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S t a t i s t i c a l T e c h n i q u e s i n

Business & Economics F i f t e e n t h E d i t i o n

Douglas A. Lind Coastal Carolina University and The University of Toledo

William G. Marchal The University of Toledo

Samuel A. Wathen Coastal Carolina University

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STATISTICAL TECHNIQUES IN BUSINESS & ECONOMICS

Published by McGraw-Hill/Irwin, a business unit of The McGraw-Hill Companies, Inc., 1221Avenue of the Americas, New York, NY, 10020. Copyright © 2012, 2010, 2008, 2005, 2002, 1999, 1996, 1993, 1990, 1986, 1982, 1978, 1974, 1970, 1967 by The McGraw-Hill Companies, Inc. All rights reserved. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning.

Some ancillaries, including electronic and print components, may not be available to customers outside the United States.

This book is printed on acid-free paper.

1 2 3 4 5 6 7 8 9 0 RJE/RJE 1 0 9 8 7 6 5 4 3 2 1

ISBN 978-0-07-340180-5 (student edition) MHID 0-07-340180-3 (student edition) ISBN 978-0-07-732701-9 (instructor’s edition) MHID 0-07-732701-2 (instructor’s edition)

Vice president and editor-in-chief: Brent Gordon Editorial director: Stewart Mattson Publisher: Tim Vertovec Executive editor: Steve Schuetz Executive director of development: Ann Torbert Senior development editor: Wanda J. Zeman Vice president and director of marketing: Robin J. Zwettler Marketing director: Brad Parkins Marketing manager: Katie White Vice president of editing, design, and production: Sesha Bolisetty Senior project manager: Diane L. Nowaczyk Senior buyer: Carol A. Bielski Interior designer: JoAnne Schopler Senior photo research coordinator: Keri Johnson Photo researcher: Teri Stratford Lead media project manager: Brian Nacik Media project manager: Ron Nelms Typeface: 9.5/11 Helvetica Neue 55 Compositor: Aptara®, Inc. Printer: R. R. Donnelley

Library of Congress Cataloging-in-Publication Data Lind, Douglas A.

Statistical techniques in business & economics / Douglas A. Lind, William G. Marchal, Samuel A. Wathen. — 15th ed.

p. cm. — (The McGraw-Hill/Irwin series operations and decision sciences) Includes index. ISBN-13: 978-0-07-340180-5 (student ed. : alk. paper) ISBN-10: 0-07-340180-3 (student ed. : alk. paper) ISBN-13: 978-0-07-732701-9 (instructor’s ed. : alk. paper) ISBN-10: 0-07-732701-2 (instructor’s ed. : alk. paper) 1. Social sciences—Statistical methods. 2. Economics—Statistical methods. 3. Commercial

statistics. I. Marchal, William G. II. Wathen, Samuel Adam. III. Title. IV. Title: Statistical techniques in business and economics. HA29.M268 2012 519.5—dc22

2010045058

www.mhhe.com

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To Jane, my wife and best friend, and our sons, their wives, and our grandchildren: Mike and Sue (Steve and Courtney), Steve and Kathryn (Kennedy and Jake), and Mark and Sarah (Jared, Drew, and Nate).

Douglas A. Lind

To John Eric Mouser, his siblings, parents, and Granny.

William G. Marchal

To my wonderful family: Isaac, Hannah, and Barb.

Samuel A. Wathen

Dedication

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Over the years, we have received many compliments on this text and understand that it’s a favorite among students. We accept that as the high- est compliment and continue to work very hard to maintain that status.

The objective of Statistical Techniques in Business and Economics is to provide students majoring in management, marketing, finance, accounting, economics, and other fields of business administration with an introductory survey of the many applications of descriptive and infer- ential statistics. We focus on business applications, but we also use many exercises and examples that relate to the current world of the col- lege student. A previous course in statistics is not necessary, and the mathematical requirement is first-year algebra.

In this text, we show beginning students every step needed to be suc- cessful in a basic statistics course. This step-by-step approach enhances performance, accelerates preparedness, and significantly improves moti- vation. Understanding the concepts, seeing and doing plenty of examples and exercises, and comprehending the application of statistical methods in business and economics are the focus of this book.

The first edition of this text was published in 1967. At that time, locat- ing relevant business data was difficult. That has changed! Today, locat- ing data is not a problem. The number of items you purchase at the gro- cery store is automatically recorded at the checkout counter. Phone companies track the time of our calls, the length of calls, and the iden- tity of the person called. Credit card companies maintain information on the number, time and date, and amount of our purchases. Medical devices automatically monitor our heart rate, blood pressure, and tem- perature from remote locations. A large amount of business information is recorded and reported almost instantly. CNN, USA Today, and MSNBC, for example, all have websites that track stock prices with a delay of less than 20 minutes.

Today, skills are needed to deal with a large volume of numerical information. First, we need to be critical consumers of information pre- sented by others. Second, we need to be able to reduce large amounts of information into a concise and meaningful form to enable us to make effective interpretations, judgments, and decisions. All students have cal- culators and most have either personal computers or access to personal computers in a campus lab. Statistical software, such as Microsoft Excel and Minitab, is available on these computers. The commands necessary to achieve the software results are available in a special section at the end of each chapter. We use screen captures within the chapters, so the student becomes familiar with the nature of the software output.

Because of the availability of computers and software, it is no Ionger necessary to dwelI on calculations. We have replaced many of the calcu- lation examples with interpretative ones, to assist the student in under- standing and interpreting the statistical results. In addition, we now place more emphasis on the conceptual nature of the statistical topics. While making these changes, we still continue to present, as best we can, the key concepts, along with supporting interesting and relevant examples.

A Note from

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the Authors

What’s New in This Fifteenth Edition? We have made changes to this edition that we think you and your stu- dents will find useful and timely.

• We have revised the learning objectives so they are more specific, added new ones, identified them in the margin, and keyed them directly to sections within the chapter.

• We have replaced the key example in Chapters 1 to 4. The new example includes more variables and more observations. It presents a realistic business situation. It is also used later in the text in Chap- ter 13.

• We have added or revised several new sections in various chapters: � Chapter 7 now includes a discussion of the exponential distribution. � Chapter 9 has been reorganized to make it more teachable and

improve the flow of the topics. � Chapter 13 has been reorganized and includes a test of hypothe-

sis for the slope of the regression coefficient. � Chapter 17 now includes a graphic test for normality and the chi-

square test for normality. • New exercises and examples use Excel 2007 screenshots and the lat-

est version of Minitab. We have also increased the size and clarity of these screenshots.

• There are new Excel 2007 software commands and updated Minitab commands at the ends of chapters.

• We have carefully reviewed the exercises within the chapters, those at the ends of chapters, and in the Review Section. We have added many new or revised exercises throughout. You can still find and assign your favorites that have worked well, or you can introduce fresh examples.

• Section numbers have been added to more clearly identify topics and more easily reference them.

• The exercises that contain data files are identified by an icon for easy identification.

• The Data Exercises at the end of each chapter have been revised. The baseball data has been updated to the most current completed season, 2009. A new business application has been added that refers to the use and maintenance of the school bus fleet of the Buena School District.

• There are many new photos throughout, with updated exercises in the chapter openers.

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How Are Chapters Organized to

3 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Explain the concept of central tendency.

LO2 Identify and compute the arithmetic mean.

LO3 Compute and interpret the weighted mean.

LO4 Determine the median.

LO5 Identify the mode.

LO6 Calculate the geometric mean.

LO7 Explain and apply mea- sures of dispersion.

LO8 Compute and explain the variance and the standard deviation.

LO9 Explain Chebyshev’s Theorem and the Empirical Rule.

LO10 Compute the mean and standard deviation of grouped data.

Describing Data: Numerical Measures

The Kentucky Derby is held the first Saturday in May at Churchill

Downs in Louisville, Kentucky. The race track is one and one-quarter

miles. The table in Exercise 82 shows the winners since 1990, their

margin of victory, the winning time, and the payoff on a $2 bet.

Determine the mean and median for the variables winning time and

payoff on a $2 bet. (See Exercise 82 and LO2 and LO4.)

Introduction to the Topic Each chapter starts with a review of the impor- tant concepts of the previous chapter and pro- vides a link to the material in the current chapter. This step-by-step approach increases com- prehension by providing continuity across the concepts.

2.1 Introduction The highly competitive automobile retailing industry in the United States has changed dramatically in recent years. These changes spurred events such as the:

• bankruptcies of General Motors and Chrysler in 2009. • elimination of well-known brands such as Pontiac and

Saturn. • closing of over 1,500 local dealerships. • collapse of consumer credit availability. • consolidation dealership groups.

Traditionally, a local family owned and operated the com- munity dealership, which might have included one or two man- ufacturers or brands, like Pontiac and GMC Trucks or Chrysler and the popular Jeep line. Recently, however, skillfully managed and well-financed companies have been acquiring local dealer-

Example

Solution

Layton Tire and Rubber Company wishes to set a minimum mileage guarantee on its new MX100 tire. Tests reveal the mean mileage is 67,900 with a stan- dard deviation of 2,050 miles and that the distribu- tion of miles follows the normal probability distrib- ution. Layton wants to set the minimum guaranteed mileage so that no more than 4 percent of the tires will have to be replaced. What minimum guaranteed mileage should Layton announce?

The facets of this case are shown in the following diagram, where X represents the minimum guaran- teed mileage.

Self-Review 3–6 The weights of containers being shipped to Ireland are (in thousands of pounds):

95 103 105 110 104 105 112 90

(a) What is the range of the weights? (b) Compute the arithmetic mean weight. (c) Compute the mean deviation of the weights.

Chapter Learning Objectives Each chapter begins with a set of learning objectives designed to provide focus for the chapter and motivate student learning. These objectives, located in the margins next to the topic, indicate what the student should be able to do after completing the chapter.

Chapter Opening Exercise A representative exercise opens the chapter and shows how the chapter content can be applied to a real-world situation.

Example/Solution After important concepts are introduced, a solved example is given to provide a how-to illustration for students and to show a relevant business or economics-based application that helps answer the question, “What will I use this for?” All examples provide a realistic scenario or application and make the math size and scale reasonable for introductory students.

Self-Reviews Self-Reviews are interspersed through- out each chapter and closely patterned after the preceding Examples. They help students monitor their progress and provide immediate reinforcement for that particular technique.

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Engage Students and Promote Learning?

Statistics in Action Statistics in Action articles are scattered through- out the text, usually about two per chapter. They provide unique and interesting applications and historical insights in the field of statistics.

Exercises Exercises are included after sections within the chapter and at the end of the chapter. Section exercises cover the material studied in the section.

The equation for the trend line is:

The slope of the trend line is .08991. This shows that over the 24 quarters the deseasonalized sales increased at a rate of 0.08991 ($ million) per quarter, or $89,910 per quarter. The value of 8.109 is the intercept of the trend line on the Y-axis (i.e., for t � 0).

Ŷ � 8.109 � .08991t

Statistics in Action

Margin Notes There are more than 300 concise notes in the margin. Each is aimed at reemphasizing the key concepts presented immediately adja- cent to it.

Definitions Definitions of new terms or terms unique to the study of statistics are set apart from the text and highlighted for easy reference and review.

Population Variance The formulas for the population variance and the sample variance are slightly different. The population variance is considered first. (Recall that a population is the totality of all observations being studied.) The population variance is found by:

Variance and standard deviation are based on squared deviations from the mean.

STANDARD DEVIATION The square root of the variance.

The variance is non-negative and is zero only if all observations are the same.

Formulas Formulas that are used for the first time are boxed and numbered for reference. In addition, a formula card is bound into the back of the text, which lists all the key formulas.

POPULATION VARIANCE [3–8]�2 � �(X � �)2

Exercises For Exercises 35–38, calculate the (a) range, (b) arithmetic mean, (c) mean deviation, and (d) interpret the values.

35. There were five customer service representatives on duty at the Electronic Super Store during last weekend’s sale. The numbers of HDTVs these representatives sold are: 5, 8, 4, 10, and 3.

36. The Department of Statistics at Western State University offers eight sections of basic statistics. Following are the numbers of students enrolled in these sections: 34, 46, 52, 29, 41, 38, 36, and 28.

Computer Output The text includes many software examples, using Excel, MegaStat®, and Minitab.

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BY CHAPTER Chapter Summary Each chapter contains a brief summary of the chapter material, including the vocabulary and the critical formulas.

How Does This Text

Chapter Summary I. A dot plot shows the range of values on the horizontal axis and the number of observa-

tions for each value on the vertical axis. A. Dot plots report the details of each observation. B. They are useful for comparing two or more data sets.

II. A stem-and-leaf display is an alternative to a histogram. A. The leading digit is the stem and the trailing digit the leaf. B. The advantages of a stem-and-leaf display over a histogram include:

Pronunciation Key This tool lists the mathematical symbol, its mean- ing, and how to pronounce it. We believe this will help the student retain the meaning of the symbol and generally enhance course communications.

Pronunciation Key SYMBOL MEANING PRONUNCIATION

Location of percentile L sub p

First quartile Q sub 1

Third quartile Q sub 3Q3

Chapter Exercises Generally, the end-of-chapter exercises are the most challenging and integrate the chapter con- cepts. The answers and worked-out solutions for all odd-numbered exercises appear at the end of the text. For exercises with more than 20 observations, the data can be found on the text’s website. These files are in Excel and Minitab formats.

Chapter Exercises 27. A sample of students attending Southeast Florida University is asked the number of social

activities in which they participated last week. The chart below was prepared from the sample data.

41 2 Activities

Data Set Exercises 44. Refer to the Real Estate data, which reports information on homes sold in the Goodyear,

Arizona, area during the last year. Prepare a report on the selling prices of the homes. Be sure to answer the following questions in your report. a. Develop a box plot. Estimate the first and the third quartiles. Are there any outliers? b. Develop a scatter diagram with price on the vertical axis and the size of the home on

the horizontal. Does there seem to be a relationship between these variables? Is the relationship direct or inverse?

c. Develop a scatter diagram with price on the vertical axis and distance from the center of the city on the horizontal axis. Does there seem to be a relationship between these variables? Is the relationship direct or inverse?

45. Refer to the Baseball 2009 data, which reports information on the 30 Major League Base- ball teams for the 2009 season. Refer to the variable team salary. a. Select the variable that refers to the year in which the stadium was built. (Hint: Subtract

the year in which the stadium was built from the current year to find the age of the stadium and work this variable.) Develop a box plot. Are there any outliers? Which sta- diums are outliers?

b. Select the variable team salary and draw a box plot. Are there any outliers? What are the quartiles? Write a brief summary of your analysis. How do the salaries of the New York Yankees compare with the other teams?

1. The Excel Commands for the descriptive statistics on page 69 are:

a. From the CD, retrieve the Applewood data. b. From the menu bar, select Data and then Data

Analysis. Select Descriptive Statistics and then click OK.

2. The Minitab commands for the descriptive summary on page 84 are:

Software Commands

Data Set Exercises The last several exercises at the end of each chapter are based on three large data sets. These data sets are printed in Appendix A in the text and are also on the text’s web- site. These data sets present the students with real-world and more complex applications.

Software Commands Software examples using Excel, MegaStat®, and Minitab are included throughout the text, but the explanations of the computer input commands for each program are placed at the end of the chapter. This allows students to focus on the sta- tistical techniques rather than on how to input data.

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Reinforce Student Learning? Answers to Self-Review The worked-out solutions to the Self-Reviews are provided at the end of each chapter.

Cases The review also includes continuing cases and several small cases that let students make decisions using tools and techniques from a variety of chapters.

Fr eq

ue nc

0 Cola-Plus Coca-Cola Pepsi

Beverage

Lemon-Lime

Chapter 2 Answers to Self-Review

2–1 a. Qualitative data, because the customers’ response to the taste test is the name of a beverage.

b. Frequency table. It shows the number of people who prefer each beverage.

c. Class frequencies. d. The largest concentration of commissions

is $1,500 up to $1,600. The smallest commission is about $1,400 and the largest is about $1,800. The typical amount earned is $15,500.

2–3 a. 26 � 64 � 73 � 128 � 27. So seven classes are recommended.

b. The interval width should be at least (488 � 320)�7 � 24. Class intervals of 25 or 30 feet are both reasonable.

c. If we use a class interval of 25 feet and begin with a lower limit of 300 feet, eight classes would be necessary. A class interval of 30 feet beginning with 300 feet is also reasonable. This alternative requires only seven classes.

2–4 a. 45 b. .250 c. .306, found by .178 � .106 � .022

2–5 a. 20

BY SECTION Section Reviews After selected groups of chapters (1–4, 5–7, 8 and 9, 10–12, 13 and 14, 15 and 16, and 17 and 18), a Sec- tion Review is included. Much like a review before an exam, these include a brief overview of the chapters, a glossary of key terms, and problems for review.

A Review of Chapters 1–4 This section is a review of the major concepts and terms introduced in Chapters 1–4. Chapter 1 began by describing the meaning and purpose of statistics. Next we described the different types of variables and the four levels of measurement. Chapter 2 was concerned with describing a set of observations by organizing it into a frequency distribution and then portraying the frequency distri- bution as a histogram or a frequency polygon. Chapter 3 began by describing measures of loca- tion, such as the mean, weighted mean, median, geometric mean, and mode. This chapter also included measures of dispersion, or spread. Discussed in this section were the range, mean devi- ation, variance, and standard deviation. Chapter 4 included several graphing techniques such as dot plots, box plots, and scatter diagrams. We also discussed the coefficient of skewness, which reports the lack of symmetry in a set of data.

Throughout this section we stressed the importance of statistical software, such as Excel and Minitab. Many computer outputs in these chapters demonstrated how quickly and effectively a large data set can be organized into a frequency distribution, several of the measures of location or measures or variation calculated, and the information presented in graphical form.

Glossary

E l d ( l f l ) d

A. Century National Bank The following case will appear in subsequent review sec- tions. Assume that you work in the Planning Department of the Century National Bank and report to Ms. Lamberg. You will need to do some data analysis and prepare a short written report. Remember, Mr. Selig is the president of the bank, so you will want to ensure that your report is complete and accurate. A copy of the data appears in Appendix A.6.

Century National Bank has offices in several cities in the Midwest and the southeastern part of the United States. Mr. Dan Selig, president and CEO, would like to know the characteristics of his checking account cus- tomers. What is the balance of a typical customer?

How many other bank services do the checking ac- count customers use? Do the customers use the ATM ser- vice and, if so, how often? What about debit cards? Who uses them, and how often are they used?

median balances for the four branches. Is there a difference among the branches? Be sure to explain the difference between the mean and the median in your report.

3. Determine the range and the standard deviation of the checking account balances. What do the first and third quartiles show? Determine the coefficient of skewness and indicate what it shows. Because Mr. Selig does not deal with statistics daily, include a brief description and interpretation of the standard devia- tion and other measures.

B. Wildcat Plumbing Supply Inc.: Do We Have Gender Differences?

Wildcat Plumbing Supply has served the plumbing needs of Southwest Arizona for more than 40 years. The company was founded by Mr. Terrence St. Julian and is run today by his son Cory. The company has grown from a handful of employees to more than 500 today. Cory is concerned about several positions within the company where he has men and women doing essentially the same job but at dif- ferent pay. To investigate, he collected the information be- low. Suppose you are a student intern in the Accounting Department and have been given the task to write a report

Cases

Practice Test The Practice Test is intended to give students an idea of content that might appear on a test and how the test might be structured. The Practice Test includes both objective questions and problems covering the material studied in the section.

Part 2—Problems 1. The Russell 2000 index of stock prices increased by the following amounts over the last three years.

18% 4% 2%

What is the geometric mean increase for the three years?

Practice Test

There is a practice test at the end of each review section. The tests are in two parts. The first part contains several ob- jective questions, usually in a fill-in-the-blank format. The second part is problems. In most cases, it should take 30 to 45 minutes to complete the test. The problems require a calculator. Check the answers in the Answer Section in the back of the book.

Part 1—Objective 1. The science of collecting, organizing, presenting, analyzing, and interpreting data to assist in making effective deci-

sions is called . 1. 2. Methods of organizing, summarizing, and presenting data in an informative way is called .

2. 3. The entire set of individuals or objects of interest or the measurements obtained from all individuals or objects of in-

terest is called the . 3. 4. List the two types of variables. 4.

5. The number of bedrooms in a house is an example of a . (discrete variable, continuous variable, qualitative variable—pick one) 5.

6. The jersey numbers of Major League Baseball players is an example of what level of measurement? 6.

7. The classification of students by eye color is an example of what level of measurement? 7. 8. The sum of the differences between each value and the mean is always equal to what value? 8. 9. A set of data contained 70 observations. How many classes would you suggest in order to construct a frequency

distribution? 9. 10. What percent of the values in a data set are always larger than the median? 10. 11. The square of the standard deviation is the . 11. 12. The standard deviation assumes a negative value when . (All the values are negative, when at least half the

values are negative, or never—pick one.) 12. 13. Which of the following is least affected by an outlier? (mean, median, or range—pick one) 13.

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What Technology Connects McGraw-Hill Connect™ Business Statistics Less Managing. More Teaching. Greater Learning. McGraw-Hill Connect Business Statistics is an online assignment and assessment solution that connects students with the tools and resources they’ll need to achieve success.

McGraw-Hill Connect Business Statistics helps prepare students for their future by enabling faster learning, more efficient studying, and higher retention of knowledge.

Features. Connect Business Statistics offers a number of powerful tools and features to make manag- ing assignments easier, so faculty can spend more time teaching. With Connect Business Statistics, students can engage with their coursework anytime and anywhere, making the learning process more accessible and efficient. Connect Business Statistics offers you the features described below.

Simple Assignment Management. With Con- nect Business Statistics, creating assignments is easier than ever, so you can spend more time teaching and less time managing. The assignment management function enables you to:

• Create and deliver assignments easily with selectable end-of-chapter questions and test bank items.

• Streamline lesson planning, student pro- gress reporting, and assignment grading to make classroom management more effi- cient than ever.

• Go paperless with the eBook and on- line submission and grading of student assignments.

Integration of Excel Data Sets. A convenient feature is the inclusion of an Excel data file link in many problems using data files in their cal- culation. This allows students to easily launch into Excel, work the problem, and return to Connect to key in the answer.

Excel Integrated Data File

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Students to Business Statistics? Smart Grading. When it comes to studying, time is precious. Connect Business Statistics helps students learn more efficiently by providing feedback and practice material when they need it, where they need it. When it comes to teaching, your time also is precious. The grading function enables you to:

• Have assignments scored automatically, giving students immediate feedback on their work and side- by-side comparisons with correct answers.

• Access and review each response; manually change grades or leave comments for students to review.

• Reinforce classroom concepts with practice tests and instant quizzes.

Instructor Library. The Connect Business Statistics Instructor Library is your repository for additional resources to improve student engagement in and out of class. You can select and use any asset that enhances your lecture. The Connect Business Statistics Instructor Library includes:

• eBook

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Student Study Center. The Connect Business Statistics Student Study Center is the place for students to access additional resources. The Student Study Center:

• Offers students quick access to lectures, practice materials, eBooks, and more.

• Provides instant practice material and study questions and is easily accessible on-the-go.

Guided Examples. These narrated video walkthroughs provide students with step-by-step guidelines for solving problems similar to those contained in the text. The student is given personalized instruction on how to solve a problem by applying the concepts presented in the chapter.

Student Progress Tracking. Connect Business Statistics keeps instructors informed about how each student, section, and class is performing, allowing for more productive use of lecture and office hours. The progress-tracking function enables you to:

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• Collect data and generate reports required by many accreditation organizations, such as AACSB.

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What Technology Connects McGraw-Hill CONNECT™ PLUS BUSINESS STATISTICS McGraw-Hill Connect Plus Business Statistics. McGraw-Hill reinvents the textbook learning experience for the modern student with Connect Plus Business Statistics. A seamless integration of an eBook and Connect Business Statistics, Connect Plus Business Statistics provides all of the Connect Business Sta- tistics features plus the following:

• An integrated eBook, allowing for anytime, anywhere access to the textbook.

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In short, Connect Business Statis- tics offers you and your students powerful tools and features that optimize your time and energies, enabling you to focus on course content, teaching, and student learning. Connect Business Statis- tics also offers a wealth of content resources for both instructors and students. This state-of-the-art, thoroughly tested system supports you in preparing students for the world that awaits. For more information about Connect, go to www.mcgrawhillconnect.com or contact your local McGraw-Hill sales representative.

Tegrity Campus: Lectures 24/7 Tegrity Campus is a service that makes class time available 24/7 by automatically capturing every lec- ture in a searchable format for students to review when they study and complete assignments. With a simple one-click start-and-stop process, you capture all computer screens and corresponding audio. Students can replay any part of any class with easy-to-use browser-based viewing on a PC or Mac.

McGraw-Hill Tegrity Campus Educators know that the more students can see, hear, and experience class resources, the better they learn. In fact, studies prove it. With Tegrity Campus, students quickly recall key moments by using Tegrity Campus’s unique search feature. This search helps students efficiently find what they need, when they need it, across an entire semester of class recordings. Help turn all your students’ study time into learn- ing moments immediately supported by your lecture.

To learn more about Tegrity, watch a two-minute Flash demo at http://tegritycampus.mhhe.com.

business statistics

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Students to Business Statistics? Assurance-of-Learning Ready Many educational institutions today are focused on the notion of assurance of learn- ing an important element of some accreditation standards. Statistical Techniques in Business & Economics is designed specifically to support your assurance-of- learning initiatives with a simple, yet powerful solution.

Each test bank question for Statistical Techniques in Business & Economics maps to a specific chapter learning outcome/objective listed in the text. You can use our test bank software, EZ Test and EZ Test Online, or Connect Business Sta- tistics to easily query for learning outcomes/objectives that directly relate to the learning objectives for your course. You can then use the reporting features of EZ Test to aggregate student results in similar fashion, making the collection and pre- sentation of assurance of learning data simple and easy.

AACSB Statement The McGraw-Hill Companies is a proud corporate member of AACSB International. Understand- ing the importance and value of AACSB accreditation, Statistical Techniques in Business & Eco- nomics recognizes the curricula guidelines detailed in the AACSB standards for business accredita- tion by connecting selected ques- tions in the text and the test bank to the six general knowledge and skill guidelines in the AACSB standards.

The statements contained in Statistical Techniques in Business & Economics are provided only as a guide for the users of this textbook. The AACSB leaves content coverage and assessment within the purview of individual schools, the mission of the school, and the faculty. While Statistical Techniques in Business & Economics and the teaching package make no claim of any specific AACSB qualification or eval- uation, we have labeled selected questions within Statistical Techniques in Business & Economics according to the six general knowledge and skills areas.

McGraw-Hill Customer Care Information At McGraw-Hill, we understand that getting the most from new technology can be challenging. That’s why our services don’t stop after you purchase our products. You can e-mail our Product Specialists 24 hours a day to get product-training online. Or you can search our knowledge bank of Frequently Asked Questions on our support website. For Customer Support, call 800-331-5094 or visit www.mhhe.com/support. One of our Technical Support Analysts will be able to assist you in a timely fashion.

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What Software Is Available with This Text? MegaStat® for Microsoft Excel®

MegaStat® by J. B. Orris of Butler University is a full-featured Excel add-in that is available on CD and on the MegaStat website at www.mhhe.com/megastat. It works with Excel 2003, 2007, and 2010. On the web- site, students have 10 days to successfully download and install MegaStat on their local computer. Once installed, MegaStat will remain active in Excel with no expiration date or time limitations. The software per- forms statistical analyses within an Excel workbook. It does basic functions, such as descriptive statistics, frequency distributions, and probability calculations as well as hypothesis testing, ANOVA, and regression.

MegaStat output is carefully formatted and ease-of-use features include Auto Expand for quick data selec- tion and Auto Label detect. Since MegaStat is easy to use, students can focus on learning statistics with- out being distracted by the software. MegaStat is always available from Excel’s main menu. Selecting a menu item pops up a dialog box. MegaStat works with all recent versions of Excel, including Excel 2007 and Excel 2010. Screencam tutorials are included that provide a walkthrough of major business statistics topics. Help files are built in, and an introductory user’s manual is also included.

Minitab®/SPSS®/JMP®

Minitab® Student Version 14, SPSS® Student Version 18.0, and JMP® Student Edition Version 8 are software tools that are available to help students solve the business statistics exercises in the text. Each can be packaged with any McGraw-Hill business statistics text.

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What Resources Are Available for Instructors? Instructor’s Resources CD-ROM (ISBN: 0077327055) This resource allows instructors to conveniently access the Instruc- tor’s Solutions Manual, Test Bank in Word and EZ Test formats, Instructor PowerPoint slides, data files, and data sets.

Online Learning Center: www.mhhe.com/lind15e The Online Learning Center (OLC) provides the instructor with a com- plete Instructor’s Manual in Word format, the complete Test Bank in both Word files and computerized EZ Test format, Instructor Power- Point slides, text art files, an introduction to ALEKS®, an introduction to McGraw-Hill Connect Business StatisticsTM, access to Visual Statistics, and more.

All test bank questions are available in an EZ Test electronic format. Included are a number of multiple- choice, true/false, and short-answer questions and problems. The answers to all questions are given, along with a rating of the level of difficulty, chapter goal the question tests, Bloom’s taxonomy question type, and the AACSB knowledge category.

WebCT/Blackboard/eCollege All of the material in the Online Learning Center is also available in portable WebCT, Blackboard, or eCollege content “cartridges” provided free to adopters of this text.

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ALEKS is an assessment and learning program that provides individualized instruction in Business Statistics, Business Math, and Accounting. Available online in partnership with McGraw- Hill/lrwin, ALEKS interacts with students much like a skilled human tutor, with the ability to assess precisely a student’s knowledge and provide instruction on the exact topics the stu- dent is most ready to learn. By providing topics to meet indi- vidual students’ needs, allowing students to move between explanation and practice, correcting and analyzing errors, and defining terms, ALEKS helps students to master course con- tent quickly and easily.

ALEKS also includes a new instructor module with powerful, assignment-driven features and extensive con- tent flexibility. ALEKS simplifies course management and allows instructors to spend less time with admin- istrative tasks and more time directing student learning. To learn more about ALEKS, visit www.aleks.com.

Online Learning Center: www.mhhe.com/lind15e The Online Learning Center (OLC) provides students with the following content:

• Quizzes • *Visual Statistics

• PowerPoint • Data sets/files

• *Narrated PowerPoint • Appendixes

• *Screencam tutorials • Chapter 20

• *Guided Examples

*Premium Content

Student Study Guide (ISBN: 007732711X) This supplement helps students master the course content. It highlights the important ideas in the text and pro- vides opportunities for students to review the worked-out solutions, review terms and concepts, and practice.

Basic Statistics Using Excel 2007 (ISBN: 0077327020) This workbook introduces students to Excel and shows how to apply it to introductory statistics. It presumes no prior familiarity with Excel or statistics and provides step-by-step directions in a how-to style using Excel 2007 with text examples and problems.

Business Statistics Center (BSC): www.mhhe.com/bstat/ The BSC contains links to statistical publications and resources, software downloads, learning aids, sta- tistical websites and databases, and McGraw-Hill/Irwin product websites and online courses.

What Resources Are Available for Students? CourseSmart CourseSmart is a convenient way to find and buy eTextbooks. CourseSmart has the largest selection of eTextbooks available anywhere, offering thousands of the most commonly adopted textbooks from a wide variety of higher-education publishers. Course Smart eTextbooks are available in one standard online reader with full text search, notes and highlighting, and e-mail tools for sharing notes between classmates. Visit www.CourseSmart.com for more information on ordering.

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Acknowledgments

Reviewers

Sung K. Ahn Washington State University–Pullman

Scott Bailey Troy University

Douglas Barrett University of North Alabama

Arnab Bisi Purdue University

Pamela A. Boger Ohio University–Athens

Emma Bojinova Canisius College

Giorgio Canarella California State University–Los Angeles

Lee Cannell El Paso Community College

James Carden University of Mississippi

Mary Coe St. Mary College of California

Anne Davey Northeastern State University

Neil Desnoyers Drexel University

Nirmal Devi Embry Riddle Aeronautical University

David Doorn University of Minnesota–Duluth

Ronald Elkins Central Washington University

Vickie Fry Westmoreland County Community College

Clifford B. Hawley West Virginia University

Lloyd R. Jaisingh Morehead State University

Mark Kesh University of Texas

Ken Kelley University of Notre Dame

Melody Kiang California State University–Long Beach

Morris Knapp Miami Dade College

Teresa Ling Seattle University

John D. McGinnis Pennsylvania State–Altoona

Mary Ruth J. McRae Appalachian State University

Jackie Miller Ohio State University

Carolyn Monroe Baylor University

Valerie Muehsam Sam Houston State University

Tariq Mughal University of Utah

Elizabeth J. T. Murff Eastern Washington University

Quinton Nottingham Virginia Polytechnic Institute and State University

René Ordonez Southern Oregon University

Robert Patterson Penn State University

Joseph Petry University of Illinois at Urbana-Champaign

Tammy Prater Alabama State University

Michael Racer University of Memphis

Darrell Radson Drexel University

Steven Ramsier Florida State University

Christopher W. Rogers Miami Dade College

Stephen Hays Russell Weber State University

Martin Sabo Community College of Denver

Farhad Saboori Albright College

Amar Sahay Salt Lake Community College and University of Utah

Abdus Samad Utah Valley University

Nina Sarkar Queensborough Community College

Roberta Schini West Chester University of Pennsylvania

Robert Smidt California Polytechnic State University

Gary Smith Florida State University

Stanley D. Stephenson Texas State University–San Marcos

Debra Stiver University of Nevada

Bedassa Tadesse University of Minnesota–Duluth

Stephen Trouard Mississippi College

Elzbieta Trybus California State University–Northridge

Daniel Tschopp Daemen College

Sue Umashankar University of Arizona

Jesus M. Valencia Slippery Rock University

Joseph Van Matre University of Alabama at Birmingham

Angie Waits Gadsden State Community College

Bin Wang St. Edwards University

Kathleen Whitcomb University of South Carolina

Blake Whitten University of Iowa

Oliver Yu San Jose State University

Zhiwei Zhu University of Louisiana

Survey and Focus Group Participants

Nawar Al-Shara American University

Charles H. Apigian Middle Tennessee State University

Nagraj Balakrishnan Clemson University

Philip Boudreaux University of Louisiana at Lafayette

Nancy Brooks University of Vermont

Qidong Cao Winthrop University

This edition of Statistical Techniques in Business and Economics is the product of many people: students, colleagues, reviewers, and the staff at McGraw-Hill/Irwin. We thank them all. We wish to express our sincere gratitude to the survey and focus group participants, and the reviewers:

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Their suggestions and thorough reviews of the previous edition and the manuscript of this edi- tion make this a better text.

Special thanks go to a number of people. Debra K. Stiver, University of Nevada–Reno, reviewed the manuscript and page proofs, checking text and exercises for accuracy. Joan McGrory, South- west Tennessee Community College, checked the Test Bank for accuracy. Professor Kathleen Whit- comb of the University of South Carolina prepared the study guide. Dr. Samuel Wathen of Coastal Carolina University prepared the quizzes and the Test Bank. Professor René Ordonez of Southern- Oregon University prepared the PowerPoint presentation, many of the screencam tutorials, and the guided examples in Connect. Ms. Denise Heban and the authors prepared the Instructor’s Manual.

We also wish to thank the staff at McGraw-Hill. This includes Steve Schuetz, Executive Edi- tor; Wanda Zeman, Senior Development Editor; Diane Nowaczyk, Senior Project Manager; and oth- ers we do not know personally, but who have made valuable contributions.

Margaret M. Capen East Carolina University

Robert Carver Stonehill College

Jan E. Christopher Delaware State University

James Cochran Louisiana Tech University

Farideh Dehkordi-Vakil Western Illinois University

Brant Deppa Winona State University

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Deborah J. Gougeon University of Scranton

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Charles F. Harrington University of Southern Indiana

Craig Heinicke Baldwin-Wallace College

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Cindy L. Hinz St. Bonaventure University

Johnny C. Ho Columbus State University

Shaomin Huang Lewis-Clark State College

J. Morgan Jones University of North Carolina at Chapel Hill

Michael Kazlow Pace University

John Lawrence California State University, Fullerton

Sheila M. Lawrence Rutgers, The State University of New Jersey

Jae Lee State University of New York at New Paltz

Rosa Lemel Kean University

Robert Lemke Lake Forest College

Francis P. Mathur California State Polytechnic University, Pomona

Ralph D. May Southwestern Oklahoma State University

Richard N. McGrath Bowling Green State University

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Carlton Scott University of California, Irvine

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Scott J. Seipel Middle Tennessee State University

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William Stein Texas A&M University

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Stuart H. Warnock Tarleton State University

Mark H. Witkowski University of Texas at San Antonio

William F. Younkin University of Miami

Shuo Zhang State University of New York, Fredonia

Zhiwei Zhu University of Louisiana at Lafayette

Acknowledgments

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Changes Made in All Chapters and Major Changes to Individual Chapters:

• Changed Goals to Learning Objectives and identified the location in the chapter where the learning objective is discussed.

• Added section numbering to each main heading.

• Identified exercises where the data file is included on the text website.

• Revised the Major League Baseball data set to reflect the latest complete season, 2009.

• Revised the Real Estate data to ensure the outcomes are more realistic to the current economy.

• Added a new data set regarding school buses in a pub- lic school system.

• Updated screens for Excel 2007, Minitab, and MegaStat.

• Revised the core example in Chapters 1–4 to reflect the current economic conditions as it relates to automobile dealers. This example is also discussed in Chapter 13 and 17.

• Added a new section in Chapter 7 describing the expo- nential distribution.

• Added a new section in Chapter 13 describing a test to determine whether the slope of the regression line dif- fers from zero.

• Added updates and clarifications throughout.

Chapter 1 What Is Statistics? • New photo and chapter opening exercise on the “Nook”

sold by Barnes and Nobel.

• Census updates on U.S. population, sales of Boeing air- craft, and Forbes data in “Statistics in Action” feature.

• New chapter exercises 17 (data on 2010 vehicle sales) and 19 (ExxonMobil sales prior to Gulf oil spill).

Chapter 2 Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation

• New data on Ohio State Lottery expenses for 2009 with new Excel 2007 screenshot.

• New exercises 45 (brides picking their wedding site) and 46 (revenue in the state of Georgia).

Chapter 3 Describing Data: Numerical Measures

• New data on averages in the introduction: average num- ber of TV sets per home, average spending on a wed- ding, and the average price of a theater ticket.

• A new description of the calculation and interpretation of the population mean using the distance between exits on I-75 through Kentucky.

• A new description of the median using the time manag- ing Facebook accounts.

• Updated example/solution on the population in Las Vegas.

• Update “Statistics in Action” on the highest batting aver- age in Major League Baseball for 2009. It was Joe Mauer of the Minnesota Twins, with an average of .365.

• New chapter exercises 22 (real estate commissions), 67 (laundry habits), 77 (public universities in Ohio), 72 (blood sugar numbers), and 82 (Kentucky Derby payoffs). Exer- cises 30 to 34 were revised to include the most recent data.

Chapter 4 Describing Data: Displaying and Exploring Data

• New exercise 22 with 2010 salary data for the New York Yankees.

• New chapter exercise 36 (American Society of Peri- Anesthesia nurses component membership).

Chapter 5 A Survey of Probability Concepts • New exercise 58 (number of hits in a Major League

Baseball game), 59 (winning a tournament), and 60 (win- ning Jeopardy).

Chapter 6 Discrete Probability Distributions • No changes.

Chapter 7 Continuous Probability Distributions • New Self-Review 7–4 and 7–5 involving coffee

temperature.

• New exercise 26 (SAT Reasoning Test).

• New exercise 29 (Hurdle Rate for economic investment).

• New section and corresponding problems on the expo- nential probability distribution.

• Several glossary updates and clarifications.

Chapter 8 Sampling Methods and the Central Limit Theorem

• No changes.

Chapter 9 Estimation and Confidence Intervals • A new Statistics in Action describing EPA fuel economy.

• New separate section on point estimates.

• Integration and application of the central limit theorem.

Enhancements to Statistical Techniques in Business & Economics, 15e

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• A revised discussion of determining the confidence interval for the population mean.

• Expanded section on calculating sample size.

• New exercise 12 (milk consumption), 33 (cost of apart- ments in Milwaukee), 47 (drug testing in the fashion industry), and 48 (survey of small-business owners regarding healthcare).

• The discussion of the finite correction factor has been relocated in the chapter.

Chapter 10 One-Sample Tests of Hypothesis • New exercises 17 (daily water consumption), 19 (number

of text messages by teenagers), 35 (household size in the United States), 49 (Super Bowl coin flip results), 54 (failure of gaming industry slot machines), 57 (study of the percentage of Americans that do not eat breakfast), and 60 (daily water usage).

Chapter 11 Two-Sample Tests of Hypothesis • New exercises 15 (2010 New York Yankee salaries), 37

(Consumer Confidence Survey), and 39 (pets as listeners).

Chapter 12 Analysis of Variance • Revised the names of airlines in the one-way ANOVA

example.

• New exercise 30 (flight times between Los Angeles and San Francisco).

Chapter 13 Correlation and Linear Regression • Rewrote the introduction section to the chapter.

• Added a new section using the Applewood Auto Group data from chapters 1 to 4.

• Added a section on testing the slope of a regression line.

• Added discussion of the regression ANOVA table with Excel examples.

• Rewrote and relocated the section on the coefficient of determination.

• Updated exercise 60 (movie box office amounts).

Chapter 14 Multiple Regression Analysis • Rewrote the section on evaluating the multiple regression

equation.

• More emphasis on the regression ANOVA table.

• Enhanced the discussion of the p-value in decision making.

• Added a separate section on qualitative variables in regression analysis.

• Moved the “Stepwise Regression” section to improve the sequence of topics.

• Added a summary problem at the end of the chapter to review the major concepts.

Chapter 15 Index Numbers • Updated census and economic data.

Chapter 16 Time Series and Forecasting • Updated economic data.

Chapter 17 Nonparametric Methods: Goodness-of-Fit Tests

• Reworked the Example/Solution on the chi-square goodness-of-fit test with equal cell frequencies (favorite meals of adults).

• Added a section and corresponding examples describing the goodness-of-fit test for testing whether sample data are from a normal population.

• Added a section and corresponding examples using graphical methods for testing whether sample data are from a normal population.

Chapter 18 Nonparametric Methods: Analysis of Ranked Data

• Revised the Example/Solution for the Kruskal-Wallis test (waiting times in the emergency room).

• Revised the Example/Solution for the Spearman coeffi- cient of rank correlation (comparison of recruiter and plant scores for trainees).

Chapter 19 Statistical Process Control and Quality Management

• Updated the section on the Malcolm Baldrige National Quality Award.

• Reworked and updated the section on Six Sigma.

Enhancements to Statistical Techniques in Business & Economics, 15e

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Brief Contents

1 What Is Statistics? 1 2 Describing Data: Frequency Tables, Frequency Distributions, and Graphic

Presentation 21

3 Describing Data: Numerical Measures 57 4 Describing Data: Displaying and Exploring Data 102 5 A Survey of Probability Concepts 144 6 Discrete Probability Distributions 186 7 Continuous Probability Distributions 222 8 Sampling Methods and the Central Limit Theorem 265 9 Estimation and Confidence Intervals 297

10 One-Sample Tests of Hypothesis 333 11 Two-Sample Tests of Hypothesis 371 12 Analysis of Variance 410 13 Correlation and Linear Regression 461 14 Multiple Regression Analysis 512 15 Index Numbers 573 16 Time Series and Forecasting 604 17 Nonparametric Methods: Goodness-of-Fit Tests 648 18 Nonparametric Methods: Analysis of Ranked Data 680 19 Statistical Process Control and Quality Management 720 20 An Introduction to Decision Theory On the website:

www.mhhe.com/lind15e

Appendixes: Data Sets, Tables, Answers 753

Photo Credits 829

Index 831

Review Section

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xxii

Contents

A Note from the Authors iv

Chapter

1 What Is Statistics? 1 1.1 Introduction 2

1.2 Why Study Statistics? 2

1.3 What Is Meant by Statistics? 4

1.4 Types of Statistics 6

Descriptive Statistics 6 Inferential Statistics 6

1.5 Types of Variables 8

1.6 Levels of Measurement 9

Nominal-Level Data 10 Ordinal-Level Data 11 Interval-Level Data 11 Ratio-Level Data 12

Exercises 14

1.7 Ethics and Statistics 14

1.8 Computer Applications 14

Chapter Summary 16

Chapter Exercises 16

Data Set Exercises 19

Answers to Self-Review 20

Chapter

2 Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 21 2.1 Introduction 22

2.2 Constructing a Frequency Table 23

Relative Class Frequencies 23 Graphic Presentation of Qualitative Data 24

Exercises 28

2.3 Constructing Frequency Distributions: Quantitative Data 29

2.4 A Software Example 34

2.5 Relative Frequency Distribution 34

Exercises 35

2.6 Graphic Presentation of a Frequency Distribution 36

Histogram 36 Frequency Polygon 38

Exercises 41

Cumulative Frequency Distributions 42

Exercises 44

Chapter Summary 46

Chapter Exercises 46

Data Set Exercises 53

Software Commands 54

Answers to Self-Review 55

Chapter

3 Describing Data: Numerical Measures 57 3.1 Introduction 58

3.2 The Population Mean 58

3.3 The Sample Mean 60

3.4 Properties of the Arithmetic Mean 61

Exercises 62

3.5 The Weighted Mean 63

Exercises 64

3.6 The Median 64

3.7 The Mode 65

Exercises 67

3.8 Software Solution 69

3.9 The Relative Positions of the Mean, Median, and Mode 69

Exercises 71

3.10 The Geometric Mean 72

Exercises 73

3.11 Why Study Dispersion? 74

3.12 Measures of Dispersion 75

Range 75 Mean Deviation 76

Exercises 79

Variance and Standard Deviation 79

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Contents xxiii

Exercises 82

3.13 Software Solution 84

Exercises 84

3.14 Interpretation and Uses of the Standard Deviation 85

Chebyshev’s Theorem 85 The Empirical Rule 86

Exercises 87

3.15 The Mean and Standard Deviation of Grouped Data 88

The Arithmetic Mean 88 Standard Deviation 89

Exercises 91

3.16 Ethics and Reporting Results 92

Chapter Summary 92

Pronunciation Key 94

Chapter Exercises 94

Data Set Exercises 99

Software Commands 100

Answers to Self-Review 100

Chapter

4 Describing Data: Displaying and Exploring Data 102 4.1 Introduction 103

4.2 Dot Plots 103

4.3 Stem-and-Leaf Displays 105

Exercises 109

4.4 Measures of Position 111

Quartiles, Deciles, and Percentiles 111

Exercises 115

Box Plots 116

Exercises 118

4.5 Skewness 119

Exercises 123

4.6 Describing the Relationship between Two Variables 124

Exercises 127

Chapter Summary 129

Pronunciation Key 129

Chapter Exercises 130

Data Set Exercises 135

Software Commands 135

Answers to Self-Review 136

A Review of Chapters 1–4 137

Glossary 137

Problems 139

Cases 141

Practice Test 142

Chapter

5 A Survey of Probability Concepts 144 5.1 Introduction 145

5.2 What Is a Probability? 146

5.3 Approaches to Assigning Probabilities 148

Classical Probability 148 Empirical Probability 149 Subjective Probability 150

Exercises 152

5.4 Some Rules for Computing Probabilities 153

Rules of Addition 153

Exercises 158

Rules of Multiplication 159

5.5 Contingency Tables 162

5.6 Tree Diagrams 164

Exercises 166

5.7 Bayes’ Theorem 167

Exercises 170

5.8 Principles of Counting 171

The Multiplication Formula 171 The Permutation Formula 172 The Combination Formula 174

Exercises 176

Chapter Summary 176

Pronunciation Key 177

Chapter Exercises 178

Data Set Exercises 182

Software Commands 183

Answers to Self-Review 184

Chapter

6 Discrete Probability Distributions 186 6.1 Introduction 187

6.2 What Is a Probability Distribution? 187

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xxiv Contents

6.3 Random Variables 189

Discrete Random Variable 190 Continuous Random Variable 190

6.4 The Mean, Variance, and Standard Deviation of a Discrete Probability Distribution 191

Mean 191 Variance and Standard Deviation 191

Exercises 193

6.5 Binomial Probability Distribution 195

How Is a Binomial Probability Computed? 196 Binomial Probability Tables 198

Exercises 201

Cumulative Binomial Probability Distributions 202

Exercises 203

6.6 Hypergeometric Probability Distribution 204

Exercises 207

6.7 Poisson Probability Distribution 207

Exercises 212

Chapter Summary 212

Chapter Exercises 213

Data Set Exercises 218

Software Commands 219

Answers to Self-Review 221

Chapter

7 Continuous Probability Distributions 222 7.1 Introduction 223

7.2 The Family of Uniform Probability Distributions 223

Exercises 226

7.3 The Family of Normal Probability Distributions 227

7.4 The Standard Normal Probability Distribution 229

Applications of the Standard Normal Distribution 231 The Empirical Rule 231

Exercises 233

Finding Areas under the Normal Curve 233

Exercises 236

Exercises 239

Exercises 241

7.5 The Normal Approximation to the Binomial 242

Continuity Correction Factor 242 How to Apply the Correction Factor 244

Exercises 245

7.6 The Family of Exponential Distributions 246

Exercises 250

Chapter Summary 251

Chapter Exercises 252

Data Set Exercises 256

Software Commands 256

Answers to Self-Review 257

A Review of Chapters 5–7 258

Glossary 259

Problems 260

Cases 261

Practice Test 263

Chapter

8 Sampling Methods and the Central Limit Theorem 265 8.1 Introduction 266

8.2 Sampling Methods 266

Reasons to Sample 266 Simple Random Sampling 267 Systematic Random Sampling 270 Stratified Random Sampling 270 Cluster Sampling 271

Exercises 272

8.3 Sampling “Error” 274

8.4 Sampling Distribution of the Sample Mean 275

Exercises 278

8.5 The Central Limit Theorem 279

Exercises 285

8.6 Using the Sampling Distribution of the Sample Mean 286

Exercises 289

Chapter Summary 289

Pronunciation Key 290

Chapter Exercises 290

Data Set Exercises 295

Software Commands 295

Answers to Self-Review 296

Lin01803_fm_i-xxx.qxd 11/15/10 3:34 PM Page xxiv

Contents xxv

Chapter

9 Estimation and Confidence Intervals 297 9.1 Introduction 298

9.2 Point Estimate for a Population Mean 298

9.3 Confidence Intervals for a Population Mean 299

Population Standard Deviation Known � 300 A Computer Simulation 304

Exercises 305

Population Standard Deviation � Unknown 306

Exercises 312

9.4 A Confidence Interval for a Proportion 313

Exercises 316

9.5 Choosing an Appropriate Sample Size 316

Sample Size to Estimate a Population Mean 317 Sample Size to Estimate a Population Proportion 318

Exercises 320

9.6 Finite-Population Correction Factor 320

Exercises 322

Chapter Summary 323

Chapter Exercises 323

Data Set Exercises 327

Software Commands 328

Answers to Self-Review 329

A Review of Chapters 8 and 9 329

Glossary 330

Problems 331

Case 332

Practice Test 332

Chapter

10 One-Sample Tests of Hypothesis 333 10.1 Introduction 334

10.2 What Is a Hypothesis? 334

10.3 What Is Hypothesis Testing? 335

10.4 Five-Step Procedure for Testing a Hypothesis 335

Step 1: State the Null Hypothesis (H0 ) and the Alternate Hypothesis (H1) 336 Step 2: Select a Level of Significance 337 Step 3: Select the Test Statistic 338 Step 4: Formulate the Decision Rule 338 Step 5: Make a Decision 339

10.5 One-Tailed and Two-Tailed Tests of Significance 340

10.6 Testing for a Population Mean: Known Population Standard Deviation 341

A Two-Tailed Test 341 A One-Tailed Test 345

10.7 p-Value in Hypothesis Testing 345

Exercises 347

10.8 Testing for a Population Mean: Population Standard Deviation Unknown 348

Exercises 352

A Software Solution 353

Exercises 355

10.9 Tests Concerning Proportions 356

Exercises 359

10.10 Type II Error 359

Exercises 362

Chapter Summary 362

Pronunciation Key 363

Chapter Exercises 364

Data Set Exercises 368

Software Commands 369

Answers to Self-Review 369

Chapter

11 Two-Sample Tests of Hypothesis 371 11.1 Introduction 372

11.2 Two-Sample Tests of Hypothesis: Independent Samples 372

Exercises 377

11.3 Two-Sample Tests about Proportions 378

Exercises 381

11.4 Comparing Population Means with Unknown Population Standard Deviations 382

Equal Population Standard Deviations 383

Exercises 386

Unequal Population Standard Deviations 388

Lin01803_fm_i-xxx.qxd 11/15/10 11:06 AM Page xxv

Exercises 391

11.5 Two-Sample Tests of Hypothesis: Dependent Samples 392

11.6 Comparing Dependent and Independent Samples 395

Exercises 398

Chapter Summary 399

Pronunciation Key 400

Chapter Exercises 400

Data Set Exercises 406

Software Commands 407

Answers to Self-Review 408

Chapter

12 Analysis of Variance 410 12.1 Introduction 411

12.2 The F Distribution 411

12.3 Comparing Two Population Variances 412

Exercises 415

12.4 ANOVA Assumptions 416

12.5 The ANOVA Test 418

Exercises 425

12.6 Inferences about Pairs of Treatment Means 426

Exercises 429

12.7 Two-Way Analysis of Variance 430

Exercises 434

12.8 Two-Way ANOVA with Interaction 435

Interaction Plots 436 Hypothesis Test for Interaction 437

Exercises 440

Chapter Summary 442

Pronunciation Key 443

Chapter Exercises 443

Data Set Exercises 451

Software Commands 452

Answers to Self-Review 454

A Review of Chapters 10–12 455

Glossary 455

Problems 456

Cases 459

Practice Test 459

Chapter

13 Correlation and Linear Regression 461 13.1 Introduction 462

13.2 What Is Correlation Analysis? 463

13.3 The Correlation Coefficient 465

Exercises 470

13.4 Testing the Significance of the Correlation Coefficient 472

Exercises 475

13.5 Regression Analysis 476

Least Squares Principle 476 Drawing the Regression Line 479

Exercises 481

13.6 Testing the Significance of the Slope 483

Exercises 486

13.7 Evaluating a Regression Equation’s Ability to Predict 486

The Standard Error of Estimate 486 The Coefficient of Determination 487

Exercises 488

Relationships among the Correlation Coefficient, the Coefficient of Determination, and the Standard Error of Estimate 488

Exercises 490

13.8 Interval Estimates of Prediction 490

Assumptions Underlying Linear Regression 490 Constructing Confidence and Prediction Intervals 492

Exercises 494

13.9 Transforming Data 495

Exercises 497

Chapter Summary 498

Pronunciation Key 499

Chapter Exercises 500

Data Set Exercises 509

Software Commands 510

Answers to Self-Review 511

Chapter

14 Multiple Regression Analysis 512 14.1 Introduction 513

14.2 Multiple Regression Analysis 513

xxvi Contents

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Contents xxvii

Exercises 517

14.3 Evaluating a Multiple Regression Equation 519

The ANOVA Table 519 Multiple Standard Error of Estimate 520 Coefficient of Multiple Determination 521 Adjusted Coefficient of Determination 522

Exercises 523

14.4 Inferences in Multiple Linear Regression 523

Global Test: Testing the Multiple Regression Model 524 Evaluating Individual Regression Coefficients 526

Exercises 530

14.5 Evaluating the Assumptions of Multiple Regression 531

Linear Relationship 532 Variation in Residuals Same for Large and Small Values 533 Distribution of Residuals 534 Multicollinearity 534 Independent Observations 537

14.6 Qualitative Independent Variables 537

14.7 Regression Models with Interaction 540

14.8 Stepwise Regression 542

Exercises 544

14.9 Review of Multiple Regression 546

Chapter Summary 551

Pronunciation Key 553

Chapter Exercises 553

Data Set Exercises 565

Software Commands 566

Answers to Self-Review 567

A Review of Chapters 13 and 14 567

Glossary 568

Problems 569

Cases 570

Practice Test 571

Chapter

15 Index Numbers 573 15.1 Introduction 574

15.2 Simple Index Numbers 574

15.3 Why Convert Data to Indexes? 577

15.4 Construction of Index Numbers 577

Exercises 578

15.5 Unweighted Indexes 579

Simple Average of the Price Indexes 579 Simple Aggregate Index 580

15.6 Weighted Indexes 581

Laspeyres Price Index 581 Paasche Price Index 582 Fisher’s Ideal Index 584

Exercises 584

15.7 Value Index 585

Exercises 586

15.8 Special-Purpose Indexes 587

Consumer Price Index 588 Producer Price Index 589 Dow Jones Industrial Average (DJIA) 589 S&P 500 Index 590

Exercises 591

15.9 Consumer Price Index 592

Special Uses of the Consumer Price Index 592

15.10 Shifting the Base 595

Exercises 597

Chapter Summary 598

Chapter Exercises 599

Software Commands 602

Answers to Self-Review 603

Chapter

16 Time Series and Forecasting 604 16.1 Introduction 605

16.2 Components of a Time Series 605

Secular Trend 605 Cyclical Variation 606 Seasonal Variation 607 Irregular Variation 608

16.3 A Moving Average 608

16.4 Weighted Moving Average 611

Exercises 614

16.5 Linear Trend 615

16.6 Least Squares Method 616

Exercises 618

16.7 Nonlinear Trends 618

Exercises 620

16.8 Seasonal Variation 621

Determining a Seasonal Index 621

Ŷ

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Exercises 626

16.9 Deseasonalizing Data 627

Using Deseasonalized Data to Forecast 628

Exercises 630

16.10 The Durbin-Watson Statistic 631

Exercises 636

Chapter Summary 636

Chapter Exercises 636

Data Set Exercise 643

Software Commands 643

Answers to Self-Review 644

A Review of Chapters 15 and 16 645

Glossary 646

Problems 646

Practice Test 647

Chapter

17 Nonparametric Methods: Goodness-of-Fit Tests 648 17.1 Introduction 649

17.2 Goodness-of-Fit Test: Equal Expected Frequencies 649

Exercises 654

17.3 Goodness-of-Fit Test: Unequal Expected Frequencies 655

17.4 Limitations of Chi-Square 657

Exercises 659

17.5 Testing the Hypothesis That a Distribution of Data Is from a Normal Population 659

17.6 Graphical and Statistical Approaches to Confirm Normality 662

Exercises 665

17.7 Contingency Table Analysis 667

Exercises 671

Chapter Summary 672

Pronunciation Key 672

Chapter Exercises 672

Data Set Exercises 677

Software Commands 678

Answers to Self-Review 679

Chapter

18 Nonparametric Methods: Analysis of Ranked Data 680 18.1 Introduction 681

18.2 The Sign Test 681

Exercises 685

Using the Normal Approximation to the Binomial 686

Exercises 688

Testing a Hypothesis about a Median 688

Exercises 689

18.3 Wilcoxon Signed-Rank Test for Dependent Samples 690

Exercises 693

18.4 Wilcoxon Rank-Sum Test for Independent Samples 695

Exercises 698

18.5 Kruskal-Wallis Test: Analysis of Variance by Ranks 698

Exercises 702

18.6 Rank-Order Correlation 704

Testing the Significance of rs 706

Exercises 707

Chapter Summary 709

Pronunciation Key 710

Chapter Exercises 710

Data Set Exercises 713

Software Commands 713

Answers to Self-Review 714

A Review of Chapters 17 and 18 716

Glossary 716

Problems 717

Cases 718

Practice Test 718

Chapter

19 Statistical Process Control and Quality Management 720 19.1 Introduction 721

19.2 A Brief History of Quality Control 721

Six Sigma 724

19.3 Causes of Variation 724

xxviii Contents

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Contents xxix

19.4 Diagnostic Charts 725

Pareto Charts 725 Fishbone Diagrams 727

Exercises 728

19.5 Purpose and Types of Quality Control Charts 729

Control Charts for Variables 729 Range Charts 733

19.6 In-Control and Out-of-Control Situations 734

Exercises 736

19.7 Attribute Control Charts 737

Percent Defective Charts 737 c-Bar Charts 740

Exercises 741

19.8 Acceptance Sampling 742

Exercises 746

Chapter Summary 746

Pronunciation Key 747

Chapter Exercises 747

Software Commands 751

Answers to Self-Review 752

On the website: www.mhhe.com/lind15e Chapter

20 An Introduction to Decision Theory 20.1 Introduction

20.2 Elements of a Decision

20.3 A Case Involving Decision Making under Conditions of Uncertainty

Payoff Table Expected Payoff

Exercises

Opportunity Loss

Exercises

Expected Opportunity Loss

Exercises

20.4 Maximin, Maximax, and Minimax Regret Strategies

20.5 Value of Perfect Information

20.6 Sensitivity Analysis

Exercises

20.7 Decision Trees

Chapter Summary

Chapter Exercises

Answers to Self-Review

Appendixes 753

Appendix A: Data Sets 754

Appendix B: Tables 764

Appendix C: Answers to Odd-Numbered Chapter Exercises and Review Exercises and Solutions to Practice Tests 782

Photo Credits 829

Index 831

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Goals When you have completed this chapter, you will be able to:

1 Organize data into a fre- quency distribution.

2 Portray a frequency distribu- tion in a histogram, frequency polygon, and cumulative fre- quency polygon.

3 Present data using such graphical techniques as line charts, bar charts, and pie charts.FPO

1 Learning Objectives When you have completed this chapter, you will be able to:

LO1 List ways that statistics is used.

LO2 Know the differences between descriptive and inferential statistics.

LO3 Understand the differ- ences between a sample and a population.

LO4 Explain the difference between qualitative and quan- titative variables.

LO5 Compare the differences between discrete and continu- ous variables.

LO6 Recognize the levels of measurement in data.

What Is Statistics?

Barnes & Noble stores recently began selling the Nook. With this

device, you can download over 1,500 books electronically and read

the book on a small monitor instead of purchasing the book. Assume

you have the number of Nooks sold each day for the last month at the

Barnes & Noble store at the Market Commons Mall in Riverside,

California. Describe a condition in which this information could be

considered a sample. Illustrate a second situation in which the same

data would be regarded as a population. (See Exercise 11 and LO3.)

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2 Chapter 1

1.1 Introduction More than 100 years ago, H. G. Wells, an English author and historian, suggested that one day quantitative reasoning will be as necessary for effective citizenship as the ability to read. He made no mention of business because the Industrial Revo- lution was just beginning. Mr. Wells could not have been more correct. While “busi-

ness experience,” some “thoughtful guesswork,” and “intuition” are key attributes of successful managers, today’s business problems tend to be too complex for this type of decision making alone.

One of the tools used to make decisions is statistics. Statistics is used not only by businesspeople; we all also apply statistical concepts in our lives. For example, to start the day you turn on the shower and let it run for a few moments. Then you put your hand in the shower to sample the temperature and decide to add more hot water or more cold water, or determine that the temperature is just right and then enter the shower. As a second example, suppose you are at Costco Wholesale and wish to buy a frozen pizza. One of the pizza makers has a stand, and they offer a small wedge of their pizza. After sampling the pizza, you

decide whether to purchase the pizza or not. In both the shower and pizza examples, you make a decision and select a course of action based on a sample.

Businesses face similar situations. The Kellogg Company must ensure that the mean amount of Raisin Bran in the 25.5-gram box meets label specifications. To do so, it sets a “target” weight somewhat higher than the amount specified on the label. Each box is then weighed after it is filled. The weighing machine reports a distribu- tion of the content weights for each hour as well as the number “kicked-out” for being under the label specification during the hour. The Quality Inspection Depart- ment also randomly selects samples from the production line and checks the qual- ity of the product and the weight of the contents of the box. If the mean product weight differs significantly from the target weight or the percent of kick-outs is too large, the process is adjusted.

As a student of business or economics, you will need basic knowledge and skills to organize, analyze, and transform data and to present the information. In this text, we will show you basic statistical techniques and methods that will develop your ability to make good personal and business decisions.

1.2 Why Study Statistics? If you look through your university catalog, you will find that statistics is required for many college programs. Why is this so? What are the differences in the sta- tistics courses taught in the Engineering College, the Psychology or Sociology Departments in the Liberal Arts College, and the College of Business? The biggest difference is the examples used. The course content is basically the same. In the College of Business we are interested in such things as profits, hours worked, and wages. Psychologists are interested in test scores, and engineers are interested in how many units are manufactured on a particular machine. However, all three are interested in what is a typical value and how much variation there is in the data. There may also be a difference in the level of mathematics required. An engi- neering statistics course usually requires calculus. Statistics courses in colleges of business and education usually teach the course at a more applied level. You should be able to handle the mathematics in this text if you have completed high school algebra.

So why is statistics required in so many majors? The first reason is that numer- ical information is everywhere. Look in the newspapers (USA Today), news maga- zines (Time, Newsweek, U.S. News and World Report), business magazines (Busi- nessWeek, Forbes), or general interest magazines (People), women’s magazines

Examples of why we study statistics

LO1 List ways that statistics is used.

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What Is Statistics? 3

(Ladies Home Journal or Elle), or sports magazines (Sports Illustrated, ESPN The Magazine), and you will be bombarded with numerical information.

Here are some examples:

• The average increase in weekly earnings, in 1982–84 dollars, from January 2009 to January 2010 was $8.32.

• In January 2010 the average amount of credit card debt per household was $7,394. This is a decrease from $7,801 in July 2009. A 2010 Federal Reserve survey found that 75 percent of U.S. households have at least one credit card.

• The following table summarizes the number of commercial aircraft manufactured by Boeing, Inc. between 2006 and 2009.

USA TODAY Snapshot

By Jae Yang and Paul Trap, USA TODAY Source: SnagAJob.com Reprinted with permission (April 29, 2010) USA TODAY.

Sales of Boeing Aircraft

Type of Aircraft

Year 737 747 767 777 787 Total

2006 733 72 8 77 160 1,050 2007 850 25 36 143 369 1,423 2008 488 4 29 54 94 669 2009 197 5 7 30 24 263

• Go to the following website: www.youtube.com/watch?v=pMcfrLYDm2U. It provides interesting numerical information about countries, business, geogra- phy, and politics.

• USA Today (www.usatoday.com) prints “Snapshots” that are the result of sur- veys conducted by various research organizations, foundations, and the federal government. The following chart summarizes what recruiters look for in hiring seasonal employees.

A second reason for taking a statistics course is that statistical techniques are used to make decisions that affect our daily lives. That is, they affect our personal welfare. Here are a few examples:

• Insurance companies use statistical analysis to set rates for home, automobile, life, and health insurance. Tables are available showing estimates that a 20-year- old female has 60.25 years of life remaining, an 87-year-old woman 4.56 years remaining, and a 50-year-old man 27.85 years remaining. Life insurance premi- ums are established based on these estimates of life expectancy. These tables are available at www.ssa.gov/OACT/STATS/table4cb.html. [This site is sensitive to capital letters.]

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• The Environmental Protection Agency is interested in the water quality of Lake Erie as well as other lakes. They periodically take water samples to establish the level of contamination and maintain the level of quality.

• Medical researchers study the cure rates for diseases using different drugs and different forms of treatment. For example, what is the effect of treating a cer- tain type of knee injury surgically or with physical therapy? If you take an aspirin each day, does that reduce your risk of a heart attack?

A third reason for taking a statistics course is that the knowledge of statistical methods will help you understand how decisions are made and give you a better understanding of how they affect you.

No matter what line of work you select, you will find yourself faced with deci- sions where an understanding of data analysis is helpful. In order to make an informed decision, you will need to be able to:

1. Determine whether the existing information is adequate or additional informa- tion is required.

2. Gather additional information, if it is needed, in such a way that it does not pro- vide misleading results.

3. Summarize the information in a useful and informative manner. 4. Analyze the available information. 5. Draw conclusions and make inferences while assessing the risk of an incorrect

conclusion.

The statistical methods presented in the text will provide you with a framework for the decision-making process.

In summary, there are at least three reasons for studying statistics: (1) data are everywhere, (2) statistical techniques are used to make many decisions that affect our lives, and (3) no matter what your career, you will make professional decisions that involve data. An understanding of statistical methods will help you make these decisions more effectively.

1.3 What Is Meant by Statistics? How do we define the word statistics? We encounter it frequently in our everyday language. It really has two meanings. In the more common usage, statistics refers to numerical information. Examples include the average starting salary of college graduates, the number of deaths due to alcoholism last year, the change in the Dow Jones Industrial Average from yesterday to today, and the number of home runs hit by the Chicago Cubs during the 2010 season. In these examples, statistics are a value or a percentage. Other examples include:

• The typical automobile in the United States travels 11,099 miles per year, the typical bus 9,353 miles per year, and the typical truck 13,942 miles per year. In Canada, the corresponding information is 10,371 miles for automobiles, 19,823 miles for buses, and 7,001 miles for trucks.

• The mean time waiting for technical support is 17 minutes. • The mean length of the business cycle since 1945 is 61 months.

The above are all examples of statistics. A collection of numerical information is called statistics (plural).

We frequently present statistical information in a graphical form. A graph is often useful for capturing reader attention and to portray a large amount of information. For example, Chart 1–1 shows Frito-Lay volume and market share for the major snack and potato chip categories in supermarkets in the United States. It requires only a quick glance to discover there were nearly 800 million pounds of potato chips sold and that Frito-Lay sold 64 percent of that total. Also note that Frito-Lay has 82 percent of the corn chip market.

4 Chapter 1

Statistics in Action

We call your atten- tion to a feature title—Statistics in Action. Read each one carefully to get an appreciation of the wide application of statistics in man- agement, economics, nursing, law enforce- ment, sports, and other disciplines.

• In 2009, Forbes published a list of the richest Ameri- cans. William Gates, founder of Microsoft Corpo- ration, is the rich- est. His net worth is estimated at $59.0 billion. (www.forbes.com)

• In 2009, the four largest American companies, ranked by revenue, were Walmart, Exxon- Mobil, Chevron, and General Electric. (www. forbes.com)

• In the United States, a typical high school gradu- ate earns $1.2 mil- lion in his or her lifetime, a typical college graduate with a bachelor’s degree earns $2.1 million, and a typical college graduate with a master’s degree earns $2.5 million. (usgovinfo.about. com/library/weekly /aa072602a.htm)

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• Research analysts for Merrill Lynch evaluate many facets of a particular stock before making a “buy” or “sell” recommendation. They collect the past sales data of the company and estimate future earnings. Other factors, such as the projected worldwide demand for the company’s products, the strength of the com- petition, and the effect of the new union–management contract, are also considered before making a recommendation.

• The marketing department at Colgate-Palmolive Co., a manufac- turer of soap products, has the responsibility of making recom- mendations regarding the potential profitability of a newly devel- oped group of face soaps having fruit smells, such as grape, orange, and pineapple. Before making a final decision, the mar- keters will test it in several markets. That is, they may advertise and sell it in Topeka, Kansas, and Tampa, Florida. On the basis of test marketing in these two regions, Colgate-Palmolive will make a decision whether to market the soaps in the entire country.

• Managers must make decisions about the quality of their product or service. For example, customers call software companies for technical advice when they are not able to resolve an issue regarding the software. One measure of the quality of customer service is the time a customer must wait for a technical consultant to answer the call. A software company might set a target of one minute as the typical response time. The company would then collect and ana- lyze data on the response time. Does the typical response time differ by day of the week or time of day? If the response times are increasing, managers might decide to increase the number of technical consultants at particular times of the day or week.

What Is Statistics? 5

STATISTICS The science of collecting, organizing, presenting, analyzing, and interpreting data to assist in making more effective decisions.

As the definition suggests, the first step in investigating a problem is to collect relevant data. They must be organized in some way and perhaps presented in a chart, such as Chart 1–1. Only after the data have been organized are we then able to analyze and interpret them. Here are some examples of the need for data collection.

Frito-Lay

Rest of Industry

0 100 200 300 400

Millions of Pounds

500 600 700 800

Potato Chips

Tortilla Chips

Pretzels

Extruded Snacks

Corn Chips

64%

75%

26%

56%

82%

CHART 1–1 Frito-Lay Volume and Share of Major Snack Chip Categories in U.S. Supermarkets

The subject of statistics, as we will explore it in this text, has a much broader meaning than just collecting and publishing numerical information. We define sta- tistics as:

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6 Chapter 1

1.4 Types of Statistics The study of statistics is usually divided into two categories: descriptive statistics and inferential statistics.

Descriptive Statistics The definition of statistics given earlier referred to “organizing, presenting, analyz- ing . . . data.” This facet of statistics is usually referred to as descriptive statistics.

DESCRIPTIVE STATISTICS Methods of organizing, summarizing, and presenting data in an informative way.

For instance, the United States government reports the population of the United States was 179,323,000 in 1960; 203,302,000 in 1970; 226,542,000 in 1980; 248,709,000 in 1990; 265,000,000 in 2000; and 308,400,000 in 2010. This informa- tion is descriptive statistics. It is descriptive statistics if we calculate the percent- age growth from one decade to the next. However, it would not be descriptive sta- tistics if we used these to estimate the population of the United States in the year 2020 or the percentage growth from 2010 to 2020. Why? The reason is these sta- tistics are not being used to summarize past populations but to estimate future pop- ulations. The following are some other examples of descriptive statistics.

• There are a total of 46,837 miles of interstate highways in the United States. The interstate system represents only 1 percent of the nation’s total roads but carries more than 20 percent of the traffic. The longest is I-90, which stretches from Boston to Seattle, a distance of 3,099 miles. The shortest is I-878 in New York City, which is 0.70 of a mile in length. Alaska does not have any interstate highways, Texas has the most interstate miles at 3,232, and New York has the most interstate routes with 28.

• The average person spent $103.00 on traditional Valentine’s Day merchandise in 2010. This is an increase of $0.50 from 2009. As in previous years, men will spend nearly twice the amount women spend on the holiday. The average man spent $135.35 to impress the people in his life while women only spent $72.28. Family pets will also feel the love, the average person spending $3.27 on their furry friends, up from $2.17 last year.

Masses of unorganized data—such as the census of population, the weekly earnings of thousands of computer programmers, and the individual responses of 2,000 registered voters regarding their choice for president of the United States— are of little value as is. However, statistical techniques are available to organize this type of data into a meaningful form. Data can be organized into a frequency dis- tribution. (This procedure is covered in Chapter 2.) Various charts may be used to describe data; several basic chart forms are also presented in Chapter 4.

Specific measures of central location, such as the mean, describe the central value of a group of numerical data. A number of statistical measures are used to describe how closely the data cluster about an average. These measures of central tendency and dispersion are discussed in Chapter 3.

Inferential Statistics The second type of statistics is inferential statistics—also called statistical infer- ence. Our main concern regarding inferential statistics is finding something about a population from a sample taken from that population. For example, a recent survey showed only 46 percent of high school seniors can solve problems involving fractions,

LO2 Know the differences between descriptive and inferential statistics.

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What Is Statistics? 7

decimals, and percentages; and only 77 percent of high school seniors correctly totaled the cost of a salad, burger, fries, and a cola on a restaurant menu. Since these are inferences about a population (all high school seniors) based on sample data, we refer to them as inferential statistics. You might think of inferential statistics as a “best guess” of a population value based on sample information.

INFERENTIAL STATISTICS The methods used to estimate a property of a population on the basis of a sample.

SAMPLE A portion, or part, of the population of interest.

POPULATION The entire set of individuals or objects of interest or the measurements obtained from all individuals or objects of interest.

Note the words population and sample in the definition of inferential statistics. We often make reference to the population of 308.8 million people living in the United States or the 1,336.1 million people living in China. However, in statistics the word population has a broader meaning. A population may consist of individuals—such as all the stu- dents enrolled at Utah State University, all the students in Accounting 201, or all the CEOs from the Fortune 500 companies. A population may also consist of objects, such as all the Cobra G/T tires produced at Cooper Tire and Rubber Company in the Find- lay, Ohio, plant; the accounts receivable at the end of October for Lorrange Plastics, Inc.; or auto claims filed in the first quarter of 2010 at the Northeast Regional Office of State Farm Insurance. The measurement of interest might be the scores on the first examination of all students in Accounting 201, the tread wear of the Cooper Tires, the dollar amount of Lorrange Plastics’s accounts receivable, or the amount of auto insur- ance claims at State Farm. Thus, a population in the statistical sense does not always refer to people.

To infer something about a population, we usually take a sample from the population.

Why take a sample instead of studying every member of the population? A sam- ple of registered voters is necessary because of the prohibitive cost of contacting millions of voters before an election. Testing wheat for moisture content destroys the wheat, thus making a sample imperative. If the wine tasters tested all the wine, none would be available for sale. It would be physically impossible for a few marine biologists to capture and tag all the seals in the ocean. (These and other reasons for sampling are discussed in Chapter 8.)

As noted, using a sample to learn something about a population is done exten- sively in business, agriculture, politics, and government, as cited in the following examples:

• Television networks constantly monitor the popularity of their programs by hir- ing Nielsen and other organizations to sample the preferences of TV viewers. For example, in a sample of 800 prime-time viewers, 320, or 40 percent, indi- cated they watched American Idol on Fox last week. These program ratings are used to set advertising rates or to cancel programs.

• Gamous and Associates, a public accounting firm, is conducting an audit of Pronto Printing Company. To begin, the accounting firm selects a random sam- ple of 100 invoices and checks each invoice for accuracy. There is at least one error on five of the invoices; hence the accounting firm estimates that 5 per- cent of the population of invoices contain at least one error.

Reasons for sampling

LO3 Understand the differences between a sample and a population.

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1.5 Types of Variables There are two basic types of variables: (1) qualitative and (2) quantitative (see Chart 1–2). When the characteristic being studied is nonnumeric, it is called a qualita- tive variable or an attribute. Examples of qualitative variables are gender, religious affiliation, type of automobile owned, state of birth, and eye color. When the data are qualitative, we are usually interested in how many or what percent fall in each category. For example, what percent of the population has blue eyes? What per- cent of the total number of cars sold last month were SUVs? Qualitative data are often summarized in charts and bar graphs (Chapter 2).

8 Chapter 1

• A random sample of 1,260 marketing graduates from four-year schools showed their mean starting salary was $42,694. We therefore estimate the mean starting salary for all marketing graduates of four-year institutions to be $42,694.

The relationship between a sample and a population is portrayed below. For example, we wish to estimate the mean miles per gallon of SUVs. Six SUVs are selected from the population. The mean MPG of the six is used to estimate MPG for the population.

Population All items

Sample Items selected

from the population

Following is a self-review problem. There are a number of them interspersed throughout each chapter. They test your comprehension of the preceding material. The answer and method of solution are given at the end of the chapter. You can find the answer to the following Self-Review on page 19. We recommend that you solve each one and then check your answer.

We strongly suggest you do the Self-Review exercise.

Self-Review 1–1 The answers are at the end of the chapter. The Atlanta-based advertising firm, Brandon and Associates, asked a sample of 1,960 consumers to try a newly developed chicken dinner by Boston Market. Of the 1,960 sampled, 1,176 said they would purchase the dinner if it is marketed. (a) What could Brandon and Associates report to Boston Market regarding acceptance of

the chicken dinner in the population? (b) Is this an example of descriptive statistics or inferential statistics? Explain.

Qualitative variable

LO4 Explain the difference between qualitative and quantitative variables.

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What Is Statistics? 9

Quantitative variable

Types of Variables

Qualitative Quantitative

ContinuousDiscrete • Brand of PC • Marital status • Hair color

• Children in a family • Strokes on a golf hole • TV sets owned

• Amount of income tax paid • Weight of a student • Yearly rainfall in Tampa, FL

CHART 1–2 Summary of the Types of Variables

When the variable studied can be reported numerically, the variable is called a quantitative variable. Examples of quantitative variables are the balance in your checking account, the ages of company presidents, the life of an automobile bat- tery (such as 42 months), and the number of children in a family.

Quantitative variables are either discrete or continuous. Discrete variables can assume only certain values, and there are “gaps” between the values. Examples of discrete variables are the number of bedrooms in a house (1, 2, 3, 4, etc.), the num- ber of cars arriving at Exit 25 on I-4 in Florida near Walt Disney World in an hour (326, 421, etc.), and the number of students in each section of a statistics course (25 in section A, 42 in section B, and 18 in section C). We count, for example, the number of cars arriving at Exit 25 on I-4, and we count the number of statistics stu- dents in each section. Notice that a home can have 3 or 4 bedrooms, but it can- not have 3.56 bedrooms. Thus, there is a “gap” between possible values. Typically, discrete variables result from counting.

Observations of a continuous variable can assume any value within a specific range. Examples of continuous variables are the air pressure in a tire and the weight of a shipment of tomatoes. Other examples are the amount of raisin bran in a box and the duration of flights from Orlando to San Diego. Grade point average (GPA) is a continuous variable. We could report the GPA of a particular student as 3.2576952. The usual practice is to round to 3 places—3.258. Typically, continuous variables result from measuring.

1.6 Levels of Measurement Data can be classified according to levels of measurement. The level of measurement of the data dictates the calculations that can be done to summarize and present the data. It will also deter- mine the statistical tests that should be performed. For example, there are six colors of candies in a bag of M&M’s. Suppose we assign brown a value of 1, yellow 2, blue 3, orange 4, green 5, and red 6. From a bag of candies, we add the assigned color val- ues and divide by the number of candies and report that the mean color is 3.56. Does this mean that the average color is blue or orange? Of course not! As a second example, in a high school track meet there are eight competitors in the 400-meter run. We

LO5 Compare the differences between discrete and continuous variables.

LO6 Recognize the levels of measurement in data.

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USA TODAY Snapshot

03/15/2007-updated 11:51 PM ET

Workers say they prefer higher salaries

to any other perks.

By Anne R. Carey and Chad Palmer, USA Today Source: hudson-index.com Reprinted with permission (March 15, 2007) USA TODAY.

10 Chapter 1

report the order of finish and that the mean finish is 4.5. What does the mean fin- ish tell us? Nothing! In both of these instances, we have not properly used the level of measurement.

There are actually four levels of measurement: nominal, ordinal, interval, and ratio. The lowest, or the most primitive, measurement is the nominal level. The high- est, or the level that gives us the most information about the observation, is the ratio level of measurement.

Nominal-Level Data For the nominal level of measurement, observations of a qualitative variable can only be classified and counted. There is no particular order to the labels. The classifica- tion of the six colors of M&M’s milk chocolate candies is an example of the nomi- nal level of measurement. We simply classify the candies by color. There is no nat- ural order. That is, we could report the brown candies first, the orange first, or any of the colors first. Gender is another example of the nominal level of measurement. Suppose we count the number of students entering a football game with a student ID and report how many are men and how many are women. We could report either the men or the women first. For the nominal level, the only measurement involved consists of counts. Sometimes, for better reader understanding, we convert these counts to percentages. The following “Snapshot” from USA Today shows the results from a survey of workers. The variable of interest is “Perks” and there are five pos- sible outcomes: “More money,” “Better healthcare,” “Better retirement,” “Work/ family balance,” and, we will assume, “Other.” The outcome “Other” is not shown on the chart, but is necessary to make the percent of respondents total 100 percent. There is no natural order to the outcomes, we could have put “Better healthcare” first instead of “More money.”

To process the data, such as the information regarding worker perks, or infor- mation on gender, employment by industry, or state of birth of a student, we often numerically code the information. That is, we assign students from Alabama a code of 1, Alaska a code of 2, Arizona as 3, and so on. Using this procedure, Wiscon- sin is coded 49 and Wyoming 50. This coding facilitates counting by a computer. However, because we have assigned numbers to the various categories, this does not give us license to manipulate the numbers. To explain, 1 � 2 does not equal 3; that is, Alabama � Alaska does not yield Arizona.

To summarize, the nominal level has the following properties:

1. The variable of interest is divided into categories or outcomes. 2. There is no natural order to the outcomes.

Statistics in Action

Where did statistics get its start? In 1662 John Graunt pub- lished an article called “Natural and Political Observa- tions Made upon Bills of Mortality.” The author’s “obser- vations” were the re- sult of a study and analysis of a weekly church publication called “Bill of Mor- tality,” which listed births, christenings, and deaths and their causes. Graunt real- ized that the Bills of Mortality repre- sented only a frac- tion of all births and deaths in London. However, he used the data to reach broad conclusions about the impact of disease, such as the plague, on the general popu- lation. His logic is an example of statistical inference. His analy- sis and interpretation of the data are thought to mark the start of statistics.

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What Is Statistics? 11

Ordinal-Level Data The next higher level of data is the ordinal level. Table 1–1 lists the student ratings of Professor James Brunner in an Introduction to Finance course. Each student in the class answered the question “Overall, how did you rate the instructor in this class?” The variable rating illustrates the use of the ordinal scale of measurement. One clas- sification is “higher” or “better” than the next one. That is, “Superior” is better than “Good,” “Good” is better than “Average,” and so on. However, we are not able to distinguish the magnitude of the differences between groups. Is the difference between “Superior” and “Good” the same as the difference between “Poor” and “Infe- rior”? We cannot tell. If we substitute a 5 for “Superior” and a 4 for “Good,” we can conclude that the rating of “Superior” is better than the rating of “Good,” but we can- not add a ranking of “Superior” and a ranking of “Good,” with the result being mean- ingful. Further we cannot conclude that a rating of “Good” (rating is 4) is necessarily twice as high as a “Poor” (rating is 2). We can only conclude that a rating of “Good” is better than a rating of “Poor.” We cannot conclude how much better the rating is.

TABLE 1–1 Rating of a Finance Professor

Rating Frequency

Superior 6 Good 28 Average 25 Poor 12 Inferior 3

Another example of ordinal-level data is the Homeland Security Advisory System. The Department of Homeland Security publishes this information regarding the risk of terrorist activity to federal, state, and local authorities and to the American people. The five risk levels from lowest to highest, including a description and color codes, are shown to the left.

This is an example of the ordinal scale because we know the order or the ranks of the risk levels—that is, orange is higher than yellow—but the amount of the difference in risk is not necessarily the same. To put it another way, the difference in the risk level between yellow and orange is not necessarily the same as between green and blue.

In summary, the properties of the ordinal level of data are:

1. Data classifications are represented by sets of labels or names (high, medium, low) that have relative values.

2. Because of the relative values, the data classified can be ranked or ordered.

Interval-Level Data The interval level of measurement is the next highest level. It includes all the characteristics of the ordinal level, but in addition, the difference between val- ues is a constant size. An example of the interval level of measurement is tem- perature. Suppose the high temperatures on three consecutive winter days in Boston are 28, 31, and 20 degrees Fahrenheit. These temperatures can be eas- ily ranked, but we can also determine the difference between temperatures. This is possible because 1 degree Fahrenheit represents a constant unit of measure- ment. Equal differences between two temperatures are the same, regardless of their position on the scale. That is, the difference between 10 degrees Fahrenheit

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12 Chapter 1

and 15 degrees is 5, the difference between 50 and 55 degrees is also 5 degrees. It is also important to note that 0 is just a point on the scale. It does not repre- sent the absence of the condition. Zero degrees Fahrenheit does not represent the absence of heat, just that it is cold! In fact 0 degrees Fahrenheit is about �18 degrees on the Celsius scale.

Another example of the interval scale of measurement is women’s dress sizes. Listed below is information on several dimensions of a standard U.S. women’s dress.

Size Bust (in) Waist (in) Hips (in)

8 32 24 35 10 34 26 37 12 36 28 39 14 38 30 41 16 40 32 43 18 42 34 45 20 44 36 47 22 46 38 49 24 48 40 51 26 50 42 53 28 52 44 55

Why is the “size” scale an interval measurement? Observe as the size changes by 2 units (say from size 10 to size 12 or from size 24 to size 26) each of the measurements increases by 2 inches. To put it another way, the intervals are the same.

There is no natural zero point for dress size. A “size 0” dress does not have “zero” material. Instead, it would have a 24-inch bust, 16-inch waist, and 27-inch hips. Moreover, the ratios are not reasonable. If you divide a size 28 by a size 14, you do not get the same answer as dividing a size 20 by 10. Neither ratio is equal to two as the “size” number would suggest. In short, if the distances between the numbers make sense, but the ratios do not, then you have an interval scale of measurement.

The properties of the interval-level data are:

1. Data classifications are ordered according to the amount of the characteristic they possess.

2. Equal differences in the characteristic are represented by equal differences in the measurements.

Ratio-Level Data Practically all quantitative data is recorded on the ratio level of measurement. The ratio level is the “highest” level of measurement. It has all the characteristics of the interval level, but in addition, the 0 point is meaningful and the ratio between two numbers is meaningful. Examples of the ratio scale of measurement include wages, units of production, weight, changes in stock prices, distance between branch offices, and height. Money is a good illustration. If you have zero dollars, then you have no money. Weight is another example. If the dial on the scale of a correctly calibrated device is at 0, then there is a complete absence of weight. The ratio of two numbers is also meaningful. If Jim earns $40,000 per year selling insur- ance and Rob earns $80,000 per year selling cars, then Rob earns twice as much as Jim.

Table 1–2 illustrates the use of the ratio scale of measurement. It shows the incomes of four father-and-son combinations.

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What Is Statistics? 13

TABLE 1–2 Father–Son Income Combinations

Name Father Son

Lahey $80,000 $ 40,000 Nale 90,000 30,000 Rho 60,000 120,000 Steele 75,000 130,000

Observe that the senior Lahey earns twice as much as his son. In the Rho fam- ily, the son makes twice as much as the father.

In summary, the properties of the ratio-level data are:

1. Data classifications are ordered according to the amount of the characteristics they possess.

2. Equal differences in the characteristic are represented by equal differences in the numbers assigned to the classifications.

3. The zero point is the absence of the characteristic and the ratio between two numbers is meaningful.

Chart 1–3 summarizes the major characteristics of the various levels of measurement.

Levels of Measurement

RatioNominal Ordinal Interval

Meaningful 0 point and ratio between values

Data may only be classified

Data are ranked Meaningful difference between values

• Jersey numbers of football players • Make of car

• Your rank in class • Team standings in the Pac-10

• Temperature • Dress size

• Number of patients seen • Number of sales calls made • Distance to class

CHART 1–3 Summary of the Characteristics for Levels of Measurement

Self-Review 1–2 What is the level of measurement reflected by the following data? (a) The age of each person in a sample of 50 adults who listen to one of the 1,230 talk

radio stations in the United States is:

35 29 41 34 44 46 42 42 37 47 30 36 41 39 44 39 43 43 44 40 47 37 41 27 33 33 39 38 43 22 44 39 35 35 41 42 37 42 38 43 35 37 38 43 40 48 42 31 51 34

(b) In a survey of 200 luxury-car owners, 100 were from California, 50 from New York, 30 from Illinois, and 20 from Ohio.

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Exercises The answers to the odd-numbered exercises are at the end of the book.

1. What is the level of measurement for each of the following variables? a. Student IQ ratings. b. Distance students travel to class. c. The jersey numbers of a sorority soccer team. d. A classification of students by state of birth. e. A summary of students by academic class—that is, freshman, sophomore, junior, and

senior. f. Number of hours students study per week.

2. What is the level of measurement for these items related to the newspaper business? a. The number of papers sold each Sunday during 2011. b. The departments, such as editorial, advertising, sports, etc. c. A summary of the number of papers sold by county. d. The number of years with the paper for each employee.

3. Look in the latest edition of USA Today or your local newspaper and find examples of each level of measurement. Write a brief memo summarizing your findings.

4. For each of the following, determine whether the group is a sample or a population. a. The participants in a study of a new cholesterol drug. b. The drivers who received a speeding ticket in Kansas City last month. c. Those on welfare in Cook County (Chicago), Illinois. d. The 30 stocks reported as a part of the Dow Jones Industrial Average.

1.7 Ethics and Statistics Following events such as Wall Street money manager Bernie Madoff’s Ponzi scheme, which swindled billions from investors, and financial misrepresentations by Enron and Tyco, business students need to understand that these events were based on the misrepresentation of business and financial data. In each case, people within each organization reported financial information to investors that indicated the companies were performing much better than the actual situation. When the true financial infor- mation was reported, the companies were worth much less than advertised. The result was many investors lost all or nearly all of the money they put into these companies.

The article “Statistics and Ethics: Some Advice for Young Statisticians,” in The American Statistician 57, no. 1 (2003), offers guidance. The authors advise us to prac- tice statistics with integrity and honesty, and urge us to “do the right thing” when collecting, organizing, summarizing, analyzing, and interpreting numerical information. The real contribution of statistics to society is a moral one. Financial analysts need to provide information that truly reflects a company’s performance so as not to mis- lead individual investors. Information regarding product defects that may be harmful to people must be analyzed and reported with integrity and honesty. The authors of The American Statistician article further indicate that when we practice statistics, we need to maintain “an independent and principled point-of-view.”

As you progress through this text, we will highlight ethical issues in the collec- tion, analysis, presentation, and interpretation of statistical information. We also hope that, as you learn about using statistics, you will become a more informed consumer of information. For example, you will question a report based on data that do not fairly represent the population, a report that does not include all relevant statistics, one that includes an incorrect choice of statistical measures, or a presentation that introduces the writer’s bias in a deliberate attempt to mislead or misrepresent.

1.8 Computer Applications Computers are now available to students at most colleges and universities. Spreadsheets, such as Microsoft Excel, and statistical software packages, such as Minitab, are available in most computer labs. The Microsoft Excel package is

14 Chapter 1

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What Is Statistics? 15

bundled with many home computers. In this text, we use both Excel and Minitab for the applications. We also use an Excel add-in called MegaStat. This add-in gives Excel the capability to produce additional statistical reports.

The following example shows the application of computers in statistical analy- sis. In Chapters 2, 3, and 4, we illustrate methods for summarizing and describing data. An example used in these chapters refers to profit, as well as other variables, on each of the 180 vehicles sold last month by the Applewood Auto Group. The following Excel output reveals, among other things, (1) there were 180 vehicles sold, the mean (average) profit per vehicle was $1,843.17, and the amount of profit ranged from $294 to $3,292.

The following output is from the Minitab system. It contains much of the same information.

Had we used a calculator to arrive at these measures and others needed to fully analyze the selling prices, hours of calculation would have been required. The

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16 Chapter 1

likelihood of an error in arithmetic is high when a large number of values are con- cerned. On the other hand, statistical software packages and spreadsheets can provide accurate information in seconds.

At the option of your instructor, and depending on the software system avail- able, we urge you to apply a computer package to the exercises in the Data Set Exercises section in each chapter. It will relieve you of the tedious calculations and allow you to concentrate on data analysis.

Chapter Summary I. Statistics is the science of collecting, organizing, presenting, analyzing, and interpreting

data to assist in making more effective decisions. II. There are two types of statistics.

A. Descriptive statistics are procedures used to organize and summarize data. B. Inferential statistics involve taking a sample from a population and making estimates

about a population based on the sample results. 1. A population is an entire set of individuals or objects of interest or the measure-

ments obtained from all individuals or objects of interest. 2. A sample is a part of the population.

III. There are two types of variables. A. A qualitative variable is nonnumeric.

1. Usually we are interested in the number or percent of the observations in each category.

2. Qualitative data are usually summarized in graphs and bar charts. B. There are two types of quantitative variables and they are usually reported numerically.

1. Discrete variables can assume only certain values, and there are usually gaps between values.

2. A continuous variable can assume any value within a specified range. IV. There are four levels of measurement.

A. With the nominal level, the data are sorted into categories with no particular order to the categories.

B. The ordinal level of measurement presumes that one classification is ranked higher than another.

C. The interval level of measurement has the ranking characteristic of the ordinal level of measurement plus the characteristic that the distance between values is a con- stant size.

D. The ratio level of measurement has all the characteristics of the interval level, plus there is a 0 point and the ratio of two values is meaningful.

Chapter Exercises 5. Explain the difference between qualitative and quantitative variables. Give an example of

qualitative and quantitative variables. 6. Explain the difference between a sample and a population. 7. Explain the difference between a discrete and a continuous variable. Give an example of

each not included in the text. 8. For the following questions, would you collect information using a sample or a popula-

tion? Why? a. Statistics 201 is a course taught at a university. Professor A. Verage has taught nearly

1,500 students in the course over the past 5 years. You would like to know the aver- age grade for the course.

b. As part of a research project, you need to report the average profitability of the num- ber one corporation in the Fortune 500 for the past 10 years.

c. You are looking forward to graduation and your first job as a salesperson for one of five large pharmaceutical corporations. Planning for your interviews, you will need to know about each company’s mission, profitability, products, and markets.

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What Is Statistics? 17

d. You are shopping for a new MP3 music player such as the Apple iPod. The manu- facturers advertise the number of music tracks that can be stored in the memory. Usu- ally, the advertisers assume relatively short, popular music to estimate the number of tracks that can be stored. You, however, like Broadway musical tunes and they are much longer. You would like to estimate how many Broadway tunes will fit on your MP3 player.

9. Exits along interstate highways were formerly numbered successively from the western or southern edge of a state. However, the Department of Transportation has recently changed most of them to agree with the numbers on the mile markers along the highway. a. What level of measurement were data on the consecutive exit numbers? b. What level of measurement are data on the milepost numbers? c. Discuss the advantages of the newer system.

10. A poll solicits a large number of college undergraduates for information on the following variables: the name of their cell phone provider (AT&T, Verizon, and so on), the numbers of minutes used last month (200, 400, for example), and their satisfaction with the ser- vice (Terrible, Adequate, Excellent, and so forth). What is the data scale for each of these three variables?

11. Barnes & Noble stores recently began selling the Nook. With this device, you can download over 1,500 books electronically and read the book on a small monitor instead of purchasing the book. Assume you have the number of Nooks sold each day for the last month at the Barnes & Noble store at the Market Commons Mall in Riverside, California. Describe a condition in which this information could be considered a sample. Illustrate a second situation in which the same data would be regarded as a population.

12. Utilize the concepts of sample and population to describe how a presidential election is unlike an “exit” poll of the electorate.

13. Place these variables in the following classification tables. For each table, summarize your observations and evaluate if the results are generally true. For example, salary is reported as a continuous quantitative variable. It is also a continuous ratio-scaled variable. a. Salary b. Gender c. Sales volume of MP3 players d. Soft drink preference e. Temperature f. SAT scores g. Student rank in class h. Rating of a finance professor i. Number of home computers

Discrete Variable Continuous Variable

Qualitative

Quantitative a. Salary

Discrete Continuous

Nominal

Ordinal

Interval

Ratio a. Salary

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18 Chapter 1

Year-to-Date Sales

Through February Through February Manufacturer 2010 2009

General Motors Corp. 287,242 252,701 Ford Motor Company 249,514 185,825 Chrysler LLC 141,592 146,207 Toyota Motor Sales USA Inc. 198,823 226,870 American Honda Motor Co. Inc. 148,150 142,606 Nissan North America Inc. 132,761 108,133 Hyundai Motor America 64,507 55,133 Mazda Motor of America Inc. 32,748 31,821

a. Compare the total sales for the eight automakers. Has there been an increase or a decrease in sales for 2010 compared to the same period in 2009?

b. Compute the market share for each of the companies. Has there been a large change in the market share for any of the companies?

c. Compare the percentage increases for each of the eight companies. What significant changes are there from 2009 to 2010 for each of the companies?

18. The following chart depicts the average amounts spent by consumers on holiday gifts.

14. Using data from such publications as the Statistical Abstract of the United States, The World Almanac, Forbes, or your local newspaper, give examples of the nominal, ordinal, interval, and ratio levels of measurement.

15. The Struthers Wells Corporation employs more than 10,000 white collar workers in its sales offices and manufacturing facilities in the United States, Europe, and Asia. A sam- ple of 300 of these workers revealed 120 would accept a transfer to a location outside the United States. On the basis of these findings, write a brief memo to Ms. Wanda Carter, Vice President of Human Services, regarding all white collar workers in the firm and their willingness to relocate.

16. AVX Stereo Equipment, Inc., recently began a “no-hassles” return policy. A sample of 500 customers who had recently returned items showed 400 thought the policy was fair, 32 thought it took too long to complete the transaction, and the rest had no opin- ion. On the basis of this information, make an inference about customer reaction to the new policy.

17. The following table reports the number of cars and light duty trucks sold by the eight largest automakers in the first two months of 2010 compared to the first two months of 2009.

Write a brief report summarizing the amounts spent during the holidays. Be sure to include the total amount spent, and the percent spent by each group.

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What Is Statistics? 19

Data Set Exercises 20. Refer to the Real Estate data at the back of the text, which report information on homes

sold in the Goodyear, Arizona, area last year. Consider the following variables: selling price, number of bedrooms, township, and distance from the center of the city. a. Which of the variables are qualitative and which are quantitative? b. Determine the level of measurement for each of the variables.

21. Refer to the Baseball 2009 data, which report information on the 30 Major League Base- ball teams for the 2009 season. Consider the following variables: number of wins, team salary, season attendance, whether the team is in the American or National League, and the number of home runs hit. a. Which of these variables are quantitative and which are qualitative? b. Determine the level of measurement for each of the variables.

22. Refer to the Buena School District bus data, which report information on the school dis- trict’s bus fleet. a. Which of the variables are qualitative and which are quantitative? b. Determine the level of measurement for each variable.

19. The following chart depicts the earnings in billions of dollars for ExxonMobil for the period 2003 until 2009. Write a brief report discussing the earnings at ExxonMobil during the period. Was one year higher than the others? Did the earnings increase, decrease, or stay the same over the period?

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20 Chapter 1

Chapter 1 Answers to Self-Review

1–1 a. On the basis of the sample of 1,960 consumers, we estimate that, if it is marketed, 60 percent of all consumers will purchase the chicken dinner (1,176/1,960) � 100 � 60 percent.

b. Inferential statistics, because a sample was used to draw a conclusion about how all consumers in the population would react if the chicken dinner were marketed.

1–2 a. Age is a ratio-scale variable. A 40-year-old is twice as old as someone 20 years old.

b. Nominal scale. We could arrange the states in any order.

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2 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Make a frequency table for a set of data.

LO2 Organize data into a bar chart.

LO3 Present a set of data in a pie chart.

LO4 Create a frequency dis- tribution for a data set.

LO5 Understand a relative frequency distribution.

LO6 Present data from a fre- quency distribution in a his- togram or frequency polygon.

LO7 Construct and interpret a cumulative frequency distribution.

Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation

Merrill Lynch recently completed a study of online investment portfolios

for a sample of clients. For the 70 participants in the study, organize

these data into a frequency distribution. (See Exercise 43 and LO4.)

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22 Chapter 2

2.1 Introduction The highly competitive automobile retailing industry in the United States has changed dramatically in recent years. These changes spurred events such as the:

• bankruptcies of General Motors and Chrysler in 2009. • elimination of well-known brands such as Pontiac and

Saturn. • closing of over 1,500 local dealerships. • collapse of consumer credit availability. • consolidation dealership groups.

ships in large regions of the country. As these groups acquire the local dealerships, they often bring standardized selling practices, common software and hardware technology platforms, and management reporting techniques. The goal of these new organizations is to provide an improved buying experience for the consumer, while increasing profitability. Megadealerships often employ over 10,000 people, generate several billion dollars in annual sales, own more than 50 franchises, and are traded on the New York Stock Exchange or NASDAQ. Today, the largest megadealership is AutoNation (ticker symbol AN). Others include Penske Auto Group (PAG and second largest), Asbury Automotive Group (ABG), and Hendrick Auto Group (which is privately held).

The Applewood Auto Group is an ownership group that includes four deal- erships. The group sells a wide range of vehicles, including the inexpensive but popular Korean brands Kia and Hyundai, BMW and Volvo sedans and luxury SUVs, and a full line of Ford and Chevrolet cars and trucks.

Ms. Kathryn Ball is a member of the senior management team at Applewood Auto Group, which has its corporate offices adjacent to Hilltop Motors. She is responsible for tracking and analyzing vehicle sales and the profitability of those vehicles. Kathryn would like to summarize the profit earned on the vehicles sold with tables, charts, and graphs that she would review monthly. She wants to know the profit per vehicle sold, as well as the lowest and highest amount of profit. She is also interested in describing the demographics of the buyers. What are their ages? How many vehicles have they previously purchased from one of the Applewood dealerships? What type of vehicle did they purchase?

The Applewood Auto Group oper- ates four dealerships:

• Tionesta Ford Lincoln Mercury sells the Ford, Lincoln, and Mercury cars and trucks.

• Olean Automotive Inc. has the Nis- san franchise as well as the General Motors brands of Chevrolet, Cadillac, and GMC Trucks.

• Sheffield Motors Inc. sells Buick, GMC trucks, Hyundai, and Kia.

• Hilltop Motors offers the Chrysler, Dodge, and Jeep line as well as BMW and Volvo.

Every month, Ms. Ball collects data from each of the four dealerships and enters it

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 23

into an Excel spreadsheet. Last month the Applewood Auto Group sold 180 vehicles at the four dealerships. A copy of the first few observations appears at the bottom of the previous page. The variables collected include:

• Profit—the amount earned by the dealership on the sale of each vehicle. • Age—the age of the buyer at the time of the purchase. • Location—the dealership where the vehicle was purchased. • Vehicle type—SUV, sedan, compact, hybrid, or truck. • Previous—the number of vehicles previously purchased at any of the four

Applewood dealerships by the consumer.

The entire data set is available at the McGraw-Hill website and in Appendix A.5 at the end of the text.

2.2 Constructing a Frequency Table Recall from Chapter 1 that techniques used to describe a set of data are called descriptive statistics. To put it another way, descriptive statistics organize data to show the general pattern of the data and where values tend to concentrate and to expose extreme or unusual data values. The first procedure we discuss is a fre- quency table.

FREQUENCY TABLE A grouping of qualitative data into mutually exclusive classes showing the number of observations in each class.

In Chapter 1, we distinguished between qualitative and quantitative variables. To review, a qualitative variable is nonnumeric, that is, it can only be classified into distinct categories. There is no particular order to these categories. Examples of qualitative data include political affiliation (Republican, Democrat, Independent), state of birth (Alabama, . . . , Wyoming), and method of payment for a purchase at Barnes & Noble (cash, check, debit, or credit). On the other hand, quantitative variables are numerical in nature. Examples of quantitative data relating to college students include the price of their textbooks, their age, and the hours they spend studying each week of the semester.

In the Applewood Auto Group data set, there are five variables for each vehicle sale: age of the buyer, amount of profit, dealer that made the sale, type of vehicle sold, and number of previous purchases by the buyer. The dealer and the type of

vehicle are qualitative variables. The amount of profit, the age of the buyer, and the number of previous purchases are quantita- tive variables.

Suppose Ms. Ball wanted to summarize last month’s sales by location. To summarize this qualitative data, we classify the vehicles sold last month according to their location: Tionesta, Olean, Sheffield, or Hilltop. We use location to develop a fre- quency table with four mutually exclusive (distinctive) classes. This means that a particular vehicle cannot belong to more than one class. Each vehicle is uniquely classified into one of the four mutually exclusive locations. This frequency table is shown in Table 2–1. The number of observations, representing the sales at each location, is called the class frequency. So the class fre- quency for vehicles sold at the Kane location is 52.

Relative Class Frequencies You can convert class frequencies to relative class frequencies to show the fraction of the total number of observations in each class. A relative frequency captures the

LO1 Make a frequency table for a set of data.

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relationship between a class total and the total number of observations. In the vehi- cle sales example, we may want to know the percentage of total cars sold at each of the four locations. To convert a frequency distribution to a relative frequency dis- tribution, each of the class frequencies is divided by the total number of observa- tions. For example, the fraction of vehicles sold last month at the Kane location is 0.289, found by 52 divided by 180. The relative frequency for each location is shown in Table 2–2.

24 Chapter 2

TABLE 2–2 Relative Frequency Table of Vehicles Sold by Type Last Month at Applewood Auto Group

Location Number of Cars Relative Frequency

Kane 52 .289 Olean 40 .222 Sheffield 45 .250 Tionesta 43 .239

Total 180 1.000

Graphic Presentation of Qualitative Data The most common graphic form to present a qualitative variable is a bar chart. In most cases, the horizontal axis shows the variable of interest. The vertical axis shows the frequency or fraction of each of the possible outcomes. A distinguishing feature of a bar chart is there is distance or a gap between the bars. That is, because the variable of interest is qualitative, the bars are not adjacent to each other. Thus, a bar chart graphically describes a frequency table using a series of uniformly wide rectangles, where the height of each rectangle is the class frequency.

BAR CHART A graph that shows qualitative classes on the horizontal axis and the class frequencies on the vertical axis. The class frequencies are proportional to the heights of the bars.

We use the Applewood Auto Group data as an example (Chart 2–1). The vari- able of interest is the location where the vehicle was sold and the number of vehi- cles sold at each location is the class frequency. We label the horizontal axis with the four locations and scale the vertical axis with the number sold. The height of the bars, or rectangles, corresponds to the number of vehicles at each location. There were 52 vehicles sold last month at the Kane location, so the height of the Kane bar is 52; the height of the bar for the Olean location is 40. The variable

LO2 Organize data into a bar chart.

TABLE 2–1 Frequency Table for Vehicles Sold Last Month at Applewood Auto Group by Location

Location Number of Cars

Kane 52 Olean 40 Sheffield 45 Tionesta 43

Total 180

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 25

location is of nominal scale, so the order of the locations on the horizontal axis does not matter. Listing this variable alphabetically or by some type of geograph- ical arrangement might also be appropriate.

Another useful type of chart for depicting qualitative information is a pie chart.

PIE CHART A chart that shows the proportion or percentage that each class represents of the total number of frequencies.

TABLE 2–3 Ohio State Lottery Expenses in 2009

Amount Percentage Use of Sales ($ million) of Sales

Prizes 1,460.0 60 Education 702.3 29 Bonuses 150.0 6 Expenses 124.3 5

Total 2,436.6 100

We explain the details of constructing a pie chart using the information in Table 2–3, which shows a breakdown of the expenses for the Ohio State Lottery in 2009.

The first step to develop a pie chart is to record the percentages 0, 5, 10, 15, and so on evenly around the circumference of a circle (see Chart 2–2). To plot the 60 percent share awarded for prizes, draw a line from the center of the circle to 0 and another line from the center of the circle to 60 percent. The area in this “slice” represents the lottery proceeds that were awarded in prizes. Next, add the 60 per- cent of expenses awarded in prizes to the 29 percent payments to education; the result is 89 percent. Draw a line from the center of the circle to 89 percent, so the area between 60 percent and 89 percent depicts the payments made to education.

Nu m

be r o

f V eh

ic le

s So

0 Kane Olean

Location

Sheffield Tionesta

CHART 2–1 Number of Vehicles Sold by Location

LO3 Present a set of data using a pie chart.

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26 Chapter 2

CHART 2–2 Pie Chart of Ohio Lottery Expenses in 2009

25%

50%

60%

75%

89%

95% 0% Operating ExpensesBonuses/

Commissions

PrizesEducation

Continuing, add the 6 percent for bonuses, which gives us a total of 95 percent. Draw a line from the center of the circle to 95, so the “slice” between 89 percent and 95 percent represents the payment of bonuses. The remainder, 5 percent, is for operating expenses.

Because each slice of the pie represents the relative share of each component, we can easily compare them:

• The largest expense of the Ohio Lottery is for prizes. • About 30 percent of the proceeds is transferred to education. • Operating expenses account for only 5 percent of the proceeds.

We can use software to quickly create a visually appealing and informative pie chart. The following chart, using the information in Table 2–3, depicts the uses of Ohio Lottery expenses in 2009.

Pie charts and bar charts serve much the same function. What are the criteria for selecting one over the other? In most cases, pie charts are the most informative

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 27

Solution

SkiLodges.com is test marketing its new website and is interested in how easy its Web page design is to navigate. It randomly selected 200 regular Internet users and asked them to perform a search task on the Web page. Each person was asked to rate the relative ease of navigation as poor, good, excellent, or awesome. The results are shown in the following table:

1. What type of measurement scale is used for ease of navigation? 2. Draw a bar chart for the survey results. 3. Draw a pie chart for the survey results.

The data are measured on an ordinal scale. That is, the scale is ranked in relative ease when moving from “poor” to “awesome.” Also, the interval between each rating is unknown so it is impossible, for example, to conclude that a rating of good is twice the value of a poor rating.

We can use a bar chart to graph the data. The vertical scale shows the rel- ative frequency and the horizontal shows the values of the ease of navigation scale.

Awesome 102 Excellent 58 Good 30 Poor 10

Re la

tiv e

Fr eq

ue nc

y %

0 Poor Good Excellent Awesome

Ease of Navigation of SkiLodges.com Web Page

Ease of Navigation

when the goal is to compare the relative difference in the percentage of observa- tions for each of the nominal scale variables. Bar charts are preferred when the goal is to compare the number of observations in each category.

A pie chart can also be used to graph this data. The pie chart emphasizes that more than half of the respondents rate the relative ease of using the website awesome.

Example

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Exercises The answers to the odd-numbered exercises are at the end of the book.

1. A pie chart shows the relative market share of cola products. The “slice” for Pepsi-Cola has a central angle of 90 degrees. What is its market share?

2. In a marketing study, 100 consumers were asked to select the best digital music player from the iPod, the iRiver, and the Magic Star MP3. To summarize the consumer responses with a frequency table, how many classes would the frequency table have?

3. A total of 1,000 residents in Minnesota were asked which season they preferred. The results were 100 liked winter best, 300 liked spring, 400 liked summer, and 200 liked fall. If the data were summarized in a frequency table, how many classes would be used? What would be the relative frequencies for each class?

4. Two thousand frequent midwestern business travelers are asked which midwestern city they prefer: Indianapolis, Saint Louis, Chicago, or Milwaukee. The results were 100 liked Indianapolis best, 450 liked Saint Louis, 1,300 liked Chicago, and the remainder preferred Milwaukee. Develop a frequency table and a relative frequency table to summarize this information.

28 Chapter 2

Poor 5%

Ease of Navigation of SkiLodges.com Web Page

Good 15%

Awesome 51% Excellent

29%

Self-Review 2–1 The answers are at the end of the chapter.

DeCenzo Specialty Food and Beverage Company has been serving a cola drink with an additional flavoring, Cola-Plus, that is very popular among its customers. The company is interested in customer preferences for Cola-Plus versus Coca-Cola, Pepsi, and a lemon- lime beverage. They ask 100 randomly sampled customers to take a taste test and select the beverage they preferred most. The results are shown in the following table:

(a) Is the data qualitative or quantitative? Why? (b) What is the table called? What does it show? (c) Develop a bar chart to depict the information. (d) Develop a pie chart using the relative frequencies.

Beverage Number

Cola-Plus 40 Coca-Cola 25 Pepsi 20 Lemon-Lime 15

Total 100

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 29

5. Wellstone Inc. produces and markets replacement covers for cell phones in a variety of colors. The company would like to allocate its production plans to five different colors: bright white, metallic black, magnetic lime, tangerine orange, and fusion red. The com- pany set up a kiosk in the Mall of America for several hours and asked randomly selected people which cover color was their favorite. The results follow:

a. What is the table called? b. Draw a bar chart for the table. c. Draw a pie chart. d. If Wellstone Inc. plans to produce 1 million cell phone covers, how many of each color

should it produce? 6. A small business consultant is investigating the performance of several companies. The

fourth quarter sales for last year (in thousands of dollars) for the selected companies were:

The consultant wants to include a chart in his report comparing the sales of the six com- panies. Use a bar chart to compare the fourth-quarter sales of these corporations and write a brief report summarizing the bar chart.

2.3 Constructing Frequency Distributions: Quantitative Data In Chapter 1 and earlier in this chapter, we distinguished between qualitative and quantitative data. In the previous section, using the Applewood Automotive Group data, we summarized a qualitative variable, location of the sale, using a frequency table, a relative frequency table, and a bar chart.

The Applewood Auto Group data also includes several quantitative variables: the age of the buyer, the profit earned on the sale of the vehicle, and the number of previous purchases. Suppose Ms. Ball wants to summarize last month’s sales by profit earned. We can describe profit using a frequency distribution.

Fourth-Quarter Sales Corporation ($ thousands)

Hoden Building Products $ 1,645.2 J & R Printing Inc. 4,757.0 Long Bay Concrete Construction 8,913.0 Mancell Electric and Plumbing 627.1 Maxwell Heating and Air Conditioning 24,612.0 Mizelle Roofing & Sheet Metals 191.9

Bright white 130 Metallic black 104 Magnetic lime 325 Tangerine orange 455 Fusion red 286

FREQUENCY DISTRIBUTION A grouping of data into mutually exclusive classes showing the number of observations in each class.

LO4 Create a frequency distribution for a data set.

How do we develop a frequency distribution? The first step is to tally the data into a table that shows the classes and the number of observations in each class. The steps in constructing a frequency distribution are best described by an example. Remember, our goal is to construct tables, charts, and graphs that will quickly reveal the concentration, extreme values, and shape of the data.

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30 Chapter 2

Example

Solution Table 2–4 shows the profits from the 180 sales. We refer to this unorganized infor- mation as raw data or ungrouped data. With a little searching, we can find the smallest profit ($294) and the highest profit ($3,292), but that is about all. It is diffi- cult to determine a typical amount of profit. It is also difficult to visualize where the profits tend to cluster. The raw data are more easily interpreted if organized into a frequency distribution.

Step 1: Decide on the number of classes. The goal is to use just enough group- ings or classes to reveal the shape of the set of observations. Some judg- ment is needed here. Too many classes or too few classes might not reveal the basic shape of the data set. In the vehicle profit example, three classes would not give much insight into the pattern of the data (see Table 2–5).

We return to the situation where Ms. Kathryn Ball of the Applewood Auto Group wants to develop tables, charts, and graphs to show the typical profit for each sale. Table 2–4 reports the profit on each of the 180 vehicles sold last month at the four Applewood locations. What is the typical profit on each sale? What is the largest profit on any sale? What is the lowest profit on any sale? Around what value did the profits tend to cluster?

TABLE 2–4 Profit on Vehicles Sold Last Month by the Applewood Auto Group

$1,387 $2,148 $2,201 $ 963 $ 820 $2,230 $3,043 $2,584 $2,370 1,754 2,207 996 1,298 1,266 2,341 1,059 2,666 2,637 1,817 2,252 2,813 1,410 1,741 3,292 1,674 2,991 1,426 1,040 1,428 323 1,553 1,772 1,108 1,807 934 2,944 1,273 1,889 352 1,648 1,932 1,295 2,056 2,063 2,147 1,529 1,166 482 2,071 2,350 1,344 2,236 2,083 1,973 3,082 1,320 1,144 2,116 2,422 1,906 2,928 2,856 2,502 1,951 2,265 1,485 1,500 2,446 1,952 1,269 2,989 783 2,692 1,323 1,509 1,549 369 2,070 1,717 910 1,538 1,206 1,761 1,638 2,348 978 2,454 1,797 1,536 2,339 1,342 1,919 1,961 2,498 1,238 1,606 1,955 1,957 2,700

443 2,357 2,127 294 1,818 1,680 2,199 2,240 2,222 754 2,866 2,430 1,115 1,824 1,827 2,482 2,695 2,597

1,621 732 1,704 1,124 1,907 1,915 2,701 1,325 2,742 870 1,464 1,876 1,532 1,938 2,084 3,210 2,250 1,837

1,174 1,626 2,010 1,688 1,940 2,639 377 2,279 2,842 1,412 1,761 2,165 1,822 2,197 842 1,220 2,626 2,434 1,809 1,915 2,231 1,897 2,646 1,963 1,401 1,501 1,640 2,415 2,119 2,389 2,445 1,461 2,059 2,175 1,752 1,821 1,546 1,766 335 2,886 1,731 2,338 1,118 2,058 2,487

The steps for organizing data into a frequency distribution.

Vehicle Profit ($) Number of Vehicles

$ 200 up to $1,400 42 1,400 up to 2,600 115 2,600 up to 3,800 23

Total 180

TABLE 2–5 An Example of Too Few Classes

Lowest

Highest

A useful recipe to determine the number of classes (k) is the “2 to the k rule.” This guide suggests you select the smallest number (k) for the number of classes such that 2k (in words, 2 raised to the power of k) is greater than the number of observations (n). In the Applewood Auto Group

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 31

example, there were 180 vehicles sold. So n � 180. If we try k � 7, which means we would use 7 classes, 27 � 128, which is less than 180. Hence, 7 is too few classes. If we let k � 8, then 28 � 256, which is greater than 180. So the recommended number of classes is 8.

Step 2: Determine the class interval or class width. Generally the class interval or class width is the same for all classes. The classes all taken together must cover at least the distance from the lowest value in the data up to the highest value. Expressing these words in a formula:

where i is the class interval, H is the highest observed value, L is the lowest observed value, and k is the number of classes.

For the Applewood Auto Group, the lowest value is $294 and the highest value is $3,292. If we need 8 classes, the interval should be:

In practice, this interval size is usually rounded up to some convenient number, such as a multiple of 10 or 100. The value of $400 is a rea- sonable choice.

In frequency distributions, equal class intervals are preferred. How- ever, unequal class intervals may be necessary in certain situations to avoid a large number of empty, or almost empty, classes. Such is the case in Table 2–6. The Internal Revenue Service used unequal-sized class intervals to report the adjusted gross income on individual tax returns. Had they used an equal-sized interval of, say, $1,000, more than 1,000 classes would have been required to describe all the incomes. A frequency distribution with 1,000 classes would be difficult to inter- pret. In this case, the distribution is easier to understand in spite of the unequal classes. Note also that the number of income tax returns or “frequencies” is reported in thousands in this particular table. This also makes the information easier to understand.

i � H � L

k �

$3,292 � $294 8

� $374.75

i � H � L

TABLE 2–6 Adjusted Gross Income for Individuals Filing Income Tax Returns

Number of Returns Adjusted Gross Income (in thousands)

No adjusted gross income 178.2 $ 1 up to $ 5,000 1,204.6

5,000 up to 10,000 2,595.5 10,000 up to 15,000 3,142.0 15,000 up to 20,000 3,191.7 20,000 up to 25,000 2,501.4 25,000 up to 30,000 1,901.6 30,000 up to 40,000 2,502.3 40,000 up to 50,000 1,426.8 50,000 up to 75,000 1,476.3 75,000 up to 100,000 338.8

100,000 up to 200,000 223.3 200,000 up to 500,000 55.2 500,000 up to 1,000,000 12.0

1,000,000 up to 2,000,000 5.1 2,000,000 up to 10,000,000 3.4

10,000,000 or more 0.6

Statistics in Action

In 1788, James Madison, John Jay, and Alexander Hamilton anony- mously published a series of essays enti- tled The Federalist. These Federalist pa- pers were an attempt to convince the peo- ple of New York that they should ratify the Constitution. In the course of history, the authorship of most of these papers became known, but 12 remained con- tested. Through the use of statistical analysis, and particu- larly the study of the frequency of the use of various words, we can now conclude that James Madison is the likely author of the 12 papers. In fact, the statistical evidence that Madi- son is the author is overwhelming.

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32 Chapter 2

Step 3: Set the individual class limits. State clear class limits so you can put each observation into only one category. This means you must avoid overlapping or unclear class limits. For example, classes such as “$1,300–$1,400” and “$1,400–$1,500” should not be used, because it is not clear whether the value $1,400 is in the first or second class. Classes stated as “$1,300–$1,400” and “$1,500–$1,600” are frequently used, but may also be confusing without the additional common convention of rounding all data at or above $1,450 up to the second class and data below $1,450 down to the first class. In this text, we will generally use the format $1,300 up to $1,400 and $1,400 up to $1,500 and so on. With this format it is clear that $1,399 goes into the first class and $1,400 in the second.

Because we round the class interval up to get a convenient class size, we cover a larger than necessary range. For example, using 8 classes with a width of $400 in the Applewood Auto Group example results in a range of 8($400) � $3,200. The actual range is $2,998, found by ($3,292 � $294). Comparing that value to $3,200, we have an excess of $202. Because we need to cover only the distance (H � L), it is nat- ural to put approximately equal amounts of the excess in each of the two tails. Of course, we should also select convenient class limits. A guideline is to make the lower limit of the first class a multiple of the class interval. Sometimes this is not possible, but the lower limit should at least be rounded. So here are the classes we could use for this data.

Classes

$ 200 up to $ 600 600 up to 1,000

1,000 up to 1,400 1,400 up to 1,800 1,800 up to 2,200 2,200 up to 2,600 2,600 up to 3,000 3,000 up to 3,400

Step 4: Tally the vehicle profit into the classes. To begin, the profit from the sale of the first vehicle in Table 2–4 is $1,387. It is tallied in the $1,000 up to $1,400 class. The second profit in the first row of Table 2–4 is $2,148. It is tallied in the $1,800 up to $2,200 class. The other profits are tallied in a sim- ilar manner. When all the profits are tallied, the table would appear as:

Step 5: Count the number of items in each class. The number of observations in each class is called the class frequency. In the $200 up to $600 class there are 8 observations, and in the $600 up to $1,000 class there are 11 observations. Therefore, the class frequency in the first class is 8 and the class frequency in the second class is 11. There are a total of

Profit Frequency

$ 200 up to $ 600 ||| 600 up to 1,000 |

1,000 up to 1,400 ||| 1,400 up to 1,800 ||| 1,800 up to 2,200 2,200 up to 2,600 || 2,600 up to 3,000 |||| 3,000 up to 3,400 ||||

Total

|||||||||||| ||||||||||||||||||||||||

|||||||||||||||||||||||||||||||||||| ||||||||||||||||||||||||||||

|||||||||||||||| ||||||||

||||

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 33

180 observations or frequencies in the entire set of data. So the sum of all the frequencies should be equal to 180.

Now that we have organized the data into a frequency distribution, we can sum- marize the pattern in the profits of the vehicles for the Applewood Auto Group. Observe the following:

1. The profit from a vehicle ranged between $200 and $3,400. 2. The profits are concentrated between $1,000 and $3,000. The profit on 157 vehi-

cles or 87 percent was within this range. 3. The largest concentration, or highest frequency, is in the $1,800 up to $2,200

class. There are 45 observations. The middle of this class is $2,000. So we say that the typical profit on selling a vehicle is $2,000.

By presenting this information to Ms. Ball, we give her a clear picture of the distri- bution of the vehicle profits for last month.

We admit that arranging the information on profits into a frequency distribution does result in the loss of some detailed information. That is, by organizing the data into a frequency distribution, we cannot pinpoint the exact profit on any vehicle, such as $1,387, $2,148, or $2,201. Further, we cannot tell that the actual lowest amount of profit for any vehicle sold is $294 or that the most profit was $3,292. However, the lower limit of the first class and the upper limit of the largest class convey essentially the same meaning. Likely, Ms. Ball will make the same judgment if she knows the lowest price is about $200 that she will if she knows the exact profit is $292. The advantages of condensing the data into a more understandable and organized form more than offset this disadvantage.

Self-Review 2–2 The commissions earned for the first quarter of last year by the 11 members of the sales staff at Master Chemical Company are:

$1,650 $1,475 $1,510 $1,670 $1,595 $1,760 $1,540 $1,495 $1,590 $1,625 $1,510

(a) What are the values such as $1,650 and $1,475 called? (b) Using $1,400 up to $1,500 as the first class, $1,500 up to $1,600 as the second class,

and so forth, organize the quarterly commissions into a frequency distribution. (c) What are the numbers in the right column of your frequency distribution called? (d) Describe the distribution of quarterly commissions, based on the frequency distribution.

What is the largest concentration of commissions earned? What is the smallest, and the largest? What is the typical amount earned?

We will use two other terms frequently: class midpoint and class interval. The midpoint is halfway between the lower limits of two consecutive classes. It is com- puted by adding the lower limits of consecutive classes and dividing the result by 2.

Profit Frequency

$ 200 up to $ 600 8 600 up to 1,000 11

1,000 up to 1,400 23 1,400 up to 1,800 38 1,800 up to 2,200 45 2,200 up to 2,600 32 2,600 up to 3,000 19 3,000 up to 3,400 4

Total 180

TABLE 2–7 Frequency Distribution of Profit for Vehicles Sold Last Month at Applewood Auto Group

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34 Chapter 2

Referring to Table 2–7, the lower class limit of the first class is $200 and the next class limit is $600. The class midpoint is $400, found by ($600 � $200)�2. The mid- point of $400 best represents, or is typical of, the profits of the vehicles in that class.

To determine the class interval, subtract the lower limit of the class from the lower limit of the next class. The class interval of the Applewood data is $400, which we find by subtracting the lower limit of the first class, $200, from the lower limit of the next class; that is, $600. ($600 � $200 � $400.) You can also determine the class interval by finding the difference between consecutive midpoints. The midpoint of the first class is $400 and the midpoint of the second class is $800. The difference is $400.

2.4 A Software Example As we mentioned in Chapter 1, there are many software packages that perform sta- tistical calculations. Throughout this text we will show the output from Microsoft Excel; from MegaStat, which is an add-in to Microsoft Excel; and from Minitab. The commands necessary to generate the outputs are given in the Software Com- mands section at the end of each chapter. By following these commands, you will be able to duplicate the output.

The following is a frequency distribution, produced by MegaStat, showing the prices of the 180 vehicles sold last month at the Applewood Auto Group. The form of the output is somewhat different than the frequency distribution of Table 2–7, but the overall conclusions are the same.

Profit Cumulative

Lower Upper Midpoint Width Frequency Percent Frequency Percent

200 � 600 400 400 8 4.4 8 4.4 600 � 1,000 800 400 11 6.1 19 10.6

1,000 � 1,400 1,200 400 23 12.8 42 23.3 1,400 � 1,800 1,600 400 38 21.1 80 44.4 1,800 � 2,200 2,000 400 45 25.0 125 69.4 2,200 � 2,600 2,400 400 32 17.8 157 87.2 2,600 � 3,000 2,800 400 19 10.6 176 97.8 3,000 � 3,400 3,200 400 4 2.2 180 100.0

180 100.0

2.5 Relative Frequency Distribution It may be desirable, as we did earlier with qualitative data, to convert class frequen- cies to relative class frequencies to show the proportion of the total number of observations in each class. In our vehicle profits, we may want to know what per- centage of the vehicle profits are in the $1,000 up to $1,400 class. In another study, we may want to know what percentage of the employees used 5 up to 10 personal leave days last year. To convert a frequency distribution to a relative frequency dis- tribution, each of the class frequencies is divided by the total number of observa- tions. From the distribution of vehicle profits, Table 2–7, the relative frequency for the $1,000 up to $1,400 class is 0.128, found by dividing 23 by 180. That is, profit

Self-Review 2–3 Barry Bonds of the San Francisco Giants established a new single-season Major League Baseball home run record by hitting 73 home runs during the 2001 season. The longest of these home runs traveled 488 feet and the shortest 320 feet. You need to construct a frequency distribution of these home run lengths. (a) How many classes would you use? (b) What class interval would you suggest? (c) What actual classes would you suggest?

LO5 Understand a relative frequency distribution.

A relative frequency converts the frequency to a percentage.

Frequency Distribution-Quantitative

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 35

TABLE 2–8 Relative Frequency Distribution of Profit for Vehicles Sold Last Month at Applewood Auto Group

Profit Frequency Relative Frequency Found by

$ 200 up to $ 600 8 .044 8�180 600 up to 1,000 11 .061 11�180

1,000 up to 1,400 23 .128 23�180 1,400 up to 1,800 38 .211 38�180 1,800 up to 2,200 45 .250 45�180 2,200 up to 2,600 32 .178 32�180 2,600 up to 3,000 19 .106 19�180 3,000 up to 3,400 4 .022 4�180

Total 180 1.000

Self-Review 2–4 Refer to Table 2–8, which shows the relative frequency distribution for the profit earned on vehicles sold last month at the Applewood Auto Group. (a) How many vehicles are in the $1,800 up to $2,200 class? (b) What proportion of the vehicles sold for a profit of between $1,800 up to $2,200? (c) What proportion of the vehicles sold for a profit of $2,200 or more?

Exercises 7. A set of data consists of 38 observations. How many classes would you recommend for

the frequency distribution? 8. A set of data consists of 45 observations between $0 and $29. What size would you rec-

ommend for the class interval? 9. A set of data consists of 230 observations between $235 and $567. What class interval

would you recommend? 10. A set of data contains 53 observations. The lowest value is 42 and the largest is 129.

The data are to be organized into a frequency distribution. a. How many classes would you suggest? b. What would you suggest as the lower limit of the first class?

11. Wachesaw Manufacturing Inc. produced the following number of units in the last 16 days. This icon indicates that the data is available at the text website: www.mhhe.com/lind15e.

You will be able to download the data directly into Excel or Minitab from this site.

27 27 27 28 27 25 25 28 26 28 26 28 31 30 26 26

The information is to be organized into a frequency distribution. a. How many classes would you recommend? b. What class interval would you suggest? c. What lower limit would you recommend for the first class? d. Organize the information into a frequency distribution and determine the relative fre-

quency distribution. e. Comment on the shape of the distribution.

12. The Quick Change Oil Company has a number of outlets in the metropolitan Seattle area. The daily number of oil changes at the Oak Street outlet in the past 20 days are:

65 98 55 62 79 59 51 90 72 56 70 62 66 80 94 79 63 73 71 85

The data are to be organized into a frequency distribution. a. How many classes would you recommend? b. What class interval would you suggest?

on 12.8 percent of the vehicles sold is between $1,000 and $1,400. The relative fre- quencies for the remaining classes are shown in Table 2–8.

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36 Chapter 2

c. What lower limit would you recommend for the first class? d. Organize the number of oil changes into a frequency distribution. e. Comment on the shape of the frequency distribution. Also determine the relative fre-

quency distribution. 13. The manager of the BiLo Supermarket in Mt. Pleasant, Rhode Island, gathered the fol-

lowing information on the number of times a customer visits the store during a month. The responses of 51 customers were:

5 3 3 1 4 4 5 6 4 2 6 6 6 7 1 1 14 1 2 4 4 4 5 6 3 5 3 4 5 6 8 4 7 6 5 9 11 3 12 4 7 6 5 15 1 1 10 8 9 2 12

a. Starting with 0 as the lower limit of the first class and using a class interval of 3, orga- nize the data into a frequency distribution.

b. Describe the distribution. Where do the data tend to cluster? c. Convert the distribution to a relative frequency distribution.

14. The food services division of Cedar River Amusement Park Inc. is studying the amount families who visit the amusement park spend per day on food and drink. A sample of 40 families who visited the park yesterday revealed they spent the following amounts:

$77 $18 $63 $84 $38 $54 $50 $59 $54 $56 $36 $26 $50 $34 $44 41 58 58 53 51 62 43 52 53 63 62 62 65 61 52 60 60 45 66 83 71 63 58 61 71

a. Organize the data into a frequency distribution, using seven classes and 15 as the lower limit of the first class. What class interval did you select?

b. Where do the data tend to cluster? c. Describe the distribution. d. Determine the relative frequency distribution.

2.6 Graphic Presentation of a Frequency Distribution Sales managers, stock analysts, hospital administrators, and other busy executives often need a quick picture of the distributions of sales, stock prices, or hospital costs. These distributions can often be depicted by the use of charts and graphs. Three charts that will help portray a frequency distribution graphically are the his- togram, the frequency polygon, and the cumulative frequency polygon.

Histogram A histogram for a frequency distribution based on quantitative data is similar to the bar chart showing the distribution of qualitative data. The classes are marked on the horizontal axis and the class frequencies on the vertical axis. The class fre- quencies are represented by the heights of the bars. However, there is one impor- tant difference based on the nature of the data. Quantitative data are usually mea- sured using scales that are continuous, not discrete. Therefore, the horizontal axis represents all possible values, and the bars are drawn adjacent to each other to show the continuous nature of the data.

LO6 Present data from a frequency distribution in a histogram or a frequency polygon.

HISTOGRAM A graph in which the classes are marked on the horizontal axis and the class frequencies on the vertical axis. The class frequencies are represented by the heights of the bars, and the bars are drawn adjacent to each other.

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 37

Example Below is the frequency distribution of the profits on vehicle sales last month at the Applewood Auto Group.

Profit Frequency

$ 200 up to $ 600 8 600 up to 1,000 11

1,000 up to 1,400 23 1,400 up to 1,800 38 1,800 up to 2,200 45 2,200 up to 2,600 32 2,600 up to 3,000 19 3,000 up to 3,400 4

Total 180

Solution

Construct a histogram. What observations can you reach based on the information presented in the histogram?

The class frequencies are scaled along the vertical axis (Y-axis) and either the class limits or the class midpoints along the horizontal axis. To illustrate the construction of the histogram, the first three classes are shown in Chart 2–3.

200 600 1000 1400

8 8

Nu m

be r o

f V eh

ic le

s (c

la ss

fr eq

ue nc

Profit $

CHART 2–3 Construction of a Histogram

From Chart 2–3 we note the profit on eight vehicles was $200 up to $600. Therefore, the height of the column for that class is 8. There are 11 vehicles sales where the profit was $600 up to $1,000. So, logically, the height of that column is 11. The height of the bar represents the number of observations in the class.

This procedure is continued for all classes. The complete histogram is shown in Chart 2–4. Note that there is no space between the bars. This is a feature of the histogram. Why is this so? Because the variable plotted on the horizontal axis is quantitative and the ratio scale of measurement. In a bar chart, the scale of mea- surement is nominal and the vertical bars are separated. These are important dis- tinctions between the histogram and the bar chart.

From Chart 2–4 we can make the following statement:

1. The profit from a vehicle ranged from about $200 up to about $3,400. 2. The profits are concentrated between $1,000 and $3,000. The profit on 157

vehicles, or 87 percent, was within this range. 3. The largest concentration, or highest frequency, is in the $1,800 up to $2,200

class. The middle of this class is $2,000. So we say that the typical profit on selling a vehicle is $2,000.

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38 Chapter 2

Thus, the histogram provides an easily interpreted visual representation of a fre- quency distribution. We should also point out that we would have made the same observations and the shape of the histogram would have been the same had we used a relative frequency distribution instead of the actual frequencies. That is, if we had used the relative frequencies of Table 2–8, we would have had a histogram of the same shape as Chart 2–4. The only difference is that the vertical axis would have been reported in percentage of vehicles instead of the number of vehicles.

We use the Microsoft Excel system to produce the histogram for the Apple- wood Auto Group vehicle sales data. Note that class midpoints are used as the labels for the classes. The software commands to create this output are given in the Software Commands section at the end of the chapter.

200 600 1000 1400

Profit $

4 8

1800 2200 2600 3000 3400

Ve hi

cl es

S ol

d (c

la ss

fr eq

ue nc

CHART 2–4 Histogram of the Profit on 180 Vehicles Sold at the Applewood Auto Group

Frequency Polygon A frequency polygon also shows the shape of a distribution and is similar to a his- togram. It consists of line segments connecting the points formed by the inter- sections of the class midpoints and the class frequencies. The construction of a frequency polygon is illustrated in Chart 2–5 (on page 39). We use the profits from the cars sold last month at the Applewood Auto Group. The midpoint of each class is scaled on the X-axis and the class frequencies on the Y-axis. Recall that the class midpoint is the value at the center of a class and represents the typical values in

Statistics in Action

Florence Nightingale is known as the founder of the nurs- ing profession. How- ever, she also saved many lives by using statistical analysis. When she encoun- tered an unsanitary condition or an un- dersupplied hospital, she improved the conditions and then used statistical data to document the im- provement. Thus, she was able to con- vince others of the need for medical re- form, particularly in the area of sanita- tion. She developed original graphs to demonstrate that, during the Crimean War, more soldiers died from unsanitary conditions than were killed in combat.

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 39

Profit Midpoint Frequency

$ 200 up to $ 600 $ 400 8 600 up to 1,000 800 11

1,000 up to 1,400 1,200 23 1,400 up to 1,800 1,600 38 1,800 up to 2,200 2,000 45 2,200 up to 2,600 2,400 32 2,600 up to 3,000 2,800 19 3,000 up to 3,400 3,200 4

Total 180

As noted previously, the $200 up to $600 class is represented by the mid- point $400. To construct a frequency polygon, move horizontally on the graph to the midpoint, $400, and then vertically to 8, the class frequency, and place a dot. The X and the Y values of this point are called the coordinates. The coordinates of the next point are X � 800 and Y � 11. The process is continued for all classes. Then the points are connected in order. That is, the point representing the lowest class is joined to the one representing the second class and so on. Note in Chart 2–5 that, to complete the frequency polygon, midpoints of $0 and $3,600 are added to the X-axis to “anchor” the polygon at zero frequencies. These two values, $0 and $3,600, were derived by subtracting the class interval of $400 from the lowest mid- point ($400) and by adding $400 to the highest midpoint ($3,200) in the frequency distribution.

Both the histogram and the frequency polygon allow us to get a quick picture of the main characteristics of the data (highs, lows, points of concentration, etc.). Although the two representations are similar in purpose, the histogram has the advantage of depicting each class as a rectangle, with the height of the rectangu- lar bar representing the number in each class. The frequency polygon, in turn, has an advantage over the histogram. It allows us to compare directly two or more

Fr eq

ue nc

4000

Profit $

800 1200 1600 2000 2400 2800 3200 3600

CHART 2–5 Frequency Polygon of Profit on 180 Vehicles Sold at Applewood Auto Group

that class. The class frequency is the number of observations in a particular class. The profit earned on the vehicles sold last month by the Applewood Auto Group is repeated below.

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40 Chapter 2

frequency distributions. Suppose Ms. Ball wants to compare the profit per vehicle sold at Applewood Auto Group with a similar auto group, Fowler Auto in Grayling, Michigan. To do this, two frequency polygons are constructed, one on top of the other, as in Chart 2–6. Two things are clear from the chart:

• The typical vehicle profit is larger at Fowler Motors—about $2,000 for Apple- wood and about $2,400 for Fowler.

• There is less dispersion in the profits at Fowler Motors than at Applewood. The lower limit of the first class for Applewood is $0 and the upper limit is $3,600. For Fowler Motors, the lower limit is $800 and the upper limit is the same: $3,600.

The total number of cars sold at the two dealerships is about the same, so a direct comparison is possible. If the difference in the total number of cars sold is large, then converting the frequencies to relative frequencies and then plotting the two distributions would allow a clearer comparison.

4000

Profit $

Fr eq

ue nc

800 1200 1600 2000 2400 2800 3200 3600

Fowler Motors Applewood

CHART 2–6 Distribution of Profit at Applewood Auto Group and Fowler Motors

Self-Review 2–5 The annual imports of a selected group of electronic suppliers are shown in the following frequency distribution.

Imports Number Imports Number ($ millions) of Suppliers ($ millions) of Suppliers

2 up to 5 6 11 up to 14 10 5 up to 8 13 14 up to 17 1 8 up to 11 20

(a) Portray the imports as a histogram. (b) Portray the imports as a relative frequency polygon. (c) Summarize the important facets of the distribution (such as classes with the highest

and lowest frequencies).

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 41

Exercises 15. Molly’s Candle Shop has several retail stores in the coastal areas of North and South

Carolina. Many of Molly’s customers ask her to ship their purchases. The following chart shows the number of packages shipped per day for the last 100 days.

Fr eq

ue nc

y Number of Packages

0 5 10 15 20 25 30 35

28 23

10 3 5

a. What is this chart called? b. What is the total number of frequencies? c. What is the class interval? d. What is the class frequency for the 10 up to 15 class? e. What is the relative frequency of the 10 up to 15 class? f. What is the midpoint of the 10 up to 15 class? g. On how many days were there 25 or more packages shipped?

16. The following chart shows the number of patients admitted daily to Memorial Hospital through the emergency room.

2 4 6 8 10 12

Fr eq

ue nc

Number of Patients

a. What is the midpoint of the 2 up to 4 class? b. How many days were 2 up to 4 patients admitted? c. Approximately how many days were studied? d. What is the class interval? e. What is this chart called?

17. The following frequency distribution reports the number of frequent flier miles, reported in thousands, for employees of Brumley Statistical Consulting Inc. during the most recent quarter.

Frequent Flier Miles Number of (000) Employees

0 up to 3 5 3 up to 6 12 6 up to 9 23 9 up to 12 8

12 up to 15 2 Total 50

a. How many employees were studied? b. What is the midpoint of the first class? c. Construct a histogram.

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42 Chapter 2

d. A frequency polygon is to be drawn. What are the coordinates of the plot for the first class?

e. Construct a frequency polygon. f. Interpret the frequent flier miles accumulated using the two charts.

18. Ecommerce.com, a large Internet retailer, is studying the lead time (elapsed time between when an order is placed and when it is filled) for a sample of recent orders. The lead times are reported in days.

Lead Time (days) Frequency

0 up to 5 6 5 up to 10 7

10 up to 15 12 15 up to 20 8 20 up to 25 7

Total 40

Cumulative Frequency Distributions Consider once again the distribution of the profits on vehicles sold by the Apple- wood Auto Group. Suppose we were interested in the number of vehicles that sold for a profit of less than $1,400 or the profit earned on the lowest selling 40 percent of the vehicles. These values can be approximated by developing a cumulative frequency distribution and portraying it graphically in a cumulative frequency polygon.

Example The frequency distribution of the profits earned at Applewood Auto Group is repeated from Table 2–7.

a. How many orders were studied? b. What is the midpoint of the first class? c. What are the coordinates of the first class for a frequency polygon? d. Draw a histogram. e. Draw a frequency polygon. f. Interpret the lead times using the two charts.

Profit Frequency

$ 200 up to $ 600 8 600 up to 1,000 11

1,000 up to 1,400 23 1,400 up to 1,800 38 1,800 up to 2,200 45 2,200 up to 2,600 32 2,600 up to 3,000 19 3,000 up to 3,400 4

Total 180

Solution

Construct a cumulative frequency polygon. Seventy-five percent of the vehicles sold earned a profit of less than what amount? Sixty of the vehicles earned a profit of less than what amount?

As the names imply, a cumulative frequency distribution and a cumulative frequency polygon require cumulative frequencies. To construct a cumulative frequency distri- bution, refer to the preceding table and note that there were eight vehicles in which

LO7 Construct and interpret a cumulative frequency distribution.

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 43

the profit earned was less than $600. Those 8 vehicles, plus the 11 in the next higher class, for a total of 19, earned a profit of less than $1,000. The cumulative frequency for the next higher class is 42, found by 8 � 11 � 23. This process is continued for all the classes. All the vehicles earned a profit of less than $3,400. (See Table 2–9.)

To plot a cumulative frequency distribution, scale the upper limit of each class along the X-axis and the corresponding cumulative frequencies along the Y-axis. To provide additional information, you can label the vertical axis on the left in units and the vertical axis on the right in percent. In the Applewood Auto Group, the vertical axis on the left is labeled from 0 to 180 and on the right from 0 to 100 percent. The value of 50 percent corresponds to 90 vehicles.

To begin, the first plot is at X � 200 and Y � 0. None of the vehicles sold for a profit of less than $200. The profit on 8 vehicles was less than $600, so the next plot is at X � 600 and Y � 8. Continuing, the next plot is X � 1,000 and Y � 19. There were 19 vehicles that sold for a profit of less than $1,000. The rest of the points are plotted and then the dots connected to form the chart below.

Profit Frequency Cumulative Frequency Found by

$ 200 up to $ 600 8 8 8 600 up to 1,000 11 19 8 � 11

1,000 up to 1,400 23 42 8 � 11 � 23 1,400 up to 1,800 38 80 8 � 11 � 23 � 30 1,800 up to 2,200 45 125 8 � 11 � 23 � 30 � 45 2,200 up to 2,600 32 157 8 � 11 � 23 � 30 � 45 � 32 2,600 up to 3,000 19 176 8 � 11 � 23 � 30 � 45 � 32 � 19 3,000 up to 3,400 4 180 8 � 11 � 23 � 30 � 45 � 32 � 19 � 4

Total 180

TABLE 2–9 Cumulative Frequency Distribution for Profit on Vehicles Sold Last Month at Applewood Auto Group

200 600 1000 1400 1800 2200 2600 3000 3400

Nu m

be r o

f V eh

ic le

s So

Pe rc

en t o

f V eh

ic le

s So

Profit $

100

120

140

160

180

CHART 2–7 Cumulative Frequency Polygon for Profit on Vehicles Sold Last Month at Applewood Auto Group

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44 Chapter 2

Exercises 19. The following chart shows the hourly wages of a sample of certified welders in the Atlanta,

Georgia area.

a. How many welders were studied? b. What is the class interval?

To find the amount of profit earned on 75 percent of the cars sold, draw a hor- izontal line from the 75 percent mark on the right-hand vertical axis over to the poly- gon, then drop down to the X-axis and read the amount of profit. The value on the X-axis is about $2,300, so we estimate that 75 percent of the vehicles sold earned a profit for the Applewood group of $2,230.

To find the profit earned on 60 vehicles, we locate the value of 60 on the left- hand vertical axis. Next, we draw a horizontal line from the value of 60 to the poly- gon and then drop down to the X-axis and read the profit. It is about $1,590, so we estimate that 60 of the vehicles sold for a profit of less than $1,590. We can also make estimates of the percentage of vehicles that sold for less than a partic- ular amount. To explain, suppose we want to estimate the percentage of vehicles that sold for a profit of less than $1,600. We begin by locating the value of $1,600 on the X-axis, move vertically to the polygon, and then horizontally to the vertical axis on the right. The value is about 56 percent, so we conclude that 56 percent of the vehicles sold for a profit of less than $1,600.

Fr eq

ue nc

Hourly Wage

Pe rc

en t

0 5 10 15 20 25 30

100

Self-Review 2–6 A sample of the hourly wages of 15 employees at Home Depot in Brunswick, Georgia, was organized into the following table.

(a) What is the table called? (b) Develop a cumulative frequency distribution and portray the distribution in a cumula-

tive frequency polygon. (c) On the basis of the cumulative frequency polygon, how many employees earn $11 an

hour or less? Half of the employees earn an hourly wage of how much or more? Four employees earn how much or less?

Hourly Wages Number of Employees

$ 8 up to $10 3 10 up to 12 7 12 up to 14 4 14 up to 16 1

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 45

c. About how many welders earn less than $10.00 per hour? d. About 75 percent of the welders make less than what amount? e. Ten of the welders studied made less than what amount? f. What percent of the welders make less than $20.00 per hour?

20. The following chart shows the selling price ($000) of houses sold in the Billings, Montana area.

Fr eq

ue nc

Pe rc

en t

200

150

100

Selling Price ($000)

500 100 150 200 250 350300

a. How many homes were studied? b. What is the class interval? c. One hundred homes sold for less than what amount? d. About 75 percent of the homes sold for less than what amount? e. Estimate the number of homes in the $150,000 up to $200,000 class. f. About how many homes sold for less than $225,000?

21. The frequency distribution representing the number of frequent flier miles accumulated by employees at Brumley Statistical Consulting Company is repeated from Exercise 17.

Frequent Flier Miles (000) Frequency

0 up to 3 5 3 up to 6 12 6 up to 9 23 9 up to 12 8

12 up to 15 2 Total 50

a. How many employees accumulated less than 3,000 miles? b. Convert the frequency distribution to a cumulative frequency distribution. c. Portray the cumulative distribution in the form of a cumulative frequency polygon. d. Based on the cumulative frequency polygon, about 75 percent of the employees accu-

mulated how many miles or less? 22. The frequency distribution of order lead time at Ecommerce.com from Exercise 18 is

repeated below.

a. How many orders were filled in less than 10 days? In less than 15 days? b. Convert the frequency distribution to a cumulative frequency distribution. c. Develop a cumulative frequency polygon. d. About 60 percent of the orders were filled in less than how many days?

Lead Time (days) Frequency

0 up to 5 6 5 up to 10 7

10 up to 15 12 15 up to 20 8 20 up to 25 7

Total 40

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Chapter Summary I. A frequency table is a grouping of qualitative data into mutually exclusive classes show-

ing the number of observations in each class. II. A relative frequency table shows the fraction of the number of frequencies in each class.

III. A bar chart is a graphic representation of a frequency table. IV. A pie chart shows the proportion each distinct class represents of the total number of

frequencies. V. A frequency distribution is a grouping of data into mutually exclusive classes showing the

number of observations in each class. A. The steps in constructing a frequency distribution are:

1. Decide on the number of classes. 2. Determine the class interval. 3. Set the individual class limits. 4. Tally the raw data into classes. 5. Count the number of tallies in each class.

B. The class frequency is the number of observations in each class. C. The class interval is the difference between the limits of two consecutive classes. D. The class midpoint is halfway between the limits of consecutive classes.

VI. A relative frequency distribution shows the percent of observations in each class. VII. There are three methods for graphically portraying a frequency distribution.

A. A histogram portrays the number of frequencies in each class in the form of a rectangle.

B. A frequency polygon consists of line segments connecting the points formed by the intersection of the class midpoint and the class frequency.

C. A cumulative frequency distribution shows the number or percent of observations below given values.

Chapter Exercises 23. Describe the similarities and differences of qualitative and quantitative variables. Be sure

to include: a. What level of measurement is required for each variable type? b. Can both types be used to describe both samples and populations?

24. Describe the similarities and differences of a frequency table and a frequency distribution. Be sure to include which requires qualitative data and which requires quantitative data.

25. Alexandra Damonte will be building a new resort in Myrtle Beach, South Carolina. She must decide how to design the resort based on the type of activities that the resort will offer to its customers. A recent poll of 300 potential customers showed the following results about customers’ preferences for planned resort activities:

46 Chapter 2

Like planned activities 63 Do not like planned activities 135 Not sure 78 No answer 24

a. What is the table called? b. Draw a bar chart to portray the survey results. c. Draw a pie chart for the survey results. d. If you are preparing to present the results to Ms. Damonte as part of a report, which

graph would you prefer to show? Why? 26. Speedy Swift is a package delivery service that serves the greater Atlanta, Georgia met-

ropolitan area. To maintain customer loyalty, one of Speedy Swift’s performance objec- tives is on-time delivery. To monitor its performance, each delivery is measured on the following scale: early (package delivered before the promised time), on-time (package delivered within 5 minutes of the promised time), late (package delivered more than 5 minutes past the promised time), lost (package never delivered). Speedy Swift’s objective

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 47

is to deliver 99 percent of all packages either early or on-time. Another objective is to never lose a package.

Speedy collected the following data for last month’s performance:

On-time On-time Early Late On-time On-time On-time On-time Late On-time Early On-time On-time Early On-time On-time On-time On-time On-time On-time Early On-time Early On-time On-time On-time Early On-time On-time On-time Early On-time On-time Late Early Early On-time On-time On-time Early On-time Late Late On-time On-time On-time On-time On-time On-time On-time On-time Late Early On-time Early On-time Lost On-time On-time On-time Early Early On-time On-time Late Early Lost On-time On-time On-time On-time On-time Early On-time Early On-time Early On-time Late On-time On-time Early On-time On-time On-time Late On-time Early On-time On-time On-time On-time On-time On-time On-time Early Early On-time On-time On-time

a. What scale is used to measure delivery performance? What kind of variable is delivery performance?

b. Construct a frequency table for delivery performance for last month. c. Construct a relative frequency table for delivery performance last month. d. Construct a bar chart of the frequency table for delivery performance for last month. e. Construct a pie chart of on-time delivery performance for last month. f. Analyze the data summaries and write an evaluation of last month’s delivery perfor-

mance as it relates to Speedy Swift’s performance objectives. Write a general recom- mendation for further analysis.

27. A data set consists of 83 observations. How many classes would you recommend for a frequency distribution?

28. A data set consists of 145 observations that range from 56 to 490. What size class inter- val would you recommend?

29. The following is the number of minutes to commute from home to work for a group of automobile executives.

28 25 48 37 41 19 32 26 16 23 23 29 36 31 26 21 32 25 31 43 35 42 38 33 28

a. How many classes would you recommend? b. What class interval would you suggest? c. What would you recommend as the lower limit of the first class? d. Organize the data into a frequency distribution. e. Comment on the shape of the frequency distribution.

30. The following data give the weekly amounts spent on groceries for a sample of households.

$271 $363 $159 $ 76 $227 $337 $295 $319 $250 279 205 279 266 199 177 162 232 303 192 181 321 309 246 278 50 41 335 116 100 151 240 474 297 170 188 320 429 294 570 342 279 235 434 123 325

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48 Chapter 2

4 6 8 7 9 6 3 7 7 6 7 1 4 7 7 4 6 4 10 2 4 6 3 4 6 8 4 3 3 6 8 8 4 6 4 6 5 5 9 6 8 8 6 5 10

Organize the above information into a frequency distribution. a. How many classes would you suggest? b. What is the most suitable class interval? c. What is the lower limit of the initial class? d. Create the frequency distribution. e. Describe the profile of the distribution.

32. David Wise handles his own investment portfolio, and has done so for many years. Listed below is the holding time (recorded to the nearest whole year) between purchase and sale for his collection of stocks.

a. How many classes would you propose? b. What class interval would you suggest? c. What quantity would you use for the lower limit of the initial class? d. Using your responses to parts (a), (b), and (c), create a frequency distribution. e. Identify the appearance of the frequency distribution.

33. You are exploring the music in your iTunes library. The total play counts over the past year for the songs on your “smart playlist” are shown below. Make a frequency dis- tribution of the counts and describe its shape. It is often claimed that a small fraction of a person’s songs will account for most of their total plays. Does this seem to be the case here?

34. The Journal of Finance made its content available on the Internet starting in July of 2005. The table below shows the number of times a monthly version was downloaded and the number of articles that were viewed during each month. Suppose you wish to make a frequency distribution of the number of downloads.

8 8 6 11 11 9 8 5 11 4 8 5 14 7 12 8 6 11 9 7 9 15 8 8 12 5 9 8 5 9 10 11 3 9 8 6

128 56 54 91 190 23 160 298 445 50 578 494 37 677 18 74 70 868 108 71 466 23 84 38 26 814 17

31. A social scientist is studying the use of iPods by college students. A sample of 45 stu- dents revealed they played the following number of songs yesterday.

312 2,753 2,595 6,057 7,624 6,624 6,362 6,575 7,760 7,085 7,272 5,967 5,256 6,160 6,238 6,709 7,193 5,631 6,490 6,682 7,829 7,091 6,871 6,230 7,253 5,507 5,676 6,974 6,915 4,999 5,689 6,143 7,086

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 49

50 60 70 80 90 100

25 20 15 10 5 0

Score

Fr eq

ue nc

a. How many students took the exam? b. What is the class interval? c. What is the class midpoint for the first class? d. How many students earned a score of less than 70?

36. The following chart summarizes the selling price of homes sold last month in the Sarasota, Florida, area.

a. What is the chart called? b. How many homes were sold during the last month? c. What is the class interval? d. About 75 percent of the houses sold for less than what amount? e. One hundred seventy-five of the homes sold for less than what amount?

37. A chain of sport shops catering to beginning skiers, headquartered in Aspen, Colorado, plans to conduct a study of how much a beginning skier spends on his or her initial purchase of equipment and supplies. Based on these figures, it wants to explore the possibility of offering combinations, such as a pair of boots and a pair of skis, to induce customers to buy more. A sample of cash register receipts revealed these initial purchases:

100

250 200 150 100 50

0 50 100 150 Selling Price ($000)

200 250 300 350

Fr eq

ue nc

Pe rc

en t

$140 $ 82 $265 $168 $ 90 $114 $172 $230 $142 86 125 235 212 171 149 156 162 118

139 149 132 105 162 126 216 195 127 161 135 172 220 229 129 87 128 126 175 127 149 126 121 118 172 126

35. The following histogram shows the scores on the first exam for a statistics class.

a. Arrive at a suggested class interval. Use six classes, and let the lower limit of the first class be $70.

b. What would be a better class interval? c. Organize the data into a frequency distribution using a lower limit of $80. d. Interpret your findings.

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50 Chapter 2

Number of Number of Shareholders Shareholders

Company (thousands) Company (thousands)

Southwest Airlines 144 Standard Oil (Indiana) 173 General Public Utilities 177 Home Depot 195 Occidental Petroleum 266 Detroit Edison 220 Middle South Utilities 133 Eastman Kodak 251 Chrysler 209 Dow Chemical 137 Standard Oil of California 264 Pennsylvania Power 150 Bethlehem Steel 160 American Electric Power 262 Long Island Lighting 143 Ohio Edison 158 RCA 246 Transamerica Corporation 162 Greyhound Corporation 151 Columbia Gas System 165 Pacific Gas & Electric 239 International Telephone & Niagara Mohawk Power 204 Telegraph 223 E. I. du Pont de Nemours 204 Union Electric 158 Westinghouse Electric 195 Virginia Electric and Power 162 Union Carbide 176 Public Service Electric & Gas 225 BankAmerica 175 Consumers Power 161 Northeast Utilities 200

Expenditure Item Amount

Fuel $ 603 Interest on car loan 279 Repairs 930 Insurance and license 646 Depreciation 492

Total $2,950

The shareholder numbers are to be organized into a frequency distribution and several graphs drawn to portray the distribution. a. Using seven classes and a lower limit of 130, construct a frequency distribution. b. Portray the distribution as a frequency polygon. c. Portray the distribution in a cumulative frequency polygon. d. According to the polygon, three out of four (75 percent) of the companies have how

many shareholders or less? e. Write a brief analysis of the number of shareholders based on the frequency distrib-

ution and graphs. 39. A recent survey showed that the typical American car owner spends $2,950 per year on

operating expenses. Below is a breakdown of the various expenditure items. Draw an appropriate chart to portray the data and summarize your findings in a brief report.

40. Midland National Bank selected a sample of 40 student checking accounts. Below are their end-of-the-month balances.

$404 $ 74 $234 $149 $279 $215 $123 $ 55 $ 43 $321 87 234 68 489 57 185 141 758 72 863

703 125 350 440 37 252 27 521 302 127 968 712 503 489 327 608 358 425 303 203

38. Following is the number of shareholders for a selected group of large companies (in thousands):

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a. Tally the data into a frequency distribution using $100 as a class interval and $0 as the starting point.

b. Draw a cumulative frequency polygon. c. The bank considers any student with an ending balance of $400 or more a “preferred

customer.” Estimate the percentage of preferred customers. d. The bank is also considering a service charge to the lowest 10 percent of the ending

balances. What would you recommend as the cutoff point between those who have to pay a service charge and those who do not?

41. Residents of the state of South Carolina earned a total of $69.5 billion in adjusted gross income. Seventy-three percent of the total was in wages and salaries; 11 percent in div- idends, interest, and capital gains; 8 percent in IRAs and taxable pensions; 3 percent in business income pensions; 2 percent in Social Security, and the remaining 3 percent from other sources. Develop a pie chart depicting the breakdown of adjusted gross income. Write a paragraph summarizing the information.

42. A recent study of home technologies reported the number of hours of personal computer usage per week for a sample of 60 persons. Excluded from the study were people who worked out of their home and used the computer as a part of their work.

9.3 5.3 6.3 8.8 6.5 0.6 5.2 6.6 9.3 4.3 6.3 2.1 2.7 0.4 3.7 3.3 1.1 2.7 6.7 6.5 4.3 9.7 7.7 5.2 1.7 8.5 4.2 5.5 5.1 5.6 5.4 4.8 2.1 10.1 1.3 5.6 2.4 2.4 4.7 1.7 2.0 6.7 1.1 6.7 2.2 2.6 9.8 6.4 4.9 5.2 4.5 9.3 7.9 4.6 4.3 4.5 9.2 8.5 6.0 8.1

$669.9 $ 7.5 $ 77.2 $ 7.5 $125.7 $516.9 $ 219.9 $645.2 301.9 235.4 716.4 145.3 26.6 187.2 315.5 89.2 136.4 616.9 440.6 408.2 34.4 296.1 185.4 526.3 380.7 3.3 363.2 51.9 52.2 107.5 82.9 63.0 228.6 308.7 126.7 430.3 82.0 227.0 321.1 403.4

39.5 124.3 118.1 23.9 352.8 156.7 276.3 23.5 31.3 301.2 35.7 154.9 174.3 100.6 236.7 171.9

221.1 43.4 212.3 243.3 315.4 5.9 1,002.2 171.7 295.7 437.0 87.8 302.1 268.1 899.5

a. Organize the data into a frequency distribution. How many classes would you suggest? What value would you suggest for a class interval?

b. Draw a histogram. Interpret your result. 43. Merrill Lynch recently completed a study regarding the size of online investment portfolios

(stocks, bonds, mutual funds, and certificates of deposit) for a sample of clients in the 40- to 50-year-old age group. Listed following is the value of all the investments in thousands of dollars for the 70 participants in the study.

a. Organize the data into a frequency distribution. How many classes would you suggest? What value would you suggest for a class interval?

b. Draw a histogram. Interpret your result. 44. A total of 5.9 percent of the prime time viewing audience watched shows on ABC, 7.6 per-

cent watched shows on CBS, 5.5 percent on Fox, 6.0 percent on NBC, 2.0 percent on Warner Brothers, and 2.2 percent on UPN. A total of 70.8 percent of the audience watched shows on other cable networks, such as CNN and ESPN. You can find the latest information on TV viewing from the following website: http://tv.zap2it.com/news/ratings. Develop a pie chart or a bar chart to depict this information. Write a paragraph summa- rizing your findings.

Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 51

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52 Chapter 2

45. Refer to the following chart, which appeared recently in the Snapshot section of USA Today.

a. What is the name given to this type of chart? b. If you studied 500 weddings, how many would you expect to take place in a house

of worship? c. Would it be reasonable to conclude that about 80 percent of weddings take place in

either a house of worship or outdoors? Cite evidence. 46. The following chart depicts the annual revenues, by type of tax, for the state of Georgia.

The chart was developed using Kids Zone, a NCES, project. Their website is: nces.ed.gov/nceskids/createagraph/.

a. What percentage of the state revenue is accounted for by sales tax and individual income tax?

b. Which category will generate more revenue, corporate taxes or license fees? c. The total annual revenue for the state of Georgia is $6.3 billion. Estimate the amount

of revenue in billions of dollars for sales taxes and for individual taxes. 47. In 2006, Canada exported $303.4 billion worth of products to the United States. The five

largest were:

a. Use a software package to develop a bar chart. b. What percentage of Canada’s total exports to the United States is represented by the

two categories “Petroleum products” and “Passenger cars”? c. Of the top five exported products, what percentage of the total do “Petroleum prod-

ucts” and “Passenger cars” represent?

Product Amount

Petroleum products $63.7 billion Passenger cars 36.6 Car parts and accessories 15.6 Aluminum 7.7 Lumber 6.6

By Michelle Healy and Veronica Salazar, USA TODAY Source: ‘BRIDES’ Magazine Reprinted with permission (January 28, 2010) USA TODAY.

Sales 44.54%Income

43.34%

Other 0.9%

License 2.9%

Corporate 8.31%

Annual Revenue State of Georgia

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 53

48. Farming has changed from the early 1900s. In the early 20th century, machinery gradu- ally replaced animal power. For example, in 1910 U.S. farms used 24.2 million horses and mules and only about 1,000 tractors. By 1960, 4.6 million tractors were used and only 3.2 million horses and mules. In 1920, there were over 6 million farms in the United States. Today there are fewer than 2 million. Listed below is the number of farms, in thou- sands, for each of the 50 states. Write a paragraph summarizing your findings.

49. One of the most popular candies in the United States is M&M’s, which are produced by the Mars Company. In the beginning M&M’s were all brown; more recently they were produced in red, green, blue, orange, brown, and yellow. You can read about the history of the prod- uct, find ideas for baking, purchase the candies in the colors of your school or favorite team, and learn the percent of each color in the standard bags at www.m-ms.com. Recently, the purchase of a 14-ounce bag of M&M’s Plain had 444 candies with the following breakdown by color: 130 brown, 98 yellow, 96 red, 35 orange, 52 blue, and 33 green. Develop a chart depicting this information and write a paragraph summarizing the results.

50. The number of families who used the Minneapolis YWCA day care service was recorded during a 30-day period. The results are as follows:

47 1 8 46 76 26 4 3 39 45 4 21 80 63 100 65 91 29 7 15 7 52 87 39 106 25 55 2 3 8

14 38 59 33 76 71 37 51 1 24 35 86 185 13 7 43 36 20 79 9

31 49 19 62 24 45 23 51 55 60 40 35 54 26 57 37 43 65 18 41 50 56 4 54 39 52 35 51 63 42

a. Construct a cumulative frequency distribution. b. Sketch a graph of the cumulative frequency polygon. c. How many days saw fewer than 30 families utilize the day care center? d. How busy were the highest 80 percent of the days?

Data Set Exercises 51. Refer to the Real Estate data at the end of the book, which reports information on homes

sold in the Goodyear, Arizona, area during the last year. Select an appropriate class inter- val and organize the selling prices into a frequency distribution. Write a brief report sum- marizing your finding. Be sure to answer the following questions in your report. a. Around what values do the data tend to cluster? b. What is the largest selling price? What is the smallest selling price? c. Draw a cumulative frequency distribution. How many homes sold for less than

$200,000? Estimate the percent of the homes that sold for more than $220,000. What percent of the homes sold for less than $125,000?

d. Refer to the variable regarding the townships. Draw a bar chart showing the number of homes sold in each township. Are there any differences or is the number of homes sold about the same in each township?

52. Refer to the Baseball 2009 data, which reports information on the 30 Major League Base- ball teams for the 2009 season. Select an appropriate class interval and organize the team salaries into a frequency distribution. a. What is a typical team salary? What is the range of salaries? b. Comment on the shape of the distribution. Does it appear that any of the team salaries

are out of line with the others? c. Draw a cumulative frequency distribution. Forty percent of the teams are paying less

than what amount in total team salary? About how many teams have total salaries of less than $80,000,000?

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53. Refer to the Buena School District bus data. Select the variable referring to the number of miles traveled last month, and then organize these data into a frequency distribution. a. What is a typical amount of miles traveled? What is the range? b. Comment on the shape of the distribution. Are there any outliers in terms of miles

driven? c. Draw a cumulative frequency distribution. Forty percent of the buses were driven fewer

than how many miles? How many buses were driven less than 850 miles? d. Refer to the variables regarding the bus type and the number of seats in each bus.

Draw a pie chart of each variable and comment on your findings.

1. The Excel commands for the pie chart on page 26 are: a. Set cell A1 as the active cell and type the words

Use of Sales. In cells A2 through A5 type Prizes, Education, Bonuses, and Expense.

b. Set B1 as the active cell and type Amount ($ Mil- lions) and in cells B2 through B5 enter the data. When finished entering data in B5, hit the Enter button.

c. From the top row of tabs, select Insert. In the Chart tool bar, select Pie. Select the top-left 2-D pie chart. A blank chart will appear.

d. In Excel’s top row, a Chart Tools tab will appear. Select the Design option. Select the Select Data option from the tool bar. A Select Data window will appear. For the Chart Data Range, using the mouse, select all cells from A1 to B5. Click OK.

e. Click on the pie chart. Right-click for the options menu. Select Add Data Labels. Click on the pie chart again. Right-click for the options menu. Select Format Data Labels, then uncheck any boxes currently checked in the dialog box. Next select Category, Percent, and Leader Lines. Then click Close.

f. Double-click on the chart title and rename Ohio State Lottery Expenses.

2. The MegaStat commands for the frequency distribu- tion on page 34 are: a. Open Excel and from the CD provided, select

Data Sets, and select the Excel format; go to Chapter 2, and select Applewood data. Click on Mega-Stat, Frequency Distribution, and select Quantitative.

b. In the dialog box, input the range from A1:A181, select Equal width intervals, use 400 as the interval width, 2000 as the lower boundary of the first interval, select Histogram, and then click OK.

3. The Excel commands for the histogram on page 38 are: a. In cell A1 indicate that the column of data is

the profit and in B1 that it is the frequency. In cells A2 to A9, insert the midpoints of the prof- its. In B2 to B9, record the class frequencies. When finished entering data in cell B9, hit the Enter key.

b. With your mouse, highlight cells B2 through B9. c. From the tabs, select Insert. From the Charts,

select Column, then 2-D column and pick the first chart type. A graph will appear.

d. When the graph area is active, a Chart Tools tab appears at the top of the screen. Select the Design tab, and then select Data. Under Hori- zontal (Category) Axis Labels, click Edit, then use the mouse to select cells A2 through A9 and click OK twice. The horizontal axis now shows the class midpoints.

e. With the Chart Tools displayed at the top, select the Design tab. Select Chart Layout. Select the layout:

f. With Chart Tools displayed at the top, select the Layout tab. Double-click on the Chart Title and type in an appropriate title. Next, under the same Layout tab, select Axis Titles. Using Primary Vertical Axis Title, name the vertical axis Fre- quency and delete the words vertical axis. Using the Primary Horizontal Axis Title, name it Profit $. Next select Legend and select None.

g. Double-click one of the columns in the graph. Then from the Tabs across the top select Lay- out. On the left-hand side of the tool bar, click the words Format Selection. A dialog box will appear. Under Series Option, change the Gap Width to 0% by moving the arrow all the way to the left, and then click the Close button at the bottom of the dialog box.

Software Commands

54 Chapter 2

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Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation 55

Fr eq

ue nc

0 Cola-Plus Coca-Cola Pepsi

Beverage

Lemon-Lime

Lemon-Lime 15%

Cola-Plus 40%

Coca-Cola 25%

Pepsi 20%

2–2 a. The raw data or ungrouped data. b.

Number of Commission Salespeople

$1,400 up to $1,500 2 1,500 up to 1,600 5 1,600 up to 1,700 3 1,700 up to 1,800 1

Total 11

Chapter 2 Answers to Self-Review

2–1 a. Qualitative data, because the customers’ response to the taste test is the name of a beverage.

b. Frequency table. It shows the number of people who prefer each beverage.

0 2 5 8 11 14 17

Imports ($ millions)

Pe rc

en t o

f T ot

c. Class frequencies. d. The largest concentration of commissions

is $1,500 up to $1,600. The smallest commission is about $1,400 and the largest is about $1,800. The typical amount earned is $15,500.

2–3 a. 26 � 64 � 73 � 128 � 27. So seven classes are recommended.

b. The interval width should be at least (488 � 320)�7 � 24. Class intervals of 25 or 30 feet are both reasonable.

2–4 a. 45 b. .250 c. .306, found by .178 � .106 � .022

2–5 a. 20

0 2

Imports ($ millions)

Nu m

be r o

f Su

pp lie

5 8 11 14 17

The plots are: (3.5, 12), (6.5, 26), (9.5, 40), (12.5, 20), and (15.5, 2).

c. The smallest annual volume of imports by a supplier is about $2 million, the largest about $17 million. The highest frequency is between $8 million and $11 million.

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56 Chapter 2

8 10 12 14 16

Cu m

ul at

iv e

Fr eq

ue nc

ie s

Hourly Wages (in dollars)

X = 12 Y = 10

X = 14 Y = 14

2–6 a. A frequency distribution.

b. Hourly Wages Cumulative Number

Less than $8 0 Less than $10 3 Less than $12 10 Less than $14 14 Less than $16 15

c. About seven employees earn $11.00 or less. About half the employees earn $11.25 or more. About four employees earn $10.25 or less.

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3 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Explain the concept of central tendency.

LO2 Identify and compute the arithmetic mean.

LO3 Compute and interpret the weighted mean.

LO4 Determine the median.

LO5 Identify the mode.

LO6 Calculate the geometric mean.

LO7 Explain and apply mea- sures of dispersion.

LO8 Compute and explain the variance and the standard deviation.

LO9 Explain Chebyshev’s Theorem and the Empirical Rule.

LO10 Compute the mean and standard deviation of grouped data.

Describing Data: Numerical Measures

The Kentucky Derby is held the first Saturday in May at Churchill

Downs in Louisville, Kentucky. The race track is one and one-quarter

miles. The table in Exercise 82 shows the winners since 1990, their

margin of victory, the winning time, and the payoff on a $2 bet.

Determine the mean and median for the variables winning time and

payoff on a $2 bet. (See Exercise 82 and LO2 and LO4.)

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58 Chapter 3

3.1 Introduction Chapter 2 began our study of descriptive statistics. To summarize raw data into a meaningful form, we organized qualitative data into a frequency table and portrayed the results in a bar chart. In a similar fashion, we organized quantitative data into a frequency distribution and portrayed the results in a histogram. We also looked at other graphical techniques such as pie charts to portray qualitative data and frequency polygons to portray quantitative data.

This chapter is concerned with two numerical ways of describing quantitative vari- ables, namely, measures of location and measures of dispersion. Measures of location are often referred to as averages. The purpose of a measure of location is to pinpoint the center of a distribution of data. An average is a measure of location that shows the central value of the data. Averages appear daily on TV, on various websites, in the newspaper, and in other journals. Here are some examples:

• The average U.S. home changes ownership every 11.8 years.

• An American receives an average of 568 pieces of mail per year.

• The average American home has more TV sets than people. There are 2.73 TV sets and 2.55 people in the typical home.

• The average American couple spends $20,398 for their wedding, while their budget is 50 percent less. This does not include the cost of a honeymoon or engagement ring.

• The average price of a theater ticket in the United States is $7.50, according to the National Association of Theatre Owners.

If we consider only measures of location in a set of data, or if we compare sev- eral sets of data using central values, we may draw an erroneous conclusion. In addi- tion to measures of location, we should consider the dispersion—often called the variation or the spread—in the data. As an illustration, suppose the average annual income of executives for Internet-related companies is $80,000, and the average income for executives in pharmaceutical firms is also $80,000. If we looked only at the average incomes, we might wrongly conclude that the distributions of the two salaries are the same. However, we need to examine the dispersion or spread of the distributions of salary. A look at the salary ranges indicates that this conclusion of equal distributions is not correct. The salaries for the executives in the Internet firms range from $70,000 to $90,000, but salaries for the marketing executives in pharmaceuticals range from $40,000 to $120,000. Thus, we conclude that although the average salaries are the same for the two industries, there is much more spread or dispersion in salaries for the pharmaceutical executives. To describe the dispersion, we will consider the range, the mean deviation, the variance, and the standard deviation.

We begin by discussing measures of location. There is not just one measure of location; in fact, there are many. We will consider five: the arithmetic mean, the weighted mean, the median, the mode, and the geometric mean. The arithmetic mean is the most widely used and widely reported measure of location. We study the mean as both a population parameter and a sample statistic.

3.2 The Population Mean Many studies involve all the values in a population. For example, there are 12 sales associates employed at the Reynolds Road outlet of Carpets by Otto. The mean amount of commission they earned last month was $1,345. This is a population

Statistics in Action

Did you ever meet the “average” Ameri- can? Well, his name is Robert (that is the nominal level of measurement), he is 31 years old (that is the ratio level), he is 69.5 inches tall (again the ratio level of measurement), weighs 172 pounds, wears a size 91⁄2 shoe, has a 34-inch waist, and wears a size 40 suit. In addition, the average man eats 4 pounds of potato chips, watches 1,456 hours of TV, and eats 26 pounds of bananas each year and also sleeps 7.7 hours per night.

The average American woman is 5� 4� tall and weighs 140 pounds, while the average Ameri- can model is 5� 11� tall and weighs 117 pounds. On any given day, almost half of the women in the United States are on a diet. Idolized in the 1950s, Marilyn Monroe would be considered over- weight by today’s standards. She fluc- tuated between a size 14 and 18 dress, and was a healthy and attractive woman.

LO1 Explain the concept of central tendency.

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Example

Describing Data: Numerical Measures 59

There are 42 exits on I-75 through the state of Kentucky. Listed below are the dis- tances between exits (in miles).

Why is this information a population? What is the mean number of miles between exits?

11 4 10 4 9 3 8 10 3 14 1 10 3 5 2 2 5 6 1 2 2 3 7 1 3 7 8 10 1 4 7 5 2 2 5 1 1 3 3 1 2 1

PARAMETER A characteristic of a population.

value, because we considered the commission of all the sales associates. Other examples of a population mean would be:

• The mean closing price for Johnson & Johnson stock for the last 5 days is $64.75. • The mean number of hours of overtime worked last week by the six welders in

the welding department of Butts Welding Inc. is 6.45 hours. • Caryn Tirsch began a website last month devoted to organic gardening. The

mean number of hits on her site for the 31 days in July was 84.36.

For raw data—that is, data that have not been grouped in a frequency distribution— the population mean is the sum of all the values in the population divided by the number of values in the population. To find the population mean, we use the fol- lowing formula.

Instead of writing out in words the full directions for computing the population mean (or any other measure), it is more convenient to use the shorthand symbols of mathematics. The mean of the population using mathematical symbols is:

POPULATION MEAN [3–1]

where: � represents the population mean. It is the Greek lowercase letter “mu.” N is the number of values in the population. X represents any particular value. ∑ is the Greek capital letter “sigma” and indicates the operation of adding. ∑X is the sum of the X values in the population.

Any measurable characteristic of a population is called a parameter. The mean of a population is an example of a parameter.

� � �X N

Population mean � Sum of all the values in the population

Number of values in the population

Solution This is a population because we are considering all the exits in Kentucky. We add the distances between each of the 42 exits. The total distance is 192 miles. To find the arithmetic mean, we divide this total by 42. So the arithmetic mean is 4.57 miles, found by 192�42. From formula (3–1):

How do we interpret the value of 4.57? It is the typical number of miles between exits. Because we considered all the exits in Kentucky, this value is a population parameter.

� � ©X N

� 11 � 4 � 10 � # # # � 1

42 �

192 42

� 4.57

LO2 Identify and compute the arithmetic mean.

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60 Chapter 3

3.3 The Sample Mean As explained in Chapter 1, we often select a sample from the population to estimate a specific characteristic of the population. Smucker’s quality assurance department needs to be assured that the amount of strawberry jam in the jar labeled as contain- ing 12 ounces actually contains that amount. It would be very expensive and time consuming to check the weight of each jar. Therefore, a sample of 20 jars is selected, the mean of the sam- ple is determined, and that value is used to estimate the amount of jam in each jar.

For raw data—that is, ungrouped data—the mean is the sum of all the sampled values divided by the total number of sampled values. To find the mean for a sample:

The mean of a sample and the mean of a population are computed in the same way, but the shorthand notation used is different. The formula for the mean of a sample is:

SAMPLE MEAN [3–2]

where: represents the sample mean. It is read “X bar.”

n is the number of values in the sample. X represents any particular value. ∑ is the Greek capital letter “sigma” and indicates the operation of adding. ∑X is the sum of the X values in the sample.

The mean of a sample, or any other measure based on sample data, is called a statistic. If the mean weight of a sample of 10 jars of Smucker’s strawberry jam is 41 ounces, this is an example of a statistic.

STATISTIC A characteristic of a sample.

X � �X n

Sample mean � Sum of all the values in the sample

Number of values in the sample Sample mean of ungrouped data.

Example SunCom is studying the number of minutes used by clients in a particular cell phone rate plan. A random sample of 12 clients showed the following number of minutes used last month.

90 77 94 89 119 112 91 110 92 100 113 83

What is the arithmetic mean number of minutes used?

Solution Using formula (3–2), the sample mean is:

The arithmetic mean number of minutes used last month by the sample of cell phone users is 97.5 minutes.

X � �X n

� 90 � 77 � # # # � 83

12 �

1170 12

� 97.5

Sample mean � Sum of all values in the sample Number of values in the sample

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Describing Data: Numerical Measures 61

3.4 Properties of the Arithmetic Mean The arithmetic mean is a widely used measure of location. It has several important properties:

1. Every set of interval- or ratio-level data has a mean. Recall from Chapter 1 that ratio-level data include such data as ages, incomes, and weights, with the distance between numbers being constant.

2. All the values are included in computing the mean. 3. The mean is unique. That is, there is only one mean in a set of data. Later in

the chapter, we will discover an average that might appear twice, or more than twice, in a set of data.

4. The sum of the deviations of each value from the mean is zero. Expressed symbolically:

As an example, the mean of 3, 8, and 4 is 5. Then:

Thus, we can consider the mean as a balance point for a set of data. To illus- trate, we have a long board with the numbers 1, 2, 3, . . . , 9 evenly spaced on it. Suppose three bars of equal weight were placed on the board at numbers 3, 4, and 8, and the balance point was set at 5, the mean of the three numbers. We would find that the board is balanced perfectly! The deviations below the mean (�3) are equal to the deviations above the mean (�3). Shown schematically:

� 0

� �2 � 3 � 1

�(X � X ) � (3 � 5) � (8 � 5) � (4 � 5)

�(X � X ) � 0

Mean as a balance point

�1

�2

1 2 3 4 5 6 7 8 9

X _

The mean does have a weakness. Recall that the mean uses the value of every item in a sample, or population, in its computation. If one or two of these values are either extremely large or extremely small compared to the majority of data, the mean might not be an appropriate average to represent the data. For example, suppose the annual incomes of a small group of stockbrokers at Merrill Lynch are $62,900, $61,600, $62,500, $60,800, and $1,200,000. The mean income is $289,560. Obvi- ously, it is not representative of this group, because all but one broker has an income in the $60,000 to $63,000 range. One income ($1.2 million) is unduly affecting the mean.

Mean unduly affected by unusually large or small values

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62 Chapter 3

Self-Review 3–1 1. The annual incomes of a sample of middle-management employees at Westinghouse are: $62,900, $69,100, $58,300, and $76,800. (a) Give the formula for the sample mean. (b) Find the sample mean. (c) Is the mean you computed in (b) a statistic or a parameter? Why? (d) What is your best estimate of the population mean?

2. All the students in advanced Computer Science 411 are a population. Their course grades are 92, 96, 61, 86, 79, and 84. (a) Give the formula for the population mean. (b) Compute the mean course grade. (c) Is the mean you computed in (b) a statistic or a parameter? Why?

Exercises The answers to the odd-numbered exercises are at the end of the book.

1. Compute the mean of the following population values: 6, 3, 5, 7, 6. 2. Compute the mean of the following population values: 7, 5, 7, 3, 7, 4. 3. a. Compute the mean of the following sample values: 5, 9, 4, 10.

b. Show that 4. a. Compute the mean of the following sample values: 1.3, 7.0, 3.6, 4.1, 5.0.

b. Show that 5. Compute the mean of the following sample values: 16.25, 12.91, 14.58. 6. Suppose you go to the grocery store and spend $61.85 for the purchase of 14 items.

What is the mean price per item?

For Exercises 7–10, (a) compute the arithmetic mean and (b) indicate whether it is a statistic or a parameter.

7. There are 10 salespeople employed by Midtown Ford. The number of new cars sold last month by the respective salespeople were: 15, 23, 4, 19, 18, 10, 10, 8, 28, 19.

8. The accounting department at a mail-order company counted the following numbers of incoming calls per day to the company’s toll-free number during the first 7 days in May: 14, 24, 19, 31, 36, 26, 17.

9. The Cambridge Power and Light Company selected a random sample of 20 residential customers. Following are the amounts, to the nearest dollar, the customers were charged for electrical service last month:

54 48 58 50 25 47 75 46 60 70 67 68 39 35 56 66 33 62 65 67

10. The Human Relations Director at Ford began a study of the overtime hours in the Inspec- tion Department. A sample of 15 workers showed they worked the following number of overtime hours last month.

13 13 12 15 7 15 5 12 6 7 12 10 9 13 12

11. AAA Heating and Air Conditioning completed 30 jobs last month with a mean revenue of $5,430 per job. The president wants to know the total revenue for the month. Based on the limited information, can you compute the total revenue? What is it?

12. A large pharmaceutical company hires business administration graduates to sell its prod- ucts. The company is growing rapidly and dedicates only one day of sales training for new salespeople. The company’s goal for new salespeople is $10,000 per month. The goal is based on the current mean sales for the entire company, which is $10,000 per month. After reviewing the retention rates of new employees, the company finds that only 1 in 10 new employees stays longer than three months. Comment on using the current mean sales per month as a sales goal for new employees. Why do new employees leave the company?

�(X � X ) � 0.

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Describing Data: Numerical Measures 63

3.5 The Weighted Mean The weighted mean is a special case of the arithmetic mean. It occurs when there are several observations of the same value. To explain, suppose the nearby Wendy’s Restaurant sold medium, large, and Biggie-sized soft drinks for $.90, $1.25, and $1.50, respectively. Of the last 10 drinks sold, 3 were medium, 4 were large, and 3 were Biggie- sized. To find the mean price of the last 10 drinks sold, we could use formula (3–2).

The mean selling price of the last 10 drinks is $1.22. An easier way to find the mean selling price is to determine the weighted mean.

That is, we multiply each observation by the number of times it happens. We will refer to the weighted mean as This is read “X bar sub w.”

In this case, the weights are frequency counts. However, any measure of importance could be used as a weight. In general, the weighted mean of a set of numbers designated X1, X2, X3, . . . , Xn with the corresponding weights w1, w2, w3, . . . , wn is computed by:

WEIGHTED MEAN [3–3]

This may be shortened to:

Note that the denominator of a weighted mean is always the sum of the weights.

Xw � �(wX )

�w

Xw � w1X1 � w2 X2 � w3 X3 �

# # # � wn Xn w1 � w2 � w3 �

# # # � wn

Xw � 3($0.90) � 4($1.25) � 3($1.50)

10 �

$12.20 10

� $1.22

Xw.

X � $12.20

10 � $1.22

X � $.90 � $.90 � $.90 � $1.25 � $1.25 � $1.25 � $1.25 � $1.50 � $1.50 � $1.50

LO3 Compute and interpret the weighted mean.

Example The Carter Construction Company pays its hourly employees $16.50, $19.00, or $25.00 per hour. There are 26 hourly employees, 14 of which are paid at the $16.50 rate, 10 at the $19.00 rate, and 2 at the $25.00 rate. What is the mean hourly rate paid the 26 employees?

Solution To find the mean hourly rate, we multiply each of the hourly rates by the number of employees earning that rate. From formula (3–3), the mean hourly rate is

The weighted mean hourly wage is rounded to $18.12.

Xw � 14($16.50) � 10($19.00) � 2($25.00)

14 � 10 � 2 �

$471.00 26

� $18.1154

Self-Review 3–2 Springers sold 95 Antonelli men’s suits for the regular price of $400. For the spring sale, the suits were reduced to $200 and 126 were sold. At the final clearance, the price was reduced to $100 and the remaining 79 suits were sold. (a) What was the weighted mean price of an Antonelli suit? (b) Springers paid $200 a suit for the 300 suits. Comment on the store’s profit per suit if

a salesperson receives a $25 commission for each one sold.

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64 Chapter 3

Exercises 13. In June, an investor purchased 300 shares of Oracle (an information technology company)

stock at $20 per share. In August, she purchased an additional 400 shares at $25 per share. In November, she purchased an additional 400 shares, but the stock declined to $23 per share. What is the weighted mean price per share?

14. The Bookstall Inc. is a specialty bookstore concentrating on used books sold via the Internet. Paperbacks are $1.00 each, and hardcover books are $3.50. Of the 50 books sold last Tuesday morning, 40 were paperback and the rest were hardcover. What was the weighted mean price of a book?

15. The Loris Healthcare System employs 200 persons on the nursing staff. Fifty are nurse’s aides, 50 are practical nurses, and 100 are registered nurses. Nurse’s aides receive $8 an hour, practical nurses $15 an hour, and registered nurses $24 an hour. What is the weighted mean hourly wage?

16. Andrews and Associates specialize in corporate law. They charge $100 an hour for researching a case, $75 an hour for consultations, and $200 an hour for writing a brief. Last week one of the associates spent 10 hours consulting with her client, 10 hours researching the case, and 20 hours writing the brief. What was the weighted mean hourly charge for her legal services?

3.6 The Median We have stressed that, for data containing one or two very large or very small val- ues, the arithmetic mean may not be representative. The center for such data can be better described by a measure of location called the median.

To illustrate the need for a measure of location other than the arithmetic mean, suppose you are seeking to buy a condominium in Palm Aire. Your real estate agent says that the typical price of the units currently available is $110,000. Would you still want to look? If you had budgeted your maximum purchase price at $75,000, you might think they are out of your price range. However, checking the prices of the individual units might change your mind. They are $60,000, $65,000, $70,000, and $80,000, and a superdeluxe penthouse costs $275,000. The arithmetic mean price is $110,000, as the real estate agent reported, but one price ($275,000) is pulling the arithmetic mean upward, causing it to be an unrepresentative average. It does seem that a price around $70,000 is a more typical or representative aver- age, and it is. In cases such as this, the median provides a more valid measure of location.

MEDIAN The midpoint of the values after they have been ordered from the smallest to the largest, or the largest to the smallest.

The median price of the units available is $70,000. To determine this, we order the prices from low ($60,000) to high ($275,000) and select the middle value ($70,000). For the median, the data must be at least an ordinal level of measurement.

Prices Ordered Prices Ordered from Low to High from High to Low

$ 60,000 $275,000 65,000 80,000 70,000 ← Median → 70,000 80,000 65,000

275,000 60,000

LO4 Determine the median.

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Describing Data: Numerical Measures 65

Median less affected by extreme values

The major properties of the median are:

1. It is not affected by extremely large or small values. Therefore, the median is a valuable measure of location when such values do occur.

2. It can be computed for ordinal-level data or higher. Recall from Chapter 1 that ordinal-level data can be ranked from low to high.

3.7 The Mode The mode is another measure of location.

The mode is especially useful in summarizing nominal-level data. As an exam- ple of its use for nominal-level data, a company has developed five bath oils. The bar chart in Chart 3–1 shows the results of a marketing survey designed to find which bath oil consumers prefer. The largest number of respondents favored Lamoure, as evidenced by the highest bar. Thus, Lamoure is the mode.

Example Facebook is a popular social networking website. Users can add friends and send them messages, and update their personal profiles to notify friends about them- selves and their activities. A sample of 10 adults revealed they spent the following number of hours last month using Facebook.

Find the median number of hours.

3 5 7 5 9 1 3 9 17 10

Solution Note that the number of adults sampled is even (10). The first step, as before, is to order the hours using Facebook from low to high. Then identify the two middle times. The arithmetic mean of the two middle observations gives us the median hours. Arranging the values from low to high:

The median is found by averaging the two middle values. The middle values are 5 hours and 7 hours, and the mean of these two values is 6. We conclude that the typical Facebook user spends 6 hours per month at the website. Notice that the median is not one of the values. Also, half of the times are below the median and half are above it.

1 3 3 5 5 7 9 9 10 17

MODE The value of the observation that appears most frequently.

The median can be determined for all levels of data but the nominal.

LO5 Identify the mode.

Note that there is the same number of prices below the median of $70,000 as above it. The median is, therefore, unaffected by extremely low or high prices. Had the highest price been $90,000, or $300,000, or even $1 million, the median price would still be $70,000. Likewise, had the lowest price been $20,000 or $50,000, the median price would still be $70,000.

In the previous illustration, there is an odd number of observations (five). How is the median determined for an even number of observations? As before, the obser- vations are ordered. Then by convention to obtain a unique value we calculate the mean of the two middle observations. So for an even number of observations, the median may not be one of the given values.

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66 Chapter 3

N um

be r

of R

es po

ns es

Bath oil

Amor Lamoure Soothing

300

200

100

400

Smell Nice Far Out

Mode

CHART 3–1 Number of Respondents Favoring Various Bath Oils

Example Recall the data regarding the distance in miles between exits on I-75 through Kentucky. The information is repeated below.

What is the modal distance?

11 4 10 4 9 3 8 10 3 14 1 10 3 5 2 2 5 6 1 2 2 3 7 1 3 7 8 10 1 4 7 5 2 2 5 1 1 3 3 1 2 1

Solution The first step is to organize the distances into a frequency table. This will help us determine the distance that occurs most frequently.

The distance that occurs most often is one mile. This happens eight times—that is, there are eight exits that are one mile apart. So the modal distance between exits is one mile.

Which of the three measures of location (mean, median, or mode) best repre- sents the central location of this data? Is the mode the best measure of location to represent the Kentucky data? No. The mode assumes only the nominal scale of

Distance in Miles between Exits Frequency

1 8 2 7 3 7 4 3 5 4 6 1 7 3 8 2 9 1

10 4 11 1 14 1

Total 42

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Describing Data: Numerical Measures 67

Self-Review 3–3 1. A sample of single persons in Towson, Texas, receiving Social Security payments revealed these monthly benefits: $852, $598, $580, $1,374, $960, $878, and $1,130. (a) What is the median monthly benefit? (b) How many observations are below the median? Above it?

2. The number of work stoppages in the automobile industry for selected months are 6, 0, 10, 14, 8, and 0. (a) What is the median number of stoppages? (b) How many observations are below the median? Above it? (c) What is the modal number of work stoppages?

In summary, we can determine the mode for all levels of data—nominal, ordi- nal, interval, and ratio. The mode also has the advantage of not being affected by extremely high or low values.

The mode does have disadvantages, however, that cause it to be used less fre- quently than the mean or median. For many sets of data, there is no mode because no value appears more than once. For example, there is no mode for this set of price data: $19, $21, $23, $20, and $18. Since every value is different, however, it could be argued that every value is the mode. Conversely, for some data sets there is more than one mode. Suppose the ages of the individuals in a stock investment club are 22, 26, 27, 27, 31, 35, and 35. Both the ages 27 and 35 are modes. Thus, this grouping of ages is referred to as bimodal (having two modes). One would ques- tion the use of two modes to represent the location of this set of age data.

Disadvantages of the mode

Exercises 17. What would you report as the modal value for a set of observations if there were a total of:

a. 10 observations and no two values were the same? b. 6 observations and they were all the same? c. 6 observations and the values were 1, 2, 3, 3, 4, and 4?

For Exercises 18–20, determine the (a) mean, (b) median, and (c) mode.

18. The following is the number of oil changes for the last 7 days at the Jiffy Lube located at the corner of Elm Street and Pennsylvania Avenue.

41 15 39 54 31 15 33

19. The following is the percent change in net income from last year to this year for a sample of 12 construction companies in Denver.

5 1 �10 �6 5 12 7 8 2 5 �1 11

20. The following are the ages of the 10 people in the video arcade at the Southwyck Shop- ping Mall at 10 A.M.

12 8 17 6 11 14 8 17 10 8

measurement and the variable miles is measured using the ratio scale. We calcu- lated the mean to be 4.57 miles. See page 59. Is the mean the best measure of location to represent this data? Probably not. There are several cases in which the distance between exits is large. These values are affecting the mean, making it too large and not representative of the distances between exits. What about the median? The median distance is 3 miles. That is, half of the distances between exits are 3 miles or less. In this case, the median of 3 miles between exits is probably a more repre- sentative measure of the distance between exits.

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68 Chapter 3

21. Several indicators of long-term economic growth in the United States are listed below.

Economic Indicator Percent Change Economic Indicator Percent Change

Inflation 4.5% Real GNP 2.9% Exports 4.7 Investment (residential) 3.6 Imports 2.3 Investment (nonresidential) 2.1 Real disposable income 2.9 Productivity (total) 1.4 Consumption 2.7 Productivity (manufacturing) 5.2

a. What is the median percent change? b. What is the modal percent change?

22. Sally Reynolds sells real estate along the coastal area of Northern California. Below is the total amount of her commissions earned since 2000. Find the mean, median, and mode of the commissions she earned for the 11 years.

23. The accounting firm of Rowatti and Koppel specializes in income tax returns for self- employed professionals, such as physicians, dentists, architects, and lawyers. The firm employs 11 accountants who prepare the returns. For last year, the number of returns prepared by each accountant was:

58 75 31 58 46 65 60 71 45 58 80

Find the mean, median, and mode for the number of returns prepared by each accoun- tant. If you could report only one, which measure of location would you recommend reporting?

24. The demand for the video games provided by Mid-Tech Video Games Inc. has exploded in the last several years. Hence, the owner needs to hire several new technical people to keep up with the demand. Mid-Tech gives each applicant a special test that Dr. McGraw, the designer of the test, believes is closely related to the ability to create video games. For the general population, the mean on this test is 100. Below are the scores on this test for the applicants.

95 105 120 81 90 115 99 100 130 10

The president is interested in the overall quality of the job applicants based on this test. Compute the mean and the median score for the ten applicants. What would you report to the president? Does it seem that the applicants are better than the general population?

Year Amount (thousands)

2000 $237.51 2001 233.80 2002 206.97 2003 248.14 2004 164.69 2005 292.16 2006 269.11 2007 225.57 2008 255.33 2009 202.67 2010 206.53

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Describing Data: Numerical Measures 69

3.9 The Relative Positions of the Mean, Median, and Mode Refer to the histogram in Chart 3–2 at the top of the following page. It is a symmetric distribution, which is also mound-shaped. This distribution has the same shape on either side of the center. If the polygon were folded in half, the two halves would be identical. For any symmetric distribution, the mode, median, and mean are located at the center and are always equal. They are all equal to 20 years in Chart 3–2. We should point out that there are symmetric distributions that are not mound-shaped.

For a symmetric, mound- shaped distribution, mean, median, and mode are equal.

Example Table 2–4 on page 30 shows the profit on the sales of 180 vehicles at Applewood Auto Group. Determine the mean and the median selling price.

Solution The mean, median, and modal amounts of profit are reported in the following Excel output (highlighted in the screen shot). (Remember: The instructions to create the output appear in the Software Commands section at the end of the chapter.) There are 180 vehicles in the study, so the calculations with a calculator would be tedious and prone to error.

The mean profit is $1,843.17 and the median is $1,882.50. These two values are less than $40 apart, so either value is reasonable. We can also see from the Excel output that there were 180 vehicles sold and their total profit was $331,770.00. We will describe the meaning of standard error, standard deviation, and other mea- sures reported on the output later in this chapter and in later chapters.

What can we conclude? The typical profit on a vehicle is about $1,850. Manage- ment at Applewood might use this value for revenue projections. For example, if the dealership could increase the number sold in a month from 180 to 200, this would result in an additional estimated $37,000 of revenue, found by 20($1,850).

3.8 Software Solution We can use a statistical software package to find many measures of location.

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70 Chapter 3

Mode $300

Fr eq

ue nc

Median $500

Mean $700

Skewed to the right (positively skewed)

Weekly Income

CHART 3–3 A Positively Skewed Distribution

Conversely, if a distribution is negatively skewed, the mean is the lowest of the three measures. The mean is, of course, influenced by a few extremely low observations. The median is greater than the arithmetic mean, and the modal value is the largest of the three measures. Again, if the distribution is highly skewed, such

Symmetric (zero skewness)

Years

Fr eq

ue nc

Mean = 20 Median = 20 Mode = 20

CHART 3–2 A Symmetric Distribution

The number of years corresponding to the highest point of the curve is the mode (20 years). Because the distribution is symmetrical, the median corresponds to the point where the distribution is cut in half (20 years). The total number of frequen- cies representing many years is offset by the total number representing few years, resulting in an arithmetic mean of 20 years. Logically, any of the three measures would be appropriate to represent the distribution’s center.

If a distribution is nonsymmetrical, or skewed, the relationship among the three measures changes. In a positively skewed distribution, the arithmetic mean is the largest of the three measures. Why? Because the mean is influenced more than the median or mode by a few extremely high values. The median is generally the next largest measure in a positively skewed frequency distribution. The mode is the small- est of the three measures.

If the distribution is highly skewed, such as the weekly incomes in Chart 3–3, the mean would not be a good measure to use. The median and mode would be more representative.

A skewed distribution is not symmetrical.

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Describing Data: Numerical Measures 71

Self-Review 3–4 The weekly sales from a sample of Hi-Tec electronic supply stores were organized into a frequency distribution. The mean of weekly sales was computed to be $105,900, the median $105,000, and the mode $104,500. (a) Sketch the sales in the form of a smoothed frequency polygon. Note the location of

the mean, median, and mode on the X-axis. (b) Is the distribution symmetrical, positively skewed, or negatively skewed? Explain.

as the distribution of tensile strengths shown in Chart 3–4, the mean should not be used to represent the data.

Mean 2,600

Fr eq

ue nc

Median 2,800

Mode 3,000

Skewed to the left (negatively skewed)

Tensile Strength

CHART 3–4 A Negatively Skewed Distribution

Exercises 25. The unemployment rate in the state of Alaska by month is given in the table below:

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

8. 7 8.8 8.7 7.8 7.3 7.8 6.6 6.5 6.5 6.8 7.3 7.6

a. What is the arithmetic mean of the Alaska unemployment rates? b. Find the median and the mode for the unemployment rates. c. Compute the arithmetic mean and median for just the winter (Dec–Mar) months. Is it

much different? 26. Big Orange Trucking is designing an information system for use in “in-cab” communica-

tions. It must summarize data from eight sites throughout a region to describe typical conditions. Compute an appropriate measure of central location for the variables wind direc- tion, temperature, and pavement.

City Wind Direction Temperature Pavement

Anniston, AL West 89 Dry Atlanta, GA Northwest 86 Wet Augusta, GA Southwest 92 Wet Birmingham, AL South 91 Dry Jackson, MS Southwest 92 Dry Meridian, MS South 92 Trace Monroe, LA Southwest 93 Wet Tuscaloosa, AL Southwest 93 Trace

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3.10 The Geometric Mean The geometric mean is useful in finding the average change of percentages, ratios, indexes, or growth rates over time. It has a wide application in business and econom- ics because we are often interested in finding the percentage changes in sales, salaries, or economic figures, such as the Gross Domestic Product, which compound or build on each other. The geometric mean of a set of n positive numbers is defined as the nth root of the product of n values. The formula for the geometric mean is written:

GEOMETRIC MEAN [3–4]

The geometric mean will always be less than or equal to (never more than) the arith- metic mean. Also, all the data values must be positive.

As an example of the geometric mean, suppose you receive a 5 percent increase in salary this year and a 15 percent increase next year. The average annual percent increase is 9.886, not 10.0. Why is this so? We begin by calculating the geometric mean. Recall, for example, that a 5 percent increase in salary is 105 per- cent. We will write it as 1.05.

This can be verified by assuming that your monthly earning was $3,000 to start and you received two increases of 5 percent and 15 percent.

Total $622.50

Your total salary increase is $622.50. This is equivalent to:

The following example shows the geometric mean of several percentages.

$622.48 is about $622.50 $3,296.58 (.09886) � 325.90

$3,000.00 (.09886) � $296.58

Raise 2 � $3,150 (.15) � 472.50

Raise 1 � $3,000 (.05) � $150.00

GM � 1(1.05)(1.15) � 1.09886

GM � n 2(X1)(X2) � � � (Xn)

72 Chapter 3

Example The return on investment earned by Atkins Construction Company for four succes- sive years was: 30 percent, 20 percent, �40 percent, and 200 percent. What is the geometric mean rate of return on investment?

Solution The number 1.3 represents the 30 percent return on investment, which is the “orig- inal” investment of 1.0 plus the “return” of 0.3. The number 0.6 represents the loss of 40 percent, which is the original investment of 1.0 less the loss of 0.4. This cal- culation assumes the total return each period is reinvested or becomes the base for the next period. In other words, the base for the second period is 1.3 and the base for the third period is (1.3)(1.2) and so forth.

Then the geometric mean rate of return is 29.4 percent, found by

The geometric mean is the fourth root of 2.808. So, the average rate of return (com- pound annual growth rate) is 29.4 percent.

Notice also that if you compute the arithmetic mean [(30 � 20 � 40 � 200)�4 � 52.5], you would have a much larger number, which would overstate the true rate of return!

GM � n 2(X1)(X2) � � � (Xn) �

4 2(1.3)(1.2)(0.6)(3.0) �

4 22.808 � 1.294

The geometric mean is never greater than the arithmetic mean.

LO6 Calculate the geometric mean.

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Describing Data: Numerical Measures 73

A second application of the geometric mean is to find an average percentage change over a period of time. For example, if you earned $30,000 in 2000 and $50,000 in 2010, what is your annual rate of increase over the period? It is 5.24 per- cent. The rate of increase is determined from the following formula.

RATE OF INCREASE [3–5]

In the above box, n is the number of periods. An example will show the details of finding the average annual percent increase.

GM � n B

Value at end of period Value at start of period

� 1OVER TIME

Self-Review 3–5 1. The percent increase in sales for the last 4 years at Combs Cosmetics were: 4.91, 5.75, 8.12, and 21.60. (a) Find the geometric mean percent increase. (b) Find the arithmetic mean percent increase. (c) Is the arithmetic mean equal to or greater than the geometric mean?

2. Production of Cablos trucks increased from 23,000 units in 2000 to 120,520 in 2010. Find the geometric mean annual percent increase.

Example During the decade of the 1990s, and into the 2000s, Las Vegas, Nevada, was the fastest-growing city in the United States. The population increased from 258,295 in 1990 to 607,876 in 2009. This is an increase of 349,581 people, or a 135.3 percent increase over the period. The population has more than doubled. What is the aver- age annual percent increase?

Solution There are 19 years between 1990 and 2009, so n � 19. Then formula (3–5) for the geometric mean as applied to this problem is:

The value of .0461 indicates that the average annual growth over the period was 4.61 percent. To put it another way, the population of Las Vegas increased at a rate of 4.61 percent per year from 1990 to 2009.

GM � B

19 Value at end of period Value at start of period � 1.0 � B

19 607,876 258,295

� 1.0 � 1.0461� 1.0 � .0461

Exercises 27. Compute the geometric mean of the following percent increases: 8, 12, 14, 26, and 5. 28. Compute the geometric mean of the following percent increases: 2, 8, 6, 4, 10, 6, 8,

and 4. 29. Listed below is the percent increase in sales for the MG Corporation over the last 5 years.

Determine the geometric mean percent increase in sales over the period.

9.4 13.8 11.7 11.9 14.7

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74 Chapter 3

Type of College 1980–81 2007–08

Public $2,550 $ 6,966 Private 5,594 13,424

3.11 Why Study Dispersion? A measure of location, such as the mean or the median, only describes the cen- ter of the data. It is valuable from that standpoint, but it does not tell us anything about the spread of the data. For example, if your nature guide told you that the river ahead averaged 3 feet in depth, would you want to wade across on foot without additional information? Probably not. You would want to know something about the variation in the depth. Is the maximum depth of the river 3.25 feet and the minimum 2.75 feet? If that is the case, you would probably agree to cross. What if you learned the river depth ranged from 0.50 feet to 5.5 feet? Your deci- sion would probably be not to cross. Before making a decision about crossing the river, you want information on both the typical depth and the dispersion in the depth of the river.

A small value for a measure of dispersion indicates that the data are clustered closely, say, around the arithmetic mean. The mean is therefore considered repre- sentative of the data. Conversely, a large measure of dispersion indicates that the mean is not reliable. Refer to Chart 3–5. The 100 employees of Hammond Iron Works Inc. a steel fabricating company, are organized into a histogram based on the number of years of employment with the company. The mean is 4.9 years, but the spread of the data is from 6 months to 16.8 years. The mean of 4.9 years is not very representative of all the employees.

A second reason for studying the dispersion in a set of data is to compare the spread in two or more distributions. Suppose, for example, that the new Vision Quest LCD computer monitor is assembled in Baton Rouge and also in Tucson. The arithmetic mean hourly output in both the Baton Rouge plant and the Tucson plant is 50. Based on the two means, you might conclude that the distributions of the hourly outputs are identical. Production records for 9 hours at the two plants, however, reveal that this conclusion is not correct (see Chart 3–6). Baton Rouge production varies from 48 to 52 assemblies per hour. Production at the Tucson plant is more erratic, ranging from 40 to 60 per hour. Therefore, the hourly output for Baton Rouge is clustered near the mean of 50; the hourly output for Tucson is more dispersed.

30. In 1996, a total of 14,968,000 taxpayers in the United States filed their individual tax returns electronically. By the year 2009, the number increased to 95,000,000. What is the geometric mean annual increase for the period?

31. The Consumer Price Index is reported monthly by the U.S. Bureau of Labor Statistics. It reports the change in prices for a market basket of goods from one period to another. The index for 2000 was 172.2. By 2009, it increased to 214.5. What was the geometric mean annual increase for the period?

32. JetBlue Airways is an American low-cost airline headquartered in New York City. Its main base is John F. Kennedy International Airport. JetBlue’s revenue in 2002 was $635.2 mil- lion. By 2009, revenue had increased to $3,290.0 million. What was the geometric mean annual increase for the period?

33. In 1985, there were 340,213 cell phone subscribers in the United States. By 2008, the number of cell phone subscribers increased to 262,700,000. What is the geometric mean annual increase for the period?

34. The information below shows the cost for a year of college in public and private colleges in 1980–81 and 2007–08. What is the geometric mean annual increase for the period for the two types of colleges? Compare the rates of increase.

Statistics in Action

The United States Postal Service has tried to become more “user friendly” in the last several years. A re- cent survey showed that customers were interested in more consistency in the time it takes to make a delivery. Under the old conditions, a local letter might take only one day to deliver, or it might take several. “Just tell me how many days ahead I need to mail the birthday card to Mom so it gets there on her birthday, not early, not late,” was a com- mon complaint. The level of consistency is measured by the stan- dard deviation of the delivery times.

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Describing Data: Numerical Measures 75

Years

Em pl

oy ee

10 20

CHART 3–5 Histogram of Years of Employment at Hammond Iron Works Inc.

LO7 Explain and apply measures of dispersion.

A measure of dispersion can be used to evaluate the reliability of two or more measures of location.

48 49 50 51 52 _ X

53 54 55 56 57 58 59 604746454443424140

Daily Production

Baton Rouge

Tucson

CHART 3–6 Hourly Production of Computer Monitors at the Baton Rouge and Tucson Plants

3.12 Measures of Dispersion We will consider several measures of dispersion. The range is based on the largest and the smallest values in the data set, that is, only two values are considered. The mean deviation, the variance, and the standard deviation use all the values in a data set and are all based on deviations from the arithmetic mean.

Range The simplest measure of dispersion is the range. It is the difference between the largest and the smallest values in a data set. In the form of an equation:

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76 Chapter 3

Mean Deviation A defect of the range is that it is based on only two values, the highest and the lowest; it does not take into consideration all of the values. The mean deviation does. It measures the mean amount by which the values in a population, or sam- ple, vary from their mean. In terms of a definition:

In terms of a formula, the mean deviation, designated MD, is computed for a sample by:

MEAN DEVIATION [3–7]

where: X is the value of each observation.

is the arithmetic mean of the values. n is the number of observations in the sample.

indicates the absolute value.

Why do we ignore the signs of the deviations from the mean? If we didn’t, the posi- tive and negative deviations from the mean would exactly offset each other, and the mean deviation would always be zero. Such a measure (zero) would be a useless statistic.

@ @

MD � � @X � X @

MEAN DEVIATION The arithmetic mean of the absolute values of the deviations from the arithmetic mean.

Example Refer to Chart 3–6 on the previous page. Find the range in the number of computer monitors produced per hour for the Baton Rouge and the Tucson plants. Interpret the two ranges.

Solution The range of the hourly production of computer monitors at the Baton Rouge plant is 4, found by the difference between the largest hourly production of 52 and the smallest of 48. The range in the hourly production for the Tucson plant is 20 com- puter monitors, found by 60 � 40. We therefore conclude that (1) there is less dis- persion in the hourly production in the Baton Rouge plant than in the Tucson plant because the range of 4 computer monitors is less than a range of 20 computer monitors, and (2) the production is clustered more closely around the mean of 50 at the Baton Rouge plant than at the Tucson plant (because a range of 4 is less than a range of 20). Thus, the mean production in the Baton Rouge plant (50 com- puter monitors) is a more representative measure of location than the mean of 50 computer monitors for the Tucson plant.

RANGE [3–6]

The range is widely used in statistical process control (SPC) applications because it is very easy to calculate and understand.

Range � Largest value � Smallest value

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Describing Data: Numerical Measures 77

Example The chart below shows the number of cap- puccinos sold at the Starbucks in the Orange County airport and the Ontario, California, air- port between 4 and 5 P.M. for a sample of five days last month.

Determine the mean, median, range, and mean deviation for each location. Comment on the similarities and differences in these measures.

Notice that all three of the measures are exactly the same. Does this indicate that there is no difference in the two sets of data? We get a clearer picture if we calcu- late the mean deviations. First, for Orange County:

MD � � @ X � X @

n �

30 � 10 � 0 � 10 � 30 5

� 80 5

� 16

Solution The mean, median, and range for each of the airport locations are reported below as part of an Excel spreadsheet.

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78 Chapter 3

Self-Review 3–6 The weights of containers being shipped to Ireland are (in thousands of pounds):

95 103 105 110 104 105 112 90

(a) What is the range of the weights? (b) Compute the arithmetic mean weight. (c) Compute the mean deviation of the weights.

The mean deviation is 16 cappuccinos. That is, the number of cappuccinos sold deviates, on average, by 16 from the mean of 50 cappuccinos.

The following shows the detail of determining the mean deviation for the num- ber of cappuccinos sold at the Ontario Airport.

So the mean, median, and range of the cappuccinos sold are the same at the two airports, but the mean deviations are different. The mean deviation at Orange County is 16, but it is 12.4 at Ontario.

Let’s interpret and compare the results of our measures for the two Starbucks airport locations. The mean and median of the two locations are exactly the same, 50 cappuccinos sold. These measures of location indicate the two distributions are the same. The range for both locations is also the same, 60. However, recall that the range provides limited information about the dispersion, because it is based on only two of the observations.

The mean deviations are not the same for the two airports. The mean deviation is based on the differences between each observation and the arithmetic mean. It shows the closeness or clustering of the data relative to the mean or center of the distribution. Compare the mean deviation for Orange County of 16 to the mean devi- ation for Ontario of 12.4. Based on the mean deviation, we conclude that the dis- persion for the sales distribution of the Ontario Starbucks is more concentrated—that is, nearer the mean of 50—than the Orange County location.

The mean deviation has two advantages. First, it uses all the values in the computation. Recall that the range uses only the highest and the lowest values. Second, it is easy to understand—it is the average amount by which values devi- ate from the mean. However, its drawback is the use of absolute values. Generally, absolute values are difficult to work with and to explain, so the mean deviation is not used as frequently as other measures of dispersion, such as the standard deviation.

MD � � @ X � X @

n �

30 � 1 � 0 � 1 � 30 5

� 62 5

� 12.4

Advantages of the mean deviation.

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Describing Data: Numerical Measures 79

Exercises For Exercises 35–38, calculate the (a) range, (b) arithmetic mean, (c) mean deviation, and (d) interpret the values.

35. There were five customer service representatives on duty at the Electronic Super Store during last weekend’s sale. The numbers of HDTVs these representatives sold are: 5, 8, 4, 10, and 3.

37. Dave’s Automatic Door installs automatic garage door openers. The following list indi- cates the number of minutes needed to install a sample of 10 door openers: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42.

38. A sample of eight companies in the aerospace industry was surveyed as to their return on investment last year. The results are (in percent): 10.6, 12.6, 14.8, 18.2, 12.0, 14.8, 12.2, and 15.6.

39. Ten randomly selected young adults living in California rated the taste of a newly devel- oped sushi pizza topped with tuna, rice, and kelp on a scale of 1 to 50, with 1 indicat- ing they did not like the taste and 50 that they did. The ratings were:

34 39 40 46 33 31 34 14 15 45

In a parallel study, 10 randomly selected young adults in Iowa rated the taste of the same pizza. The ratings were:

28 25 35 16 25 29 24 26 17 20

As a market researcher, compare the potential markets for sushi pizza. 40. A sample of the personnel files of eight employees at the Pawnee location of Acme Car-

pet Cleaners Inc. revealed that during the last six-month period they lost the following number of days due to illness:

2 0 6 3 10 4 1 2

A sample of eight employees during the same period at the Chickpee location of Acme Carpets revealed they lost the following number of days due to illness.

2 0 1 0 5 0 1 0

As the director of human relations, compare the two locations. What would you recommend?

Variance and Standard Deviation The variance and standard deviation are also based on the deviations from the mean. However, instead of using the absolute value of the deviations, the variance and the standard deviation square the deviations.

VARIANCE The arithmetic mean of the squared deviations from the mean.

LO8 Compute and explain the variance and the standard deviation.

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80 Chapter 3

POPULATION VARIANCE [3–8]

where: �2 is the population variance (� is the lowercase Greek letter sigma). It is read

as “sigma squared.” X is the value of an observation in the population. � is the arithmetic mean of the population. N is the number of observations in the population.

Note the process of computing the variance.

1. Begin by finding the mean. 2. Find the difference between each observation and the mean, and square that

difference. 3. Sum all the squared differences. 4. Divide the sum of the squared differences by the number of items in the

population.

So you might think of the population variance as the mean of the squared differ- ence between each value and the mean. For populations whose values are near the mean, the variance will be small. For populations whose values are dispersed from the mean, the population variance will be large.

The variance overcomes the weakness of the range by using all the values in the population, whereas the range uses only the largest and the smallest. We over- come the issue where (X � �) � 0 by squaring the differences, instead of using the absolute values. Squaring the differences will always result in non-negative values.

�

�2 � �(X � �)2

Variance and standard deviation are based on squared deviations from the mean.

Example The number of traffic citations issued last year by month in Beaufort County, South Carolina, is reported below.

Determine the population variance.

Month January February March April May June July August September October November December Citations 19 17 22 18 28 34 45 39 38 44 34 10

Solution Because we are studying all the citations for a year, the data comprise a population. To determine the population variance, we use formula (3-8). The table below details the calculations.

STANDARD DEVIATION The square root of the variance.

The variance is non-negative and is zero only if all observations are the same.

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Describing Data: Numerical Measures 81

Citations Month (X ) X � � (X � �)2

January 19 �10 100 February 17 �12 144 March 22 �7 49 April 18 �11 121 May 28 �1 1 June 34 5 25 July 45 16 256 August 39 10 100 September 38 9 81 October 44 15 225 November 34 5 25 December 10 �19 361

Total 348 0 1,488

1. We begin by determining the arithmetic mean of the population. The total num- ber of citations issued for the year is 348, so the mean number issued per month is 29.

2. Next we find the difference between each observation and the mean. This is shown in the third column of the table. Recall that earlier in the chapter (page 61) we indicated that the sum of the differences between each value and the mean is 0. From the spreadsheet, the sum of the differences between the mean and the number of citations each month is 0.

3. The next step is to square the difference between each monthly value. That is shown in the fourth column of the table. By squaring the differences, we convert both the positive and the negative values to a plus sign. Hence, each difference will be positive.

4. The squared differences are totaled. The total of the fourth column is 1,488. That is the term �(X � �)2.

5. Finally, we divide the squared differences by N, the number of observations in the population.

So, the population variation for the number of citations is 124.

Like the range and the mean deviation, the variance can be used to compare dispersion in two or more sets of observations. For example, the variance for the number of citations issued in Beaufort County was just computed to be 124. If the variance in the number of citations issued in Marlboro County, South Carolina, is 342.9, we conclude that (1) there is less dispersion in the distribution of the num- ber of citations issued in Beaufort County than in Marlboro County (because 124 is less than 342.9); and (2) the number of citations in Beaufort County is more closely clustered around the mean of 29 than for the number of citations issued in Marlboro County. Thus the mean number of citations issued in Beaufort County is a more representative measure of location than the mean number of citations in Marlboro County.

2 � ©(X � �)2

N �

1488 12

� 124

� � �X N

� 19 � 17 � . . . � 10

12 �

348 12

� 29

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82 Chapter 3

Population Standard Deviation Both the range and the mean deviation are easy to interpret. The range is the difference between the high and low values of a set of data, and the mean deviation is the mean of the deviations from the mean. However, the variance is difficult to interpret for a single set of observations. The variance of 124 for the number of citations issued is not in terms of citations, but citations squared.

There is a way out of this difficulty. By taking the square root of the population variance, we can transform it to the same unit of measurement used for the origi- nal data. The square root of 124 citations-squared is 11.14 citations. The units are now simply citations. The square root of the population variance is the population standard deviation.

POPULATION STANDARD DEVIATION [3–9]� � B

�(X � �)2

Variance is difficult to interpret because the units are squared.

Standard deviation is in the same units as the data.

Self-Review 3–7 The Philadelphia office of Price Waterhouse Coopers LLP hired five accounting trainees this year. Their monthly starting salaries were: $3,536; $3,173; $3,448; $3,121; and $3,622. (a) Compute the population mean. (b) Compute the population variance. (c) Compute the population standard deviation. (d) The Pittsburgh office hired six trainees. Their mean monthly salary was $3,550, and the

standard deviation was $250. Compare the two groups.

Exercises 41. Consider these five values a population: 8, 3, 7, 3, and 4.

a. Determine the mean of the population. b. Determine the variance.

42. Consider these six values a population: 13, 3, 8, 10, 8, and 6. a. Determine the mean of the population. b. Determine the variance.

43. The annual report of Dennis Industries cited these primary earnings per common share for the past 5 years: $2.68, $1.03, $2.26, $4.30, and $3.58. If we assume these are pop- ulation values, what is: a. The arithmetic mean primary earnings per share of common stock? b. The variance?

44. Referring to Exercise 43, the annual report of Dennis Industries also gave these returns on stockholder equity for the same five-year period (in percent): 13.2, 5.0, 10.2, 17.5, and 12.9. a. What is the arithmetic mean return? b. What is the variance?

45. Plywood Inc. reported these returns on stockholder equity for the past 5 years: 4.3, 4.9, 7.2, 6.7, and 11.6. Consider these as population values. a. Compute the range, the arithmetic mean, the variance, and the standard deviation. b. Compare the return on stockholder equity for Plywood Inc. with that for Dennis

Industries cited in Exercise 44. 46. The annual incomes of the five vice presidents of TMV Industries are: $125,000; $128,000;

$122,000; $133,000; and $140,000. Consider this a population. a. What is the range? b. What is the arithmetic mean income? c. What is the population variance? The standard deviation? d. The annual incomes of officers of another firm similar to TMV Industries were also

studied. The mean was $129,000 and the standard deviation $8,612. Compare the means and dispersions in the two firms.

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Describing Data: Numerical Measures 83

Sample Variance The formula for the population mean is We just changed the symbols for the sample mean; that is, Unfortunately, the conversion from the population variance to the sample variance is not as direct. It requires a change in the denominator. Instead of substituting n (number in the sample) for N (number in the population), the denominator is n � 1. Thus the formula for the sample variance is:

SAMPLE VARIANCE [3–10]

where: s2 is the sample variance. X is the value of each observation in the sample.

is the mean of the sample. n is the number of observations in the sample.

Why is this change made in the denominator? Although the use of n is logical since is used to estimate �, it tends to underestimate the population variance, 2. The use of (n � 1) in the denominator provides the appropriate correction for this tendency. Because the primary use of sample statistics like s2 is to estimate population parameters like 2, (n � 1) is preferred to n in defining the sample variance. We will also use this convention when computing the sample standard deviation.

s2 � �( X � X )2

n � 1

X � �X�n. � � �X�N.

Example The hourly wages for a sample of part-time employees at Home Depot are: $12, $20, $16, $18, and $19. What is the sample variance?

Solution The sample variance is computed by using formula (3–10).

X � �X n

� $85

5 � $17

� 10 in dollars squared

s2 � �(X � X)2

n � 1 �

40 5 � 1

Hourly Wage (X ) (X � ) 2

$12 �$5 25 20 3 9 16 �1 1 18 1 1 19 2 4

$85 0 40

XX � X

Sample Standard Deviation The sample standard deviation is used as an esti- mator of the population standard deviation. As noted previously, the population stan- dard deviation is the square root of the population variance. Likewise, the sample

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Exercises For Exercises 47–52, do the following:

a. Compute the sample variance. b. Determine the sample standard deviation.

47. Consider these values a sample: 7, 2, 6, 2, and 3. 48. The following five values are a sample: 11, 6, 10, 6, and 7.

84 Chapter 3

3.13 Software Solution On page 69, we used Excel to determine the mean and median of the Applewood Auto Group data. You will also note that it lists the sample standard deviation. Excel, like most other statistical software, assumes the data are from a sample.

Another software package that we will use in this text is Minitab. This pack- age uses a spreadsheet format, much like Excel, but produces a wider variety of sta- tistical information. The information for the profit on the sales of 180 vehicles last month at Applewood Auto Group follows.

Self-Review 3–8 The years of service for a sample of seven employees at a State Farm Insurance claims office in Cleveland, Ohio, are: 4, 2, 5, 4, 5, 2, and 6. What is the sample variance? Com- pute the sample standard deviation.

standard deviation is the square root of the sample variance. The sample standard deviation is most easily determined by:

SAMPLE STANDARD DEVIATION [3–11]s � A

�(X � X )2

n � 1

Example The sample variance in the previous example involving hourly wages was computed to be 10. What is the sample standard deviation?

Solution The sample standard deviation is $3.16, found by . Note again that the sam- ple variance is in terms of dollars squared, but taking the square root of 10 gives us $3.16, which is in the same units (dollars) as the original data.

110

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49. Dave’s Automatic Door, referred to in Exercise 37, installs automatic garage door open- ers. Based on a sample, following are the times, in minutes, required to install 10 door openers: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42.

50. The sample of eight companies in the aerospace industry, referred to in Exercise 38, was surveyed as to their return on investment last year. The results are: 10.6, 12.6, 14.8, 18.2, 12.0, 14.8, 12.2, and 15.6.

51. The Houston, Texas, Motel Owner Association conducted a survey regarding weekday motel rates in the area. Listed below is the room rate for business-class guests for a sample of 10 motels.

$101 $97 $103 $110 $78 $87 $101 $80 $106 $88

52. A consumer watchdog organization is concerned about credit card debt. A survey of 10 young adults with credit card debt of more than $2,000 showed they paid an average of just over $100 per month against their balances. Listed below are the amounts each young adult paid last month.

$110 $126 $103 $93 $99 $113 $87 $101 $109 $100

3.14 Interpretation and Uses of the Standard Deviation The standard deviation is commonly used as a measure to compare the spread in two or more sets of observations. For example, the standard deviation of the biweekly amounts invested in the Dupree Paint Company profit-sharing plan is com- puted to be $7.51. Suppose these employees are located in Georgia. If the stan- dard deviation for a group of employees in Texas is $10.47, and the means are about the same, it indicates that the amounts invested by the Georgia employees are not dispersed as much as those in Texas (because $7.51 < $10.47). Since the amounts invested by the Georgia employees are clustered more closely about the mean, the mean for the Georgia employees is a more reliable measure than the mean for the Texas group.

Chebyshev’s Theorem We have stressed that a small standard deviation for a set of values indicates that these values are located close to the mean. Conversely, a large standard deviation reveals that the observations are widely scattered about the mean. The Russian mathematician P. L. Chebyshev (1821–1894) developed a theorem that allows us to determine the minimum proportion of the values that lie within a specified number of standard deviations of the mean. For example, according to Chebyshev’s theorem, at least three of four values, or 75 percent, must lie between the mean plus two standard deviations and the mean minus two standard deviations. This relationship applies regardless of the shape of the distribution. Further, at least eight of nine val- ues, or 88.9 percent, will lie between plus three standard deviations and minus three standard deviations of the mean. At least 24 of 25 values, or 96 percent, will lie between plus and minus five standard deviations of the mean.

Chebyshev’s theorem states:

Statistics in Action

Most colleges report the “average class size.” This informa- tion can be mislead- ing because average class size can be found in several ways. If we find the number of students in each class at a particular university, the result is the mean number of students per class. If we compile a list of the class sizes for each student and find the mean class size, we might find the mean to be quite different. One school found the mean number of students in each of its 747 classes to be 40. But when it found the mean from a list of the class sizes of each student, it was 147. Why the disparity? Because there are few

(continued)

Describing Data: Numerical Measures 85

CHEBYSHEV’S THEOREM For any set of observations (sample or population), the proportion of the values that lie within k standard deviations of the mean is at least 1 � 1/k2, where k is any constant greater than 1.

LO9 Explain Chebyshev’s Theorem and the Empirical Rule.

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students in the small classes and a larger number of students in the larger classes, which has the effect of increasing the mean class size when it is calculated this way. A school could reduce this mean class size for each stu- dent by reducing the number of students in each class. That is, cut out the large fresh- man lecture classes.

86 Chapter 3

The Empirical Rule Chebyshev’s theorem is concerned with any set of values; that is, the distribution of values can have any shape. However, for a symmetrical, bell-shaped distribution such as the one in Chart 3–7, we can be more precise in explaining the dispersion about the mean. These relationships involving the standard deviation and the mean are described by the Empirical Rule, sometimes called the Normal Rule.

EMPIRICAL RULE For a symmetrical, bell-shaped frequency distribution, approximately 68 percent of the observations will lie within plus and minus one standard deviation of the mean; about 95 percent of the observations will lie within plus and minus two standard deviations of the mean; and practically all (99.7 percent) will lie within plus and minus three standard deviations of the mean.

These relationships are portrayed graphically in Chart 3–7 for a bell-shaped distri- bution with a mean of 100 and a standard deviation of 10.

908070 110 120 130100 68% 95%

99.7%

CHART 3–7 A Symmetrical, Bell-Shaped Curve Showing the Relationships between the Standard Deviation and the Observations

The Empirical Rule applies only to symmetrical, bell- shaped distributions.

It has been noted that if a distribution is symmetrical and bell-shaped, practi- cally all of the observations lie between the mean plus and minus three standard deviations. Thus, if and s � 10, practically all the observations lie between 100 � 3(10) and 100 � 3(10), or 70 and 130. The estimated range is therefore 60, found by 130 � 70.

X � 100

Example The arithmetic mean biweekly amount contributed by the Dupree Paint employees to the company’s profit-sharing plan is $51.54, and the standard deviation is $7.51. At least what percent of the contributions lie within plus 3.5 standard deviations and minus 3.5 standard deviations of the mean?

Solution About 92 percent, found by

1 � 1 k2

� 1 � 1

(3.5)2 � 1 �

1 12.25

� 0.92

(continued from p. 85)

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Describing Data: Numerical Measures 87

Example A sample of the rental rates at University Park Apartments approximates a sym- metrical, bell-shaped distribution. The sample mean is $500; the standard deviation is $20. Using the Empirical Rule, answer these questions:

1. About 68 percent of the monthly rentals are between what two amounts? 2. About 95 percent of the monthly rentals are between what two amounts? 3. Almost all of the monthly rentals are between what two amounts?

Solution 1. About 68 percent are between $480 and $520, found by 2. About 95 percent are between $460 and $540, found by 3. Almost all (99.7 percent) are between $440 and $560, found by

3($20). X � 3s � $500 �

X � 2s � $500 � 2($20). X � 1s � $500 � 1($20).

Conversely, if we know that the range is 60, we can approximate the standard deviation by dividing the range by 6. For this illustration: range � 6 � 60 � 6 � 10, the standard deviation.

Self-Review 3–9 The Pitney Pipe Company is one of several domestic manufacturers of PVC pipe. The quality control department sampled 600 10-foot lengths. At a point 1 foot from the end of the pipe, they measured the outside diameter. The mean was 14.0 inches and the standard deviation 0.1 inches. (a) If the shape of the distribution is not known, at least what percent of the observations

will be between 13.85 inches and 14.15 inches? (b) If we assume that the distribution of diameters is symmetrical and bell-shaped, about

95 percent of the observations will be between what two values?

Exercises 53. According to Chebyshev’s theorem, at least what percent of any set of observations will

be within 1.8 standard deviations of the mean? 54. The mean income of a group of sample observations is $500; the standard deviation is

$40. According to Chebyshev’s theorem, at least what percent of the incomes will lie between $400 and $600?

55. The distribution of the weights of a sample of 1,400 cargo containers is symmetric and bell-shaped. According to the Empirical Rule, what percent of the weights will lie: a. Between and ? b. Between and ? Below ?

56. The following graph portrays the distribution of the number of Biggie-sized soft drinks sold at a nearby Wendy’s for the last 141 days. The mean number of drinks sold per day is 91.9 and the standard deviation is 4.67.

X � 2sX � 2sX X � 2sX � 2s

10090 Sales

If we use the Empirical Rule, sales will be between what two values on 68 percent of the days? Sales will be between what two values on 95 percent of the days?

Statistics in Action

Joe Mauer of the Minnesota Twins had the highest batting

(continued)

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average, at .365, dur- ing the 2009 Major League Baseball sea- son. Tony Gwynn hit .394 in the strike- shortened season of 1994, and Ted Williams hit .406 in 1941. No one has hit over .400 since 1941. The mean batting average has remained constant at about .260 for more than 100 years, but the standard deviation declined from .049 to .031. This indicates less dispersion in the batting averages today and helps explain the lack of any .400 hit- ters in recent times.

88 Chapter 3

3.15 The Mean and Standard Deviation of Grouped Data In most instances, measures of location, such as the mean, and measures of dis- persion, such as the standard deviation, are determined by using the individual values. Statistical software packages make it easy to calculate these values, even for large data sets. However, sometimes we are given only the frequency distrib- ution and wish to estimate the mean or standard deviation. In the following dis- cussion, we show how we can estimate the mean and standard deviation from data organized into a frequency distribution. We should stress that a mean or a standard deviation from grouped data is an estimate of the corresponding actual values.

The Arithmetic Mean To approximate the arithmetic mean of data organized into a frequency distribution, we begin by assuming the observations in each class are represented by the mid- point of the class. The mean of a sample of data organized in a frequency distrib- ution is computed by:

ARITHMETIC MEAN OF GROUPED DATA [3–12]

where: is the designation for the sample mean.

M is the midpoint of each class. f is the frequency in each class. fM is the frequency in each class times the midpoint of the class.

is the sum of these products. n is the total number of frequencies. �fM

X � �fM

Example The computations for the arithmetic mean of data grouped into a frequency distri- bution will be shown based on the Applewood Auto Group profit data. Recall in Chapter 2, in Table 2–7 on page 33, we constructed a frequency distribution for the vehicle profit. The information is repeated below. Determine the arithmetic mean profit per vehicle.

Profit Frequency

$ 200 up to $ 600 8 600 up to 1,000 11

1,000 up to 1,400 23 1,400 up to 1,800 38 1,800 up to 2,200 45 2,200 up to 2,600 32 2,600 up to 3,000 19 3,000 up to 3,400 4

Total 180

L10 Compute the mean and standard deviation of grouped data.

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Describing Data: Numerical Measures 89

Solution The mean vehicle selling price can be estimated from data grouped into a frequency distribution. To find the estimated mean, assume the midpoint of each class is rep- resentative of the data values in that class. Recall that the midpoint of a class is halfway between the lower class limits of two consecutive classes. To find the mid- point of a particular class, we add the lower limits of two consecutive classes and divide by 2. Hence, the midpoint of the first class is $400, found by ($200 � $600)�2. We assume the value of $400 is representative of the eight values in that class. To put it another way, we assume the sum of the eight values in this class is $3,200, found by 8($400). We continue the process of multiplying the class midpoint by the class frequency for each class and then sum these products. The results are sum- marized in Table 3–1.

Solving for the arithmetic mean using formula (3–12), we get:

We conclude that the mean profit per vehicle is about $1,851.

X � ©fM

n �

$333,200 180

� $1,851.11

Profit Frequency (f ) Midpoint (M ) fM

$ 200 up to $ 600 8 $ 400 $ 3,200 600 up to 1,000 11 800 8,800

1,000 up to 1,400 23 1,200 27,600 1,400 up to 1,800 38 1,600 60,800 1,800 up to 2,200 45 2,000 90,000 2,200 up to 2,600 32 2,400 76,800 2,600 up to 3,000 19 2,800 53,200 3,000 up to 3,400 4 3,200 12,800

Total 180 $333,200

TABLE 3–1 Profit on 180 Vehicles Sold Last Month at Applewood Auto Group

Standard Deviation To calculate the standard deviation of data grouped into a frequency distribution, we need to adjust formula (3–11) slightly. We weight each of the squared differences by the number of frequencies in each class. The formula is:

STANDARD DEVIATION, GROUPED DATA [3–13]

where: s is the symbol for the sample standard deviation. M is the midpoint of the class. f is the class frequency. n is the number of observations in the sample.

is the designation for the sample mean.X

s � B

�f(M � X)2

n � 1

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90 Chapter 3

Self-Review 3–10 The net incomes of a sample of large importers of antiques were organized into the following table:

Net Income Number of Net Income Number of ($ millions) Importers ($ millions) Importers

2 up to 6 1 14 up to 18 3 6 up to 10 4 18 up to 22 2

10 up to 14 10

Example Refer to the frequency distribution for the Applewood Auto Group profit data reported in Table 3–1. Compute the standard deviation of the vehicle selling prices.

Solution Following the same practice used earlier for computing the mean of data groupedinto a frequency distribution, f is the class frequency, M the class midpoint, and n the number of observations.

To find the standard deviation:

Step 1: Subtract the mean from the class midpoint. That is, find �(M � X ) ($400 � $1,851 � �$1,451) for the first class, for the second class ($800 � $1,851 � �$1,051), and so on.

Step 2: Square the difference between the class midpoint and the mean. For the first class, it would be ($400 � $1,851)2 � 2,105,401 for the second class ($800 � $1,851)2 � 1,104,601, and so on.

Step 3: Multiply the squared difference between the class midpoint and the mean by the class frequency. For the first class, the value is 8($400 � $1,851)2 � 16,843,208; for the second, 11($800 � $1,851)2 � 12,150,611, and so on.

Step 4: Sum the The total is 76,169,920. To find the standard deviation, we insert these values in formula (3–13).

The mean and the standard deviation calculated from the data grouped into a frequency distribution are usually close to the values calculated from raw data. The grouped data result in some loss of information. For the vehicle profit example, the mean profit reported in the Excel output on page 69 is $1,843.17 and the standard deviation is $643.63. The respective values estimated from data grouped into a fre- quency distribution are $1,851.11 and $652.33. The difference in the means is $7.94, or about 0.4 percent. The standard deviations differ by $8.70, or 1.4 percent. Based on the percentage difference, the estimates are very close to the actual values.

s � B

©f(M � X)2

n � 1 � B

76,169,780 180 � 1

� 652.33

f(M � X )2.

Profit Frequency (f ) Midpoint (M ) fM ( ) ( )2 f ( )2

$ 200 up to $ 600 8 400 3,200 �1,451 2,105,401 16,843,208 600 up to 1,000 11 800 8,800 �1,051 1,104,601 12,150,611

1,000 up to 1,400 23 1,200 27,600 �651 423,801 9,747,423 1,400 up to 1,800 38 1,600 60,800 �251 63,001 2,394,038 1,800 up to 2,200 45 2,000 90,000 149 22,201 999,045 2,200 up to 2,600 32 2,400 76,800 549 301,401 9,644,832 2,600 up to 3,000 19 2,800 53,200 949 900,601 17,111,419 3,000 up to 3,400 4 3,200 12,800 1,349 1,819,801 7,279,204

Total 180 333,200 76,169,780

M � XM � XM � X

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Describing Data: Numerical Measures 91

Exercises 57. When we compute the mean of a frequency distribution, why do we refer to this as an

estimated mean? 58. Determine the mean and the standard deviation of the following frequency distribution.

Class Frequency

0 up to 5 2 5 up to 10 7

10 up to 15 12 15 up to 20 6 20 up to 25 3

59. Determine the mean and the standard deviation of the following frequency distribution.

Class Frequency

20 up to 30 7 30 up to 40 12 40 up to 50 21 50 up to 60 18 60 up to 70 12

60. SCCoast, an Internet provider in the Southeast, developed the following frequency dis- tribution on the age of Internet users. Find the mean and the standard deviation.

Age (years) Frequency

10 up to 20 3 20 up to 30 7 30 up to 40 18 40 up to 50 20 50 up to 60 12

(a) What is the table called? (b) Based on the distribution, what is the estimate of the arithmetic mean net income? (c) Based on the distribution, what is the estimate of the standard deviation?

61. The IRS was interested in the number of individual tax forms prepared by small account- ing firms. The IRS randomly sampled 50 public accounting firms with 10 or fewer employ- ees in the Dallas–Fort Worth area. The following frequency table reports the results of the study. Estimate the mean and the standard deviation.

Number of Clients Frequency

20 up to 30 1 30 up to 40 15 40 up to 50 22 50 up to 60 8 60 up to 70 4

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92 Chapter 3

3.16 Ethics and Reporting Results In Chapter 1, we discussed the ethical and unbiased reporting of statistical results. While you are learning about how to organize, summarize, and interpret data using statistics, it is also important to understand statistics so that you can be an intelli- gent consumer of information.

In this chapter, we learned how to compute numerical descriptive statistics. Specifically, we showed how to compute and interpret measures of location for a data set: the mean, median, and mode. We also discussed the advantages and disadvantages for each statistic. For example, if a real estate developer tells a client that the average home in a particular subdivision sold for $150,000, we assume that $150,000 is a representative selling price for all the homes. But sup- pose that the client also asks what the median sales price is, and the median is $60,000. Why was the developer only reporting the mean price? This information is extremely important to a person’s decision making when buying a home. Knowing the advantages and disadvantages of the mean, median, and mode is important as we report statistics and as we use statistical information to make decisions.

We also learned how to compute measures of dispersion: range, mean devia- tion, and standard deviation. Each of these statistics also has advantages and dis- advantages. Remember that the range provides information about the overall spread of a distribution. However, it does not provide any information about how the data is clustered or concentrated around the center of the distribution. As we learn more about statistics, we need to remember that when we use statistics we must maintain an independent and principled point of view. Any statistical report requires objective and honest communication of the results.

62. Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 60 manufacturing companies located in the Southwest. Estimate the mean and the standard deviation of advertising expenses.

Advertising Expenditure Number of ($ millions) Companies

25 up to 35 5 35 up to 45 10 45 up to 55 21 55 up to 65 16 65 up to 75 8

Total 60

Chapter Summary I. A measure of location is a value used to describe the center of a set of data.

A. The arithmetic mean is the most widely reported measure of location. 1. It is calculated by adding the values of the observations and dividing by the total

number of observations. a. The formula for a population mean of ungrouped or raw data is

[3–1]� � �X N

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Describing Data: Numerical Measures 93

b. The formula for the mean of a sample is

[3–2]

c. The formula for the sample mean of data in a frequency distribution is

[3–12]

2. The major characteristics of the arithmetic mean are: a. At least the interval scale of measurement is required. b. All the data values are used in the calculation. c. A set of data has only one mean. That is, it is unique. d. The sum of the deviations from the mean equals 0.

B. The weighted mean is found by multiplying each observation by its corresponding weight. 1. The formula for determining the weighted mean is

[3–3]

2. It is a special case of the arithmetic mean. C. The median is the value in the middle of a set of ordered data.

1. To find the median, sort the observations from smallest to largest and identify the middle value.

2. The major characteristics of the median are: a. At least the ordinal scale of measurement is required. b. It is not influenced by extreme values. c. Fifty percent of the observations are larger than the median. d. It is unique to a set of data.

D. The mode is the value that occurs most often in a set of data. 1. The mode can be found for nominal-level data. 2. A set of data can have more than one mode.

E. The geometric mean is the nth root of the product of n positive values. 1. The formula for the geometric mean is

[3–4]

2. The geometric mean is also used to find the rate of change from one period to another.

[3–5]

3. The geometric mean is always equal to or less than the arithmetic mean. II. The dispersion is the variation or spread in a set of data.

A. The range is the difference between the largest and the smallest value in a set of data. 1. The formula for the range is

Range � Largest value � Smallest value [3–6]

2. The major characteristics of the range are: a. Only two values are used in its calculation. b. It is influenced by extreme values. c. It is easy to compute and to understand.

B. The mean absolute deviation is the sum of the absolute values of the deviations from the mean divided by the number of observations. 1. The formula for computing the mean absolute deviation is

[3–7]

2. The major characteristics of the mean absolute deviation are: a. It is not unduly influenced by large or small values. b. All observations are used in the calculation. c. The absolute values are somewhat difficult to work with.

MD � © 0X � X 0

GM � n B

Value at end of period Value at beginning of period

� 1

GM � n2(X1)(X2)(X3) � � � (Xn)

X w � w1X1 � w2X2 � w3X3 � � � � � wnXn

w1 � w2 � w3 � � � � � wn

X � ©fM

X � ©X n

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94 Chapter 3

C. The variance is the mean of the squared deviations from the arithmetic mean. 1. The formula for the population variance is

[3–8]

2. The formula for the sample variance is

[3–10]

3. The major characteristics of the variance are: a. All observations are used in the calculation. b. It is not unduly influenced by extreme observations. c. The units are somewhat difficult to work with; they are the original units squared.

D. The standard deviation is the square root of the variance. 1. The major characteristics of the standard deviation are:

a. It is in the same units as the original data. b. It is the square root of the average squared distance from the mean. c. It cannot be negative. d. It is the most widely reported measure of dispersion.

2. The formula for the sample standard deviation is

[3–11]

3. The formula for the standard deviation of grouped data is

[3–13]

III. We interpret the standard deviation using two measures. A. Chebyshev’s theorem states that regardless of the shape of the distribution, at least

1 � 1/k2 of the observations will be within k standard deviations of the mean, where k is greater than 1.

B. The Empirical Rule states that for a bell-shaped distribution about 68 percent of the values will be within one standard deviation of the mean, 95 percent within two, and virtually all within three.

Pronunciation Key SYMBOL MEANING PRONUNCIATION

� Population mean mu

� Operation of adding sigma �X Adding a group of values sigma X

Sample mean X bar

Weighted mean X bar sub w

GM Geometric mean G M

�fM Adding the product of the frequencies and the class midpoints sigma f M

2 Population variance sigma squared

Population standard deviation sigma

Chapter Exercises 63. The accounting firm of Crawford and Associates has five senior partners. Yesterday the

senior partners saw six, four, three, seven, and five clients, respectively. a. Compute the mean and median number of clients seen by the partners. b. Is the mean a sample mean or a population mean? c. Verify that (X � �) � 0.�

s � B

©f(M � X)2

n � 1

s � B

©(X � X)2

n � 1

s2 � ©(X � X )2

n � 1

2 � ©(X � �)2

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Describing Data: Numerical Measures 95

64. Owens Orchards sells apples in a large bag by weight. A sample of seven bags con- tained the following numbers of apples: 23, 19, 26, 17, 21, 24, 22. a. Compute the mean and median number of apples in a bag. b. Verify that

65. A sample of households that subscribe to United Bell Phone Company for land line phone service revealed the following number of calls received per household last week. Deter- mine the mean and the median number of calls received.

66. The Citizens Banking Company is studying the number of times the ATM located in a Loblaws Supermarket at the foot of Market Street is used per day. Following are the num- ber of times the machine was used daily over each of the last 30 days. Determine the mean number of times the machine was used per day.

67. A recent study of the laundry habits of Americans included the time in minutes of the wash cycle. A sample of 40 observations follows. Determine the mean and the median of a typical wash cycle.

68. Trudy Green works for the True-Green Lawn Company. Her job is to solicit lawn-care business via the telephone. Listed below is the number of appointments she made in each of the last 25 hours of calling. What is the arithmetic mean number of appointments she made per hour? What is the median number of appointments per hour? Write a brief report summarizing the findings.

69. The Split-A-Rail Fence Company sells three types of fence to homeowners in suburban Seattle, Washington. Grade A costs $5.00 per running foot to install, Grade B costs $6.50 per running foot, and Grade C, the premium quality, costs $8.00 per running foot. Yes- terday, Split-A-Rail installed 270 feet of Grade A, 300 feet of Grade B, and 100 feet of Grade C. What was the mean cost per foot of fence installed?

70. Rolland Poust is a sophomore in the College of Business at Scandia Tech. Last semes- ter he took courses in statistics and accounting, 3 hours each, and earned an A in both. He earned a B in a five-hour history course and a B in a two-hour history of jazz course. In addition, he took a one-hour course dealing with the rules of basketball so he could get his license to officiate high school basketball games. He got an A in this course. What was his GPA for the semester? Assume that he receives 4 points for an A, 3 for a B, and so on. What measure of location did you just calculate?

71. The table below shows the percent of the labor force that is unemployed and the size of the labor force for three counties in Northwest Ohio. Jon Elsas is the Regional Direc- tor of Economic Development. He must present a report to several companies that are

9 5 2 6 5 6 4 4 7 2 3 6 3 4 4 7 8 4 4 5 5 4 8 3 3

35 37 28 37 33 38 37 32 28 29 39 33 32 37 33 35 36 44 36 34 40 38 46 39 37 39 34 39 31 33 37 35 39 38 37 32 43 31 31 35

83 64 84 76 84 54 75 59 70 61 63 80 84 73 68 52 65 90 52 77 95 36 78 61 59 84 95 47 87 60

52 43 30 38 30 42 12 46 39 37 34 46 32 18 41 5

©(X � X) � 0.

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96 Chapter 3

considering locating in Northwest Ohio. What would be an appropriate unemployment rate to show for the entire region?

72. The American Diabetes Association recommends a blood glucose reading of less than 130 for those with Type 2 diabetes. Blood glucose measures the amount of sugar in the blood. Below are the readings for February for a person recently diagnosed with Type 2 diabetes.

a. What is the arithmetic mean glucose reading? b. What is the median glucose reading? c. What is the modal glucose reading?

73. The metropolitan area of Los Angeles–Long Beach, California, is the area expected to show the largest increase in the number of jobs between 1989 and 2010. The number of jobs is expected to increase from 5,164,900 to 6,286,800. What is the geometric mean expected yearly rate of increase?

74. A recent article suggested that, if you earn $25,000 a year today and the inflation rate continues at 3 percent per year, you’ll need to make $33,598 in 10 years to have the same buying power. You would need to make $44,771 if the inflation rate jumped to 6 percent. Confirm that these statements are accurate by finding the geometric mean rate of increase.

75. The ages of a sample of Canadian tourists flying from Toronto to Hong Kong were: 32, 21, 60, 47, 54, 17, 72, 55, 33, and 41. a. Compute the range. b. Compute the mean deviation. c. Compute the standard deviation.

76. The weights (in pounds) of a sample of five boxes being sent by UPS are: 12, 6, 7, 3, and 10. a. Compute the range. b. Compute the mean deviation. c. Compute the standard deviation.

77. The enrollments of the 13 public universities in the state of Ohio are listed below.

College Enrollment

University of Akron 25,942 Bowling Green State University 18,989 Central State University 1,820 University of Cincinnati 36,415 Cleveland State University 15,664 Kent State University 34,056 Miami University 17,161 Ohio State University 59,091 Ohio University 20,437 Shawnee State University 4,300 University of Toledo 20,775 Wright State University 18,786 Youngstown State University 14,682

112 122 116 103 112 96 115 98 106 111 106 124 116 127 116 108 112 112 121 115 124 116 107 118 123 109 109 106

County Percent Unemployed Size of Workforce

Wood 4.5 15,300 Ottawa 3.0 10,400 Lucas 10.2 150,600

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Describing Data: Numerical Measures 97

a. Is this a sample or a population? b. What is the mean enrollment? c. What is the median enrollment? d. What is the range of the enrollments? e. Compute the standard deviation.

78. Health issues are a concern of managers, especially as they evaluate the cost of med- ical insurance. A recent survey of 150 executives at Elvers Industries, a large insurance and financial firm located in the Southwest, reported the number of pounds by which the executives were overweight. Compute the mean and the standard deviation.

79. The Apollo space program lasted from 1967 until 1972 and included 13 missions. The missions lasted from as little as 7 hours to as long as 301 hours. The duration of each flight is listed below.

a. Explain why the flight times are a population. b. Find the mean and median of the flight times. c. Find the range and the standard deviation of the flight times.

80. Creek Ratz is a very popular restaurant located along the coast of northern Florida. They serve a variety of steak and seafood dinners. During the summer beach season, they do not take reservations or accept “call ahead” seating. Management of the restaurant is concerned with the time a patron must wait before being seated for dinner. Listed below is the wait time, in minutes, for the 25 tables seated last Saturday night.

a. Explain why the times are a population. b. Find the mean and median of the times. c. Find the range and the standard deviation of the times.

81. A sample of 25 undergraduates reported the following dollar amounts of entertainment expenses last year:

a. Find the mean, median, and mode of this information. b. What are the range and standard deviation? c. Use the Empirical Rule to establish an interval which includes about 95 percent of the

observations.

684 710 688 711 722 698 723 743 738 722 696 721 685 763 681 731 736 771 693 701 737 717 752 710 697

28 39 23 67 37 28 56 40 28 50 51 45 44 65 61 27 24 61 34 44 64 25 24 27 29

9 195 241 301 216 260 7 244 192 147 10 295 142

Pounds Overweight Frequency

0 up to 6 14 6 up to 12 42

12 up to 18 58 18 up to 24 28 24 up to 30 8

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98 Chapter 3

82. The Kentucky Derby is held the first Saturday in May at Churchill Downs in Louisville, Kentucky. The race track is one and one-quarter miles. The following table shows the winners since 1990, their margin of victory, the winning time, and the payoff on a $2 bet.

a. Determine the mean and median for the variables winning time and payoff on a $2 bet. b. Determine the range and standard deviation of the variables time and payoff. c. Refer to the variable winning margin. What is the level of measurement? What mea-

sure of location would be most appropriate? 83. The manager of the local Walmart Supercenter is studying the number of items purchased

by customers in the evening hours. Listed below is the number of items for a sample of 30 customers.

a. Find the mean and the median of the number of items. b. Find the range and the standard deviation of the number of items. c. Organize the number of items into a frequency distribution. You may want to review

the guidelines in Chapter 2 for establishing the class interval and the number of classes.

d. Find the mean and the standard deviation of the data organized into a frequency distribution. Compare these values with those computed in part (a). Why are they different?

84. The following frequency distribution reports the electricity cost for a sample of 50 two-bedroom apartments in Albuquerque, New Mexico, during the month of May last year.

15 8 6 9 9 4 18 10 10 12 12 4 7 8 12 10 10 11 9 13

5 6 11 14 5 6 6 5 13 5

Winning Margin Winning Time Payoff on a Year Winner (lengths) (minutes) $2 Win Bet

1990 Unbridled 3.5 2.03333 10.80 1991 Strike the Gold 1.75 2.05000 4.80 1992 Lil E. Tee 1 2.05000 16.80 1993 Sea Hero 2.5 2.04000 12.90 1994 Go For Gin 2 2.06000 9.10 1995 Thunder Gulch 2.25 2.02000 24.50 1996 Grindstone nose 2.01667 5.90 1997 Silver Charm head 2.04000 4.00 1998 Real Quiet 0.5 2.03667 8.40 1999 Charismatic neck 2.05333 31.30 2000 Fusaichi Pegasus 1.5 2.02000 2.30 2001 Monarchos 4.75 1.99950 10.50 2002 War Emblem 4 2.01883 20.50 2003 Funny Cide 1.75 2.01983 12.80 2004 Smarty Jones 2.75 2.06767 4.10 2005 Giacomo 0.5 2.04583 50.30 2006 Barbaro 6.5 2.02267 6.10 2007 Street Sense 2.25 2.03617 4.90 2008 Big Brown 4.75 2.03033 6.80 2009 Mine That Bird 6.75 2.04433 103.20 2010 Super Saver 2.50 2.07417 18.00

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Describing Data: Numerical Measures 99

a. Estimate the mean cost. b. Estimate the standard deviation. c. Use the Empirical Rule to estimate the proportion of costs within two standard devi-

ations of the mean. What are these limits? 85. Bidwell Electronics Inc. recently surveyed a sample of employees to determine how far

they lived from corporate headquarters. The results are shown below. Compute the mean and the standard deviation.

Data Set Exercises 86. Refer to the Real Estate data, which reports information on homes sold in the Goodyear,

Arizona, area during the last year. Prepare a report on the selling prices of the homes. Be sure to answer the following questions in your report. a. Around what values do the data tend to cluster? What is the mean selling price? What

is the median selling price? Is one measure more representative of the typical selling prices than the others?

b. What is the range of selling prices? What is the standard deviation? About 95 per- cent of the selling prices are between what two values?

87. Refer to the Baseball 2009 data, which reports information on the 30 Major League Base- ball teams for the 2009 season. Refer to the variable team salary. a. Prepare a report on the team salaries. Be sure to answer the following questions in

your report. 1. Around what values do the data tend to cluster? Specifically what is the mean

team salary? What is the median team salary? Is one measure more representative of the typical team salary than the others?

2. What is the range of the team salaries? What is the standard deviation? About 95 percent of the salaries are between what two values?

b. Refer to the information on the average salary for each year. In 1989 the average player salary was $512,930. By 2009 the average player salary had increased to $3,240,000. What was the rate of increase over the period?

88. Refer to the Buena School District bus data. Prepare a report on the maintenance cost for last month. Be sure to answer the following questions in your report. a. Around what values do the data tend to cluster? Specifically what was the mean main-

tenance cost last month? What is the median cost? Is one measure more represen- tative of the typical cost than the others?

b. What is the range of maintenance costs? What is the standard deviation? About 95 percent of the maintenance costs are between what two values?

Distance (miles) Frequency M

0 up to 5 4 2.5 5 up to 10 15 7.5

10 up to 15 27 12.5 15 up to 20 18 17.5 20 up to 25 6 22.5

Electricity Cost Frequency

$ 80 up to $100 3 100 up to 120 8 120 up to 140 12 140 up to 160 16 160 up to 180 7 180 up to 200 4

Total 50

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100 Chapter 3

1. The Excel Commands for the descriptive statistics on page 69 are:

a. From the CD, retrieve the Applewood data. b. From the menu bar, select Data and then Data

Analysis. Select Descriptive Statistics and then click OK.

c. For the Input Range, type C1:C181, indicate that the data are grouped by column and that the labels are in the first row. Click on Output Range, indicate that the output should go in G1 (or any place you wish), click on Summary sta- tistics, then click OK.

d. After you get your results, double-check the count in the output to be sure it contains the correct number of items.

2. The Minitab commands for the descriptive summary on page 84 are:

a. From the CD retrieve the Applewood data. b. Select Stat, Basic Statistics, and then Display

Descriptive Statistics. In the dialog box, select Profit as the variable and then click OK.

Software Commands

Chapter 3 Answers to Self-Review

3–1 1. a.

c. Statistic, because it is a sample value. d. $66,775. The sample mean is our best

estimate of the population mean.

2. a.

c. Parameter, because it was computed using all the population values.

� � 498 6

� 83

� � ©X N

X � $267,100

4 � $66,775

X � ©X n

3–2 a. $237, found by:

b. The profit per suit is $12, found by $237 � $200 cost � $25 commission. The total profit for the 300 suits is $3,600, found by 300 � $12.

3–3 1. a. $878 b. 3, 3

2. a. 7, found by (6 � 8)/2 � 7 b. 3, 3 c. 0

(95 � $400) � (126 � $200) � (79 � $100) 95 � 126 � 79

� $237.00

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Describing Data: Numerical Measures 101

3–4 a.

b. Positively skewed, because the mean is the largest average and the mode is the smallest.

3–5 1. a. About 9.9 percent, found by , then 1.099 � 1.00 � .099

b. About 10.095 percent c. Greater than, because 10.095 9.9

2. 8.63 percent, found by

3–6 a. 22 thousands of pounds, found by 112 � 90

b. thousands of pounds

X � 824 8

� 103

1.0863 � 1

120,520 23,000

� 1 �

1 4 1.458602236

Fr eq

ue nc

M od

e M

ed ia

n M

ea n

Weekly Sales

d. There is more variation in the Pittsburgh office because the standard deviation is larger. The mean is also larger in the Pittsburgh office.

3–8 2.33, found by:

X � �X n

� 28 7

� 4

X | | Absolute Deviation

95 @ �8 @ 8 103 @ 0 @ 0 105 @ �2 @ 2 110 @ �7 @ 7 104 @ �1 @ 1 105 @ �2 @ 2 112 @ �9 @ 9 90 @�13 @ 13

Total 42

X � X

X X � (X � )2

4 0 0 2 �2 4 5 1 1 4 0 0 5 1 1 2 �2 4 6 2 4

28 0 14

3–9 a.

b. 13.8 and 14.2 3–10 a. Frequency distribution.

1 � 1

(1.5)2 � 1 � .44 � .56

k � 13.85 � 14.0

.10 � �1.5

k � 14.15 � 14.00

.10 � 1.5

s � 12.33 � 1.53 � 2.33

� 14

7 � 1

s2 � ©(X � X)2

n � 1

thousands of pounds

3–7 a.

c. � 139,490.8 � 198.72

� 197,454

5 � 39,490.8

� � (�259)2 � (242)2

(156)2 � (�207)2 � (68)2

2 � (3536 � 3380)2 � � � � � (3622 � 3380)2

� � $16,900

5 � $3,380

MD � 42 8

� 5.25

f M fM ( ) f ( )2

1 4 4 �8.2 67.24 4 8 32 �4.2 70.56

10 12 120 �0.2 0.40 3 16 48 3.8 43.32 2 20 40 7.8 121.68

20 244 303.20

M � XM � X

c. s � B

303.20 20 � 1

� $3.99

X � ©fM M

� $244 20

� $12.20

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4 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Construct and interpret a dot plot.

LO2 Construct and describe a stem-and-leaf display.

LO3 Identify and compute measures of position.

LO4 Construct and analyze a box plot.

LO5 Compute and describe the coefficient of skewness.

LO6 Create and interpret a scatter diagram.

LO7 Develop and explain a contingency table.

Describing Data: Displaying and Exploring Data

McGivern Jewelers recently ran an advertisement in the local newspaper

reporting the shape, size, price, and cut grade for 33 of its diamonds in

stock. Develop a box plot of the variable price and comment on the

result. (See Exercise 37 and LO4.)

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Describing Data: Displaying and Exploring Data 103

4.1 Introduction Chapter 2 began our study of descriptive statistics. In order to transform raw or ungrouped data into a meaningful form, we organize the data into a frequency dis- tribution. We present the frequency distribution in graphic form as a histogram or a frequency polygon. This allows us to visualize where the data tends to cluster, the largest and the smallest values, and the general shape of the data.

In Chapter 3, we first computed several measures of location, such as the mean and the median. These measures of location allow us to report a typical value in the set of observations. We also computed several measures of dispersion, such as the range and the standard deviation. These measures of dispersion allow us to describe the variation or the spread in a set of observations.

We continue our study of descriptive statistics in this chapter. We study (1) dot plots, (2) stem-and-leaf displays, (3) percentiles, and (4) box plots. These charts and statistics give us additional insight into where the values are concentrated as well as the general shape of the data. Then we consider bivariate data. In bivariate data, we observe two variables for each individual or observation selected. Examples include: the number of hours a student studied and the points earned on an exam- ination; whether a sampled product is acceptable or not and the shift on which it is manufactured; and the amount of electricity used in a month by a homeowner and the mean daily high temperature in the region for the month.

4.2 Dot Plots Recall for the Applewood Auto Group data, we summarized the profit earned on the 180 vehicles sold into eight classes. When we organized the data into the eight classes, we lost the exact value of the observations. A dot plot, on the other hand, groups the data as little as possible, and we do not lose the identity of an individ- ual observation. To develop a dot plot, we simply display a dot for each observa- tion along a horizontal number line indicating the possible values of the data. If there are identical observations or the observations are too close to be shown individu- ally, the dots are “piled” on top of each other. This allows us to see the shape of the distribution, the value about which the data tend to cluster, and the largest and smallest observations. Dot plots are most useful for smaller data sets, whereas his- tograms tend to be most useful for large data sets. An example will show how to construct and interpret dot plots.

Dot plots give a visual idea of the spread and concentration of the data.

LO1 Construct and interpret a dot plot.

Example The service departments at Tionesta Ford Lincoln Mercury and Sheffield Motors Inc., two of the four Applewood Auto Group dealerships, were both open 24 work- ing days last month. Listed below is the number of vehicles serviced last month at the two dealerships. Construct dot plots and report summary statistics to compare the two dealerships.

Tionesta Ford Lincoln Mercury

Monday Tuesday Wednesday Thursday Friday Saturday

23 33 27 28 39 26 30 32 28 33 35 32 29 25 36 31 32 27 35 32 35 37 36 30

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104 Chapter 4

Solution The Minitab system provides a dot plot and outputs the mean, median, maximum, and minimum values, and the standard deviation for the number of cars serviced at both of the dealerships over the last 24 working days.

The dot plots, shown in the center of the software output, graphically illustrate the distributions for both dealerships. The plots show the difference in the location and dispersion of the observations. By looking at the dot plots, we can see that the number of vehicles serviced at the Sheffield dealership is more widely dispersed and has a larger mean than at the Tionesta dealership. Several other features of the number of vehicles serviced are:

• Tionesta serviced the fewest cars in any day, 23. • Sheffield serviced 26 cars during their slowest day, which is 4 cars less than

the next lowest day. • Tionesta serviced exactly 32 cars on four different days. • The numbers of cars serviced cluster around 36 for Sheffield and 32 for

Tionesta.

From the descriptive statistics, we see that Sheffield serviced a mean of 35.83 vehi- cles per day. Tionesta serviced a mean of 31.292 vehicles per day during the same period. So Sheffield typically services 4.54 more vehicles per day. There is also more dispersion, or variation, in the daily number of vehicles serviced at Sheffield than at Tionesta. How do we know this? The standard deviation is larger at Sheffield (4.96 vehicles per day) than at Tionesta (4.112 cars per day).

Sheffield Motors Inc.

Monday Tuesday Wednesday Thursday Friday Saturday

31 35 44 36 34 37 30 37 43 31 40 31 32 44 36 34 43 36 26 38 37 30 42 33

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Describing Data: Displaying and Exploring Data 105

4.3 Stem-and-Leaf Displays In Chapter 2, we showed how to organize data into a frequency distribution so we could summarize the raw data into a meaningful form. The major advantage to orga- nizing the data into a frequency distribution is that we get a quick visual picture of the shape of the distribution without doing any further calculation. To put it another way, we can see where the data are concentrated and also determine whether there are any extremely large or small values. There are two disadvantages, however, to organizing the data into a frequency distribution: (1) we lose the exact identity of each value and (2) we are not sure how the values within each class are distributed. To explain, the following frequency distribution shows the number of advertising spots purchased by the 45 members of the Greater Buffalo Automobile Dealers Association in the year 2010. We observe that 7 of the 45 dealers purchased at least 90 but less than 100 spots. However, are the spots purchased within this class clustered about 90, spread evenly throughout the class, or clustered near 99? We cannot tell.

Number of Spots Purchased Frequency

80 up to 90 2 90 up to 100 7

100 up to 110 6 110 up to 120 9 120 up to 130 8 130 up to 140 7 140 up to 150 3 150 up to 160 3

Total 45

One technique that is used to display quantitative information in a condensed form is the stem-and-leaf display. An advantage of the stem-and-leaf display over a frequency distribution is that we do not lose the identity of each observation. In the above example, we would not know the identity of the values in the 90 up to 100 class. To illustrate the construction of a stem-and-leaf display using the num- ber of advertising spots purchased, suppose the seven observations in the 90 up to 100 class are: 96, 94, 93, 94, 95, 96, and 97. The stem value is the leading digit or digits, in this case 9. The leaves are the trailing digits. The stem is placed to the left of a vertical line and the leaf values to the right.

The values in the 90 up to 100 class would appear as follows:

It is also customary to sort the values within each stem from smallest to largest. Thus, the second row of the stem-and-leaf display would appear as follows:

With the stem-and-leaf display, we can quickly observe that there were two deal- ers that purchased 94 spots and that the number of spots purchased ranged from 93 to 97. A stem-and-leaf display is similar to a frequency distribution with more information, that is, the identity of the observations is preserved.

STEM-AND-LEAF DISPLAY A statistical technique to present a set of data. Each numerical value is divided into two parts. The leading digit(s) becomes the stem and the trailing digit the leaf. The stems are located along the vertical axis, and the leaf values are stacked against each other along the horizontal axis.

9 @ 3 4 4 5 6 6 7

9 @ 6 4 3 4 5 6 7

LO2 Construct and describe a stem-and-leaf display.

Statistics in Action

John W. Tukey (1915–2000) received a PhD in mathemat- ics from Princeton in 1939. However, when he joined the Fire Control Research Office during World War II, his interest in abstract mathematics shifted to applied sta- tistics. He developed effective numerical and graphical meth- ods for studying pat- terns in data. Among the graphics he de- veloped are the stem- and-leaf diagram and the box-and-whisker plot or box plot. From 1960 to 1980, Tukey headed the statistical division of NBC’s election night vote projection team. He became renowned in 1960 for preventing an early call of vic- tory for Richard Nixon in the presi- dential election won by John F. Kennedy.

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Example Listed in Table 4–1 is the number of 30-second radio advertising spots purchased by each of the 45 members of the Greater Buffalo Automobile Dealers Association last year. Organize the data into a stem-and-leaf display. Around what values do the number of advertising spots tend to cluster? What is the fewest number of spots purchased by a dealer? The largest number purchased?

Solution From the data in Table 4–1, we note that the smallest number of spots purchased is 88. So we will make the first stem value 8. The largest number is 156, so we will have the stem values begin at 8 and continue to 15. The first number in Table 4–1 is 96, which will have a stem value of 9 and a leaf value of 6. Moving across the top row, the second value is 93 and the third is 88. After the first 3 data values are considered, your chart is as follows.

96 93 88 117 127 95 113 96 108 94 148 156 139 142 94 107 125 155 155 103 112 127 117 120 112 135 132 111 125 104 106 139 134 119 97 89 118 136 125 143 120 103 113 124 138

TABLE 4–1 Number of Advertising Spots Purchased by Members of the Greater Buffalo Automobile Dealers Association

Stem Leaf

8 8 9 6 3

10 11 12 13 14 15

Stem Leaf

8 8 9 9 6 3 5 6 4 4 7

10 8 7 3 4 6 3 11 7 3 2 7 2 1 9 8 3 12 7 5 7 0 5 5 0 4 13 9 5 2 9 4 6 8 14 8 2 3 15 6 5 5

Organizing all the data, the stem-and-leaf chart looks as follows.

The usual procedure is to sort the leaf values from the smallest to largest. The last line, the row referring to the values in the 150s, would appear as:

15 @ 5 5 6

The final table would appear as follows, where we have sorted all of the leaf values.

The following example will explain the details of developing a stem-and-leaf display.

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You can draw several conclusions from the stem-and-leaf display. First, the mini- mum number of spots purchased is 88 and the maximum is 156. Two dealers pur- chased less than 90 spots, and three purchased 150 or more. You can observe, for example, that the three dealers who purchased more than 150 spots actually pur- chased 155, 155, and 156 spots. The concentration of the number of spots is between 110 and 130. There were nine dealers who purchased between 110 and 119 spots and eight who purchased between 120 and 129 spots. We can also tell that within the 120 to 129 group the actual number of spots purchased was spread evenly through- out. That is, two dealers purchased 120 spots, one dealer purchased 124 spots, three dealers purchased 125 spots, and two purchased 127 spots.

We can also generate this information on the Minitab software system. We have named the variable Spots. The Minitab output is below. You can find the Minitab commands that will produce this output at the end of the chapter.

The Minitab solution provides some additional information regarding cumu- lative totals. In the column to the left of the stem values are numbers such as 2, 9, 15, and so on. The number 9 indicates that there are 9 observations that have occurred before the value of 100. The number 15 indicates that 15 obser- vations have occurred prior to 110. About halfway down the column the number 9 appears in parentheses. The parentheses indicate that the middle value or median appears in that row and that there are nine values in this group. In this case, we describe the middle value as the value below which half of the obser- vations occur. There are a total of 45 observations, so the middle value, if the data were arranged from smallest to largest, would be the 23rd observation; its value is 118. After the median, the values begin to decline. These values repre- sent the “more than” cumulative totals. There are 21 observations of 120 or more, 13 of 130 or more, and so on.

Stem Leaf

8 8 9 9 3 4 4 5 6 6 7

10 3 3 4 6 7 8 11 1 2 2 3 3 7 7 8 9 12 0 0 4 5 5 5 7 7 13 2 4 5 6 8 9 9 14 2 3 8 15 5 5 6

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Which is the better choice, a dot plot or a stem-and-leaf chart? This is really a matter of personal choice and convenience. For presenting data, especially with a large number of observations, you will find dot plots are more frequently used. You will see dot plots in analytical literature, marketing reports, and occasionally in annual reports. If you are doing a quick analysis for yourself, stem-and-leaf tallies are handy and easy, particularly on a smaller set of data.

Self-Review 4–1

1. The number of employees at each of the 142 Home Depot Stores in the Southeast region is shown in the following dot plot.

(a) What are the maximum and minimum numbers of employees per store? (b) How many stores employ 91 people? (c) Around what values does the number of employees per store tend to cluster?

2. The rate of return for 21 stocks is:

8.3 9.6 9.5 9.1 8.8 11.2 7.7 10.1 9.9 10.8 10.2 8.0 8.4 8.1 11.6 9.6 8.8 8.0 10.4 9.8 9.2

Organize this information into a stem-and-leaf display. (a) How many rates are less than 9.0? (b) List the rates in the 10.0 up to 11.0 category. (c) What is the median? (d) What are the maximum and the minimum rates of return?

100 10484 88 92 Number of employees

9680

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Describing Data: Displaying and Exploring Data 109

Exercises 1. Describe the differences between a histogram and a dot plot. When might a dot plot be

better than a histogram? 2. Describe the differences between a histogram and a stem-and-leaf display. 3. Consider the following chart.

a. What is this chart called? b. How many observations are in the study? c. What are the maximum and the minimum values? d. Around what values do the observations tend to cluster?

4. The following chart reports the number of cell phones sold at Radio Shack for the last 26 days.

a. What are the maximum and the minimum number of cell phones sold in a day? b. What is a typical number of cell phones sold?

5. The first row of a stem-and-leaf chart appears as follows: 62 @ 1 3 3 7 9. Assume whole number values. a. What is the “possible range” of the values in this row? b. How many data values are in this row? c. List the actual values in this row of data.

6. The third row of a stem-and-leaf chart appears as follows: 21 @ 0 1 3 5 7 9. Assume whole number values. a. What is the “possible range” of the values in this row? b. How many data values are in this row? c. List the actual values in this row of data.

7. The following stem-and-leaf chart from the Minitab software shows the number of units produced per day in a factory.

a. How many days were studied? b. How many observations are in the first class? c. What are the minimum value and the maximum value?

1 3 8 1 4 2 5 6 9 6 0133559

(7) 7 0236778 9 8 59 7 9 00156 2 10 36

199 144

6 72 3 4 51

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110 Chapter 4

d. List the actual values in the fourth row. e. List the actual values in the second row. f. How many values are less than 70?

g. How many values are 80 or more? h. What is the median? i. How many values are between 60 and 89, inclusive?

8. The following stem-and-leaf chart reports the number of movies rented per day at Video Connection on the corner of Fourth and Main Streets.

a. How many days were studied? b. How many observations are in the last class? c. What are the maximum and the minimum values in the entire set of data? d. List the actual values in the fourth row. e. List the actual values in the next to the last row. f. On how many days were less than 160 movies rented?

g. On how many days were 220 or more movies rented? h. What is the middle value? i. On how many days were between 170 and 210 movies rented?

9. A survey of the number of cell phone calls made by a sample of Verizon subscribers last week revealed the following information. Develop a stem-and-leaf chart. How many calls did a typical subscriber make? What were the maximum and the minimum number of calls made?

10. Aloha Banking Co. is studying ATM use in suburban Honolulu. A sample of 30 ATMs showed they were used the following number of times yesterday. Develop a stem-and- leaf chart. Summarize the number of times each ATM was used. What was the typical, minimum, and maximum number of times each ATM was used?

83 64 84 76 84 54 75 59 70 61 63 80 84 73 68 52 65 90 52 77 95 36 78 61 59 84 95 47 87 60

52 43 30 38 30 42 12 46 39 37 34 46 32 18 41 5

3 12 689 6 13 123

10 14 6889 13 15 589 15 16 35 20 17 24568 23 18 268 (5) 19 13456 22 20 034679 16 21 2239 12 22 789

9 23 00179 4 24 8 3 25 13 1 26 1 27 0

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Describing Data: Displaying and Exploring Data 111

4.4 Measures of Position The standard deviation is the most widely used measure of dispersion. However, there are other ways of describing the variation or spread in a set of data. One method is to determine the location of values that divide a set of observations into equal parts. These measures include quartiles, deciles, and percentiles.

Quartiles divide a set of observations into four equal parts. To explain further, think of any set of values arranged from smallest to largest. In Chapter 3, we called the middle value of a set of data arranged from smallest to largest the median. That is, 50 percent of the observations are larger than the median and 50 percent are smaller. The median is a measure of location because it pinpoints the center of the data. In a similar fashion, quartiles divide a set of observations into four equal parts. The first quartile, usually labeled Q1, is the value below which 25 percent of the observations occur, and the third quartile, usually labeled Q3, is the value below which 75 percent of the observations occur. Logically, Q2 is the median. Q1 can be thought of as the “median” of the lower half of the data and Q3 the “median” of the upper half of the data.

Similarly, deciles divide a set of observations into 10 equal parts and per- centiles into 100 equal parts. So if you found that your GPA was in the 8th decile at your university, you could conclude that 80 percent of the students had a GPA lower than yours and 20 percent had a higher GPA. A GPA in the 33rd percentile means that 33 percent of the students have a lower GPA and 67 percent have a higher GPA. Percentile scores are frequently used to report results on such national standardized tests as the SAT, ACT, GMAT (used to judge entry into many master of business administration programs), and LSAT (used to judge entry into law school).

Quartiles, Deciles, and Percentiles To formalize the computational procedure, let Lp refer to the location of a desired percentile. So if we want to find the 33rd percentile we would use L33 and if we wanted the median, the 50th percentile, then L50. The number of observations is n, so if we want to locate the median, its position is at (n � 1)/2, or we could write this as (n � 1)(P/100), where P is the desired percentile.

LOCATION OF A PERCENTILE [4–1]

An example will help to explain further.

Lp � (n � 1) P

100

Example Listed below are the commissions earned last month by a sample of 15 brokers at Salomon Smith Barney’s Oakland, California office. Salomon Smith Barney is an investment company with offices located throughout the United States.

$2,038 $1,758 $1,721 $1,637 $2,097 $2,047 $2,205 $1,787 $2,287 1,940 2,311 2,054 2,406 1,471 1,460

Locate the median, the first quartile, and the third quartile for the commissions earned.

Quartiles divide a set of data into four parts.

LO3 Identify and compute measures of position.

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In the above example, the location formula yielded a whole number. That is, we wanted to find the first quartile and there were 15 observations, so the location for- mula indicated we should find the fourth ordered value. What if there were 20 obser- vations in the sample, that is n � 20, and we wanted to locate the first quartile? From the location formula (4–1):

We would locate the fifth value in the ordered array and then move .25 of the distance between the fifth and sixth values and report that as the first quartile. Like the median, the quartile does not need to be one of the actual values in the data set.

To explain further, suppose a data set contained the six values: 91, 75, 61, 101, 43, and 104. We want to locate the first quartile. We order the values from smallest to largest: 43, 61, 75, 91, 101, and 104. The first quartile is located at

The position formula tells us that the first quartile is located between the first and the second value and that it is .75 of the distance between the first and the sec- ond values. The first value is 43 and the second is 61. So the distance between these two values is 18. To locate the first quartile, we need to move .75 of the dis- tance between the first and second values, so .75(18) � 13.5. To complete the pro- cedure, we add 13.5 to the first value and report that the first quartile is 56.5.

L25 � (n � 1) P

100 � (6 � 1)

25 100

� 1.75

L25 � (n � 1) P

100 � (20 � 1)

25 100

� 5.25

Solution The first step is to sort the data from the smallest commission to the largest.

$1,460 $1,471 $1,637 $1,721 $1,758 $1,787 $1,940 $2,038 2,047 2,054 2,097 2,205 2,287 2,311 2,406

The median value is the observation in the center. The center value or L50 is located at (n � 1) (50/100), where n is the number of observations. In this case,

that is position number 8, found by (15 � 1)(50/100). The eighth largest commission is $2,038. So we conclude this is the median and that half the brokers earned commissions more than $2,038 and half earned less than $2,038.

Recall the definition of a quartile. Quartiles divide a set of observations into four equal parts. Hence 25 percent of the obser- vations will be less than the first quartile. Seventy-five percent of the observations will be less than the third quartile. To locate the first quartile, we use formula (4–1), where n � 15 and P � 25:

and to locate the third quartile, n � 15 and P � 75:

Therefore, the first and third quartile values are located at positions 4 and 12, respectively. The fourth value in the ordered array is $1,721 and the twelfth is $2,205. These are the first and third quartiles.

L75 � (n � 1) P

100 � (15 � 1)

75 100

� 12

L25 � (n � 1) P

100 � (15 � 1)

25 100

� 4

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Describing Data: Displaying and Exploring Data 113

We can extend the idea to include both deciles and percentiles. To locate the 23rd percentile in a sample of 80 observations, we would look for the 18.63 position.

To find the value corresponding to the 23rd percentile, we would locate the 18th value and the 19th value and determine the distance between the two values. Next, we would multiply this difference by 0.63 and add the result to the smaller value. The result would be the 23rd percentile.

With a statistical software package, it is easy to sort the data from smallest to largest and to locate percentiles and deciles. Both Minitab and Excel provide sum- mary statistics. Listed below is the output from the Minitab system for the Smith Barney commission data. Included are the first and third quartiles, mean, median, and standard deviation. We conclude that 25 percent of the commissions earned were less than $1,721 and 75 percent were less than $2,205. The same values were reported in the Example on the previous page.

L23 � (n � 1) P

100 � (80 � 1)

23 100

� 18.63

Excel and MegaStat, which is based on Excel, will also calculate quartiles and output the results. However, the method of solution used is slightly different. To simplify the issues, assume the data set contains an odd number of values. The method described in the Example, and supported by Minitab, for the first quartile is:

1. Find the median of the set of n observations. 2. Focus only on the observations below the median and find the median of these

values. That is, do not consider the median as part of the new data set. 3. Report this value as the first quartile.

So in our Smith Barney commissions data, the median commission is the 8th observation in the set of 15 observations. This commission is $2,038, so we focus on the seven observations less than $2,038. The median of these seven observa- tions is located in position 4 and that value is $1,721, the value found in our Exam- ple and in the Minitab output.

Below is the Excel spreadsheet. Also shown are the first and third quartiles for the Smith Barney commission data. Notice the results differ. Again, to simplify the situation, assume there are an odd number of values. Excel finds the median accord- ing to the following method:

1. Find the median of the set of n observations. 2. Focus on all the observation equal to or less than the median. That is, include

the median in the new subset of data. 3. Find the median of this set of values. 4. Report this value as the first quartile.

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114 Chapter 4

In our Smith Barney commission data, the median of the original 15 observa- tions is $2,038. So our new set of values is the eight ordered observations between $1,460 and $2,038. The median is halfway between $1,721 and $1,758, or $1,739 as reported by Excel.

So the essential difference between the two methods is:

• In the Minitab method, the median is not included in the subset of data. • In the Excel method, the median is included in the subset of data.

In this example, there was an odd number of observations. What happens in the Excel method if there is an even number of observations? Instead of using for- mula 4–1 to find the location, it uses 0.25n � 0.75 to locate the position of the first quartile and 0.75n � 0.25 to locate the position of the third quartile.

Is this difference important? No, usually it is just a nuisance. Statisticians usu- ally prefer the first method discussed. When the sample is large, the difference in the results from the two methods is small. For example, recall the Applewood Auto Group data in which the profit data on the sale of 180 vehicles is reported. Below are the Minitab and Excel results. Not much difference, only $7.00 over 180 vehi- cles! Reporting either value would make little difference in the interpretation.

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Self-Review 4–2 The Quality Control department of Plainsville Peanut Company is responsible for checking the weight of the 8-ounce jar of peanut butter. The weights of a sample of nine jars pro- duced last hour are:

(a) What is the median weight? (b) Determine the weights corresponding to the first and third quartiles.

7.69 7.72 7.8 7.86 7.90 7.94 7.97 8.06 8.09

Exercises 11. Determine the median and the values corresponding to the first and third quartiles in the

following data.

12. Determine the median and the values corresponding to the first and third quartiles in the following data.

13. The Thomas Supply Company Inc. is a distributor of gas-powered generators. As with any business, the length of time customers take to pay their invoices is important. Listed below, arranged from smallest to largest, is the time, in days, for a sample of The Thomas Supply Company Inc. invoices.

13 13 13 20 26 27 31 34 34 34 35 35 36 37 38 41 41 41 45 47 47 47 50 51 53 54 56 62 67 82

a. Determine the first and third quartiles. b. Determine the second decile and the eighth decile. c. Determine the 67th percentile.

5.24 6.02 6.67 7.30 7.59 7.99 8.03 8.35 8.81 9.45 9.61 10.37 10.39 11.86 12.22 12.71 13.07 13.59 13.89 15.42

46 47 49 49 51 53 54 54 55 55 59

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14. Kevin Horn is the national sales manager for National Textbooks Inc. He has a sales staff of 40 who visit college professors all over the United States. Each Saturday morning he requires his sales staff to send him a report. This report includes, among other things, the number of professors visited during the previous week. Listed below, ordered from smallest to largest, are the number of visits last week.

38 40 41 45 48 48 50 50 51 51 52 52 53 54 55 55 55 56 56 57 59 59 59 62 62 62 63 64 65 66 66 67 67 69 69 71 77 78 79 79

a. Determine the median number of calls. b. Determine the first and third quartiles. c. Determine the first decile and the ninth decile. d. Determine the 33rd percentile.

Box Plots A box plot is a graphical display, based on quartiles, that helps us picture a set of data. To construct a box plot, we need only five statistics: the minimum value, Q1 (the first quartile), the median, Q3 (the third quartile), and the maximum value. An example will help to explain.

LO4 Construct and analyze a box plot.

Example Alexander’s Pizza offers free delivery of its pizza within 15 miles. Alex, the owner, wants some information on the time it takes for delivery. How long does a typical delivery take? Within what range of times will most deliveries be completed? For a sample of 20 deliveries, he determined the following information:

Develop a box plot for the delivery times. What conclusions can you make about the delivery times?

Maximum value � 30 minutes

Q3 � 22 minutes

Median � 18 minutes

Q1 � 15 minutes

Minimum value � 13 minutes

Solution The first step in drawing a box plot is to create an appropriate scale along the horizontal axis. Next, we draw a box that starts at Q1 (15 minutes) and ends at Q3 (22 minutes). Inside the box we place a vertical line to represent the median (18 minutes). Finally, we extend horizontal lines from the box out to the minimum value (13 minutes) and the maximum value (30 minutes). These horizontal lines outside of the box are sometimes called “whiskers” because they look a bit like a cat’s whiskers.

12 14 16 18 20 22 24 26 28 30 32

Q1 Median

Minimum value

Maximum value

Minutes

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Describing Data: Displaying and Exploring Data 117

The box plot also reveals that the distribution of delivery times is positively skewed. Recall from page 70 in Chapter 3 that we defined skewness as the lack of symme- try in a set of data. How do we know this distribution is positively skewed? In this case, there are actually two pieces of information that suggest this. First, the dashed line to the right of the box from 22 minutes (Q3) to the maximum time of 30 minutes is longer than the dashed line from the left of 15 minutes (Q1) to the minimum value of 13 minutes. To put it another way, the 25 percent of the data larger than the third quartile is more spread out than the 25 percent less than the first quartile. A second indication of positive skewness is that the median is not in the center of the box. The distance from the first quartile to the median is smaller than the distance from the median to the third quartile. We know that the number of delivery times between 15 minutes and 18 minutes is the same as the number of delivery times between 18 minutes and 22 minutes.

Example Refer to the Applewood Auto Group data. Develop a box plot for the variable age of the buyer. What can we conclude about the distribution of the age of the buyer?

Solution The Minitab statistical software system was used to develop the following chart and summary statistics.

The box plot shows that the middle 50 percent of the deliveries take between 15 minutes and 22 minutes. The distance between the ends of the box, 7 min- utes, is the interquartile range. The interquartile range is the distance between the first and the third quartile. It shows the spread or dispersion of the majority of deliveries.

The median age of the purchaser was 46 years, 25 percent of the purchasers were less than 40 years of age, and 25 percent were more than 52.75 years of age. Based on the summary information and the box plot, we conclude:

• Fifty percent of the purchasers were between the ages of 40 and 52.75 years. • The distribution of ages is symmetric. There are two reasons for this conclu-

sion. The length of the whisker above 52.75 years (Q3) is about the same length as the whisker below 40 years (Q1). Also, the area in the box between 40 years and the median of 46 years is about the same as the area between the median and 52.75.

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118 Chapter 4

Self-Review 4–3 The following box plot shows the assets in millions of dollars for credit unions in Seattle, Washington.

What are the smallest and largest values, the first and third quartiles, and the median? Would you agree that the distribution is symmetrical? Are there any outliers?

0 10 20 30 40 50 60 70 80 90 100

Exercises 15. The box plot below shows the amount spent for books and supplies per year by stu-

dents at four-year public colleges.

a. Estimate the median amount spent. b. Estimate the first and third quartiles for the amount spent. c. Estimate the interquartile range for the amount spent. d. Beyond what point is a value considered an outlier?

$1,750

1,400

1,050

700

350

There are three asterisks (*) above 70 years. What do they indicate? In a box plot, an asterisk identifies an outlier. An outlier is a value that is inconsistent with the rest of the data. It is defined as a value that is more than 1.5 times the interquar- tile range smaller than Q1 or larger than Q3. In this example, an outlier would be a value larger than 71.875 years, found by:

An outlier would also be a value less than 20.875 years.

From the box plot, we conclude that there are three purchasers 72 years of age or older and none less than 21 years of age. Technical note: In some cases, a single asterisk may represent more than one observation, because of the limi- tations of the software and space available. It is a good idea to check the actual data. In this instance, there are three purchasers 72 years old or older; two are 72 and one is 73.

Outlier 6 Q1 � 1.5(Q3 � Q1) � 40 � 1.5(52.75 � 40) � 20.875

Outlier 7 Q3 � 1.5(Q3 � Q1) � 52.75 � 1.5(52.75 � 40) � 71.875

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Describing Data: Displaying and Exploring Data 119

e. Identify any outliers and estimate their value. f. Is the distribution symmetrical or positively or negatively skewed?

16. The box plot shows the undergraduate in-state charge per credit hour at four-year public colleges.

a. Estimate the median. b. Estimate the first and third quartiles. c. Determine the interquartile range. d. Beyond what point is a value considered an outlier? e. Identify any outliers and estimate their value. f. Is the distribution symmetrical or positively or negatively skewed?

17. In a study of the gasoline mileage of model year 2011 automobiles, the mean miles per gallon was 27.5 and the median was 26.8. The smallest value in the study was 12.70 miles per gallon, and the largest was 50.20. The first and third quartiles were 17.95 and 35.45 miles per gallon, respectively. Develop a box plot and comment on the distribution. Is it a symmetric distribution?

18. A sample of 28 time shares in the Orlando, Florida, area revealed the following daily charges for a one-bedroom suite. For convenience, the data are ordered from smallest to largest. Construct a box plot to represent the data. Comment on the distribution. Be sure to identify the first and third quartiles and the median.

4.5 Skewness In Chapter 3, we described measures of central location for a set of observations by reporting the mean, median, and mode. We also described measures that show the amount of spread or variation in a set of data, such as the range and the stan- dard deviation.

Another characteristic of a set of data is the shape. There are four shapes com- monly observed: symmetric, positively skewed, negatively skewed, and bimodal. In a symmetric set of observations the mean and median are equal and the data values are evenly spread around these values. The data values below the mean and median are a mirror image of those above. A set of values is skewed to the right or posi- tively skewed if there is a single peak and the values extend much further to the right of the peak than to the left of the peak. In this case, the mean is larger than the median. In a negatively skewed distribution there is a single peak but the observa- tions extend further to the left, in the negative direction, than to the right. In a nega- tively skewed distribution, the mean is smaller than the median. Positively skewed

$116 $121 $157 $192 $207 $209 $209 229 232 236 236 239 243 246 260 264 276 281 283 289 296 307 309 312 317 324 341 353

$1,500

1,200

900

600

300

LO5 Compute and describe the coefficient of skewness.

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120 Chapter 4

distributions are more common. Salaries often follow this pattern. Think of the salaries of those employed in a small company of about 100 people. The president and a few top executives would have very large salaries relative to the other workers and hence the distribution of salaries would exhibit positive skewness. A bimodal distribution will have two or more peaks. This is often the case when the values are from two or more populations. This information is summarized in Chart 4–1.

M ed

ia n

M ea

45 X

Fr eq

ue nc

Fr eq

ue nc

Fr eq

ue nc

Fr eq

ue nc

Years

Ages

Symmetric

Monthly Salaries

Positively Skewed

$3,000 $4,000

M ed

ia n

M ea

Test Scores

Negatively Skewed

75 80 Score

M ea

Outside Diameter

Bimodal

.98 1.04 Inches$

There are several formulas in statistical literature used to calculate skewness. The simplest, developed by Professor Karl Pearson (1857–1936), is based on the difference between the mean and the median.

CHART 4–1 Shapes of Frequency Polygons

PEARSON’S COEFFICIENT OF SKEWNESS [4–2]sk � 3(X � Median)

Using this relationship, coefficient of skewness can range from �3 up to 3. A value near �3, such as �2.57, indicates considerable negative skewness. A value such as 1.63 indicates moderate positive skewness. A value of 0, which will occur when the mean and median are equal, indicates the distribution is symmetrical and that there is no skewness present.

In this text, we present output from the statistical software packages Minitab and Excel. Both of these software packages compute a value for the coeffi- cient of skewness that is based on the cubed deviations from the mean. The formula is:

SOFTWARE COEFFICIENT OF SKEWNESS [4–3]sk � n

(n � 1)(n � 2) ca a

X � X s b

Formula (4–3) offers an insight into skewness. The right-hand side of the for- mula is the difference between each value and the mean, divided by the standard deviation. That is the portion of the formula. This idea is called stan- dardizing. We will discuss the idea of standardizing a value in more detail in Chap- ter 7 when we describe the normal probability distribution. At this point, observe that the result is to report the difference between each value and the mean in units

(X � X )/s

Statistics in Action

The late Stephen Jay Gould (1941–2002) was a professor of zoology and professor of geology at Harvard University. In 1982, he was diagnosed with cancer and had an expected survival time of eight months. However, never to be discouraged, his re- search showed that the distribution of survival time is dra- matically skewed to the right and showed that not only do 50 percent of similar cancer patients survive more than 8 months, but that the survival time could be years rather than months! Based on his experience, he wrote a widely pub- lished essay titled, “The Median Is not the Message.”

Skewness shows the lack of symmetry in a set of observations.

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Describing Data: Displaying and Exploring Data 121

of the standard deviation. If this difference is positive, the particular value is larger than the mean; if the value is negative, the standardized quantity is smaller than the mean. When we cube these values, we retain the information on the direction of the difference. Recall that in the formula for the standard deviation [see formula (3–11)] we squared the difference between each value and the mean, so that the result was all non-negative values.

If the set of data values under consideration is symmetric, when we cube the standardized values and sum over all the values, the result would be near zero. If there are several large values, clearly separate from the others, the sum of the cubed differences would be a large positive value. Several values much smaller will result in a negative cubed sum.

An example will illustrate the idea of skewness.

Example Following are the earnings per share for a sample of 15 software companies for the year 2010. The earnings per share are arranged from smallest to largest.

Compute the mean, median, and standard deviation. Find the coefficient of skew- ness using Pearson’s estimate and the software methods. What is your conclusion regarding the shape of the distribution?

Solution These are sample data, so we use formula (3–2) to determine the mean

The median is the middle value in a set of data, arranged from smallest to largest. In this case, the middle value is $3.18, so the median earnings per share is $3.18.

We use formula (3–11) on page 84 to determine the sample standard deviation.

Pearson’s coefficient of skewness is 1.017, found by

This indicates there is moderate positive skewness in the earnings per share data. We obtain a similar, but not exactly the same, value from the software method.

The details of the calculations are shown in Table 4–2. To begin, we find the differ- ence between each earnings per share value and the mean and divide this result by the standard deviation. Recall that we referred to this as standardizing. Next, we cube, that is, raise to the third power, the result of the first step. Finally, we sum the cubed values. The details for the first company, that is, the company with an earn- ings per share of $0.09, are:

a X � X

s b

� a 0.09 � 4.95

5.22 b

� (�0.9310)3 � �0.8070

sk � 3(X � Median)

s �

3($4.95 � $3.18) $5.22

� 1.017

s � C

©(X � X )2

n � 1 � C

($0.09 � $4.95)2 � . . . � ($16.40 � $4.95)2

15 � 1 � $5.22

X � ©X n

� $74.26

15 � $4.95

$0.09 $0.13 $0.41 $0.51 $ 1.12 $ 1.20 $ 1.49 $3.18 3.50 6.36 7.83 8.92 10.13 12.99 16.40

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122 Chapter 4

When we sum the 15 cubed values, the result is 11.8274. That is, the term � 11.8274. To find the coefficient of skewness, we use formula (4–3),

with n � 15.

sk � n

(n � 1)(n � 2) a a

X � X s b

� 15

(15 � 1)(15 � 2) (11.8274) � 0.975

�[(X � X )/s]3

We conclude that the earnings per share values are somewhat positively skewed. The following chart, from Minitab, reports the descriptive measures, such as the mean, median, and standard deviation of the earnings per share data. Also included are the coefficient of skewness and a histogram with a bell-shaped curve superimposed.

TABLE 4–2 Calculation of the Coefficient of Skewness

Earnings per Share

0.09 �0.9310 �0.8070 0.13 �0.9234 �0.7873 0.41 �0.8697 �0.6579 0.51 �0.8506 �0.6154 1.12 �0.7337 �0.3950 1.20 �0.7184 �0.3708 1.49 �0.6628 �0.2912 3.18 �0.3391 �0.0390 3.50 �0.2778 �0.0214 6.36 0.2701 0.0197 7.83 0.5517 0.1679 8.92 0.7605 0.4399

10.13 0.9923 0.9772 12.99 1.5402 3.6539 16.40 2.1935 10.5537

11.8274

a X � X

s b

3(X � X ) s

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Describing Data: Displaying and Exploring Data 123

Self-Review 4–4 A sample of five data entry clerks employed in the Horry County Tax Office revised the following number of tax records last hour: 73, 98, 60, 92, and 84. (a) Find the mean, median, and the standard deviation. (b) Compute the coefficient of skewness using Pearson’s method. (c) Calculate the coefficient of skewness using the software method. (d) What is your conclusion regarding the skewness of the data?

Exercises For Exercises 19–22:

a. Determine the mean, median, and the standard deviation. b. Determine the coefficient of skewness using Pearson’s method. c. Determine the coefficient of skewness using the software method.

19. The following values are the starting salaries, in $000, for a sample of five accounting graduates who accepted positions in public accounting last year.

20. Listed below are the salaries, in $000, for a sample of 15 chief financial officers in the electronics industry.

21. Listed below are the commissions earned ($000) last year by the sales representatives at Furniture Patch Inc.

22. Listed below are the salaries in $000 of the 25 players on the opening day roster of the 2010 New York Yankees Major League Baseball team.

$ 3.9 $ 5.7 $ 7.3 $10.6 $13.0 $13.6 $15.1 $15.8 $17.1 17.4 17.6 22.3 38.6 43.2 87.7

$516.0 $548.0 $566.0 $534.0 $586.0 $529.0 546.0 523.0 538.0 523.0 551.0 552.0 486.0 558.0 574.0

36.0 26.0 33.0 28.0 31.0

Player Salary ($000) Position

Pena, Ramiro 412.1 Infielder Pettitte, Andy 11,750.0 Pitcher Posada, Jorge 13,100.0 Catcher Rivera, Mariano 15,000.0 Pitcher Robertson, David 426.7 Pitcher Rodriguez, Alex 33,000.0 Third Baseman Sabathia, CC 24,285.7 Pitcher Swisher, Nick 6,850.0 Outfielder Teixeira, Mark 20,625.0 First Baseman Thames, Marcus 900.0 Outfielder Vazquez, Javier 11,500.0 Pitcher Winn, Randy 1,100.0 Outfielder

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124 Chapter 4

4.6 Describing the Relationship between Two Variables

In Chapter 2 and the first section of this chapter we presented graphical techniques to summarize the distribution of a single variable. We used a his- togram in Chapter 2 to summarize the profit on vehi- cles sold by the Applewood Auto Group. Earlier in this chapter we used dot plots and stem-and-leaf dis- plays to visually summarize a set of data. Because we are studying a single variable, we refer to this as univariate data.

There are situations where we wish to study and visually portray the relationship between two vari- ables. When we study the relationship between two variables, we refer to the data as bivariate. Data analysts frequently wish to understand the rela- tionship between two variables. Here are some examples:

• Tybo and Associates is a law firm that advertises extensively on local TV. The partners are considering increasing their advertising budget. Before doing so, they would like to know the relationship between the amount spent per month on advertising and the total amount of billings for that month. To put it another way, will increasing the amount spent on advertising result in an increase in billings?

• Coastal Realty is studying the selling prices of homes. What variables seem to be related to the selling price of homes? For example, do larger homes sell for more than smaller ones? Probably. So Coastal might study the relationship between the area in square feet and the selling price.

• Dr. Stephen Givens is an expert in human development. He is studying the rela- tionship between the height of fathers and the height of their sons. That is, do tall fathers tend to have tall children? Would you expect Shaquille O’Neal, the 7�1�, 335-pound professional basketball player, to have relatively tall sons?

One graphical technique we use to show the relationship between variables is called a scatter diagram.

To draw a scatter diagram we need two variables. We scale one variable along the horizontal axis (X-axis) of a graph and the other variable along the vertical axis (Y-axis). Usually one variable depends to some degree on the other. In the third exam- ple above, the height of the son depends on the height of the father. So we scale the height of the father on the horizontal axis and that of the son on the vertical axis.

We can use statistical software, such as Excel, to perform the plotting function for us. Caution: You should always be careful of the scale. By changing the scale of either the vertical or the horizontal axis, you can affect the apparent visual strength of the relationship.

Following are three scatter diagrams (Chart 4–2). The one on the left shows a rather strong positive relationship between the age in years and the maintenance cost last year for a sample of 10 buses owned by the city of Cleveland, Ohio. Note that as the age of the bus increases, the yearly maintenance cost also increases. The example in the center, for a sample of 20 vehicles, shows a rather strong indi- rect relationship between the odometer reading and the auction price. That is, as the number of miles driven increases, the auction price decreases. The example on the right depicts the relationship between the height and yearly salary for a sample of 15 shift supervisors. This graph indicates there is little relationship between their height and yearly salary.

LO6 Create and interpret a scatter diagram.

A scatter diagram is used as a way to understand the relationship between two variables.

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Describing Data: Displaying and Exploring Data 125

CHART 4–2 Three Examples of Scatter Diagrams.

$5,600 5,200 4,800 4,400 4,000A

uc tio

n pr

ic e

10,000 30,000 50,000 Odometer

Auction Price versus Odometer $10,000

8,000 6,000 4,000 2,000

Co st

(a nn

ua l)

0 1 2 3 4 5 6 Age (years)

Age of Buses and Maintenance Cost Height versus Salary

125 120 115 110 105 100 95 90S

al ar

y ($

00 0)

54 55 56 57 58 59 60 61 62 63 Height (inches)

Example In the introduction to Chapter 2, we presented data from the Applewood Auto Group. We gathered information concerning several variables, including the profit earned from the sale of 180 vehicles sold last month. In addition to the amount of profit on each sale, one of the other variables is the age of the purchaser. Is there a relationship between the profit earned on a vehicle sale and the age of the pur- chaser? Would it be reasonable to conclude that more profit is made on vehicles purchased by older buyers?

Solution We can investigate the relationship between vehicle profit and the age of the buyer with a scatter diagram. We scale age on the horizontal, or X-axis, and the profit on the vertical, or Y-axis. We use Microsoft Excel to develop the scatter diagram. The Excel commands necessary for the output are shown in the Software Commands section at the end of the chapter.

The scatter diagram shows a rather weak positive relationship between the two variables. It does not appear there is much relationship between the vehicle profit and the age of the buyer. In Chapter 13, we will study the relationship between vari- ables more extensively, even calculating several numerical measures to express the relationship between variables.

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126 Chapter 4

In the preceding example, there is a weak positive, or direct, relationship between the variables. There are, however, many instances where there is a relationship between the variables, but that relationship is inverse or negative. For example:

• The value of a vehicle and the number of miles driven. As the number of miles increases, the value of the vehicle decreases.

• The premium for auto insurance and the age of the driver. Auto rates tend to be the highest for young adults and less for older people.

• For many law enforcement personnel, as the number of years on the job increases, the number of traffic citations decreases. This may be because personnel become more liberal in their interpretations or they may be in supervisor posi- tions and not in a position to issue as many citations. But in any event, as age increases, the number of citations decreases.

A scatter diagram requires that both of the variables be at least interval scale. In the Applewood Auto Group example, both age and vehicle profit are ratio scale variables. Height is also ratio scale as used in the discussion of the relationship between the height of fathers and the height of their sons. What if we wish to study the relationship between two variables when one or both are nominal or ordinal scale? In this case, we tally the results in a contingency table.

A contingency table is a cross-tabulation that simultaneously summarizes two variables of interest. For example:

• Students at a university are classified by gender and class rank. • A product is classified as acceptable or unacceptable and by the shift (day,

afternoon, or night) on which it is manufactured. • A voter in a school bond referendum is classified as to party affiliation (Demo-

crat, Republican, other) and the number of children that voter has attending school in the district (0, 1, 2, etc.).

CONTINGENCY TABLE A table used to classify observations according to two identifiable characteristics.

Example There are four dealerships in the Applewood Auto Group. Suppose we want to com- pare the profit earned on each vehicle sold by the particular dealership. To put it another way, is there a relationship between the amount of profit earned and the dealership?

Solution The level of measurement for the variable dealership is nominal and ratio for the variable profit. To effectively use a contingency table, both variables need to be either of the nominal or ordinal scale. To make the variables compatible, we classify the vari- able profit into two categories, those cases where the profit earned is more than the median and those cases where it is less. On page 69 we calculated the median profit for all sales last month at Applewood Auto Group to be $1,882.50.

Contingency Table Showing the Relationship between Profit and Dealership

Above/Below Median Profit Kane Olean Sheffield Tionesta Total

Above 25 20 19 26 90 Below 27 20 26 17 90

Total 52 40 45 43 180

LO7 Develop and explain a contingency table.

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Describing Data: Displaying and Exploring Data 127

Self-Review 4–5 The rock group Blue String Beans is touring the United States. The following chart shows the relationship between concert seating capacity and revenue in $000 for a sample of concerts.

5800 6300 6800 Seating Capacity

Am ou

nt ($

00 0)

7300

(a) What is the diagram called? (b) How many concerts were studied? (c) Estimate the revenue for the concert with the largest seating capacity. (d) How would you characterize the relationship between revenue and seating capacity?

Is it strong or weak, direct or inverse?

We will return to the study of contingency tables in Chapter 5 during the study of probability and in Chapter 17 during the study of nonparametric methods of analysis.

Exercises 23. Develop a scatter diagram for the following sample data. How would you describe the

relationship between the values?

X-Value Y-Value X-Value Y-Value

10 6 11 6 8 2 10 5 9 6 7 2

11 5 7 3 13 7 11 7

By organizing the information into a contingency table, we can compare the profit at the four dealerships. We observe the following:

• From the Total column on the right, 90 of the 180 cars sold had a profit above the median and half below. From the definition of the median, this is expected.

• For the Kane dealership 25 out of the 52, or 48 percent, of the cars sold were sold for a profit more than the median.

• The percentage of profits above the median for the other dealerships are 50 per- cent for Olean, 42 percent for Sheffield, and 60 percent for Tionesta.

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128 Chapter 4

24. Silver Springs Moving and Storage Inc. is studying the relationship between the num- ber of rooms in a move and the number of labor hours required for the move. As part of the analysis, the CFO of Silver Springs developed the following scatter diagram.

a. How many moves are in the sample? b. Does it appear that more labor hours are required as the number of rooms increases,

or do labor hours decrease as the number of rooms increases? 25. The Director of Planning for Devine Dining Inc. wishes to study the relationship between

the gender of a guest and whether the guest orders dessert. To investigate the relation- ship, the manager collected the following information on 200 recent customers.

a. What is the level of measurement of the two variables? b. What is the above table called? c. Does the evidence in the table suggest men are more likely to order dessert than

women? Explain why. 26. Ski Resorts of Vermont Inc. is considering a merger with Gulf Shores Beach Resorts Inc.

of Alabama. The board of directors surveyed 50 stockholders concerning their position on the merger. The results are reported below.

Opinion

Number of Shares Held Favor Oppose Undecided Total

Under 200 8 6 2 16 200 up to 1,000 6 8 1 15 Over 1,000 6 12 1 19

Total 20 26 4 50

1 2 3 Rooms

0 Ho

ur s

Gender

Dessert Ordered Male Female Total

Yes 32 15 47 No 68 85 153

Total 100 100 200

a. What level of measurement is used in this table? b. What is this table called? c. What group seems most strongly opposed to the merger?

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Describing Data: Displaying and Exploring Data 129

Chapter Summary I. A dot plot shows the range of values on the horizontal axis and the number of observa-

tions for each value on the vertical axis. A. Dot plots report the details of each observation. B. They are useful for comparing two or more data sets.

II. A stem-and-leaf display is an alternative to a histogram. A. The leading digit is the stem and the trailing digit the leaf. B. The advantages of a stem-and-leaf display over a histogram include:

1. The identity of each observation is not lost. 2. The digits themselves give a picture of the distribution. 3. The cumulative frequencies are also shown.

III. Measures of location also describe the shape of a set of observations. A. Quartiles divide a set of observations into four equal parts.

1. Twenty-five percent of the observations are less than the first quartile, 50 per- cent are less than the second quartile, and 75 percent are less than the third quartile.

2. The interquartile range is the difference between the third quartile and the first quartile. B. Deciles divide a set of observations into ten equal parts and percentiles into 100 equal

parts. C. A box plot is a graphic display of a set of data.

1. A box is drawn enclosing the regions between the first quartile and the third quartile. a. A line is drawn inside the box at the median value. b. Dotted line segments are drawn from the third quartile to the largest value to

show the highest 25 percent of the values and from the first quartile to the small- est value to show the lowest 25 percent of the values.

2. A box plot is based on five statistics: the maximum and minimum values, the first and third quartiles, and the median.

IV. The coefficient of skewness is a measure of the symmetry of a distribution. A. There are two formulas for the coefficient of skewness.

1. The formula developed by Pearson is:

[4–2]

2. The coefficient of skewness computed by statistical software is:

[4–3]

V. A scatter diagram is a graphic tool to portray the relationship between two variables. A. Both variables are measured with interval or ratio scales. B. If the scatter of points moves from the lower left to the upper right, the variables under

consideration are directly or positively related. C. If the scatter of points moves from the upper left to the lower right, the variables are

inversely or negatively related. VI. A contingency table is used to classify nominal-scale observations according to two

characteristics.

Pronunciation Key SYMBOL MEANING PRONUNCIATION

Location of percentile L sub p

First quartile Q sub 1

Third quartile Q sub 3Q3

sk � n

(n � 1)(n � 2) c a a

X � X s b

sk � 3(X � Median)

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130 Chapter 4

Chapter Exercises 27. A sample of students attending Southeast Florida University is asked the number of social

activities in which they participated last week. The chart below was prepared from the sample data.

41 2 Activities

5020 30 Patients

4010

Location

Georgetown

Monk Corners

Aynor

a. What is the name given to this chart? b. How many students were in the study? c. How many students reported attending no social activities?

28. Doctor’s Care is a walk-in clinic, with locations in Georgetown, Monks Corners, and Aynor, at which patients may receive treatment for minor injuries, colds, and flu, as well as physical examinations. The following charts report the number of patients treated in each of the three locations last month.

Describe the number of patients served at the three locations each day. What are the maximum and minimum numbers of patients served at each of the locations?

29. The screen size for 23 LCD televisions is given below. Make a stem-and-leaf display of this variable.

30. The top 25 companies (by market capitalization) operating in the Washington, DC, area along with the year they were founded and the number of employees are given below. Make a stem-and-leaf display of each of these variables and write a short description of your findings.

46 52 46 40 42 46 40 37 46 40 52 32 37 32 52 40 32 52 40 52 46 46 52

Company Name Year Founded Employees

AES Corp. 1981 30000 American Capital Strategies Ltd. 1986 484 AvalonBay Communities Inc. 1978 1767 Capital One Financial Corp. 1995 31800 Constellation Energy Group Inc. 1816 9736 Coventry Health Care Inc. 1986 10250 Danaher Corp. 1984 45000 Dominion Resources Inc. 1909 17500 Fannie Mae 1938 6450 Freddie Mac 1970 5533

(continued )

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Describing Data : Displaying and Exploring Data 131

31. In recent years, due to low interest rates, many homeowners refinanced their home mortgages. Linda Lahey is a mortgage officer at Down River Federal Savings and Loan. Below is the amount refinanced for 20 loans she processed last week. The data are reported in thousands of dollars and arranged from smallest to largest.

Company Name Year Founded Employees

Gannett Co. 1906 49675 General Dynamics Corp. 1952 81000 Genworth Financial Inc. 2004 7200 Harman International Industries Inc. 1980 11246 Host Hotels & Resorts Inc. 1927 229 Legg Mason Inc. 1899 3800 Lockheed Martin Corp. 1995 140000 Marriott International Inc. 1927 151000 MedImmune Inc. 1988 2516 NII Holdings Inc. 1996 7748 Norfolk Southern Corp. 1982 30594 Pepco Holdings Inc. 1896 5057 Sallie Mae 1972 11456 Sprint Nextel Corp. 1899 64000 T. Rowe Price Group Inc. 1937 4605 The Washington Post Co. 1877 17100

59.2 59.5 61.6 65.5 66.6 72.9 74.8 77.3 79.2 83.7 85.6 85.8 86.6 87.0 87.1 90.2 93.3 98.6

100.2 100.7

a. Find the median, first quartile, and third quartile. b. Find the 26th and 83rd percentiles. c. Draw a box plot of the data.

32. A study is made by the recording industry in the United States of the number of music CDs owned by senior citizens and young adults. The information is reported below.

Seniors

28 35 41 48 52 81 97 98 98 99 118 132 133 140 145 147 153 158 162 174 177 180 180 187 188

Young Adults

81 107 113 147 147 175 183 192 202 209 233 251 254 266 283 284 284 316 372 401 417 423 490 500 507 518 550 557 590 594

a. Find the median and the first and third quartiles for the number of CDs owned by senior citizens. Develop a box plot for the information.

b. Find the median and the first and third quartiles for the number of CDs owned by young adults. Develop a box plot for the information.

c. Compare the number of CDs owned by the two groups.

33. The corporate headquarters of Bank.com, a new Internet company that performs all bank- ing transactions via the Internet, is located in downtown Philadelphia. The director of human resources is making a study of the time it takes employees to get to work. The city is planning to offer incentives to each downtown employer if they will encourage their

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132 Chapter 4

employees to use public transportation. Below is a listing of the time to get to work this morning according to whether the employee used public transportation or drove a car.

Public Transportation

23 25 25 30 31 31 32 33 35 36 37 42

Private

32 32 33 34 37 37 38 38 38 39 40 44

0 20 40 60 80 10 0

****

Number of Newspapers

a. Find the median and the first and third quartiles for the time it took employees using public transportation. Develop a box plot for the information.

b. Find the median and the first and third quartiles for the time it took employees who drove their own vehicle. Develop a box plot for the information.

c. Compare the times of the two groups. 34. The following box plot shows the number of daily newspapers published in each state

and the District of Columbia. Write a brief report summarizing the number published. Be sure to include information on the values of the first and third quartiles, the median, and whether there is any skewness. If there are any outliers, estimate their value.

35. Walter Gogel Company is an industrial supplier of fasteners, tools, and springs. The amounts of its invoices vary widely, from less than $20.00 to more than $400.00. During the month of January the company sent out 80 invoices. Here is a box plot of these invoices. Write a brief report summarizing the invoice amounts. Be sure to include infor- mation on the values of the first and third quartiles, the median, and whether there is any skewness. If there are any outliers, approximate the value of these invoices.

0 50 10 0

15 0

20 0

25 0

Invoice Amount

36. The American Society of PeriAnesthesia Nurses (ASPAN; www.aspan.org) is a national organization serving nurses practicing in ambulatory surgery preanesthesia and postanes- thesia care. The organization consists of 40 components, which are listed below.

State/Region Membership

Alabama 95 Arizona 399 Maryland, Delaware, DC 531 Connecticut 239 Florida 631 Georgia 384 Hawaii 73

State/Region Membership

Illinois 562 Indiana 270 Iowa 117 Kentucky 197 Louisiana 258 Michigan 411 Massachusetts 480

(continued )

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Describing Data : Displaying and Exploring Data 133

Shape Size (carats) Price Cut Grade Shape Size (carats) Price Cut Grade

Princess 5.03 $44,312 Ideal cut Round 0.77 $2,828 Ultra ideal cut Round 2.35 20,413 Premium cut Oval 0.76 3,808 Premium cut Round 2.03 13,080 Ideal cut Princess 0.71 2,327 Premium cut Round 1.56 13,925 Ideal cut Marquise 0.71 2,732 Good cut Round 1.21 7,382 Ultra ideal cut Round 0.70 1,915 Premium cut Round 1.21 5,154 Average cut Round 0.66 1,885 Premium cut Round 1.19 5,339 Premium cut Round 0.62 1,397 Good cut Emerald 1.16 5,161 Ideal cut Round 0.52 2,555 Premium cut Round 1.08 8,775 Ultra ideal cut Princess 0.51 1,337 Ideal cut Round 1.02 4,282 Premium cut Round 0.51 1,558 Premium cut Round 1.02 6,943 Ideal cut Round 0.45 1,191 Premium cut Marquise 1.01 7,038 Good cut Princess 0.44 1,319 Average cut Princess 1.00 4,868 Premium cut Marquise 0.44 1,319 Premium cut Round 0.91 5,106 Premium cut Round 0.40 1,133 Premium cut Round 0.90 3,921 Good cut Round 0.35 1,354 Good cut Round 0.90 3,733 Premium cut Round 0.32 896 Premium cut Round 0.84 2,621 Premium cut

a. Develop a box plot of the variable price and comment on the result. Are there any out- liers? What is the median price? What is the value of the first and the third quartile?

b. Develop a box plot of the variable size and comment on the result. Are there any out- liers? What is the median price? What is the value of the first and the third quartile?

c. Develop a scatter diagram between the variables price and size. Be sure to put price on the vertical axis and size on the horizontal axis. Does there seem to be an asso- ciation between the two variables? Is the association direct or indirect? Does any point seem to be different from the others?

d. Develop a contingency table for the variables shape and cut grade. What is the most common cut grade? What is the most common shape? What is the most common combination of cut grade and shape?

Use statistical software to answer the following questions. a. Find the mean, median, and standard deviation of the number of members per

component. b. Find the coefficient of skewness, using the software. What do you conclude about the

shape of the distribution of component size? c. Determine the first and third quartiles. Do not use the method described by Excel. d. Develop a box plot. Are there any outliers? Which components are outliers? What are

the limits for outliers?

37. McGivern Jewelers is located in the Levis Square Mall just south of Toledo, Ohio. Recently it ran an advertisement in the local newspaper reporting the shape, size, price, and cut grade for 33 of its diamonds currently in stock. The information is reported below.

State/Region Membership

Maine 97 Minnesota, Dakotas 289 Missouri, Kansas 282 Mississippi 90 Nebraska 115 North Carolina 542 Nevada 106 New Jersey, Bermuda 517 Alaska, Idaho, Montana,

Oregon, Washington 708 New York 891 Ohio 708 Oklahoma 171 Arkansas 68

State/Region Membership

California 1,165 New Mexico 79 Pennsylvania 575 Rhode Island 53 Colorado 409 South Carolina 237 Texas 1,026 Tennessee 167 Utah 67 Virginia 414 Vermont,

New Hampshire 144 Wisconsin 311 West Virginia 62

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134 Chapter 4

42. Wendy’s offers eight different condiments (mustard, catsup, onion, mayonnaise, pickle, lettuce, tomato, and relish) on hamburgers. A store manager collected the following infor- mation on the number of condiments ordered and the age group of the customer. What can you conclude regarding the information? Who tends to order the most or least num- ber of condiments?

41. An auto insurance company reported the following information regarding the age of a dri- ver and the number of accidents reported last year. Develop a scatter diagram for the data and write a brief summary.

Age Accidents Age Accidents

16 4 23 0 24 2 27 1 18 5 32 1 17 4 22 3

Age

Number of Condiments Under 18 18 up to 40 40 up to 60 60 or older

0 12 18 24 52 1 21 76 50 30 2 39 52 40 12

3 or more 71 87 47 28

34 39 44 Months

$130

120

110

100

M on

th ly

M ai

nt en

an ce

C os

980.9 1,036.5 1,099.5 1,153.9 1,409.0 1,456.4 1,718.4 1,721.2

39. Listed below is the number of car thefts in a large city over the last week. Calculate the coefficient of skewness using both methods. Hint: Use of a spreadsheet will expedite the calculations.

38. Listed below is the amount of commissions earned last month for the eight members of the sales staff at Best Electronics. Calculate the coefficient of skewness using both meth- ods. Hint: Use of a spreadsheet will expedite the calculations.

40. The manager of Information Services at Wilkin Investigations, a private investigation firm, is studying the relationship between the age (in months) of a combination printer, copy, and fax machine and its monthly maintenance cost. For a sample of 15 machines, the manager developed the following chart. What can the manager conclude about the rela- tionship between the variables?

3 12 13 7 8 3 8

43. Listed at the top of the next page is a table showing the number of employed and unem- ployed workers 20 years or older by gender in the United States.

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Describing Data: Displaying and Exploring Data 135

a. How many workers were studied? b. What percent of the workers were unemployed? c. Compare the percent unemployed for the men and the women.

Data Set Exercises 44. Refer to the Real Estate data, which reports information on homes sold in the Goodyear,

the horizontal. Does there seem to be a relationship between these variables? Is the relationship direct or inverse?

the year in which the stadium was built from the current year to find the age of the stadium and work this variable.) Develop a box plot. Are there any outliers? Which sta- diums are outliers?

c. Draw a scatter diagram with the number of games won on the vertical axis and the team salary on the horizontal axis. What are your conclusions?

d. Select the variable wins. Draw a dot plot. What can you conclude from this plot? 46. Refer to the Buena School District bus data.

a. Refer to the maintenance cost variable. Develop a box plot. What are the first and third quartiles? Are there any outliers?

b. Determine the median maintenance cost. Based on the median, develop a contingency table with bus manufacturer as one variable and whether the maintenance cost was above or below the median as the other variable. What are your conclusions?

Number of Workers (000)

Gender Employed Unemployed

Men 70,415 4,209 Women 61,402 3,314

1. The Minitab commands for the dot plot on page 104 are: a. Enter the number of vehicles serviced at Tionesta

Ford Lincoln Mercury in column C1 and Sheffield Motors in C2. Name the variables accordingly.

b. Select Graph and Dotplot. In the first dialog box, select Multiple Y’s, Simple in the lower left corner, and click OK. In the next dialog box select Tionesta and Sheffield as the variables to Graph, click on Labels and write an appro- priate title. Then click OK.

c. To calculate the descriptive statistics shown in the output, select Stat, Basic statistics, and then Display Descriptive statistics. In the dialog box, select Tionesta and Sheffield as the Variables, click on Statistics, select the

desired statistics to be output, and finally click OK twice.

Software Commands

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136 Chapter 4

b. Select Graph and then Boxplot. In the dialog box, select Simple in the upper left corner and click OK. Select Age as the Graph Variable, click on Labels and include an appropriate heading, and then click OK.

6. The Minitab commands for the descriptive sum- mary on page 122 are: a. Enter the data in the first column. In the cell

below C1, enter the variable Earnings. b. Select Stat, Basic Statistics, and then click on

Graphical Summary. Select Earnings as the variable, and then click OK.

7. The Excel commands for the scatter diagram on page 125 are: a. Retrieve the Applewood Auto data. b. Using the mouse, highlight the column of age

and profit. Include the first row. c. Select the Insert tab. Select Scatter from the

Chart options. Select the top left option. The scatter plot will appear.

d. With Chart Tools displayed at the top, select the Layout tab. Select Chart Title and type in a title for the plot. Next, under the same Layout tab, select AxisTitles. Using Primary Vertical Axis Title, name the vertical axis Profit. Using the Pri- mary Horizontal Axis Title, name the horizon- tal axis Age. Next, select Legend and select None.

2. The Minitab commands for the stem-and-leaf dis- play on page 107 are: a. Import the data for Table 4–1. b. Select Graph, and click on Stem-and-Leaf. c. Select the variable Spots, enter 10 for the Incre-

ment, and then click OK.

3. The Minitab commands for the descriptive sum- mary on page 113 are: a. Input the data on the Smith Barney commissions

from the Example on page 111. b. From the toolbar, select Stat, Basic Statistics,

and Display Descriptive Statistics. In the dia- log box, select Commissions as the Variable, and then click OK.

4. The Excel commands for the descriptive statistics on page 114 are: a. Input the data on the Smith Barney commissions

from the Example on page 111. b. In cell C4 write Quartile 1 and in C6 write Quar-

tile 3. c. In cell D4 write “ �QUARTILE(A1:A16,1) ” and hit

Enter. In cell D6 write “ �QUARTILE(A1:A16,1) ” and hit Enter.

5. The Minitab commands for the box plot on page 117 are: a. Import the Applewood Auto Group data.

Chapter 4 Answers to Self-Review

4–1 1. a. 79, 105 b. 15 c. From 88 to 97; 75 percent of the stores are

in this range.

a. 8 b. 10.1, 10.2, 10.4, 10.8 c. 9.5 d. 11.6, 7.7

4–2 a. 7.9 b.

4–3 The smallest value is 10 and the largest 85; the first quartile is 25 and the third 60. About 50 percent of the values are between 25 and 60. The median value is 40. The distribution is positively skewed. There are no outliers.

Q3 � 8.015Q1 � 7.76,7 7 8 0013488 9 1256689

10 1248 11 26

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A Review of Chapters 1–4 137

4–4 a.

sk � 3(81.4 � 84.0)

15.19 � �0.51

s � A

923.2 5 � 1

� 15.19

Median � 84X � 407

5 � 81.4,

d. The distribution is somewhat negatively skewed.

4–5 a. Scatter diagram b. 16 c. $7,500 d. Strong and direct

� �0.5481

sk � 5

(4)(3) [�1.3154]

73 �0.5530 �0.1691 98 1.0928 1.3051 60 �1.4088 �2.7962 92 0.6978 0.3398 84 0.1712 0.0050

�1.3154

c X � X

s d

3X � X s

Glossary

Chapter 1 Descriptive statistics The techniques used to describe the important characteristics of a set of data. This includes organizing the data values into a frequency distribution, computing measures of location, and computing mea- sures of dispersion and skewness. Inferential statistics, also called statistical inference This facet of statistics deals with estimating a population para- meter based on a sample statistic. For example, if a sample of 10 TI-36X solar calculators revealed 2 to be defective, we might infer that 20 percent of the production is defective. Interval measurement If one observation is greater than another by a certain amount, and the zero point is arbi- trary, the measurement is on an interval scale. For exam- ple, the difference between temperatures of 70 degrees and 80 degrees is 10 degrees. Likewise, a temperature of

90 degrees is 10 degrees more than a temperature of 80 degrees, and so on. Nominal measurement The “lowest” level of measure- ment. If data are classified into categories and the order of those categories is not important, it is the nominal level of measurement. Examples are gender (male, female) and political affiliation (Republican, Democrat, Independent, all others). If it makes no difference whether male or female is listed first, the data are nominal level. Ordinal measurement Data that can be ranked are referred to as ordinal measures. For example, consumer response to the sound of a new speaker might be excel- lent, very good, fair, or poor. Population The collection, or set, of all individuals, objects, or measurements whose properties are being studied.

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Ratio measurement If the distance between numbers is a constant size, there is a true zero point, and the ratio of two values is meaningful, then the data are ratio scale. For example, the distance between $200 and $300 is $100, and in the case of money there is a true zero point. If you have zero dollars, there is an absence of money (you have none). Also the ratio between $200 and $300 is meaningful. Sample A portion, or subset, of the population being studied. Statistics The science of collecting, organizing, analyz- ing, and interpreting numerical data for the purpose of making more effective decisions.

Chapter 2 Charts Special graphical formats used to portray a fre- quency distribution, including histograms, frequency poly- gons, and cumulative frequency polygons. Other graphical devices used to portray data are bar charts and pie charts. Class The interval in which the data are tallied. For exam- ple, $4 up to $7 is a class; $7 up to $11 is another class. Class frequency The number of observations in each class. If there are 16 observations in the $4 up to $6 class, 16 is the class frequency. Exhaustive Each observation must fall into one of the categories. Frequency distribution A grouping of data into classes showing the number of observations in each of the mutu- ally exclusive classes. Histogram A graphical display of a frequency or relative frequency distribution. The horizontal axis shows the classes. The vertical height of adjacent bars shows the fre- quency or relative frequency of each class. Midpoint The value that divides the class into two equal parts. For the classes $10 up to $20 and $20 up to $30, the midpoints are $15 and $25, respectively. Mutually exclusive A property of a set of categories such that an individual, object, or measurement is in- cluded in only one category. Relative frequency distribution A frequency distribution that shows the fraction or proportion of the total observa- tions in each class.

Chapter 3 Arithmetic mean The sum of the values divided by the number of values. The symbol for the mean of a sample is

and the symbol for a population mean is �. Geometric mean The nth root of the product of all the values. It is especially useful for averaging rates of change and index numbers. It minimizes the importance of ex- treme values. A second use of the geometric mean is to find the mean annual percent change over a period of time. For example, if gross sales were $245 million in 1990 and $692 million in 2010, the average annual rate of return is 5.33 percent. Mean deviation The mean of the deviations from the mean, disregarding signs. It is identified as MD. Measure of dispersion A value that shows the spread of a data set. The range, variance, and standard deviation are measures of dispersion.

Measure of location A single value that is typical of the data. It pinpoints the center of a distribution. The arith- metic mean, weighted mean, median, mode, and geomet- ric mean are measures of location. Median The value of the middle observation after all the observations have been arranged from low to high. For ex- ample, if observations 6, 9, 4 are rearranged to read 4, 6, 9, the median is 6, the middle value. Mode The value that appears most frequently in a set of data. For grouped data, it is the midpoint of the class con- taining the largest number of values. Range It is a measure of dispersion. The range is found by subtracting the minimum value from the maximum value. Standard deviation The square root of the variance. Variance A measure of dispersion based on the average squared differences from the arithmetic mean. Weighted mean Each value is weighted according to its relative importance. For example, if 5 shirts cost $10 each and 20 shirts cost $8 each, the weighted mean price is $8.40:

Chapter 4 Box plot A graphic display that shows the general shape of a variable’s distribution. It is based on five descriptive statistics: the maximum and minimum values, the first and third quartiles, and the median. Coefficient of skewness A measure of the lack of sym- metry in a distribution. For a symmetric distribution there is no skewness, so the coefficient of skewness is zero. Otherwise, it is either positive or negative, with the limits of �3.0. Contingency table A table used to classify observations according to two characteristics. Deciles Values of an ordered (minimum to maximum) data set that divide the data into ten equal parts. Dot plot A dot plot summarizes the distribution of one variable by stacking dots at points on a number line that shows the values of the variable. A dot plot shows all values. Interquartile range The absolute numerical difference between the first and third quartiles. Fifty percent of a dis- tribution’s values occur in this range. Outlier A data point that is usually far from the others. An accepted rule is to classify an observation as an out- lier if it is 1.5 times the interquartile range above the third quartile or below the first quartile. Percentiles Values of an ordered (minimum to maxi- mum) data set that divide the data into one hundred intervals. Quartiles Values of an ordered (minimum to maximum) data set that divide the data into four intervals. Scatter diagram Graphical technique used to show the relationship between two variables measured with interval or ratio scales. Stem-and-leaf display A method to display a variable’s distribution using every value. Values are classified by the data’s leading digit. For example, if a data set contains val- ues between 13 and 84, eight classes based on the 10s digit would be used for the stems. The 1s digits would be the leaves.

[(5 $10) � (20 $8)]/25 � $210/25 � $8.40.

138 Chapter 4

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Problems

1. A sample of the funds deposited in First Federal Savings Bank’s MCA (miniature check- ing account) revealed the following amounts.

$124 $14 $150 $289 $52 $156 $203 $82 $27 $248 39 52 103 58 136 249 110 298 251 157

186 107 142 185 75 202 119 219 156 78 116 152 206 117 52 299 58 153 219 148 145 187 165 147 158 146 185 186 149 140

Use a statistical software package such as Excel or Minitab to help answer the follow- ing questions. a. Determine the mean, median, and standard deviation. b. Determine the first and third quartiles. c. Develop a box plot. Are there any outliers? Do the amounts follow a symmetric dis-

tribution or are they skewed? Justify your answer. d. Organize the distribution of funds into a frequency distribution. e. Write a brief summary of the results in parts a to d.

2. Listed below are the 44 U.S. presidents and their age as they began their terms in office.

A Review of Chapters 1–4 139

Number Name Age

1 Washington 57 2 J. Adams 61 3 Jefferson 57 4 Madison 57 5 Monroe 58 6 J. Q. Adams 57 7 Jackson 61 8 Van Buren 54 9 W. H. Harrison 68

10 Tyler 51 11 Polk 49 12 Taylor 64 13 Fillmore 50 14 Pierce 48 15 Buchanan 65 16 Lincoln 52 17 A. Johnson 56 18 Grant 46 19 Hayes 54 20 Garfield 49 21 Arthur 50 22 Cleveland 47

Number Name Age

23 B. Harrison 55 24 Cleveland 55 25 McKinley 54 26 T. Roosevelt 42 27 Taft 51 28 Wilson 56 29 Harding 55 30 Coolidge 51 31 Hoover 54 32 F. D. Roosevelt 51 33 Truman 60 34 Eisenhower 62 35 Kennedy 43 36 L. B. Johnson 55 37 Nixon 56 38 Ford 61 39 Carter 52 40 Reagan 69 41 G.H.W. Bush 64 42 Clinton 46 43 G. W. Bush 54 44 Obama 47

tribution or are they skewed? Justify your answer. d. Organize the distribution of ages into a frequency distribution. e. Write a brief summary of the results in parts a to d.

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Home Size Home Size (thousands of Selling Price (thousands of Selling Price square feet) ($ thousands) square feet) ($ thousands)

1.4 100 1.3 110 1.3 110 0.8 85 1.2 105 1.2 105 1.1 120 0.9 75 1.4 80 1.1 70 1.0 105 1.1 95

Use a statistical software package such as Excel or Minitab to help answer the follow- ing questions. Determine the first and third quartiles. a. Determine the mean, median, and standard deviation. b. Determine the first and third quartiles. c. Develop a box plot. Are there any outliers? Do the amounts follow a symmetric dis-

tribution or are they skewed? Justify your answer. d. Organize the distribution of funds into a frequency distribution. e. Write a brief summary of the results in parts a to d.

4. A sample of 12 homes sold last week in St. Paul, Minnesota, revealed the following infor- mation. Draw a scatter diagram. Can we conclude that, as the size of the home (reported below in thousands of square feet) increases, the selling price (reported in $ thousands) also increases?

140 Chapter 4

State Amount

Alabama $30,894 Alaska 38,138 Arizona 31,936 Arkansas 28,473 California 39,626 Colorado 39,491 Connecticut 50,762 Delaware 39,131 DC 57,746 Florida 36,720 Georgia 32,095 Hawaii 37,023 Idaho 29,920 Illinois 38,409 Indiana 32,288 Iowa 33,038 Kansas 34,799 Kentucky 29,729 Louisiana 31,821 Maine 32,095 Maryland 43,788 Massachusetts 46,299 Michigan 33,788 Minnesota 38,859 Mississippi 27,028 Missouri 32,789

State Amount

Montana $30,790 Nebraska 34,440 Nevada 38,994 New Hampshire 39,753 New Jersey 46,763 New Mexico 29,929 New York 44,027 North Carolina 32,247 North Dakota 32,763 Ohio 33,320 Oklahoma 32,391 Oregon 33,299 Pennsylvania 36,825 Rhode Island 37,523 South Carolina 29,767 South Dakota 32,030 Tennessee 32,172 Texas 35,166 Utah 29,406 Vermont 34,871 Virginia 39,540 Washington 38,212 West Virginia 28,206 Wisconsin 34,405 Wyoming 40,655

3. Listed below is the per capita income for the 50 states and the District of Columbia.

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A Review of Chapters 1–4 141

5. Refer to the following diagram.

0 40 80 120 160 200

* *

a. What is the graph called? b. What are the median, and first and third quartile values? c. Is the distribution positively skewed? Tell how you know. d. Are there any outliers? If yes, estimate these values. e. Can you determine the number of observations in the study?

How many other bank services do the checking ac- count customers use? Do the customers use the ATM ser- vice and, if so, how often? What about debit cards? Who uses them, and how often are they used?

To better understand the customers, Mr. Selig asked Ms. Wendy Lamberg, director of planning, to se- lect a sample of customers and prepare a report. To be- gin, she has appointed a team from her staff. You are the head of the team and responsible for preparing the report. You select a random sample of 60 customers. In addition to the balance in each account at the end of last month, you determine: (1) the number of ATM (auto- matic teller machine) transactions in the last month; (2) the number of other bank services (a savings account, a certificate of deposit, etc.) the customer uses; (3) whether the customer has a debit card (this is a bank service in which charges are made directly to the cus- tomer’s account); and (4) whether or not interest is paid on the checking account. The sample includes cus- tomers from the branches in Cincinnati, Ohio; Atlanta, Georgia; Louisville, Kentucky; and Erie, Pennsylvania. 1. Develop a graph or table that portrays the checking

balances. What is the balance of a typical customer? Do many customers have more than $2,000 in their accounts? Does it appear that there is a difference in the distribution of the accounts among the four branches? Around what value do the account bal- ances tend to cluster?

2. Determine the mean and median of the checking account balances. Compare the mean and the

median balances for the four branches. Is there a difference among the branches? Be sure to explain the difference between the mean and the median in your report.

B. Wildcat Plumbing Supply Inc.: Do We Have Gender Differences?

To kick off the project, Mr. Cory St. Julian held a meeting with his staff and you were invited. At this meet- ing, it was suggested that you calculate several mea- sures of location, draw charts, such as a cumulative

Cases

Yearly Salary ($000) Women Men

Less than 30 2 0 30 up to 40 3 1 40 up to 50 17 4 50 up to 60 17 24 60 up to 70 8 21 70 up to 80 3 7 80 or more 0 3

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Part 2—Problems 1. The Russell 2000 index of stock prices increased by the following amounts over the last three years.

18% 4% 2%

What is the geometric mean increase for the three years?

Practice Test

Part 1—Objective 1. The science of collecting, organizing, presenting, analyzing, and interpreting data to assist in making effective deci-

sions is called . 1. 2. Methods of organizing, summarizing, and presenting data in an informative way is called .

2. 3. The entire set of individuals or objects of interest or the measurements obtained from all individuals or objects of in-

terest is called the . 3. 4. List the two types of variables. 4.

5. The number of bedrooms in a house is an example of a . (discrete variable, continuous variable, qualitative variable—pick one) 5.

6. The jersey numbers of Major League Baseball players is an example of what level of measurement? 6.

values are negative, or never—pick one.) 12. 13. Which of the following is least affected by an outlier? (mean, median, or range—pick one) 13.

142 Chapter 4

frequency distribution, and determine the quartiles for both men and women. Develop the charts and write the report summarizing the yearly salaries of employees at Wildcat Plumbing Supply. Does it appear that there are pay differences based on gender?

C. Kimble Products: Is There a Difference In the Commissions?

At the January national sales meeting, the CEO of Kimble Products was questioned extensively regarding the com- pany policy for paying commissions to its sales repre- sentatives. The company sells sporting goods to two major markets. There are 40 sales representatives who

Commissions Earned by Sales Representatives Calling on Large Retailers ($)

1,116 681 1,294 12 754 1,206 1,448 870 944 1,255 1,213 1,291 719 934 1,313 1,083 899 850 886 1,556

886 1,315 1,858 1,262 1,338 1,066 807 1,244 758 918

Commissions Earned by Sales Representatives Calling on Athletic Departments ($)

354 87 1,676 1,187 69 3,202 680 39 1,683 1,106 883 3,140 299 2,197 175 159 1,105 434 615 149

1,168 278 579 7 357 252 1,602 2,321 4 392 416 427 1,738 526 13 1,604 249 557 635 527

call directly on large volume customers, such as the ath- letic departments at major colleges and universities and professional sports franchises. There are 30 sales repre- sentatives who represent the company to retail stores lo- cated in shopping malls and large discounters such as Kmart and Target.

Upon his return to corporate headquarters, the CEO asked the sales manager for a report comparing the com- missions earned last year by the two parts of the sales team. The information is reported below. Write a brief re- port. Would you conclude that there is a difference? Be sure to include information in the report on both the cen- tral tendency and dispersion of the two groups.

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Selling Price ($000) Frequency

120.0 up to 150.0 4 150.0 up to 180.0 18 180.0 up to 210.0 30 210.0 up to 240.0 20 240.0 up to 270.0 17 270.0 up to 300.0 10 300.0 up to 330.0 6

a. What is the class interval? b. How many homes were sold in 2010? c. How many homes sold for less than $210,000? d. What is the relative frequency of the 210 up to 240 class? e. What is the midpoint of the 150 up to 180 class? f. The selling prices range between what two amounts?

3. A sample of eight college students revealed they owned the following number of CDs.

52 76 64 79 80 74 66 69

a. What is the mean number of CDs owned? b. What is the median number of CDs owned? c. What is the 40th percentile? d. What is the range of the number of CDs owned? e. What is the standard deviation of the number of CDs owned?

4. An investor purchased 200 shares of the Blair Company for $36 each in July of 2010, 300 shares at $40 each in Sep- tember 2010, and 500 shares at $50 each in January 2011. What is the investor’s weighted mean price per share?

5. During the 2008 Super Bowl, 30 million pounds of snack food was eaten. The chart below depicts this information.

2. The information below refers to the selling prices ($000) of homes sold in Warren, PA, during 2010.

A Review of Chapters 1–4 143

Snack Nuts 8%

Potato Chips 37%

Tortilla Chips 28%

Pretzels 14%

Popcorn 13%

a. What is the name given to this graph? b. Estimate, in millions of pounds, the amount of potato chips eaten during the game. c. Estimate the relationship of potato chips to popcorn. (twice as much, half as much, three times, none of these—

pick one) d. What percent of the total do potato chips and tortilla chips comprise?

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5 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Explain the terms experi- ment, event, and outcome.

LO2 Identify and apply the appropriate approach to as- signing probabilities.

LO3 Calculate probabilities using the rules of addition.

LO4 Define the term joint probability.

LO5 Calculate probabilities using the rules of multiplication.

LO6 Define the term condi- tional probability.

LO7 Compute probabilities using a contingency table.

LO8 Calculate probabilities using Bayes’ theorem.

LO9 Determine the number of outcomes using the appro- priate principle of counting.

A Survey of Probability Concepts

It was found that 60 percent of the tourists to China visited the

Forbidden City, the Temple of Heaven, the Great Wall, and other

historical sites in or near Beijing. Forty percent visited Xi’an and its

magnificent terracotta soldiers, horses, and chariots, which lay buried

for over 2,000 years. Thirty percent of the tourists went to both Beijing

and Xi’an. What is the probability that a tourist visited at least one of

these places? (See Exercise 76 and LO4.)

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A Survey of Probability Concepts 145

5.1 Introduction The emphasis in Chapters 2, 3, and 4 is on descriptive statistics. In Chapter 2, we organize the profits on 180 vehicles sold by the Applewood Auto Group into a fre- quency distribution. This frequency distribution shows the smallest and the largest profits and where the largest concentration of data occurs. In Chapter 3, we use numerical measures of location and dispersion to locate a typical profit on vehicle sales and to examine the variation in the profit of a sale. We describe the variation in the profits with such measures of dispersion as the range and the standard devi- ation. In Chapter 4, we develop charts and graphs, such as a scatter diagram, to further describe the data graphically.

Descriptive statistics is concerned with summarizing data collected from past events. We now turn to the second facet of statistics, namely, computing the chance that something will occur in the future. This facet of statistics is called statistical inference or inferential statistics.

Seldom does a decision maker have complete information to make a decision. For example:

• Toys and Things, a toy and puzzle man- ufacturer, recently developed a new game based on sports trivia. It wants to know whether sports buffs will purchase the game. “Slam Dunk” and “Home Run” are two of the names under consideration. One way to minimize the risk of making an incorrect decision is to hire a market research firm to select a sample of 2,000 consumers from the population and ask each respondent for a reaction to the new game and its proposed titles. Using the

sample results, the company can estimate the proportion of the population that will purchase the game.

• The quality assurance department of a Bethlehem Steel mill must assure man- agement that the quarter-inch wire being produced has an acceptable ten- sile strength. Obviously, not all the wire produced can be tested for tensile strength because testing requires the wire to be stretched until it breaks—thus destroying it. So a random sample of 10 pieces is selected and tested. Based on the test results, all the wire produced is deemed to be either acceptable or unacceptable.

• Other questions involving uncertainty are: Should the daytime drama Days of Our Lives be discontinued immediately? Will a newly developed mint-flavored cereal be profitable if marketed? Will Charles Linden be elected to county auditor in Batavia County?

Statistical inference deals with conclusions about a population based on a sam- ple taken from that population. (The populations for the preceding illustrations are: all consumers who like sports trivia games, all the quarter-inch steel wire produced, all television viewers who watch soaps, all who purchase breakfast cereal, and so on.)

Because there is uncertainty in decision making, it is important that all the known risks involved be scientifically evaluated. Helpful in this evaluation is probability the- ory, which has often been referred to as the science of uncertainty. The use of prob- ability theory allows the decision maker with only limited information to analyze the risks and minimize the gamble inherent, for example, in marketing a new product or accepting an incoming shipment possibly containing defective parts.

Because probability concepts are so important in the field of statistical inference (to be discussed starting with Chapter 8), this chapter introduces the basic language of probability, including such terms as experiment, event, subjective probability, and addition and multiplication rules.

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5.2 What Is a Probability? No doubt you are familiar with terms such as probability, chance, and likelihood. They are often used interchangeably. The weather forecaster announces that there is a 70 percent chance of rain for Super Bowl Sunday. Based on a survey of consumers who tested a newly developed pickle with a banana taste, the probability is .03 that, if marketed, it will be a financial success. (This means that the chance of the banana- flavor pickle being accepted by the public is rather remote.) What is a probability? In general, it is a number that describes the chance that something will happen.

146 Chapter 5

PROBABILITY A value between zero and one, inclusive, describing the relative possibility (chance or likelihood) an event will occur.

EXPERIMENT A process that leads to the occurrence of one and only one of several possible observations.

OUTCOME A particular result of an experiment.

A probability is frequently expressed as a decimal, such as .70, .27, or .50. How- ever, it may be given as a fraction such as 7/10, 27/100, or 1/2. It can assume any number from 0 to 1, inclusive. If a company has only five sales regions, and each re- gion’s name or number is written on a slip of paper and the slips put in a hat, the prob- ability of selecting one of the five regions is 1. The probability of selecting from the hat a slip of paper that reads “Pittsburgh Steelers” is 0. Thus, the probability of 1 repre- sents something that is certain to happen, and the probability of 0 represents some- thing that cannot happen.

The closer a probability is to 0, the more improbable it is the event will happen. The closer the probability is to 1, the more sure we are it will happen. The relationship is shown in the following diagram along with a few of our personal beliefs. You might, however, select a different probability for Slo Poke’s chances to win the Kentucky Derby or for an increase in federal taxes.

Cannot Sure to happen happen

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

Probability our sun will disappear this year

Chance Slo Poke will

win the Kentucky

Derby

Chance of a head in

single toss of a coin

Chance of an

increase in federal

taxes

Chance of rain in Florida

this year

Three key words are used in the study of probability: experiment, outcome, and event. These terms are used in our everyday language, but in statistics they have specific meanings.

This definition is more general than the one used in the physical sciences, where we picture someone manipulating test tubes or microscopes. In reference to prob- ability, an experiment has two or more possible results, and it is uncertain which will occur.

LO1 Explain the terms experiment, event, and outcome.

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A Survey of Probability Concepts 147

For example, the tossing of a coin is an experiment. You may observe the toss of the coin, but you are unsure whether it will come up “heads” or “tails.” Similarly, ask- ing 500 college students whether they would purchase a new Dell computer system at a particular price is an experiment. If the coin is tossed, one particular outcome is a “head.” The alternative outcome is a “tail.” In the computer purchasing experiment, one possible outcome is that 273 students indicate they would purchase the com- puter. Another outcome is that 317 students would purchase the computer. Still an- other outcome is that 423 students indicate that they would purchase it. When one or more of the experiment’s outcomes are observed, we call this an event.

EVENT A collection of one or more outcomes of an experiment.

Examples to clarify the definitions of the terms experiment, outcome, and event are presented in the following figure.

In the die-rolling experiment, there are six possible outcomes, but there are many possible events. When counting the number of members of the board of directors for Fortune 500 companies over 60 years of age, the number of possible outcomes can be anywhere from zero to the total number of members. There are an even larger num- ber of possible events in this experiment.

Experiment

All possible outcomes

Some possible events

Roll a die

Observe a 1

Observe a 2

Observe a 3

Observe a 4

Observe a 5

Observe a 6

Observe an even number Observe a number greater than 4

Observe a number 3 or less

Count the number of members of the board of directors

for Fortune 500 companies who are over 60 years of age

None are over 60

One is over 60

Two are over 60

...

29 are over 60

...

48 are over 60

...

More than 13 are over 60 Fewer than 20 are over 60

Self-Review 5–1 Video Games Inc. recently developed a new video game. Its playability is to be tested by 80 veteran game players. (a) What is the experiment? (b) What is one possible outcome? (c) Suppose 65 players tried the new game and said they liked it. Is 65 a probability? (d) The probability that the new game will be a success is computed to be �1.0. Comment. (e) Specify one possible event.

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5.3 Approaches to Assigning Probabilities Two approaches to assigning probabilities to an event will be discussed, namely, the objective and the subjective viewpoints. Objective probability is subdivided into (1) classical probability and (2) empirical probability.

Classical Probability Classical probability is based on the assumption that the outcomes of an experi- ment are equally likely. Using the classical viewpoint, the probability of an event happening is computed by dividing the number of favorable outcomes by the num- ber of possible outcomes:

CLASSICAL PROBABILITY [5–1]Probability

of an event �

Number of favorable outcomes Total number of possible outcomes

148 Chapter 5

The mutually exclusive concept appeared earlier in our study of frequency distribu- tions in Chapter 2. Recall that we create classes so that a particular value is included in only one of the classes and there is no overlap between classes. Thus, only one of several events can occur at a particular time.

MUTUALLY EXCLUSIVE The occurrence of one event means that none of the other events can occur at the same time.

The variable “gender” presents mutually exclusive outcomes, male and female. An employee selected at random is either male or female but cannot be both. A manu- factured part is acceptable or unacceptable. The part cannot be both acceptable and unacceptable at the same time. In a sample of manufactured parts, the event of selecting an unacceptable part and the event of selecting an acceptable part are mu- tually exclusive.

LO2 Identify and apply the appropriate approach to assigning probabilities.

Example Consider an experiment of rolling a six-sided die. What is the probability of the event “an even number of spots appear face up”?

Solution The possible outcomes are:

There are three “favorable” outcomes (a two, a four, and a six) in the collection of six equally likely possible outcomes. Therefore:

Probability of an even number

� .5

Number of favorable outcomes Total number of possible outcomesd

d �

3 6

a one-spot

a two-spot

a three-spot

a four-spot

a five-spot

a six-spot

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A Survey of Probability Concepts 149

If an experiment has a set of events that includes every possible outcome, such as the events “an even number” and “an odd number” in the die-tossing experiment, then the set of events is collectively exhaustive. For the die-tossing experiment, every outcome will be either even or odd. So the set is collectively exhaustive.

COLLECTIVELY EXHAUSTIVE At least one of the events must occur when an experiment is conducted.

If the set of events is collectively exhaustive and the events are mutually exclu- sive, the sum of the probabilities is 1. Historically, the classical approach to proba- bility was developed and applied in the 17th and 18th centuries to games of chance, such as cards and dice. It is unnecessary to do an experiment to determine the prob- ability of an event occurring using the classical approach because the total number of outcomes is known before the experiment. The flip of a coin has two possible out- comes; the roll of a die has six possible outcomes. We can logically arrive at the prob- ability of getting a tail on the toss of one coin or three heads on the toss of three coins.

The classical approach to probability can also be applied to lotteries. In South Carolina, one of the games of the Education Lottery is “Pick 3.” A person buys a lot- tery ticket and selects three numbers between 0 and 9. Once per week, the three numbers are randomly selected from a machine that tumbles three containers each with balls numbered 0 through 9. One way to win is to match the numbers and the or- der of the numbers. Given that 1,000 possible outcomes exist (000 through 999), the probability of winning with any three-digit number is 0.001, or 1 in 1,000.

Empirical Probability Empirical or relative frequency is the second type of objective probability. It is based on the number of times an event occurs as a proportion of a known number of trials.

LAW OF LARGE NUMBERS Over a large number of trials, the empirical probability of an event will approach its true probability.

EMPIRICAL PROBABILITY The probability of an event happening is the fraction of the time similar events happened in the past.

In terms of a formula:

The empirical approach to probability is based on what is called the law of large numbers. The key to establishing probabilities empirically is that more observations will provide a more accurate estimate of the probability.

Empirical probability � Number of times the event occurs

Total number of observations

To explain the law of large numbers, suppose we toss a fair coin. The result of each toss is either a head or a tail. With just one toss of the coin the empirical proba- bility for heads is either zero or one. If we toss the coin a great number of times, the probability of the outcome of heads will approach .5. The following table reports the results of an experiment of flipping a fair coin 1, 10, 50, 100, 500, 1,000, and 10,000 times and then computing the relative frequency of heads. Note as we increase the number of trials the empirical probability of a head appearing approaches .5, which is its value based on the classical approach to probability.

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Number of Trials Number of Heads Relative Frequency of Heads

1 0 .00 10 3 .30 50 26 .52

100 52 .52 500 236 .472

1,000 494 .494 10,000 5,027 .5027

What have we demonstrated? Based on the classical definition of probability, the likeli- hood of obtaining a head in a single toss of a fair coin is .5. Based on the empirical or relative frequency approach to probability, the probability of the event happening approaches the same value based on the classical definition of probability.

This reasoning allows us to use the empirical or relative frequency approach to finding a probability. Here are some examples.

• Last semester, 80 students registered for Business Statistics 101 at Scandia University. Twelve students earned an A. Based on this information and the em- pirical approach to assigning a probability, we estimate the likelihood a student will earn an A is .15.

• Kobe Bryant of the Los Angeles Lakers made 403 out of 491 free throw attempts during the 2009–10 NBA season. Based on the empirical rule of probability, the likelihood of him making his next free throw attempt is .821.

Life insurance companies rely on past data to determine the acceptability of an applicant as well as the premium to be charged. Mortality tables list the likelihood a person of a particular age will die within the upcoming year. For example, the likeli- hood a 20-year-old female will die within the next year is .00105.

The empirical concept is illustrated with the following example.

150 Chapter 5

SUBJECTIVE CONCEPT OF PROBABILITY The likelihood (probability) of a particular event happening that is assigned by an individual based on whatever information is available.

Subjective Probability If there is little or no experience or information on which to base a probability, it may be arrived at subjectively. Essentially, this means an individual evaluates the available opinions and information and then estimates or assigns the probability. This probability is aptly called a subjective probability.

Example On February 1, 2003, the Space Shuttle Columbia exploded. This was the second disaster in 113 space missions for NASA. On the basis of this information, what is the probability that a future mission is successfully completed?

Solution To simplify, letters or numbers may be used. P stands for probability, and in this case P(A) stands for the probability a future mission is successfully completed.

We can use this as an estimate of probability. In other words, based on past experi- ence, the probability is .98 that a future space shuttle mission will be safely completed.

P(A) � 111 113

� .98

Probability of a successful flight � Number of successful flights

Total number of flights

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A Survey of Probability Concepts 151

Illustrations of subjective probability are:

1. Estimating the likelihood the New England Patriots will play in the Super Bowl next year.

2. Estimating the likelihood you will be married before the age of 30. 3. Estimating the likelihood the U.S. budget deficit will be reduced by half in the

next 10 years.

The types of probability are summarized in Chart 5–1. A probability statement always assigns a likelihood to an event that has not yet occurred. There is, of course, a considerable latitude in the degree of uncertainty that surrounds this probability, based primarily on the knowledge possessed by the individual concerning the underlying process. The individual possesses a great deal of knowledge about the toss of a die and can state that the probability that a one-spot will appear face up on the toss of a true die is one-sixth. But we know very little concerning the accep- tance in the marketplace of a new and untested product. For example, even though a market research director tests a newly developed product in 40 retail stores and states that there is a 70 percent chance that the product will have sales of more than 1 million units, she has limited knowledge of how consumers will react when it is marketed nationally. In both cases (the case of the person rolling a die and the testing of a new product), the individual is assigning a probability value to an event of interest, and a difference exists only in the predictor’s confidence in the preci- sion of the estimate. However, regardless of the viewpoint, the same laws of proba- bility (presented in the following sections) will be applied.

Approaches to Probability

SubjectiveObjective

Empirical ProbabilityClassical Probability Based on available information

Based on equally likely outcomes

Based on relative frequencies

CHART 5–1 Summary of Approaches to Probability

Self-Review 5–2 1. One card will be randomly selected from a standard 52-card deck. What is the proba- bility the card will be a queen? Which approach to probability did you use to answer this question?

2. The Center for Child Care reports on 539 children and the marital status of their par- ents. There are 333 married, 182 divorced, and 24 widowed parents. What is the prob- ability a particular child chosen at random will have a parent who is divorced? Which approach did you use?

3. What is the probability that the Dow Jones Industrial Average will exceed 12,000 dur- ing the next 12 months? Which approach to probability did you use to answer this question?

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Exercises 1. Some people are in favor of reducing federal taxes to increase consumer spending and oth-

ers are against it. Two persons are selected and their opinions are recorded. Assuming no one is undecided, list the possible outcomes.

2. A quality control inspector selects a part to be tested. The part is then declared acceptable, repairable, or scrapped. Then another part is tested. List the possible outcomes of this ex- periment regarding two parts.

3. A survey of 34 students at the Wall College of Business showed the following majors:

Accounting 10 Finance 5 Economics 3 Management 6 Marketing 10

Suppose you select a student and observe his or her major. a. What is the probability he or she is a management major? b. Which concept of probability did you use to make this estimate?

4. A large company that must hire a new president prepares a final list of five candidates, all of whom are equally qualified. Two of these candidates are members of a minority group. To avoid bias in the selection of the candidate, the company decides to select the president by lottery. a. What is the probability one of the minority candidates is hired? b. Which concept of probability did you use to make this estimate?

5. In each of the following cases, indicate whether classical, empirical, or subjective probabil- ity is used. a. A baseball player gets a hit in 30 out of 100 times at bat. The probability is .3 that

he gets a hit in his next at bat. b. A seven-member committee of students is formed to study environmental issues. What

is the likelihood that any one of the seven is chosen as the spokesperson? c. You purchase one of 5 million tickets sold for Lotto Canada. What is the likelihood you

will win the $1 million jackpot? d. The probability of an earthquake in northern California in the next 10 years above 5.0

on the Richter Scale is .80. 6. A firm will promote two employees out of a group of six men and three women.

a. List the chances of this experiment if there is particular concern about gender equity. b. Which concept of probability would you use to estimate these probabilities?

7. A sample of 40 oil industry executives was selected to test a questionnaire. One question about environmental issues required a yes or no answer. a. What is the experiment? b. List one possible event. c. Ten of the 40 executives responded yes. Based on these sample responses, what is

the probability that an oil industry executive will respond yes? d. What concept of probability does this illustrate? e. Are each of the possible outcomes equally likely and mutually exclusive?

8. A sample of 2,000 licensed drivers revealed the following number of speeding violations.

Number of Violations Number of Drivers

0 1,910 1 46 2 18 3 12 4 9 5 or more 5

Total 2,000

a. What is the experiment? b. List one possible event.

152 Chapter 5

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Example

A Survey of Probability Concepts 153

c. What is the probability that a particular driver had exactly two speeding violations? d. What concept of probability does this illustrate?

9. Bank of America customers select their own three-digit personal identification number (PIN) for use at ATMs. a. Think of this as an experiment and list four possible outcomes. b. What is the probability Mr. Jones and Mrs. Smith select the same PIN? c. Which concept of probability did you use to answer (b)?

10. An investor buys 100 shares of AT&T stock and records its price change daily. a. List several possible events for this experiment. b. Estimate the probability for each event you described in (a). c. Which concept of probability did you use in (b)?

5.4 Some Rules for Computing Probabilities Now that we have defined probability and described the different approaches to prob- ability, we turn our attention to computing the probability of two or more events by applying rules of addition and multiplication.

Rules of Addition There are two rules of addition, the special rule of addition and the general rule of addition. We begin with the special rule of addition.

Special Rule of Addition To apply the special rule of addition, the events must be mutually exclusive. Recall that mutually exclusive means that when one event oc- curs, none of the other events can occur at the same time. An illustration of mutually exclusive events in the die-tossing experiment is the events “a number 4 or larger” and “a number 2 or smaller.” If the outcome is in the first group {4, 5, and 6}, then it cannot also be in the second group {1 and 2}. Another illustration is a product coming off the assembly line cannot be defective and satisfactory at the same time.

If two events A and B are mutually exclusive, the special rule of addition states that the probability of one or the other event’s occurring equals the sum of their probabilities. This rule is expressed in the following formula:

SPECIAL RULE OF ADDITION [5–2]

For three mutually exclusive events designated A, B, and C, the rule is written:

An example will help to show the details.

P(A or B or C) � P(A) � P(B) � P(C)

P(A or B) � P(A) � P(B)

LO3 Calculate probabilities using the rules of addition.

A machine fills plastic bags with a mixture of beans, broccoli, and other vegetables. Most of the bags contain the correct weight, but because of the variation in the size

of the beans and other vegetables, a package might be underweight or overweight. A check of 4,000 packages filled in the past month revealed:

Number of Probability of Weight Event Packages Occurrence

Underweight A 100 .025 Satisfactory B 3,600 .900 Overweight C 300 .075

4,000 1.000

d 100

4,000

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English logician J. Venn (1834–1923) developed a diagram to portray graphically the outcome of an experiment. The mutually exclusive concept and various other rules for combining probabilities can be illustrated using this device. To construct a Venn di- agram, a space is first enclosed representing the total of all possible outcomes. This space is usually in the form of a rectangle. An event is then represented by a circular area which is drawn inside the rectangle proportional to the probability of the event. The following Venn diagram represents the mutually exclusive concept. There is no overlapping of events, meaning that the events are mutually exclusive. In the follow- ing diagram, assume the events A, B, and C are about equally likely.

154 Chapter 5

Event A

Event B

Event C

Complement Rule The probability that a bag of mixed vegetables selected is underweight, P(A), plus the probability that it is not an underweight bag, written P and read “not A,” must logically equal 1. This is written:

This can be revised to read:

COMPLEMENT RULE [5–3]

This is the complement rule. It is used to determine the probability of an event occurring by subtracting the probability of the event not occurring from 1. This rule is useful because sometimes it is easier to calculate the probability of an event hap- pening by determining the probability of it not happening and subtracting the result from 1. Notice that the events A and are mutually exclusive and collectively exhaustive. Therefore, the probabilities of A and sum to 1. A Venn diagram illus- trating the complement rule is shown as:

�A �A

P(A) � 1 � P(�A)

P(A) � P(�A) � 1

(�A)

Event A

� A

What is the probability that a particular package will be either underweight or overweight?

Solution The outcome “underweight” is the event A. The outcome “overweight” is the event C. Applying the special rule of addition:

Note that the events are mutually exclusive, meaning that a package of mixed veg- etables cannot be underweight, satisfactory, and overweight at the same time. They are also collectively exhaustive; that is, a selected package must be either under- weight, satisfactory, or overweight.

P(A or C) � P(A) � P(C) � .025 � .075 � .10

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Example Recall the probability a bag of mixed vegetables is underweight is .025 and the prob- ability of an overweight bag is .075. Use the complement rule to show the probability of a satisfactory bag is .900. Show the solution using a Venn diagram.

Solution The probability the bag is unsatisfactory equals the probability the bag is overweight plus the probability it is underweight. That is, P(A or C) � P(A) � P(C) � .025 � .075 � .100. The bag is satisfactory if it is not underweight or overweight, so P(B) � 1 � [P(A) � P(C)] � 1 � [.025 � .075] � 0.900. The Venn diagram portraying this situation is:

A Survey of Probability Concepts 155

A .025

not A or C .90

C .075

The General Rule of Addition The outcomes of an experiment may not be mutually exclusive. Suppose, for illustration, that the Florida Tourist Commission selected a sample of 200 tourists who visited the state during the year. The survey revealed that 120 tourists went to Disney World and 100 went to Busch Gardens near Tampa. What is the probability that a person selected visited either Disney World or Busch Gardens? If the special rule of addition is used, the probability of selecting a tourist who went to Disney World is .60, found by 120/200. Similarly, the probability of a tourist going to Busch Gardens is .50. The sum of these probabili- ties is 1.10. We know, however, that this probability cannot be greater than 1. The explanation is that many tourists visited both attractions and are being counted twice! A check of the survey responses revealed that 60 out of 200 sampled did, in fact, visit both attractions.

To answer our question, “What is the probability a selected person visited either Disney World or Busch Gardens?” (1) add the probability that a tourist visited Disney

Self-Review 5–3 A sample of employees of Worldwide Enterprises is to be surveyed about a new health care plan. The employees are classified as follows:

Classification Event Number of Employees

Supervisors A 120 Maintenance B 50 Production C 1,460 Management D 302 Secretarial E 68

(a) What is the probability that the first person selected is: (i) either in maintenance or a secretary? (ii) not in management?

(b) Draw a Venn diagram illustrating your answers to part (a). (c) Are the events in part (a)(i) complementary or mutually exclusive or both?

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World and the probability he or she visited Busch Gardens, and (2) subtract the prob- ability of visiting both. Thus:

When two events both occur, the probability is called a joint probability. The prob- ability that a tourist visits both attractions (.30) is an example of a joint probability.

The following Venn diagram shows two events that are not mutually exclusive. The two events overlap to illustrate the joint event that some people have visited both attractions.

� .60 � .50 � .30 � .80

P(Disney or Busch) � P(Disney) � P(Busch) � P(both Disney and Busch)

156 Chapter 5

P (Disney) = .60 P (Busch) = .50

P (Disney and Busch) = .30

JOINT PROBABILITY A probability that measures the likelihood two or more events will happen concurrently.

This rule for two events designated A and B is written:

GENERAL RULE OF ADDITION [5–4]P(A or B) � P(A) � P(B) � P(A and B)

LO4 Define the term joint probability.

Statistics in Action

If you wish to get some attention at the next gathering you attend, announce that you believe that at least two people present were born on the same date—that is, the same day of the year but not nec- essarily the same year. If there are 30 people in the room, the probability of a duplicate is .706. If there are 60 people in the room, the probability is .994 that at least two peo- ple share the same birthday. With as few as 23 people the chances are even, that is .50, that at least two people share the same birth- day. Hint: To com- pute this, find the probability everyone was born on a differ- ent day and use the complement rule. Try this in your class.

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Example What is the probability that a card chosen at random from a standard deck of cards will be either a king or a heart?

Solution We may be inclined to add the probability of a king and the probability of a heart. But this creates a problem. If we do that, the king of hearts is counted with the kings and also with the hearts. So, if we simply add the probability of a king (there are 4 in a deck of 52 cards) to the probability of a heart (there are 13 in a deck of 52 cards) and report that 17 out of 52 cards meet the requirement, we have counted the king of hearts twice. We need to subtract 1 card from the 17 so the king of hearts is counted only once. Thus, there are 16 cards that are either hearts or kings. So the probability is 16/52 � .3077.

Card Probability Explanation

King P (A ) � 4/52 4 kings in a deck of 52 cards Heart P(B) � 13/52 13 hearts in a deck of 52 cards King of Hearts P(A and B) � 1/52 1 king of hearts in a deck of 52 cards

From formula (5–4):

A Venn diagram portrays these outcomes, which are not mutually exclusive.

� 16/52, or .3077

� 4/52 � 13/52 � 1/52

P(A or B) � P(A) � P(B) � P(A and B)

A Survey of Probability Concepts 157

Kings

Hearts

Both

A B A

and

For the expression P(A or B), the word or suggests that A may occur or B may occur. This also includes the possibility that A and B may occur. This use of or is some- times called an inclusive. You could also write P(A or B or both) to emphasize that the union of the events includes the intersection of A and B.

If we compare the general and special rules of addition, the important differ- ence is determining if the events are mutually exclusive. If the events are mutually exclusive, then the joint probability P(A and B) is 0 and we could use the special rule of addition. Otherwise, we must account for the joint probability and use the general rule of addition.

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Exercises 11. The events A and B are mutually exclusive. Suppose P(A) � .30 and P(B) � .20. What

is the probability of either A or B occurring? What is the probability that neither A nor B will happen?

12. The events X and Y are mutually exclusive. Suppose P(X) � .05 and P(Y) � .02. What is the probability of either X or Y occurring? What is the probability that neither X nor Y will happen?

13. A study of 200 advertising firms revealed their income after taxes:

Income after Taxes Number of Firms

Under $1 million 102 $1 million to $20 million 61 $20 million or more 37

a. What is the probability an advertising firm selected at random has under $1 million in income after taxes?

b. What is the probability an advertising firm selected at random has either an income between $1 million and $20 million, or an income of $20 million or more? What rule of probability was applied?

14. The chair of the board of directors says, “There is a 50 percent chance this company will earn a profit, a 30 percent chance it will break even, and a 20 percent chance it will lose money next quarter.” a. Use an addition rule to find the probability the company will not lose money next

quarter. b. Use the complement rule to find the probability it will not lose money next quarter.

15. Suppose the probability you will get an A in this class is .25 and the probability you will get a B is .50. What is the probability your grade will be above a C?

16. Two coins are tossed. If A is the event “two heads” and B is the event “two tails,” are A and B mutually exclusive? Are they complements?

17. The probabilities of the events A and B are .20 and .30, respectively. The probability that both A and B occur is .15. What is the probability of either A or B occurring?

18. Let P(X) � .55 and P(Y) � .35. Assume the probability that they both occur is .20. What is the probability of either X or Y occurring?

19. Suppose the two events A and B are mutually exclusive. What is the probability of their joint occurrence?

20. A student is taking two courses, history and math. The probability the student will pass the history course is .60, and the probability of passing the math course is .70. The prob- ability of passing both is .50. What is the probability of passing at least one?

21. A survey of grocery stores in the Southeast revealed 40 percent had a pharmacy, 50 per- cent had a floral shop, and 70 percent had a deli. Suppose 10 percent of the stores have all three departments, 30 percent have both a pharmacy and a deli, 25 percent have both a floral shop and deli, and 20 percent have both a pharmacy and floral shop. a. What is the probability of selecting a store at random and finding it has both a phar-

macy and a floral shop? b. What is the probability of selecting a store at random and finding it has both a phar-

macy and a deli?

158 Chapter 5

Self-Review 5–4 Routine physical examinations are conducted annually as part of a health service program for General Concrete Inc. employees. It was discovered that 8 percent of the employees need corrective shoes, 15 percent need major dental work, and 3 percent need both corrective shoes and major dental work. (a) What is the probability that an employee selected at random will need either corrective

shoes or major dental work? (b) Show this situation in the form of a Venn diagram.

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Example A survey by the American Automobile Association (AAA) revealed 60 percent of its members made airline reservations last year. Two members are selected at random. What is the probability both made airline reservations last year?

Solution The probability the first member made an airline reservation last year is .60, written P(R1) � .60, where R1 refers to the fact that the first member made a reservation.

A Survey of Probability Concepts 159

c. Are the events “select a store with a deli” and “select a store with a pharmacy” mutu- ally exclusive?

d. What is the name given to the event of “selecting a store with a pharmacy, a floral shop, and a deli?”

e. What is the probability of selecting a store that does not have all three departments? 22. A study by the National Park Service revealed that 50 percent of vacationers going

to the Rocky Mountain region visit Yellowstone Park, 40 percent visit the Tetons, and 35 percent visit both. a. What is the probability a vacationer will visit at least one of these attractions? b. What is the probability .35 called? c. Are the events mutually exclusive? Explain.

Rules of Multiplication When we used the rules of addition in the previous section, we found the likelihood of combining two events. In this section, we find the likelihood that two events both happen. For example, a marketing firm may want to estimate the likelihood that a person is 21 years old or older and buys a Hummer. Venn diagrams illustrate this as the intersection of two events. To find the likelihood of two events happening we use the rules of multiplication. There are two rules of multiplication, the special rule and the general rule.

Special Rule of Multiplication The special rule of multiplication requires that two events A and B are independent. Two events are independent if the occurrence of one event does not alter the probability of the occurrence of the other event.

INDEPENDENCE The occurrence of one event has no effect on the probability of the occurrence of another event.

One way to think about independence is to assume that events A and B occur at different times. For example, when event B occurs after event A occurs, does A have any effect on the likelihood that event B occurs? If the answer is no, then A and B are independent events. To illustrate independence, suppose two coins are tossed. The outcome of a coin toss (head or tail) is unaffected by the outcome of any other prior coin toss (head or tail).

For two independent events A and B, the probability that A and B will both occur is found by multiplying the two probabilities. This is the special rule of multiplication and is written symbolically as:

SPECIAL RULE OF MULTIPLICATION [5–5]

For three independent events, A, B, and C, the special rule of multiplication used to determine the probability that all three events will occur is:

P(A and B and C) � P(A)P(B)P(C)

P(A and B) � P(A)P(B)LO5 Calculate probabilities using the rules of multiplication.

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General Rule of Multiplication If two events are not independent, they are referred to as dependent. To illustrate dependency, suppose there are 10 cans of soda in a cooler, 7 are regular and 3 are diet. A can is selected from the cooler. The probability of selecting a can of diet soda is 3/10, and the probability of selecting a can of regular soda is 7/10. Then a second can is selected from the cooler, without returning the first. The probability the second is diet depends on whether the first one selected was diet or not. The probability that the second is diet is:

2/9, if the first can is diet. (Only two cans of diet soda remain in the cooler.) 3/9, if the first can selected is regular. (All three diet sodas are still in the

cooler.)

The fraction 2/9 (or 3/9) is aptly called a conditional probability because its value is conditional on (dependent on) whether a diet or regular soda was the first selec- tion from the cooler.

160 Chapter 5

Self-Review 5–5 From experience, Teton Tire knows the probability is .95 that a particular XB-70 tire will last 60,000 miles before it becomes bald or fails. An adjustment is made on any tire that does not last 60,000 miles. You purchase four XB-70s. What is the probability all four tires will last at least 60,000 miles?

CONDITIONAL PROBABILITY The probability of a particular event occurring, given that another event has occurred.

LO6 Define the term conditional probability.

The probability that the second member selected made a reservation is also .60, so P(R2) � .60. Since the number of AAA members is very large, you may assume that R1 and R2 are independent. Consequently, using formula (5–5), the probability they both make a reservation is .36, found by:

All possible outcomes can be shown as follows. R means a reservation is made, and NR means no reservation was made.

With the probabilities and the complement rule, we can compute the joint prob- ability of each outcome. For example, the probability that neither member makes a reservation is .16. Further, the probability of the first or the second member (spe- cial addition rule) making a reservation is .48 (.24 � .24). You can also observe that the outcomes are mutually exclusive and collectively exhaustive. Therefore, the probabilities sum to 1.00.

Outcomes Joint Probability

R1 R2 (.60)(.60) � .36 R1 NR2 (.60)(.40) � .24 NR1 R2 (.40)(.60) � .24 NR1 NR2 (.40)(.40) � .16

Total 1.00

P(R1 and R2) � P(R1)P(R2) � (.60)(.60) � .36

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A Survey of Probability Concepts 161

Incidentally, it is assumed that this experiment was conducted without replacement. That is, the first shirt was not laundered and put back in the closet before the sec- ond was selected. So the outcome of the second event is conditional or dependent on the outcome of the first event.

We can extend the general rule of multiplication to more than two events. For three events A, B, and C, the formula is:

In the case of the golf shirt example, the probability of selecting three white shirts without replacement is:

So the likelihood of selecting three shirts without replacement and all being white is .38.

P(W1 and W2 and W3) � P(W1)P(W2 ƒ W1)P(W3 ƒ W1 and W2) � a 9 12 b a

8 11 b a

7 10 b � .38

P(A and B and C) � P(A)P(B ƒ A)P(C ƒ A and B)

We use the general rule of multiplication to find the joint probability of two events when the events are not independent. For example, when event B occurs after event A occurs, and A has an effect on the likelihood that event B occurs, then A and B are not independent.

The general rule of multiplication states that for two events, A and B, the joint probability that both events will happen is found by multiplying the probability that event A will happen by the conditional probability of event B occurring given that A has occurred. Symbolically, the joint probability, P(A and B), is found by:

GENERAL RULE OF MULTIPLICATION [5–6]P(A and B) � P(A)P(B ƒ A)

Example A golfer has 12 golf shirts in his closet. Suppose 9 of these shirts are white and the others blue. He gets dressed in the dark, so he just grabs a shirt and puts it on. He plays golf two days in a row and does not do laundry. What is the likelihood both shirts selected are white?

Solution The event that the first shirt selected is white is W1. The probability is P(W1) � 9/12 because 9 of the 12 shirts are white. The event that the second shirt selected is also white is identified as W2. The conditional probability that the second shirt selected is white, given that the first shirt selected is also white, is P(W2 ƒW1) � 8/11. Why is this so? Because after the first shirt is selected there are only 11 shirts remaining in the closet and 8 of these are white. To determine the probability of 2 white shirts being selected, we use formula (5–6).

So the likelihood of selecting two shirts and finding them both to be white is .55.

P(W1 and W2) � P(W1)P(W2 ƒ W1) � a 9 12 b a

8 11 b � .55

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162 Chapter 5

A contingency table is a cross-tabulation that simultaneously summarizes two vari- ables of interest and their relationship. The level of measurement can be nominal. Below are several examples.

• A survey of 150 adults classified each as to gender and the number of movies attended last month. Each respondent is classified according to two criteria— the number of movies attended and gender.

Gender

Movies Attended Men Women Total

0 20 40 60 1 40 30 70 2 or more 10 10 20

Total 70 80 150

• The American Coffee Producers Association reports the following information on age and the amount of coffee consumed in a month.

Coffee Consumption

Age (Years) Low Moderate High Total

Under 30 36 32 24 92 30 up to 40 18 30 27 75 40 up to 50 10 24 20 54 50 and over 26 24 29 79

Total 90 110 100 300

According to this table, each of the 300 respondents is classified according to two criteria: (1) age and (2) the amount of coffee consumed.

5.5 Contingency Tables Often we tally the results of a survey in a two-way table and use the results of this tally to determine various probabilities. We described this idea beginning on page 126 in Chapter 4. To review, we refer to a two-way table as a contingency table.

CONTINGENCY TABLE A table used to classify sample observations according to two or more identifiable characteristics.

Self-Review 5–6 The board of directors of Tarbell Industries consists of eight men and four women. A four- member search committee is to be chosen at random to conduct a nationwide search for a new company president. (a) What is the probability all four members of the search committee will be women? (b) What is the probability all four members will be men? (c) Does the sum of the probabilities for the events described in parts (a) and (b) equal 1?

Explain.

Statistics in Action

In 2000 George W. Bush won the U.S. presidency by the slimmest of margins. Many election stories resulted, some in- volving voting irreg- ularities, others raising interesting election questions. In a local Michigan elec- tion, there was a tie be- tween two candidates for an elected posi- tion. To break the tie, the candidates drew a slip of paper from a box that contained two slips of paper, one marked “Winner” and the other un- marked. To determine which candidate drew first, election officials flipped a coin. The winner of the coin flip also drew the winning slip of paper. But was the coin flip really necessary? No, be- cause the two events are independent. Winning the coin flip did not alter the prob- ability of either candi- date drawing the winning slip of paper.

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Example A sample of executives were surveyed about loyalty to their company. One of the ques- tions was, “If you were given an offer by another company equal to or slightly bet- ter than your present position, would you remain with the company or take the other position?” The responses of the 200 executives in the survey were cross-classified with their length of service with the company. (See Table 5–1.)

TABLE 5–1 Loyalty of Executives and Length of Service with Company

A Survey of Probability Concepts 163

The following example shows how the rules of addition and multiplication are used when we employ contingency tables.

Length of Service

Less than 1–5 6–10 More than 1 Year, Years, Years, 10 Years,

Loyalty B1 B2 B3 B4 Total

Would remain, A1 10 30 5 75 120 Would not remain, A2 25 15 10 30 80

35 45 15 105 200

LO7 Compute probabilities using a contingency table.

What is the probability of randomly selecting an executive who is loyal to the com- pany (would remain) and who has more than 10 years of service?

Solution Note that two events occur at the same time—the executive would remain with the company, and he or she has more than 10 years of service.

1. Event A1 happens if a randomly selected executive will remain with the company despite an equal or slightly better offer from another company. To find the proba- bility that event A1 will happen, refer to Table 5–1. Note there are 120 executives out of the 200 in the survey who would remain with the company, so P(A1) � 120/200, or .60.

2. Event B4 happens if a randomly selected executive has more than 10 years of ser- vice with the company. Thus, P(B4 ƒ A1) is the conditional probability that an exec- utive with more than 10 years of service would remain with the company despite an equal or slightly better offer from another company. Referring to the contin- gency table, Table 5–1, 75 of the 120 executives who would remain have more than 10 years of service, so P(B4 ƒA1) � 75/120.

Solving for the probability that an executive randomly selected will be one who would remain with the company and who has more than 10 years of service with the com- pany, using the general rule of multiplication in formula (5–6), gives:

To find the probability of selecting an executive who would remain with the com- pany or has less than 1 year of experience, we use the general rule of addition, for- mula (5–4).

1. Event A1 refers to executives that would remain with the company. So P(A1) � 120/200 � .60.

2. Event B1 refers to executives that have been with the company less than 1 year. The probability of B1 is P(B1) � 35/200 � .175.

3. The events A1 and B1 are not mutually exclusive. That is, an executive can both be willing to remain with the company and have less than 1 year of experience.

P(A1 and B4) � P(A1)P(B4 ƒ A1) � a 120 200 b a

75 120 b �

9,000 24,000

� .375

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5.6 Tree Diagrams The tree diagram is a graph that is helpful in organizing calculations that involve several stages. Each segment in the tree is one stage of the problem. The branches of a tree diagram are weighted by probabilities. We will use the data in Table 5–1 to show the construction of a tree diagram.

1. To construct a tree diagram, we begin by drawing a heavy dot on the left to represent the root of the tree (see Chart 5–2).

2. For this problem, two main branches go out from the root, the upper one repre- senting “would remain” and the lower one “would not remain.” Their probabilities are written on the branches, namely, 120/200 and 80/200. These probabilities could also be denoted P(A1) and P(A2).

3. Four branches “grow” out of each of the two main branches. These branches represent the length of service—less than 1 year, 1–5 years, 6–10 years, and more than 10 years. The conditional probabilities for the upper branch of the tree, 10/120, 30/120, 5/120, and so on are written on the appropriate branches. These are and where refers to less than 1 year of service, 1 to 5 years, 6 to 10 years, and more than 10 years. Next, write the conditional probabilities for the lower branch.

4. Finally, joint probabilities, that the events and or the events and will occur together, are shown on the right side. For example, the joint probability of randomly selecting an executive who would remain with the company and who has less than 1 year of service, from formula (5–6), is:

Because the joint probabilities represent all possible outcomes (would remain, 6–10 years service; would not remain, more than 10 years of service; etc.), they must sum to 1.00 (see Chart 5–2).

P(A1 and B1) � P(A1)P(B1 ƒ A1) � a 120 200 b a

10 120 b � .05

BiA2BiA1

B4B3B2

B1P(B4 ƒ A1),P(B1 ƒ A1), P(B2 ƒ A1), P(B3 ƒ A1),

164 Chapter 5

Self-Review 5–7 Refer to Table 5–1 on page 163 to find the following probabilities. (a) What is the probability of selecting an executive with more than 10 years of service? (b) What is the probability of selecting an executive who would not remain with the com-

pany, given that he or she has more than 10 years of service? (c) What is the probability of selecting an executive with more than 10 years of service

or one who would not remain with the company?

We write this probability, which is called the joint probability, as P(A1 and B1). There are 10 executives who would both stay with the company and have less than 1 year of service, so P(A1 and B1) � 10/200 � .05. These 10 people are in both groups, those who would remain with the company and those with less than 1 year with the company. They are actually being counted twice, so we need to subtract out this value.

4. We insert these values in formula (5–4) and the result is as follows.

So the likelihood that a selected executive would either remain with the company or has been with the company less than 1 year is .725.

� .60 � .175 � .05 � .725

P(A1 or B1) � P(A1) � P(B1) � P(A1 and B1)

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A Survey of Probability Concepts 165

Self-Review 5–8 Consumers were surveyed on the relative number of visits to a Sears store (often, occa- sional, and never) and if the store was located in an enclosed mall (yes and no). When variables are measured nominally, such as these data, the results are usually summarized in a contingency table.

Enclosed Mall

Visits Yes No Total

Often 60 20 80 Occasional 25 35 60 Never 5 50 55

90 105 195

(a) Are the number of visits and enclosed mall variables independent? Why? Interpret your conclusion.

(b) Draw a tree diagram and determine the joint probabilities.

10 ____ 120 Less than 1 year

1–5 years

6–10 years

Over 10 years

120 ____ 200

� � .050

� � .150

� � .025

� � .375

10 ____ 120

30 ____ 120 120 ____ 200

30 ____ 120

5 ____ 120

120 ____ 200

5 ____ 120

75 ____ 120

120 ____ 200

75 ____ 120

25 ___ 80 Less than 1 year

1–5 years

6–10 years

Over 10 years

80 ____ 200

� � .125

� � .075

� � .050

� � .150

25 ___ 80

15 ___ 80 80 ____ 200

15 ___ 80

10 ___ 80

80 ____ 200

10 ___ 80

30 ___ 80

80 ____ 200

30 ___ 80

1.000Must total 1.00

Joint probabilities

Conditional probabilities

ServiceLoyalty

Would remain

Would not remain

120 ____ 200

80 ____ 200

CHART 5–2 Tree Diagram Showing Loyalty and Length of Service

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166 Chapter 5

Exercises 23. Suppose and What is the joint probability of A and B? 24. Suppose and What is the joint probability of and ? 25. A local bank reports that 80 percent of its customers maintain a checking account, 60 per-

cent have a savings account, and 50 percent have both. If a customer is chosen at ran- dom, what is the probability the customer has either a checking or a savings account? What is the probability the customer does not have either a checking or a savings account?

26. All Seasons Plumbing has two service trucks that frequently need repair. If the probabil- ity the first truck is available is .75, the probability the second truck is available is .50, and the probability that both trucks are available is .30, what is the probability neither truck is available?

27. Refer to the following table.

First Event

Second Event Total

2 1 3 6 1 2 1 4

Total 3 3 4 10

a. Determine P(A1). b. Determine c. Determine

28. Three defective electric toothbrushes were accidentally shipped to a drugstore by Clean- brush Products along with 17 nondefective ones. a. What is the probability the first two electric toothbrushes sold will be returned to the

drugstore because they are defective? b. What is the probability the first two electric toothbrushes sold will not be defective?

29. Each salesperson at Puchett, Sheets, and Hogan Insurance Agency is rated either below average, average, or above average with respect to sales ability. Each salesperson is also rated with respect to his or her potential for advancement—either fair, good, or excellent. These traits for the 500 salespeople were cross-classified into the following table.

Potential for Advancement

Sales Ability Fair Good Excellent

Below average 16 12 22 Average 45 60 45

Above average 93 72 135

a. What is this table called? b. What is the probability a salesperson selected at random will have above average

sales ability and excellent potential for advancement? c. Construct a tree diagram showing all the probabilities, conditional probabilities, and

joint probabilities. 30. An investor owns three common stocks. Each stock, independent of the others, has

equally likely chances of (1) increasing in value, (2) decreasing in value, or (3) remaining the same value. List the possible outcomes of this experiment. Estimate the probability at least two of the stocks increase in value.

31. The board of directors of a small company consists of five people. Three of those are “strong leaders.” If they buy an idea, the entire board will agree. The other “weak” mem- bers have no influence. Three salespeople are scheduled, one after the other, to make sales presentations to a board member of the salesperson’s choice. The salespeople are convincing but do not know who the “strong leaders” are. However, they will know who the previous salespeople spoke to. The first salesperson to find a strong leader will win the account. Do the three salespeople have the same chance of winning the account? If not, find their respective probabilities of winning.

P(B2 and A3). P(B1 ƒ A2).

A3A2A1

Y2X1P(Y2 @ X1) � .40.P(X1) � .75 P(B @ A) � .30.P(A) � .40

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A Survey of Probability Concepts 167

The prior probability a person is not afflicted with the disease is therefore .95, or found by

There is a diagnostic technique to detect the disease, but it is not very accu- rate. Let B denote the event “test shows the disease is present.” Assume that his- torical evidence shows that if a person actually has the disease, the probability that the test will indicate the presence of the disease is .90. Using the conditional prob- ability definitions developed earlier in this chapter, this statement is written as:

Assume the probability is .15 that for a person who actually does not have the dis- ease the test will indicate the disease is present.

Let’s randomly select a person from Umen and perform the test. The test results indicate the disease is present. What is the probability the person actually has the disease? In symbolic form, we want to know which is interpreted as: P(has the disease ƒ the test results are positive). The probability is called a pos- terior probability.

P(A1 ƒ B) P(A1 ƒ B),

P(B ƒ A2) � .15

P(B ƒ A1) � .90

1 � .05.P(A2) � .95,

32. If you ask three strangers about their birthdays, what is the probability: (a) All were born on Wednesday? (b) All were born on different days of the week? (c) None were born on Saturday?

5.7 Bayes’ Theorem In the 18th century, Reverend Thomas Bayes, an English Presbyterian minister, pon- dered this question: Does God really exist? Being interested in mathematics, he attempted to develop a formula to arrive at the probability God does exist based on evidence available to him on earth. Later Pierre-Simon Laplace refined Bayes’ work and gave it the name “Bayes’ theorem.” In a workable form, Bayes’ theorem is:

BAYES’ THEOREM [5–7]

Assume in formula 5–7 that the events and are mutually exclusive and col- lectively exhaustive, and refers to either event or Hence and are in this case complements. The meaning of the symbols used is illustrated by the fol- lowing example.

Suppose 5 percent of the population of Umen, a fictional Third World country, have a disease that is peculiar to that country. We will let refer to the event “has the dis- ease” and refer to the event “does not have the disease.” Thus, we know that if we select a person from Umen at random, the probability the individual chosen has the disease is .05, or This probability, is called the prior probability. It is given this name because the probability is assigned before any empirical data are obtained.

P(A1) � P(has the disease) � .05,P(A1) � .05.

A2A1A2.A1Ai

A2A1

P(Ai ƒ B) � P(Ai )P(B ƒ Ai )

P(A1)P(B ƒ A1) � P(A2)P(B ƒ A2)

PRIOR PROBABILITY The initial probability based on the present level of information.

POSTERIOR PROBABILITY A revised probability based on additional information.

With the help of Bayes’ theorem, formula (5–7), we can determine the posterior probability.

LO8 Calculate probabilities using Bayes’ theorem.

Statistics in Action

A recent study by the National Collegiate Athletic Association (NCAA) reported that of 150,000 se- nior boys playing on their high school basketball team, 64 would make a profes- sional team. To put it another way, the odds of a high school senior basketball player making a pro- fessional team are 1 in 2,344. From the same study:

1. The odds of a high school senior basketball player playing some col- lege basketball are about 1 in 40.

2. The odds of a high school senior playing college basketball as a senior in college are about 1 in 60.

3. If you play basket- ball as a senior in college, the odds of making a pro- fessional team are about 1 in 37.5.

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Solution

Example A manufacturer of DVD players purchases a particular microchip, called the LS-24, from three suppliers: Hall Electronics, Schuller Sales,

and Crawford Components. Thirty percent of the LS-24 chips are purchased from Hall Electronics, 20 percent from Schuller Sales, and the remaining 50 percent from Crawford Components. The manufacturer has exten- sive histories on the three suppliers and knows that 3 percent of the LS-24 chips from Hall Electronics are

defective, 5 percent of chips from Schuller Sales are defective, and 4 percent of the chips purchased from

Crawford Components are defective. When the LS-24 chips arrive at the manufacturer, they

are placed directly in a bin and not inspected or otherwise identified by supplier. A worker selects a chip for installation

in a DVD player and finds it defective. What is the probability that it was manufactured by Schuller Sales?

168 Chapter 5

So the probability that a person has the disease, given that he or she tested posi- tive, is .24. How is the result interpreted? If a person is selected at random from the population, the probability that he or she has the disease is .05. If the person is tested and the test result is positive, the probability that the person actually has the disease is increased about fivefold, from .05 to .24.

In the preceding problem, we had only two mutually exclusive and collectively exhaustive events, and If there are n such events, Bayes’ the- orem, formula (5–7), becomes

With the preceding notation, the calculations for the Umen problem are sum- marized in the following table.

P(Ai ƒ B) � P(Ai)P(B ƒ Ai)

P(A1)P(B ƒ A1) � P(A2)P(B ƒ A2) � � � � � P(An)P(B ƒ An)

A1, A2, � � � , An,A2.A1

� (.05)(.90)

(.05)(.90) � (.95)(.15) �

.0450

.1875 � .24

P(A1 ƒ B) � P(A1)P(B ƒ A1)

P(A1)P(B ƒ A1) � P(A2)P(B ƒ A2)

Prior Conditional Joint Posterior Event, Probability, Probability, Probability, Probability,

Disease, .05 .90 .0450 .0450/.1875 � .24A1

P(Ai ƒ B)P(Ai and B)P(B ƒ Ai)P(Ai)Ai

Another illustration of Bayes’ theorem follows.

No disease, .95 .15 .1425 .1425/.1875 � .76A2 P(B ) � .1875 1.00

As a first step, let’s summarize some of the information given in the problem statement.

• There are three mutually exclusive and collectively exhaustive events, that is, three suppliers.

The LS-24 was purchased from Hall Electronics. The LS-24 was purchased from Schuller Sales. The LS-24 was purchased from Crawford Components.A3

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• The prior probabilities are:

The probability the LS-24 was manufactured by Hall Electronics. The probability the LS-24 was manufactured by Schuller Sales. The probability the LS-24 was manufactured by Crawford Components.

• The additional information can be either:

The LS-24 appears defective, or The LS-24 appears not to be defective.

• The following conditional probabilities are given.

The probability that an LS-24 chip produced by Hall Electronics is defective. The probability that an LS-24 chip produced by Schuller Sales is defective. The probability that an LS-24 chip produced by Crawford Components is defective.

• A chip is selected from the bin. Because the chips are not identified by sup- plier, we are not certain which supplier manufactured the chip. We want to deter- mine the probability that the defective chip was purchased from Schuller Sales. The probability is written

Look at Schuller’s quality record. It is the worst of the three suppliers. Now that we have found a defective LS-24 chip, we suspect that is greater than That is, we expect the revised probability to be greater than .20. But how much greater? Bayes’ theorem can give us the answer. As a first step, consider the tree diagram in Chart 5–3.

The events are dependent, so the prior probability in the first branch is multi- plied by the conditional probability in the second branch to obtain the joint proba- bility. The joint probability is reported in the last column of Chart 5–3. To construct the tree diagram of Chart 5–3, we used a time sequence that moved from the sup- plier to the determination of whether the chip was acceptable or unacceptable.

What we need to do is reverse the time process. That is, instead of moving from left to right in Chart 5–3, we need to move from right to left. We have a defec- tive chip, and we want to determine the likelihood that it was purchased from Schuller Sales. How is that accomplished? We first look at the joint probabilities as relative frequencies out of 1,000 cases. For example, the likelihood of a defective LS-24 chip that was produced by Hall Electronics is .009. So of 1,000 cases, we would expect to find 9 defective chips produced by Hall Electronics. We observe that in 39 of 1,000 cases the LS-24 chip selected for assembly will be defective, found by Of these 39 defective chips, 10 were produced by Schuller Sales. Thus, the probability that the defective LS-24 chip was purchased from Schuller Sales is We have now determined the revised probability of Before we found the defective chip, the likelihood that it was purchased from Schuller Sales was .20. This likelihood has been increased to .2564.

This information is summarized in the following table.

Prior Conditional Joint Posterior Event, Probability, Probability, Probability, Probability,

Hall .30 .03 .009 Schuller .20 .05 .010 Crawford .50 .04 .020

1.0000P (B1) � .039

.020/.039 � .5128

.010/.039 � .2564

.009/.039 � .2308

P (Ai ƒ B1)P(Ai and B1)P (B1 ƒ Ai)P(Ai)Ai

P(A2 ƒ B1).10/39 � .2564.

9 � 10 � 20.

P(A2).P(A2 ƒ B1)

P(A2 ƒ B1).

P(B1 ƒ A3) � .04

P(B1 ƒ A2) � .05

P(B1 ƒ A1) � .03

P(A3) � .50 P(A2) � .20 P(A1) � .30

A Survey of Probability Concepts 169

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170 Chapter 5

The probability the defective LS-24 chip came from Schuller Sales can be formally found by using Bayes’ theorem. We compute where A2 refers to Schuller Sales and B1 to the fact that the selected LS-24 chip was defective.

This is the same result obtained from Chart 5–3 and from the conditional probability table.

� (.20)(.05)

(.30)(.03) � (.20)(.05) � (.50)(.04) �

.010

.039 � .2564

P(A2 ƒ B1) � P(A2)P(B1 ƒ A2)

P(A1)P(B1 ƒ A1) � P(A2)P(B1 ƒ A2) � P(A3)(B1 ƒ A3)

P(A2 ƒ B1),

B1 = Defective

B2 = Good

B1 = Defective

B2 = Good

B1 = Defective

B2 = Good

P (B1| A1) = .03

P (B2| A1) = .97

P (B1| A2) = .05

P (B2| A2) = .95

P (B1| A3) = .04

P (B2| A3) = .96

A2 = Schuller P (A2) = .20

A1 = Hall P (A1) = .30

A3 = Crawford P (A3) = .50

Joint probability Conditional probability

Prior probability

P (A1 and B1) = P (A1) P (B1| A1) = (.30) (.03) = .009

P (A1 and B2) = P (A1) P (B2|A1) = (.30) (.97) = .291

P (A2 and B1) = P (A2) P (B1|A2) = (.20) (.05) = .010

P (A2 and B2) = P (A2) P (B2|A2) = (.20) (.95) = .190

P (A3 and B1) = P (A3) P (B1|A3) = (.50) (.04) = .020

P (A3 and B2) = P (A3) P (B2|A3) = (.50) (.96) = .480

Total 1.000

CHART 5–3 Tree Diagram of DVD Manufacturing Problem

Self-Review 5–9 Refer to the preceding example and solution. (a) Design a formula to find the probability the part selected came from Crawford Com-

ponents, given that it was a good chip. (b) Compute the probability using Bayes’ theorem.

Exercises 33. and Use Bayes’ theorem to deter-

mine 34. and Use

Bayes’ theorem to determine P(A3 @B1). P(B1 @A3) � .10.P(A3) � .40, P(B1 0A1) � .25, P(B1 0 A2) � .05,P(A1) � .20, P(A2) � .40,

P(A1 @ B1). P(B1 @A2) � .10.P(A1) � .60, P(A2) � .40, P(B1 @ A1) � .05,

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A Survey of Probability Concepts 171

35. The Ludlow Wildcats baseball team, a minor league team in the Cleveland Indians organi- zation, plays 70 percent of their games at night and 30 percent during the day. The team wins 50 percent of their night games and 90 percent of their day games. According to today’s newspaper, they won yesterday. What is the probability the game was played at night?

36. Dr. Stallter has been teaching basic statistics for many years. She knows that 80 percent of the students will complete the assigned problems. She has also determined that among those who do their assignments, 90 percent will pass the course. Among those students who do not do their homework, 60 percent will pass. Mike Fishbaugh took sta- tistics last semester from Dr. Stallter and received a passing grade. What is the proba- bility that he completed the assignments?

37. The credit department of Lion’s Department Store in Anaheim, California, reported that 30 percent of their sales are cash or check, 30 percent are paid with a credit card and 40 percent with a debit card. Twenty percent of the cash or check purchases, 90 per- cent of the credit card purchases, and 60 percent of the debit card purchases are for more than $50. Ms. Tina Stevens just purchased a new dress that cost $120. What is the prob- ability that she paid cash or check?

38. One-fourth of the residents of the Burning Ridge Estates leave their garage doors open when they are away from home. The local chief of police estimates that 5 percent of the garages with open doors will have something stolen, but only 1 percent of those closed will have something stolen. If a garage is robbed, what is the probability the doors were left open?

5.8 Principles of Counting If the number of possible outcomes in an experiment is small, it is relatively easy to count them. There are six possible outcomes, for example, resulting from the roll of a die, namely:

If, however, there are a large number of possible outcomes, such as the number of heads and tails for an experiment with 10 tosses, it would be tedious to count all the possibilities. They could have all heads, one head and nine tails, two heads and eight tails, and so on. To facilitate counting, we discuss three formulas: the multiplication formula (not to be confused with the multiplication rule described earlier in the chap- ter), the permutation formula, and the combination formula.

The Multiplication Formula We begin with the multiplication formula.

MULTIPLICATION FORMULA If there are m ways of doing one thing and n ways of doing another thing, there are ways of doing both.m � n

In terms of a formula:

MULTIPLICATION FORMULA Total number of arrangements [5–8]

This can be extended to more than two events. For three events m, n, and o:

Total number of arrangements � (m)(n)(o)

� (m)(n)

LO9 Determine the number of outcomes using the appropriate principle of counting.

Example An automobile dealer wants to advertise that for $29,999 you can buy a convertible, a two-door sedan, or a four-door model with your choice of either wire wheel covers or solid wheel covers. How many different arrangements of models and wheel cov- ers can the dealer offer?

Solution Of course the dealer could determine the total number of arrangements by picturing and counting them. There are six.

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It was not difficult to count all the possible model and wheel cover combinations in this example. Suppose, however, that the dealer decided to offer eight models and six types of wheel covers. It would be tedious to picture and count all the possible alternatives. Instead, the multiplication formula can be used. In this case, there are

possible arrangements. Note in the preceding applications of the multiplication formula that there were

two or more groupings from which you made selections. The automobile dealer, for example, offered a choice of models and a choice of wheel covers. If a home builder offered you four different exterior styles of a home to choose from and three inte- rior floor plans, the multiplication formula would be used to find how many differ- ent arrangements were possible. There are 12 possibilities.

(m)(n) � (8)(6) � 48

172 Chapter 5

We can employ the multiplication formula as a check (where m is the number of models and n the wheel cover type). From formula (5–8):

Total possible arrangements � (m)(n) � (3)(2) � 6

Convertible with wire wheels

Convertible with solid wheels

Four-door with wire wheels

Two-door with solid wheels

Four-door with solid wheels

Two-door with wire wheels

Self-Review 5–10 1. The Women’s Shopping Network on cable TV offers sweaters and slacks for women. The sweaters and slacks are offered in coordinating colors. If sweaters are available in five colors and the slacks are available in four colors, how many different outfits can be advertised?

2. Pioneer manufactures three models of stereo receivers, two MP3 docking stations, four speakers, and three CD carousels. When the four types of components are sold together, they form a “system.” How many different systems can the electronics firm offer?

The Permutation Formula As noted, the multiplication formula is applied to find the number of possible arrangements for two or more groups. The permutation formula is applied to find the possible number of arrangements when there is only one group of objects. Illus- trations of this type of problem are:

• Three electronic parts are to be assembled into a plug-in unit for a television set. The parts can be assembled in any order. How many different ways can the three parts be assembled?

• A machine operator must make four safety checks before starting his machine. It does not matter in which order the checks are made. In how many different ways can the operator make the checks?

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A Survey of Probability Concepts 173

One order for the first illustration might be: the transistor first, the LEDs second, and the synthesizer third. This arrangement is called a permutation.

PERMUTATION Any arrangement of r objects selected from a single group of n possible objects.

Note that the arrangements a b c and b a c are different permutations. The formula to count the total number of different permutations is:

PERMUTATION FORMULA [5–9]

where: n is the total number of objects. r is the number of objects selected.

Before we solve the two problems illustrated, note that permutations and combi- nations (to be discussed shortly) use a notation called n factorial. It is written n! and means the product of For instance,

Many of your calculators have a button with x! that will perform this calculation for you. It will save you a great deal of time. For example the Texas Instrument TI-36X calculator has the following key:

It is the “third function” so check your users’ manual or the Internet for instructions. The factorial notation can also be canceled when the same number appears in

both the numerator and the denominator, as shown below.

By definition, zero factorial, written 0!, is 1. That is, 0! � 1.

6!3! 4!

� 6 # 5 # 4 # 3 # 2 # 1(3 # 2 # 1)

4 # 3 # 2 # 1 � 180

1 � 120. 5! � 5 # 4 # 3 # 2 # n(n � 1)(n � 2)(n � 3) p (1).

n Pr � n!

(n � r)!

10x

LOG

x !

Example Referring to the group of three electronic parts that are to be assembled in any order, in how many different ways can they be assembled?

Solution There are three electronic parts to be assembled, so Because all three are to be inserted in the plug-in unit, Solving using formula (5–9) gives:

We can check the number of permutations arrived at by using the permutation formula. We determine how many “spaces” have to be filled and the possibilities for each “space.” In the problem involving three electronic parts, there are three loca- tions in the plug-in unit for the three parts. There are three possibilities for the first place, two for the second (one has been used up), and one for the third, as follows:

6 permutations

The six ways in which the three electronic parts, lettered A, B, C, can be arranged are:

ABC BAC CAB ACB BCA CBA

(3)(2)(1) �

n Pr � n!

(n � r)! �

3! (3 � 3)!

� 3! 0!

� 3! 1

� 6

r � 3. n � 3.

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In the previous example, we selected and arranged all the objects, that is In many cases, only some objects are selected and arranged from the n possible objects. We explain the details of this application in the following example.

n � r.

174 Chapter 5

The Combination Formula If the order of the selected objects is not important, any selection is called a combi- nation. The formula to count the number of r object combinations from a set of n objects is:

COMBINATION FORMULA [5–10]

For example, if executives Able, Baker, and Chauncy are to be chosen as a com- mittee to negotiate a merger, there is only one possible combination of these three; the committee of Able, Baker, and Chauncy is the same as the committee of Baker, Chauncy, and Able. Using the combination formula:

nCr � n!

r!(n � r)! �

3 # 2 # 1 3 # 2 # 1(1) � 1

nCr � n!

r!(n � r)!

Example Betts Machine Shop Inc. has eight screw machines but only three spaces avail- able in the production area for the machines. In how many different ways can the eight machines be arranged in the three spaces available?

Solution There are eight possibilities for the first available space in the production area, seven for the second space (one has been used up), and six for the third space. Thus:

that is, there are a total of 336 different possible arrangements. This could also be found by using formula (5–9). If machines, and spaces available, the formula leads to

n Pr � n!

(n � r)! �

8! (8 � 3)!

� 8! 5!

� (8)(7)(6)5!

5! � 336

r � 3n � 8

(8)(7)(6) � 336,

Example The marketing department has been given the assignment of designing color codes for the 42 different lines of compact disks sold by Goody Records. Three colors are to be used on each CD, but a combination of three colors used for one CD cannot be rearranged and used to identify a different CD. This means that if green, yellow, and violet were used to identify one line, then yellow, green, and violet (or any other combination of these three colors) cannot be used to identify another line. Would seven colors taken three at a time be adequate to color-code the 42 lines?

Solution According to formula (5–10), there are 35 combinations, found by

The seven colors taken three at a time (i.e., three colors to a line) would not be ade- quate to color-code the 42 different lines because they would provide only 35 com- binations. Eight colors taken three at a time would give 56 different combinations. This would be more than adequate to color-code the 42 different lines.

7C3 � n!

r!(n � r)! �

7! 3!(7 � 3)!

� 7!

3!4! � 35

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A Survey of Probability Concepts 175

When the number of permutations or combinations is large, the calculations are tedious. Computer software and handheld calculators have “functions” to compute these numbers. The Excel output for the location of the eight screw machines in the production area of Betts Machine Shop Inc. is shown below. There are a total of 336 arrangements.

Below is the output for the color codes at Goody Records. Three colors are chosen from among seven possible. The number of combinations possible is 35.

Self-Review 5–11 1. A musician wants to write a score based on only five chords: B-flat, C, D, E, and G. However, only three chords out of the five will be used in succession, such as C, B-flat, and E. Repetitions, such as B-flat, B-flat, and E, will not be permitted. (a) How many permutations of the five chords, taken three at a time, are possible? (b) Using formula (5–9), how many permutations are possible?

2. The 10 numbers 0 through 9 are to be used in code groups of four to identify an item of clothing. Code 1083 might identify a blue blouse, size medium; the code group 2031 might identify a pair of pants, size 18; and so on. Repetitions of numbers are not per- mitted. That is, the same number cannot be used twice (or more) in a total sequence. For example, 2256, 2562, or 5559 would not be permitted. How many different code groups can be designed?

3. In the above example involving Goody Records, we said that eight colors taken three at a time would give 56 different combinations. (a) Use formula (5–10) to show this is true. (b) As an alternative plan for color-coding the 42 different lines, it has been suggested

that only two colors be placed on a disk. Would 10 colors be adequate to color-code

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Exercises 39. Solve the following:

a. 40!/35! b. c.

40. Solve the following: a. 20!/17! b. c.

41. A pollster randomly selected 4 of 10 available people. How many different groups of 4 are possible?

42. A telephone number consists of seven digits, the first three representing the exchange. How many different telephone numbers are possible within the 537 exchange?

43. An overnight express company must include five cities on its route. How many different routes are possible, assuming that it does not matter in which order the cities are included in the routing?

44. A representative of the Environmental Protection Agency (EPA) wants to select samples from 10 landfills. The director has 15 landfills from which she can collect samples. How many different samples are possible?

45. A national pollster has developed 15 questions designed to rate the performance of the president of the United States. The pollster will select 10 of these questions. How many different arrangements are there for the order of the 10 selected questions?

46. A company is creating three new divisions and seven managers are eligible to be appointed head of a division. How many different ways could the three new heads be appointed? Hint: Assume the division assignment makes a difference.

7 C2 9 P3

5 C2 7 P4

176 Chapter 5

Chapter Summary I. A probability is a value between 0 and 1 inclusive that represents the likelihood a

particular event will happen. A. An experiment is the observation of some activity or the act of taking some

measurement. B. An outcome is a particular result of an experiment. C. An event is the collection of one or more outcomes of an experiment.

II. There are three definitions of probability. A. The classical definition applies when there are n equally likely outcomes to an

experiment. B. The empirical definition occurs when the number of times an event happens is divided

by the number of observations. C. A subjective probability is based on whatever information is available.

III. Two events are mutually exclusive if by virtue of one event happening the other cannot happen.

IV. Events are independent if the occurrence of one event does not affect the occurrence of another event.

V. The rules of addition refer to the union of events.

the 42 different lines? (Again, a combination of two colors could be used only once— that is, if pink and blue were coded for one line, blue and pink could not be used to identify a different line.)

4. In a lottery game, three numbers are randomly selected from a tumbler of balls num- bered 1 through 50. (a) How many permutations are possible? (b) How many combinations are possible?

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A. The special rule of addition is used when events are mutually exclusive.

[5–2]

B. The general rule of addition is used when the events are not mutually exclusive.

[5–4]

C. The complement rule is used to determine the probability of an event happening by subtracting the probability of the event not happening from 1.

[5–3]

VI. The rules of multiplication refer to the product of events. A. The special rule of multiplication refers to events that are independent.

[5–5]

B. The general rule of multiplication refers to events that are not independent.

[5–6]

C. A joint probability is the likelihood that two or more events will happen at the same time.

D. A conditional probability is the likelihood that an event will happen, given that another event has already happened.

E. Bayes’ theorem is a method of revising a probability, given that additional information is obtained. For two mutually exclusive and collectively exhaustive events:

[5–7]

VII. There are three counting rules that are useful in determining the number of outcomes in an experiment. A. The multiplication rule states that if there are m ways one event can happen and n ways

another event can happen, then there are mn ways the two events can happen.

Number of arrangements [5–8]

B. A permutation is an arrangement in which the order of the objects selected from a specific pool of objects is important.

[5–9]

C. A combination is an arrangement where the order of the objects selected from a spe- cific pool of objects is not important.

[5–10]

Pronunciation Key SYMBOL MEANING PRONUNCIATION

Probability of A P of A

P(�A) Probability of not A P of not A P(A and B) Probability of A and B P of A and B

P(A or B) Probability of A or B P of A or B

P Probability of A given B has happened P of A given B

Permutation of n items selected r at a time Pnr

nCr Combination of n items selected r at a time Cnr nPr

(A @B)

P(A)

n Cr � n!

r!(n � r)!

n Pr � n!

(n � r)!

� (m)(n)

P(A1 ƒ B) � P(A1)P(B ƒ A1)

P(A1)P(B ƒ A1) � P(A2)P(B ƒ A2)

P(A and B) � P(A)P(B @ A)

P(A and B) � P(A)P(B)

P(A) � 1 � P(�A)

P(A or B) � P(A) � P(B) � P(A and B)

P(A or B) � P(A) � P(B)

A Survey of Probability Concepts 177

Statistics in Action

Government statistics show there are about 1.7 automobile- caused fatalities for every 100,000,000 vehicle-miles. If you drive 1 mile to the store to buy your lot- tery ticket and then return home, you have driven 2 miles. Thus the probability that you will join this statistical group on your next 2 mile round trip is

This can also be stated as “One in 29,411,765.” Thus if you drive to the store to buy your Powerball ticket, your chance of being killed (or killing someone else) is more than 4 times greater than the chance that you will win the Powerball Jackpot, one chance in 120,526,770. http://www. durangobill.com/ Powerball Odds.html

0.000000034. 100,000,000 �

2 � 1.7/

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Chapter Exercises 47. The marketing research department at Pepsico plans to survey teenagers about a newly

developed soft drink. Each will be asked to compare it with his or her favorite soft drink. a. What is the experiment? b. What is one possible event?

48. The number of times a particular event occurred in the past is divided by the number of occurrences. What is this approach to probability called?

49. The probability that the cause and the cure for all cancers will be discovered before the year 2020 is .20. What viewpoint of probability does this statement illustrate?

50. Berdine’s Chicken Factory has several stores in the Hilton Head, South Carolina, area. When interviewing applicants for server positions, the owner would like to include information on the amount of tip a server can expect to earn per check (or bill). A study of 500 recent checks indicated the server earned the following amounts in tips per 8-hour shift.

Amount of Tip Number

$0 up to $ 20 200 20 up to 50 100 50 up to 100 75

100 up to 200 75 200 or more 50

Total 500

a. What is the probability of a tip of $200 or more? b. Are the categories “$0 up to $20,” “$20 up to $50,” and so on considered mutually

exclusive? c. If the probabilities associated with each outcome were totaled, what would that total be? d. What is the probability of a tip of up to $50? e. What is the probability of a tip of less than $200?

51. Winning all three “Triple Crown” races is considered the greatest feat of a pedigree race- horse. After a successful Kentucky Derby, Big Brown is a 1 to 2 favorite to win the Preak- ness Stakes. a. If he is a 1 to 2 favorite to win the Belmont Stakes as well, what is his probability of

winning the Triple Crown? b. What do his chances for the Preakness Stakes have to be in order for him to be “even

money” to earn the Triple Crown? 52. The first card selected from a standard 52-card deck is a king.

a. If it is returned to the deck, what is the probability that a king will be drawn on the second selection?

b. If the king is not replaced, what is the probability that a king will be drawn on the second selection?

c. What is the probability that a king will be selected on the first draw from the deck and another king on the second draw (assuming that the first king was not replaced)?

53. Armco, a manufacturer of traffic light systems, found that under accelerated-life tests, 95 percent of the newly developed systems lasted 3 years before failing to change signals properly. a. If a city purchased four of these systems, what is the probability all four systems would

operate properly for at least 3 years? b. Which rule of probability does this illustrate? c. Using letters to represent the four systems, write an equation to show how you arrived

at the answer to part (a). 54. Refer to the following picture.

178 Chapter 5

�B

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a. What is the picture called? b. What rule of probability is illustrated? c. B represents the event of choosing a family that receives welfare payments. What

does equal? 55. In a management trainee program at Claremont Enterprises, 80 percent of the trainees

are female and 20 percent male. Ninety percent of the females attended college, and 78 percent of the males attended college. a. A management trainee is selected at random. What is the probability that the person

selected is a female who did not attend college? b. Are gender and attending college independent? Why? c. Construct a tree diagram showing all the probabilities, conditional probabilities, and

joint probabilities. d. Do the joint probabilities total 1.00? Why?

56. Assume the likelihood that any flight on Delta Airlines arrives within 15 minutes of the scheduled time is .90. We select four flights from yesterday for study. a. What is the likelihood all four of the selected flights arrived within 15 minutes of the

scheduled time? b. What is the likelihood that none of the selected flights arrived within 15 minutes of

the scheduled time? c. What is the likelihood at least one of the selected flights did not arrive within 15 min-

utes of the scheduled time? 57. There are 100 employees at Kiddie Carts International. Fifty-seven of the employees are

production workers, 40 are supervisors, 2 are secretaries, and the remaining employee is the president. Suppose an employee is selected: a. What is the probability the selected employee is a production worker? b. What is the probability the selected employee is either a production worker or a supervisor? c. Refer to part (b). Are these events mutually exclusive? d. What is the probability the selected employee is neither a production worker nor a

supervisor? 58. Joe Mauer of the Minnesota Twins had the highest batting average in the 2009 Major

League Baseball season. His average was .365. So assume the probability of getting a hit is .365 for each time he batted. In a particular game, assume he batted three times. a. This is an example of what type of probability? b. What is the probability of getting three hits in a particular game? c. What is the probability of not getting any hits in a game? d. What is the probability of getting at least one hit?

59. Four sports teams remain in a single-elimination playoff competition. If one team is favored in its semi-final match by odds of 2 to 1 and another squad is favored in its con- test by odds of 3 to 1. What is the probability that: a. Both favored teams win their games? b. Neither favored team wins its game? c. At least one of the favored teams wins its game?

60. There are three clues labeled “daily double” on the game show Jeopardy. If three equally matched contenders play, what is the probability that: a. A single contestant finds all three “daily doubles”? b. The returning champion gets all three of the “daily doubles”? c. Each of the players selects precisely one of the “daily doubles”?

61. Brooks Insurance Inc. wishes to offer life insurance to men age 60 via the Internet. Mortality tables indicate the likelihood of a 60-year-old man surviving another year is .98. If the policy is offered to five men age 60: a. What is the probability all five men survive the year? b. What is the probability at least one does not survive?

62. Forty percent of the homes constructed in the Quail Creek area include a security sys- tem. Three homes are selected at random: a. What is the probability all three of the selected homes have a security system? b. What is the probability none of the three selected homes have a security system? c. What is the probability at least one of the selected homes has a security system? d. Did you assume the events to be dependent or independent?

63. Refer to Exercise 62, but assume there are 10 homes in the Quail Creek area and 4 of them have a security system. Three homes are selected at random: a. What is the probability all three of the selected homes have a security system? b. What is the probability none of the three selected homes have a security system? c. What is the probability at least one of the selected homes has a security system? d. Did you assume the events to be dependent or independent?

P(B) � P(�B)

A Survey of Probability Concepts 179

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64. There are 20 families living in the Willbrook Farms Development. Of these families, 10 pre- pared their own federal income taxes for last year, 7 had their taxes prepared by a local professional, and the remaining 3 by H&R Block. a. What is the probability of selecting a family that prepared their own taxes? b. What is the probability of selecting two families, both of which prepared their own taxes? c. What is the probability of selecting three families, all of which prepared their own taxes? d. What is the probability of selecting two families, neither of which had their taxes pre-

pared by H&R Block? 65. The board of directors of Saner Automatic Door Company consists of 12 members, 3 of

whom are women. A new policy and procedures manual is to be written for the com- pany. A committee of 3 is randomly selected from the board to do the writing. a. What is the probability that all members of the committee are men? b. What is the probability that at least 1 member of the committee is a woman?

66. A recent survey reported in BusinessWeek dealt with the salaries of CEOs at large cor- porations and whether company shareholders made money or lost money.

If a company is randomly selected from the list of 20 studied, what is the probability: a. The CEO made more than $1 million? b. The CEO made more than $1 million or the shareholders lost money? c. The CEO made more than $1 million given the shareholders lost money? d. Of selecting 2 CEOs and finding they both made more than $1 million?

67. Althoff and Roll, an investment firm in Augusta, Georgia, advertises extensively in the Augusta Morning Gazette, the newspaper serving the region. The Gazette marketing staff estimates that 60 percent of Althoff and Roll’s potential market read the newspaper. It is further estimated that 85 percent of those who read the Gazette remember the Althoff and Roll advertisement. a. What percent of the investment firm’s potential market sees and remembers the

advertisement? b. What percent of the investment firm’s potential market sees, but does not remember

the advertisement? 68. An Internet company located in Southern California has season tickets to the Los Angeles

Lakers basketball games. The company president always invites one of the four vice pres- idents to attend games with him, and claims he selects the person to attend at random. One of the four vice presidents has not been invited to attend any of the last five Lakers home games. What is the likelihood this could be due to chance?

69. A computer-supply retailer purchased a batch of 1,000 CD-R disks and attempted to for- mat them for a particular application. There were 857 perfect CDs, 112 CDs were usable but had bad sectors, and the remainder could not be used at all. a. What is the probability a randomly chosen CD is not perfect? b. If the disk is not perfect, what is the probability it cannot be used at all?

70. An investor purchased 100 shares of Fifth Third Bank stock and 100 shares of Santee Electric Cooperative stock. The probability the bank stock will appreciate over a year is .70. The probability the electric utility will increase over the same period is .60. a. What is the probability both stocks appreciate during the period? b. What is the probability the bank stock appreciates but the utility does not? c. What is the probability at least one of the stocks appreciates?

71. Flashner Marketing Research Inc. specializes in providing assessments of the prospects for women’s apparel shops in shopping malls. Al Flashner, president, reports that he assesses the prospects as good, fair, or poor. Records from previous assessments show that 60 per- cent of the time the prospects were rated as good, 30 percent of the time fair, and 10 per- cent of the time poor. Of those rated good, 80 percent made a profit the first year; of those rated fair, 60 percent made a profit the first year; and of those rated poor, 20 percent made a profit the first year. Connie’s Apparel was one of Flashner’s clients. Connie’s Apparel made a profit last year. What is the probability that it was given an original rating of poor?

72. Two boxes of men’s Old Navy shirts were received from the factory. Box 1 contained 25 mesh polo shirts and 15 Super-T shirts. Box 2 contained 30 mesh polo shirts and 10

180 Chapter 5

CEO Paid More CEO Paid Less Than $1 Million Than $1 Million Total

Shareholders made money 2 11 13 Shareholders lost money 4 3 7

Total 6 14 20

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Super-T shirts. One of the boxes was selected at random, and a shirt was chosen at ran- dom from that box to be inspected. The shirt was a mesh polo shirt. Given this infor- mation, what is the probability that the mesh polo shirt came from box 1?

73. With each purchase of a large pizza at Tony’s Pizza, the customer receives a coupon that can be scratched to see if a prize will be awarded. The odds of winning a free soft drink are 1 in 10, and the odds of winning a free large pizza are 1 in 50. You plan to eat lunch tomorrow at Tony’s. What is the probability: a. That you will win either a large pizza or a soft drink? b. That you will not win a prize? c. That you will not win a prize on three consecutive visits to Tony’s? d. That you will win at least one prize on one of your next three visits to Tony’s?

74. For the daily lottery game in Illinois, participants select three numbers between 0 and 9. A number cannot be selected more than once, so a winning ticket could be, say, 307 but not 337. Purchasing one ticket allows you to select one set of numbers. The winning numbers are announced on TV each night. a. How many different outcomes (three-digit numbers) are possible? b. If you purchase a ticket for the game tonight, what is the likelihood you will win? c. Suppose you purchase three tickets for tonight’s drawing and select a different number

for each ticket. What is the probability that you will not win with any of the tickets? 75. Several years ago, Wendy’s Hamburgers advertised that there are 256 different ways to

order your hamburger. You may choose to have, or omit, any combination of the follow- ing on your hamburger: mustard, ketchup, onion, pickle, tomato, relish, mayonnaise, and lettuce. Is the advertisement correct? Show how you arrive at your answer.

76. It was found that 60 percent of the tourists to China visited the Forbidden City, the Tem- ple of Heaven, the Great Wall, and other historical sites in or near Beijing. Forty percent visited Xi’an with its magnificent terracotta soldiers, horses, and chariots, which lay buried for over 2,000 years. Thirty percent of the tourists went to both Beijing and Xi’an. What is the probability that a tourist visited at least one of these places?

77. A new chewing gum has been developed that is helpful to those who want to stop smok- ing. If 60 percent of those people chewing the gum are successful in stopping smoking, what is the probability that in a group of four smokers using the gum at least one quits smoking?

78. Reynolds Construction Company has agreed not to erect all “look-alike” homes in a new subdivision. Five exterior designs are offered to potential home buyers. The builder has stan- dardized three interior plans that can be incorporated in any of the five exteriors. How many different ways can the exterior and interior plans be offered to potential home buyers?

79. A new sports car model has defective brakes 15 percent of the time and a defective steering mechanism 5 percent of the time. Let’s assume (and hope) that these problems occur independently. If one or the other of these problems is present, the car is called a “lemon.” If both of these problems are present, the car is a “hazard.” Your instructor pur- chased one of these cars yesterday. What is the probability it is: a. A lemon? b. A hazard?

80. The state of Maryland has license plates with three numbers followed by three letters. How many different license plates are possible?

81. There are four people being considered for the position of chief executive officer of Dalton Enterprises. Three of the applicants are over 60 years of age. Two are female, of which only one is over 60. a. What is the probability that a candidate is over 60 and female? b. Given that the candidate is male, what is the probability he is less than 60? c. Given that the person is over 60, what is the probability the person is female?

82. Tim Bleckie is the owner of Bleckie Investment and Real Estate Company. The company recently purchased four tracts of land in Holly Farms Estates and six tracts in Newburg Woods. The tracts are all equally desirable and sell for about the same amount. a. What is the probability that the next two tracts sold will be in Newburg Woods? b. What is the probability that of the next four sold at least one will be in Holly Farms? c. Are these events independent or dependent?

83. A computer password consists of four characters. The characters can be one of the 26 letters of the alphabet. Each character may be used more than once. How many differ- ent passwords are possible?

84. A case of 24 cans contains 1 can that is contaminated. Three cans are to be chosen randomly for testing. a. How many different combinations of 3 cans could be selected? b. What is the probability that the contaminated can is selected for testing?

A Survey of Probability Concepts 181

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85. A puzzle in the newspaper presents a matching problem. The names of 10 U.S. presi- dents are listed in one column, and their vice presidents are listed in random order in the second column. The puzzle asks the reader to match each president with his vice president. If you make the matches randomly, how many matches are possible? What is the probability all 10 of your matches are correct?

86. Two components, A and B, operate in series. Being in series means that for the system to operate, both components A and B must work. Assume the two components are indepen- dent. What is the probability the system works under these conditions? The probability A works is .90 and the probability B functions is also .90.

87. Horwege Electronics Inc. purchases TV picture tubes from four different suppliers. Tyson Wholesale supplies 20 percent of the tubes, Fuji Importers 30 percent, Kirkpatricks 25 per- cent, and Parts Inc. 25 percent. Tyson Wholesale tends to have the best quality, as only 3 percent of its tubes arrive defective. Fuji Importers’ tubes are 4 percent defective, Kirkpatricks’ 7 percent, and Parts Inc. are 6.5 percent defective. a. What is the overall percent defective? b. A defective picture tube was discovered in the latest shipment. What is the probabil-

ity that it came from Tyson Wholesale? 88. ABC Auto Insurance classifies drivers as good, medium, or poor risks. Drivers who apply

to them for insurance fall into these three groups in the proportions 30 percent, 50 per- cent, and 20 percent, respectively. The probability a “good” driver will have an accident is .01, the probability a “medium” risk driver will have an accident is .03, and the prob- ability a “poor” driver will have an accident is .10. The company sells Mr. Brophy an insur- ance policy and he has an accident. What is the probability Mr. Brophy is: a. A “good” driver? b. A “medium” risk driver? c. A “poor” driver?

89. You take a trip by air that involves three independent flights. If there is an 80 percent chance each specific leg of the trip is done on time, what is the probability all three flights arrive on time?

90. The probability a HP network server is down is .05. If you have three independent servers, what is the probability that at least one of them is operational?

91. Twenty-two percent of all liquid crystal displays (LCDs) are manufactured by Samsung. What is the probability that in a collection of three independent LCD purchases, at least one is a Samsung?

Data Set Exercises 92. Refer to the Real Estate data, which reports information on homes sold in the Goodyear,

Arizona, area during the last year. a. Sort the data into a table that shows the number of homes that have a pool versus

the number that don’t have a pool in each of the five townships. If a home is selected at random, compute the following probabilities. 1. The home is in Township 1 or has a pool. 2. Given that it is in Township 3, that it has a pool. 3. The home has a pool and is in Township 3.

b. Sort the data into a table that shows the number of homes that have a garage attached versus those that don’t in each of the five townships. If a home is selected at random, compute the following probabilities: 1. The home has a garage attached. 2. The home does not have a garage attached, given that it is in Township 5. 3. The home has a garage attached and is in Township 3. 4. The home does not have a garage attached or is in Township 2.

93. Refer to the Baseball 2009 data, which reports information on the 30 Major League Base- ball teams for the 2009 season. Set up three variables: • Divide the teams into two groups, those that had a winning season and those that

did not. That is, create a variable to count the teams that won 81 games or more, and those that won 80 or less.

• Create a new variable for attendance, using three categories: attendance less than 2.0 mil- lion, attendance of 2.0 million up to 3.0 million, and attendance of 3.0 million or more.

• Create a variable that shows the teams that play in a stadium less than 15 years old versus one that is 15 years old or more.

182 Chapter 5

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Answer the following questions. a. Create a table that shows the number of teams with a winning season versus those

with a losing season by the three categories of attendance. If a team is selected at random, compute the following probabilities: 1. The team had a winning season. 2. The team had a winning season or attendance of more than 3.0 million. 3. The team had a winning season given attendance was more than 3.0 million. 4. The team has a winning season and attracted fewer than 2.0 million fans.

b. Create a table that shows the number of teams with a winning season versus those that play in new or old stadiums. If a team is selected at random, compute the fol- lowing probabilities: 1. Selecting a team with a winning season. 2. The likelihood of selecting a team with a winning record and playing in a new stadium. 3. The team had a winning record or played in a new stadium.

94. Refer to the data on the school buses in the Buena School District. Set up a variable that divides the age of the buses into three groups: new (less than 5 year old), medium (5 but less than 10 years), and old (10 or more years). The median maintenance cost is $456. Based on this value, create a variable for those less than the median (low mainte- nance) and those more than the median (high maintenance). Finally, develop a table to show the relationship between maintenance cost and age of the bus. a. What percentage of the buses are new? b. What percentage of the new buses have low maintenance? c. What percentage of the old buses have high maintenance? d. Does maintenance cost seem to be related to the age of the bus? Hint: Compare the

maintenance cost of the old buses with the cost of the new buses? Would you con- clude maintenance cost is independent of the age?

A Survey of Probability Concepts 183

Software Commands 1. The Excel Commands to determine the number of

permutations shown on page 175 are: a. Click on the Formulas tab in the top menu, then,

on the far left, select Insert Function fx.

b. In the Insert Function box, select Statistical as the category, then scroll down to PERMUT in the Select a function list. Click OK.

c. In the PERM box after Number, enter 8 and in the Number_chosen box enter 3. The correct answer of 336 appears twice in the box.

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184 Chapter 5

2. The Excel Commands to determine the number of combinations shown on page 175 are:

a. Click on the Formulas tab in the top menu, then, on the far left, select Insert Function fx.

b. In the Insert Function box, select Math & Trig as the category, then scroll down to COMBIN in the Select a function list. Click OK.

c. In the COMBIN box after Number, enter 7 and in the Number_chosen box enter 3. The correct answer of 35 appears twice in the box.

Chapter 5 Answers to Self-Review

5–1 a. Count the number who think the new game is playable.

b. Seventy-three players found the game playable. Many other answers are possible.

c. No. Probability cannot be greater than 1. The probability that the game, if put on the market, will be successful is 65/80, or .8125.

d. Cannot be less than 0. Perhaps a mistake in arithmetic.

e. More than half of the players testing the game liked it. (Of course, other answers are possible.)

5–2 1.

Classical.

2. Empirical.

3. The author’s view when writing the text of the chance that the DJIA will climb to 12,000 is .25. You may be more optimistic or less optimistic. Subjective.

182 539

� .338

4 queens in deck 52 cards total

� 4

52 � .0769

5–3 a. i.

ii.

1 � 302

2,000 � .849

(50 � 68) 2,000

� .059

B E

� D

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B .15

A .08

Both .03

A Survey of Probability Concepts 185

c. They are not complementary, but are mutually exclusive.

5–4 a. Need for corrective shoes is event A. Need for major dental work is event B.

b. One possibility is:

� .20 � .08 � .15 � .03

P(A or B) � P(A) � P(B) � P(A and B)

5–5 5–6 a. .002, found by:

b. .14, found by:

c. No, because there are other possibilities, such as three women and one man.

5–7 a.

5–8 a. Independence requires that One possibility is:

Does No, the two variables are not independent.

Therefore, any joint probability in the table must be computed by using the general rule of multiplication.

60/90 � 80/195?

P(visit often) P(visit often @ yes enclosed mall) �

P(A @B) � P(A).

P(A2 or B4) � 80 200

� 105 200

� 30 200

� 155 200

� .775

P(A2 @ B4) � 30 105

� .286

P(B4) � 105 200

� .525

a 8

12 b a

7 11 b a

6 10 b a

5 9 b �

1,680 11,880

� .1414

a 4

12 b a

3 11 b a

2 10 b a

1 9 b �

24 11,880

� .002

(.95)(.95)(.95)(.95) � .8145

5–9 a.

5–10 1. 2.

5–11 1. a. 60, found by (5)(4)(3). b. 60, found by:

2. 5,040, found by:

3. a. 56 is correct, found by:

b. Yes. There are 45 combinations, found by:

4. a.

b. 50C3 � 50!

3! (50 � 3)! � 19,600

50P3 � 50!

(50 � 3)! � 117,600

10C2 � n!

r!(n � r)! �

10! 2!(10 � 2)!

� 45

8C3 � n!

r!(n � r)! �

8! 3!(8 � 3)!

� 56

10! (10 � 4)!

� 10 # 9 # 8 # 7 # 6 # 5 # 4 # 3 # 2 # 1

6 # 5 # 4 # 3 # 2 # 1

5! (5 � 3)!

� 5 # 4 # 3 # 2 # 1

2 # 1

(3)(2)(4)(3) � 72 (5)(4) � 20

� .480 .961

� .499

� (.50)(.96)

(.30)(.97) � (.20)(.95) � (.50)(.96)

P(A3)P(B2 ƒ A3) P(A1)P(B2 ƒ A1) � P(A2)P(B2 ƒ A2) � P(A3)P(B2 ƒ A3)

�P(A3 ƒ B2)

Often

Occasional

Joint probabilities 60/90

25/90 90/195

105/195

5/90

20/105

35/105

50/105

.31

.13

.03

.10

.18

.25

Never

Occasional

Visits

Enclosed

yes

Never

Often

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6 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Identify the characteristics of a probability distribution.

LO2 Distinguish between a discrete and a continuous random variable.

LO3 Compute the mean of a probability distribution.

LO4 Compute the variance and standard deviation of a probability distribution.

LO5 Describe and compute probabilities for a binomial distribution.

LO6 Describe and compute probabilities for a hypergeo- metric distribution.

LO7 Describe and compute probabilities for a Poisson distribution.

Discrete Probability Distributions

Recent statistics suggest that 15 percent of those who visit a retail site on

the Web make a purchase. A retailer wished to verify this claim. To do so,

she selected a sample of 16 “hits” to her site and found that 4 had actually

made a purchase. What is the likelihood of exactly four purchases? How

many purchases should she expect? What is the likelihood that four or

more “hits” result in a purchase? (See Exercise 49 and LO5.)

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Example Suppose we are interested in the number of heads showing face up on three tosses of a coin. This is the experiment. The possible results are: zero heads, one head, two heads, and three heads. What is the probability distribution for the number of heads?

Solution There are eight possible outcomes. A tail might appear face up on the first toss, another tail on the second toss, and another tail on the third toss of the coin. Or we might get a tail, tail, and head, in that order. We use the multiplication formula for counting outcomes (5–8). There are (2)(2)(2) or 8 possible results. These results are on the next page.

Discrete Probability Distributions 187

6.1 Introduction Chapters 2 through 4 are devoted to descriptive statistics. We describe raw data by organizing it into a frequency distribution and portraying the distribution in tables, graphs, and charts. Also, we compute a measure of location—such as the arith- metic mean, median, or mode—to locate a typical value near the center of the dis- tribution. The range and the standard deviation are used to describe the spread in the data. These chapters focus on describing something that has already happened.

Starting with Chapter 5, the emphasis changes—we begin examining something that would probably happen. We note that this facet of statistics is called statistical inference. The objective is to make inferences (statements) about a population based on a number of observations, called a sample, selected from the population. In Chap- ter 5, we state that a probability is a value between 0 and 1 inclusive, and we exam- ine how probabilities can be combined using rules of addition and multiplication.

This chapter will begin the study of probability distributions. A probability distri- bution gives the entire range of values that can occur based on an experiment. A prob- ability distribution is similar to a relative frequency distribution. However, instead of describing the past, it describes a likely future event. For example, a drug manufacturer may claim a treatment will cause weight loss for 80 percent of the population. A con- sumer protection agency may test the treatment on a sample of six people. If the man- ufacturer’s claim is true, it is almost impossible to have an outcome where no one in the sample loses weight and it is most likely that five out of the six do lose weight.

In this chapter, we discuss the mean, variance, and standard deviation of a prob- ability distribution. We also discuss three frequently occurring probability distribu- tions: the binomial, hypergeometric, and Poisson.

6.2 What Is a Probability Distribution? A probability distribution shows the possible outcomes of an experiment and the probability of each of these outcomes.

PROBABILITY DISTRIBUTION A listing of all the outcomes of an experiment and the probability associated with each outcome.

CHARACTERISTICS OF A PROBABILITY DISTRIBUTION 1. The probability of a particular outcome is between 0 and 1 inclusive. 2. The outcomes are mutually exclusive events. 3. The list is exhaustive. So the sum of the probabilities of the various events

is equal to 1.

Below are the major characteristics of a probability distribution.

How can we generate a probability distribution? The following example will explain.

LO1 Identify the characteristics of a probability distribution.

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Coin TossPossible Number of Result First Second Third Heads

1 T T T 0 2 T T H 1 3 T H T 1 4 T H H 2 5 H T T 1 6 H T H 2 7 H H T 2 8 H H H 3

188 Chapter 6

Number of Probability Heads, of Outcome,

x P (x)

Total 8 8

� 1 .000

1 8

� .125

3 8

� .375

3 8

� .375

1 8

� .125

CHART 6–1 Graphical Presentation of the Number of Heads Resulting from Three Tosses of a Coin and the Corresponding Probability

Pr ob

ab ili

0 1 2 3

Number of heads

3 8

2 8

1 8

P (x )

TABLE 6–1 Probability Distribution for the Events of Zero, One, Two, and Three Heads Showing Face Up on Three Tosses of a Coin

Note that the outcome “zero heads” occurred only once, “one head” occurred three times, “two heads” occurred three times, and the outcome “three heads” occurred only once. That is, “zero heads” happened one out of eight times. Thus, the probability of zero heads is one-eighth, the probability of one head is three-eighths, and so on. The probability distribution is shown in Table 6–1. Because one of these outcomes must happen, the total of the probabilities of all possible events is 1.000. This is always true. The same information is shown in Chart 6–1.

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Discrete Probability Distributions 189

Refer to the coin-tossing example in Table 6–1. We write the probability of x as P(x). So the probability of zero heads is P(0 heads) � .125, and the probability of one head is P(1 head) � .375, and so forth. The sum of these mutually exclusive probabilities is 1; that is, from Table 6–1, 0.125 � 0.375 � 0.375 � 0.125 � 1.00.

6.3 Random Variables In any experiment of chance, the outcomes occur randomly. So it is often called a random variable. For example, rolling a single die is an experiment: any one of six possible outcomes can occur. Some experiments result in outcomes that are quan- titative (such as dollars, weight, or number of children), and others result in quali- tative outcomes (such as color or religious preference). Each value of the random variable is associated with a probability to indicate the chance of a particular outcome. A few examples will further illustrate what is meant by a random variable.

• If we count the number of employees absent from the day shift on Monday, the number might be 0, 1, 2, 3, . . . The number absent is the random variable.

• If we weigh four steel ingots, the weights might be 2,492 pounds, 2,497 pounds, 2,506 pounds, and so on. The weight is the random variable.

• If we toss two coins and count the number of heads, there could be zero, one, or two heads. Because the number of heads resulting from this experiment is due to chance, the number of heads appearing is the random variable.

• Other random variables might be the number of defective lightbulbs produced in an hour at the Cleveland Company Inc., the grade level (9, 10, 11, or 12) of the members of the St. James girls’ basketball team, the number of runners in the Boston Marathon for the 2010 race, and the daily number of drivers charged with driving under the influence of alcohol in Texas.

RANDOM VARIABLE A quantity resulting from an experiment that, by chance, can assume different values.

The following diagram illustrates the terms experiment, outcome, event, and ran- dom variable. First, for the experiment where a coin is tossed three times, there are eight possible outcomes. In this experiment, we are interested in the event that one head occurs in the three tosses. The random variable is the number of heads. In terms of probability, we want to know the probability of the event that the random variable equals 1. The result is P(1 head in 3 tosses) � 0.375.

TTT TTH THT HTT

THH HTH HHT

HHH

Possible outcomes for three coin tosses

The event {one head} occurs and the random variable x � 1.

A random variable may be either discrete or continuous.

Self-Review 6–1 The possible outcomes of an experiment involving the roll of a six-sided die are a one- spot, a two-spot, a three-spot, a four-spot, a five-spot, and a six-spot. (a) Develop a probability distribution for the number of possible spots. (b) Portray the probability distribution graphically. (c) What is the sum of the probabilities?

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Discrete Random Variable A discrete random variable can assume only a certain number of separated values. If there are 100 employees, then the count of the number absent on Monday can only be 0, 1, 2, 3, . . . , 100. A discrete random variable is usually the result of count- ing something.

DISCRETE RANDOM VARIABLE A random variable that can assume only certain clearly separated values.

A discrete random variable can, in some cases, assume fractional or decimal values. These values must be separated—that is, have distance between them. As an example, the scores awarded by judges for technical competence and artistic form in figure skating are decimal values, such as 7.2, 8.9, and 9.7. Such values are discrete because there is distance between scores of, say, 8.3 and 8.4. A score cannot be 8.34 or 8.347, for example.

Continuous Random Variable On the other hand, if the random variable is continuous, then the distribution is a continuous probability distribution. If we measure something such as the width of a room, the height of a person, or the pressure in an automobile tire, the variable is a continuous random variable. It can assume one of an infinitely large number of values, within certain limitations. As examples:

• The times of commercial flights between Atlanta and Los Angeles are 4.67 hours, 5.13 hours, and so on. The random variable is the time in hours.

• Tire pressure, measured in pounds per square inch (psi), for a new Chevy Trail- blazer might be 32.78 psi, 31.62 psi, 33.07 psi, and so on. In other words, any values between 28 and 35 could reasonably occur. The random variable is the tire pressure.

Logically, if we organize a set of possible values from a random variable into a probability distribution, the result is a probability distribution. So what is the dif- ference between a probability distribution and a random variable? A random vari- able reports the particular outcome of an experiment. A probability distribution reports all the possible outcomes as well as the corresponding probability.

The tools used, as well as the probability interpretations, are different for dis- crete and continuous probability distributions. This chapter is limited to the discus- sion and interpretation of discrete distributions. In the next chapter, we discuss continuous distributions. How do you tell the difference between the two types of distributions? Usually a discrete distribution is the result of counting something, such as:

• The number of heads appearing when a coin is tossed 3 times. • The number of students earning an A in this class. • The number of production employees absent from the second shift today. • The number of 30-second commercials on NBC from 8 to 11 P.M. tonight.

Continuous distributions are usually the result of some type of measurement, such as:

• The length of each song on the latest Linkin Park CD. • The weight of each student in this class.

190 Chapter 6

LO2 Distinguish between a discrete and a continuous random variable.

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Discrete Probability Distributions 191

• The temperature outside as you are reading this book. • The amount of money earned by each of the more than 750 players currently

on Major League Baseball team rosters.

6.4 The Mean, Variance, and Standard Deviation of a Discrete Probability Distribution In Chapter 3, we discussed measures of location and variation for a frequency distribution. The mean reports the central location of the data, and the variance describes the spread in the data. In a similar fashion, a probability distribution is summarized by its mean and variance. We identify the mean of a probability distri- bution by the lowercase Greek letter mu (�) and the standard deviation by the lower- case Greek letter sigma (�).

Mean The mean is a typical value used to represent the central location of a probability distribution. It also is the long-run average value of the random variable. The mean of a probability distribution is also referred to as its expected value. It is a weighted average where the possible values of a random variable are weighted by their cor- responding probabilities of occurrence.

The mean of a discrete probability distribution is computed by the formula:

MEAN OF A PROBABILITY DISTRIBUTION [6–1]

where P(x) is the probability of a particular value x. In other words, multiply each x value by its probability of occurrence, and then add these products.

Variance and Standard Deviation As noted, the mean is a typical value used to summarize a discrete probability distribution. However, it does not describe the amount of spread (variation) in a dis- tribution. The variance does this. The formula for the variance of a probability distribution is:

VARIANCE OF A PROBABILITY DISTRIBUTION [6–2]

The computational steps are

1. Subtract the mean from each value, and square this difference. 2. Multiply each squared difference by its probability. 3. Sum the resulting products to arrive at the variance.

The standard deviation, �, is found by taking the positive square root of �2; that is,

An example will help explain the details of the calculation and interpretation of the mean and standard deviation of a probability distribution.

� � 2�2.

�2 � ©[(x � �)2P(x)]

� � ©[xP(x)]LO3 Compute the mean of a probability distribution.

LO4 Compute the variance and standard deviation of a probability distribution.

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Example

Solution 1. This is a discrete probability distribution for the random variable called “number of cars sold.” Note that John expects to sell only within a certain range of cars; he does not expect to sell 5 cars or 50 cars. Further, he cannot sell half a car. He can sell only 0, 1, 2, 3, or 4 cars. Also, the outcomes are mutually exclusive— he cannot sell a total of both 3 and 4 cars on the same Saturday. The sum of the possible outcomes total 1. Hence, these circumstance qualify as a probability distribution.

2. The mean number of cars sold is computed by weighting the number of cars sold by the probability of selling that number and adding or summing the prod- ucts, using formula (6–1):

These calculations are summarized in the following table.

Number of Cars Sold, Probability,

x P(x ) x � P (x )

0 .10 0.00 1 .20 0.20 2 .30 0.60 3 .30 0.90 4 .10 0.40

Total 1.00 � � 2.10

How do we interpret a mean of 2.1? This value indicates that, over a large number of Saturdays, John Ragsdale expects to sell a mean of 2.1 cars a day. Of course, it is not possible for him to sell exactly 2.1 cars on any par- ticular Saturday. However, the expected value can be used to predict the arithmetic mean number of cars sold on Saturdays in the long run. For example,

� 2.1

� 0(.10) � 1(.20) � 2(.30) � 3(.30) � 4(.10)

� � ©[xP(x)]

192 Chapter 6

John Ragsdale sells new cars for Pelican Ford. John usually sells the largest number of cars on Saturday. He has developed the following prob- ability distribution for the number of cars he expects to sell on a particular Saturday.

1. What type of distribution is this? 2. On a typical Saturday, how many cars does

John expect to sell? 3. What is the variance of the distribution?

Number of Probability, Cars Sold, x P (x)

0 .10 1 .20 2 .30 3 .30 4 .10

Total 1.00

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Discrete Probability Distributions 193

if John works 50 Saturdays during a year, he can expect to sell (50)(2.1) or 105 cars just on Saturdays. Thus, the mean is sometimes called the expected value.

3. Again, a table is useful for systemizing the computations for the variance, which is 1.290.

Number of Cars Sold, Probability,

x P(x) (x � �) (x � �)2 (x � �)2P(x )

0 .10 0 � 2.1 4.41 0.441 1 .20 1 � 2.1 1.21 0.242 2 .30 2 � 2.1 0.01 0.003 3 .30 3 � 2.1 0.81 0.243 4 .10 4 � 2.1 3.61 0.361

�2 � 1.290

Recall that the standard deviation, �, is the positive square root of the variance. In this example, . How do we interpret a standard deviation of 1.136 cars? If salesperson Rita Kirsch also sold a mean of 2.1 cars on Saturdays, and the standard deviation in her sales was 1.91 cars, we would conclude that there is more variability in the Saturday sales of Ms. Kirsch than in those of Mr. Ragsdale (because 1.91 � 1.136).

1�2 � 11.290 � 1.136 cars

Self-Review 6–2 The Pizza Palace offers three sizes of cola—small, medium, and large—to go with its pizza. The colas are sold for $0.80, $0.90, and $1.20, respectively. Thirty percent of the orders are for small, 50 percent are for medium, and 20 percent are for the large sizes. Organize the size of the colas and the probability of a sale into a probability distribution. (a) Is this a discrete probability distribution? Indicate why or why not. (b) Compute the mean amount charged for a cola. (c) What is the variance in the amount charged for a cola? The standard deviation?

Exercises 1. Compute the mean and variance of the following discrete probability distribution.

x P(x)

0 .2 1 .4 2 .3 3 .1

2. Compute the mean and variance of the following discrete probability distribution.

x P(x)

2 .5 8 .3

10 .2

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4. Which of these variables are discrete and which are continuous random variables? a. The number of new accounts established by a salesperson in a year. b. The time between customer arrivals to a bank ATM. c. The number of customers in Big Nick’s barber shop. d. The amount of fuel in your car’s gas tank. e. The number of minorities on a jury. f. The outside temperature today.

5. The information below is the number of daily emergency service calls made by the vol- unteer ambulance service of Walterboro, South Carolina, for the last 50 days. To explain, there were 22 days on which there were 2 emergency calls, and 9 days on which there were 3 emergency calls.

3. Compute the mean and variance of the following probability distribution.

x P(x)

5 .1 10 .3 15 .2 20 .4

Number of Calls Frequency

0 8 1 10 2 22 3 9 4 1

Total 50

a. Convert this information on the number of calls to a probability distribution. b. Is this an example of a discrete or continuous probability distribution? c. What is the mean number of emergency calls per day? d. What is the standard deviation of the number of calls made daily?

6. The director of admissions at Kinzua University in Nova Scotia estimated the distribution of student admissions for the fall semester on the basis of past experience. What is the expected number of admissions for the fall semester? Compute the variance and the standard deviation of the number of admissions.

Admissions Probability

1,000 .6 1,200 .3 1,500 .1

7. Belk Department Store is having a special sale this weekend. Customers charging pur- chases of more than $50 to their Belk credit card will be given a special Belk Lottery card. The customer will scratch off the card, which will indicate the amount to be taken off the total amount of the purchase. Listed below are the amount of the prize and the per- cent of the time that amount will be deducted from the total amount of the purchase.

Prize Amount Probability

$ 10 .50 25 .40 50 .08

100 .02

a. What is the mean amount deducted from the total purchase amount? b. What is the standard deviation of the amount deducted from the total purchase?

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a. Convert the information on the number of hours parked to a probability distribution. Is this a discrete or a continuous probability distribution?

b. Find the mean and the standard deviation of the number of hours parked. How would you answer the question: How long is a typical customer parked?

c. Find the mean and the standard deviation of the amount charged.

6.5 Binomial Probability Distribution The binomial probability distribution is a widely occurring discrete probability dis- tribution. One characteristic of a binomial distribution is that there are only two possi-

ble outcomes on a particular trial of an experiment. For example, the statement in a true/false question is either true or false. The outcomes are mutually exclusive, mean- ing that the answer to a true/false question cannot be both true and false at the same time. As other examples, a product is clas- sified as either acceptable or not accept- able by the quality control department, a worker is classified as employed or unem- ployed, and a sales call results in the cus-

tomer either purchasing the product or not purchasing the product. Frequently, we classify the two possible outcomes as “success” and “failure.” However, this clas- sification does not imply that one outcome is good and the other is bad.

Another characteristic of the binomial distribution is that the random variable is the result of counts. That is, we count the number of successes in the total num- ber of trials. We flip a fair coin five times and count the number of times a head appears, we select 10 workers and count the number who are over 50 years of age, or we select 20 boxes of Kellogg’s Raisin Bran and count the number that weigh more than the amount indicated on the package.

A third characteristic of a binomial distribution is that the probability of a suc- cess remains the same from one trial to another. Two examples are:

• The probability you will guess the first question of a true/false test correctly (a success) is one-half. This is the first “trial.” The probability that you will guess correctly on the second question (the second trial) is also one-half, the proba- bility of success on the third trial is one-half, and so on.

• If past experience revealed the swing bridge over the Intracoastal Waterway in Socastee was raised one out of every 20 times you approach it, then the prob- ability is one-twentieth that it will be raised (a “success”) the next time you approach it, one-twentieth the following time, and so on.

Discrete Probability Distributions 195

8. The Downtown Parking Authority of Tampa, Florida, reported the following information for a sample of 250 customers on the number of hours cars are parked and the amount they are charged.

LO5 Describe and compute probabilities for a binomial distribution.

Number of Hours Frequency Amount Charged

1 20 $ 3.00 2 38 6.00 3 53 9.00 4 45 12.00 5 40 14.00 6 13 16.00 7 5 18.00 8 36 20.00

250

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How Is a Binomial Probability Computed? To construct a particular binomial probability, we use (1) the number of trials and (2) the probability of success on each trial. For example, if an examination at the conclusion of a management seminar consists of 20 multiple-choice questions, the number of trials is 20. If each question has five choices and only one choice is cor- rect, the probability of success on each trial is .20. Thus, the probability is .20 that a person with no knowledge of the subject matter will guess the answer to a ques- tion correctly. So the conditions of the binomial distribution just noted are met.

A binomial probability is computed by the formula:

BINOMIAL PROBABILITY FORMULA P(x) � nCx � x(1 � �)n � x [6–3]

where: C denotes a combination. n is the number of trials. x is the random variable defined as the number of successes. � is the probability of a success on each trial.

We use the Greek letter � (pi) to denote a binomial population parameter. Do not confuse it with the mathematical constant 3.1416.

BINOMIAL PROBABILITY EXPERIMENT 1. An outcome on each trial of an experiment is classified into one of two

mutually exclusive categories—a success or a failure. 2. The random variable counts the number of successes in a fixed number

of trials. 3. The probability of success and failure stay the same for each trial. 4. The trials are independent, meaning that the outcome of one trial does not

affect the outcome of any other trial.

The final characteristic of a binomial probability distribution is that each trial is independent of any other trial. Independent means that there is no pattern to the trials. The outcome of a particular trial does not affect the outcome of any other trial. Two examples are:

• A young family has two children, both boys. The probability of a third birth being a boy is still .50. That is, the gender of the third child is independent of the other two.

• Suppose 20 percent of the patients served in the emergency room at Wacca- maw Hospital do not have insurance. If the second patient served on the after- noon shift today did not have insurance, that does not affect the probability the third, the tenth, or any of the other patients will or will not have insurance.

Example There are five flights daily from Pittsburgh via US Airways into the Bradford Regional Airport in Bradford, Pennsylvania. Suppose the probability that any flight arrives late is .20. What is the probability that none of the flights are late today? What is the probability that exactly one of the flights is late today?

Solution We can use Formula (6–3). The probability that a particular flight is late is .20, so let � � .20. There are five flights, so n � 5, and x, the random variable, refers to

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the number of successes. In this case, a “success” is a flight that arrives late. Because there are no late arrivals, x � 0.

P(0) � nCx(�) x(1 � �)n � x

� 5C0(.20) 0(1 � .20)5 � 0 � (1)(1)(.3277) � .3277

The probability that exactly one of the five flights will arrive late today is .4096, found by

The entire binomial probability distribution with � � .20 and n � 5 is shown in the following bar chart. We can observe that the probability of exactly 3 late flights is .0512 and from the bar chart that the distribution of the number of late arrivals is positively skewed.

� 5C1(.20) 1(1 � .20)5�1 � (5)(.20)(.4096) � .4096

P(1) � nCx(�) x(1 � �)n�x

Discrete Probability Distributions 197

The mean (�) and the variance (�2) of a binomial distribution can be computed in a “shortcut” fashion by:

MEAN OF A BINOMIAL DISTRIBUTION [6–4]

VARIANCE OF A BINOMIAL DISTRIBUTION [6–5]

For the example regarding the number of late flights, recall that � � .20 and n � 5. Hence:

�2 � n�(1 � �) � 5(.20)(1 � .20) � .80

� � n� � (5)(.20) � 1.0

�2 � n�(1 � �)

� � n�

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TABLE 6–2 Binomial Probabilities for n � 6 and Selected Values of �

n � 6 Probability

x \� .05 .1 .2 .3 .4 .5 .6 .7 .8 .9 .95

0 .735 .531 .262 .118 .047 .016 .004 .001 .000 .000 .000 1 .232 .354 .393 .303 .187 .094 .037 .010 .002 .000 .000 2 .031 .098 .246 .324 .311 .234 .138 .060 .015 .001 .000 3 .002 .015 .082 .185 .276 .313 .276 .185 .082 .015 .002 4 .000 .001 .015 .060 .138 .234 .311 .324 .246 .098 .031 5 .000 .000 .002 .010 .037 .094 .187 .303 .393 .354 .232 6 .000 .000 .000 .001 .004 .016 .047 .118 .262 .531 .735

Binomial Probability Tables Formula (6–3) can be used to build a binomial probability distribution for any value of n and �. However, for a larger n, the calculations take more time. For conve- nience, the tables in Appendix B.9 show the result of using the formula for various values of n and �. Table 6–2 shows part of Appendix B.9 for n � 6 and various val- ues of �.

Number of Late Flights,

x P(x) xP(x) x � � (x � �)2 (x � �)2P(x)

0 0.3277 0.0000 �1 1 0.3277 1 0.4096 0.4096 0 0 0 2 0.2048 0.4096 1 1 0.2048 3 0.0512 0.1536 2 4 0.2048 4 0.0064 0.0256 3 9 0.0576 5 0.0003 0.0015 4 16 0.0048

� � 1.0000 �2 � 0.7997

The mean of 1.0 and the variance of .80 can be verified from formulas (6–1) and (6–2). The probability distribution from the Excel output on the previous page and the details of the calculations are shown below.

Example Five percent of the worm gears produced by an automatic, high-speed Carter-Bell milling machine are defective. What is the probability that out of six gears selected at random none will be defective? Exactly one? Exactly two? Exactly three? Exactly four? Exactly five? Exactly six out of six?

Solution The binomial conditions are met: (a) there are only two possible outcomes (a partic- ular gear is either defective or acceptable), (b) there is a fixed number of trials (6), (c) there is a constant probability of success (.05), and (d) the trials are independent.

Refer to Table 6–2 above for the probability of exactly zero defective gears. Go down the left margin to an x of 0. Now move horizontally to the column headed by a � of .05 to find the probability. It is .735.

The probability of exactly one defective in a sample of six worm gears is .232. The complete binomial probability distribution for n � 6 and � � .05 is:

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Of course, there is a slight chance of getting exactly five defective gears out of six random selections. It is .00000178, found by inserting the appropriate values in the binomial formula:

For six out of the six, the exact probability is .000000016. Thus, the probability is very small that five or six defective gears will be selected in a sample of six.

We can compute the mean or expected value of the distribution of the number defective:

�2 � n�(1 � �) � 6(.05)(.95) � 0.285

� � n� � (6)(.05) � 0.30

P(5) � 6C5(.05) 5(.95)1 � (6)(.05)5(.95) � .00000178

Discrete Probability Distributions 199

MegaStat software will also compute the probabilities for a binomial distribution. Below is the output for the previous example. In MegaStat, p is used to represent the probability of success rather than �. The cumulative probability, expected value, variance, and standard deviation are also reported.

Number of Number of Defective Probability of Defective Probability of

Gears, Occurrence, Gears, Occurrence, x P(x) x P (x)

0 .735 4 .000 1 .232 5 .000 2 .031 6 .000 3 .002

Self-Review 6–3 Eighty percent of the employees at the J. M. Smucker Company plant on Laskey Road have their bimonthly wages sent directly to their bank by electronic funds transfer. This is also called direct deposit. Suppose we select a random sample of seven employees. (a) Does this situation fit the assumptions of the binomial distribution? (b) What is the probability that all seven employees use direct deposit? (c) Use formula (6–3) to determine the exact probability that four of the seven sampled

employees use direct deposit. (d) Use Appendix B.9 to verify your answers to parts (b) and (c).

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2. If �, the probability of success, remains the same but n becomes larger, the shape of the binomial distribution becomes more symmetrical. Chart 6–3 shows a situation where � remains constant at .10 but n increases from 7 to 40.

TABLE 6–3 Probability of 0, 1, 2, . . . Successes for a � of .05, .10, .20, .50, and .70 and an n of 10

x\� .05 .1 .2 .3 .4 .5 .6 .7 .8 .9 .95

0 .599 .349 .107 .028 .006 .001 .000 .000 .000 .000 .000 1 .315 .387 .268 .121 .040 .010 .002 .000 .000 .000 .000 2 .075 .194 .302 .233 .121 .044 .011 .001 .000 .000 .000 3 .010 .057 .201 .267 .215 .117 .042 .009 .001 .000 .000 4 .001 .011 .088 .200 .251 .205 .111 .037 .006 .000 .000 5 .000 .001 .026 .103 .201 .246 .201 .103 .026 .001 .000 6 .000 .000 .006 .037 .111 .205 .251 .200 .088 .011 .001 7 .000 .000 .001 .009 .042 .117 .215 .267 .201 .057 .010 8 .000 .000 .000 .001 .011 .044 .121 .233 .302 .194 .075 9 .000 .000 .000 .000 .002 .010 .040 .121 .268 .387 .315

10 .000 .000 .000 .000 .000 .001 .006 .028 .107 .349 .599

Appendix B.9 is limited. It gives probabilities for n values from 1 to 15 and � values of .05, .10, . . . , .90, and .95. A software program can generate the proba- bilities for a specified number of successes, given n and �. The Excel output to the left shows the probability when n � 40 and � � .09. Note that the number of suc- cesses stops at 15 because the probabilities for 16 to 40 are very close to 0. The instructions are detailed in the Software Commands section on page 219.

Several additional points should be made regarding the binomial probability distribution.

1. If n remains the same but � increases from .05 to .95, the shape of the distri- bution changes. Look at Table 6–3 and Chart 6–2. The distribution for a � of .05 is positively skewed. As � approaches .50, the distribution becomes symmet- rical. As � goes beyond .50 and moves toward .95, the probability distribution becomes negatively skewed. Table 6–3 highlights probabilities for n � 10 and � of .05, .10, .20, .50, and .70. The graphs of these probability distributions are shown in Chart 6–2.

P (x )

.60

.50

.40

.30

.20

.10

.00

� = .05 n = 10

� = .10 n = 10

� = .20 n = 10

� = .50 n = 10

� = .70 n = 10

0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10

x Successes

CHART 6–2 Graphing the Binomial Probability Distribution for a � of .05, .10, .20, .50, and .70 and an n of 10

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Discrete Probability Distributions 201

P(x )

.50

.40

.30

.20

.10

.00

n = 7 n = 12 n = 20 n = 40

Number of Successes (x )

0 1 2 3 4 5 6 7 8 9100 1 2 3 4 5 6 7 80 1 2 3 4 5 6 70 1 2 3 4

CHART 6–3 Chart Representing the Binomial Probability Distribution for a � of .10 and an n of 7, 12, 20, and 40

Exercises 9. In a binomial situation, n � 4 and � � .25. Determine the probabilities of the following

events using the binomial formula. a. x � 2 b. x � 3

10. In a binomial situation, n � 5 and � � .40. Determine the probabilities of the following events using the binomial formula. a. x � 1 b. x � 2

11. Assume a binomial distribution where n � 3 and � � .60. a. Refer to Appendix B.9, and list the probabilities for values of x from 0 to 3. b. Determine the mean and standard deviation of the distribution from the general defi-

nitions given in formulas (6–1) and (6–2). 12. Assume a binomial distribution where n � 5 and � � .30.

a. Refer to Appendix B.9, and list the probabilities for values of x from 0 to 5. b. Determine the mean and standard deviation of the distribution from the general defi-

nitions given in formulas (6–1) and (6–2). 13. An American Society of Investors survey found 30 percent of individual investors have

used a discount broker. In a random sample of nine individuals, what is the probability: a. Exactly two of the sampled individuals have used a discount broker? b. Exactly four of them have used a discount broker? c. None of them have used a discount broker?

14. The United States Postal Service reports 95 percent of first class mail within the same city is delivered within two days of the time of mailing. Six letters are randomly sent to different locations.

a. What is the probability that all six arrive within two days? b. What is the probability that exactly five arrive within two days? c. Find the mean number of letters that will arrive within two days. d. Compute the variance and standard deviation of the number that will arrive within two

days. 15. Industry standards suggest that 10 percent of new vehicles require warranty service within

the first year. Jones Nissan in Sumter, South Carolina, sold 12 Nissans yesterday.

a. What is the probability that none of these vehicles requires warranty service? b. What is the probability exactly one of these vehicles requires warranty service?

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c. Determine the probability that exactly two of these vehicles require warranty service. d. Compute the mean and standard deviation of this probability distribution.

16. A telemarketer makes six phone calls per hour and is able to make a sale on 30 percent of these contacts. During the next two hours, find:

a. The probability of making exactly four sales. b. The probability of making no sales. c. The probability of making exactly two sales. d. The mean number of sales in the two-hour period.

17. A recent survey by the American Accounting Association revealed 23 percent of students graduating with a major in accounting select public accounting. Suppose we select a sample of 15 recent graduates.

a. What is the probability two select public accounting? b. What is the probability five select public accounting? c. How many graduates would you expect to select public accounting?

18. It is reported that 16 percent of American households use a cell phone exclusively for their telephone service. In a sample of eight households, find the probability that:

a. None use a cell phone as their exclusive service. b. At least one uses the cell exclusively. c. At least five use the cell phone.

Cumulative Binomial Probability Distributions We may wish to know the probability of correctly guessing the answers to 6 or more true/false questions out of 10. Or we may be interested in the probability of selecting less than two defectives at random from production during the previous hour. In these cases, we need cumulative frequency distributions similar to the ones developed in Chapter 2. See page 42. The following example will illustrate.

Example A study by the Illinois Department of Transportation concluded that 76.2 percent of front seat occupants used seat belts. That means that both occupants of the front seat were using their seat belts. Suppose we decide to compare that information with current usage. We select a sample of 12 vehicles.

1. What is the probability the front seat occupants in exactly 7 of the 12 vehicles selected are wearing seat belts?

2. What is the probability the front seat occupants in at least 7 of the 12 vehicles are wearing seat belts?

Solution This situation meets the binomial requirements. • In a particular vehicle, both the front seat occupants are either wearing seat belts

or they are not. There are only two possible outcomes. • There are a fixed number of trials, 12 in this case, because 12 vehicles are checked. • The probability of a “success” (occupants wearing seat belts) is the same from

one vehicle to the next: 76.2 percent. • The trials are independent. If the fourth vehicle selected in the sample has all

the occupants wearing their seat belts, this does not have any effect on the results for the fifth or tenth vehicle.

To find the likelihood the occupants of exactly 7 of the sampled vehicles are wear- ing seat belts, we use formula 6-3. In this case, n � 12 and � � .762.

So we conclude the likelihood that the occupants of exactly 7 of the 12 sampled vehicles will be wearing their seat belts is about 9 percent. We often use, as we

� 12C7(.762) 7(1 � .762)12�7 � 792(.149171)(.000764) � .0902

P(x � 7 ƒ n � 12 and � � .762)

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did in this equation, a bar “�” to mean “given that.” So in this equation we want to know the probability that x is equal to 7 “given that the number of trials is 12 and the probability of a success is .762.”

To find the probability that the occupants in 7 or more of the vehicles will be wearing seat belts, we use formula (6–3) from this chapter as well as the special rule of addition from the previous chapter. See formula (5–2) on page 153.

Because the events are mutually exclusive (meaning that a particular sample of 12 vehicles cannot have both a total of 7 and a total of 8 vehicles where the occupants are wearing seat belts), we find the probability of 7 vehicles where the occupants are wearing seat belts, the probability of 8, and so on up to the proba- bility that occupants of all 12 sample vehicles are wearing seat belts. The probability of each of these outcomes is then totaled.

So the probability of selecting 12 cars and finding that the occupants of 7 or more vehicles were wearing seat belts is .9562. This information is shown on the following Excel spreadsheet. There is a slight difference in the software answer due to rounding. The Excel commands are similar to those detailed in the Soft- ware Commands section on page 219, number 2.

� .9562

� .0902 � .1805 � .2569 � .2467 � .1436 � .0383

� P(x � 7) � P(x � 8) � P(x � 9) � P(x � 10) � P(x � 11) � P(x � 12)

P(x 7 ƒ n � 12 and � � .762)

Discrete Probability Distributions 203

Self-Review 6–4 For a case where n � 4 and � � .60, determine the probability that: (a) x � 2. (b) x 2. (c) x � 2.

Exercises 19. In a binomial distribution, n � 8 and � � .30. Find the probabilities of the following events.

a. x � 2. b. x 2 (the probability that x is equal to or less than 2). c. x 3 (the probability that x is equal to or greater than 3).

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20. In a binomial distribution, n � 12 and � � .60. Find the following probabilities. a. x � 5. b. x 5. c. x 6.

21. In a recent study, 90 percent of the homes in the United States were found to have large- screen TVs. In a sample of nine homes, what is the probability that: a. All nine have large-screen TVs? b. Less than five have large-screen TVs? c. More than five have large-screen TVs? d. At least seven homes have large-screen TVs?

22. A manufacturer of window frames knows from long experience that 5 percent of the pro- duction will have some type of minor defect that will require an adjustment. What is the probability that in a sample of 20 window frames: a. None will need adjustment? b. At least one will need adjustment? c. More than two will need adjustment?

23. The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 70 percent of the cases. Suppose the 15 cases reported today are representative of all complaints.

a. How many of the problems would you expect to be resolved today? What is the stan- dard deviation?

b. What is the probability 10 of the problems can be resolved today? c. What is the probability 10 or 11 of the problems can be resolved today? d. What is the probability more than 10 of the problems can be resolved today?

24. It is asserted that 80 percent of the cars approaching an individual toll both in New Jer- sey are equipped with an E-ZPass transponder. Find the probability that in a sample of six cars:

a. All six will have the transponder. b. At least three will have the transponder. c. None will have a transponder.

6.6 Hypergeometric Probability Distribution For the binomial distribution to be applied, the probability of a success must stay the same for each trial. For example, the probability of guessing the correct answer to a true/false question is .50. This probability remains the same for each question on an examination. Likewise, suppose that 40 percent of the registered voters in a precinct are Republicans. If 27 registered voters are selected at random, the probability of choosing a Republican on the first selection is .40. The chance of choosing a Republican on the next selection is also .40, assuming that the sam- pling is done with replacement, meaning that the person selected is put back in the population before the next person is selected.

Most sampling, however, is done without replacement. Thus, if the population is small, the probability for each observation will change. For example, if the pop- ulation consists of 20 items, the probability of selecting a particular item from that population is 1/20. If the sampling is done without replacement, after the first selec- tion there are only 19 items remaining; the probability of selecting a particular item on the second selection is only 1/19. For the third selection, the probability is 1/18, and so on. This assumes that the population is finite—that is, the number in the population is known and relatively small in number. Examples of a finite population are 2,842 Republicans in the precinct, 9,241 applications for medical school, and the 18 2010 Dakota 4x4 Crew Cabs at Helfman Dodge Chrysler Jeep in Houston, TX.

Recall that one of the criteria for the binomial distribution is that the probability of success remains the same from trial to trial. Because the probability of success does not remain the same from trial to trial when sampling is from a relatively small population without replacement, the binomial distribution should not be used. Instead, the hypergeometric distribution is applied. Therefore, (1) if a sample is

LO6 Describe and compute probabilities for a hypergeometric distribution.

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Example PlayTime Toys Inc. employs 50 people in the Assembly Department. Forty of the employees belong to a union and ten do not. Five employees are selected at ran- dom to form a committee to meet with management regarding shift starting times. What is the probability that four of the five selected for the committee belong to a union?

Solution The population in this case is the 50 Assem- bly Department employees. An employee can be selected for the committee only once. Hence, the sampling is done without replacement. Thus, the probability of select- ing a union employee, for example, changes from one trial to the next. The hyperge- ometric distribution is appropriate for determining the probability. In this problem,

N is 50, the number of employees. S is 40, the number of union employees. x is 4, the number of union employees selected. n is 5, the number of employees selected.

Discrete Probability Distributions 205

selected from a finite population without replacement and (2) if the size of the sam- ple n is more than 5 percent of the size of the population N, then the hypergeometric distribution is used to determine the probability of a specified number of successes or failures. It is especially appropriate when the size of the population is small.

The formula for the hypergeometric distribution is:

HYPERGEOMETRIC DISTRIBUTION [6–6]P(x) � (SCx)(N�SCn�x)

NCn

where: N is the size of the population. S is the number of successes in the population. x is the number of successes in the sample. It may be 0, 1, 2, 3, . . . . n is the size of the sample or the number of trials. C is the symbol for a combination.

In summary, a hypergeometric probability distribution has these characteristics:

HYPERGEOMETRIC PROBABILITY EXPERIMENT 1. An outcome on each trial of an experiment is classified into one of two mutu-

ally exclusive categories—a success or a failure. 2. The random variable is the number of successes in a fixed number of

trials. 3. The trials are not independent. 4. We assume that we sample from a finite population without replacement and

n/N � 0.05. So, the probability of a success changes for each trial.

The following example illustrates the details of determining a probability using the hypergeometric distribution.

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Table 6–4 shows the hypergeometric probabilities of finding 0, 1, 2, 3, 4, and 5 union members on the committee.

TABLE 6–4 Hypergeometric Probabilities (n � 5, N � 50, and S � 40) for the Number of Union Members on the Committee

Union Members Probability

0 .000 1 .004 2 .044 3 .210 4 .431 5 .311

1.000

TABLE 6–5 Hypergeometric and Binomial Probabilities for PlayTime Toys Inc. Assembly Department

Number of Union Hypergeometric Binomial Probability Members on Committee Probability, P(x) (n � 5 and � � .80)

0 .000 .000 1 .004 .006 2 .044 .051 3 .210 .205 4 .431 .410 5 .311 .328

1.000 1.000

In order for you to compare the two probability distributions, Table 6–5 shows the hypergeometric and binomial probabilities for the PlayTime Toys Inc. example. Because 40 of the 50 Assembly Department employees belong to the union, we let � � .80 for the binomial distribution. The binomial probabilities for Table 6–5 come from the binomial distribution with n � 5 and � � .80.

When the binomial requirement of a constant probability of success cannot be met, the hypergeometric distribution should be used. However, as Table 6–5 shows, under certain conditions the results of the binomial distribution can be used to approximate the hypergeometric. This leads to a rule of thumb:

If selected items are not returned to the population, the binomial distribu- tion can be used to closely approximate the hypergeometric distribution when n � .05N. In words, the binomial will suffice if the sample is less than 5 percent of the population.

A hypergeometric distribution can be created using Excel. See the output on the left. The necessary steps are given in the Software Commands section on page 219 at the end of the chapter.

We wish to find the probability 4 of the 5 committee members belong to a union. Inserting these values into formula (6–6):

Thus, the probability of selecting 5 assembly workers at random from the 50 workers and finding 4 of the 5 are union members is .431.

P(4) � (40C4)(50�40C5�4)

50C5 �

(91,390)(10) 2,118,760

� .431

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Discrete Probability Distributions 207

Self-Review 6–5 Horwege Discount Brokers plans to hire 5 new financial analysts this year. There is a pool of 12 approved applicants, and George Horwege, the owner, decides to randomly select those who will be hired. There are 8 men and 4 women among the approved applicants. What is the probability that 3 of the 5 hired are men?

Exercises 25. A CD contains 10 songs; 6 are classical and 4 are rock and roll. In a sample of 3 songs,

what is the probability that exactly 2 are classical? Assume the samples are drawn with- out replacement.

26. A population consists of 15 items, 10 of which are acceptable. In a sample of 4 items, what is the probability that exactly 3 are acceptable? Assume the samples are drawn without replacement.

27. Kolzak Appliance Outlet just received a shipment of 10 DVD players. Shortly after they were received, the manufacturer called to report that he had inadvertently shipped 3 defective units. Ms. Kolzak, the owner of the outlet, decided to test 2 of the 10 DVD players she received. What is the probability that neither of the 2 DVD players tested is defective? Assume the samples are drawn without replacement.

28. The Computer Systems Department has 8 faculty, 6 of whom are tenured. Dr. Vonder, the chairman, wants to establish a committee of 3 department faculty members to review the curriculum. If she selects the committee at random: a. What is the probability all members of the committee are tenured? b. What is the probability that at least one member is not tenured? (Hint: For this ques-

tion, use the complement rule.) 29. Keith’s Florists has 15 delivery trucks, used mainly to deliver flowers and flower arrange-

ments in the Greenville, South Carolina, area. Of these 15 trucks, 6 have brake problems. A sample of 5 trucks is randomly selected. What is the probability that 2 of those tested have defective brakes?

30. The game called Lotto sponsored by the Louisiana Lottery Commission pays its largest prize when a contestant matches all 6 of the 40 possible numbers. Assume there are 40 ping-pong balls each with a single number between 1 and 40. Any number appears only once, and the winning balls are selected without replacement. a. The commission reports that the probability of matching all the numbers are 1 in

3,838,380. What is this in terms of probability? b. Use the hypergeometric formula to find this probability. The lottery commission also pays if a contestant matches 4 or 5 of the 6 winning num- bers. Hint: Divide the 40 numbers into two groups, winning numbers and nonwinning numbers. c. Find the probability, again using the hypergeometric formula, for matching 4 of the 6

winning numbers. d. Find the probability of matching 5 of the 6 winning numbers.

6.7 Poisson Probability Distribution The Poisson probability distribution describes the number of times some event occurs during a specified interval. The interval may be time, distance, area, or volume.

The distribution is based on two assumptions. The first assumption is that the probability is proportional to the length of the interval. The second assumption is that the intervals are independent. To put it another way, the longer the interval, the larger the probability, and the number of occurrences in one interval does not affect the other intervals. This distribution is also a limiting form of the binomial distribution when the probability of a success is very small and n is large. It is often referred to as the “law of improbable events,” meaning that the probability, �, of a particular

LO7 Describe and compute probabilities for a Poisson distribution.

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Statistics in Action

Near the end of World War II, the Germans developed rocket bombs, which were fired at the city of London. The Al- lied military com- mand didn’t know whether these bombs were fired at random or whether they had an aiming device. To investigate, the city of London was divided into 586 square regions. The distribution of hits in each square was recorded as follows:

Regions 229 221 93 35 7 1

To interpret, the above chart indicates that 229 regions were not hit with one of the bombs. Seven re- gions were hit four times. Using the Poisson distribution, with a mean of 0.93 hits per region, the expected number of hits is as follows:

Hits 0 1 2 3 4 5 or

more Regions 231.2 215.0 100.0 31.0 7.2 1.6

Because the actual number of hits was close to the expected number of hits, the military command concluded that the

(continued)

event’s happening is quite small. The Poisson distribution is a discrete probability distribution because it is formed by counting.

In summary, a Poisson probability distribution has these characteristics:

POISSON PROBABILITY EXPERIMENT 1. The random variable is the number of times some event occurs during a

defined interval. 2. The probability of the event is proportional to the size of the interval. 3. The intervals do not overlap and are independent.

This distribution has many applications. It is used as a model to describe the distrib- ution of errors in data entry, the number of scratches and other imperfections in newly painted car panels, the number of defective parts in outgoing shipments, the number of customers waiting to be served at a restaurant or waiting to get into an attraction at Disney World, and the number of accidents on I–75 during a three-month period.

The Poisson distribution can be described mathematically by the formula:

POISSON DISTRIBUTION [6–7]P(x) � �xe��

where: � (mu) is the mean number of occurrences (successes) in a particular interval.

e is the constant 2.71828 (base of the Napierian logarithmic system). x is the number of occurrences (successes).

P(x) is the probability for a specified value of x.

The mean number of successes, �, can be determined by n�, where n is the total number of trials and � the probability of success.

MEAN OF A POISSON DISTRIBUTION [6–8]� � n�

The variance of the Poisson is also equal to its mean. If, for example, the proba- bility that a check cashed by a bank will bounce is .0003, and 10,000 checks are cashed, the mean and the variance for the number of bad checks is 3.0, found by � � n� � 10,000(.0003) � 3.0.

Recall that for a binomial distribution there is a fixed number of trials. For exam- ple, for a four-question multiple-choice test there can only be zero, one, two, three, or four successes (correct answers). The random variable, x, for a Poisson distribution, however, can assume an infinite number of values—that is, 0, 1, 2, 3, 4, 5, . . . How- ever, the probabilities become very small after the first few occurrences (successes).

To illustrate the Poisson probability computation, assume baggage is rarely lost by Delta Airlines. Most flights do not experience any mishandled bags; some have one bag lost; a few have two bags lost; rarely a flight will have three lost bags; and so on. Suppose a random sample of 1,000 flights shows a total of 300 bags were lost. Thus, the arithmetic mean number of lost bags per flight is 0.3, found by 300/1,000. If the number of lost bags per flight follows a Poisson distribution with � � 0.3, we can compute the various probabilities using formula (6–7):

For example, the probability of not losing any bags is:

P(0) � (0.3)0(e�0.3)

0! � 0.7408

P(x) � �xe��

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bombs were falling at random. The Ger- mans had not devel- oped a bomb with an aiming device.

Example Recall from the previous illustration that the number of lost bags follows a Poisson distribution with a mean of 0.3. Use Appendix B.5 to find the probability that no bags will be lost on a particular flight. What is the probability exactly one bag will be lost on a particular flight? When should the supervisor become suspicious that a flight is having too many lost bags?

Solution Part of Appendix B.5 is repeated as Table 6–6. To find the probability of no lost bags, locate the column headed “0.3” and read down that column to the row labeled “0.” The probability is .7408. That is the probability of no lost bags. The probability of one lost bag is .2222, which is in the next row of the table, in the same column. The probability of two lost bags is .0333, in the row below; for three lost bags, it is .0033; and for four lost bags, it is .0003. Thus, a supervisor should not be surprised to find one lost bag but should expect to see more than one lost bag infrequently.

Discrete Probability Distributions 209

�

x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

0 0.9048 0.8187 0.7408 0.6703 0.6065 0.5488 0.4966 0.4493 0.4066 1 0.0905 0.1637 0.2222 0.2681 0.3033 0.3293 0.3476 0.3595 0.3659 2 0.0045 0.0164 0.0333 0.0536 0.0758 0.0988 0.1217 0.1438 0.1647 3 0.0002 0.0011 0.0033 0.0072 0.0126 0.0198 0.0284 0.0383 0.0494 4 0.0000 0.0001 0.0003 0.0007 0.0016 0.0030 0.0050 0.0077 0.0111 5 0.0000 0.0000 0.0000 0.0001 0.0002 0.0004 0.0007 0.0012 0.0020 6 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0002 0.0003 7 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

TABLE 6–6 Poisson Table for Various Values of � (from Appendix B.5)

These probabilities can also be found using the Minitab system. The com- mands necessary are reported at the end of the chapter.

In other words, 74 percent of the flights will have no lost baggage. The probability of exactly one lost bag is:

Thus, we would expect to find exactly one lost bag on 22 percent of the flights. Poisson probabilities can also be found in the table in Appendix B.5.

P(1) � (0.3)1(e�0.3)

1! � 0.2222

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Earlier in this section, we mentioned that the Poisson probability distribution is a lim- iting form of the binomial. That is, we could estimate a binomial probability using the Poisson.

The Poisson probability distribution is characterized by the number of times an event happens during some interval or continuum. Examples include:

• The number of misspelled words per page in a newspaper. • The number of calls per hour received by Dyson Vacuum Cleaner Company. • The number of vehicles sold per day at Hyatt Buick GMC in Durham, North

Carolina. • The number of goals scored in a college soccer game.

In each of these examples, there is some type of continuum—misspelled words per page, calls per hour, vehicles per day, or goals per game.

In the previous example, we investigated the number of bags lost per flight, so the continuum was a “flight.” We knew the mean number of bags of luggage lost per flight, but we did not know the number of passengers or the probability of a bag being lost. We suspected the number of passengers was fairly large and the probability of a passenger losing his or her bag of luggage was small. In the fol- lowing example, we use the Poisson distribution to estimate a binomial probability when n, the number of trials, is large and �, the probability of a success, small.

Example Coastal Insurance Company underwrites insurance for beachfront properties along the Virginia, North and South Carolina, and Georgia coasts. It uses the estimate that the probability of a named Category III hurricane (sustained winds of more than 110 miles per hour) or higher striking a particular region of the coast (for example, St. Simons Island, Georgia) in any one year is .05. If a homeowner takes a 30-year mortgage on a recently purchased property in St. Simons, what is the likelihood that the owner will experience at least one hurricane during the mort- gage period?

Solution To use the Poisson probability distribution, we begin by determining the mean or expected number of storms meeting the criterion hitting St. Simons during the 30-year period. That is:

where: n is the number of years, 30 in this case. � is the probability a hurricane meeting the strength criteria comes ashore. � is the mean or expected number of storms in a 30-year period.

To find the probability of at least one storm hitting St. Simons Island, Georgia, we first find the probability of no storms hitting the coast and subtract that value from 1.

We conclude that the likelihood a hurricane meeting the strength criteria will strike the beachfront property at St. Simons during the 30-year period when the mort- gage is in effect is .7769. To put it another way, the probability St. Simons will be hit by a Category III or higher hurricane during the 30-year period is a little more than 75 percent.

We should emphasize that the continuum, as previously described, still exists. That is, there are expected to be 1.5 storms hitting the coast per 30-year period. The continuum is the 30-year period.

P(x 1) � 1 � P(x � 0) � 1 � �0e�1.5

0! � 1 � .2231 � .7769

� � n� � 30(.05) � 1.5

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Discrete Probability Distributions 211

The Poisson probability distribution is always positively skewed and the random variable has no specific upper limit. The Poisson distribution for the lost bags illus- tration, where � � 0.3, is highly skewed. As � becomes larger, the Poisson distrib- ution becomes more symmetrical. For example, Chart 6–4 shows the distributions of the number of transmission services, muffler replacements, and oil changes per day at Avellino’s Auto Shop. They follow Poisson distributions with means of 0.7, 2.0, and 6.0, respectively.

CHART 6–4 Poisson Probability Distributions for Means of 0.7, 2.0, and 6.0

11 12 13

� = 0.7 � = 2.0 � = 6.0

0 1 2 3 4 5 6 7 8 9 100 1 2 3 4 5 6 70 1 2 3 4

Pr ob

ab ili

ty o

f o cc

ur re

nc e

.30

.20

.10

.00

.40

.50

Transmission services

Muffler replacements

Oil changes

Number of occurrences

P (x )

In summary, the Poisson distribution is actually a family of discrete distributions. All that is needed to construct a Poisson probability distribution is the mean number of defects, errors, and so on—designated as �.

Only � needed to construct Poisson distribution

In the preceding case, we are actually using the Poisson distribution as an estimate of the binomial. Note that we’ve met the binomial conditions outlined on page 196.

• There are only two possible outcomes: a hurricane hits the St. Simons area or it does not.

• There is a fixed number of trials, in this case 30 years. • There is a constant probability of success; that is, the probability of a hurricane

hitting the area is .05 each year. • The years are independent. That means if a named storm strikes in the fifth

year, that has no effect on any other year.

To find the probability of at least one storm striking the area in a 30-year period using the binomial distribution:

The probability of at least one hurricane hitting the St. Simons area during the 30-year period using the binomial distribution is .7854.

Which answer is correct? Why should we look at the problem both ways? The binomial is the more “technically correct” solution. The Poisson can be thought of as an approximation for the binomial, when n, the number of trials is large, and �, the probability of a success, is small. We look at the problem using both distributions to emphasize the convergence of the two discrete distributions. In some instances, using the Poisson may be the quicker solution, and as you see there is little practical dif- ference in the answers. In fact, as n gets larger and � smaller, the differences between the two distributions gets smaller.

P(x 1) � 1 � P(x � 0) � 1 � 30C0(.05) 0 (.95)30 � 1 � (1)(1)(.2146) � .7854

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Self-Review 6–6 From actuary tables, Washington Insurance Company determined the likelihood that a man age 25 will die within the next year is .0002. If Washington Insurance sells 4,000 policies to 25-year-old men this year, what is the probability they will pay on exactly one policy?

Exercises 31. In a Poisson distribution � � 0.4.

a. What is the probability that x � 0? b. What is the probability that x � 0?

32. In a Poisson distribution � � 4. a. What is the probability that x � 2? b. What is the probability that x 2? c. What is the probability that x � 2?

33. Ms. Bergen is a loan officer at Coast Bank and Trust. From her years of experience, she estimates that the probability is .025 that an applicant will not be able to repay his or her installment loan. Last month she made 40 loans. a. What is the probability that 3 loans will be defaulted? b. What is the probability that at least 3 loans will be defaulted?

34. Automobiles arrive at the Elkhart exit of the Indiana Toll Road at the rate of two per minute. The distribution of arrivals approximates a Poisson distribution. a. What is the probability that no automobiles arrive in a particular minute? b. What is the probability that at least one automobile arrives during a particular minute?

35. It is estimated that 0.5 percent of the callers to the Customer Service department of Dell Inc. will receive a busy signal. What is the probability that of today’s 1,200 callers at least 5 received a busy signal?

36. In the past, schools in Los Angeles County have closed an average of three days each year for weather emergencies. What is the probability that schools in Los Angeles County will close for four days next year?

Chapter Summary I. A random variable is a numerical value determined by the outcome of an experiment.

II. A probability distribution is a listing of all possible outcomes of an experiment and the probability associated with each outcome. A. A discrete probability distribution can assume only certain values. The main features

are: 1. The sum of the probabilities is 1.00. 2. The probability of a particular outcome is between 0.00 and 1.00. 3. The outcomes are mutually exclusive.

B. A continuous distribution can assume an infinite number of values within a specific range. III. The mean and variance of a probability distribution are computed as follows.

A. The mean is equal to:

[6–1]

B. The variance is equal to:

[6–2]

IV. The binomial distribution has the following characteristics. A. Each outcome is classified into one of two mutually exclusive categories. B. The distribution results from a count of the number of successes in a fixed number of

trials.

�2 � ©[(x � �)2P(x)]

� � ©[xP(x)]

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Discrete Probability Distributions 213

C. The probability of a success remains the same from trial to trial. D. Each trial is independent. E. A binomial probability is determined as follows:

[6–3]

F. The mean is computed as:

[6–4]

G. The variance is

[6–5]

V. The hypergeometric distribution has the following characteristics. A. There are only two possible outcomes. B. The probability of a success is not the same on each trial. C. The distribution results from a count of the number of successes in a fixed number of

trials. D. It is used when sampling without replacement from a finite population. E. A hypergeometric probability is computed from the following equation:

[6–6]

VI. The Poisson distribution has the following characteristics. A. It describes the number of times some event occurs during a specified interval. B. The probability of a “success” is proportional to the length of the interval. C. Nonoverlapping intervals are independent. D. It is a limiting form of the binomial distribution when n is large and � is small. E. A Poisson probability is determined from the following equation:

[6–7]

F. The mean and the variance are:

[6–8]

Chapter Exercises 37. What is the difference between a random variable and a probability distribution? 38. For each of the following indicate whether the random variable is discrete or continuous.

a. The length of time to get a haircut. b. The number of cars a jogger passes each morning while running. c. The number of hits for a team in a high school girls’ softball game. d. The number of patients treated at the South Strand Medical Center between 6 and

10 P.M. each night. e. The distance your car traveled on the last fill-up. f. The number of customers at the Oak Street Wendy’s who used the drive-through

facility. g. The distance between Gainesville, Florida, and all Florida cities with a population of

at least 50,000. 39. An investment will be worth $1,000, $2,000, or $5,000 at the end of the year. The prob-

abilities of these values are .25, .60, and .15, respectively. Determine the mean and vari- ance of the worth of the investment.

40. The personnel manager of Cumberland Pig Iron Company is studying the number of on- the-job accidents over a period of one month. He developed the following probability distribution. Compute the mean, variance, and standard deviation of the number of acci- dents in a month.

�2 � n�

� � n�

P(x) � �xe��

P(x) � (SCx)(N�SCn�x)

(NCn)

�2 � n�(1 � �)

� � n�

P(x) � nCx � x(1 � �)n�x

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214 Chapter 6

Number of Accidents Probability

0 .40 1 .20 2 .20 3 .10 4 .10

41. Croissant Bakery Inc. offers special decorated cakes for birthdays, weddings, and other occasions. It also has regular cakes available in its bakery. The following table gives the total number of cakes sold per day and the corresponding probability. Compute the mean, variance, and standard deviation of the number of cakes sold per day.

Number of Cakes Sold in a Day Probability

12 .25 13 .40 14 .25 15 .10

42. The payouts for the Powerball lottery and their corresponding odds and probabilities of occurrence are shown below. The price of a ticket is $1.00. Find the mean and standard deviation of the payout. Hint: Don’t forget to include the cost of the ticket and its corre- sponding probability.

Divisions Payout Odds Probability

Five plus Powerball $50,000,000 146,107,962 0.000000006844 Match 5 200,000 3,563,609 0.000000280614 Four plus Powerball 10,000 584,432 0.000001711060 Match 4 100 14,255 0.000070145903 Three plus Powerball 100 11,927 0.000083836351 Match 3 7 291 0.003424657534 Two plus Powerball 7 745 0.001340482574 One plus Powerball 4 127 0.007812500000 Zero plus Powerball 3 69 0.014285714286

43. In a recent survey, 35 percent indicated chocolate was their favorite flavor of ice cream. Suppose we select a sample of ten people and ask them to name their favorite flavor of ice cream. a. How many of those in the sample would you expect to name chocolate? b. What is the probability exactly four of those in the sample name chocolate? c. What is the probability four or more name chocolate?

44. Thirty percent of the population in a southwestern community are Spanish-speaking Amer- icans. A Spanish-speaking person is accused of killing a non-Spanish-speaking American and goes to trial. Of the first 12 potential jurors, only 2 are Spanish-speaking Americans, and 10 are not. The defendant’s lawyer challenges the jury selection, claiming bias against her client. The government lawyer disagrees, saying that the probability of this particular jury composition is common. Compute the probability and discuss the assumptions.

45. An auditor for Health Maintenance Services of Georgia reports 40 percent of policyhold- ers 55 years or older submit a claim during the year. Fifteen policyholders are randomly selected for company records. a. How many of the policyholders would you expect to have filed a claim within the last

year? b. What is the probability that 10 of the selected policyholders submitted a claim last year?

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Discrete Probability Distributions 215

c. What is the probability that 10 or more of the selected policyholders submitted a claim last year?

d. What is the probability that more than 10 of the selected policyholders submitted a claim last year?

46. Tire and Auto Supply is considering a 2-for-1 stock split. Before the transaction is finalized, at least two-thirds of the 1,200 company stockholders must approve the proposal. To evaluate the likelihood the proposal will be approved, the CFO selected a sample of 18 stockholders. He contacted each and found 14 approved of the proposed split. What is the likelihood of this event, assuming two-thirds of the stock- holders approve?

47. A federal study reported that 7.5 percent of the U.S. workforce has a drug problem. A drug enforcement official for the State of Indiana wished to investigate this state- ment. In her sample of 20 employed workers: a. How many would you expect to have a drug problem? What is the standard deviation? b. What is the likelihood that none of the workers sampled has a drug problem? c. What is the likelihood at least one has a drug problem?

48. The Bank of Hawaii reports that 7 percent of its credit card holders will default at some time in their life. The Hilo branch just mailed out 12 new cards today. a. How many of these new cardholders would you expect to default? What is the stan-

dard deviation? b. What is the likelihood that none of the cardholders will default? c. What is the likelihood at least one will default?

49. Recent statistics suggest that 15 percent of those who visit a retail site on the World Wide Web make a purchase. A retailer wished to verify this claim. To do so, she selected a sample of 16 “hits” to her site and found that 4 had actually made a purchase. a. What is the likelihood of exactly four purchases? b. How many purchases should she expect? c. What is the likelihood that four or more “hits” result in a purchase?

50. In Chapter 19, we discuss the acceptance sample. Acceptance sampling is used to mon- itor the quality of incoming raw materials. Suppose a purchaser of electronic components allows 1 percent of the components to be defective. To ensure the quality of incoming parts, a purchaser or manufacturer normally samples 20 parts and allows 1 defect. a. What is the likelihood of accepting a lot that is 1 percent defective? b. If the quality of the incoming lot was actually 2 percent, what is the likelihood of

accepting it? c. If the quality of the incoming lot was actually 5 percent, what is the likelihood of

accepting it? 51. Colgate-Palmolive Inc. recently developed a new toothpaste flavored with honey. It

tested a group of ten people. Six of the group said they liked the new flavor, and the remaining four indicated they definitely did not. Four of the ten are selected to participate in an in-depth interview. What is the probability that of those selected for the in-depth interview two liked the new flavor and two did not?

52. Dr. Richmond, a psychologist, is studying the daytime television viewing habits of college students. She believes 45 percent of college students watch soap operas during the after- noon. To further investigate, she selects a sample of 10. a. Develop a probability distribution for the number of students in the sample who watch

soap operas. b. Find the mean and the standard deviation of this distribution. c. What is the probability of finding exactly four watch soap operas? d. What is the probability less than half of the students selected watch soap operas?

53. A recent study conducted by Penn, Shone, and Borland, on behalf of LastMinute.com, revealed that 52 percent of business travelers plan their trips less than two weeks before departure. The study is to be replicated in the tri-state area with a sample of 12 frequent business travelers.

a. Develop a probability distribution for the number of travelers who plan their trips within two weeks of departure.

b. Find the mean and the standard deviation of this distribution. c. What is the probability exactly 5 of the 12 selected business travelers plan their trips

within two weeks of departure? d. What is the probability 5 or fewer of the 12 selected business travelers plan their trips

within two weeks of departure?

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54. Suppose the Internal Revenue Service is studying the category of charitable contribu- tions. A sample of 25 returns is selected from young couples between the ages of 20 and 35 who had an adjusted gross income of more than $100,000. Of these 25 returns, five had charitable contributions of more than $1,000. Suppose four of these returns are selected for a comprehensive audit. a. Explain why the hypergeometric distribution is appropriate. b. What is the probability exactly one of the four audited had a charitable deduction of

more than $1,000? c. What is the probability at least one of the audited returns had a charitable contribu-

tion of more than $1,000? 55. The law firm of Hagel and Hagel is located in downtown Cincinnati. There are 10 partners

in the firm; 7 live in Ohio and 3 in northern Kentucky. Ms. Wendy Hagel, the managing partner, wants to appoint a committee of 3 partners to look into moving the firm to northern Kentucky. If the committee is selected at random from the 10 partners, what is the probability that: a. One member of the committee lives in northern Kentucky and the others live in Ohio? b. At least 1 member of the committee lives in northern Kentucky?

56. Recent information published by the U.S. Environmental Protection Agency indicates that Honda is the manufacturer of four of the top nine vehicles in terms of fuel economy. a. Determine the probability distribution for the number of Hondas in a sample of three

cars chosen from the top nine. b. What is the likelihood that in the sample of three at least one Honda is included?

57. The position of chief of police in the city of Corry, Pennsylvania, is vacant. A search committee of Corry residents is charged with the responsibility of recommending a new chief to the city council. There are 12 applicants, 4 of which are either female or members of a minority. The search committee decides to interview all 12 of the appli- cants. To begin, they randomly select four applicants to be interviewed on the first day, and none of the four is female or a member of a minority. The local newspaper, the Corry Press, suggests discrimination in an editorial. What is the likelihood of this occurrence?

58. Listed below is the population by state for the 15 states with the largest population. Also included is whether that state’s border touches the Gulf of Mexico, the Atlantic Ocean, or the Pacific Ocean (coastline).

Rank State Population Coastline

1 California 36,553,215 Yes 2 Texas 23,904,380 Yes 3 New York 19,297,729 Yes 4 Florida 18,251,243 Yes 5 Illinois 12,852,548 No 6 Pennsylvania 12,432,792 No 7 Ohio 11,466,917 No 8 Michigan 10,071,822 No 9 Georgia 9,544,750 Yes

10 North Carolina 9,061,032 Yes 11 New Jersey 8,685,920 Yes 12 Virginia 7,712,091 Yes 13 Washington 6,468,424 Yes 14 Massachusetts 6,449,755 Yes 15 Indiana 6,345,289 No

Note that 5 of the 15 states do not have any coastline. Suppose three states are selected at random. What is the probability that: a. None of the states selected have any coastline? b. Exactly one of the selected states has a coastline? c. At least one of the selected states has a coastline?

216 Chapter 6

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Discrete Probability Distributions 217

59. The sales of Lexus automobiles in the Detroit area follow a Poisson distribution with a mean of 3 per day. a. What is the probability that no Lexus is sold on a particular day? b. What is the probability that for five consecutive days at least one Lexus is sold?

60. Suppose 1.5 percent of the antennas on new Nokia cell phones are defective. For a ran- dom sample of 200 antennas, find the probability that: a. None of the antennas is defective. b. Three or more of the antennas are defective.

61. A study of the checkout lines at the Safeway Supermarket in the South Strand area revealed that between 4 and 7 P.M. on weekdays there is an average of four customers waiting in line. What is the probability that you visit Safeway today during this period and find: a. No customers are waiting? b. Four customers are waiting? c. Four or fewer are waiting? d. Four or more are waiting?

62. An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of two emails per hour. Assume the arrival of these emails is approximated by the Poisson distribution. a. What is the probability Linda Lahey, company president, received exactly 1 email

between 4 P.M. and 5 P.M. yesterday? b. What is the probability she received 5 or more emails during the same period? c. What is the probability she did not receive any emails during the period?

63. Recent crime reports indicate that 3.1 motor vehicle thefts occur each minute in the United States. Assume that the distribution of thefts per minute can be approximated by the Poisson probability distribution. a. Calculate the probability exactly four thefts occur in a minute. b. What is the probability there are no thefts in a minute? c. What is the probability there is at least one theft in a minute?

64. New Process Inc. a large mail-order supplier of women’s fashions, advertises same- day service on every order. Recently, the movement of orders has not gone as planned, and there were a large number of complaints. Bud Owens, director of customer ser- vice, has completely redone the method of order handling. The goal is to have fewer than five unfilled orders on hand at the end of 95 percent of the working days. Fre- quent checks of the unfilled orders at the end of the day reveal that the distribution of the unfilled orders follows a Poisson distribution with a mean of two orders. a. Has New Process Inc. lived up to its internal goal? Cite evidence. b. Draw a histogram representing the Poisson probability distribution of unfilled orders.

65. The National Aeronautics and Space Administration (NASA) has experienced two dis- asters. The Challenger exploded over the Atlantic Ocean in 1986, and the Columbia disintegrated on reentry over East Texas in 2003. Based on the first 113 missions, and assuming failures occur at the same rate, consider the next 23 missions. What is the probability of exactly two failures? What is the probability of no failures?

66. According to the “January theory,” if the stock market is up for the month of January, it will be up for the year. If it is down in January, it will be down for the year. According to an article in The Wall Street Journal, this theory held for 29 out of the last 34 years. Suppose there is no truth to this theory; that is, the probability it is either up or down is .50. What is the probability this could occur by chance? You will probably need a software package such as Excel or Minitab.

67. During the second round of the 1989 U.S. Open golf tournament, four golfers scored a hole in one on the sixth hole. The odds of a professional golfer making a hole in one are estimated to be 3,708 to 1, so the probability is 1/3,709. There were 155 golfers parti- cipating in the second round that day. Estimate the probability that four golfers would score a hole in one on the sixth hole.

68. Suppose the National Hurricane Center forecasts that hurricanes will hit the strike area with a .95 probability. Answer the following questions: a. What probability distribution does this follow? b. What is the probability that 10 hurricanes reach landfall in the strike area? c. What is the probability at least one of 10 hurricanes reaches land outside the strike

area?

Lin01803_ch06_186-221.qxd 10/29/10 12:58 PM Page 217

Pa.

Va. N.C.

S.C.

N.J. Del. Md.

CUBA

Atlantic Ocean

0 300 mi

7080 W

Potential area of

movement

STORM CONTINUES NORTHWEST Position : 27.8 N, 71.4 W Movement: NNW at 8 mph Sustained winds: 105 mph As of 11 p.m. EDT Tuesday

Hurricane watch Tropical storm watch

69. A recent CBS News survey reported that 67 percent of adults felt the U.S. Treasury should continue making pennies.

Suppose we select a sample of 15 adults. a. How many of the 15 would we expect to indicate that the Treasury should continue

making pennies? What is the standard deviation? b. What is the likelihood that exactly 8 adults would indicate the Treasury should con-

tinue making pennies? c. What is the likelihood at least 8 adults would indicate the Treasury should continue

making pennies?

A penny saved is... Those who always take

pennies in change46%

56%

71%

18–29

30–44

45–64

65–over

40%

Data Set Exercises 70. Refer to the Real Estate data, which report information on homes sold in the Goodyear,

Arizona, area last year. a. Create a probability distribution for the number of bedrooms. Compute the mean and

the standard deviation of this distribution. b. Create a probability distribution for the number of bathrooms. Compute the mean and

the standard deviation of this distribution. 71. Refer to the Baseball 2009 data. Compute the mean number of home runs per game. To

do this, first find the mean number of home runs per team for 2009. Next, divide this value by 162 (a season comprises 162 games). Then multiply by 2, because there are two teams in each game. Use the Poisson distribution to estimate the number of home runs that will be hit in a game. Find the probability that: a. There are no home runs in a game. b. There are two home runs in a game. c. There are at least four home runs in a game.

218 Chapter 6

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Discrete Probability Distributions 219

Software Commands 1. The MegaStat commands to create the binomial

probability distribution on page 199 are: a. Select the Add-Ins tab in the top menu. On the

far left, select the MegaStat pull-down menu.

b. In the dialog box select the Binomial tab. The number of trials is 6, the probability of a success is .05. If you wish to see a graph, click on dis- play graph.

2. The Excel commands necessary to determine the binomial probability distribution on page 200 are: a. On a blank Excel worksheet, write the word Suc-

cess in cell A1 and the word Probability in B1. In cells A2 through A17, write the integers 0 to 15. Click on B2 as the active cell.

b. Click on the Formulas tab in the top menu, then, on the far left, select Insert Function fx.

c. In the first dialog box, select Statistical in the function category and BINOMDIST in the func- tion name category, then click OK.

d. In the second dialog box, enter the four items necessary to compute a binomial probability. 1. Enter 0 for the number of successes. 2. Enter 40 for the number of trials. 3. Enter .09 for the probability of a success. 4. Enter the word false or the number 0 for the

individual probabilities and click on OK.

5. Excel will compute the probability of 0 suc- cesses in 40 trials, with a .09 probability of success. The result, .02299618, is stored in cell B2.

e. To complete the probability distribution for suc- cesses of 1 through 15, double click on cell B2. The binomial function should appear. Replace the 0 to the right of the open parentheses with the cell reference A2.

f. Move the mouse to the lower right corner of cell B2 till a solid black � symbol appears, then click and hold and highlight the B column to cell B17. The probability of a success for the various val- ues of the random variable will appear.

3. The Excel commands necessary to determine the hypergeometric distribution on page 206 are: a. On a blank Excel worksheet, write the word

Union Members in cell A1 and the word Proba- bility in B1. In cells A2 through A7, enter the num- bers 0 through 5. Click on cell B2.

b. Click the Formulas tab in the top menu, then, on the far left, select Insert Function fx.

Click on probability, and Discrete Probability Distributions. Enter n, number of trials, and p, probability of occurrence, and click OK.

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c. In the first dialog box, select Statistical and HYPGEOMDIST, and then click OK.

d. In the second dialog box, enter the four items nec- essary to compute a hypergeometric probability. 1. Enter 0 for the number of successes. 2. Enter 5 for the number of trials. 3. Enter 40 for the number of successes in the

population. 4. Enter 50 for the size of the population and

click OK. 5. Excel will compute the probability of 0 suc-

cesses in 5 trials (.000118937) and store that result in cell F9.

e. To complete the probability distribution for suc- cesses of 1 through 5, double click on cell B2. The hypergeometric function should appear. Replace the 0 to the right of the open paren- theses with the cell reference A2.

f. Move the mouse to the lower right corner of cell F9 till a solid black � symbol appears, then click and hold and highlight the F column to cell F14. The probability of a success for the various out- comes will appear.

4. The Minitab commands to generate the Poisson distribution on page 209 are: a. Label column C1 as Successes and C2 as Prob-

ability. Enter the integers 0 though 5 in the first column.

b. Select Calc, then Probability Distributions, and Poisson.

c. In the dialog box, click on Probability, set the mean equal to .3, and select C1 as the Input col- umn. Designate C2 as Optional storage, and then click OK.

220 Chapter 6

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Discrete Probability Distributions 221

Chapter 6 Answers to Self-Review

6–1 a.

c. or 1.

6–2 a. It is discrete, because the value $0.80, $0.90, and $1.20 are clearly separated from each other. Also the sum of the probabilities is 1.00, and the outcomes are mutually exclusive.

6 6

The mean is 93 cents. c.

The variance is .02010, and the standard deviation is 14 cents.

6–3 a. It is reasonable because each employee either uses direct deposit or does not; employees are independent; the probability of using direct deposit is .80 for all; and we count the number using the service out of 7.

b. P(7) � 7C7 (.80) 7 (.20)0 � .2097

c. P(4) � 7C4 (.80) 4 (.20)3 � .1147

d. Answers are in agreement. 6–4 n � 4, � � .60

a. P(x � 2) � .346 b. P(x 2) � .526 c. P(x � 2) � 1 � .526 � .474

6–5

6–6 � � 4,000(.0002) � 0.8

P(1) � 0.81e�0.8

1! � .3595

� (56)(6) 792

� .424

P(3) � 8 C3 4C2 12C5

�

a 8!

3!5! b a

4! 2!2! b

12! 5!7!

Number of Spots Probability

Total 6 6

� 1.00

1 6

Pr ob

ab ili

Number of spots

1 6

1 2 3 4 5 6

x P(x) xP(x)

$ .80 .30 0.24 .90 .50 0.45

1.20 .20 0.24

0.93

x P (x) (x � �) (x � �)2P (x)

$0.80 .30 �0.13 .00507 0.90 .50 �0.03 .00045 1.20 .20 0.27 .01458

.02010

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7 Learning Objectives When you have completed this chapter, you will be able to:

LO1 List the characteristics of the uniform distribution.

LO2 Compute probabilities using the uniform distribution.

LO3 List the characteristics of the normal distribution.

LO4 Convert a normal distribution to the standard normal distribution.

LO5 Find the probability that a normally distributed random variable is between two values.

LO6 Find probabilities using the Empirical Rule.

LO7 Approximate the binomial distribution using the normal distribution.

LO8 Describe the character- istics and compute probabili- ties using the exponential distribution.

Continuous Probability Distributions

Cruise ships of the Royal Viking line report that 80 percent of their

rooms are occupied during September. For a cruise ship having 800

rooms, what is the probability that 665 or more are occupied in

September? (See Exercise 60 and LO7.)

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Continuous Probability Distributions 223

7.1 Introduction Chapter 6 began our study of probability distributions. We consider three discrete probability distributions: binomial, hypergeometric, and Poisson. These distributions are based on discrete random variables, which can assume only clearly separated values. For example, we select for study 10 small businesses that began operations during the year 2000. The number still operating in 2011 can be 0, 1, 2, . . . , 10. There cannot be 3.7, 12, or �7 still operating in 2011. In this example, only cer- tain outcomes are possible and these outcomes are represented by clearly sepa- rated values. In addition, the result is usually found by counting the number of suc- cesses. We count the number of the businesses in the study that are still in operation in 2011.

In this chapter, we continue our study of probability distributions by examining continuous probability distributions. A continuous probability distribution usually results from measuring something, such as the distance from the dormitory to the classroom, the weight of an individual, or the amount of bonus earned by CEOs. Suppose we select five students and find the distance, in miles, they travel to attend class as 12.2, 8.9, 6.7, 3.6, and 14.6. When examining a continuous distribution we are usually interested in information such as the percent of students who travel less than 10 miles or the percent who travel more than 8 miles. In other words, for a continuous distribution we may wish to know the percent of observations that occur within a certain range. It is important to realize that a continuous random variable has an infinite number of values within a particular range. So you think of the probability a variable will have a value within a specified range, rather than the probability for a specific value.

We consider three families of continuous probability distributions, the uniform probability distribution, the normal probability distribution, and the exponential probability distribution.

7.2 The Family of Uniform Probability Distributions The uniform probability distribution is perhaps the simplest distribution for a con- tinuous random variable. This distribution is rectangular in shape and is defined by minimum and maximum values. Here are some examples that follow a uniform distribution.

• The time to fly via a commercial airliner from Orlando, Florida, to Atlanta, Georgia, ranges from 60 minutes to 120 minutes. The random variable is the flight time within this interval. The variable of interest, flight time in minutes, is continuous in the interval from 60 minutes to 120 minutes.

• Volunteers at the Grand Strand Public Library prepare fed- eral income tax forms. The time to prepare form 1040-EZ follows a uniform distribution over the interval between 10 minutes and 30 minutes. The random variable is the number of minutes to complete the form, and it can assume any value between 10 and 30.

A uniform distribution is shown in Chart 7–1. The distribution’s shape is rectan- gular and has a minimum value of a and a maximum of b. Also notice in Chart 7–1 the height of the distribution is constant or uniform for all values between a and b.

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224 Chapter 7

P (x )

1 b � a

The mean of a uniform distribution is located in the middle of the interval between the minimum and maximum values. It is computed as:

MEAN OF THE UNIFORM DISTRIBUTION [7–1]

The standard deviation describes the dispersion of a distribution. In the uniform distribution, the standard deviation is also related to the interval between the maxi- mum and minimum values.

STANDARD DEVIATION [7–2]OF THE UNIFORM DISTRIBUTION

The equation for the uniform probability distribution is:

UNIFORM DISTRIBUTION if and 0 elsewhere [7–3]

As described in Chapter 6, probability distributions are useful for making prob- ability statements concerning the values of a random variable. For distributions describing a continuous random variable, areas within the distribution represent probabilities. In the uniform distribution, its rectangular shape allows us to apply the area formula for a rectangle. Recall that we find the area of a rectangle by multi- plying its length by its height. For the uniform distribution, the height of the rectan- gle is P(x), which is 1/(b � a). The length or base of the distribution is b � a. So if we multiply the height of the distribution by its entire range to find the area, the result is always 1.00. To put it another way, the total area within a continuous prob- ability distribution is equal to 1.00. In general

So if a uniform distribution ranges from 10 to 15, the height is 0.20, found by 1/(15 � 10). The base is 5, found by 15 � 10. The total area is:

An example will illustrate the features of a uniform distribution and how we calcu- late probabilities using it.

Area � (height)(base) � 1

(15 � 10) (15 � 10) � 1.00

Area � (height)(base) � 1

(b � a) (b � a) � 1.00

a � x � bP(x) � 1

b � a

� � B

(b � a)2

� � a � b

CHART 7–1 A Continuous Uniform Distribution

LO1 List the characteristics of the uniform distribution.

The total area under the curve is always 1.

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Continuous Probability Distributions 225

Example

Solution

Southwest Arizona State University provides bus service to students while they are on campus. A bus arrives at the North Main Street and College Drive stop every 30 minutes between 6 A.M. and 11 P.M. during weekdays. Students arrive at the bus stop at random times. The time that a student waits is uniformly distributed from 0 to 30 minutes.

1. Draw a graph of this distribution. 2. Show that the area of this uniform distribution is 1.00. 3. How long will a student “typically” have to wait for a bus? In other words, what

is the mean waiting time? What is the standard deviation of the waiting times? 4. What is the probability a student will wait more than 25 minutes? 5. What is the probability a student will wait between 10 and 20 minutes?

In this case, the random variable is the length of time a student must wait. Time is measured on a continuous scale, and the wait times may range from 0 minutes up to 30 minutes.

1. The graph of the uniform distribution is shown in Chart 7–2. The horizontal line is drawn at a height of .0333, found by 1/(30 � 0). The range of this distribu- tion is 30 minutes.

CHART 7–2 Uniform Probability Distribution of Student Waiting Times

2. The times students must wait for the bus is uniform over the interval from 0 minutes to 30 minutes, so in this case a is 0 and b is 30.

3. To find the mean, we use formula (7–1).

The mean of the distribution is 15 minutes, so the typical wait time for bus ser- vice is 15 minutes.

To find the standard deviation of the wait times, we use formula (7–2).

The standard deviation of the distribution is 8.66 minutes. This measures the variation in the student wait times.

4. The area within the distribution for the interval 25 to 30 represents this partic- ular probability. From the area formula:

P(25 6 wait time 6 30) � (height)(base) � 1

(30 � 0) (5) � .1667

� � B

(b � a)2

12 � B

(30 � 0)2

12 � 8.66

� � a � b

2 �

0 � 30 2

� 15

Area � (height)(base) � 1

(30 � 0) (30 � 0) � 1.00

.060

.0333

0 10 40

Length of Wait (minutes)

Pr ob

ab ili

3020

LO2 Compute probabilities using the uniform distribution.

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So the probability a student waits between 25 and 30 minutes is .1667. This conclusion is illustrated by the following graph.

226 Chapter 7

P (x )

.0333

10 20

Area � .3333

μ � 15 30

Self-Review 7–1 Australian sheepdogs have a relatively short life. The length of their life follows a uniform distribution between 8 and 14 years. (a) Draw this uniform distribution. What are the height and base values? (b) Show the total area under the curve is 1.00. (c) Calculate the mean and the standard deviation of this distribution. (d) What is the probability a particular dog lives between 10 and 14 years? (e) What is the probability a dog will live less than 9 years?

Exercises 1. A uniform distribution is defined over the interval from 6 to 10.

a. What are the values for a and b? b. What is the mean of this uniform distribution? c. What is the standard deviation? d. Show that the total area is 1.00. e. Find the probability of a value more than 7. f. Find the probability of a value between 7 and 9.

2. A uniform distribution is defined over the interval from 2 to 5. a. What are the values for a and b? b. What is the mean of this uniform distribution? c. What is the standard deviation? d. Show that the total area is 1.00. e. Find the probability of a value more than 2.6. f. Find the probability of a value between 2.9 and 3.7.

3. The closing price of Schnur Sporting Goods Inc. common stock is uniformly distributed between $20 and 30 per share. What is the probability that the stock price will be: a. More than $27? b. Less than or equal to $24?

5. The area within the distribution for the interval 10 to 20 represents the probability.

We can illustrate this probability as follows.

P(10 6 wait time 6 20) � (height)(base) � 1

(30 � 0) (10) � .3333

P (x )

.0333

10 20 25

Area � .1667

μ � 15 30

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Continuous Probability Distributions 227

4. According to the Insurance Institute of America, a family of four spends between $400 and $3,800 per year on all types of insurance. Suppose the money spent is uniformly distributed between these amounts. a. What is the mean amount spent on insurance? b. What is the standard deviation of the amount spent? c. If we select a family at random, what is the probability they spend less than $2,000

per year on insurance per year? d. What is the probability a family spends more than $3,000 per year?

5. The April rainfall in Flagstaff, Arizona, follows a uniform distribution between 0.5 and 3.00 inches. a. What are the values for a and b? b. What is the mean amount of rainfall for the month? What is the standard deviation? c. What is the probability of less than an inch of rain for the month? d. What is the probability of exactly 1.00 inch of rain? e. What is the probability of more than 1.50 inches of rain for the month?

6. Customers experiencing technical difficulty with their Internet cable hookup may call an 800 number for technical support. It takes the technician between 30 seconds to 10 minutes to resolve the problem. The distribution of this support time follows the uniform distribution. a. What are the values for a and b in minutes? b. What is the mean time to resolve the problem? What is the standard deviation of the time? c. What percent of the problems take more than 5 minutes to resolve? d. Suppose we wish to find the middle 50 percent of the problem-solving times. What

are the end points of these two times?

7.3 The Family of Normal Probability Distributions Next we consider the normal probability distribution. Unlike the uniform distribution [see formula (7–3)] the normal probability distribution has a very complex formula.

NORMAL PROBABILITY DISTRIBUTION [7–4]

However, do not be bothered by how complex this formula looks. You are already familiar with many of the values. The symbols � and � refer to the mean and the standard deviation, as usual. The Greek symbol � is a natural mathematical con- stant and its value is approximately 22/7 or 3.1416. The letter e is also a mathe- matical constant. It is the base of the natural log system and is equal to 2.718. X is the value of a continuous random variable. So a normal distribution is based on— that is, it is defined by—its mean and standard deviation.

You will not need to make any calculations from formula (7–4). Instead you will be using a table, which is given as Appendix B.1, to look up the various probabilities.

The normal probability distribution has the following major characteristics:

• It is bell-shaped and has a single peak at the center of the distribution. The arithmetic mean, median, and mode are equal and located in the center of the distribution. The total area under the curve is 1.00. Half the area under the nor- mal curve is to the right of this center point and the other half to the left of it.

• It is symmetrical about the mean. If we cut the normal curve vertically at the center value, the two halves will be mirror images.

• It falls off smoothly in either direction from the central value. That is, the distri- bution is asymptotic: The curve gets closer and closer to the X-axis but never actually touches it. To put it another way, the tails of the curve extend indefi- nitely in both directions.

• The location of a normal distribution is determined by the mean, �. The disper- sion or spread of the distribution is determined by the standard deviation, �.

These characteristics are shown graphically in Chart 7–3.

� c (X � �)2

2� 2 dP(x) �

1 �12�

LO3 List the characteristics of the normal distribution.

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228 Chapter 7

Chart 7–5 shows the distribution of box weights of three different cereals. The weights follow a normal distribution with different means but identical standard deviations.

Unequal means, equal standard deviations

CHART 7–5 Normal Probability Distributions Having Different Means but Equal Standard Deviations

Sugar Yummies

σ = 1.6 grams

Alphabet Gems

σ = 1.6 grams

Weight Droppers

σ = 1.6 grams

� 283

grams

� 301

grams

� 321

grams

CHART 7–3 Characteristics of a Normal Distribution

Normal curve is symmetrical Two halves identical

Mean, median, and mode are

equal

Theoretically, curve extends to – ∞

Theoretically, curve extends to + ∞

Tail Tail

There is not just one normal probability distribution, but rather a “family” of them. For example, in Chart 7–4 the probability distributions of length of employee service in three different plants can be compared. In the Camden plant, the mean is 20 years and the standard deviation is 3.1 years. There is another normal prob- ability distribution for the length of service in the Dunkirk plant, where � � 20 years and � � 3.9 years. In the Elmira plant, � � 20 years and � � 5.0 years. Note that the means are the same but the standard deviations are different.

CHART 7–4 Normal Probability Distributions with Equal Means but Different Standard Deviations

Equal means, unequal standard deviations

0 4 7 10 13 16 19 22 25 28 μ � 20 years of service

σ � 3.1 years, Camden plant

σ � 3.9 years, Dunkirk plant

σ � 5.0 years, Elmira plant

37 403431

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Continuous Probability Distributions 229

Finally, Chart 7–6 shows three normal distributions having different means and standard deviations. They show the distribution of tensile strengths, measured in pounds per square inch (psi), for three types of cables.

Unequal means, unequal standard deviations

In Chapter 6, recall that discrete probability distributions show the specific like- lihood a discrete value will occur. For example, in Section 6.5 on page 196, the binomial distribution is used to calculate the probability that none of the five flights arriving at the Bradford Pennsylvania Regional Airport would be late.

With a continuous probability distribution, areas below the curve define proba- bilities. The total area under the normal curve is 1.0. This accounts for all possible outcomes. Because a normal probability distribution is symmetric, the area under the curve to the left of the mean is 0.5, and the area under the curve to the right of the mean is 0.5. Apply this to the distribution of Sugar Yummies in Chart 7–5. It is nor- mally distributed with a mean of 283 grams. Therefore, the probability of filling a box with more than 283 grams is 0.5 and the probability of filling a box with less than 283 grams is 0.5. We can also determine the probability that a box weighs between 280 and 286 grams. However, to determine this probability we need to know about the standard normal probability distribution.

7.4 The Standard Normal Probability Distribution The number of normal distributions is unlimited, each having a different mean (�), standard deviation (�), or both. While it is possible to provide probability tables for discrete distributions such as the binomial and the Poisson, providing tables for the infinite number of normal distributions is impossible. Fortunately, one member of the family can be used to determine the probabilities for all normal probability distribu- tions. It is called the standard normal probability distribution, and it is unique because it has a mean of 0 and a standard deviation of 1.

Any normal probability distribution can be converted into a standard normal probability distribution by subtracting the mean from each observation and dividing this difference by the standard deviation. The results are called z values or z scores.

CHART 7–6 Normal Probability Distributions with Different Means and Standard Deviations

μ 2,000

psi

σ � 41 psi σ � 52 psi

σ � 26 psi

μ 2,107

psi

μ 2,186

psi

z VALUE The signed distance between a selected value, designated X, and the mean, �, divided by the standard deviation, �.

So, a z value is the distance from the mean, measured in units of the standard deviation.

In terms of a formula:

STANDARD NORMAL VALUE [7–5]z � X � �

�

LO4 Convert a normal distribution to the standard normal distribution.

There is only one standard normal distribution. It has a mean of 0 and a standard deviation of 1.

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230 Chapter 7

.4332

285.4 1.50

� = 283 Grams z Values 0

TABLE 7–1 Areas under the Normal Curve

To explain, suppose we wish to compute the probability that boxes of Sugar Yummies weigh between 283 and 285.4 grams. From Chart 7–5, we know that the box weight of Sugar Yummies follows the normal distribution with a mean of 283 grams and a standard deviation of 1.6 grams. We want to know the probability or area under the curve between the mean, 283 grams, and 285.4 grams. We can also express this problem using probability notation, similar to the style used in the previous chapter: P(283 weight 285.4). To find the probability, it is necessary to convert both 283 grams and 285.4 grams to z values using formula (7–5). The z value corresponding to 283 is 0, found by (283 � 283)/1.6. The z value corresponding to 285.4 is 1.50 found by (285.4 � 283)/1.6. Next we go to the table in Appendix B.1. A portion of the table is repeated as Table 7–1. Go down the column of the table headed by the letter z to 1.5. Then move horizontally to the right and read the probability under the column headed 0.00. It is 0.4332. This means the area under the curve between 0.00 and 1.50 is 0.4332. This is the probability that a randomly selected box of Sugar Yummies will weigh between 283 and 285.4 grams. This is illustrated in the following graph.

z 0.00 0.01 0.02 0.03 0.04 0.05 . . .

1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744

where: X is the value of any particular observation or measurement. � is the mean of the distribution. � is the standard deviation of the distribution.

As noted in the preceding definition, a z value expresses the distance or difference between a particular value of X and the arithmetic mean in units of the standard deviation. Once the normally distributed observations are standardized, the z values are normally distributed with a mean of 0 and a standard deviation of 1. So the z distribution has all the characteristics of any normal probability distribution. These characteristics are listed on page 227. The table in Appendix B.1 (also on the inside back cover) lists the probabilities for the standard normal probability distribution. A small portion of this table follows.

Statistics in Action

An individual’s skills depend on a combi- nation of many hereditary and envi- ronmental factors, each having about the same amount of weight or influence on the skills. Thus, much like a bino- mial distribution with a large number of trials, many skills and attributes follow the normal distribu- tion. For example, scores on the Scholastic Aptitude Test (SAT) are nor- mally distributed with a mean of 1000 and a standard devi- ation of 140.

LO5 Find the probability that a normally distributed random variable is between two values.

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Continuous Probability Distributions 231

Applications of the Standard Normal Distribution What is the area under the curve between the mean and X for the z values in Table 7–2 below? Check your answers against those given. You will need to use Appendix B.1 or the table located on the inside back cover of the text.

Selected z Values Area

2.84 .4977 1.00 .3413 0.49 .1879

Now we will compute the z value given the population mean, �, the population standard deviation, �, and a selected X.

Self-Review 7–2 Using the information in the preceding example (� � $1,000, � � $100), convert: (a) The weekly income of $1,225 to a z value. (b) The weekly income of $775 to a z value.

The Empirical Rule Before examining further applications of the standard normal probability distribution, we will consider three areas under the normal curve that will be used extensively in the fol- lowing chapters. These facts were called the Empirical Rule in Chapter 3 (see page 86).

1. About 68 percent of the area under the normal curve is within one standard deviation of the mean. This can be written as � � 1�.

2. About 95 percent of the area under the normal curve is within two standard deviations of the mean, written � � 2�.

3. Practically all of the area under the normal curve is within three standard devi- ations of the mean, written � � 3�.

LO6 Find probabilities using the Empirical Rule.

TABLE 7–2 Areas for Selected Values of z

Example

Solution

The weekly incomes of shift foremen in the glass industry follow the normal prob- ability distribution with a mean of $1,000 and a standard deviation of $100. What is the z value for the income, let’s call it X, of a foreman who earns $1,100 per week? For a foreman who earns $900 per week?

Using formula (7–5), the z values for the two X values ($1,100 and $900) are:

The z of 1.00 indicates that a weekly income of $1,100 is one standard devia- tion above the mean, and a z of �1.00 shows that a $900 income is one standard deviation below the mean. Note that both incomes ($1,100 and $900) are the same distance ($100) from the mean.

� �1.00 � 1.00

� $900 � $1,000

$100 �

$1,100 � $1,000 $100

z � X � �

� z �

X � � �

For X � $900: For X � $1,100:

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232 Chapter 7

Transforming measurements to standard normal deviates changes the scale. The conversions are also shown in the graph. For example, � � 1� is converted to a z value of 1.00. Likewise, � � 2� is transformed to a z value of �2.00. Note that the center of the z distribution is zero, indicating no deviation from the mean, �.

converts to μ

– 3 – 2 – 1 1 2 30

μ – 3σ μ – 2σ μ – 1σ μ + 3σμ + 2σμ + 1σ

68%

95%

Practically all

Scale of z

Scale of X

This information is summarized in the following graph.

Example

Solution

As part of its quality assurance program, the Autolite Battery Company conducts tests on battery life. For a particular D-cell alkaline battery, the mean life is 19 hours. The useful life of the battery follows a normal distribution with a standard deviation of 1.2 hours. Answer the following questions.

1. About 68 percent of the batteries failed between what two values? 2. About 95 percent of the batteries failed between what two values? 3. Virtually all of the batteries failed between what two values?

We can use the results of the Empirical Rule to answer these questions.

1. About 68 percent of the batteries will fail between 17.8 and 20.2 hours, found by 19.0 1(1.2) hours.

2. About 95 percent of the batteries will fail between 16.6 and 21.4 hours, found by 19.0 2(1.2) hours.

3. Practically all failed between 15.4 and 22.6 hours, found by 19.0 3(1.2) hours.

This information is summarized on the following chart.

μμ – 3σ μ – 2σ μ – 1σ μ + 3σμ + 2σμ + 1σ 15.4 16.6 17.8 20.2 21.4 22.619.0

68% 95%

Practically all

Scale of hours

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Continuous Probability Distributions 233

Exercises 7. Explain what is meant by this statement: “There is not just one normal probability distri-

bution but a ‘family’ of them.” 8. List the major characteristics of a normal probability distribution. 9. The mean of a normal probability distribution is 500; the standard deviation is 10.

a. About 68 percent of the observations lie between what two values? b. About 95 percent of the observations lie between what two values? c. Practically all of the observations lie between what two values?

10. The mean of a normal probability distribution is 60; the standard deviation is 5. a. About what percent of the observations lie between 55 and 65? b. About what percent of the observations lie between 50 and 70? c. About what percent of the observations lie between 45 and 75?

11. The Kamp family has twins, Rob and Rachel. Both Rob and Rachel graduated from college 2 years ago, and each is now earning $50,000 per year. Rachel works in the retail industry, where the mean salary for executives with less than 5 years’ experi- ence is $35,000 with a standard deviation of $8,000. Rob is an engineer. The mean salary for engineers with less than 5 years’ experience is $60,000 with a standard devi- ation of $5,000. Compute the z values for both Rob and Rachel and comment on your findings.

12. A recent article in the Cincinnati Enquirer reported that the mean labor cost to repair a heat pump is $90 with a standard deviation of $22. Monte’s Plumbing and Heating Service completed repairs on two heat pumps this morning. The labor cost for the first was $75 and it was $100 for the second. Assume the distribution of labor costs fol- lows the normal probability distribution. Compute z values for each and comment on your findings.

Finding Areas under the Normal Curve The next application of the standard normal distribution involves finding the area in a normal distribution between the mean and a selected value, which we identify as X. The following example will illustrate the details.

Self-Review 7–3 The distribution of the annual incomes of a group of middle-management employees at Compton Plastics approximates a normal distribution with a mean of $47,200 and a stan- dard deviation of $800. (a) About 68 percent of the incomes lie between what two amounts? (b) About 95 percent of the incomes lie between what two amounts? (c) Virtually all of the incomes lie between what two amounts? (d) What are the median and the modal incomes? (e) Is the distribution of incomes symmetrical?

Example

Solution

Recall in an earlier example (see page 231) we reported that the mean weekly income of a shift foreman in the glass industry is normally distributed with a mean of $1,000 and a standard deviation of $100. That is, � � $1,000 and � � $100. What is the likelihood of selecting a foreman whose weekly income is between $1,000 and $1,100? We write this question in probability notation as: P($1,000 weekly income $1,100).

We have already converted $1,100 to a z value of 1.00 using formula (7–5). To repeat:

z � X � �

� �

$1,100 � $1,000 $100

� 1.00

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234 Chapter 7

In the example just completed, we are interested in the probability between the mean and a given value. Let’s change the question. Instead of wanting to know the probability of selecting at random a foreman who earned between $1,000 and $1,100, suppose we wanted the probability of selecting a foreman who earned less than $1,100. In probability notation, we write this statement as P(weekly income $1,100). The method of solution is the same. We find the probability of selecting a foreman who earns between $1,000, the mean, and $1,100. This probability is .3413. Next, recall that half the area, or probability, is above the mean and half is below. So the probability of selecting a foreman earning less than $1,000 is .5000. Finally, we add the two probabilities, so .3413 � .5000 � .8413. About 84 percent of the foremen in the glass industry earn less than $1,100 per week. See the following diagram.

Scale of z1.0

$1,000 Scale of dollars

$1,100

.3413

The probability associated with a z of 1.00 is available in Appendix B.1. A portion of Appendix B.1 follows. To locate the probability, go down the left column to 1.0, and then move horizontally to the column headed .00. The value is .3413.

z 0.00 0.01 0.02 � � � �

� � � �

0.7 .2580 .2611 .2642 0.8 .2881 .2910 .2939 0.9 .3159 .3186 .3212 1.0 .3413 .3438 .3461 1.1 .3643 .3665 .3686

� � � �

The area under the normal curve between $1,000 and $1,100 is .3413. We could also say 34.13 percent of the shift foremen in the glass industry earn between $1,000 and $1,100 weekly, or the likelihood of selecting a foreman and finding his or her income is between $1,000 and $1,100 is .3413.

This information is summarized in the following diagram.

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Continuous Probability Distributions 235

Scale of z1.0

$1,000 Scale of dollars

$1,100

.3413.5000

Excel will calculate this probability. The necessary commands are in the Soft- ware Commands section at the end of the chapter. The answer is .8413, the same as we calculated.

Statistics in Action

Many processes, such as filling soda bottles and canning fruit, are normally distributed. Manufacturers must guard against both over- and underfill- ing. If they put too much in the can or bottle, they are giving away their product. If they put too little in, the customer may feel cheated and the government may question the label description. “Control charts,” with limits drawn three standard deviations above and below the mean, are routinely used to monitor this type of production process.

Example

Solution

Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribu- tion, with a mean of $1,000 and a standard deviation of $100. What is the proba- bility of selecting a shift foreman in the glass industry whose income is:

1. Between $790 and $1,000? 2. Less than $790?

We begin by finding the z value corresponding to a weekly income of $790. From formula (7–5):

See Appendix B.1. Move down the left margin to the row 2.1 and across that row to the column headed 0.00. The value is .4821. So the area under the standard nor- mal curve corresponding to a z value of 2.10 is .4821. However, because the nor- mal distribution is symmetric, the area between 0 and a negative z value is the same as that between 0 and the corresponding positive z value. The likelihood of finding a foreman earning between $790 and $1,000 is .4821. In probability notation, we write P($790 weekly income $1000) � .4821.

z � X � �

s �

$790 � $1,000 $100

� �2.10

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236 Chapter 7

Exercises 13. A normal population has a mean of 20.0 and a standard deviation of 4.0.

a. Compute the z value associated with 25.0. b. What proportion of the population is between 20.0 and 25.0? c. What proportion of the population is less than 18.0?

14. A normal population has a mean of 12.2 and a standard deviation of 2.5. a. Compute the z value associated with 14.3. b. What proportion of the population is between 12.2 and 14.3? c. What proportion of the population is less than 10.0?

Self-Review 7–4 The temperature of coffee sold at the Coffee Bean Cafe follows the normal probability distribution, with a mean of 150 degrees. The standard deviation of this distribution is 5 degrees. (a) What is the probability that the coffee temperature is between 150 degrees and

154 degrees? (b) What is the probability that the coffee temperature is more than 164 degrees?

.5000

Scale of z0 Scale of dollars$1,000

–2.10 $790

.0179

.4821

z 0.00 0.01 0.02

. � � �

. � � � 2.0 .4772 .4778 .4783 2.1 .4821 .4826 .4830 2.2 .4861 .4864 .4868 2.3 .4893 .4896 .4898 . � � � . � � � . � � �

The mean divides the normal curve into two identical halves. The area under the half to the left of the mean is .5000, and the area to the right is also .5000. Because the area under the curve between $790 and $1,000 is .4821, the area below $790 is .0179, found by .5000 � .4821. In probability notation, we write P(weekly income $790) � .0179.

This means that 48.21 percent of the foremen have weekly incomes between $790 and $1,000. Further, we can anticipate that 1.79 percent earn less than $790 per week. This information is summarized in the following diagram.

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Continuous Probability Distributions 237

15. A recent study of the hourly wages of maintenance crew members for major airlines showed that the mean hourly salary was $20.50, with a standard deviation of $3.50. Assume the distribution of hourly wages follows the normal probability distribution. If we select a crew member at random, what is the probability the crew member earns: a. Between $20.50 and $24.00 per hour? b. More than $24.00 per hour? c. Less than $19.00 per hour?

16. The mean of a normal probability distribution is 400 pounds. The standard deviation is 10 pounds. a. What is the area between 415 pounds and the mean of 400 pounds? b. What is the area between the mean and 395 pounds? c. What is the probability of selecting a value at random and discovering that it has a

value of less than 395 pounds?

Another application of the normal distribution involves combining two areas, or probabilities. One of the areas is to the right of the mean and the other to the left.

Another application of the normal distribution involves determining the area between values on the same side of the mean.

Example

Solution

Recall the distribution of weekly incomes of shift foremen in the glass industry. The weekly incomes follow the normal probability distribution, with a mean of $1,000 and a standard deviation of $100. What is the area under this normal curve between $840 and $1,200?

The problem can be divided into two parts. For the area between $840 and the mean of $1,000:

For the area between the mean of $1,000 and $1,200:

The area under the curve for a z of �1.60 is .4452 (from Appendix B.1). The area under the curve for a z of 2.00 is .4772. Adding the two areas: .4452 � .4772 � .9224. Thus, the probability of selecting an income between $840 and $1,200 is .9224. In probability notation, we write P($840 weekly income $1,200) � .4452 � .4772 � .9224. To summarize, 92.24 percent of the foremen have weekly incomes between $840 and $1,200. This is shown in a diagram:

z � $1,200 � $1,000

$100 �

$200 $100

� 2.00

z � $840 � $1,000

$100 �

�$160 $100

� �1.60

Scale of z2.00

.4772.4452

�1.6 Scale of dollars$1,200$1,000$840

What is this probability?

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238 Chapter 7

In brief, there are four situations for finding the area under the standard normal probability distribution.

1. To find the area between 0 and z or (�z), look up the probability directly in the table.

2. To find the area beyond z or (�z), locate the probability of z in the table and subtract that probability from .5000.

3. To find the area between two points on different sides of the mean, determine the z values and add the corresponding probabilities.

4. To find the area between two points on the same side of the mean, determine the z values and subtract the smaller probability from the larger.

Self-Review 7–5 Refer to Self-Review 7–4. The temperature of coffee sold at the Coffee Bean Cafe follows the normal probability distribution with a mean of 150 degrees. The standard deviation of this distribution is 5 degrees. (a) What is the probability the coffee temperature is between 146 degrees and 156 degrees? (b) What is the probability the coffee temperature is more than 156 but less than 162 degrees?

Example

Solution

Returning to the weekly income distribution of shift foremen in the glass industry (� � $1,000, � � $100), what is the area under the normal curve between $1,150 and $1,250?

The situation is again separated into two parts, and formula (7–5) is used. First, we find the z value associated with a weekly salary of $1,250:

Next we find the z value for a weekly salary of $1,150:

From Appendix B.1, the area associated with a z value of 2.50 is .4938. So the probability of a weekly salary between $1,000 and $1,250 is .4938. Similarly, the area associated with a z value of 1.50 is .4332, so the probability of a weekly salary between $1,000 and $1,150 is .4332. The probability of a weekly salary between $1,150 and $1,250 is found by subtracting the area associated with a z value of 1.50 (.4332) from that associated with a z of 2.50 (.4938). Thus, the probability of a weekly salary between $1,150 and $1,250 is .0606. In probability notation, we write P($1,150 weekly income $1,250) � .4938 � .4332 � .0606.

z � $1,150 � $1,000

$100 � 1.50

z � $1,250 � $1,000

$100 � 2.50

Scale of incomes Scale of z

$1,250 2.50

$1,000 0

.0606

$1,150 1.50

.4332

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Continuous Probability Distributions 239

Exercises 17. A normal distribution has a mean of 50 and a standard deviation of 4.

a. Compute the probability of a value between 44.0 and 55.0. b. Compute the probability of a value greater than 55.0. c. Compute the probability of a value between 52.0 and 55.0.

18. A normal population has a mean of 80.0 and a standard deviation of 14.0. a. Compute the probability of a value between 75.0 and 90.0. b. Compute the probability of a value 75.0 or less. c. Compute the probability of a value between 55.0 and 70.0.

19. According to the Internal Revenue Service, the mean tax refund for the year 2007 was $2,708. Assume the standard deviation is $650 and that the amounts refunded follow a normal probability distribution. a. What percent of the refunds are more than $3,000? b. What percent of the refunds are more than $3,000 but less than $3,500? c. What percent of the refunds are more than $2,500 but less than $3,500?

20. The number of viewers of American Idol has a mean of 29 million with a standard devi- ation of 5 million. Assume this distribution follows a normal distribution. What is the prob- ability that next week’s show will: a. Have between 30 and 34 million viewers? b. Have at least 23 million viewers? c. Exceed 40 million viewers?

21. WNAE, an all-news AM station, finds that the distribution of the lengths of time listeners are tuned to the station follows the normal distribution. The mean of the distribution is 15.0 minutes and the standard deviation is 3.5 minutes. What is the probability that a particular listener will tune in: a. More than 20 minutes? b. For 20 minutes or less? c. Between 10 and 12 minutes?

22. Among U.S. cities with a population of more than 250,000, the mean one-way commute time to work is 24.3 minutes. The longest one-way travel time is New York City, where the mean time is 38.3 minutes. Assume the distribution of travel times in New York City follows the normal probability distribution and the standard deviation is 7.5 minutes. a. What percent of the New York City commutes are for less than 30 minutes? b. What percent are between 30 and 35 minutes? c. What percent are between 30 and 40 minutes?

Previous examples require finding the percent of the observations located between two observations or the percent of the observations above, or below, a particular obser- vation X. A further application of the normal distribution involves finding the value of the observation X when the percent above or below the observation is given.

Example

Solution

The facets of this case are shown in the following diagram, where X represents the minimum guaran- teed mileage.

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240 Chapter 7

Excel will also find the mileage value. See the following output. The necessary com- mands are given in the Software Commands section at the end of the chapter.

Scale of milesμ 67,900

.5000

4% or .0400

X ?

Tire replaced if it does not get this amount of mileage

.4600

Inserting these values in formula (7–5) for z gives:

Notice that there are two unknowns, z and X. To find X, we first find z, and then solve for X. Notice the area under the normal curve to the left of � is .5000. The area between � and X is .4600, found by .5000 � .0400. Now refer to Appendix B.1. Search the body of the table for the area closest to .4600. The closest area is .4599. Move to the margins from this value and read the z value of 1.75. Because the value is to the left of the mean, it is actually �1.75. These steps are illustrated in Table 7–3.

TABLE 7–3 Selected Areas under the Normal Curve

z … .03 .04 .05 .06

� � � � �

1.5 .4370 .4382 .4394 .4406 1.6 .4484 .4495 .4505 .4515 1.7 .4582 .4591 .4599 .4608 1.8 .4664 .4671 .4678 .4686

Knowing that the distance between � and X is �1.75� or z � �1.75, we can now solve for X (the minimum guaranteed mileage):

So Layton can advertise that it will replace for free any tire that wears out before it reaches 64,312 miles, and the company will know that only 4 percent of the tires will be replaced under this plan.

X � 67,900 � 1.75(2,050) � 64,312

�1.75(2,050) � X � 67,900

�1.75 � X � 67,900

2,050

z � X � 67,900

2,050

z � X � �

� �

X � 67,900 2,050

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Continuous Probability Distributions 241

Exercises 23. A normal distribution has a mean of 50 and a standard deviation of 4. Determine the

value below which 95 percent of the observations will occur. 24. A normal distribution has a mean of 80 and a standard deviation of 14. Determine the

value above which 80 percent of the values will occur. 25. Assume that the mean hourly cost to operate a commercial airplane follows the normal

distribution with a mean of $2,100 per hour and a standard deviation of $250. What is the operating cost for the lowest 3 percent of the airplanes?

26. The SAT Reasoning Test (formerly called the Scholastic Aptitude Test) is perhaps the most widely used standardized test for college admissions in the United States. Scores are based on a normal distribution with a mean of 1500 and a standard deviation of 300. Clinton College would like to offer an honors scholarship to students who score in the top 10 percent of this test. What is the minimum score that qualifies for the scholarship?

27. According to media research, the typical American listened to 195 hours of music in the last year. This is down from 290 hours four years earlier. Dick Trythall is a big country and western music fan. He listens to music while working around the house, reading, and riding in his truck. Assume the number of hours spent listening to music follows a normal probability distribution with a standard deviation of 8.5 hours. a. If Dick is in the top 1 percent in terms of listening time, how many hours does he listen

per year? b. Assume that the distribution of times four years earlier also follows the normal prob-

ability distribution with a standard deviation of 8.5 hours. How many hours did the 1 percent who listen to the least music actually listen?

28. For the most recent year available, the mean annual cost to attend a private university in the United States was $26,889. Assume the distribution of annual costs follows the normal probability distribution and the standard deviation is $4,500. Ninety-five percent of all students at private universities pay less than what amount?

29. In economic theory, a “hurdle rate” is the minimum return that a person requires before they will make an investment. A research report says that annual returns from a specific class of common equities are distributed according to a normal distribution with a mean of 12 percent and a standard deviation of 18 percent. A stock screener would like to identify a hurdle rate such that only 1 in 20 equities is above that value. Where should the hurdle rate be set?

Self-Review 7–6 An analysis of the final test scores for Introduction to Business reveals the scores follow the normal probability distribution. The mean of the distribution is 75 and the standard deviation is 8. The professor wants to award an A to students whose score is in the high- est 10 percent. What is the dividing point for those students who earn an A and those earning a B?

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242 Chapter 7

30. The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,200. The distribution of pages printed per cartridge closely follows the normal probability distribution and the standard deviation is 820 pages. The manufacturer wants to provide guidelines to potential customers as to how long they can expect a cartridge to last. How many pages should the manufacturer adver- tise for each cartridge if it wants to be correct 99 percent of the time?

7.5 The Normal Approximation to the Binomial Chapter 6 describes the binomial probability distribution, which is a discrete distri- bution. The table of binomial probabilities in Appendix B.9 goes successively from an n of 1 to an n of 15. If a problem involved taking a sample of 60, generating a binomial distribution for that large a number would be very time consuming. A more efficient approach is to apply the normal approximation to the binomial.

We can use the normal distribution (a continuous distribution) as a substitute for a binomial distribution (a discrete distribution) for large values of n because, as n increases, a binomial distribution gets closer and closer to a normal distribution. Chart 7–7 depicts the change in the shape of a binomial distribution with � � .50 from an n of 1, to an n of 3, to an n of 20. Notice how the case where n � 20 approximates the shape of the normal distribution. That is, compare the case where n � 20 to the normal curve in Chart 7–3 on page 228.

n = 1

P (x

)

.40

.30

.20

0 1

Number of occurrences

.10

.50 n = 3

.40

.30

.20

0 1

Number of occurrences

.10

2 x3

n = 20

.20

.15

.10

Number of occurrences

.05

2 x4 6 8 10 12 14 16 18 20

CHART 7–7 Binomial Distributions for an n of 1, 3, and 20, Where � � .50

When can we use the normal approximation to the binomial? The normal prob- ability distribution is a good approximation to the binomial probability distribution when n� and n(1 � �) are both at least 5. However, before we apply the normal approximation, we must make sure that our distribution of interest is in fact a bino- mial distribution. Recall from Chapter 6 that four criteria must be met:

1. There are only two mutually exclusive outcomes to an experiment: a “success” and a “failure.”

2. The distribution results from counting the number of successes in a fixed num- ber of trials.

3. The probability of a success, �, remains the same from trial to trial. 4. Each trial is independent.

Continuity Correction Factor To show the application of the normal approximation to the binomial and the need for a correction factor, suppose the management of the Santoni Pizza Restaurant found that 70 percent of its new customers return for another meal. For a week

When to use the normal approximation

LO7 Approximate the binomial distribution using the normal distribution.

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We can find the probability of 60 or more returning by summing .063 � .048 � . . . � .001, which is .197. However, a look at the plot on the following page shows the similarity of this distribution to a normal distribution. All we need do is “smooth out” the discrete probabilities into a continuous distribution. Further- more, working with a normal distribution will involve far fewer calculations than working with the binomial.

The trick is to let the discrete probability for 56 customers be represented by an area under the continuous curve between 55.5 and 56.5. Then let the probability for 57 customers be represented by an area between 56.5 and 57.5, and so on. This is just the opposite of rounding off the numbers to a whole number.

Continuous Probability Distributions 243

in which 80 new (first-time) customers dined at Santoni’s, what is the probability that 60 or more will return for another meal?

Notice the binomial conditions are met: (1) There are only two possible outcomes—a customer either returns for another meal or does not return. (2) We can count the number of successes, meaning, for example, that 57 of the 80 cus- tomers return. (3) The trials are independent, meaning that if the 34th person returns for a second meal, that does not affect whether the 58th person returns. (4) The probability of a customer returning remains at .70 for all 80 customers.

Therefore, we could use the binomial formula (6–3) described in Section 6.5 on page 196.

To find the probability 60 or more customers return for another pizza, we need to first find the probability exactly 60 customers return. That is:

Next we find the probability that exactly 61 customers return. It is:

We continue this process until we have the probability that all 80 customers return. Finally, we add the probabilities from 60 to 80. Solving the above prob- lem in this manner is tedious. We can also use a computer software package such as Minitab or Excel to find the various probabilities. Listed below are the binomial probabilities for n � 80, � � .70, and x, the number of customers return- ing, ranging from 43 to 68. The probability of any number of customers less than 43 or more than 68 returning is less than .001. We can assume these probabili- ties are 0.000.

P(x � 61) � 80C61 (.70) 61 (1 � .70)19 � .048

P(x � 60) � 80C60 (.70) 60 (1 � .70)20 � .063

P(x) � nCx (�) x (1 � �)n�x

Number Number Returning Probability Returning Probability

43 .001 56 .097 44 .002 57 .095 45 .003 58 .088 46 .006 59 .077 47 .009 60 .063 48 .015 61 .048 49 .023 62 .034 50 .033 63 .023 51 .045 64 .014 52 .059 65 .008 53 .072 66 .004 54 .084 67 .002 55 .093 68 .001

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244 Chapter 7

.10

.09

.08

.07

.06

.05

.04

.03

.02

.01

43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68

Customers

Pr ob

ab ili

Because we use the normal distribution to determine the binomial probability of 60 or more successes, we must subtract, in this case, .5 from 60. The value .5 is called the continuity correction factor. This small adjustment must be made because a continuous distribution (the normal distribution) is being used to approximate a discrete distribution (the binomial distribution). Subtracting, 60 � .5 � 59.5.

CONTINUITY CORRECTION FACTOR The value .5 subtracted or added, depending on the question, to a selected value when a discrete probability distribution is approximated by a continuous probability distribution.

How to Apply the Correction Factor Only four cases may arise. These cases are:

1. For the probability at least X occurs, use the area above (X � .5). 2. For the probability that more than X occurs, use the area above (X � .5). 3. For the probability that X or fewer occurs, use the area below (X � .5). 4. For the probability that fewer than X occurs, use the area below (X � .5).

To use the normal distribution to approximate the probability that 60 or more first-time Santoni customers out of 80 will return, follow the procedure shown below.

Step 1: Find the z corresponding to an X of 59.5 using formula (7–5), and formulas (6–4) and (6–5) for the mean and the variance of a binomial distribution:

Step 2: Determine the area under the normal curve between a � of 56 and an X of 59.5. From step 1, we know that the z value corresponding to 59.5 is 0.85. So we go to Appendix B.1 and read down the left margin to 0.8, and then we go horizontally to the area under the column headed by .05. That area is .3023.

z � X � �

� �

59.5 � 56 4.10

� 0.85

� � 116.8 � 4.10

�2 � n�(1 � �) � 80(.70)(1 � .70) � 16.8

� � n� � 80(.70) � 56

Statistics in Action

Many variables are approximately, nor- mally distributed, such as IQ scores, life expectancies, and adult height. This implies that nearly all observa- tions occur within 3 standard deviations of the mean. On the other hand, observa- tions that occur beyond 3 standard deviations from the mean are extremely rare. For example, the mean adult male height is 68.2 inches (about 5 feet 8 inches) with a stan- dard deviation of 2.74. This means that almost all males are between 60.0 inches (5 feet) and 76.4 inches (6 feet 4 inches). Shaquille O’Neal, a profes- sional basketball player with the Phoenix Suns, is 86 inches or 7 feet 2 inches, which is clearly beyond 3 standard deviations from the mean. The height of a standard doorway is 6 feet 8 inches, and should be high enough for almost all adult males, except for a rare person like Shaquille O’Neal.

(continued)

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No doubt you will agree that using the normal approximation to the binomial is a more efficient method of estimating the probability of 60 or more first-time cus- tomers returning. The result compares favorably with that computed on page 243, using the binomial distribution. The probability using the binomial distribution is .197, whereas the probability using the normal approximation is .1977.

Continuous Probability Distributions 245

Step 3: Calculate the area beyond 59.5 by subtracting .3023 from .5000 (.5000 � .3023 � .1977). Thus, .1977 is the probability that 60 or more first-time Santoni customers out of 80 will return for another meal. In probability notation, P(customers � 59.5) � .5000 � .3023 � .1977. The facets of this problem are shown graphically:

As another exam- ple, the driver’s seat in most vehicles is set to comfortably fit a person who is at least 159 cm (62.5 inches) tall. The dis- tribution of heights of adult women is approximately a nor- mal distribution with a mean of 161.5 cm and a standard devia- tion of 6.3 cm. Thus about 35 percent of adult women will not fit comfortably in the driver’s seat.

Scale of X Scale of z

56 0

.5000

59.5 .85

Probability is .1977 that 60 or more out of

80 will return to Santoni’s

.3023

.1977

Self-Review 7–7 A study by Great Southern Home Insurance revealed that none of the stolen goods were recovered by the homeowners in 80 percent of reported thefts. (a) During a period in which 200 thefts occurred, what is the probability that no stolen

goods were recovered in 170 or more of the robberies? (b) During a period in which 200 thefts occurred, what is the probability that no stolen

goods were recovered in 150 or more robberies?

Exercises 31. Assume a binomial probability distribution with n � 50 and � � .25. Compute the

following: a. The mean and standard deviation of the random variable. b. The probability that X is 15 or more. c. The probability that X is 10 or less.

32. Assume a binomial probability distribution with n � 40 and � � .55. Compute the following: a. The mean and standard deviation of the random variable. b. The probability that X is 25 or greater. c. The probability that X is 15 or less. d. The probability that X is between 15 and 25, inclusive.

33. Dottie’s Tax Service specializes in federal tax returns for professional clients, such as physicians, dentists, accountants, and lawyers. A recent audit by the IRS of the returns she prepared indicated that an error was made on 7 percent of the returns she prepared last year. Assuming this rate continues into this year and she prepares 80 returns, what is the probability that she makes errors on: a. More than six returns? b. At least six returns? c. Exactly six returns?

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246 Chapter 7

34. Shorty’s Muffler advertises it can install a new muffler in 30 minutes or less. However, the work standards department at corporate headquarters recently conducted a study and found that 20 percent of the mufflers were not installed in 30 minutes or less. The Maumee branch installed 50 mufflers last month. If the corporate report is correct: a. How many of the installations at the Maumee branch would you expect to take more

than 30 minutes? b. What is the likelihood that fewer than eight installations took more than 30 minutes? c. What is the likelihood that eight or fewer installations took more than 30 minutes? d. What is the likelihood that exactly 8 of the 50 installations took more than 30 minutes?

35. A study conducted by the nationally known Taurus Health Club revealed that 30 percent of its new members are significantly overweight. A membership drive in a metropolitan area resulted in 500 new members. a. It has been suggested that the normal approximation to the binomial be used to deter-

mine the probability that 175 or more of the new members are significantly overweight. Does this problem qualify as a binomial problem? Explain.

b. What is the probability that 175 or more of the new members are significantly overweight? c. What is the probability that 140 or more new members are significantly overweight?

36. A recent issue of Bride Magazine suggested that couples planning their wedding should expect two-thirds of those who are sent an invitation to respond that they will attend. Rich and Stacy are planning to be married later this year. They plan to send 197 invitations. a. How many guests would you expect to accept the invitation? b. What is the standard deviation? c. What is the probability 140 or more will accept the invitation? d. What is the probability exactly 140 will accept the invitation?

7.6 The Family of Exponential Distributions So far in this chapter, we have considered two continuous probability distributions, the uniform and the normal. The next continuous distribution we consider is the expo- nential distribution. This continuous probability distribution usually describes times between events in a sequence. The actions occur independently at a constant rate per unit of time or length. Because time is never negative, an exponential random variable is always positive. The exponential distribution usually describes situations such as:

• The service times in a system (e.g., how long it takes to serve a customer). • The time between “hits” on a website. • The lifetime of an electrical component. • The time until the next phone call arrives in a customer service center.

The exponential probability distribution is positively skewed. That differs from the uniform and normal distributions, which were both symmetric. Moreover, the distrib- ution is described by only one parameter, which we will identify as . is often referred to as the “rate” parameter. The following chart shows the change in the shape of the exponential distribution as we vary the value of from 1/3 to 1 to 2. Observe that as we decrease , the shape of the distribution changes to become “less skewed.”

2.5

1.5

0.5

0 0 1 2 3 4

Three Exponential Distributions

λ � 0.33 λ � 1.0 λ � 2.0

Exponential distribution is positively skewed

LO8 Describe the characteristics and compute probabilities using the exponential distribution.

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Continuous Probability Distributions 247

Another feature of the exponential distribution is its close relationship to the Poisson distribution. The Poisson is a discrete probability distribution and also has a single parameter, �. We described the Poisson distribution in Section 6.7 in Chapter 6. It too is a positively skewed distribution. To explain the relationship between the Poisson and the exponential distributions, suppose the rate customers arrive at a family restaurant during the dinner hour is 6 per hour. We use the Poisson distribution to find the probability in any particular dinner hour that 2 customers arrive, or that 7 arrive, and so on. So we have a Poisson distribution with a mean of 6. But suppose instead of studying the number

of customers arriving in an hour, we wished to study the time between their arrivals. The time between arrivals is a continuous distribution, because time is measured as a continuous random variable. If customers arrive at a rate of 6 per hour, then logi- cally the typical or mean time between arrivals is 1/6 of an hour, or 10 minutes. We need to be careful here to be consistent with our units, so let’s stay with 1/6 of an hour. So in general, if we know customers arrive at a certain rate per hour, which we call �, then we can expect the mean time between arrivals to be 1/�. The rate para- meter is equal to 1/�. So in our restaurant arrival example, � 1/6.

The graph of the exponential distribution starts at the value of when the ran- dom variable’s (X ) value is 0. The distribution declines steadily as we move to the right with increasing values of X. Formula (7–6) describes the exponential probability distribution with as rate parameter. As we described with the Poisson distribution in Section 6.7 on page 207, e is a mathematical constant equal to 2.71828. It is the basis of the Napierian logarithmic system. It is a pleasant surprise that both the mean and the standard deviation of the exponential probability distribution are equal to 1� .

EXPONENTIAL DISTRIBUTION [7–6]P( x ) � le�l x

FINDING A PROBABILITY USING THE EXPONENTIAL DISTRIBUTION

With continuous distributions, we do not address the probability that a distinct value will occur. Instead, areas or regions below the graph of the probability distrib- ution between two specified values give the probability the random variable is in that interval. A table, such as Appendix B.1 for the normal distribution, is not necessary for the exponential distribution. The area under the exponential density function is found by a simple formula and the necessary calculations can be accomplished with a handheld calculator that has an ex key. Most statistical software packages will also calculate exponential probabilities by inputting the rate parameter, , only. The prob- ability of obtaining an arrival value less than a particular value of x is:

[7–7]P (Arrival time 6 x ) � 1 � e�l x

Example

Solution

Orders for prescriptions arrive at a pharmacy website according to an exponential probability distribution at a mean of one every 20 seconds. Find the probability the next order arrives in less than 5 seconds, in more than 40 seconds, or between 5 and 40 seconds.

To begin, we determine the rate parameter , which in this case is 1/20. To find the probability, we insert 1/20 for and 5 for x in formula (7–7).

P ( Arrival time 6 5) � 1 � e �

1 20

(5) � 1 � e�0.25 � 1 � .7788 � .2212

The mean and standard deviation of the exponential distribution are both equal to 1� .

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248 Chapter 7

0 10 20 30 40 50 60 70 80 90 100 0

0.01

0.02

0.03

0.04

0.05

0.06 Exponential, λ λ � 1/20

What if you wish to determine the probability that it will take more than 5 sec- onds but less than 40 seconds for the next order to arrive? Use formula (7–7) with an x value of 40 and then subtract the value of formula (7–7) when x is 5.

So we conclude there is a 22 percent chance the next order will arrive in less than five seconds. The region is identified as the rust-colored area under the curve.

0 10 20 30 40 50 60 70 80 90 100 0

0.01

0.02

0.03

0.04

0.05

0.06 Exponential, λ λ � 1/20

The above computations addressed the area in the left-tail area of the expo- nential distribution with � 1/20 and the area between 0 and 5—that is, the area which is below 5 seconds. What if you are interested in the right-tail area? It is found using the complement rule. See formula (5–3) in Section 5.4 on page 154 in Chap- ter 5. To put it another way, to find the probability the next order will arrive in more than 40 seconds, we find the probability the order arrives in less than 40 seconds and subtract the result from 1.00. We show this in two steps.

1. Find the probability an order is received in less than 40 seconds.

2. Find the probability an order is received in more than 40 seconds.

We conclude that the likelihood that it will be 40 seconds or more before the next order is received at the pharmacy is 13.5 percent.

P ( Arrival 7 40 ) � 1 � P ( Arrival 6 40 ) � 1 � .8647 � .1353

P (Arrival 6 40) � 1 � e �

1 20

(40) � 1 � .1353 � .8647

As you probably observed, there is some redundancy in this example. In general, if we wish to find the likelihood of a time greater than some value X, such as 40 in the above equations. Let

P(Arrival X ) � 1 � P(Arrival � X ) � (1 � e�kx) � e�kx

In other words, subtract formula [7–7] from the number 1, and the area in the right tail is e�kx. Thus, the probability that more than 40 seconds go by before the next order arrives is computed directly, without the aid of the complement rule as follows:

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Continuous Probability Distributions 249

Previous examples require finding the percentage of the observations located between two values or the percentage of the observations above or below a par- ticular value, x. We can also use formula (7–7) in “reverse” to find the value of the observation x when the percentage above or below the observation is given. The following example illustrates this situation.

In symbols, you can write this as:

We conclude that 64 percent of the time, the time between orders will be between 5 seconds and 40 seconds.

� (1 � e �

1 20

(40) ) � (1 � e

� 1 20

(5) ) � .8647 � .2212 � .6435

P ( 5 � x � 40 ) � P ( Arrival � 40 ) � P( Arrival � 5 )

0 10 20 30 40 50 60 70 80 90 100 0

0.01

0.02

0.03

0.04

0.05

0.06 Exponential, λ λ � 1/20

Example

Solution

Compton Computers wishes to set a minimum lifetime guarantee on its new power supply unit. Quality testing shows the time to failure follows an exponential distrib- ution with a mean of 4,000 hours. Compton wants a warranty period such that only 5 percent of the power supply units fail during that period. What value should they set for the warranty period?

Note that 4,000 hours is a mean and not a rate. Therefore, we must compute as 1/4,000, or 0.00025 failures per hour. A diagram of the situation is shown below, where x represents the minimum guaranteed lifetime.

We use formula (7–7) and essentially work backward for the solution. In this case, the rate parameter is 4,000 hours and we want the area, as shown in the diagram, to be .05.

.05 � 1� e �

1 4,000

(x)

P ( Arrival time 6 x ) � 1 � e(�l x )

0 2000 4000 6000 8000 10000 12000 0

0.00005

0.0001

0.00015

0.0002

0.00025

0.0003 Exponential, λ λ � 0.00025

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Self-Review 7–8 The time between ambulance arrivals at the Methodist Hospital emergency room follows an exponential distribution with a mean of 10 minutes. (a) What is the likelihood the next ambulance will arrive in 15 minutes or less? (b) What is the likelihood the next ambulance will arrive in more than 25 minutes? (c) What is the likelihood the next ambulance will arrive in more than 15 minutes but less

than 25? (d) Find the 80th percentile for the time between ambulance arrivals. (This means only

20 percent of the runs are longer than this time.)

250 Chapter 7

Exercises 37. Waiting times to receive food after placing an order at the local Subway sandwich shop

follow an exponential distribution with a mean of 60 seconds. Calculate the probability a customer waits: a. Less than 30 seconds. b. More than 120 seconds. c. Between 45 and 75 seconds. d. Fifty percent of the patrons wait less than how many seconds? What is the median?

38. The lifetime of plasma and LCD TV sets follows an exponential distribution with a mean of 100,000 hours. Compute the probability a television set: a. Fails in less than 10,000 hours. b. Lasts more than 120,000 hours. c. Fails between 60,000 and 100,000 hours of use. d. Find the 90th percentile. So 10 percent of the TV sets last more than what length

of time? 39. The Bureau of Labor Statistics’ American Time Use Survey showed that the amount of

time spent using a computer for leisure varied greatly by age. Individuals age 75 and over averaged 0.3 hour (18 minutes) per day using a computer for leisure. Individuals ages 15 to 19 spend 1.0 hour per day using a computer for leisure. If these times fol- low an exponential distribution, find the proportion of each group that spends: a. Less than 15 minutes per day using a computer for leisure. b. More than two hours. c. Between 30 minutes and 90 minutes using a computer for leisure. d. Find the 20th percentile. Eighty percent spend more than what amount of time?

40. The cost per item at a supermarket follows an exponential distribution. There are many inexpensive items and a few relatively expensive ones. The mean cost per item is $3.50. What is the percentage of items that cost: a. Less than $1? b. More than $4? c. Between $2 and $3? d. Find the 40th percentile. Sixty percent of the supermarket items cost more than what

amount?

Next, we solve this equation for x. So, we subtract 1 from both sides of the equation and multiply by �1 to simplify the signs. The result is:

Next, we take the natural log of both sides and solve for x:

In this case, x � 205.17. Hence, Compton can set the warranty period at 205 hours and expect about 5 percent of the power supply units to be returned.

x � 205.17

�(.051293294) � � 1

4,000 x

ln (.95 ) � � 1

4,000 x

.95 � e �

1 4,000

(x)

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Continuous Probability Distributions 251

Chapter Summary I. The uniform distribution is a continuous probability distribution with the following charac-

teristics. A. It is rectangular in shape. B. The mean and the median are equal. C. It is completely described by its minimum value a and its maximum value b. D. It is also described by the following equation for the region from a to b:

[7–3]

E. The mean and standard deviation of a uniform distribution are computed as follows:

[7–1]

[7–2]

II. The normal probability distribution is a continuous distribution with the following charac- teristics. A. It is bell-shaped and has a single peak at the center of the distribution. B. The distribution is symmetric. C. It is asymptotic, meaning the curve approaches but never touches the X-axis. D. It is completely described by its mean and standard deviation. E. There is a family of normal probability distributions.

1. Another normal probability distribution is created when either the mean or the stan- dard deviation changes.

2. The normal probability distribution is described by the following formula:

[7–4]

III. The standard normal probability distribution is a particular normal distribution. A. It has a mean of 0 and a standard deviation of 1. B. Any normal probability distribution can be converted to the standard normal probabil-

ity distribution by the following formula.

[7–5]

C. By standardizing a normal probability distribution, we can report the distance of a value from the mean in units of the standard deviation.

IV. The normal probability distribution can approximate a binomial distribution under certain conditions. A. n� and n(1 � �) must both be at least 5.

1. n is the number of observations. 2. � is the probability of a success.

B. The four conditions for a binomial probability distribution are: 1. There are only two possible outcomes. 2. � remains the same from trial to trial. 3. The trials are independent. 4. The distribution results from a count of the number of successes in a fixed number

of trials. C. The mean and variance of a binomial distribution are computed as follows:

D. The continuity correction factor of .5 is used to extend the continuous value of X one- half unit in either direction. This correction compensates for approximating a discrete distribution by a continuous distribution.

�2 � n�(1 � �)

� � n�

z � X � �

�

P(x) � 1

�12� e�c

(x � �)2

2� 2 d

� � B

(b � a)2

� � (a � b)

P(x) � 1

b � a

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252 Chapter 7

Chapter Exercises 41. The amount of cola in a 12-ounce can is uniformly distributed between 11.96 ounces

and 12.05 ounces. a. What is the mean amount per can? b. What is the standard deviation amount per can? c. What is the probability of selecting a can of cola and finding it has less than 12 ounces? d. What is the probability of selecting a can of cola and finding it has more than 11.98

ounces? e. What is the probability of selecting a can of cola and finding it has more than 11.00

ounces? 42. A tube of Listerine Tartar Control toothpaste contains 4.2 ounces. As people use the

toothpaste, the amount remaining in any tube is random. Assume the amount of tooth- paste left in the tube follows a uniform distribution. From this information, we can determine the following information about the amount remaining in a toothpaste tube without invading anyone’s privacy. a. How much toothpaste would you expect to be remaining in the tube? b. What is the standard deviation of the amount remaining in the tube? c. What is the likelihood there is less than 3.0 ounces remaining in the tube? d. What is the probability there is more than 1.5 ounces remaining in the tube?

43. Many retail stores offer their own credit cards. At the time of the credit application, the customer is given a 10 percent discount on the purchase. The time required for the credit application process follows a uniform distribution with the times ranging from 4 minutes to 10 minutes. a. What is the mean time for the application process? b. What is the standard deviation of the process time? c. What is the likelihood a particular application will take less than 6 minutes? d. What is the likelihood an application will take more than 5 minutes?

44. The time patrons at the Grande Dunes Hotel in the Bahamas spend waiting for an elevator follows a uniform distribution between 0 and 3.5 minutes. a. Show that the area under the curve is 1.00. b. How long does the typical patron wait for elevator service? c. What is the standard deviation of the waiting time? d. What percent of the patrons wait for less than a minute? e. What percent of the patrons wait more than 2 minutes?

45. The net sales and the number of employees for aluminum fabricators with similar char- acteristics are organized into frequency distributions. Both are normally distributed. For the net sales, the mean is $180 million and the standard deviation is $25 million. For the number of employees, the mean is 1,500 and the standard deviation is 120. Clarion Fab- ricators had sales of $170 million and 1,850 employees. a. Convert Clarion’s sales and number of employees to z values. b. Locate the two z values. c. Compare Clarion’s sales and number of employees with those of the other fabricators.

46. The accounting department at Weston Materials Inc., a national manufacturer of unat- tached garages, reports that it takes two construction workers a mean of 32 hours and a standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times follow the normal distribution. a. Determine the z values for 29 and 34 hours. What percent of the garages take between

32 hours and 34 hours to erect? b. What percent of the garages take between 29 hours and 34 hours to erect?

V. The exponential probability distribution describes times between events in a sequence. A. The actions occur independently at a constant rate per unit of time or length. B. The probability density is given by the formula:

[7–6]

C. It is nonnegative, positively skewed, declines steadily to the right, and is asymptotic. D. The area under the curve is given by the formula

[7–7]

E. Both the mean and standard deviation are 1/ .

P ( Arrival time 6 x ) � 1 � e�l x

P ( x ) � le�l x

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Continuous Probability Distributions 253

c. What percent of the garages take 28.7 hours or less to erect? d. Of the garages, 5 percent take how many hours or more to erect?

47. A recent report in USA Today indicated a typical family of four spends $490 per month on food. Assume the distribution of food expenditures for a family of four follows the nor- mal distribution, with a mean of $490 and a standard deviation of $90. a. What percent of the families spend more than $30 but less than $490 per month on

food? b. What percent of the families spend less than $430 per month on food? c. What percent spend between $430 and $600 per month on food? d. What percent spend between $500 and $600 per month on food?

48. A study of long-distance phone calls made from General Electric revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 4.2 minutes and the standard deviation was 0.60 minutes. a. What fraction of the calls last between 4.2 and 5 minutes? b. What fraction of the calls last more than 5 minutes? c. What fraction of the calls last between 5 and 6 minutes? d. What fraction of the calls last between 4 and 6 minutes? e. As part of her report to the president, the director of communications would

like to report the length of the longest (in duration) 4 percent of the calls. What is this time?

49. Shaver Manufacturing Inc. offers dental insurance to its employees. A recent study by the human resource director shows the annual cost per employee per year followed the normal probability distribution, with a mean of $1,280 and a standard deviation of $420 per year. a. What fraction of the employees cost more than $1,500 per year for dental

expenses? b. What fraction of the employees cost between $1,500 and $2,000 per year? c. Estimate the percent that did not have any dental expense. d. What was the cost for the 10 percent of employees who incurred the highest dental

expense? 50. The annual commissions earned by sales representatives of Machine Products Inc., a

manufacturer of light machinery, follow the normal probability distribution. The mean yearly amount earned is $40,000 and the standard deviation is $5,000. a. What percent of the sales representatives earn more than $42,000 per year? b. What percent of the sales representatives earn between $32,000 and $42,000? c. What percent of the sales representatives earn between $32,000 and $35,000? d. The sales manager wants to award the sales representatives who earn the largest

commissions a bonus of $1,000. He can award a bonus to 20 percent of the repre- sentatives. What is the cutoff point between those who earn a bonus and those who do not?

51. According to the South Dakota Department of Health, the mean number of hours of TV viewing per week is higher among adult women than men. A recent study showed women spent an average of 34 hours per week watching TV and men 29 hours per week. Assume that the distribution of hours watched follows the normal distribution for both groups, and that the standard deviation among the women is 4.5 hours and is 5.1 hours for the men. a. What percent of the women watch TV less than 40 hours per week? b. What percent of the men watch TV more than 25 hours per week? c. How many hours of TV do the one percent of women who watch the most TV per

week watch? Find the comparable value for the men. 52. According to a government study among adults in the 25- to 34-year age group, the

mean amount spent per year on reading and entertainment is $1,994. Assume that the distribution of the amounts spent follows the normal distribution with a standard devia- tion of $450. a. What percent of the adults spend more than $2,500 per year on reading and

entertainment? b. What percent spend between $2,500 and $3,000 per year on reading and

entertainment? c. What percent spend less than $1,000 per year on reading and entertainment?

53. Management at Gordon Electronics is considering adopting a bonus system to increase production. One suggestion is to pay a bonus on the highest 5 percent of production based on past experience. Past records indicate weekly production follows the normal distribution. The mean of this distribution is 4,000 units per week and the standard

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254 Chapter 7

deviation is 60 units per week. If the bonus is paid on the upper 5 percent of produc- tion, the bonus will be paid on how many units or more?

54. Fast Service Truck Lines uses the Ford Super Duty F-750 exclusively. Management made a study of the maintenance costs and determined the number of miles traveled during the year followed the normal distribution. The mean of the distribution was 60,000 miles and the standard deviation 2,000 miles. a. What percent of the Ford Super Duty F-750s logged 65,200 miles or more? b. What percent of the trucks logged more than 57,060 but less than 58,280 miles? c. What percent of the Fords traveled 62,000 miles or less during the year? d. Is it reasonable to conclude that any of the trucks were driven more than 70,000 miles?

Explain. 55. Best Electronics Inc. offers a “no hassle” returns policy. The number of items returned

per day follows the normal distribution. The mean number of customer returns is 10.3 per day and the standard deviation is 2.25 per day. a. In what percent of the days are there 8 or fewer customers returning items? b. In what percent of the days are between 12 and 14 customers returning items? c. Is there any chance of a day with no returns?

56. A recent report in BusinessWeek indicated that 20 percent of all employees steal from their company each year. If a company employs 50 people, what is the probability that: a. Fewer than 5 employees steal? b. More than 5 employees steal? c. Exactly 5 employees steal? d. More than 5 but fewer than 15 employees steal?

57. The Orange County Register, as part of its Sunday health supplement, reported that 64 per- cent of American men over the age of 18 consider nutrition a top priority in their lives. Suppose we select a sample of 60 men. What is the likelihood that: a. 32 or more consider nutrition important? b. 44 or more consider nutrition important? c. More than 32 but fewer than 43 consider nutrition important? d. Exactly 44 consider diet important?

58. It is estimated that 10 percent of those taking the quantitative methods portion of the CPA examination fail that section. Sixty students are taking the exam this Saturday. a. How many would you expect to fail? What is the standard deviation? b. What is the probability that exactly two students will fail? c. What is the probability at least two students will fail?

59. The Georgetown, South Carolina, Traffic Division reported 40 percent of high-speed chases involving automobiles result in a minor or major accident. During a month in which 50 high-speed chases occur, what is the probability that 25 or more will result in a minor or major accident?

60. Cruise ships of the Royal Viking line report that 80 percent of their rooms are occupied during September. For a cruise ship having 800 rooms, what is the probability that 665 or more are occupied in September?

61. The goal at U.S. airports handling international flights is to clear these flights within 45 min- utes. Let’s interpret this to mean that 95 percent of the flights are cleared in 45 minutes, so 5 percent of the flights take longer to clear. Let’s also assume that the distribution is approximately normal. a. If the standard deviation of the time to clear an international flight is 5 minutes, what

is the mean time to clear a flight? b. Suppose the standard deviation is 10 minutes, not the 5 minutes suggested in part

(a). What is the new mean? c. A customer has 30 minutes from the time her flight lands to catch her limousine.

Assuming a standard deviation of 10 minutes, what is the likelihood that she will be cleared in time?

62. The funds dispensed at the ATM machine located near the checkout line at the Kroger’s in Union, Kentucky, follows a normal probability distribution with a mean of $4,200 per day and a standard deviation of $720 per day. The machine is programmed to notify the nearby bank if the amount dispensed is very low (less than $2,500) or very high (more than $6,000). a. What percent of the days will the bank be notified because the amount dispensed is

very low? b. What percent of the time will the bank be notified because the amount dispensed is high? c. What percent of the time will the bank not be notified regarding the amount of funds

dispersed?

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Continuous Probability Distributions 255

63. The weights of canned hams processed at Henline Ham Company follow the normal dis- tribution, with a mean of 9.20 pounds and a standard deviation of 0.25 pounds. The label weight is given as 9.00 pounds. a. What proportion of the hams actually weigh less than the amount claimed on the label? b. The owner, Glen Henline, is considering two proposals to reduce the proportion of

hams below label weight. He can increase the mean weight to 9.25 and leave the standard deviation the same, or he can leave the mean weight at 9.20 and reduce the standard deviation from 0.25 pounds to 0.15. Which change would you recommend?

64. The Cincinnati Enquirer, in its Sunday business supplement, reported that the mean num- ber of hours worked per week by those employed full time is 43.9. The article further indi- cated that about one-third of those employed full time work less than 40 hours per week. a. Given this information and assuming that number of hours worked follows the normal

distribution, what is the standard deviation of the number of hours worked? b. The article also indicated that 20 percent of those working full time work more than

49 hours per week. Determine the standard deviation with this information. Are the two estimates of the standard deviation similar? What would you conclude?

65. Most four-year automobile leases allow up to 60,000 miles. If the lessee goes beyond this amount, a penalty of 20 cents per mile is added to the lease cost. Suppose the dis- tribution of miles driven on four-year leases follows the normal distribution. The mean is 52,000 miles and the standard deviation is 5,000 miles. a. What percent of the leases will yield a penalty because of excess mileage? b. If the automobile company wanted to change the terms of the lease so that 25 per-

cent of the leases went over the limit, where should the new upper limit be set? c. One definition of a low-mileage car is one that is 4 years old and has been driven

less than 45,000 miles. What percent of the cars returned are considered low-mileage? 66. The price of shares of Bank of Florida at the end of trading each day for the last year

followed the normal distribution. Assume there were 240 trading days in the year. The mean price was $42.00 per share and the standard deviation was $2.25 per share. a. What percent of the days was the price over $45.00? How many days would you

estimate? b. What percent of the days was the price between $38.00 and $40.00? c. What was the stock’s price on the highest 15 percent of days?

67. The annual sales of romance novels follow the normal distribution. However, the mean and the standard deviation are unknown. Forty percent of the time sales are more than 470,000, and 10 percent of the time sales are more than 500,000. What are the mean and the standard deviation?

68. In establishing warranties on HDTV sets, the manufacturer wants to set the limits so that few will need repair at the manufacturer’s expense. On the other hand, the warranty period must be long enough to make the purchase attractive to the buyer. For a new HDTV, the mean number of months until repairs are needed is 36.84 with a standard devi- ation of 3.34 months. Where should the warranty limits be set so that only 10 percent of the HDTVs need repairs at the manufacturer’s expense?

69. DeKorte Tele-Marketing Inc. is considering purchasing a machine that randomly selects and automatically dials telephone numbers. DeKorte Tele-Marketing makes most of its calls during the evening, so calls to business phones are wasted. The manufacturer of the machine claims that its programming reduces the calling to business phones to 15 per- cent of all calls. To test this claim, the director of purchasing at DeKorte programmed the machine to select a sample of 150 phone numbers. What is the likelihood that more than 30 of the phone numbers selected are those of businesses, assuming the manufacturer’s claim is correct?

70. A carbon monoxide detector in the Wheelock household activates once every 200 days on average. Assume this activation follows the exponential distribution. What is the prob- ability that: a. There will be an alarm within the next 60 days? b. At least 400 days will pass before the next alarm? c. It will be between 150 and 250 days until the next warning? d. Find the median time until the next activation.

71. “Boot Time” (the time between the appearance of the Bios screen to the first file that is loaded in Windows) on Eric Mouser’s personal computer follows an exponential distrib- ution with a mean of 27 seconds. What is the probability his “boot” will require: a. Less than 15 seconds? b. More than 60 seconds?

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256 Chapter 7

c. Between 30 and 45 seconds? d. What is the point below which only 10 percent of the boots occur?

72. The time between visits to a U.S. emergency room for a member of the general popula- tion follows an exponential distribution with a mean of 2.5 years. What proportion of the population will visit an emergency room: a. Within the next six months? b. Not visit the ER over the next six years? c. Next year, but not this year? d. Find the first and third quartiles of this distribution.

73. The times between failures on a personal computer follow an exponential distribution with a mean of 300,000 hours. What is the probability of: a. A failure in less than 100,000 hours? b. No failure in the next 500,000 hours? c. The next failure occurring between 200,000 and 350,000 hours? d. What are the mean and standard deviation of the time between failures?

Data Set Exercises 74. Refer to the Real Estate data, which report information on homes sold in the Goodyear,

Arizona, area during the last year. a. The mean selling price (in $ thousands) of the homes was computed earlier to be

$221.10, with a standard deviation of $47.11. Use the normal distribution to estimate the percentage of homes selling for more than $280.0. Compare this to the actual results. Does the normal distribution yield a good approximation of the actual results?

b. The mean distance from the center of the city is 14.629 miles, with a standard devia- tion of 4.874 miles. Use the normal distribution to estimate the number of homes 18 or more miles but less than 22 miles from the center of the city. Compare this to the actual results. Does the normal distribution yield a good approximation of the actual results?

75. Refer to the Baseball 2009 data, which report information on the 30 Major League Base- ball teams for the 2009 season. a. The mean attendance per team for the season was 2.448 million, with a standard devi-

ation of 0.698 million. Use the normal distribution to estimate the number of teams with attendance of more than 3.5 million. Compare that estimate with the actual num- ber. Comment on the accuracy of your estimate.

b. The mean team salary was $88.51 million, with a standard deviation of $33.90 million. Use the normal distribution to estimate the number of teams with a team salary of more than $50 million. Compare that estimate with the actual number. Comment on the accuracy of the estimate.

76. Refer to the Buena School District bus data. a. Refer to the maintenance cost variable. The mean maintenance cost for last year is

$450.29, with a standard deviation of 53.69. Estimate the number of buses with a cost of more than $500. Compare that with the actual number.

b. Refer to the variable on the number of miles driven. The mean is 830.11 and the standard deviation is 42.19 miles. Estimate the number of buses traveling more than 900 miles. Compare that number with the actual value.

1. The Excel commands necessary to produce the out- put on page 235 are: a. Click on the Formulas tab in the top menu, then,

on the far left, select Insert Function fx. Then from the category box, select Statistical and below that NORMDIST and click OK.

b. In the dialog box, put 1100 in the box for X, 1000 for the Mean, 100 for the Standard_dev, True in the Cumulative box, and click OK.

c. The result will appear in the dialog box. If you click OK, the answer appears in your spread- sheet.

Software Commands

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Continuous Probability Distributions 257

2. The Excel Commands necessary to produce the out- put on page 241 are: a. Click the Formulas tab in the top menu, then, on

the far left, select Insert Function fx. Then from the category box, select Statistical and below that NORMINV and click OK.

b. In the dialog box, set the Probability to .04, the Mean to 67900, and the Standard_dev to 2050.

c. The results will appear in the dialog box. Note that the answer is different from page 240 because of rounding error. If you click OK, the answer also appears in your spreadsheet.

d. Try entering a Probability of .04, a Mean of 0, and a Standard_dev of 1. The z value will be computed.

Chapter 7 Answers to Self-Review

7–1 a.

c. � � � � � 11

� �

� 1.73 d.

7–2 a. 2.25, found by:

b. �2.25, found by:

7–3 a. $46,400 and $48,000, found by $47,200 1($800).

b. $45,600 and $48,800, found by $47,200 2($800).

c. $44,800 and $49,600, found by $47,200 3($800).

d. $47,200. The mean, median, and mode are equal for a normal distribution.

e. Yes, a normal distribution is symmetrical. 7–4 a. Computing z:

Referring to Appendix B.1, the area is .2881. So P(150 temp 154) � .2881.

b. Computing z:

Referring to Appendix B.1, the area is .4974. So P(164 � temp) � .5000 � .4974 � .0026.

7–5 a. Computing the z-values:

and

P(146 temp 156) � P(�0.80 z 1.20) � .2881 � .3948 � .6829.

b. Computing the z-values:

and

P(156 temp 162) � P(1.20 z 2.40) � .4918 � .3849 � .1069

7–6 85.24 (instructor would no doubt make it 85). The closest area to .4000 is .3997; z is 1.28. Then:

7–7 a. .0465, found by � � n� � 200(.80) � 160, and �2 � n� (1 � �) � 200(.80)(1 � .80) � 32. Then,

z � 169.5 � 160

5.66 � 1.68

� � 132 � 5.66

X � 85.24

10.24 � X � 75

1.28 � X � 75

z � 156 � 150

5 � 1.20z �

162 � 150 5

� 2.40

z � 156 � 150

5 � 1.20z �

146 � 150 5

� �0.80

z � 164 � 150

5 � 2.80

z � 154 � 150

5 � 0.80

z � $775 � $1,000

$100 �

�$225 $100

� �2.25

z � $1,225 � $1,000

$100 �

$225 $100

� 2.25

� 0.167

� a 1

14 � 8 b (9 � 8)

P(x 6 9) � (height)(base) � .667

� 1 6

(4)

� a 1

14 � 8 b (14 � 10)

P(10 6 x 6 14) � (height)(base)

(b � a)2

12 � B

(14 � 8)2

12 � B

36 12

� 23

22 2

14 � 8 2

a � b 2

� a 1 6 b (6) � 1.00

� a 1

14 � 8 b (14 � 8)

P(x) � (height)(base)

P (x )

.167

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258 Chapter 7

Area from Appendix B.1 is .4535. Subtracting from .5000 gives .0465.

b. .9686, found by .4686 � .5000. First calculate z:

Area from Appendix B.1 is .4686. 7–8 a. .7769, found by:

b. .0821, found by:

c. .1410, found by

d. 16.09 minutes, found by:

x � �(�1.609)(10) � 1.609(10) � 16.09

�ln 0.20 � 1

10 x

.80 � 1 � e �

1 10

(x)

� .9179 � .7769 � .1410 P (15 6 x 6 25) � P ( Arrival 6 25 ) � P ( Arrival 15)

P ( Arrival 6 25 ) � e �

1 10

(15) � .0821

� 1 � .2231 � .7769 P ( Arrival 6 15 ) � 1 � e

� 1 10

(15)

z � 149.5 � 160

5.66 � �1.86

A Review of Chapters 5–7 This section is a review of the major concepts, terms, symbols, and equations introduced in Chap- ters 5, 6, and 7. These three chapters are concerned with methods of dealing with uncertainty. As an example of the uncertainty in business, consider the role of the quality assurance department in most mass-production firms. Usually, the department has neither the personnel nor the time to check, say, all 200 plug-in modules produced during a two-hour period. Standard operating proce- dure may call for selecting a sample of 5 modules and shipping all 200 modules if the 5 operate correctly. However, if 1 or more in the sample are defective, all 200 are checked. Assuming that all 5 function correctly, quality assurance personnel cannot be absolutely certain that their action (allow- ing shipment of the modules) will prove to be correct. The study of probability allows us to mea- sure the uncertainty of shipping defective modules. Also, probability as a measurement of uncer- tainty comes into play when SurveyUSA, The Gallop Poll, Zogby, and other pollsters measure public opinion on issues such as taxes and health care.

Chapter 5 introduces the concept of probability. A probability is a value between 0 and 1 that expresses the likelihood a particular event will occur. A weather forecaster states the probability of rain tomorrow is .20. The project director of a firm bidding on a subway station in Bangkok assesses the firm’s chance of being awarded the contract at .70. In this chapter, we looked at methods for combining probabilities using rules of addition and multiplication, presented some principles of counting, and described situations for using Bayes’ theorem.

Chapter 6 describes discrete probability distributions. Probability distributions are listings of the possible outcomes of an experiment and the probability associated with each outcome. In this chap- ter, we describe three discrete probability distributions: the binomial distribution, the hypergeomet- ric distribution, and the Poisson distribution.

Chapter 7 describes three continuous probability distributions: the uniform probability distribution, the normal probability distribution, and the exponential distribution.

The uniform probability distribution is rectangular in shape and is defined by a minimum and maximum value. The mean and the median of a uniform probability distribution are equal, and it does not have a mode. A normal probability distribution is used to describe phenomena such as the weights of newborn babies, the time it takes to assemble products, or the scores students earn on an examination. Actually, there is a family of normal probability distributions—each with its own mean and standard deviation. So there is a normal distribution with a mean of 100 and a standard deviation of 5, another for a mean of 149 and a standard deviation of 5.26, and so on.

A normal probability distribution is symmetrical about its mean, and the tails of the normal curve extend in either direction infinitely. There are an unlimited number of normal probability distributions, so the number of tables such as B.1 would be very large. Instead of using a large number of tables, we convert a normal distribution to a standard normal probability distribution by computing the z value. The standard normal probability distribution has a mean of 0 and a standard deviation of 1. It is useful because the probability for any event from a normal probability distribution can be com- puted using standard normal probability tables.

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A Review of Chapters 5–7 259

Glossary

Chapter 5 Bayes’ theorem Developed by Reverend Bayes in the 1700s, it is designed to find the probability of one event, A, occurring, given that another event, B, has already occurred. Classical probability Probability based on the assump- tion that each of the outcomes is equally likely. According to this concept of probability, if there are n possible out- comes, the probability of a particular outcome is 1/n. Thus, on the toss of a coin, the probability of a head is 1/n � 1/2. Combination formula A formula to count the number of possible outcomes when the order of the outcomes is not im- portant. If the order a, b, c is considered the same as b, a, c, or c, b, a, and so on, the number of arrangements is found by:

Conditional probability The likelihood that an event will occur given that another event has already occurred. Empirical probability A concept of probability based on past experience. For example, Metropolitan Life Insurance Company reported that, during the year, 100.2 of every 100,000 persons in Wyoming died of accidental causes (motor vehicle accidents, falls, drowning, firearms, etc.). On the basis of this experience, Metropolitan can estimate the probability of accidental death for a particular person in Wyoming: 100.2/100,000 � .001002. Event A collection of one or more outcomes of an ex- periment. For example, an event is the collection of even numbers in the roll of a fair die. Experiment An activity that is either observed or mea- sured. An experiment may be counting the number of cor- rect responses to a question, for example. General rule of addition Used to find the probabilities of complex events made up of A or B when the events are not mutually exclusive.

General rule of multiplication Used to find the proba- bilities of events A and B both happening when the events are not independent. Example: It is known that there are 3 defective radios in a box containing 10 radios. What is the probability of selecting 2 defective radios on the first two selections from the box?

where is the conditional probability and means “the probability of B occurring given that A has already occurred.”

P(B ƒ A)

P(A and B) � P(A)P(B ƒ A) � 3

10 �

2 9

� 6 90

� .067

P(A or B) � P(A) � P(B) � P(A and B)

nCr � n!

r!(n � r)!

Independent The probability of one event has no effect on the probability of another event. Multiplication formula One of the formulas used to count the number of possible outcomes of an experiment. It states that if there are m ways of doing one thing and n ways of doing another, there are m � n ways of doing both. Example: A men’s clothier offers two sport coats and three contrasting pants for $400. How many different out- fits can there be? Answer: m � n � 2 � 3 � 6. Mutually exclusive The occurrence of one event means that none of the other events can occur at the same time. Outcome A particular observation or measurement of an experiment. Permutation formula A formula to count the number of possible outcomes when the order of the outcomes is im- portant. If a, b, c is one arrangement, b, a, c another, c, a, b another, and so on, the total number of arrangements is determined by

Probability A value between 0 and 1, inclusive, that re- ports the likelihood that a specific event will occur. Special rule of addition For this rule to apply, the events must be mutually exclusive. For two events, the probabil- ity of A or B occurring is found by:

Example: The probability of a one-spot or a two-spot oc- curring on the toss of one die.

Special rule of multiplication If two events are not related—that is, they are independent—this rule is applied to determine the probability of their joint occurrence.

Example: The probability of two heads on two tosses of a coin is:

Subjective probability The chance of an event hap- pening based on whatever information is available— hunches, personal opinion, opinions of others, rumors, and so on.

P(A and B) � P(A)P(B) � 1 2

� 1 2

� 1 4

P(A and B) � P(A)P(B)

P(A or B) � 1 6

� 1 6

� 2 6

� 1 3

P(A or B) � P(A) � P(B)

nPr � n!

(n � r)!

The exponential distribution describes the time between events in a sequence. These events occur independently at a constant rate per unit of time or length. The exponential probability dis- tribution is positively skewed, with as “rate” parameter. The mean and standard deviation are equal and are the reciprocal of . If the mean life of the television set is 8 years, then the annual rate of failure is 1/8 and the standard deviation of the failure rate is also 1/8.

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260 Chapter 7

Problems

1. It is claimed that Proactine, a new medicine for acne, is 80 percent effective. It is applied to the affected area of a sample of 15 people. What is the probability that: a. All 15 will show significant improvement? b. Fewer than 9 of 15 will show significant improvement? c. That 12 or more people will show significant improvement?

2. First National Bank thoroughly investigates its applicants for small home-improvement loans. Its default record is very impressive: The probability that a homeowner will default is only .005. The bank has approved 400 small home-improvement loans. Assuming the Poisson probability distribution applies to this problem: a. What is the probability that no homeowners out of the 400 will default? b. How many of the 400 are expected not to default? c. What is the probability that 3 or more homeowners will default on their small home-

improvement loans? 3. A study of the attendance at the University of Alabama’s basketball games revealed that

the distribution of attendance is normally distributed with a mean of 10,000 and a stan- dard deviation of 2,000. a. What is the probability a particular game has an attendance of 13,500 or more? b. What percent of the games have an attendance between 8,000 and 11,500? c. Ten percent of the games have an attendance of how many or less?

4. Daniel-James Insurance Company will insure an offshore Mobil Oil production plat- form against weather losses for one year. The president of Daniel-James estimates

Chapter 6 Binomial probability distribution A probability distribu- tion based on a discrete random variable. Its major char- acteristics are:

1. Each outcome can be classified into one of two mu- tually exclusive categories.

2. The distribution is the result of counting the number of successes.

3. Each trial is independent, meaning that the answer to trial 1 (correct or wrong) in no way affects the answer to trial 2.

4. The probability of a success stays the same from trial to trial.

Continuous random variable A random variable that may assume an infinite number of values within a given range. Discrete random variable A random variable that can assume only certain separate values. Hypergeometric probability distribution A probability distribution based on a discrete random variable. Its major characteristics are:

1. There is a fixed number of trials. 2. The probability of success is not the same from trial

to trial. 3. There are only two possible outcomes.

Poisson distribution A distribution often used to ap- proximate binomial probabilities when n is large and � is small. What is considered “large” or “small” is not pre- cisely defined, but a general rule is that n should be equal to or greater than 20 and � equal to or less than .05. Probability distribution A listing of the possible out- comes of an experiment and the probability associated with each outcome.

Random variable A quantity obtained from an experi- ment that may, by chance, result in different values. For example, a count of the number of accidents (the experi- ment) on I-75 during a week might be 10, or 11, or 12, or some other number.

Chapter 7 Continuity correction factor Used to improve the accu- racy of the approximation of a discrete distribution by a continuous distribution. Exponential probability distribution A positively skewed continuous probability distribution that is fully described by a single “rate” parameter ( ). Its likelihood is at the starting value 0 and declines steadily while extending in- definitely to the right or positive direction. Both the mean and standard deviation are the reciprocal of the rate para- meter . Normal probability distribution A continuous distribu- tion that is bell-shaped, with the mean dividing the distrib- ution into two equal parts. Further, the normal curve ex- tends indefinitely in either direction, and it never touches the X-axis. The distribution is defined by its mean and standard deviation. Uniform probability distribution A continuous proba- bility distribution that is rectangular in shape. It is com- pletely described by using the minimum and maximum values of the distribution to compute the mean and stan- dard deviation. Also, minimum and maximum values are used to compute the probability for any event. z value The distance between a selected value and the mean measured in units of the standard deviation.

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A Review of Chapters 5–7 261

the following losses for that platform (in millions of dollars) with the accompanying probabilities:

Amount of Loss Probability ($ millions) of Loss

0 .98 40 .016

300 .004

a. What is the expected amount Daniel-James will have to pay to Mobil in claims? b. What is the likelihood that Daniel-James will actually lose less than the expected

amount? c. Given that Daniel-James suffers a loss, what is the likelihood that it is for $300 million? d. Daniel-James has set the annual premium at $2.0 million. Does that seem like a fair

premium? Will it cover its risk?

5. The distribution of the number of school-age children per family in the Whitehall Estates area of Boise, Idaho, is:

Number of children 0 1 2 3 4 Percent of families 40 30 15 10 5

a. Determine the mean and standard deviation of the number of school-age children per family in Whitehall Estates.

b. A new school is planned in Whitehall Estates. An estimate of the number of school- age children is needed. There are 500 family units. How many children would you esti- mate?

c. Some additional information is needed about only the families having children. Con- vert the preceding distribution to one for families with children. What is the mean num- ber of children among families that have children?

6. The following table shows a breakdown of the 110th U.S. Congress by party affiliation.

Party

Democrats Republicans Total

House 236 199 435 Senate 48 52 100

Total 284 251 535

a. A member of Congress is selected at random. What is the probability of selecting a Republican?

b. Given that the person selected is a member of the House of Representatives, what is the probability he or she is a Republican?

c. What is the probability of selecting a member of the House of Representatives or a Democrat?

Cases

A. Century National Bank Refer to the Century National Bank data. Is it reasonable that the distribution of checking account balances approx- imates a normal probability distribution? Determine the mean and the standard deviation for the sample of 60 cus- tomers. Compare the actual distribution with the theoreti- cal distribution. Cite some specific examples and comment on your findings.

Divide the account balances into three groups, of about 20 each, with the smallest third of the balances in the first group, the middle third in the second group, and those with the largest balances in the third group. Next, develop a table that shows the number in each of the cat- egories of the account balances by branch. Does it appear that account balances are related to the branch? Cite some examples and comment on your findings.

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262 Chapter 7

Avondale Midway Kingston

9 19 38 14 22 39 11 23 41 8 14 39

14 22 41 6 17 39

10 15 39 13 20 8 18 8 9

12 7

C. Geoff “Applies” His Education Geoff Brown is the manager for a small telemarketing firm and is evaluating the sales rate of experienced workers in order to set minimum standards for new hires. During the past few weeks, he has recorded the number of success- ful calls per hour for the staff. These data appear next

B. Elections Auditor An item such as an increase in taxes, recall of elected offi- cials, or an expansion of public services can be placed on the ballot if a required number of valid signatures are col- lected on the petition. Unfortunately, many people will sign the petition even though they are not registered to vote in that particular district, or they will sign the petition more than once.

Sara Ferguson, the elections auditor in Venango County, must certify the validity of these signatures after the petition is officially presented. Not surprisingly, her staff is overloaded, so she is considering using statistical methods to validate the pages of 200 signatures, instead of validating each individual signature. At a recent profes- sional meeting, she found that, in some communities in the state, election officials were checking only five signa- tures on each page and rejecting the entire page if two or more signatures were invalid. Some people are concerned that five may not be enough to make a good decision. They suggest that you should check 10 signatures and re- ject the page if three or more are invalid.

In order to investigate these methods, Sara asks her staff to pull the results from the last election and sample 30 pages. It happens that the staff selected 14 pages from the Avondale district, 9 pages from the Midway district, and 7 pages from the Kingston district. Each page had 200 signatures, and the data below show the number of invalid signatures on each.

Use the data to evaluate Sara’s two proposals. Cal- culate the probability of rejecting a page under each of the approaches. Would you get about the same results by ex- amining every single signature? Offer a plan of your own, and discuss how it might be better or worse than the two plans proposed by Sara.

along with some summary statistics he worked out with a statistical software package. Geoff has been a student at the local community college and has heard of many dif- ferent kinds of probability distributions (binomial, normal, hypergeometric, Poisson, etc.). Could you give Geoff some advice on which distribution to use to fit these data as well as possible and how to decide when a probation- ary employee should be accepted as having reached full production status? This is important because it means a pay raise for the employee, and there have been some probationary employees in the past who have quit be- cause of discouragement that they would never meet the standard.

Successful sales calls per hour during the week of August 14:

4 2 3 1 4 5 5 2 3 2 2 4 5 2 5 3 3 0 1 3 2 8 4 5 2 2 4 1 5 5 4 5 1 2 4

Descriptive statistics:

N MEAN MEDIAN TRMEAN STDEV SEMEAN 35 3.229 3.000 3.194 1.682 0.284

MIN MAX Q1 Q3 0.0 8.000 2.000 5.000

Which distribution do you think Geoff should use for his analysis?

D. CNP Bank Card Before banks issue a credit card, they usually rate or score the customer in terms of his or her projected probability of being a profitable customer. A typical scoring table ap- pears below.

Age Under 25 25–29 30–34 35� (12 pts.) (5 pts.) (0 pts.) (18 pts.)

Time at �1 yr. 1–2 yrs. 3–4 yrs. 5� yrs. same (9 pts.) (0 pts.) (13 pts.) (20 pts.) address

Auto age None 0–1yr. 2–4 yrs. 5� yrs. (18 pts.) (12 pts.) (13 pts.) (3 pts.)

Monthly None $1–$99 $100–$299 $300� car (15 pts.) (6 pts.) (4 pts.) (0 pts.) payment

Housing $1–$199 $200–$399 Owns Lives with cost (0 pts.) (10 pts.) (12 pts.) relatives

(24 pts.)

Checking/ Both Checking Savings Neither savings only only accounts (15 pts.) (3 pts.) (2 pts.) (0 pts.)

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A Review of Chapters 5–7 263

Sushi’s score of 44 would translate into a probability of be- ing profitable of approximately .81. In other words, 81 per- cent of customers like Sushi will make money for the bank card operations.

Here are the interview results for three potential customers.

David Edward Ann Name Born Brendan McLaughlin Age 42 23 33 Time at same address 9 2 5 Auto age 2 3 7 Monthly car payment $140 $99 $175 Housing cost $300 $200 Owns clear Checking/savings accounts Both Checking only Neither

1. Score each of these customers and estimate their probability of being profitable.

2. What is the probability that all three are profitable? 3. What is the probability that none of them are profitable? 4. Find the entire probability distribution for the number

of profitable customers among this group of three. 5. Write a brief summary of your findings.

Score 30 40 50 60 70 80 90 Probability .70 .78 .85 .90 .94 .95 .96

The score is the sum of the points on the six items. For ex- ample, Sushi Brown is under 25 years old (12 pts.), has lived at the same address for 2 years (0 pts.), owns a 4- year-old car (13 pts.), with car payments of $75 (6 pts.), housing cost of $200 (10 pts.), and a checking account (3 pts.). She would score 44.

A second chart is then used to convert scores into the probability of being a profitable customer. A sample chart of this type appears below.

Practice Test

Part 1—Objective 1. Under what conditions will a probability be greater than 1 or 100 percent? 1. 2. An is the observation of some activity or the act of taking some type of measurement.

2. 3. An is the collection of one or more outcomes to an experiment. 3. 4. A probability is the likelihood that two or more events will happen at the same time.

4. 5. In a (5a) , the order in which the events are counted is important, but in a (5b) , it is not impor-

tant. 5. a. 5. b.

6. In a discrete probability distribution, the sum of the possible outcomes is equal to . 6.

7. Which of the following is NOT a requirement of the binomial distribution? (constant probability of success, three or more outcomes, the result of counts) 7.

8. How many normal distributions are there? (1, 10, 30, 1,000, or infinite—pick one) 8. 9. How many standard normal distributions are there? (1, 10, 30, 1,000, or infinite—pick one) 9.

10. What is the probability of finding a z value between 0 and �0.76? 10. 11. What is the probability of finding a z value greater than 1.67? 11. 12. Two events are if the occurrence of one event does not affect the occurrence of another event.

12. 13. Two events are if by virtue of one event happening the other cannot happen. 13. 14. Which of the following is not true regarding the normal probability distribution? (asymptotic, family of distributions,

only two outcomes, 50 percent of the observations greater than the mean) 14. 15. Which of the following statements best describes the shape of a normal probability distribution? (bell-shaped, uni-

form, V-shaped, no constant shape) 15.

Part 2—Problems 1. Fred Friendly, CPA, has 20 tax returns to prepare before the April 15th deadline. It is late at night so he decides

to do two more before going home. In his stack of accounts, 12 are personal, 5 businesses, and 3 are for charitable organizations. If he selects the two returns at random, what is the probability: a. Both are business? b. At least one is business?

2. The IRS reports that 15 percent of returns where the adjusted gross income is more than $1,000,000 will be sub- ject to a computer audit. For the year 2008, Fred Friendly, CPA, completed 16 returns where the adjusted gross income was more than $1,000,000. a. What is the probability exactly one of these returns will be audited? b. What is the probability at least one will be audited?

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264 Chapter 7

3. Fred works in a tax office with five other CPAs. There are five parking spots beside the office. In how many dif- ferent ways can the cars belonging to the CPAs be arranged in the five spots? Assume they all drive to work.

4. Fred decided to study the number of exemptions claimed on personal tax returns he prepared in 2007. The data are summarized in the following table.

Exemptions Percent

1 20 2 50 3 20 4 10

a. What is the mean number of exemptions per return? b. What is the variance of the number of exemptions per return?

5. In a memo to all those involved in tax preparation, the IRS indicated that the mean amount of refund was $1,600 with a standard deviation of $850. Assume the distribution of the amounts returned follows the normal distribution. a. What percent of the refunds were between $1,600 and $2,000? b. What percent of the refunds were between $900 and $2,000? c. According to the above information, what percent of the refunds were less than $0. That is, the taxpayer owed

the IRS. 6. For the year 2008, Fred Friendly completed a total of 80 returns. He developed the following table summarizing

the relationship between number of dependents and whether or not the client received a refund.

a. What is the name given to this table? b. What is the probability of selecting a client who received a refund? c. What is the probability of selecting a client who received a refund or had one dependent? d. Given that the client received a refund, what is the probability he or she had one dependent? e. What is the probability of selecting a client who did not receive a refund and had one dependent?

7. The IRS offers taxpayers the choice of allowing the IRS to compute the amount of their tax refund. During the busy filing season, the number of returns received at the Springfield Service Center which request this service follows a Poisson distribution with a mean of three per day. What is the probability that on a particular day: a. There are no requests? b. Exactly three requests appear? c. Five or more requests take place? d. There are no requests on two consecutive days?

Dependents

Refund 1 2 3 or more Total

Yes 20 20 10 50 No 10 20 0 30

Total 30 40 10 80

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8 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Explain why a sample is often the only feasible way to learn something about a population.

LO2 Describe methods to select a sample.

LO3 Define sampling error.

LO4 Describe the sampling distribution of the sample mean.

LO5 Explain the central limit theorem.

LO6 Define the standard error of the mean.

LO7 Apply the central limit theorem to find probabilities of selecting possible sample means from a specified population.

Sampling Methods and the Central Limit Theorem

The Nike annual report says that the average American buys 6.5 pairs

of sports shoes per year. Suppose the population standard deviation

is 2.1 and that a sample of 81 customers will be examined next year.

What is the standard error of the mean in this experiment? (See

Exercise 45 and LO5.)

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266 Chapter 8

8.1 Introduction Chapters 2 through 4 emphasize techniques to describe data. To illustrate these tech- niques, we organize the profits for the sale of 180 vehicles by the four dealers included in the Applewood Auto Group into a frequency distribution and compute various mea- sures of location and dispersion. Such measures as the mean and the standard devi- ation describe the typical profit and the spread in the profits. In these chapters, the emphasis is on describing the condition of the data. That is, we describe something that has already happened.

Chapter 5 starts to lay the foundation for statistical inference with the study of probability. Recall that in statistical inference our goal is to determine some- thing about a population based only on the sample. The population is the entire group of individuals or objects under consideration, and the sample is a part or subset of that population. Chapter 6 extends the probability concepts by describ- ing three discrete probability distributions: the binomial, the hypergeometric, and the Poisson. Chapter 7 describes three continuous probability distributions: the uniform, normal, and exponential. Probability distributions encompass all possi- ble outcomes of an experiment and the probability associated with each outcome. We use probability distributions to evaluate the likelihood something occurs in the future.

This chapter begins our study of sampling. Sampling is a tool to infer something about a population. We begin this chapter by discussing methods of selecting a sam- ple from a population. Next, we construct a distribution of the sample mean to under- stand how the sample means tend to cluster around the population mean. Finally, we show that for any population the shape of this sampling distribution tends to follow the normal probability distribution.

8.2 Sampling Methods In Chapter 1, we said the purpose of inferential statistics is to find something about a population based on a sample. A sample is a portion or part of the population of interest. In many cases, sampling is more feasible than studying the entire popula- tion. In this section, we discuss the major reasons for sampling, and then several methods for selecting a sample.

Reasons to Sample When studying characteristics of a population, there are many practical reasons why we prefer to select portions or samples of a population to observe and measure. Here are some of the reasons for sampling:

1. To contact the whole population would be time consuming. A candidate for a national office may wish to determine her chances for election. A sample poll using the regular staff and field interviews of a professional polling firm would take only 1 or 2 days. Using the same staff and interviewers and working 7 days a week, it would take nearly 200 years to contact all the voting population! Even if a large staff of interviewers could be assembled, the benefit of contacting all of the voters would probably not be worth the time.

2. The cost of studying all the items in a population may be prohibitive. Pub- lic opinion polls and consumer testing organizations, such as Harris International, CBS News Polls, and Zogby International, usually contact fewer than 2,000 of the nearly 60 million families in the United States. One consumer panel–type organization charges about $40,000 to mail samples and tabulate responses in order to test a product (such as breakfast cereal, cat food, or perfume). The same product test using all 60 million families would cost about $1 billion.

Statistics in Action

With the significant role played by infer- ential statistics in all branches of science, the availability of large sources of ran- dom numbers has become a necessity. The first book of random numbers, containing 41,600 random digits gener- ated by L. Tippett, was published in 1927. In 1938, R. A. Fisher and F. Yates published 15,000 random digits generated using two decks of cards. In 1955, RAND Corpo- ration published a million random dig- its, generated by the random frequency pulses of an elec- tronic roulette wheel. By 1970, applications of sam- pling required bil- lions of random numbers. Methods have since been developed for gener- ating, using a com- puter, digits that are “almost” random and hence are called pseudo-random. The question of whether a computer program can be used to gener- ate numbers that are truly random remains a debatable issue.

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Sampling Methods and the Central Limit Theorem 267

3. The physical impossibility of checking all items in the population. Some populations are infinite. It would be impossible to check all the water in Lake Erie for bacterial levels, so we select samples at various locations. The popu- lations of fish, birds, snakes, mosquitoes, and the like are large and are con- stantly moving, being born, and dying. Instead of even attempting to count all the ducks in Canada or all the fish in Lake Pontchartrain, we make estimates using various techniques—such as counting all the ducks on a pond picked at random, making creel checks, or setting nets at predetermined places in the lake.

4. The destructive nature of some tests. If the wine tasters at the Sutter Home Winery in California drank all the wine to evaluate the vintage, they would consume the entire crop, and none would be available for sale. In the area of industrial production, steel plates, wires, and similar products must have a certain minimum tensile strength. To ensure that the product meets the minimum standard, the Quality Assurance Depart- ment selects a sample from the current production. Each piece is stretched until it breaks and the breaking point (usually mea- sured in pounds per square inch) recorded. Obviously, if all the wire or all the plates were tested for tensile strength, none would be available for sale or use. For the same reason, only a few seeds are tested for germination by Burpee Seeds Inc. prior to the planting season.

5. The sample results are adequate. Even if funds were available, it is doubtful the additional accuracy of a 100 percent sample—that is, studying the entire population—is essential in most problems. For example, the federal government uses a sample of grocery stores scattered throughout the United States to determine the monthly index of food prices. The prices of bread, beans, milk, and other major food items are included in the index. It is unlikely that the inclu- sion of all grocery stores in the United States would significantly affect the index, since the prices of milk, bread, and other major foods usually do not vary by more than a few cents from one chain store to another.

Simple Random Sampling The most widely used type of sampling is a simple random sample.

To illustrate simple random sampling and selection, suppose a population con- sists of 845 employees of Nitra Industries. A sample of 52 employees is to be selected from that population. One way of ensuring that every employee in the pop- ulation has the same chance of being chosen is to first write the name of each employee on a small slip of paper and deposit all of the slips in a box. After they have been thoroughly mixed, the first selection is made by drawing a slip out of the box without looking at it. This process is repeated until the sample of 52 employees is chosen.

A more convenient method of selecting a random sample is to use the iden- tification number of each employee and a table of random numbers such as the one in Appendix B.6. As the name implies, these numbers have been generated by a random process (in this case, by a computer). For each digit of a number, the probability of 0, 1, 2, . . . , 9 is the same. Thus, the probability that employee

SIMPLE RANDOM SAMPLE A sample selected so that each item or person in the population has the same chance of being included.

A table of random numbers is an efficient way to select members of the sample.

LO1 Explain why a sample is often the only feasible way to learn something about a population.

LO2 Describe methods to select a sample.

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Statistics in Action

Is discrimination tak- ing a bite out of your paycheck? Before you answer, consider a recent article in Per- sonnel Journal. These findings indicate that attractive men and women earn about 5 percent more than av- erage lookers, who in turn earn about 5 per- cent more than their plain counterparts. This is true for both men and women. It is also true for a wide range of occupations, from construction work to auto repair to telemarketing, occu- pations for which it would seem that looks would not matter.

number 011 will be selected is the same as for employee 722 or employee 382. By using random numbers to select employees, bias is eliminated from the selec- tion process.

A portion of a table of random numbers is shown in the following illustration. To select a sample of employees, you first choose a starting point in the table. Any starting point will do. Suppose the time is 3:04. You might look at the third column and then move down to the fourth set of numbers. The number is 03759. Since there are only 845 employees, we will use the first three digits of a five-digit ran- dom number. Thus, 037 is the number of the first employee to be a member of the sample. Another way of selecting the starting point is to close your eyes and point at a number in the table. To continue selecting employees, you could move in any direction. Suppose you move right. The first three digits of the number to the right of 03759 are 447—the number of the employee selected to be the second member of the sample. The next three-digit number to the right is 961. You skip 961 because there are only 845 employees. You continue to the right and select employee 784, then 189, and so on.

5 0 5 2 5 5 7 4 5 4 2 8 4 5 5 6 8 2 2 6 3 4 6 5 6 3 8 8 8 4 3 9 0 1 8 7 2 5 0 7 5 3 3 8 0 5 3 8 2 7 4 2 4 8 6 5 4 4 6 5 7 1 8 1 9 9 1 1 9 9 3 4 9 8 6 7 4 2 9 7 0 0 1 4 4 3 8 6 7 6 8 9 9 6 7 9 8 8 6 9 3 9 7 4 4 6 8 8 5 1 2 7 3 0 5 0 3 7 5 9 4 4 7 2 3 9 6 1 0 8 7 8 4 8 9 1 8 9 1 0 0 6 7 3 8 6 2 8 7 9 0 3 9 1 0 1 7 3 5 0 4 9 1 6 9 0 3 8 5 0 1 8 9 1 0 1 1 4 4 8 1 0 7 3 4 0 5 8 3 7 2 4 3 9 7 1 0 4 2 0 1 6 7 1 2 9 4 4 9 6

Starting Second Third Fourth point employee employee employee

Most statistical software packages have available a routine that will select a simple random sample. The following example uses the Excel System to select a random sample.

268 Chapter 8

Example Jane and Joe Miley operate the Foxtrot Inn, a bed and breakfast in Tryon, North Carolina. There are eight rooms available for rent at this B&B. Listed below is the number of these eight rooms rented each day during June 2011. Use Excel to select a sample of five nights during the month of June.

Solution Excel will select the random sample and report the results. On the first sampled date, four of the eight rooms were rented. On the second sampled date in June, seven of the eight rooms were rented. The information is reported in column D of the Excel

June Rentals

1 0 2 2 3 3 4 2 5 3 6 4 7 2 8 3 9 4

10 7

June Rentals

11 3 12 4 13 4 14 4 15 7 16 0 17 5 18 3 19 6 20 2

June Rentals

21 3 22 2 23 3 24 6 25 0 26 4 27 1 28 1 29 3 30 3

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Sampling Methods and the Central Limit Theorem 269

Self-Review 8–1 The following class roster lists the students enrolling in an introductory course in business statistics. Three students are to be randomly selected and asked various questions regarding course content and method of instruction. (a) The numbers 00 through 45 are handwritten on slips of paper and placed in a bowl.

The three numbers selected are 31, 7, and 25. Which students would be included in the sample?

(b) Now use the table of random digits, Appendix B.6, to select your own sample. (c) What would you do if you encountered the number 59 in the table of random digits?

CSPM 264 01 BUSINESS & ECONOMIC STAT 8:00 AM 9:40 AM MW ST 118 LIND D

RANDOM CLASS NUMBER NAME RANK 00 ANDERSON, RAYMOND SO 01 ANGER, CHERYL RENEE SO 02 BALL, CLAIRE JEANETTE FR 03 BERRY, CHRISTOPHER G FR 04 BOBAK, JAMES PATRICK SO 05 BRIGHT, M. STARR JR 06 CHONTOS, PAUL JOSEPH SO 07 DETLEY, BRIAN HANS JR 08 DUDAS, VIOLA SO 09 DULBS, RICHARD ZALFA JR 10 EDINGER, SUSAN KEE SR 11 FINK, FRANK JAMES SR 12 FRANCIS, JAMES P JR 13 GAGHEN, PAMELA LYNN JR 14 GOULD, ROBYN KAY SO 15 GROSENBACHER, SCOTT ALAN SO 16 HEETFIELD, DIANE MARIE SO 17 KABAT, JAMES DAVID JR 18 KEMP, LISA ADRIANE FR 19 KILLION, MICHELLE A SO 20 KOPERSKI, MARY ELLEN SO 21 KOPP, BRIDGETTE ANN SO 22 LEHMANN, KRISTINA MARIE JR

RANDOM CLASS NUMBER NAME RANK 23 MEDLEY, CHERYL ANN SO 24 MITCHELL, GREG R FR 25 MOLTER, KRISTI MARIE SO 26 MULCAHY, STEPHEN ROBERT SO 27 NICHOLAS, ROBERT CHARLES JR 28 NICKENS, VIRGINIA SO 29 PENNYWITT, SEAN PATRICK SO 30 POTEAU, KRIS E JR 31 PRICE, MARY LYNETTE SO 32 RISTAS, JAMES SR 33 SAGER, ANNE MARIE SO 34 SMILLIE, HEATHER MICHELLE SO 35 SNYDER, LEISHA KAY SR 36 STAHL, MARIA TASHERY SO 37 ST. JOHN, AMY J SO 38 STURDEVANT, RICHARD K SO 39 SWETYE, LYNN MICHELE SO 40 WALASINSKI, MICHAEL SO 41 WALKER, DIANE ELAINE SO 42 WARNOCK, JENNIFER MARY SO 43 WILLIAMS, WENDY A SO 44 YAP, HOCK BAN SO 45 YODER, ARLAN JAY JR

spreadsheet. The Excel steps are listed in the Software Commands section at the end of the chapter. The Excel system performs the sampling with replacement. This means it is possible for the same day to appear more than once in a sample.

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Systematic Random Sampling The simple random sampling procedure may be awkward in some research situa- tions. For example, suppose the sales division of Computer Graphic Inc. needs to quickly estimate the mean dollar revenue per sale during the past month. It finds that 2,000 sales invoices were recorded and stored in file drawers, and decides to select 100 invoices to estimate the mean dollar revenue. Simple random sampling requires the numbering of each invoice before using the random number table to select the 100 invoices. The numbering process would be a very time-consuming task. Instead, we use systematic random sampling.

First, k is calculated as the population size divided by the sample size. For Com- puter Graphic Inc., we would select every 20th (2,000/100) invoice from the file drawers; in so doing, the numbering process is avoided. If k is not a whole number, then round down.

Random sampling is used in the selection of the first invoice. For example, a number from a random number table between 1 and k, or 20, would be selected. Say the random number was 18. Then, starting with the 18th invoice, every 20th invoice (18, 38, 58, etc.) would be selected as the sample.

Before using systematic random sampling, we should carefully observe the phys- ical order of the population. When the physical order is related to the population char- acteristic, then systematic random sampling should not be used. For example, if the invoices in the example were filed in order of increasing sales, systematic random sam- pling would not guarantee a random sample. Other sampling methods should be used.

Stratified Random Sampling When a population can be clearly divided into groups based on some characteristic, we may use stratified random sampling. It guarantees each group is represented in the sample. The groups are also called strata. For example, college students can be grouped as full time or part time, male or female, or traditional or nontraditional. Once the strata are defined, we can apply simple random sampling within each group or stratum to collect the sample.

For instance, we might study the advertising expenditures for the 352 largest com- panies in the United States. Suppose the objective of the study is to determine whether firms with high returns on equity (a measure of profitability) spent more of each sales dollar on advertising than firms with a low return or deficit. To make sure that the sam- ple is a fair representation of the 352 companies, the companies are grouped on percent return on equity. Table 8–1 shows the strata and the relative frequencies. If simple random sampling was used, observe that firms in the 3rd and 4th strata have a high chance of selection (probability of 0.87) while firms in the other strata have a low chance of selection (probability of 0.13). We might not select any firms in stratum 1 or 5 simply by chance. However, stratified random sampling will guarantee that at least one firm in strata 1 and 5 are represented in the sample. Let’s say that 50 firms are selected for intensive study. Then 1 (0.02 � 50) firm from stratum 1 would be ran- domly selected, 5 (0.10 � 50) firms from stratum 2 would be randomly selected, and so on. In this case, the number of firms sampled from each stratum is proportional to the stratum’s relative frequency in the population. Stratified sampling has the advantage, in

STRATIFIED RANDOM SAMPLE A population is divided into subgroups, called strata, and a sample is randomly selected from each stratum.

SYSTEMATIC RANDOM SAMPLE A random starting point is selected, and then every kth member of the population is selected.

270 Chapter 8

Statistics in Action

Random and unbi- ased sampling meth- ods are extremely important to make valid statistical infer- ences. In 1936, a straw vote to predict the outcome of the presidential race be- tween Franklin Roosevelt and Alfred Landon was done. Ten million ballots in the form of return- able postcards were sent to addresses taken from telephone directories and auto- mobile registrations. A high proportion of the ballots were returned, with 59 percent in favor of Landon and 41 percent favoring Roosevelt. On Elec- tion Day, Roosevelt won with 61 percent of the vote. Landon had 39 percent. In the mid-1930s peo- ple who had tele- phones and drove automobiles clearly did not represent American voters!

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Sampling Methods and the Central Limit Theorem 271

Profitability Number of Relative Number Stratum (return on equity) Firms Frequency Sampled

1 30 percent and over 8 0.02 1* 2 20 up to 30 percent 35 0.10 5* 3 10 up to 20 percent 189 0.54 27 4 0 up to 10 percent 115 0.33 16 5 Deficit 5 0.01 1

Total 352 1.00 50

CLUSTER SAMPLE A population is divided into clusters using naturally occurring geographic or other boundaries. Then, clusters are randomly selected and a sample is collected by randomly selecting from each cluster.

Suppose you want to determine the views of residents in Oregon about state and federal environmental protection policies. Selecting a random sample of resi- dents in Oregon and personally contacting each one would be time consuming and very expensive. Instead, you could employ cluster sampling by subdividing the state into small units—either counties or regions. These are often called primary units.

Suppose you divided the state into 12 primary units, then selected at random four regions—2, 7, 4, and 12—and concentrated your efforts in these primary units. You could take a random sample of the residents in each of these regions and interview them. (Note that this is a combination of cluster sampling and simple random sampling.)

some cases, of more accurately reflecting the characteristics of the population than does simple random or systematic random sampling.

TABLE 8–1 Number Selected for a Proportional Stratified Random Sample

*0.02 of 50 � 1, 0.10 of 50 � 5, etc.

Cluster Sampling Another common type of sampling is cluster sampling. It is often employed to reduce the cost of sampling a population scattered over a large geographic area.

4 5

Self-Review 8–2 Refer to Self-Review 8–1 and the class roster on page 269. Suppose a systematic random sample will select every ninth student enrolled in the class. Initially, the fourth student on the list was selected at random. That student is numbered 03. Remembering that the ran- dom numbers start with 00, which students will be chosen to be members of the sample?

Many other sampling methods

The discussion of sampling methods in the preceding sections did not include all the sampling methods available to a researcher. Should you become involved in a major research project in marketing, finance, accounting, or other areas, you would need to consult books devoted solely to sample theory and sample design.

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Exercises 1. The following is a list of Marco’s Pizza stores in Lucas County. Also noted is whether the

store is corporate-owned (C) or manager-owned (M). A sample of four locations is to be selected and inspected for customer convenience, safety, cleanliness, and other features.

ID No. Address Type ID No. Address Type

00 2607 Starr Av C 12 2040 Ottawa River Rd C 01 309 W Alexis Rd C 13 2116 N Reynolds Rd C 02 2652 W Central Av C 14 3678 Rugby Dr C 03 630 Dixie Hwy M 15 1419 South Av C 04 3510 Dorr St C 16 1234 W Sylvania Av C 05 5055 Glendale Av C 17 4624 Woodville Rd M 06 3382 Lagrange St M 18 5155 S Main M 07 2525 W Laskey Rd C 19 106 E Airport Hwy C 08 303 Louisiana Av C 20 6725 W Central M 09 149 Main St C 21 4252 Monroe C 10 835 S McCord Rd M 22 2036 Woodville Rd C 11 3501 Monroe St M 23 1316 Michigan Av M

a. The random numbers selected are 08, 18, 11, 54, 02, 41, and 54. Which stores are selected?

b. Use the table of random numbers to select your own sample of locations. c. A sample is to consist of every seventh location. The number 03 is the starting point.

Which locations will be included in the sample? d. Suppose a sample is to consist of three locations, of which two are corporate-owned

and one is manager-owned. Select a sample accordingly. 2. The following is a list of hospitals in the Cincinnati (Ohio) and Northern Kentucky Region.

Also included is whether the hospital is a general medical/surgical hospital (M/S) or a specialty hospital (S). We are interested in estimating the average number of full- and part-time nurses employed in the area hospitals. a. A sample of five hospitals is to be randomly selected. The random numbers are 09,

16, 00, 49, 54, 12, and 04. Which hospitals are included in the sample? b. Use a table of random numbers to develop your own sample of five hospitals.

272 Chapter 8

ID Number Name Address Type

00 Bethesda North 10500 Montgomery M/S Cincinnati, Ohio 45242

01 Ft. Hamilton-Hughes 630 Eaton Avenue M/S Hamilton, Ohio 45013

02 Jewish Hospital- 4700 East Galbraith Rd. M/S Kenwood Cincinnati, Ohio 45236

03 Mercy Hospital- 3000 Mack Road M/S Fairfield Fairfield, Ohio 45014

04 Mercy Hospital- 100 Riverfront Plaza M/S Hamilton Hamilton, Ohio 45011

05 Middletown 105 McKnight Drive M/S Regional Middletown, Ohio 45044

06 Clermont Mercy 3000 Hospital Drive M/S Hospital Batavia, Ohio 45103

07 Mercy Hospital- 7500 State Road M/S Anderson Cincinnati, Ohio 45255

08 Bethesda Oak 619 Oak Street M/S Hospital Cincinnati, Ohio 45206

09 Children’s Hospital 3333 Burnet Avenue M/S Medical Center Cincinnati, Ohio 45229

ID Number Name Address Type

10 Christ Hospital 2139 Auburn Avenue M/S Cincinnati, Ohio 45219

11 Deaconess 311 Straight Street M/S Hospital Cincinnati, Ohio 45219

12 Good Samaritan 375 Dixmyth Avenue M/S Hospital Cincinnati, Ohio 45220

13 Jewish Hospital 3200 Burnet Avenue M/S Cincinnati, Ohio 45229

14 University Hospital 234 Goodman Street M/S Cincinnati, Ohio 45267

15 Providence Hospital 2446 Kipling Avenue M/S Cincinnati, Ohio 45239

16 St. Francis- 3131 Queen City Avenue M/S St. George Hospital Cincinnati, Ohio 45238

17 St. Elizabeth Medical 401 E. 20th Street M/S Center, North Unit Covington, Kentucky 41014

18 St. Elizabeth Medical One Medical Village M/S Center, South Unit Edgewood, Kentucky 41017

19 St. Luke’s Hospital 7380 Turfway Drive M/S West Florence, Kentucky 41075

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Sampling Methods and the Central Limit Theorem 273

c. A sample is to consist of every fifth location. We select 02 as the starting point. Which hospitals will be included in the sample?

d. A sample is to consist of four medical and surgical hospitals and one specialty hos- pital. Select an appropriate sample.

3. Listed below are the 35 members of the Metro Toledo Automobile Dealers Association. We would like to estimate the mean revenue from dealer service departments.

ID ID ID Number Dealer Number Dealer Number Dealer

00 Dave White Acura 11 Thayer Chevrolet/Toyota 23 Kistler Ford, Inc. 01 Autofair Nissan 12 Spurgeon Chevrolet Motor Sales, Inc. 24 Lexus of Toledo 02 Autofair Toyota-Suzuki 13 Dunn Chevrolet 25 Mathews Ford Oregon, Inc. 03 George Ball’s Buick GMC Truck 14 Don Scott Chevrolet-Pontiac 26 Northtowne Chevrolet 04 Yark Automotive Group 15 Dave White Chevrolet Co. 27 Quality Ford Sales, Inc. 05 Bob Schmidt Chevrolet 16 Dick Wilson Pontiac 28 Rouen Chrysler Jeep Eagle 06 Bowling Green Lincoln Mercury 17 Doyle Pontiac Buick 29 Saturn of Toledo

Jeep Eagle 18 Franklin Park Lincoln Mercury 30 Ed Schmidt Pontiac Jeep Eagle 07 Brondes Ford 19 Genoa Motors 31 Southside Lincoln Mercury 08 Brown Honda 20 Great Lakes Ford Nissan 32 Valiton Chrysler 09 Brown Mazda 21 Grogan Towne Chrysler 33 Vin Divers 10 Charlie’s Dodge 22 Hatfield Motor Sales 34 Whitman Ford

a. We want to select a random sample of five dealers. The random numbers are: 05, 20, 59, 21, 31, 28, 49, 38, 66, 08, 29, and 02. Which dealers would be included in the sample?

b. Use the table of random numbers to select your own sample of five dealers. c. A sample is to consist of every seventh dealer. The number 04 is selected as the start-

ing point. Which dealers are included in the sample? 4. Listed below are the 27 Nationwide Insurance agents in the Toledo, Ohio, metropolitan

area. We would like to estimate the mean number of years employed with Nationwide.

ID ID ID Number Agent Number Agent Number Agent

00 Bly Scott 3332 W Laskey Rd 10 Heini Bernie 7110 W Centra 19 Riker Craig 2621 N Reynolds Rd 01 Coyle Mike 5432 W Central Av 11 Hinckley Dave 20 Schwab Dave 572 W Dussel Dr 02 Denker Brett 7445 Airport Hwy 14 N Holland Sylvania Rd 21 Seibert John H 201 S Main 03 Denker Rollie 7445 Airport Hwy 12 Joehlin Bob 3358 Navarre Av 22 Smithers Bob 229 Superior St 04 Farley Ron 1837 W Alexis Rd 13 Keisser David 3030 W Sylvania Av 23 Smithers Jerry 229 Superior St 05 George Mark 7247 W Central Av 14 Keisser Keith 5902 Sylvania Av 24 Wright Steve 105 S Third St 06 Gibellato Carlo 6616 Monroe St 15 Lawrence Grant 342 W Dussel Dr 25 Wood Tom 112 Louisiana Av 07 Glemser Cathy 5602 Woodville Rd 16 Miller Ken 2427 Woodville Rd 26 Yoder Scott 6 Willoughby Av 08 Green Mike 4149 Holland Sylvania Rd 17 O’Donnell Jim 7247 W Central Av 09 Harris Ev 2026 Albon Rd 18 Priest Harvey 5113 N Summit St

ID Number Name Address Type

20 St. Luke’s Hospital 85 North Grand Avenue M/S East Ft. Thomas, Kentucky 41042

21 Care Unit Hospital 3156 Glenmore Avenue S Cincinnati, Ohio 45211

22 Emerson Behavioral 2446 Kipling Avenue S Science Cincinnati, Ohio 45239

23 Pauline Warfield 1101 Summit Road S Lewis Center for Cincinnati, Ohio 45237 Psychiatric Treat.

24 Children’s Psychiatric 502 Farrell Drive S No. Kentucky Covington, Kentucky 41011

ID Number Name Address Type

25 Drake Center Rehab— 151 W. Galbraith Road S Long Term Cincinnati, Ohio 45216

26 No. Kentucky Rehab 201 Medical Village S Hospital—Short Term Edgewood, Kentucky

27 Shriners Burns 3229 Burnet Avenue S Institute Cincinnati, Ohio 45229

28 VA Medical Center 3200 Vine S Cincinnati, Ohio 45220

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SAMPLING ERROR The difference between a sample statistic and its corresponding population parameter.

a. We want to select a random sample of four agents. The random numbers are: 02, 59, 51, 25, 14, 29, 77, 69, and 18. Which dealers would be included in the sample?

b. Use the table of random numbers to select your own sample of four agents. c. A sample is to consist of every seventh dealer. The number 04 is selected as the start-

ing point. Which agents will be included in the sample?

8.3 Sampling “Error” In the previous section, we discussed sampling methods that are used to select a sam- ple that is a fair and unbiased representation of the population. In each method, the selection of every possible sample of a specified size from a population has a known chance or probability. This is another way to describe an unbiased sampling method.

Samples are used to estimate population characteristics. For example, the mean of a sample is used to estimate the population mean. However, since the sample is a part or portion of the population, it is unlikely that the sample mean would be exactly equal to the population mean. Similarly, it is unlikely that the sample stan- dard deviation would be exactly equal to the population standard deviation. We can therefore expect a difference between a sample statistic and its corresponding pop- ulation parameter. This difference is called sampling error.

274 Chapter 8

The following example clarifies the idea of sampling error.

LO3 Define sampling error.

Example Refer to the previous example on page 268, where we studied the number of rooms rented at the Foxtrot Inn bed and breakfast in Tryon, North Carolina. The popula- tion is the number of rooms rented each of the 30 days in June 2011. Find the mean of the population. Use Excel or other statistical software to select three random samples of five days. Calculate the mean of each sample and compare it to the population mean. What is the sampling error in each case?

During the month, there were a total of 94 rentals. So the mean number of units rented per night is 3.13. This is the population mean. Hence we designate this value with the Greek letter �.

The first random sample of five nights resulted in the following number of rooms rented: 4, 7, 4, 3, and 1. The mean of this sample of five nights is 3.8 rooms, which we designate as . The bar over the X reminds us that it is a sample mean and the subscript 1 indicates it is the mean of the first sample.

The sampling error for the first sample is the difference between the population mean (3.13) and the first sample mean (3.80). Hence, the sampling error is ( ). The second random sample of five days from the population of all 30 days in June revealed the following number of rooms rented: 3, 3, 2, 3, and 6. The mean of these five values is 3.4, found by

The sampling error is ( ).X2 � � � 3.4 � 3.13 � 0.27

X2 � ©X n

� 3 � 3 � 2 � 3 � 6

5 � 3.4

X1 � � � 3.80 � 3.13 � 0.67

X1 � ©X n

� 4 � 7 � 4 � 3 � 1

5 �

19 5

� 3.80

� � ©X N

� 0 � 2 � 3 � . . . � 3

30 �

94 30

� 3.13

Solution

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Sampling Methods and the Central Limit Theorem 275

8.4 Sampling Distribution of the Sample Mean Now that we have discovered the possibility of a sampling error when sample results are used to estimate a population parameter, how can we make an accurate prediction about the possible success of a newly developed toothpaste or other product, based only on sample results? How can the quality-assurance department in a mass- production firm release a shipment of microchips based only on a sample of 10 chips? How can the CNN/USA Today or ABC News/Washington Post polling organizations make an accurate prediction about a presidential race based on a sample of 1,200 registered voters out of a voting population of nearly 90 million? To answer these questions, we first develop a sampling distribution of the sample mean.

The sample means in the previous example varied from one sample to the next. The mean of the first sample of 5 days was 3.80 rooms, and the second sample was 3.40 rooms. The population mean was 3.13 rooms. If we organized the means of all possible samples of 5 days into a probability distribution, the result is called the sampling distribution of the sample mean.

Sample means vary from sample to sample.

SAMPLING DISTRIBUTION OF THE SAMPLE MEAN A probability distribution of all possible sample means of a given sample size.

In the third random sample, the mean was 1.8 and the sampling error was �1.33. Each of these differences, 0.67, 0.27, and �1.33, is the sampling error made in

estimating the population mean. Sometimes these errors are positive values, indi- cating that the sample mean overestimated the population mean; other times they are negative values, indicating the sample mean was less than the population mean.

In this case, where we have a population of 30 values and samples of 5 values, there is a very large number of possible samples—142,506 to be exact! To find this value, use the combination formula (5–10) in Section 5.8 on page 174. Each of the 142,506 different samples has the same chance of being selected. Each sample may have a different sample mean and therefore a different sampling error. The value of the sampling error is based on the particular one of the 142,506 different possible samples selected. Therefore, the sampling errors are random and occur by chance. If you were to determine the sum of these sampling errors over a large number of samples, the result would be very close to zero. This is true because the sample mean is an unbiased estimator of the population mean.

LO4 Describe the sampling distribution of the sample mean.

A sample mean is an unbiased estimate of the population mean.

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The following example illustrates the construction of a sampling distribution of the sample mean.

276 Chapter 8

Example

Solution

Employee Hourly Earnings Employee Hourly Earnings

Joe $7 Jan $7 Sam 7 Art 8 Sue 8 Ted 9 Bob 8

Hourly Hourly Sample Employees Earnings Sum Mean Sample Employees Earnings Sum Mean

1 Joe, Sam $7, $7 $14 $7.00 12 Sue, Bob $8, $8 $16 $8.00 2 Joe, Sue 7, 8 15 7.50 13 Sue, Jan 8, 7 15 7.50 3 Joe, Bob 7, 8 15 7.50 14 Sue, Art 8, 8 16 8.00 4 Joe, Jan 7, 7 14 7.00 15 Sue, Ted 8, 9 17 8.50 5 Joe, Art 7, 8 15 7.50 16 Bob, Jan 8, 7 15 7.50 6 Joe, Ted 7, 9 16 8.00 17 Bob, Art 8, 8 16 8.00 7 Sam, Sue 7, 8 15 7.50 18 Bob, Ted 8, 9 17 8.50 8 Sam, Bob 7, 8 15 7.50 19 Jan, Art 7, 8 15 7.50 9 Sam, Jan 7, 7 14 7.00 20 Jan, Ted 7, 9 16 8.00

10 Sam, Art 7, 8 15 7.50 21 Art, Ted 8, 9 17 8.50 11 Sam, Ted 7, 9 16 8.00

Tartus Industries has seven production employees (considered the population). The hourly earnings of each employee are given in Table 8–2.

TABLE 8–2 Hourly Earnings of the Production Employees of Tartus Industries

1. What is the population mean? 2. What is the sampling distribution of the sample mean for samples of size 2? 3. What is the mean of the sampling distribution? 4. What observations can be made about the population and the sampling

distribution?

Here are the solutions to the questions.

1. The population mean is $7.71, found by:

We identify the population mean with the Greek letter �. Our policy, stated in Chapters 1, 3, and 4, is to identify population parameters with Greek letters.

2. To arrive at the sampling distribution of the sample mean, we need to select all possible samples of 2 without replacement from the population, then compute the mean of each sample. There are 21 possible samples, found by using for- mula (5–10) in Section 5.8 on page 174.

where N � 7 is the number of items in the population and n � 2 is the num- ber of items in the sample.

The 21 sample means from all possible samples of 2 that can be drawn from the population are shown in Table 8–3. These 21 sample means are used to construct a probability distribution. This is the sampling distribution of the sample mean, and it is summarized in Table 8–4.

TABLE 8–3 Sample Means for All Possible Samples of 2 Employees

NCn � N!

n!(N � n)! �

7! 2!(7 � 2)!

� 21

� � ©X N

� $7 � $7 � $8 � $8 � $7 � $8 � $9

7 � $7.71

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Sampling Methods and the Central Limit Theorem 277

In summary, we took all possible random samples from a population and for each sample calculated a sample statistic (the mean amount earned). This example illustrates important relationships between the population distribution and the sam- pling distribution of the sample mean:

1. The mean of the sample means is exactly equal to the population mean. 2. The dispersion of the sampling distribution of sample means is narrower than

the population distribution. 3. The sampling distribution of sample means tends to become bell-shaped and

to approximate the normal probability distribution.

TABLE 8–4 Sampling Distribution of the Sample Mean for n � 2

3. The mean of the sampling distribution of the sample mean is obtained by sum- ming the various sample means and dividing the sum by the number of sam- ples. The mean of all the sample means is usually written The � reminds us that it is a population value because we have considered all possible samples. The subscript indicates that it is the sampling distribution of the sample mean.

4. Refer to Chart 8–1, which shows both the population distribution and the distri- bution of the sample mean. These observations can be made: a. The mean of the distribution of the sample mean ($7.71) is equal to the mean

of the population: . b. The spread in the distribution of the sample mean is less than the spread

in the population values. The sample mean ranges from $7.00 to $8.50, while the population values vary from $7.00 up to $9.00. Notice, as we increase the size of the sample, the spread of the distribution of the sample mean becomes smaller.

c. The shape of the sampling distribution of the sample mean and the shape of the frequency distribution of the population values are different. The distri- bution of the sample mean tends to be more bell-shaped and to approximate the normal probability distribution.

� � �X

� $162

21 � $7.71

�X � Sum of all sample means Total number of samples

� $7.00 � $7.50 � # # # � $8.50

�X.

Population mean is equal to the mean of the sample means

Population distribution

.40

.30

.20

.10

.40

.30

.20

.10

Hourly earnings8 9μ

Distribution of sample mean

7 Sample mean of hourly earnings

8 97.5 8.5 X Xμ

Pr ob

ab ili

Pr ob

ab ili

CHART 8–1 Distributions of Population Values and Sample Mean

Sample Mean Number of Means Probability

$7.00 3 .1429 7.50 9 .4285 8.00 6 .2857 8.50 3 .1429

21 1.0000

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Self-Review 8–3 The lengths of service of all the executives employed by Standard Chemicals are:

(a) Using the combination formula, how many samples of size 2 are possible? (b) List all possible samples of 2 executives from the population and compute their means. (c) Organize the means into a sampling distribution. (d) Compare the population mean and the mean of the sample means. (e) Compare the dispersion in the population with that in the distribution of the sample mean. (f) A chart portraying the population values follows. Is the distribution of population val-

ues normally distributed (bell-shaped)?

(g) Is the distribution of the sample mean computed in part (c) starting to show some ten- dency toward being bell-shaped?

278 Chapter 8

Name Years

Mr. Snow 20 Ms. Tolson 22 Mr. Kraft 26 Ms. Irwin 24 Mr. Jones 28

Given a bell-shaped or normal probability distribution, we will be able to apply concepts from Chapter 7 to determine the probability of selecting a sample with a specified sample mean. In the next section, we will show the importance of sam- ple size as it relates to the sampling distribution of sample means.

0 20 22 24 26 28

Length of service

Fr eq

ue nc

Exercises 5. A population consists of the following four values: 12, 12, 14, and 16.

a. List all samples of size 2, and compute the mean of each sample. b. Compute the mean of the distribution of the sample mean and the population mean.

Compare the two values. c. Compare the dispersion in the population with that of the sample mean.

6. A population consists of the following five values: 2, 2, 4, 4, and 8. a. List all samples of size 2, and compute the mean of each sample. b. Compute the mean of the distribution of sample means and the population mean.

Compare the two values. c. Compare the dispersion in the population with that of the sample means.

7. A population consists of the following five values: 12, 12, 14, 15, and 20. a. List all samples of size 3, and compute the mean of each sample. b. Compute the mean of the distribution of sample means and the population mean.

Compare the two values. c. Compare the dispersion in the population with that of the sample means.

8. A population consists of the following five values: 0, 0, 1, 3, 6. a. List all samples of size 3, and compute the mean of each sample. b. Compute the mean of the distribution of sample means and the population mean.

Compare the two values. c. Compare the dispersion in the population with that of the sample means.

9. In the law firm Tybo and Associates, there are six partners. Listed next is the number of cases each associate actually tried in court last month.

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Sampling Methods and the Central Limit Theorem 279

Associate Number of Cases

Ruud 3 Wu 6 Sass 3 Flores 3 Wilhelms 0 Schueller 1

a. How many different samples of 3 are possible? b. List all possible samples of size 3, and compute the mean number of cases in each

sample. c. Compare the mean of the distribution of sample means to the population mean. d. On a chart similar to Chart 8–1, compare the dispersion in the population with that of

the sample means. 10. There are five sales associates at Mid-Motors Ford. The five representatives and the num-

ber of cars they sold last week are:

Sales Representative Cars Sold

Peter Hankish 8 Connie Stallter 6 Juan Lopez 4 Ted Barnes 10 Peggy Chu 6

a. How many different samples of size 2 are possible? b. List all possible samples of size 2, and compute the mean of each sample. c. Compare the mean of the sampling distribution of sample means with that of the

population. d. On a chart similar to Chart 8–1, compare the dispersion in sample means with that

of the population.

8.5 The Central Limit Theorem In this section, we examine the central limit theorem. Its application to the sam- pling distribution of the sample mean, introduced in the previous section, allows us to use the normal probability distribution to create confidence intervals for the pop- ulation mean (described in Chapter 9) and perform tests of hypothesis (described in Chapter 10). The central limit theorem states that, for large random samples, the shape of the sampling distribution of the sample mean is close to the normal prob- ability distribution. The approximation is more accurate for large samples than for small samples. This is one of the most useful conclusions in statistics. We can rea- son about the distribution of the sample mean with absolutely no information about the shape of the population distribution from which the sample is taken. In other words, the central limit theorem is true for all distributions.

A formal statement of the central limit theorem follows.

CENTRAL LIMIT THEOREM If all samples of a particular size are selected from any population, the sampling distribution of the sample mean is approximately a normal distribution. This approximation improves with larger samples.

If the population follows a normal probability distribution, then for any sample size the sampling distribution of the sample mean will also be normal. If the population dis- tribution is symmetrical (but not normal), you will see the normal shape of the distrib- ution of the sample mean emerge with samples as small as 10. On the other hand, if you start with a distribution that is skewed or has thick tails, it may require samples of 30 or more to observe the normality feature. This concept is summarized in Chart 8–2

LO5 Explain the central limit theorem.

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for various population shapes. Observe the convergence to a normal distribution regardless of the shape of the population distribution. Most statisticians consider a sample of 30 or more to be large enough for the central limit theorem to be employed.

The idea that the distribution of the sample means from a population that is not normal will converge to normality is illustrated in Charts 8–3, 8–4, and 8–5. We will discuss this example in more detail shortly, but Chart 8–3 is a graph of a discrete probability distribution that is positively skewed. There are many possible samples of 5 that might be selected from this population. Suppose we randomly select 25 samples of size 5 each and compute the mean of each sample. These results are shown in Chart 8–4. Notice that the shape of the distribution of sample means has changed from the shape of the original population even though we selected only 25 of the many possible samples. To put it another way, we selected 25 random samples of 5 each from a population that is positively skewed and found the dis- tribution of sample means has changed from the shape of the population. As we take larger samples, that is, n � 20 instead of n � 5, we will find the distribution of the sample mean will approach the normal distribution. Chart 8–5 shows the results of 25 random samples of 20 observations each from the same population. Observe the clear trend toward the normal probability distribution. This is the point of the central limit theorem. The following example will underscore this condition.

280 Chapter 8

n � 2

n � 6

n � 30

n � 2

n � 6

n � 30

n � 2

n � 6

n � 30

n � 2

n � 6

n � 30

x x x x

_ x

Populations

Sampling Distributions

CHART 8–2 Results of the Central Limit Theorem for Several Populations

Any sampling distribution of the sample mean will move toward a normal distribution as we increase the sample size.

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Sampling Methods and the Central Limit Theorem 281

Example

Solution Chart 8–3 shows the distribution of the years of experience for the population of 40 current employees. This distribution of lengths of service is positively skewed because there are a few employees who have worked at Spence Sprockets for a longer period of time. Specifically, six employees have been with the company 10 years or more. However, because the business has grown, the number of employees has increased in the last few years. Of the 40 employees, 18 have been with the com- pany two years or less.

Years of service

Fr eq

ue nc

105 15 20

CHART 8–3 Length of Service for Spence Sprockets Inc. Employees

Let’s consider the first of Ed Spence’s problems. He would like to form a committee of five employees to look into the health care question and suggest what type of health care coverage would be most appropriate for the majority of workers. How should he select the committee? If he selects the committee ran- domly, what might he expect in terms of mean length of service for those on the committee?

Ed Spence began his sprocket business 20 years ago. The business has grown over the years and now employs 40 people. Spence Sprockets Inc. faces some major decisions regarding health care for these employees. Before making a final decision on what health care plan to purchase, Ed decides to form a committee of five rep- resentative employees. The committee will be asked to study the health care issue carefully and make a recommendation as to what plan best fits the employees’ needs. Ed feels the views of newer employees toward health care may differ from those of more experienced employees. If Ed randomly selects this committee, what can he expect in terms of the mean years with Spence Sprockets for those on the committee? How does the shape of the distribution of years of experience of all employees (the population) compare with the shape of the sampling distribution of the mean? The lengths of service (rounded to the nearest year) of the 40 employ- ees currently on the Spence Sprockets Inc. payroll are as follows.

11 4 18 2 1 2 0 2 2 4 3 4 1 2 2 3 3 19 8 3 7 1 0 2 7 0 4 5 1 14

16 8 9 1 1 2 5 10 2 3

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282 Chapter 8

To begin, Ed writes the length of service for each of the 40 employees on pieces of paper and puts them into an old baseball hat. Next, he shuffles the pieces of paper around and randomly selects five slips of paper. The lengths of service for these five employees are 1, 9, 0, 19, and 14 years. Thus, the mean length of service for these five sampled employees is 8.60 years. How does that compare with the population mean? At this point, Ed does not know the population mean, but the num- ber of employees in the population is only 40, so he decides to calculate the mean length of service for all his employees. It is 4.8 years, found by adding the lengths of service for all the employees and dividing the total by 40.

The difference between the sample mean ( ) and the population mean (�) is called sampling error. In other words, the difference of 3.80 years between the pop- ulation mean of 4.80 and the sample mean of 8.60 is the sampling error. It is due to chance. Thus, if Ed selected these five employees to constitute the committee, their mean length of service would be larger than the population mean.

What would happen if Ed put the five pieces of paper back into the baseball hat and selected another sample? Would you expect the mean of this second sample to be exactly the same as the previous one? Suppose he selects another sample of five employees and finds the lengths of service in this sample to be 7, 4, 4, 1, and 3. This sample mean is 3.80 years. The result of selecting 25 samples of five employees each is shown in Table 8–5 and Chart 8–4. There are actually 658,008 possible samples of 5 from the population of 40 employees, found by the combination formula (5–10) for 40 things taken 5 at a time. Notice the difference in the shape of the population and

� � 11 � 4 � 18 � # # # � 2 � 3

40 � 4.80

Sample Sample I.D. Sample Data Mean

A 1 9 0 19 14 8.6 B 7 4 4 1 3 3.8 C 8 19 8 2 1 7.6 D 4 18 2 0 11 7.0 E 4 2 4 7 18 7.0 F 1 2 0 3 2 1.6 G 2 3 2 0 2 1.8 H 11 2 9 2 4 5.6 I 9 0 4 2 7 4.4 J 1 1 1 11 1 3.0 K 2 0 0 10 2 2.8 L 0 2 3 2 16 4.6 M 2 3 1 1 1 1.6 N 3 7 3 4 3 4.0 O 1 2 3 1 4 2.2 P 19 0 1 3 8 6.2 Q 5 1 7 14 9 7.2 R 5 4 2 3 4 3.6 S 14 5 2 2 5 5.6 T 2 1 1 4 7 3.0 U 3 7 1 2 1 2.8 V 0 1 5 1 2 1.8 W 0 3 19 4 2 5.6 X 4 2 3 4 0 2.6 Y 1 1 2 3 2 1.8

TABLE 8–5 Twenty-Five Random Samples of Five Employees

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Sampling Methods and the Central Limit Theorem 283

CHART 8–4 Histogram of Mean Lengths of Service for 25 Samples of Five Employees

the distribution of these sample means. The population of the lengths of service for employees (Chart 8–3) is positively skewed, but the distribution of these 25 sample means does not reflect the same positive skew. There is also a difference in the range of the sample means versus the range of the population. The population ranged from 0 to 19 years, whereas the sample means range from 1.6 to 8.6 years.

Table 8–6 reports the result of selecting 25 samples of 20 employees each and computing their sample means. These sample means are shown graphically in Chart 8–5. Compare the shape of this distribution to the population (Chart 8–3) and to the distribution of sample means where the sample is n � 5 (Chart 8–4). You should observe two important features:

TABLE 8–6 Random Samples and Sample Means of 25 Samples of 20 Spence Sprockets Inc. Employees

Sample Sample Number Sample Data (Length of Service) Mean

A 3 8 3 0 2 1 2 3 11 5 1 3 4 2 7 1 1 2 4 16 3.95 B 2 3 8 2 1 5 2 0 3 1 0 7 1 4 3 11 4 4 3 1 3.25 C 14 5 0 3 2 14 11 9 2 2 1 2 19 1 0 1 4 2 19 8 5.95 D 9 2 1 1 4 10 0 8 4 3 2 1 0 8 1 14 5 10 1 3 4.35 E 18 1 2 2 4 3 2 8 2 1 0 19 4 19 0 1 4 0 3 14 5.35 F 10 4 4 18 3 3 1 0 0 2 2 4 7 10 2 0 3 4 2 1 4.00 G 5 7 11 8 11 18 1 1 16 2 2 16 2 3 2 16 2 2 2 4 6.55 H 3 0 2 0 5 4 5 3 8 3 2 5 1 1 2 9 8 3 16 5 4.25 I 0 0 18 2 1 7 4 1 3 0 3 2 11 7 2 8 5 1 2 3 4.00 J 2 7 2 4 1 3 3 2 5 10 0 1 1 2 9 3 2 19 3 2 4.05 K 7 4 5 3 3 0 18 2 0 4 2 7 2 7 4 2 10 1 1 2 4.20 L 0 3 10 5 9 2 1 4 1 2 1 8 18 1 4 3 3 2 0 4 4.05 M 4 1 2 1 7 3 9 14 8 19 4 4 1 2 0 3 1 2 1 2 4.40 N 3 16 1 2 4 4 4 2 1 5 2 3 5 3 4 7 16 1 11 1 4.75 O 2 19 2 0 2 2 16 2 3 11 9 2 8 0 8 2 7 3 2 2 5.10 P 2 18 16 5 2 2 19 0 1 2 11 4 2 2 1 4 2 0 4 3 5.00 Q 3 2 3 11 10 1 1 5 19 16 7 10 3 1 1 1 2 2 3 1 5.10 R 2 3 1 2 7 4 3 19 9 2 2 1 1 2 2 2 1 8 0 2 3.65 S 2 14 19 1 19 2 8 4 2 2 14 2 8 16 4 7 2 9 0 7 7.10 T 0 1 3 3 2 2 3 1 1 0 3 2 3 5 2 10 14 4 2 0 3.05 U 1 0 1 2 16 1 1 2 5 1 4 1 2 2 2 2 2 8 9 3 3.25 V 1 9 4 4 2 8 7 1 14 18 1 5 10 11 19 0 3 7 2 11 6.85 W 8 1 9 19 3 19 0 5 2 1 5 3 3 4 1 5 3 1 8 7 5.35 X 4 2 0 3 1 16 1 11 3 3 2 18 2 0 1 5 0 7 2 5 4.30 Y 1 2 1 2 0 2 7 2 4 8 19 2 5 3 3 0 19 2 1 18 5.05

0 1 2 3 4 5 6 7 8

Mean length of service

Fr eq

ue nc

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284 Chapter 8

What should we conclude from this example? The central limit theorem indi- cates that, regardless of the shape of the population distribution, the sampling distribution of the sample mean will move toward the normal probability distribution. The larger the number of observations in each sample, the stronger the conver- gence. The Spence Sprockets Inc. example shows how the central limit theorem works. We began with a positively skewed population (Chart 8–3). Next, we selected 25 random samples of 5 observations, computed the mean of each sample, and finally organized these 25 sample means into a graph (Chart 8–4). We observe a change in the shape of the sampling distribution of the sample mean from that of the population. The movement is from a positively skewed distribution to a distrib- ution that has the shape of the normal probability distribution.

To further illustrate the effects of the central limit theorem, we increased the number of observations in each sample from 5 to 20. We selected 25 samples of 20 observations each and calculated the mean of each sample. Finally, we orga- nized these sample means into a graph (Chart 8–5). The shape of the histogram in Chart 8–5 is clearly moving toward the normal probability distribution.

1. The shape of the distribution of the sample mean is different from that of the population. In Chart 8–3, the distribution of all employees is positively skewed. However, as we select random samples from this population, the shape of the distribution of the sample mean changes. As we increase the size of the sam- ple, the distribution of the sample mean approaches the normal probability dis- tribution. This illustrates the central limit theorem.

2. There is less dispersion in the sampling distribution of sample means than in the population distribution. In the population, the lengths of service ranged from 0 to 19 years. When we selected samples of 5, the sample means ranged from 1.6 to 8.6 years, and when we selected samples of 20, the means ranged from 3.05 to 7.10 years.

CHART 8–5 Histogram of Mean Lengths of Service for 25 Samples of 20 Employees

We can also compare the mean of the sample means to the population mean. The mean of the 25 samples of 20 employees reported in Table 8–6 is 4.676 years.

We use the symbol to identify the mean of the distribution of the sample mean. The subscript reminds us that the distribution is of the sample mean. It is read “mu sub X bar.” We observe that the mean of the sample means, 4.676 years, is very close to the population mean of 4.80.

�X

�X � 3.95 � 3.25 � # # # � 4.30 � 5.05

25 � 4.676

2.5 3.5 4.5 5.5 6.5

Mean length of service

Fr eq

ue nc

7.5

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Sampling Methods and the Central Limit Theorem 285

If you go back to Chapter 6 where several binomial distributions with a “success” proportion of .10 are shown in Chart 6–3 in Section 6.5 on page 201, you can see yet another demonstration of the central limit theorem. Observe as n increases from 7 through 12 and 20 up to 40 that the profile of the probability distributions moves closer and closer to a normal probability distribution. Chart 8–5 on page 284 also shows the convergence to normality as n increases. This again reinforces the fact that, as more observations are sampled from any population distribution, the shape of the sampling distribution of the sample mean will get closer and closer to a normal distribution.

The central limit theorem itself (reread the definition on page 279) does not say anything about the dispersion of the sampling distribution of the sample mean or about the comparison of the mean of the sampling distribution of the sample mean to the mean of the population. However, in our Spence Sprockets example, we did observe that there was less dispersion in the distribution of the sample mean than in the population distribution by noting the difference in the range in the population and the range of the sample means. We observe that the mean of the sample means is close to the mean of the population. It can be demonstrated that the mean of the sampling distribution is the population mean (i.e., ), and if the standard deviation in the population is �, the standard deviation of the sample means is where n is the number of observations in each sample. We refer to as the standard error of the mean. Its longer name is actually the standard deviation of the sampling distribution of the sample mean.

STANDARD ERROR OF THE MEAN [8–1]

In this section, we also came to other important conclusions.

1. The mean of the distribution of sample means will be exactly equal to the pop- ulation mean if we are able to select all possible samples of the same size from a given population. That is:

Even if we do not select all samples, we can expect the mean of the distribu- tion of sample means to be close to the population mean.

2. There will be less dispersion in the sampling distribution of the sample mean than in the population. If the standard deviation of the population is �, the stan- dard deviation of the distribution of sample means is . Note that when we increase the size of the sample, the standard error of the mean decreases.

��1n

� � �X

� X �

�

��1n ��1n,

�X � �

Self-Review 8–4 Refer to the Spence Sprockets Inc. data on page 281. Select 10 random samples of 5 employees each. Use the methods described earlier in the chapter and the Table of Ran- dom Numbers (Appendix B.6) to find the employees to include in the sample. Compute the mean of each sample and plot the sample means on a chart similar to Chart 8–3. What is the mean of your 10 sample means?

Exercises 11. Appendix B.6 is a table of random numbers. Hence, each digit from 0 to 9 has the same

likelihood of occurrence.

a. Draw a graph showing the population distribution. What is the population mean?

LO6 Define the standard error of the mean.

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286 Chapter 8

b. Following are the first 10 rows of five digits from Appendix B.6. Assume that these are 10 random samples of five values each. Determine the mean of each sample and plot the means on a chart similar to Chart 8–3. Compare the mean of the sampling distribution of the sample means with the population mean.

0 2 7 1 1 9 4 8 7 3 5 4 9 2 1 7 7 6 4 0 6 1 5 4 5 1 7 1 4 7 1 3 7 4 8 8 7 4 5 5 0 8 9 9 9 7 8 8 0 4

12. Scrapper Elevator Company has 20 sales representatives who sell its product through- out the United States and Canada. The number of units sold last month by each repre- sentative is listed below. Assume these sales figures to be the population values.

2 3 2 3 3 4 2 4 3 2 2 7 3 4 5 3 3 3 3 5

a. Draw a graph showing the population distribution. b. Compute the mean of the population. c. Select five random samples of 5 each. Compute the mean of each sample. Use the

methods described in this chapter and Appendix B.6 to determine the items to be included in the sample.

d. Compare the mean of the sampling distribution of the sample means to the popula- tion mean. Would you expect the two values to be about the same?

e. Draw a histogram of the sample means. Do you notice a difference in the shape of the distribution of sample means compared to the shape of the population distribution?

13. Consider all of the coins (pennies, nickels, quarters, etc.) in your pocket or purse as a population. Make a frequency table beginning with the current year and counting back- ward to record the ages (in years) of the coins. For example, if the current year is 2009, then a coin with 2007 stamped on it is 2 years old. a. Draw a histogram or other graph showing the population distribution. b. Randomly select five coins and record the mean age of the sampled coins. Repeat

this sampling process 20 times. Now draw a histogram or other graph showing the distribution of the sample means.

c. Compare the shapes of the two histograms. 14. Consider the digits in the phone numbers on a randomly selected page of your local

phone book a population. Make a frequency table of the final digit of 30 randomly selected phone numbers. For example, if a phone number is 555-9704, record a 4. a. Draw a histogram or other graph of this population distribution. Using the uniform dis-

tribution, compute the population mean and the population standard deviation. b. Also record the sample mean of the final four digits (9704 would lead to a mean of 5).

Now draw a histogram or other graph showing the distribution of the sample means. c. Compare the shapes of the two histograms.

8.6 Using the Sampling Distribution of the Sample Mean The previous discussion is important because most business decisions are made on the basis of sampling results. Here are some examples.

1. Arm and Hammer Company wants to ensure that its laundry detergent actually contains 100 fluid ounces, as indicated on the label. Historical summaries from

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Sampling Methods and the Central Limit Theorem 287

the filling process indicate the mean amount per container is 100 fluid ounces and the standard deviation is 2 fluid ounces. The quality technician in her 10 A.M. check of 40 containers finds the mean amount per container is 99.8 fluid ounces. Should the technician shut down the filling operation or is the sampling error reasonable?

2. A. C. Nielsen Company provides information to organizations advertising on tele- vision. Prior research indicates that adult Americans watch an average of 6.0 hours per day of television. The standard deviation is 1.5 hours. For a sample of 50 adults in the greater Boston area, would it be reasonable that we could randomly select a sample and find that they watch an average of 6.5 hours of television per day?

3. Haughton Elevator Company wishes to develop specifications for the number of people who can ride in a new oversized elevator. Suppose the mean weight for an adult is 160 pounds and the standard deviation is 15 pounds. However, the distribution of weights does not follow the normal probability distribution. It is positively skewed. What is the likelihood that for a sample of 30 adults their mean weight is 170 pounds or more?

In each of these situations, we have a population about which we have some infor- mation. We take a sample from that population and wish to conclude whether the difference between the population parameter and the sample statistic is random sampling error due to chance. Or is the difference not random sampling error and, therefore, a statistically significant real difference?

Using ideas discussed in the previous section, we can compute the probability that a sample mean will fall within a certain range. We know that the sampling distribution of the sample mean will follow the normal probability distribution under two conditions:

1. When the samples are taken from populations known to follow the normal dis- tribution. In this case, the size of the sample is not a factor.

2. When the shape of the population distribution is not known or the shape is known to be nonnormal, but our sample contains at least 30 observations. We should point out that the number 30 is a guideline that has evolved over the years. In this case, the central limit theorem guarantees the sampling distribu- tion of the mean follows a normal distribution.

We can use formula (7–5) from Section 7.5 in the previous chapter to convert any normal distribution to the standard normal distribution. We also refer to this as a z value. Then we can use the standard normal table, Appendix B.1, to find the probability of selecting an observation that would fall within a specific range. The formula for finding a z value is:

In this formula, X is the value of the random variable, � is the population mean, and � the population standard deviation.

However, most business decisions refer to a sample—not just one observation. So we are interested in the distribution of , the sample mean, instead of X, the value of one observation. That is the first change we make in formula (7–5). The second is that we use the standard error of the mean of n observations instead of the population standard deviation. That is, we use in the denominator rather than �. Therefore, to find the likelihood of a sample mean with a specified range, we first use the following formula to find the corresponding z value. Then we use Appendix B.1 to locate the probability.

FINDING THE z VALUE OF X WHEN THE [8–2] POPULATION STANDARD DEVIATION IS KNOWN

z � X � � ��1n

��1n

z � X � �

�

LO7 Apply the central limit theorem to find probabilities of selecting possible sample means from a specified population.

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288 Chapter 8

The following example will show the application.

Example

Solution

The Quality Assurance Department for Cola Inc. maintains records regarding the amount of cola in its Jumbo bottle. The actual amount of cola in each bottle is crit- ical, but varies a small amount from one bottle to the next. Cola Inc. does not wish to underfill the bottles, because it will have a problem with truth in labeling. On the other hand, it cannot overfill each bottle, because it would be giving cola away, hence reducing its profits. Its records indicate that the amount of cola follows the normal probability distribution. The mean amount per bottle is 31.2 ounces and the population standard deviation is 0.4 ounces. At 8 A.M. today the quality technician randomly selected 16 bottles from the filling line. The mean amount of cola con- tained in the bottles is 31.38 ounces. Is this an unlikely result? Is it likely the process is putting too much soda in the bottles? To put it another way, is the sampling error of 0.18 ounces unusual?

We can use the results of the previous section to find the likelihood that we could select a sample of 16 (n) bottles from a normal population with a mean of 31.2 (�) ounces and a population standard deviation of 0.4 (�) ounces and find the sample mean to be 31.38( ). We use formula (8–2) to find the value of z.

The numerator of this equation, is the sampling error. The denominator, is the standard error of the sam- pling distribution of the sample mean. So the z values express the sampling error in standard units—in other words, the standard error.

Next, we compute the likelihood of a z value greater than 1.80. In Appendix B.1, locate the probability corresponding to a z value of 1.80. It is .4641. The likelihood of a z value greater than 1.80 is .0359, found by .5000 � .4641.

What do we conclude? It is unlikely, less than a 4 percent chance, we could select a sample of 16 observations from a normal population with a mean of 31.2 ounces and a population standard deviation of 0.4 ounces and find the sample mean equal to or greater than 31.38 ounces. We conclude the process is putting too much cola in the bottles. The quality technician should see the production supervisor about reducing the amount of soda in each bottle. This information is summarized in Chart 8–6.

31.20 31.38 Ounces (X –

)

0 1.80 z value

.4641

.0359

��1n � 0.4�116 � 0.1, X � � � 31.38 � 31.20 � .18,

z � X � � ��1n

� 31.38 � 31.20

0.4�116 � 1.80

CHART 8–6 Sampling Distribution of the Mean Amount of Cola in a Jumbo Bottle

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Sampling Methods and the Central Limit Theorem 289

Self-Review 8–5 Refer to the Cola Inc. information. Suppose the quality technician selected a sample of 16 Jumbo bottles that averaged 31.08 ounces. What can you conclude about the filling process?

Exercises 15. A normal population has a mean of 60 and a standard deviation of 12. You select a

random sample of 9. Compute the probability the sample mean is: a. Greater than 63. b. Less than 56. c. Between 56 and 63.

16. A normal population has a mean of 75 and a standard deviation of 5. You select a sample of 40. Compute the probability the sample mean is: a. Less than 74. b. Between 74 and 76. c. Between 76 and 77. d. Greater than 77.

17. The rent for a one-bedroom apartment in Southern California follows the normal distrib- ution with a mean of $2,200 per month and a standard deviation of $250 per month. The distribution of the monthly costs does not follow the normal distribution. In fact, it is pos- itively skewed. What is the probability of selecting a sample of 50 one-bedroom apart- ments and finding the mean to be at least $1,950 per month?

18. According to an IRS study, it takes a mean of 330 minutes for taxpayers to prepare, copy, and electronically file a 1040 tax form. This distribution of times follows the normal dis- tribution and the standard deviation is 80 minutes. A consumer watchdog agency selects a random sample of 40 taxpayers. a. What is the standard error of the mean in this example? b. What is the likelihood the sample mean is greater than 320 minutes? c. What is the likelihood the sample mean is between 320 and 350 minutes? d. What is the likelihood the sample mean is greater than 350 minutes?

Chapter Summary I. There are many reasons for sampling a population.

A. The results of a sample may adequately estimate the value of the population parameter, thus saving time and money.

B. It may be too time consuming to contact all members of the population. C. It may be impossible to check or locate all the members of the population. D. The cost of studying all the items in the population may be prohibitive. E. Often testing destroys the sampled item and it cannot be returned to the population.

II. In an unbiased or probability sample, all members of the population have a chance of being selected for the sample. There are several probability sampling methods. A. In a simple random sample, all members of the population have the same chance of

being selected for the sample. B. In a systematic sample, a random starting point is selected, and then every kth item

thereafter is selected for the sample. C. In a stratified sample, the population is divided into several groups, called strata, and

then a random sample is selected from each stratum. D. In cluster sampling, the population is divided into primary units, then samples are drawn

from the primary units.

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290 Chapter 8

III. The sampling error is the difference between a population parameter and a sample statistic.

IV. The sampling distribution of the sample mean is a probability distribution of all possible sample means of the same sample size. A. For a given sample size, the mean of all possible sample means selected from a pop-

ulation is equal to the population mean. B. There is less variation in the distribution of the sample mean than in the population

distribution. C. The standard error of the mean measures the variation in the sampling distribution of

the sample mean. The standard error is found by:

[8–1]

D. If the population follows a normal distribution, the sampling distribution of the sample mean will also follow the normal distribution for samples of any size. Assume the pop- ulation standard deviation is known. To determine the probability that a sample mean falls in a particular region, use the following formula.

[8–2]

Pronunciation Key SYMBOL MEANING PRONUNCIATION

Mean of the sampling distribution mu sub X bar of the sample mean

Population standard error of sigma sub X bar the sample mean

Chapter Exercises 19. The retail stores located in the North Towne Square Mall are:

00 Elder-Beerman 09 Lion Store 18 County Seat 01 Sears 10 Bootleggers 19 Kid Mart 02 Deb Shop 11 Formal Man 20 Lerner 03 Frederick’s of Hollywood 12 Leather Ltd. 21 Coach House Gifts 04 Petries 13 B Dalton Bookseller 22 Spencer Gifts 05 Easy Dreams 14 Pat’s Hallmark 23 CPI Photo Finish 06 Summit Stationers 15 Things Remembered 24 Regis Hairstylists 07 E. B. Brown Opticians 16 Pearle Vision Express 08 Kay-Bee Toy & Hobby 17 Dollar Tree

a. If the following random numbers are selected, which retail stores should be contacted for a survey? 11, 65, 86, 62, 06, 10, 12, 77, and 04

b. Select a random sample of four retail stores. Use Appendix B.6. c. A systematic sampling procedure is to be used. The first store is to be contacted and

then every third store. Which stores will be contacted? 20. Medical Mutual Insurance is investigating the cost of a routine office visit to

family-practice physicians in the Rochester, New York, area. The following is a list of family-practice physicians in the region. Physicians are to be randomly selected and contacted regarding their charges. The 39 physicians have been coded from 00 to 38. Also noted is whether they are in practice by themselves (S), have a partner (P), or are in a group practice (G).

�X

z � X � � ��1n

� X �

�

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Sampling Methods and the Central Limit Theorem 291

Type of Type of Number Physician Practice Number Physician Practice

00 R. E. Scherbarth, M.D. S 20 Gregory Yost, M.D. P 01 Crystal R. Goveia, M.D. P 21 J. Christian Zona, M.D. P 02 Mark D. Hillard, M.D. P 22 Larry Johnson, M.D. P 03 Jeanine S. Huttner, M.D. P 23 Sanford Kimmel, M.D. P 04 Francis Aona, M.D. P 24 Harry Mayhew, M.D. S 05 Janet Arrowsmith, M.D. P 25 Leroy Rodgers, M.D. S 06 David DeFrance, M.D. S 26 Thomas Tafelski, M.D. S 07 Judith Furlong, M.D. S 27 Mark Zilkoski, M.D. G 08 Leslie Jackson, M.D. G 28 Ken Bertka, M.D. G 09 Paul Langenkamp, M.D. S 29 Mark DeMichiei, M.D. G 10 Philip Lepkowski, M.D. S 30 John Eggert, M.D. P 11 Wendy Martin, M.D. S 31 Jeanne Fiorito, M.D. P 12 Denny Mauricio, M.D. P 32 Michael Fitzpatrick, M.D. P 13 Hasmukh Parmar, M.D. P 33 Charles Holt, D.O. P 14 Ricardo Pena, M.D. P 34 Richard Koby, M.D. P 15 David Reames, M.D. P 35 John Meier, M.D. P 16 Ronald Reynolds, M.D. G 36 Douglas Smucker, M.D. S 17 Mark Steinmetz, M.D. G 37 David Weldy, M.D. P 18 Geza Torok, M.D. S 38 Cheryl Zaborowski, M.D. P 19 Mark Young, M.D. P

a. The random numbers obtained from Appendix B.6 are: 31, 94, 43, 36, 03, 24, 17, and 09. Which physicians should be contacted?

b. Select a random sample of four physicians using the random numbers of Appendix B.6. c. A sample is to consist of every fifth physician. The number 04 is selected as the

starting point. Which physicians will be contacted? d. A sample is to consist of two physicians in solo practice (S), two in partnership

(P), and one in group practice (G). Select a sample accordingly. Explain your procedure.

21. A population consists of the following three values: 1, 2, and 3. a. List all possible samples of size 2 (including possible repeats) and compute the mean

of every sample. b. Find the means of the distribution of the sample mean and the population mean. Com-

pare the two values. c. Compare the dispersion of the population with that of the sample mean. d. Describe the shapes of the two distributions.

22. In the Department of Education at UR University, student records suggest that the population of students spends an average of 5.5 hours per week playing organized sports. The population’s standard deviation is 2.2 hours per week. Based on a sample of 121 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates. a. Compute the standard error of the sample mean. b. What is the chance HLI will find a sample mean between 5 and 6 hours? c. Calculate the probability that the sample mean will be between 5.3 and 5.7 hours. d. How strange would it be to obtain a sample mean greater than 6.5 hours?

23. The manufacturer of eMachines, an economy-priced computer, recently completed the design for a new laptop model. eMachine’s top management would like some assis- tance in pricing the new laptop. Two market research firms were contacted and asked to prepare a pricing strategy. Marketing-Gets-Results tested the new eMachines laptop with 50 randomly selected consumers, who indicated they plan to purchase a laptop within the next year. The second marketing research firm, called Marketing- Reaps-Profits, test-marketed the new eMachines laptop with 200 current laptop owners. Which of the marketing research companies’ test results will be more useful? Discuss why.

24. Answer the following questions in one or two well-constructed sentences. a. What happens to the standard error of the mean if the sample size is increased?

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b. What happens to the distribution of the sample means if the sample size is increased? c. When using the distribution of sample means to estimate the population mean, what

is the benefit of using larger sample sizes? 25. There are 25 motels in Goshen, Indiana. The number of rooms in each motel follows:

90 72 75 60 75 72 84 72 88 74 105 115 68 74 80 64 104 82 48 58 60 80 48 58 100

Technician Shutdowns Technician Shutdowns

Taylor 4 Rousche 3 Hurley 3 Huang 2 Gupta 5

a. Using a table of random numbers (Appendix B.6), select a random sample of five motels from this population.

b. Obtain a systematic sample by selecting a random starting point among the first five motels and then select every fifth motel.

c. Suppose the last five motels are “cut-rate” motels. Describe how you would select a random sample of three regular motels and two cut-rate motels.

26. As a part of their customer-service program, United Airlines randomly selected 10 pas- sengers from today’s 9 A.M. Chicago–Tampa flight. Each sampled passenger is to be interviewed in depth regarding airport facilities, service, and so on. To identify the sam- ple, each passenger was given a number on boarding the aircraft. The numbers started with 001 and ended with 250. a. Select 10 usable numbers at random using Appendix B.6. b. The sample of 10 could have been chosen using a systematic sample. Choose the

first number using Appendix B.6, and then list the numbers to be interviewed. c. Evaluate the two methods by giving the advantages and possible disadvantages. d. In what other way could a random sample be selected from the 250 passengers?

27. Suppose your statistics instructor gave six examinations during the semester. You received the following grades (percent correct): 79, 64, 84, 82, 92, and 77. Instead of averaging the six scores, the instructor indicated he would randomly select two grades and compute the final percent correct based on the two percents. a. How many different samples of two test grades are possible? b. List all possible samples of size two and compute the mean of each. c. Compute the mean of the sample means and compare it to the population mean. d. If you were a student, would you like this arrangement? Would the result be different

from dropping the lowest score? Write a brief report. 28. At the downtown office of First National Bank, there are five tellers. Last week, the tellers

made the following number of errors each: 2, 3, 5, 3, and 5. a. How many different samples of 2 tellers are possible? b. List all possible samples of size 2 and compute the mean of each. c. Compute the mean of the sample means and compare it to the population mean.

29. The Quality Control Department employs five technicians during the day shift. Listed below is the number of times each technician instructed the production foreman to shut down the manufacturing process last week.

Sales Number Sales Number Representative Sold Representative Sold

Zina Craft 54 Jan Niles 48 Woon Junge 50 Molly Camp 50 Ernie DeBrul 52 Rachel Myak 52

a. How many different samples of two technicians are possible from this population? b. List all possible samples of two observations each and compute the mean of each

sample. c. Compare the mean of the sample means with the population mean. d. Compare the shape of the population distribution with the shape of the distribution of

the sample means. 30. The Appliance Center has six sales representatives at its North Jacksonville outlet. Listed

below is the number of refrigerators sold by each last month.

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Sampling Methods and the Central Limit Theorem 293

a. How many samples of size 2 are possible? b. Select all possible samples of size 2 and compute the mean number sold. c. Organize the sample means into a frequency distribution. d. What is the mean of the population? What is the mean of the sample means? e. What is the shape of the population distribution? f. What is the shape of the distribution of the sample mean?

31. Mattel Corporation produces a remote-controlled car that requires three AA batteries. The mean life of these batteries in this product is 35.0 hours. The distribution of the battery lives closely follows the normal probability distribution with a standard deviation of 5.5 hours. As a part of its testing program, Sony tests samples of 25 batteries. a. What can you say about the shape of the distribution of the sample mean? b. What is the standard error of the distribution of the sample mean? c. What proportion of the samples will have a mean useful life of more than 36 hours? d. What proportion of the sample will have a mean useful life greater than 34.5 hours? e. What proportion of the sample will have a mean useful life between 34.5 and 36.0

hours? 32. CRA CDs Inc. wants the mean lengths of the “cuts” on a CD to be 135 seconds (2 min-

utes and 15 seconds). This will allow the disk jockeys to have plenty of time for com- mercials within each 10-minute segment. Assume the distribution of the length of the cuts follows the normal distribution with a population standard deviation of 8 seconds. Suppose we select a sample of 16 cuts from various CDs sold by CRA CDs Inc. a. What can we say about the shape of the distribution of the sample mean? b. What is the standard error of the mean? c. What percent of the sample means will be greater than 140 seconds? d. What percent of the sample means will be greater than 128 seconds? e. What percent of the sample means will be greater than 128 but less than 140 seconds?

33. Recent studies indicate that the typical 50-year-old woman spends $350 per year for per- sonal-care products. The distribution of the amounts spent follows a normal distribution with a standard deviation of $45 per year. We select a random sample of 40 women. The mean amount spent for those sampled is $335. What is the likelihood of finding a sample mean this large or larger from the specified population?

34. Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000. This distribution follows the normal distribution with a standard deviation of $40,000. a. If we select a random sample of 50 households, what is the standard error of the mean? b. What is the expected shape of the distribution of the sample mean? c. What is the likelihood of selecting a sample with a mean of at least $112,000? d. What is the likelihood of selecting a sample with a mean of more than $100,000? e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less

than $112,000. 35. The mean age at which men in the United States marry for the first time follows the

normal distribution with a mean of 24.8 years. The standard deviation of the distribution is 2.5 years. For a random sample of 60 men, what is the likelihood that the age at which they were married for the first time is less than 25.1 years?

36. A recent study by the Greater Los Angeles Taxi Drivers Association showed that the mean fare charged for service from Hermosa Beach to Los Angeles International Airport is $21 and the standard deviation is $3.50. We select a sample of 15 fares. a. What is the likelihood that the sample mean is between $20 and $23? b. What must you assume to make the above calculation?

37. Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 6,000 pounds and the standard deviation is 150 pounds. Assume that the population follows the normal distribution. Forty trucks are randomly selected and weighed. Within what limits will 95 percent of the sample means occur?

38. The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 50 customers, answer the following questions. a. What is the likelihood the sample mean is at least $25.00? b. What is the likelihood the sample mean is greater than $22.50 but less than $25.00? c. Within what limits will 90 percent of the sample means occur?

39. The mean SAT score for Division I student-athletes is 947 with a standard deviation of 205. If you select a random sample of 60 of these students, what is the probability the mean is below 900?

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40. Suppose we roll a fair die two times. a. How many different samples are there? b. List each of the possible samples and compute the mean. c. On a chart similar to Chart 8–1, compare the distribution of sample means with the

distribution of the population. d. Compute the mean and the standard deviation of each distribution and compare

them. 41. Following is a list of the 50 states with the numbers 0 through 49 assigned to them.

Number State Number State

0 Alabama 25 Montana 1 Alaska 26 Nebraska 2 Arizona 27 Nevada 3 Arkansas 28 New Hampshire 4 California 29 New Jersey 5 Colorado 30 New Mexico 6 Connecticut 31 New York 7 Delaware 32 North Carolina 8 Florida 33 North Dakota 9 Georgia 34 Ohio 10 Hawaii 35 Oklahoma 11 Idaho 36 Oregon 12 Illinois 37 Pennsylvania 13 Indiana 38 Rhode Island 14 Iowa 39 South Carolina 15 Kansas 40 South Dakota 16 Kentucky 41 Tennessee 17 Louisiana 42 Texas 18 Maine 43 Utah 19 Maryland 44 Vermont 20 Massachusetts 45 Virginia 21 Michigan 46 Washington 22 Minnesota 47 West Virginia 23 Mississippi 48 Wisconsin 24 Missouri 49 Wyoming

a. You wish to select a sample of eight from this list. The selected random numbers are 45, 15, 81, 09, 39, 43, 90, 26, 06, 45, 01, and 42. Which states are included in the sample?

b. You wish to use a systematic sample of every sixth item and the digit 02 is chosen as the starting point. Which states are included?

42. Human Resource Consulting (HRC) surveyed a random sample of 60 Twin Cities con- struction companies to find information on the costs of their health care plans. One of the items being tracked is the annual deductible that employees must pay. The Minnesota Department of Labor reports that historically the mean deductible amount per employee is $502 with a standard deviation of $100. a. Compute the standard error of the sample mean for HRC. b. What is the chance HRC finds a sample mean between $477 and $527? c. Calculate the likelihood that the sample mean is between $492 and $512. d. What is the probability the sample mean is greater than $550?

43. Over the past decade, the mean number of members of the Information Systems Secu- rity Association who have experienced a denial-of-service attack each year is 510, with a standard deviation of 14.28 attacks. Suppose nothing in this environment changes. a. What is the likelihood this group will suffer an average of more than 600 attacks in

the next 10 years? b. Compute the probability the mean number of attacks over the next 10 years is

between 500 and 600. c. What is the possibility they will experience an average of less than 500 attacks over

the next 10 years?

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Sampling Methods and the Central Limit Theorem 295

44. The Oil Price Information Center of greater Houston reports the mean price per gallon of regular gasoline is $3.00 with a population standard deviation of $0.18. Assume a ran- dom sample of 40 gasoline stations is selected and their mean cost for regular gasoline is computed. a. What is the standard error of the mean in this experiment? b. What is the probability that the sample mean is between $2.98 and $3.02? c. What is the probability that the difference between the sample mean and the popu-

lation mean is less than 0.01? d. What is the likelihood the sample mean is greater than $3.08?

45. Nike’s annual report says that the average American buys 6.5 pairs of sports shoes per year. Suppose the population standard deviation is 2.1 and that a sample of 81 cus- tomers will be examined next year. a. What is the standard error of the mean in this experiment? b. What is the probability that the sample mean is between 6 and 7 pairs of sports shoes? c. What is the probability that the difference between the sample mean and the popu-

lation mean is less than 0.25 pairs? d. What is the likelihood the sample mean is greater than 7 pairs?

Data Set Exercises 46. Refer to the Real Estate data, which report information on the homes sold in the

Goodyear, Arizona, area last year. Use statistical software to compute the mean and the standard deviation of the selling prices. Assume this to be the population. Select a sam- ple of 10 homes. Compute the mean and the standard deviation of the sample. Deter- mine the likelihood of a sample mean this large or larger from the population.

47. Refer to the Baseball 2009 data, which report information on the 30 Major League Base- ball teams for the 2009 season. Over the last decade, the mean attendance per team followed a normal distribution with a mean of 2.25 million per team and a standard devi- ation of 0.70 million. Use statistical software to compute the mean attendance per team for the 2009 season. Determine the likelihood of a sample mean this large or larger from the population.

48. Refer to the Buena School District bus data. Information provided by manufacturers of school buses suggests the mean maintenance cost per month is $455 per bus. Use statistical software to find the mean and the standard deviation for the Buena buses. Does the Buena data seem to be in line with that reported by the manufacturer? Specif- ically, what is the probability of the sample mean being less than Buena’s, given the manufacturer’s data?

1. The Excel commands to select a simple random sample on page 269 are: a. Select the Data tab on the top of the menu.

Then on the far right select Data Analysis, then Sampling and OK.

b. For Input Range, insert B1:B31. Since the col- umn is named, click the Labels box. Select Random, and enter the sample size for the Number of Samples, in this case 5. Click on Output Range and indicate the place in the spreadsheet where you want the sample infor- mation. Note that your sample results will differ from those in the text. Also recall that Excel samples with replacement, so it is possible for a population value to appear more than once in the sample.

Software Commands

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296 Chapter 8

Chapter 8 Answers to Self-Review

Mean Number Probability

21 1 .10 22 1 .10 23 2 .20 24 2 .20 25 2 .20 26 1 .10 27 1 .10

10 1.00

Sample Service Mean

Snow, Tolson 20, 22 21 Snow, Kraft 20, 26 23 Snow, Irwin 20, 24 22 Snow, Jones 20, 28 24 Tolson, Kraft 22, 26 24 Tolson, Irwin 22, 24 23 Tolson, Jones 22, 28 25 Kraft, Irwin 26, 24 25 Kraft, Jones 26, 28 27 Irwin, Jones 24, 28 26

8–1 a. Students selected are Price, Detley, and Molter. b. Answers will vary. c. Skip it and move to the next random number.

8–2 The students selected are Berry, Francis, Kopp, Poteau, and Swetye.

8–3 a. 10, found by:

d. Identical: population mean, �, is 24, and mean of sample means, , is also 24.

e. Sample means range from 21 to 27. Population values go from 20 to 28.

f. Nonnormal. g. Yes.

�X

5C2 � 5!

2!(5 � 2)!

8–4 The answers will vary. Here is one solution.

Sample Number

1 2 3 4 5 6 7 8 9 10

8 2 2 19 3 4 0 4 1 2 19 1 14 9 2 5 8 2 14 4

8 3 4 2 4 4 1 14 4 1 0 3 2 3 1 2 16 1 2 3 2 1 7 2 19 18 18 16 3 7

Total 37 10 29 35 29 33 43 37 24 17 7.4 2 5.8 7.0 5.8 6.6 8.6 7.4 4.8 3.4

Mean of the 10 sample means is 5.88.

8–5

The probability that z is greater than �1.20 is .5000 � .3849 � .8849. There is more than an 88 percent chance the filling operation will produce bottles with at least 31.08 ounces.

z � 31.08 � 31.20

0.4�116 � �1.20

2 3 5 7 4 6 8 Mean years

Fr eq

ue nc

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9 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Define point estimate.

LO2 Define level of confidence.

LO3 Compute a confidence interval for the population mean when the population standard deviation is known.

LO4 Compute a confidence interval for the population mean when the population standard deviation is unknown.

LO5 Compute a confidence interval for a population proportion.

LO6 Calculate the required sample size to estimate a population proportion or population mean.

LO7 Adjust a confidence interval for finite populations.

Estimation and Confidence Intervals

The American Restaurant Association collected information on the

number of meals eaten outside the home per week by young

married couples. A survey of 60 couples showed the sample

mean number of meals eaten outside the home was 2.76 meals

per week, with a standard deviation of 0.75 meals per week.

Construct a 97 percent confidence interval for the population mean.

(See Exercise 36 and LO4.)

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The following examples illustrate point estimates of population means.

1. Tourism is a major source of income for many Caribbean countries, such as Barbados. Suppose the Bureau of Tourism for Barbados wants an estimate of the mean amount spent by tourists visiting the country. It would not be feasi- ble to contact each tourist. Therefore, 500 tourists are randomly selected as they depart the country and asked in detail about their spending while visiting the island. The mean amount spent by the sample of 500 tourists is an esti- mate of the unknown population parameter. That is, we let the sample mean serve as a point estimate of the population mean.

298 Chapter 9

9.1 Introduction The previous chapter began our discussion of statistical inference. It introduced both the reasons for, and the methods of, sampling. The reasons for sampling were:

• Contacting the entire population is too time consuming. • Studying all the items in the population is often too expensive. • The sample results are usually adequate. • Certain tests are destructive. • Checking all the items is physically impossible.

There are several methods of sampling. Simple random sampling is the most widely used method. With this type of sampling, each member of the population has the same chance of being selected to be a part of the sample. Other methods of sam- pling include systematic sampling, stratified sampling, and cluster sampling.

Chapter 8 assumes information about the population, such as the mean, the stan- dard deviation, or the shape of the population. In most business situations, such infor- mation is not available. In fact, the purpose of sampling may be to estimate some of these values. For example, you select a sample from a population and use the mean of the sample to estimate the mean of the population.

This chapter considers several important aspects of sampling. We begin by studying point estimates. A point estimate is a single value (point) derived from a sample and used to estimate a population value. For example, suppose we select a sample of 50 junior executives and ask how many hours they worked last week. Compute the mean of this sample of 50 and use the value of the sample mean as a point estimate of the unknown population mean. However, a point estimate is a single value. A more informative approach is to present a range of values in which we expect the population parameter to occur. Such a range of values is called a confidence interval.

Frequently in business we need to determine the size of a sample. How many vot- ers should a polling organization contact to forecast the election outcome? How many products do we need to examine to ensure our quality level? This chapter also devel- ops a strategy for determining the appropriate number of observations in the sample.

9.2 Point Estimate for a Population Mean A point estimate is a single statistic used to estimate a population parameter. Sup- pose Best Buy Inc. wants to estimate the mean age of people who purchase HD LCD televisions. They select a random sample of 50 recent purchases, determine the age of each buyer, and compute the mean age of the buyers in the sample. The mean of this sample is a point estimate of the population mean.

Statistics in Action

On all new cars, a fuel economy estimate is prominently dis- played on the window sticker as required by the Environmental Protection Agency (EPA). Often, fuel economy is a factor in a consumer’s choice of a new car because of fuel costs or envi- ronmental concerns. The fuel estimates for a 2010 BMW 328i Sedan (6-cylinder, au- tomatic) are 18 miles per gallon (mpg) in the city and 28 on the highway. The EPA recognizes that actual fuel economy may differ from the esti- mates by noting, “No test can simulate all possible combina- tions of conditions and climate, driver behavior, and car care habits. Actual mileage depends on how, when, and where the vehicle is driven. EPA has found that the mpg obtained by most drivers will be within a few mpg of the estimates.” In fact, the window sticker also includes an in- terval estimate for fuel economy: 14 to 22 mpg in the city and 23 to 33 mpg on the highway.

POINT ESTIMATE The statistic, computed from sample information that estimates the population parameter.

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Estimation and Confidence Intervals 299

2. Litchfield Home Builders Inc. builds homes in the southeastern region of the United States. One of the major concerns of new buyers is the date when the home will be completed. Recently, Litchfield has been telling customers, “Your home will be completed 45 working days from the date we begin installing dry- wall.” The customer relations department at Litchfield wishes to compare this pledge with recent experience. A sample of 50 homes completed this year revealed that the point estimate of the population mean is 46.7 working days from the start of drywall to the completion of the home. Is it reasonable to con- clude that the population mean is still 45 days and that the difference between the sample mean (46.7 days) and the proposed population mean (45 days) is sampling error? In other words, is the sample mean significantly different from the population mean?

3. Recent medical studies indicate that exercise is an impor- tant part of a person’s overall health. The director of human resources at OCF, a large glass manufacturer, wants an esti- mate of the number of hours per week employees spend exercising. A sample of 70 employees reveals the mean number of hours of exercise last week is 3.3. The point esti- mate of 3.3 hours estimates the unknown population mean.

The sample mean, , is not the only point estimate of a population parameter. For example, p, a sample proportion, is a point estimate of �, the population proportion; and s, the sam- ple standard deviation, is a point estimate of �, the population standard deviation.

9.3 Confidence Intervals for a Population Mean A point estimate, however, tells only part of the story. While we expect the point esti- mate to be close to the population parameter, we would like to measure how close it really is. A confidence interval serves this purpose. For example, we estimate the mean yearly income for construction workers in the New York–New Jersey area is $85,000. The range of this estimate might be from $81,000 to $89,000. We can de- scribe how confident we are that the population parameter is in the interval by making a probability statement. We might say, for instance, that we are 90 percent sure that the mean yearly income of construction workers in the New York–New Jersey area is between $81,000 and $89,000.

CONFIDENCE INTERVAL A range of values constructed from sample data so that the population parameter is likely to occur within that range at a specified probability. The specified probability is called the level of confidence.

To compute a confidence interval, we will consider two situations:

• We use sample data to estimate � with and the population standard devia- tion (�) is known.

• We use sample data to estimate � with , and the population standard devia- tion is unknown. In this case, we substitute the sample standard deviation (s) for the population standard deviation (�).

There are important distinctions in the assumptions between these two situations. We first consider the case where � is known.

LO1 Define point estimate.

LO2 Define level of confidence.

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Population Standard Deviation Known � A confidence interval is computed using two statistics: the sample mean, , and the standard deviation. From previous chapters, you know that the standard devia- tion is an important statistic because it measures the dispersion, or width, of a pop- ulation or sample distribution. In computing a confidence interval, the standard devi- ation is used to compute the range of the confidence interval.

To demonstrate the idea of a confidence interval, we start with one simplifying assumption. That assumption is that we know the value of the population standard deviation, �. Knowing � allows us to simplify the development of a confidence inter- val because we can use the standard normal distribution from Chapter 8.

Recall that the sampling distribution of the sample mean is the distribution of all sample means, , of sample size n from a population. The population standard deviation, �, is known. From this information, and the central limit theorem, we know that the sampling distribution follows the normal probability distribution with a mean of � and a standard deviation . Also recall that this value is called the stan- dard error.

The results of the central limit theorem allow us to make the following general confidence interval statements using z-statistics:

1. Ninety-five percent of the sample means selected from a population will be within 1.96 standard errors of the population mean �.

2. Ninety-nine percent of the sample means will lie within 2.58 standard errors of the population mean.

These confidence interval statements provide examples of levels of confidence and are called a 95 percent confidence interval and a 99 percent confidence inter- val. The 95 percent and 99 percent are the levels of confidence and refer to the percentage of similarly constructed intervals that would include the parameter being estimated—in this case, �.

How are the values of 1.96 and 2.58 obtained? For a 95 percent confidence interval, see the following diagram and use Appendix B.1 to find the appropriate z values. Locate .4750 in the body of the table. Read the corresponding row and col- umn values. The value is 1.96. Thus, the probability of finding a z value between 0 and 1.96 is .4750. Likewise, the probability of being in the interval between �1.96 and 0 is also .4750. When we combine these two, the probability of being in the interval �1.96 to 1.96 is .9500. At the top of the next page is a portion of Appendix B.1. The z value for the 90 percent level of confidence is determined in a similar manner. It is 1.65. For a 99 percent level of confidence, the z value is 2.58.

��1n

z value

.025

1.96–1.96

.025 .4750.4750

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Estimation and Confidence Intervals 301

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

o o o o o o o o o o 1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441 1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633 1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706 1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767

2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817 2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857 2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890 2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916 2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936

How do we determine a 95 percent confidence interval? The width of the interval is determined by the level of confidence and the size of the standard error of the mean. We described earlier how to find the z value for a particular level of confidence. Recall from the previous chapter (see formula 8–1 on page 285) the standard error of the mean reports the variation in the distribution of sample means. It is really the standard deviation of the distribution of sample means. The formula is repeated below:

where: is the symbol for the standard error of the mean. We use a Greek letter because it is a population value, and the subscript reminds us that it refers to a sampling distribution of the sample means.

� is the population standard deviation. n is the number of observations in the sample.

The size of the standard error is affected by two values. The first is the stan- dard deviation of the population. The larger the population standard deviation, �, the larger . If the population is homogenous, resulting in a small population standard deviation, the standard error will also be small. However, the standard error is also affected by the number of observations in the sample. A large number of observations in the sample will result in a small standard error of estimate, indicat- ing that there is less variability in the sample means.

To explain these ideas, consider the following example. Del Monte Foods dis- tributes diced peaches in 4-ounce plastic cups. To be sure each cup contains at

least the required amount, Del Monte sets the filling operation to dispense 4.01 ounces of peaches and gel in each cup. So 4.01 is the population mean. Of course not every cup will contain exactly 4.01 ounces of peaches and gel. Some cups will have more and others less. Suppose the population standard deviation of the process is 0.04 ounces. Let’s also assume that the process follows the normal probability distribution. Now, we select a random sam- ple of 64 cups and determine the sample mean. It is 4.015 ounces of peaches and gel. The 95 percent confidence interval for the population mean of this particular sample is:

The 95 percent confidence interval is between 4.0052 and 4.0248. Of course in this case we observe that the population mean of 4.01 ounces is in this interval. But that will not always be the case. Theoretically, if we selected 100 samples of 64 cups

4.015 � 1.96(.04�164) � 4.015 � .0098

�/1n

x� �x�

�x� � �

LO3 Compute a confidence interval for the population mean when the population standard deviation is known.

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from the population, calculated the sample mean, and developed a confidence interval based on each sample mean, we would expect to find the population mean in about 95 of the 100 intervals.

We can summarize the following calculations for a 95 percent confidence inter- val in the following formula:

Similarly, a 99 percent confidence interval is computed as follows.

As we discussed earlier, the values 1.96 and 2.58 are z values corresponding to the middle 95 percent and the middle 99 percent of the observations, respectively.

We are not restricted to the 95 and 99 percent levels of confidence. We can select any confidence level between 0 and 100 percent and find the corresponding value for z. In general, a confidence interval for the population mean when the population fol- lows the normal distribution and the standard deviation is known is computed by:

X � 2.58 �

X � 1.96 �

CONFIDENCE INTERVAL FOR POPULATION MEAN WITH � KNOWN [9–1]

X � z �

where z depends on the level of confidence. Thus for a 92 percent level of confi- dence, the value of z in formula (9–1) is 1.75. The value of z is from Appendix B.1. This table is based on half the normal distribution, so .9200/2 � .4600. The closest value in the body of the table is .4599 and the corresponding z value is 1.75.

Frequently, we also use the 90 percent level of confidence. In this case, we want the area between 0 and z to be .4500, found by .9000/2. To find the z value for this level of confidence, move down the left column of Appendix B.1 to 1.6 and then over to the columns headed 0.04 and 0.05. The area corresponding to a z value of 1.64 is .4495, and for 1.65 it is .4505. To be conservative, we use 1.65. Try looking up the following levels of confidence and check your answers with the corresponding z values given on the right.

Confidence Nearest Half z Level Probability Value

80 percent .3997 1.28 94 percent .4699 1.88 96 percent .4798 2.05

The following example shows the details for calculating a confidence interval and interpreting the result.

Example

Solution

The American Management Association wishes to have information on the mean income of store managers in the retail industry. A random sample of 256 managers reveals a sample mean of $45,420. The standard deviation of this population is $2,050. The association would like answers to the following questions:

1. What is the population mean? 2. What is a reasonable range of values for the population mean? 3. How do we interpret these results?

Generally, distributions of salary and income are positively skewed, because a few indi- viduals earn considerably more than others, thus skewing the distribution in the pos- itive direction. Fortunately, the central limit theorem says that the sampling distribution

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of the mean becomes a normal distribution as sample size increases. In this instance, a sample of 256 store managers is large enough that we can assume that the sam- pling distribution will follow the normal distribution. Now to answer the questions posed in the example.

1. What is the population mean? In this case, we do not know. We do know the sample mean is $45,420. Hence, our best estimate of the unknown population value is the corresponding sample statistic. Thus the sample mean of $45,420 is a point estimate of the unknown population mean.

2. What is a reasonable range of values for the population mean? The asso- ciation decides to use the 95 percent level of confidence. To determine the cor- responding confidence interval, we use formula (9–1).

The usual practice is to round these endpoints to $45,169 and $45,671. These endpoints are called the confidence limits. The degree of confidence or the level of confidence is 95 percent and the confidence interval is from $45,169 to $45,671. The �$251 is often referred to as the margin of error.

3. How do we interpret these results? Suppose we select many samples of 256 store managers, perhaps several hundred. For each sample, we compute the mean and then construct a 95 percent confidence interval, such as we did in the previous section. We could expect about 95 percent of these confidence inter- vals to contain the population mean. About 5 percent of the intervals would not contain the population mean annual income, which is �. However, a particular confidence interval either contains the population parameter or it does not. The following diagram shows the results of selecting samples from the population of store managers in the retail industry, computing the mean of each and then, using formula (9–1), determining a 95 percent confidence interval for the population mean. Note that not all intervals include the population mean. Both the endpoints of the fifth sample are less than the population mean. We attribute this to sam- pling error, and it is the risk we assume when we select the level of confidence.

X � z �

1n � $45,420 � 1.96

$2,050 1256

� $45,420 � $251

μμ – 1.96 �

√n μ + 1.96 �

√n

Sample 1 of size 256. It includes the population mean. Sample 2 of size 256. It includes the population mean.

Sample 3 of size 256. It includes the population mean.

Sample 4 of size 256. It includes the population mean.

Sample 6 of size 256. It includes the population mean.

Sample 5 of size 256. It does not include the population mean.

Population mean

Scale of X

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A Computer Simulation With the aid of a computer, we can randomly select samples from a population, quickly compute the confidence interval, and show how confidence intervals usually, but not always, include the population parameter. The following example will help to explain.

Example

Solution

From many years in the automobile leasing business, Town Bank knows the mean dis- tance driven on a four-year lease is 50,000 miles and the standard deviation is 5,000. These are population values. Suppose, using the Minitab statistical software system, we want to find what proportion of the 95 percent confidence intervals will include the population mean of 50,000. To make the calculations easier to understand, we’ll con- duct the study in thousands of miles, instead of miles. We select 60 random samples of 30 observations from a population with a mean of 50 and a standard deviation of 5.

The results of 60 random samples of 30 automobiles each are summarized in the computer output below. Of the 60 confidence intervals with a 95 percent confidence level, 2, or 3.33 percent, did not include the population mean of 50. The intervals (C3 and C59) that do not include the population mean are highlighted. 3.33 percent is close to the estimate that 5 percent of the intervals will not include the popula- tion mean, and the 58 of 60, or 96.67 percent, is close to 95 percent.

To explain the first calculation in more detail: Minitab began by selecting a ran- dom sample of 30 observations from a population with a mean of 50 and a stan- dard deviation of 5. The mean of these 30 observations is 50.053. The sampling error is 0.053, found by . The endpoints of the confidence inter- val are 48.264 and 51.842. These endpoints are determined by using formula (9–1):

X � 1.96 �

1n � 50.053 � 1.96

5 130

� 50.053 � 1.789

X � � � 50.053 � 50.000

One-Sample Z:

The assumed sigma = 5

Variable N Mean StDev SE Mean 95.0% CI C1 30 50.053 5.002 0.913 ( 48.264, 51.842) C2 30 49.025 4.450 0.913 ( 47.236, 50.815) C3 30 52.023 5.918 0.913 ( 50.234, 53.812) C4 30 50.056 3.364 0.913 ( 48.267, 51.845) C5 30 49.737 4.784 0.913 ( 47.948, 51.526) C6 30 51.074 5.495 0.913 ( 49.285, 52.863) C7 30 50.040 5.930 0.913 ( 48.251, 51.829) C8 30 48.910 3.645 0.913 ( 47.121, 50.699) C9 30 51.033 4.918 0.913 ( 49.244, 52.822) C10 30 50.692 4.571 0.913 ( 48.903, 52.482) C11 30 49.853 4.525 0.913 ( 48.064, 51.642) C12 30 50.286 3.422 0.913 ( 48.497, 52.076) C13 30 50.257 4.317 0.913 ( 48.468, 52.046) C14 30 49.605 4.994 0.913 ( 47.816, 51.394) C15 30 51.474 5.497 0.913 ( 49.685, 53.264) C16 30 48.930 5.317 0.913 ( 47.141, 50.719) C17 30 49.870 4.847 0.913 ( 48.081, 51.659) C18 30 50.739 6.224 0.913 ( 48.950, 52.528) C19 30 50.979 5.520 0.913 ( 49.190, 52.768) C20 30 48.848 4.130 0.913 ( 47.059, 50.638) C21 30 49.481 4.056 0.913 ( 47.692, 51.270) C22 30 49.183 5.409 0.913 ( 47.394, 50.973) C23 30 50.084 4.522 0.913 ( 48.294, 51.873) C24 30 50.866 5.142 0.913 ( 49.077, 52.655) C25 30 48.768 5.582 0.913 ( 46.979, 50.557) C26 30 50.904 6.052 0.913 ( 49.115, 52.694) C27 30 49.481 5.535 0.913 ( 47.691, 51.270) C28 30 50.949 5.916 0.913 ( 49.160, 52.739)

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Exercises 1. A sample of 49 observations is taken from a normal population with a standard devia-

tion of 10. The sample mean is 55. Determine the 99 percent confidence interval for the population mean.

2. A sample of 81 observations is taken from a normal population with a standard devia- tion of 5. The sample mean is 40. Determine the 95 percent confidence interval for the population mean.

3. A sample of 250 observations is selected from a normal population for which the popu- lation standard deviation is known to be 25. The sample mean is 20. a. Determine the standard error of the mean. b. Explain why we can use formula (9–1) to determine the 95 percent confidence interval. c. Determine the 95 percent confidence interval for the population mean.

4. Suppose you know � and you want an 85 percent confidence level. What value would you use to multiply the standard error of the mean by?

5. A research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. They found the distribution of amounts spent per week

Self-Review 9–1 The Bun-and-Run is a franchise fast-food restaurant located in the Northeast specializing in half-pound hamburgers, fish sandwiches, and chicken sandwiches. Soft drinks and French fries are also available. The Planning Department of Bun-and-Run Inc. reports that the distri- bution of daily sales for restaurants follows the normal distribution and that the population standard deviation is $3,000. A sample of 40 showed the mean daily sales to be $20,000. (a) What is the population mean? (b) What is the best estimate of the population mean? What is this value called? (c) Develop a 99 percent confidence interval for the population mean. (d) Interpret the confidence interval.

Variable N Mean StDev SE Mean 95.0% CI C29 30 49.106 4.641 0.913 ( 47.317, 50.895) C30 30 49.994 5.853 0.913 ( 48.205, 51.784) C31 30 49.601 5.064 0.913 ( 47.811, 51.390) C32 30 51.494 5.597 0.913 ( 49.705, 53.284) C33 30 50.460 4.393 0.913 ( 48.671, 52.249) C34 30 50.378 4.075 0.913 ( 48.589, 52.167) C35 30 49.808 4.155 0.913 ( 48.019, 51.597) C36 30 49.934 5.012 0.913 ( 48.145, 51.723) C37 30 50.017 4.082 0.913 ( 48.228, 51.806) C38 30 50.074 3.631 0.913 ( 48.285, 51.863) C39 30 48.656 4.833 0.913 ( 46.867, 50.445) C40 30 50.568 3.855 0.913 ( 48.779, 52.357) C41 30 50.916 3.775 0.913 ( 49.127, 52.705) C42 30 49.104 4.321 0.913 ( 47.315, 50.893) C43 30 50.308 5.467 0.913 ( 48.519, 52.097) C44 30 49.034 4.405 0.913 ( 47.245, 50.823) C45 30 50.399 4.729 0.913 ( 48.610, 52.188) C46 30 49.634 3.996 0.913 ( 47.845, 51.424) C47 30 50.479 4.881 0.913 ( 48.689, 52.268) C48 30 50.529 5.173 0.913 ( 48.740, 52.318) C49 30 51.577 5.822 0.913 ( 49.787, 53.366) C50 30 50.403 4.893 0.913 ( 48.614, 52.192) C51 30 49.717 5.218 0.913 ( 47.927, 51.506) C52 30 49.796 5.327 0.913 ( 48.007, 51.585) C53 30 50.549 4.680 0.913 ( 48.760, 52.338) C54 30 50.200 5.840 0.913 ( 48.410, 51.989) C55 30 49.138 5.074 0.913 ( 47.349, 50.928) C56 30 49.667 3.843 0.913 ( 47.878, 51.456) C57 30 49.603 5.614 0.913 ( 47.814, 51.392) C58 30 49.441 5.702 0.913 ( 47.652, 51.230) C59 30 47.873 4.685 0.913 ( 46.084, 49.662) C60 30 51.087 5.162 0.913 ( 49.297, 52.876)

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followed the normal distribution with a population standard deviation of $5. A sample of 49 steady smokers revealed that . a. What is the point estimate of the population mean? Explain what it indicates. b. Using the 95 percent level of confidence, determine the confidence interval for �. Ex-

plain what it indicates. 6. Refer to the previous exercise. Suppose that 64 smokers (instead of 49) were sampled.

Assume the sample mean remained the same. a. What is the 95 percent confidence interval estimate of �? b. Explain why this confidence interval is narrower than the one determined in the pre-

vious exercise. 7. Bob Nale is the owner of Nale’s Quick Fill. Bob would like to estimate the mean number of

gallons of gasoline sold to his customers. Assume the number of gallons sold follows the normal distribution with a population standard deviation of 2.30 gallons. From his records, he selects a random sample of 60 sales and finds the mean number of gallons sold is 8.60. a. What is the point estimate of the population mean? b. Develop a 99 percent confidence interval for the population mean. c. Interpret the meaning of part (b).

8. Dr. Patton is a professor of English. Recently she counted the number of misspelled words in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a population standard deviation of 2.44 words per essay. For her 10 A.M. section of 40 students, the mean number of misspelled words was 6.05. Construct a 95 percent confidence interval for the mean number of misspelled words in the population of student essays.

Population Standard Deviation � Unknown In the previous section, we assumed the population standard deviation was known. In the case involving Del Monte 4-ounce cups of peaches, there would likely be a long history of measurements in the filling process. Therefore, it is reasonable to assume the standard deviation of the population is available. However, in most sam- pling situations the population standard deviation (�) is not known. Here are some examples where we wish to estimate the population means and it is unlikely we would know the population standard deviations. Suppose each of these studies involves students at West Virginia University.

• The Dean of the Business College wants to estimate the mean number of hours full-time students work at paying jobs each week. He selects a sample of 30 stu- dents, contacts each student and asks them how many hours they worked last week. From the sample information, he can calculate the sample mean, but it is not likely he would know or be able to find the population (�) standard devia- tion required in formula (9–1). He could calculate the standard deviation of the sample and use that as an estimate, but he would not likely know the population standard deviation.

• The Dean of Students wants to estimate the distance the typical commuter stu- dent travels to class. She selects a sample of 40 commuter students, contacts each, and determines the one-way distance from each student’s home to the center of campus. From the sample data, she calculates the mean travel dis- tance, that is . It is unlikely the standard deviation of the population would be known or available, again making formula (9–1) unusable.

• The Director of Student Loans wants to know the mean amount owed on stu- dent loans at the time of his/her graduation. The director selects a sample of 20 graduating students and contacts each to find the information. From the sample information, he can estimate the mean amount. However, to develop a confidence interval using formula (9–1), the population standard deviation is necessary. It is not likely this information is available.

Fortunately we can use the sample standard deviation to estimate the population standard deviation. That is, we use s, the sample standard deviation, to estimate �,

X � $20

LO4 Compute a confidence interval for the population mean when the population standard deviation is unknown.

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Estimation and Confidence Intervals 307

the population standard deviation. But in doing so, we cannot use formula (9–1). Because we do not know �, we cannot use the z distribution. However, there is a remedy. We use the sample standard deviation and replace the z distribution with the t distribution.

The t distribution is a continuous probability distribution, with many similar char- acteristics to the z distribution. William Gosset, an English brewmaster, was the first to study the t distribution.

He was particularly concerned with the exact behavior of the distribution of the following statistic:

where s is an estimate of �. He was especially worried about the discrepancy between s and � when s was calculated from a very small sample. The t distribu- tion and the standard normal distribution are shown graphically in Chart 9–1. Note particularly that the t distribution is flatter, more spread out, than the standard nor- mal distribution. This is because the standard deviation of the t distribution is larger than the standard normal distribution.

t � X � � s�1n

CHART 9–1 The Standard Normal Distribution and Student’s t Distribution

z distribution

t distribution

The following characteristics of the t distribution are based on the assumption that the population of interest is normal, or nearly normal.

• It is, like the z distribution, a continuous distribution. • It is, like the z distribution, bell-shaped and symmetrical. • There is not one t distribution, but rather a family of t distributions. All t distri-

butions have a mean of 0, but their standard deviations differ according to the sample size, n. There is a t distribution for a sample size of 20, another for a sample size of 22, and so on. The standard deviation for a t distribution with 5 observations is larger than for a t distribution with 20 observations.

• The t distribution is more spread out and flatter at the center than the standard normal distribution (see Chart 9–1). As the sample size increases, however, the t distribution approaches the standard normal distribution, because the errors in using s to estimate � decrease with larger samples.

Because Student’s t distribution has a greater spread than the z distribution, the value of t for a given level of confidence is larger in magnitude than the corre- sponding z value. Chart 9–2 shows the values of z for a 95 percent level of confi- dence and of t for the same level of confidence when the sample size is n � 5. How we obtained the actual value of t will be explained shortly. For now, observe that for the same level of confidence the t distribution is flatter or more spread out than the standard normal distribution.

Statistics in Action

William Gosset was born in England in 1876 and died there in 1937. He worked for many years at Arthur Guinness, Sons and Company. In fact, in his later years he was in charge of the Guinness Brewery in London. Guinness preferred its employees to use pen names when publishing papers, so in 1908, when Gosset wrote “The Probable Error of a Mean,” he used the name “Student.” In this pa- per, he first described the properties of the t distribution.

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To develop a confidence interval for the population mean using the t distribution, we adjust formula (9–1) as follows.

CONFIDENCE INTERVAL FOR THE POPULATION MEAN, � UNKNOWN [9–2]

To determine a confidence interval for the population mean with an unknown standard deviation, we:

1. Assume the sampled population is either normal or approximately normal. From the central limit theorem, we know that this assumption is questionable for small sample sizes, and becomes more valid with larger sample sizes.

2. Estimate the population standard deviation (�) with the sample standard deviation (s).

3. Use the t distribution rather than the z distribution.

We should be clear at this point. We base the decision on whether to use the t or the z on whether or not we know �, the population standard deviation. If we know the population standard deviation, then we use z. If we do not know the population standard deviation, then we must use t. Chart 9–3 summarizes the decision-making process.

The following example will illustrate a confidence interval for a population mean when the population standard deviation is unknown and how to find the appropriate value of t in a table.

X � t s 1n

CHART 9–2 Values of z and t for the 95 Percent Level of Confidence

Distribution of z

.025

1.96 Scale of z

Distribution of t

.025

.025 .95

.95 .025

2.776

1.96

2.776 Scale of t

n = 5

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Estimation and Confidence Intervals 309

CHART 9–3 Determining When to Use the z Distribution or the t Distribution

Use the t distribution Use the z distribution

Is the population standard

deviation known?

Assume the population is

normal

YesNo

Example

Solution

A tire manufacturer wishes to investigate the tread life of its tires. A sample of 10 tires driven 50,000 miles revealed a sample mean of 0.32 inches of tread remaining with a standard deviation of 0.09 inches. Construct a 95 percent confidence interval for the population mean. Would it be reasonable for the manufacturer to conclude that after 50,000 miles the population mean amount of tread remaining is 0.30 inches?

To begin, we assume the population distribution is normal. In this case, we don’t have a lot of evidence, but the assumption is probably reasonable. We know the sample standard deviation is .09 inches. We use formula (9–2):

From the information given, , s � 0.09, and n � 10. To find the value of t, we use Appendix B.2, a portion of which is reproduced in Table 9–1. Appendix B.2 is also reproduced on the inside back cover of the text. The first step for locating

X � 0.32

X � t s 1n

TABLE 9–1 A Portion of the t Distribution

Confidence Intervals

80% 90% 95% 98% 99%

Level of Significance for One-Tailed Test

df 0.10 0.05 0.025 0.010 0.005

Level of Significance for Two-Tailed Test

0.20 0.10 0.05 0.02 0.01

1 3.078 6.314 12.706 31.821 63.657 2 1.886 2.920 4.303 6.965 9.925 3 1.638 2.353 3.182 4.541 5.841 4 1.533 2.132 2.776 3.747 4.604 5 1.476 2.015 2.571 3.365 4.032 6 1.440 1.943 2.447 3.143 3.707

7 1.415 1.895 2.365 2.998 3.499 8 1.397 1.860 2.306 2.896 3.355 9 1.383 1.833 2.262 2.821 3.250

10 1.372 1.812 2.228 2.764 3.169

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Here is another example to clarify the use of confidence intervals. Suppose an ar- ticle in your local newspaper reported that the mean time to sell a residential property in the area is 60 days. You select a random sample of 20 homes sold in the last year and find the mean selling time is 65 days. Based on the sample data, you develop a 95 percent confidence interval for the population mean. You find that the endpoints of the confidence interval are 62 days and 68 days. How do you interpret this result? You can be reasonably confident the population mean is within this range. The value pro- posed for the population mean, that is, 60 days, is not included in the interval. It is not likely that the population mean is 60 days. The evidence indicates the statement by the local newspaper may not be correct. To put it another way, it seems unreasonable to obtain the sample you did from a population that had a mean selling time of 60 days.

The following example will show additional details for determining and inter- preting a confidence interval. We used Minitab to perform the calculations.

t is to move across the columns identified for “Confidence Intervals” to the level of confidence requested. In this case, we want the 95 percent level of confidence, so we move to the column headed “95%.” The column on the left margin is identified as “df.” This refers to the number of degrees of freedom. The number of degrees of freedom is the number of observations in the sample minus the number of sam- ples, written n � 1. In this case, it is 10 � 1 � 9. Why did we decide there were 9 degrees of freedom? When sample statistics are being used, it is necessary to determine the number of values that are free to vary.

To illustrate the meaning of degrees of freedom: Assume that the mean of four numbers is known to be 5. The four numbers are 7, 4, 1, and 8. The deviations of these numbers from the mean must total 0. The deviations of �2, �1, �4, and �3 do total 0. If the deviations of �2, �1, and �4 are known, then the value of �3 is fixed (restricted) in order to satisfy the condition that the sum of the deviations must equal 0. Thus, 1 degree of freedom is lost in a sampling problem involving the standard deviation of the sample because one number (the arithmetic mean) is known. For a 95 percent level of confidence and 9 degrees of freedom, we select the row with 9 degrees of freedom. The value of t is 2.262.

To determine the confidence interval, we substitute the values in formula (9–2).

The endpoints of the confidence interval are 0.256 and 0.384. How do we interpret this result? If we repeated this study 200 times, calculating the 95 percent confi- dence interval with each sample’s mean and the standard deviation, 190 of the inter- vals would include the population mean. Ten of the intervals would not include the population mean. This is the effect of sampling error. A further interpretation is to conclude that the population mean is in this interval. The manufacturer can be rea- sonably sure (95 percent confident) that the mean remaining tread depth is between 0.256 and 0.384 inches. Because the value of 0.30 is in this interval, it is possible that the mean of the population is 0.30.

X � t s 1n

� 0.32 � 2.262 0.09 110

� 0.32 � .064

Example The manager of the Inlet Square Mall, near Ft. Myers, Florida, wants to estimate the mean amount spent per shopping visit by customers. A sample of 20 customers reveals the following amounts spent.

$48.16 $42.22 $46.82 $51.45 $23.78 $41.86 $54.86 37.92 52.64 48.59 50.82 46.94 61.83 61.69 49.17 61.46 51.35 52.68 58.84 43.88

What is the best estimate of the population mean? Determine a 95 percent confidence interval. Interpret the result. Would it be reasonable to conclude that the population mean is $50? What about $60?

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Solution

Estimation and Confidence Intervals 311

The calculations to construct a confidence interval are also available in Excel. The output follows. Note that the sample mean ($49.35) and the sample standard deviation ($9.01) are the same as those in the Minitab calculations. In the Excel information, the last line of the output also includes the margin of error, which is the amount that is added and subtracted from the sample mean to form the endpoints of the confidence interval. This value is found from

t s 1n

� 2.093 $9.01 120

� $4.22

The mall manager assumes that the population of the amounts spent follows the normal distribution. This is a rea- sonable assumption in this case. Additionally, the confidence interval technique is quite powerful and tends to commit any errors on the conservative side if the population is not nor- mal. We should not make the normality assumption when the population is severely skewed or when the distribution has “thick tails.” In Chapter 18, we present methods for han- dling this problem if we cannot make the normality assump- tion. In this case, the normality assumption is reasonable.

The population standard deviation is not known. Hence, it is appropriate to use the t distribution and formula (9–2) to find the confidence interval. We use the Minitab system to find the mean and standard deviation of this sample. The results are shown below.

The mall manager does not know the population mean. The sample mean is the best estimate of that value. From the pictured Minitab output, the mean is $49.35, which is the best estimate, the point estimate, of the unknown population mean.

We use formula (9–2) to find the confidence interval. The value of t is available from Appendix B.2. There are n � 1 � 20 � 1 � 19 degrees of freedom. We move across the row with 19 degrees of freedom to the column for the 95 percent confidence level. The value at this intersection is 2.093. We substitute these values into formula (9–2) to find the confidence interval.

The endpoints of the confidence interval are $45.13 and $53.57. It is reasonable to conclude that the population mean is in that interval.

The manager of Inlet Square wondered whether the population mean could have been $50 or $60. The value of $50 is within the confidence interval. It is reasonable that the population mean could be $50. The value of $60 is not in the confidence interval. Hence, we conclude that the population mean is unlikely to be $60.

X � t s 1n

� $49.35 � 2.093 $9.01 120

� $49.35 � $4.22

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Exercises 9. Use Appendix B.2 to locate the value of t under the following conditions.

a. The sample size is 12 and the level of confidence is 95 percent. b. The sample size is 20 and the level of confidence is 90 percent. c. The sample size is 8 and the level of confidence is 99 percent.

10. Use Appendix B.2 to locate the value of t under the following conditions. a. The sample size is 15 and the level of confidence is 95 percent. b. The sample size is 24 and the level of confidence is 98 percent. c. The sample size is 12 and the level of confidence is 90 percent.

11. The owner of Britten’s Egg Farm wants to estimate the mean number of eggs laid per chicken. A sample of 20 chickens shows they laid an average of 20 eggs per month with a standard deviation of 2 eggs per month. a. What is the value of the population mean? What is the best estimate of this value? b. Explain why we need to use the t distribution. What assumption do you need to

make? c. For a 95 percent confidence interval, what is the value of t? d. Develop the 95 percent confidence interval for the population mean. e. Would it be reasonable to conclude that the population mean is 21 eggs? What about

25 eggs?

Self-Review 9–2 Dottie Kleman is the “Cookie Lady.” She bakes and sells cookies at 50 different locations in the Philadelphia area. Ms. Kleman is concerned about absenteeism among her workers. The information below reports the number of days absent for a sample of 10 workers during the last two-week pay period.

4 1 2 2 1 2 2 1 0 3

(a) Determine the mean and the standard deviation of the sample. (b) What is the population mean? What is the best estimate of that value? (c) Develop a 95 percent confidence interval for the population mean. (d) Explain why the t distribution is used as a part of the confidence interval. (e) Is it reasonable to conclude that the typical worker does not miss any

days during a pay period?

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12. The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of 16 people reveals the mean yearly consumption to be 60 gallons with a standard deviation of 20 gallons. a. What is the value of the population mean? What is the best estimate of this value? b. Explain why we need to use the t distribution. What assumption do you need to make? c. For a 90 percent confidence interval, what is the value of t? d. Develop the 90 percent confidence interval for the population mean. e. Would it be reasonable to conclude that the population mean is 63 gallons?

13. Merrill Lynch Securities and Health Care Retirement Inc. are two large employers in down- town Toledo, Ohio. They are considering jointly offering child care for their employees. As a part of the feasibility study, they wish to estimate the mean weekly child-care cost of their employees. A sample of 10 employees who use child care reveals the following amounts spent last week.

$107 $92 $97 $95 $105 $101 $91 $99 $95 $104

Develop a 90 percent confidence interval for the population mean. Interpret the result. 14. The Greater Pittsburgh Area Chamber of Commerce wants to estimate the mean time

workers who are employed in the downtown area spend getting to work. A sample of 15 workers reveals the following number of minutes spent traveling.

29 38 38 33 38 21 45 34 40 37 37 42 30 29 35

Develop a 98 percent confidence interval for the population mean. Interpret the result.

9.4 A Confidence Interval for a Proportion The material presented so far in this chapter uses the ratio scale of measurement. That is, we use such variables as incomes, weights, distances, and ages. We now want to consider situations such as the following:

• The career services director at Southern Tech- nical Institute reports that 80 percent of its graduates enter the job market in a position related to their field of study.

• A company representative claims that 45 per- cent of Burger King sales are made at the drive-through window.

• A survey of homes in the Chicago area indi- cated that 85 percent of the new construc- tion had central air conditioning.

• A recent survey of married men between the ages of 35 and 50 found that 63 percent felt that both partners should earn a living.

These examples illustrate the nominal scale of measurement. When we measure with a nom-

inal scale, an observation is classified into one of two or more mutually exclusive groups. For example, a graduate of Southern Tech either entered the job market in a position related to his or her field of study or not. A particular Burger King customer either made a purchase at the drive-through window or did not make a purchase at the drive-through window. There are only two possibilities, and the outcome must be classified into one of the two groups.

LO5 Compute a confidence interval for a population proportion.

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314 Chapter 9

PROPORTION The fraction, ratio, or percent indicating the part of the sample or the population having a particular trait of interest.

As an example of a proportion, a recent survey indicated that 92 out of 100 surveyed favored the continued use of daylight savings time in the summer. The sample proportion is 92/100, or .92, or 92 percent. If we let p represent the sam- ple proportion, X the number of “successes,” and n the number of items sampled, we can determine a sample proportion as follows.

CONFIDENCE INTERVAL FOR A POPULATION PROPORTION

[9–4]p � z C

p(1� p) n

SAMPLE PROPORTION [9–3]p � X n

The population proportion is identified by �. Therefore, � refers to the percent of successes in the population. Recall from Chapter 6 that � is the proportion of “successes” in a binomial distribution. This continues our practice of using Greek letters to identify population parameters and Roman letters to identify sample statistics.

To develop a confidence interval for a proportion, we need to meet the follow- ing assumptions.

1. The binomial conditions, discussed in Chapter 6, have been met. Briefly, these conditions are: a. The sample data is the result of counts. b. There are only two possible outcomes. (We usually label one of the outcomes

a “success” and the other a “failure.”) c. The probability of a success remains the same from one trial to the next. d. The trials are independent. This means the outcome on one trial does not

affect the outcome on another. 2. The values n� and n(1 � �) should both be greater than or equal to 5. This

condition allows us to invoke the central limit theorem and employ the standard normal distribution, that is, z, to complete a confidence interval.

Developing a point estimate for a population proportion and a confidence inter- val for a population proportion is similar to doing so for a mean. To illustrate, John Gail is running for Congress from the third district of Nebraska. From a random sample of 100 voters in the district, 60 indicate they plan to vote for him in the upcoming election. The sample proportion is .60, but the population proportion is unknown. That is, we do not know what proportion of voters in the population will vote for Mr. Gail. The sample value, .60, is the best estimate we have of the unknown population parameter. So we let p, which is .60, be an estimate of �, which is not known.

To develop a confidence interval for a population proportion, we use:

An example will help to explain the details of determining a confidence interval and the result.

Statistics in Action

Many survey results reported in newspa- pers, in news maga- zines, and on TV use confidence intervals. For example, a recent survey of 800 TV viewers in Toledo, Ohio, found 44 per- cent watched the evening news on the local CBS affiliate. The article went on to indicate the margin of error was 3.4 per- cent. The margin of error is actually the amount that is added and subtracted from the point estimate to find the endpoints of a confidence interval. From formula (9–4) and the 95 percent level of confidence:

� 0.034

� 1.96 C

.44(1 � .44) 800

z C

p ( 1 � p) n

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To review the interpretation of the confidence interval: If the poll was conducted 100 times with 100 different samples, the confidence intervals constructed from 95 of the samples would contain the true population proportion. In addition, the inter- pretation of a confidence interval can be very useful in decision making and play a very important role especially on election night. For example, Cliff Obermeyer is run- ning for Congress from the 6th District of New Jersey. Suppose 500 voters are con- tacted upon leaving the polls and 275 indicate they voted for Mr. Obermeyer. We will assume that the exit poll of 500 voters is a random sample of those voting in the 6th District. That means that 55 percent of those in the sample voted for Mr. Obermeyer. Based on formula (9–3):

Now, to be assured of election, he must earn more than 50 percent of the votes in the population of those voting. At this point, we know a point estimate, which is .55, of the population of voters that will vote for him. But we do not know the per- cent in the population that will ultimately vote for the candidate. So the question is: Could we take a sample of 500 voters from a population where 50 percent or less of the voters support Mr. Obermeyer and find that 55 percent of the sample support him? To put it another way, could the sampling error, which is p � � � .55 � .50 � .05 be due to chance, or is the population of voters who support Mr. Obermeyer greater than .50. If we develop a confidence interval for the sample proportion and find that .50 is not in the interval, then we conclude that the proportion of voters supporting Mr. Obermeyer is greater than .50. What does that mean? Well, it means he should be elected! What if .50 is in the interval? Then we conclude that it is possible that 50 percent or less of the voters support his candidacy and we cannot conclude that he will be elected based on the sample information. In this case, using the 95 per- cent significance level and formula (9–4):

p � z C

p(1 � p) n

� .55 � 1.96 C

.55(1 � .55) 500

� .55 � .044

p � X n

� 275 500

� .55

Example The union representing the Bottle Blowers of America (BBA) is considering a pro- posal to merge with the Teamsters Union. According to BBA union bylaws, at least three-fourths of the union membership must approve any merger. A random sample of 2,000 current BBA members reveals 1,600 plan to vote for the merger proposal. What is the estimate of the population proportion? Develop a 95 percent confidence interval for the population proportion. Basing your decision on this sample informa- tion, can you conclude that the necessary proportion of BBA members favor the merger? Why?

Solution First, calculate the sample proportion from formula (9–3). It is .80, found by

Thus, we estimate that 80 percent of the population favor the merger proposal. We determine the 95 percent confidence interval using formula (9–4). The z value cor- responding to the 95 percent level of confidence is 1.96.

The endpoints of the confidence interval are .782 and .818. The lower endpoint is greater than .75. Hence, we conclude that the merger proposal will likely pass because the interval estimate includes values greater than 75 percent of the union membership.

p � z C

p(1 � p) n

� .80 � 1.96 C

.80(1 � .80) 2,000

� .80 � .018

p � X n

� 1,600 2,000

� .80

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316 Chapter 9

So the endpoints of the confidence interval are .55 � .044 � .506 and .55 � .044 � .594. The value of .50 is not in this interval. So we conclude that probably more than 50 percent of the voters support Mr. Obermeyer and that is enough to get him elected.

Is this procedure ever used? Yes! It is exactly the procedure used by television networks, news magazines, and polling organizations on election night.

Self-Review 9–3 A market survey was conducted to estimate the proportion of homemakers who would recognize the brand name of a cleanser based on the shape and the color of the con- tainer. Of the 1,400 homemakers sampled, 420 were able to identify the brand by name. (a) Estimate the value of the population proportion. (b) Develop a 99 percent confidence interval for the population proportion. (c) Interpret your findings.

Exercises 15. The owner of the West End Kwick Fill Gas Station wishes to determine the proportion of

customers who use a credit card or debit card to pay at the pump. He surveys 100 cus- tomers and finds that 80 paid at the pump. a. Estimate the value of the population proportion. b. Develop a 95 percent confidence interval for the population proportion. c. Interpret your findings.

16. Ms. Maria Wilson is considering running for mayor of the town of Bear Gulch, Montana. Before completing the petitions, she decides to conduct a survey of voters in Bear Gulch. A sample of 400 voters reveals that 300 would support her in the November election. a. Estimate the value of the population proportion. b. Develop a 99 percent confidence interval for the population proportion. c. Interpret your findings.

17. The Fox TV network is considering replacing one of its prime-time crime investigation shows with a new family-oriented comedy show. Before a final decision is made, net- work executives commission a sample of 400 viewers. After viewing the comedy, 250 indicated they would watch the new show and suggested it replace the crime investiga- tion show. a. Estimate the value of the population proportion. b. Develop a 99 percent confidence interval for the population proportion. c. Interpret your findings.

18. Schadek Silkscreen Printing Inc. purchases plastic cups on which to print logos for sporting events, proms, birthdays, and other special occasions. Zack Schadek, the owner, received a large shipment this morning. To ensure the quality of the shipment, he selected a random sample of 300 cups. He found 15 to be defective. a. What is the estimated proportion defective in the population? b. Develop a 95 percent confidence interval for the proportion defective. c. Zack has an agreement with his supplier that he is to return lots that are 10 percent

or more defective. Should he return this lot? Explain your decision.

9.5 Choosing an Appropriate Sample Size When working with confidence intervals, one important variable is sample size. However, in practice, sample size is not a variable. It is a decision we make so that our estimate of a population parameter is a good one. Our decision is based on three variables:

1. The margin of error the researcher will tolerate. 2. The level of confidence desired, for example, 95 percent. 3. The variation or dispersion of the population being studied.

LO6 Calculate the required sample size to estimate a population proportion or population mean.

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The first variable is the margin of error. It is designated as E and is the amount that is added and subtracted to the sample mean (or sample proportion) to deter- mine the endpoints of the confidence interval. For example, in a study of wages, we may decide that we want to estimate the population average wage with a mar- gin of error of plus or minus $1,000. Or in an opinion poll, we may decide that we want to estimate the population proportion with a margin of error of plus or minus 5 percent. The margin of error is the amount of error we are willing to tolerate in estimating a population parameter. You may wonder why we do not choose small margins of error. There is a trade-off between the margin of error and sample size. A small margin of error will require a larger sample and more money and time to collect the sample. A larger margin of error will permit a smaller sample and a wider confidence interval.

The second choice is the level of confidence. In working with confidence inter- vals, we logically choose relatively high levels of confidence such as 95 percent and 99 percent. To compute the sample size, we need the z-statistic that corresponds to the chosen level of confidence. The 95 percent level of confidence corresponds to a z value of 1.96, and a 99 percent level of confidence corresponds to a z value of 2.58. Notice that larger sample sizes (and more time and money to collect the sample) correspond with higher levels of confidence. Also, notice that we use a z- statistic.

The third choice to determine the sample size is the population standard devi- ation. If the population is widely dispersed, a large sample is required. On the other hand, if the population is concentrated (homogeneous), the required sample size will be smaller. Often, we do not know the population standard deviation. Here are three suggestions for finding a value for the population standard deviation.

1. Conduct a pilot study. This is the most common method. Suppose we want an estimate of the number of hours per week worked by students enrolled in the College of Business at the University of Texas. To test the validity of our questionnaire, we use it on a small sample of students. From this small sam- ple, we compute the standard deviation of the number of hours worked and use this value as the population standard deviation.

2. Use a comparable study. Use this approach when there is an estimate of the standard deviation from another study. Suppose we want to estimate the num- ber of hours worked per week by refuse workers. Information from certain state or federal agencies that regularly study the workforce may provide a reliable value to use for the population standard deviation.

3. Use a range-based approach. To use this approach, we need to know or have an estimate of the largest and smallest values in the population. Recall from Chapter 3, the Empirical Rule states that virtually all the observations could be expected to be within plus or minus 3 standard deviations of the mean, assum- ing that the distribution follows the normal distribution. Thus, the distance between the largest and the smallest values is 6 standard deviations. We can estimate the standard deviation as one-sixth of the range. For example, the director of operations at University Bank wants to estimate the number of checks written per month by college students. She believes that the distribu- tion of the number of checks written follows the normal distribution. The mini- mum and maximum numbers written per month are 2 and 50, so the range is 48, found by (50 � 2). Then 8 checks per month, 48/6, would be the value we would use for the population standard deviation.

Sample Size to Estimate a Population Mean To estimate a population mean, we can express the interaction among these three factors and the sample size in the following formula. Notice that this formula is the

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318 Chapter 9

margin of error used to calculate the endpoints of confidence intervals to estimate a population mean!

Solving this equation for n yields the following result.

E � z �

SAMPLE SIZE FOR ESTIMATING THE POPULATION MEAN

where: n is the size of the sample. z is the standard normal value corresponding to the desired level of confidence. � is the population standard deviation. E is the maximum allowable error.

The result of this calculation is not always a whole number. When the outcome is not a whole number, the usual practice is to round up any fractional result to the next whole number. For example, 201.21 would be rounded up to 202.

[9–5]n � a z� E b

Example A student in public administration wants to determine the mean amount members of city councils in large cities earn per month as remuneration for being a council mem- ber. The error in estimating the mean is to be less than $100 with a 95 percent level of confidence. The student found a report by the Department of Labor that reported a standard deviation of $1,000. What is the required sample size?

Solution The maximum allowable error, E, is $100. The value of z for a 95 percent level of confidence is 1.96, and the value of the standard deviation is $1,000. Substituting these values into formula (9–5) gives the required sample size as:

The computed value of 384.16 is rounded up to 385. A sample of 385 is required to meet the specifications. If the student wants to increase the level of confidence, for example to 99 percent, this will require a larger sample. The z value corre- sponding to the 99 percent level of confidence is 2.58.

We recommend a sample of 666. Observe how much the change in the confidence level changed the size of the sample. An increase from the 95 percent to the 99 per- cent level of confidence resulted in an increase of 281 observations or 73 percent [(666/385)*100]. This would greatly increase the cost of the study, both in terms of time and money. Hence, the level of confidence should be considered carefully.

n � a z� E b

� a (2.58)($1,000)

$100 b

� (25.8)2 � 665.64

n � a z� E b

� a (1.96)($1,000)

$100 b

� (19.6)2 � 384.16

Sample Size to Estimate a Population Proportion To determine the sample size for a proportion, the same three variables need to be specified:

1. The margin of error. 2. The desired level of confidence. 3. The variation or dispersion of the population being studied.

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SAMPLE SIZE FOR THE POPULATION PROPORTION

where: n is the size of the sample. z is the standard normal value corresponding to the desired level of confidence. � is the population proportion. E is the maximum allowable error.

The choices for the z-statistic and the margin of error, E, are the same as the choices for estimating the population mean. However, in this case the population standard deviation for a binomial distribution is represented by �(1 � �). To find a value of the population proportion, we would find a comparable study or conduct a pilot study. If a reliable value cannot be found, then a value of .50 should be used for �. Note that �(1 � �) has the largest value using 0.50 and, therefore, without a good estimate of the population proportion, overstates the sample size. This differ- ence will not hurt the estimate of the population proportion.

Self-Review 9–4 The registrar wants to estimate the arithmetic mean grade point average (GPA) of all grad- uating seniors during the past 10 years. GPAs range between 2.0 and 4.0. The mean GPA is to be estimated within plus or minus .05 of the population mean. Based on prior expe- rience, the population standard deviation is 0.279. Use the 99 percent level of confidence. Will you assist the college registrar in determining how many transcripts to study?

Example The study in the previous example also estimates the proportion of cities that have private refuse collectors. The student wants the margin of error to be within .10 of the population proportion, the desired level of confidence is 90 percent, and no estimate is available for the population proportion. What is the required sample size?

Solution The estimate of the population proportion is to be within .10, so E � .10. The desired level of confidence is .90, which corresponds to a z value of 1.65. Because no esti- mate of the population proportion is available, we use .50. The suggested number of observations is

The student needs a random sample of 69 cities.

n � (.5)(1 � .5)a 1.65 .10 b

� 68.0625

[9–6]n � �(1 � �)a z E b

For the binomial distribution, the margin of error is:

Solving this equation for n yields the following equation

E � z C

�(1 � �) n

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320 Chapter 9

Exercises 19. A population is estimated to have a standard deviation of 10. We want to estimate the

population mean within 2, with a 95 percent level of confidence. How large a sample is required?

20. We want to estimate the population mean within 5, with a 99 percent level of confidence. The population standard deviation is estimated to be 15. How large a sample is required?

21. The estimate of the population proportion is to be within plus or minus .05, with a 95 per- cent level of confidence. The best estimate of the population proportion is .15. How large a sample is required?

22. The estimate of the population proportion is to be within plus or minus .10, with a 99 per- cent level of confidence. The best estimate of the population proportion is .45. How large a sample is required?

23. A survey is being planned to determine the mean amount of time corporation executives watch television. A pilot survey indicated that the mean time per week is 12 hours, with a standard deviation of 3 hours. It is desired to estimate the mean viewing time within one-quarter hour. The 95 percent level of confidence is to be used. How many execu- tives should be surveyed?

24. A processor of carrots cuts the green top off each carrot, washes the carrots, and inserts six to a package. Twenty packages are inserted in a box for shipment. To test the weight of the boxes, a few were checked. The mean weight was 20.4 pounds; the standard deviation, 0.5 pounds. How many boxes must the processor sample to be 95 percent confident that the sample mean does not differ from the population mean by more than 0.2 pounds?

25. Suppose the U.S. president wants an estimate of the proportion of the population who support his current policy toward revisions in the health care system. The president wants the estimate to be within .04 of the true proportion. Assume a 95 percent level of confi- dence. The president’s political advisors estimated the proportion supporting the current policy to be .60. a. How large of a sample is required? b. How large of a sample would be necessary if no estimate were available for the pro-

portion supporting current policy? 26. Past surveys reveal that 30 percent of tourists going to Las Vegas to gamble spend more

than $1,000. The Visitor’s Bureau of Las Vegas wants to update this percentage. a. The new study is to use the 90 percent confidence level. The estimate is to be within

1 percent of the population proportion. What is the necessary sample size? b. The Bureau feels the sample size determined above is too large. What can be done to

reduce the sample? Based on your suggestion, recalculate the sample size.

9.6 Finite-Population Correction Factor The populations we have sampled so far have been very large or infinite. What if the sampled population is not very large? We need to make some adjustments in the way we compute the standard error of the sample means and the standard error of the sample proportions.

A population that has a fixed upper bound is finite. For example, there are 12,179 students enrolled at Eastern Illinois University, there are 40 employees at Spence Sprockets, Chrysler assembled 917 Jeep Wranglers at the Alexis Avenue plant yesterday, or there were 65 surgical patients at St. Rose Memorial Hospital in Sarasota yesterday. A finite population can be rather small; it could be all the stu- dents registered for this class. It can also be very large, such as all senior citizens living in Florida.

For a finite population, where the total number of objects or individuals is N and the number of objects or individuals in the sample is n, we need to adjust the standard errors in the confidence interval formulas. To put it another way, to find the confidence interval for the mean, we adjust the standard error of the mean in formulas (9–1) and (9–2). If we are determining the confidence interval for a

LO7 Adjust a confidence interval for finite populations.

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TABLE 9–2 Finite-Population Correction Factor for Selected Samples When the Population Is 1,000

Sample Fraction of Correction Size Population Factor

10 .010 .9955 25 .025 .9879 50 .050 .9752

100 .100 .9492 200 .200 .8949 500 .500 .7075

So if we wished to develop a confidence interval for the mean from a finite popu- lation and the population standard deviation was unknown, we would adjust formula (9–2) as follows:

We would make a similar adjustment to formula (9–3) in the case of a proportion. The following example summarizes the steps to find a confidence interval for

the mean.

X � t s 1n ° C

N � n N � 1

Example There are 250 families in Scandia, Pennsylvania. A random sample of 40 of these families revealed the mean annual church contribution was $450 and the standard deviation of this was $75. Could the population mean be $445 or $425?

1. What is the population mean? What is the best estimate of the population mean?

2. Develop a 90 percent confidence interval for the population mean. What are the endpoints of the confidence interval?

3. Interpret the confidence interval.

proportion, then we need to adjust the standard error of the proportion in for- mula (9–3).

This adjustment is called the finite-population correction factor. It is often shortened to FPC and is:

Why is it necessary to apply a factor, and what is its effect? Logically, if the sam- ple is a substantial percentage of the population, the estimate is more precise. Note the effect of the term (N � n)/(N � 1). Suppose the population is 1,000 and the sam- ple is 100. Then this ratio is (1,000 � 100)/(1,000 � 1), or 900/999. Taking the square root gives the correction factor, .9492. Multiplying this correction factor by the stan- dard error reduces the standard error by about 5 percent (1 � .9492 � .0508). This reduction in the size of the standard error yields a smaller range of values in esti- mating the population mean or the population proportion. If the sample is 200, the correction factor is .8949, meaning that the standard error has been reduced by more than 10 percent. Table 9–2 shows the effects of various sample sizes.

FPC � C

N � n N � 1

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Exercises 27. Thirty-six items are randomly selected from a population of 300 items. The sample mean

is 35 and the sample standard deviation 5. Develop a 95 percent confidence interval for the population mean.

28. Forty-nine items are randomly selected from a population of 500 items. The sample mean is 40 and the sample standard deviation 9. Develop a 99 percent confidence interval for the population mean.

29. The attendance at the Savannah Colts minor league baseball game last night was 400. A random sample of 50 of those in attendance revealed that the mean number of soft drinks consumed per person was 1.86, with a standard deviation of 0.50. Develop a 99 percent confidence interval for the mean number of soft drinks consumed per person.

30. There are 300 welders employed at Maine Shipyards Corporation. A sample of 30 welders revealed that 18 graduated from a registered welding course. Construct the 95 percent confidence interval for the proportion of all welders who graduated from a registered welding course.

Self-Review 9–5 The same study of church contributions in Scandia revealed that 15 of the 40 families sampled attend church regularly. Construct the 95 percent confidence interval for the proportion of families attending church regularly.

Solution First, note the population is finite. That is, there is a limit to the number of people in Scandia, in this case 250.

1. We do not know the population mean. This is the value we wish to estimate. The best estimate we have of the population mean is the sample mean, which is $450.

2. The formula to find the confidence interval for a population mean follows.

In this case, we know , s � 75, N � 250, and n � 40. We do not know the population standard deviation, so we use the t distribution. To find the appropriate value of t, we use Appendix B.2, and move across the top row to the column headed 90 percent. The degrees of freedom is df � n � 1 � 40 � 1 � 39, so we move to the cell where the df row of 39 intersects with the column headed 90 percent. The value is 1.685. Inserting these values in the formula:

The endpoints of the confidence interval are $431.65 and $468.35. 3. It is likely that the population mean is more than $431.65 but less than $468.35.

To put it another way, could the population mean be $445? Yes, but it is not likely that it is $425. Why is this so? Because the value $445 is within the con- fidence interval and $425 is not within the confidence interval.

� $450 � 1.685 $75 140

° C

250 � 40 250 � 1

¢ � $450 � $19.981.8434 � $450 � $18.35

X � t s 1n ° C

N � n N � 1

X � 450

X � t s 1n ° C

N � n N � 1

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Chapter Summary I. A point estimate is a single value (statistic) used to estimate a population value (parameter).

II. A confidence interval is a range of values within which the population parameter is expected to occur. A. The factors that determine the width of a confidence interval for a mean are:

1. The number of observations in the sample, n. 2. The variability in the population, usually estimated by the sample standard deviation, s. 3. The level of confidence.

a. To determine the confidence limits when the population standard deviation is known, we use the z distribution. The formula is

[9–1]

b. To determine the confidence limits when the population standard deviation is unknown, we use the t distribution. The formula is

[9–2]

III. The major characteristics of the t distribution are: A. It is a continuous distribution. B. It is mound-shaped and symmetrical. C. It is flatter, or more spread out, than the standard normal distribution. D. There is a family of t distributions, depending on the number of degrees of freedom.

IV. A proportion is a ratio, fraction, or percent that indicates the part of the sample or pop- ulation that has a particular characteristic. A. A sample proportion is found by X, the number of successes, divided by n, the number

of observations. B. We construct a confidence interval for a sample proportion from the following formula.

[9–4]

V. We can determine an appropriate sample size for estimating both means and proportions. A. There are three factors that determine the sample size when we wish to estimate the mean.

1. The margin of error, E. 2. The desired level of confidence. 3. The variation in the population. 4. The formula to determine the sample size for the mean is

[9–5]

B. There are three factors that determine the sample size when we wish to estimate a proportion. 1. The margin of error, E. 2. The desired level of confidence. 3. A value for � to calculate the variation in the population. 4. The formula to determine the sample size for a proportion is

[9–6]

VI. For a finite population, the standard error is adjusted by the factor .

Chapter Exercises 31. A random sample of 85 group leaders, supervisors, and similar personnel at General Motors

revealed that, on average, they spent 6.5 years on the job before being promoted. The standard deviation of the sample was 1.7 years. Construct a 95 percent confidence interval.

N � n N � 1

n � �(1 � �)a z E b

n � a z� E b

p � z C

p(1 � p) n

X � t s 1n

X � z �

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32. A state meat inspector in Iowa has been given the assignment of estimating the mean net weight of packages of ground chuck labeled “3 pounds.” Of course, he realizes that the weights cannot be precisely 3 pounds. A sample of 36 packages reveals the mean weight to be 3.01 pounds, with a standard deviation of 0.03 pounds. a. What is the estimated population mean? b. Determine a 95 percent confidence interval for the population mean.

33. As part of their business promotional package, the Milwaukee Chamber of Commerce would like an estimate of the mean cost per month of a one-bedroom apartment. A ran- dom sample of 40 apartments currently available for lease showed the mean cost per month was $323. The standard deviation of the sample was $25. a. Develop a 98 percent confidence interval for the population mean. b. Would it be reasonable to conclude that the population mean is $350 per month?

34. A recent survey of 50 executives who were laid off during a recent recession revealed it took a mean of 26 weeks for them to find another position. The standard deviation of the sample was 6.2 weeks. Construct a 95 percent confidence interval for the popula- tion mean. Is it reasonable that the population mean is 28 weeks? Justify your answer.

35. Marty Rowatti recently assumed the position of director of the YMCA of South Jersey. He would like some current data on how long current members of the YMCA have been members. To investigate, suppose he selects a random sample of 40 current members. The mean length of membership of those included in the sample is 8.32 years and the standard deviation is 3.07 years. a. What is the mean of the population? b. Develop a 90 percent confidence interval for the population mean. c. The previous director, in the summary report she prepared as she retired, indicated

the mean length of membership was now “almost 10 years.” Does the sample infor- mation substantiate this claim? Cite evidence.

36. The American Restaurant Association collected information on the number of meals eaten outside the home per week by young married couples. A survey of 60 couples showed the sample mean number of meals eaten outside the home was 2.76 meals per week, with a standard deviation of 0.75 meals per week. Construct a 97 percent confidence interval for the population mean.

37. The National Collegiate Athletic Association (NCAA) reported that the mean number of hours spent per week on coaching and recruiting by college football assistant coaches during the season was 70. A random sample of 50 assistant coaches showed the sam- ple mean to be 68.6 hours, with a standard deviation of 8.2 hours. a. Using the sample data, construct a 99 percent confidence interval for the popula-

tion mean. b. Does the 99 percent confidence interval include the value suggested by the NCAA?

Interpret this result. c. Suppose you decided to switch from a 99 to a 95 percent confidence interval. With-

out performing any calculations, will the interval increase, decrease, or stay the same? Which of the values in the formula will change?

38. The Human Relations Department of Electronics Inc. would like to include a dental plan as part of the benefits package. The question is: How much does a typical employee and his or her family spend per year on dental expenses? A sample of 45 employees reveals the mean amount spent last year was $1,820, with a standard deviation of $660. a. Construct a 95 percent confidence interval for the population mean. b. The information from part (a) was given to the president of Electronics Inc. He indi-

cated he could afford $1,700 of dental expenses per employee. Is it possible that the population mean could be $1,700? Justify your answer.

39. A student conducted a study and reported that the 95 percent confidence interval for the mean ranged from 46 to 54. He was sure that the mean of the sample was 50, that the standard deviation of the sample was 16, and that the sample was at least 30, but could not remember the exact number. Can you help him out?

40. A recent study by the American Automobile Dealers Association revealed the mean amount of profit per car sold for a sample of 20 dealers was $290, with a standard devi- ation of $125. Develop a 95 percent confidence interval for the population mean.

41. A study of 25 graduates of four-year colleges by the American Banker’s Association revealed the mean amount owed by a student in student loans was $14,381. The stan- dard deviation of the sample was $1,892. Construct a 90 percent confidence interval for the population mean. Is it reasonable to conclude that the mean of the population is actu- ally $15,000? Tell why or why not.

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Estimation and Confidence Intervals 325

42. An important factor in selling a residential property is the number of people who look through the home. A sample of 15 homes recently sold in the Buffalo, New York, area revealed the mean number looking through each home was 24 and the standard deviation of the sample was 5 people. Develop a 98 percent confidence interval for the population mean.

43. Warren County Telephone Company claims in its annual report that “the typical customer spends $60 per month on local and long-distance service.” A sample of 12 subscribers revealed the following amounts spent last month.

$64 $66 $64 $66 $59 $62 $67 $61 $64 $58 $54 $66

a. What is the point estimate of the population mean? b. Develop a 90 percent confidence interval for the population mean. c. Is the company’s claim that the “typical customer” spends $60 per month reason-

able? Justify your answer. 44. The manufacturer of a new line of ink-jet printers would like to include as part of its

advertising the number of pages a user can expect from a print cartridge. A sample of 10 cartridges revealed the following number of pages printed.

2,698 2,028 2,474 2,395 2,372 2,475 1,927 3,006 2,334 2,379

a. What is the point estimate of the population mean? b. Develop a 95 percent confidence interval for the population mean.

45. Dr. Susan Benner is an industrial psychologist. She is currently studying stress among executives of Internet companies. She has developed a questionnaire that she believes measures stress. A score above 80 indicates stress at a dangerous level. A random sam- ple of 15 executives revealed the following stress level scores.

94 78 83 90 78 99 97 90 97 90 93 94 100 75 84

a. Find the mean stress level for this sample. What is the point estimate of the popula- tion mean?

b. Construct a 95 percent confidence level for the population mean. c. Is it reasonable to conclude that Internet executives have a mean stress level in the

dangerous level, according to Dr. Benner’s test? 46. As a condition of employment, Fashion Industries applicants must pass a drug test. Of the

last 220 applicants, 14 failed the test. Develop a 99 percent confidence interval for the pro- portion of applicants that fail the test. Would it be reasonable to conclude that more than 10 percent of the applicants are now failing the test?

47. Fashion Industries randomly tests its employees throughout the year. Last year in the 400 random tests conducted, 14 employees failed the test. Would it be reasonable to conclude that less than 5 percent of the employees are not able to pass the random drug test? Explain.

48. During a national debate on changes to health care, a cable news service performs an opin- ion poll of 500 small-business owners. It shows that 65 percent of small-business owners do not approve of the changes. Develop a 95 percent confidence interval for the proportion opposing health care changes. Comment on the result.

49. There are 20,000 eligible voters in York County, South Carolina. A random sample of 500 York County voters revealed 350 plan to vote to return Louella Miller to the state senate. Construct a 99 percent confidence interval for the proportion of voters in the county who plan to vote for Ms. Miller. From this sample information, can you confirm she will be re-elected?

50. In a poll to estimate presidential popularity, each person in a random sample of 1,000 voters was asked to agree with one of the following statements: 1. The president is doing a good job. 2. The president is doing a poor job. 3. I have no opinion.

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A total of 560 respondents selected the first statement, indicating they thought the pres- ident was doing a good job. a. Construct a 95 percent confidence interval for the proportion of respondents who feel

the president is doing a good job. b. Based on your interval in part (a), is it reasonable to conclude that a majority (more

than half ) of the population believes the president is doing a good job? 51. Police Chief Edward Wilkin of River City reports 500 traffic citations were issued last

month. A sample of 35 of these citations showed the mean amount of the fine was $54, with a standard deviation of $4.50. Construct a 95 percent confidence interval for the mean amount of a citation in River City.

52. The First National Bank of Wilson has 650 checking account customers. A recent sam- ple of 50 of these customers showed 26 have a Visa card with the bank. Construct the 99 percent confidence interval for the proportion of checking account customers who have a Visa card with the bank.

53. It is estimated that 60 percent of U.S. households subscribe to cable TV. You would like to verify this statement for your class in mass communications. If you want your esti- mate to be within 5 percentage points, with a 95 percent level of confidence, how large of a sample is required?

54. You need to estimate the mean number of travel days per year for outside salespeople. The mean of a small pilot study was 150 days, with a standard deviation of 14 days. If you must estimate the population mean within 2 days, how many outside salespeople should you sample? Use the 90 percent confidence level.

55. You are to conduct a sample survey to determine the mean family income in a rural area of central Florida. The question is, how many families should be sampled? In a pilot sam- ple of 10 families, the standard deviation of the sample was $500. The sponsor of the survey wants you to use the 95 percent confidence level. The estimate is to be within $100. How many families should be interviewed?

56. Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,979. The standard deviation of the sample was $1,000. a. Based on this sample information, develop a 90 percent confidence interval for the

population mean yearly premium. b. How large a sample is needed to find the population mean within $250 at 99 percent

confidence? 57. Passenger comfort is influenced by the amount of pressurization in an airline cabin.

Higher pressurization permits a closer-to-normal environment and a more relaxed flight. A study by an airline user group recorded the corresponding air pressure on 30 randomly chosen flights. The study revealed a mean equivalent pressure of 8,000 feet with a stan- dard deviation of 300 feet. a. Develop a 99 percent confidence interval for the population mean equivalent pressure. b. How large a sample is needed to find the population mean within 25 feet at 95 percent

confidence? 58. A random sample of 25 people employed by the Florida state authority established they

earned an average wage (including benefits) of $65.00 per hour. The sample standard deviation was $6.25 per hour. a. What is the population mean? What is the best estimate of the population mean? b. Develop a 99 percent confidence interval for the population mean wage (including

benefits) for these employees. c. How large a sample is needed to assess the population mean with an allowable error

of $1.00 at 95 percent confidence? 59. A film alliance used a random sample of 50 U.S. citizens to estimate that the typical

American spent 78 hours watching videos and DVDs last year. The standard deviation of this sample was 9 hours. a. Develop a 95 percent confidence interval for the population mean number of hours

spent watching videos and DVDs last year. b. How large a sample should be used to be 90 percent confident the sample mean is

within 1.0 hour of the population mean? 60. Dylan Jones kept careful records of the fuel efficiency of his new car. After the first nine

times he filled up the tank, he found the mean was 23.4 miles per gallon (mpg) with a sample standard deviation of 0.9 mpg. a. Compute the 95 percent confidence interval for his mpg. b. How many times should he fill his gas tank to obtain a margin of error below 0.1 mpg?

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61. A survey of 36 randomly selected “iPhone” owners showed that the purchase price has a mean of $416 with a sample standard deviation of $180. a. Compute the standard error of the sample mean. b. Compute the 95 percent confidence interval for the mean. c. How large a sample is needed to estimate the population mean within $10?

62. You plan to conduct a survey to find what proportion of the workforce has two or more jobs. You decide on the 95 percent confidence level and state that the estimated pro- portion must be within 2 percent of the population proportion. A pilot survey reveals that 5 of the 50 sampled hold two or more jobs. How many in the workforce should be inter- viewed to meet your requirements?

63. The proportion of public accountants who have changed companies within the last three years is to be estimated within 3 percent. The 95 percent level of confidence is to be used. A study conducted several years ago revealed that the percent of public accoun- tants changing companies within three years was 21. a. To update this study, the files of how many public accountants should be studied? b. How many public accountants should be contacted if no previous estimates of the

population proportion are available? 64. As part of an annual review of its accounts, a discount brokerage selects a random

sample of 36 customers. Their accounts are reviewed for total account valuation, which showed a mean of $32,000, with a sample standard deviation of $8,200. What is a 90 percent confidence interval for the mean account valuation of the population of customers?

65. The National Weight Control Registry tries to mine secrets of success from people who have lost at least 30 pounds and kept it off for at least a year. It reports that out of 2,700 registrants, 459 were on low-carbohydrate diets (less than 90 grams a day). a. Develop a 95 percent confidence interval for this fraction. b. Is it possible that the population percentage is 18 percent? c. How large a sample is needed to estimate the proportion within 0.5 percent?

66. Near the time of an election, a cable news service performs an opinion poll of 1,000 probable voters. It shows that the Republican contender has an advantage of 52 percent to 48 percent. a. Develop a 95 percent confidence interval for the proportion favoring the Republican

candidate. b. Estimate the probability that the Democratic candidate is actually leading. c. Repeat the above analysis based on a sample of 3,000 probable voters.

67. A sample of 352 subscribers to Wired magazine shows the mean time spent using the Internet is 13.4 hours per week, with a sample standard deviation of 6.8 hours. Find the 95 percent confidence interval for the mean time Wired subscribers spend on the Internet.

68. The Tennessee Tourism Institute (TTI) plans to sample information center visitors enter- ing the state to learn the fraction of visitors who plan to camp in the state. Current esti- mates are that 35 percent of visitors are campers. How large a sample would you take to estimate at a 95 percent confidence level the population proportion with an allowable error of 2 percent?

Data Set Exercises 69. Refer to the Real Estate data, which report information on homes sold in the Goodyear,

Arizona, area during the last year. a. Develop a 95 percent confidence interval for the mean selling price of the homes. b. Develop a 95 percent confidence interval for the mean distance the home is from the

center of the city. c. Develop a 95 percent confidence interval for the proportion of homes with an attached

garage. d. To report your findings, write a business style memo to Gary Loftus, the president of

the Goodyear Chamber of Commerce. 70. Refer to the Baseball 2009 data, which report information on the 30 Major League Base-

ball teams for the 2009 season. a. Develop a 95 percent confidence interval for the mean number of home runs per team. b. Develop a 95 percent confidence interval for the mean number of errors committed

by each team. c. Develop a 95 percent confidence interval for the mean number of stolen bases for

each team.

Estimation and Confidence Intervals 327

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328 Chapter 9

71. Refer to the Buena School District bus data. a. Develop a 95 percent confidence interval for the mean bus maintenance. b. Develop a 95 percent confidence interval for the mean bus miles. c. Write a business memo to the state transportation official to report your results.

Software Commands 1. The Minitab commands for the 60 columns of 30

random numbers used in the example/solution on page 304 are: a. Select Calc, Random Data, and then click on

Normal. b. From the dialog box, click in the Generate box

and type 30 for the number of rows of data, Store in column(s) is C1-C60, Mean is 50, Standard deviation is 5.0, and finally click OK.

2. The Minitab commands for the 60 confidence inter- vals on page 304 follow. a. Select Stat, Basic Statistics, and then click on

1-Sample Z. b. In the dialog box, indicate that the Variables are

C1-C60 and that Standard Duration is 5. Next, click on Options in the lower right corner, and in the next dialog box, indicate that the Confi- dence level is 95, and then click OK. Click OK in the main dialog box.

3. The Minitab commands for the descriptive statistics on page 311 follow. Enter the data in the first col- umn and label this column Amount. On the Toolbar, select Stat, Basic Statistics, and Display Descrip- tive Statistics. In the dialog box, select Amount as the Variable and click OK.

4. The Minitab commands for the confidence interval for the amount spent at the Inlet Square Mall on page 311 are: a. Enter the 20 amounts spent in column C1 and

name the variable Amount, or locate the data on the student data disk. It is named Shopping and is found in the folder for Chapter 9.

b. On the Toolbar, select Stat, Basic Statistics, and click on 1-Sample t.

c. Select Samples in columns: and select Amount and click OK.

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A Review of Chapters 8 and 9 329

5. The Excel commands for the confidence interval for the amounts spent at the Inlet Square Mall on page 312 are: a. Select the Data tab on the top menu. Then, on

the far right, select Data Analysis, and then Descriptive Statistics, and click OK.

b. For the Input Range, type A1:A21, click on Labels in first row, type C1 as the Output Range, click on Summary statistics and Confi- dence Level for Mean, and then click on OK.

Chapter 9 Answers to Self-Review

9–1 a. Unknown. This is the value we wish to estimate. b. $20,000, point estimate.

d. The endpoints of the confidence interval are $18,776 and $21,224. About 99 percent of the intervals similarly constructed would include the population mean.

9–2 a.

b. The population mean is not known. The best estimate is the sample mean, 1.8 days.

The endpoints are 0.99 and 2.61 d. t is used because the population standard

deviation is unknown. e. The value of 0 is not in the interval. It is

unreasonable to conclude that the mean

1.80 � 2.262 1.1353 110

� 1.80 � 0.81

X � 18 10

� 1.8 s � C

11.6 10 � 1

� 1.1353

$20,000 � 2.58 $3,000 140

� $20,000 � $1,224

number of days of work missed is 0 per employee.

9–3 a.

b. .30 � 2.58(.0122) � .30 � .03 c. The interval is between .27 and .33. About 99

percent of the similarly constructed intervals would include the population mean.

9–4 . The sample should

be rounded to 208.

9–5

.375 � 1.96(.0765)(.9184) � .375 � .138

.375 � 1.96 C

.375(1 � .375) 40

250 � 40 250 � 1

�

n � a 2.58(.279)

.05 b

� 207.26

p � 420

1,400 � .30

A Review of Chapters 8 and 9 We began Chapter 8 by describing the reasons sampling is necessary. We sample because it is often impossible to study every item, or individual, in some populations. It would be too expensive and time consuming, for example, to contact and record the annual incomes of all U.S. bank offi- cers. Also, sampling often destroys the product. A drug manufacturer cannot test the properties of each vitamin tablet manufactured, because there would be none left to sell. To estimate a popula- tion parameter, therefore, we select a sample from the population. A sample is a part of the popu- lation. Care must be taken to ensure that every member of our population has a chance of being

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330 Chapter 9

selected; otherwise, the conclusions might be biased. A number of probability-type sampling meth- ods can be used, including simple random, systematic, stratified, and cluster sampling.

Regardless of the sampling method selected, a sample statistic is seldom equal to the corre- sponding population parameter. For example, the mean of a sample is seldom exactly the same as the mean of the population. The difference between this sample statistic and the population para- meter is the sampling error.

In Chapter 8, we demonstrated that, if we select all possible samples of a specified size from a population and calculate the mean of these samples, the result will be exactly equal to the pop- ulation mean. We also showed that the dispersion in the distribution of the sample means is equal to the population standard deviation divided by the square root of the sample size. This result is called the standard error of the mean. There is less dispersion in the distribution of the sample means than in the population. In addition, as we increase the number of observations in each sam- ple, we decrease the variation in the sampling distribution.

The central limit theorem is the foundation of statistical inference. It states that, if the popula- tion from which we select the samples follows the normal probability distribution, the distribution of the sample means will also follow the normal distribution. If the population is not normal, it will approach the normal probability distribution as we increase the size of the sample.

Our focus in Chapter 9 was point estimates and interval estimates. A point estimate is a sin- gle value used to estimate a population parameter. An interval estimate is a range of values within which we expect the population parameter to occur. For example, based on a sample, we estimate that the mean annual income of all professional house painters in Atlanta, Georgia (the population), is $45,300. That estimate is called a point estimate. If we state that the population mean is proba- bly in the interval between $45,200 and $45,400, that estimate is called an interval estimate. The two endpoints ($45,200 and $45,400) are the confidence limits for the population mean. We also described procedures for establishing a confidence interval for a population mean when the population stan- dard deviation is not known and for a population proportion. In this chapter, we also provided a method to determine the necessary sample size based on the dispersion in the population, the level of confidence desired, and the desired precision of the estimate or margin of error.

Glossary

Bias A possible consequence if certain members of the population are denied the chance to be selected for the sample. As a result, the sample may not be representative of the population. Central limit theorem The sampling distribution of the sample mean will approach a normal distribution regard- less of the shape of the population as sample size increases. Cluster sampling A method often used to lower the cost of sampling if the population is dispersed over a wide geo- graphic area. The area is divided into smaller units (coun- ties, precincts, blocks, etc.) called primary units. Then a few primary units are chosen, and a random sample is se- lected from each primary unit. Finite-population correction factor (FPC) When sam- pling without replacement from a finite population, a cor- rection term is used to reduce the standard error of the mean according to the relative size of the sample to the size of the population. The correction factor is used when the sample is more than 5 percent of a finite population. Interval estimate The interval within which a population parameter probably lies, based on sample information. Example: According to sample data, the population mean is in the interval between 1.9 and 2.0 pounds. Point estimate A single value computed from a sample and used to estimate a population parameter. Example: If the sample mean is 1,020, it is the best estimate of the population mean.

Probability sample A sample of items or individuals chosen so that each member of the population has a chance of being included in the sample. Sampling distribution of the sample mean A probabil- ity distribution consisting of all possible means of samples of a given size selected from the population. Sampling error The difference between a sample statistic and the corresponding population parameter. Example: The sample mean income is $22,100; the population mean is $22,000. The sampling error is $22,100 � $22,000 � $100. This error can be attributed to sampling—that is, chance. Simple random sampling A sampling scheme such that each member of the population has the same chance of being selected as part of the sample. Stratified random sampling A population is first divided into subgroups called strata. A sample is then chosen from each stratum. If, for example, the population of interest consisted of all undergraduate students, the sample de- sign might call for sampling 62 freshmen, 51 sophomores, 40 juniors, and 39 seniors. Systematic random sampling Assuming the popula- tion is arranged in some way, such as alphabetically, by height, or in a file drawer, a random starting point is se- lected, then every kth item becomes a member of the sample. If a sample design called for interviewing every ninth household on Main Street starting with 932 Main, the sample would consist of households at 932 Main, 941 Main, 950 Main, and so on.

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A Review of Chapters 8 and 9 331

Problems

1. A recent study indicated that women took an average of 8.6 weeks of unpaid leave from their jobs after the birth of a child. Assume that this distribution follows the normal prob- ability distribution with a standard deviation of 2.0 weeks. We select a sample of 35 women who recently returned to work after the birth of a child. What is the likelihood that the mean of this sample is at least 8.8 weeks?

2. The manager of Tee Shirt Emporium reports that the mean number of shirts sold per week is 1,210, with a standard deviation of 325. The distribution of sales follows the nor- mal distribution. What is the likelihood of selecting a sample of 25 weeks and finding the sample mean to be 1,100 or less?

3. The owner of the Gulf Stream Café wished to estimate the mean number of lunch cus- tomers per day. A sample of 40 days revealed a mean of 160 per day, with a standard deviation of 20 per day. Develop a 98 percent confidence interval for the mean number of customers per day.

4. The manager of the local Hamburger Express wishes to estimate the mean time cus- tomers spend at the drive-through window. A sample of 20 customers experienced a mean waiting time of 2.65 minutes, with a standard deviation of 0.45 minutes. Develop a 90 percent confidence interval for the mean waiting time.

5. The office manager for a large company is studying the usage of its copy machines. A random sample of six copy machines revealed the following number of copies (reported in 000s) made yesterday.

826 931 1,126 918 1,011 1,101

Develop a 95 percent confidence interval for the mean number of copies per machine. 6. John Kleman is the host of KXYZ Radio 55 AM drive-time news in Denver. During his

morning program, John asks listeners to call in and discuss current local and national news. This morning, John was concerned with the number of hours children under 12 years of age watch TV per day. The last 5 callers reported that their children watched the following number of hours of TV last night.

3.0 3.5 4.0 4.5 3.0

Would it be reasonable to develop a confidence interval from these data to show the mean number of hours of TV watched? If yes, construct an appropriate confidence interval and interpret the result. If no, why would a confidence interval not be appropriate?

7. Historically, Widgets Manufacturing Inc. produces 250 widgets per day. Recently the new owner bought a new machine to produce more widgets per day. A sample of 16 days’ production revealed a mean of 240 units with a standard deviation of 35. Construct a confidence interval for the mean number of widgets produced per day. Does it seem rea- sonable to conclude that the mean daily widget production has increased? Justify your conclusion.

8. A manufacturer of cell phone batteries wants to estimate the useful life of its battery (in thousands of hours). The estimate is to be within 0.10 (100) hours. Assume a 95 per- cent level of confidence and that the standard deviation of the useful life of the battery is 0.90 (900 hours). Determine the required sample size.

9. The manager of a home improvement store wishes to estimate the mean amount of money spent in the store. The estimate is to be within $4.00 with a 95 percent level of confidence. The manager does not know the standard deviation of the amounts spent. However, he does estimate that the range is from $5.00 up to $155.00. How large of a sample is needed?

10. In a sample of 200 residents of Georgetown County, 120 reported they believed the county real estate taxes were too high. Develop a 95 percent confidence interval for the proportion of residents who believe the tax rate is too high. Would it be reasonable to conclude that the majority of the taxpayers feel that the taxes are too high?

11. In recent times, the percent of buyers purchasing a new vehicle via the Internet has been large enough that local automobile dealers are concerned about its impact on their business. The information needed is an estimate of the proportion of purchases via the

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332 Chapter 9

Internet. How large of a sample of purchasers is necessary for the estimate to be within 2 percentage points with a 98 percent level of confidence? Current thinking is that about 8 percent of the vehicles are purchased via the Internet.

12. Historically, the proportion of adults over the age of 24 who smoke has been .30. In recent years, much information has been published and aired on radio and TV that smoking is not good for one’s health. A sample of 500 adults revealed only 25 per- cent of those sampled smoked. Develop a 98 percent confidence interval for the pro- portion of adults who currently smoke. Would you agree that the proportion is less than 30 percent?

13. The auditor of the State of Ohio needs an estimate of the proportion of residents who regularly play the state lottery. Historically, about 40 percent regularly play, but the audi- tor would like some current information. How large a sample is necessary for the esti- mate to be within 3 percentage points, with a 98 percent level of confidence?

Century National Bank Refer to the description of Century National Bank at the end of the Review of Chapters 1–4 on page 141. When Mr. Selig took over as president of Century several years ago, the use of debit cards was just beginning. He would like an update

on the use of these cards. Develop a 95 percent confidence interval for the proportion of customers using these cards. On the basis of the confidence interval, is it reasonable to conclude that more than half of the customers use a debit card? Write a brief report interpreting the results.

Case

Practice Test

Part 1—Objective 1. If each item in the population has the same chance of being selected, this is called a .

1. 2. The difference between the population mean and the sample mean is called the . 2. 3. The is the standard deviation of the distribution of sample means. 3. 4. If the sample size is increased, the variance of the sample means will . (become smaller, become larger, not

change) 4. 5. A single value used to estimate a population parameter is called a . 5. 6. A range of values within which the population parameter is expected to occur is called a .

6. 7. Which of the following do not affect the width of a confidence interval? (sample size, variation in the population, level

of confidence, size of population) 7. 8. The fraction of a population that has a particular characteristic is called a . 8. 9. Which of the following is not a characteristic of the t distribution? (positively skewed, continuous, mean of zero,

based on degrees of freedom) 9. 10. To determine the required sample size of a proportion when no estimate of the population proportion is available,

what value is used? 10.

Part 2—Problems 1. Americans spend an average (mean) of 12.2 minutes (per day) in the shower. The distribution of times follows

the normal distribution with a population standard deviation of 2.3 minutes. What is the likelihood that the mean time per day for a sample of 12 Americans was 11 minutes or less?

2. A recent study of 26 Conway, SC, residents revealed they had lived at their current address an average of 9.3 years. The standard deviation of the sample was 2 years. a. What is the population mean? b. What is the best estimate of the population mean? c. What is the standard error of estimate? d. Develop a 90 percent confidence interval for the population mean.

3. A recent federal report indicated that 27 percent of children ages 2 to 5 ate a vegetable at least 5 times a week. How large a sample is needed to estimate the true population proportion within 2 percent with a 98 percent level of confi- dence? Be sure to use the information contained in the federal report.

4. The Philadelphia Area Transit Authority wishes to estimate the proportion of central city workers that use public trans- portation to get to work. A sample of 100 workers revealed that 64 used public transportation. Develop a 95 percent confidence interval for the population proportion.

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10 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Define a hypothesis.

LO2 Explain the five-step hypothesis-testing procedure.

LO3 Describe Type I and Type II errors.

LO4 Define the term test statistic and explain how it is used.

LO5 Distinguish between a one-tailed and a two-tailed test of hypothesis.

LO6 Conduct a test of hypothesis about a population mean.

LO7 Compute and interpret a p-value.

LO8 Conduct a test of hypothesis about a population proportion.

LO9 Compute the probability of a Type II error.

One-Sample Tests of Hypothesis

Dole Pineapple Inc. is concerned that the 16-ounce can of sliced

pineapple is being overfilled. Assume the standard deviation of the

process is .03 ounces. The quality control department took a random

sample of 50 cans and found that the arithmetic mean weight was

16.05 ounces. At the 5 percent level of significance, can we conclude

that the mean weight is greater than 16 ounces? Determine the

p-value. (See Exercise 32 and LO6.)

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10.1 Introduction Chapter 8 began our study of statistical inference. We described how we could select a random sample and from this sample estimate the value of a population parameter. For example, we selected a sample of five employees at Spence Sprock- ets, found the number of years of service for each sampled employee, computed the mean years of service, and used the sample mean to estimate the mean years of service for all employees. In other words, we estimated a population parameter from a sample statistic.

Chapter 9 continued the study of statistical inference by developing a confi- dence interval. A confidence interval is a range of values within which we expect the population parameter to occur. In this chapter, rather than develop a range of values within which we expect the population parameter to occur, we develop a procedure to test the validity of a statement about a population parameter. Some examples of statements we might want to test are:

• The mean speed of automobiles passing milepost 150 on the West Virginia Turnpike is 68 miles per hour.

• The mean number of miles driven by those leasing a Chevy TrailBlazer for three years is 32,000 miles.

• The mean time an American family lives in a par- ticular single-family dwelling is 11.8 years.

• The 2010 mean starting salary for a graduate of a four-year college is $47,673.

• Thirty-five percent of retirees in the upper Mid- west sell their home and move to a warm climate within 1 year of their retirement.

• Eighty percent of those who regularly play the state lotteries never win more than $100 in any one play.

This chapter and several of the following chapters are concerned with statistical hypothesis testing. We begin by defining what we mean by a statistical hypothesis and statistical hypothesis testing. Next, we outline the steps in statistical hypothe- sis testing. Then we conduct tests of hypothesis for means and proportions. In the last section of the chapter, we describe possible errors due to sampling in hypothesis testing.

10.2 What Is a Hypothesis? A hypothesis is a statement about a population. Data are then used to check the reasonableness of the statement. To begin, we need to define the word hypothesis. In the United States legal system, a person is innocent until proven guilty. A jury hypothesizes that a person charged with a crime is innocent and subjects this hypothesis to verification by reviewing the evidence and hearing testimony before reaching a verdict. In a similar sense, a patient goes to a physician and reports var- ious symptoms. On the basis of the symptoms, the physician will order certain diag- nostic tests, then, according to the symptoms and the test results, determine the treatment to be followed.

In statistical analysis, we make a claim—that is, state a hypothesis—collect data, and then use the data to test the assertion. We define a statistical hypothesis as follows.

A hypothesis is a statement about a population parameter.

HYPOTHESIS A statement about a population parameter subject to verification.LO1 Define a hypothesis.

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One-Sample Tests of Hypothesis 335

In most cases, the population is so large that it is not feasible to study all the items, objects, or persons in the population. For example, it would not be possible to contact every systems analyst in the United States to find his or her monthly income. Likewise, the quality assurance department at Cooper Tire cannot check each tire produced to determine whether it will last more than 60,000 miles.

As noted in Chapter 8, an alternative to measuring or interviewing the entire population is to take a sample from the population. We can, therefore, test a state- ment to determine whether the sample does or does not support the statement con- cerning the population.

10.3 What Is Hypothesis Testing? The terms hypothesis testing and testing a hypothesis are used interchangeably. Hypothesis testing starts with a statement, or assumption, about a population parameter—such as the population mean. This statement is referred to as a hypothesis. A hypothesis might be that the mean monthly commission of sales associates in retail electronics stores, such as Circuit City, is $2,000. We cannot contact all these sales associates to ascertain that the mean is in fact $2,000. The cost of locating and interviewing every electronics sales associate in the United States would be exorbitant. To test the validity of the assumption (� � $2,000), we must select a sample from the population of all electronics sales associates, calculate sample statistics, and based on certain decision rules accept or reject the hypothesis. A sample mean of $1,000 for the electronics sales associates would certainly cause rejection of the hypothesis. However, suppose the sample mean is $1,995. Is that close enough to $2,000 for us to accept the assumption that the population mean is $2,000? Can we attribute the difference of $5 between the two means to sampling error, or is that difference statistically significant?

HYPOTHESIS TESTING A procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement.

10.4 Five-Step Procedure for Testing a Hypothesis There is a five-step procedure that systematizes hypothesis testing; when we get to step 5, we are ready to reject or not reject the hypothesis. However, hypothe- sis testing as used by statisticians does not provide proof that something is true, in the manner in which a mathematician “proves” a statement. It does provide a kind of “proof beyond a reasonable doubt,” in the manner of the court system. Hence, there are specific rules of evidence, or procedures, that are followed. The steps are shown in the following diagram. We will discuss in detail each of the steps.

Step 1 Step 2 Step 3 Step 4 Step 5 Do not reject H0

or reject H0

and accept H1

Take a sample, arrive at decision

Formulate a decision

rule

Select a level of

significance

Identify the test

statistic

State null and alternate

hypotheses

Statistics in Action

LASIK is a 15-minute surgical procedure that uses a laser to reshape an eye’s cornea with the goal of improving eye- sight. Research shows that about 5 percent of all surgeries involve complications such as glare, corneal haze, over-correction or under-correction of vision, and loss of vision. In a statistical sense, the research tests a null hypothesis that the surgery will not improve eyesight with the alternative hypothe- sis that the surgery will improve eyesight. The sample data of LASIK surgery shows that 5 per- cent of all cases result in complications. The 5 percent represents a Type I error rate. When a person decides to have the surgery, he or she ex- pects to reject the null hypothesis. In 5 percent of future cases, this ex- pectation will not be met. (Source: American Academy of Ophthal- mology Journal, San Francisco, Vol. 16, no. 43.)

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Step 1: State the Null Hypothesis (H0) and the Alternate Hypothesis (H1) The first step is to state the hypothesis being tested. It is called the null hypothesis, designated H0, and read “H sub zero.” The capital letter H stands for hypothesis, and the subscript zero implies “no difference.” There is usually a “not” or a “no” term in the null hypothesis, meaning that there is “no change.” For example, the null hypothesis is that the mean number of miles driven on the steel-belted tire is not dif- ferent from 60,000. The null hypothesis would be written H0: � � 60,000. Generally speaking, the null hypothesis is developed for the purpose of testing. We either reject or fail to reject the null hypothesis. The null hypothesis is a statement that is not rejected unless our sample data provide convincing evidence that it is false.

We should emphasize that, if the null hypothesis is not rejected on the basis of the sample data, we cannot say that the null hypothesis is true. To put it another way, failing to reject the null hypothesis does not prove that H0 is true, it means we have failed to disprove H0. To prove without any doubt the null hypothesis is true, the population parameter would have to be known. To actually determine it, we would have to test, survey, or count every item in the population. This is usually not feasible. The alternative is to take a sample from the population.

It should also be noted that we often begin the null hypothesis by stating, “There is no significant difference between . . . ,” or “The mean impact strength of the glass is not significantly different from. . . .” When we select a sample from a population, the sample statistic is usually numerically different from the hypothesized population parameter. As an illustration, suppose the hypothesized impact strength of a glass plate is 70 psi, and the mean impact strength of a sample of 12 glass plates is 69.5 psi. We must make a decision about the difference of 0.5 psi. Is it a true difference, that is, a significant difference, or is the difference between the sample statistic (69.5) and the hypothesized population parameter (70.0) due to chance (sampling)? To answer this question, we conduct a test of significance, commonly referred to as a test of hypothesis. To define what is meant by a null hypothesis:

State the null hypothesis and the alternative hypothesis.

NULL HYPOTHESIS A statement about the value of a population parameter developed for the purpose of testing numerical evidence.

ALTERNATE HYPOTHESIS A statement that is accepted if the sample data provide sufficient evidence that the null hypothesis is false.

The alternate hypothesis describes what you will conclude if you reject the null hypothesis. It is written H1 and is read “H sub one.” It is also referred to as the research hypothesis. The alternate hypothesis is accepted if the sample data provide us with enough statistical evidence that the null hypothesis is false.

The following example will help clarify what is meant by the null hypothesis and the alternate hypothesis. A recent article indicated the mean age of U.S. commer- cial aircraft is 15 years. To conduct a statistical test regarding this statement, the first step is to determine the null and the alternate hypotheses. The null hypothesis represents the current or reported condition. It is written H0: � � 15. The alternate hypothesis is that the statement is not true, that is, H1: � � 15. It is important to remember that no matter how the problem is stated, the null hypothesis will always contain the equal sign. The equal sign (�) will never appear in the alternate hypoth- esis. Why? Because the null hypothesis is the statement being tested, and we need a specific value to include in our calculations. We turn to the alternate hypothesis only if the data suggests the null hypothesis is untrue.

LO2 Explain the five- step hypothesis-testing procedure.

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Step 2: Select a Level of Significance After setting up the null hypothesis and alternate hypothesis, the next step is to state the level of significance.

Select a level of significance or risk.

LEVEL OF SIGNIFICANCE The probability of rejecting the null hypothesis when it is true.

The level of significance is designated �, the Greek letter alpha. It is also some- times called the level of risk. This may be a more appropriate term because it is the risk you take of rejecting the null hypothesis when it is really true.

There is no one level of significance that is applied to all tests. A decision is made to use the .05 level (often stated as the 5 percent level), the .01 level, the .10 level, or any other level between 0 and 1. Traditionally, the .05 level is selected for consumer research projects, .01 for quality assurance, and .10 for political polling. You, the researcher, must decide on the level of significance before formulating a decision rule and collecting sample data.

To illustrate how it is possible to reject a true hypothesis, suppose a firm manu- facturing personal computers uses a large number of printed circuit boards. Suppliers

bid on the boards, and the one with the lowest bid is awarded a sizable contract. Suppose the contract specifies that the com- puter manufacturer’s quality-assurance department will sample all incoming shipments of circuit boards. If more than 6 percent of the boards sampled are substandard, the shipment will be rejected. The null hypothesis is that the incoming shipment of boards contains 6 percent or less substandard boards. The alter- nate hypothesis is that more than 6 percent of the boards are defective.

A sample of 50 circuit boards received July 21 from Allied Elec- tronics revealed that 4 boards, or 8 percent, were substandard. The shipment was rejected because it exceeded the maximum of 6 per- cent substandard printed circuit boards. If the shipment was actu-

ally substandard, then the decision to return the boards to the supplier was correct. However, suppose the 4 substandard printed circuit boards selected in the sample of 50 were the only substandard boards in the shipment of 4,000 boards. Then only 1�10 of 1 percent were defective (4�4,000 � .001). In that case, less than 6 percent of the entire shipment was substandard and rejecting the shipment was an error. In terms of hypothesis testing, we rejected the null hypothesis that the shipment was not substandard when we should have accepted the null hypothesis. By rejecting a true null hypothesis, we committed a Type I error. The probability of committing a Type I error is �.

TYPE I ERROR Rejecting the null hypothesis, H0, when it is true.

The probability of committing another type of error, called a Type II error, is des- ignated by the Greek letter beta (�).

TYPE II ERROR Accepting the null hypothesis when it is false.

The firm manufacturing personal computers would commit a Type II error if, unknown to the manufacturer, an incoming shipment of printed circuit boards from Allied Electronics contained 15 percent substandard boards, yet the shipment

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338 Chapter 10

Researcher

Null Does Not Reject Rejects Hypothesis H0 H0

H0 is true Correct Type I decision error

H0 is false Type II Correct error decision

was accepted. How could this happen? Suppose 2 of the 50 boards in the sample (4 percent) tested were substandard, and 48 of the 50 were good boards. Accord- ing to the stated procedure, because the sample contained less than 6 percent sub- standard boards, the shipment was accepted. It could be that by chance the 48 good boards selected in the sample were the only acceptable ones in the entire shipment consisting of thousands of boards!

In retrospect, the researcher cannot study every item or individual in the pop- ulation. Thus, there is a possibility of two types of error—a Type I error, wherein the null hypothesis is rejected when it should have been accepted, and a Type II error, wherein the null hypothesis is not rejected when it should have been rejected.

We often refer to the probability of these two possible errors as alpha, �, and beta, �. Alpha (�) is the probability of making a Type I error, and beta (�) is the prob- ability of making a Type II error.

The following table summarizes the decisions the researcher could make and the possible consequences.

Step 3: Select the Test Statistic There are many test statistics. In this chapter, we use both z and t as the test sta- tistic. In later chapters, we will use such test statistics as F and �2, called chi-square.

TEST STATISTIC A value, determined from sample information, used to determine whether to reject the null hypothesis.

In hypothesis testing for the mean (�) when � is known, the test statistic z is com- puted by:

TESTING A MEAN, � KNOWN [10–1]z � X � � 1n

The z value is based on the sampling distribution of which follows the normal distribution with a mean equal to �, and a standard deviation , which is equal to . We can thus determine whether the difference between and � is statistically significant by finding the number of standard deviations is from �, using formula (10–1).

Step 4: Formulate the Decision Rule A decision rule is a statement of the specific conditions under which the null hypoth- esis is rejected and the conditions under which it is not rejected. The region or area of rejection defines the location of all those values that are so large or so small that the probability of their occurrence under a true null hypothesis is rather remote.

Chart 10–1 portrays the rejection region for a test of significance that will be conducted later in the chapter.

X X��1n

�x(�x ) X,

The decision rule states the conditions when H0 is rejected.

LO3 Define Type I and Type II errors.

LO4 Define the term test statistic and explain how it is used.

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One-Sample Tests of Hypothesis 339

Scale of z0 1.65 Critical value

Region of rejection

Do not reject Ho

Probability = .95 Probability = .05

CHART 10–1 Sampling Distribution of the Statistic z, a Right-Tailed Test, .05 Level of Significance

Note in the chart that:

• The area where the null hypothesis is not rejected is to the left of 1.65. We will explain how to get the 1.65 value shortly.

• The area of rejection is to the right of 1.65. • A one-tailed test is being applied. (This will also be explained later.) • The .05 level of significance was chosen. • The sampling distribution of the statistic z follows the normal probability

distribution. • The value 1.65 separates the regions where the null hypothesis is rejected and

where it is not rejected. • The value 1.65 is the critical value.

CRITICAL VALUE The dividing point between the region where the null hypothesis is rejected and the region where it is not rejected.

Step 5: Make a Decision The fifth and final step in hypothesis testing is computing the test statistic, com- paring it to the critical value, and making a decision to reject or not to reject the null hypothesis. Referring to Chart 10–1, if, based on sample information, z is computed to be 2.34, the null hypothesis is rejected at the .05 level of significance. The decision to reject H0 was made because 2.34 lies in the region of rejection, that is, beyond 1.65. We would reject the null hypothesis, reasoning that it is highly improb- able that a computed z value this large is due to sampling error (chance).

Had the computed value been 1.65 or less, say 0.71, the null hypothesis would not be rejected. It would be reasoned that such a small computed value could be attributed to chance, that is, sampling error.

As noted, only one of two decisions is possible in hypothesis testing—either accept or reject the null hypothesis. Instead of “accepting” the null hypothesis, H0, some researchers prefer to phrase the decision as: “Do not reject H0,” “We fail to reject H0,” or “The sample results do not allow us to reject H0.”

It should be reemphasized that there is always a possibility that the null hypothe- sis is rejected when it should not be rejected (a Type I error). Also, there is a definable chance that the null hypothesis is accepted when it should be rejected (a Type II error).

Statistics in Action

During World War II, allied military plan- ners needed estimates of the number of German tanks. The information provided by traditional spying methods was not reli- able, but statistical methods proved to be valuable. For exam- ple, espionage and reconnaissance led analysts to estimate that 1,550 tanks were produced during June 1941. However, using the serial numbers of captured tanks and statistical analysis, military planners estimated 244. The actual num- ber produced, as determined from German production records, was 271. The estimate using statisti- cal analysis turned out to be much more accurate. A similar type of analysis was used to estimate the number of Iraqi tanks destroyed during Desert Storm.

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SUMMARY OF THE STEPS IN HYPOTHESIS TESTING 1. Establish the null hypothesis (H0 ) and the alternate hypothesis (H1). 2. Select the level of significance, that is, �. 3. Select an appropriate test statistic. 4. Formulate a decision rule based on steps 1, 2, and 3 above. 5. Make a decision regarding the null hypothesis based on the sample infor-

mation. Interpret the results of the test.

10.5 One-Tailed and Two-Tailed Tests of Significance Refer to Chart 10–1. It depicts a one-tailed test. The region of rejection is only in the right (upper) tail of the curve. To illustrate, suppose that the packaging depart- ment at General Foods Corporation is concerned that some boxes of Grape Nuts are significantly overweight. The cereal is packaged in 453-gram boxes, so the null hypothesis is H0: � � 453. This is read, “the population mean (�) is equal to or less than 453.” The alternate hypothesis is, therefore, H1: � � 453. This is read, “� is greater than 453.” Note that the inequality sign in the alternate hypothesis (�) points to the region of rejection in the upper tail. (See Chart 10–1.) Also note that the null hypothesis includes the equal sign. That is, H0: � � 453. The equality condition always appears in H0, never in H1.

Chart 10–2 portrays a situation where the rejection region is in the left (lower) tail of the standard normal distribution. As an illustration, consider the problem of auto- mobile manufacturers, large automobile leasing companies, and other organizations that purchase large quantities of tires. They want the tires to average, say, 60,000 miles of wear under normal usage. They will, therefore, reject a shipment of tires if tests reveal that the mean life of the tires is significantly below 60,000 miles. They gladly accept a shipment if the mean life is greater than 60,000 miles! They are not concerned with this possibility, however. They are concerned only if they have sample evidence to con- clude that the tires will average less than 60,000 miles of useful life. Thus, the test is set up to satisfy the concern of the automobile manufacturers that the mean life of

Scale of z0–1.65 Critical value

Do not reject H0

Region of rejection

CHART 10–2 Sampling Distribution for the Statistic z, Left-Tailed Test, .05 Level of Significance

Before actually conducting a test of hypothesis, we will differentiate between a one-tailed test of significance and a two-tailed test.

LO5 Distinguish between a one-tailed and a two-tailed test of hypothesis.

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the tires is no less than 60,000 miles. This statement appears in the alternate hypoth- esis. The null and alternate hypotheses in this case are written H0: � 60,000 and H1: � � 60,000.

One way to determine the location of the rejection region is to look at the direction in which the inequality sign in the alternate hypothesis is pointing (either � or �). In this problem, it is pointing to the left, and the rejection region is therefore in the left tail.

In summary, a test is one-tailed when the alternate hypothesis, H1, states a direction, such as:

H0: The mean income of women stockbrokers is less than or equal to $65,000 per year.

H1: The mean income of women stockbrokers is greater than $65,000 per year.

If no direction is specified in the alternate hypothesis, we use a two-tailed test. Changing the previous problem to illustrate, we can say:

H0: The mean income of women stockbrokers is $65,000 per year. H1: The mean income of women stockbrokers is not equal to $65,000 per year.

If the null hypothesis is rejected and H1 accepted in the two-tailed case, the mean income could be significantly greater than $65,000 per year or it could be significantly less than $65,000 per year. To accommodate these two possibilities, the 5 percent area of rejection is divided equally into the two tails of the sampling distribution (2.5 percent each). Chart 10–3 shows the two areas and the critical values. Note that the total area in the normal distribution is 1.0000, found by .9500 � .0250 � .0250.

If H1 states a direction, test is one-tailed.

Scale of z0–1.96

Do not reject H0

Critical value

Region of rejection

.025

1.96 Critical value

Region of rejection

.025

.95

CHART 10–3 Regions of Nonrejection and Rejection for a Two-Tailed Test, .05 Level of Significance

10.6 Testing for a Population Mean: Known Population Standard Deviation A Two-Tailed Test An example will show the details of the five-step hypothesis testing procedure. We also wish to use a two-tailed test. That is, we are not concerned whether the sam- ple results are larger or smaller than the proposed population mean. Rather, we are interested in whether it is different from the proposed value for the population mean. We begin, as we did in the previous chapter, with a situation in which we have his- torical information about the population and in fact know its standard deviation.

Test is one-tailed if H1 states � � or � �.

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Example Jamestown Steel Company manufactures and assembles desks and other office equip- ment at several plants in western New York state. The weekly production of the Model A325 desk at the Fredonia Plant follows a normal probability distribution with a mean of 200 and a standard deviation of 16. Recently, because of market expansion, new produc- tion methods have been introduced and new employees hired. The vice president of man- ufacturing would like to investigate whether

there has been a change in the weekly production of the Model A325 desk. Is the mean number of desks produced at the Fredonia Plant different from 200 at the .01 significance level?

Solution

Sample mean Population mean

Sample sizeStandard deviation of population

z = X _

– μ σ n√

Step 4: Formulate the decision rule. The decision rule is formulated by finding the critical values of z from Appendix B.1. Since this is a two-tailed test, half of .01, or .005, is placed in each tail. The area where H0 is not rejected, located between the two tails, is therefore .99. Appendix B.1 is based on half of the area under the curve, or .5000. Then, .5000 .0050 is .4950, so .4950 is the area between 0 and the critical value. Locate

Formula for the test statistic

[10–1]

In this example, we know two important pieces of information: (1) the population of weekly production follows the normal distribution, and (2) the standard deviation of this normal distribution is 16 desks per week. So it is appropriate to use the z statistic for this problem. We use the statistical hypothesis testing procedure to investigate whether the production rate has changed from 200 per week.

Step 1: State the null hypothesis and the alternate hypothesis. The null hypothesis is “The population mean is 200.” The alternate hypothesis is “The mean is different from 200” or “The mean is not 200.” These two hypotheses are written:

This is a two-tailed test because the alternate hypothesis does not state a direction. In other words, it does not state whether the mean produc- tion is greater than 200 or less than 200. The vice president wants only to find out whether the production rate is different from 200.

Step 2: Select the level of significance. As we indicated in the Problem, the significance level is .01. This is �, the probability of committing a Type I error, and it is the probability of rejecting a true null hypothesis.

Step 3: Select the test statistic. The test statistic is z when the population standard deviation is known. Transforming the production data to stan- dard units (z values) permits their use not only in this problem but also in other hypothesis-testing problems. Formula (10–1) for z is repeated below with the various letters identified.

H1: � � 200

H0: � � 200

LO6 Conduct a test of hypothesis about a population mean.

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Scale of z0

.5000.5000

2.58

Region of rejection

–2.58

α _ 2

.4950

.01 ___ 2

= = .005 α _ 2

.01 ___ 2

= = .005

.4950

Critical value

H0 not rejected Region of rejection

Critical value

H0: μ = 200 H1: μ ≠ 200

CHART 10–4 Decision Rule for the .01 Significance Level

The decision rule is, therefore: Reject the null hypothesis and accept the alternate hypothesis (which states that the population mean is not 200) if the computed value of z is not between 2.58 and �2.58. Do not reject the null hypothesis if z falls between 2.58 and �2.58.

Step 5: Make a decision and interpret the result. Take a sample from the pop- ulation (weekly production), compute z, apply the decision rule, and arrive at a decision to reject H0 or not to reject H0. The mean number of desks produced last year (50 weeks, because the plant was shut down 2 weeks for vacation) is 203.5. The standard deviation of the population is 16 desks per week. Computing the z value from formula (10–1):

Because 1.55 does not fall in the rejection region, H0 is not rejected. We conclude that the population mean is not different from 200. So we would report to the vice president of manufacturing that the sample evidence does not show that the production rate at the Fredonia Plant has changed from 200 per week. The difference of 3.5 units between the historical weekly pro- duction rate and that last year can reasonably be attributed to sampling error. This information is summarized in the following chart.

z � X � ��1n

� 203.5 200

16�150 � 1.55

z scale0 2.581.55–2.58

Reject H0 Reject H0

Do not reject H0

Computed value of z

.4950 in the body of the table. The value nearest to .4950 is .4951. Then read the critical value in the row and column corresponding to .4951. It is 2.58. For your convenience, Appendix B.1, Areas under the Normal Curve, is repeated in the inside back cover.

All the facets of this problem are shown in the diagram in Chart 10–4.

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Did we prove that the assembly rate is still 200 per week? Not really. What we did, technically, was fail to disprove the null hypothesis. Failing to disprove the hypothesis that the population mean is 200 is not the same thing as proving it to be true. As we suggested in the chapter introduction, the conclusion is analogous to the American judicial system. To explain, suppose a person is accused of a crime but is acquitted by a jury. If a person is acquitted of a crime, the conclusion is that there was not enough evidence to prove the person guilty. The trial did not prove that the individual was innocent, only that there was not enough evidence to prove the defendant guilty. That is what we do in statistical hypothesis testing when we do not reject the null hypothesis. The correct interpretation is that we have failed to disprove the null hypothesis.

We selected the significance level, .01 in this case, before setting up the deci- sion rule and sampling the population. This is the appropriate strategy. The signifi- cance level should be set by the investigator, but it should be determined before gathering the sample evidence and not changed based on the sample evidence.

How does the hypothesis testing procedure just described compare with that of confidence intervals discussed in the previous chapter? When we conducted the test of hypothesis regarding the production of desks, we changed the units from desks per week to a z value. Then we compared the computed value of the test statistic (1.55) to that of the critical values ( 2.58 and 2.58). Because the computed value was in the region where the null hypothesis was not rejected, we concluded that the population mean could be 200. To use the confidence inter- val approach, on the other hand, we would develop a confidence interval, based on formula (9–1). See page 302. The interval would be from 197.66 to 209.34, found by Note that the proposed population value, 200, is within this interval. Hence, we would conclude that the population mean could reasonably be 200.

In general, H0 is rejected if the confidence interval does not include the hypoth- esized value. If the confidence interval includes the hypothesized value, then H0 is not rejected. So the “do not reject region” for a test of hypothesis is equivalent to the proposed population value occurring in the confidence interval. The primary dif- ference between a confidence interval and the “do not reject” region for a hypothesis test is whether the interval is centered around the sample statistic, such as , as in the confidence interval, or around 0, as in the test of hypothesis.

203.5 � 2.58(16�150).

Comparing confidence intervals and hypothesis testing.

Heinz, a manufacturer of ketchup, uses a particular machine to dispense 16 ounces of its ketchup into containers. From many years of experience with the particular dispensing machine, Heinz knows the amount of product in each container follows a normal distribution with a mean of 16 ounces and a standard deviation of 0.15 ounce. A sample of 50 containers filled last hour revealed the mean amount per container was 16.017 ounces. Does this evidence suggest that the mean amount dispensed is different

from 16 ounces? Use the .05 significance level. (a) State the null hypothesis and the alternate hypothesis. (b) What is the probability of a Type I error? (c) Give the formula for the test statistic. (d) State the decision rule. (e) Determine the value of the test statistic. (f ) What is your decision regarding the null hypothesis? (g) Interpret, in a single sentence, the result of the statistical test.

Self-Review 10–1

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One-Sample Tests of Hypothesis 345

A One-Tailed Test In the previous example, we emphasized that we were concerned only with report- ing to the vice president whether there had been a change in the mean number of desks assembled at the Fredonia Plant. We were not concerned with whether the change was an increase or a decrease in the production.

To illustrate a one-tailed test, let’s change the problem. Suppose the vice president wants to know whether there has been an increase in the number of units assembled. Can we conclude, because of the improved production meth- ods, that the mean number of desks assembled in the last 50 weeks was more than 200? Look at the difference in the way the problem is formulated. In the first case, we wanted to know whether there was a difference in the mean number assembled, but now we want to know whether there has been an increase. Because we are investigating different questions, we will set our hypotheses dif- ferently. The biggest difference occurs in the alternate hypothesis. Before, we stated the alternate hypothesis as “different from”; now we want to state it as “greater than.” In symbols:

The critical values for a one-tailed test are different from a two-tailed test at the same significance level. In the previous example, we split the significance level in half and put half in the lower tail and half in the upper tail. In a one-tailed test, we put all the rejection region in one tail. See Chart 10–5.

For the one-tailed test, the critical value is 2.33, found by: (1) subtracting .01 from .5000 and (2) finding the z value corresponding to .4900.

H1: � 7 200H1: � � 200

H0: � � 200H0: � � 200

A one-tailed test:A two-tailed test:

Scale of z0

.005 Region of rejection

–2.58 Critical value

H0 is not rejected

H0: μ = 200 H1: μ ≠ 200

.99

.005 Region of rejection

2.58 Critical value

H0 is not rejected

H0: μ ≤ 200 H1: μ > 200

.99

.01 Region of rejection

2.33 Critical value

One-tailed testTwo-tailed test

CHART 10–5 Rejection Regions for Two-Tailed and One-Tailed Tests, � � .01

10.7 p-Value in Hypothesis Testing In testing a hypothesis, we compare the test statistic to a critical value. A decision is made to either reject the null hypothesis or not to reject it. So, for example, if the critical value is 1.96 and the computed value of the test statistic is 2.19, the deci- sion is to reject the null hypothesis.

In recent years, spurred by the availability of computer software, additional infor- mation is often reported on the strength of the rejection or acceptance. That is, how confident are we in rejecting the null hypothesis? This approach reports the

LO5 Distinguish between a one-tailed and a two-tailed test of hypothesis.

LO7 Compute and interpret a p-value.

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346 Chapter 10

p-VALUE The probability of observing a sample value as extreme as, or more extreme than, the value observed, given that the null hypothesis is true.

probability (assuming that the null hypothesis is true) of getting a value of the test statistic at least as extreme as the value actually obtained. This process compares the probability, called the p-value, with the significance level. If the p-value is smaller than the significance level, H0 is rejected. If it is larger than the significance level, H0 is not rejected.

Determining the p-value not only results in a decision regarding H0, but it gives us additional insight into the strength of the decision. A very small p-value, such as .0001, indicates that there is little likelihood the H0 is true. On the other hand, a p-value of .2033 means that H0 is not rejected, and there is little likelihood that it is false.

How do we compute the p-value? To illustrate, we will use the example in which we tested the null hypothesis that the mean number of desks produced per week at Fredonia was 200. We did not reject the null hypothesis, because the z value of 1.55 fell in the region between 2.58 and 2.58. We agreed not to reject the null hypothesis if the computed value of z fell in this region. The probability of finding a z value of 1.55 or more is .0606, found by .5000 .4394. To put it another way, the probability of obtaining an greater than 203.5 if � � 200 is .0606. To com- pute the p-value, we need to be concerned with the region less than 1.55 as well as the values greater than 1.55 (because the rejection region is in both tails). The two-tailed p-value is .1212, found by 2(.0606). The p-value of .1212 is greater than the significance level of .01 decided upon initially, so H0 is not rejected. The details are shown in the following graph. In general, the area is doubled in a two-sided test. Then the p-value can easily be compared with the significance level. The same deci- sion rule is used as in the one-sided test.

A p-value is a way to express the likelihood that H0 is false. But how do we interpret a p-value? We have already said that if the p-value is less than the signif- icance level, then we reject H0; if it is greater than the significance level, then we do not reject H0. Also, if the p-value is very large, then it is likely that H0 is true. If the p-value is small, then it is likely that H0 is not true. The following box will help to interpret p-values.

Scale of z0

Rejection region Rejection region

p-value +

–2.58 –1.55 2.581.55

α _ 2

.01 ___ 2

= = .005α _ 2

.01 ___ 2

= = .005

.0606.0606

INTERPRETING THE WEIGHT OF EVIDENCE AGAINST H0 If the p-value is less than (a) .10, we have some evidence that H0 is not true. (b) .05, we have strong evidence that H0 is not true. (c) .01, we have very strong evidence that H0 is not true. (d) .001, we have extremely strong evidence that H0 is not true.

Statistics in Action

There is a difference between statistically significant and prac- tically significant. To explain, suppose we develop a new diet pill and test it on 100,000 people. We conclude that the typical person taking the pill for two years lost one pound. Do you think many peo- ple would be inter- ested in taking the pill to lose one pound? The results of using the new pill were statistically sig- nificant but not prac- tically significant.

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One-Sample Tests of Hypothesis 347

Self-Review 10–2 Refer to Self-Review 10–1. (a) Suppose the next to the last sentence is changed to read: Does this evidence suggest

that the mean amount dispensed is more than 16 ounces? State the null hypothesis and the alternate hypothesis under these conditions.

(b) What is the decision rule under the new conditions stated in part (a)? (c) A second sample of 50 filled containers revealed the mean to be 16.040 ounces. What

is the value of the test statistic for this sample? (d) What is your decision regarding the null hypothesis? (e) Interpret, in a single sentence, the result of the statistical test. (f ) What is the p-value? What is your decision regarding the null hypothesis based on the

p-value? Is this the same conclusion reached in part (d)?

Exercises For Exercises 1–4, answer the questions: (a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding H0? (e) What is the p-value? Interpret it.

1. A sample of 36 observations is selected from a normal population. The sample mean is 49, and the population standard deviation is 5. Conduct the following test of hypothesis using the .05 significance level.

2. A sample of 36 observations is selected from a normal population. The sample mean is 12, and the population standard deviation is 3. Conduct the following test of hypothesis using the .02 significance level.

3. A sample of 36 observations is selected from a normal population. The sample mean is 21, and the population standard deviation is 5. Conduct the following test of hypothesis using the .05 significance level.

4. A sample of 64 observations is selected from a normal population. The sample mean is 215, and the population standard deviation is 15. Conduct the following test of hypoth- esis using the .03 significance level.

For Exercises 5–8: (a) State the null hypothesis and the alternate hypothesis. (b) State the decision rule. (c) Compute the value of the test statistic. (d) What is your decision regarding H0? (e) What is the p-value? Interpret it.

5. The manufacturer of the X-15 steel-belted radial truck tire claims that the mean mileage the tire can be driven before the tread wears out is 60,000 miles. Assume the mileage wear follows the normal distribution and the standard deviation of the distribution is 5,000 miles. Crosset Truck Company bought 48 tires and found that the mean mileage for its trucks is 59,500 miles. Is Crosset’s experience different from that claimed by the manufacturer at the .05 significance level?

6. The waiting time for customers at MacBurger Restaurants follows a normal distribution with a mean of 3 minutes and a standard deviation of 1 minute. At the Warren Road MacBurger, the quality-assurance department sampled 50 customers and found that the mean waiting time was 2.75 minutes. At the .05 significance level, can we conclude that the mean waiting time is less than 3 minutes?

7. A recent national survey found that high school students watched an average (mean) of 6.8 DVDs per month with a population standard deviation of 0.5 hours. The distribution of

H1: � 6 220 H0: � 220

H1: � 7 20 H0: � � 20

H1: � 7 10 H0: � � 10

H1: � � 50 H0: � � 50

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times follows the normal distribution. A random sample of 36 college students revealed that the mean number of DVDs watched last month was 6.2. At the .05 significance level, can we conclude that college students watch fewer DVDs a month than high school students?

8. At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average $80 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $3.24. Over the first 35 days she was employed at the restaurant, the mean daily amount of her tips was $84.85. At the .01 significance level, can Ms. Brigden conclude that her daily tips average more than $80?

10.8 Testing for a Population Mean: Population Standard Deviation Unknown In the preceding example, we knew �, the population standard deviation, and that the population followed the normal distribution. In most cases, however, the population standard deviation is unknown. Thus, � must be based on prior studies or estimated by the sample standard deviation, s. The population standard deviation in the follow- ing example is not known, so the sample standard deviation is used to estimate �.

To find the value of the test statistic, we use the t distribution and revise for- mula [10–1] as follows:

TESTING A MEAN, � UNKNOWN [10–2]

with n 1 degrees of freedom, where: is the sample mean.

� is the hypothesized population mean. s is the sample standard deviation. n is the number of observations in the sample.

We encountered this same situation when constructing confidence intervals in the previous chapter. See pages 306–312 in Chapter 9. We summarized this problem in Chart 9–3 on page 309. Under these conditions, the correct statistical procedure is to replace the standard normal distribution with the t distribution. To review, the major characteristics of the t distribution are:

• It is a continuous distribution. • It is bell-shaped and symmetrical. • There is a family of t distributions. Each time the degrees of freedom change,

a new distribution is created. • As the number of degrees of freedom increases, the shape of the t distribution

approaches that of the standard normal distribution. • The t distribution is flatter, or more spread out, than the standard normal

distribution.

The following example shows the details.

t � X � s�1n

Example The McFarland Insurance Company Claims Department reports the mean cost to process a claim is $60. An industry comparison showed this amount to be larger than most other insurance companies, so the company instituted cost-cutting mea- sures. To evaluate the effect of the cost-cutting measures, the Supervisor of the Claims Department selected a random sample of 26 claims processed last month. The sample information is reported in the following.

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Solution We will use the five-step hypothesis testing procedure. Step 1: State the null hypothesis and the alternate hypothesis. The null

hypothesis is that the population mean is at least $60. The alternate hypothesis is that the population mean is less than $60. We can express the null and alternate hypotheses as follows:

The test is one-tailed because we want to determine whether there has been a reduction in the cost. The inequality in the alternate hypothesis points to the region of rejection in the left tail of the distribution.

Step 2: Select the level of significance. We decided on the .01 significance level. Step 3: Select the test statistic. The test statistic in this situation is the t distribu-

tion. Why? First it is reasonable to conclude that the distribution of the cost per claim follows the normal distribution. We can confirm this from the his- togram in the center of the following Minitab output. Observe the normal distribution superimposed on the frequency distribution.

H1: � 6 $60

H0: � $60

$45 $49 $62 $40 $43 $61 48 53 67 63 78 64 48 54 51 56 63 69 58 51 58 59 56 57 38 76

At the .01 significance level, is it reasonable to conclude that the mean cost to process a claim is now less than $60?

We do not know the standard deviation of the population. So we substi- tute the sample standard deviation. The value of the test statistic is com- puted by formula (10–2):

t � X � s�1n

LO6 Conduct a test of hypothesis about a population mean.

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Step 4: Formulate the decision rule. The critical values of t are given in Appendix B.2, a portion of which is shown in Table 10–1. Appendix B.2 is also repeated in the back inside cover of the text. The far left column of the table is labeled “df ” for degrees of freedom. The number of degrees of freedom is the total number of observations in the sample minus the number of populations sampled, written n 1. In this case, the number of observations in the sam- ple is 26, and we sampled 1 population, so there are 26 1 � 25 degrees of freedom. To find the critical value, first locate the row with the appropri- ate degrees of freedom. This row is shaded in Table 10–1. Next, determine whether the test is one-tailed or two-tailed. In this case, we have a one- tailed test, so find the portion of the table that is labeled “one-tailed.” Locate the column with the selected significance level. In this example, the signifi- cance level is .01. Move down the column labeled “0.01” until it intersects the row with 25 degrees of freedom. The value is 2.485. Because this is a one-sided test and the rejection region is in the left tail, the critical value is negative. The decision rule is to reject H0 if the value of t is less than 2.485.

Step 5: Make a decision and interpret the result. From the Minitab output, to the right of the histogram, the mean cost per claim for the sample of 26 observations is $56.42. The standard deviation of this sample is $10.04. We insert these values in formula (10–2) and compute the value of t:

Because 1.818 lies in the region to the right of the critical value of the null hypothesis is not rejected at the .01 significance level.

We have not demonstrated that the cost-cutting measures reduced the mean cost per claim to less than $60. To put it another way, the difference of $3.58 ($56.42 $60) between the sample mean and the population mean could be due to sampling error. The computed value of t is shown in Chart 10–6. It is in the region where the null hypothesis is not rejected.

2.485,

t � X � s�1n

� $56.42 $60 $10.04�126

� 1.818

Confidence Intervals

80% 90% 95% 98% 99% 99.9%

Level of Significance for One-Tailed Test, �

df 0.10 0.05 0.025 0.01 0.005 0.0005

Level of Significance for Two-Tailed Test, �

0.20 0.10 0.05 0.02 0.01 0.001

o o o o o o o 21 1.323 1.721 2.080 2.518 2.831 3.819 22 1.321 1.717 2.074 2.508 2.819 3.792 23 1.319 1.714 2.069 2.500 2.807 3.768 24 1.318 1.711 2.064 2.492 2.797 3.745 25 1.316 1.708 2.060 2.485 2.787 3.725

26 1.315 1.706 2.056 2.479 2.779 3.707 27 1.314 1.703 2.052 2.473 2.771 3.690 28 1.313 1.701 2.048 2.467 2.763 3.674 29 1.311 1.699 2.045 2.462 2.756 3.659 30 1.310 1.697 2.042 2.457 2.750 3.646

TABLE 10–1 A Portion of the t Distribution Table

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0–1.818 Computed value of t

Region of rejection

–2.485 Critical value

Scale of t

α = .01

H0: μ ≥ $60 H1: μ < $60

df = 26 − 1 = 25

CHART 10–6 Rejection Region, t Distribution, .01 Significance Level

Example The mean length of a small counterbalance bar is 43 millimeters. The production supervisor is concerned that the adjustments of the machine producing the bars have changed. He asks the Engineering Department to investigate. Engineering selects a random sample of 12 bars and measures each. The results are reported below in millimeters.

42 39 42 45 43 40 39 41 40 42 43 42

Is it reasonable to conclude that there has been a change in the mean length of the bars? Use the .02 significance level.

Solution We begin by stating the null hypothesis and the alternate hypothesis.

The alternate hypothesis does not state a direction, so this is a two-tailed test. There are 11 degrees of freedom, found by n 1 � 12 1 � 11. The t value is 2.718, found by referring to Appendix B.2 for a two-tailed test, using the .02 significance level, with 11 degrees of freedom. The decision rule is: Reject the null hypothesis if the computed t is to the left of 2.718 or to the right of 2.718. This information is summarized in Chart 10–7.

H1: � � 43 H0: � � 43

CHART 10–7 Regions of Rejection, Two-Tailed Test, Student’s t Distribution, � � .02

Region of rejection

–2.718 Critical value

Scale of t

H0: μ = 43 H1: μ ≠ 43

df = 11 H0 is not rejected

Region of rejection

2.718 Critical value

α _ 2

= .01 α _ 2

= .01

In the previous example, the mean and the standard deviation were computed using Minitab. The following example shows the details when the sample mean and sample standard deviation are calculated from sample data.

LO6 Conduct a test of hypothesis about a population mean.

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Exercises 9. Given the following hypothesis:

A random sample of 10 observations is selected from a normal population. The sample mean was 12 and the sample standard deviation 3. Using the .05 significance level: a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis?

10. Given the following hypothesis:

H1: � � 400 H0: � � 400

H1: � 7 10 H0: � � 10

Self-Review 10–3 The mean life of a battery used in a digital clock is 305 days. The lives of the batteries follow the normal distribution. The battery was recently modified to last longer. A sample of 20 of the modified batteries had a mean life of 311 days with a standard deviation of 12 days. Did the modification increase the mean life of the battery? (a) State the null hypothesis and the alternate hypothesis. (b) Show the decision rule graphically. Use the .05 significance level. (c) Compute the value of t. What is your decision regarding the null hypothesis? Briefly

summarize your results.

We calculate the standard deviation of the sample using formula (3–11). The mean, , is 41.5 millimeters, and the standard deviation, s, is 1.784 millimeters. The details are shown in Table 10–2.

TABLE 10–2 Calculations of the Sample Standard Deviation

X (mm) X � (X � )2

42 0.5 0.25 39 2.5 6.25 42 0.5 0.25 45 3.5 12.25 43 1.5 2.25 40 1.5 2.25 39 2.5 6.25 41 0.5 0.25 40 1.5 2.25 42 0.5 0.25 43 1.5 2.25 42 0.5 0.25

498 0 35.00

s � B

�(X X )2

n 1 � B

35 12 1

� 1.784

X � 498 12

� 41.5 mm

Now we are ready to compute the value of t, using formula (10–2).

The null hypothesis that the population mean is 43 millimeters is rejected because the computed t of 2.913 lies in the area to the left of 2.718. We accept the alternate hypothesis and conclude that the population mean is not 43 millime- ters. The machine is out of control and needs adjustment.

t � X � s�1n

� 41.5 43.0 1.784�112

� 2.913

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A random sample of 12 observations is selected from a normal population. The sample mean was 407 and the sample standard deviation 6. Using the .01 significance level: a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis?

11. The Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook pub- lishing company, claims that the sales representatives make an average of 40 sales calls per week on professors. Several reps say that this estimate is too low. To investigate, a random sample of 28 sales representatives reveals that the mean number of calls made last week was 42. The standard deviation of the sample is 2.1 calls. Using the .05 significance level, can we conclude that the mean number of calls per salesperson per week is more than 40?

12. The management of White Industries is considering a new method of assembling its golf cart. The present method requires 42.3 minutes, on the average, to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes. Using the .10 level of significance, can we conclude that the assembly time using the new method is faster?

13. The mean income per person in the United States is $40,000, and the distribution of incomes follows a normal distribution. A random sample of 10 residents of Wilmington, Delaware, had a mean of $50,000 with a standard deviation of $10,000. At the .05 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average?

14. Most air travelers now use e-tickets. Electronic ticketing allows passengers to not worry about a paper ticket, and it costs the airline companies less to handle than paper tick- eting. However, in recent times the airlines have received complaints from passengers regarding their e-tickets, particularly when connecting flights and a change of airlines were involved. To investigate the problem, an independent watchdog agency contacted a ran- dom sample of 20 airports and collected information on the number of complaints the airport had with e-tickets for the month of March. The information is reported below.

At the .05 significance level, can the watchdog agency conclude the mean number of complaints per airport is less than 15 per month? a. What assumption is necessary before conducting a test of hypothesis? b. Plot the number of complaints per airport in a frequency distribution or a dot plot. Is

it reasonable to conclude that the population follows a normal distribution? c. Conduct a test of hypothesis and interpret the results.

A Software Solution The Minitab statistical software system, used in earlier chapters and the previous section, provides an efficient way of conducting a one-sample test of hypothesis for a population mean. The steps to generate the following output are shown in the Software Commands section at the end of the chapter.

14 14 16 12 12 14 13 16 15 14 12 15 15 14 13 13 12 13 10 13

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–2.913 2.913 Scale of t –2.718 2.718

TABLE 10–3 A Portion of Student’s t Distribution

Confidence Intervals

80% 90% 95% 98% 99% 99.9%

Level of Significance for One-Tailed Test, �

df 0.10 0.05 .0025 0.01 0.005 0.0005

Level of Significance for Two-Tailed Test, �

0.20 0.10 0.05 0.02 0.01 0.001

o o o o o o o 9 1.383 1.833 2.262 2.821 3.250 4.781

10 1.372 1.812 2.228 2.764 3.169 4.587

11 1.363 1.796 2.201 2.718 3.106 4.437 12 1.356 1.782 2.179 2.681 3.055 4.318 13 1.350 1.771 2.160 2.650 3.012 4.221 14 1.345 1.761 2.145 2.624 2.977 4.140 15 1.341 1.753 2.131 2.602 2.947 4.073

In the preceding example, the alternate hypothesis was two-sided, so there were rejection areas in both the lower (left) tail and the upper (right) tail. To determine the p-value, it was necessary to determine the area to the left of 2.913 for a t distri- bution with 11 degrees of freedom and add to it the value of the area to the right of 2.913, also with 11 degrees of freedom.

What if we were conducting a one-sided test, so that the entire rejection region would be in either the upper or the lower tail? In that case, we would report the area from only the one tail. In the counterbalance example, if H1 were stated as � � 43, the inequality would point to the left. Thus, we would have reported the p-value as the area to the left of 2.913. This value is .007, found by .014�2. Thus, the p-value for a one-tailed test would be .007.

How can we estimate a p-value without a computer? To illustrate, recall that, in the example regarding the length of a counterbalance, we rejected the null

An additional feature of most statistical software packages is to report the p-value, which gives additional information on the null hypothesis. The p-value is the proba- bility of a t value as extreme as that computed, given that the null hypothesis is true. Using the data from the previous counterbalance bar example, the p-value of .014 is the likelihood of a t value of 2.91 or less plus the likelihood of a t value of 2.91 or larger, given a population mean of 43. Thus, comparing the p-value to the significance level tells us whether the null hypothesis was close to being rejected, barely rejected, and so on.

To explain further, refer to the diagram below. The p-value of .014 is the darker or shaded area and the significance level is the total shaded area. Because the p-value of .014 is less than the significance level of .02, the null hypothesis is rejected. Had the p-value been larger than the significance level—say, .06, .19, or .57—the null hypothesis would not be rejected. If the significance level had initially been selected as .01, the null hypothesis would not be rejected.

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Self-Review 10–4 A machine is set to fill a small bottle with 9.0 grams of medicine. A sample of eight bot- tles revealed the following amounts (grams) in each bottle.

At the .01 significance level, can we conclude that the mean weight is less than 9.0 grams? (a) State the null hypothesis and the alternate hypothesis. (b) How many degrees of freedom are there? (c) Give the decision rule. (d) Compute the value of t. What is your decision regarding the null hypothesis? (e) Estimate the p-value.

9.2 8.7 8.9 8.6 8.8 8.5 8.7 9.0

hypothesis that � � 43 and accepted the alternate hypothesis that � � 43. The significance level was .02, so logically the p-value is less than .02. To estimate the p-value more accurately, go to Appendix B.2 and find the row with 11 degrees of freedom. The computed t value of 2.913 is between 2.718 and 3.106. (A portion of Appendix B.2 is reproduced as Table 10–3.) The two-tailed significance level corre- sponding to 2.718 is .02, and for 3.106 it is .01. Therefore, the p-value is between .01 and .02. The usual practice is to report that the p-value is less than the larger of the two significance levels. So we would report, “the p-value is less than .02.”

Exercises 15. Given the following hypothesis:

A random sample of five resulted in the following values: 18, 15, 12, 19, and 21. Assume a normal population. Using the .01 significance level, can we conclude the population mean is less than 20? a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis? d. Estimate the p-value.

16. Given the following hypothesis:

A random sample of six resulted in the following values: 118, 105, 112, 119, 105, and 111. Assume a normal population. Using the .05 significance level, can we conclude the mean is different from 100? a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis? d. Estimate the p-value.

17. The amount of water consumed each day by a healthy adult follows a normal distribu- tion with a mean of 1.4 liters. A health campaign promotes the consumption of at least 2.0 liters per day. A sample of 10 adults after the campaign shows the following con- sumption in liters:

At the .01 significance level, can we conclude that water consumption has increased? Calculate and interpret the p-value.

18. The liquid chlorine added to swimming pools to combat algae has a relatively short shelf life before it loses its effectiveness. Records indicate that the mean shelf life of a 5-gallon jug of chlorine is 2,160 hours (90 days). As an experiment, Holdlonger was added to the

1.5 1.6 1.5 1.4 1.9 1.4 1.3 1.9 1.8 1.7

H1: � � 100 H0: � � 100

H1: � 6 20 H0: � 20

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chlorine to find whether it would increase the shelf life. A sample of nine jugs of chlorine had these shelf lives (in hours):

At the .025 level, has Holdlonger increased the shelf life of the chlorine? Estimate the p-value.

19. A Washington, D.C., “think tank” announces the typical teenager sent 50 text messages per day in 2009. To update that estimate, you phone a sample of teenagers and ask them how many text messages they sent the previous day. Their responses were:

At the .05 level, can you conclude that the mean number is greater than 50? Estimate the p-value and describe what it tells you.

20. Hugger Polls contends that an agent conducts a mean of 53 in-depth home surveys every week. A streamlined survey form has been introduced, and Hugger wants to eval- uate its effectiveness. The number of in-depth surveys conducted during a week by a random sample of agents are:

At the .05 level of significance, can we conclude that the mean number of interviews conducted by the agents is more than 53 per week? Estimate the p-value.

10.9 Tests Concerning Proportions In the previous chapter, we discussed confidence intervals for proportions. See Section 9.4 on pages 313–316. We can also conduct a test of hypothesis for a proportion. Recall that a proportion is the ratio of the number of successes to the number of observations. We let X refer to the number of successes and n the num- ber of observations, so the proportion of successes in a fixed number of trials is X�n. Thus, the formula for computing a sample proportion, p, is p � X�n. Consider the following potential hypothesis-testing situations.

• Historically, General Motors reports that 70 percent of leased vehicles are returned with less than 36,000 miles. A recent sample of 200 vehicles returned at the end of their lease showed 158 had less than 36,000 miles. Has the pro- portion increased?

• The American Association of Retired Persons (AARP) reports that 60 percent of retired people under the age of 65 would return to work on a full-time basis if a suitable job were available. A sample of 500 retirees under 65 revealed 315 would return to work. Can we conclude that more than 60 percent would return to work?

• Able Moving and Storage Inc. advises its clients for long-distance residential moves that their household goods will be delivered in 3 to 5 days from the time they are picked up. Able’s records show it is successful 90 percent of the time with this claim. A recent audit revealed it was successful 190 times out of 200. Can the company conclude its success rate has increased?

Some assumptions must be made and conditions met before testing a popula- tion proportion. To test a hypothesis about a population proportion, a random sam- ple is chosen from the population. It is assumed that the binomial assumptions dis- cussed in Chapter 6 are met: (1) the sample data collected are the result of counts; (2) the outcome of an experiment is classified into one of two mutually exclusive categories—a “success” or a “failure”; (3) the probability of a success is the same for each trial; and (4) the trials are independent, meaning the outcome of one trial does not affect the outcome of any other trial. The test we will conduct shortly is appropriate when both n� and n(1 � �) are at least 5. n is the sample size, and � is the population proportion. It takes advantage of the fact that a binomial distrib- ution can be approximated by the normal distribution.

53 57 50 55 58 54 60 52 59 62 60 60 51 59 56

51 175 47 49 44 54 145 203 21 59 42 100

2,159 2,170 2,180 2,179 2,160 2,167 2,171 2,181 2,185

356 Chapter 10

LO8 Conduct a test of hypothesis about a population proportion.

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One-Sample Tests of Hypothesis 357

Example Suppose prior elections in a certain state indicated it is necessary for a candidate for governor to receive at least 80 percent of the vote in the northern section of the state to be elected. The incumbent governor is interested in assessing his chances of returning to office and plans to conduct a survey of 2,000 registered voters in the northern section of the state.

Using the hypothesis-testing procedure, assess the governor’s chances of reelection.

Solution This situation regarding the governor’s reelection meets the binomial conditions. • There are only two possible outcomes. That is, a sampled voter will either vote

or not vote for the governor. • The probability of a success is the same for each trial. In this case, the likeli-

hood a particular sampled voter will support reelection is .80. • The trials are independent. This means, for example, the likelihood the 23rd voter

sampled will support reelection is not affected by what the 24th or 52nd voter does. • The sample data is the result of counts. We are going to count the number of

voters who support reelection in the sample of 2,000.

We can use the normal approximation to the binomial distribution, discussed in Chapter 7, because both n� and n(1 �) exceed 5. In this case, n � 2,000 and � � .80 (� is the proportion of the vote in the northern part of the state, or 80 per- cent, needed to be elected). Thus, n� � 2,000(.80) � 1,600 and n(1 �) � 2,000 (1 .80) � 400. Both 1,600 and 400 are greater than 5.

Step 1: State the null hypothesis and the alternate hypothesis. The null hypothesis, H0, is that the population proportion is .80 or larger. The alternate hypothesis, H1, is that the proportion is less than .80. From a practical standpoint, the incumbent governor is concerned only when the proportion is less than .80. If it is equal to or greater than .80, he will have no problem; that is, the sample data would indicate he will probably be reelected. These hypotheses are written symbolically as:

H1 states a direction. Thus, as noted previously, the test is one-tailed with the inequality sign pointing to the tail of the distribution containing the region of rejection.

Step 2: Select the level of significance. The level of significance is .05. This is the likelihood that a true hypothesis will be rejected.

Step 3: Select the test statistic. z is the appropriate statistic, found by:

H1: � 6 .80 H0: � .80

�

Finding the critical value

TEST OF HYPOTHESIS, ONE PROPORTION [10–3] z �

p �

�(1 �) n

where: � is the population proportion. p is the sample proportion. n is the sample size.

Step 4: Formulate the decision rule. The critical value or values of z form the di- viding point or points between the regions where H0 is rejected and where it is not rejected. Since the alternate hypothesis states a direction, this is a one-tailed test. The sign of the inequality points to the left, so only the left side of the curve is used. (See Chart 10–8.) The significance level was

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358 Chapter 10

.4500 .5000

H0: � ≥ .80 H1: � < .80

0–1.65 Critical value

Scale of z

.05 Region of rejection

H0 is not rejected

CHART 10–8 Rejection Region for the .05 Level of Significance, One-Tailed Test

given as .05 in step 2. This probability is in the left tail and determines the region of rejection. The area between zero and the critical value is .4500, found by .5000 .0500. Referring to Appendix B.1 and searching for .4500, we find the critical value of z is 1.65. The decision rule is, therefore: Reject the null hypothesis and accept the alternate hypothesis if the com- puted value of z falls to the left of 1.65; otherwise do not reject H0.

Step 5: Make a decision and interpret the result. Select a sample and make a decision about H0. A sample survey of 2,000 potential voters in the north- ern part of the state revealed that 1,550 planned to vote for the incum- bent governor. Is the sample proportion of .775 (found by 1,550�2,000) close enough to .80 to conclude that the difference is due to sampling error? In this case:

p is .775, the proportion in the sample who plan to vote for the governor.

n is 2,000, the number of voters surveyed. � is .80, the hypothesized population proportion. z is a normally distributed test statistic when the hypothesis is

true and the other assumptions are true.

Using formula (10–3) and computing z gives

The computed value of z ( 2.80) is in the rejection region, so the null hypothesis is rejected at the .05 level. The difference of 2.5 percentage points between the sample percent (77.5 percent) and the hypothesized population percent in the northern part of the state necessary to carry the state (80 percent) is statistically significant. It is probably not due to sam- pling variation. To put it another way, the evidence at this point does not support the claim that the incumbent governor will return to the gover- nor’s mansion for another four years.

The p-value is the probability of finding a z value less than 2.80. From Appendix B.1, the probability of a z value between zero and 2.80 is .4974. So the p-value is .0026, found by .5000 .4974. The governor cannot be confident of reelection because the p-value is less than the significance level.

z � p �

�(1 �) n

�

1,550 2,000

.80

.80(1 .80) 2,000

� .775 .80 1.00008

� 2.80

Select a sample and make a decision regarding .H0

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One-Sample Tests of Hypothesis 359

Self-Review 10–5 A recent insurance industry report indicated that 40 percent of those persons involved in minor traffic accidents this year have been involved in a least one other traffic accident in the last five years. An advisory group decided to investigate this claim, believing it was too large. A sample of 200 traffic accidents this year showed 74 persons were also involved in another accident within the last five years. Use the .01 significance level. (a) Can we use z as the test statistic? Tell why or why not. (b) State the null hypothesis and the alternate hypothesis. (c) Show the decision rule graphically. (d) Compute the value of z and state your decision regarding the null hypothesis. (e) Determine and interpret the p-value.

Exercises 21. The following hypotheses are given.

A sample of 100 observations revealed that p � .75. At the .05 significance level, can the null hypothesis be rejected? a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis?

22. The following hypotheses are given.

A sample of 120 observations revealed that p � .30. At the .05 significance level, can the null hypothesis be rejected? a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis?

Note: It is recommended that you use the five-step hypothesis-testing procedure in solving the following problems.

23. The National Safety Council reported that 52 percent of American turnpike drivers are men. A sample of 300 cars traveling southbound on the New Jersey Turnpike yesterday revealed that 170 were driven by men. At the .01 significance level, can we conclude that a larger proportion of men were driving on the New Jersey Turnpike than the national statistics indicate?

24. A recent article in USA Today reported that a job awaits only one in three new college graduates. The major reasons given were an overabundance of college graduates and a weak economy. A survey of 200 recent graduates from your school revealed that 80 stu- dents had jobs. At the .02 significance level, can we conclude that a larger proportion of students at your school have jobs?

25. Chicken Delight claims that 90 percent of its orders are delivered within 10 minutes of the time the order is placed. A sample of 100 orders revealed that 82 were delivered within the promised time. At the .10 significance level, can we conclude that less than 90 percent of the orders are delivered in less than 10 minutes?

26. Research at the University of Toledo indicates that 50 percent of students change their major area of study after their first year in a program. A random sample of 100 students in the College of Business revealed that 48 had changed their major area of study after their first year of the program. Has there been a significant decrease in the proportion of students who change their major after the first year in this program? Test at the .05 level of significance.

10.10 Type II Error Recall that the level of significance, identified by the symbol �, is the probability that the null hypothesis is rejected when it is true. This is called a Type I error. The most common levels of significance are .05 and .01 and are set by the researcher at the outset of the test.

H1: � � .40 H0: � � .40

H1: � 7 .70 H0: � � .70

LO9 Compute the probability of a Type II error.

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360 Chapter 10

In a hypothesis-testing situation there is also the possibility that a null hypothesis is not rejected when it is actually false. That is, we accept a false null hypothesis. This is called a Type II error. The probability of a Type II error is identified by the Greek let- ter beta (�). The following examples illustrate the details of determining the value of �.

Example Western Wire Products purchases steel bars to make cotter pins. Past experience indicates that the mean tensile strength of all incoming shipments is 10,000 psi and that the standard deviation, �, is 400 psi.

To make a decision about incoming shipments of steel bars, Western Wire Prod- ucts set up this rule for the quality-control inspector to follow: “Take a sample of 100 steel bars. At the .05 significance level, if the sample mean strength falls between 9,922 psi and 10,078 psi, accept the lot. Otherwise, the lot is to be rejected.” Refer to Chart 10–9, Region A. It shows the region where each lot is rejected and where it is not rejected. The mean of this distribution is designated �0. The tails of the curve represent the probability of making a Type I error, that is, rejecting the incoming lot of steel bars when in fact it is a good lot, with a mean of 10,000 psi.

(X)

Solution The probability of committing a Type II error, as represented by the shaded area in Chart 10–9, Region B, can be computed by determining the area under the nor- mal curve that lies above 9,922 pounds. The calculation of the areas under the normal curve was discussed in Chapter 7. Reviewing briefly, it is necessary first to determine the probability of the sample mean falling between 9,900 and 9,922. Then

CHART 10–9 Charts Showing Type I and Type II Errors

9,922

Reject lot

10,07810,000 –1.96 � x– 1.96 �x–

Region A

Probability of �

psi

�

9,900 9,922

.2912.2088.5000

�1 X c

Region B

Suppose the unknown population mean of an incoming lot, designated �1, is really 9,900 psi. What is the probability that the quality-control inspector will fail to reject the shipment (a Type II error)?

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One-Sample Tests of Hypothesis 361

9,922

Rejection region

10,07810,000 –1.96 �– 1.96 ��

�

xx –

2 � .025�

10,120 10,078

Probability of making a Type II

error Probability

of not making a Type II

error

Probability of not making

a Type II error

� 1 �

Region A

Region C

psi

CHART 10–10 Type I and Type II Errors (Another Example)

this probability is subtracted from .5000 (which represents all the area beyond the mean of 9,900) to arrive at the probability of making a Type II error in this case.

The number of standard units (z value) between the mean of the incoming lot (9,900), designated by �1, and , representing the critical value for 9,922, is computed by:

TYPE II ERROR [10–4]

With n � 100 and � � 400, the value of z is 0.55:

The area under the curve between 9,900 and 9,922 (a z value of 0.55) is .2088. The area under the curve beyond 9,922 pounds is .5000 .2088, or .2912; this is the probability of making a Type II error—that is, accepting an incoming lot of steel bars when the population mean is 9,900 psi.

Another illustration, in Chart 10–10, Region C, depicts the probability of accept- ing a lot when the population mean is 10,120. To find the probability:

The probability that z is less than 1.05 is .1469, found by .5000 .3531. There- fore, �, or the probability of a Type II error, is .1469.

z � Xc �1 ��1n

� 10,078 10,120

400�1100 � 1.05

z � Xc �1 ��1n

� 9,922 9,900

400�1100 �

22 40

� 0.55

z � Xc �1

��1n

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362 Chapter 10

TABLE 10–4 Probabilities of a Type II Error for �0 � 10,000 Pounds and Selected Alternative Means, .05 Level of Significance

Selected Probability Probability of Alternative of Type II Not Making a

Mean Error Type II Error (pounds) (�) (1 � �)

9,820 .0054 .9946 9,880 .1469 .8531 9,900 .2912 .7088 9,940 .6736 .3264 9,980 .9265 .0735

10,000 — * — 10,020 .9265 .0735 10,060 .6736 .3264 10,100 .2912 .7088 10,120 .1469 .8531 10,180 .0054 .9946

*It is not possible to make a Type II error when � � �0.

Using the methods illustrated by Charts 10–9 Region B and 10-10 Region C, the probability of accepting a hypothesis as true when it is actually false can be determined for any value of �1.

Type II error probabilities are shown in the center column of Table 10–4 for selected values of �, given in the left column. The right column gives the probabil- ity of not making a Type II error, which is also known as the power of a test.

Self-Review 10–6 Refer to the previous Example. Suppose the true mean of an incoming lot of steel bars is 10,180 psi. What is the probability that the quality control inspector will accept the bars as having a mean of 10,000 psi? (It sounds implausible that steel bars will be rejected if the tensile strength is higher than specified. However, it may be that the cotter pin has a dual function in an outboard motor. It may be designed not to shear off if the motor hits a small object, but to shear off if it hits a rock. Therefore, the steel should not be too strong.)

The light area in Chart 10–10, Region C, represents the probability of falsely accept- ing the hypothesis that the mean tensile strength of the incoming steel is 10,000 psi. What is the probability of committing a Type II error?

Chapter Summary I. The objective of hypothesis testing is to verify the validity of a statement about a popu-

lation parameter. II. The steps to conduct a test of hypothesis are:

A. State the null hypothesis (H0) and the alternate hypothesis (H1). B. Select the level of significance.

1. The level of significance is the likelihood of rejecting a true null hypothesis. 2. The most frequently used significance levels are .01, .05, and .10, but any value

between 0 and 1.00 is possible.

Exercises 27. Refer to Table 10–4 and the example just completed. With n � 100, � � 400,

and �1 � 9,880, verify that the probability of a Type II error is .1469. 28. Refer to Table 10–4 and the example just completed. With n � 100, � � 400,

and �1 � 9,940, verify that the probability of a Type II error is .6736. Xc � 9,922,

Xc � 9,922,

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One-Sample Tests of Hypothesis 363

C. Select the test statistic. 1. A test statistic is a value calculated from sample information used to determine

whether to reject the null hypothesis. 2. Two test statistics were considered in this chapter.

a. The standard normal distribution (the z distribution) is used when the population follows the normal distribution and the population standard deviation is known.

b. The t distribution is used when the population follows the normal distribution and the population standard deviation is unknown.

D. State the decision rule. 1. The decision rule indicates the condition or conditions when the null hypothesis is

rejected. 2. In a two-tailed test, the rejection region is evenly split between the upper and lower

tails. 3. In a one-tailed test, all of the rejection region is in either the upper or the lower tail.

E. Select a sample, compute the value of the test statistic, make a decision regarding the null hypothesis, and interpret the results.

III. A p-value is the probability that the value of the test statistic is as extreme as the value computed, when the null hypothesis is true.

IV. When testing a hypothesis about a population mean: A. If the population standard deviation, �, is known, the test statistic is the standard nor-

mal distribution and is determined from:

[10–1]

B. If the population standard deviation is not known, s is substituted for �. The test sta- tistic is the t distribution, and its value is determined from:

[10–2]

The major characteristics of the t distribution are: 1. It is a continuous distribution. 2. It is mound-shaped and symmetrical. 3. It is flatter, or more spread out, than the standard normal distribution. 4. There is a family of t distributions, depending on the number of degrees of freedom.

V. When testing about a population proportion: A. The binomial conditions must be met. B. Both n� and n(1 � �) must be at least 5. C. The test statistic is

[10–3]

VI. There are two types of errors that can occur in a test of hypothesis. A. A Type I error occurs when a true null hypothesis is rejected.

1. The probability of making a Type I error is equal to the level of significance. 2. This probability is designated by the Greek letter �.

B. A Type II error occurs when a false null hypothesis is not rejected. 1. The probability of making a Type II error is designated by the Greek letter �. 2. The likelihood of a Type II error is found by

[10–4]

Pronunciation Key SYMBOL MEANING PRONUNCIATION

H0 Null hypothesis H sub zero

H1 Alternate hypothesis H sub one

�/2 Two-tailed significance level Alpha over 2

c Limit of the sample mean X bar sub c

�0 Assumed population mean mu sub zero

z � Xc � �1

��1n

z � p � �

�(1 � �) n

t � X � � s�1n

z � X � � ��1n

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364 Chapter 10

45.6 47.7 47.6 46.3 46.2 47.4 49.2 55.8 47.5 48.5

Chapter Exercises 29. According to the local union president, the mean gross income of plumbers in the Salt Lake

City area follows the normal probability distribution with a mean of $45,000 and a standard deviation of $3,000. A recent investigative reporter for KYAK TV found, for a sample of 120 plumbers, the mean gross income was $45,500. At the .10 significance level, is it reason- able to conclude that the mean income is not equal to $45,000? Determine the p-value.

30. Rutter Nursery Company packages its pine bark mulch in 50-pound bags. From a long history, the production department reports that the distribution of the bag weights follows the normal distribution and the standard deviation of this process is 3 pounds per bag. At the end of each day, Jeff Rutter, the production manager, weighs 10 bags and com- putes the mean weight of the sample. Below are the weights of 10 bags from today’s production.

a. Can Mr. Rutter conclude that the mean weight of the bags is less than 50 pounds? Use the .01 significance level.

b. In a brief report, tell why Mr. Rutter can use the z distribution as the test statistic. c. Compute the p-value.

31. A new weight-watching company, Weight Reducers International, advertises that those who join will lose, on the average, 10 pounds the first two weeks with a standard deviation of 2.8 pounds. A random sample of 50 people who joined the new weight reduction pro- gram revealed the mean loss to be 9 pounds. At the .05 level of significance, can we conclude that those joining Weight Reducers on average will lose less than 10 pounds? Determine the p-value.

32. Dole Pineapple Inc. is concerned that the 16-ounce can of sliced pineapple is being overfilled. Assume the standard deviation of the process is .03 ounces. The quality- control department took a random sample of 50 cans and found that the arithmetic mean weight was 16.05 ounces. At the 5 percent level of significance, can we conclude that the mean weight is greater than 16 ounces? Determine the p-value.

33. According to a recent survey, Americans get a mean of 7 hours of sleep per night. A ran- dom sample of 50 students at West Virginia University revealed the mean number of hours slept last night was 6 hours and 48 minutes (6.8 hours). The standard deviation of the sample was 0.9 hours. Is it reasonable to conclude that students at West Virginia sleep less than the typical American? Compute the p-value.

34. A statewide real estate sales agency, Farm Associates, specializes in selling farm property in the state of Nebraska. Its records indicate that the mean selling time of farm property is 90 days. Because of recent drought conditions, the agency believes that the mean selling time is now greater than 90 days. A statewide survey of 100 farms sold recently revealed that the mean selling time was 94 days, with a standard deviation of 22 days. At the .10 significance level, has there been an increase in selling time?

35. According to the Census Bureau, 3.13 people reside in the typical American household. A sample of 25 households in Arizona retirement communities showed the mean number of residents per household was 2.86 residents. The standard deviation of this sample was 1.20 residents. At the .05 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.13 persons?

36. A recent article in Vitality magazine reported that the mean amount of leisure time per week for American men is 40.0 hours. You believe this figure is too large and decide to conduct your own test. In a random sample of 60 men, you find that the mean is 37.8 hours of leisure per week and that the standard deviation of the sample is 12.2 hours. Can you conclude that the information in the article is untrue? Use the .05 sig- nificance level. Determine the p-value and explain its meaning.

37. In recent years, the interest rate on home mortgages has declined to less than 6.0 per- cent. However, according to a study by the Federal Reserve Board, the rate charged on credit card debt is more than 14 percent. Listed below is the interest rate charged on a sample of 10 credit cards.

14.6 16.7 17.4 17.0 17.8 15.4 13.1 15.8 14.3 14.5

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4.8 5.3 6.5 4.8 6.1 5.8 6.2 5.6

3.1 3.3 3.5 2.6 2.6 4.3 4.4 3.8 3.1 4.1 3.1 3.2

25 27 25 26 25 28 28 27 24 26 25 29 25 27 24

0 4 3 4 9 4 5 9 1 6 7 10

2.98 2.40 2.70 2.25 3.23 3.17 2.93 3.18 2.80 2.38 3.75 3.20 3.27 2.52 2.58 4.45 2.45

Is it reasonable to conclude the mean rate charged is greater than 14 percent? Use the .01 significance level.

38. A recent article in The Wall Street Journal reported that the 30-year mortgage rate is now less than 6 percent. A sample of eight small banks in the Midwest revealed the follow- ing 30-year rates (in percent):

At the .01 significance level, can we conclude that the 30-year mortgage rate for small banks is less than 6 percent? Estimate the p-value.

39. According to the Coffee Research Organization (http://www.coffeeresearch.org) the typical American coffee drinker consumes an average of 3.1 cups per day. A sample of 12 senior citizens revealed they consumed the following amounts of coffee, reported in cups, yesterday.

At the .05 significance level, does this sample data suggest there is a difference between the national average and the sample mean from senior citizens?

40. The postanesthesia care area (recovery room) at St. Luke’s Hospital in Maumee, Ohio, was recently enlarged. The hope was that with the enlargement the mean number of patients per day would be more than 25. A random sample of 15 days revealed the following numbers of patients.

At the .01 significance level, can we conclude that the mean number of patients per day is more than 25? Estimate the p-value and interpret it.

41. www.golfsmith.com receives an average of 6.5 returns per day from online shoppers. For a sample of 12 days, it received the following number of returns.

At the .01 significance level, can we conclude the mean number of returns is less than 6.5?

42. During recent seasons, Major League Baseball has been criticized for the length of the games. A report indicated that the average game lasts 3 hours and 30 minutes. A sam- ple of 17 games revealed the following times to completion. (Note that the minutes have been changed to fractions of hours, so that a game that lasted 2 hours and 24 minutes is reported at 2.40 hours.)

Can we conclude that the mean time for a game is less than 3.50 hours? Use the .05 significance level.

43. Watch Corporation of Switzerland claims that its watches on average will neither gain nor lose time during a week. A sample of 18 watches provided the following gains or losses in seconds per week.(�)

(�)

Is it reasonable to conclude that the mean gain or loss in time for the watches is 0? Use the .05 significance level. Estimate the p-value.

44. Listed below is the rate of return for one year (reported in percent) for a sample of 12 mutual funds that are classified as taxable money market funds.

�0.38 �0.20 �0.38 �0.32 �0.32 �0.23 �0.30 �0.25 �0.10 �0.37 �0.61 �0.48 �0.47 �0.64 �0.04 �0.20 �0.68 �0.05

4.63 4.15 4.76 4.70 4.65 4.52 4.70 5.06 4.42 4.51 4.24 4.52

One-Sample Tests of Hypothesis 365

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366 Chapter 10

120 108 120 114 118 91 118 92 104 104 112 97 118 108 117

$321 $286 $290 $330 $310 $250 $270 $280 $299 $265 $291 $275 $281

Is it reasonable to conclude that the mean number of customers using the self-checkout system is more than 100 per day? Use the .05 significance level.

46. For a recent year, the mean fare to fly from Charlotte, North Carolina, to Seattle, Wash- ington, on a discount ticket was $267. A random sample of round-trip discount fares on this route last month gives:

At the .01 significance level, can we conclude that the mean fare has increased? What is the p-value?

47. The publisher of Celebrity Living claims that the mean sales for personality magazines that feature people such as Angelina Jolie or Paris Hilton are 1.5 million copies per week. A sample of 10 comparable titles shows a mean weekly sales last week of 1.3 million copies with a standard deviation of 0.9 million copies. Does this data contradict the pub- lisher’s claim? Use the 0.01 significance level.

48. A United Nations report shows the mean family income for Mexican migrants to the United States is $27,000 per year. A FLOC (Farm Labor Organizing Committee) evalua- tion of 25 Mexican family units reveals a mean to be $30,000 with a sample standard deviation of $10,000. Does this information disagree with the United Nations report? Apply the 0.01 significance level.

49. A coin toss is used to decide which team gets the ball first in most sports. It involves little effort and is believed to give each side the same chance. In 43 Super Bowl games, the National Football Conference has won the coin-flip 29 times. Meanwhile, the American Football Conference has won only 14 times. Use the five-step hypothesis-testing procedure at the .01 significance level to test whether this data suggests a fair coin flip. a. Why can you employ z as the test statistic? b. State the null and alternate hypotheses. c. Make a diagram of the decision rule. d. Evaluate the test statistic and make the decision. e. What is the p-value and what does that imply?

50. According to a study by the American Pet Food Dealers Association, 63 percent of U.S. households own pets. A report is being prepared for an editorial in the San Francisco Chronicle. As a part of the editorial, a random sample of 300 households showed 210 own pets. Does this data disagree with the Pet Food Dealers Association data? Use a .05 level of significance.

51. Tina Dennis is the comptroller for Meek Industries. She believes that the current cash- flow problem at Meek is due to the slow collection of accounts receivable. She believes that more than 60 percent of the accounts are in arrears more than three months. A ran- dom sample of 200 accounts showed that 140 were more than three months old. At the .01 significance level, can she conclude that more than 60 percent of the accounts are in arrears for more than three months?

52. The policy of the Suburban Transit Authority is to add a bus route if more than 55 per- cent of the potential commuters indicate they would use the particular route. A sample of 70 commuters revealed that 42 would use a proposed route from Bowman Park to the downtown area. Does the Bowman-to-downtown route meet the STA criterion? Use the .05 significance level.

53. Past experience at the Crowder Travel Agency indicated that 44 percent of those persons who wanted the agency to plan a vacation for them wanted to go to Europe. During the most recent busy season, a sampling of 1,000 plans was selected at random from the

Using the .05 significance level, is it reasonable to conclude that the mean rate of return is more than 4.50 percent?

45. Many grocery stores and large retailers such as Walmart and Kmart have installed self- checkout systems so shoppers can scan their own items and cash out themselves. How do customers like this service and how often do they use it? Listed below is the num- ber of customers using the service for a sample of 15 days at the Walmart on Highway 544 in Surfside, South Carolina.

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One-Sample Tests of Hypothesis 367

files. It was found that 480 persons wanted to go to Europe on vacation. Has there been a significant shift upward in the percentage of persons who want to go to Europe? Test at the .05 significance level.

54. Research in the gaming industry showed that 10 percent of all slot machines in the United States stop working each year. Short’s Game Arcade has 60 slot machines and only 3 failed last year. Use the five-step hypothesis-testing procedure at the .05 significance level to test whether this data contradicts the research report. a. Why can you employ z as the test statistic? b. State the null and alternate hypotheses. c. Evaluate the test statistic and make the decision. d. What is the p-value and what does that imply?

55. An urban planner claims that, nationally, 20 percent of all families renting condominiums move during a given year. A random sample of 200 families renting condominiums in the Dallas Metroplex revealed that 56 had moved during the past year. At the .01 significance level, does this evidence suggest that a larger proportion of condominium owners moved in the Dallas area? Determine the p-value.

56. The cost of weddings in the United States has skyrocketed in recent years. As a result, many couples are opting to have their weddings in the Caribbean. A Caribbean vacation resort recently advertised in Bride Magazine that the cost of a Caribbean wedding was less than $10,000. Listed below is a total cost in $000 for a sample of 8 Caribbean weddings.

9.7 9.4 11.7 9.0 9.1 10.5 9.1 9.8

At the .05 significance level is it reasonable to conclude the mean wedding cost is less than $10,000 as advertised?

57. According to an ABC News survey, 40 percent of Americans do not eat breakfast. A sam- ple of 30 college students found 16 had skipped breakfast that day. Use the .01 signif- icance level to check whether college students are more likely to skip breakfast.

58. After a losing season, there is a great uproar to fire the head football coach. In a ran- dom sample of 200 college alumni, 80 favor keeping the coach. Test at the .05 level of significance whether the proportion of alumni who support the coach is less than 50 percent.

59. During the 1990s, the fatality rate for lung cancer was 80 per 100,000 people. After the turn of the century and the establishment of newer treatments and adjustment in public health advertising, a random sample of 10,000 people exhibits only six deaths due to lung cancer. Test at the .05 significance level whether that data is proof of a reduced fatality rate for lung cancer.

60. The American Water Works Association reports that the per capita water use in a single- family home is 69 gallons per day. Legacy Ranch is a relatively new housing development consisting of 100 homes. The builders installed more efficient water fixtures, such as low- flush toilets, and subsequently conducted a survey of the residences. Thirty-six homes responded, and the sample mean water use per day was 64 gallons with a standard devi- ation of 8.8 gallons per day. At the .10 level of significance, is that enough evidence to conclude that residents of Legacy Ranch use less water on average?

61. A cola-dispensing machine is set to dispense 9.00 ounces of cola per cup, with a stan- dard deviation of 1.00 ounces. The manufacturer of the machine would like to set the control limit in such a way that, for samples of 36, 5 percent of the sample means will be greater than the upper control limit, and 5 percent of the sample means will be less than the lower control limit. a. At what value should the control limit be set? b. What is the probability that if the population mean shifts to 8.9, this change will not

be detected? c. What is the probability that if the population mean shifts to 9.3, this change will not

be detected? 62. The owners of the Westfield Mall wished to study customer shopping habits. From earlier

studies, the owners were under the impression that a typical shopper spends 0.75 hours at the mall, with a standard deviation of 0.10 hours. Recently the mall owners added some specialty restaurants designed to keep shoppers in the mall longer. The consulting firm, Brunner and Swanson Marketing Enterprises, was hired to evaluate the effects of

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368 Chapter 10

the restaurants. A sample of 45 shoppers by Brunner and Swanson revealed that the mean time spent in the mall had increased to 0.80 hours. a. Develop a test of hypothesis to determine if the mean time spent in the mall is more

than 0.75 hours. Use the .05 significance level. b. Suppose the mean shopping time actually increased from 0.75 hours to 0.77 hours.

What is the probability this increase would not be detected? c. When Brunner and Swanson reported the information in part (b) to the mall own-

ers, the owners were upset with the statement that a survey could not detect a change from 0.75 to 0.77 hours of shopping time. How could this probability be reduced?

63. The following null and alternate hypotheses are given.

Suppose the population standard deviation is 10. The probability of a Type I error is set at .01 and the probability of a Type II error at .30. Assume that the population mean shifts from 50 to 55. How large a sample is necessary to meet these requirements?

64. An insurance company, based on past experience, estimates the mean damage for a nat- ural disaster in its area is $5,000. After introducing several plans to prevent loss, it ran- domly samples 200 policyholders and finds the mean amount per claim was $4,800 with a standard deviation of $1,300. Does it appear the prevention plans were effective in reducing the mean amount of a claim? Use the .05 significance level.

65. A national grocer’s magazine reports the typical shopper spends eight minutes in line waiting to check out. A sample of 24 shoppers at the local Farmer Jack’s showed a mean of 7.5 minutes with a standard deviation of 3.2 minutes. Is the waiting time at the local Farmer Jack’s less than that reported in the national magazine? Use the .05 significance level.

Data Set Exercises 66. Refer to the Real Estate data, which report information on the homes sold in Goodyear,

Arizona, last year. a. A recent article in the Arizona Republic indicated that the mean selling price of the

homes in the area is more than $220,000. Can we conclude that the mean selling price in the Goodyear, AZ, area is more than $220,000? Use the .01 significance level. What is the p-value?

b. The same article reported the mean size was more than 2,100 square feet. Can we conclude that the mean size of homes sold in the Goodyear, AZ, area is more than 2,100 square feet? Use the .01 significance level. What is the p-value?

c. Determine the proportion of homes that have an attached garage. At the .05 signifi- cance level, can we conclude that more than 60 percent of the homes sold in the Goodyear, AZ, area had an attached garage? What is the p-value?

d. Determine the proportion of homes that have a pool. At the .05 significance level, can we conclude that more than 60 percent of the homes sold in the Goodyear, AZ, area had a pool? What is the p-value?

67. Refer to the Baseball 2009 data, which report information on the 30 Major League Base- ball teams for the 2009 season. a. Conduct a test of hypothesis to determine whether the mean salary of the teams was

different from $80.0 million. Use the .05 significance level. b. Conduct a test of hypothesis to determine whether the mean attendance was more

than 2,000,000 per team. 68. Refer to the Buena School District bus data.

a. Select the variable for the number of miles traveled last month. Conduct a test of hypothesis to determine whether the mean number of miles traveled is equal to 840. Use the .01 significance level. Find the p-value and explain what it means.

b. Using the maintenance cost variable, conduct a test of hypothesis to determine whether the mean maintenance cost is less than $500 at the .05 significance level. Determine the p-value and interpret the result.

c. Suppose we consider a bus “old” if it is more than eight years old. At the .01 significance level, can we conclude that less than 40 percent of the buses are old? Report the p-value.

H1 : � 7 50 H0 : � � 50

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One-Sample Tests of Hypothesis 369

1. The Minitab commands for the histogram and the de- scriptive statistics on page 349 are: a. Enter the 26 sample observations in column C1

and name the variable Cost. b. From the menu bar, select Stat, Basic Statistics,

and Graphical Summary. In the dialog box, select Cost as the variable and click OK.

Software Commands 2. The Minitab commands for the one-sample t test on

page 353 are: a. Enter the sample data into column C1 and name

the variable Length. b. From the menu bar, select Stat, Basic Statistics,

and 1-Sample t, and then hit Enter. c. Select Length as the variable, select Test mean,

insert the number 43, and click OK.

10–1 a. b. .05

d. Reject H0 if .

f. Do not reject H0. g. We cannot conclude the mean amount

dispensed is different from 16.0 ounces. 10–2 a.

b. Reject H0 if .

d. Reject H0. e. The mean amount dispensed is more than 16.0

ounces. f. p-value The p-value

is less than � (.05), so H0 is rejected. It is the same conclusion as in part (d).

� .5000 .4706 � .0294.

z � 16.040 16.0

0.15�150 �

.0400

.0212 � 1.89

z 7 1.65 H0: � � 16.0; H1: � 7 16.0

z � 16.017 16.0

0.15�150 �

0.0170 0.0212

� 0.80

z 6 1.96 or z 7 1.96

z � X � ��1n

H0: � � 16.0; H1: � � 16.0

Chapter 10 Answers to Self-Review

10–3 a. b. The decision rule is to reject H0 if t � 1.729.

df � n 1 � 20 1 � 19 H0: � � 305; H1: � 7 305.

0 1.729 t Critical value

Do not reject H0

Region of rejection

α � .05

Reject H0 because 2.236 � 1.729. The modification increased the mean battery life to more than 305 days.

t � X � s�1n

� 311 305

12�120 � 2.236

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370 Chapter 10

10–4 a. ; . b. 7, found by . c. Reject H0 if t 6 2.998.

n 1 � 8 1 � 7 H1: � 6 9.0H0: � 9.0

d. , found by:

Then

Since lies to the right of H0 is not rejected. We have not shown that the mean is less than 9.0. e. The p-value is between .025 and .010.

10–5 a. Yes, because both and exceed 5: , and

.n(1 �) � 200(.60) � 120 n� � 200(.40) � 80

n(1 �)n�

2.998, 2.494

t � 8.8 9.0

0.2268�18 � 2.494

X � 70.4

8 � 8.8

s � A

0.36 8 1

� 0.2268

t � 2.494

c. Reject H0 if z � 2.33. H1 : � 6 .40 H0 : � .40

–2.998 0 Scale of t Critical value

Do not reject H0

Region of rejection

Do not reject e. The p-value is .1922, found by

10–6 .0054, found by determining the area under the curve between 10,078 and 10,180.

The area under the curve for a z of is .4946 (Appendix B.1), and ..5000 .4946 � .0054

2.55

� 10,078 10,180

400�1100 � 2.55

z � Xc �1 ��1n

.5000 .3078. H0.

z � .37 .40

.40(1 .40) 200

� .03 1.0012

� 0.87

z � 0.87, found by:

Region of rejection � � .01

0 2.33 Critical value

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11 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Test a hypothesis that two independent population means with known population standard deviations are equal.

LO2 Carry out a hypothesis test that two population proportions are equal.

LO3 Conduct a test of a hypothesis that two independent population means are equal, assuming equal but unknown population standard deviations.

LO4 Conduct a test of a hypothesis that two independent population means are equal, assuming unequal but unknown population standard deviations.

LO5 Explain the difference between dependent and independent samples.

LO6 Carry out a test of a hypothesis about the mean difference between paired and dependent observations.

Two-Sample Tests of Hypothesis

The Damon family owns a large grape vineyard in western New York

along Lake Erie. The grapevines must be sprayed at the beginning of the

growing season to protect against various insects and diseases. Two new

insecticides have just been marketed, Pernod 5 and Action. When the

grapes ripened, 400 of the vines treated with Pernod 5 were checked for

infestation, and 400 of the vines treated with Action were checked.

Referring to the table in the text, at the .05 significance level, can we

conclude that there is a difference in the proportion of vines infested

using Pernod 5 as opposed to Action? (See Exercise 9 and LO2.)

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11.1 Introduction Chapter 10 began our study of hypothesis testing. We described the nature of hypothesis testing and conducted tests of a hypothesis in which we compared the results of a single sample to a population value. That is, we selected a single ran- dom sample from a population and conducted a test of whether the proposed pop- ulation value was reasonable. Recall in Chapter 10 that we selected a sample of the number of desks assembled per week at Jamestown Steel Company to determine whether there was a change in the production rate. Similarly, we sampled voters in

one area of a particular state to determine whether the population proportion that would support the governor for reelection was less than .80. In both of these cases, we compared the results of a single sam- ple statistic to a population parameter.

In this chapter, we expand the idea of hypothesis testing to two samples. That is, we select random samples from two different populations to determine whether the popu- lation means or proportions are equal. Some questions we might want to test are:

1. Is there a difference in the mean value of residential real estate sold by male agents and female agents in south Florida?

2. Is there a difference in the mean number of defects produced on the day and the afternoon shifts at Kimble Products?

3. Is there a difference in the mean number of days absent between young work- ers (under 21 years of age) and older workers (more than 60 years of age) in the fast-food industry?

4. Is there a difference in the proportion of Ohio State University graduates and University of Cincinnati graduates who pass the state Certified Public Accoun- tant Examination on their first attempt?

5. Is there an increase in the production rate if music is piped into the production area?

We begin this chapter with the case in which we select random samples from two independent populations and wish to investigate whether these populations have the same mean.

11.2 Two-Sample Tests of Hypothesis: Independent Samples A city planner in Florida wishes to know whether there is a difference in the mean hourly wage rate of plumbers and electricians in central Florida. A financial accoun- tant wishes to know whether the mean rate of return for high yield mutual funds is different from the mean rate of return on global mutual funds. In each of these cases, there are two independent populations. In the first case, the plumbers represent one population and the electricians the other. In the second case, high-yield mutual funds are one population and global mutual funds the other.

In each of these cases, to investigate the question, we would select a random sample from each population and compute the mean of the two samples. If the two population means are the same, that is, the mean hourly rate is the same for the plumbers and the electricians, we would expect the difference between the two sam- ple means to be zero. But what if our sample results yield a difference other than

372 Chapter 11

Statistics in Action

The U.S. presidential election of 2000 turned out to be one of the closest in his- tory. The news media were unable to pro- ject a winner, and the final decision, in- cluding recounts and court decisions, took more than five weeks. This was not the only election in which there was controversy. Shortly before the 1936 presidential election, The New York Times carried the headline: “Digest Poll Gives Landon 32 States: Landon Leads 4–3.” However, Alfred Landon of Kansas was not elected president. In fact, Roosevelt won by more than 11 million votes and received 523 Electoral College votes. How could the headline have been so wrong?

The Literary Digest collected a sample of voters from lists of telephone numbers, automobile registrations, and Di- gest readers. In 1936 not many people could afford a tele- phone or an automo- bile. In addition, those who read the Digest tended to be

(continued)

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Two-Sample Tests of Hypothesis 373

wealthier and vote Republican. Thus, the population that was sampled did not represent the popula- tion of voters. A sec- ond problem was with the nonre- sponses. More than 10 million people were sent surveys, and more than 2.3 million responded. However, no attempt was made to see whether those responding repre- sented a cross-section of all the voters.

With modern com- puters and survey methods, samples are carefully selected and checked to ensure they are representa- tive. What happened to the Literary Di- gest? It went out of business shortly after the 1936 election.

zero? Is that difference due to chance or is it because there is a real difference in the hourly earnings? A two-sample test of means will help to answer this question.

We do need to return to the results of Chapter 8. Recall that we showed that a distribution of sample means would tend to approximate the normal distribution. We need to again assume that a distribution of sample means will follow the nor- mal distribution. It can be shown mathematically that the distribution of the differ- ences between sample means for two normal distributions is also normal.

We can illustrate this theory in terms of the city planner in Tampa, Florida. To begin, let’s assume some information that is not usually available. Suppose that the population of plumbers has a mean of $30.00 per hour and a standard deviation of $5.00 per hour. The population of electricians has a mean of $29.00 and a standard deviation of $4.50. Now, from this information it is clear that the two population means are not the same. The plumbers actually earn $1.00 per hour more than the electricians. But we cannot expect to uncover this difference each time we sample the two populations.

Suppose we select a random sample of 40 plumbers and a random sample of 35 electricians and compute the mean of each sample. Then, we determine the dif- ference between the sample means. It is this difference between the sample means that holds our interest. If the populations have the same mean, then we would expect the difference between the two sample means to be zero. If there is a differ- ence between the population means, then we expect to find a difference between the sample means.

To understand the theory, we need to take several pairs of samples, compute the mean of each, determine the difference between the sample means, and study the distribution of the differences in the sample means. Because of our study of the distribution of sample means in Chapter 8, we know that the distribution of the sam- ple means follows the normal distribution. If the two distributions of sample means follow the normal distribution, then we can reason that the distribution of their dif- ferences will also follow the normal distribution. This is the first hurdle.

The second hurdle refers to the mean of this distribution of differences. If we find the mean of this distribution is zero, that implies that there is no difference in the two populations. On the other hand, if the mean of the distribution of differ- ences is equal to some value other than zero, either positive or negative, then we conclude that the two populations do not have the same mean.

To report some concrete results, let’s return to the city planner in Tampa, Florida. Table 11–1 shows the result of selecting 20 different samples of 40 plumbers and 35 electricians, computing the mean of each sample, and finding the difference between the two sample means. In the first case, the sample of 40 plumbers has a mean of $29.80, and for the 35 electricians the mean is $28.76. The difference between the sample means is $1.04. This process was repeated 19 more times. Observe that in 17 of the 20 cases the mean of the plumbers is larger than the mean of the electricians.

Our final hurdle is that we need to know something about the variability of the distribution of differences. To put it another way, what is the standard deviation of this distribution of differences? Statistical theory shows that when we have inde- pendent populations, such as the case here, the distribution of the differences has a variance (standard deviation squared) equal to the sum of the two individual vari- ances. This means that we can add the variances of the two sampling distribu- tions. To put it another way, the variance of the difference in sample means is equal to the sum of the variance for the plumbers and the variance for the electricians.

VARIANCE OF THE DISTRIBUTION [11–1]�2 X1�X2

� �21 n1

� �22 n2

(X1 � X2)

OF DIFFERENCES IN MEANS

LO1 Test a hypothesis that two independent population means with known population standard deviations are equal.

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374 Chapter 11

Sample Plumbers Electricians Difference

1 $29.80 $28.76 $1.04 2 30.32 29.40 0.92 3 30.57 29.94 0.63 4 30.04 28.93 1.11 5 30.09 29.78 0.31 6 30.02 28.66 1.36 7 29.60 29.13 0.47 8 29.63 29.42 0.21 9 30.17 29.29 0.88

10 30.81 29.75 1.06 11 30.09 28.05 2.04 12 29.35 29.07 0.28 13 29.42 28.79 0.63 14 29.78 29.54 0.24 15 29.60 29.60 0.00 16 30.60 30.19 0.41 17 30.79 28.65 2.14 18 29.14 29.95 �0.81 19 29.91 28.75 1.16 20 28.74 29.21 �0.47

TABLE 11–1 The Means of Random Samples of Plumbers and Electricians

The term looks complex but need not be difficult to interpret. The por- tion reminds us that it is a variance, and the subscript that it is a distribu- tion of differences in the sample means.

We can put this equation in a more usable form by taking the square root, so that we have the standard deviation of the distribution or “standard error” of the dif- ferences. Finally, we standardize the distribution of the differences. The result is the following equation.

TWO-SAMPLE TEST OF [11–2]

Before we present an example, let’s review the assumptions necessary for using formula (11–2).

• The two populations follow normal distributions. • The two samples must be unrelated, that is, independent. • The standard deviations for both populations must be known.

The following example shows the details of the test of hypothesis for two pop- ulation means.

z � X1 � X2

�21 n1

� �22 n2

X1 � X2

�2� 2 X1�X2

MEANS—KNOWN �

Example Customers at the FoodTown Supermarket have a choice when paying for their gro- ceries. They may check out and pay using the standard cashier-assisted checkout, or they may use the new Fast Lane procedure. In the standard procedure, a Food- Town employee scans each item, puts it on a short conveyor where another employee puts it in a bag and then into the grocery cart. In the Fast Lane procedure, the customer scans each item, bags it, and places the bags in the cart themselves.

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Two-Sample Tests of Hypothesis 375

Solution We use the five-step hypothesis testing procedure to investigate the question. Step 1: State the null hypothesis and the alternate hypothesis. The null

hypothesis is that the mean checkout times for the two groups are equal. In other words, the difference of 0.20 minutes between the mean checkout time for the standard method and the mean checkout time for Fast Lane is due to chance. The alternate hypothesis is that the mean checkout time is larger for those using the standard method. We will let refer to the mean checkout time for the population of stan- dard customers and the mean checkout time for the Fast Lane cus- tomers. The null and alternative hypotheses are:

Step 2: Select the level of significance. The significance level is the probability that we reject the null hypothesis when it is actually true. This likelihood is determined prior to selecting the sample or performing any calcula- tions. The .05 and .01 significance levels are the most common, but other values, such as .02 and .10, are also used. In theory, we may select any value between 0 and 1 for the significance level. In this case, we selected the .01 significance level.

Step 3: Determine the test statistic. In Chapter 10, we used the standard nor- mal distribution (that is z) and t as test statistics. In this case, we use the z distribution as the test statistic because we assume the two population distributions are both normal and the standard deviations of both popu- lations are known.

Step 4: Formulate a decision rule. The decision rule is based on the null and the alternate hypotheses (i.e., one-tailed or two-tailed test), the level of sig- nificance, and the test statistic used. We selected the .01 significance level and the z distribution as the test statistic, and we wish to determine whether the mean checkout time is longer using the standard method. We set the alternate hypothesis to indicate that the mean checkout time is longer for those using the standard method than the Fast Lane method. Hence, the rejection region is in the upper tail of the standard normal distribution (a one-tailed test). To find the critical value, place .01 of the total area in the upper tail. This means that .4900 ( ) of the area is located between the z value of 0 and the critical value. Next,

.5000 � .0100

H1: �s 7 �f H0: �s � �f

�f

�s

The Fast Lane procedure is designed to reduce the time a cus- tomer spends in the checkout line.

The Fast Lane facility was recently installed at the Byrne Road FoodTown location. The store manager would like to know if the mean checkout time using the standard checkout method is longer than using the Fast Lane. She gathered the following sample information. The time is measured from when the customer enters the line until their bags are in the cart. Hence the time includes both waiting in line and checking out. What is the p-value?

Population Standard Customer Type Sample Mean Deviation Sample Size

Standard 5.50 minutes 0.40 minutes 50 Fast Lane 5.30 minutes 0.30 minutes 100

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376 Chapter 11

In summary, the criteria for using formula (11–2) are:

1. The samples are from independent populations. This means the checkout time for the Fast Lane customers is unrelated to the checkout time for the other cus- tomers. For example, Mr. Smith’s checkout time does not affect any other customer’s checkout time.

2. Both populations follow the normal distribution. In the FoodTown example, this means the population of times in both the standard checkout line and the Fast Lane follow the normal distribution.

3. Both population standard deviations are known. In the FoodTown example, the population standard deviation of the Fast Lane times was 0.30 minutes. The standard deviation of the standard checkout times was 0.40 minutes.

Statistics in Action

Do you live to work or work to live? A recent poll of 802 working Americans revealed that, among those who considered their work as a career, the mean number of hours worked per day was 8.7. Among those who considered their work as a job, the mean number of hours worked per day was 7.6.

we search the body of Appendix B.1 for a value located near .4900. It is 2.33, so our decision rule is to reject if the value computed from the test statistic exceeds 2.33. Chart 11–1 depicts the decision rule.

Region of rejection

.01

Scale of z

.5000 .4900

2.33 Critical value

H0: μs ≤ μf H1: μs > μf

CHART 11–1 Decision Rule for One-Tailed Test at .01 Significance Level

Step 5: Make the decision regarding H0 and interpret the result. We use formula (11–2) to compute the value of the test statistic.

The computed value of 3.13 is larger than the critical value of 2.33. Our decision is to reject the null hypothesis and accept the alternate hypoth- esis. The difference of .20 minutes between the mean checkout time using the standard method is too large to have occurred by chance. To put it another way, we conclude the Fast Lane method is faster.

What is the p-value for the test statistic? Recall that the p-value is the probability of finding a value of the test statistic this extreme when the null hypothesis is true. To calculate the p-value, we need the probability of a z value larger than 3.13. From Appendix B.1 we can- not find the probability associated with 3.13. The largest value available is 3.09. The area corresponding to 3.09 is .4990. In this case, we can report that the p-value is less than .0010, found by We conclude that there is very little likelihood that the null hypothesis is true!

.5000 � .4990.

z � Xs � Xf

� 2s

ns �

� 2f nf

� 5.5 � 5.3

0.402

50 �

0.302

100

� 0.2

0.064 � 3.13

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Two-Sample Tests of Hypothesis 377

Self-Review 11–1 Tom Sevits is the owner of the Appliance Patch. Recently Tom observed a difference in the dollar value of sales between the men and women he employs as sales associates. A sample of 40 days revealed the men sold a mean of $1,400 worth of appliances per day. For a sample of 50 days, the women sold a mean of $1,500 worth of appliances per day. Assume the population standard deviation for men is $200 and for women $250. At the .05 significance level, can Mr. Sevits conclude that the mean amount sold per day is larger for the women? (a) State the null hypothesis and the alternate hypothesis. (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding the null hypothesis? (e) What is the p-value? (f ) Interpret the result.

Exercises 1. A sample of 40 observations is selected from one population with a population standard

deviation of 5. The sample mean is 102. A sample of 50 observations is selected from a second population with a population standard deviation of 6. The sample mean is 99. Conduct the following test of hypothesis using the .04 significance level.

a. Is this a one-tailed or a two-tailed test? b. State the decision rule. c. Compute the value of the test statistic. d. What is your decision regarding ? e. What is the p-value?

2. A sample of 65 observations is selected from one population with a population standard deviation of 0.75. The sample mean is 2.67. A sample of 50 observations is selected from a second population with a population standard deviation of 0.66. The sample mean is 2.59. Conduct the following test of hypothesis using the .08 significance level.

a. Is this a one-tailed or a two-tailed test? b. State the decision rule. c. Compute the value of the test statistic. d. What is your decision regarding H0? e. What is the p-value?

Note: Use the five-step hypothesis testing procedure to solve the following exercises.

3. Gibbs Baby Food Company wishes to compare the weight gain of infants using its brand versus its competitor’s. A sample of 40 babies using the Gibbs products revealed a mean weight gain of 7.6 pounds in the first three months after birth. For the Gibbs brand, the population standard deviation of the sample is 2.3 pounds. A sample of 55 babies using the competitor’s brand revealed a mean increase in weight of 8.1 pounds. The population standard deviation is 2.9 pounds. At the .05 significance level, can we con- clude that babies using the Gibbs brand gained less weight? Compute the p-value and interpret it.

4. As part of a study of corporate employees, the director of human resources for PNC Inc. wants to compare the distance traveled to work by employees at its office in downtown Cincinnati with the distance for those in downtown Pittsburgh. A sample of 35 Cincinnati employees showed they travel a mean of 370 miles per month. A sample of 40 Pittsburgh employees showed they travel a mean of 380 miles per month. The population standard deviation for the Cincinnati and Pittsburgh employees

H1: �1 7 �2 H0: �1 � �2

H1: �1 � �2

H0: �1 � �2

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378 Chapter 11

are 30 and 26 miles, respectively. At the .05 significance level, is there a difference in the mean number of miles traveled per month between Cincinnati and Pittsburgh employees?

5. Women’s height is a suspected factor for difficult deliveries, that is, shorter women are more likely to have Caesarean sections. A medical researcher found in a sample of 45 women who had a normal delivery that their mean height was 61.4 inches. A second sample of 39 women who had a Caesarean section had a mean height of 60.6 inches. Assume that the population of heights of normal deliveries has a population standard deviation of 1.2 inches. Also assume that the heights of the population of women who had Caesarean section births has a standard deviation of 1.1 inches. Are those who had a Caesarean section shorter? Use the .05 significance level. Find the p-value and explain what it means.

6. Mary Jo Fitzpatrick is the vice president for Nursing Services at St. Luke’s Memorial Hospital. Recently she noticed in the job postings for nurses that those that are union- ized seem to offer higher wages. She decided to investigate and gathered the following information.

Population Group Mean Wage Standard Deviation Sample Size

Union $20.75 $2.25 40 Nonunion $19.80 $1.90 45

OF PROPORTIONS

z � p1 � p2

pc(1 � pc) n1

� pc(1 � pc)

LO2 Carry out a hypothesis test that two population proportions are equal.

Would it be reasonable for her to conclude that union nurses earn more? Use the .02 significance level. What is the p-value?

11.3 Two-Sample Tests about Proportions In the previous section, we considered a test involving population means. However, we are often interested also in whether two sample proportions come from popula- tions that are equal. Here are several examples.

• The vice president of human resources wishes to know whether there is a dif- ference in the proportion of hourly employees who miss more than five days of work per year at the Atlanta and the Houston plants.

• General Motors is considering a new design for the Chevy Malibu. The design is shown to a group of potential buyers under 30 years of age and another group over 60 years of age. General Motors wishes to know whether there is a difference in the proportion of the two groups who like the new design.

• A consultant to the airline industry is investigating the fear of flying among adults. Specifically, the company wishes to know whether there is a difference in the proportion of men versus women who are fearful of flying.

In the above cases, each sampled item or individual can be classified as a “suc- cess” or a “failure.” That is, in the Chevy Malibu example each potential buyer is clas- sified as “liking the new design” or “not liking the new design.” We then compare the proportion in the under 30 group with the proportion in the over 60 group who indicated they liked the new design. Can we conclude that the differences are due to chance? In this study, there is no measurement obtained, only classifying the indi- viduals or objects.

To conduct the test, we assume each sample is large enough that the normal distribution will serve as a good approximation of the binomial distribution. The test statistic follows the standard normal distribution. We compute the value of z from the following formula:

TWO-SAMPLE TEST [11–3]

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Two-Sample Tests of Hypothesis 379

Example Manelli Perfume Company recently developed a new fragrance that it plans to market under the name Heavenly. A number of market stud- ies indicate that Heavenly has very good mar- ket potential. The Sales Department at Manelli is particularly interested in whether there is a difference in the proportions of younger and older women who would purchase Heavenly if it were marketed. There are two independent populations, a population consisting of the younger women and a population consisting of the older women. Each sampled woman will be asked to smell Heavenly and indicate whether she likes the fragrance well enough to purchase a bottle.

Solution We will use the usual five-step hypothesis-testing procedure. Step 1: State H0 and H1. In this case, the null hypothesis is: “There is no dif-

ference in the proportion of young women and older women who prefer Heavenly.” We designate 1 as the proportion of young women who would purchase Heavenly and 2 as the proportion of older women who would purchase it. The alternate hypothesis is that the two proportions are not equal.

Step 2: Select the level of significance. We choose the .05 significance level in this example.

Step 3: Determine the test statistic. The test statistic follows the standard nor- mal distribution. The value of the test statistic can be computed from formula (11–3).

Step 4: Formulate the decision rule. Recall that the alternate hypothesis from Step 1 does not state a direction, so this is a two-tailed test. To determine the critical value, we divide the significance level in half and place this amount in each tail of the z distribution. Next, we subtract this amount from the total area to the right of zero. That is .5000 � .0250 � .4750.

H1: 1 � 2 H0: 1 � 2

Formula (11–3) is formula (11–2) with the respective sample proportions replacing the sample means and pc(1 � pc) replacing the two variances. In addition:

n1 is the number of observations in the first sample. n2 is the number of observations in the second sample. p1 is the proportion in the first sample possessing the trait. p2 is the proportion in the second sample possessing the trait. pc is the pooled proportion possessing the trait in the combined samples. It is

called the pooled estimate of the population proportion and is computed from the following formula.

POOLED PROPORTION [11–4]

where: X1 is the number possessing the trait in the first sample. X2 is the number possessing the trait in the second sample.

The following example will illustrate the two-sample test of proportions.

pc � X1 � X2 n1 � n2

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380 Chapter 11

–1.96

H0: π1 � π2 H1: π1 � π2

H0 is not rejected

H0 is rejected

.025 .95

.025

H0 is rejected

1.96 Scale of z

CHART 11–2 Decision Rules for Heavenly Fragrance Test, .05 Significance Level

Step 5: Select a sample and make a decision. A random sample of 100 young women revealed 19 liked the Heavenly fragrance well enough to pur- chase it. Similarly, a sample of 200 older women revealed 62 liked the fragrance well enough to make a purchase. We let p1 refer to the young women and p2 to the older women.

The research question is whether the difference of .12 in the two sample proportions is due to chance or whether there is a difference in the pro- portion of younger and older women who like the Heavenly fragrance.

Next, we combine or pool the sample proportions. We use formula (11–4).

Note that the pooled proportion is closer to .31 than to .19 because more older women than younger women were sampled.

We use formula (11–3) to find the value of the test statistic.

The computed value of �2.21 is in the area of rejection; that is, it is to the left of �1.96. Therefore, the null hypothesis is rejected at the .05 significance level. To put it another way, we reject the null hypothesis that the proportion of young women who would purchase Heavenly is equal to the proportion of older women who would purchase Heavenly. It is unlikely that the difference between the two sample pro- portions is due to chance. To find the p-value, go to Appendix B.1 and find the likelihood of a z value less than �2.21 or greater than 2.21. The z value corresponding

� �2.21� .19 � .31

.27(1 � .27) 100

� .27(1 � .27)

200

z � p1 � p2

pc(1 � pc) n1

� pc(1 � pc )

pc � X1 � X2 n1 � n2

� 19 � 62

100 � 200 �

81 300

� 0.27

p1 � X1 n1

� 19

100 � .19 p2 �

X2 n2

� 62

200 � .31

Finally, we search the body of the z table (Appendix B.1) for the closest value. It is 1.96. The critical values are �1.96 and �1.96. As before, if the computed z value falls in the region between �1.96 and �1.96, the null hypothesis is not rejected. If that does occur, it is assumed that any difference between the two sample proportions is due to chance varia- tion. This information is summarized in Chart 11–2.

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Two-Sample Tests of Hypothesis 381

Notice the Minitab output includes the two sample proportions, the value of z, and the p-value.

to 2.21 is .4864. So the likelihood of finding the value of the test statistic to be less than �2.21 or greater than 2.21 is:

The p-value of .0272 is less than the significance level of .05, so our decision is to reject the null hypothesis. Again, we conclude that there is a difference in the proportion of younger and older women who would purchase Heavenly.

The Minitab system has a procedure to quickly determine the value of the test statistic and compute the p-value. The results follow.

p-value � 2(.5000 � .4864) � 2(.0136) � .0272

Self-Review 11–2 Of 150 adults who tried a new peach-flavored Peppermint Pattie, 87 rated it excellent. Of 200 children sampled, 123 rated it excellent. Using the .10 level of significance, can we conclude that there is a significant difference in the proportion of adults and the propor- tion of children who rate the new flavor excellent? (a) State the null hypothesis and the alternate hypothesis. (b) What is the probability of a Type I error? (c) Is this a one-tailed or a two-tailed test? (d) What is the decision rule? (e) What is the value of the test statistic? (f) What is your decision regarding the null hypothesis? (g) What is the p-value? Explain what it means in terms of this problem.

Exercises 7. The null and alternate hypotheses are:

A sample of 100 observations from the first population indicated that X1 is 70. A sample of 150 observations from the second population revealed X2 to be 90. Use the .05 sig- nificance level to test the hypothesis. a. State the decision rule. b. Compute the pooled proportion. c. Compute the value of the test statistic. d. What is your decision regarding the null hypothesis?

8. The null and alternate hypotheses are:

H1: 1 � 2

H0: 1 � 2

H1: 1 7 2 H0: 1 � 2

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382 Chapter 11

Number of Vines Checked Number of

Insecticide (sample size) Infested Vines

Pernod 5 400 24 Action 400 40

At the .05 significance level, can we conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action?

10. GfK Custom Research North America conducted identical surveys five years apart. One question asked of women was “Are most men basically kind, gentle, and thoughtful?” The earlier survey revealed that, of the 3,000 women surveyed, 2,010 said that they were. The later revealed 1,530 of the 3,000 women surveyed thought that men were kind, gentle, and thoughtful. At the .05 level, can we conclude that women think men are less kind, gentle, and thoughtful in the later survey compared with the earlier one?

11. A nationwide sample of influential Republicans and Democrats was asked as a part of a comprehensive survey whether they favored lowering environmental standards so that high-sulfur coal could be burned in coal-fired power plants. The results were:

A sample of 200 observations from the first population indicated that X1 is 170. A sam- ple of 150 observations from the second population revealed X2 to be 110. Use the .05 significance level to test the hypothesis. a. State the decision rule. b. Compute the pooled proportion. c. Compute the value of the test statistic. d. What is your decision regarding the null hypothesis?

Note: Use the five-step hypothesis-testing procedure in solving the following exercises.

9. The Damon family owns a large grape vineyard in western New York along Lake Erie. The grapevines must be sprayed at the beginning of the growing season to protect against various insects and diseases. Two new insecticides have just been marketed: Pernod 5 and Action. To test their effectiveness, three long rows were selected and sprayed with Pernod 5, and three others were sprayed with Action. When the grapes ripened, 400 of the vines treated with Pernod 5 were checked for infestation. Likewise, a sample of 400 vines sprayed with Action were checked. The results are:

At the .02 level of significance, can we conclude that there is a larger proportion of Democrats in favor of lowering the standards? Determine the p-value.

12. The research department at the home office of New Hampshire Insurance conducts on- going research on the causes of automobile accidents, the characteristics of the drivers, and so on. A random sample of 400 policies written on single persons revealed 120 had at least one accident in the previous three-year period. Similarly, a sample of 600 poli- cies written on married persons revealed that 150 had been in at least one accident. At the .05 significance level, is there a significant difference in the proportions of single and married persons having an accident during a three-year period? Determine the p-value.

11.4 Comparing Population Means with Unknown Population Standard Deviations In the previous two sections, we described conditions where the standard normal dis- tribution, that is z, is used as the test statistic. In one case, we were working with a variable (calculating the mean) and in the second an attribute (calculating a propor- tion). In the first case, we wished to compare two sample means from independent populations to determine if they came from the same or equal populations. In that

Republicans Democrats

Number sampled 1,000 800 Number in favor 200 168

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Two-Sample Tests of Hypothesis 383

instance, we assumed the population followed the normal probability distribution and that we knew the standard deviation of the population. In many cases, in fact in most cases, we do not know the population standard deviation. We can overcome this problem, as we did in the one sample case in the previous chapter, by substituting the sample standard deviation (s) for the population standard deviation (�). See for- mula (10–2) on page 348.

Equal Population Standard Deviations This section describes another method for comparing the sample means of two independent populations to determine if the sampled populations could reasonably have the same mean. The method described does not require that we know the standard deviations of the populations. This gives us a great deal more flexibility when investigating the difference in sample means. There are two major differences in this test and the previous test described earlier in this chapter.

1. We assume the sampled populations have equal but unknown standard devia- tions. Because of this assumption, we combine or “pool” the sample standard deviations.

2. We use the t distribution as the test statistic.

The formula for computing the value of the test statistic t is similar to (11–2), but an additional calculation is necessary. The two sample standard deviations are pooled to form a single estimate of the unknown population standard deviation. In essence, we compute a weighted mean of the two sample standard deviations and use this value as an estimate of the unknown population standard deviation. The weights are the degrees of freedom that each sample provides. Why do we need to pool the sample standard deviations? Because we assume that the two popula- tions have equal standard deviations, the best estimate we can make of that value is to combine or pool all the sample information we have about the value of the population standard deviation.

The following formula is used to pool the sample standard deviations. Notice that two factors are involved: the number of observations in each sample and the sample standard deviations themselves.

POOLED VARIANCE [11–5]

where: is the variance (standard deviation squared) of the first sample.

is the variance of the second sample.

The value of t is computed from the following equation.

TWO-SAMPLE TEST OF MEANS—UNKNOWN �’S [11–6]

where: is the mean of the first sample. is the mean of the second sample.

n1 is the number of observations in the first sample. n2 is the number of observations in the second sample.

is the pooled estimate of the population variance.s2p

t � X1 � X2

B s2pa

1 n1

� 1 n2 b

s22

s21

s2p � (n1 � 1)s

2 1 � (n2 � 1)s

2 2

n1 � n2 � 2

LO3 Conduct a test of a hypothesis that two independent population means are equal, assuming equal but unknown population standard deviations.

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384 Chapter 11

The number of degrees of freedom in the test is the total number of items sampled minus the total number of samples. Because there are two samples, there are n1 � n2 � 2 degrees of freedom.

To summarize, there are three requirements or assumptions for the test.

1. The sampled populations follow the normal distribution. 2. The sampled populations are independent. 3. The standard deviations of the two populations are equal.

The following example/solution explains the details of the test.

Example Owens Lawn Care Inc. manufactures and assembles lawnmowers that are shipped to dealers throughout the United States and Canada. Two different pro- cedures have been proposed for mounting the engine on the frame of the lawn- mower. The question is: Is there a difference in the mean time to mount the engines on the frames of the lawnmowers? The first procedure was developed by longtime Owens employee Herb Welles (designated as procedure 1), and the other procedure was developed by Owens Vice President of Engineering William Atkins (designated as procedure 2). To evaluate the two methods, it was decided to con- duct a time and motion study. A sample of five employees was timed using the Welles method and six using the Atkins method. The results, in minutes, are shown below. Is there a difference in the mean mounting times? Use the .10 significance level.

Solution Following the five steps to test a hypothesis, the null hypothesis states that there is no difference in mean mounting times between the two procedures. The alternate hypothesis indicates that there is a difference.

The required assumptions are:

• The observations in the Welles sample are independent of the observations in the Atkins sample.

• The two populations follow the normal distribution. • The two populations have equal standard deviations.

Is there a difference between the mean assembly times using the Welles and the Atkins methods? The degrees of freedom are equal to the total number of items sampled minus the number of samples. In this case, that is n1 � n2 � 2. Five assemblers used the Welles method and six the Atkins method. Thus, there are 9 degrees of freedom, found by 5 � 6 � 2. The critical values of t, from Appen- dix B.2 for df � 9, a two-tailed test, and the .10 significance level, are �1.833 and 1.833. The decision rule is portrayed graphically in Chart 11–3. We do not reject the null hypothesis if the computed value of t falls between �1.833 and 1.833.

H1: �1 � �2

H0: �1 � �2

Welles Atkins (minutes) (minutes)

2 3 4 7 9 5 3 8 2 4

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Two-Sample Tests of Hypothesis 385

Region of rejection

.05

Region of rejection

.05

–1.833 Critical value

Scale of t

Do not reject H0

1.833 Critical value

H0: μ1 = μ2 H1: μ1 ≠ μ2

CHART 11–3 Regions of Rejection, Two-Tailed Test, df � 9, and .10 Significance Level

We use three steps to compute the value of t.

Step 1: Calculate the sample standard deviations. To compute the sample standard deviations, we use formula (3–11) from page 84. See the details below.

Step 2: Pool the sample variances. We use formula (11–5) to pool the sam- ple variances (standard deviations squared).

Step 3: Determine the value of t. The mean mounting time for the Welles method is 4.00 minutes, found by . The mean mounting time for the Atkins method is 5.00 minutes, found by . We use for- mula (11–6) to calculate the value of t.

The decision is not to reject the null hypothesis, because �0.662 falls in the region between �1.833 and 1.833. We conclude that there is no difference in the mean times to mount the engine on the frame using the two methods.

t � X1 � X2

A s2pa

1 n1

� 1 n2 b

� 4.00 � 5.00

A 6.2222a

1 5

� 1 6 b

� �0.662

X2 � 30�6 X1 � 20�5

s2p � (n1 � 1)s

2 1 � (n2 � 1)s

2 2

n1 � n2 � 2 �

(5 � 1)(2.9155)2 � (6 � 1)(2.0976)2

5 � 6 � 2 � 6.2222

s1 �C ©(X1 � X1)2

n1 � 1 � C

34 5 � 1

� 2.9155 s2 �C ©(X2 � X2)2

n2 � 1 � C

22 6 � 1

� 2.0976

X2 � ©X2 n2

� 30 6

� 5X1 � ©X1 n1

� 20 5

� 4

Welles Method Atkins Method

X1 X2 2 (2 � 4)2 � 4 3 (3 � 5)2 � 4 4 (4 � 4)2 � 0 7 (7 � 5)2 � 4 9 (9 � 4)2 � 25 5 (5 � 5)2 � 0 3 (3 � 4)2 � 1 8 (8 � 5)2 � 9 2 (2 � 4)2 � 4 4 (4 � 5)2 � 1

20 34 3 (3 � 5)2 � 4

30 22

(X2 � X 2) 2(X1 � X 1)

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Excel has a procedure called “t-Test: Two Sample Assuming Equal Variances” that will perform the calculations of formulas (11–5) and (11–6) as well as find the sample means and sample variances. The details of the procedure are provided in the Software Commands section at the end of the chapter. The data are input in the first two columns of the Excel spreadsheet. They are labeled “Welles” and “Atkins.” The output follows. The value of t, called the “t Stat,” is �0.662, and the two-tailed p-value is .525. As we would expect, the p-value is larger than the significance level of .10. The conclusion is not to reject the null hypothesis.

Self-Review 11–3 The production manager at Bellevue Steel, a manufacturer of wheelchairs, wants to com- pare the number of defective wheelchairs produced on the day shift with the number on the afternoon shift. A sample of the production from 6 day shifts and 8 afternoon shifts revealed the following number of defects.

Day 5 8 7 6 9 7 Afternoon 8 10 7 11 9 12 14 9

At the .05 significance level, is there a difference in the mean number of defects per shift? (a) State the null hypothesis and the alternate hypothesis. (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding the null hypothesis? (e) What is the p-value? (f) Interpret the result. (g) What are the assumptions necessary for this test?

We can also estimate the p-value using Appendix B.2. Locate the row with 9 degrees of freedom, and use the two-tailed test column. Find the t value, with- out regard to the sign, which is closest to our computed value of 0.662. It is 1.383, corresponding to a significance level of .20. Thus, even had we used the 20 per- cent significance level, we would not have rejected the null hypothesis of equal means. We can report that the p-value is greater than .20.

Exercises For Exercises 13 and 14: (a) state the decision rule, (b) compute the pooled estimate of the population variance, (c) compute the test statistic, (d) state your decision about the null hypothesis, and (e) estimate the p-value.

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Two-Sample Tests of Hypothesis 387

13. The null and alternate hypotheses are:

A random sample of 10 observations from one population revealed a sample mean of 23 and a sample deviation of 4. A random sample of 8 observations from another population revealed a sample mean of 26 and a sample standard deviation of 5. At the .05 significance level, is there a difference between the population means?

14. The null and alternate hypotheses are:

A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the .10 significance level, is there a difference in the population means?

Note: Use the five-step hypothesis testing procedure for the following exercises.

15. Listed below are the salaries in $000 of the 25 players on the opening-day roster of the 2010 New York Yankees Major League baseball team. These data also appear in Chap- ter 4, exercise 22.

Sort the players into two groups, pitchers and nonpitchers (position players). Assume equal population variances for the position players and the pitchers. Test the hypothesis that mean salaries between position players and pitchers are the same versus the alter- nate hypothesis that they are not the same. Use the .01 significance level.

16. A recent study compared the time spent together by single- and dual-earner couples. According to the records kept by the wives during the study, the mean amount of time spent together watching television among the single-earner couples was 61 minutes per

Player Salary ($000) Position

Aceves, Alfredo 435.7 Pitcher Burnett, A.J. 16,500.0 Pitcher Cano, Robinson 9,000.0 Second Baseman Cervelli, Francisco 410.8 Catcher Chamberlain, Joba 488.0 Pitcher Gardner, Brett 452.5 Outfielder Granderson, Curtis 5,500.0 Outfielder Hughes, Phil 447.0 Pitcher Jeter, Derek 22,600.0 Shortstop Johnson, Nick 5,500.0 First Baseman Marte, Damaso 4,000.0 Pitcher Mitre, Sergio 850.0 Pitcher Park, Chan Ho 1,200.0 Pitcher Pena, Ramiro 412.1 Infielder Pettitte, Andy 11,750.0 Pitcher Posada, Jorge 13,100.0 Catcher Rivera, Mariano 15,000.0 Pitcher Robertson, David 426.7 Pitcher Rodriguez, Alex 33,000.0 Third Baseman Sabathia, CC 24,285.7 Pitcher Swisher, Nick 6,850.0 Outfielder Teixeira, Mark 20,625.0 First Baseman Thames, Marcus 900.0 Outfielder Vazquez, Javier 11,500.0 Pitcher Winn, Randy 1,100.0 Outfielder

H1: �1 � �2

H0: �1 � �2

H1: �1 � �2

H0: �1 � �2

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IN MEANS, UNEQUAL VARIANCES

UNEQUAL VARIANCE TEST

388 Chapter 11

Unequal Population Standard Deviations In the previous sections, it was necessary to assume that the populations had equal standard deviations. To put it another way, we did not know the population stan- dard deviations but we assumed they were equal. In many cases, this is a reason- able assumption, but what if it is not? In the next chapter, we present a formal method for testing this equal variance assumption.

If it is not reasonable to assume the population standard deviations are equal, then we use a statistic very much like formula [11–2]. The sample standard deviations, s1 and s2, are used in place of the respective population standard deviations. In addition, the degrees of freedom are adjusted downward by a rather complex approximation formula. The effect is to reduce the number of degrees of freedom in the test, which will require a larger value of the test statistic to reject the null hypothesis.

The formula for the t statistic is:

TEST STATISTIC FOR NO DIFFERENCE [11–7]

The degrees of freedom statistic is found by:

DEGREES OF FREEDOM FOR [11–8]

where n1 and n2 are the respective sample sizes and s1 and s2 are the respective sam- ple standard deviations. If necessary, this fraction is rounded down to an integer value. An example will explain the details.

df � [(s21/n1) � (s

2 2/n2)]

(s21/n1) 2

n1 � 1 �

(s22/n2) 2

n2 � 1

t � X1 � X2

s21 n1

� s22 n2

day, with a standard deviation of 15.5 minutes. For the dual-earner couples, the mean number of minutes spent watching television was 48.4 minutes, with a standard devia- tion of 18.1 minutes. At the .01 significance level, can we conclude that the single-earner couples on average spend more time watching television together? There were 15 single- earner and 12 dual-earner couples studied.

17. Ms. Lisa Monnin is the budget director for Nexus Media Inc. She would like to compare the daily travel expenses for the sales staff and the audit staff. She collected the fol- lowing sample information.

At the .10 significance level, can she conclude that the mean daily expenses are greater for the sales staff than the audit staff? What is the p-value?

18. The Tampa Bay (Florida) Area Chamber of Commerce wanted to know whether the mean weekly salary of nurses was larger than that of school teachers. To investigate, they col- lected the following information on the amounts earned last week by a sample of school teachers and nurses.

Is it reasonable to conclude that the mean weekly salary of nurses is higher? Use the .01 significance level. What is the p-value?

School Teachers ($) 845 826 827 875 784 809 802 820 829 830 842 832 Nurses ($) 841 890 821 771 850 859 825 829

Sales ($) 131 135 146 165 136 142 Audit ($) 130 102 129 143 149 120 139

LO4 Conduct a test of a hypothesis that two independent population means are equal, assuming unequal but unknown population standard deviations.

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Two-Sample Tests of Hypothesis 389

Example Personnel in a consumer testing laboratory are evaluating the absorbency of paper towels. They wish to compare a set of store brand towels to a similar group of name brand ones. For each brand they dip a ply of the paper into a tub of fluid, allow the paper to drain back into the vat for two minutes, and then eval- uate the amount of liquid the paper has taken up from the vat. A random sam- ple of 9 store brand paper towels absorbed the following amounts of liquid in milliliters.

8 8 3 1 9 7 5 5 12

An independent random sample of 12 name brand towels absorbed the following amounts of liquid in milliliters:

12 11 10 6 8 9 9 10 11 9 8 10

Use the .10 significance level and test if there is a difference in the mean amount of liquid absorbed by the two types of paper towels.

Solution To begin, let’s assume that the amounts of liquid absorbed follow the normal prob- ability distribution for both the store brand and the name brand towels. We do not know either of the population standard deviations, so we are going to use the t dis- tribution as the test statistic. The assumption of equal population standard devia- tions does not appear reasonable. The amount of absorption in the store brand ranges from 1 ml to 12 ml. For the name brand, the amount of absorption ranges from 6 ml to 12 ml. That is, there is considerably more variation in the amount of absorption in the store brand than in the name brand. We observe the difference in the variation in the following dot plot provided by Minitab. The software commands to create a Minitab dot plot are given on page 135.

So we decide to use the t distribution and assume that the population standard deviations are not the same.

In the five-step hypothesis testing procedure, the first step is to state the null hypothesis and the alternate hypothesis. The null hypothesis is that there is no dif- ference in the mean amount of liquid absorbed between the two types of paper towels. The alternate hypothesis is that there is a difference.

H1: �1 � �2

H0: �1 � �2

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390 Chapter 11

The significance level is .10 and the test statistic follows the t distribution. Because we do not wish to assume equal population standard deviations, we adjust the degrees of freedom using formula (11–8). To do so, we need to find the sample standard deviations. We can use the Minitab system to quickly find these results. We will also find the mean absorption rate, which we will use shortly. The respec- tive samples sizes are n1 � 9 and n2 � 12 and the respective standard deviations are 3.32 ml and 1.621 ml.

Descriptive Statistics: Store, Name

Variable N Mean StDev Store 9 6.44 3.32 Name 12 9.417 1.621

Inserting this information into formula (11–8):

The usual practice is to round down to the integer, so we use 10 degrees of free- dom. From Appendix B.2 with 10 degrees of freedom, a two-tailed test, and the .10 significance level, the critical t values are �1.812 and 1.812. Our decision rule is to reject the null hypothesis if the computed value of t is less than �1.812 or greater than 1.812.

To find the value of the test statistic, we use formula (11–7). Recall from the Minitab output above that the mean amount of absorption for the store paper towels is 6.44 ml and 9.417 ml for the brand.

The computed value of t is less than the lower critical value, so our decision is to reject the null hypothesis. We conclude that the mean absorption rate for the two towels is not the same. The Minitab output for this example follows.

t � X1 � X2

s21 n1

� s22 n2

� 6.44 � 9.417

3.322

9 �

1.6212

� �2.478

df � [(s21/n1) � (s22/n2)]2

(s21/n1)2

n1 � 1 �

(s22/n2) 2

n2 � 1

� [(3.322�9) � (1.6212�12)]2

(3.322�9)2

9 � 1 �

(1.6212�12)2

12 � 1

� 1.44362

.1875 � .0043 � 10.88

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Two-Sample Tests of Hypothesis 391

Self-Review 11–4 It is often useful for companies to know who their customers are and how they became customers. A credit card company is interested in whether the owner of the card applied for the card on their own or was contacted by a telemarketer. The com- pany obtained the following sample information regarding end-of-the-month balances for the two groups.

Is it reasonable to conclude the mean balance is larger for the credit card holders that were contacted by telemarketers than for those who applied on their own for the card? Assume the population standard deviations are not the same. Use the .05 significance level. (a) State the null hypothesis and the alternate hypothesis. (b) How many degrees of freedom are there? (c) What is the decision rule? (d) What is the value of the test statistic? (e) What is your decision regarding the null hypothesis? (f) Interpret the result.

Source Mean Standard Deviation Sample Size

Applied $1,568 $356 10 Contacted 1,967 857 8

Exercises For exercises 19 and 20, assume the sample populations do not have equal standard devia- tions and use the .05 significance level: (a) determine the number of degrees of freedom, (b) state the decision rule, (c) compute the value of the test statistic, and (d) state your deci- sion about the null hypothesis.

19. The null and alternate hypotheses are:

A random sample of 15 items from the first population showed a mean of 50 and a stan- dard deviation of 5. A sample of 12 items for the second population showed a mean of 46 and a standard deviation of 15.

20. The null and alternate hypotheses are:

A random sample of 20 items from the first population showed a mean of 100 and a standard deviation of 15. A sample of 16 items for the second population showed a mean of 94 and a standard deviation of 8. Use the .05 significant level.

21. A recent article in The Wall Street Journal compared the cost of adopting children from China with that of Russia. For a sample of 16 adoptions from China, the mean cost was $11,045, with a standard deviation of $835. For a sample of 18 adoptions from Russia, the mean cost was $12,840, with a standard deviation of $1,545. Can we conclude the mean cost is larger for adopting children from Russia? Assume the two population stan- dard deviations are not the same. Use the .05 significance level.

22. Suppose you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne attire with those of Calvin Klein. The following is the amount ($000) earned per month by a sample of Claiborne models:

H1: �1 7 �2 H0: �1 � �2

H1: �1 � �2

H0: �1 � �2

$5.0 $4.5 $3.4 $3.4 $6.0 $3.3 $4.5 $4.6 $3.5 $5.2 4.8 4.4 4.6 3.6 5.0

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392 Chapter 11

$3.1 $3.7 $3.6 $4.0 $3.8 $3.8 $5.9 $4.9 $3.6 $3.6 2.3 4.0

The following is the amount ($000) earned by a sample of Klein models.

Is it reasonable to conclude that Claiborne models earn more? Use the .05 significance level and assume the population standard deviations are not the same.

11.5 Two-Sample Tests of Hypothesis: Dependent Samples On page 383, we tested the difference between the means from two independent samples. We compared the mean time required to mount an engine using the Welles method to the time to mount the engine using the Atkins method. The samples were independent, meaning that the sample of assembly times using the Welles method was in no way related to the sample of assembly times using the Atkins method.

There are situations, however, in which the samples are not independent. To put it another way, the samples are dependent or related. As an example, Nickel Sav- ings and Loan employs two firms, Schadek Appraisals and Bowyer Real Estate, to appraise the value of the real estate properties on which it makes loans. It is impor- tant that these two firms be similar in their appraisal values. To review the consis- tency of the two appraisal firms, Nickel Savings randomly selects 10 homes and has both Schadek Appraisals and Bowyer Real Estate appraise the value of the selected homes. For each home, there will be a pair of appraisal values. That is, for

each home there will be an appraised value from both Schadek Appraisals and Bowyer Real Estate. The appraised values depend on, or are related to, the home selected. This is also referred to as a paired sample.

For hypothesis testing, we are interested in the distribution of the differences in the appraised value of each home. Hence, there is only one sample. To put it more formally, we are investigating whether the mean of the distribution of differences in the appraised values is 0. The sample is made up of the differences between the appraised values determined by Schadek Appraisals and the values from Bowyer Real Estate. If the two appraisal firms are reporting similar estimates, then sometimes Schadek

Appraisals will be the higher value and sometimes Bowyer Real Estate will have the higher value. However, the mean of the distribution of differences will be 0. On the other hand, if one of the firms consistently reports the larger appraisal values, then the mean of the distribution of the differences will not be 0.

We will use the symbol �d to indicate the population mean of the distribution of differences. We assume the distribution of the population of differences follows the normal distribution. The test statistic follows the t distribution and we calculate its value from the following formula:

PAIRED t TEST [11–9]

There are degrees of freedom and

is the mean of the difference between the paired or related observations. is the standard deviation of the differences between the paired or related observations.

n is the number of paired observations.

n � 1

t � d

sd�1n

LO5 Explain the difference between dependent and independent samples.

LO6 Carry out a test of a hypothesis about the mean difference between paired and dependent observations.

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Two-Sample Tests of Hypothesis 393

The standard deviation of the differences is computed by the familiar formula for the standard deviation (see formula 3–11), except d is substituted for X. The formula is:

The following example illustrates this test.

sd �C ©(d � d )2

n � 1

Example Recall that Nickel Savings and Loan wishes to compare the two companies it uses to appraise the value of residential homes. Nickel Savings selected a sample of 10 residential properties and scheduled both firms for an appraisal. The results, reported in $000, are:

Home Schadek Bowyer

1 235 228 2 210 205 3 231 219 4 242 240 5 205 198 6 230 223 7 231 227 8 210 215 9 225 222

10 249 245

Solution The first step is to state the null and the alternate hypotheses. In this case, a two- tailed alternative is appropriate because we are interested in determining whether there is a difference in the appraised values. We are not interested in showing whether one particular firm appraises property at a higher value than the other. The question is whether the sample differences in the appraised values could have come from a population with a mean of 0. If the population mean of the differences is 0, then we conclude that there is no difference in the appraised values. The null and alternate hypotheses are:

There are 10 homes appraised by both firms, so n � 10, and df � n � 1 � 10 � 1 � 9. We have a two-tailed test, and the significance level is .05. To determine the critical value, go to Appendix B.2, move across the row with 9 degrees of freedom to the column for a two-tailed test and the .05 significance level. The value at the inter- section is 2.262. This value appears in the box in Table 11–2. The decision rule is to reject the null hypothesis if the computed value of t is less than �2.262 or greater than 2.262. Here are the computational details.

H1: �d � 0 H0: �d � 0

At the .05 significance level, can we conclude there is a difference in the mean appraised values of the homes?

Home Schadek Bowyer Difference, d (d � ) (d � )2

1 235 228 7 2.4 5.76 2 210 205 5 0.4 0.16 3 231 219 12 7.4 54.76 4 242 240 2 �2.6 6.76

(continued )

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394 Chapter 11

Home Schadek Bowyer Difference, d (d � ) (d � )2

5 205 198 7 2.4 5.76 6 230 223 7 2.4 5.76 7 231 227 4 �0.6 0.36 8 210 215 �5 �9.6 92.16 9 225 222 3 �1.6 2.56

10 249 245 4 �0.6 0.36

46 0 174.40

Using formula (11–9), the value of the test statistic is 3.305, found by

Because the computed t falls in the rejection region, the null hypothesis is rejected. The population distribution of differences does not have a mean of 0. We conclude that there is a difference in the mean appraised values of the homes. The largest difference of $12,000 is for Home 3. Perhaps that would be an appropriate place to begin a more detailed review.

To find the p-value, we use Appendix B.2 and the section for a two-tailed test. Move along the row with 9 degrees of freedom and find the values of t that are closest to our calculated value. For a .01 significance level, the value of t is 3.250. The computed value is larger than this value, but smaller than the value of 4.781 corresponding to the .001 significance level. Hence, the p-value is less than .01. This information is highlighted in Table 11–2.

t � d

sd�1n �

4.6 4.402�110

� 4.6

1.3920 � 3.305

sd �C ©(d � d)2

n � 1 � C

174.4 10 � 1

� 4.402

d � ©d n

� 46 10

� 4.60

TABLE 11–2 A Portion of the t Distribution from Appendix B.2

Confidence Intervals

80% 90% 95% 98% 99% 99.9%

Level of Significance for One-Tailed Test

df 0.10 0.05 0.025 0.01 0.005 0.0005

Level of Significance for Two-Tailed Test

0.20 0.10 0.05 0.02 0.01 0.001

1 3.078 6.314 12.706 31.821 63.657 636.619 2 1.886 2.920 4.303 6.965 9.925 31.599 3 1.638 2.353 3.182 4.541 5.841 12.924 4 1.533 2.132 2.776 3.747 4.604 8.610 5 1.476 2.015 2.571 3.365 4.032 6.869 6 1.440 1.943 2.447 3.143 3.707 5.959 7 1.415 1.895 2.365 2.998 3.499 5.408 8 1.397 1.860 2.306 2.896 3.355 5.041 9 1.383 1.833 2.262 2.821 3.250 4.781

10 1.372 1.812 2.228 2.764 3.169 4.587

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Two-Sample Tests of Hypothesis 395

Excel has a procedure called “t-Test: Paired Two-Sample for Means” that will perform the calculations of formula (11–9). The output from this procedure is given below.

The computed value of t is 3.305, and the two-tailed p-value is .009. Because the p-value is less than .05, we reject the hypothesis that the mean of the dis- tribution of the differences between the appraised values is zero. In fact, this p-value is between .01 and .001. There is a small likelihood that the null hypoth- esis is true.

11.6 Comparing Dependent and Independent Samples Beginning students are often confused by the difference between tests for inde- pendent samples [formula (11–6)] and tests for dependent samples [formula (11–9)]. How do we tell the difference between dependent and independent samples? There are two types of dependent samples: (1) those characterized by a measurement, an intervention of some type, and then another measurement; and (2) a matching or pairing of the observations. To explain further:

1. The first type of dependent sample is characterized by a measurement followed by an intervention of some kind and then another measurement. This could be called a “before” and “after” study. Two examples will help to clarify. Suppose we want to show that, by placing speakers in the production area and playing soothing music, we are able to increase production. We begin by selecting a sample of workers and measuring their output under the current conditions. The speakers are then installed in the production area, and we again measure the output of the same workers. There are two measurements, before placing the speakers in the production area and after. The intervention is placing speakers in the production area.

A second example involves an educational firm that offers courses designed to increase test scores and reading ability. Suppose the firm wants to offer a course that will help high school juniors increase their SAT scores. To begin, each student takes the SAT in the junior year in high school. During the sum- mer between the junior and senior year, they participate in the course that gives

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The mean of the appraised value of the 10 properties by Schadek is $226,800, and the standard deviation is $14,500. For Bowyer Real Estate, the mean appraised value is $222,200, and the standard deviation is $14,290. To make the calculations easier, we use $000 instead of $. The value of the pooled estimate of the variance from formula (11–5) is

s2p � (n1 � 1)s

2 1 � (n2 � 1)s

2 2

n1 � n2 � 2 �

(10 � 1)(14.452) � (10 � 1)(14.29)2

10 � 10 � 2 � 206.50

them tips on taking tests. Finally, during the fall of their senior year in high school, they retake the SAT. Again, the procedure is characterized by a mea- surement (taking the SAT as a junior), an intervention (the summer workshops), and another measurement (taking the SAT during their senior year).

2. The second type of dependent sample is characterized by matching or pairing observations. Nickel Savings in the previous example is a dependent sample of this type. It selected a property for appraisal and then had two appraisals on the same property. As a second example, suppose an industrial psychologist wishes to study the intellectual similarities of newly married couples. She selects a sample of newlyweds. Next, she administers a standard intelligence test to both the man and woman to determine the difference in the scores. Notice the matching that occurred: comparing the scores that are paired or matched by marriage.

Why do we prefer dependent samples to independent samples? By using depen- dent samples, we are able to reduce the variation in the sampling distribution. To illustrate, we will use the Nickel Savings and Loan example just completed. Sup- pose we assume that we have two independent samples of real estate property for appraisal and conduct the following test of hypothesis, using formula (11–6). The null and alternate hypotheses are:

There are now two independent samples of 10 each. So the number of degrees of freedom is From Appendix B.2, for the .05 significance level, H0 is rejected if t is less than or greater than 2.101.

We use the same Excel commands as on page 100 in Chapter 3 to find the mean and the standard deviation of the two independent samples. We use the Excel commands on page 408 of this chapter to find the pooled variance and the value of the “t Stat.” These values are highlighted in yellow.

�2.101 10 � 10 � 2 � 18.

H1: �1 � �2

H0: �1 � �2

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Two-Sample Tests of Hypothesis 397

From formula (11–6), t is 0.716.

The computed t (0.716) is less than 2.101, so the null hypothesis is not rejected. We cannot show that there is a difference in the mean appraisal value. That is not the same conclusion that we got before! Why does this happen? The numer- ator is the same in the paired observations test (4.6). However, the denominator is smaller. In the paired test, the denominator is 1.3920 (see the calculations on page 394). In the case of the independent samples, the denominator is 6.4265. There is more variation or uncertainty. This accounts for the difference in the t values and the difference in the statistical decisions. The denominator measures the standard error of the statistic. When the samples are not paired, two kinds of variation are present: differences between the two appraisal firms and the difference in the value of the real estate. Properties numbered 4 and 10 have relatively high values, whereas number 5 is relatively low. These data show how different the values of the property are, but we are really interested in the difference between the two appraisal firms.

The trick is to pair the values to reduce the variation among the properties. The paired test uses only the difference between the two appraisal firms for the same property. Thus, the paired or dependent statistic focuses on the variation between Schadek Appraisals and Bowyer Real Estate. Thus, its standard error is always smaller. That, in turn, leads to a larger test statistic and a greater chance of reject- ing the null hypothesis. So whenever possible you should pair the data.

There is a bit of bad news here. In the paired observations test, the degrees of freedom are half of what they are if the samples are not paired. For the real estate example, the degrees of freedom drop from 18 to 9 when the observa- tions are paired. However, in most cases, this is a small price to pay for a bet- ter test.

t � X1 � X2

A s2pa

1 n1

� 1 n2 b

� 226.8 � 222.2

A 206.50a

1 10

� 1 10 b

� 4.6

6.4265 � 0.716

Self-Review 11–5 Advertisements by Sylph Fitness Center claim that completing its course will result in losing weight. A random sample of eight recent participants showed the following weights before and after completing the course. At the .01 significance level, can we conclude the students lost weight?

(a) State the null hypothesis and the alternate hypothesis. (b) What is the critical value of t? (c) What is the computed value of t? (d) Interpret the result. What is the p-value? (e) What assumption needs to be made about the distribution of the differences?

Name Before After

Hunter 155 154 Cashman 228 207 Mervine 141 147 Massa 162 157 Creola 211 196 Peterson 164 150 Redding 184 170 Poust 172 165

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398 Chapter 11

Day

1 2 3 4

Day shift 10 12 15 19 Afternoon shift 8 9 12 15

Day

May June July August September

Officer Dhondt 30 22 25 19 26 Officer Meredith 26 19 20 15 19

At the .05 significance level, can we conclude there are more defects produced on the afternoon shift?

24. The null and alternate hypotheses are:

The following paired observations show the number of traffic citations given for speed- ing by Officer Dhondt and Officer Meredith of the South Carolina Highway Patrol for the last five months.

H1: �d � 0 H0: �d � 0

Exercises 23. The null and alternate hypotheses are:

The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month.

H1: �d 7 0 H0: �d � 0

Salesperson Before After

Sid Mahone $320 $340 Carol Quick 290 285 Tom Jackson 421 475 Andy Jones 510 510 Jean Sloan 210 210 Jack Walker 402 500 Peg Mancuso 625 631 Anita Loma 560 560 John Cuso 360 365 Carl Utz 431 431 A. S. Kushner 506 525 Fern Lawton 505 619

At the .05 significance level, is there a difference in the mean number of citations given by the two officers?

Note: Use the five-step hypothesis testing procedure to solve the following exercises.

25. The management of Discount Furniture, a chain of discount furniture stores in the North- east, designed an incentive plan for salespeople. To evaluate this innovative plan, 12 salespeople were selected at random, and their weekly incomes before and after the plan were recorded.

Was there a significant increase in the typical salesperson’s weekly income due to the innovative incentive plan? Use the .05 significance level. Estimate the p-value, and interpret it.

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Chapter Summary I. In comparing two population means, we wish to know whether they could be equal.

A. We are investigating whether the distribution of the difference between the means could have a mean of 0.

B. The test statistic follows the standard normal distribution if the population standard deviations are known. 1. No assumption about the shape of either population is required. 2. The samples are from independent populations. 3. The formula to compute the value of z is

[11–2]

II. We can also test whether two samples came from populations with an equal proportion of successes. A. The two sample proportions are pooled using the following formula:

[11–4]

B. We compute the value of the test statistic from the following formula:

[11–3]

III. The test statistic to compare two means is the t distribution if the population standard deviations are not known. A. Both populations must follow the normal distribution. B. The populations must have equal standard deviations. C. The samples are independent. D. Finding the value of t requires two steps.

1. The first step is to pool the standard deviations according to the following formula:

[11–5]

2. The value of t is computed from the following formula:

[11–6]

3. The degrees of freedom for the test are n1 � n2 � 2.

t � X1 � X2

B s2pa

1 n1

� 1 n2 b

s2p � (n1 � 1)s21 � (n2 � 1)s22

n1 � n2 � 2

z � p1 � p2

pc(1 � pc) n1

� pc(1 � pc)

pc � X1 � X2 n1 � n2

z � X1 � X2

�21 n1

� �22 n2

Two-Sample Tests of Hypothesis 399

26. The federal government recently granted funds for a special program designed to reduce crime in high-crime areas. A study of the results of the program in eight high-crime areas of Miami, Florida, yielded the following results.

Number of Crimes by Area

A B C D E F G H

Before 14 7 4 5 17 12 8 9 After 2 7 3 6 8 13 3 5

Has there been a decrease in the number of crimes since the inauguration of the pro- gram? Use the .01 significance level. Estimate the p-value.

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400 Chapter 11

IV. If we cannot assume the population standard deviations are equal, we adjust the degrees of freedom and the formula for finding t. A. We determine the degrees of freedom based on the following formula.

[11–8]

B. The value of the test statistic is computed from the following formula.

[11–7]

V. For dependent samples, we assume the distribution of the paired differences between the populations has a mean of 0. A. We first compute the mean and the standard deviation of the sample differences. B. The value of the test statistic is computed from the following formula:

[11–9]

Pronunciation Key SYMBOL MEANING PRONUNCIATION

pc Pooled proportion p sub c

Pooled sample variance s sub p squared

Mean of the first sample X bar sub 1

Mean of the second sample X bar sub 2

Mean of the difference between d bar dependent observations

sd Standard deviation of the difference s sub d between dependent observations

Chapter Exercises 27. A recent study focused on the number of times men and women who live alone buy

take-out dinner in a month. The information is summarized below.

At the .01 significance level, is there a difference in the mean number of times men and women order take-out dinners in a month? What is the p-value?

28. Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shift than on the day shift. Assume the population standard deviation for the number of units produced on the day shift is 21 and is 28 on the night shift. A sample of 54 day-shift workers showed that the mean number of units produced was 345. A sample of 60 night-shift workers showed that the mean number of units produced was 351. At the .05 significance level, is the number of units produced on the night shift larger?

29. Fry Brothers Heating and Air Conditioning Inc. employs Larry Clark and George Murnen to make service calls to repair furnaces and air conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service

Statistic Men Women

Sample mean 24.51 22.69 Population standard deviation 4.48 3.86 Sample size 35 40

s2p

t � d

sd�1n

t � X1 � X2

s21 n1

� s22 n2

df � [(s21�n1) � (s22 �n2)]

(s21 �n1) 2

n1 � 1 �

(s22 �n2) 2

n2 � 1

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Two-Sample Tests of Hypothesis 401

calls they make per day. Assume the population standard deviation for Larry Clark is 1.05 calls per day and 1.23 calls per day for George Murnen. A random sample of 40 days last year showed that Larry Clark made an average of 4.77 calls per day. For a sample of 50 days George Murnen made an average of 5.02 calls per day. At the .05 signifi- cance level, is there a difference in the mean number of calls per day between the two employees? What is the p-value?

30. A coffee manufacturer is interested in whether the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers. Assume the population standard deviation for those drinking regular coffee is 1.20 cups per day and 1.36 cups per day for those drinking decaffeinated coffee. A random sample of 50 regular-coffee drinkers showed a mean of 4.35 cups per day. A sample of 40 decaffeinated-coffee drinkers showed a mean of 5.84 cups per day. Use the .01 significance level. Compute the p-value.

31. A cell phone company offers two plans to its subscribers. At the time new subscribers sign up, they are asked to provide some demographic information. The mean yearly income for a sample of 40 subscribers to Plan A is $57,000 with a standard deviation of $9,200. This distribution is positively skewed; the actual coefficient of skewness is 2.11. For a sample of 30 subscribers to Plan B, the mean income is $61,000 with a standard deviation of $7,100. The distribution of Plan B subscribers is also positively skewed, but not as severely. The coefficient of skewness is 1.54. At the .05 significance level, is it reasonable to conclude the mean income of those selecting Plan B is larger? What is the p-value? Do the coefficients of skewness affect the results of the hypothesis test? Why?

32. A computer manufacturer offers a help line that purchasers can call for help 24 hours a day, 7 days a week. Clearing these calls for help in a timely fashion is important to the company’s image. After telling the caller that resolution of the problem is important, the caller is asked whether the issue is software or hardware related. The mean time it takes a technician to resolve a software issue is 18 minutes with a standard deviation of 4.2 min- utes. This information was obtained from a sample of 35 monitored calls. For a study of 45 hardware issues, the mean time for the technician to resolve the problem was 15.5 minutes with a standard deviation of 3.9 minutes. This information was also obtained from monitored calls. At the .05 significance level, does it take longer to resolve software issues? What is the p-value?

33. Suppose the manufacturer of Advil, a common headache remedy, recently developed a new formulation of the drug that is claimed to be more effective. To evaluate the new drug, a sample of 200 current users is asked to try it. After a one-month trial, 180 indi- cated the new drug was more effective in relieving a headache. At the same time, a sam- ple of 300 current Advil users is given the current drug but told it is the new formulation. From this group, 261 said it was an improvement. At the .05 significance level, can we conclude that the new drug is more effective?

34. Each month the National Association of Purchasing Managers publishes the NAPM index. One of the questions asked on the survey to purchasing agents is: Do you think the economy is contracting? Last month, of the 300 responses, 160 answered yes to the question. This month, 170 of the 290 responses indicated they felt the economy was contracting. At the .05 significance level, can we conclude that a larger proportion of the agents believe the economy is contracting this month?

35. As part of a recent survey among dual-wage-earner couples, an industrial psycholo- gist found that 990 men out of the 1,500 surveyed believed the division of household duties was fair. A sample of 1,600 women found 970 believed the division of house- hold duties was fair. At the .01 significance level, is it reasonable to conclude that the proportion of men who believe the division of household duties is fair is larger? What is the p-value?

36. There are two major Internet providers in the Colorado Springs, Colorado, area, one called HTC and the other Mountain Communications. We want to investigate whether there is a difference in the proportion of times a customer is able to access the Internet. During a one-week period, 500 calls were placed at random times throughout the day and night to HTC. A connection was made to the Internet on 450 occasions. A similar one-week study with Mountain Communications showed the Internet to be available on 352 of 400 trials. At the .01 significance level, is there a difference in the percent of time that access to the Internet is successful?

37. The Consumer Confidence Survey is a monthly review that measures consumer confi- dence in the U.S. economy. It is based on a typical sample of 5,000 U.S. households. Last month 9.1 percent of consumers said conditions were “good.” In the prior month, only 8.5 percent held they were “good.” Use the five-step hypothesis-testing method at

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the .05 level of significance to see whether you can determine there is an increase in the share asserting conditions are “good.” Find the p-value and explain what it means.

38. A study was conducted to determine if there was a difference in the humor content in British and American trade magazine advertisements. In an independent random sample of 270 American trade magazine advertisements, 56 were humorous. An independent random sample of 203 British trade magazines contained 52 humorous ads. Does this data provide evidence at the .05 significance level that there is a difference in the pro- portion of humorous ads in British versus American trade magazines?

39. The AP-Petside.com poll contacted 300 married women and 200 married men. All owned pets. One hundred of the women and 36 of the men replied that their pets are better lis- teners than their spouses. At the .05 significance level, is there a difference between the responses of women and men?

40. The National Basketball Association had 39 black top executives (presidents or vice pres- idents) among its 388 senior managers. Meanwhile, Major League Baseball had only 11 blacks among its 307 top administrators. Test at the .05 significance level if this reveals the NBA has significantly more black participation in higher levels of management.

41. One of the music industry’s most pressing questions is: Can paid download stores contend nose-to-nose with free peer-to-peer download services? Data gathered over the last 12 months show Apple’s iTunes was used by an average of 1.65 million households with a sample standard deviation of 0.56 million family units. Over the same 12 months WinMX (a no-cost P2P download service) was used by an average of 2.2 million fam- ilies with a sample standard deviation of 0.30 million. Assume the population standard deviations are not the same. Using a significance level of 0.05, test the hypothesis of no difference in the mean number of households picking either variety of service to download songs.

42. Businesses, particularly those in the food preparation industry such as General Mills, Kellogg, and Betty Crocker regularly use coupons as a brand allegiance builder to stim- ulate their retailing. There is uneasiness that the users of paper coupons are different from the users of e-coupons (coupons disseminated by means of the Internet). One sur- vey recorded the age of each person who redeemed a coupon along with the type (either electronic or paper). The sample of 35 e-coupon users had a mean age of 33.6 years with a standard deviation of 10.9, while a similar sample of 25 traditional paper-coupon clippers had a mean age of 39.5 with a standard deviation of 4.8. Assume the popula- tion standard deviations are not the same. Using a significance level of 0.01, test the hypothesis of no difference in the mean ages of the two groups of coupon clients.

43. The owner of Bun ‘N’ Run Hamburgers wishes to compare the sales per day at two loca- tions. The mean number sold for 10 randomly selected days at the Northside site was 83.55, and the standard deviation was 10.50. For a random sample of 12 days at the Southside location, the mean number sold was 78.80 and the standard deviation was 14.25. At the .05 significance level, is there a difference in the mean number of ham- burgers sold at the two locations? What is the p-value?

44. The Engineering Department at Sims Software Inc. recently developed two chemical solutions designed to increase the usable life of computer disks. A sample of disks treated with the first solution lasted 86, 78, 66, 83, 84, 81, 84, 109, 65, and 102 hours. Those treated with the second solution lasted 91, 71, 75, 76, 87, 79, 73, 76, 79, 78, 87, 90, 76, and 72 hours. Assume the population standard deviations are not the same. At the .10 significance level, can we conclude that there is a difference in the length of time the two types of treatment lasted?

45. The Willow Run Outlet Mall has two Haggar Outlet Stores, one located on Peach Street and the other on Plum Street. The two stores are laid out differently, but both store man- agers claim their layout maximizes the amounts customers will purchase on impulse. A sample of 10 customers at the Peach Street store revealed they spent the following amounts more than planned: $17.58, $19.73, $12.61, $17.79, $16.22, $15.82, $15.40, $15.86, $11.82, and $15.85. A sample of 14 customers at the Plum Street store revealed they spent the following amounts more than they planned: $18.19, $20.22, $17.38, $17.96, $23.92, $15.87, $16.47, $15.96, $16.79, $16.74, $21.40, $20.57, $19.79, and $14.83. At the .01 significance level, is there a difference in the mean amounts purchased on impulse at the two stores?

46. Grand Strand Family Medical Center is specifically set up to treat minor medical emer- gencies for visitors to the Myrtle Beach area. There are two facilities, one in the Little

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Two-Sample Tests of Hypothesis 403

River Area and the other in Murrells Inlet. The Quality Assurance Department wishes to compare the mean waiting time for patients at the two locations. Samples of the waiting times, reported in minutes, follow:

Assume the population standard deviations are not the same. At the .05 significance level, is there a difference in the mean waiting time?

47. Commercial Bank and Trust Company is studying the use of its automatic teller machines (ATMs). Of particular interest is whether young adults (under 25 years) use the machines more than senior citizens. To investigate further, samples of customers under 25 years of age and customers over 60 years of age were selected. The number of ATM transactions last month was determined for each selected individual, and the results are shown below. At the .01 significance level, can bank management conclude that younger customers use the ATMs more?

48. Two boats, the Prada (Italy) and the Oracle (U.S.A.), are competing for a spot in the upcoming America’s Cup race. They race over a part of the course several times. Below are the sample times in minutes. Assume the population standard deviations are not the same. At the .05 significance level, can we conclude that there is a difference in their mean times?

49. The manufacturer of an MP3 player wanted to know whether a 10 percent reduction in price is enough to increase the sales of its product. To investigate, the owner randomly selected eight outlets and sold the MP3 player at the reduced price. At seven randomly selected outlets, the MP3 player was sold at the regular price. Reported below is the number of units sold last month at the sampled outlets. At the .01 significance level, can the manufacturer conclude that the price reduction resulted in an increase in sales?

50. A number of minor automobile accidents occur at various high-risk intersections in Teton County despite traffic lights. The Traffic Department claims that a modification in the type of light will reduce these accidents. The county commissioners have agreed to a pro- posed experiment. Eight intersections were chosen at random, and the lights at those intersections were modified. The numbers of minor accidents during a six-month period before and after the modifications were:

At the .01 significance level, is it reasonable to conclude that the modification reduced the number of traffic accidents?

Number of Accidents

A B C D E F G H

Before modification 5 7 6 4 8 9 8 10 After modification 3 7 7 0 4 6 8 2

Regular price 138 121 88 115 141 125 96 Reduced price 128 134 152 135 114 106 112 120

Boat Time (minutes)

Prada (Italy) 12.9 12.5 11.0 13.3 11.2 11.4 11.6 12.3 14.2 11.3 Oracle (U.S.A.) 14.1 14.1 14.2 17.4 15.8 16.7 16.1 13.3 13.4 13.6 10.8 19.0

Under 25 10 10 11 15 7 11 10 9 Over 60 4 8 7 7 4 5 1 7 4 10 5

Location Waiting Time

Little River 31.73 28.77 29.53 22.08 29.47 18.60 32.94 25.18 29.82 26.49 Murrells Inlet 22.93 23.92 26.92 27.20 26.44 25.62 30.61 29.44 23.09 23.10 26.69 22.31

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404 Chapter 11

51. Lester Hollar is vice president for human resources for a large manufacturing company. In recent years, he has noticed an increase in absenteeism that he thinks is related to the general health of the employees. Four years ago, in an attempt to improve the situation, he began a fitness program in which employees exercise during their lunch hour. To eval- uate the program, he selected a random sample of eight participants and found the num- ber of days each was absent in the six months before the exercise program began and in the last six months. Below are the results. At the .05 significance level, can he con- clude that the number of absences has declined? Estimate the p-value.

52. The president of the American Insurance Institute wants to compare the yearly costs of auto insurance offered by two leading companies. He selects a sample of 15 families, some with only a single insured driver, others with several teenage drivers, and pays each family a stipend to contact the two companies and ask for a price quote. To make the data comparable, certain features, such as the deductible amount and limits of liability, are standardized. The sample information is reported below. At the .10 significance level, can we conclude that there is a difference in the amounts quoted?

53. Fairfield Homes is developing two parcels near Pigeon Fork, Tennessee. In order to test different advertising approaches, it uses different media to reach potential buyers. The mean annual family income for 15 people making inquiries at the first development is $150,000, with a standard deviation of $40,000. A corresponding sample of 25 people at the second development had a mean of $180,000, with a standard deviation of $30,000. Assume the population standard deviations are the same. At the .05 significance level, can Fairfield conclude that the population means are different?

54. The following data resulted from a taste test of two different chocolate bars. The first number is a rating of the taste, which could range from 0 to 5, with a 5 indicating the person liked the taste. The second number indicates whether a “secret ingredient” was present. If the ingredient was present, a code of 1 was used and a 0 otherwise. Assume

Progressive GEICO Family Car Insurance Mutual Insurance

Becker $2,090 $1,610 Berry 1,683 1,247 Cobb 1,402 2,327 Debuck 1,830 1,367 DuBrul 930 1,461 Eckroate 697 1,789 German 1,741 1,621 Glasson 1,129 1,914 King 1,018 1,956 Kucic 1,881 1,772 Meredith 1,571 1,375 Obeid 874 1,527 Price 1,579 1,767 Phillips 1,577 1,636 Tresize 860 1,188

Employee Before After

1 6 5 2 6 2 3 7 1 4 7 3 5 4 3 6 3 6 7 5 3 8 6 7

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Two-Sample Tests of Hypothesis 405

the population standard deviations are the same. At the .05 significance level, do these data show a difference in the taste ratings?

55. An investigation of the effectiveness of an antibacterial soap in reducing operating room contamination resulted in the accompanying table. The new soap was tested in a sam- ple of eight operating rooms in the greater Seattle area during the last year.

At the 0.05 significance level, can we conclude the contamination measurements are lower after use of the new soap?

56. The following data on annual rates of return were collected from five stocks listed on the New York Stock Exchange (“the big board”) and five stocks listed on NASDAQ. Assume the population standard deviations are the same. At the .10 significance level, can we con- clude that the annual rates of return are higher on the big board?

57. The city of Laguna Beach operates two public parking lots. The one on Ocean Drive can accommodate up to 125 cars and the one on Rio Rancho can accommodate up to 130 cars. City planners are considering both increasing the size of the lots and changing the fee structure. To begin, the Planning Office would like some information on the num- ber of cars in the lots at various times of the day. A junior planner officer is assigned the task of visiting the two lots at random times of the day and evening and counting the num- ber of cars in the lots. The study lasted over a period of one month. Below is the number of cars in the lots for 25 visits of the Ocean Drive lot and 28 visits of the Rio Rancho lot. Assume the population standard deviations are equal.

NYSE NASDAQ

17.16 15.80 17.08 16.28 15.51 16.21

8.43 17.97 25.15 7.77

Operating Room

A B C D E F G H

Before 6.6 6.5 9.0 10.3 11.2 8.1 6.3 11.6 After 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2.0

Rating With/Without Rating With/Without

3 1 1 1 1 1 4 0 0 0 4 0 2 1 2 1 3 1 3 0 1 1 4 0

Ocean Drive 89 115 93 79 113 77 51 75 118 105 106 91 54 63 121 53 81 115 67 53 69 95 121 88 64

Rio Rancho 128 110 81 126 82 114 93 40 94 45 84 71 74

92 66 69 100 114 113 107 62 77 80 107 90 129 105 124

Is it reasonable to conclude that there is a difference in the mean number of cars in the two lots? Use the .05 significance level.

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406 Chapter 11

59–60. Use this information to do exercises 59 and 60. The drivers, ages, odds against winning, row of their start- ing position, and car number for the 2008 Indianapolis 500 auto race are listed below. Use the .01 signif- icance level.

58. The amount of income spent on housing is an important component of the cost of liv- ing. The total costs of housing for homeowners might include mortgage payments, prop- erty taxes, and utility costs (water, heat, electricity). An economist selected a sample of 20 homeowners in New England and then calculated these total housing costs as a per- cent of monthly income, five years ago and now. The information is reported below. Is it reasonable to conclude the percent is less now than five years ago?

Homeowner Five Years Ago Now Homeowner Five Years Ago Now

1 17% 10% 11 35% 32% 2 20 39 12 16 32 3 29 37 13 23 21 4 43 27 14 33 12 5 36 12 15 44 40 6 43 41 16 44 42 7 45 24 17 28 22 8 19 26 18 29 19 9 49 28 19 39 35

10 49 26 20 22 12

Car Driver Age Odds Row Number

Dixon 27 4 1 9 Wheldon 29 4 1 10 Briscoe 26 4 1 6 Castroneves 33 4 2 3 Patrick 26 8 2 7 Kanaan 33 4 2 11 Andretti, M 21 8 3 26 Meira 31 25 3 4 Mutoh 25 20 3 27 Carpenter 27 50 4 20 Scheckter 27 45 4 12 Bell 33 200 4 99 Rahal 19 40 5 6 Manning 33 100 5 14 Junqueira 31 75 5 18 Wilson 29 50 6 2 Rice 32 50 6 15

Car Driver Age Odds Row Number

Hamilton 45 100 6 22 Lloyd 23 200 7 16 Hunter-Reay 27 100 7 17 Andretti, J 45 100 7 24 Fisher 27 200 8 67 Power 27 100 8 8 Simmons 31 200 8 41 Servia 33 150 9 5 Viso 23 200 9 33 Duno 36 200 9 23 Moraes 19 200 10 19 Bernoldi 29 200 10 36 Camara 27 200 10 34 Foyt 24 150 11 2 Lazier 40 150 11 91 Roth 49 300 11 25

59. Is it reasonable to conclude that starting in the first five rows significantly increases the odds of winning, in contrast to the last four rows?

60. Does having a car number of 20 or below significantly change the odds of winning?

Data Set Exercises 61. Refer to the Real Estate data, which report information on the homes sold in Goodyear,

Arizona, last year. a. At the .05 significance level, can we conclude that there is a difference in the mean

selling price of homes with a pool and homes without a pool? b. At the .05 significance level, can we conclude that there is a difference in the mean sell-

ing price of homes with an attached garage and homes without an attached garage?

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Two-Sample Tests of Hypothesis 407

1. The Minitab commands for the two-sample test of proportions on page 381 are: a. From the toolbar, select Stat, Basic Statistics, and

then 2 Proportions. b. In the next dialog box, select Summarized data, in

the row labeled First enter 100 for Trials and 19 for Events. In the row labeled Second, put 200 for Tri- als and 62 for Events. Then, click on Options and select Use pooled estimate of p for test, and click OK twice.

2. The Excel commands for the two-sample t-test on page 386 are: a. Enter the data into columns A and B (or any other

columns) in the spreadsheet. Use the first row of each column to enter the variable name.

b. Select the Data tab on the top menu. Then, on the far right, select Data Analysis. Select t-Test: Two Sample Assuming Equal Variances, and then click OK.

c. In the dialog box, indicate that the range of Vari- able 1 is from A1 to A6 and Variable 2 from B1 to B7, the Hypothesized Mean Difference is 0, click Labels, Alpha is 0.05, and the Output Range is D1. Click OK.

Software Commands

c. At the .05 significance level, can we conclude that there is a difference in the mean selling price of homes in Township 1 and Township 2?

d. Find the median selling price of the homes. Divide the homes into two groups, those that sold for more than (or equal to) the median price and those that sold for less. Is there a difference in the proportion of homes with a pool for those that sold at or above the median price versus those that sold for less than the median price? Use the .05 significance level.

e. Write a summary report on your findings to parts (a), (b), (c), and (d). Address the report to all real estate agents who sell property in Goodyear.

62. Refer to the Baseball 2009 data, which report information on the 30 Major League Base- ball teams for the 2009 season. a. At the .05 significance level, can we conclude that there is a difference in the mean

salary of teams in the American League versus teams in the National League? b. At the .05 significance level, can we conclude that there is a difference in the mean

home attendance of teams in the American League versus teams in the National League?

c. Compute the mean and the standard deviation of the number of wins for the 10 teams with the highest salaries. Do the same for the 10 teams with the lowest salaries. At the .05 significance level, is there a difference in the mean number of wins for the two groups?

63. Refer to the Buena School District bus data. Is there a difference in the mean mainte- nance cost for the diesel versus the gasoline buses? Use the .05 significance level.

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408 Chapter 11

3. The Minitab commands for the two-sample t-test on page 390 are: a. Put the amount absorbed by the Store brand in C1

and the amount absorbed by the Name brand paper towel in C2.

b. From the toolbar, select Stat, Basic Statistics, and then 2-Sample, and click OK.

c. In the next dialog box, select Samples in different columns, select C1 Store for the First column and C2 Name of the Second and click OK.

4. The Excel commands for the paired t-test on page 395 are: a. Enter the data into columns B and C (or any other

two columns) in the spreadsheet, with the variable names in the first row.

b. Select the Data tab on the top menu. Then, on the far right, select Data Analysis. Select t-Test: Paired Two Sample for Means, and then click OK.

c. In the dialog box, indicate that the range of Vari- able 1 is from B1 to B11 and Variable 2 from C1 to C11, the Hypothesized Mean Difference is 0, click Labels, Alpha is .05, and the Output Range is E1. Click OK.

11–1 a.

The subscript W refers to the women and M to the men.

b. Reject H0 if z � 1.65.

d. Reject the null hypothesis. e. p-value � .5000 � .4826 � .0174 f. The mean amount sold per day is larger for

women. 11–2 a.

b. .10 c. Two-tailed d. Reject H0 if z is less than �1.65 or greater

than 1.65.

H1: �1 � �2

H0: �1 � �2

z � $1,500 � $1,400

($250)2

50 �

($200)2

� 2.11

H1: �W 7 �M H0: �W �M

Chapter 11 Answers to Self-Review

f. Do not reject H0. g. p-value � 2(.5000 � .2454) � .5092

There is no difference in the proportion of adults and children that liked the proposed flavor.

11–3 a.

b. df � 6 � 8 � 2 � 12 Reject H0 if t is less than �2.179 or t is greater than 2.179.

H1: �d � �a H0: �d � �a

z � .58 � .615

.60(.40) 150

� .60(.40)

200

� �0.66

p1 � 87

150 � .58 p2 �

123 200

� .615

pc � 87 � 123

150 � 200 �

210 350

� .60

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Two-Sample Tests of Hypothesis 409

d. Reject H0 because �2.837 is less than the critical value.

e. The p-value is less than .02. f. The mean number of defects is not the same on

the two shifts. g. Independent populations, populations follow

the normal distribution, populations have equal standard deviations.

11–4 a.

so df � 8 c. Reject H0 if t � �1.860.

e. Do not reject H0. f. There is no difference in the mean account

balance of those who applied for their card or were contacted by a telemarketer.

t � $1,568 � $1,967

3562

10 �

8572

� �399.00 323.23

� �1.234

df � [(3562�10) � (8572�8)]2

(3562�10)2

10 � 1 �

(8572�8)2

8 � 1

� 8.93

H0: �c �a H1: �c 6 �a

t � 7.00 � 10.00

A 3.8333a

1 6

� 1 8 b

� �2.837

� 3.8333

s2p � (6 � 1)(1.4142)2 � (8 � 1)(2.2678)2

6 � 8 � 2

X2 � 80 8

� 10.00 s2 � A 36

8 � 1 � 2.2678

X1 � 42 6

� 7.00 s1 � A 10

6 � 1 � 1.4142

11–5 a. . b. Reject H0 if t 2.998. c.

d. Do not reject H0. We cannot conclude that the students lost weight. The p-value is less than .025 but larger than .01.

e. The distribution of the differences must follow a normal distribution.

t � 8.875

8.774�18 � 2.861

sd � A

538.875 8 � 1

� 8.774

d � 71 8

� 8.875

H0: �d � 0, H1: �d 7 0

Name Before After d (d � ) (d � d )2

Hunter 155 154 1 �7.875 62.0156 Cashman 228 207 21 12.125 147.0156 Mervine 141 147 �6 �14.875 221.2656 Massa 162 157 5 �3.875 15.0156 Creola 211 196 15 6.125 37.5156 Peterson 164 150 14 5.125 26.2656 Redding 184 170 14 5.125 26.2656 Poust 172 165 7 �1.875 3.5156

71 538.8750

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12 Learning Objectives When you have completed this chapter, you will be able to:

LO1 List the characteristics of the F distribution and locate values in an F table.

LO2 Perform a test of hypothesis to determine whether the variances of two populations are equal.

LO3 Describe the ANOVA approach for testing differences in sample means.

LO4 Organize data into appropriate ANOVA tables for analysis.

LO5 Conduct a test of hypothesis among three or more treatment means and describe the results.

LO6 Develop confidence intervals for the differences between treatment means and interpret the results.

LO7 Carry out a test of hypothesis among treatment means using a blocking variable and understand the results.

LO8 Perform a two-way ANOVA with interaction and describe the results.

Analysis of Variance

A computer manufacturer is about to unveil a new, faster personal

computer. The new machine clearly is faster, but initial tests indicate

there is more variation in the processing time, which depends on

the program being run, and the amount of input and output data.

A sample of 16 computer runs, covering a range of production jobs,

showed that the standard deviation of the processing time was 22

(hundredths of a second) for the new machine and 12 (hundredths of

a second) for the current machine. At the .05 significance level, can

we conclude that there is more variation in the processing time of the

new machine? (See Exercise 24 and LO2.)

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Analysis of Variance 411

12.1 Introduction In this chapter, we continue our discussion of hypothesis testing. Recall that in Chap- ters 10 and 11 we examined the general theory of hypothesis testing. We described the case where a sample was selected from the population. We used the z distri- bution (the standard normal distribution) or the t distribution to determine whether it was reasonable to conclude that the population mean was equal to a specified value. We tested whether two population means are the same. We also conducted both one- and two-sample tests for population proportions, using the standard normal dis- tribution as the distribution of the test statistic. In this chapter, we expand our idea of hypothesis tests. We describe a test for variances and then a test that simulta- neously compares several means to determine if they came from equal populations.

12.2 The F Distribution The probability distribution used in this chapter is the F distribution. It was named to honor Sir Ronald Fisher, one of the founders of modern-day statistics. The test statistic for several situations follows this probability distribution. It is used to test whether two samples are from populations having equal variances, and it is also applied when we want to compare several population means simultaneously. The simultaneous comparison of several population means is called analysis of vari- ance (ANOVA). In both of these situations, the populations must follow a normal distribution, and the data must be at least interval-scale.

What are the characteristics of the F distribution?

1. There is a family of F distributions. A particular member of the family is deter- mined by two parameters: the degrees of freedom in the numerator and the degrees of freedom in the denominator. The shape of the distribution is illus- trated by the following graph. There is one F distribution for the combination of 29 degrees of freedom in the numerator (df ) and 28 degrees of freedom in the denominator. There is another F distribution for 19 degrees in the numer- ator and 6 degrees of freedom in the denominator. The final distribution shown has 6 degrees of freedom in the numerator and 6 degrees of freedom in the denominator. We will describe the concept of degrees of freedom later in the chapter. Note that the shapes of the distributions change as the degrees of freedom change.

0 1 2 3 4 5

Re la

tiv e

fre qu

en cy

df � (29, 28)

df � (19, 6)

df � (6, 6)

2. The F distribution is continuous. This means that it can assume an infinite number of values between zero and positive infinity.

3. The F distribution cannot be negative. The smallest value F can assume is 0.

LO1 List the characteristics of the F distribution and locate values in an F Table.

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412 Chapter 12

4. It is positively skewed. The long tail of the distribution is to the right-hand side. As the number of degrees of freedom increases in both the numerator and denominator, the distribution approaches a normal distribution.

5. It is asymptotic. As the values of X increase, the F distribution approaches the X-axis but never touches it. This is similar to the behavior of the normal prob- ability distribution, described in Chapter 7.

12.3 Comparing Two Population Variances The first application of the F distribution that we describe occurs when we test the hypothesis that the variance of one normal population equals the variance of another normal population. The following examples will show the use of the test:

• Two Barth shearing machines are set to produce steel bars of the same length. The bars, therefore, should have the same mean length. We want to ensure that in addition to having the same mean length they also have similar variation.

• The mean rate of return on two types of common stock may be the same, but there may be more variation in the rate of return in one than the other. A sample of 10 technology and 10 utility stocks shows the same mean rate of return, but there is likely more variation in the technology stocks.

• A study by the marketing department for a large newspaper found that men and women spent about the same amount of time per day surfing the Net. However, the same report indicated

there was nearly twice as much variation in time spent per day among the men than the women.

The F distribution is also used to test assumptions for some statistical tests. Recall that in the previous chapter we used the t test to investigate whether the means of two independent populations differed. To employ that test, we sometimes assume that the variances of two normal populations are the same. See this list of assump- tions in Section 11.4 on page 384. The F distribution is used to test if the variances of two normal populations are equal.

Regardless of whether we want to determine whether one population has more variation than another population or validate an assumption for a statistical test, we first state the null hypothesis. The null hypothesis is that the variance of one normal population, equals the variance of the other normal population, The alternate hypothesis could be that the variances differ. In this instance, the null hypothesis and the alternate hypothesis are:

To conduct the test, we select a random sample of observations from one pop- ulation, and a random sample of observations from the second population. The test statistic is defined as follows.

H1: � 2 1 � �

2 2

H0: � 2 1 � �

2 2

�22.� 2 1,

TEST STATISTIC FOR COMPARING [12–1] TWO VARIANCES

F � s21 s22

LO2 Perform a test of hypothesis to determine whether the variances of two populations are equal.

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Analysis of Variance 413

The terms s21 and s 2 2 are the respective sample variances. If the null hypothesis is true,

the test statistic follows the F distribution with n1 � 1 and n2 � 1 degrees of freedom. In order to reduce the size of the table of critical values, the larger sample variance is placed in the numerator; hence, the tabled F ratio is always larger than 1.00. Thus, the right-tail critical value is the only one required. The critical value of F for a two- tailed test is found by dividing the significance level in half and then referring to the appropriate degrees of freedom in Appendix B.4. An example will illustrate.

(��2)

Example Lammers Limos offers limousine service from the city hall in Toledo, Ohio, to Metro Airport in Detroit. Sean Lammers, president of the company, is considering two routes. One is via U.S. 25 and the other via I-75. He wants to study the time it takes to drive to the airport using each route and then compare the results. He collected the fol- lowing sample data, which is reported in minutes. Using the .10 significance level, is there a difference in the variation in the dri- ving times for the two routes?

U.S. Route 25 Interstate 75

52 59 67 60 56 61 45 51 70 56 54 63 64 57

Solution The mean driving times along the two routes are nearly the same. The mean time is 58.29 minutes for the U.S. 25 route and 59.0 minutes along the I-75 route. How- ever, in evaluating travel times, Mr. Lammers is also concerned about the variation in the travel times. The first step is to compute the two sample variances. We’ll use formula (3–11) to compute the sample standard deviations. To obtain the sample variances, we square the standard deviations.

U.S. Route 25

Interstate 75

There is more variation, as measured by the standard deviation, in the U.S. 25 route than in the I-75 route. This is consistent with his knowledge of the two routes; the U.S. 25 route contains more stoplights, whereas I-75 is a limited-access interstate

X � ©X n

� 472 8

� 59.00 s � B

©(X � X)2

n � 1 � B

134 8 � 1

� 4.3753

X � ©X n

� 408

7 � 58.29 s �

©(X � X )2

n � 1 � B

485.43 7 � 1

� 8.9947

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414 Chapter 12

highway. However, the I-75 route is several miles longer. It is important that the service offered be both timely and consistent, so he decides to conduct a statisti- cal test to determine whether there really is a difference in the variation of the two routes.

The usual five-step hypothesis-testing procedure will be employed.

Step 1: We begin by stating the null hypothesis and the alternate hypothesis. The test is two-tailed because we are looking for a difference in the variation of the two routes. We are not trying to show that one route has more variation than the other.

Step 2: We selected the .10 significance level. Step 3: The appropriate test statistic follows the F distribution. Step 4: The critical value is obtained from Appendix B.4, a portion of which is re-

produced as Table 12–1. Because we are conducting a two-tailed test, the tabled significance level is .05, found by There are

degrees of freedom in the numerator, and degrees of freedom in the denominator. To find the critical

value, move horizontally across the top portion of the F table (Table 12–1 or Appendix B.4) for the .05 significance level to 6 degrees of freedom in the numerator. Then move down that column to the critical value oppo- site 7 degrees of freedom in the denominator. The critical value is 3.87. Thus, the decision rule is: Reject the null hypothesis if the ratio of the sample variances exceeds 3.87.

8 � 1 � 7 n2 � 1 �n1 � 1 � 7 � 1 � 6

��2 � .10�2 � .05.

H1: � 2 1 � �

2 2

H0: � 2 1 � �

2 2

Step 5: The final step is to take the ratio of the two sample variances, determine the value of the test statistic, and make a decision regarding the null hy- pothesis. Note that formula (12–1) refers to the sample variances but we calculated the sample standard deviations. We need to square the stan- dard deviations to determine the variances.

The decision is to reject the null hypothesis, because the computed F value (4.23) is larger than the critical value (3.87). We conclude that there is a difference in the variation of the travel times along the two routes.

F � s21 s22

� (8.9947)2

(4.3753)2 � 4.23

TABLE 12–1 Critical Values of the F Distribution, � � .05

Degrees of Freedom for NumeratorDegrees of Freedom for Denominator 5 6 7 8

1 230 234 237 239 2 19.3 19.3 19.4 19.4 3 9.01 8.94 8.89 8.85 4 6.26 6.16 6.09 6.04 5 5.05 4.95 4.88 4.82 6 4.39 4.28 4.21 4.15 7 3.97 3.87 3.79 3.73 8 3.69 3.58 3.50 3.44 9 3.48 3.37 3.29 3.23

10 3.33 3.22 3.14 3.07

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Analysis of Variance 415

As noted, the usual practice is to determine the F ratio by putting the larger of the two sample variances in the numerator. This will force the F ratio to be at least 1.00. This allows us to always use the right tail of the F distribution, thus avoiding the need for more extensive F tables.

A logical question arises: Is it possible to conduct one-tailed tests? For exam- ple, suppose in the previous example we suspected that the variance of the times using the U.S. 25 route is larger than the variance of the times along the I-75 route. We would state the null and the alternate hypothesis as

The test statistic is computed as Notice that we labeled the population with the suspected largest variance as population 1. So appears in the numerator. The F ratio will be larger than 1.00, so we can use the upper tail of the F distribution. Under these conditions, it is not necessary to divide the significance level in half. Because Appendix B.4 gives us only the .05 and .01 significance levels, we are restricted to these levels for one-tailed tests and .10 and .02 for two-tailed tests unless we consult a more complete table or use statistical software to compute the F statistic.

The Excel software system has a procedure to perform a test of variances. Below is the output. The computed value of F is the same as that determined by using formula (12–1).

s21

s21/s 2 2.

H1: � 2 1 7 �22

H0: � 2 1 � �

2 2

Self-Review 12–1 Steele Electric Products Inc. assembles electrical components for cell phones. For the last 10 days, Mark Nagy has averaged 9 rejects, with a standard deviation of 2 rejects per day. Debbie Richmond averaged 8.5 rejects, with a standard deviation of 1.5 rejects, over the same period. At the .05 significance level, can we conclude that there is more varia- tion in the number of rejects per day attributed to Mark?

Exercises 1. What is the critical F value for a sample of six observations in the numerator and four in

the denominator? Use a two-tailed test and the .10 significance level. 2. What is the critical F value for a sample of four observations in the numerator and seven

in the denominator? Use a one-tailed test and the .01 significance level.

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416 Chapter 12

3. The following hypotheses are given.

A random sample of eight observations from the first population resulted in a standard deviation of 10. A random sample of six observations from the second population resulted in a standard deviation of 7. At the .02 significance level, is there a difference in the vari- ation of the two populations?

4. The following hypotheses are given.

A random sample of five observations from the first population resulted in a standard deviation of 12. A random sample of seven observations from the second population showed a standard deviation of 7. At the .01 significance level, is there more variation in the first population?

5. Arbitron Media Research Inc. conducted a study of the iPod listening habits of men and women. One facet of the study involved the mean listening time. It was discovered that the mean listening time for men was 35 minutes per day. The standard deviation of the sample of the 10 men studied was 10 minutes per day. The mean listening time for the 12 women studied was also 35 minutes, but the standard deviation of the sample was 12 minutes. At the .10 significance level, can we conclude that there is a difference in the variation in the listening times for men and women?

6. A stockbroker at Critical Securities reported that the mean rate of return on a sample of 10 oil stocks was 12.6 percent with a standard deviation of 3.9 percent. The mean rate of return on a sample of 8 utility stocks was 10.9 percent with a standard deviation of 3.5 percent. At the .05 significance level, can we conclude that there is more variation in the oil stocks?

12.4 ANOVA Assumptions Another use of the F distribution is the analysis of variance (ANOVA) technique in which we compare three or more population means to determine whether they could be equal. To use ANOVA, we assume the following:

1. The populations follow the normal distribution. 2. The populations have equal standard deviations 3. The populations are independent.

When these conditions are met, F is used as the distribution of the test statistic. Why do we need to study ANOVA? Why can’t we just use the test of differ-

ences in population means discussed in the previous chapter? We could compare the population means two at a time. The major reason is the unsatisfactory buildup of Type I error. To explain further, suppose we have four different methods (A, B, C, and D) of training new recruits to be firefighters. We randomly assign each of the 40 recruits in this year’s class to one of the four methods. At the end of the train- ing program, we administer to the four groups a common test to measure under- standing of firefighting techniques. The question is: Is there a difference in the mean test scores among the four groups? An answer to this question will allow us to compare the four training methods.

Using the t distribution to compare the four population means, we would have to conduct six different t tests. That is, we would need to compare the mean scores for the four methods as follows: A versus B, A versus C, A versus D, B versus C, B versus D, and C versus D. If we set the significance level at .05, the probability of a correct statistical decision is .95, found by Because we conduct six1 � .05.

(�).

H1: � 2 1 7 �22

H0: � 2 1 � �

2 2

H1: � 2 1 � �

2 2

H0: � 2 1 � �

2 2

Using the t distribution leads to a buildup of Type I error.

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Example Joyce Kuhlman manages a regional financial center. She wishes to compare the pro- ductivity, as measured by the number of customers served, among three employ- ees. Four days are randomly selected and the number of customers served by each employee is recorded. The results are:

Solution Is there a difference in the mean number of customers served? Chart 12–1 illus- trates how the populations would appear if there were a difference in the treatment means. Note that the populations follow the normal distribution and the variation in each population is the same. However, the means are not the same.

Analysis of Variance 417

separate (independent) tests, the probability that we do not make an incorrect decision due to sampling error in any of the six independent tests is:

To find the probability of at least one error due to sampling, we subtract this result from 1. Thus, the probability of at least one incorrect decision due to sampling is

To summarize, if we conduct six independent tests using the t dis- tribution, the likelihood of rejecting a true null hypothesis because of sampling error is increased from .05 to an unsatisfactory level of .265. It is obvious that we need a better method than conducting six t tests. ANOVA will allow us to compare the treatment means simultaneously and avoid the buildup of Type I error.

ANOVA was first developed for applications in agriculture, and many of the terms related to that context remain. In particular, the term treatment is used to iden- tify the different populations being examined. For example, treatment refers to how a plot of ground was treated with a particular type of fertilizer. The following illus- tration will clarify the term treatment and demonstrate an application of ANOVA.

1 � .735 � .265.

P(All correct) � (.95)(.95)(.95)(.95)(.95)(.95) � .735

Wolfe White Korosa

55 66 47 54 76 51 59 67 46 56 71 48

Em pl

oy ee

Customers served

White

Korosa

Wolfe

μ1 μ2 μ3

CHART 12–1 Case Where Treatment Means Are Different

LO3 Describe the ANOVA approach for testing differences in sample means.

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418 Chapter 12

12.5 The ANOVA Test How does the ANOVA test work? Recall that we want to determine whether the var- ious sample means came from a single population or populations with different means. We actually compare these sample means through their variances. To explain, on page 416 we listed the assumptions required for ANOVA. One of those assumptions was that the standard deviations of the various normal populations had to be the same. We take advantage of this requirement in the ANOVA test. The underlying strategy is to estimate the population variance (standard deviation squared) two ways and then find the ratio of these two estimates. If this ratio is about 1, then logically the two estimates are the same, and we conclude that the population means are the same. If the ratio is quite different from 1, then we con- clude that the population means are not the same. The F distribution serves as a referee by indicating when the ratio of the sample variances is too much greater than 1 to have occurred by chance.

Refer to the financial center example in the previous section. The manager wants to determine whether there is a difference in the mean number of customers served. To begin, find the overall mean of the 12 observations. It is 58, found by

Next, for each of the 12 observations find the difference between the particular value and the overall mean. Each of these differences is squared and these squares summed. This term is called the total variation.

(55 � 54 � . . . � 48)�12.

Suppose the populations are the same. That is, there is no difference in the (treatment) means. This is shown in Chart 12–2. This would indicate that the pop- ulation means are the same. Note again that the populations follow the normal distribution and the variation in each of the populations is the same.

Em plo

ye e

Customers served

White

Korosa

Wolfe

µ1 µ2 µ3� �

CHART 12–2 Case Where Treatment Means Are the Same

TOTAL VARIATION The sum of the squared differences between each observation and the overall mean.

In our example, the total variation is 1,082, found by

Next, break this total variation into two components: that which is due to the treatments and that which is random. To find these two components, determine

(48 � 58)2. (55 � 58)2 � (54 � 58)2 � . . . �

LO5 Conduct a test of hypothesis among three or more treatment means and describe the results.

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Analysis of Variance 419

In the example, the variation due to the treatments is the sum of the squared differences between the mean number of customers served by each employee and the overall mean. This term is 992. To calculate it, we first find the mean of each of the three treatments. The mean for Wolfe is 56 customers, found by

The other means are 70 and 48, respectively. The sum of the squares due to the treatments is:

� 992

(56 � 58)2 � (56 � 58)2 � . . . � (48 � 58)2 � 4(56 � 58)2 � 4(70 � 58)2 � 4(48 � 58)2

(55 � 54 � 59 � 56)�4.

RANDOM VARIATION The sum of the squared differences between each observation and its treatment mean.

In the example, this term is the sum of the squared differences between each value and the mean for that particular employee. The error variation is 90.

We determine the test statistic, which is the ratio of the two estimates of the population variance, from the following equation.

Our first estimate of the population variance is based on the treatments, that is, the difference between the means. It is 992�2. Why did we divide by 2? Recall from Chapter 3, to find a sample variance [see formula (3–11)], we divide by the number of observations minus one. In this case, there are three treatments, so we divide by 2. Our first estimate of the population variance is 992�2.

The variance estimate within the treatments is the random variation divided by the total number of observations less the number of treatments—that is, Hence, our second estimate of the population variance is 90�9. This is actually a generalization of formula (11–5), where we pooled the sample variances from two populations.

The last step is to take the ratio of these two estimates.

Because this ratio is quite different from 1, we can conclude that the treatment means are not the same. There is a difference in the mean number of customers served by the three employees.

Here’s another example, which deals with samples of different sizes.

F � 992�2 90�9

� 49.6

90 (12 � 3).

F �

Estimate of the population variance based on the differences among the sample means

Estimate of the population variance based on the variation within the sample

(55 � 56)2 � (54 � 56)2 � . . . � (48 � 48)2 � 90

TREATMENT VARIATION The sum of the squared differences between each treatment mean and the grand or overall mean.

If there is considerable variation among the treatment means, it is logical that this term will be large. If the treatment means are similar, this term will be a small value. The smallest possible value would be zero. This would occur when all the treatment means are the same.

The other source of variation is referred to as the random component, or the error component.

the mean of each of the treatments. The first source of variation is due to the treatments.

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Example Recently airlines have cut services, such as meals and snacks during flights, and started charging extra for some services, such as accommodating overweight lug- gage, last-minute flight changes, and pets traveling in the cabin. However, they are still concerned about service. Recently, a group of four carriers hired Brunner Mar- keting Research Inc. to survey passengers regarding their level of satisfaction with a recent flight. The survey included questions on ticketing, boarding, in-flight service, baggage handling, pilot communication, and so forth. Twenty-five questions offered a range of possible answers: excellent, good, fair, or poor. A response of excellent was given a score of 4, good a 3, fair a 2, and poor a 1. These responses were then totaled, so the total score was an indication of the satisfaction with the flight. The greater the score, the higher the level of satisfaction with the service. The highest pos- sible score was 100.

Brunner randomly selected and surveyed passengers from the four airlines. Below is the sample information. Is there a difference in the mean satisfaction level among the four airlines? Use the .01 significance level.

Solution We will use the five-step hypothesis-testing procedure. Step 1: State the null hypothesis and the alternate hypothesis. The null

hypothesis is that the mean scores are the same for the four airlines.

The alternate hypothesis is that the mean scores are not all the same for the four airlines.

The mean scores are not all equal.

We can also think of the alternate hypothesis as “at least two mean scores are not equal.”

If the null hypothesis is not rejected, we conclude that there is no dif- ference in the mean scores for the four airlines. If is rejected, we con- clude that there is a difference in at least one pair of mean scores, but at this point we do not know which pair or how many pairs differ.

Step 2: Select the level of significance. We selected the .01 significance level. Step 3: Determine the test statistic. The test statistic follows the F distribution. Step 4: Formulate the decision rule. To determine the decision rule, we need the

critical value. The critical value for the F statistic is found in Appendix B.4. The critical values for the .05 significance level are found on the first page and the .01 significance level on the second page. To use this table, we need to know the degrees of freedom in the numerator and the denominator. The degrees of freedom in the numerator equals the number of treatments, des- ignated as k, minus 1. The degrees of freedom in the denominator is the total number of observations, n, minus the number of treatments. For this problem, there are four treatments and a total of 22 observations.

Degrees of freedom in the denominator � n � k � 22 � 4 � 18

Degrees of freedom in the numerator � k � 1 � 4 � 1 � 3

H1:

H0: �1 � �2 � �3 � �4

420 Chapter 12

Northern WTA Pocono Branson

94 75 70 68 90 68 73 70 85 77 76 72 80 83 78 65

88 80 74 68 65 65

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Analysis of Variance 421

Refer to Appendix B.4 and the .01 significance level. Move horizontally across the top of the page to 3 degrees of freedom in the numerator. Then move down that column to the row with 18 degrees of freedom. The value at this intersection is 5.09. So the decision rule is to reject if the computed value of F exceeds 5.09.

Step 5: Select the sample, perform the calculations, and make a decision. It is convenient to summarize the calculations of the F statistic in an ANOVA table. The format for an ANOVA table is as follows. Statistical software packages also use this format.

There are three values, or sum of squares, used to compute the test statistic F. You can determine these values by obtaining SS total and SSE, then finding SST by subtraction. The SS total term is the total variation, SST is the variation due to the treatments, and SSE is the variation within the treatments or the random error.

We usually start the process by finding SS total. This is the sum of the squared differences between each observation and the overall mean. The formula for finding SS total is:

[12–2]

where: X is each sample observation.

is the overall or grand mean.

Next determine SSE or the sum of the squared errors. This is the sum of the squared differences between each observation and its respective treatment mean. The formula for finding SSE is:

[12–3]

where: is the sample mean for treatment c.

The detailed calculations of SS total and SSE for this example follow. To deter- mine the values of SS total and SSE we start by calculating the overall or grand mean. There are 22 observations and the total is 1,664, so the grand mean is 75.64.

XG � 1,664

22 � 75.64

SSE � ©(X � Xc)2

SS total � ©(X � XG)2

ANOVA Table

Source of Variation Sum of Squares Degrees of Freedom Mean Square F

Treatments SST k � 1 SST�(k � 1) � MST MST�MSE Error SSE n � k SSE�(n � k) � MSE

Total SS total n � 1

Northern WTA Pocono Branson Total

94 75 70 68 90 68 73 70 85 77 76 72 80 83 78 65

88 80 74 68 65 65

Column total 349 391 510 414 1,664 n 4 5 7 6 22 Mean 87.25 78.20 72.86 69.00 75.64

LO4 Organize data into appropriate ANOVA tables for analysis.

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422 Chapter 12

Next we find the deviation of each observation from the grand mean, square those deviations, and sum this result for all 22 observations. For example, the first sampled passenger had a score of 94 and the overall or grand mean is 75.64. So For the last passenger,

The calculations for all other passengers follow.�10.64. (X � XG) � 65 � 75.64 �(X � XG) � 94 � 75.64 � 18.36.

Northern WTA Pocono Branson

18.36 14.36

9.36 1.36 0.36 4.36 7.36 2.36

12.36 4.36

�10.64 �10.64�7.64 �1.64

�10.64 �3.64 �5.64�2.64�7.64 �7.64�5.64�0.64

Then square each of these differences and sum all the values. Thus, for the first passenger:

Finally, sum all the squared differences as formula (12–2) directs. Our SS total value is 1,485.10.

(X � XG) 2 � (94 � 75.64)2 � (18.36)2 � 337.09

Northern WTA Pocono Branson Total

337.09 0.41 31.81 58.37 206.21 58.37 6.97 31.81

87.61 1.85 0.13 13.25 19.01 54.17 5.57 113.21

152.77 19.01 2.69 58.37 113.21

113.21 Total 649.92 267.57 235.07 332.54 1,485.10

To compute the term SSE, find the deviation between each observation and its treat- ment mean. In the example, the mean of the first treatment (that is, the passengers on Northern Airlines) is 87.25, found by The subscript N refers to Northern Airlines.

The first passenger rated Northern a 94, so The first passenger in the WTA group responded with a total score of 75, so

The detail for all the passengers follows.(X � XW) � (75 � 78.20) � �3.2.

(X � XN) � (94 � 87.25) � 6.75.

XN � 349�4.

Northern WTA Pocono Branson

6.75 �3.2 �2.86 �1 2.75 �10.2 0.14 1

�2.25 �1.2 3.14 3 �7.25 4.8 5.14 �4

9.8 7.14 5 �4.86 �4 �7.86

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Analysis of Variance 423

As noted in the previous example, the calculations are tedious if the number of observations in each treatment is large. There are many software packages that will perform the calculations and output the results. Following is the Excel output in the form of an ANOVA table for the previous example involving airlines and passenger

Northern WTA Pocono Branson Total

45.5625 10.24 8.18 1 7.5625 104.04 0.02 1 5.0625 1.44 9.86 9

52.5625 23.04 26.42 16 96.04 50.98 25

23.62 16 61.78

Total 110.7500 234.80 180.86 68 594.41

So the SSE value is 594.41. That is Finally, we determine SST, the sum of the squares due to the treatments, by

subtraction.

[12–4]

For this example:

To find the computed value of F, work your way across the ANOVA table. The degrees of freedom for the numerator and the denominator are the same as in step 4 on page 420 when we were finding the critical value of F. The term mean square is another expression for an estimate of the variance. The mean square for treat- ments is SST divided by its degrees of freedom. The result is the mean square for treatments and is written MST. Compute the mean square error in a similar fash- ion. To be precise, divide SSE by its degrees of freedom. To complete the process and find F, divide MST by MSE.

Insert the particular values of F into an ANOVA table and compute the value of F as follows.

SST � SS total � SSE � 1,485.10 � 594.41 � 890.69.

SST � SS total � SSE

©(X � Xc)2 � 594.41.

Source of Variation Sum of Squares Degrees of Freedom Mean Square F

Treatments 890.69 3 296.90 8.99 Error 594.41 18 33.02

Total 1,485.10 21

The computed value of F is 8.99, which is greater than the critical value of 5.09, so the null hypothesis is rejected. We conclude the population means are not all equal. The mean scores are not the same for the four airlines. It is likely that the passenger scores are related to the particular airline. At this point, we can only con- clude there is a difference in the treatment means. We cannot determine which treat- ment groups differ or how many treatment groups differ.

Each of these values is squared and then summed for all 22 observations. The values are shown in the following table.

Statistics in Action

Have you ever waited in line for a tele- phone and it seemed like the person using the phone talked on and on? There is evi- dence that people actually talk longer on public telephones when someone is waiting. In a recent survey, researchers measured the length of time that 56 shop- pers in a mall spent on the phone (1) when they were alone, (2) when a person was using the adjacent phone, and (3) when a person was using an adja- cent phone and someone was waiting to use the phone. The study, using the one-way ANOVA technique, showed that the mean time using the telephone was significantly less when the person was alone.

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424 Chapter 12

Notice Excel uses the term “Between Groups” for treatments and “Within Groups” for error. However, they have the same meanings. The p-value is .0007. This is the probability of finding a value of the test statistic this large or larger when the null hypothesis is true. To put it another way, it is the likelihood of calculating an F value larger than 8.99 with 3 degrees of freedom in the numerator and 18 degrees of free- dom in the denominator. So when we reject the null hypothesis in this instance, there is a very small likelihood of committing a Type I error!

Following is the Minitab output from the airline passenger ratings example, which is similar to the Excel output. The output is also in the form of an ANOVA table. In addition, Minitab provides information about the differences between means. This is discussed in the next section.

The Minitab system uses the term “Factor” instead of treatment, with the same intended meaning.

ratings. There are some slight differences between the software output and the pre- vious calculations. These differences are due to rounding.

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Analysis of Variance 425

Exercises 7. The following is sample information. Test the hypothesis that the treatment means are

equal. Use the .05 significance level.

Self-Review 12–2 Citrus Clean is a new all-purpose cleaner being test-marketed by placing displays in three different locations within various supermarkets. The number of 12-ounce bottles sold from each location within the supermarket is reported below.

At the .05 significance level, is there a difference in the mean number of bottles sold at the three locations? (a) State the null hypothesis and the alternate hypothesis. (b) What is the decision rule? (c) Compute the values of SS total, SST, and SSE. (d) Develop an ANOVA table. (e) What is your decision regarding the null hypothesis?

Near bread 18 14 19 17 Near beer 12 18 10 16 Other cleaners 26 28 30 32

Treatment 1 Treatment 2 Treatment 3

8 3 3 6 2 4

10 4 5 9 3 4

Treatment 1 Treatment 2 Treatment 3

9 13 10 7 20 9

11 14 15 9 13 14

12 15 10

a. State the null hypothesis and the alternate hypotheses. b. What is the decision rule? c. Compute SST, SSE, and SS total. d. Complete an ANOVA table. e. State your decision regarding the null hypothesis.

8. The following is sample information. Test the hypothesis at the .05 significance level that the treatment means are equal.

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426 Chapter 12

9. A real estate developer is considering investing in a shopping mall on the outskirts of Atlanta, Georgia. Three parcels of land are being evaluated. Of particular importance is the income in the area surrounding the proposed mall. A random sample of four families is selected near each proposed mall. Following are the sample results. At the .05 significance level, can the developer conclude there is a difference in the mean income? Use the usual five-step hypothesis testing procedure.

Southwyck Area Franklin Park Old Orchard ($000) ($000) ($000)

64 74 75 68 71 80 70 69 76 60 70 78

10. The manager of a computer software company wishes to study the number of hours senior executives by type of industry spend at their desktop computers. The manager selected a sample of five executives from each of three industries. At the .05 significance level, can she conclude there is a difference in the mean number of hours spent per week by industry?

Banking Retail Insurance

12 8 10 10 8 8 10 6 6 12 8 8 10 10 10

12.6 Inferences about Pairs of Treatment Means Suppose we carry out the ANOVA procedure and make the decision to reject the null hypothesis. This allows us to conclude that all the treatment means are not the same. Sometimes we may be satisfied with this conclusion, but in other instances we may want to know which treatment means differ. This section provides the details for such a test.

Recall that in the Brunner Research example regarding airline passenger ratings, there was a difference in the treatment means. That is, the null hypothesis was rejected and the alternate hypothesis accepted. If the passenger ratings do differ, the question is: Between which groups do the treatment means differ?

Several procedures are available to answer this question. The simplest is through the use of confidence intervals, that is, formula (9–2). From the computer output of the previous example (see page 424), note that the sample mean score for those passengers rating Northern’s service is 87.25, and for those rating Branson’s service, the sample mean score is 69.00. Is there enough disparity to justify the con- clusion that there is a significant difference in the mean satisfaction scores of the two airlines?

The t distribution, described in Chapters 10 and 11, is used as the basis for this test. Recall that one of the assumptions of ANOVA is that the population variances are the same for all treatments. This common population value is the

LO6 Develop confidence intervals for the differences between treatment means and interpret the results.

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DIFFERENCE IN TREATMENT MEANS

Analysis of Variance 427

mean square error, or MSE, and is determined by A confidence inter- val for the difference between two populations is found by:

CONFIDENCE INTERVAL FOR THE [12–5]

where: is the mean of the first sample. is the mean of the second sample.

t is obtained from Appendix B.2. The degrees of freedom is equal to MSE is the mean square error term obtained from the ANOVA table [SSE (n � k)].

n1 is the number of observations in the first sample. n2 is the number of observations in the second sample.

How do we decide whether there is a difference in the treatment means? If the confidence interval includes zero, there is not a difference between the treatment means. For example, if the left endpoint of the confidence interval has a negative sign and the right endpoint has a positive sign, the interval includes zero and the two means do not differ. So if we develop a confidence interval from formula (12–5) and find the difference in the sample means was 5.00—that is, if and

—the confidence interval would range from �7.00 up to

17.00. To put it in symbols:

Note that zero is included in this interval. Therefore, we conclude that there is no significant difference in the selected treatment means.

On the other hand, if the endpoints of the confidence interval have the same sign, this indicates that the treatment means differ. For example, if and

the confidence interval would range from �0.60 up to

�0.10. Because �0.60 and �0.10 have the same sign, both negative, zero is not in the interval and we conclude that these treatment means differ.

Using the previous airline example, let us compute the confidence interval for the difference between the mean scores of passengers on Northern and Branson. With a 95 percent level of confidence, the endpoints of the confidence interval are 10.46 and 26.04.

where: is 87.25. is 69.00.

t is 2.101: from Appendix B.2 with (n � k) � 22 � 4 � 18 degrees of freedom. MSE is 33.0: from the ANOVA table with SSE (n � k) � 594.4 18.

nA is 4. nUS is 6.

The 95 percent confidence interval ranges from 10.46 up to 26.04. Both endpoints are positive; hence, we can conclude these treatment means differ significantly.

��

XUS

� 18.25 � 7.79

(XA � XUS) � t B

MSEa 1 nA

� 1

nUS b � (87.25 � 69.00) � 2.101

B 33.0a

1 4

� 1 6 b

t B

MSEa 1 n1

� 1 n2 b � 0.25,

X1 � X2 � �0.35

t B

MSEa 1 n1

� 1 n2 b � 12

X1 � X2 � 5

� n � k.

(X1 � X2) � tB MSEa

1 n1

� 1 n2 b

SSE�(n � k).

(X1 � X2) � tBMSEa 1 n1

� 1 n2 b � 5.00 � 12.00 � �7.00 up to 17.00

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428 Chapter 12

On the right side of the printout is a confidence interval for each treatment mean. The asterisk (*) indicates the location of the treatment mean and the open parenthesis and close parenthesis, the endpoints of the confidence interval. In those instances where the intervals overlap, the treatment means may not differ. If there is no common area in the confidence intervals, that pair of means differ.

The endpoints of a 95 percent confidence interval for mean passenger scores for Pocono are about 69 and 77. For Branson, the endpoints of the 95 percent con- fidence interval for the mean passenger score are about 64 and 73. There is com- mon area between these points, so we conclude this pair of means does not differ. In other words, there is no significant difference between the mean passenger ratings for Pocono and Branson Airlines. The difference in the mean scores is due to chance.

There are two pairs of means that differ. The mean scores of passengers on Northern Airlines are significantly different from the mean ratings of passengers on Pocono Airlines and on Branson Airlines. There is no common area between these two pairs of confidence intervals.

We should emphasize that this investigation is a step-by-step process. The initial step is to conduct the ANOVA test. Only if the null hypothesis that the treat- ment means are equal is rejected should any analysis of the individual treatment means be attempted.

Self-Review 12–3 The following data are the semester tuition charges ($000) for a sample of private colleges in various regions of the United States. At the .05 significance level, can we conclude there is a difference in the mean tuition rates for the various regions?

(a) State the null and the alternate hypotheses. (b) What is the decision rule? (c) Develop an ANOVA table. What is the value of the test statistic? (d) What is your decision regarding the null hypothesis? (e) Could there be a significant difference between the mean tuition in the Northeast and

that of the West? If so, develop a 95 percent confidence interval for that difference.

Northeast Southeast West ($000) ($000) ($000)

10 8 7 11 9 8 12 10 6 10 8 7 12 6

That is, passengers on Northern Airlines rated service significantly different from those on Branson Airlines.

Approximate results can also be obtained directly from the Minitab output. Fol- lowing is the lower portion of the output from page 424. On the left side is the number of observations, the mean, and the standard deviation for each treatment. Seven pas- sengers on Pocono rated the service as 72.857 with a standard deviation of 5.490.

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Analysis of Variance 429

a. State the null hypothesis and the alternate hypothesis. b. What is the decision rule? c. Compute SST, SSE, and SS total. d. Complete an ANOVA table. e. State your decision regarding the null hypothesis. f. If is rejected, can we conclude that treatment 1 and treatment 2 differ? Use the

95 percent level of confidence. 12. Given the following sample information, test the hypothesis that the treatment means are

equal at the .05 significance level.

Treatment 1 Treatment 2 Treatment 3

8 3 3 11 2 4 10 1 5

3 4 2

Treatment 1 Treatment 2 Treatment 3

3 9 6 2 6 3 5 5 5 1 6 5 3 8 5 1 5 4

4 1 7 5 6 4

95 percent level of confidence. 13. A senior accounting major at Midsouth State University has job offers from four CPA

firms. To explore the offers further, she asked a sample of recent trainees how many months each worked for the firm before receiving a raise in salary. The sample informa- tion is submitted to Minitab with the following results:

Analysis of Variance Source DF SS MS F P Factor 3 32.33 10.78 2.36 0.133 Error 10 45.67 4.57 Total 13 78.00

At the .05 level of significance, is there a difference in the mean number of months before a raise was granted among the four CPA firms?

14. A stock analyst wants to determine whether there is a difference in the mean rate of return for three types of stock: utility, retail, and banking stocks. The following output is obtained:

Exercises 11. Given the following sample information, test the hypothesis that the treatment means are

equal at the .05 significance level.

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Example WARTA, the Warren Area Regional Transit Authority, is expanding bus service from the suburb of Starbrick into the central business district of Warren. There are four routes being considered from Starbrick to downtown Warren: (1) via U.S. 6, (2) via the West End, (3) via the Hickory Street Bridge, and (4) via Route 59. WARTA conducted several tests to determine whether there was a difference in the mean travel times along the four routes. Because there will be many different drivers, the test was set up

so each driver drove along each of the four routes. Below is the travel time, in minutes, for each driver–route combination.

430 Chapter 12

a. Using the .05 level of significance, is there a difference in the mean rate of return among the three types of stock?

b. Suppose the null hypothesis is rejected. Can the analyst conclude there is a difference between the mean rates of return for the utility and the retail stocks? Explain.

12.7 Two-Way Analysis of Variance In the airline passenger ratings example, we divided the total variation into two cat- egories: the variation between the treatments and the variation within the treatments. We also called the variation within the treatments the error or the random variation. To put it another way, we considered only two sources of variation, that due to the treatments and the random differences. In the airline passenger ratings example, there may be other causes of variation. These factors might include, for example, the sea- son of the year, the particular airport, or the number of passengers on the flight.

The benefit of considering other factors is that we can reduce the error vari- ance. That is, if we can reduce the denominator of the F statistic (reducing the error variance or, more directly, the SSE term), the value of F will be larger, causing us to reject the hypothesis of equal treatment means. In other words, if we can explain more of the variation, then there is less “error.” An example will clarify the reduc- tion in the error variance.

Analysis of Variance Source DF SS MS F P Factor 2 86.49 43.25 13.09 0.001 Error 13 42.95 3.30 Total 15 129.44

Individual 95% CIs For Mean Based on Pooled StDev

Level N Mean StDev ----------+----------+----------+------ Utility 5 17.400 1.916 (------*------) Retail 5 11.620 0.356 (------*------) Banking 6 15.400 2.356 (------*------)

----------+----------+----------+------ Pooled StDev � 1.818 12.0 15.0 18.0

LO7 Carry out a test of hypothesis among treatment means using a blocking variable and understand the results.

Travel Time from Starbrick to Warren (minutes)

Driver U.S. 6 West End Hickory St. Rte. 59

Deans 18 17 21 22 Snaverly 16 23 23 22 Ormson 21 21 26 22 Zollaco 23 22 29 25 Filbeck 25 24 28 28

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Solution To begin, we conduct a test of hypothesis using a one-way ANOVA. That is, we consider only the four routes. Under this condition, the variation in travel times is either due to the treatments or it is random. The null hypothesis and the alternate hypothesis for comparing the mean travel time along the four routes are:

Not all treatment means are the same.

There are four routes, so for the numerator the degrees of freedom is k � 1 � 4 � 1 � 3. There are 20 observations, so the degrees of freedom in the denominator is n � k � 20 � 4 � 16. From Appendix B.4, with the .05 significance level, the critical value of F is 3.24. The decision rule is to reject the null hypothesis if the computed value of F is greater than 3.24.

We use Excel to perform the calculations and output the results. The computed value of F is 2.482, so our decision is to not reject the null hypothesis. We con- clude there is no difference in the mean travel time along the four routes. There is no reason to select one of the routes as faster than the other.

H1: H0: 1 � 2 � 3 � 4

Analysis of Variance 431

In the above example, we considered the variation due to the treatments (routes) and took all the remaining variation to be random. If we could consider the effect of the several drivers, this would allow us to reduce the SSE term, which would lead to a larger value of F. The second treatment variable, the drivers in this case, is referred to as a blocking variable.

At the .05 significance level, is there a difference in the mean travel time along the four routes? If we remove the effect of the drivers, is there a difference in the mean travel time?

From the above Excel output, the mean travel times along the routes were: 20.6 minutes along U.S. 6, 21.4 minutes along the West End route, 25.4 minutes using Hickory Street, and 23.8 minutes using Route 59. We conclude these differences could reasonably be attributed to chance. From the ANOVA table, we note: SST is 72.8, SSE is 156.4, and SS total is 229.2.

BLOCKING VARIABLE A second treatment variable that when included in the ANOVA analysis will have the effect of reducing the SSE term.

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432 Chapter 12

In this case, we let the drivers be the blocking variable, and removing the effect of the drivers from the SSE term will change the F ratio for the treatment variable. First, we need to determine the sum of squares due to the blocks.

In a two-way ANOVA, the sum of squares due to blocks is found by the fol- lowing formula.

[12–6]

where: k is the number of treatments. b is the number of blocks.

is the sample mean of block b. is the overall or grand mean.

From the calculations below, the means for the respective drivers are 19.5 minutes, 21 minutes, 22.5 minutes, 24.75 minutes, and 26.25 minutes. The overall mean is 22.8 minutes, found by adding the travel time for all 20 drives (456 minutes) and dividing by 20.

SSB � k©(Xb � XG)2

Travel Time from Starbrick to Warren (minutes)

Driver U.S. 6 West End Hickory St. Rte. 59 Driver Sums Driver Means

Deans 18 17 21 22 78 19.5 Snaverly 16 23 23 22 84 21 Ormson 21 21 26 22 90 22.5 Zollaco 23 22 29 25 99 24.75 Filbeck 25 24 28 28 105 26.25

Source of Variation Sum of Squares Degrees of Freedom Mean Square F

Treatments SST k � 1 SST�(k � 1) � MST MST�MSE Blocks SSB b � 1 SSB�(b � 1) � MSB MSB�MSE Error SSE (k � 1)(b � 1) SSE�(k � 1)(b � 1) � MSE

Total SS total n � 1

Substituting this information into formula (12–6) we determine SSB, the sum of squares due to the drivers (the blocking variable), is 119.7.

The same format is used in the two-way ANOVA table as in the one-way case, except there is an additional row for the blocking variable. SS total and SST are calculated as before, and SSB is found from formula (12–6). The SSE term is found by subtraction.

SUM OF SQUARES ERROR, TWO-WAY [12–7]

The values for the various components of the ANOVA table are computed as follows.

SSE � SS total � SST � SSB

� 119.7

� 4(24.75 � 22.8)2 � 4(26.25 � 22.8)2 � 4(19.5 � 22.8)2 � 4(21.0 � 22.8)2 � 4(22.5 � 22.8)2

SSB � k©(Xb � XG)2

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Analysis of Variance 433

SSE is found by formula (12–7).

SSE � SS total � SST � SSB � 229.2 � 72.8 � 119.7 � 36.7

(1) (2) (3) Source of Sum of Degrees of Mean Square Variation Squares Freedom (1) (2)

Treatments 72.8 3 24.27 Blocks 119.7 4 29.93 Error 36.7 12 3.06

Total 229.2 19

�

There is disagreement at this point. If the purpose of the blocking variable (the drivers in this example) was only to reduce the error variation, we should not con- duct a test of hypothesis for the difference in block means. That is, if our goal was to reduce the MSE term, then we should not test a hypothesis regarding the block- ing variable. On the other hand, we may wish to give the blocks the same status as the treatments and conduct a test of hypothesis. In the latter case, when the blocks are important enough to be considered as a second factor, we refer to this as a two-factor experiment. In many cases the decision is not clear. In our example we are concerned about the difference in the travel time for the different drivers, so we will conduct the test of hypothesis. The two sets of hypotheses are:

1. The treatment means are the same The treatment means are not the same.

2. The block means are the same The block means are not the same.

First, we will test the hypothesis concerning the treatment means. There are k � 1 � 4 � 1 � 3 degrees of freedom in the numerator and (b � 1)(k � 1) � (5 � 1) (4 � 1) � 12 degrees of freedom in the denominator. Using the .05 significance level, the critical value of F is 3.49. The null hypothesis that the mean times for the four routes are the same is rejected if the F ratio exceeds 3.49.

The null hypothesis is rejected and the alternate accepted. We conclude that the mean travel time is not the same for all routes. WARTA will want to conduct some tests to determine which treatment means differ.

Next, we test to find whether the travel time is the same for the various drivers. The degrees of freedom in the numerator for blocks is b � 1 � 5 � 1 � 4. The degrees of freedom for the denominator are the same as before: (b � 1)(k � 1) � (5 � 1)(4 � 1) � 12. The null hypothesis that the block means are the same is rejected if the F ratio exceeds 3.26.

The null hypothesis is rejected, and the alternate is accepted. The mean time is not the same for the various drivers. Thus, WARTA management can conclude, based on the sample results, that there is a difference in the routes and in the drivers.

The Excel spreadsheet has a two-factor ANOVA procedure. The output for the WARTA example just completed follows. The results are the same as reported ear- lier. In addition, the Excel output reports the p-values. The p-value for the null hypoth- esis regarding the drivers is .001 and .004 for the routes. These p-values confirm that the null hypotheses for treatments and blocks should both be rejected because the p-value is less than the significance level.

F � MSB MSE

� 29.93 3.06

� 9.78

F � MST MSE

� 24.27 3.06

� 7.93

H1: ( 1 � 2 � 3 � 4 � 5).H0:

H1: ( 1 � 2 � 3 � 4).H0:

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434 Chapter 12

Exercises For exercises 15 and 16, conduct a test of hypothesis to determine whether the block or the treatment means differ. Using the .05 significance level: (a) state the null and alternate hypotheses for treatments; (b) state the decision rule for treatments; and (c) state the null and alternate hypotheses for blocks. Also, state the decision rule for blocks, then: (d) compute SST, SSB, SS total, and SSE; (e) complete an ANOVA table; and (f ) give your decision regard- ing the two sets of hypotheses.

15. The following data are given for a two-factor ANOVA.

Self-Review 12–4 Rudduck Shampoo sells three shampoos, one each for dry, normal, and oily hair. Sales, in millions of dollars, for the past five months are given in the following table. Using the .05 significance level, test whether the mean sales differ for the three types of shampoo or by month.

Sales ($ million)

Month Dry Normal Oily

June 7 9 12 July 11 12 14 August 13 11 8 September 8 9 7 October 9 10 13

Treatment

Block 1 2

A 46 31 B 37 26 C 44 35

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Analysis of Variance 435

16. The following data are given for a two-factor ANOVA.

Treatment

Block 1 2 3

A 12 14 8 B 9 11 9 C 7 8 8

Units Produced

Employee Day Afternoon Night

Skaff 31 25 35 Lum 33 26 33 Clark 28 24 30 Treece 30 29 28 Morgan 28 26 27

Number of Surgeries Performed

Day St. Luke’s St. Vincent Mercy

Monday 14 18 24 Tuesday 20 24 14 Wednesday 16 22 14 Thursday 18 20 22 Friday 20 28 24

17. Chapin Manufacturing Company operates 24 hours a day, five days a week. The work- ers rotate shifts each week. Management is interested in whether there is a difference in the number of units produced when the employees work on various shifts. A sample of five workers is selected and their output recorded on each shift. At the .05 significance level, can we conclude there is a difference in the mean production rate by shift or by employee?

18. There are three hospitals in the Tulsa, Oklahoma, area. The following data show the num- ber of outpatient surgeries performed at each hospital last week. At the .05 significance level, can we conclude there is a difference in the mean number of surgeries performed by hospital or by day of the week?

12.8 Two-Way ANOVA with Interaction In the previous section, we studied the separate or independent effects of two variables, routes into the city and drivers, on mean travel time. The sample results indicated differences in mean time among the routes. Perhaps this is simply related to differences in the distance among the routes. The results also indicated differences in the mean drive time among the several drivers. Perhaps this differ- ence is explained by differing average speeds by the drivers regardless of the route. There is another effect that may influence travel time. This is called an inter- action effect between route and driver on travel time. For example, is it possible that one of the drivers is especially good driving one or more of the routes? Per- haps one driver knows how to effectively time the traffic lights or how to avoid heavily congested intersections for one or more of the routes. In this case, the combined effect of driver and route may also explain differences in mean travel time. To measure interaction effects, it is necessary to have at least two observa- tions in each cell.

LO8 Perform a two-way ANOVA with interaction and describe the results.

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436 Chapter 12

When we use a two-way ANOVA to study interaction, instead of using the terms treatments and blocks, we now call the two variables factors. So in this method there is a route factor, a driver factor, and an interaction of the two factors. That is, there is an effect for the routes, for the driver, and for the interaction of drivers and routes.

Interaction occurs if the combination of two factors has some effect on the vari- able under study, in addition to each factor alone. We refer to the variable being studied as the response variable. An everyday illustration of interaction is the effect of diet and exercise on weight. It is generally agreed that a person’s weight (the response variable) can be controlled with two factors, diet and exercise. Research shows that weight is affected by diet alone and that weight is affected by exercise alone. However, the general recommended method to control weight is based on the combined or interaction effect of diet and exercise.

INTERACTION The effect of one factor on a response variable differs depending on the value of another factor.

Interaction Plots One way to study interaction is by plotting factor means in a graph called an inter- action plot. Consider the bus driver example in the previous section. WARTA, the War- ren Area Regional Transit Authority, wants to study the mean travel time for different routes and drivers. To complete the study, we should also explore the prospect of interaction between driver and route. We begin by graphing the mean travel times on each route for each driver and connect the points. We compute Deans’ mean travel times for each route and plot them in a graph of mean travel times versus route. We repeat this process for each of the drivers. The interaction plot follows.

This plot helps us understand the interaction between the effects of drivers and routes on travel time. If the line segments for the drivers appear essentially parallel, then there is probably no interaction. On the other hand, if the line segments do not

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Analysis of Variance 437

seem to be parallel or cross, this suggests an interaction between the factors. The above plot suggests interaction because:

• The line segments for Zollaco and Filbeck cross over each other. • Snaverly’s line segment from U.S. 6 to West End crosses over three other line

segments.

These observations suggest an interaction between driver and route.

Hypothesis Test for Interaction The next step is to conduct statistical tests to further investigate the possible inter- action effects. In summary, our study of travel times has several questions:

• Is there an interaction between routes and drivers? • Are the travel times for drivers the same? • Are the travel times for the routes the same?

Of the three questions, we are most interested in the test for interactions. We can investigate these questions statistically by extending the two-way

ANOVA procedure presented in the previous section. We add another source of vari- ation, namely, the interaction. However, in order to estimate the “error” sum of squares, we need at least two measurements for each driver/route combination. So suppose the experiment reported on page 430 is repeated by measuring two more travel times for each driver and route combination. Let’s replicate the experiment. Now we have three observations for each driver/route combination. Using the mean of three travel times for each driver/route combination, we get a more reliable mea- sure of the mean travel time. The results of replicating the experiment are in the fol- lowing Excel table. Note that the data should be entered in this exact format so we can use statistical software.

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438 Chapter 12

To explain the spreadsheet, consider the “20, 21, 22” for the rows labeled “Deans” and column labeled “Hickory St.” These are the three travel time measurements for Deans to drive the Hickory Street route. Specifically, Deans drove the Hickory Street route the first time in 20 minutes, 21 minutes on the second trip, and 22 minutes on the third trip.

The ANOVA now has three sets of hypotheses to test:

1. H0: There is no interaction between drivers and routes. H1: There is interaction between drivers and routes.

2. H0: The driver means are the same. H1: The driver means are not the same.

3. H0: The route means are the same. H1: The route means are not the same.

Note that we will label the route effect as Factor A and the driver effect as Factor B.

Each of these hypotheses is tested using the familiar F statistic. We can use a decision rule for each of the above tests or we can use the p-values for each test. In this case, we will use the .05 significance level and compare the p-value gener- ated by the statistical software with the significance level. So the various null hypotheses are rejected if the computed p-value is less than .05. Instead of com- puting treatment and block sum of squares, we compute factor and interaction sum of squares. The computations for the factor sum of squares are very similar to the SST and SSB as computed before. See formulas (12–4) and (12–6). The sum of squares due to possible interaction is:

[12–8]

where: i is a subscript or label representing a route. j is a subscript or label representing a driver. k is the number of Factor A (route effect) levels. b is the number of Factor B (driver effect) levels. n is the number of observations.

is the mean travel time on a route, i, for driver, j. Note these are the means that we plotted in the graph on page 436. is the mean travel time for route i. Note the dot shows that the mean is cal- culated over all drivers. These are the route means that we compared on page 434. is the mean travel time for driver j. Note the dot shows that the mean is cal- culated over all routes. These are the driver means that we compared on page 434. is the grand mean.

Once you have SSI, then SSE is found as:

[12–9]

The complete ANOVA table including interactions is:

SSE � SS total � SS Factor A � SS Factor B � SSI

X. j

Xi.

Xij

SSI � n�bk [©© (Xij � Xi. � X. j � XG)2]

Source Sum of Squares df Mean Square F

Route Factor A k � 1 SSA�(k � 1) � MSA MSA�MSE Driver Factor B b � 1 SSB�(b � 1) � MSB MSB�MSE Interaction SSI (k � 1)(b � 1) SSI�[(k � 1)(b � 1)] � MSI MSI�MSE Error SSE n � kb SSE�(n � kb) � MSE Total SS total n � 1

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Analysis of Variance 439

The p-value for interactions of 0.036 (noted in yellow), is less than our significance level of 0.05. So our decision is to reject the null hypothesis of no interaction and conclude that the combination of route and driver has a significant effect on the response variable travel time.

Interaction effects provide information about the combined effects of variables. If interaction is present, then you should conduct a one-way ANOVA to test differences in the factor means for each level of the other factor. This analysis requires some time and work to complete but the results are usually enlightening.

We will continue the analysis by conducting a one-way ANOVA for each driver testing the hypothesis: H0: Route travel times are equal. The results follow.

Deans: H0: Route travel times are equal.

Source DF SS MS F P Deans RTE 3 51.00 17.00 2.43 0.140 Error 8 56.00 7.00 Total 11 107.00

Ormson: H0: Route travel times are equal.

Source DF SS MS F P Ormson RTE 3 51.00 17.00 3.78 0.059 Error 8 36.00 4.50 Total 11 87.00

Filbeck: H0: Route travel times are equal.

Source DF SS MS F P Filbeck RTE 3 38.25 12.75 6.38 0.016 Error 8 16.00 2.00 Total 11 54.25

Snaverly: H0: Route travel times are equal.

Source DF SS MS F P SN RTE 3 102.00 34.00 7.16 0.012 Error 8 38.00 4.75 Total 11 140.00

Zollaco: H0: Route travel times are equal.

Source DF SS MS F P Z-RTE 3 86.25 28.75 8.85 0.006 Error 8 26.00 3.25 Total 11 112.25

Recall the results from the two-way ANOVA without interaction on page 433. In that analysis, the results clearly showed that the factor “route” had a significant effect on travel time. However, now that we include the interaction effect, the results show that conclusion is not generally true. As we review the p-values for the five one-way ANOVAs

The resulting Excel output shows the summary descriptive statistics for each driver and an ANOVA table.

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above (reject the null if the p-value is less than 0.05), we know that route mean travel times are different for three drivers: Filbeck, Snaverly, and Zollaco. However, for Deans and Ormson, their mean route travel times are not significantly different.

Now that we know this new and interesting information, we would want to know why these differences exist. Further investigation of the driving habits of the five drivers is required.

In review, our presentation of two-way ANOVA with interaction shows the power of statistical analysis. In this analysis, we were able to test the combined effect of driver and route on travel time, and to show that different drivers evidently behave differently as they travel their routes. Gaining understanding of interaction effects is extremely important in many applications, from scientific areas such as agriculture and quality control to managerial fields like human resource management and gender equality in salary and performance ratings.

Self-Review 12–5 See the following ANOVA table.

ANOVA

Source of Variation SS df MS F p-value

Factor A 6.41 3 2.137 3.46 0.0322 Factor B 5.01 2 2.507 4.06 0.0304 Interaction 33.15 6 5.525 8.94 0.0000 Error 14.83 24 0.618

Total 59.41 35

Use the .05 significance level to answer the following questions. (a) How many levels does Factor A have? Is there a significant difference among the Fac-

tor A means? How do you know? (b) How many levels does Factor B have? Is there a significant difference among the Fac-

tor B means? How do you know? (c) How many observations are there in each cell? Is there a significant interaction between

Factor A and Factor B on the response variable? How do you know?

Exercises 19. Consider the following sample data for a two-factor ANOVA experiment:

Size

Small Medium Large

23 20 11 Heavy 21 32 20

25 26 20 Weight

13 20 11 Light 32 17 23

17 15 8

Use the .05 significance level to answer the following questions. a. Is there a difference in the Size means? b. Is there a difference in the Weight means? c. Is there a significant interaction between Weight and Size?

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Analysis of Variance 441

20. Consider the following partially completed two-way ANOVA table. Suppose there are four levels of Factor A and three levels of Factor B. The number of replications per cell is 5. Complete the table and test to determine if there is a significant difference in Factor A means, Factor B means, or the interaction means. Use the .05 significance level. (Hint: estimate the values from the F table.)

ANOVA

Source SS df MS F

Factor A 75 Factor B 25 Interaction 300 Error 600

Total 1000

Position/Machine J-1000 D-320 UV-57

Inside 33, 40, 30, 31 29, 28, 33, 33 47, 39, 39, 45 Outside 43, 36, 41, 40 48, 45, 40, 44 37, 32, 36, 35

Area/Gender Female Male

Manufacturing 1016, 1007, 875, 968 978, 1056, 982, 748 Marketing 1045, 895, 848, 904 1154, 1091, 878, 876 Research and Development 770, 733, 844, 771 926, 1055, 1066, 1088

21. The distributor of the Wapakoneta Daily News, a regional newspaper serving western Ohio, is considering three types of dispensing machines or “racks.” Management wants to know if the different machines affect sales. These racks are designated as J-1000, D-320, and UV-57. Management also wants to know if the placement of the racks either inside or outside supermarkets affects sales. Each of six similar stores was randomly assigned a machine and location combination. The data below is the number of papers sold over four days.

a. Draw the interaction graph. Based on your observations, is there an interaction effect? Based on the graph, describe the interaction effect of machine and position.

b. Use the 0.05 level to test for position, machine, and interaction effects on sales. Report the statistical results.

c. Compare the inside and outside mean sales for each machine using statistical tech- niques. What do you conclude?

22. A large company is organized into three functional areas: manufacturing, marketing, and research and development. The employees claim that the company pays women less than men for similar jobs. The company randomly selected four males and four females in each area and recorded their weekly salaries in dollars.

a. Draw the interaction graph. Based on your observations, is there an interaction effect? Based on the graph, describe the interaction effect of gender and area on salary.

b. Use the 0.05 level to test for gender, area, and interaction effects on salary. Report the statistical results.

c. Compare the male and female mean sales for each area using statistical techniques. What do you recommend to the distributor?

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Chapter Summary I. The characteristics of the F distribution are:

A. It is continuous. B. Its values cannot be negative. C. It is positively skewed. D. There is a family of F distributions. Each time the degrees of freedom in either the

numerator or the denominator changes, a new distribution is created. II. The F distribution is used to test whether two population variances are the same.

A. The sampled populations must follow the normal distribution. B. The larger of the two sample variances is placed in the numerator, forcing the ratio to

be at least 1.00. C. The value of F is computed using the following equation:

[12–1]

III. A one-way ANOVA is used to compare several treatment means. A. A treatment is a source of variation. B. The assumptions underlying ANOVA are:

1. The samples are from populations that follow the normal distribution. 2. The populations have equal standard deviations. 3. The samples are independent.

C. The information for finding the value of F is summarized in an ANOVA table. 1. The formula for SS total, the sum of squares total, is:

[12–2]

2. The formula for SSE, the sum of squares error, is:

[12–3]

3. The formula for the SST, the sum of squares treatment, is found by subtraction.

[12–4]

4. This information is summarized in the following table and the value of F deter- mined.

SST � SS total � SSE

SSE � ©(X � Xc)2

SS total � ©(X � XG)2

F � s21 s22

IV. If a null hypothesis of equal treatment means is rejected, we can identify the pairs of means that differ from the following confidence interval.

[12–5]

V. In a two-way ANOVA, we consider a second treatment variable. A. The second treatment variable is called the blocking variable. B. It is determined using the following equation:

[12–6]

C. The SSE term, or sum of squares error, is found from the following equation.

[12–7]SSE � SS total � SST � SSB

SSB � k©(Xb � XG)2

(X1 � X2) � tB MSEa

1 n1

� 1 n2 b

Source of Variation Sum of Squares Degrees of Freedom Mean Square F

Treatments SST k � 1 SST�(k � 1) � MST MST�MSE Error SSE n � k SSE�(n � k) � MSE

Total SS total n � 1

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Analysis of Variance 443

VI. In a two-way ANOVA with repeated observations, we consider two treatment variables and the possible interaction between the variables. A. The sum of squares due to possible interactions is found by:

[12–8]

B. The SSE term is found by subtraction.

[12–9]

C. The complete ANOVA table including interactions is:

SSI � n�bk [©© (Xij � Xi. � X.j � XG)2]

Sum of Source Squares df Mean Square F

Factor A SSA k � 1 SSA�(k � 1) � MSA MSA�MSE Factor B SSB b � 1 SSB�(b � 1) � MSB MSB�MSE Interaction SSI (k � 1)(b � 1) SSI�[(k � 1)(b � 1)] � MSI MSI�MSE Error SSE n � kb SSE�(n � kb) � MSE

Total SS total n � 1

Pronunciation Key SYMBOL MEANING PRONUNCIATION

SS total Sum of squares total S S total

SST Sum of squares treatment S S T

SSE Sum of squares error S S E

MSE Mean square error M S E

SSB Block sum of squares S S B

SSI Sum of squares interaction S S I

Chapter Exercises 23. A real estate agent in the coastal area of Georgia wants to compare the variation in the

selling price of homes on the oceanfront with those one to three blocks from the ocean. A sample of 21 oceanfront homes sold within the last year revealed the standard devia- tion of the selling prices was $45,600. A sample of 18 homes, also sold within the last year, that were one to three blocks from the ocean revealed that the standard deviation was $21,330. At the .01 significance level, can we conclude that there is more variation in the selling prices of the oceanfront homes?

24. A computer manufacturer is about to unveil a new, faster personal computer. The new machine clearly is faster, but initial tests indicate there is more variation in the process- ing time. The processing time depends on the particular program being run, the amount of input data, and the amount of output. A sample of 16 computer runs, covering a range of production jobs, showed that the standard deviation of the processing time was 22 (hundredths of a second) for the new machine and 12 (hundredths of a second) for the

Source of Variation Sum of Squares Degrees of Freedom Mean Square F

Treatments SST k � 1 SST�(k � 1) � MST MST�MSE Blocks SSB b � 1 SSB�(b � 1) � MSB MSB�MSE Error SSE [(k � 1)(b � 1)] SSE�[(k � 1)(b � 1)] � MSE

Total SS total n � 1

D. The F statistics for the treatment variable and the blocking variable are determined in the following table.

SSE � SS total � SSA � SSB � SSI

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current machine. At the .05 significance level can we conclude that there is more variation in the processing time of the new machine?

25. There are two Chevrolet dealers in Jamestown, New York. The mean monthly sales at Sharkey Chevy and Dave White Chevrolet are about the same. However, Tom Sharkey, the owner of Sharkey Chevy, believes his sales are more consistent. Below is the num- ber of new cars sold at Sharkey in the last seven months and for the last eight months at Dave White. Do you agree with Mr. Sharkey? Use the .01 significance level.

Sum of Mean Source Squares df Square F

Treatment 2 Error 20

Total 500 11

26. Random samples of five were selected from each of three populations. The sum of squares total was 100. The sum of squares due to the treatments was 40. a. Set up the null hypothesis and the alternate hypothesis. b. What is the decision rule? Use the .05 significance level. c. Complete the ANOVA table. What is the value of F? d. What is your decision regarding the null hypothesis?

27. In an ANOVA table, MSE was equal to 10. Random samples of six were selected from each of four populations, where the sum of squares total was 250. a. Set up the null hypothesis and the alternate hypothesis. b. What is the decision rule? Use the .05 significance level. c. Complete the ANOVA table. What is the value of F? d. What is your decision regarding the null hypothesis?

28. The following is a partial ANOVA table.

Discount Variety Department

$12 $15 $19 13 17 17 14 14 16 12 18 20 15 17 19

Complete the table and answer the following questions. Use the .05 significance level. a. How many treatments are there? b. What is the total sample size? c. What is the critical value of F? d. Write out the null and alternate hypotheses. e. What is your conclusion regarding the null hypothesis?

29. A consumer organization wants to know whether there is a difference in the price of a particular toy at three different types of stores. The price of the toy was checked in a sample of five discount stores, five variety stores, and five department stores. The results are shown below. Use the .05 significance level.

30. Jacob Lee is a frequent traveler between Los Angeles and San Francisco. For the past month, he wrote down the flight times on three different airlines. The results are:

Sharkey 98 78 54 57 68 64 70 Dave White 75 81 81 30 82 46 58 101

Goust Jet Red Cloudtran

51 50 52 51 53 55 52 52 60

(continued )

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Analysis of Variance 445

a. Use the .05 significance level and the five-step hypothesis-testing process to check if there is a difference in the mean flight times among the three airlines.

b. Develop a 95 percent confidence interval for the difference in the means between Goust and Cloudtran.

31. The City of Maumee comprises four districts. Chief of Police Andy North wants to deter- mine whether there is a difference in the mean number of crimes committed among the four districts. He recorded the number of crimes reported in each district for a sample of six days. At the .05 significance level, can the chief of police conclude there is a dif- ference in the mean number of crimes?

Number of Crimes

Rec Center Key Street Monclova Whitehouse

13 21 12 16 15 13 14 17 14 18 15 18 15 19 13 15 14 18 12 20 15 19 15 18

32. A study of the effect of television commercials on 12-year-old children measured their attention span, in seconds. The commercials were for clothes, food, and toys. At the .05 significance level, is there a difference in the mean attention span of the children for the various commercials? Are there significant differences between pairs of means? Would you recommend dropping one of the three commercial types?

Clothes Food Toys

26 45 60 21 48 51 43 43 43 35 53 54 28 47 63 31 42 53 17 34 48 31 43 58 20 57 47

47 51 44 51 54

33. When only two treatments are involved, ANOVA and the Student t test (Chapter 11) result in the same conclusions. Also, t2 � F. As an example, suppose that 14 randomly selected students were divided into two groups, one consisting of 6 students and the other of 8. One group was taught using a combination of lecture and programmed instruction, the

Goust Jet Red Cloudtran

42 62 64 51 53 61 57 49 49 47 50 49 47 49 50 58 60 54 54 51 49 49 48 49 48 50

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446 Chapter 12

a. Using analysis of variance techniques, test H0 that the two mean test scores are equal; � � .05.

b. Using the t test from Chapter 11, compute t. c. Interpret the results.

34. There are four auto body shops in Bangor, Maine, and all claim to promptly serve cus- tomers. To check if there is any difference in service, customers are randomly selected from each repair shop and their waiting times in days are recorded. The output from a statistical software package is:

ANOVA

Source of Variation SS df MS F p-value

Between Groups 23.37321 3 7.791069 9.612506 0.001632 Within Groups 9.726167 12 0.810514

Total 33.09938 15

Summary

Groups Count Sum Average Variance

Body Shop A 3 15.4 5.133333 0.323333 Body Shop B 4 32 8 1.433333 Body Shop C 5 25.2 5.04 0.748 Body Shop D 4 25.9 6.475 0.595833

Is there evidence to suggest a difference in the mean waiting times at the four body shops? Use the .05 significance level.

35. The fuel efficiencies for a sample of 27 compact, midsize, and large cars are entered into a statistical software package. Analysis of variance is used to investigate if there is a difference in the mean mileage of the three cars. What do you conclude? Use the .01 significance level.

Summary

Groups Count Sum Average Variance

Compact 12 268.3 22.35833 9.388106 Midsize 9 172.4 19.15556 7.315278 Large 6 100.5 16.75 7.303

Lecture and Programmed Lecture and Instruction Television

19 32 17 28 23 31 22 26 17 23 16 24

27 25

other using a combination of lecture and television. At the end of the course, each group was given a 50-item test. The following is a list of the number correct for each of the two groups.

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Analysis of Variance 447

a. Use a .01 level of significance to test if there is a difference in the mean production of the three assembly lines.

b. Develop a 99 percent confidence interval for the difference in the means between Line B and Line C.

37. The postal service groups first-class mail as letters, cards, flats, or parcels. Over a period of three weeks, one item of each kind was sent from a particular postal administrative center. The total time in transit was recorded. A statistical software package was then used to perform the analysis. The results follow.

Source DF SS MS F P Factor 3 13.82 4.61 2.72 0.051 Error 68 115.17 1.69 Total 71 128.99

S = 1.301 R-Sq = 10.71% R-Sq(adj) = 6.77%

Individual 95% CIs for Mean Based on Pooled StDev

Level N Mean StDev ------+---------+---------+---------+--- Letters 18 1.444 1.097 (---------*---------) Cards 18 1.667 1.455 (---------*---------) Flats 18 2.444 1.617 (---------*---------) Parcels 18 2.389 0.916 (---------*---------)

------+---------+---------+---------+--- 1.20 1.80 2.40 3.00

Use the .05 significance level to test if this evidence suggests a difference in the means for the different types of first-class mail.

ANOVA

Source of Variation SS df MS F p-value

Between Groups 12.33333 2 6.166667 11.32653 0.001005 Within Groups 8.166667 15 0.544444

Total 20.5 17

ANOVA

Source of Variation SS df MS F p-value

Between Groups 136.4803 2 68.24014 8.258752 0.001866 Within Groups 198.3064 24 8.262766

Total 334.7867 26

36. Three assembly lines are used to produce a certain component for an airliner. To exam- ine the production rate, a random sample of six hourly periods is chosen for each assem- bly line and the number of components produced during these periods for each line is recorded. The output from a statistical software package is:

Summary

Groups Count Sum Average Variance

Line A 6 250 41.66667 0.266667 Line B 6 260 43.33333 0.666667 Line C 6 249 41.5 0.7

Additional results are shown below.

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448 Chapter 12

Color of Ad

Size of Ad Red Blue Orange Green

Small 2 3 3 8 Medium 3 5 6 7 Large 6 7 8 8

Restaurant

Week Metro Interstate University River

1 124 160 320 190 2 234 220 340 230 3 430 290 290 240 4 105 245 310 170 5 240 205 280 180 6 310 260 270 205

Is there a difference in the effectiveness of an advertisement by color and by size? Use the .05 level of significance.

40. There are four McBurger restaurants in the Columbus, Georgia, area. The numbers of burgers sold at the respective restaurants for each of the last six weeks are shown below. At the .05 significance level, is there a difference in the mean number sold among the four restaurants when the factor of week is considered?

a. Is there a difference in the treatment means? b. Is there a difference in the block means?

38. For your email, you use a filter to block spam from your inbox. The number of items blocked by day of week is recorded and a statistical software system is used to perform the analysis that follows. Here are the results:

Source DF SS MS F P Factor 6 1367.8 228.0 5.72 0.000 Error 48 1913.2 39.9 Total 54 3281.0

S = 6.313 R-Sq = 41.69% R-Sq(adj) = 34.40%

Individual 95% CIs for Mean Based on Pooled StDev

Level N Mean StDev ----+---------+---------+---------+----- Monday 10 74.000 6.164 (-----*------) Tuesday 9 66.111 7.288 (------*------) Wednesday 7 74.143 2.268 (-------*-------) Thursday 8 62.375 5.041 (-------*------) Friday 8 75.125 4.454 (------*-------) Saturday 5 63.200 7.259 (--------*---------) Sunday 8 72.375 9.164 (-------*------)

----+---------+---------+---------+----- 60.0 66.0 72.0 78.0

Use the .05 significance level to test if this evidence suggests a difference in the means for the different days of the week.

39. Shank’s Inc. a nationwide advertising firm, wants to know whether the size of an adver- tisement and the color of the advertisement make a difference in the response of mag- azine readers. A random sample of readers is shown ads of four different colors and three different sizes. Each reader is asked to give the particular combination of size and color a rating between 1 and 10. Assume that the ratings follow the normal distribution. The rating for each combination is shown in the following table (for example, the rating for a small red ad is 2).

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Using the .05 level of significance: a. Is there a difference among types of gasoline? b. Is there a difference in the cars?

43. A research firm wants to compare the miles per gallon of unleaded regular, mid-grade, and super premium gasolines. Because of differences in the performance of different automobiles, seven different automobiles were selected and treated as blocks. Therefore, each brand of gasoline was tested with each type of automobile. The results of the trials, in miles per gallon, are shown in the following table. At the .05 significance level, is there a difference in the gasolines or automobiles?

Distance (miles)

Types of Gasoline Car A Car B Car C

Regular 22.4 20.8 21.5 Super regular 17.0 19.4 20.7 Unleaded 19.2 20.2 21.2 Premium unleaded 20.3 18.6 20.4

Automobile Regular Mid-grade Super Premium

1 21 23 26 2 23 22 25 3 24 25 27 4 24 24 26 5 26 26 30 6 26 24 27 7 28 27 32

44. Three supermarket chains in the Denver area each claim to have the lowest overall prices. As part of an investigative study on supermarket advertising, the Denver Daily News con- ducted a study. First, a random sample of nine grocery items was selected. Next, the price of each selected item was checked at each of the three chains on the same day.

Analysis of Variance 449

Assessor

Home Zawodny Norman Cingle Holiday

A $53.0 $55.0 $49.0 $45.0 B 50.0 51.0 52.0 53.0 C 48.0 52.0 47.0 53.0 D 70.0 68.0 65.0 64.0 E 84.0 89.0 92.0 86.0

a. Is there a difference in the treatment means? b. Is there a difference in the block means?

42. Martin Motors has in stock three cars of the same make and model. The president would like to compare the gas consumption of the three cars (labeled car A, car B, and car C) using four different types of gasoline. For each trial, a gallon of gasoline was added to an empty tank, and the car was driven until it ran out of gas. The following table shows the number of miles driven in each trial.

41. The city of Tucson, Arizona, employs people to assess the value of homes for the purpose of establishing real estate tax. The city manager sends each assessor to the same five homes and then compares the results. The information is given below, in thousands of dol- lars. Can we conclude that there is a difference in the assessors, at � � .05?

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450 Chapter 12

46. There are four radio stations in Midland. The stations have different formats (hard rock, classical, country/western, and easy listening), but each is concerned with the number of minutes of music played per hour. From a sample of 10 hours from each station, the following sample means were offered.

a. Determine SST. b. Determine SSE. c. Complete an ANOVA table. d. At the .05 significance level, is there a difference in the treatment means? e. Is there a difference in the mean amount of music time between station 1 and station

4? Use the .05 significance level.

We recommend you complete the following exercises using a statistical software pack- age such as Excel, MegaStat, or Minitab.

47. The American Accounting Association recently conducted a study to compare the weekly wages of men and women employed in either the public or private sector of accounting.

SS total � 650.75

X4 � 50.85X3 � 47.2X2 � 44.64X1 � 51.32

Red Orange Yellow Brown Tan Green

0.946 0.902 0.929 0.896 0.845 0.935 1.107 0.943 0.960 0.888 0.909 0.903 0.913 0.916 0.938 0.906 0.873 0.865 0.904 0.910 0.933 0.941 0.902 0.822 0.926 0.903 0.932 0.838 0.956 0.871 0.926 0.901 0.899 0.892 0.959 0.905 1.006 0.919 0.907 0.905 0.916 0.905 0.914 0.901 0.906 0.824 0.822 0.852 0.922 0.930 0.930 0.908 0.965 1.052 0.883 0.952 0.833 0.898 0.903 0.939 0.895 0.940

0.882 0.906

Item Super$ Ralph’s Lowblaws

1 $1.12 $1.02 $1.07 2 1.14 1.10 1.21 3 1.72 1.97 2.08 4 2.22 2.09 2.32 5 2.40 2.10 2.30 6 4.04 4.32 4.15 7 5.05 4.95 5.05 8 4.68 4.13 4.67 9 5.52 5.46 5.86

45. Listed below are the weights (in grams) of a sample of M&M’s Plain candies, classified according to color. Use a statistical software system to determine whether there is a difference in the mean weights of candies of different colors. Use the .05 significance level.

At the .05 significance level, is there a difference in the mean prices at the supermarkets or for the items?

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At the .05 significance level: a. Draw an interaction plot of men and women means by sector. b. Test the interaction effect of gender and sector on wages. c. Based on your results in part (b), conduct the appropriate tests of hypotheses for dif-

ferences in factor means. d. Interpret the results in a brief report.

48. Robert Altoff is vice president for engineering for a manufacturer of household washing machines. As part of new product development, he wishes to determine the optimal length of time for the washing cycle. A part of the development is to study the relationship between the detergent used (four brands) and the length of the washing cycle (18, 20, 22, or 24 min- utes). In order to run the experiment, 32 standard household laundry loads (having equal amounts of dirt and the same total weights) are randomly assigned to the 16 detergent– washing cycle combinations. The results (in pounds of dirt removed) are shown below.

Cycle Time (min)

Detergent Brand 18 20 22 24

A 0.13 0.12 0.19 0.15 0.11 0.11 0.17 0.18

B 0.14 0.15 0.18 0.20 0.10 0.14 0.17 0.18

C 0.16 0.15 0.18 0.19 0.17 0.14 0.19 0.21

D 0.09 0.12 0.16 0.15 0.13 0.13 0.16 0.17

At the .05 significance level: a. Draw an interaction plot of the detergent means by cycle time. b. Test the interaction effect of brand and cycle time on “dirt removed.” c. Based on your results in part (b), conduct the appropriate tests of hypotheses for dif-

ferences in factor means. d. Interpret the results in a brief report.

Data Set Exercises 49. Refer to the Real Estate data, which report information on the homes sold in the

Goodyear, Arizona, area last year. a. At the .02 significance level, is there a difference in the variability of the selling prices

of the homes that have a pool versus those that do not have a pool?

Analysis of Variance 451

Sector

Gender Public Private

Men $ 978 $1,335 1,035 1,167

964 1,236 996 1,317

1,117 1,192 Women $ 863 $1,079

975 1,160 999 1,063

1,019 1,110 1,037 1,093

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452 Chapter 12

b. At the .02 significance level, is there a difference in the variability of the selling prices of the homes with an attached garage versus those that do not have an attached garage?

c. At the .05 significance level, is there a difference in the mean selling price of the homes among the five townships?

50. Refer to the Baseball 2009 data, which report information on the 30 Major League Base- ball teams for the 2009 season. a. At the .10 significance level, is there a difference in the variation in team salary among

the American and National League teams? b. Create a variable that classifies a team’s total attendance into three groups: less than

2.0 (million), 2.0 up to 3.0, and 3.0 or more. At the .05 significance level, is there a difference in the mean number of games won among the three groups? Use the .01 significance level.

c. Using the same attendance variable developed in part (b), is there a difference in the mean number of home runs hit per team? Use the .01 significance level.

d. Using the same attendance variable developed in part (b), is there a difference in the mean salary of the three groups? Use the .01 significance level.

51. Refer to the Buena School District bus data. a. Conduct a test of hypothesis to reveal whether the mean maintenance cost is equal

for each of the bus producers. Use the .01 significance level. b. Conduct a test of hypothesis to determine whether the mean miles traveled is equal

for each make of bus. Use the .05 significance level. c. Develop a 95 percent confidence interval for the disparity in the average maintenance

cost between buses made by Bluebird and Thompson.

1. The Excel commands for the test of variances on page 415 are: a. Enter the data for U.S. 25 in column A and for

I-75 in column B. Label the two columns. b. Select the Data tab on the top menu. Then, on

the far right, select Data Analysis. Select F-Test: Two-Sample for Variances, then click OK.

c. The range of the first variable is A1:A8, and B1:B9 for the second. Click on Labels, enter 0.05 for Alpha, select D1 for the Output Range, and click OK.

Software Commands

2. The Excel commands for the one-way ANOVA on page 424 are: a. Key in data into four columns labeled: Northern,

WTA, Pocono, and Branson. b. Select the Data tab on the top menu. Then, on

the far right, select Data Analysis. Select ANOVA: Single Factor, then click OK.

c. In the subsequent dialog box, make the input range A1:D8, click on Grouped by Columns, click on Labels in first row, the Alpha text box is .05, and finally select Output Range as F1 and click OK.

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Analysis of Variance 453

3. The Minitab commands for the one-way ANOVA on page 424 are: a. Input the data into four columns and identify

the columns as Northern, WTA, Pocono, and Branson.

b. Select Stat, ANOVA, and One-way (Unstacked), select the data in columns C1 to C4, check on Select in the lower left, and then click OK.

4. The Excel commands for the two-way ANOVA on page 434 are: a. In the first row of the first column, write the word

Driver, then list the five drivers in the first column. In the first row of the next four columns, enter the name of the routes. Enter the data under each route name.

b. Select the Data tab on the top menu. Then, on the far right, select Data Analysis. Select ANOVA: Two-Factor Without Replication, then click OK.

c. In the dialog box, the Input Range is A3:E8, click on Labels, select G3 for the Output Range, and then click OK.

5. The Excel commands for the two-way ANOVA with interaction on page 439 are: a. Enter the data into Excel as shown on page 437. b. Select the Data tab on the top menu. Then, on

the far right, select Data Analysis. Select ANOVA: Two-Factor With Replication, then click OK.

c. In the dialog box, enter the Input Range as B2:F17, enter Rows per sample as 3, select New Worksheet Ply, and then click OK.

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Chapter 12 Answers to Self-Review

12–1 Let Mark’s assemblies be population 1, then and

also equals 9. is rejected if

is not rejected. The variation is the same for both employees.

12–2 a. At least one treatment mean is different.

b. Reject if .

SST � 578 � 74 � 504 � 74

(32 � 29)2 SSE � (18 � 17)2 � (14 � 17)2 � . . . �

� 578 SS total � (18 � 20)2 � . . . � (32 � 20)2

X � 240 12

� 20

F 7 4.26H0 H1:

1 � 2 � 3H0:

F � (2.0)2

(1.5)2 � 1.78

F 7 3.18.H0df2 df1 � 10 � 1 � 9;�

2 1 7 �22;H1:�21 � �22;H0:

d. is rejected. The treatment means differ. e. (11.0 � 6.8) � 2.201

These treatment means differ because both endpoints of the confidence interval are of the same sign—positive in this problem.

12–4 For types:

The treatment means are not equal.

Reject if For months:

The block means are not equal.

Reject if The analysis of variance table is as follows:

F 7 3.84.H0

H1: 1 � 2 � 3 � 4 � 5H0:

F 7 4.46.H0

H1: 1 � 2 � 3H0:

4.2 � 1.30 � 2.90 and 5.50 20.8682(15 �

1 5) �

Sum of Degrees of Mean Source Squares Freedom Square F

Treatment 504 2 252 30.65 Error 74 9 8.22

Total 578 11

e. is rejected. There is a difference in the mean number of bottles sold at the various locations.

12–3 a. Not all means are equal.

b. is rejected if c.

SSE � 9.55 SST � 44.16

SS total � 53.71 X3 � 6.8X2 � 8.75,X1 � 11,XG � 8.86,

F 7 3.98.H0 H1:

1 � 2 � 3H0:

Source df SS MS F

Types 2 3.60 1.80 0.39 Months 4 31.73 7.93 1.71 Error 8 37.07 4.63

Total 14 72.40

The null hypotheses cannot be rejected for either types or months. There is no difference in the mean sales among types or months.

12–5 a. There are four levels of Factor A. The p-value is less than .05, so Factor A means differ.

b. There are three levels of Factor B. The p-value is less than .05, so the Factor B means differ.

c. There are three observations in each cell. There is an interaction between Factor A and Factor B means, because the p-value is less than .05.

454 Chapter 12

Sum of Mean Source Squares df Square F

Treatment 44.16 2 22.08 25.43 Error 9.55 11 0.8682

Total 53.71 13

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A Review of Chapters 10–12 455

A Review of Chapters 10–12 This section is a review of the major concepts and terms introduced in Chapters 10, 11, and 12. Chapter 10 began our study of hypothesis testing. A hypothesis is a statement about the value of a population parameter. In statistical hypothesis testing, we begin by making a statement about the value of the population parameter in the null hypothesis. We establish the null hypothesis for the purpose of testing. When we complete the testing, our decision is either to reject or to fail to reject the null hypoth- esis. If we reject the null hypothesis, we conclude that the alternate hypothesis is true. The alternate hypothesis is “accepted” only if we show that the null hypothesis is false. We also refer to the alternate hypothesis as the research hypothesis. Most of the time we want to prove the alternate hypothesis.

In Chapter 10, we selected random samples from a single population and tested whether it was reasonable that the population parameter under study equaled a particular value. For example, we wish to investigate whether the mean tenure time of those holding the position of CEO in large firms is 12 years. We select a sample of CEOs, compute the sample mean, and compare the mean of the sample to the population. The single population under consideration is the length of tenure of CEOs of large firms. We described methods for conducting the test when the population standard deviation was available and when it was not available. Also, in Chapter 10 we conducted tests of hypothesis about a population proportion. A proportion is the fraction of individuals or objects pos- sessing a certain characteristic. For example, industry records indicate that 70 percent of gasoline sales for automobiles are for the regular grade of gasoline. A sample of the 100 sales from last month at the Pantry in Conway revealed 76 were for the regular grade. Can the owners conclude that more than 70 percent of their customers purchase the regular grade?

In Chapter 11, we extended the idea of hypothesis testing to whether two independent ran- dom samples came from populations having the same or equal population means. For example, St. Mathews Hospital operates an urgent care facility on both the north and south sides of Knoxville, Tennessee. The research question is: Is the mean waiting time for patients visiting the two facilities the same? To investigate, we select a random sample from each of the facilities and compute the sample means. We test the null hypothesis that the mean waiting time is the same at the two facili- ties. The alternate hypothesis is that the mean waiting time is not the same for the two facilities. If the population standard deviations are known, we use the z distribution as the distribution of the test sta- tistic. If the population standard deviations are not known, the test statistic follows the t distribution.

Our discussion in Chapter 11 also concerned dependent samples. The test statistic is the t dis- tribution and we assume that the distribution of differences follows the normal distribution. One typ- ical paired sample problem calls for recording an individual’s blood pressure before administering medication and then again afterward in order to evaluate the effectiveness of the medication. We also considered the case of testing two population proportions. For example, the production man- ager wished to compare the proportion of defects on the day shift with that of the second shift.

Chapter 11 dealt with the difference between two population means. Chapter 12 presented tests for variances and a procedure called the analysis of variance, or ANOVA. ANOVA is used to simultaneously determine whether several independent normal populations have the same mean. This is accomplished by comparing the variances of the random samples selected from these pop- ulations. We apply the usual hypothesis-testing procedure, but we use the F distribution as the test statistic. Often the calculations are tedious, so a software package is recommended.

As an example of analysis of variance, a test could be conducted to resolve whether there is any difference in effectiveness among five fertilizers on the weight of popcorn ears. This type of analysis is referred to as one-factor ANOVA because we are able to draw conclusions about only one factor, called a treatment. If we want to draw conclusions about the simultaneous effects of more than one factor or variable, we use the two-factor ANOVA technique. Both the one-factor and two-factor tests use the F distribution as the distribution of the test statistic. The F distribution is also the distribution of the test statistic used to find whether one normal population has more variation than another.

Two-factor analysis of variance is further complicated by the possibility that interactions may exist between the factors. There is an interaction if the response to one of the factors depends on the level of the other factor. Fortunately, the ANOVA technique is easily extended to include a test for interactions.

Glossary

Chapter 10 Alpha The probability of a Type I error or the level of sig- nificance. Its symbol is the Greek letter �.

Alternate hypothesis The conclusion we accept when we demonstrate that the null hypothesis is false. It is also called the research hypothesis.

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456 Chapter 12

Problems

For problems 1–8, state: (a) the null and the alternate hypothesis, (b) the decision rule, and (c) the decision regarding the null hypothesis, (d) then interpret the result.

1. A machine is set to produce tennis balls so the mean bounce is 36 inches when the ball is dropped from a platform of a certain height. The production supervisor suspects that the mean bounce has changed and is less than 36 inches. As an experiment, a sam- ple of 12 balls was dropped from the platform and the mean height of the bounce was

Critical value A value that is the dividing point between the region where the null hypothesis is not rejected and the region where it is rejected. Degrees of freedom The number of items in a sample that are free to vary. Suppose there are two items in a sam- ple, and we know the mean. We are free to specify only one of the two values, because the other value is auto- matically determined (since the two values total twice the mean). Example: If the mean is $6, we are free to choose only one value. Choosing $4 makes the other value $8 be- cause So there is 1 degree of freedom in this illustration. We can determine the degrees of free- dom by If n is 4, then there are 3 degrees of freedom, found by Hypothesis A statement or claim about the value of a population parameter. Examples: 40.7 percent of all per- sons 65 years old or older live alone. The mean number of people in a car is 1.33. Hypothesis testing A statistical procedure, based on sample evidence and probability theory, used to determine whether the statement about the population parameter is reasonable. Null hypothesis A statement about the value of a popu- lation parameter that is developed for testing in the face of numerical evidence. It is written as H0. One-tailed test Used when the alternate hypothesis states a direction, such as read “the popula- tion mean is greater than 40.” Here the rejection region is only in one tail (the right tail). Proportion A fraction or percentage of a sample or a population having a particular trait. If 5 out of 50 in a sam- ple liked a new cereal, the proportion is 5�50, or .10. p-value The probability of computing a value of the test statistic at least as extreme as the one found in the sample data when the null hypothesis is true. Significance level The probability of rejecting the null hypothesis when it is true. Two-tailed test Used when the alternate hypothesis does not state a direction, such as read “the population mean is not equal to 75.” There is a region of rejection in each tail. Type I error Occurs when a true is rejected. Type II error Occurs when a false is accepted.

Chapter 11 Dependent samples Dependent samples are charac- terized by a measurement, then some type of intervention, followed by another measurement. Paired samples are also dependent because the same individual or item is a member of both samples. Example: Ten participants in a marathon were weighed prior to and after competing in the race. We wish to study the mean amount of weight loss. Independent samples The samples chosen at random are not related to each other. We wish to study the mean

H1: � � 75,

H1: � 7 40,

n � 1 � 4 � 1 � 3. n � 1 � 2 � 1 � 1.

$4 � $8 � 2($6).

age of the inmates at the Auburn and Allegheny prisons. We select a random sample of 28 inmates from the Auburn prison and a sample of 19 inmates at the Allegheny prison. A person cannot be an inmate in both prisons. The sam- ples are independent, that is, unrelated. Pooled estimate of the population variance A weighted average of and used to estimate the common variance,

when using small samples to test the difference between two population means. t distribution Investigated and reported by William S. Gossett in 1908 and published under the pseudonym Student. It is similar to the standard normal distribution presented in Chapter 7. The major characteristics of t are:

1. It is a continuous distribution. 2. It can assume values between minus infinity and plus

infinity. 3. It is symmetrical about its mean of zero. However, it is

more spread out and flatter at the apex than the stan- dard normal distribution.

4. It approaches the standard normal distribution as n gets larger.

5. There is a family of t distributions. One t distribution exists for a sample of 15 observations, another for 25, and so on.

Chapter 12 Analysis of variance (ANOVA) A technique used to test simultaneously whether the means of several populations are equal. It uses the F distribution as the distribution of the test statistic. Block A second source of variation, in addition to treatments. F distribution It is used as the test statistic for ANOVA problems, as well as others. The major characteristics of the F distribution are:

1. It is never negative. 2. It is a continuous distribution approaching the X-axis

but never touching it. 3. It is positively skewed. 4. It is based on two sets of degrees of freedom. 5. Like the t distribution, there is a family of F distribu-

tions. There is one distribution for 17 degrees of free- dom in the numerator and 9 degrees of freedom in the denominator, there is another F distribution for 7 de- grees of freedom in the numerator and 12 degrees of freedom in the denominator, and so on.

Interaction Two variables interact if the effect that one factor has on the variable being studied is different for dif- ferent levels of the other factor. Treatment A source of variation. It identifies the several populations being examined.

�2, s22s

2 1

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A Review of Chapters 10–12 457

Using the .05 significance level, determine whether the additive is effective in prolonging the life of the paint.

6. A Buffalo, New York, cola distributor is featuring a super-special sale on 12-packs. She wonders where in the grocery store to place the cola for maximum attention. Should it be near the front door of the grocery stores, in the cola section, at the checkout regis- ters, or near the milk and other dairy products? Four stores with similar total sales coop- erated in an experiment. In one store, the 12-packs were stacked near the front door, in

Number of Hours by Sample

A B C D E F G H I J

Without additive 325 313 320 340 318 312 319 330 333 319 With additive 323 313 326 343 310 320 313 340 330 315

35.5 inches, with a standard deviation of 0.9 inches. At the .05 significance level, can the supervisor conclude that the mean bounce height is less than 36 inches?

2. Research by First Bank of Illinois revealed that 8 percent of its customers wait more than five minutes to do their banking when not using the drive-through facility. Management considers this reasonable and will not add more tellers unless the proportion becomes larger than 8 percent. The branch manager at the Litchfield Branch believes that the wait is longer than the standard at her branch and requested additional part-time tellers. To support her request, she found that, in a sample of 100 customers, 10 waited more than five minutes. At the .01 significance level, is it reasonable to conclude that more than 8 per- cent of the customers wait more than five minutes?

3. It was hypothesized that road construction workers do not engage in productive work 20 minutes on the average out of every hour. Some claimed the nonproductive time is greater than 20 minutes. An actual study was conducted at a construction site, using a stopwatch and other ways of checking the work habits. A random check of workers revealed the following unproductive times, in minutes, during a one-hour period (exclu- sive of regularly scheduled breaks):

10 25 17 20 28 30 18 23 18

Using the .05 significance level, is it reasonable to conclude the mean unproductive time is greater than 20 minutes?

4. A test is to be conducted involving the mean holding power of two glues designed for plas- tic. First, a small plastic hook was coated at one end with Epox glue and fastened to a sheet of plastic. After it dried, weight was added to the hook until it separated from the sheet of plastic. The weight was then recorded. This was repeated until 12 hooks were tested. The same procedure was followed for Holdtite glue, but only 10 hooks were used. The sample results, in pounds, were:

Epox Holdtite

Sample mean 250 252 Sample standard deviation 5 8 Sample size 12 10

At the .01 significance level, is there a difference between the mean holding power of Epox and that of Holdtite?

5. Pittsburgh Paints wishes to test an additive formulated to increase the life of paints used in the hot and arid conditions of the Southwest. The top half of a piece of wood was painted using the regular paint. The bottom half was painted with the paint including the additive. The same procedure was followed for a total of 10 pieces. Then each piece was subjected to brilliant light. The data, the number of hours each piece lasted before it faded beyond a certain point, follow:

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458 Chapter 12

Cola at the Door In Soft Drink Section Near Registers Dairy Section

$6 $ 5 $ 7 $10 8 10 10 9 3 12 9 6 7 4 4 11

9 5 7

The Buffalo distributor wants to find out whether there is a difference in the mean sales for cola stacked at the four locations in the store. Use the .05 significance level.

7. Williams Corporation is investigating the effects of educational background on employee performance. A potential relevant variable in this case is the self-rated social status of the employee. The company has recorded the annual sales volumes (in $000) achieved by sales employees in each of the categories below. Perform a complete two-way analy- sis of variance (including the possibility of interactions) on the data and describe what your results suggest.

School Type

Self-Rated Social Status Ivy League State-supported Small Private

Low 62, 61 68, 64 70, 70 Medium 68, 64 74, 68 62, 65 High 70, 71 57, 60 57, 56

8. A school supervisor is reviewing initial wages of former students (in $000). Samples were taken over three years for four different majors (accounting, administration, finance, and marketing).

Major/Year 2003 2004 2005

Accounting 75.4, 69.8, 62.3 73.9, 78.8, 62.0 64.2, 80.8, 68.2 Administration 61.5, 59.9, 62.1 63.9, 57.6, 66.5 74.2, 67.5, 58.1 Finance 63.6, 70.2, 72.2 69.2, 72.5, 67.2 74.7, 66.4, 77.9 Marketing 71.3, 69.2, 66.4 74.0, 67.6, 61.7 60.0, 61.3, 62.5

a. Here is an interaction plot of the information. What does it reveal?

another they were placed near the checkout registers, and so on. Sales were checked at specified times in each store for exactly four minutes. The results were:

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A Review of Chapters 10–12 459

b. Write out all of the pairs of null and alternative hypotheses you would apply for a two- way ANOVA.

c. Here is the statistical software output. Use the 0.05 level to check for interactions.

d. If proper, test the other hypotheses at the 0.05 significance level. If it is not appro- priate, describe why you should not do the tests.

A. Century National Bank Refer to the description of Century National Bank at the end of the Review of Chapters 1–4 on page 141.

With many other options available, customers no longer let their money sit in a checking account. For many years the mean checking balance has been $1,600. Does the sample data indicate that the mean account balance has declined from this value?

Recent years have also seen an increase in the use of ATM machines. When Mr. Selig took over the bank, the mean number of transactions per month per customer was 8; now he believes it has increased to more than 10. In fact, the advertising agency that prepares TV commer- cials for Century would like to use this on the new com- mercial being designed. Is there sufficient evidence to conclude that the mean number of transactions per cus- tomer is more than 10 per month? Could the advertising agency say the mean is more than 9 per month?

The bank has branch offices in four different cities: Cincinnati, Ohio; Atlanta, Georgia; Louisville, Kentucky; and Erie, Pennsylvania. Mr. Selig would like to know whether there is a difference in the mean checking account balances among the four branches. If there are differences, between which branches do these differences occur?

Mr. Selig is also interested in the bank’s ATMs. Is there a difference in ATM use among the branches? Also, do customers who have debit cards tend to use ATMs dif- ferently from those who do not have debit cards? Is there a difference in ATM use by those with checking accounts that pay interest versus those that do not? Prepare a report for Mr. Selig answering these questions.

B. Bell Grove Medical Center Ms. Gene Dempsey manages the emergency care center at Bell Grove Medical Center. One of her responsibilities is

Date Day Patients

9-29-06 Monday 38 9-30-06 Tuesday 28 10-1-06 Wednesday 28 10-2-06 Thursday 30 10-3-06 Friday 35 10-6-06 Monday 35 10-7-06 Tuesday 25 10-8-06 Wednesday 22 10-9-06 Thursday 21 10-10-06 Friday 32 10-13-06 Monday 37 10-14-06 Tuesday 29 10-15-06 Wednesday 27 10-16-06 Thursday 28 10-17-06 Friday 35 10-20-06 Monday 37 10-21-06 Tuesday 26 10-22-06 Wednesday 28 10-23-06 Thursday 23 10-24-06 Friday 33

to have enough nurses so that incoming patients needing service can be handled promptly. It is stressful for patients to wait a long time for emergency care even when their care needs are not life threatening. Ms. Dempsey gathered the following information regarding the number of patients over the last several weeks. The center is not open on weekends. Does it appear that there are any differences in the number of patients served by the day of the week? If there are differences, which days seem to be the busiest? Write a brief report summarizing your findings.

Source DF SS MS F P Major 3 329.20 109.732 3.39 0.034 Year 2 7.32 3.659 0.11 0.894 Interaction 6 183.57 30.595 0.94 0.482 Error 24 777.29 32.387 Total 35 1297.37

Cases

Practice Test

Part 1—Objective 1. A statement about the value of a population parameter that always includes the equal sign is called the .

1. 2. The likelihood of rejecting a true null hypothesis is called the . 2.

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3. When conducting a test of hypothesis about a population proportion, the value n� should be at least . 3.

4. When conducting a test of hypothesis about a single population mean, the z distribution is used as the test statistic only when the is known. 4.

5. In a two-sample test of hypothesis for means, the population standard deviations are not known and we must assume what about the shape of the populations? 5.

6. A value calculated from sample information used to determine whether to reject the null hypothesis is called the . 6.

7. In a two-tailed test, the rejection region is . (all in the upper tail, all in the lower tail, split evenly between the two tails, none of these—pick one) 7.

8. Which of the following is not a characteristic of the F distribution? (continuous, positively skewed, range from �q to q, family of distributions) 8.

9. To perform a one-way ANOVA, the treatments must be . (independent, mutually exclusive, continuous) 9.

10. In a one-way ANOVA, there are four treatments and six observations in each treatment. What are the degrees of free- dom for the F distribution? 10.

Part 2—Problems For problems 1 and 2, state the null and alternate hypotheses and the decision rule, make a decision regarding the null hy- pothesis, and interpret the result. 1. The Park Manager at Fort Fisher State Park in North Carolina believes the typical summer visitor spends more than

90 minutes in the park. A sample of 18 visitors during the months of June, July, and August of 2008 revealed the mean time in the park for visitors was 96 minutes, with a standard deviation of 12 minutes. At the .01 significance level, is it reasonable to conclude the mean time in the park is greater than 90 minutes?

2. Is there a difference in the mean miles traveled per week by each of two taxi cab companies operating in the Grand Strand area? The Sun News, the local paper, is investigating and obtained the following sample information. At the .05 significance level, is it reasonable to conclude there is a difference in the mean miles traveled? Assume equal pop- ulation variances.

460 Chapter 12

ANOVA

Source of Variation SS df MS F

Between groups 6.892202 2 3.446101 4.960047 Within groups 12.50589 18 0.694772

Total 19.3981 20

Horse and Buggy Variable Yellow Cab Cab Company

Mean miles 837 797 Standard deviation 30 40 Sample size 14 12

3. The results of a one-way ANOVA are reported below. Use the .05 significance level.

Answer the following questions. a. How many treatments are in the study? b. What is the total sample size? c. What is the critical value of F? d. Write out the null hypothesis and the alternate hypothesis. e. What is your decision regarding the null hypothesis? f. Can we conclude the treatment means differ?

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13 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Define the terms independent variable and dependent variable.

LO2 Calculate, test, and interpret the relationship between two variables using the correlation coefficient.

LO3 Apply regression analysis to estimate the linear relationship between two variables.

LO4 Interpret the regression analysis.

LO5 Evaluate the significance of the slope of the regression equation.

LO6 Evaluate a regression equation to predict the dependent variable.

LO7 Calculate and interpret the coefficient of determination.

LO8 Calculate and interpret confidence and prediction intervals.

Correlation and Linear Regression

Exercise 61 lists the movies with the largest world box office sales and

their world box office budget. Is there a correlation between the world

box office sales for a movie and the total amount spent making the

movie? Comment on the association between the two variables.

(See Exercise 61 and LO2.)

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462 Chapter 13

13.1 Introduction Chapters 2 through 4 presented descriptive statistics. We orga- nized raw data into a frequency distribution and computed several measures of location and measures of dispersion to describe the major characteristics of the distribution. In chapters 5 through 7, we described probability, and from probability statements, we created probability distributions. In Chapter 8, we began the study of statistical inference, where we collected a sample to estimate a population parameter such as the population mean or population proportion. In addition, we used the sample data to test an inference or hypothesis about a population mean or a population proportion, the difference between two population means, or the equality of several population means. Each of these tests involved just one interval- or ratio-level variable, such as the profit made on a car sale, the income of bank presidents, or the number of patients admitted each month to a particular hospital.

In this chapter, we shift the emphasis to the study of rela- tionships between two interval- or ratio-level variables. In all busi- ness fields, identifying and studying relationships between vari-

ables can provide information on ways to increase profits, methods to decrease costs, or variables to predict demand. In marketing products, many firms use price reductions through coupons and discount pricing to increase sales. In this exam- ple, we are interested in the relationship between two variables: price reductions and sales. To collect the data, a company can test-market a variety of price reduc- tion methods and observe sales. We hope to confirm a relationship that decreasing price leads to increased sales. In economics, you will find many relationships between two variables that are the basis of economics, such as supply and demand and demand and price.

As another familiar example, recall in Section 4.6 in Chapter 4 we used the Applewood Auto Group data to show the relationship between two variables with a scatter diagram. We plotted the profit for each vehicle sold on the vertical axis and the age of the buyer on the horizontal axis. See the statistical software output on page 125. In that diagram, we observed that as the age of the buyer increased, the profit for each vehicle also increased.

Other examples of relationships between two variables are:

• Does the amount Healthtex spends per month on training its sales force affect its monthly sales?

• Is the number of square feet in a home related to the cost to heat the home in January?

• In a study of fuel efficiency, is there a relationship between miles per gallon and the weight of a car?

• Does the number of hours that students study for an exam influence the exam score?

In this chapter, we carry this idea further. That is, we develop numerical measures to express the relationship between two variables. Is the relationship strong or weak? Is it direct or inverse? In addition, we develop an equation to express the relationship between variables. This will allow us to estimate one variable on the basis of another.

To begin our study of relationships between two variables, we examine the meaning and purpose of correlation analysis. We continue by developing an equa- tion that will allow us to estimate the value of one variable based on the value of another. This is called regression analysis. We will also evaluate the ability of the equation to accurately make estimations.

Statistics in Action

The space shuttle Challenger exploded on January 28, 1986. An investigation of the cause examined four contractors: Rockwell Interna- tional for the shuttle and engines, Lockheed Martin for ground support, Mar- tin Marietta for the external fuel tanks, and Morton Thiokol for the solid fuel booster rockets. After several months, the investigation blamed the explosion on de- fective O-rings pro- duced by Morton Thiokol. A study of the contractor’s stock

(continued)

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prices showed an interesting happen- stance. On the day of the crash, Morton Thiokol stock was down 11.86% and the stock of the other three lost only 2 to 3%. Can we con- clude that financial markets predicted the outcome of the investigation?

Correlation and Linear Regression 463

13.2 What Is Correlation Analysis? When we study the relationship between two interval- or ratio-scale variables, we often start with a scatter diagram. This procedure provides a visual representation of the relationship between the variables. The next step is usually to calculate the cor- relation coefficient. It provides a quantitative measure of the strength of the rela- tionship between two variables. As an example, the sales manager of Copier Sales of America, which has a large sales force throughout the United States and Canada, wants to determine whether there is a relationship between the number of sales calls made in a month and the number of copiers sold that month. The manager selects a random sample of 10 representatives and determines the number of sales calls each representative made. This information is reported in Table 13–1.

TABLE 13–1 Number of Sales Calls and Copiers Sold for 10 Salespeople

Number of Number of Sales Representative Sales Calls Copiers Sold

Tom Keller 20 30 Jeff Hall 40 60 Brian Virost 20 40 Greg Fish 30 60 Susan Welch 10 30 Carlos Ramirez 10 40 Rich Niles 20 40 Mike Kiel 20 50 Mark Reynolds 20 30 Soni Jones 30 70

By reviewing the data, we observe that there does seem to be some relation- ship between the number of sales calls and the number of units sold. That is, the salespeople who made the most sales calls sold the most units. However, the rela- tionship is not “perfect” or exact. For example, Soni Jones made fewer sales calls than Jeff Hall, but she sold more units.

In addition to the graphical techniques in Chapter 4, we will develop numerical measures to precisely describe the relationship between the two variables, sales calls and copiers sold. This group of statistical techniques is called correlation analysis.

CORRELATION ANALYSIS A group of techniques to measure the relationship between two variables.

The basic idea of correlation analysis is to report the relationship between two variables. The usual first step is to plot the data in a scatter diagram. An example will show how a scatter diagram is used.

Example Copier Sales of America sells copiers to businesses of all sizes throughout the United States and Canada. Ms. Marcy Bancer was recently promoted to the position of national sales manager. At the upcoming sales meeting, the sales representatives from all over the country will be in attendance. She would like to impress upon them the importance of making that extra sales call each day. She decides to gather some information on the relationship between the number of sales calls and the number of copiers sold. She selects a random sample of 10 sales representatives and deter- mines the number of sales calls they made last month and the number of copiers they sold. The sample information is reported in Table 13–1. What observations can you make about the relationship between the number of sales calls and the number of copiers sold? Develop a scatter diagram to display the information.

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464 Chapter 13

Solution Based on the information in Table 13–1, Ms. Bancer suspects there is a relation- ship between the number of sales calls made in a month and the number of copiers sold. Soni Jones sold the most copiers last month, and she was one of three representatives making 30 or more sales calls. On the other hand, Susan Welch and Carlos Ramirez made only 10 sales calls last month. Ms. Welch, along with two others, had the lowest number of copiers sold among the sampled representatives.

The implication is that the number of copiers sold is related to the number of sales calls made. As the number of sales calls increases, it appears the number of copiers sold also increases. We refer to number of sales calls as the independent variable and number of copiers sold as the dependent variable.

The independent variable provides the basis for estimation. It is the predictor variable. For example, we would like to predict the expected number of copiers sold if a salesperson makes 20 sales calls. Notice that we choose this value. The inde- pendent variable is not a random number.

The dependent variable is the variable that is being predicted or estimated. It can also be described as the result or outcome for a known value of the indepen- dent variable. The dependent variable is random. That is, for a given value of the independent variable, there are many possible outcomes for the dependent variable. In this example, notice that five different sales representatives made 20 sales calls. The result or outcome of making 20 sales calls is three different values of the depen- dent variable.

It is common practice to scale the dependent variable (copiers sold) on the verti- cal or Y-axis and the independent variable (number of sales calls) on the horizontal or X-axis. To develop the scatter diagram of the Copier Sales of America sales infor- mation, we begin with the first sales representative, Tom Keller. Tom made 20 sales calls last month and sold 30 copiers, so and To plot this point, move along the horizontal axis to then go vertically to and place a dot at the intersection. This process is continued until all the paired data are plotted, as shown in Chart 13–1.

Y � 30X � 20, Y � 30.X � 20

Sales calls

Co pi

er s

so ld

0 10 20 30 40 50

80 70 60 50 40 30 20 10 0

CHART 13–1 Scatter Diagram Showing Sales Calls and Copiers Sold

The scatter diagram shows graphically that the sales representatives who make more calls tend to sell more copiers. It is reasonable for Ms. Bancer, the national sales manager at Copier Sales of America, to tell her salespeople that, the more sales calls they make, the more copiers they can expect to sell. Note that, while there appears to be a positive relationship between the two variables, all the points do not fall on a line. In the following section, you will measure the strength and direction of this relationship between two variables by determining the correlation coefficient.

LO1 Define the terms independent variable and dependent variable.

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Correlation and Linear Regression 465

13.3 The Correlation Coefficient Originated by Karl Pearson about 1900, the correlation coefficient describes the strength of the relationship between two sets of interval-scaled or ratio-scaled variables. Designated r, it is often referred to as Pearson’s r and as the Pearson product-moment correlation coefficient. It can assume any value from to �1.00 inclusive. A correlation coefficient of or indicates perfect correlation. For example, a correlation coefficient for the preceding example com- puted to be would indicate that the number of sales calls and the number of copiers sold are perfectly related in a positive linear sense. A computed value of

reveals that sales calls and the number of copiers sold are perfectly related in an inverse linear sense. How the scatter diagram would appear if the relationship between the two sets of data were linear and perfect is shown in Chart 13–2.

�1.00

�1.00�1.00 �1.00

If there is absolutely no relationship between the two sets of variables, Pearson’s r is zero. A correlation coefficient r close to 0 (say, .08) shows that the linear rela- tionship is quite weak. The same conclusion is drawn if Coefficients of

and have equal strength; both indicate very strong correlation between the two variables. Thus, the strength of the correlation does not depend on the direction (either � or �).

Scatter diagrams for r � 0, a weak r (say, �.23), and a strong r (say, are shown in Chart 13–3. Note that, if the correlation is weak, there is considerable scatter about a line drawn through the center of the data. For the scatter diagram represent- ing a strong relationship, there is very little scatter about the line. This indicates, in the example shown on the chart, that hours studied is a good predictor of exam score.

The following drawing summarizes the strength and direction of the correlation coefficient.

�.87)

�.91�.91 r � �.08.

Y Perfect negative correlation Perfect positive correlation

Line has negative slope

Line has positive slope

r = –1.00

r = +1.00

CHART 13–2 Scatter Diagrams Showing Perfect Negative Correlation and Perfect Positive Correlation

Positive correlation

.50 1.00–.50–1.00 0

Negative correlation

Moderate negative

correlation

Perfect negative

correlation

Weak negative

correlation

Strong negative

correlation

Moderate positive

correlation

No correlation

Perfect positive

correlation

Weak positive

correlation

Strong positive

correlation

LO2 Calculate, test, and interpret the relationship between two variables using the correlation coefficient.

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The characteristics of the correlation coefficient are summarized below.

CORRELATION COEFFICIENT A measure of the strength of the linear relationship between two variables.

CHARACTERISTICS OF THE CORRELATION COEFFICIENT 1. The sample correlation coefficient is identified by the lowercase letter r. 2. It shows the direction and strength of the linear relationship between two

interval- or ratio-scale variables. 3. It ranges from up to and including �1. 4. A value near 0 indicates there is little relationship between the variables. 5. A value near 1 indicates a direct or positive relationship between the variables. 6. A value near �1 indicates inverse or negative relationship between the variables.

�1

How is the value of the correlation coefficient determined? We will use the Copier Sales of America data, which are reported in Table 13–2, as an example. We begin

TABLE 13–2 Sales Calls and Copiers Sold for 10 Salespeople

Sales Copiers Calls, Sold,

Sales Representative (X ) (Y )

Tom Keller 20 30 Jeff Hall 40 60 Brian Virost 20 40 Greg Fish 30 60 Susan Welch 10 30 Carlos Ramirez 10 40 Rich Niles 20 40 Mike Kiel 20 50 Mark Reynolds 20 30 Soni Jones 30 70

Total 220 450

Examples of degrees of correlation

CHART 13–3 Scatter Diagrams Depicting Zero, Weak, and Strong Correlation

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Correlation and Linear Regression 467

with a scatter diagram, similar to Chart 13–2. Draw a vertical line through the data values at the mean of the X-values and a horizontal line at the mean of the Y-values. In Chart 13–4, we’ve added a vertical line at 22.0 calls and a horizontal line at 45.0 copiers These lines pass through the “center” of the data and divide the scatter diagram into four quadrants. Think of moving the origin from (0, 0) to (22, 45).

(Y � ©Y�n � 450�10 � 45.0). (X � ©X�n � 220�10 � 22)

Co pi

er s

so ld

(Y )

0 10 20 30 40 50

80 70 60 50 40 30 20 10

Sales calls (X )

X = 22

Y = 45

III II

CHART 13–4 Computation of the Correlation Coefficient

TABLE 13–3 Deviations from the Mean and Their Products

Sales Representative Calls, X Sales, Y X � Y � (X � )(Y � )

Tom Keller 20 30 �2 �15 30 Jeff Hall 40 60 18 15 270 Brian Virost 20 40 �2 �5 10 Greg Fish 30 60 8 15 120 Susan Welch 10 30 �12 �15 180 Carlos Ramirez 10 40 �12 �5 60 Rich Niles 20 40 �2 �5 10 Mike Kiel 20 50 �2 5 �10 Mark Reynolds 20 30 �2 �15 30 Soni Jones 30 70 8 25 200

900

YXYX

Two variables are positively related when the number of copiers sold is above the mean and the number of sales calls is also above the mean. These points appear in the upper-right quadrant (labeled Quadrant I) of Chart 13–4. Similarly, when the number of copiers sold is less than the mean, so is the number of sales calls. These points fall in the lower-left quadrant of Chart 13–4 (labeled Quadrant III). For example, the last person on the list in Table 13–2, Soni Jones, made 30 sales calls and sold 70 copiers. These values are above their respective means, so this point is located in Quadrant I which is in the upper-right quadrant. She made 8 more sales calls than the mean and sold 25 more copiers than the mean. Tom Keller, the first name on the list in Table 13–2, made 20 sales calls and sold 30 copiers. Both of these values are less than their respective mean; hence this point is in the lower-left quadrant. Tom made 2 less sales calls and sold 15 less copiers than the respective means. The deviations from the mean number of sales calls and for the mean number of copiers sold are summarized in Table 13–3 for the 10 sales rep- resentatives. The sum of the products of the deviations from the respective means is 900. That is, the term

In both the upper-right and the lower-left quadrants, the product of is positive because both of the factors have the same sign. In our example, this

(X � X )(Y � Y ) ©(X � X )(Y � Y ) � 900.

(Y � Y � 70 � 45) (X � X � 30 � 22)

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468 Chapter 13

happens for all sales representatives except Mike Kiel. We can therefore expect the correlation coefficient to have a positive value.

If the two variables are inversely related, one variable will be above the mean and the other below the mean. Most of the points in this case occur in the upper- left and lower-right quadrants, that is, Quadrant II and IV. Now and will have opposite signs, so their product is negative. The resulting correlation coef- ficient is negative.

What happens if there is no linear relationship between the two variables? The points in the scatter diagram will appear in all four quadrants. The negative products of offset the positive products, so the sum is near zero. This leads to a correlation coefficient near zero. So, the term drives the strength as well as the sign of the relationship between the two variables.

The correlation coefficient also needs to be unaffected by the units of the two variables. For example, if we had used hundreds of copiers sold instead of the number sold, the correlation coefficient would be the same. The correlation coeffi- cient is independent of the scale used if we divide the term by the sample standard deviations. It is also made independent of the sample size and bounded by the values and if we divide by

This reasoning leads to the following formula: (n � 1).�1.00�1.00

©(X � X )(Y � Y )

g (X � X )(Y � Y ) (X � X )(Y � Y )

(Y � Y )(X � X )

CORRELATION COEFFICIENT [13–1]r � ©(X � X )(Y � Y )

(n � 1)sx sy

To compute the correlation coefficient, we use the standard deviations of the sample of 10 sales calls and 10 copiers sold. We could use formula (3–12) to cal- culate the sample standard deviations or we could use a software package. For the specific Excel and Minitab commands, see the Software Commands section at the end of Chapter 3. The following is the Excel output. The standard deviation of the number of sales calls is 9.189 and of the number of copiers sold 14.337.

We now insert these values into formula (13–1) to determine the correlation coefficient:

How do we interpret a correlation of 0.759? First, it is positive, so we con- clude there is a direct relationship between the number of sales calls and the num- ber of copiers sold. This confirms our reasoning based on the scatter diagram,

r � ©(X � X )(Y � Y )

(n � 1)sx sy �

900 (10 � 1)(9.189)(14.337)

� 0.759

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Correlation and Linear Regression 469

Chart 13–4. The value of 0.759 is fairly close to 1.00, so we conclude that the association is strong.

We must be careful with the interpretation. The correlation of 0.759 indicates a strong positive association between the variables. Ms. Bancer would be correct to encourage the sales personnel to make that extra sales call, because the number of sales calls made is related to the number of copiers sold. However, does this mean that more sales calls cause more sales? No, we have not demonstrated cause and effect here, only that the two variables—sales calls and copiers sold—are related.

If there is a strong relationship (say, .91) between two variables, we are tempted to assume that an increase or decrease in one variable causes a change in the other variable. For example, it can be shown that the consumption of Georgia peanuts and the consumption of aspirin have a strong correlation. However, this does not indicate that an increase in the consumption of peanuts caused the consumption of aspirin to increase. Likewise, the incomes of professors and the number of inmates in mental institutions have increased proportionately. Further, as the popu- lation of donkeys has decreased, there has been an increase in the number of doc- toral degrees granted. Relationships such as these are called spurious correlations. What we can conclude when we find two variables with a strong correlation is that there is a relationship or association between the two variables, not that a change in one causes a change in the other.

Example The Applewood Auto Group’s marketing department believes younger buyers pur- chase vehicles on which lower profits are earned and the older buyers purchase vehicles on which higher profits are earned. They would like to use this information as part of an upcoming advertising campaign to try to attract older buyers on which the profits tend to be higher. Develop a scatter diagram depicting the relationship between vehicle profits and age of the buyer. Use statistical software to determine the correlation coefficient. Would this be a useful advertising feature?

The scatter diagram suggests that a positive relationship does exist between age and profit; however, that relationship does not appear strong.

The next step is to calculate the correlation coefficient to evaluate the relative strength of the relationship. Statistical software provides an easy way to calculate the value of the correlation coefficient. The Excel output follows.

Solution Using the Applewood Auto Group example, the first step is to graph the data using a scatter plot. It is shown in Chart 13-5.

3500

3000

2500

2000

1500

1000

500

0 0 20 40 60 80

Age

Scatter Plot of Profit vs. Age

Pr of

CHART 13–5 Scatter Diagram of Applewood Auto Group Data

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470 Chapter 13

Self-Review 13–1 Haverty’s Furniture is a family business that has been selling to retail customers in the Chicago area for many years. The company advertises extensively on radio, TV, and the Internet, emphasizing low prices and easy credit terms. The owner would like to review the relationship between sales and the amount spent on advertising. Below is information on sales and advertising expense for the last four months.

Advertising Expense Sales Revenue Month ($ million) ($ million)

July 2 7 August 1 3 September 3 8 October 4 10

(a) The owner wants to forecast sales on the basis of advertising expense. Which variable is the dependent variable? Which variable is the independent variable?

(b) Draw a scatter diagram. (c) Determine the correlation coefficient. (d) Interpret the strength of the correlation coefficient.

Exercises 1. The following sample observations were randomly selected.

Determine the correlation coefficient and interpret the relationship between X and Y.

X 4 5 3 6 10 y 4 6 5 7 7

For this data, r � 0.262. To evaluate the relationship between a buyer’s age and the profit on a car sale:

1. The relationship is positive or direct. Why? Because the sign of the correlation coefficient is positive. This confirms that as the age of the buyer increases, the profit on a car sale also increases.

2. The relationship between the two variables is weak. For a positive relationship, values of the correlation coefficient close to one indicate stronger relationships. In this case, r � 0.262. It is closer to zero, and we would observe that the rela- tionship is not very strong.

It is not recommended that Applewood use this information as part of an adver- tising campaign to attract older more profitable buyers.

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Correlation and Linear Regression 471

X 5 3 6 3 4 4 6 8 Y 13 15 7 12 13 11 9 5

2. The following sample observations were randomly selected.

Determine the correlation coefficient and interpret the relationship between X and Y. 3. Bi-lo Appliance Super-Store has outlets in several large metropolitan areas in New England.

The general sales manager aired a commercial for a digital camera on selected local TV stations prior to a sale starting on Saturday and ending Sunday. She obtained the infor- mation for Saturday–Sunday digital camera sales at the various outlets and paired it with the number of times the advertisement was shown on the local TV stations. The purpose is to find whether there is any relationship between the number of times the advertise- ment was aired and digital camera sales. The pairings are:

Location of Number of Saturday–Sunday Sales TV Station Airings ($ thousands)

Providence 4 15 Springfield 2 8 New Haven 5 21 Boston 6 24 Hartford 3 17

One-Hour Number of Production Assemblers (units)

2 15 4 25 1 10 5 40 3 30

a. What is the dependent variable? b. Draw a scatter diagram. c. Determine the correlation coefficient. d. Interpret these statistical measures.

4. The production department of Celltronics International wants to explore the relationship between the number of employees who assemble a subassembly and the number produced. As an experiment, two employees were assigned to assemble the subassemblies. They pro- duced 15 during a one-hour period. Then four employees assembled them. They produced 25 during a one-hour period. The complete set of paired observations follows.

The dependent variable is production; that is, it is assumed that different levels of pro- duction result from a different number of employees. a. Draw a scatter diagram. b. Based on the scatter diagram, does there appear to be any relationship between the

number of assemblers and production? Explain. c. Compute the correlation coefficient.

5. The city council of Pine Bluffs is considering increasing the number of police in an effort to reduce crime. Before making a final decision, the council asked the chief of police to survey other cities of similar size to determine the relationship between the number

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472 Chapter 13

City Police Number of Crimes City Police Number of Crimes

Oxford 15 17 Holgate 17 7 Starksville 17 13 Carey 12 21 Danville 25 5 Whistler 11 19 Athens 27 7 Woodville 22 6

a. Which variable is the dependent variable and which is the independent variable? Hint: If you were the Chief of Police, which variable would you decide? Which variable is the random variable?

b. Draw a scatter diagram. c. Determine the correlation coefficient. d. Interpret the correlation coefficient. Does it surprise you that the correlation coefficient

is negative? 6. The owner of Maumee Ford-Mercury-Volvo wants to study the relationship between the

age of a car and its selling price. Listed below is a random sample of 12 used cars sold at the dealership during the last year.

Car Age (years) Selling Price ($000) Car Age (years) Selling Price ($000)

1 9 8.1 7 8 7.6 2 7 6.0 8 11 8.0 3 11 3.6 9 10 8.0 4 12 4.0 10 12 6.0 5 8 5.0 11 6 8.6 6 7 10.0 12 6 8.0

a. Draw a scatter diagram. b. Determine the correlation coefficient. c. Interpret the correlation coefficient. Does it surprise you that the correlation coefficient

is negative?

13.4 Testing the Significance of the Correlation Coefficient Recall that the sales manager of Copier Sales of America found the correlation between the number of sales calls and the number of copiers sold was 0.759. This indicated a strong positive association between the two variables. However, only 10 salespeople were sampled. Could it be that the correlation in the population is actually 0? This would mean the correlation of 0.759 was due to chance. The pop- ulation in this example is all the salespeople employed by the firm.

Resolving this dilemma requires a test to answer the obvious question: Could there be zero correlation in the population from which the sample was selected? To put it another way, did the computed r come from a population of paired observations with zero correlation? To continue our convention of allowing Greek letters to represent a population parameter, we will let represent the correlation in the population. It is pronounced “rho.”

�

Could the correlation in the population be zero?

of police and the number of crimes reported. The chief gathered the following sample information.

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We will continue with the illustration involving sales calls and copiers sold. We employ the same hypothesis testing steps described in Chapter 10. The null hypoth- esis and the alternate hypothesis are:

(The correlation in the population is zero.) (The correlation in the population is different from zero.)

From the way is stated, we know that the test is two-tailed. The formula for t is:

H1: � � 0 H0: � � 0

with degrees of freedom [13–2] n � 2t � r1n � 2 11 � r 2

t TEST FOR THE CORRELATION COEFFICIENT

Using the .05 level of significance, the decision rule in this instance states that if the computed t falls in the area between plus 2.306 and minus 2.306, the null hypothesis is not rejected. To locate the critical value of 2.306, refer to Appendix B.2 for See Chart 13–6.df � n � 2 � 10 � 2 � 8.

–2.306 0 2.306 Scale of t

Region of rejection

(there is correlation) .025

Region of rejection

(there is correlation) .025

H0 not rejected

(no correlation in population)

CHART 13–6 Decision Rule for Test of Hypothesis at .05 Significance Level and 8 df

Applying formula (13–2) to the example regarding the number of sales calls and units sold:

The computed t is in the rejection region. Thus, is rejected at the .05 significance level. Hence we conclude the correlation in the population is not zero. From a prac- tical standpoint, it indicates to the sales manager that there is correlation with respect to the number of sales calls made and the number of copiers sold in the population of salespeople.

We can also interpret the test of hypothesis in terms of p-values. A p-value is the likelihood of finding a value of the test statistic more extreme than the one com- puted, when is true. To determine the p-value, go to the t distribution in Appen- dix B.2 and find the row for 8 degrees of freedom. The value of the test statistic is 3.297, so in the row for 8 degrees of freedom and a two-tailed test, find the value closest to 3.297. For a two-tailed test at the .02 significance level, the critical value is 2.896, and the critical value at the .01 significance level is 3.355. Because 3.297 is between 2.896 and 3.355, we conclude that the p-value is between .01 and .02.

Both Minitab and Excel will report the correlation between two variables. In addition to the correlation, Minitab reports the p-value for the test of hypothesis that the correlation in the population between the two variables is 0. The Minitab output is at the top of the following page.

t � r1n � 2 11 � r 2

� .759110 � 2 11 � .7592

� 3.297

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Example In the Example on page 470, we found that the correlation coefficient between the profit on the sale of a vehicle by the Applewood Auto Group and the age of the per- son that purchased the vehicle was 0.262. Because the sign of the correlation coef- ficient was positive, we concluded there was a direct relationship between the two variables. However, because the amount of correlation was low—that is, near zero— we concluded that an advertising campaign directed toward the older buyers, where there is a large profit, was not warranted. Does this mean we should conclude that there is no relationship between the two variables? Use the .05 significance level.

Solution To begin to answer the question in the last sentence above, we need to clarify the sample and population issues. Let’s assume that the data collected on the 180 vehi- cles sold by the Applewood Group is a sample from the population of all vehicles sold over many years by the Applewood Auto Group. The Greek letter � is the cor- relation coefficient in the population and r the correlation coefficient in the sample.

Our next step is to set up the null hypothesis and the alternate hypothesis. We test the null hypothesis that the correlation coefficient is equal to zero. The alter- nate hypothesis is that there is positive correlation between the two variables.

(The correlation in the population is zero.) (The correlation in the population is positive)

This is a one-tailed test because we are interested in confirming a positive asso- ciation between the variables. The test statistic follows the t distribution with n � 2 degrees of freedom, so the degrees of freedom is 180 � 2 � 178. However, 178 degrees of freedom is not in Appendix B.2. The closest value is 180, so we will use that value. Our decision rule is to reject the null hypothesis if the computed value of the test statistic is greater than 1.653.

We use formula 13–2 to find the value of the test statistic.

Comparing the value of our test statistic of 3.622 to the critical value of 1.653, we reject the null hypothesis. We conclude that the sample correlation coefficient of 0.262 is too large to have come from a population with no correlation. To put our results another way, there is a positive correlation between profits and age in the population.

t � r2n � 2

21 � r 2 �

0.2622180 � 2

21 � 0.2622 � 3.622

H1: � 7 0 H0: � � 0

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This result is confusing and seems contradictory. On one hand, we observed that the correlation coefficient did not indicate a very strong relationship and that the Applewood Auto Group marketing department should not use this information for its promotion and advertising decisions. On the other hand, the hypothesis test indicated that the correlation coefficient is not equal to zero and that a positive rela- tionship between age and profit exists. How can this be? We must be very careful about the interpretation of the hypothesis test results. The conclusion is that the correlation coefficient is not equal to zero and that there is a positive relationship between the amount of profit earned and the age of the buyer. The result of the hypothesis test only shows that a relationship exists. The hypothesis test makes no claims regarding the strength of the relationship.

Self-Review 13–2 A sample of 25 mayoral campaigns in medium-sized cities with populations between 50,000 and 250,000 showed that the correlation between the percent of the vote received and the amount spent on the campaign by the candidate was .43. At the .05 significance level, is there a positive association between the variables?

Exercises 7. The following hypotheses are given.

A random sample of 12 paired observations indicated a correlation of .32. Can we con- clude that the correlation in the population is greater than zero? Use the .05 signifi- cance level.

8. The following hypotheses are given.

A random sample of 15 paired observations have a correlation of �.46. Can we conclude that the correlation in the population is less than zero? Use the .05 significance level.

9. Pennsylvania Refining Company is studying the relationship between the pump price of gasoline and the number of gallons sold. For a sample of 20 stations last Tuesday, the correlation was .78. At the .01 significance level, is the correlation in the population greater than zero?

10. A study of 20 worldwide financial institutions showed the correlation between their assets and pretax profit to be .86. At the .05 significance level, can we conclude that there is positive correlation in the population?

11. The Airline Passenger Association studied the relationship between the number of passen- gers on a particular flight and the cost of the flight. It seems logical that more pas- sengers on the flight will result in more weight and more luggage, which in turn will result in higher fuel costs. For a sample of 15 flights, the correlation between the num- ber of passengers and total fuel cost was .667. Is it reasonable to conclude that there is positive association in the population between the two variables? Use the .01 sig- nificance level.

12. The Student Government Association at Middle Carolina University wanted to demon- strate the relationship between the number of beers a student drinks and their blood alcohol content (BAC). A random sample of 18 students participated in a study in which each participating student was randomly assigned a number of 12-ounce cans of beer to drink. Thirty minutes after consuming their assigned number of beers a member of the

H1: � 6 0 H0: � � 0

H1: � 7 0 H0: � � 0

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13.5 Regression Analysis In the previous sections of this chapter, we evaluated the direction and the signifi- cance of the linear relationship between two variables by finding the correlation coefficient. If the correlation coefficient is significantly different from zero, then the next step is to develop an equation to express the linear relationship between the two variables. Using this equation, we will be able to estimate the value of the dependent

variable Y based on a selected value of the independent variable X. The technique used to develop the equation and provide the estimates is called regression analysis.

In Table 13–1, we reported the number of sales calls and the num- ber of units sold for a sample of 10 sales representatives employed by Copier Sales of America. Chart 13–1 portrayed this information in a scatter diagram. Recall that we tested the significance of the correlation coefficient (r � 0.759) and concluded that a significant relationship exists between the two variables. Now we want to develop a linear equation that expresses the relationship between the number of sales calls, the independent variable, and the number of units sold, the dependent variable. The equation for the line used to estimate Y on the basis of X is referred to as the regression equation.

local sheriff’s office measured their blood alcohol content. The sample information is reported below.

Student Beers BAC Student Beers BAC

1 6 0.10 10 3 0.07 2 7 0.09 11 3 0.05 3 7 0.09 12 7 0.08 4 4 0.10 13 1 0.04 5 5 0.10 14 4 0.07 6 3 0.07 15 2 0.06 7 3 0.10 16 7 0.12 8 6 0.12 17 2 0.05 9 6 0.09 18 1 0.02

REGRESSION EQUATION An equation that expresses the linear relationship between two variables.

Use a statistical software package to answer the following questions. a. Develop a scatter diagram for the number of beers consumed and BAC. Comment on

the relationship. Does it appear to be strong or weak? Does it appear to be positive or inverse?

b. Determine the correlation coefficient. c. At the .01 significance level, is it reasonable to conclude that there is a positive rela-

tionship in the population between the number of beers consumed and the BAC? What is the p-value?

Least Squares Principle In regression analysis, our objective is to use the data to position a line that best represents the relationship between the two variables. Our first approach is to use a scatter diagram to visually position the line.

The scatter diagram in Chart 13–1 is reproduced in Chart 13–7, with a line drawn with a ruler through the dots to illustrate that a line would probably fit the data.

476 Chapter 13

LO3 Apply regression analysis to estimate the linear relationship between two variables.

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Correlation and Linear Regression 477

However, the line drawn using a straight edge has one disadvantage: Its position is based in part on the judgment of the person drawing the line. The hand-drawn lines in Chart 13–8 represent the judgments of four people. All the lines except line A seem to be reasonable. That is, each line is centered among the graphed data. However, each would result in a different estimate of units sold for a particular num- ber of sales calls.

Co pi

er s

so ld

0 10 20 30 40 50

80 70 60 50 40 30 20 10 0

Sales calls

Co pi

er s

so ld

0 10 20 30 40 50

80 70 60 50 40 30 20 10 0

Sales calls

CHART 13–7 Sales Calls and Copiers Sold for 10 Sales Representatives

CHART 13–8 Four Lines Superimposed on the Scatter Diagram

However, we would prefer a method that results in a single, best regression line. This method is called the least squares principle. It gives what is commonly referred to as the “best-fitting” line.

LEAST SQUARES PRINCIPLE A mathematical procedure that uses the data to position a line with the objective of minimizing the sum of the squares of the vertical distances between the actual Y values and the predicted values of Y.

To illustrate this concept, the same data are plotted in the three charts that fol- low. The dots are the actual values of Y, and the asterisks are the predicted values of Y for a given value of X. The regression line in Chart 13–9 was determined using the least squares method. It is the best-fitting line because the sum of the squares of the vertical deviations about it is at a minimum. The first plot devi- ates by 2 from the line, found by The deviation squared is 4. The squared deviation for the plot is 16. The squared deviation for the plot

is 4. The sum of the squared deviations is 24, found by Assume that the lines in Charts 13–10 and 13–11 were drawn with a straight edge.

The sum of the squared vertical deviations in Chart 13–10 is 44. For Chart 13–11,

4 � 16 � 4.Y � 16 X � 5,Y � 18X � 4,

10 � 8. (X � 3, Y � 8)

CHART 13–11 Different Line Drawn with a Straight Edge

2 3 4 5 6

Ac hi

ev em

en t s

co re

Years of service with company

6 2 3 4 5 6

Ac hi

ev em

en t s

co re

Years of service with company

6 2 3 4 5 6

Ac hi

ev em

en t s

co re

Years of service with company

2 8

CHART 13–10 Line Drawn with a Straight Edge

CHART 13–9 The Least Squares Line

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478 Chapter 13

it is 132. Both sums are greater than the sum for the line in Chart 13–9, found by using the least squares method.

The equation of a line has the form

GENERAL FORM OF LINEAR REGRESSION EQUATION [13–3]

where: read Y hat, is the estimated value of the Y variable for a selected X value.

a is the Y-intercept. It is the estimated value of Y when Another way to put it is: a is the estimated value of Y where the regression line crosses the Y-axis when X is zero.

b is the slope of the line, or the average change in for each change of one unit (either increase or decrease) in the independent variable X.

X is any value of the independent variable that is selected.

The general form of the linear regression equation is exactly the same form as the equation of any line. a is the Y intercept and b is the slope. The purpose of regression analysis is to calculate the values of a and b to develop a linear equa- tion that best fits the data.

The formulas for a and b are:

SLOPE OF THE REGRESSION LINE [13–4]

where: r is the correlation coefficient.

is the standard deviation of Y (the dependent variable). is the standard deviation of X (the independent variable).

Y-INTERCEPT [13–5]

where: is the mean of Y (the dependent variable). is the mean of X (the independent variable).X

a � Y � bX

b � r sy sx

Ŷ

X � 0. Ŷ,

Ŷ � a � bX

Example Recall the example involving Copier Sales of America. The sales manager gathered information on the number of sales calls made and the number of copiers sold for a random sample of 10 sales representatives. As a part of her presentation at the upcoming sales meeting, Ms. Bancer, the sales manager, would like to offer spe- cific information about the relationship between the number of sales calls and the number of copiers sold. Use the least squares method to determine a linear equa- tion to express the relationship between the two variables. What is the expected number of copiers sold by a representative who made 20 calls?

Solution The first step in determining the regression equation is to find the slope of the least squares regression line. That is, we need the value of b. On page 468, we determined the correlation coefficient r (.759). In the Excel output on the same page, we determined the standard deviation of the independent variable X (9.189) and the standard deviation of the dependent variable Y (14.337). The values are inserted in formula (13–4).

b � r a sy sx b � .759 a

14.337 9.189

b � 1.1842

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Correlation and Linear Regression 479

Next we need to find the value of a. To do this, we use the value for b that we just calculated as well as the means for the number of sales calls and the number of copiers sold. These means are also available in the Excel printout on page 468. From formula (13–5):

Thus, the regression equation is So if a salesperson makes 20 calls, he or she can expect to sell 42.6316 copiers, found by

The b value of 1.1842 indicates that for each additional sales call made the sales representative can expect to increase the number of copiers sold by about 1.2. To put it another way, five addi- tional sales calls in a month will result in about six more copiers being sold, found by

The a value of 18.9476 is the point where the equation crosses the Y-axis. A literal translation is that if no sales calls are made, that is, 18.9476 copiers will be sold. Note that is outside the range of values included in the sample and, therefore, should not be used to estimate the number of copiers sold. The sales calls ranged from 10 to 40, so estimates should be limited to that range.

X � 0 X � 0,

1.1842(5) � 5.921.

Ŷ � 18.9476 � 1.1842X � 18.9476 � 1.1842(20).

Ŷ � 18.9476 � 1.1842X.

a � Y � bX � 45 � 1.1842(22) � 18.9476

Drawing the Regression Line The least squares equation, , can be drawn on the scat- ter diagram. The first sales representative in the sample is Tom Keller. He made 20 calls. His estimated number of copiers sold is Ŷ � 18.9476 � 1.1842(20) �

Ŷ � 18.9476 � 1.1842X

The plot and is located by moving to 20 on the X-axis and then going vertically to 42.6316. The other points on the regression equation can be determined by substituting the particular value of X into the regression equation. All the points are connected to give the line. See Chart 13–12.

Ŷ � 42.6316X � 2042.6316.

Sales Sales Calls Estimated Sales Sales Sales Calls Estimated Sales Representative (X ) ( ) Representative (X ) ( )

Tom Keller 20 42.6316 Carlos Ramirez 10 30.7896 Jeff Hall 40 66.3156 Rich Niles 20 42.6316 Brian Virost 20 42.6316 Mike Kiel 20 42.6316 Greg Fish 30 54.4736 Mark Reynolds 20 42.6316 Susan Welch 10 30.7896 Soni Jones 30 54.4736

ŶŶ

(X = 20, Y = 42.6316)

(X = 40, Y = 66.3156)

Co pi

er s

so ld

0 10 20 30 40 50

80 70 60 50 40 30 20 10 0

Sales calls

CHART 13–12 The Line of Regression Drawn on the Scatter Diagram

Statistics in Action

In finance, investors are interested in the trade-off between returns and risk. One technique to quantify risk is a regression analysis of a com- pany’s stock price (dependent variable) and an average measure of the stock market (independent variable). Often the Standard and Poor’s (S&P) 500 Index is used to estimate the market. The regres- sion coefficient, called beta in finance, shows the change in a company’s stock price for a one-unit change in the S&P Index. For

(continued)

LO4 Interpret the regression analysis.

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The least squares regression line has some interesting and unique features. First, it will always pass through the point To show this is true, we can use the mean number of sales calls to predict the number of copiers sold. In this example, the mean number of sales calls is 22.0, found by The mean number of copiers sold is 45.0, found by If we let and then use the regression equation to find the estimated value for the result is:

The estimated number of copiers sold is exactly equal to the mean number of copiers sold. This simple example shows the regression line will pass through the point rep- resented by the two means. In this case, the regression equation will pass through the point and

Second, as we discussed earlier in this section, there is no other line through the data where the sum of the squared deviations is smaller. To put it another way, the term is smaller for the least squares regression equation than for any other equation. We use the Excel system to demonstrate this condition.

©(Y � Ŷ )2

Y � 45.X � 22

Ŷ � 18.9476 � 1.1842(22) � 45

Ŷ, X � 22Y � 450�10 � 45.

X � 220�10.

(X, Y ).

In Columns A, B, and C in the Excel spreadsheet above, we duplicated the sample information on sales and copiers sold from Table 13–1. In column D, we provide the estimated sales values, the values, as calculated above.

In column E, we calculate the residuals, or the error values. This is the differ- ence between the actual values and the predicted values. That is, column E is

For Soni Jones,

Her actual value is 70. So the residual, or error of estimate, is

This value reflects the amount the predicted value of sales is “off” from the actual sales value.

Next, in Column F, we square the residuals for each of the sales representatives and total the result. The total is 784.2105.

This is the sum of the squared differences or the least squares value. There is no other line through these 10 data points where the sum of the squared differences is smaller.

We can demonstrate the least squares criterion by choosing two arbitrary equa- tions that are close to the least squares equation and determining the sum of the

©(Y � Ŷ )2 � 159.5573 � 39.8868 � . . . � 241.0691 � 784.2105

(Y � Ŷ ) � (70 � 54.4736) � 15.5264

Ŷ � 18.9476 � 1.1842(30) � 54.4736

(Y � Ŷ ).

Ŷ

example, if a stock has a beta of 1.5, then when the S&P index increases by 1%, the stock price will increase by 1.5%. The opposite is also true. If the S&P decreases by 1%, the stock price will decrease by 1.5%. If the beta is 1.0, then a 1% change in the index should show a 1% change in a stock price. If the beta is less than 1.0, then a 1% change in the index shows less than a 1% change in the stock price.

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Correlation and Linear Regression 481

squared differences for these equations. In column G, we use the equation to find the predicted value. Notice this equation is very similar to

the least squares equation. In Column H, we determine the residuals and square these residuals. For the first sales representative, Tom Keller,

This procedure is continued for the other nine sales representatives and the squared residuals totaled. The result is 786. This is a larger value (786 versus 784.2105) than the residuals for the least squares line.

In columns I and J on the output, we repeat the above process for yet another equation Again, this equation is similar to the least squares equation. The details for Tom Keller are:

This procedure is continued for the other nine sales representatives and the residuals totaled. The result is 900, which is also larger than the least squares values.

What have we shown with the example? The sum of the squared residuals for the least squares equation is smaller than for other selected lines.

The bottom line is you will not be able to find a line passing through these data points where the sum of the squared residuals is smaller.

[©(Y � Ŷ )2]

(Y � Y **)2 � (30 � 40)2 � 100

Y ** � 20 � X � 20 � 20 � 40

Y ** � 20 � X.

(Y � Y *)2 � (43 � 30)2 � 169

Y * � 19 � 1.2(20) � 43

Y * � 19 � 1.2X

Self-Review 13–3 Refer to Self-Review 13–1, where the owner of Haverty’s Furniture Company was studying the relationship between sales and the amount spent on advertising. The sales information for the last four months is repeated below.

(a) Determine the regression equation. (b) Interpret the values of a and b. (c) Estimate sales when $3 million is spent on advertising.

Advertising Expense Sales Revenue Month ($ million) ($ million)

July 2 7 August 1 3 September 3 8 October 4 10

Exercises 13. The following sample observations were randomly selected.

X: 4 5 3 6 10 Y: 4 6 5 7 7

a. Determine the regression equation. b. Determine the value of when X is 7.

14. The following sample observations were randomly selected. Ŷ

X: 5 3 6 3 4 4 6 8 Y: 13 15 7 12 13 11 9 5

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482 Chapter 13

a. Determine the regression equation. b. Determine the number of kilowatt-hours, in thousands, for a six-room house.

16. Mr. James McWhinney, president of Daniel-James Financial Services, believes there is a relationship between the number of client contacts and the dollar amount of sales. To document this assertion, Mr. McWhinney gathered the following sample information. The X column indicates the number of client contacts last month, and the Y column shows the value of sales ($ thousands) last month for each client sampled.

a. Determine the regression equation. b. Determine the value of when X is 7.

15. Bradford Electric Illuminating Company is studying the relationship between kilowatt- hours (thousands) used and the number of rooms in a private single-family residence. A random sample of 10 homes yielded the following.

Ŷ

Number of Sales Number of Sales Contacts, ($ thousands), Contacts, ($ thousands),

X Y X Y

14 24 23 30 12 14 48 90 20 28 50 85 16 30 55 120 46 80 50 110

a. Determine the regression equation. b. Determine the estimated sales if 40 contacts are made.

17. A recent article in BusinessWeek listed the “Best Small Companies.” We are interested in the current results of the companies’ sales and earnings. A random sample of 12 companies was selected and the sales and earnings, in millions of dollars, are reported below.

Sales Earnings Sales Earnings Company ($ millions) ($ millions) Company ($ millions) ($ millions)

Papa John’s International $89.2 $4.9 Checkmate Electronics $17.5 $ 2.6 Applied Innovation 18.6 4.4 Royal Grip 11.9 1.7 Integracare 18.2 1.3 M-Wave 19.6 3.5 Wall Data 71.7 8.0 Serving-N-Slide 51.2 8.2 Davidson & Associates 58.6 6.6 Daig 28.6 6.0 Chico’s FAS 46.8 4.1 Cobra Golf 69.2 12.8

Let sales be the independent variable and earnings be the dependent variable. a. Draw a scatter diagram. b. Compute the correlation coefficient. c. Determine the regression equation. d. For a small company with $50.0 million in sales, estimate the earnings.

18. We are studying mutual bond funds for the purpose of investing in several funds. For this particular study, we want to focus on the assets of a fund and its five-year performance. The question is: Can the five-year rate of return be estimated based on the assets of the

Number of Kilowatt-Hours Number of Kilowatt-Hours Rooms (thousands) Rooms (thousands)

12 9 8 6 9 7 10 8

14 10 10 10 6 5 5 4

10 8 7 7

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a. Draw a scatter diagram. b. Compute the correlation coefficient. c. Write a brief report of your findings for parts (b) and (c). d. Determine the regression equation. Use assets as the independent variable. e. For a fund with $400.0 million in sales, determine the five-year rate of return (in

percent). 19. Refer to Exercise 5.

a. Determine the regression equation. b. Estimate the number of crimes for a city with 20 police officers. c. Interpret the regression equation.

20. Refer to Exercise 6. a. Determine the regression equation. b. Estimate the selling price of a 10-year-old car. c. Interpret the regression equation.

13.6 Testing the Significance of the Slope In the prior section, we showed how to find the equation of the regression line that best fits the data. The method for finding the equation is based on the least squares principle. The purpose of the regression equation is to quantify a linear relationship between two variables.

The next step is to analyze the regression equation by conducting a test of hypothesis to see if the slope of the regression line is different from zero. Why is this important? If we can show that the slope of the line in the population is differ- ent from zero, then we can conclude that using the regression equation adds to our ability to predict or forecast the dependent variable based on the independent vari- able. If we cannot demonstrate that this slope is different from zero, then we con- clude there is no merit to using the independent variable as a predictor. To put it another way, if we cannot show the slope of the line is different from zero, we might as well use the mean of the dependent variable as a predictor, rather than use the regression equation.

Following from the hypothesis-testing procedure in Chapter 10, the null and alternative hypotheses are:

We use (the Greek letter beta) to represent the population slope for the regres- sion equation. This is consistent with our policy to identify population parameters by Greek letters. We assumed the information regarding Copier Sales of America, Table 13–2, and the Example for the Applewood Auto Group are samples. Be care- ful here. Remember, this is a single sample, but when we selected a particular sales- person we identified two pieces of information, how many customers they called on and how many copiers they sold. It is still a single sample, however.

H1: � 0 H0: � 0

Assets Return Assets Return Fund ($ millions) (%) Fund ($ millions) (%)

AARP High Quality Bond $622.2 10.8 MFS Bond A $494.5 11.6 Babson Bond L 160.4 11.3 Nichols Income 158.3 9.5 Compass Capital Fixed Income 275.7 11.4 T. Rowe Price Short-term 681.0 8.2 Galaxy Bond Retail 433.2 9.1 Thompson Income B 241.3 6.8 Keystone Custodian B-1 437.9 9.2

fund? Nine mutual funds were selected at random, and their assets and rates of return are shown below.

LO5 Evaluate the significance of the slope of the regression equation.

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484 Chapter 13

We identified the slope value as b. So our computed slope “b” is based on a sample and is an estimate of the population’s slope, identified as “�.” The null hypothesis is that the slope of the regression equation in the population is zero. If this is the case, the regression line is horizontal and there is no relationship between the independent variable, X, and the dependent variable, Y. In other words, the value of the dependent variable is the same for any value of the independent variable and does not offer us any help in estimating the value of the dependent variable.

What if the null hypothesis is rejected? If the null hypothesis is rejected and the alternate hypothesis accepted, this indicates that the slope of the regression line for the population is not equal to zero. That is, knowing the value of the independent variable allows us to make a better estimate of the dependent variable. To put it another way, a significant relationship exists between the two variables.

Before we test the hypothesis, we use statistical software to determine the needed regression statistics. We continue to use the Copier Sales of America data from Table 13–2 and use Excel to perform the necessary calculations. The follow- ing spreadsheet shows three tables to the right of the sample data.

1. Starting on the top are the Regression Statistics. We will use this information later in the chapter, but notice that the “Multiple R” value is familiar. It is .759, which is the correlation coefficient we calculated in Section 13.2 using formula (13–1).

2. Next is an ANOVA table. This is a useful table for summarizing regression infor- mation. We will refer to it later in this chapter and use it extensively in the next chapter when we study multiple regression.

3. At the bottom, highlighted in blue, is the information needed to conduct our test of hypothesis regarding the slope of the line. It includes the value of the slope, which is 1.18421, and the intercept, which is 18.9474. (Note that these values for the slope and the intercept are slightly different from those computed on pages 478 and 479. These small differences are due to rounding.) In the column to the right of the regression coefficient is a column labeled “Standard Error.” This is a value similar to the standard error of the mean. Recall that the stan- dard error of the mean reports the variation in the sample means. In a similar fashion, these standard errors report the possible variation in slope and inter- cept values. The standard error of the slope coefficient is 0.35914.

To test the null hypothesis, we use the t-distribution with (n � 2) and the fol- lowing formula.

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Correlation and Linear Regression 485

where: b is the estimate of the regression line’s slope calculated from the sample

information. sb is the standard error of the slope estimate, also determined from sample

information.

Our first step is to set the null and the alternative hypotheses. They are:

Notice that we have a one-tailed test. If we do not reject the null hypothesis, we conclude that the slope of the regression line in the population could be zero. This means the independent variable is of no value in improving our estimate of the dependent variable. In our case, this means that knowing the number of sales calls made by a representative does not help us predict the sales.

If we reject the null hypothesis and accept the alternative, we conclude the slope of the line is greater than zero. Hence, the independent variable is an aid in predicting the dependent variable. Thus, if we know the number of sales calls made by a representative, this will help us forecast that representative’s sales. We also know, because we have demonstrated that the slope of the line is greater than zero—that is, positive—that more sales calls will result in the sale of more copiers.

The t-distribution is the test statistic; there are 8 degrees of freedom, found by n � 2 � 10 � 2. We use the .05 significance level. From Appendix B.2, the critical value is 1.860. Our decision rule is to reject the null hypothesis if the value com- puted from formula (13–6) is greater than 1.860. We apply formula (13–6) to find t.

The computed value of 3.297 exceeds our critical value of 1.860, so we reject the null hypothesis and accept the alternative hypothesis. We conclude that the slope of the line is greater than zero. The independent variable referring to the number of sales calls is useful for obtaining a better estimate of sales.

The table also provides us information on the p-value of this test. This cell is highlighted in purple. So we could select a significance level, say .05, and compare that value with the p-value. In this case, the calculated p-value in the table is .01090, so our decision is to reject the null hypothesis. An important caution is that the p- values reported in the statistical software are usually for a two-tailed test.

Before moving on, here is an interesting note. Observe that on page 473, when we conducted a test of hypothesis regarding the correlation coefficient for these same data using formula (13–2), we obtained the same value of the t-statistic, t � 3.297. Actually, the two-tests are equivalent and will always yield exactly the same values of t and the same p-values.

t � b � 0

sb �

1.18421 � 0 0.35814

� 3.297

H1: � 7 0 H0: � � 0

TEST FOR THE SLOPE with n � 2 degrees of freedom [13–6]t � b � 0

Self-Review 13–4 Refer to Self-Review 13–1, where the owner of Haverty’s Furniture Company studied the relationship between the amount spent on advertising in a month and sales revenue for that month. The amount of sales is the dependent variable, and advertising expense the independent variable. The regression equation in that study was for a sample of five months. Conduct a test of hypothesis to show there is a positive relationship between advertising and sales. From statistical software, the standard error of the regression coeffi- cient is 0.42. Use the .05 significance level.

Ŷ � 1.5 � 2.2X

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13.7 Evaluating a Regression Equation’s Ability to Predict The Standard Error of Estimate The results of the regression analysis for Copier Sales of America show a sig- nificant relationship between number of sales calls and the number of sales made. By substituting the names of the variables into the equation, it can be written as:

Number of copiers sold � 18.9476 � 1.1842 (Number of sales calls)

The equation can be used to estimate the number of copiers sold for any given “number of sales calls” within the range of the data. For example, if the number of sales calls is 30, then we can predict the number of copiers sold. It is 54.4736, found by 18.9476 � 1.1842(30). However, the data show two sales representatives with sales of 60 and 70 copiers sold. Is the regression equation a good predictor of “Number of copiers sold”?

Perfect prediction, which is finding the exact outcome, in economics and busi- ness is practically impossible. For example, the revenue for the year from gaso- line sales (Y) based on the number of automobile registrations (X) as of a certain date could no doubt be approximated fairly closely, but the prediction would not be exact to the nearest dollar, or probably even to the nearest thousand dollars. Even predictions of tensile strength of steel wires based on the outside diameters of the wires are not always exact, because of slight differences in the composi- tion of the steel.

What is needed, then, is a measure that describes how precise the prediction of Y is based on X or, conversely, how inaccurate the estimate might be. This mea- sure is called the standard error of estimate. The standard error of estimate is symbolized by , The subscript, , is interpreted as the standard error of y for a given value of x. It is the same concept as the standard deviation discussed in Chapter 3. The standard deviation measures the dispersion around the mean. The standard error of estimate measures the dispersion about the regression line for a given value of X.

y # xsy # x

STANDARD ERROR OF ESTIMATE A measure of the dispersion, or scatter, of the observed values around the line of regression for a given value of X.

LO6 Evaluate a regression equation to predict the dependent variable.

Exercises 21. Refer to Exercise 5. The regression equation is , the sample size is 8,

and the standard error of the slope is 0.22. Use the .05 significance level. Can we con- clude that the slope of the regression line is less than zero?

22. Refer to Exercise 6. The regression equation is , the sample size is 12, and the standard error of the slope is 0.23. Use the .05 significance level. Can we conclude that the slope of the regression line is less than zero?

23. Refer to Exercise 17. The regression equation is , the sample size is 12, and the standard error of the slope is 0.03. Use the .05 significance level. Can we con- clude that the slope of the regression line is different from zero?

24. Refer to Exercise 18. The regression equation is , the sample size is 9, and the standard error of the slope is 0.0032. Use the .05 significance level. Can we conclude that the slope of the regression line is less than zero?

Ŷ � 9.9198 � 0.00039X

Ŷ � 1.85 � .08X

Ŷ � 11.18 � 0.49X

Ŷ � 29.29 � 0.96X

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Correlation and Linear Regression 487

The coefficient of determination is easy to compute. It is the correlation coeffi- cient squared. Therefore, the term R-square is also used. With the Copier Sales of America, the correlation coefficient for the relationship between the number of copiers sold and the number of sales calls is 0.759. If we compute (0.759)2, the coefficient of determination is 0.576. See the blue (Multiple R) and green (R-square) highlighted cells in the spreadsheet on page 484. To better interpret the coeffi- cient of determination, convert it to a percentage. Hence, we say that 57.6 percent of the variation in the number of copiers sold is explained, or accounted for, by the variation in the number of sales calls.

How well can the regression equation predict number of copiers sold with num- ber of sales calls made? If it were possible to make perfect predictions, the coeffi- cient of determination would be 100 percent. That would mean that the indepen- dent variable, number of sales calls, explains or accounts for all the variation in the number of copiers sold. A coefficient of determination of 100 percent is associated with a correlation coefficient of �1.0 or �1.0. Refer to Chart 13–2, which shows that a perfect prediction is associated with a perfect linear relationship where all the data points form a perfect line in a scatter diagram. Our analysis shows that only 57.6 percent of the variation in copiers sold is explained by the number of sales

The standard error of estimate is found using formula (13–7).

STANDARD ERROR OF ESTIMATE [13–7]

The calculation of the standard error of estimate requires the sum of the squared differences between each observed value of Y and the predicted value of Y, which is identified as in the numerator. This calculation is illustrated in the spreadsheet on page 484. See cell G13 in the spreadsheet. It is a very important value. It is the numerator in the calculation of the standard error of the estimate.

This calculation can be eliminated by using statistical software such as Excel. The standard error of the estimate is included in Excel’s regression analysis and high- lighted in yellow on page 484. Its value is 9.901.

If the standard error of estimate is small, this indicates that the data are relatively close to the regression line and the regression equation can be used to predict Y with little error. If the standard error of estimate is large, this indicates that the data are widely scattered around the regression line, and the regression equation will not provide a precise estimate of Y.

The Coefficient of Determination Using the standard error of the estimate provides a relative measure of a regression equation’s ability to predict. We will use it to provide more specific information about a prediction in the next section. In this section, another statistic is explained that will provide a more interpretable measure of a regression equation’s ability to pre- dict. It is called the coefficient of determination, or R-square.

sy # x � B ©(Y � Ŷ )2

n � 2 � B

784.211 10 � 2

� 9.901

Ŷ

sy # x � B

�(Y � Ŷ )2

n � 2

COEFFICIENT OF DETERMINATION The proportion of the total variation in the dependent variable Y that is explained, or accounted for, by the variation in the independent variable X.

LO7 Calculate and interpret the coefficient of determination.

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488 Chapter 13

calls. Clearly, this data does not form a perfect line. Instead, the data are scat- tered around the best-fitting, least squares regression line, and there will be error in the predictions. In the next section, the standard error of the estimate is used to provide more specific information regarding the error associated with using the regression equation to make predictions.

Exercises (You may wish to use a software package such as Excel to assist in your calculations.)

25. Refer to Exercise 5. Determine the standard error of estimate and the coefficient of deter- mination. Interpret the coefficient of determination.

26. Refer to Exercise 6. Determine the standard error of estimate and the coefficient of deter- mination. Interpret the coefficient of determination.

27. Refer to Exercise 15. Determine the standard error of estimate and the coefficient of determination. Interpret the coefficient of determination.

28. Refer to Exercise 16. Determine the standard error of estimate and the coefficient of determination. Interpret the coefficient of determination.

Relationships among the Correlation Coefficient, the Coefficient of Determination, and the Standard Error of Estimate In Section 13.7, we described the standard error of estimate. Recall that it measures how close the actual values are to the regression line. When the standard error is small, it indicates that the two variables are closely related. In the calculation of the standard error, the key term is

If the value of this term is small, then the standard error will also be small. The correlation coefficient measures the strength of the linear association

between two variables. When the points on the scatter diagram appear close to the line, we note that the correlation coefficient tends to be large. Therefore, the corre- lation coefficient and the standard error of the estimate are inversely related. As the strength of a linear relationship between two variables increases, the correlation coefficient increases and the standard error of the estimate decreases.

We also noted that the square of the correlation coefficient is the coefficient of determination. The coefficient of determination measures the percentage of the vari- ation in Y that is explained by the variation in X.

A convenient vehicle for showing the relationship among these three measures is an ANOVA table. See the yellow highlighted portion of the spreadsheet on page 489. This table is similar to the analysis of variance table developed in Chapter 12. In that chapter, the total variation was divided into two components: variation due to the treatments and that due to random error. The concept is similar in regression analysis. The total variation is divided into two components: (1) variation explained

�(Y � Ŷ )2

Self-Review 13–5 Refer to Self-Review 13–1, where the owner of Haverty’s Furniture Company studied the relationship between the amount spent on advertising in a month and sales revenue for that month. The amount of sales is the dependent variable, and advertising expense is the independent variable. (a) Determine the standard error of estimate. (b) Determine the coefficient of determination. (c) Interpret the coefficient of determination.

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Correlation and Linear Regression 489

by the regression (explained by the independent variable) and (2) the error, or resid- ual. This is the unexplained variation. These three categories are identified in the first column of the spreadsheet ANOVA table. The column headed “df” refers to the degrees of freedom associated with each category. The total number of degrees of freedom is n � 1. The number of degrees of freedom in the regression is 1, because there is only one independent variable. The number of degrees of freedom associ- ated with the error term is n � 2. The term “SS” located in the middle of the ANOVA table refers to the sum of squares. You should note that the total degrees of freedom is equal to the sum of the regression and residual (error) degrees of freedom, and the total sum of squares is equal to the sum of the regression and residual (error) sum of squares. This is true for any ANOVA table.

The ANOVA sum of squares are computed as follows:

Regression Sum of Squares � SSR � � 1065.789

Residual or Error Sum of Squares � SSE � � 784.211

Total Sum of Squares � SS Total � � 1850.00

Recall that the coefficient of determination is defined as the percentage of the total variation (SS Total) explained by the regression equation (SSR). Using the ANOVA table, the reported value of R-square can be validated.

�(Y � Y )2 �(Y � Ŷ )2 �(Ŷ � Y )2

Using the values from the ANOVA table, the coefficient of determination is 1065.789/1850.00 � 0.576. Therefore, the more variation of the dependent variable (SS Total) explained by the independent variable (SSR), the higher the coefficient of determination.

We can also express the coefficient of determination in terms of the error or residual variation:

In this case, the coefficient of determination and the residual or error sum of squares are inversely related. The higher the unexplained or error variation as a percentage of the total variation, the lower is the coefficient of determination. In this case, 42.4 per- cent of the total variation in the dependent variable is error or residual variation.

r 2 � 1 � SSE

SS Total � 1 � 784.211 1850.00

� 1 � 0.424 � 0.576

COEFFICIENT OF DETERMINATION [13–8]r 2 � SSR

SS Total � 1 �

SSE SS Total

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490 Chapter 13

The final observation that relates the correlation coefficient, the coefficient of determination, and the standard error of the estimate is to show the relationship between the standard error of the estimate and SSE. By substituting [SSE Residual or Error Sum of Squares � SSE � ] into the formula for the standard error of the estimate, we find:

�(Y � Ŷ )2

In sum, regression analysis provides two statistics to evaluate the predictive ability of a regression equation, the standard error of the estimate and the coeffi- cient of determination. When reporting the results of a regression analysis, the find- ings must be clearly explained, especially when using the results to make predic- tions of the dependent variable. The report must always include a statement regarding the coefficient of determination so that the relative precision of the prediction is known to the reader of the report. Objective reporting of statistical analysis is required so that the readers can make their own decisions.

Exercises 29. Given the following ANOVA table:

a. Determine the coefficient of determination. b. Assuming a direct relationship between the variables, what is the correlation

coefficient? c. Determine the standard error of estimate.

30. On the first statistics exam, the coefficient of determination between the hours studied and the grade earned was 80 percent. The standard error of estimate was 10. There were 20 students in the class. Develop an ANOVA table for the regression analysis of hours studied as a predictor of the grade earned on the first statistics exam.

13.8 Interval Estimates of Prediction The standard error of the estimate and the coefficient of determination are two statis- tics that provide an overall evaluation of the ability of a regression equation to predict a dependent variable. Another way to report the ability of a regression equation to predict is specific to a stated value of the independent variable. For example, we can predict the number of copiers sold (Y ) for a selected value of number of sales calls made (X ). In fact, we can calculate a confidence interval for the predicted value of the dependent variable for a selected value of the independent variable.

Assumptions Underlying Linear Regression Before we present the confidence intervals, the assumptions for properly applying linear regression should be reviewed. Chart 13–13 illustrates these assumptions.

1. For each value of X, there are corresponding Y values. These Y values follow the normal distribution.

2. The means of these normal distributions lie on the regression line.

STANDARD ERROR OF ESTIMATE [13–9]sy # x � B

SSE n � 2

Source DF SS MS F

Regression 1 1000.0 1000.0 26.00 Error 13 500.0 38.46

Total 14 1500.0

Statistics in Action

Studies indicate that for both men and women, those who are considered good looking earn higher wages than those who are not. In addition, for men there is a correlation between

(continued)

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height and salary. For each additional inch of height, a man can expect to earn an additional $250 per year. So a man tall receives a $3,000 “stature” bonus over his counterpart. Being overweight or underweight is also related to earnings, particularly among women. A study of young women showed the heaviest 10 percent earned about 6 percent less than their lighter counterparts.

5 6–

6 6–

Correlation and Linear Regression 491

3. The standard deviations of these normal distributions are all the same. The best estimate we have of this common standard deviation is the standard error of estimate

4. The Y values are statistically independent. This means that in selecting a sam- ple, a particular X does not depend on any other value of X. This assumption is particularly important when data are collected over a period of time. In such situations, the errors for a particular time period are often correlated with those of other time periods.

Recall from Chapter 7 that if the values follow a normal distribution, then the mean plus or minus one standard deviation will encompass 68 percent of the obser- vations, the mean plus or minus two standard deviations will encompass 95 percent of the observations, and the mean plus or minus three standard deviations will encompass virtually all of the observations. The same relationship exists between the predicted values and the standard error of estimate

1. will include the middle 68 percent of the observations.Ŷ � sy # x

(sy # x).Ŷ

(sy # x).

X10

Regression Equation

Y-Intercept

X2 X3

Each of these distributions 1. follows the normal distribution, 2. has a mean on the regression line, 3. has the same standard error of estimate (sy . x ), and 4. is independent of the others.

CHART 13–13 Regression Assumptions Shown Graphically

2. will include the middle 95 percent of the observations.Ŷ � 2sy # x 3. will include virtually all the observations.

We can now relate these assumptions to Copier Sales of America, where we studied the relationship between the number of sales calls and the number of copiers sold. Assume that we took a much larger sample than but that the standard error of estimate was still 9.901. If we drew a parallel line 9.901 units above the regression line and another 9.901 units below the regression line, about 68 per- cent of the points would fall between the two lines. Similarly, a line 19.802

units above the regression line and another 19.802 units below the regression line should include about 95 percent of the data values.

As a rough check, refer to column E in the Excel spreadsheet in Section 13.5 on page 480. Three of the 10 deviations exceed one standard error of estimate. That is, the deviation of for Tom Keller, for Mark Reynolds, and for Soni Jones all exceed the value of 9.901, which is one standard error from the regression line. All of the values are within 19.802 units of the regression line. To put

�15.5264�12.6316�12.6316

[2sy # x � 2(9.901)]

n � 10,

Ŷ � 3sy # x

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492 Chapter 13

it another way, 7 of the 10 deviations in the sample are within one standard error of the regression line and all are within two—a good result for a relatively small sample.

Constructing Confidence and Prediction Intervals When using a regression equation, two different predictions can be made for a selected value of the independent variable. The differences are subtle but very important and are related to the assumptions stated in the last section. Recall that for any selected value of the independent variable (X), the dependent variable (Y ) is a random variable that is normally distributed with a mean, . Each distribution of Y has a standard devi- ation equal to the regression analysis’ standard error of the estimate.

The first interval estimate is called a confidence interval. This is used when the regression equation is used to predict the mean value of Y for a given value of X. For example, we would use a confidence interval to estimate the mean salary of all executives in the retail industry based on their years of experience. To determine the confidence interval for the mean value of Y for a given X, the formula is:

Ŷ

CONFIDENCE INTERVAL FOR THE MEAN OF Y, GIVEN X

[13–10]Ŷ � t(sy # x) B

1 n

� (X � X )2

©(X � X )2

PREDICTION INTERVAL FOR Y, GIVEN X

[13–11]Ŷ � tsy # x B

1 � 1 n

� (X � X )2

©(X � X )2

The second interval estimate is called a prediction interval. This is used when the regression equation is used to predict an individual Y (n � 1) for a given value of X. For example, we would estimate the salary of a particular retail executive who has 20 years of experience. To determine the prediction interval for an estimate of an individual for a given X, the formula is:

LO8 Calculate and interpret confidence and prediction intervals.

Example We return to the Copier Sales of America illustration. Determine a 95 percent confi- dence interval for all sales representatives who make 25 calls, and determine a pre- diction interval for Sheila Baker, a West Coast sales representative who made 25 calls.

Solution We use formula 13–10 to determine a confidence level. Table 13–4 includes the necessary totals and a repeat of the information of Table 13–2 on page 466.

TABLE 13–4 Calculations Needed for Determining the Confidence Interval and Prediction Interval

Sales Copier Sales Representative Calls, (X ) Sales, (Y ) (X � ) (X � )2

Tom Keller 20 30 �2 4 Jeff Hall 40 60 18 324 Brian Virost 20 40 �2 4 Greg Fish 30 60 8 64 Susan Welch 10 30 �12 144 Carlos Ramirez 10 40 �12 144 Rich Niles 20 40 �2 4 Mike Kiel 20 50 �2 4 Mark Reynolds 20 30 �2 4 Soni Jones 30 70 8 64

0 760

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Correlation and Linear Regression 493

The first step is to determine the number of copiers we expect a sales repre- sentative to sell if he or she makes 25 calls. It is 48.5526, found by

To find the t value, we need to first know the number of degrees of freedom. In this case, the degrees of freedom is n � 2 � 10 � 2 � 8. We set the confidence level at 95 percent. To find the value of t, move down the left-hand column of Appen- dix B.2 to 8 degrees of freedom, then move across to the column with the 95 per- cent level of confidence. The value of t is 2.306.

In the previous section, we calculated the standard error of estimate to be 9.901. We let X � 25, and from Table 13–4 Inserting these values in formula (13–10), we can determine the confidence interval.

Thus, the 95 percent confidence interval for all sales representatives who make 25 calls is from 40.9170 up to 56.1882. To interpret, let’s round the values. If a sales representative makes 25 calls, he or she can expect to sell 48.6 copiers. It is likely those sales will range from 40.9 to 56.2 copiers.

Suppose we want to estimate the number of copiers sold by Sheila Baker, who made 25 sales calls. The 95 percent prediction interval is determined as follows:

Thus, the interval is from 24.478 up to 72.627 copiers. We conclude that the num- ber of copiers sold will be between about 24 and 73 for a particular sales repre- sentative who makes 25 calls. This interval is quite large. It is much larger than the confidence interval for all sales representatives who made 25 calls. It is logical, how- ever, that there should be more variation in the sales estimate for an individual than for a group.

� 48.5526 � 24.0746

� 48.5526 � 2.306(9.901) B

1 � 1 10

� (25 � 22)2

760

Prediction Interval � Ŷ � tsy # x B

1 � 1 n

� (X � X )2

©(X � X )2

� 48.5526 � 7.6356

� 48.5526 � 2.306(9.901) B

1 10

� (25 � 22)2

760

Confidence Interval � Ŷ � tsy # x B

1 n

� (X � X )2

©(X � X )2

©(X � X )2 � 760.X � ©X�n � 220�10 � 22,

1.1842X � 18.9476 � 1.1842(25). Ŷ � 18.9476 �

The following Minitab graph shows the relationship between the regression line (in the center), the confidence interval (shown in crimson), and the prediction inter- val (shown in green). The bands for the prediction interval are always further from the regression line than those for the confidence interval. Also, as the values of X move away from the mean number of calls (22) in either the positive or the nega- tive direction, the confidence interval and prediction interval bands widen. This is caused by the numerator of the right-hand term under the radical in formulas (13–10) and (13–11). That is, as the term increases, the widths of the confidence interval and the prediction interval also increase. To put it another way, there is less precision in our estimates as we move away, in either direction, from the mean of the independent variable.

(X � X )2

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494 Chapter 13

We wish to emphasize again the distinction between a confidence interval and a prediction interval. A confidence interval refers to all cases with a given value of X and is computed by formula (13–10). A prediction interval refers to a particular case for a given value of X and is computed using formula (13–11). The prediction interval will always be wider because of the extra 1 under the radical in the second equation.

Exercises 31. Refer to Exercise 13.

a. Determine the .95 confidence interval for the mean predicted when X � 7. b. Determine the .95 prediction interval for an individual predicted when X � 7.

32. Refer to Exercise 14. a. Determine the .95 confidence interval for the mean predicted when X � 7. b. Determine the .95 prediction interval for an individual predicted when X � 7.

33. Refer to Exercise 15. a. Determine the .95 confidence interval, in thousands of kilowatt-hours, for the mean of

all six-room homes. b. Determine the .95 prediction interval, in thousands of kilowatt-hours, for a particular

six-room home.

Self-Review 13–6 Refer to the sample data in Self-Review 13–1, where the owner of Haverty’s Furniture was studying the relationship between sales and the amount spent on advertising. The sales information for the last four months is repeated below.

Advertising Expense Sales Revenue Month ($ million) ($ million)

July 2 7 August 1 3 September 3 8 October 4 10

The regression equation was computed to be and the standard error 0.9487. Both variables are reported in millions of dollars. Determine the 90 percent confi- dence interval for the typical month in which $3 million was spent on advertising.

Ŷ � 1.5 � 2.2X,

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Correlation and Linear Regression 495

34. Refer to Exercise 16. a. Determine the .95 confidence interval, in thousands of dollars, for the mean of all sales

personnel who make 40 contacts. b. Determine the .95 prediction interval, in thousands of dollars, for a particular sales-

person who makes 40 contacts.

13.9 Transforming Data The correlation coefficient describes the strength of the linear relation- ship between two variables. It could be that two variables are closely related, but their relationship is not linear. Be cautious when you are interpreting the correlation coefficient. A value of r may indicate there is no linear relationship, but it could be there is a relationship of some other nonlinear or curvilinear form.

To explain, below is a listing of 22 professional golfers, the num- ber of events in which they participated, the amount of their winnings, and their mean score. In golf, the objective is to play 18 holes in the least number of strokes. So, we would expect that those golfers with the lower mean scores would have the larger winnings. In other words, score and winnings should be inversely related.

Phil Mickelson played 22 events, earned $5,784,823, and had a mean score per round of 69.16. Fred Couples played in 16 events, earned $1,396,109, and had a mean score per round of 70.92. The data for the 22 golfers follows.

Player Events Winnings Score

Vijay Singh 29 $10,905,166 68.84 Ernie Els 16 5,787,225 68.98 Phil Mickelson 22 5,784,823 69.16 Tiger Woods 19 5,365,472 69.04 Davis Love III 24 3,075,092 70.13 Chris DiMarco 27 2,971,842 70.28 John Daly 22 2,359,507 70.82 Charles Howell III 30 1,703,485 70.77 Kirk Triplett 24 1,566,426 70.31 Fred Couples 16 1,396,109 70.92 Tim Petrovic 32 1,193,354 70.91 Briny Baird 30 1,156,517 70.79 Hank Kuehne 30 816,889 71.36 J. L. Lewis 32 807,345 71.21 Aaron Baddeley 27 632,876 71.61 Craig Perks 27 423,748 71.75 David Frost 26 402,589 71.75 Rich Beem 28 230,499 71.76 Dicky Pride 23 230,329 72.91 Len Mattiace 25 213,707 72.03 Esteban Toledo 36 115,185 72.36 David Gossett 25 21,250 75.01

The correlation between the variables Winnings and Score is �0.782. This is a fairly strong inverse relationship. However, when we plot the data on a scatter diagram the relationship does not appear to be linear; it does not seem to follow a line. See the scatter diagram on the right-hand side of the following Minitab output. The data points for the lowest score and the highest score seem to be well away from the regression line. In addition, for the scores between 70 and 72, the winnings are

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496 Chapter 13

below the regression line. If the relationship were linear, we would expect these points to be both above and below the line.

What can we do to explore other (nonlinear) relationships? One possibility is to transform one of the variables. For example, instead of using Y as the dependent variable, we might use its log, reciprocal, square, or square root. Another possibil- ity is to transform the independent variable in the same way. There are other trans- formations, but these are the most common.

In the golf winnings example, changing the scale of the dependent variable is effec- tive. We determine the log of each golfer’s winnings and then find the correlation between the log of winnings and score. That is, we find the log to the base 10 of Tiger Woods’ earnings of $5,365,472, which is 6.72961. Next we find the log to the base 10 of each golfer’s winnings and then determine the correlation between log of winnings and the score. The correlation coefficient increases from �0.782 to �0.969. This means that the coefficient of determination is .939 [r2 � (�0.969)2 � .939]. That is, 93.9 percent of the variation in the log of winnings is accounted for by the independent variable score.

We have determined an equation that fits the data more closely than the line did. Clearly, as the mean score for a golfer increases, he can expect his winnings to decrease. It no longer appears that some of the data points are different from the regression line, as we found when using winnings instead of the log of winnings as the dependent variable. Also note the points between 70 and 72 in particular are now randomly distributed above and below the regression line.

We can also estimate the amount of winnings based on the score. Following is the Minitab regression output using score as the independent variable and the log of winnings as the dependent variable.

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Correlation and Linear Regression 497

To compute the earnings for a golfer with a mean score of 70, we first use the regression equation to compute the log of earnings.

The value 6.4372 is the log to the base 10 of winnings. The antilog of 6.4372 is 2,736,528. So a golfer that had a mean score of 70 could expect to earn $2,736,528. We can also evaluate the change in scores. The above golfer had a mean score of 70 and estimated earnings of $2,736,528. How much less would a golfer expect to win if his mean score was 71? Again solving the regression equation:

The antilog of this value is $994,855. So based on the regression analysis, there is a large financial incentive for a professional golfer to reduce his mean score by even one stroke. Those of you that play golf or know a golfer understand how difficult that change would be! That one stroke is worth over $1,700,000.

Ŷ � 37.198 � .43944X � 37.198 � .43944(71) � 5.99776

Ŷ � 37.198 � .43944X � 37.198 � .43944(70) � 6.4372

Exercises 35. Given the following sample observations, develop a scatter diagram. Compute the cor-

relation coefficient. Does the relationship between the variables appear to be linear? Try squaring the X-variable and then determine the correlation coefficient.

Demand Price 2 $120.0 5 90.0 8 80.0 12 70.0 16 50.0 21 45.0 27 31.0 35 30.0 45 25.0 60 21.0

36. According to basic economics, as the demand for a product increases, the price will decrease. Listed below is the number of units demanded and the price.

a. Determine the correlation between price and demand. Plot the data in a scatter dia- gram. Does the relationship seem to be linear?

b. Transform the price to a log to the base 10. Plot the log of the price and the demand. Determine the correlation coefficient. Does this seem to improve the relationship between the variables?

X �8 �16 12 2 18 Y 58 247 153 3 341

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498 Chapter 13

Chapter Summary I. A scatter diagram is a graphic tool to portray the relationship between two variables.

A. The dependent variable is scaled on the Y-axis and is the variable being estimated. B. The independent variable is scaled on the X-axis and is the variable used as the

predictor. II. The correlation coefficient measures the strength of the linear association between two

variables. A. Both variables must be at least the interval scale of measurement. B. The correlation coefficient can range from �1.00 to 1.00. C. If the correlation between the two variables is 0, there is no association between them. D. A value of 1.00 indicates perfect positive correlation, and a value of �1.00 indicates

perfect negative correlation. E. A positive sign means there is a direct relationship between the variables, and a neg-

ative sign means there is an inverse relationship. F. It is designated by the letter r and found by the following equation:

[13–1]

G. The following equation is used to determine whether the correlation in the population is different from 0.

with degrees of freedom [13–2]

III. In regression analysis, we estimate one variable based on another variable. A. The variable being estimated is the dependent variable. B. The variable used to make the estimate or predict the value is the independent

variable. 1. The relationship between the variables is linear. 2. Both the independent and the dependent variables must be interval or ratio scale. 3. The least squares criterion is used to determine the regression equation.

IV. The least squares regression line is of the form A. is the estimated value of Y for a selected value of X. B. a is the constant or intercept.

1. It is the value of when X � 0. 2. a is computed using the following equation.

[13–5]

C. b is the slope of the fitted line. 1. It shows the amount of change in for a change of one unit in X. 2. A positive value for b indicates a direct relationship between the two variables. A

negative value indicates an inverse relationship. 3. The sign of b and the sign of r, the correlation coefficient, are always the same. 4. b is computed using the following equation.

[13–4]

D. X is the value of the independent variable. V. For a regression equation, the slope is tested for significance.

A. We test the hypothesis that the slope of the line in the population is 0. 1. If we do not reject the null hypothesis, we conclude there is no relationship

between the two variables. 2. The test is equivalent to the test for the correlation coefficient.

B. When testing the null hypothesis about the slope, the test statistic is with n � 2 degrees of freedom:

[13–6]t � b � 0

b � ra sy sx b

Ŷ

a � Y � bX

Ŷ

Ŷ Ŷ � a � bX.

n � 2t � r1n � 2 11 � r 2

r � ©(X � X )(Y � Y)

(n � 1)sxsy

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Correlation and Linear Regression 499

VI. The standard error of estimate measures the variation around the regression line. A. It is in the same units as the dependent variable. B. It is based on squared deviations from the regression line. C. Small values indicate that the points cluster closely about the regression line. D. It is computed using the following formula.

[13–7]

VII. The coefficient of determination is the proportion of the variation of a dependent variable explained by the independent variable. A. It ranges from 0 to 1.0. B. It is the square of the correlation coefficient. C. It is found from the following formula.

[13–8]

VIII. Inference about linear regression is based on the following assumptions. A. For a given value of X, the values of Y are normally distributed about the line of

regression. B. The standard deviation of each of the normal distributions is the same for all values

of X and is estimated by the standard error of estimate. C. The deviations from the regression line are independent, with no pattern to the size

or direction. IX. There are two types of interval estimates.

A. In a confidence interval, the mean value of Y is estimated for a given value of X. 1. It is computed from the following formula.

[13–10]

2. The width of the interval is affected by the level of confidence, the size of the stan- dard error of estimate, and the size of the sample, as well as the value of the inde- pendent variable.

B. In a prediction interval, the individual value of Y is estimated for a given value of X. 1. It is computed from the following formula.

[13–11]

2. The difference between formulas (13–10) and (13–11) is the 1 under the radical. a. The prediction interval will be wider than the confidence interval. b. The prediction interval is also based on the level of confidence, the size of the

standard error of estimate, the size of the sample, and the value of the inde- pendent variable.

Pronunciation Key SYMBOL MEANING PRONUNCIATION

Sum of the products of X and Y Sum X Y

Correlation coefficient in Rho the population

Estimated value of Y Y hat

Standard error of estimate s sub y dot x

Coefficient of determination r squarer2 sy # x

Ŷ

�

©XY

Ŷ � tsy # x B

1 � 1 n

� (X � X )2

©(X � X )2

Ŷ � t(sy # x) B

1 n

� (X � X )2

©(X � X )2

r 2 � SSR

SS Total � 1 �

SSE SS Tatal

sy # x � B

©(Y � Ŷ )2

n � 2

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500 Chapter 13

Chapter Exercises 37. A regional commuter airline selected a random sample of 25 flights and found that the

correlation between the number of passengers and the total weight, in pounds, of lug- gage stored in the luggage compartment is 0.94. Using the .05 significance level, can we conclude that there is a positive association between the two variables?

38. A sociologist claims that the success of students in college (measured by their GPA) is related to their family’s income. For a sample of 20 students, the correlation coefficient is 0.40. Using the 0.01 significance level, can we conclude that there is a positive cor- relation between the variables?

39. An Environmental Protection Agency study of 12 automobiles revealed a correlation of 0.47 between engine size and emissions. At the .01 significance level, can we conclude that there is a positive association between these variables? What is the p-value? Interpret.

40. A suburban hotel derives its gross income from its hotel and restaurant operations. The owners are interested in the relationship between the number of rooms occupied on a nightly basis and the revenue per day in the restaurant. Below is a sample of 25 days (Monday through Thursday) from last year showing the restaurant income and number of rooms occupied.

Use a statistical software package to answer the following questions. a. Does the breakfast revenue seem to increase as the number of occupied rooms

increases? Draw a scatter diagram to support your conclusion. b. Determine the correlation coefficient between the two variables. Interpret the value. c. Is it reasonable to conclude that there is a positive relationship between revenue and

occupied rooms? Use the .10 significance level. d. What percent of the variation in revenue in the restaurant is accounted for by the num-

ber of rooms occupied? 41. The table below shows the number of cars (in millions) sold in the United States for var-

ious years and the percent of those cars manufactured by GM.

Day Income Occupied Day Income Occupied

1 $1,452 23 14 $1,425 27 2 1,361 47 15 1,445 34 3 1,426 21 16 1,439 15 4 1,470 39 17 1,348 19 5 1,456 37 18 1,450 38 6 1,430 29 19 1,431 44 7 1,354 23 20 1,446 47 8 1,442 44 21 1,485 43 9 1,394 45 22 1,405 38

10 1,459 16 23 1,461 51 11 1,399 30 24 1,490 61 12 1,458 42 25 1,426 39 13 1,537 54

Year Cars Sold (millions) Percent GM Year Cars Sold (millions) Percent GM

1950 6.0 50.2 1980 11.5 44.0 1955 7.8 50.4 1985 15.4 40.1 1960 7.3 44.0 1990 13.5 36.0 1965 10.3 49.9 1995 15.5 31.7 1970 10.1 39.5 2000 17.4 28.6 1975 10.8 43.1 2005 16.9 26.9

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Correlation and Linear Regression 501

Use a statistical software package to answer the following questions. a. Is the number of cars sold directly or indirectly related to GM’s percent of the mar-

ket? Draw a scatter diagram to show your conclusion. b. Determine the correlation coefficient between the two variables. Interpret the value. c. Is it reasonable to conclude that there is a negative association between the two vari-

ables? Use the .01 significance level. d. How much of the variation in GM’s market share is accounted for by the variation in

cars sold? 42. For a sample of 32 large U.S. cities, the correlation between the mean number of square

feet per office worker and the mean monthly rental rate in the central business district is �.363. At the .05 significance level, can we conclude that there is a negative associa- tion in the population between the two variables?

43. What is the relationship between the amount spent per week on recreation and the size of the family? Do larger families spend more on recreation? A sample of 10 families in the Chicago area revealed the following figures for family size and the amount spent on recreation per week.

Family Amount Spent Family Amount Spent Size on Recreation Size on Recreation

3 $ 99 3 $111 6 104 4 74 5 151 4 91 6 129 5 119 6 142 3 91

a. Compute the correlation coefficient. b. Determine the coefficient of determination. c. Can we conclude that there is a positive association between the amount spent on

recreation and family size? Use the .05 significance level. 44. A sample of 12 homes sold last week in St. Paul, Minnesota, is selected. Can we con-

clude that, as the size of the home (reported below in thousands of square feet) increases, the selling price (reported in $ thousands) also increases?

Home Size Home Size (thousands of Selling Price (thousands of Selling Price square feet) ($ thousands) square feet) ($ thousands)

1.4 100 1.3 110 1.3 110 0.8 85 1.2 105 1.2 105 1.1 120 0.9 75 1.4 80 1.1 70 1.0 105 1.1 95

a. Compute the correlation coefficient. b. Determine the coefficient of determination. c. Can we conclude that there is a positive association between the size of the home

and the selling price? Use the .05 significance level. 45. The manufacturer of Cardio Glide exercise equipment wants to study the relationship

between the number of months since the glide was purchased and the length of time the equipment was used last week.

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502 Chapter 13

a. Plot the information on a scatter diagram. Let hours of exercise be the dependent variable. Comment on the graph.

b. Determine the correlation coefficient. Interpret. c. At the .01 significance level, can we conclude that there is a negative association

between the variables? 46. The following regression equation was computed from a sample of 20 observations:

SSE was found to be 100 and SS total 400. a. Determine the standard error of estimate. b. Determine the coefficient of determination. c. Determine the correlation coefficient. (Caution: Watch the sign!)

47. City planners believe that larger cities are populated by older residents. To investi- gate the relationship, data on population and median age in ten large cities were collected.

Ŷ � 15 � 5X

Person Months Owned Hours Exercised Person Months Owned Hours Exercised

Rupple 12 4 Massa 2 8 Hall 2 10 Sass 8 3 Bennett 6 8 Karl 4 8 Longnecker 9 5 Malrooney 10 2 Phillips 7 5 Veights 5 5

Population City (in millions) Median age

Chicago, IL 2.833 31.5 Dallas, TX 1.233 30.5 Houston, TX 2.144 30.9 Los Angeles, CA 3.849 31.6 New York, NY 8.214 34.2 Philadelphia, PA 1.448 34.2 Phoenix, AZ 1.513 30.7 San Antonio, TX 1.297 31.7 San Diego, CA 1.257 32.5 San Jose, CA 0.930 32.6

a. Plot this data on a scatter diagram with median age as the dependent variable. b. Find the correlation coefficient. c. A regression analysis was performed and the resulting regression equation is Median

age � 31.4 � 0.272 population. Interpret the meaning of the slope. d. Estimate the median age for a city of 2.5 million people. e. Here is a portion of the regression software output. What does it tell you?

Predictor Coef SE Coef T P Constant 31.3672 0.6158 50.94 0.000 Population 0.2722 0.1901 1.43 0.190

f. Using the .10 significance level, test the significance of the slope. Interpret the result. Is there a significant relationship between the two variables?

48. Emily Smith decides to buy a fuel-efficient used car. Here are several vehicles she is con- sidering, with the estimated cost to purchase and the age of the vehicle.

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Correlation and Linear Regression 503

a. Determine the regression equation. Interpret the equation. Do more bidders tend to increase or decrease the amount of the winning bid?

b. Estimate the amount of the winning bid if there were seven bidders. c. A new entrance is to be constructed on the Ohio Turnpike. There are seven bidders

on the project. Develop a 95 percent prediction interval for the winning bid. d. Determine the coefficient of determination. Interpret its value.

50. Mr. William Profit is studying companies going public for the first time. He is particularly interested in the relationship between the size of the offering and the price per share. A sample of 15 companies that recently went public revealed the following information.

a. Plot this data on a scatter diagram with estimated cost as the dependent variable. b. Find the correlation coefficient. c. A regression analysis was performed and the resulting regression equation is Esti-

mated Cost � 18358 � 1534 age. Interpret the meaning of the slope. d. Estimate the cost of a five-year-old car. e. Here is a portion of the regression software output. What does it tell you?

Predictor Coef SE Coef T P Constant 18358 1817 10.10 0.000 Age -1533.6 306.3 -5.01 0.000

f. Using the .10 significance level, test the significance of the slope. Interpret the result. Is there a significant relationship between the two variables?

49. The National Highway Association is studying the relationship between the number of bidders on a highway project and the winning (lowest) bid for the project. Of particu- lar interest is whether the number of bidders increases or decreases the amount of the winning bid.

Winning Bid Winning Bid Number of ($ millions), Number of ($ millions),

Project Bidders, X Y Project Bidders, X Y

1 9 5.1 9 6 10.3 2 9 8.0 10 6 8.0 3 3 9.7 11 4 8.8 4 10 7.8 12 7 9.4 5 5 7.7 13 7 8.6 6 10 5.5 14 7 8.1 7 7 8.3 15 6 7.8 8 11 5.5

Vehicle Estimated Cost Age

Honda Insight $5,555 8 Toyota Prius $17,888 3 Toyota Prius $9,963 6 Toyota Echo $6,793 5 Honda Civic Hybrid $10,774 5 Honda Civic Hybrid $16,310 2 Chevrolet Prizm $2,475 8 Mazda Protege $2,808 10 Toyota Corolla $7,073 9 Acura Integra $8,978 8 Scion xB $11,213 2 Scion xA $9,463 3 Mazda3 $15,055 2 Mini Cooper $20,705 2

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504 Chapter 13

Size Price Size Price ($ millions), per Share, ($ millions), per Share,

Company X Y Company X Y

1 9.0 10.8 9 160.7 11.3 2 94.4 11.3 10 96.5 10.6 3 27.3 11.2 11 83.0 10.5 4 179.2 11.1 12 23.5 10.3 5 71.9 11.1 13 58.7 10.7 6 97.9 11.2 14 93.8 11.0 7 93.5 11.0 15 34.4 10.8 8 70.0 10.7

a. Determine the regression equation. b. Conduct a test to determine whether the slope of the regression line is positive. c. Determine the coefficient of determination. Do you think Mr. Profit should be satisfied

with using the size of the offering as the independent variable? 51. Bardi Trucking Co., located in Cleveland, Ohio, makes deliveries in the Great Lakes

region, the Southeast, and the Northeast. Jim Bardi, the president, is studying the rela- tionship between the distance a shipment must travel and the length of time, in days, it takes the shipment to arrive at its destination. To investigate, Mr. Bardi selected a random sample of 20 shipments made last month. Shipping distance is the inde- pendent variable, and shipping time is the dependent variable. The results are as follows:

Distance Shipping Time Distance Shipping Time Shipment (miles) (days) Shipment (miles) (days)

1 656 5 11 862 7 2 853 14 12 679 5 3 646 6 13 835 13 4 783 11 14 607 3 5 610 8 15 665 8 6 841 10 16 647 7 7 785 9 17 685 10 8 639 9 18 720 8 9 762 10 19 652 6

10 762 9 20 828 10

a. Draw a scatter diagram. Based on these data, does it appear that there is a relation- ship between how many miles a shipment has to go and the time it takes to arrive at its destination?

b. Determine the correlation coefficient. Can we conclude that there is a positive corre- lation between distance and time? Use the .05 significance level.

c. Determine and interpret the coefficient of determination. d. Determine the standard error of estimate. e. Would you recommend using the regression equation to predict shipping time? Why

or why not. 52. Super Markets Inc. is considering expanding into the Scottsdale, Arizona, area. You as

director of planning, must present an analysis of the proposed expansion to the operating committee of the board of directors. As a part of your proposal, you need to include infor- mation on the amount people in the region spend per month for grocery items. You would also like to include information on the relationship between the amount spent for grocery items and income. Your assistant gathered the following sample information. The data are available on the data disk supplied with the text.

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Correlation and Linear Regression 505

a. Calculate the regression equation using selling price based on the annual dividend. b. Test the significance of the slope. c. Determine the coefficient of determination. Interpret its value. d. Determine the correlation coefficient. Can you conclude that it is greater than 0 using

the .05 significance level? 54. A highway employee performed a regression analysis of the relationship between the

number of construction work-zone fatalities and the number of unemployed people in a state. The regression equation is (Unemp). Some additional output is:

Fatalities � 12.7 � 0.000114

a. Let the amount spent be the dependent variable and monthly income the indepen- dent variable. Create a scatter diagram, using a software package.

b. Determine the regression equation. Interpret the slope value. c. Determine the correlation coefficient. Can you conclude that it is greater than 0?

53. Below is information on the price per share and the dividend for a sample of 30 companies. The sample data are available on the data disk supplied with the text.

Company Price per Share Dividend

1 $20.00 $ 3.14 2 22.01 3.36 . . .. . .. . .

29 77.91 17.65 30 80.00 17.36

Predictor Coef SE Coef T P Constant 12.726 8.115 1.57 0.134 Unemp 0.00011386 0.00002896 3.93 0.001

Analysis of Variance Source DF SS MS F P Regression 1 10354 10354 15.46 0.001 Residual Error 18 12054 670 Total 19 22408

a. How many states were in the sample? b. Determine the standard error of estimate. c. Determine the coefficient of determination. d. Determine the correlation coefficient. e. At the .05 significance level, does the evidence suggest there is a positive association

between fatalities and the number unemployed? 55. A regression analysis relating the current market value in dollars to the size in square feet

of homes in Greene County, Tennessee, follows. The regression equation is: Value � �37,186 � 65.0 Size.

Predictor Coef SE Coef T P Constant �37186 4629 �8.03 0.000 Size 64.993 3.047 21.33 0.000

Analysis of Variance Source DF SS MS F P Regression 1 13548662082 13548662082 454.98 0.000 Residual Error 33 982687392 29778406 Total 34 14531349474

Household Amount Spent Monthly Income

1 $ 555 $4,388 2 489 4,558

39 1,206 9,862 40 1,145 9,883

.. .

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506 Chapter 13

a. How many homes were in the sample? b. Compute the standard error of estimate. c. Compute the coefficient of determination. d. Compute the correlation coefficient. e. At the .05 significance level does the evidence suggest a positive association between

the market value of homes and the size of the home in square feet? 56. The following table shows the mean annual percent return on capital (profitability) and the

mean annual percentage sales growth for eight aerospace and defense companies.

a. Compute the correlation coefficient. Conduct a test of hypothesis to determine if it is reasonable to conclude that the population correlation is greater than zero. Use the .05 significance level.

b. Develop the regression equation for profitability based on growth. Can we conclude that the slope of the regression line is negative?

c. Use a software package to determine the residual for each observation. Which com- pany has the largest residual?

57. The following data show the retail price for 12 randomly selected laptop computers along with their corresponding processor speeds in gigahertz.

Company Profitability Growth

Alliant Techsystems 23.1 8.0 Boeing 13.2 15.6 General Dynamics 24.2 31.2 Honeywell 11.1 2.5 L-3 Communications 10.1 35.4 Northrop Grunmman 10.8 6.0 Rockwell Collins 27.3 8.7 United Technologies 20.1 3.2

a. Develop a linear equation that can be used to describe how the price depends on the processor speed.

b. Based on your regression equation, is there one machine that seems particularly over- or underpriced?

c. Compute the correlation coefficient between the two variables. At the .05 significance level, conduct a test of hypothesis to determine if the population correlation is greater than zero.

58. A consumer buying cooperative tested the effective heating area of 20 different electric space heaters with different wattages. Here are the results.

Computers Speed Price Computers Speed Price

1 2.0 $2,017 7 2.0 $2,197 2 1.6 922 8 1.6 1,387 3 1.6 1,064 9 2.0 2,114 4 1.8 1,942 10 1.6 2,002 5 2.0 2,137 11 1.0 937 6 1.2 1,012 12 1.4 869

Heater Wattage Area Heater Wattage Area

1 1,500 205 11 1,250 116 2 750 70 12 500 72 3 1,500 199 13 500 82 4 1,250 151 14 1,500 206 5 1,250 181 15 2,000 245 6 1,250 217 16 1,500 219 7 1,000 94 17 750 63 8 2,000 298 18 1,500 200 9 1,000 135 19 1,250 151

10 1,500 211 20 500 44

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Correlation and Linear Regression 507

a. Compute the correlation between the wattage and heating area. Is there a direct or an indirect relationship?

b. Conduct a test of hypothesis to determine if it is reasonable that the coefficient is greater than zero. Use the .05 significance level.

c. Develop the regression equation for effective heating based on wattage. d. Which heater looks like the “best buy” based on the size of the residual?

59. A dog trainer is exploring the relationship between the size of the dog (weight in pounds) and its daily food consumption (measured in standard cups). Below is the result of a sample of 18 observations.

Dog Weight Consumption Dog Weight Consumption

1 41 3 10 91 5 2 148 8 11 109 6 3 79 5 12 207 10 4 41 4 13 49 3 5 85 5 14 113 6 6 111 6 15 84 5 7 37 3 16 95 5 8 111 6 17 57 4 9 41 3 18 168 9

Answer the following questions. a. Write out the regression equation. Is there a direct or indirect relationship between the

distance from the fire station and the amount of fire damage? b. How much damage would you estimate for a fire 5 miles from the nearest fire

station? c. Determine and interpret the coefficient of determination. d. Determine the correlation coefficient. Interpret its value. How did you determine the

sign of the correlation coefficient? e. Conduct a test of hypothesis to determine if there is a significant relationship between

the distance from the fire station and the amount of damage. Use the .01 significance level and a two-tailed test.

a. Compute the correlation coefficient. Is it reasonable to conclude that the correlation in the population is greater than zero? Use the .05 significance level.

b. Develop the regression equation for cups based on the dog’s weight. How much does each additional cup change the estimated weight of the dog?

c. Is one of the dogs a big undereater or overeater? 60. Waterbury Insurance Company wants to study the relationship between the amount of

fire damage and the distance between the burning house and the nearest fire station. This information will be used in setting rates for insurance coverage. For a sample of 30 claims for the last year, the director of the actuarial department determined the distance from the fire station (X ) and the amount of fire damage, in thousands of dollars (Y ). The MegaStat output is reported below. (You can find the actual data in the data set on the CD as prb13-60.)

ANOVA table Source SS df MS F Regression 1,864.5782 1 1,864.5782 38.83 Residual 1,344.4934 28 48.0176 Total 3,209.0716 29

Regression output Variables Coefficients Std. Error t(df = 28) Intercept 12.3601 3.2915 3.755 Distance–X 4.7956 0.7696 6.231

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508 Chapter 13

61. Listed below are the movies with the largest world box office sales and their world box office budget (total amount available to spend making the picture).

World Box Office Adjusted Budget Rank Movie Year ($ million) ($ million)

1 Avatar 2009 2,729.7 237.0 2 Titanic 1997 1,835.0 789.3 3 LOTR: The Return of the King 2003 1,129.2 377.0 4 Pirates of the Caribbean: Dead Man's Chest 2006 1,060.6 321.4 5 Alice in Wonderland 2010 1,017.3 200.0 6 The Dark Knight 2008 1,001.9 185.0 7 Harry Potter and the Sorcerer's Stone 2001 968.7 338.3 8 Pirates of the Caribbean: At World's End 2007 958.4 308.9 9 Harry Potter and the Order of The Phoenix 2007 937.0 306.3

10 Harry Potter and the Half-Blood Prince 2009 934.0 382.2 11 Star Wars: Episode I—The Phantom Menace 1999 925.5 511.7 12 The Lord of the Rings: The Two Towers 2002 920.5 354.0 13 Jurassic Park 1993 920.0 513.8 14 Shrek 2 2004 912.0 436.5 15 Harry Potter and the Goblet of Fire 2005 892.2 300.8 16 Ice Age: Dawn of the Dinosaurs 2009 886.7 380.4 17 Spider-Man 3 2007 885.4 354.0 18 Harry Potter and the Chamber of Secrets 2002 866.4 272.4 19 The Lord of the Rings: The Fellowship of the Ring 2001 860.7 334.3 20 Finding Nemo 2003 853.2 339.7 21 Star Wars: Episode III—Revenge of the Sith 2005 848.5 278.0 22 Independence Day 1996 813.1 417.5 23 Spider-Man 2002 806.7 419.7 24 Star Wars 1977 797.9 1,084.3 25 Harry Potter and the Prisoner of Azkaban 2004 789.8 249.4 26 Spider-Man 2 2004 784.0 373.4 27 The Lion King 1994 771.9 446.2 28 E.T. 1982 757.0 860.6 29 The Matrix: Reloaded 2003 735.7 281.5 30 Forrest Gump 1994 680.0 470.2 31 The Sixth Sense 1999 661.5 348.4 32 Pirates of the Caribbean 2003 653.2 305.4 33 Star Wars: Episode II—Attack of the Clones 2002 648.3 323.0 34 The Incredibles 2004 631.2 261.4 35 The Lost World 1997 614.4 301.0 36 The Passion of the Christ 2004 611.8 370.3 37 Men In Black 1997 587.2 328.6 38 Return of the Jedi 1983 573.0 563.1 39 Mission: Impossible 2 2000 545.4 241.0 40 The Empire Strikes Back 1980 533.9 586.8 41 Home Alone 1990 533.8 401.6 42 Monsters, Inc. 2001 524.2 272.6 43 Ghost 1990 517.6 306.6 44 Meet the Fockers 2004 511.9 279.2 45 Aladdin 1992 502.4 311.7 46 Twister 1996 495.0 329.7 47 Toy Story 2 1999 485.7 291.8 48 Saving Private Ryan 1998 479.3 278.1 49 Jaws 1975 471.0 782.7 50 Shrek 2001 469.7 285.1

Find the correlation between the world box office budget and world box office sales. Comment on the association between the two variables. Does it appear that the movies with large budgets result in large box office revenues?

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Correlation and Linear Regression 509

Data Set Exercises 62. Refer to the Real Estate data, which reports information on homes sold in Goodyear,

Arizona, last year. a. Let selling price be the dependent variable and size of the home the independent

variable. Determine the regression equation. Estimate the selling price for a home with an area of 2,200 square feet. Determine the 95 percent confidence interval and the 95 percent prediction interval for the selling price of a home with 2,200 square feet.

b. Let selling price be the dependent variable and distance from the center of the city the independent variable. Determine the regression equation. Estimate the selling price of a home 20 miles from the center of the city. Determine the 95 percent confidence interval and the 95 percent prediction interval for homes 20 miles from the center of the city.

c. Can you conclude that the independent variables “distance from the center of the city” and “selling price” are negatively correlated and that the area of the home and the selling price are positively correlated? Use the .05 significance level. Report the p-value of the test. Summarize your results in a brief report.

63. Refer to the Baseball 2009 data, which reports information on the 2009 Major League Baseball season. Let the games won be the dependent variable and total team salary, in millions of dollars, be the independent variable. Determine the regression equation and answer the following questions. a. Draw a scatter diagram. From the diagram, does there seem to be a direct relation-

ship between the two variables? b. How many wins would you estimate with a salary of $100.0 million? c. How many additional wins will an additional $5 million in salary bring? d. At the .05 significance level, can we conclude that the slope of the regression line is

positive? Conduct the appropriate test of hypothesis. e. What percentage of the variation in wins is accounted for by salary? f. Determine the correlation between wins and team batting average and between wins

and team ERA. Which is stronger? Conduct an appropriate test of hypothesis for each set of variables.

64. Refer to the Buena School bus data. Develop a regression equation that expresses the relationship between age of the bus and maintenance. The age of the bus is the inde- pendent variable. a. Draw a scatter diagram. What does this diagram suggest as to the relationship

between the two variables? Is it direct or indirect? Does it appear to be strong or weak?

b. Develop a regression equation. How much does an additional year add to the main- tenance cost. What is the estimated maintenance cost for a 10-year-old bus?

c. Conduct a test of hypothesis to determine whether the slope of the regression line is greater than zero. Use the .05 significance level. Interpret your findings from parts (a), (b), and (c) in a brief report.

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510 Chapter 13

2. The computer commands for the Excel output on page 487 are: a. Enter the variable names in row 1 of columns A,

B, and C. Enter the data in rows 2 through 11 in the same columns.

b. Select the Data tab on the top of the menu. Then, on the far right, select Data Analysis. Select Regression, then click OK.

c. For our spreadsheet, we have Calls in column B and Sales in column C. The Input Y-Range is C1:C11 and the Input X-Range is B1:B11. Click on Labels, select E1 as the Output Range, and click OK.

3. The Minitab commands to the confidence intervals and prediction intervals on page 494 are: a. Select Stat, Regression, and Fitted line plot. b. In the next dialog box, the Response (Y) is Sales

and Predictor (X) is Calls. Select Linear for the type of regression model and then click on Options.

c. In the Options dialog box, click on Display con- fidence and prediction bands, use the 95.0 for confidence level, type an appropriate heading in the Title box, then click OK and then OK again.

1. The Minitab commands for the output showing the correlation coefficient on page 474 are: a. Enter the sales representative’s name in C1, the

number of calls in C2, and the sales in C3. b. Select Stat, Basic Statistics, and Correlation. c. Select Calls and Units Sold as the variables,

click on Display p-values, and then click OK.

Software Commands

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Correlation and Linear Regression 511

Chapter 13 Answers to Self-Review

13–1 a. Advertising expense is the independent variable, and sales revenue is the dependent variable.

Advertising expense

Sa le

s re

ve nu

0 1 2 3 4

d. There is a strong correlation between the advertising expense and sales.

13–2 is rejected if

is rejected. There is a positive correlation between the percent of the vote received and the amount spent on the campaign.

t � .43125 � 2 11 � (.43)2

� 2.284

t 7 1.714.H0H1: � 7 0.H0: � � 0,

� 0.9648

r � ©(X�X )(Y� Y )

(n � 1)sx sy �

11 (4 �1)(1.2909944)(2.9439203)

sy � B

26 3

� 2.9439203

sx � B

5 3

� 1.2909944

Y � 28 4

� 7X � 10 4

� 2.5

13–3 a. See the calculations in Self-Review 13–1, part (c).

b. The slope is 2.2. This indicates that an increase of $1 million in advertising will result in an increase of $2.2 million in sales. The intercept is 1.5. If there was no expenditure for advertising, sales would be $1.5 million.

13–4 , reject H0 if t � 3.182.

Reject H0. The slope of the line is greater than 0. 13–5 a.

t � 2.2 � 0

0.42 � 5.238

H0: 1 � 0; H1: 7 0 Ŷ � 1.5 � 2.2(3) � 8.1

a � 28 4

� 2.2a 10 4 b � 7 � 5.5 � 1.5

b � rsy sx

� (0.9648)(2.9439)

1.2910 � 2.2

� B

1.80 4 � 2

� .9487

sy # x � B

©(Y � Ŷ )2

n � 2

b. r2 � (.9487)2 � .90 c. Ninety percent of the variation in sales is

accounted for by advertising expense. 13–6 6.58 and 9.62, since for an X of 3 is 8.1, found by

then and

t from Appendix B.2 for 4 � 2 � 2 degrees of freedom at the .10 level is 2.920.

� 6.58 and 9.62 (in $ millions)

� 8.1 � 2.920(0.9487)(0.5477)

� 8.1 � 2.920(0.9487) B

1 4

� (3 � 2.5)2

Ŷ � t(sy # x) B

1 n

� (X � X )2

©(X � X )2

©(X � X )2 � 5. X � 2.5Ŷ � 1.5 � 2.2(3) � 8.1,

Ŷ

X Y (X � ) (X � )2 (Y � ) (Y � )2 (X � )(Y � )

2 7 �0.5 .25 0 0 0 1 3 �1.5 2.25 �4 16 6 3 8 0.5 .25 1 1 0.5 4 10 1.5 2.25 3 9 4.5

10 28 5.00 26 11.0

YXYYXX

Y (Y � ) (Y � )2

7 5.9 1.1 1.21 3 3.7 �0.7 .49 8 8.1 �0.1 .01

10 10.3 �0.3 .09 1.80

ŶŶŶ

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14 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Describe the relationship between several independent variables and a dependent variable using multiple regression analysis.

LO2 Set up, interpret, and apply an ANOVA table.

LO3 Compute and interpret measures of association in multiple regression.

LO4 Conduct a hypothesis test to determine whether a set of regression coefficients differ from zero.

LO5 Conduct a hypothesis test of each regression coefficient.

LO6 Use residual analysis to evaluate the assumptions of multiple regression analysis.

LO7 Evaluate the effects of correlated independent variables.

LO8 Evaluate and use qualitative independent variables.

LO9 Explain the possible interaction among independent variables.

LO10 Explain stepwise regression.

Multiple Regression Analysis

The mortgage department of the Bank of New England is studying data

from recent loans. Of particular interest is how such factors as the value of

the home being purchased, education level of the head of the household,

age of the head of the household, current monthly mortgage payment, and

gender of the head of the household relate to the family income. Are the

proposed variables effective predictors of the dependent variable family

income? (See the Example/Solution in Section 14.9 and LO1.)

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Multiple Regression Analysis 513

GENERAL MULTIPLE REGRESSION EQUATION

where: a is the intercept, the value of Y when all the X’s are zero.

is the amount by which Y changes when that particular increases by one unit, with the values of all other independent variables held constant. The subscript j is simply a label that helps to identify each independent variable; it is not used in any calculations. Usually the subscript is an integer value between 1 and k, which is the number of independent variables. However, the subscript can also be a short or abbreviated label. For example, age could be used as a subscript.

In Chapter 13, the regression analysis described and tested the relationship between a dependent variable, , and a single independent variable, X. The relationship between and X was graphically portrayed by a line. When there are two inde- pendent variables, the regression equation is

Because there are two independent variables, this relationship is graphically por- trayed as a plane and is shown in Chart 14–1. The chart shows the residuals as the difference between the actual Y and the fitted on the plane. If a multiple regres- sion analysis includes more than two independent variables, we cannot use a graph to illustrate the analysis since graphs are limited to three dimensions.

To illustrate the interpretation of the intercept and the two regression coefficients, suppose a vehicle’s mileage per gallon of gasoline is directly related to the octane rat- ing of the gasoline being used and inversely related to the weight of the automobile

Assume that the regression equation, calculated using statistical software, is:

Ŷ � 6.3 � 0.2X1 � 0.001X2

(X2). (X1)

Ŷ

Ŷ � a � b1X1 � b2X2

Ŷ Ŷ

Xjbj

Ŷ � a � b1X1 � b2X2 � b3X3 � . . . � bkXk [14–1]

14.1 Introduction In Chapter 13, we described the relationship between a pair of interval- or ratio-scaled variables. We began the chapter by studying the correlation coefficient, which mea- sures the strength of the relationship. A coefficient near plus or minus 1.00 or .78, for example) indicates a very strong linear relationship, whereas a value near 0 ( or .18, for example) means that the relationship is weak. Next we developed a procedure to determine a linear equation to express the relationship between the two variables. We referred to this as a regression line. This line describes the rela- tionship between the variables. It also describes the overall pattern of a dependent variable (Y ) to a single independent or explanatory variable (X ).

In multiple linear correlation and regression, we use additional independent vari- ables (denoted . . . , and so on) that help us better explain or predict the dependent variable (Y ). Almost all of the ideas we saw in simple linear correlation and regression extend to this more general situation. However, the additional inde- pendent variables do lead to some new considerations. Multiple regression analysis can be used either as a descriptive or as an inferential technique.

14.2 Multiple Regression Analysis The general descriptive form of a multiple linear equation is shown in formula (14–1). We use k to represent the number of independent variables. So k can be any positive integer.

X1, X2,

�.12

(�.88

LO1 Describe the relationship between several independent variables and a dependent variable using multiple regression analysis.

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The intercept value of 6.3 indicates the regression equation intersects the Y-axis at 6.3 when both and are zero. Of course, this does not make any physical sense to own an automobile that has no (zero) weight and to use gasoline with no octane. It is important to keep in mind that a regression equation is not generally used outside the range of the sample values.

The of 0.2 indicates that for each increase of 1 in the octane rating of the gasoline, the automobile would travel 2/10 of a mile more per gallon, regardless of the weight of the vehicle. The value of �0.001 reveals that for each increase of one pound in the vehicle’s weight, the number of miles traveled per gallon decreases by 0.001, regardless of the octane of the gasoline being used.

As an example, an automobile with 92-octane gasoline in the tank and weighing 2,000 pounds would travel an average 22.7 miles per gallon, found by:

The values for the coefficients in the multiple linear equation are found by using the method of least squares. Recall from the previous chapter that the least squares method makes the sum of the squared differences between the fitted and actual values of Y as small as possible, that is, the term is minimized. The calculations are very tedious, so they are usually performed by a statistical software package, such as Excel or Minitab.

In the following example, we show a multiple regression analysis using three independent variables employing Excel and Minitab. Both packages report a stan- dard set of statistics and reports. However, Minitab also provides advanced regres- sion analysis techniques that we will use later in the chapter.

�(Y � Ŷ )2

Ŷ � a � b1X1 � b2X2 � 6.3 � 0.2(92) � 0.001(2,000) � 22.7

X2X1

514 Chapter 14

Estimated point (Y )

Observed point (Y )

Plane formed through the sample points � a � b1 X1 � b2 X2

CHART 14–1 Regression Plane with 10 Sample Points

Example Salsberry Realty sells homes along the east coast of the United States. One of the questions most frequently asked by prospective buyers is: If we purchase this home, how much can we expect to pay to heat it during the winter? The research department at Salsberry has been asked to develop some guidelines regarding heating costs for single-family homes. Three variables are thought to relate to the heating costs: (1) the mean daily outside temperature, (2) the number of inches of insulation in the attic, and (3) the age in years of the furnace. To investigate, Salsberry’s research department selected a random sample of 20 recently sold homes. It determined the cost to heat each home last January, as well as the January outside temperature in the region, the number of inches of insulation in the attic, and the age of the furnace. The sample information is reported in Table 14–1.

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Multiple Regression Analysis 515

Solution We begin the analysis by defining the dependent and independent variables. The dependent variable is the January heating cost. It is represented by Y. There are three independent variables:

• The mean outside temperature in January, represented by . • The number of inches of insulation in the attic, represented by . • The age in years of the furnace, represented by .

Given these definitions, the general form of the multiple regression equation follows. The value is used to estimate the value of Y.

Now that we have defined the regression equation, we are ready to use either Excel or Minitab to compute all the statistics needed for the analysis. The outputs from the two software systems are shown below.

To use the regression equation to predict the January heating cost, we need to know the values of the regression coefficients, . These are highlighted in the software reports. Note that the software used the variable names or labels associated with each independent variable. The regression equation intercept, a, is labeled as “constant” in the Minitab output and “intercept” in the Excel output.

Ŷ � a � b1X1 � b2X2 � b3X3

Ŷ

X3 X2

The data in Table 14–1 is available in both Excel and Minitab formats at the textbook website, www.mhhe.com/lind15e. The basic instructions for using Excel and Minitab for this data are in the Software Commands section at the end of this chapter.

Determine the multiple regression equation. Which variables are the indepen- dent variables? Which variable is the dependent variable? Discuss the regression coefficients. What does it indicate if some coefficients are positive and some coef- ficients are negative? What is the intercept value? What is the estimated heating cost for a home if the mean outside temperature is 30 degrees, there are 5 inches of insulation in the attic, and the furnace is 10 years old?

TABLE 14–1 Factors in January Heating Cost for a Sample of 20 Homes

Statistics in Action

Many studies indi- cate a woman will earn about 70 per- cent of what a man would for the same work. Researchers at the University of Michigan Institute for Social Research found that about one-third of the difference can be explained by such social factors as dif- ferences in educa- tion, seniority, and work interruptions. The remaining two- thirds is not ex- plained by these social factors.

Heating Cost Mean Outside Attic Insulation Age of Furnace Home ($) Temperature (�F) (inches) (years)

1 $250 35 3 6 2 360 29 4 10 3 165 36 7 3 4 43 60 6 9 5 92 65 5 6 6 200 30 5 5 7 355 10 6 7 8 290 7 10 10 9 230 21 9 11

10 120 55 2 5 11 73 54 12 4 12 205 48 5 1 13 400 20 5 15 14 320 39 4 7 15 72 60 8 6 16 272 20 5 8 17 94 58 7 3 18 190 40 8 11 19 235 27 9 8 20 139 30 7 5

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516 Chapter 14

In this case, the estimated regression equation is:

We can now estimate or predict the January heating cost for a home if we know the mean outside temperature, the inches of insulation, and the age of the furnace. For an example home, the mean outside temperature for the month is 30 degrees

there are 5 inches of insulation in the attic and the furnace is 10 years old . By substituting the values for the independent variables:

The estimated January heating cost is $276.56.

Ŷ � 427.194 � 4.583(30) � 14.831(5) � 6.101(10) � 276.56

(X3) (X2),(X1),

Ŷ � 427.194 � 4.583X1 � 14.831X2 � 6.101X3

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Multiple Regression Analysis 517

The regression coefficients, and their algebraic signs, also provide information about their individual relationships with the January heating cost. The regression coefficient for mean outside temperature is �4.583. The coefficient is negative and shows an inverse relationship between heating cost and temperature. This is not sur- prising. As the outside temperature increases, the cost to heat the home decreases. The numeric value of the regression coefficient provides more information. If we increase temperature by 1 degree and hold the other two independent variables con- stant, we can estimate a decrease of $4.583 in monthly heating cost. So if the mean temperature in Boston is 25 degrees and it is 35 degrees in Philadelphia, all other things being the same (insulation and age of furnace), we expect the heating cost would be $45.83 less in Philadelphia.

The attic insulation variable also shows an inverse relationship: the more insu- lation in the attic, the less the cost to heat the home. So the negative sign for this coefficient is logical. For each additional inch of insulation, we expect the cost to heat the home to decline $14.83 per month, holding the outside temperature and the age of the furnace constant.

The age of the furnace variable shows a direct relationship. With an older fur- nace, the cost to heat the home increases. Specifically, for each additional year older the furnace is, we expect the cost to increase $6.10 per month.

Self-Review 14–1 There are many restaurants in northeastern South Carolina. They serve beach vacationers in the summer, golfers in the fall and spring, and snowbirds in the winter. Bill and Joyce Tuneall manage several restaurants in the North Jersey area and are considering moving to Myrtle Beach, SC, to open a new restaurant. Before making a final decision, they wish to investigate existing restaurants and what variables seem to be related to profitability. They gather sample information where profit (reported in $000) is the dependent variable and the independent variables are:

X1 the number of parking spaces near the restaurant. X2 the number of hours the restaurant is open per week. X3 the distance from Peaches Corner, a landmark in Myrtle Beach. X4 the number of servers employed. X5 the number of years the current owner has owned the restaurant.

The following is part of the output obtained using statistical software.

Predictor Coef SE Coef T Constant 2.50 1.50 1.667 X1 3.00 1.500 2.000 X2 4.00 3.000 1.333 X3 �3.00 0.20 �15.00 X4 0.20 .05 4.00 X5 1.00 1.50 0.667

(a) What is the amount of profit for a restaurant with 40 parking spaces that is open 72 hours per week, is 10 miles from Peaches Corner, has 20 servers, and has been open 5 years?

(b) Interpret the values of and in the multiple regression equation.b3b2

Exercises 1. The director of marketing at Reeves Wholesale Products is studying monthly sales. Three

independent variables were selected as estimators of sales: regional population, per capita income, and regional unemployment rate. The regression equation was computed to be (in dollars):

Ŷ � 64,100 � 0.394X1 � 9.6X2 � 11,600X3

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a. What is the full name of the equation? b. Interpret the number 64,100. c. What are the estimated monthly sales for a particular region with a population of

796,000, per capita income of $6,940, and an unemployment rate of 6.0 percent? 2. Thompson Photo Works purchased several new, highly sophisticated processing

machines. The production department needed some guidance with respect to qualifica- tions needed by an operator. Is age a factor? Is the length of service as an operator (in years) important? In order to explore further the factors needed to estimate performance on the new processing machines, four variables were listed:

X1 � Length of time an employee was in the industry X2 � Mechanical aptitude test score X3 � Prior on-the-job rating X4 � Age

Performance on the new machine is designated Y. Thirty employees were selected at random. Data were collected for each, and their

performances on the new machines were recorded. A few results are:

518 Chapter 14

The equation is:

a. What is this equation called? b. How many dependent variables are there? Independent variables? c. What is the number 0.286 called? d. As age increases by one year, how much does estimated performance on the new

machine increase? e. Carl Knox applied for a job at Photo Works. He has been in the business for six

years, and scored 280 on the mechanical aptitude test. Carl’s prior on-the-job per- formance rating is 97, and he is 35 years old. Estimate Carl’s performance on the new machine.

3. A sample of General Mills employees was studied to determine their degree of satis- faction with their quality of life. A special index, called the index of satisfaction, was used to measure satisfaction. Six factors were studied, namely, age at the time of first marriage annual income number of children living value of all assets status of health in the form of an index and the average number of social activi- ties per week—such as bowling and dancing Suppose the multiple regression equation is:

a. What is the estimated index of satisfaction for a person who first married at 18, has an annual income of $26,500, has three children living, has assets of $156,000, has an index of health status of 141, and has 2.5 social activities a week on the average?

b. Which would add more to satisfaction, an additional income of $10,000 a year or two more social activities a week?

4. Cellulon, a manufacturer of home insulation, wants to develop guidelines for builders and consumers on how the thickness of the insulation in the attic of a home and the outdoor temperature affect natural gas consumption. In the laboratory, it varied the insulation thickness and temperature. A few of the findings are:

Ŷ � 16.24 � 0.017X1 � 0.0028X2 � 42X3 � 0.0012X4 � 0.19X5 � 26.8X6

(X6). (X5),

(X4),(X3),(X2),(X1),

Ŷ � 11.6 � 0.4X1 � 0.286X2 � 0.112X3 � 0.002X4

Performance Length of Mechanical Prior on New Time in Aptitude On-the-Job

Machine, Industry, Score, Performance, Age, Name Y X1 X2 X3 X4 Mike Miraglia 112 12 312 121 52 Sue Trythall 113 2 380 123 27

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Multiple Regression Analysis 519

On the basis of the sample results, the regression equation is:

a. How much natural gas can homeowners expect to use per month if they install 6 inches of insulation and the outdoor temperature is 40 degrees F?

b. What effect would installing 7 inches of insulation instead of 6 have on the monthly natural gas consumption (assuming the outdoor temperature remains at 40 degrees F)?

c. Why are the regression coefficients and negative? Is this logical?

14.3 Evaluating a Multiple Regression Equation Many statistics and statistical methods are used to evaluate the relationship between a dependent variable and more than one independent variable. Our first step was to write the relationship in terms of a multiple regression equation. The next step follows on the concepts presented in Chapter 13 by using the information in an ANOVA table to evaluate how well the equation fits the data.

The ANOVA Table As in Chapter 13, the statistical analysis of a multiple regression equation is sum- marized in an ANOVA table. To review, the total variation of the dependent variable, Y, is divided into two components: (1) regression, or the variation of Y explained by all the independent variables and (2) the error or residual, or unexplained variation of Y. These two categories are identified in the first column of an ANOVA table below. The column headed “df” refers to the degrees of freedom associated with each cat- egory. The total number of degrees of freedom is n � 1. The number of degrees of freedom in the regression is equal to the number of independent variables in the multiple regression equation. We call the regression degrees of freedom k. The num- ber of degrees of freedom associated with the error term is equal to the total degrees of freedom minus the regression degrees of freedom. In multiple regression, the degrees of freedom are n � (k � 1).

b2b1

Ŷ � 62.65 � 1.86X1 � 0.52X2

Monthly Natural Thickness of Outdoor Gas Consumption Insulation Temperature

(cubic feet), (inches), (�F), Y X1 X2

30.3 6 40 26.9 12 40 22.1 8 49

Source df SS MS F

Regression k SSR MSR � SSR�k MSR�MSE Residual or error n � (k � 1) SSE MSE � SSE�[n � (k � 1)]

Total n � 1 SS total

The term “SS” located in the middle of the ANOVA table refers to the sum of squares. Notice that there is a sum of squares for each source of variation. The sum of squares column shows the amount of variation attributable to each source. The total variation of the dependent variable, Y, is summarized in SS total. You should

LO2 Set up, interpret, and apply an ANOVA table.

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520 Chapter 14

note that this is simply the numerator of the usual formula to calculate any variation— in other words, the sum of the squared deviations from the mean. It is computed as:

As we have seen, the total sum of squares is the sum of the regression and residual sum of squares. The regression sum of squares is the sum of the squared differences between the estimated or predicted values, , and the overall mean of Y. The regres- sion sum of squares is found by:

The residual sum of squares is the sum of the squared differences between the observed values of the dependent variable, Y, and their corresponding estimated or predicted values, . Notice that this difference is the error of estimating or predict- ing the dependent variable with the multiple regression equation. It is calculated as:

We will use the ANOVA table information from the previous example to evalu- ate the regression equation to estimate January heating costs.

Residual or Error Sum of Squares � SSE � ©(Y � Ŷ )2

Ŷ

Regression Sum of Squares � SSR � ©(Ŷ � Y )2

Ŷ

Total Sum of Squares � SS total � ©(Y � Y )2

Multiple Standard Error of Estimate We begin with the multiple standard error of estimate. Recall that the standard error of estimate is comparable to the standard deviation. To explain the details of the standard error of estimate, refer to the first sampled home in Table 14–1 in the previous example on page 515. The actual heating cost for the first observation, Y, is $250, the outside temperature, is 35 degrees, the depth of insulation, is 3 inches, and the age of the furnace, is 6 years. Using the regression equation developed in the previous section, the estimated heating cost for this home is:

So we would estimate that a home with a mean January outside temperature of 35 degrees, 3 inches of insulation, and a 6-year-old furnace would cost $258.90 to heat. The actual heating cost was $250, so the residual—which is the difference between the actual value and the estimated value—is

This difference of $8.90 is the random or unexplained error for the first home sampled. Our next step is to square this difference—that is, find (Y � Ŷ )2 � (250 � 258.90)2 � (�8.90)2 � 79.21.

250 � 258.90 � �8.90. Y � Ŷ �

� 258.90

� 427.194 � 4.583(35) � 14.831(3) � 6.101(6)

Ŷ � 427.194 � 4.583X1 � 14.831X2 � 6.101X3

X3, X2,X1,

LO3 Compute and interpret measures of association in multiple regression.

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Multiple Regression Analysis 521

where: Y is the actual observation.

is the estimated value computed from the regression equation. n is the number of observations in the sample. k is the number of independent variables. SSR is the Residual Sum of Squares from an ANOVA table.

There is still more information in the ANOVA table that can be used to compute the multiple standard error of the estimate. Note that the next column in the ANOVA table is labeled MS, or mean square. For the regression and residual sources of vari- ation, the mean squares are calculated as the sum of squares divided by its corre- sponding degrees of freedom. In the case of the multiple standard error of the mean, the multiple standard error of the estimate is the square root of the residual mean square.

How do we interpret the standard error of estimate of 51.05? It is the typical “error” when we use this equation to predict the cost. First, the units are the same as the dependent variable, so the standard error is in dollars, $51.05. Second, we expect the residuals to be approximately normally distributed, so about 68 percent of the residuals will be within and about 95 percent within or As before with similar measures of dispersion, such as the standard error of estimate in Chapter 13, a smaller multiple standard error indicates a bet- ter or more effective predictive equation.

Coefficient of Multiple Determination Next, let’s look at the coefficient of multiple determination. Recall from the previous chapter the coefficient of determination is defined as the percent of variation in the dependent variable explained, or accounted for, by the independent variable. In the multiple regression case, we extend this definition as follows.

�$102.10. �2(51.05)�$51.05

sY.123...K � 2MSE � 22605.995 � $51.05

Ŷ

MULTIPLE STANDARD ERROR OF ESTIMATE

[14–2]sY.123...k � B

©(Y � Ŷ )2

n � (k � 1) � B

SSR n � (k � 1)

If we repeat this calculation for the other 19 observations and sum all 20 squared differences, the total will be the residual or error sum of squares from the ANOVA table. Using this information, we can calculate the multiple standard error of the estimate as:

COEFFICIENT OF MULTIPLE DETERMINATION The percent of variation in the dependent variable, explained by the set of independent variables, . . . .XkX3,X2,X1,Y,

The characteristics of the coefficient of multiple determination are:

1. It is symbolized by a capital R squared. In other words, it is written as because it behaves like the square of a correlation coefficient.

2. It can range form 0 to 1. A value near 0 indicates little association between the set of independent variables and the dependent variable. A value near 1 means a strong association.

3. It cannot assume negative values. Any number that is squared or raised to the second power cannot be negative.

4. It is easy to interpret. Because is a value between 0 and 1, it is easy to interpret, compare, and understand.

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522 Chapter 14

We can calculate the coefficient of determination from the information found in the ANOVA table. We look in the sum of squares column, which is labeled SS in the Excel output, and use the regression sum of squares, SSR, then divide by the total sum of squares, SS total.

Using the residual and total sum of squares from the ANOVA table, we can use formula (14–3) to calculate the coefficient of multiple determination.

How do we interpret this value? We conclude that the independent variables (out- side temperature, amount of insulation, and age of furnace) explain, or account for, 80.4 percent of the variation in heating cost. To put it another way, 19.6 percent of the variation is due to other sources, such as random error or variables not included in the analysis. Using the ANOVA table, 19.6 percent is the error sum of squares divided by the total sum of squares. Knowing that the SSR � SSE � SS total, the following relationship is true.

Adjusted Coefficient of Determination The number of independent variables in a multiple regression equation makes the coefficient of determination larger. Each new independent variable causes the pre- dictions to be more accurate. That, in turn, makes SSE smaller and SSR larger. Hence, increases only because of the total number of independent variables and not because the added independent variable is a good predictor of the dependent variable. In fact, if the number of variables, k, and the sample size, n, are equal, the coefficient of determination is 1.0. In practice, this situation is rare and would also be ethically questionable. To balance the effect that the number of independent vari- ables has on the coefficient of multiple determination, statistical software packages use an adjusted coefficient of multiple determination.

1 � R2 � 1 � SSR

SS total �

SSE SS total

� 41,695

212,916 � .196

R2 � SSR

SS total �

171,220 212,916

� .804

ADJUSTED COEFFICIENT OF DETERMINATION [14–4]R2adj � 1 �

SSE n � (k � 1)

SS total n � 1

The error and total sum of squares are divided by their degrees of freedom. Notice especially the degrees of freedom for the error sum of squares includes k, the number of independent variables. For the cost of heating example, the adjusted coefficient of determination is:

If we compare the (0.80) to the adjusted (0.77), the difference in this case is small.

R2R2

R2adj � 1 �

41,695 20 � (3 � 1)

212,916 20 � 1

� 1 � 2,606

11,206.0 � 1 � .23 � .77

COEFFICIENT OF MULTIPLE DETERMINATION [14–3]R2 � SSR

SS total

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Multiple Regression Analysis 523

Exercises 5. Consider the ANOVA table that follows.

Analysis of Variance Source DF SS MS F P Regression 2 77.907 38.954 4.14 0.021 Residual Error 62 583.693 9.414 Total 64 661.600

Analysis of Variance Source DF SS MS F Regression 5 3710.00 742.00 12.89 Residual Error 46 2647.38 57.55 Total 51 6357.38

a. Determine the standard error of estimate. About 95 percent of the residuals will be between what two values?

b. Determine the coefficient of multiple determination. Interpret this value. c. Determine the coefficient of multiple determination, adjusted for the degrees of

freedom. 6. Consider the ANOVA table that follows.

a. Determine the standard error of estimate. About 95 percent of the residuals will be between what two values?

b. Determine the coefficient of multiple determination. Interpret this value. c. Determine the coefficient of multiple determination, adjusted for the degrees of freedom.

14.4 Inferences in Multiple Linear Regression Thus far, multiple regression analysis has been viewed only as a way to describe the relationship between a dependent variable and several independent vari- ables. However, the least squares method also has the ability to draw inferences or generalizations about the relationship for an entire population. Recall that when you create confidence intervals or perform hypothesis tests as a part of inferential statistics, you view the data as a random sample taken from some population.

In the multiple regression setting, we assume there is an unknown popula- tion regression equation that relates the dependent variable to the k independent

Self-Review 14–2 Refer to Self-Review 14–1 on the subject of restaurants in Myrtle Beach. The ANOVA portion of the regression output is presented below.

Analysis of Variance Source DF SS MS Regression 5 100 20 Residual Error 20 40 2 Total 25 140

(a) How large was the sample? (b) How many independent variables are there? (c) How many dependent variables are there? (d) Compute the standard error of estimate. About 95 percent of the residuals will be

between what two values? (e) Determine the coefficient of multiple determination. Interpret this value. ( f ) Find the coefficient of multiple determination, adjusted for the degrees of freedom.

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524 Chapter 14

variables. This is sometimes called a model of the relationship. In symbols we write:

This equation is analogous to formula (14–1) except the coefficients are now reported as Greek letters. We use the Greek letters to denote population parameters. Then under a certain set of assumptions, which will be discussed shortly, the computed values of a and are sample statistics. These sample statistics are point estimates of the corresponding population parameters and For example, the sample regression coefficient is a point estimate of the population parameter The sam- pling distribution of these point estimates follows the normal probability distribution. These sampling distributions are each centered at their respective parameter values. To put it another way, the means of the sampling distributions are equal to the pa- rameter values to be estimated. Thus, by using the properties of the sampling distri- butions of these statistics, inferences about the population parameters are possible.

Global Test: Testing the Multiple Regression Model We can test the ability of the independent variables to explain the behavior of the dependent variable Y. To put this in question form: Can the depen- dent variable be estimated without relying on the independent variables? The test used is referred to as the global test. Basically, it investigates whether it is possi- ble all the independent variables have zero regression coefficients.

To relate this question to the heating cost example, we will test whether the independent variables (amount of insulation in the attic, mean daily outside tem- perature, and age of furnace) effectively estimate home heating costs. In testing a hypothesis, we first state the null hypothesis and the alternate hypothesis. In the heating cost example, there are three independent variables. Recall that are sample regression coefficients. The corresponding coefficients in the population are given the symbols We now test whether the regression coefficients in the population are all zero. The null hypothesis is:

The alternate hypothesis is:

Not all the �i’s are 0.

If the null hypothesis is true, it implies the regression coefficients are all zero and, logically, are of no use in estimating the dependent variable (heating cost). Should that be the case, we would have to search for some other independent variables— or take a different approach—to predict home heating costs.

To test the null hypothesis that the multiple regression coefficients are all zero, we employ the F distribution introduced in Chapter 12. We will use the .05 level of significance. Recall these characteristics of the F distribution:

1. There is a family of F distributions. Each time the degrees of freedom in either the numerator or the denominator changes, a new F distribution is created.

2. The F distribution cannot be negative. The smallest possible value is 0. 3. It is a continuous distribution. The distribution can assume an infinite number

of values between 0 and positive infinity. 4. It is positively skewed. The long tail of the distribution is to the right-hand side.

As the number of degrees of freedom increases in both the numerator and the denominator, the distribution approaches the normal probability distribution. That is, the distribution will move toward a symmetric distribution.

5. It is asymptotic. As the values of X increase, the F curve will approach the hor- izontal axis, but will never touch it.

The F-statistic to test the global hypothesis follows. As in Chapter 12, it is the ratio of two variances. In this case, the numerator is the regression sum of squares

H1:

H0: �1 � �2 � �3 � 0

�1, �2, and �3.

b1, b2, and b3

X1, X2, . . . , Xk

�2.b2 �j.�

Ŷ � � � �1X1 � �2 X2 � . . . � �k Xk

LO4 Conduct a hypothesis test to determine whether a set of regression coefficients differ from zero.

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divided by its degrees of freedom, k. The denominator is the residual sum of squares divided by its degrees of freedom, n � (k � 1). The formula follows.

The critical value of F is found in Appendix B.4. Using the table for the .05 sig- nificance level, move horizontally to 3 degrees of freedom in the numerator, then down to 16 degrees of freedom in the denominator, and read the critical value. It is 3.24. The region where is not rejected and the region where is rejected are shown in the following diagram.

H0H0

Using the ANOVA table, the F-statistic is

Remember that the F-statistic tests the basic null hypothesis that two variances or, in this case, two mean squares are equal. Also remember that we always put the larger of the two variances in the numerator. In our global multiple regression hypothesis test, we will reject the null hypothesis, H0, that all regression coefficients are zero when the regression mean square is larger in comparison to the residual mean square. If this is true, the F-statistic will be relatively large and in the far right tail of the F-distribution, and the p-value will be small, that is, less than our choice of our significance level of 0.05. Thus, we will reject the null hypothesis.

As with other hypothesis-testing methods, the decision rule can be based on either of two methods: (1) comparing the test statistic to a critical value or (2) cal- culating a p-value based on the test statistic and comparing the p-value to the sig- nificance level. Using the critical value method, we first find the critical value of F that requires three pieces of information: (1) the numerator degrees of freedom, (2) the denominator degrees of freedom, and (3) the significance level. The degrees of freedom for the numerator and the denominator are reported in the Excel ANOVA table that follows. The ANOVA output is highlighted in light green. The top number in the column marked “df” is 3, indicating there are 3 degrees of freedom in the numerator. This value corresponds to the number of independent variables. The mid- dle number in the “df” column (16) indicates that there are 16 degrees of freedom in the denominator. The number 16 is found by n � (k � 1) � 20 � (3 � 1) � 16.

F � SSR �k

SSE�[n � (k � 1)] �

MSR MSE

� 21.90

[14–5]F � SSR �k

SSE �[n � (k � 1)] GLOBAL TEST

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526 Chapter 14

Continuing with the global test, the decision rule is: Do not reject the null hypothesis, , that all the regression coefficients are 0 if the computed value of F is less than or equal to 3.24. If the computed F is greater than 3.24, reject and accept the alternate hypothesis,

The computed value of F is 21.90, which is in the rejection region. The null hypoth- esis that all the multiple regression coefficients are zero is therefore rejected. This means that at least one of the independent variables has the ability to explain the variation in the dependent variable (heating cost). We expected this decision. Logically, the out- side temperature, the amount of insulation, or age of the furnace have a great bear- ing on heating costs. The global test assures us that they do.

Testing the null hypothesis can also be based on a p-value, which is reported in the computer software output for all hypothesis tests. In the case of the F-statistic, the p-value is defined as the probability of observing an F-value as large or larger than the F test statistic, assuming the null hypothesis is true. If the p-value is less than our selected significance level, then we decide to reject the null hypothesis. The ANOVA shows the F-statistic’s p-value is equal to 0.000. It is clearly less than our significance level of 0.05. Therefore, we decide to reject the global null hypothesis and conclude that at least one of the regression coefficients is not equal to zero.

The decision is the same as when we used the critical value approach. The advan- tage to using the p-value approach is that the p-value gives us a “flavor” of the deci- sion. The computed p-value is much smaller than our significance level (.000 versus .05). We reject the null hypothesis that all the regression coefficients are 0 and, on the basis of the p-value, conclude that there is little likelihood this hypothesis is true.

Evaluating Individual Regression Coefficients So far we have shown that at least one, but not necessarily all, of the regression coefficients are not equal to zero and thus useful for predictions. The next step is to test the independent variables individually to determine which regression coeffi- cients may be 0 and which are not.

Why is it important to know if any of the equal 0? If a could equal 0, it implies that this particular independent variable is of no value in explaining any vari- ation in the dependent value. If there are coefficients for which cannot be rejected, we may want to eliminate them from the regression equation.

We will now conduct three separate tests of hypothesis—for temperature, for insulation, and for the age of the furnace.

For temperature: For insulation: For furnace age:

H1: �3 � 0H1: �2 � 0H1: �1 � 0

H0: �3 � 0H0: �2 � 0H0: �1 � 0

��i’s

H1. H0

3.24 Scale of F

F distribution

df = (3, 16)

Region of rejection (.05 level)

Region where H0 is not rejected

LO5 Conduct a hypothesis test of each regression coefficient.

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Multiple Regression Analysis 527

We will test the hypotheses at the .05 level. Note that these are two-tailed tests. The test statistic follows Student’s t distribution with degrees of

freedom. The number of sample observations is n. There are 20 homes in the study, so The number of independent variables is k, which is 3. Thus, there are

degrees of freedom. The critical value for t is in Appendix B.2. For a two-tailed test with 16 degrees

of freedom using the .05 significance level, is rejected if t is less than �2.120 or greater than 2.120.

Refer to the Excel output in the previous section. (See page 525.) The column highlighted in yellow, headed Coefficients, shows the values for the multiple regres- sion equation:

Interpreting the term in the equation: For each degree the temperature increases, it is expected that the heating cost will decrease about $4.58, holding the two other variables constant.

The column in the Excel output labeled “Standard Error” shows the standard error of the sample regression coefficients. Recall that Salsberry Realty selected a sample of 20 homes along the East Coast of the United States. If Salsberry Realty selected a second random sample and computed the regression coefficients for that sample, the values would not be exactly the same. If the sampling process was repeated many times, we could construct a sampling distribution for each of these regression coefficients. The column labeled “Standard Error” estimates the variabil- ity for each of these regression coefficients. The sampling distributions of the coef- ficients follow the t distribution with degrees of freedom. Hence, we are able to test the independent variables individually to determine whether the net regression coefficients differ from zero. The formula is:

n � (k � 1)

�4.583X1

Ŷ � 427.194 � 4.583X1 � 14.831X2 � 6.101X3

n � (k � 1) � 20 � (3 � 1) � 16 n � 20.

n � (k � 1)

TESTING INDIVIDUAL REGRESSION COEFFICIENTS

[14–6]t � bi � 0

sbi

The refers to any one of the regression coefficients, and refers to the stan- dard deviation of that distribution of the regression coefficient. We include 0 in the equation because the null hypothesis is

To illustrate this formula, refer to the test of the regression coefficient for the independent variable temperature. From the computer output on page 525, the regression coefficient for temperature is �4.583. The standard deviation of the sam- pling distribution of the regression coefficient for the independent variable temper- ature is 0.772. Inserting these values in formula (14–6):

Applying the formula, the computed t ratio is �5.937 for temperature (the small dif- ference between the computed value and that shown on the Excel output is due to rounding) and �3.119 for insulation. Both of these t-values are in the rejection region to the left of �2.120. Thus, we conclude that the regression coefficients for the tem- perature and insulation variables are not zero. The computed t for the age of the fur- nace is 1.521, so we conclude that could equal 0. The independent variable age of the furnace is not a significant predictor of heating cost. It can be dropped from the analysis.

We can also use p-values to test the individual regression coefficients. Again, these are commonly reported in computer software output. The computed t ratio for temperature on the Excel output is �5.934 and has a p-value of 0.000. Because the p-value is less than 0.05, the regression coefficient for the independent variable

t � b1 � 0

sb1 �

�4.583 � 0 0.772

� �5.937

�i � 0.

sbibi

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528 Chapter 14

temperature is not equal to zero and should be included in the equation to predict heating costs. For insulation, the t ratio is �3.119 and has a p-value of 0.007. As with temperature, the p-value is less than 0.05, so we conclude that the insulation regression coefficient is not equal to zero and should be included in the equation to predict heating cost. In contrast to temperature and insulation, the p-value to test the “age of the furnace” regression coefficient is 0.148. It is clearly greater than 0.05, so we conclude that the “age of furnace” regression coefficient could equal 0. Further, as an independent variable it is not a significant predictor of heating cost. Thus, age of furnace should not be included in the equation to predict heating costs.

At this point, we need to develop a strategy for deleting independent variables. In the Salsberry Realty case, there were three independent variables, and one (the age of the furnace) had a regression coefficient that did not differ from 0. It is clear that we should drop that variable and rerun the regression equation. Below is the Minitab output where heating cost is the dependent variable and outside temper- ature and amount of insulation are the independent variables.

Summarizing the results from this new Minitab output:

1. The new regression equation is:

Notice that the regression coefficients for outside temperature and amount of insulation are similar to but not exactly the same as when we included the independent variable age of the furnace. Compare the above equation to that in the Excel output on page 525. Both of the regression coefficients are negative as in the earlier equation.

2. The details of the global test are as follows:

H0: �1 � �2 � 0

(X2) (X1)

Ŷ � 490.29 � 5.1499X1 � 14.718X2

Not all of the �i’s � 0

The F distribution is the test statistic and there are degrees of freedom in the numerator and degrees of freedom in the denominator. Using the .05 significance level and Appendix B.4, the decision

n � (k � 1) � 20 � (2 � 1) � 17 k � 2

H1:

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rule is to reject if F is greater than 3.59. We compute the value of F as follows:

Because the computed value of F (29.42) is greater than the critical value (3.59), the null hypothesis is rejected and the alternate accepted. We conclude that at least one of the regression coefficients is different from 0.

Using the p-value, the F test statistic (29.42) has a p-value (0.000) that is clearly less than 0.05. Therefore, we reject the null hypothesis and accept the alternate. We conclude that at least one of the regression coefficients is differ- ent from 0.

3. The next step is to conduct a test of the regression coefficients individually. We want to find out if one or both of the regression coefficients are different from 0. The null and alternate hypotheses for each of the independent variables are:

Outside Temperature Insulation

The test statistic is the t distribution with degrees of freedom. Using the .05 significance level and Appendix B.2, the decision rule is to reject if the computed value of t is less than �2.110 or greater than 2.110.

Outside Temperature Insulation

In both tests, we reject and accept We conclude that each of the regression coefficients is different from 0. Both outside temperature and amount of insulation are useful variables in explaining the variation in heating costs.

Using p-values, the p-value for the temperature t-statistic is 0.000 and the p- value for the insulation t-statistic is 0.008. Both p-values are less than 0.05, so in both tests we reject the null hypothesis and conclude that each of the regression coefficients is different from 0. Both outside temperature and amount of insulation are useful variables in explaining the variation in heating costs.

In the heating cost example, it was clear which independent variable to delete. However, in some instances which variable to delete may not be as clear-cut. To explain, suppose we develop a multiple regression equation based on five indepen- dent variables. We conduct the global test and find that some of the regression coef- ficients are different from zero. Next, we test the regression coefficients individually and find that three are significant and two are not. The preferred procedure is to drop the single independent variable with the smallest absolute t value or largest p-value and rerun the regression equation with the four remaining variables, then, on the new regression equation with four independent variables, conduct the individual tests. If there are still regression coefficients that are not significant, again drop the variable with the smallest absolute t value or the largest, nonsignificant p-value. To describe the process in another way, we should delete only one variable at a time. Each time we delete a variable, we need to rerun the regression equation and check the remain- ing variables.

This process of selecting variables to include in a regression model can be auto- mated, using Excel, Minitab, MegaStat, or other statistical software. Most of the soft- ware systems include methods to sequentially remove and/or add independent variables and at the same time provide estimates of the percentage of variation

H1.H0

t � b2 � 0

sb2 �

�14.718 � 0 4.934

� �2.98t � b1 � 0

sb1 �

�5.1499 � 0 0.7019

� �7.34

n � (k � 1) � 20 � (2 � 1) � 17

H1: �2 � 0H1: �1 � 0 H0: �2 � 0H0: �1 � 0

F � SSR �k

SSE �[n � (k � 1)] �

165,195�2 47,721�[20 � (2 � 1)]

� 29.42

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530 Chapter 14

Self-Review 14–3 The regression output about eating places in Myrtle Beach is repeated below (see earlier self-reviews).

explained (the R-square term). Two of the common methods are stepwise regres- sion and best subset regression. It may take a long time, but in the extreme we could compute every regression between the dependent variable and any possible subset of the independent variables.

Unfortunately, on occasion, the software may work “too hard” to find an equa- tion that fits all the quirks of your particular data set. The suggested equation may not represent the relationship in the population. A judgment is needed to choose among the equations presented. Consider whether the results are logical. They should have a simple interpretation and be consistent with your knowledge of the application under study.

Predictor Coef SE Coef T p-value Constant 2.50 1.50 1.667 — X1 3.00 1.500 2.000 0.056 X2 4.00 3.000 1.333 0.194 X3 �3.00 0.20 �15.00 0.000 X4 0.20 .05 4.00 0.000 X5 1.00 1.50 0.667 0.511

Analysis of Variance Source DF SS MS F p-value Regression 5 100 20 10 0.000 Residual Error 20 40 2 Total 25 140

Predictor Coef SE Coef T P Constant 84.998 1.863 45.61 0.000 X1 2.391 1.200 1.99 0.051 X2 �0.4086 0.1717 �2.38 0.020

Analysis of Variance Source DF SS MS F P Regression 2 77.907 38.954 4.14 0.021 Residual Error 62 583.693 9.414 Total 64 661.600

(a) Perform a global test of hypothesis to check if any of the regression coefficients are different from 0. What do you decide? Use the .05 significance level.

(b) Do an individual test of each independent variable. Which variables would you consider eliminating? Use the .05 significance level.

Exercises 7. Given the following regression output,

answer the following questions: a. Write the regression equation. b. If is 4 and is 11, what is the value of the dependent variable? c. How large is the sample? How many independent variables are there? d. Conduct a global test of hypothesis to see if any of the set of regression coefficients

could be different from 0. Use the .05 significance level. What is your conclusion? e. Conduct a test of hypothesis for each independent variable. Use the .05 significance

level. Which variable would you consider eliminating? f. Outline a strategy for deleting independent variables in this case.

X2X1

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Multiple Regression Analysis 531

8. The following regression output was obtained from a study of architectural firms. The dependent variable is the total amount of fees in millions of dollars.

is the number of architects employed by the company. is the number of engineers employed by the company. is the number of years involved with health care projects. is the number of states in which the firm operates. is the percent of the firm’s work that is health care–related.

a. Write out the regression equation. b. How large is the sample? How many independent variables are there? c. Conduct a global test of hypothesis to see if any of the set of regression coefficients

could be different from 0. Use the .05 significance level. What is your conclusion? d. Conduct a test of hypothesis for each independent variable. Use the .05 significance

level. Which variable would you consider eliminating first? e. Outline a strategy for deleting independent variables in this case.

14.5 Evaluating the Assumptions of Multiple Regression In the previous section, we described the methods to statistically evaluate the multiple regression equation. The results of the test let us know if at least one of the coefficients was not equal to zero and we described a procedure of eval- uating each regression coefficient. We also discussed the decision-making pro- cess for including and excluding independent variables in the multiple regression equation.

It is important to know that the validity of the statistical global and individ- ual tests rely on several assumptions. That is, if the assumptions are not true, the results might be biased or misleading. However, strict adherence to the fol- lowing assumptions is not always possible. Fortunately, the statistical techniques discussed in this chapter work well even when one or more of the assumptions are violated. Even if the values in the multiple regression equation are “off” slightly, our estimates using a multiple regression equation will be closer than any that could be made otherwise. Usually the statistical procedures are robust enough to over- come violations of some assumptions.

In Chapter 13, we listed the necessary assumptions for regression when we con- sidered only a single independent variable. (See Section 13.8 on page 490.) The assumptions for multiple regression are similar.

1. There is a linear relationship. That is, there is a straight-line relationship between the dependent variable and the set of independent variables.

X5 X4 X3 X2 X1

Predictor Coef SE Coef T p-value Constant 7.987 2.967 2.69 — X1 0.12242 0.03121 3.92 0.000 X2 �0.12166 0.05353 �2.27 0.028 X3 �0.06281 0.03901 �1.61 0.114 X4 0.5235 0.1420 3.69 0.001 X5 �0.06472 0.03999 �1.62 0.112

Analysis of Variance Source DF SS MS F p-value Regression 5 3710.00 742.00 12.89 0.000 Residual Error 46 2647.38 57.55 Total 51 6357.38

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2. The variation in the residuals is the same for both large and small values of To put it another way, is unrelated to whether is large or small.

3. The residuals follow the normal probability distribution. Recall the residual is the difference between the actual value of Y and the estimated value So the term is computed for every observation in the data set. These resid- uals should approximately follow a normal probability distribution. In addition, the mean of the residuals should be 0.

4. The independent variables should not be correlated. That is, we would like to select a set of independent variables that are not themselves correlated.

5. The residuals are independent. This means that successive observations of the dependent variable are not correlated. This assumption is often violated when time is involved with the sampled observations.

In this section, we present a brief discussion of each of these assumptions. In addi- tion, we provide methods to validate these assumptions and indicate the conse- quences if these assumptions cannot be met. For those interested in additional discussion, Kutner, Nachtscheim, Neter, and Li, Applied Linear Statistical Models, 5th ed., (McGraw-Hill: 2005), is an excellent reference.

Linear Relationship Let’s begin with the linearity assumption. The idea is that the relationship between the set of independent variables and the dependent variable is linear. If we are considering two independent variables, we can visualize this assumption. The two independent variables and the dependent variable would form a three- dimensional space. The regression equation would then form a plane as shown on page 514. We can evaluate this assumption with scatter diagrams and residual plots.

Using Scatter Diagrams The evaluation of a multiple regression equation should always include a scatter diagram that plots the dependent variable against each independent variable. These graphs help us to visualize the relationships and provide some initial information about the direction (positive or negative), linearity, and strength of the relationship. For example, the scatter diagrams for the home heat- ing example follow. The plots suggest a fairly strong negative, linear relationship between heating cost and temperature, and a negative relationship between heating cost and insulation.

Using Residual Plots Recall that a residual can be computed using the multiple regression equation for each observation in a data set. In Chapter 13, we

(Y � Ŷ )

(Y � Ŷ ) Ŷ.

Ŷ(Y � Ŷ )Ŷ.

532 Chapter 14

LO6 Use residual analysis to evaluate the assumptions of multiple regression analysis.

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discussed the idea that the best regression line passed through the center of the data in a scatter plot. In this case, you would find a good number of the observa- tions above the regression line (these residuals would have a positive sign), and a good number of the observations below the line (these residuals would have a neg- ative sign). Further, the observations would be scattered above and below the line over the entire range of the independent variable.

The same concept is true for multiple regression, but we cannot graphically por- tray the multiple regression. However, plots of the residuals can help us evaluate the linearity of the multiple regression equation. To investigate, the residuals are plot- ted on the vertical axis against the predictor variable, The graph on the left below shows the residual plots for the home heating cost example. Notice the following:

• The residuals are plotted on the vertical axis and are centered around zero. There are both positive and negative residuals.

• The residual plots show a random distribution of positive and negative values across the entire range of the variable plotted on the horizontal axis.

• The points are scattered and there is no obvious pattern, so there is no reason to doubt the linearity assumption.

This plot supports the assumption of linearity.

Ŷ.

If there is a pattern to the points in the scatter plot, further investigation is necessary. The points in the graph on the right above show nonrandom residuals. See that the residual plot does not show a random distribution of positive and negative values across the entire range of the variable plotted on the horizontal axis. In fact, the graph shows a curvature to the residual plots. This indicates the relationship may not be linear. In this case, we would evaluate different transfor- mations of the equation as discussed in Chapter 13.

Variation in Residuals Same for Large and Small Values This requirement indicates that the variation about the predicted values is con- stant, regardless of whether the predicted values are large or small. To cite a specific example, which may violate the assumption, suppose we use the single independent variable age to explain the variation in income. We suspect that as age increases so does salary, but it also seems reasonable that as age increases there may be more variation around the regression line. That is, there will likely be more variation in income for a 50-year-old person than for a 35-year-old

Ŷ

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person. The requirement for constant variation around the regression line is called homoscedasticity.

HOMOSCEDASTICITY The variation around the regression equation is the same for all of the values of the independent variables.

To check for homoscedasticity the residuals are plotted against the fitted values of Y. This is the same graph that we used to evaluate the assumption of linearity. (See page 533.) Based on the scatter diagram in that software output, it is reasonable to conclude that this assumption has not been violated.

Distribution of Residuals To be sure that the inferences we make in the global and individual hypotheses tests are valid, we evaluate the distribution of residuals. Ideally, the residuals should follow a normal probability distribution.

To evaluate this assumption, we can organize the residuals into a frequency dis- tribution. The Minitab histogram of the residuals is shown following on the left for the home heating cost example. Although it is difficult to show that the residuals follow a normal distribution with only 20 observations, it does appear the normality assumption is reasonable.

Both Minitab and Excel offer another graph that helps to evaluate the assump- tion of normally distributed residuals. It is a called a normal probability plot and is shown to the right of the histogram. We describe this graph further in Section 17.6 starting on page 663. Basically, the normal probability plot supports the assump- tion of normally distributed residuals if the plotted points are fairly close to a straight line drawn from the lower left to the upper right of the graph.

In this case, both graphs support the assumption that the residuals follow the normal probability distribution. Therefore, the inferences that we made based on the global and individual hypothesis tests are supported with the results of this evaluation.

Multicollinearity Multicollinearity exists when independent variables are correlated. Correlated inde- pendent variables make it difficult to make inferences about the individual regression coefficients and their individual effects on the dependent variable. In practice, it is

LO7 Evaluate the effects of correlated independent variables.

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Example Refer to the data in Table 14–1, which relates the heating cost to the independent variables outside temperature, amount of insulation, and age of furnace. Develop a correlation matrix for all the independent variables. Does it appear there is a prob- lem with multicollinearity? Find and interpret the variance inflation factor for each of the independent variables.

nearly impossible to select variables that are completely unrelated. To put it another way, it is nearly impossible to create a set of independent variables that are not correlated to some degree. However, a general understanding of the issue of multi- collinearity is important.

First, we should point out that multicollinearity does not affect a multiple regres- sion equation’s ability to predict the dependent variable. However, when we are interested in evaluating the relationship between each independent variable and the dependent variable, multicollinearity may show unexpected results.

For example, if we use two highly multicollinear variables, high school GPA and high school class rank, to predict the GPA of incoming college freshmen (dependent variable), we would expect that both independent variables would be positively related to the dependent variable. However, because the independent variables are highly correlated, one of the independent variables may have an unexpected and inexplicable negative sign. In essence, these two independent variables are redundant in that they explain the same variation in the dependent variable.

A second reason for avoiding correlated independent variables is they may lead to erroneous results in the hypothesis tests for the individual independent variables. This is due to the instability of the standard error of estimate. Several clues that indicate problems with multicollinearity include the following:

1. An independent variable known to be an important predictor ends up having a regression coefficient that is not significant.

2. A regression coefficient that should have a positive sign turns out to be nega- tive, or vice versa.

3. When an independent variable is added or removed, there is a drastic change in the values of the remaining regression coefficients.

In our evaluation of a multiple regression equation, an approach to reducing the effects of multicollinearity is to carefully select the independent variables that are included in the regression equation. A general rule is if the correlation between two independent variables is between �0.70 and 0.70, there likely is not a prob- lem using both of the independent variables. A more precise test is to use the variance inflation factor. It is usually written VIF. The value of VIF is found as follows:

The term refers to the coefficient of determination, where the selected independent variable is used as a dependent variable and the remaining inde- pendent variables are used as independent variables. A VIF greater than 10 is considered unsatisfactory, indicating that the independent variable should be removed from the analysis. The following example will explain the details of finding the VIF.

R2j

VARIANCE INFLATION FACTOR [14–7]VIF � 1

1 � R2j

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536 Chapter 14

Solution We begin by using the Minitab system to find the correlation matrix for the depen- dent variable and the three independent variables. A portion of that output follows:

Cost Temp Insul Temp �0.812 Insul �0.257 �0.103 Age 0.537 �0.486 0.064

Cell Contents: Pearson correlation

The highlighted area indicates the correlation among the independent variables. None of the correlations among the independent variables exceed �.70 or .70, so we do not suspect problems with multicollinearity. The largest correlation among the independent variables is �0.486 between age and temperature.

To confirm this conclusion, we compute the VIF for each of the three indepen- dent variables. We will consider the independent variable temperature first. We use Minitab to find the multiple coefficient of determination with temperature as the dependent variable and amount of insulation and age of the furnace as indepen- dent variables. The relevant Minitab output follows.

Regression Analysis: Temp versus Insul, Age

The regression equation is Temp � 58.0 � 0.51 Insul � 2.51 Age

Predictor Coef SE Coef T P Constant 57.99 12.35 4.70 0.000 Insul �0.509 1.488 �0.34 0.737 Age �2.509 1.103 �2.27 0.036

S � 16.0311 R-Sq � 24.1% R-Sq(adj) � 15.2%

Analysis of Variance Source DF SS MS F P Regression 2 1390.3 695.1 2.70 0.096 Residual Error 17 4368.9 257.0 Total 19 5759.2

The coefficient of determination is .241, so inserting this value into the VIF formula:

The VIF value of 1.32 is less than the upper limit of 10. This indicates that the inde- pendent variable temperature is not strongly correlated with the other independent variables.

Again, to find the VIF for insulation we would develop a regression equation with insulation as the dependent variable and temperature and age of furnace as independent variables. For this equation, we would determine the coefficient of determination. This would be the value for . We would substitute this value in equation 14–7, and solve for VIF.

Fortunately, Minitab will generate the VIF values for each of the independent variables. These values are reported in the right-hand column under the heading VIF in the following Minitab output. All these values are less than 10. Hence, we con- clude there is not a problem with multicollinearity in this example.

The regression equation is Cost � 427 � 4.58 Temp � 14.8 Insul � 6.10 Age

Predictor Coef SE Coef T P VIF Constant 427.19 59.60 7.17 0.000 Temp �4.5827 0.7723 �5.93 0.000 1.318 Insul �14.831 4.754 �3.12 0.007 1.011 Age �6.101 4.012 1.52 0.148 1.310

R22

VIF � 1

1 � R21 �

1 1 � .241

� 1.32

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Multiple Regression Analysis 537

Independent Observations The fifth assumption about regression and correlation analysis is that successive residuals should be independent. This means that there is not a pattern to the residuals, the residuals are not highly correlated, and there are not long runs of positive or negative residuals. When successive residuals are correlated, we refer to this condition as autocorrelation.

Autocorrelation frequently occurs when the data are collected over a period of time. For example, we wish to predict yearly sales of Ages Software Inc. based on the time and the amount spent on advertising. The dependent variable is yearly sales and the independent variables are time and amount spent on advertising. It is likely that for a period of time the actual points will be above the regression plane (remem- ber there are two independent variables) and then for a period of time the points will be below the regression plane. The graph below shows the residuals plotted on the vertical axis and the fitted values on the horizontal axis. Note the run of residuals above the mean of the residuals, followed by a run below the mean. A scatter plot such as this would indicate possible autocorrelation.

Ŷ

Re si

du al

s (Y

– Y

Fitted values ̂Y

There is a test for autocorrelation, called the Durbin-Watson. We present the details of this test in Chapter 16, Section 16.10.

14.6 Qualitative Independent Variables In the previous example regarding heating cost, the two independent variables out- side temperature and insulation were quantitative; that is, numerical in nature. Fre- quently we wish to use nominal-scale variables—such as gender, whether the home has a swimming pool, or whether the sports team was the home or the visiting team—in our analysis. These are called qualitative variables because they describe a particular quality, such as male or female. To use a qualitative variable in regression analysis, we use a scheme of dummy variables in which one of the two possible conditions is coded 0 and the other 1.

DUMMY VARIABLE A variable in which there are only two possible outcomes. For analysis, one of the outcomes is coded a 1 and the other a 0.

For example, we might be interested in estimating an executive’s salary on the basis of years of job experience and whether he or she graduated from college. “Graduation from college” can take on only one of two conditions: yes or no. Thus, it is considered a qualitative variable.

Suppose in the Salsberry Realty example that the independent variable “garage” is added. For those homes without an attached garage, 0 is used; for homes with an attached garage, a 1 is used. We will refer to the “garage” variable as The data from Table 14–2 are entered into the Minitab system.

X4.

LO8 Evaluate and use qualitative independent variables.

Statistics in Action

In recent years, mul- tiple regression has been used in a vari- ety of legal proceed- ings. It is particularly useful in cases alleg- ing discrimination

(continued)

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by gender or race. As an example, suppose that a woman alleges that Company X’s wage rates are unfair to women. To sup- port the claim, the plaintiff produces data showing that, on the average, women earn less than men. In re- sponse, Company X argues that its wage rates are based on experience, training, and skill and that its female employees, on the average, are younger and less ex- perienced than the male employees. In fact, the company might further argue that the current situ- ation is actually due to its recent success- ful efforts to hire more women.

538 Chapter 14

The output from Minitab is:

Cost, Temperature, Insulation, Garage, Y X1 X2 X4

$250 35 3 0 360 29 4 1 165 36 7 0 43 60 6 0 92 65 5 0

200 30 5 0 355 10 6 1 290 7 10 1 230 21 9 0 120 55 2 0 73 54 12 0

205 48 5 1 400 20 5 1 320 39 4 1 72 60 8 0

272 20 5 1 94 58 7 0

190 40 8 1 235 27 9 0 139 30 7 0

TABLE 14–2 Home Heating Costs, Temperature, Insulation, and Presence of a Garage for a Sample of 20 Homes

What is the effect of the garage variable? Should it be included in the analy- sis? To show the effect of the variable, suppose we have two houses exactly alike next to each other in Buffalo, New York; one has an attached garage, and the other

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Multiple Regression Analysis 539

does not. Both homes have 3 inches of insulation, and the mean January temper- ature in Buffalo is 20 degrees. For the house without an attached garage, a 0 is substituted for in the regression equation. The estimated heating cost is $280.90, found by:

For the house with an attached garage, a 1 is substituted for in the regression equation. The estimated heating cost is $358.30, found by:

The difference between the estimated heating costs is $77.40 Hence, we can expect the cost to heat a house with an attached garage to be $77.40 more than the cost for an equivalent house without a garage.

We have shown the difference between the two types of homes to be $77.40, but is the difference significant? We conduct the following test of hypothesis.

The information necessary to answer this question is on the Minitab output above. The net regression coefficient for the independent variable garage is 77.43, and the standard deviation of the sampling distribution is 22.78. We identify this as the fourth independent variable, so we use a subscript of 4. Finally, we insert these values in formula (14–6).

There are three independent variables in the analysis, so there are degrees of freedom. The critical value from Appendix B.2 is

2.120. The decision rule, using a two-tailed test and the .05 significance level, is to reject if the computed t is to the left of �2.120 or to the right of 2.120. Since the computed value of 3.40 is to the right of 2.120, the null hypothesis is rejected. It is concluded that the regression coefficient is not zero. The independent variable garage should be included in the analysis.

Using the p-value approach, the computed t value of 3.40 has a p-value of 0.004. This value is less than the .05 significance level. Therefore, we reject the null hypothesis. We conclude that the regression coefficient is not zero and the inde- pendent variable garage should be included in the analysis.

Is it possible to use a qualitative variable with more than two possible out- comes? Yes, but the coding scheme becomes more complex and will require a series of dummy variables. To explain, suppose a company is studying its sales as they relate to advertising expense by quarter for the last 5 years. Let sales be the dependent variable and advertising expense be the first independent variable, To include the qualitative information regarding the quarter, we use three additional independent variables. For the variable the five observations referring to the first quarter of each of the 5 years are coded 1 and the other quarters 0. Similarly, for

the five observations referring to the second quarter are coded 1 and the other quarters 0. For , the five observations referring to the third quarter are coded 1 and the other quarters 0. An observation that does not refer to any of the first three quarters must refer to the fourth quarter, so a distinct independent variable refer- ring to this quarter is not necessary.

X4 X3

X2,

X1.

20 � (3 � 1) � 16 n � (k � 1) �

t � b4 � 0

sb4 �

77.43 � 0 22.78

� 3.40

H1: �4 � 0

H0: �4 � 0

$280.90).($358.30 �

� 394 � 3.96(20) � 11.3(3) � 77.4(1) � 358.30

Ŷ � 394 � 3.96X1 � 11.3X2 � 77.4X4

� 394 � 3.96(20) � 11.3(3) � 77.4(0) � 280.90

Ŷ � 394 � 3.96X1 � 11.3X2 � 77.4X4

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540 Chapter 14

Self-Review 14–4 A study by the American Realtors Association investigated the relationship between the commissions earned by sales associates last year and the number of months since the associates earned their real estate licenses. Also of interest in the study is the gender of the sales associate. Below is a portion of the regression output. The dependent variable is commissions, which is reported in $000, and the independent variables are months since the license was earned and gender and male � 0).(female � 1

(a) Write out the regression equation. How much commission would you expect a female agent to make who earned her license 30 months ago?

(b) Do the female agents on the average make more or less than the male agents? How much more?

(c) Conduct a test of hypothesis to determine if the independent variable gender should be included in the analysis. Use the 0.05 significance level. What is your conclusion?

Regression Analysis R2 0.642

Adjusted R2 0.600 n 20 R 0.801 k 2

Std. Error 3.219 Dep. Var. Commissions

ANOVA table Source SS df MS F p-value Regression 315.9291 2 157.9645 15.25 .0002 Residual 176.1284 17 10.3605 Total 492.0575 19

Regression output Variables coefficients std. error t (df � 17) p-value 95% lower 95% upper Intercept 15.7625 3.0782 5.121 .0001 9.2680 22.2570

Months 0.4415 0.0839 5.263 .0001 0.2645 0.6186 Gender 3.8598 1.4724 2.621 .0179 0.7533 6.9663

14.7 Regression Models with Interaction In Chapter 12, we discussed interaction among independent variables. To explain, sup- pose we are studying weight loss and assume, as the current literature suggests, that diet and exercise are related. So the dependent variable is amount of change in weight and the independent variables are: diet (yes or no) and exercise (none, moderate, sig- nificant). We are interested in whether there is interaction among the independent vari- ables. That is, if those studied maintain their diet and exercise significantly, will that increase the mean amount of weight lost? Is total weight loss more than the sum of the loss due to the diet effect and the loss due to the exercise effect?

We can expand on this idea. Instead of having two nominal-scale variables, diet and exercise, we can examine the effect (interaction) of several ratio-scale variables. For example, suppose we want to study the effect of room temperature (68, 72, 76, or 80 degrees Fahrenheit) and noise level (60, 70, or 80 decibels) on the number of units produced. To put it another way, does the combination of noise level in the room and the temperature of the room have an effect on the productivity of the workers? Would the workers in a quiet, cool room produce more units than those in a hot, noisy room?

In regression analysis, interaction can be examined as a separate independent variable. An interaction prediction variable can be developed by multiplying the data values in one independent variable by the values in another independent variable, thereby creating a new independent variable. A two-variable model that includes an interaction term is:

Y � � � �1X1 � �2 X2 � �3X1X2

LO9 Explain the possible interaction among independent variables.

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Multiple Regression Analysis 541

The term is the interaction term. We create this variable by multiplying the values of and to create the third independent variable. We then develop a regression equation using the three independent variables and test the significance of the third independent variable using the individual test for independent variables, described earlier in the chapter. An example will illustrate the details.

X2X1 X1X2

Example Refer to the heating cost example and the data in Table 14–1. Is there an interac- tion between the outside temperature and the amount of insulation? If both vari- ables are increased, is the effect on heating cost greater than the sum of savings from warmer temperature and the savings from increased insulation separately?

Solution The information from Table 14–1 for the independent variables temperature and insu- lation is repeated below. We create the interaction variable by multiplying the temper- ature variable by the insulation. For the first sampled home, the value temperature is 35 degrees and insulation is 3 inches so the value of the interaction variable is 35 � 3 � 105. The values of the other interaction products are found in a similar fashion.

We find the multiple regression using temperature, insulation, and the interaction of temperature and insulation as independent variables. The regression equation is reported below.

The question we wish to answer is whether the interaction variable is significant. We will use the .05 significance level. In terms of a hypothesis:

There is degrees of freedom. Using the .05 signif- icance level and a two-tailed test, the critical values of t are and We reject the null hypothesis if t is less than or t is greater than 2.120. From the output, and To find the value of t, we use formula (14–6).

Because the computed value of 1.324 is less than the critical value of 2.120, we do not reject the null hypothesis. In addition, the p-value of 0.204 exceeds 0.05. We con- clude that there is not a significant interaction between temperature and insulation.

t � b3 � 0

sb3 �

0.385 � 0 0.291

� 1.324

sb3 � 0.291.b3 � 0.385 �2.120

2.120.�2.120 n � (k � 1) � 20 � (3 � 1) � 16

H1: �3 � 0 H0: �3 � 0

Ŷ � 598.070 � 7.811 X1 � 30.161X2 � 0.385X1X2

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542 Chapter 14

STEPWISE REGRESSION A step-by-step method to determine a regression equation that begins with a single independent variable and adds or deletes independent variables one by one. Only independent variables with nonzero regression coefficients are included in the regression equation.

There are other situations that can occur when studying interaction among inde- pendent variables.

1. It is possible to have a three-way interaction among the independent variables. In our heating example, we might have considered the three-way interaction between temperature, insulation, and age of the furnace.

2. It is possible to have an interaction where one of the independent variables is nominal scale. In our heating cost example, we could have studied the inter- action between temperature and garage.

Studying all possible interactions can become very complex. However, careful con- sideration to possible interactions among independent variables can often provide useful insight into the regression models.

14.8 Stepwise Regression In our heating cost example (see sample information in Table 14–1 and Table 14–2), we considered four independent variables: the mean outside temperature, the amount of insulation in the home, the age of the furnace, and whether or not there was an attached garage. To obtain the equation, we first ran a global or “all at once” test to determine if any of the regression coefficients were significant. When we found at least one to be significant, we tested the regression coefficients individually to deter- mine which were important. We left out the independent variables that did not have significant regression coefficients and kept the others. By retaining the independent variables with significant coefficients, we found the regression equation that used the fewest independent variables. This made the regression equation easy to interpret and explained as much variation in the dependent variable as possible.

We are now going to describe a technique called stepwise regression, which is more efficient in building the regression equation.

In the stepwise method, we develop a sequence of equations. The first equation con- tains only one independent variable. However, this independent variable is the one from the set of proposed independent variables that explains the most variation in the depen- dent variable. Stated differently, if we compute all the simple correlations between each independent variable and the dependent variable, the stepwise method first selects the independent variable with the strongest correlation with the dependent variable.

Next, the stepwise method looks at the remaining independent variables and then selects the one that will explain the largest percentage of the variation yet unex- plained. We continue this process until all the independent variables with significant regression coefficients are included in the regression equation. The advantages to the stepwise method are:

1. Only independent variables with significant regression coefficients are entered into the equation.

2. The steps involved in building the regression equation are clear. 3. It is efficient in finding the regression equation with only significant regression

coefficients. 4. The changes in the multiple standard error of estimate and the coefficient of

determination are shown.

LO10 Explain stepwise regression.

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Multiple Regression Analysis 543

The stepwise Minitab output for the heating cost problem follows. Note that the final equation, which is reported in the column labeled 3 includes the independent variables temperature, garage, and insulation. These are the same independent vari- ables that were included in our equation using the global test and the test for indi- vidual independent variables. (See page 538.) The independent variable age, for age of the furnace, is not included because it is not a significant predictor of cost.

Reviewing the steps and interpreting output:

1. The stepwise procedure selects the independent variable temperature first. This variable explains more of the variation in heating cost than any of the other three proposed independent variables. Temperature explains 65.85 percent of the variation in heating cost. The regression equation is:

There is an inverse relationship between heating cost and temperature. For each degree the temperature increases, heating cost is reduced by $4.93.

2. The next independent variable to enter the regression equation is garage. When this variable is added to the regression equation, the coefficient of determination is increased from 65.85 percent to 80.46 percent. That is, by adding garage as an independent variable, we increase the coefficient of determination by 14.61 per- cent. The regression equation after step 2 is:

Usually the regression coefficients will change from one step to the next. In this case, the coefficient for temperature retained its negative sign, but it changed from to This change is reflective of the added influence of the independent variable garage. Why did the stepwise method select the inde- pendent variable garage instead of either insulation or age? The increase in the coefficient of determination, is larger if garage is included rather than either of the other two variables.

3. At this point, there are two unused variables remaining, insulation and age. Notice on the third step the procedure selects insulation and then stops. This indicates the variable insulation explains more of the remaining variation in heating cost than the age variable does. After the third step, the regression equation is:

Ŷ � 393.7 � 3.96 X1 � 77.0 X2 � 11.3 X3

R2,

�3.56.�4.93

Ŷ � 300.3 � 3.56X1 � 93.0X2

Ŷ � 388.8 � 4.93X1

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544 Chapter 14

Exercises 9. The production manager of High Point Sofa and Chair, a large furniture manufacturer

located in North Carolina, is studying the job performance ratings of a sample of 15 elec- trical repairmen employed by the company. An aptitude test is required by the human resources department to become an electrical repairman. The production manager was able to get the score for each repairman in the sample. In addition, he determined which of the repairmen were union members and which were not The sam-(code � 0).(code � 1)

Job Performance Worker Score Aptitude Test Score Union Membership

Abbott 58 5 0 Anderson 53 4 0 Bender 33 10 0 Bush 97 10 0 Center 36 2 0 Coombs 83 7 0 Eckstine 67 6 0 Gloss 84 9 0 Herd 98 9 1

(continued )

At this point, 86.98 percent of the variation in heating cost is explained by the three independent variables temperature, garage, and insulation. This is the same value and regression equation we found on page 538 except for round- ing differences.

4. Here, the stepwise procedure stops. This means the independent variable age does not add significantly to the coefficient of determination.

The stepwise method developed the same regression equation, selected the same independent variables, and found the same coefficient of determination as the global and individual tests described earlier in the chapter. The advantages to the step- wise method is that it is more direct than using a combination of the global and individual procedures.

Other methods of variable selection are available. The stepwise method is also called the forward selection method because we begin with no independent vari- ables and add one independent variable to the regression equation at each itera- tion. There is also the backward elimination method, which begins with the entire set of variables and eliminates one independent variable at each iteration.

The methods described so far look at one variable at a time and decide whether to include or eliminate that variable. Another approach is the best-subset regression. With this method, we look at the best model using one independent variable, the best model using two independent variables, the best model with three and so on. The cri- terion is to find the model with the largest value, regardless of the number of inde- pendent variables. Also, each independent variable does not necessarily have a nonzero regression coefficient. Since each independent variable could either be included or not included, there are possible models, where k refers to the number of indepen- dent variables. In our heating cost example, we considered four independent variables so there are 15 possible regression models, found by We would examine all regression models using one independent variable, all combinations using two variables, all combinations using three independent variables, and the possibility of using all four independent variables. The advantages to the best-subset method is it may examine combinations of independent variables not considered in the stepwise method. The process is available in Minitab and MegaStat.

24 � 1 � 16 � 1 � 15.

2k � 1

ple information is reported below.

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Multiple Regression Analysis 545

Job Performance Worker Score Aptitude Test Score Union Membership

Householder 45 2 1 Iori 97 8 1 Lindstrom 90 6 1 Mason 96 7 1 Pierse 66 3 1 Rohde 82 6 1

a. Use a statistical software package to develop a multiple regression equation using the job performance score as the dependent variable, and aptitude test score and union membership as independent variables.

b. Comment on the regression equation. Be sure to include the coefficient of determi- nation and the effect of union membership. Are these two variables effective in explaining the variation in job performance?

c. Conduct a test of hypothesis to determine if union membership should be included as an independent variable.

d. Repeat the analysis considering possible interaction terms. 10. Cincinnati Paint Company sells quality brands of paints through hardware stores through-

out the United States. The company maintains a large sales force whose job it is to call on existing customers as well as look for new business. The national sales manager is inves- tigating the relationship between the number of sales calls made and the miles driven by the sales representative. Also, do the sales representatives who drive the most miles and make the most calls necessarily earn the most in sales commissions? To investigate, the vice president of sales selected a sample of 25 sales representatives and determined:

• The amount earned in commissions last month (Y ) • The number of miles driven last month (X1) • The number of sales calls made last month (X2) The information is reported below.

Commissions ($000) Calls Driven

22 139 2,371 13 132 2,226 33 144 2,731

25 127 2,671 43 154 2,988 34 147 2,829

ooo

Develop a regression equation including an interaction term. Is there a significant inter- action between the number of sales calls and the miles driven?

11. An art collector is studying the relationship between the selling price of a painting and two independent variables. The two independent variables are the number of bidders at the particular auction and the age of the painting, in years. A sample of 25 paintings revealed the following sample information.

Painting Auction Price Bidders Age

1 3,470 10 67 2 3,500 8 56 3 3,700 7 73

23 4,660 5 94 24 4,710 3 88 25 4,880 1 84

oooo

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546 Chapter 14

a. Develop a multiple regression equation using the independent variables number of bidders and age of painting to estimate the dependent variable auction price. Discuss the equation. Does it surprise you that there is an inverse relationship between the number of bidders and the price of the painting?

b. Create an interaction variable and include it in the regression equation. Explain the meaning of the interaction. Is this variable significant?

c. Use the stepwise method and the independent variables for the number of bidders, the age of the painting, and the interaction between the number of bidders and the age of the painting. Which variables would you select?

12. A real estate developer wishes to study the relationship between the size of home a client will purchase (in square feet) and other variables. Possible independent variables include the family income, family size, whether there is a senior adult parent living with the fam- ily (1 for yes, 0 for no), and the total years of education beyond high school for the hus- band and wife. The sample information is reported below.

Square Income Family Senior Family Feet (000s) Size Parent Education

1 2,240 60.8 2 0 4 2 2,380 68.4 2 1 6 3 3,640 104.5 3 0 7 4 3,360 89.3 4 1 0 5 3,080 72.2 4 0 2 6 2,940 114 3 1 10 7 4,480 125.4 6 0 6 8 2,520 83.6 3 0 8 9 4,200 133 5 0 2

10 2,800 95 3 0 6

Develop an appropriate multiple regression equation. Which independent variables would you include in the final regression equation? Use the stepwise method.

14.9 Review of Multiple Regression We described many topics involving multiple regression in this chapter. In this sec- tion of the chapter, we focus on a single example with a solution that reviews the procedure and guides your application of multiple regression analysis.

Example The Bank of New England is a large financial institution serving the New England states as well as New York and New Jersey. The mortgage department of the Bank of New England is studying data from recent loans. Of particular interest is how such factors as the value of the home being purchased ($000), education level of the head of the household (number of years, beginning with first grade), age of the head of the household, current monthly mortgage payment (in dollars), and gender of the head of the household (male � 1, female � 0) relate to the family income. The mortgage department would like to know whether these variables are effective predictors of family income.

Solution To begin, consider a random sample of 25 loan applications submitted to the Bank of New England last month. A portion of the sample information is shown in Table 14–3. The entire data set is available at the website (www.mhhe.com/lind15e) and is iden- tified as Bank of New England.

Next, we develop a correlation matrix. It will show the relationship between each of the independent variables and the dependent variable. This will help identify the

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Multiple Regression Analysis 547

Loan Income ($000) Value ($000) Education Age Mortgage Gender

1 100.7 190 14 53 230 1 2 99.0 121 15 49 370 1 3 102.0 161 14 44 397 1

23 102.3 163 14 46 142 1 24 100.2 150 15 50 343 0 25 96.3 139 14 45 373 0

ooooooo

Income ($000) Value ($000) Education Age Mortgage Gender

Income ($000) 1.0000 Value ($000) 0.7197 1.0000 Education 0.1880 �0.1437 1.0000 Age 0.2426 0.2195 0.6209 1.0000 Mortgage 0.1157 0.3579 �0.2103 �0.0379 1.0000 Gender 0.4856 0.1841 0.0619 0.1558 �0.1290 1.0000

TABLE 14–3 Information on Sample of 25 Loans by the Bank of New England

independent variables that are more closely related to the dependent variable (family income.) The correlation matrix will also reveal independent variables that are highly correlated and possibly redundant. The correlation matrix is below.

What can we learn from this correlation matrix?

1. The family income is strongly related to the value of the home. There is also a moderate correlation between the gender of the person seeking the loan and family income. These two correlations are highlighted in yellow in the correla- tion matrix.

2. The amount of the mortgage has a weak correlation with family income. This correlation is indentified in red.

3. All possible correlations among the independent variables are identified in blue type. Our standard is to look for correlations that exceed an absolute value of .700. None of the independent variables are strongly correlated with each other. This indicates that multicollinearity is not likely.

Next, we compute the multiple regression equation using all the independent variables. The software output follows.

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548 Chapter 14

The coefficients of determination, that is, both R2 and adjusted R2, are reported at the top of the summary output and highlighted in yellow. The R2 value is 75.0 percent, so the five independent variables account for three-quarters of the varia- tion in family income. The adjusted R2 measures the strength of the relationship between the set of independent variables and family income and also accounts for the number of variables in the regression equation. The adjusted R2 indicates that the five variables account for 68.4 percent of the variance of family income. Both of these suggest that the proposed independent variables are useful in predicting family income.

The output also includes the regression equation.

Be careful in this interpretation. Both income and the value of the home are in thou- sands of dollars. Here is a summary:

1. An increase of $1,000 dollars in the value of the home suggests an increase of $70 in family income. An increase of one year of education increases income by $1,620, another year older reduces income by $120, and an increase of $1,000 in the mortgage reduces income by $1.

2. If a male is head of the household, the value of family income will increase by $1,807. Remember that “female” was coded 0 and “male” was coded 1, so a male head of household is positively related to home value.

3. The age of the head of household and monthly mortgage payment are inversely related to family income. This is true because the sign of the regression coef- ficient is negative.

Next we conduct the global hypothesis test. Here we check to see if any of the regression coefficients are different from 0. We use the .05 significance level.

H1: Not all the �s are 0.

The p-value from the table (cell F12) is 0.000. Because the p-value is less than the significance level, we reject the null hypothesis and conclude that at least one of the regression coefficients is not equal to zero.

Next we evaluate the individual regression coefficients. Refer to software output p-values to test each regression coefficient. They are reported in cells E18 to E22. The null hypothesis and the alternate hypothesis are:

The subscript i represents any particular independent variable. Again using .05 sig- nificance levels, the p-values for the regression coefficients for home value, years of education, and gender are all less than .05. We conclude that these regression coefficients are not equal to zero and are significant predictors of family income. For age and mortgage amount, the p-values are greater than .05, the significance level, so we do not reject the null hypothesis. The regression coefficients for these two variables are not different from zero and are not related to family income.

Based on the results of testing each of the regression coefficients, we conclude that the variables age and mortgage amount are not effective predictors of family income. Thus, they should be removed from the multiple regression equation. Remem- ber that we must remove one independent variable at a time and redo the analysis to evaluate the overall effect of removing the variable. Our strategy is to remove the variable with the smallest t-statistic or the largest p-value. This variable is mortgage amount. The result of the regression analysis without the mortgage variable follows.

H1: �i Z 0 H0: �i � 0

H0: �1 � �2 � �3 � �4 � �5 � 0

� .001(Mortgage) � 1.807(Gender) Ŷ � 70.61 � .07(Value) � 1.62(Education) � 0.12(Age)

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Multiple Regression Analysis 549

Observe that the R2 and adjusted R2 change very little without the mortgage variable. Also observe that the p-value associated with age is greater than the .05 sig- nificance level. So next we remove the age variable and redo the analysis. The regression output with the variables age and mortgage amount removed follows.

From this output, we conclude:

1. The R2 and adjusted R2 values have declined but only slightly. Using all five independent variables, the R2 value was .750. With the two nonsignificant vari- ables removed, the R2 and adjusted R2 values are .716 and .676, respectively. We prefer the equation with the fewer number of independent variables. It is easier to interpret.

2. In ANOVA, we observe that the p-value is less than .05. Hence, at least one of the regression coefficients is not equal to zero.

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3. Reviewing the significance of the individual coefficients, the p-values associated with each of the remaining independent variables are less than .05. We con- clude that all the regression coefficients are different from zero. Each independent variable is a useful predictor of family income.

Our final step is to examine the regression assumptions, listed in Section 14.5 beginning on page 531, with our regression model. The first assumption is that there is a linear relationship between each independent variable and the dependent vari- able. It is not necessary to review the dummy variable Gender, because there are only two possible outcomes. Below are the scatter plots of family income versus home value and family income versus years of education.

The scatter plot of income versus home value shows a general increasing trend. As the home value increases, so does family income. The points appear to be linear. That is, there is no observable nonlinear pattern in the data. The scatter plot on the right, of income versus years of education, shows that the data are measured to the nearest year. The measurement is to the nearest year and is a discrete variable. Given the measurement method, it is difficult to make an observation that the rela- tionship is linear.

A plot of the residuals is also useful to evaluate the overall assumption of linearity. Recall that a residual is (Y � ), the difference between the actual value of the depen-Ŷ dent variable (Y ) and the predicted value of the dependent variable ( ). Assuming a lin- ear relationship, the distribution of the residuals should show about an equal proportion of negative residuals (points above the line) and positive residuals (points below the line) centered on zero. There should be no observable pattern to the plots. The graph follows.

Ŷ

550 Chapter 14

There is no discernable pattern to the plot, so we conclude that the linearity assumption is reasonable.

If the linearity assumption is valid, then the distribution of residuals should fol- low the normal probability distribution with a mean of zero. To evaluate this assump- tion, we will use a histogram and a normal probability plot.

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Multiple Regression Analysis 551

In general, the histogram on the left shows the major characteristics of a normal distribution, that is, a majority of observations in the middle and centered on the mean of zero, with lower frequencies in the tails of the distribution. The normal prob- ability plot on the right is based on a cumulative normal probability distribution. The blue line shows the standardized cumulative normal distribution. The red points show the cumulative distribution of the residuals. To confirm the normal distribution of the residuals, the red dots should be close to the blue line. This is true for most of the plot. However, we would note that there are departures and even perhaps a nonlin- ear pattern in the residuals in the lower part of the graph. As before, we are looking for serious departures from linearity and these are not indicated in these graphs.

The final assumption refers to multicollinearity. This means that the independent variables should not be highly correlated. We suggested a rule of thumb that mul- ticollinearity would be a concern if the correlations among independent variables were close to 0.7 or �0.7. There are no violations of this guideline.

There is a statistical test to more precisely evaluate multicollinearity, the variance inflation factor (VIF). We use Minitab to calculate the VIF’s below. The standard is that the VIF should be less than 10. Note that all the VIFs are clearly less than 10, so multicollinearity is not a concern.

To summarize, the multiple regression equation is

This equation explains 71.6 percent of the variation in family income. There are no major departures from the multiple regression assumptions of linearity, normally dis- tributed residuals, and multicollinearity.

Ŷ � 74.527 � .0634(Value) � 1.0158(Education) � 1.7697(Gender)

The regression equation is Income ($000) � 74.5 � 0.0634 Value ($000) � 1.02 Education � 1.77 Gender

Predictor Coef SE Coef T P VIF Constant 74.527 6.870 10.85 0.000 Value ($000) 0.06336 0.01092 5.80 0.000 1.062 Education 1.0158 0.4492 2.26 0.034 1.030 Gender 1.7697 0.6163 2.87 0.009 1.044

Chapter Summary I. The general form of a multiple regression equation is:

[14–1]

where a is the Y-intercept when all X’s are zero, refers to the sample regression coefficients, and refers to the value of the various independent variables.Xj

Ŷ � a � b1X1 � b2X2 � . . . � bkXk

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552 Chapter 14

A. There can be any number of independent variables. B. The least squares criterion is used to develop the regression equation. C. A statistical software package is needed to perform the calculations.

II. An ANOVA table summarizes the multiple regression analysis. A. It reports the total amount of the variation in the dependent variable and divides this

variation into that explained by the set of independent variables and that not explained. B. It reports the degrees of freedom associated with the independent variables, the error

variation, and the total variation. III. There are two measures of the effectiveness of the regression equation.

A. The multiple standard error of estimate is similar to the standard deviation. 1. It is measured in the same units as the dependent variable. 2. It is based on squared deviations from the regression equation. 3. It ranges from 0 to plus infinity. 4. It is calculated from the following equation.

[14–2]

B. The coefficient of multiple determination reports the percent of the variation in the dependent variable explained by the set of independent variables. 1. It may range from 0 to 1. 2. It is also based on squared deviations from the regression equation. 3. It is found by the following equation.

[14–3]

4. When the number of independent variables is large, we adjust the coefficient of determination for the degrees of freedom as follows.

[14–4]

IV. A global test is used to investigate whether any of the independent variables have significant regression coefficients. A. The null hypothesis is: All the regression coefficients are zero. B. The alternate hypothesis is: At least one regression coefficient is not zero. C. The test statistic is the F distribution with k (the number of independent variables)

degrees of freedom in the numerator and degrees of freedom in the denominator, where n is the sample size.

D. The formula to calculate the value of the test statistic for the global test is:

[14–5]

V. The test for individual variables determines which independent variables have nonzero regression coefficients. A. The variables that have zero regression coefficients are usually dropped from the

analysis. B. The test statistic is the t distribution with degrees of freedom. C. The formula to calculate the value of the test statistic for the individual test is:

[14–6]

VI. There are five assumptions to use multiple regression analysis. A. The relationship between the dependent variable and the set of independent vari-

ables must be linear. 1. To verify this assumption, develop a scatter diagram and plot the residuals on the

vertical axis and the fitted values on the horizontal axis. 2. If the plots appear random, we conclude the relationship is linear.

B. The variation is the same for both large and small values of 1. Homoscedasticity means the variation is the same for all fitted values of the

dependent variable.

Ŷ.

t � bi � 0

sbi

n � (k � 1)

F � SSR�k

SSE�[n � (k � 1)]

n � (k � 1)

R2adj � 1 �

SSE n � (k � 1)

SS total n � 1

R2 � SSR

SS total

sY.123...k � B

©(Y � Ŷ )2

n � (k � 1)

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2. This condition is checked by developing a scatter diagram with the residuals on the vertical axis and the fitted values on the horizontal axis.

3. If there is no pattern to the plots—that is, they appear random—the residuals meet the homoscedasticity requirement.

C. The residuals follow the normal probability distribution. 1. This condition is checked by developing a histogram of the residuals to see if they

follow a normal distribution. 2. The mean of the distribution of the residuals is 0.

D. The independent variables are not correlated. 1. A correlation matrix will show all possible correlations among independent vari-

ables. Signs of trouble are correlations larger than 0.70 or less than 2. Signs of correlated independent variables include when an important predictor

variable is found insignificant, when an obvious reversal occurs in signs in one or more of the independent variables, or when a variable is removed from the solu- tion, there is a large change in the regression coefficients.

3. The variance inflation factor is used to identify correlated independent variables.

[14–7]

E. Each residual is independent of other residuals. 1. Autocorrelation occurs when successive residuals are correlated. 2. When autocorrelation exists, the value of the standard error will be biased and will

return poor results for tests of hypothesis regarding the regression coefficients. VII. Several techniques help build a regression model.

A. A dummy or qualitative independent variable can assume one of two possible outcomes. 1. A value of 1 is assigned to one outcome and 0 the other. 2. Use formula (14–6) to determine if the dummy variable should remain in the equation.

B. Interaction is the case in which one independent variable (such as affects the rela- tionship with another independent variable and the dependent variable (Y ).

C. Stepwise regression is a step-by-step process to find the regression equation. 1. Only independent variables with nonzero regression coefficients enter the equation. 2. Independent variables are added one at a time to the regression equation.

Pronunciation Key SYMBOL MEANING PRONUNCIATION

Regression coefficient for the first b sub 1 independent variable

Regression coefficient for any b sub k independent variable

Multiple standard error of estimate s sub Y dot 1, 2, 3 . . . k

Chapter Exercises 13. A multiple regression equation yields the following partial results.

sY.123...k

(X1) X2)

VIF � 1

1 � R2j

�0.70.

Source Sum of Squares df

Regression 750 4 Error 500 35

a. What is the total sample size? b. How many independent variables are being considered? c. Compute the coefficient of determination. d. Compute the standard error of estimate. e. Test the hypothesis that none of the regression coefficients is equal to zero. Let

� � .05.

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554 Chapter 14

14. In a multiple regression equation, two independent variables are considered, and the sam- ple size is 25. The regression coefficients and the standard errors are as follows.

Conduct a test of hypothesis to determine whether either independent variable has a coefficient equal to zero. Would you consider deleting either variable from the regression equation? Use the .05 significance level.

15. The following output was obtained.

sb2 � 0.71b2 � �0.880

sb1 � 0.56b1 � 2.676

a. What is the sample size? b. Compute the value of c. Compute the multiple standard error of estimate. d. Conduct a global test of hypothesis to determine whether any of the regression coef-

ficients are significant. Use the .05 significance level. e. Test the regression coefficients individually. Would you consider omitting any vari-

able(s)? If so, which one(s)? Use the .05 significance level. 16. In a multiple regression equation, k � 5 and the MSE value is 5.10, and SS total

is 519.68. At the .05 significance level, can we conclude that any of the regression coef- ficients are not equal to 0?

17. The district manager of Jasons, a large discount electronics chain, is investigating why certain stores in her region are performing better than others. She believes that three fac- tors are related to total sales: the number of competitors in the region, the population in the surrounding area, and the amount spent on advertising. From her district, consisting of several hundred stores, she selects a random sample of 30 stores. For each store, she gathered the following information.

sales last year (in $ thousands) of competitors in the region

of the region (in millions) expense (in $ thousands)

The sample data were run on Minitab, with the following results.

X3 � advertising X2 � population X1 � number Y � total

n � 20,

R2.

Analysis of variance

SOURCE DF SS MS Regression 5 100 20 Error 20 40 2 Total 25 140

Predictor Coef StDev t-ratio Constant 3.00 1.50 2.00

X1 4.00 3.00 1.33 X2 3.00 0.20 15.00 X3 0.20 0.05 4.00 X4 �2.50 1.00 �2.50 X5 3.00 4.00 0.75

Analysis of variance

SOURCE DF SS MS Regression 3 3050.00 1016.67 Error 26 2200.00 84.62 Total 29 5250.00

Predictor Coef StDev t-ratio Constant 14.00 7.00 2.00

X1 �1.00 0.70 �1.43 X2 30.00 5.20 5.77 X3 0.20 0.08 2.50

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a. What are the estimated sales for the Bryne store, which has four competitors, a regional population of 0.4 (400,000), and an advertising expense of 30 ($30,000)?

b. Compute the value. c. Compute the multiple standard error of estimate. d. Conduct a global test of hypothesis to determine whether any of the regression coef-

ficients are not equal to zero. Use the .05 level of significance. e. Conduct tests of hypotheses to determine which of the independent variables have

significant regression coefficients. Which variables would you consider eliminating? Use the .05 significance level.

18. Suppose that the sales manager of a large automotive parts distributor wants to esti- mate as early as April the total annual sales of a region. On the basis of regional sales, the total sales for the company can also be estimated. If, based on past experience, it is found that the April estimates of annual sales are reasonably accurate, then in future years the April forecast could be used to revise production schedules and maintain the correct inventory at the retail outlets.

Several factors appear to be related to sales, including the number of retail outlets in the region stocking the company’s parts, the number of automobiles in the region reg- istered as of April 1, and the total personal income for the first quarter of the year. Five independent variables were finally selected as being the most important (according to the sales manager). Then the data were gathered for a recent year. The total annual sales for that year for each region were also recorded. Note in the following table that for region 1 there were 1,739 retail outlets stocking the company’s automotive parts, there were 9,270,000 registered automobiles in the region as of April 1, and so on. The sales for that year were $37,702,000.

Number of Average Annual Number of Automobiles Personal Age of Sales Retail Registered Income Automobiles Number of

($ millions), Outlets, (millions), ($ billions), (years), Supervisors, Y X1 X2 X3 X4 X5

37.702 1,739 9.27 85.4 3.5 9.0 24.196 1,221 5.86 60.7 5.0 5.0 32.055 1,846 8.81 68.1 4.4 7.0

3.611 120 3.81 20.2 4.0 5.0 17.625 1,096 10.31 33.8 3.5 7.0 45.919 2,290 11.62 95.1 4.1 13.0 29.600 1,687 8.96 69.3 4.1 15.0

8.114 241 6.28 16.3 5.9 11.0 20.116 649 7.77 34.9 5.5 16.0 12.994 1,427 10.92 15.1 4.1 10.0

a. Consider the following correlation matrix. Which single variable has the strongest cor- relation with the dependent variable? The correlations between the independent vari- ables outlets and income and between cars and outlets are fairly strong. Could this be a problem? What is this condition called?

sales outlets cars income age outlets 0.899 cars 0.605 0.775 income 0.964 0.825 0.409 age �0.323 �0.489 �0.447 �0.349 bosses 0.286 0.183 0.395 0.155 0.291

b. The output for all five variables is on the following page. What percent of the varia- tion is explained by the regression equation?

Multiple Regression Analysis 555

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556 Chapter 14

The regression equation is Sales � �19.7 � 0.00063 outlets � 1.74 cars � 0.410 income

� 2.04 age � 0.034 bosses

Predictor Coef SE Coef T P Constant �19.672 5.422 �3.63 0.022 outlets �0.000629 0.002638 �0.24 0.823 cars 1.7399 0.5530 3.15 0.035 income 0.40994 0.04385 9.35 0.001 age 2.0357 0.8779 2.32 0.081 bosses �0.0344 0.1880 �0.18 0.864

Analysis of Variance SOURCE DF SS MS F P Regression 5 1593.81 318.76 140.36 0.000 Residual Error 4 9.08 2.27 Total 9 1602.89

c. Conduct a global test of hypothesis to determine whether any of the regression coef- ficients are not zero. Use the .05 significance level.

d. Conduct a test of hypothesis on each of the independent variables. Would you con- sider eliminating “outlets” and “bosses”? Use the .05 significance level.

e. The regression has been rerun below with “outlets” and “bosses” eliminated. Compute the coefficient of determination. How much has changed from the previous analysis?R2

The regression equation is Sales � �18.9 � 1.61 cars � 0.400 income � 1.96 age

Predictor Coef SE Coef T P Constant �18.924 3.636 �5.20 0.002 cars 1.6129 0.1979 8.15 0.000 income 0.40031 0.01569 25.52 0.000 age 1.9637 0.5846 3.36 0.015

Analysis of Variance SOURCE DF SS MS F P Regression 3 1593.66 531.22 345.25 0.000 Residual Error 6 9.23 1.54 Total 9 1602.89

g. Following is a plot of the fitted values of Y (i.e., and the residuals. Do you see any violations of the assumptions?

Ŷ )

f. Following is a histogram and a stem-and-leaf chart of the residuals. Does the nor- mality assumption appear reasonable?

Histogram of residual N � 10 Stem-and-leaf of residual N � 10 Leaf Unit � 0.10

Midpoint Count �1.5 1 * 1 �1 7 �1.0 1 * 2 �1 2 �0.5 2 ** 2 �0 �0.0 2 ** 5 �0 440 0.5 2 ** 5 0 24 1.0 1 * 3 0 68 1.5 1 * 1 1

1 1 7

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a. Consider the following correlation matrix. Which variable has the strongest correlation with the dependent variable? Some of the correlations among the independent vari- ables are strong. Does this appear to be a problem?

b. Consider the following output. Compute the coefficient of multiple determination.

19. The administrator of a new paralegal program at Seagate Technical College wants to esti- mate the grade point average in the new program. He thought that high school GPA, the verbal score on the Scholastic Aptitude Test (SAT), and the mathematics score on the SAT would be good predictors of paralegal GPA. The data on nine students are:

High School SAT SAT Paralegal Student GPA Verbal Math GPA

1 3.25 480 410 3.21 2 1.80 290 270 1.68 3 2.89 420 410 3.58 4 3.81 500 600 3.92 5 3.13 500 490 3.00 6 2.81 430 460 2.82 7 2.20 320 490 1.65 8 2.14 530 480 2.30 9 2.63 469 440 2.33

legal gpa verbal gpa 0.911 verbal 0.616 0.609 math 0.487 0.636 0.599

1.2

–1.2

8 16 24 32 40

Re si

du al

(Y –

Y )

Fitted ^

The regression equation is Legal � �0.411 � 1.20 GPA � 0.00163 Verbal � 0.00194 Math

Predictor Coef SE Coef T P Constant �0.4111 0.7823 �0.53 0.622 GPA 1.2014 0.2955 4.07 0.010 Verbal 0.001629 0.002147 0.76 0.482 Math �0.001939 0.002074 �0.94 0.393

Analysis of Variance SOURCE DF SS MS F P Regression 3 4.3595 1.4532 10.33 0.014 Residual Error 5 0.7036 0.1407 Total 8 5.0631

SOURCE DF Seq SS GPA 1 4.2061 Verbal 1 0.0303 Math 1 0.1231

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558 Chapter 14

f. Following are a histogram and a stem-and-leaf diagram of the residuals. Does the nor- mality assumption for the residuals seem reasonable?

The regression equation is Legal � �0.454 � 1.16 GPA

Predictor Coef SE Coef T P Constant �0.4542 0.5542 �0.82 0.439 GPA 1.1589 0.1977 5.86 0.001

Analysis of Variance SOURCE DF SS MS F P Regression 1 4.2061 4.2061 34.35 0.001 Residual Error 7 0.8570 0.1224 Total 8 5.0631

Histogram of residual N � 9

Midpoint Count �0.4 1 * �0.2 3 *** 0.0 3 *** 0.2 1 * 0.4 0 0.6 1 *

Stem-and-leaf of residual N � 9 Leaf unit � 0.10 1 �0 4 2 �0 2

(3) �0 110 4 0 00 2 0 1 0 1 0 6

g. Following is a plot of the residuals and the values. Do you see any violation of the assumptions?

Ŷ

0.70

0.35

0.00

–0.35

1.50 2.00 2.50 3.00 3.50 4.00

Re si

du al

s (Y

– Y

)

Ŷ

c. Conduct a global test of hypothesis from the preceding output. Does it appear that any of the regression coefficients are not equal to zero?

d. Conduct a test of hypothesis on each independent variable. Would you consider elimi- nating the variables “verbal” and “math”? Let

e. The analysis has been rerun without “verbal” and “math.” See the following output. Compute the coefficient of determination. How much has changed from the pre- vious analysis?

� � .05.

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Salary Years of Principal’s Master’s ($ thousands), Experience, Rating, Degree,*

Y X1 X2 X3

31.1 8 35 0 33.6 5 43 0 29.3 2 51 1

30.7 4 62 0 32.8 2 80 1 42.8 8 72 0

oooo

*1 � yes, 0 � no.

a. Develop a correlation matrix. Which independent variable has the strongest correla- tion with the dependent variable? Does it appear there will be any problems with multicollinearity?

b. Determine the regression equation. What salary would you estimate for a teacher with five years’ experience, a rating by the principal of 60, and no master’s degree?

c. Conduct a global test of hypothesis to determine whether any of the regression coef- ficients differ from zero. Use the .05 significance level.

d. Conduct a test of hypothesis for the individual regression coefficients. Would you con- sider deleting any of the independent variables? Use the .05 significance level.

e. If your conclusion in part (d) was to delete one or more independent variables, run the analysis again without those variables.

f. Determine the residuals for the equation of part (e). Use a stem-and-leaf chart or a histogram to verify that the distribution of the residuals is approximately normal.

g. Plot the residuals computed in part (f) in a scatter diagram with the residuals on the Y-axis and the values on the X-axis. Does the plot reveal any violations of the assumptions of regression?

21. A consumer analyst collected the following data on the screen sizes of popular LCD tele- visions sold recently at a large retailer:

Ŷ

Manufacturer Screen Price

Sharp 46 $1473.00 Samsung 52 2300.00 Samsung 46 1790.00 Sony 40 1250.00 Sharp 42 1546.50 Samsung 46 1922.50 Samsung 40 1372.00 Sharp 37 1149.50 Sharp 46 2000.00 Sony 40 1444.50 Sony 52 2615.00 Samsung 32 747.50

Manufacturer Screen Price

Sharp 37 $1314.50 Sharp 32 853.50 Sharp 52 2778.00 Samsung 40 1749.50 Sharp 32 1035.00 Samsung 52 2950.00 Sony 40 1908.50 Sony 52 3103.00 Sony 46 2606.00 Sony 46 2861.00 Sony 52 3434.00

20. Mike Wilde is president of the teachers’ union for Otsego School District. In preparing for upcoming negotiations, he would like to investigate the salary structure of classroom teachers in the district. He believes there are three factors that affect a teacher’s salary: years of experience, a rating of teaching effectiveness given by the principal, and whether the teacher has a master’s degree. A random sample of 20 teachers resulted in the fol- lowing data.

Multiple Regression Analysis 559

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560 Chapter 14

a. Does there appear to be a linear relationship between the screen size and the price? b. Which variable is the “dependent” variable? c. Using statistical software, determine the regression equation. Interpret the value of the

slope in the regression equation. d. Include the manufacturer in a multiple linear regression analysis using a “dummy”

variable. Does it appear that some manufacturers can command a premium price? Hint: You will need to use a set of indicator variables.

e. Test each of the individual coefficients to see if they are significant. f. Make a plot of the residuals and comment on whether they appear to follow a nor-

mal distribution. g. Plot the residuals versus the fitted values. Do they seem to have the same amount of

variation? 22. A regional planner is studying the demographics in a region of a particular state. She has

gathered the following data on nine counties.

a. Is there a linear relationship between the median income and median age? b. Which variable is the “dependent” variable? c. Use statistical software to determine the regression equation. Interpret the value of

the slope in a simple regression equation. d. Include the aspect that the county is “coastal” or not in a multiple linear regression

analysis using a “dummy” variable. Does it appear to be a significant influence on incomes?

e. Test each of the individual coefficients to see if they are significant. f. Make a plot of the residuals and comment on whether they appear to follow a nor-

mal distribution. g. Plot the residuals versus the fitted values. Do they seem to have the same amount of

variation? 23. Great Plains Roofing and Siding Company Inc. sells roofing and siding products to home

repair retailers, such as Lowe’s and Home Depot, and commercial contractors. The owner is interested in studying the effects of several variables on the value of shingles sold ($000). The marketing manager is arguing that the company should spend more money on adver- tising, while a market researcher suggests it should focus more on making its brand and product more distinct from its competitors.

The company has divided the United States into 26 marketing districts. In each dis- trict, it collected information on the following variables: volume of sales (in thousands of dollars), advertising dollars (in thousands), number of active accounts, number of compet- ing brands, and a rating of district potential.

Advertising Sales Dollars Number of Number of Market (000s) (000s) Accounts Competitors Potential

79.3 5.5 31 10 8 200.1 2.5 55 8 6 163.2 8.0 67 12 9

(continued )

County Median Income Median Age Coastal

A $48,157 57.7 1 B 48,568 60.7 1 C 46,816 47.9 1 D 34,876 38.4 0 E 35,478 42.8 0 F 34,465 35.4 0 G 35,026 39.5 0 H 38,599 65.6 0 J 33,315 27.0 0

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Advertising Sales Dollars Number of Number of Market (000s) (000s) Accounts Competitors Potential

200.1 3.0 50 7 16 146.0 3.0 38 8 15 177.7 2.9 71 12 17

93.5 4.2 26 8 3 259.0 4.5 75 8 19 331.2 5.6 71 4 9

ooooo

Conduct a multiple regression analysis to find the best predictors of sales. a. Draw a scatter diagram comparing sales volume with each of the independent vari-

ables. Comment on the results. b. Develop a correlation matrix. Do you see any problems? Does it appear there are any

redundant independent variables? c. Develop a regression equation. Conduct the global test. Can we conclude that some

of the independent variables are useful in explaining the variation in the dependent variable?

d. Conduct a test of each of the independent variables. Are there any that should be dropped?

e. Refine the regression equation so the remaining variables are all significant. f. Develop a histogram of the residuals and a normal probability plot. Are there any

problems? g. Determine the variance inflation factor for each of the independent variables. Are there

any problems? 24. The Times-Observer is a daily newspaper in Metro City. Like many city newspapers, the

Times-Observer is suffering through difficult financial times. The circulation manager is studying other papers in similar cities in the United States and Canada. She is particu- larly interested in what variables relate to the number of subscriptions to the paper. She is able to obtain the following sample information on 25 newspapers in similar cities. The following notation is used:

Income � The median family income in the metropolitan area (in $ thousands) Adv � The advertising budget of the paper (in $ hundreds)

Popul � The metropolitan population (in thousands) Sub � Number of subscriptions (in thousands)

Paper Sub Popul Adv Income

1 37.95 588.9 13.2 35.1 2 37.66 585.3 13.2 34.7 3 37.55 566.3 19.8 34.8

23 38.83 629.6 22.0 35.3 24 38.33 680.0 24.2 34.7 25 40.24 651.2 33.0 35.8

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a. Determine the regression equation. b. Conduct a global test of hypothesis to determine whether any of the regression coef-

ficients are not equal to zero. c. Conduct a test for the individual coefficients. Would you consider deleting any

coefficients? d. Determine the residuals and plot them against the fitted values. Do you see any

problems? e. Develop a histogram of the residuals. Do you see any problems with the normality

assumption?

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562 Chapter 14

Family Food Income Size

1 $5.04 $ 73.98 4 2 4.08 54.90 2 3 5.76 94.14 4

23 4.56 38.16 3 24 5.40 43.74 7 25 4.80 48.42 5

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a. Determine the regression equation, using salary as the dependent variable and the other four variables as independent variables.

b. What is the value of Comment on this value. c. Conduct a global test of hypothesis to determine whether any of the independent vari-

ables are different from 0. d. Conduct an individual test to determine whether any of the independent variables can

be dropped. e. Rerun the regression equation, using only the independent variables that are signifi-

cant. How much more does a man earn per month than a woman? Does it make a difference whether the employee has a technical or a clerical job?

26. Many regions along the coast in North and South Carolina and Georgia have experienced rapid population growth over the last 10 years. It is expected that the growth will continue over the next 10 years. This has motivated many of the large grocery store chains to build new stores in the region. The Kelley’s Super Grocery Stores Inc. chain is no exception. The director of planning for Kelley’s Super Grocery Stores wants to study adding more stores in this region. He believes there are two main factors that indicate the amount fam- ilies spend on groceries. The first is their income and the other is the number of people in the family. The director gathered the following sample information.

R2?

25. Fred G. Hire is the manager of human resources at Crescent Tool and Die Inc. As part of his yearly report to the CEO, he is required to present an analysis of the salaried employ- ees. Because there are over 1,000 employees, he does not have the staff to gather infor- mation on each salaried employee, so he selects a random sample of 30. For each employee, he records monthly salary; service at Crescent, in months; gender

and whether the employee has a technical or clerical job. Those working technical jobs are coded 1, and those who are clerical 0. female);

(1 � male, 0 �

Food and income are reported in thousands of dollars per year, and the variable size refers to the number of people in the household. a. Develop a correlation matrix. Do you see any problems with multicollinearity? b. Determine the regression equation. Discuss the regression equation. How much does

an additional family member add to the amount spent on food? c. What is the value of Can we conclude that this value is greater than 0? d. Would you consider deleting either of the independent variables? e. Plot the residuals in a histogram. Is there any problem with the normality assumption? f. Plot the fitted values against the residuals. Does this plot indicate any problems with

homoscedasticity?

R2?

Sampled Monthly Length of Employee Salary Service Age Gender Job

1 $1,769 93 42 1 0 2 1,740 104 33 1 0 3 1,941 104 42 1 1

28 1,791 131 56 0 1 29 2,001 95 30 1 1 30 1,874 98 47 1 0

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27. An investment advisor is studying the relationship between a common stock’s price to earnings (P/E) ratio and factors that she thinks would influence it. She has the following data on the earnings per share (EPS) and the dividend percentage (Yield) for a sample of 20 stocks.

a. Develop a multiple regression equation with the amount of tips as the dependent vari- able and the amount of the bill and the number of diners as independent variables. Write out the regression equation. How much does another diner add to the amount of the tips?

b. Conduct a global test of hypothesis to determine if at least one of the independent variables is significant. What is your conclusion?

c. Conduct an individual test on each of the variables. Should one or the other be deleted?

d. Use the equation developed in part (c) to determine the coefficient of determination. Interpret the value.

e. Plot the residuals. Is it reasonable to assume they follow the normal distribution? f. Plot the residuals against the fitted values. Is it reasonable to conclude they are

random?

Stock P/E EPS Yield

1 20.79 $2.46 1.42 2 3.03 2.69 4.05 3 44.46 �0.28 4.16

18 30.21 1.71 3.07 19 32.88 0.35 2.21 20 15.19 5.02 3.50

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a. Develop a multiple linear regression with P/E as the dependent variable. b. Are either of the two independent variables an effective predictor of P/E? c. Interpret the regression coefficients. d. Do any of these stocks look particularly undervalued? e. Plot the residuals and check the normality assumption. Plot the fitted values against

the residuals. f. Does there appear to be any problems with homoscedasticity? g. Develop a correlation matrix. Do any of the correlations indicate multicollinearity?

28. The Conch Café, located in Gulf Shores, Alabama, features casual lunches with a great view of the Gulf of Mexico. To accommodate the increase in business during the summer vacation season, Fuzzy Conch, the owner, hires a large number of servers as seasonal help. When he interviews a prospective server, he would like to provide data on the amount a server can earn in tips. He believes that the amount of the bill and the number of diners are both related to the amount of the tip. He gathered the following sample information.

Amount Amount Number of Customer of Tip of Bill Diners

1 $7.00 $48.97 5 2 4.50 28.23 4 3 1.00 10.65 1

28 2.50 26.25 2 29 9.25 56.81 5 30 8.25 50.65 5

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564 Chapter 14

29. The president of Blitz Sales Enterprises sells kitchen products through television com- mercials, often called infomercials. He gathered data from the last 15 weeks of sales to determine the relationship between sales and the number of infomercials.

Infomercials Sales ($000s) Infomercials Sales ($000s)

20 3.2 22 2.5 15 2.6 15 2.4 25 3.4 25 3.0 10 1.8 16 2.7 18 2.2 12 2.0 18 2.4 20 2.6 15 2.4 25 2.8 12 1.5

a. Determine the regression equation. Are the sales predictable from the number of commercials?

b. Determine the residuals and plot a histogram. Does the normality assumption seem reasonable?

30. The director of special events for Sun City believed that the amount of money spent on fireworks displays on the 4th of July was predictive of attendance at the Fall Festival held in October. She gathered the following data to test her suspicion.

4th of July ($000) Fall Festival (000) 4th of July ($000) Fall Festival (000)

10.6 8.8 9.0 9.5 8.5 6.4 10.0 9.8

12.5 10.8 7.5 6.6 9.0 10.2 10.0 10.1 5.5 6.0 6.0 6.1

12.0 11.1 12.0 11.3 8.0 7.5 10.5 8.8 7.5 8.4

Determine the regression equation. Is the amount spent on fireworks related to atten- dance at the Fall Festival? Conduct a hypothesis test to determine if there is a problem with autocorrelation.

31. You are a new hire at Laurel Woods Real Estate which specializes in selling foreclosed homes via public auction. Your boss has asked you to use the following data (mortgage balance, monthly payments, payments made before default, and final auction price) on a random sample of recent sales in order to estimate what the actual auction price will be.

Monthly Payments Auction Loan Payments Made Price

$ 85,600 $ 985.87 1 $16,900 115,300 902.56 33 75,800 103,100 736.28 6 43,900

119,400 1021.23 58 69,000 90,600 836.46 3 35,600

104,500 1056.37 22 63,000

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a. Carry out a global test of hypothesis to verify if any of the regression coefficients are different from zero.

b. Do an individual test of the independent variables. Would you remove any of the variables?

c. If it seems one or more of the independent variables is not needed, remove it and work out the revised regression equation.

32. Think about the figures from the previous exercise. Add a new variable that describes the potential interaction between the loan amount and the number of payments made. Then do a test of hypothesis to check if the interaction is significant.

Data Set Exercises 33. Refer to the Real Estate data, which report information on homes sold in Goodyear, Arizona.

Use the selling price of the home as the dependent variable and determine the regres- sion equation with number of bedrooms, size of the house, center of the city, and num- ber of bathrooms as independent variables. a. Use a statistical software package to determine the multiple regression equation. Dis-

cuss each of the variables. For example, are you surprised that the regression coeffi- cient for distance from the center of the city is negative? How much does a garage or a swimming pool add to the selling price of a home?

b. Determine the value of the Intercept. c. Develop a correlation matrix. Which independent variables have strong or weak corre-

lations with the dependent variable? Do you see any problems with multicollinearity? d. Conduct the global test on the set of independent variables. Interpret. e. Conduct a test of hypothesis on each of the independent variables. Would you con-

sider deleting any of the variables? If so, which ones? f. Rerun the analysis until only significant regression coefficients remain in the analysis.

Identify these variables. g. Develop a histogram or a stem-and-leaf display of the residuals from the final regres-

sion equation developed in part (f). Is it reasonable to conclude that the normality assumption has been met?

h. Plot the residuals against the fitted values from the final regression equation developed in part (f). Plot the residuals on the vertical axis and the fitted values on the hori- zontal axis.

34. Refer to the Baseball 2009 data, which report information on the 30 Major League Base- ball teams for the 2009 season. Let the number of games won be the dependent variable and the following variables be independent variables: team batting average, number of stolen bases, number of errors committed, team ERA, number of home runs, and whether the team plays in the American or the National League. a. Use a statistical software package to determine the multiple regression equation. Dis-

cuss each of the variables. For example, are you surprised that the regression coeffi- cient for ERA is negative? Is the number of wins affected by whether the team plays in the National or the American League?

b. Find the coefficient of determination for this set of independent variables. c. Develop a correlation matrix. Which independent variables have strong or weak corre-

lations with the dependent variable? Do you see any problems with multicollinearity? d. Conduct a global test on the set of independent variables. Interpret. e. Conduct a test of hypothesis on each of the independent variables. Would you con-

sider deleting any of the variables? If so, which ones? f. Rerun the analysis until only significant net regression coefficients remain in the analy-

sis. Identify these variables. g. Develop a histogram or a stem-and-leaf display of the residuals from the final regres-

sion equation developed in part (f). Is it reasonable to conclude that the normality assumption has been met?

h. Plot the residuals against the fitted values from the final regression equation developed in part (f). Plot the residuals on the vertical axis and the fitted values on the hori- zontal axis.

35. Refer to the Buena School District bus data. First, add a variable to change the type of bus (diesel or gasoline) to a qualitative variable. If the bus type is diesel, then set the qual- itative variable to 0. If the bus type is gasoline, then set the qualitative variable to 1. Develop

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566 Chapter 14

a regression equation using statistical software with maintenance as the dependent vari- able and age, miles, and bus type as the independent variables. a. Write out the multiple regression equation analysis. Discuss each of the variables. b. Determine the value of R2. Interpret. c. Develop a correlation matrix. Which independent variables have strong or weak corre-

sider deleting any of the variables? If so, which ones? f. Rerun the analysis until only significant regression coefficients remain in the analysis.

Identify these variables. g. Develop a histogram or a stem-and-leaf display of the residuals from the final regres-

sion equation developed in part (f). Is it reasonable to conclude that the normality assumption has been met?

h. Plot the residuals against the fitted values from the final regression equation developed in part (f) against the fitted values of Y. Plot the residuals on the vertical axis and the fitted values on the horizontal axis.

Note: We do not show steps for all the statistical soft- ware used in this chapter. Below are the first two, which show the basic steps.

1. The Minitab commands for the multiple regression output on page 516 are: a. Import the data from the CD. The file name is

Tbl14–1. b. Select Stat, Regression, and then click on

Regression. c. Select Cost as the Response variable, and

Temp, Insul, and Age as the Predictors, then click on OK.

Software Commands

2. The Excel commands to produce the multiple regression output on page 516 are: a. Import the data from the CD. The file name is

Tbl14. b. Select the Data tab on the top menu. Then on

the far right, select Data analysis. Select Regression and click OK.

c. Make the Input Y Range A1:A21, the Input X Range B1:D21, check the Labels box, the Out- put Range is G1, then click OK.

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A Review of Chapters 13 and 14 567

Chapter 14 Answers to Self-Review

14–1 a. $389,500 or 389.5 (in $000); found by

b. The of 4 shows profit will go up $4,000 for each extra hour the restaurant is open (if none of the other variables change). The of implies profit will fall $3,000 for each added mile away from the central area (if none of the other variables change).

14–2 a. The total degrees of freedom is 25. So the sample size is 26.

b. There are 5 independent variables. c. There is only 1 dependent variable (profit). d. found by . Ninety-five

percent of the residuals will be between �2.828 and 2.828, found by

e. found by 100�140. 71.4% of the deviation in profit is accounted for by these five variables.

f. found by

14–3 a. Not all of the �’s are 0.

The decision rule is to reject if The computed value of F is 10, found by 20/2. So, you reject which indicates at least one of the regression coefficients is different from zero.

Based on p-values, the decision rule is to reject the null hypothesis if the p-value is less than 0.05. The computed value of F is 10, found by 20/2, and has a p-value of 0.000. Thus, we reject the null hypothesis, which indicates that at least one of the regression coefficients is different from zero.

b. For variable 1: and The decision rule is: Reject if or

Since 2.000 does not go beyond either of those limits, we fail to reject the null hypothesis. This regression coefficient could be zero. We can consider dropping this variable. By parallel logic the null hypothesis is rejected for variables 3 and 4.

For variable 1, the decision rule is to reject H0: �1 � 0 if the p-value is less than 0.05. Because the p-value is 0.056, we cannot reject the null hypothesis. This regression coefficient could be zero. Therefore, we can consider dropping this variable. By parallel logic, we reject the null hypothesis for variables 3 and 4.

c. We should consider dropping variables 1, 2, and 5. Variable 5 has the smallest absolute value of t or largest p-value. So delete it first and refigure the regression analysis.

14–4 a.

b. Female agents make $3,860 more than male agents.

df � 17, reject H0 if or

The t-statistic exceeds the critical value of 2.110. Also, the p-value � 0.0179 and is less than 0.05. Reject H0. Gender should be included in the regression equation.

t � 3.8598 � 0

1.4724 � 2.621

t 7 2.110t 6 �2.110 �3 � 0H1: �3 � 0H0:

� 32.87 Ŷ � 15.7625 � 0.4415(30) � 3.8598(1) Ŷ � 15.7625 � 0.4415X1 � 3.8598X2

t 7 2.086. t 6 �2.086H0

�1 � 0H1:�1 � 0H0:

H0,

F 7 2.71.H0 H1:

�1 � �2 � �3 � �4 � �5 � 0H0:

1 � c 40

(26 � (5 � 1)) d / c 140(26 � 1) d

R2adj � .643,

R2 � .714, �2(1.414).

12SY.12345 � 1.414,

(n � 1)

�3b3

b2 � 3895 2.5 � 3(40) � 4(72) � 3(10) � .2(20) � 1(5)

A Review of Chapters 13 and 14 This section is a review of the major concepts and terms introduced in Chapters 13 and 14. Chapter 13 noted that the strength of the relationship between the independent variable and the dependent variable can be measured by the correlation coefficient. The correlation coef- ficient is designated by the letter r. It can assume any value between and inclu- sive. Coefficients of and indicate a perfect relationship, and 0 indicates no rela- tionship. A value near 0, such as or .14, indicates a weak relationship. A value near or such as or �.90, indicates a strong relationship. Squaring r gives the coefficient of determination, also called It indicates the proportion of the total variation in the depen- dent variable explained by the independent variable.

Likewise, the strength of the relationship between several independent variables and a dependent variable is measured by the coefficient of multiple determination, It measures the proportion of the variation in Y explained by two or more independent variables.

R2.

r 2. �.90�1,

�1�.14 �1.00�1.00

�1.00�1.00

Simple regression and correlation examine the relationship between two variables.

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568 Chapter 14

The linear relationship in the simple case involving one independent variable and one dependent variable is described by the equation For three independent variables,Ŷ � a � bx.

and the same multiple regression equation is

Solving for . . . , would involve tedious calculations. Fortunately, this type of problem can be quickly solved using one of the many statistical software packages and spreadsheet packages. Various measures, such as the coefficient of determination, the mul- tiple standard error of estimate, the results of the global test, and the test of the individual variables, are reported in the output of most computer software programs.

Glossary

bkb2,b1,

Ŷ � a � b1X1 � b2X2 � . . . � bkXk

X3,X2,X1,

Chapter 13 Correlation coefficient A measure of the strength of as- sociation between two variables. Coefficient of determination The proportion of the total variation in the dependent variable that is explained by the in- dependent variable. It can assume any value between 0 and

inclusive. This coefficient is computed by squaring the correlation coefficient, r. Correlation analysis A group of statistical techniques used to measure the strength of the relationship between two variables. Dependent variable The variable that is being predicted or estimated. Independent variable A variable that provides the basis for estimation. Least squares method A technique used to arrive at the regression equation by minimizing the sum of the squares of the vertical distances between the actual Y values and the predicted Y values. Linear regression equation A mathematical equation that defines the relationship between two variables. It has the form It is used to predict Y based on a selected X value. Y is the dependent variable and X the in- dependent variable. Scatter diagram A chart that visually depicts the rela- tionship between two variables. Standard error of estimate Measures the dispersion of the actual Y values about the regression line. It is reported in the same units as the dependent variable. t test of significance of r A formula to answer the ques- tion: Is the correlation in the population from which the sample was selected zero? The test statistic is t, and the number of degrees of freedom is

[13–2]

Chapter 14 Autocorrelation Correlation of successive residuals. This condition frequently occurs when time is involved in the analysis.

t � r1n � 2 11 � r2

n � 2.

Ŷ � a � bX.

�1.00

Correlation matrix A listing of all possible simple coeffi- cients of correlation. A correlation matrix includes the cor- relations between each of the independent variables and the dependent variable, as well as those among all the independent variables. Dummy variable A qualitative variable. It can assume only one of two possible outcomes. Global test A test used to determine if any of the set of independent variables have regression coefficients differ- ent from zero. Homoscedasticity The standard error of estimate is the same for all fitted values of the dependent variable. Individual test A test to determine if a particular inde- pendent variable has a regression coefficient different from zero. Interaction The case in which one independent variable (such as affects the relationship between another inde- pendent variable and the dependent variable (Y ). Multicollinearity A condition that occurs in multiple regression analysis if the independent variables are them- selves correlated. Multiple regression equation The relationship in the form of a mathematical equation between several inde- pendent variables and a dependent variable. The general form is It is used to estimate Y given k independent variables, Xi. Qualitative variables A nominal-scale variable that is coded to assume only one of two possible outcomes. For example, a person is considered either employed or unemployed. Residual The difference between the actual value of the dependent variable and the estimated value of the depen- dent variable, that is, Stepwise regression A step-by-step process for finding the regression equation. Only independent variables with nonzero regression coefficients enter the regression equa- tion. Independent variables are added one at a time to the regression equation. Variance inflation factor A test used to detect correla- tion among independent variables.

Y � Ŷ.

Ŷ � a � b1X1 � b2X2 � b3X3 � . . . � bkXk.

(X1) X2)

Multiple regression and correlation is concerned with the relationship between two or more independent variables and the dependent variable.

Computers are invaluable in multiple regression and correlation.

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Problems

1. The accounting department at Crate and Barrel wishes to estimate the profit for each of the chain’s many stores on the basis of the number of employees in the store, overhead costs, average markup, and theft loss. A few statistics from the stores are:

Net Number Overhead Average Theft Profit of Cost Markup Loss

Store ($ thousands) Employees ($ thousands) (percent) ($ thousands)

1 $846 143 $79 69% $52 2 513 110 64 50 45

a. The dependent variable is _________. b. The general equation for this problem is _________. c. The multiple regression equation was computed to be

What are the predicted sales for a store with 112 employees, an overhead cost of $65,000, a markup rate of 50 percent, and a loss from theft of $50,000?

d. Suppose was computed to be .86. Explain. e. Suppose that the multiple standard error of estimate was 3 (in $ thousands). Explain

what this means in this problem. 2. Quick-print firms in a large downtown business area spend most of their advertising dol-

lars on displays on bus benches. A research project involves predicting monthly sales based on the annual amount spent on placing ads on bus benches. A sample of quick- print firms revealed these advertising expenses and sales:

3X4. Ŷ � 67 � 8X1 � 10X2 � 0.004X3 �

Annual Bus Bench Monthly Advertising Sales

Firm ($ thousands) ($ thousands)

A 2 10 B 4 40 C 5 30 D 7 50 E 3 20

a. Draw a scatter diagram. b. Determine the correlation coefficient. c. What is the coefficient of determination? d. Compute the regression equation. e. Estimate the monthly sales of a quick-print firm that spends $4,500 on bus bench

advertisements. f. Summarize your findings.

3. The following ANOVA output is given.

SOURCE Sum of Squares DF MS Regression 1050.8 4 262.70 Error 83.8 20 4.19 Total 1134.6 24

Predictor Coef St.Dev. t-ratio Constant 70.06 2.13 32.89

X1 0.42 0.17 2.47 X2 0.27 0.21 1.29 X3 0.75 0.30 2.50 X4 0.42 0.07 6.00

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570 Chapter 14

Cases

A. The Century National Bank Refer to the Century National Bank data. Using checking account balance as the dependent variable and using as independent variables the number of ATM transactions, the number of other services used, whether the individual has a debit card, and whether interest is paid on the par- ticular account, write a report indicating which of the vari- ables seem related to the account balance and how well they explain the variation in account balances. Should all of the independent variables proposed be used in the analysis or can some be dropped?

B. Terry and Associates: The Time to Deliver Medical Kits

Terry and Associates is a specialized medical testing cen- ter in Denver, Colorado. One of the firm’s major sources of revenue is a kit used to test for elevated amounts of lead in the blood. Workers in auto body shops, those in the lawn care industry, and commercial house painters are exposed to large amounts of lead and thus must be ran- domly tested. It is expensive to conduct the test, so the kits are delivered on demand to a variety of locations throughout the Denver area.

Kathleen Terry, the owner, is concerned about setting appropriate costs for each delivery. To investigate, Ms. Terry gathered information on a random sample of 50 recent deliveries. Factors thought to be related to the cost of delivering a kit were:

Prep The time in minutes between when the cus- tomized order is phoned into the company and when it is ready for delivery.

Delivery The actual travel time in minutes from Terry’s plant to the customer.

Mileage The distance in miles from Terry’s plant to the customer.

Sample Number Cost Prep Delivery Mileage

1 $32.60 10 51 20 2 23.37 11 33 12 3 31.49 6 47 19 4 19.31 9 18 8 5 28.35 8 88 17 6 22.63 9 20 11 7 22.63 9 39 11 8 21.53 10 23 10 9 21.16 13 20 8

10 21.53 10 32 10 11 28.17 5 35 16

Sample Number Cost Prep Delivery Mileage

12 $20.42 7 23 9 13 21.53 9 21 10 14 27.55 7 37 16 15 23.37 9 25 12 16 17.10 15 15 6 17 27.06 13 34 15 18 15.99 8 13 4 19 17.96 12 12 4 20 25.22 6 41 14 21 24.29 3 28 13 22 22.76 4 26 10 23 28.17 9 54 16 24 19.68 7 18 8 25 25.15 6 50 13 26 20.36 9 19 7 27 21.16 3 19 8 28 25.95 10 45 14 29 18.76 12 12 5 30 18.76 8 16 5 31 24.29 7 35 13 32 19.56 2 12 6 33 22.63 8 30 11 34 21.16 5 13 8 35 21.16 11 20 8 36 19.68 5 19 8 37 18.76 5 14 7 38 17.96 5 11 4 39 23.37 10 25 12 40 25.22 6 32 14 41 27.06 8 44 16 42 21.96 9 28 9 43 22.63 8 31 11 44 19.68 7 19 8 45 22.76 8 28 10 46 21.96 13 18 9 47 25.95 10 32 14 48 26.14 8 44 15 49 24.29 8 34 13 50 24.35 3 33 12

1. Develop a multiple linear regression equation that describes the relationship between the cost of delivery and the other variables. Do these three variables explain a reasonable amount of the variation in the

a. Compute the coefficient of determination. b. Compute the multiple standard error of estimate. c. Conduct a test of hypothesis to determine whether any of the regression coefficients

are different from zero. d. Conduct a test of hypothesis on the individual regression coefficients. Can any of the

variables be deleted?

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A Review of Chapters 13 and 14 571

dependent variable? Estimate the delivery cost for a kit that takes 10 minutes for preparation, takes 30 minutes to deliver, and must cover a distance of 14 miles.

2. Test to determine that at least one net regression coef- ficient differs from zero. Also test to see whether any of

the variables can be dropped from the analysis. If some of the variables can be dropped, rerun the regression equation until only significant variables are included.

3. Write a brief report interpreting the final regression equation.

Practice Test

Part 1—Objective 1. In a scatter diagram, the dependent variable is always scaled on which axis? 1. 2. What level of measurement is required to compute the correlation coefficient? 2. 3. If there is no correlation between two variables, what is the value of the correlation coefficient?

3. 4. Which of the following values indicates the strongest correlation between two variables? (0.65, �0.77, 0, �.12)

4. 5. Under what conditions will the coefficient of determination assume a value greater than 1? 5.

Given the following regression equation, and that the coefficient of determination is 0.81, answer questions 7, 8, and 9.

6. At what point does the regression equation cross the Y-axis? 6. 7. An increase of 1 unit in the independent variable will result in what amount of an increase or decrease in the depen-

dent variable? 7. 8. What is the correlation coefficient? (Be careful of the sign.) 8. 9. If all the data points in a scatter diagram were on the regression line, what would be the value of the standard error

of estimate? 9. 10. In a multiple regression equation, what is the maximum number of independent variables allowed? (2, 10, 30, unlimited)

10. 11. In multiple regression analysis, we assume what type of relationship between the dependent variable and the set of

independent variables? (linear, multiple, curved, none of these) 11. 12. The difference between Y and is called a . 12. 13. For a particular dummy variable, such as gender, how many different outcomes are possible?

13. 14. What is the term given to a table that shows all possible correlation coefficients between the dependent variable and

all the independent variables and among all the independent variables? 14. 15. If there is a linear relationship between the dependent variable and the set of independent variables, a graph of the

residuals will show what type of distribution? 15.

Part 2—Problems 1. Given the following software output:

Ŷ

Ŷ � 7 � .5X,

ANOVA Table

Source SS df MS F p-value

Regression 129.7275 1 129.7275 14.50 .0007 Residual 250.4391 28 8.9443

Total 380.1667 29

Regression Output

Variables Coefficients Standard Error t (df � 28)

Intercept 90.6190 1.5322 59.141 Slope �0.9401 0.2468 �3.808

Regression Analysis

a. What is the sample size? b. Write out the regression equation. Interpret the slope and intercept values. c. If the value of the independent variable is 10, what is the value of the dependent variable?

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572 Chapter 14

d. Calculate the coefficient of determination. Interpret this value. e. Calculate the correlation coefficient. Conduct a test of hypothesis to determine if there is a significant nega-

tive association between the variables. 2. Given the following software output.

a. How large is the sample size? b. How many independent variables are in the study? c. Determine the coefficient of determination. d. Conduct a global test of hypothesis. Can you conclude at least one of the independent variables does not

equal zero? Use the .01 significance level. e. Conduct an individual test of hypothesis on each of the independent variables. Would you consider dropping

any of the independent variables? If so, which variable or variables would you drop? Use the .01 significance level.

ANOVA Table

Source SS df MS F p-value

Regression 227.0928 4 56.7732 9.27 0.000 Residual 153.0739 25 6.1230

Total 380.1667 29

Regression Output

Variables Coefficients Standard Error t (df � 25) p-value

Intercept 68.3366 8.9752 7.614 0.000 X1 0.8595 0.3087 2.784 0.010 X2 �0.3380 0.8381 �0.403 0.690 X3 �0.8179 0.2749 �2.975 0.006 X4 �0.5824 0.2541 �2.292 0.030

Regression Analysis

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15 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Compute and interpret a simple index.

LO2 Describe the difference between a weighted and an unweighted index.

LO3 Compute and interpret a Laspeyres price index.

LO4 Compute and interpret a Paasche price index.

LO5 Compute and interpret a value index.

LO6 Explain how the Consumer Price Index is constructed and interpreted.

Index Numbers

Information on prices and quantities for margarine, shortening, milk, and

potato chips for the years 2000 and 2009 is provided in Exercise 27.

Compute a simple price index for each of the four items, using 2000

as the base period. (See Exercise 27 and LO1.)

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15.1 Introduction In this chapter, we will examine a useful descriptive tool called an index. An index expresses the relative change in a value from one period to another. No doubt you are familiar with indexes such as the Consumer Price Index, which is released monthly by the U.S. Department of Labor. There are many other indexes, such as the Dow Jones Industrial Average (DJIA), Nasdaq, NIKKEI 225, and Standard & Poor’s 500 Stock Average. Indexes are published on a regular basis by the fed- eral government, by business publications such as BusinessWeek and Forbes, in most daily newspapers, and on the Internet.

Of what importance is an index? Why is the Consumer Price Index so important and so widely reported? As the name implies, it measures the change in the price of a large group of items consumers purchase. The Federal Reserve Board, consumer groups, unions, management, senior citizens orga- nizations, and others in business and economics are very con- cerned about changes in prices. These groups closely monitor the Consumer Price Index as well as the Producer Price Index, which measures price fluctuations at all stages of pro- duction. To combat sharp price increases, the Federal Reserve often raises the interest rate to “cool down” the economy. Likewise, the Dow Jones Industrial Average, which is updated continuously during the business day, describes the overall change in common stock prices of 30 large companies.

A few stock market indexes appear daily in the financial section of most newspapers. Many are reported in real time such as in the business section of USA Today’s website (http://www.usatoday.com/money/default.htm). Shown below are the Dow Jones Industrial Average, Nasdaq, and S&P 500 from the USA Today website.

15.2 Simple Index Numbers What is an index number? An index or index number measures the change in a par- ticular item (typically a product or service) between two time periods.

INDEX NUMBER A number that expresses the relative change in price, quantity, or value compared to a base period.

If the index number is used to measure the relative change in just one variable, such as hourly wages in manufacturing, we refer to this as a simple index. It is the ratio of two variables converted to a percentage. The following four examples will serve to illustrate the use of index numbers. As noted in the definition, the main use of an index number in business is to show the percent change in one or more items from one time period to another.

LO1 Compute and interpret a simple index.

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Index Numbers 575

Example According to the Bureau of Labor Statistics, in 2000 the average hourly earnings of production workers was $14.02. In 2009, it was $18.62. What is the index of hourly earnings of production workers for 2009 based on 2000 data?

Solution It is 132.81, found by:

Thus, the hourly earnings in 2009 compared to 2000 were 132.81 percent. This means there was a 32.81 percent increase in hourly earnings during the period, found by 132.81 � 100.0 � 32.81.

P � Average hourly earnings in 2009 Average hourly earnings in 2000 (100) �

$18.62 $14.02

(100) � 132.81

Example An index can also compare one item with another. The population of the Canadian province of British Columbia in 2010 was 4,494,232, and for Ontario it was 13,069,182. What is the population index of British Columbia compared to Ontario?

Solution The index of population for British Columbia is 34.4, found by:

This indicates that the population of British Columbia is 34.4 percent (about one- third) of the population of Ontario, or the population of British Columbia is 65.6 per- cent less than the population of Ontario (100 � 34.4 � 65.6).

P � Population of British Columbia

Population of Ontario (100) � 4,494,232 13,069,182

(100) � 34.4

You can check the latest information on wages, the Consumer Price Indexes, and other business-related values at the Bureau of Labor Statistics (BLS) website, http://www.bls.gov. The following chart shows some statistics from the BLS.

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Example The following Excel output shows the number of passengers (in millions) for the five busiest airports in the United States in 2009. What is the index for Atlanta, Chicago, Los Angeles, and Dallas/Ft. Worth compared to Denver?

Solution To find the four indexes, we divide the passengers for Atlanta, Chicago, Los Angeles, and Dallas/Ft. Worth by the number at Denver. We conclude that Atlanta had 75.1 percent more passengers than Denver, Chicago 34.8 percent more, Los Angeles 15.8 percent more, and Dallas/Ft. Worth 10.9 percent more.

Number of Passengers Airport (millions) Index Found by

Hartsfield-Jackson Atlanta International Airport (ATL) 90.0 175.1 (90.0�51.4)(100)

O’Hare International Airport (ORD) 69.3 134.8 (69.3�51.4)(100) Los Angeles

International Airport (LAX) 59.5 115.8 (59.5�51.4)(100) Dallas-Fort Worth

International Airport (DFW) 57.0 110.9 (57.0�51.4)(100) Denver International Airport (DEN) 51.4 100.0 (51.4�51.4)(100)

Note from the previous discussion that:

1. The index of average hourly earnings of production workers (132.81) is a per- centage, but the percent symbol is usually omitted.

2. Each index has a base period. In the example regarding the average hourly earn- ings of production workers, we used 2000 as the base period. The base period for the Consumer Price Index is 1993–95. The parity ratio, which is the ratio of the prices received by farmers to the prices paid by farmers, still has 1910–14 as the base period.

3. Most business and economic indexes are computed to the nearest whole num- ber, such as 214 or 96, or to the nearest tenth of a percent, such as 83.4 or 118.7.

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Index Numbers 577

15.3 Why Convert Data to Indexes? Compiling index numbers is not a recent innovation. An Italian, G. R. Carli, is credited with originating index numbers in 1764. They were incorporated in a report he made regarding price fluctuations in Europe from 1500 to 1750. No systematic approach to collecting and reporting data in index form was evident in the United States until about 1900. The cost-of-living index (now called the Consumer Price Index) was introduced in 1913, and a long list of indexes has been compiled since then.

Why convert data to indexes? An index is a convenient way to express a change in a diverse group of items. The Consumer Price Index (CPI), for example, encompasses about 400 items—including golf balls, lawn mowers, hamburgers, funeral services, and dentists’ fees. Prices are expressed in dollars per pound, box, yard, and many other different units. Only by converting the prices of these many diverse goods and services to one index number can the federal government and others concerned with inflation keep informed of the overall movement of consumer prices.

Converting data to indexes also makes it easier to assess the trend in a series composed of exceptionally large numbers. For example, the estimate of U.S. retail e-commerce sales for the fourth quarter of 2010, adjusted for seasonal variation, was $36,200,000. e-Commerce sales for the fourth quarter of 2009 was $30,700,000. This is an increase of $5,500,000. If e-commerce sales for the fourth quarter of 2010 are expressed as an index based on the fourth quarter of 2009, the increase is 17.9 percent.

15.4 Construction of Index Numbers We already discussed the construction of a simple price index. The price in a selected year (such as 2010) is divided by the price in the base year. The base- period price is designated as p0, and a price other than the base period is often referred to as the given period or selected period and designated pt. To calculate the simple price index P using 100 as the base value for any given period, use the formula:

4th qtr 2010 e-commerce sales 4th qtr 2009 e-commerce sales (100) �

$36,200,000 $30,700,000

(100) � 117.9

Indexes allow us to express a change in price, quantity, or value as a percent.

SIMPLE INDEX [15–1]P � pt p0

� 100

Suppose the price of a fall weekend package (including lodging and all meals) at Tryon Mountain Lodge in western North Carolina in 2000 was $450. The price rose to $795 in 2010. What is the price index for 2010 using 2000 as the base period and 100 as the base value? It is 176.7, found by:

Interpreting this result, the price of the fall weekend package increased 76.7 per- cent from 2000 to 2010.

The base period need not be a single year. Note in Table 15–1 that if we use 2000–01 � 100, the base price for the stapler would be $21 [found by deter- mining the mean price of 2000 and 2001, ($20 � $22)�2 � $21]. The prices $20, $22, and $23 are averaged if 2000–02 is selected as the base. The mean price would be $21.67. The indexes constructed using the three different base periods

P � pt p0

(100) � $795 $450

(100) � 176.7

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578 Chapter 15

are presented in Table 15–1. (Note that when 2000–02 � 100, the index numbers for 2000, 2001, and 2002 average 100.0, as we would expect.) Logically, the index numbers for 2010 using the three different bases are not the same.

Price of Price Index Price Index Price Index Year Stapler (2000 � 100) (2000–01 � 100) (2000–02 � 100)

1995 $18 90.0

2000 20 100.0

2001 22 110.0

2002 23 115.0

2010 38 190.0 38

21.67 � 100 � 175.4

38 21

� 100 � 181.0

23 21.67

� 100 � 106.1 23 21

� 100 � 109.5

22 21.67

� 100 � 101.5 22 21

� 100 � 104.8

20 21.67

� 100 � 92.3 20 21

� 100 � 95.2

18 21.67

� 100 � 83.1 18 21

� 100 � 85.7

TABLE 15–1 Prices of a Benson Automatic Stapler, Model 3, Converted to Indexes Using Three Different Base Periods

Self-Review 15–1 1. Listed below are the top steel-producing regions for the year 2009. Express the amount produced by China, the European Union, Japan, and Russia as an index, using the United States as a base. What percent more steel does China produce than the United States?

2. The average hourly earnings of production workers for January of selected years are given below.

(a) Using 1995 as the base period and 100 as the base value, determine the indexes for the other years. Interpret the index.

(b) Use the average of 1995 and 2000 as the base and determine indexes for the other years. Interpret the index.

Year Average Hourly Earnings

1995 $11.65 2000 14.02 2005 16.13 2010 (May) 19.01

Amount Region (millions of tons)

People’s Republic of China 500.5 European Union 198.0 Japan 118.7 United States 91.4 Russia 68.5

Exercises 1. PNC Bank Inc., which has its headquarters in Pittsburgh, reported $17,446 (million) in

commercial loans in 1995, $19,989 in 1997, $21,468 in 1999, $21,685 in 2000, $15,922 in 2002, $18,375 for 2004, and $54,818 in 2009. Using 1995 as the base, develop a simple

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Index Numbers 579

index for the change in the amount of commercial loans for the years 1997, 1999, 2000, 2002, 2004, and 2009, based on 1995.

2. The table below reports the earnings per share of common stock for Home Depot Inc. for recent years. Develop an index, with 2001 as the base, for the change in earnings per share over the period.

Year Earnings per Share

2001 $1.29 2002 1.56 2003 1.88 2004 2.26 2005 2.72

Year Earnings per Share

2006 $2.63 2007 2.55 2008 2.27 2009 0.71 2010 1.70

Year Sales (millions) Year Sales (millions)

1997 $486.6 2002 $568.5 1998 506.8 2003 581.9 1999 522.2 2004 496.1 2000 574.6 2005 456.6 2001 580.7 2006 433.3

3. Listed below are the net sales for Blair Corporation, a mail-order retailer located in Warren, Pennsylvania, for the years 1997 to 2006. In 2007, Blair became a subsidiary of Appleseed’s Topco. Its website is www.blair.com. Use the mean sales for the earliest three years to determine a base and then find the index for 2003 and 2006. By how much have net sales increased from the base period?

4. In January 1994, the price for a whole fresh chicken was $0.899 per pound. In April 2010, the price for the same chicken was $1.230 per pound. Use the January 1994 price as the base period and 100 as the base value to develop a simple index. By what percent has the cost of chicken increased?

15.5 Unweighted Indexes In many situations, we wish to combine several items and develop an index to com- pare the cost of this aggregation of items in two different time periods. For exam- ple, we might be interested in an index for items that relate to the expense of oper- ating and maintaining an automobile. The items in the index might include tires, oil changes, and gasoline prices. Or we might be interested in a college student index. This index might include the cost of books, tuition, housing, meals, and entertain- ment. There are several ways we can combine the items to determine the index.

Simple Average of the Price Indexes Table 15–2 on the next page reports the prices for several food items for the years 1999 and 2009. We would like to develop an index for this group of food items for 2009, using 1999 as the base. This is written in the abbreviated code 1999 � 100.

We could begin by computing a simple average of the price indexes for each item, using 1999 as the base year and 2009 as the given year. The simple index for bread is 147.1, found by using formula (15–1).

We compute the simple index for the other items in Table 15–2 similarly. The largest price increase was for eggs, 106.7 percent, and bread was second with 47.1 percent.

P � pt p0

(100) � 1.28 .87

(100) � 147.1

LO2 Describe the difference between a weighted and an unweighted index.

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The price of coffee increased 7.3 percent, found by 107.3 � 100 � 7.3. Then it would be natural to average the simple indexes. The formula is:

SIMPLE AVERAGE OF THE PRICE RELATIVES [15–2]

where Pi refers to the simple index for each of the items and n the number of items. In our example, the index is 145.5, found by:

This indicates that the mean of the group of indexes increased 45.5 percent from 1999 to 2009.

A positive feature of the simple average of price indexes is that we would obtain the same value for the index regardless of the units of measure. In the above index, if apples were priced in tons, instead of pounds, the impact of apples on the com- bined index would not change. That is, the commodity “apples” represents one of six items in the index, so the impact of the item is not related to the units. A nega- tive feature of this index is that it fails to consider the relative importance of the items included in the index. For example, milk and eggs receive the same weight, even though a typical family might spend far more over the year on milk than on eggs.

Simple Aggregate Index A second possibility is to sum the prices (rather than the indexes) for the two peri- ods and then determine the index based on the totals. The formula is:

SIMPLE AGGREGATE INDEX [15–3]

This is called a simple aggregate index. The index for the above food items is found by summing the prices in 1999 and 2009. The sum of the prices for the base period is $10.90 and for the given period it is $14.70. The simple aggregate index is 134.9. This means that the aggregate group of prices had increased 34.9 percent in the 10-year period.

Because the value of a simple aggregate index can be influenced by the units of measurement, it is not used frequently. In our example, the value of the index would differ significantly if we were to report the price of apples in tons rather than pounds.

P � ©pt ©p0

(100) � $14.70 $10.90

(100) � 134.9

P � ©pt ©p0

� 100

P � ©Pi n

� 147.1 � . . . � 107.3

6 �

872.7 6

� 145.5

P � ©Pi n

TABLE 15–2 Computation of Index for Food Price 2009, 1999 � 100

Item 1999 Price 2009 Price Simple Index

Bread, white, cost per pound $ 0.87 $ 1.28 147.1 Eggs, dozen 1.05 2.17 206.7 Milk, gallon, white 2.94 3.87 131.6 Apples, Red Delicious, 1 pound 0.86 1.16 134.9 Orange Juice, 12 oz concentrate 1.75 2.54 145.1 Coffee, 100% ground roast, 1 pound 3.43 3.68 107.3

Total $10.90 $14.70

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Example The prices for the six food items from Table 15–2 are repeated below in Table 15–3. Also included is the number of units of each consumed by a typical family in 1999 and 2009.

Determine a weighted price index using the Laspeyres method. Interpret the result.

Solution First we determine the total amount spent for the six items in the base period, 1999. To find this value, we multiply the base period price for bread ($0.87) by the base period quantity of 50. The result is $43.50. This indicates that a total of $43.50 was spent in the base period on bread. We continue that for all items and total the results. The base period total is $507.64. The current period total is computed in a similar fashion. For the first item, bread, we multiply the quantity in 1999 by the

Index Numbers 581

1999 1999 2009 2009 Item Price Quantity Price Quantity

Bread, white, cost per pound $0.87 50 $1.28 55 Eggs, dozen 1.05 26 2.17 20 Milk, gallon, white 2.94 102 3.87 130 Apples, Red Delicious, 1 pound 0.86 30 1.16 40 Orange Juice, 12 oz concentrate 1.75 40 2.54 41 Coffee, 100% ground roast, 1 pound 3.43 12 3.68 12

TABLE 15–3 Price and Quantity of Food Items in 1999 and 2009

Also, note the effect of coffee on the total index. For both the current year and the base year, coffee is a significant contributor to the total index, so a change in the price of coffee will drive the index much more than any other item. Therefore, we need a way to appropriately “weight” the items according to their relative importance.

15.6 Weighted Indexes Two methods of computing a weighted price index are the Laspeyres method and the Paasche method. They differ only in the period used for weighting. The Laspeyres method uses base-period weights; that is, the original prices and quan- tities of the items bought are used to find the percent change over a period of time in either price or quantity consumed, depending on the problem. The Paasche method uses current-year weights.

Laspeyres Price Index Etienne Laspeyres developed a method in the latter part of the 18th century to deter- mine a weighted price index using base-period quantities as weights. Applying his method, a weighted price index is computed by:

LASPEYRES PRICE INDEX [15–4]P � ©ptq0 ©p0q0

� 100

where: P is the price index. pt is the current price. p0 is the price in the base period. q0 is the quantity used in the base period.

LO3 Compute and interpret a Laspeyres price index.

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Paasche Price Index The major disadvantage of the Laspeyres index is it assumes that the base-period quantities are still realistic in the given period. That is, the quantities used for the six items are about the same in 1999 as 2009. In this case, notice that the quantity of eggs purchased declined by 23 percent, the quantity of milk increased by nearly 28 percent, and the quantity of apples increased by 33 percent.

The Paasche index is an alternative. The procedure is similar, but instead of using base-period quantities as weights, we use current-period quantities as weights. We use the sum of the products of the 1999 prices and the 2009 quantities. This has the advantage of using the more recent quantities. If there has been a change in the quan- tities consumed since the base period, such a change is reflected in the Paasche index.

The weighted price index for 2009 is 137.0, found by

Based on this analysis, we conclude that the price of this group of items has increased 37.0 percent in the ten-year period. The advantage of this method over the simple aggregate index is that the weight of each of the items is considered. In the simple aggregate index, coffee had about 40 percent of the weight in determining the index. In the Laspeyres index, the item with the most weight is milk, because the prod- uct of the price and the units sold is the largest.

P � ©pt q0 ©p0q0

(100) � $695.72 $507.64

(100) � 137.0

price of bread in 2009, that is, $1.28(50). The result is $64.00. We make the same calculation for each item and total the result. The total is $695.72. Because of the repetitive nature of these calculations, a spreadsheet is effective for carrying out the calculations. Following is a copy of the Excel output.

PAASCHE PRICE INDEX [15–5]P � ©ptqt ©p0qt

� 100

582 Chapter 15

LO4 Compute and interpret a Paasche price index.

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Index Numbers 583

Example Use the information from Table 15–3 to determine the Paasche index. Discuss which of the indexes should be used.

Solution Again, because of the repetitive nature of the calculations, Excel is used to perform the calculations. The results are shown in the following output.

The Paasche index is 135.6, found by

This result indicates that there has been an increase of 35.6 percent in the price of this “market basket” of goods between 1999 and 2009. That is, it costs 35.6 per- cent more to purchase these items in 2009 than it did in 1999. The Laspeyres index is more widely used because there are fewer pieces of data to update each period. The Consumer Price Index, the most widely reported index, is an example of a Laspeyres index.

P � ©ptqt ©p0qt

(100) � $811.60 $598.36

(100) � 135.6

How do we decide which index to use? When is Laspeyres most appropriate and when is Paasche the better choice?

Laspeyres Advantages Requires quantity data from only the base period. This allows

a more meaningful comparison over time. The changes in the index can be attributed to changes in the price.

Disadvantages Does not reflect changes in buying patterns over time. Also, it may overweight goods whose prices increase.

Paasche Advantages Because it uses quantities from the current period, it reflects

current buying habits. Disadvantages It requires quantity data for the current year. Because different

quantities are used each year, it is impossible to attribute changes in the index to changes in price alone. It tends to overweight the goods whose prices have declined. It requires the prices to be recomputed each year.

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Fisher’s Ideal Index As noted earlier, Laspeyres’ index tends to overweight goods whose prices have increased. Paasche’s index, on the other hand, tends to overweight goods whose prices have gone down. In an attempt to offset these shortcomings, Irving Fisher, in his book The Making of Index Numbers, published in 1922, proposed an index called Fisher’s ideal index. It is the geometric mean of the Laspeyres and Paasche indexes. We described the geometric mean in Chapter 3. It is determined by tak- ing the kth root of the product of k positive numbers.

Self-Review 15–2 An index of clothing prices for 2009 based on 2000 is to be constructed. The clothing items considered are shoes and dresses. The prices and quantities for both years are given below. Use 2000 as the base period and 100 as the base value.

2000 2009

Item Price Quantity Price Quantity

Dress (each) $75 500 $85 520 Shoes (pair) 40 1,200 45 1,300

(a) Determine the simple average of the price indexes. (b) Determine the aggregate price index for the two years. (c) Determine Laspeyres’ price index. (d) Determine the Paasche price index. (e) Determine Fisher’s ideal index.

[15–6]Fisher’s ideal index � 2(Laspeyres’ index)(Paasche’s index)

Fisher’s index seems to be theoretically ideal because it combines the best fea- tures of both Laspeyres’ and Paasche’s. That is, it balances the effects of the two indexes. However, it is rarely used in practice because it has the same basic set of problems as the Paasche index. It requires that a new set of quantities be deter- mined for each period.

Exercises For exercises 5–8: a. Determine the simple price indexes. d. Determine the Paasche b. Determine the simple aggregate price index for price index.

the two years. e. Determine Fisher’s ideal index. c. Determine Laspeyres’ price index.

584 Chapter 15

Example Determine Fisher’s ideal index for the data in Table 15–3.

Solution Fisher’s ideal index is 136.3.

� 1(137.0)(135.6) � 136.3

Fisher’s ideal index � 2(Laspeyres’ index)(Paasche’s index)

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Index Numbers 585

15.7 Value Index A value index measures changes in both the price and quantities involved. A value index, such as the index of department store sales, considers the base-year prices, the base-year quantities, the present-year prices, and the present-year quantities for its construction. Its formula is:

5. Below are the prices of toothpaste (9 oz), shampoo (7 oz), cough tablets (package of 100), and antiperspirant (2 oz) for August 2000 and August 2009. Also included are the quantity purchased. Use August 2000 as the base.

August 2000 August 2009

Item Price Quantity Price Quantity

Toothpaste $2.49 6 $3.35 6 Shampoo 3.29 4 4.49 5 Cough tablets 1.59 2 4.19 3 Antiperspirant 1.79 3 2.49 4

6. Fruit prices and the amounts consumed for 2000 and 2009 are below. Use 2000 as the base.

2000 2009

Fruit Price Quantity Price Quantity

Bananas (pound) $0.23 100 $0.69 120 Grapefruit (each) 0.29 50 1.00 55 Apples (pound) 0.35 85 1.89 85 Strawberries (basket) 1.02 8 3.79 10 Oranges (bag) 0.89 6 2.99 8

7. The prices and the numbers of various items produced by a small machine and stamp- ing plant are reported below. Use 2000 as the base.

2000 2009

Item Price Quantity Price Quantity

Washer $0.07 17,000 $0.10 20,000 Cotter pin 0.04 125,000 0.03 130,000 Stove bolt 0.15 40,000 0.15 42,000 Hex nut 0.08 62,000 0.10 65,000

8. Following are the quantities and prices for the years 2000 and 2009 for Kinzua Valley Geriatrics. Use 2000 as the base period.

Value index measures percent change in value

2000 2009

Item Price Quantity Price Quantity

Syringes (dozen) $ 6.10 1,500 $ 6.83 2,000 Thermometers 8.10 10 9.35 12 Advil (bottle) 4.00 250 4.62 250 Patient record forms (box) 6.00 1,000 6.85 900 Computer paper (box) 12.00 30 13.65 40

VALUE INDEX [15–7]V � ©pt qt ©p0 q0

� 100LO5 Compute and interpret a value index.

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Self-Review 15–3 The number of items produced by Houghton Products for 1996 and 2009 and the whole- sale prices for the two periods are:

Price Number Produced

Item Produced 1996 2009 1996 2009

Shear pins (box) $ 3 $4 10,000 9,000 Cutting compound

(pound) 1 5 600 200 Tie rods (each) 10 8 3,000 5,000

(a) Find the value index of production for 2009 using 1996 as the base period. (b) Interpret the index.

Exercises 9. The prices and production of grains for August 1995 and August 2009 are listed on the

following page.

586 Chapter 15

Example The prices and quantities sold at the Waleska Clothing Emporium for various items of apparel for May 2000 and May 2009 are:

Solution What is the index of value for May 2009 using May 2000 as the base period? Total sales in May 2009 were $10,600,000, and the comparable figure for 2000 is $9,000,000. (See Table 15–4.) Thus, the value index for May 2009 using 2000 � 100 is 117.8. The value of apparel sales in 2009 was 117.8 percent of the 2000 sales. To put it another way, the value of apparel sales increased 17.8 percent from May 2000 to May 2009.

V � ©pt qt ©p0 q0

(100) � $10,600,000 $9,000,000

(100) � 117.8

2000 2009 Quantity Quantity

2000 Sold 2009 Sold Price, (thousands), Price, (thousands),

Item p0 q0 pt qt Ties (each) $ 1 1,000 $ 2 900 Suits (each) 30 100 40 120 Shoes (pair) 10 500 8 500

TABLE 15–4 Construction of a Value Index for 2009 (2000 � 100)

2000 2009 Quantity Quantity

2000 Sold 2009 Sold Price, (thousands), p0q0 Price, (thousands), ptqt

Item p0 q0 ($ thousands) pt qt ($ thousands)

Ties (each) $ 1 1,000 $1,000 $ 2 900 $ 1,800 Suits (each) 30 100 3,000 40 120 4,800 Shoes (pair) 10 500 5,000 8 500 4,000

$9,000 $10,600

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Example The Seattle Chamber of Commerce wants to develop a measure of general busi- ness activity for the northwest portion of the United States. The director of eco- nomic development has been assigned to develop the index. It will be called the General Business Activity Index of the Northwest.

Solution After considerable thought and research, the director has concluded that four fac- tors should be considered: the regional department store sales (which are reported in $ millions), the regional employment index (which has a 2000 base and is reported by the State of Washington), the freight car loadings (reported in millions), and exports for the Seattle Harbor (reported in thousands of tons). Recent information on these variables is reported in Table 15–5.

TABLE 15–5 Data for the Computation of the General Business Activity Index of the Northwest

Department Index of Freight Car Year Store Sales Employment Loadings Exports

1999 20 100 50 500 2004 41 110 30 900 2009 44 125 18 700

Using 1995 as the base period, find the value index of grains produced for August 2009. 10. Johnson Wholesale Company manufactures a variety of products. The prices and quan-

tities produced for April 1994 and April 2009 are:

1995 2009 Quantity Quantity Produced Produced

1995 (millions of 2009 (millions of Grain Price bushels) Price bushels)

Oats $1.52 200 $5.95 214 Wheat 2.10 565 9.80 489 Corn 1.48 291 6.00 203 Barley 3.05 87 3.29 106

1994 2009 1994 2009 Quantity Quantity

Product Price Price Produced Produced

Small motor (each) $23.60 $28.80 1,760 4,259 Scrubbing compound (gallon) 2.96 3.08 86,450 62,949 Nails (pound) 0.40 0.48 9,460 22,370

Using April 1994 as the base period, find the value index of goods produced for April 2009.

15.8 Special-Purpose Indexes Many important indexes are prepared and published by private organizations. J. D. Power & Associates surveys automobile purchasers to determine how satisfied cus- tomers are with their vehicle after one year of ownership. This special index is called the Consumer Satisfaction Index. Financial institutions, utility companies, and uni- versity research centers often prepare indexes on employment, factory hours and wages, and retail sales for the regions they serve. Many trade associations prepare indexes of price and quantity that are vital to their particular area of interest. How are these special indexes prepared? An example, simplified of course, will help to explain some of the details.

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As we stated at the start of the section, there are many special-purpose indexes. Here are a few examples.

Consumer Price Index The U.S. Bureau of Labor Statistics reports this index monthly. It describes the changes in prices from one period to another for a “market basket” of goods and services. We discuss its history in detail and present some applications in the next section. You can access this information by going to www.bls.gov, then under Inflation & Prices select Consumer Price Index, then select CPI databases, then select All Urban Con- sumers (Current Series), Top Picks, and then click on U.S. all items 1982–84 � 100. You may elect to include different periods. Following is a recent summary report.

588 Chapter 15

The General Business Activity Index of the Northwest for 2004 is 157.0 and for 2009 it is 157.1. Interpreting, business activity has increased 57.0 percent from 1999 to 2004 and 57.1 percent from the base period of 1999 to 2009.

After review and consultation, the director assigned weights of 40 percent to depart- ment store sales, 30 percent to employment, 10 percent to freight car loadings, and 20 percent to exports.

To develop the General Business Activity Index of the Northwest for 2009 using 1999 � 100, each 2009 value is expressed as a percentage, with the base-period value as the denominator. For illustration, department store sales for 2009 are con- verted to a percentage by ($44�$20)(100) � 220. This means that department store sales have increased 120 percent in the period. This percentage is then multiplied by the appropriate weight. For the department store sales, this is (220)(.40) � 88.0. The details of the calculations for the years 2004 and 2009 are shown below.

2004 2009

Department store sales ($41�$20)(100)(.40) � 82.0 ($44�$20)(100)(.40) � 88.0 Employment (110�100)(100)(.30) � 33.0 (125�100)(100)(.30) � 37.5 Freight car loadings (30�50)(100)(.10) � 6.0 (18�50)(100)(.10) � 3.6 Exports (900�500)(100)(.20) � 36.0 (700�500)(100)(.20) � 28.0

Total 157.0 157.1

LO6 Explain how the Consumer Price Index is constructed and interpreted.

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Index Numbers 589

Dow Jones Industrial Average (DJIA) This is an index of stock prices, but perhaps it would be better to say it is an “indicator” rather than an index. It is supposed to be the mean price of 30 specific industrial stocks. However, summing the 30 stock prices and dividing by 30 does not calculate its value. This is because of stock splits, mergers, and stocks being added or dropped. When changes occur, adjustments are made in the denominator used with the average. Today the DJIA is more of a psychological indicator than a representation of the general price movement on the New York Stock Exchange. The lack of representativeness of the stocks on the DJIA is one of the reasons for the development of the New York Stock Exchange Index. This index was developed as an average price of all stocks on the New York Stock Exchange. You can find more information about the DJIA by going to the website: www.dowjones.com and select The Company, then select about Dow Jones; finally, under Enterprise Media Group, select Dow Jones Indexes. You can find its current value as well as the 30 stocks that are now a part of its calculation. The chart below summarizes the DJIA for one day. It can be located at the Merrill Lynch website: www.ml.com.

Producer Price Index Formerly called the Wholesale Price Index, it dates back to 1890 and is also pub- lished by the U.S. Bureau of Labor Statistics. It reflects the prices of over 3,400 com- modities. Price data are collected from the sellers of the commodities, and it usually refers to the first large-volume transaction for each commodity. It is a Laspeyres-type index. To access this information, go to www.bls.gov, then Inflation & Prices. Select Producer Price Indexes, then select PPI databases, then select Commodity Data, Top Picks, and finally select Finished Goods. You may select to include different periods. Below is a recent output.

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S&P 500 Index The full name of this index is the Standard and Poor’s Composite Index of Stock Prices. It is an aggregate price index of 500 common stocks. It, too, is probably a better reflection of the market than is the DJIA. You can access information about the S&P 500 from the Merrill Lynch website. Below is a recent summary.

There are many other indexes that track business and economic behavior, such as the Nasdaq, the Russell 2000, and the Wilshire 5000.

Self-Review 15–4 As an intern in the Fulton County Economic Development Office, you have been asked to develop a special-purpose index for your county. Three economic series seem to hold promise as the basis of an index. These data are the price of cotton (per pound), the number of new automobiles sold in the county, and the rate of money turnover (published by the local bank). After discussing the project with your supervisor and the director, you decide that money turnover should have a weight of .60, the number of new automobiles sold a weight of .30, and the cotton price .10. The base period is 1999.

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Index Numbers 591

Year Cotton Price Automobiles Sold Money Turnover

1999 $0.20 1,000 80 2004 0.25 1,200 90 2009 0.50 900 75

(a) Construct the index for 2004 and 2009. (b) Interpret the index for 2004 and 2009.

Exercises 11. The index of leading economic indicators, compiled and published by the U.S. National

Bureau of Economic Research, is composed of 12 time series, such as the average work hours of production in manufacturing, manufacturers’ new orders, and money supply. This index and similar indexes are designed to move up or down before the economy begins to move the same way. Thus, an economist has statistical evidence to forecast future trends.

You want to construct a leading indicator for Erie County in upstate New York. The index is to be based on 2000 data. Because of the time and work involved, you decide to use only four time series. As an experiment, you select these four series: unemploy- ment in the county, a composite index of county stock prices, the County Price Index, and retail sales. Here are the figures for 2000 and 2009.

2000 2009

Unemployment rate (percent) 5.3 6.8 Composite county stocks 265.88 362.26 County Price Index (1982 � 100) 109.6 125.0 Retail sales ($ millions) 529,917.0 622,864.0

The weights you assign are: unemployment rate 20 percent, stock prices 40 percent, County Price Index 25 percent, and retail sales 15 percent. a. Using 2000 as the base period, construct a leading economic indicator for 2009. b. Interpret your leading index.

12. You are employed by the state bureau of economic development. There is a demand for a leading economic index to review past economic activity and to forecast future eco- nomic trends in the state. You decide that several key factors should be included in the index: number of new businesses started during the year, number of business failures, state income tax receipts, college enrollment, and the state sales tax receipts. Here are the data for 2000 and 2009.

2000 2009

New businesses 1,088 1,162 Business failures 627 520 State income tax receipts ($ millions) 191.7 162.6 College student enrollment 242,119 290,841 State sales tax ($ millions) 41.6 39.9

a. Decide on the weights to be applied to each item in the leading index. b. Compute the leading economic indicator for 2009. c. Interpret the indexes.

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15.9 Consumer Price Index Frequent mention has been made of the Consumer Price Index (CPI) in the pre- ceding pages. It measures the change in price of a fixed market basket of goods and services from one period to another. In January 1978, the Bureau of Labor Sta- tistics began publishing CPIs for two groups of the population. One index, called the Consumer Price Index—All Urban Consumers, covers about 87 percent of the total population. The other index is for urban wage earners and clerical workers and covers about 32 percent of the population.

In brief, the CPI serves several major functions. It allows consumers to deter- mine the degree to which their purchasing power is being eroded by price increases. In that respect, it is a yardstick for revising wages, pensions, and other income pay- ments to keep pace with changes in prices. Equally important, it is an economic indicator of the rate of inflation in the United States.

The index includes about 400 items, and about 250 agents collect price data monthly. Prices are collected from more than 21,000 retail establishments and 60,000 housing units in 91 urban areas across the country. The prices of baby cribs, bread, beer, cigars, gasoline, haircuts, mortgage interest rates, physicians’ fees, taxes, and operating-room charges are just a few of the items included in what is often termed a typical “market basket” of goods and services that you purchase.

The CPI originated in 1913 and has been published regularly since 1921. The standard reference period (the base period) has been updated periodically. The cur- rent base period is 1982–84. The earlier base periods were: 1967, 1957–59, 1947–49, 1935–39, and 1925–29. Why is it necessary to change the base? Our expen- diture patterns change dramatically, and these changes must be reflected in the index. The most recent revision includes consumer items, such as VCRs, home com- puters, and cell phones. Earlier versions of the CPI did not include these items. By changing the base, the CPI captures the most recent expenditure patterns. You may want to go to www.bls.gov, click on the Consumer Price Index, and read more about it.

The CPI is actually not just one index. There are Consumer Price Indexes for New York, Chicago, Seattle, and Atlanta, as well as a number of other large cities. There are also price indexes for food, apparel, medical care, and other items. A few of them are shown below, 1982–84 � 100, for December 2009.

There are two consumer price indexes.

A perusal of this listing shows that a weighted index of all items has increased 115.949 percent since 1982–84; medical care has increased the most, 279.516 per- cent; and apparel went up the least, 19.357 percent.

Special Uses of the Consumer Price Index In addition to measuring changes in the prices of goods and services, both con- sumer price indexes have a number of other applications. The CPI is used to deter- mine real disposable personal income, to deflate sales or other variables, to find the

Item CPI-U

All items 215.949 Food and beverage 218.049 Apparel 119.357 Transportation 188.318 Medical care 379.516 Housing 215.523

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Does it seem that prices only increase? The Consumer Price Index, computed and reported by the U.S. Department of La- bor, is a relative mea- sure of price changes. It shows interesting price information for categories of products and services. For ex- ample, did you know that the CPI shows a decrease from 2008 to 2009 in the rela- tive prices of personal computers and pe- ripheral equipment? In fact, using a base of 1982–1984 � 100, the CPI for comput- ers and peripherals is 77.960. This means that relative prices for computers and peripherals have decreased about 22 percent from the 1982–1984 base.

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Index Numbers 593

Real income

Money income

The concept of real income is sometimes called deflated income, and the CPI is called the deflator. Also, a popular term for deflated income is income expressed in constant dollars. Thus, in Table 15–6, to determine whether Ms. Watts’ standard of living changed, her money income was converted to constant dollars. We found that her purchasing power, expressed in 1982–84 dollars (constant dollars), remained at $20,000.

Deflated income and real income are the same.

Self-Review 15–5 The take-home pay of Jon Greene and the CPI for 2000 and 2009 are:

Take-Home CPI Year Pay (1982–84 � 100)

2000 $25,000 170.8 2009 41,200 216.6

(a) What was Jon’s real income in 2000? (b) What was his real income in 2009? (c) Interpret your findings.

purchasing power of the dollar, and to establish cost-of-living increases. We first discuss the use of the CPI in determining real income.

Real Income As an example of the meaning and computation of real income, assume the Consumer Price Index is presently 200 with 1982–84 � 100. Also, assume that Ms. Watts earned $20,000 per year in the base period of 1982, 1983, and 1984. She has a current income of $40,000. Note that although her money income has doubled since the base period of 1982–84, the prices she paid for food, gasoline, clothing, and other items have also doubled. Thus, Ms. Watts’ standard of living has remained the same from the base period to the present time. Price increases have exactly offset an increase in income, so her present buying power (real income) is still $20,000. (See Table 15–6 for computations.) In general:

REAL INCOME [15–8]Real income � Money income CPI

� 100

Consumer Annual Price Index Computation

Year Money Income (1982–84 � 100) of Real Income Real Income

1982–84 $20,000 100 $20,000

Present year 40,000 200 20,000 $40,000

200 (100)

$20,000 100

(100)

TABLE 15–6 Computation of Real Income for 1982–84 and Present Year

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Deflating Sales A price index can also be used to “deflate” sales or similar money series. Deflated sales are determined byDeflated sales important

for showing the trend in “real” sales

Purchasing Power of the Dollar The Consumer Price Index is also used to de- termine the purchasing power of the dollar.

What has happened to the purchasing power of your dollar?

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Example The sales of Hill Enterprises, a small injection molding company in upstate New York, increased from $875,000 in 1982 to $1,482,000 in 1995, $1,491,000 in 2000, $1,502,000 in 2004, $1,515,000 in 2007, and $1,596,000 in 2009. The owner, Harry Hill, realizes that the price of raw materials used in the process has also increased over the period, so Mr. Hill wants to deflate sales to account for the increase in raw material prices. What are the deflated sales for 1995, 2000, 2004, 2007, and 2009 based on 1982 dollars? That is, what are sales for 1995, 2000, 2004, 2007, and 2009 expressed in constant 1982 dollars?

Solution The Producer Price Index (PPI) is an index released every month and published in the Monthly Labor Review and is also available at the Bureau of Labor Statistics website. The prices included in the PPI reflect the prices charged the manufacturer for the metals, rubber, and other items purchased. So the PPI seems an appro- priate index to use to deflate the manufacturer’s sales. The manufacturer’s sales are listed in the second column of Table 15–7, and the PPI for each year is in the third column. The next column shows sales divided by the PPI. The right-hand column details the calculations.

Sales increased from 1995 through 2009, but if we compare the sales in con- stant dollars, sales declined during the period. That is, deflated sales were $1,072,661.87 in 2000 but declined to $1,011,477.81 in 2004. Deflated sales declined even further to $909,363.75 in 2007. This is so because the prices Hill Enterprises paid for raw materials grew more rapidly than sales. Then, in 2009 deflated sales increased from the 2007 level.

TABLE 15–7 Calculation of Deflated Sales for Hill Enterprises

Year Sales PPI Constant Dollars Found By

1982 $ 875,000.00 100.0 $ 875,000.00 ($ 875,000.00/100.0)*100.0 1995 1,482,000.00 127.9 1,158,717.75 ($1,482,000.00/127.9)*100.0 2000 1,491,000.00 139.0 1,072,661.87 ($1,491,000.00/139.0)*100.0 2004 1,502,000.00 148.5 1,011,447.81 ($1,502,000.00/148.4)*100.0 2007 1,515,000.00 166.6 909,363.75 ($1,515,000.00/166.6)*100.0 2009 1,596,000.00 172.5 925,217.39 ($1,596,000.00/172.5)*100.0

USING AN INDEX TO FIND PURCHASING POWER [15–10]

Purchasing power of dollar � $1 CPI

� 100

USING AN INDEX AS A DEFLATOR

[15–9]Deflated sales � Actual sales An appropriate index � 100

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Index Numbers 595

Example Suppose the Consumer Price Index this month is 200.0 (1982–84 � 100). What is the purchasing power of the dollar?

Solution From formula (15–10), it is 50 cents, found by:

The CPI of 200.0 indicates that prices have doubled from the years 1982–84 to this month. Thus, the purchasing power of a dollar has been cut in half. That is, a 1982–84 dollar is worth only 50 cents this month. To put it another way, if you lost $1,000 in the period 1982–84 and just found it, the $1,000 could only buy half of what it could have bought in the years 1982, 1983, and 1984.

Purchasing power of dollar � $1

200.0 (100) � $0.50

Cost-of-Living Adjustments The Consumer Price Index (CPI) is also the basis for cost-of-living adjustments, or COLA, in many management–union contracts. The specific clause in the contract is often referred to as the “escalator clause.” About 31 million Social Security beneficiaries, 2.5 million retired military and federal civil service employees and survivors, and 600,000 postal workers have their incomes or pensions pegged to the CPI.

The CPI is also used to adjust alimony and child support payments; attorneys’ fees; workers’ compensation payments; rentals on apartments, homes, and office buildings; welfare payments; and so on. In brief, say a retiree receives a pension of $500 a month and the CPI increases 5 points from 165 to 170. Suppose for each point that the CPI increases, the pension benefits increase 1.0 percent, so the monthly increase in benefits will be $25, found by $500 (5 points)(.01). Now the retiree will receive $525 per month.

CPI used to adjust wages, pensions, and so on

15.10 Shifting the Base If two or more time series have the same base period, they can be compared directly. As an example, suppose we are interested in the trend in the prices of food and beverages, housing, medical care, and so on since the base period, 1982–84. Note in Table 15–8 that all of the consumer price indexes use the same base. Hence,

All Food and Apparel and Medical Year Items Beverages Housing Upkeep Care

1982–84 100.0 100.0 100.0 100.0 100.0 1990 130.7 132.1 128.5 124.1 162.8 1995 152.4 148.9 148.5 132.0 220.5 2000 172.2 168.4 169.6 129.6 260.8 2004 188.9 186.6 189.5 120.4 310.1 2005 195.3 191.2 195.7 119.5 323.2 2007 207.392 203.300 209.586 118.257 369.302 2009 214.537 218.249 217.057 120.078 375.613

TABLE 15–8 Trend in Consumer Prices to 2009 (1982–84 � 100)

Self-Review 15–6 Suppose the Consumer Price Index for the latest month is 195.4 (1982–84 � 100). What is the purchasing power of the dollar? Interpret.

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we conclude that the price of all consumer items combined increased 114.537 per- cent from the base period (1982–84) to the year 2009. (Beginning with January 2007, the CPI is reported to three decimal places instead of one.) Likewise, housing prices increased 117.057 percent, medical care 375.613 percent, and so on.

A problem arises, however, when two or more series being compared do not have the same base period. The following example compares the two most widely reported business indexes, the DJIA and Nasdaq.

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Example We want to compare the price changes in the Dow Jones Industrial Average (DJIA) with the Nasdaq. The two indexes for selected periods since 1995 follow. The infor- mation is reported on July 1 of each year.

Date DJIA Nasdaq 1-Jul-95 4,708.47 1,001.21 1-Jul-00 10,521.98 3,766.99 1-Jul-01 10,522.81 2,027.13 1-Jul-02 8,736.59 1,328.26 1-Jul-03 9,233.80 1,735.02 1-Jul-04 10,139.71 1,887.36 1-Jul-05 10,640.91 2,184.83 1-Jul-06 11,228.02 2,190.43 1-Jul-07 13,535.43 2,632.30 1-Jul-08 11,382.26 1,875.42 1-Jul-09 8,504.06 1,845.72

Solution From the information given, we are not sure the base periods are the same. Hence, a direct comparison is not appropriate. Because we want to compare the changes in the two business indexes, the logical approach is to let a particular year, say 1995, be the base for both indexes. For the DJIA, the base is 4,708.47, and for the Nasdaq it is 1,001.21.

The calculation of the index for the DJIA in 2005 is:

The following table reports the complete set of indexes.

Index � 10,640.91 4,708.47

(100) � 226.0

We conclude that both indexes have increased over the period. The DJIA has in- creased 187.5 percent and the Nasdaq 162.9 percent for the period from July 1, 1995, until July 1, 2007. Notice that both indexes reached a peak in 2000, declined to their lowest points in 2002 and increased up to 2007. Both indexes declined in 2008 and 2009.

The following chart, obtained from the financial section of Yahoo!, is a line graph of the DJIA and Nasdaq. The vertical axis shows the percent change from the base period of June 2003 for both indexes. From this graph we conclude that both indexes reached their largest percent increase in late 2007, then declined in 2008 and 2009. Of course, if we select different periods as the base, the results may not be exactly the same.

Comparison of DJIA and NASDAQ Date DJIA Index NASDAQ Index

1-Jul-95 4,708.47 100.0 1,001.21 100.0 1-Jul-00 10,521.98 223.5 3,766.99 376.2 1-Jul-01 10,522.81 223.5 2,027.13 202.5 1-Jul-02 8,736.59 185.6 1,328.26 132.7 1-Jul-03 9,233.80 196.1 1,735.02 173.3 1-Jul-04 10,139.71 215.4 1,887.36 188.5 1-Jul-05 10,640.91 226.0 2,184.83 218.2 1-Jul-06 11,228.02 238.5 2,190.43 218.8 1-Jul-07 13,535.43 287.5 2,632.30 262.9 1-Jul-08 11,382.26 241.7 2,304.97 230.2 1-Jul-09 8,504.06 180.6 1,845.72 184.3

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Index Numbers 597

Exercises 13. In April 2008, the mean salary for a nurse manager with a bachelor’s degree was $89,673.

The Consumer Price Index for April 2009 was 213.240 (1982–84 � 100). The mean annual salary for a nurse in the base period of 1982–84 was $19,800. What was the real income of the nurse in April 2009? How much had the mean salary increased?

14. The Trade Union Association of Orlando, Florida, maintains indexes on the hourly wages for a number of the trades. Unfortunately, the indexes do not all have the same base periods. Listed below is information on plumbers and electricians. Shift the base periods to 2000 and compare the hourly wage increases for the period from 2000 to 2009.

Self-Review 15–7 (a) From the preceding example, verify that the DJIA price index for 2004, using 1995 as the base period, is 215.4.

(b) The changes in industrial production and in the prices manufacturers paid for raw mate- rials since 1982 are to be compared. Unfortunately, the index of industrial production, which measures changes in production, and the Producer Price Index for crude mate- rials, have different base periods. The production index has a 2002 base period, and the Producer Price Index uses 1982 as the base period. Shift the base to 2002 and make the two series comparable. Interpret.

Industrial Producer Production Price

Index Index Year (2002 � 100) (1982 � 100)

2004 103.8 159.1 2005 107.2 182.3 2006 109.7 185.0 2007 111.3 206.9 2008 108.8 251.0

Year Plumbers (1995 � 100) Electricians (1998 � 100)

2000 133.8 126.0 2009 159.4 158.7

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Chapter Summary I. An index number measures the relative change from one period to another.

A. The major characteristics of an index are: 1. It is a percentage, but the percent sign is usually omitted. 2. It has a base period. 3. Most indexes are reported to the nearest tenth of a percent, such as 153.1. 4. The base of most indexes is 100.

B. The reasons for computing an index are: 1. It facilitates the comparison of unlike series. 2. If the numbers are very large, often it is easier to comprehend the change of the

index than the actual numbers. II. There are two types of price indexes, unweighted and weighted.

A. In an unweighted index, we do not consider the quantities. 1. In a simple index, we compare the base period to the given period.

[15–1]

where pt refers to the price in the current period, and p0 is the price in the base period. 2. In the simple average of price indexes, we add the simple indexes for each item

and divide by the number of items.

[15–2]

3. In a simple aggregate price index, the price of the items in the group are totaled for both periods and compared.

[15–3]

B. In a weighted index, the quantities are considered. 1. In the Laspeyres method, the base period quantities are used in both the base period

and the given period.

[15–4]

2. In the Paasche method, current period quantities are used.

[15–5]P � ©ptqt ©p0qt

� 100

P � ©pt q0 ©p0 q0

� 100

P � ©pt ©p0

� 100

P � ©Pi n

P � pt p0

� 100

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In the 1920s, whole- sale prices in Germany increased dramatically. In 1920, wholesale prices in- creased about 80 per- cent, in 1921 the rate of increase was 140 percent, and in 1922 it was a whopping 4,100 percent! Be- tween December 1922 and November 1923, wholesale prices increased an- other 4,100 percent. By that time govern- ment printing presses could not keep up, even by printing notes as

(continued)

15. In 1995, the mean salary of classroom teachers in Tinora School District was $28,650. By 2004, the mean salary increased to $33,972, and further increased in 2009 to $37,382. The American Federation of Classroom Teachers maintains information on the trends throughout the United States in classroom teacher salaries. Its index, which has a base period of 1995, was 122.5 for 2004 and 136.9 for 2009. Compare the Tinora teachers to the national trends.

16. Sam Steward is a freelance Web page designer. Listed below are his yearly wages for several years between 2002 and 2008. Also included is an industry index for Web page designers that reports the rate of wage inflation in the industry. This index has a base period of 1995.

Year Wage ($000) Index (1995 � 100)

2002 134.8 160.6 2004 145.2 173.6 2006 156.6 187.9 2008 168.8 203.3

Compute Sam’s real income for the selected years during the six-year period. Did his wages keep up with inflation, or did he lose ground?

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large as 500 million marks. Stories are told that workers were paid daily, then twice daily, so their wives could shop for necessities before the wages became too devalued.

Index Numbers 599

Domestic International Total Sales Sales Sales Employees

Year ($ million) ($ million) ($ million) (thousands)

2000 17,316 11,856 29,172 100.9 2001 19,825 12,492 32,317 101.8 2002 22,455 13,843 36,298 108.3 2003 25,274 16,588 41,862 110.6 2004 27,770 19,578 47,348 109.9 2005 28,377 22,137 50,514 115.6 2006 29,775 23,549 53,324 122.2 2007 32,444 28,651 61,095 119.2 2008 32,309 31,438 63,747 118.7 2009 30,889 31,008 61,897 115.5

Revenue Employees Year ($ million) (000)

2004 134 325 2005 152 307 2006 157 316

Revenue Employees Year ($ million) (000)

2007 168 319 2008 177 327 2009 183 323

3. Fisher’s ideal index is the geometric mean of Laspeyres’ index and Paasche’s index.

[15–6]

C. A value index uses both base period and current period prices and quantities.

[15–7]

III. The most widely reported index is the Consumer Price Index (CPI). A. It is often used to show the rate of inflation in the United States. B. It is reported monthly by the U.S. Bureau of Labor Statistics. C. The current base period is 1982–84. D. It is used by the Social Security system, so when the CPI changes, retirement benefits

also change.

Chapter Exercises The following information was taken from Johnson & Johnson annual reports. The principal office of Johnson & Johnson is in New Brunswick, New Jersey. Its common stock is listed on the New York Stock Exchange, using the symbol JNJ.

V � ©ptqt ©p0 q0

� 100

Fisher’s ideal index � 2(Laspeyres’ index)(Paasche’s index)

17. Refer to the Johnson & Johnson data. Use 2000 as the base period and compute a sim- ple index of domestic sales for each year from 2000 until 2009. Interpret the trend in domestic sales.

18. Refer to the Johnson & Johnson data. Use the period 2000–02 as the base period and compute a simple index of domestic sales for each year from 2003 to 2009.

19. Refer to the Johnson & Johnson data. Use 2000 as the base period and compute a sim- ple index of international sales for each year from 2001 until 2009. Interpret the trend in international sales.

20. Refer to the Johnson & Johnson data. Use the period 2000–02 as the base period and compute a simple index of international sales for each year from 2003 to 2009.

21. Refer to the Johnson & Johnson data. Use 2000 as the base period and compute a sim- ple index of the number of employees for each year from 2001 until 2009. Interpret the trend in the number of employees.

22. Refer to the Johnson & Johnson data. Use the period 2000–02 as the base period and com- pute a simple index of the number of employees for each year from 2003 to 2009.

The following information is from the General Electric Corporation annual reports.

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600 Chapter 15

2000 2009

Item Price Quantity Price Quantity

Margarine (pound) $0.81 18 $2.00 27 Shortening (pound) 0.84 5 1.88 9 Milk (1⁄2 gallon) 1.44 70 2.89 65 Potato chips 2.91 27 3.99 33

23. Compute a simple index for the revenue of GE. Use 2004 as the base period. What can you conclude about the change in revenue over the period?

24. Compute a simple index for the revenue of GE using the period 2004–06 as the base. What can you conclude about the change in revenue over the period?

25. Compute a simple index for the number of employees for GE. Use 2004 as the base period. What can you conclude about the change in the number of employees over the period?

26. Compute a simple index for the number of employees for GE using the period 2004–06 as the base. What can you conclude about the change in the number of employees over the period?

Below is information on food items for the years 2000 and 2009.

Price Quantity

Part 2000 2009 2000 2009

RC-33 $0.50 $0.60 320 340 SM-14 1.20 0.90 110 130 WC50 0.85 1.00 230 250

27. Compute a simple price index for each of the four items. Use 2000 as the base period.

28. Compute a simple aggregate price index. Use 2000 as the base period. 29. Compute Laspeyres’ price index for 2009 using 2000 as the base period. 30. Compute Paasche’s index for 2009 using 2000 as the base period. 31. Determine Fisher’s ideal index using the values for the Laspeyres and Paasche indexes

computed in the two previous problems. 32. Determine a value index for 2009 using 2000 as the base period.

Betts Electronics purchases three replacement parts for robotic machines used in its manu- facturing process. Information on the price of the replacement parts and the quantity purchased is given below.

33. Compute a simple price index for each of the three items. Use 2000 as the base period.

34. Compute a simple aggregate price index for 2009. Use 2000 as the base period. 35. Compute Laspeyres’ price index for 2009 using 2000 as the base period. 36. Compute Paasche’s index for 2009 using 2000 as the base period. 37. Determine Fisher’s ideal index using the values for the Laspeyres and Paasche indexes

computed in the two previous problems. 38. Determine a value index for 2009 using 2000 as the base period.

Prices for selected foods for 2000 and 2009 are given in the following table.

Price Quantity

Item 2000 2009 2000 2009

Cabbage (pound) $0.06 $0.05 2,000 1,500 Carrots (bunch) 0.10 0.12 200 200 Peas (quart) 0.20 0.18 400 500 Endive (bunch) 0.15 0.15 100 200

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Index Numbers 601

Price Quantity

Item 1990 2009 1990 2009

Aluminum (cents per pound) $ 0.287 $ 0.76 1,000 1,200 Natural gas (1,000 cu. ft.) 0.17 2.50 5,000 4,000 Petroleum (barrel) 3.18 26.00 60,000 60,000 Platinum (troy ounce) 133.00 490.00 500 600

39. Compute a simple price index for each of the four items. Use 2000 as the base period.

40. Compute a simple aggregate price index. Use 2000 as the base period. 41. Compute Laspeyres’ price index for 2009 using 2000 as the base period. 42. Compute Paasche’s index for 2009 using 2000 as the base period. 43. Determine Fisher’s ideal index using the values for the Laspeyres and Paasche indexes

computed in the two previous problems. 44. Determine a value index for 2009 using 2000 as the base period.

The prices of selected items for 1990 and 2009 follow. Production figures for those two periods are also given.

45. Compute a simple price index for each of the four items. Use 1990 as the base period.

46. Compute a simple aggregate price index. Use 1990 as the base period. 47. Compute Laspeyres’ price index for 2009 using 1990 as the base period. 48. Compute Paasche’s index for 2009 using 1990 as the base period. 49. Determine Fisher’s ideal index using the values for the Laspeyres and Paasche indexes

computed in the two previous problems. 50. Determine a value index for 2009 using 1990 as the base period. 51. A special-purpose index is to be designed to monitor the overall economy of the South-

west. Four key series were selected. After considerable deliberation, it was decided to weight retail sales 20 percent, total bank deposits 10 percent, industrial production in the area 40 percent, and nonagricultural employment 30 percent. The data for 1996 and 2009 are:

Bank Industrial Retail Sales Deposits Production

Year ($ millions) ($ billions) (1990 � 100) Employment

1996 1,159.0 87 110.6 1,214,000 2009 1,971.0 91 114.7 1,501,000

Construct a special-purpose index for 2009 using 1996 as the base period and interpret. 52. We are making a historical study of the American economy from 1950 to 1980. Data on

prices, the labor force, productivity, and the GNP were collected. Note in the following table that the CPI has a base period of 1967, employment is in millions of persons, and so on. A direct comparison, therefore, is not feasible. a. Make whatever calculations are necessary to compare the trend in the four series from

1950 to 1980. b. Interpret.

Consumer Total Index of Gross Price Labor Productivity National Index Force in Manufacturing Product

Year (1967 � 100) (millions) (1967 � 100) ($ billions)

1950 72.1 64 64.9 286.2 1967 100.0 81 100.0 789.6 1971 121.3 87 110.3 1,063.4 1975 161.2 95 114.9 1,516.3 1980 246.8 107 146.6 2,626.0

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53. The management of Ingalls Super Discount stores, with several stores in the Oklahoma City area, wants to construct an index of economic activity for the metropolitan area. Management contends that, if the index reveals that the economy is slowing down, inven- tory should be kept at a low level.

Three series seem to hold promise as predictors of economic activity—area retail sales, bank deposits, and employment. All of these data can be secured monthly from the U.S. government. Retail sales is to be weighted 40 percent, bank deposits 35 percent, and employment 25 percent. Seasonally adjusted data for the first three months of the year are:

Retail Sales Bank Deposits Employment Month ($ millions) ($ billions) (thousands)

January 8.0 20 300 February 6.8 23 303 March 6.4 21 297

Consumer Price Index Mr. Martin’s Monthly Year (1982–84 � 100) Take-Home Pay

1982–84 100.0 $ 600 2009 214.537 2,000

Construct an index of economic activity for each of the three months, using January as the base period.

54. The following table gives information on the Consumer Price Index and the monthly take- home pay of Bill Martin, an employee at Jeep Corporation.

Year Producer Price Index Sales

1995 127.9 $2,400,000 2009 172.5 3,500,000

a. What is the purchasing power of the dollar in 2009, based on the period 1982–84? b. Determine Mr. Martin’s “real” monthly income for 2009.

55. Suppose that the Producer Price Index and the sales of Hoskin’s Wholesale Distributors for 1995 and 2009 are:

What are Hoskin’s real sales (also called deflated sales) for the two years?

Software Commands appear. This is the product of the price of bread ($0.87) and the quantity of bread (50) sold in 1999.

c. With cell E3 still highlighted, move the cursor to the bottom right corner of cell E3, hold the left mouse button and drag the cell down to cell E8. The remaining products should appear.

d. Move to cell E9, click on on the menu and hit Enter. The value of 507.64 will appear. This is the denominator for the Laspeyres price index. The other products and column totals are deter- mined in a similar manner. The other Excel out- put in the chapter is computed similarly.

�

1. The Excel commands for the spreadsheet on page 582 are: a. Enter the data for the prices and quantities. We

entered the label Item in cell A2, and the item names in cells A3 through A8. The label 1999 Price was entered in B2, and the price data for 1999 in cells B3 through B8. The label 1999 Quantity was entered in cell C2, with the 1999 quantities in cells C3 through C8. Cell E2 was labeled 1999 Price*1999 Quantity.

b. To determine the product of the 1999 prices and quantities, highlight the cell E3. Type � B2*C2 in cell E3 and hit Enter. The value of 43.5 should

602 Chapter 15

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Index Numbers 603

Chapter 15 Answers to Self-Review

15–1 1. Region AMT Index

China 500.5 547.59 EU 198.0 216.63 Japan 118.7 129.87 US 91.4 100.00 Russia 68.5 74.95

Average (a) (b) Year Hourly Wage Index Index

1995 11.65 100.0 90.8 2000 14.02 120.3 109.2 2005 16.13 138.5 125.7 2010 (May) 19.01 163.2 148.1

b. Business activity increased 16 percent from 1999 to 2004. It increased 8.25 percent from 1999 to 2009.

15–5 a. $14,637, found by ($25,000�170.8)(100). b. $19,021, found by ($41,200�216.6)(100). c. In terms of the base period, Jon’s salary was

$14,637 in 2000 and $19,021 in 2009. This indicates his take-home pay increased at a faster rate than the price paid for food, transportation, etc.

15–6 $0.51, found by ($1.00�195.4)(100). The purchasing power has declined by $0.49.

15–7 a. 215.4, found by (10,139.71�4,708.47)(100). b. With 2004 as the base period for both series:

From the base of 2004, the producer price index for crude materials increased at a faster rate (57.76 percent) than industrial production (4.82 percent).

China produces 447.6 percent more steel than the United States.

15–2 a.

15–3 a.

b. The value of sales has gone up 27.1 percent from 1996 to 2009.

15–4 a.

� $77,000 60,600

(100) � 127.1

P � $4(9,000) � $5(200) � $8(5,000)

$3(10,000) � $1(600) � $10(3,000) (100)

P � 2(112.9)(112.9) � 112.9

� $102,700 $91,000

(100) � 112.9

P � $85(520) � $45(1,300) $75(520) � $40(1,300)

(100)

� $96,500 85,500

(100) � 112.9

P � $85(500) � $45(1,200) $75(500) � $40(1,200)

(100)

P � ($130�$115)(100) � 113.0 P � (113.3 � 112.5)�2 � 112.9 P2 � ($45�$40)(100) � 112.5 P1 � ($85�$75)(100) � 113.3

For 2009

Item Weight

Cotton ($0.50�$0.20)(100)(.10) � 25.00 Autos (900�1,000)(100)(.30) � 27.00 Money turnover (75�80)(100)(.60) � 56.25

108.25

Industrial Production Producer

Index Price Index

2004 1.0000 1.0000 2005 1.0328 1.1458 2006 1.0568 1.1628 2007 1.0723 1.3004 2008 1.0482 1.5776

For 2004

Item Weight

Cotton ($0.25�$0.20)(100)(.10) � 12.5 Autos (1,200�1,000)(100)(.30) � 36.0 Money turnover (90�80)(100)(.60) � 67.5

116.0

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16 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Define the components of a time series.

LO2 Compute a moving average.

LO3 Determine a linear trend equation.

LO4 Use a trend equation to compute forecasts.

LO5 Compute a nonlinear trend equation.

LO6 Determine and interpret a set of seasonal indexes.

LO7 Deseasonalize data using a seasonal index.

LO8 Calculate seasonally adjusted forecasts.

LO9 Test for autocorrelation.

Time Series and Forecasting

Team Sports Inc. sells sporting goods to high schools and colleges via

a nationally distributed catalog. Management at Team Sports

estimates it will sell 2,000 Wilson Model A2000 catcher’s mitts next

year. The deseasonalized sales are projected to be the same for each

of the four quarters next year. The seasonal factor for the second

quarter is 145. Determine the seasonally adjusted sales for the second

quarter of next year. (See Exercise 12 and LO8.)

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Time Series and Forecasting 605

16.1 Introduction The emphasis in this chapter is on time series analysis and forecasting. A time series is a collection of data recorded over a period of time—weekly, monthly, quarterly,

or yearly. Two examples of time series are Microsoft Corporation sales by quarter since 1985 and the annual production of sulfuric acid since 1970.

An analysis of history—a time series— is used by management to make current decisions and plans based on long-term forecasting. We usually assume past pat- terns will continue into the future. Long-term forecasts extend more than 1 year into the future; 2-, 5-, and 10-year projections are common. Long-range predictions are essen-

tial to allow sufficient time for the procurement, manufacturing, sales, finance, and other departments of a company to develop plans for possible new plants, financing, development of new products, and new methods of assembling.

Forecasting the level of sales, both short-term and long-term, is practically dic- tated by the very nature of business organizations in the United States. Competi- tion for the consumer’s dollar, stress on earning a profit for the stockholders, a desire to procure a larger share of the market, and the ambitions of executives are some of the prime motivating forces in business. Thus, a forecast (a statement of the goals of management) is necessary to have the raw materials, production facilities, and staff to meet the projected demand.

This chapter deals with the use of data to forecast future events. First, we look at the components of a time series. Then, we examine some of the techniques used in analyzing data. Finally, we forecast future events.

16.2 Components of a Time Series There are four components to a time series: the trend, the cyclical variation, the sea- sonal variation, and the irregular variation.

Secular Trend The long-term trends of sales, employment, stock prices, and other business and eco- nomic series follow various patterns. Some move steadily upward, others decline, and still others stay the same over time.

What is a time series?

SECULAR TREND The smooth long-term direction of a time series.

The following are several examples of a secular trend.

• Home Depot was founded in 1978 and is the world’s largest home improve- ment retailer. The following chart shows the number of employees at Home Depot Inc. You can see this number has increased rapidly over the last 15 years. In 1993, there were just over 50,000 associates and by 2006 that number increased to 364,000. Since then, the number of associates has declined to 317,000 in 2010.

LO1 Define the components of a time series.

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606 Chapter 16

• The number of manufactured homes shipped in the United States showed a steady increase from 1990 to 1996, then remained about the same until 1999, when the number began to decline. By the year 2002, the number shipped was less than it had been in 1990 and continued to decline to 2009. This informa- tion is shown in the following chart.

Cyclical Variation The second component of a time series is cyclical variation. A typical business cycle consists of a period of prosperity followed by periods of recession, depression, and then recovery. There are sizable fluctuations unfolding over more than one year in time above and below the secular trend. In a recession, for example, employment, pro- duction, the Dow Jones Industrial Average, and many other business and economic series are below the long-term trend lines. Conversely, in periods of prosperity they are above their long-term trend lines.

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Time Series and Forecasting 607

CYCLICAL VARIATION The rise and fall of a time series over periods longer than one year.

CHART 16–1 Batteries Sold by National Battery Retailers Inc. from 1991 to 2010

Year 1991 1996 2001 2006 2011

Prosperity

Recovery

Contraction

Long-term secular trend

Depression

Ba tte

rie s

So ld

(0 00

)

Chart 16–1 shows the annual unit sales of batteries sold by National Battery Retailers Inc. from 1991 through 2010. The cyclical nature of business is highlighted. There are periods of recovery, followed by prosperity, then contraction, and finally the cycle bottoms out with depression.

Seasonal Variation The third component of a time series is the seasonal variation. Many sales, pro- duction, and other series fluctuate with the seasons. The unit of time reported is either quarterly or monthly.

SEASONAL VARIATION Patterns of change in a time series within a year. These patterns tend to repeat themselves each year.

Almost all businesses tend to have recurring seasonal patterns. Men’s and boy’s clothing, for example, have extremely high sales just prior to Christmas and relatively low sales just after Christmas and during the summer. Toy sales is another example with an extreme seasonal pattern. More than half of the business for the year is usu- ally done in the months of November and December. The lawn care business is sea- sonal in the northeast and north-central states. Many businesses try to even out the seasonal effects by engaging in an offsetting seasonal business. In the Northeast, you will see the operator of a lawn care business with a snowplow on the front of the truck in an effort to earn income in the off-season. At ski resorts throughout the coun- try, you will often find golf courses nearby. The owners of the lodges try to rent to skiers in the winter and golfers in the summer. This is an effective method of spreading their fixed costs over the entire year rather than a few months.

Chart 16–2 shows the quarterly sales, in millions of dollars, of Hercher Sport- ing Goods Inc. The Chicago area sporting goods company specializes in selling baseball and softball equipment to high schools, colleges, and youth leagues. It also has several retail outlets in some of the larger shopping malls. There is a distinct seasonal pattern to its business. Most of its sales are in the first and second quar- ters of the year, when schools and organizations are purchasing equipment for the upcoming season. During the early summer, it keeps busy by selling replacement equipment. It does some business during the holidays (fourth quarter). The late summer (third quarter) is its slow season.

Statistics in Action

Statisticians, econo- mists, and business executives are con- stantly looking for variables that will forecast the country’s economy. The pro- duction of crude oil, price of gold on world markets, and the Dow Jones aver- age, as well as many published govern- ment indexes are variables that have been used with some success. Variables such as the length of hemlines and the winner of the Super Bowl have also been tried. The variable that seems overall to be the most success- ful is the price of scrap metal. Why? Scrap metal is the beginning of the manufacturing chain. When its de- mand increases, this is an indication that manufacturing is also increasing.

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608 Chapter 16

Irregular Variation Many analysts prefer to subdivide the irregular variation into episodic and residual variations. Episodic fluctuations are unpredictable, but they can be identified. The ini- tial impact on the economy of a major strike or a war can be identified, but a strike or war cannot be predicted. After the episodic fluctuations have been removed, the remaining variation is called the residual variation. The residual fluctuations, often called chance fluctuations, are unpredictable, and they cannot be identified. Of course, neither episodic nor residual variation can be projected into the future.

16.3 A Moving Average A moving average is useful in smoothing a time series to see its trend. It is also the basic method used in measuring the seasonal fluctuation, described later in the chapter. In contrast to the least squares method, which expresses the trend in terms of a mathematical equation ( � a � bt), the moving-average method merely smooths the fluctuations in the data. This is accomplished by “moving” the arithmetic mean values through the time series.

To apply the moving average to a time series, the data should follow a fairly lin- ear trend and have a definite rhythmic pattern of fluctuations (repeating, say, every three years). The data in the following example have three components—trend, cycle, and irregular, abbreviated T, C, and I. There is no seasonal variation, because the data are recorded annually. What the moving average accomplishes is to average out C and I. What is left is the trend.

If the duration of the cycles is constant, and if the amplitudes of the cycles are equal, the cyclical and irregular fluctuations are removed entirely using the moving average. The result is a line. For example, in the following time series the cycle repeats itself every seven years, and the amplitude of each cycle is 4; that is, there are exactly four units from the trough (lowest time period) to the peak. The seven- year moving average, therefore, averages out the cyclical and irregular fluctuations perfectly, and the residual is a linear trend.

The first step in computing the seven-year moving average is to determine the seven-year moving totals. The total sales for the first seven years (1984–90 inclusive) are $22 million, found by 1 � 2 � 3 � 4 � 5 � 4 � 3. (See Table 16–1.) The total of $22 million is divided by 7 to determine the arithmetic mean sales per year. The seven-year total (22) and the seven-year mean (3.143) are positioned opposite the middle year for that group of seven, namely, 1987, as shown in Table 16–1. Then

Ŷ

CHART 16–2 Sales of Baseball and Softball Equipment, Hercher Sporting Goods, 2008–2010 by Quarter

Q1 Q2 Q3 Q4 Q1 Q2 Q3 Q4Q1 Q2 Q3 Q4 2008 2009 2010

Sa le

s ($

m ill

io n)

20.0

10.0

Moving-average method smooths out fluctuations

Compute mean of first seven years

LO2 Compute a moving average.

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Time Series and Forecasting 609

Sales Seven-Year Seven-Year Year ($ millions) Moving Total Moving Average

1984 $1 1985 2 1986 3 1987 4 22 3.143 1988 5 23 3.286 1989 4 24 3.429 1990 3 25 3.571 1991 2 26 3.714 1992 3 27 3.857 1993 4 28 4.000 1994 5 29 4.143 1995 6 30 4.286 1996 5 31 4.429 1997 4 32 4.571 1998 3 33 4.714 1999 4 34 4.857 2000 5 35 5.000 2001 6 36 5.143 2002 7 37 5.286 2003 6 38 5.429 2004 5 39 5.571 2005 4 40 5.714 2006 5 41 5.857 2007 6 2008 7 2009 8

TABLE 16–1 The Computations for the Seven-Year Moving Average

the total sales for the next seven years (1985–91 inclusive) are determined. (A con- venient way of doing this is to subtract the sales for 1984 [$1 million] from the first seven-year total [$22 million] and add the sales for 1991 [$2 million], to give the new total of $23 million.) The mean of this total, $3.286 million, is positioned oppo- site the middle year, 1988. The sales data and seven-year moving average are shown graphically in Chart 16–3.

CHART 16–3 Sales and Seven-Year Moving Average

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610 Chapter 16

The number of data values to include in a moving average depends on the character of the data collected. If the data are quarterly, since there are four quarters in a year, then four terms might be typical. If the data are daily, since there are seven days in a week, then seven terms might be appropriate. You might also use trial and error to determine a number that best levels out the chance fluctuations.

A moving average can be easily computed in Excel. In fact, it requires only one command. If the original data are in locations D3 to D20 and you wish a three- period moving average, you can go to position E4 and type � (D3�D4�D5)�3 and then copy that same formula down to position E19.

A three-year and a five-year moving average for some production data are shown in Table 16–2 and depicted in Chart 16–4.

Three-Year Three-Year Five-Year Five-Year Production, Moving Moving Moving Moving

Year Y Total Average Total Average

1991 5 1992 6 19 6.3 1993 8 24 8.0 34 6.8 1994 10 23 7.7 32 6.4 1995 5 18 6.0 33 6.6 1996 3 15 5.0 35 7.0 1997 7 20 6.7 37 7.4 1998 10 29 9.7 43 8.6 1999 12 33 11.0 49 9.8 2000 11 32 10.7 55 11.0 2001 9 33 11.0 60 12.0 2002 13 37 12.3 66 13.2 2003 15 46 15.3 70 14.0 2004 18 48 16.0 72 14.4 2005 15 44 14.7 73 14.6 2006 11 40 13.3 75 15.0 2007 14 42 14.0 79 15.8 2008 17 53 17.7 2009 22

TABLE 16–2 A Three-Year Moving Average and a Five-Year Moving Average

Sales, production, and other economic and business series usually do not have (1) periods of oscillation that are of equal length or (2) oscillations that have identi- cal amplitudes. Thus, in practice, the application of a moving average does not result precisely in a line. For example, the production series in Table 16–2 repeats about every five years, but the amplitude of the data varies from one oscillation to another. The trend appears to be upward and somewhat linear. Both moving averages—the three-year and the five-year—seem to adequately describe the trend in production since 1991.

Four-year, six-year, and other even-numbered-year moving averages present one minor problem regarding the centering of the moving totals and moving aver- ages. Note in Table 16–3 on the next page that there is no center time period, so the moving totals are positioned between two time periods. The total for the first four years ($42) is positioned between 2002 and 2003. The total for the next four years is $43. The averages of the first four years and the second four years ($10.50 and $10.75, respectively) are averaged, and the resulting figure is centered on 2003. This procedure is repeated until all possible four-year averages are computed.

Determining a moving average for an even- numbered period, such as four years

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Time Series and Forecasting 611

CHART 16–4 A Three-Year and Five-Year Moving Average, 1991 to 2009

Centered Four-Year Four-Year Four-Year

Sales, Moving Moving Moving Year Y Total Average Average

2001 $ 8

2002 11 $42 (8 � 11 � 9 � 14) $10.50 ($42 � 4)

2003 9 10.625 43 (11 � 9 � 14 � 9) 10.75 ($43 � 4)

2004 14 10.625 42 10.50

2005 9 10.625 43 10.75

2006 10 10.000 37 9.25

2007 10 9.625 40 10.00

2008 8

2009 12

TABLE 16–3 A Four-Year Moving Average

16.4 Weighted Moving Average A moving average uses the same weight for each observation. For example, a three- year moving total is divided by the value 3 to yield the three-year moving average. To put it another way, each data value has a weight of one-third in this case. Similarly, you can see that for a five-year moving average each data value has a weight of one-fifth.

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612 Chapter 16

A natural extension of the weighted mean discussed in Chapter 3 is to compute a weighted moving average. This involves selecting a possibly different weight for each data value and then computing a weighted average of the most recent n values as the smoothed value. In the majority of applications, we use the smoothed value as a fore- cast of the future. So, the most recent observation receives the most weight, and the weight decreases for older data values. Notice that for both the simple moving aver- age and for the weighted moving average the sum of the weights is equal to 1.

Suppose, for example, we compute a two-year weighted moving average for the data in Table 16–3 giving twice as much weight to the most recent value. In other words, give a weight of 2�3 to the last year and 1�3 to the value immediately before that. Then “forecast” sales for 2003 would be found by (1�3)($8) � (2�3)($11) � $10. The next moving average would be computed as (1�3)($11) � (2�3)($9) � $9.667. Pro- ceeding in the same fashion, the final or 2010 weighted moving average would be (1�3)($8) � (2�3)($12) � $10.667. To summarize the technique of using moving aver- ages, its purpose is to help identify the long-term trend in a time series (because it smooths out short-term fluctuations). It is used to reveal any cyclical and seasonal fluctuations.

Example Cedar Fair operates seven amusement parks and five separately gated water parks. Its combined attendance (in thousands) for the last 17 years is given in the following table. A partner asks you to study the trend in attendance. Compute a three-year moving average and a three-year weighted moving average with weights of 0.2, 0.3, and 0.5 for successive years.

Solution The three-year moving average is:

Attendance Year (000)

1993 5,761 1994 6,148 1995 6,783 1996 7,445 1997 7,405 1998 11,450 1999 11,224 2000 11,703 2001 11,890 2002 12,380 2003 12,181 2004 12,557 2005 12,700 2006 19,300 2007 22,100 2008 22,720 2009 21,136

Attendance Year (000) Moving Average Found by

1993 5,761 1994 6,148 6,231 (5,761 � 6,148 � 6,783)�3 1995 6,783 6,792 (6,148 � 6,783 � 7,445)�3 1996 7,445 7,211 (6,783 � 7,445 � 7,405)�3 1997 7,405 8,767 (7,445 � 7,405 � 11,450)�3 1998 11,450 10,026 (7,405 � 11,450 � 11,224)�3 1999 11,224 11,459 (11,450 � 11,224 � 11,703)�3 2000 11,703 11,606 (11,224 � 11,703 � 11,890)�3

(continued )

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Time Series and Forecasting 613

The three-year weighted moving average is:

Attendance Year (000) Moving Average Found by

2001 11,890 11,991 (11,703 � 11,890 � 12,380)�3 2002 12,380 12,150 (11,890 � 12,380 � 12,181)�3 2003 12,181 12,373 (12,380 � 12,181 � 12,557)�3 2004 12,557 12,479 (12,181 � 12,557 � 12,700)�3 2005 12,700 14,852 (12,557 � 12,700 � 19,300)�3 2006 19,300 18,033 (12,700 � 19,300 � 22,100)�3 2007 22,100 21,373 (19,300 � 22,100 � 22,720)/3 2008 22,720 21,985 (22,100 � 22,720 � 21,136)/3 2009 21,136

Attendance Weighted Year (000) Moving Average Found by

1993 5,761 1994 6,148 6,388 .2(5,761) � .3(6,148) � .5(6,783) 1995 6,783 6,987 .2(6,148) � .3(6,783) � .5(7,445) 1996 7,445 7,293 .2(6,783) � .3(7,445) � .5(7,405) 1997 7,405 9,436 .2(7,445) � .3(7,405) � .5(11,450) 1998 11,450 10,528 .2(7,405) � .3(11,450) � .5(11,224) 1999 11,224 11,509 .2(11,450) � .3(11,224) � .5(11,703) 2000 11,703 11,701 .2(11,224) � .3(11,703) � .5(11,890) 2001 11,890 12,098 .2(11,703) � .3(11,890) � .5(12,380) 2002 12,380 12,183 .2(11,890) � .3(12,380) � .5(12,181) 2003 12,181 12,409 .2(12,380) � .3(12,181) � .5(12,557) 2004 12,557 12,553 .2(12,181) � .3(12,557) � .5(12,700) 2005 12,700 15,971 .2(12,557) � .3(12,700) � .5(19,300) 2006 19,300 19,380 .2(12,700) � .3(19,300) � .5(22,100) 2007 22,100 21,850 .2(19,300) � .3(22,100) � .5(22,720) 2008 22,720 21,804 .2(22,100) � .3(22,720) � .5(21,136) 2009 21,136

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Self-Review 16–1 Determine a three-year moving average for the sales of Waccamaw Machine Tool Inc. Plot both the original data and the moving average.

Number Produced Number Produced Year (thousands) Year (thousands)

2005 2 2008 5 2006 6 2009 3 2007 4 2010 10

Exercises 1. Calculate a four-quarter weighted moving average for the number of shares outstanding

for the Boxley Box Company for the nine quarters of data. The data are reported in thou- sands. Apply weights of .1, .2, .3, and .4, respectively, for the quarters. In a few words, describe the trend in the number of subscribers.

1st Quarter 2008 28,766 2nd Quarter 2008 30,057 3rd Quarter 2008 31,336 4th Quarter 2008 33,240 1st Quarter 2009 34,610 2nd Quarter 2009 35,102 3rd Quarter 2009 35,308 4th Quarter 2009 35,203 1st Quarter 2010 34,386

Study the graph carefully. You will see that the attendance trend is evenly upward with 360,000 added visitors each year. However, there is a “hop” of approximately 3 mil- lion per year between 1997 and 1998. That probably reflects the fact Cedar Fair acquired Knott’s Berry Farm in late 1997, leading to a boost in attendance. A similar boost occurred in 2006 with the purchase of King’s Island near Cincinnati. The weighted moving average follows the data more closely than the moving average. This reflects the additional influence given to the most recent period. In other words, the weighted method, where the most recent period is given the largest weight, won’t be quite as smooth. However, it may be more accurate as a forecasting tool.

1998 8.61 2005 6.61 1999 8.14 2006 5.58 2000 7.67 2007 5.87 2001 6.59 2008 5.94 2002 7.37 2009 5.49 2003 6.88 2010 5.43 2004 6.71

2. Listed below is the number of movie tickets sold at the Library Cinema-Complex, in thou- sands, for the period from 1998 to 2010. Compute a five-year weighted moving average using weights of .1, .1, .2, .3, and .3, respectively. Describe the trend in yield.

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Time Series and Forecasting 615

16.5 Linear Trend The long-term trend of many business series, such as sales, exports, and production, often approximates a straight line. If so, the equation to describe this growth is:

LINEAR TREND EQUATION [16–1]

where: read Y hat, is the projected value of the Y variable for a selected value of t.

a is the Y-intercept. It is the estimated value of Y when t � 0. Another way to put it is: a is the estimated value of Y where the line crosses the Y-axis when t is zero.

b is the slope of the line, or the average change in for each increase of one unit in t.

t is any value of time that is selected.

To illustrate the meaning of , a, b, and t in a time-series problem, a line has been drawn in Chart 16–5 to represent the typical trend of sales. Assume that this company started in business in 2001. This beginning year (2001) has been arbitrar- ily designated as year 1. Note that sales increased $2 million on the average every year; that is, based on the straight line drawn through the sales data, sales increased from $3 million in 2001 to $5 million in 2002, to $7 million in 2003, to $9 million in 2004, and so on. The slope, or b, is therefore 2. Note too that the line intercepts the Y-axis (when t � 0) at $1 million. This point is a. Another way of determining b is to locate the starting place of the straight line in year (1). It is 3 for 2001 in this problem. Then locate the value on the straight line for the last year. It is 19 for 2009. Sales went up $19 million � $3 million � $16 million, in eight years (2001 to 2009). Thus, 16 � 8 � 2, which is the slope of the line, or b.

Ŷ

Ŷ � a � bt

Slope of trend line is b

CHART 16–5 A Straight Line Fitted to Sales Data

2001 2002 2003 2004 2005 2006 2007 2008 2009 Years 1 2 3 4 5 6 7 8 9 Codes

Sa le

s ($

m ill

io ns

)

Trend

Sales

LO3 Determine a linear trend equation.

The equation for the line in Chart 16–5 is:

where: is sales in millions of dollars.

1 is the intercept with the Y-axis. It is also the sales in millions of dollars for year 0, or 2000.

t refers to the yearly increase in sales.

Ŷ

Ŷ � 1 � 2t

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616 Chapter 16

In Chapter 13, we drew a line through points on a scatter diagram to approxi- mate the regression line. We stressed, however, that this method for determining the regression equation has a serious drawback—namely, the position of the line depends on the judgment of the individual who drew the line. Three people would probably draw three different lines through the scatter plots. Likewise, the line we drew through the sales data in Chart 16–5 might not be the best-fitting line. Because of the subjective judgment involved, this method should be used only when a quick approximation of the straight-line equation is needed, or to check the reasonableness of the least squares line, which is discussed next.

16.6 Least Squares Method In the discussion of simple linear regression in Chapter 13, we showed how the least squares method is used to find the best linear relationship between two vari- ables. In forecasting methods, time is the independent variable and the value of the time series is the dependent variable. Furthermore, we often code the independent variable time to make the equations easier to interpret. In other words, we let t be 1 for the first year, 2 for the second, and so on. If a time series includes the sales of General Electric for five years starting in 2002 and continuing through 2006, we would code the year 2002 as 1, 2003 as 2, and 2006 as 5.

LO4 Use a trend equation to compute forecasts.

Example The sales of Jensen Foods, a small grocery chain located in southwest Texas, since 2005 are:

Solution To determine the trend equation, we could use formula (13–4) to find the slope, or b value, and formula (13–5) to locate the intercept, or a value. We would substitute t, the coded values for the year, for X in these equations. Another approach is to use a software package, such as Minitab or Excel. Chart 16–6 is output from

Sales Year ($ million)

2005 7 2006 10 2007 9 2008 11 2009 13

Determine the regression equation. How much are sales increasing each year? What is the sales forecast for 2012?

CHART 16–6 Sales and Trend Line, 2005–2009

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Time Series and Forecasting 617

Minitab. The values for year, coded year, sales, and fitted sales are shown in the lower right portion of the output. The left half is a scatter plot of the data and the fitted regression line.

From the output, the trend equation is � 6.1 � 1.3t. How do we interpret this equation? The sales are in millions of dollars. So the value 1.3 tells us that sales increased at a rate of 1.3 million dollars per year. The value 6.1 is the estimated value of sales in the year 0. That is the estimate for 2004, which is called the base year. For example, to determine the point on the line for 2008, insert the t value of 4 in the equation. Then � 6.1 � 1.3(4) � 11.3.

If sales, production, or other data approximate a linear trend, the equation devel- oped by the least squares technique can be used to estimate future values. It is reasonable that the sales for Jensen Foods follow a linear trend. So we can use the trend equation to forecast future sales.

See Table 16–4. The year 2005 is coded 1, the year 2007 is coded 3, and 2009 is coded 5. Logically, we code 2011 as 7 and 2012 as 8. So we substitute 8 into the trend equation and solve for .

Thus, on the basis of past sales, the estimate for 2012 is $16.5 million.

Ŷ � 6.1 � 1.3t � 6.1 � 1.3(8) � 16.5

Ŷ

Statistics in Action

Investors frequently use regression analy- sis to study the rela- tionship between a particular stock and the general condition of the market. The dependent variable is the monthly percent- age change in the value of the stock, and the independent variable is the monthly percentage change in a market index, such as the Standard & Poor’s 500 Composite Index. The value of b in the regression equation is the particular stock’s beta coefficient or just the beta. If b is greater than 1, the implica- tion is that the stock is sensitive to market changes. If b is be- tween 0 and 1, the implication is that the stock is not sensitive to market changes.

Sales ($ millions),

Year Y t Found by

2005 7 1 7.4 6.1 � 1.3(1) 2006 10 2 8.7 6.1 � 1.3(2) 2007 9 3 10 6.1 � 1.3(3) 2008 11 4 11.3 6.1 � 1.3(4) 2009 13 5 12.6 6.1 � 1.3(5)

Ŷ

TABLE 16–4 Calculations for Determining the Points on the Least Squares Line Using the Coded Values

In this time series example, there were five years of sales data. Based on those five sales figures, we estimated sales for 2012. Many researchers suggest that we do not project sales, production, and other business and economic series more than n�2 time periods into the future where n is the number of data points. If, for example, there are 10 years of data, we would make estimates only up to 5 years into the future (n�2 � 10�2 � 5). Others suggest the forecast may be for no longer than 2 years, especially in rapidly changing economic times.

Self-Review 16–2 Annual production of king-size rockers by Wood Products Inc. since 2002 follows.

(a) Plot the production data. (b) Determine the least squares equation using a software package. (c) Determine the points on the line for 2002 and 2009. Connect the two points to arrive

at the line. (d) Based on the linear trend equation, what is the estimated production for 2012?

Production Production Year (thousands) Year (thousands)

2002 4 2006 11 2003 8 2007 9 2004 5 2008 11 2005 8 2009 14

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Year Net Sales Year Net Sales Year Net Sales

1993 $ 9,239 1999 $38,434 2005 $81,511 1994 12,477 2000 45,738 2006 90,837 1995 15,470 2001 53,553 2007 77,349 1996 19,535 2002 58,247 2008 71,300 1997 24,156 2003 64,816 2009 66,200 1998 30,219 2004 73,094

Determine the least squares equation. According to this information, what is the esti- mated number of rentals for 2010?

4. Listed below is the net sales in $ million for Home Depot Inc. and its subsidiaries from 1993 to 2009.

Exercises 3. Listed below is the number of rooms rented at Plantation Resorts of Georgia for the years

from 1999 to 2009.

Year Rental Year Rental Year Rental

1999 6,714 2003 9,762 2007 6,162 2000 7,991 2004 10,180 2008 6,897 2001 9,075 2005 8,334 2009 8,285 2002 9,775 2006 8,272

Determine the least squares equation. On the basis of this information, what are the esti- mated sales for 2010 and 2011?

5. The following table lists the annual amounts of glass cullet produced by Kimble Glass Works Inc.

Scrap Year Code (tons)

2006 1 2 2007 2 4 2008 3 3

Scrap Year Code (tons)

2009 4 5 2010 5 6

Determine the least squares trend equation. Estimate the amount of scrap for the year 2012. 6. The sales by Walker’s Milk and Dairy Products in millions of dollars for the period from

2004 to 2010 are reported in the following table.

Year Code Sales ($ millions)

2004 1 17.5 2005 2 19.0 2006 3 21.0 2007 4 22.7

Year Code Sales ($ millions)

2008 5 24.5 2009 6 26.7 2010 7 27.3

Determine the least squares regression trend equation. Estimate the sales for 2012.

16.7 Nonlinear Trends The emphasis in the previous discussion was on a time series whose growth or decline approximated a straight line. A linear trend equation is used to represent the time series when it is believed that the data are increasing (or decreasing) by equal amounts, on the average, from one period to another.

Data that increase (or decrease) by increasing amounts over a period of time appear curvilinear when plotted on an arithmetic scale. To put it another way, data

LO5 Compute a nonlinear trend equation.

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Time Series and Forecasting 619

that increase (or decrease) by equal percents or proportions over a period of time appear curvilinear. (See Chart 16–7.)

The trend equation for a time series that does approximate a curvilinear trend, such as the one portrayed in Chart 16–7, is computed by using the logarithms of the data and the least squares method. The general equation for the logarithmic trend equation is:

LOG TREND EQUATION [16–2]

The logarithmic trend equation can be determined for the Gulf Shores Importers data in Chart 16–7 using Excel. The first step is to enter the data, then find the log base 10 of each year’s imports. Finally, use the regression procedure to find the least squares equation. To put it another way, we take the log of each year’s data, then use the logs as the dependent variable and the coded year as the independent variable.

log Ŷ � log a � log b(t)

1996 0.0

2000.0

4000.0

6000.0

8000.0

10000.0

12000.0

14000.0

16000.0

18000.0

20000.0

’97 ’98 ’99 ’00 ’01 ’02 ’03 ’04 ’05 ’06 ’07 ’08 ’09 ’10

Sa le

s ($

00 0)

Year

CHART 16–7 Sales for Gulf Shores Importers from 1996–2010

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620 Chapter 16

The regression equation is � 2.053805 � 0.153357t. This equation is the log form. We now have a trend equation in terms of percent of change. That is, the value 0.153357 is the percent of change in for each unit increase in t. This value is similar to the geometric mean described in Section 3.10 in Chapter 3.

The log of b is 0.153357 and its antilog or inverse is 1.423498. If we subtract 1 from this value, as we did in Chapter 3, the value 0.423498 indicates the geo- metric mean annual rate of increase from 1996 to 2010. We conclude that imports increased at a rate of 42.35 percent annually during the period.

We can also use the logarithmic trend equation to make estimates of future val- ues. Suppose we want to estimate the imports in the year 2014. The first step is to determine the code for the year 2014. It is 19. To explain, the year 2010 has a code of 15 and the year 2014 is four years later, so 15 � 4 � 19. The log of imports for the year 2014 is

To find the estimated imports for the year 2014, we need the antilog of 4.967588. It is 92,809. This is our estimate of the number of imports for 2014. Recall that the data were in thousands of dollars, so the estimate is $92,809,000.

Ŷ � 2.053805 � 0.153357t � 2.053805 � 0.153357(19) � 4.967588

Ŷ

Self-Review 16–3 Sales at Tomlin Manufacturing since 2006 are:

(a) Determine the logarithmic trend equation for the sales data. (b) Sales increased by what percentage annually from 2006–2010? (c) What is the projected sales amount for 2011?

Sales Year ($ millions)

2006 2.13 2007 18.10 2008 39.80 2009 81.40 2010 112.00

Exercises 7. Sally’s Software Inc. is a rapidly growing supplier of computer software to the Sarasota

area. Sales for the last five years are given on the left.

a. Determine the logarithmic trend equation. b. By what percent did sales increase, on the average, during the period? c. Estimate sales for the year 2010.

8. It appears that the imports of carbon black have been increasing by about 10 percent annually.

Sales Year ($ millions)

2003 1.1 2004 1.5 2005 2.0 2006 2.4 2007 3.1 Imports Imports

of Carbon Black of Carbon Black Year (thousands of tons) Year (thousands of tons)

2000 92.0 2004 135.0 2001 101.0 2005 149.0 2002 112.0 2006 163.0 2003 124.0 2007 180.0

a. Determine the logarithmic trend equation. b. By what percent did imports increase, on the average, during the period? c. Estimate imports for the year 2010.

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Time Series and Forecasting 621

16.8 Seasonal Variation We mentioned that seasonal variation is another of the components of a time series. Business series, such as automobile sales, shipments of soft-drink bottles, and residential construction, have periods of above- average and below-average activity each year. In the area of production, one of the reasons for analyzing seasonal fluctuations is to have a suf- ficient supply of raw materials on hand to meet the varying seasonal demand. The glass container division of a large glass company, for example, manufactures nonreturnable beer bottles, iodine bottles, aspirin bottles, bottles for rubber cement, and so on. The production scheduling department must know how many bottles to produce and when to produce each kind. A run of too many bottles of one kind may cause a serious storage problem. Production cannot be based entirely on orders on hand, because many orders are telephoned in for immediate shipment. Since the demand for many of the bottles varies according to the season, a forecast a year or two in advance, by month, is essen- tial to good scheduling.

An analysis of seasonal fluctuations over a period of years can also help in evaluating current sales. The typical sales of department stores in the United States, excluding mail-order sales, are expressed as indexes in Table 16–5. Each index represents the average sales for a period of sev- eral years. The actual sales for some months were above average (which

is represented by an index over 100.0), and the sales for other months were below average. The index of 126.8 for December indicates that, typically, sales for December are 26.8 percent above an average month; the index of 86.0 for July indicates that department store sales for July are typically 14 percent below an average month.

January 87.0 July 86.0 February 83.2 August 99.7 March 100.5 September 101.4 April 106.5 October 105.8 May 101.6 November 111.9 June 89.6 December 126.8

TABLE 16–5 Typical Seasonal Indexes for U.S. Department Store Sales, Excluding Mail-Order Sales

Suppose an enterprising store manager, in an effort to stimulate sales during December, introduced a number of unique promotions, including bands of carol- ers strolling through the store singing holiday songs, large mechanical exhibits, and clerks dressed in Santa Claus costumes. When the index of sales was com- puted for that December, it was 150.0. Compared with the typical December sales of 126.8, it was concluded that the promotional program was a huge success.

Determining a Seasonal Index A typical set of monthly indexes consists of 12 indexes that are representative of the data for a 12-month period. Logically, there are four typical seasonal indexes for data reported quarterly. Each index is a percent, with the average for the year equal to 100.0; that is, each monthly index indicates the level of sales, production, or another variable in relation to the annual average of 100.0. A typical index of 96.0 for January indicates that sales (or whatever the variable is) are usually 4 percent below the average for the year. An index of 107.2 for October means that the variable is typically 7.2 percent above the annual average.

Several methods have been developed to measure the typical seasonal fluctu- ation in a time series. The method most commonly used to compute the typical

LO6 Determine and interpret a set of seasonal indexes.

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622 Chapter 16

seasonal pattern is called the ratio-to-moving-average method. It eliminates the trend, cyclical, and irregular components from the original data (Y ). In the following discussion, T refers to trend, C to cyclical, S to seasonal, and I to irregular varia- tion. The numbers that result are called the typical seasonal index.

We will discuss in detail the steps followed in arriving at typical seasonal indexes using the ratio-to-moving-average method. The data of interest might be monthly or quarterly. To illustrate, we have chosen the quarterly sales of Toys International. First, we will show the steps needed to arrive at a set of typical quarterly indexes. Then we use MegaStat Excel and Minitab software to calculate the seasonal indexes.

Example Table 16–6 shows the quarterly sales for Toys International for the years 2004 through 2009. The sales are reported in millions of dollars. Determine a quarterly seasonal index using the ratio-to-moving-average method.

Solution Chart 16–8 depicts the quarterly sales for Toys International over the six-year period.Notice the seasonal nature of the sales. For each year, the fourth-quarter sales are the largest and the second-quarter sales are the smallest. Also, there is a moderate increase in the sales from one year to the next. To observe this feature, look only at the six fourth-quarter sales values. Over the six-year period, the sales in the fourth quarter increased. If you connect these points in your mind, you can visualize fourth- quarter sales increasing for 2010.

Year Winter Spring Summer Fall

2004 6.7 4.6 10.0 12.7 2005 6.5 4.6 9.8 13.6 2006 6.9 5.0 10.4 14.1 2007 7.0 5.5 10.8 15.0 2008 7.1 5.7 11.1 14.5 2009 8.0 6.2 11.4 14.9

TABLE 16–6 Quarterly Sales of Toys International ($ millions)

CHART 16–8 Quarterly Sales of Toys International 2004–2009

There are six steps to determining the quarterly seasonal indexes.

Step 1: For the following discussion, refer to Table 16–7. The first step is to de- termine the four-quarter moving total for 2004. Starting with the winter quarter of 2004, we add $6.7, $4.6, $10.0, and $12.7. The total is $34.0 (million). The four-quarter total is “moved along” by adding the spring,

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Time Series and Forecasting 623

(1) (2) (3) (4) (5) Four-Quarter Centered

Sales Four-Quarter Moving Moving Specific Year Quarter ($ millions) Total Average Average Seasonal

2004 Winter 6.7

Spring 4.6 34.0 8.500

Summer 10.0 8.475 1.180 33.8 8.450

Fall 12.7 8.450 1.503 33.8 8.450

2005 Winter 6.5 8.425 0.772 33.6 8.400

Spring 4.6 8.513 0.540 34.5 8.625

Summer 9.8 8.675 1.130 34.9 8.725

Fall 13.6 8.775 1.550 35.3 8.825

2006 Winter 6.9 8.900 0.775 35.9 8.975

Spring 5.0 9.038 0.553 36.4 9.100

Summer 10.4 9.113 1.141 36.5 9.125

Fall 14.1 9.188 1.535 37.0 9.250

2007 Winter 7.0 9.300 0.753 37.4 9.350

Spring 5.5 9.463 0.581 38.3 9.575

Summer 10.8 9.588 1.126 38.4 9.600

Fall 15.0 9.625 1.558 38.6 9.650

2008 Winter 7.1 9.688 0.733 38.9 9.725

Spring 5.7 9.663 0.590 38.4 9.600

Summer 11.1 9.713 1.143 39.3 9.825

Fall 14.5 9.888 1.466 39.8 9.950

2009 Winter 8.0 9.888 0.801 40.1 10.025

Spring 6.2 10.075 0.615 40.5 10.125

Summer 11.4

Fall 14.9

TABLE 16–7 Computations Needed for the Specific Seasonal Indexes

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624 Chapter 16

summer, and fall sales of 2004 to the winter sales of 2005. The total is $33.8 (million), found by 4.6 � 10.0 � 12.7 � 6.5. This procedure is continued for the quarterly sales for each of the six years. Column 2 of Table 16–7 shows all of the moving totals. Note that the moving total 34.0 is positioned between the spring and summer sales of 2004. The next moving total, 33.8, is positioned between sales for summer and fall 2004, and so on. Check the totals frequently to avoid arithmetic errors.

Step 2: Each quarterly moving total in column 2 is divided by 4 to give the four- quarter moving average. (See column 3.) All the moving averages are still positioned between the quarters. For example, the first moving average (8.500) is positioned between spring and summer 2004.

Step 3: The moving averages are then centered. The first centered moving aver- age is found by (8.500 � 8.450)�2 � 8.475 and centered opposite sum- mer 2004. The second moving average is found by (8.450 � 8.450)�2 � 8.450. The others are found similarly. Note in column 4 that a centered moving average is positioned on a particular quarter.

Step 4: The specific seasonal index for each quarter is then computed by di- viding the sales in column 1 by the centered moving average in column 4. The specific seasonal index reports the ratio of the original time series value to the moving average. To explain further, if the time series is repre- sented by TSCI and the moving average by TC, then, algebraically, if we compute TSCI�TC, the result is the specified seasonal component SI. The specific seasonal index for the summer quarter of 2004 is 1.180, found by 10.0�8.475.

Step 5: The specific seasonal indexes are organized in Table 16–8. This table will help us locate the specific seasonals for the corresponding quarters. The values 1.180, 1.130, 1.141, 1.126, and 1.143 all represent estimates of the typical seasonal index for the summer quarter. A reasonable method to find a typical seasonal index is to average these values in order to eliminate the irregular component. So we find the typical index for the summer quarter by (1.180 � 1.130 � 1.141 � 1.126 � 1.143)�5 � 1.144. We used the arithmetic mean, but the median or a modified mean can also be used.

Year Winter Spring Summer Fall

2004 1.180 1.503 2005 0.772 0.540 1.130 1.550 2006 0.775 0.553 1.141 1.535 2007 0.753 0.581 1.126 1.558 2008 0.733 0.590 1.143 1.466 2009 0.801 0.615

Total 3.834 2.879 5.720 7.612 Mean 0.767 0.576 1.144 1.522 4.009 Adjusted 0.765 0.575 1.141 1.519 4.000 Index 76.5 57.5 114.1 151.9

TABLE 16–8 Calculations Needed for Typical Quarterly Indexes

Step 6: The four quarterly means (0.767, 0.576, 1.144, and 1.522) should theo- retically total 4.00 because the average is set at 1.0. The total of the four quarterly means may not exactly equal 4.00 due to rounding. In this prob- lem, the total of the means is 4.009. A correction factor is therefore ap- plied to each of the four means to force them to total 4.00.

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Time Series and Forecasting 625

In this example,

The adjusted winter quarterly index is, therefore, .767(.997755) � .765. Each of the means is adjusted downward so that the total of the four quarterly means is 4.00. Usually indexes are reported as percentages, so each value in the last row of Table 16–8 has been multiplied by 100. So the index for the winter quarter is 76.5 and for the fall it is 151.9. How are these values interpreted? Sales for the fall quar- ter are 51.9 percent above the typical quarter, and for winter they are 23.5 below the typical quarter (100.0 � 76.5). These findings should not surprise you. The period prior to Christmas (the fall quarter) is when toy sales are brisk. After Christmas (the winter quarter), sales of the toys decline drastically.

Correction factor � 4.00 4.009

� 0.997755

As we noted earlier, there is software that will perform the calculations and output the results. The MegaStat Excel output is shown below. Use of software will greatly reduce the computational time and the chance of an error in arithmetic, but you should understand the steps in the process, as outlined earlier. There can be slight differences in the answers, due to the number of digits carried in the calculations.

Centered Moving Average and Deseasonalization

Centered Moving Ratio to Seasonal Sales

t Year Quarter Sales Average CMA Indexes Deseasonalized

1 2004 1 6.70 0.765 8.759 2 2004 2 4.60 0.575 8.004 3 2004 3 10.00 8.475 1.180 1.141 8.761 4 2004 4 12.70 8.450 1.503 1.519 8.361

5 2005 1 6.50 8.425 0.772 0.765 8.498 6 2005 2 4.60 8.513 0.540 0.575 8.004 7 2005 3 9.80 8.675 1.130 1.141 8.586 8 2005 4 13.60 8.775 1.550 1.519 8.953

9 2006 1 6.90 8.900 0.775 0.765 9.021 10 2006 2 5.00 9.038 0.553 0.575 8.700 11 2006 3 10.40 9.113 1.141 1.141 9.112 12 2006 4 14.10 9.188 1.535 1.519 9.283

13 2007 1 7.00 9.300 0.753 0.765 9.151 14 2007 2 5.50 9.463 0.581 0.575 9.570 15 2007 3 10.80 9.588 1.126 1.141 9.462 16 2007 4 15.00 9.625 1.558 1.519 9.875

17 2008 1 7.10 9.688 0.733 0.765 9.282 18 2008 2 5.70 9.663 0.590 0.575 9.918 19 2008 3 11.10 9.713 1.143 1.141 9.725 20 2008 4 14.50 9.888 1.466 1.519 9.546

21 2009 1 8.00 9.988 0.801 0.765 10.459 22 2009 2 6.20 10.075 0.615 0.575 10.788 23 2009 3 11.40 1.141 9.988 24 2009 4 14.90 1.519 9.809

CORRECTION FACTOR FOR ADJUSTING QUARTERLY MEANS

Correction factor � 4.00

Total of four means [16–3]

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Calculation of Seasonal Indexes

1 2 3 4 2004 1.180 1.503 2005 0.772 0.540 1.130 1.550 2006 0.775 0.553 1.141 1.535 2007 0.753 0.581 1.126 1.558 2008 0.733 0.590 1.143 1.466 2009 0.801 0.615

Mean: 0.767 0.576 1.144 1.522 4.009 Adjusted: 0.765 0.575 1.141 1.519 4.000

Now we briefly summarize the reasoning underlying the preceding calculations. The original data in column 1 of Table 16–7 contain trend (T ), cyclical (C), seasonal (S), and irregular (I) components. The ultimate objective is to remove seasonal (S) from the original sales valuation.

Columns 2 and 3 in Table 16–7 are concerned with deriving the centered mov- ing average given in column 4. Basically, we “average out” the seasonal and irreg- ular fluctuations from the original data in column 1. Thus, in column 4 we have only trend and cyclical (TC).

Next, we divide the sales data in column 1 (TCSI) by the centered fourth-quarter moving average in column 4 (TC) to arrive at the specific seasonals in column 5 (SI). In terms of letters, TCSI�TC � SI. We multiply SI by 100.0 to express the typical seasonal in index form.

Finally, we take the mean of all the winter typical indexes, all the spring indexes, and so on. This averaging eliminates most of the irregular fluctuations from the specific seasonals, and the resulting four indexes indicate the typical seasonal sales pattern.

Self-Review 16–4 Teton Village, Wyoming, near Grand Teton Park and Yellowstone Park, contains shops, restaurants, and motels. The village has two peak seasons—winter, for skiing on the 10,000-foot slopes, and summer, for tourists visiting the parks. The number of visitors (in thousands) by quarter for five years follows.

(a) Develop the typical seasonal pattern for Teton Village using the ratio-to-moving-average method.

(b) Explain the typical index for the winter season.

Quarter

Year Winter Spring Summer Fall

2006 117.0 80.7 129.6 76.1 2007 118.6 82.5 121.4 77.0 2008 114.0 84.3 119.9 75.0 2009 120.7 79.6 130.7 69.6 2010 125.2 80.2 127.6 72.0

626 Chapter 16

Exercises 9. Victor Anderson, the owner of Anderson Belts Inc., is studying absenteeism among

his employees. His workforce is small, consisting of only five employees. For the last three years, he recorded the following number of employee absences, in days, for each quarter.

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Time Series and Forecasting 627

Determine a typical seasonal index for each of the four quarters. 10. Appliance Center sells a variety of electronic equipment and home appliances. For the

last four years, the following quarterly sales (in $ millions) were reported.

Quarter

Year I II III IV

2008 4 10 7 3 2009 5 12 9 4 2010 6 16 12 4

16.9 Deseasonalizing Data A set of typical indexes is very useful in adjusting a sales series, for example, for seasonal fluctuations. The resulting sales series is called deseasonalized sales or seasonally adjusted sales. The reason for deseasonalizing the sales series is to remove the seasonal fluctuations so that the trend and cycle can be studied. To illustrate the procedure, the quarterly sales totals of Toys International from Table 16–6 are repeated in column 1 of Table 16–9.

Quarter

Year I II III IV

2007 5.3 4.1 6.8 6.7 2008 4.8 3.8 5.6 6.8 2009 4.3 3.8 5.7 6.0 2010 5.6 4.6 6.4 5.9

Determine a typical seasonal index for each of the four quarters.

(1) (2) (3) Seasonal Deseasonalized

Year Quarter Sales Index Sales

2004 Winter 6.7 0.765 8.76 Spring 4.6 0.575 8.00 Summer 10.0 1.141 8.76 Fall 12.7 1.519 8.36

2005 Winter 6.5 0.765 8.50 Spring 4.6 0.575 8.00 Summer 9.8 1.141 8.59 Fall 13.6 1.519 8.95

2006 Winter 6.9 0.765 9.02 Spring 5.0 0.575 8.70 Summer 10.4 1.141 9.11 Fall 14.1 1.519 9.28

2007 Winter 7.0 0.765 9.15 Spring 5.5 0.575 9.57 Summer 10.8 1.141 9.47 Fall 15.0 1.519 9.87

(continued )

TABLE 16–9 Actual and Deseasonalized Sales for Toys International

LO7 Deseasonalize data using a seasonal index.

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628 Chapter 16

To remove the effect of seasonal variation, the sales amount for each quarter (which contains trend, cyclical, irregular, and seasonal effects) is divided by the sea- sonal index for that quarter, that is, TSCI�S. For example, the actual sales for the first quarter of 2004 were $6.7 million. The seasonal index for the winter quarter is 76.5 percent, using the MegaStat results on page 626. The index of 76.5 indicates that sales for the first quarter are typically 23.5 percent below the average for a typical quarter. By dividing the actual sales of $6.7 million by 76.5 and multiplying the result by 100, we find the deseasonalized sales value—that is, removed the seasonal effect on sales—for the first quarter of 2004. It is $8,758,170, found by ($6,700,000�76.5)100. We continue this process for the other quarters in column 3 of Table 16–9, with the results reported in millions of dollars. Because the seasonal component has been removed (divided out) from the quarterly sales, the deseasonalized sales figure con- tains only the trend (T ), cyclical (C), and irregular (I ) components. Scanning the desea- sonalized sales in column 3 of Table 16–9, we see that the sales of toys showed a moderate increase over the six-year period. Chart 16–9 shows both the actual sales and the deseasonalized sales. It is clear that removing the seasonal factor allows us to focus on the overall long-term trend of sales. We will also be able to determine the regression equation of the trend data and use it to forecast future sales.

CHART 16–9 Actual and Deseasonalized Sales for Toys International from 2004–2009

(1) (2) (3) Seasonal Deseasonalized

Year Quarter Sales Index Sales

2008 Winter 7.1 0.765 9.28 Spring 5.7 0.575 9.91 Summer 11.1 1.141 9.73 Fall 14.5 1.519 9.55

2009 Winter 8.0 0.765 10.46 Spring 6.2 0.575 10.79 Summer 11.4 1.141 9.99 Fall 14.9 1.519 9.81

Using Deseasonalized Data to Forecast The procedure for identifying trend and the seasonal adjustments can be combined to yield seasonally adjusted forecasts. To identify the trend, we determine the least squares trend equation on the deseasonalized historical data. Then we project this trend into future periods, and finally we adjust these trend values to account for the seasonal factors. The following example will help to clarify.

LO8 Calculate seasonally adjusted forecasts.

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Time Series and Forecasting 629

Example Toys International would like to forecast its sales for each quarter of 2010. Use the information in Table 16–9 to determine the forecast.

Solution The deseasonalized data depicted in Chart 16–9 seems to follow a straight line. Hence, it is reasonable to develop a linear trend equation based on these data. The deseasonalized trend equation is:

where: is the estimated trend value for Toys International sales for the period t.

a is the intercept of the trend line at time 0. b is the slope of the line. t is the coded time period.

The winter quarter of 2004 is the first quarter, so it is coded 1, the spring quarter of 2004 is coded 2, and so on. The last quarter of 2009 is coded 24. A portion of these coded values are shown in the data section of the Minitab output associated with Chart 16–9.

We use Minitab to find the regression equation. The output follows. The output includes a scatter diagram of the coded time periods and the deseasonalized sales as well as the regression line.

The equation for the trend line is:

Ŷ � 8.109 � .08991t

Ŷ

Ŷ � a � bt

The Minitab system also outputs the coefficient of determination. This value, called R2, is 78.6 percent. It is shown in the upper right of the Minitab output. We can use this value as an indication of the fit of the data. Because this is not sam- ple information, technically we should not use R2 for judging a regression equation. However, it will serve to quickly evaluate the fit of the deseasonalized sales data. In this instance, because R2 is rather large, we conclude the deseasonalized sales of Toys International are effectively explained by a linear trend equation.

If we assume that the past 24 periods are a good indicator of future sales, we can use the trend equation to estimate future sales. For example, for the winter quarter of 2010 the value of t is 25. Therefore, the estimated sales of that period is 10.35675, found by

The estimated deseasonalized sales for the winter quarter of 2010 are $10,356,750. This is the sales forecast, before we consider the effects of seasonality.

Ŷ � 8.109 � .08991t � 8.109 � .08991(25) � 10.35675

Statistics in Action

Forecasts are not al- ways correct. The re- ality is that a forecast may just be a best guess as to what will happen. What are the reasons forecasts are not correct? One expert lists eight common errors: (1) Failure to carefully examine the assump- tions, (2) Limited expertise, (3) Lack of imagination, (4) Neglect of constraints, (5) Exces- sive optimism, (6) Re- liance on mechanical extrapolation, (7) Premature clo- sure, and (8) Overspecification.

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630 Chapter 16

Self-Review 16–5 Westberg Electric Company sells electric motors to customers in the Jamestown, New York, area. The monthly trend equation, based on five years of monthly data, is

The seasonal factor for the month of January is 120, and it is 95 for February. Determine the seasonally adjusted forecast for January and February of the sixth year.

Ŷ � 4.4 � 0.5t

Exercises 11. The planning department of Padget and Kure Shoes, the manufacturer of an exclusive

brand of women’s shoes, developed the following trend equation, in millions of pairs, based on five years of quarterly data.

The following table gives the seasonal factors for each quarter.

Ŷ � 3.30 � 1.75t

We use the same procedure and an Excel spreadsheet to determine a forecast for each of the four quarters of 2010. A partial Excel output follows.

Now that we have the forecasts for the four quarters of 2010, we can season- ally adjust them. The index for a winter quarter is 0.765. So we can seasonally adjust the forecast for the winter quarter of 2010 by 10.35675(0.765) � 7.923. The esti- mates for each of the four quarters of 2010 are in the right-hand column of the Excel output. Notice how the seasonal adjustments drastically increase the sales estimates for the last two quarters of the year.

Quarter

I II III IV

Index 110.0 120.0 80.0 90.0

Determine the seasonally adjusted forecast for each of the four quarters of the sixth year. 12. Team Sports Inc. sells sporting goods to high schools and colleges via a nationally dis-

tributed catalog. Management at Team Sports estimates it will sell 2,000 Wilson Model A2000 catcher’s mitts next year. The deseasonalized sales are projected to be the same for each of the four quarters next year. The seasonal factor for the second quarter is 145. Determine the seasonally adjusted sales for the second quarter of next year.

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Time Series and Forecasting 631

13. Refer to Exercise 9 regarding the absences at Anderson Belts Inc. Use the seasonal indexes you computed to determine the deseasonalized absences. Determine the linear trend equation based on the quarterly data for the three years. Forecast the seasonally adjusted absences for 2011.

14. Refer to Exercise 10, regarding sales at Appliance Center. Use the seasonal indexes you computed to determine the deseasonalized sales. Determine the linear trend equation based on the quarterly data for the four years. Forecast the seasonally adjusted sales for 2011.

16.10 The Durbin-Watson Statistic Time series data or observations collected successively over a period of time pre- sent a particular difficulty when you use the technique of regression. One of the assumptions traditionally used in regression is that the successive residuals are independent. This means that there is not a pattern to the residuals, the residu- als are not highly correlated, and there are not long runs of positive or negative residuals. In Chart 16–10, the residuals are scaled on the vertical axis and the values along the horizontal axis. Notice there are “runs” of residuals above and below the 0 line. If we computed the correlation between successive residuals, it is likely the correlation would be strong.

Ŷ

CHART 16–10 Correlated Residuals

This condition is called autocorrelation or serial correlation.

AUTOCORRELATION Successive residuals are correlated.

Re si

du al

(0) e1

e2 e3

e4 e5 e6

e7 e8 e9

e10 e11

Successive residuals are correlated in time series data because an event in one time period often influences the event in the next period. To explain, the owner of a furniture store decides to have a sale this month and spends a large amount of money advertising the event. We would expect a correlation between sales and advertising expense, but all the results of the increase in advertising are not experienced this month. It is likely that some of the effect of the advertising carries over into next month. Therefore, we expect correlation among the residuals.

The regression relationship in a time series is written

where the subscript t is used in place of i to suggest the data were collected over time. If the residuals are correlated, problems occur when we try to conduct tests

of hypotheses about the regression coefficients. Also, a confidence interval or a

Yt � � � �1Xt � et

LO9 Test for autocorrelation.

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632 Chapter 16

prediction interval, where the multiple standard error of estimate is used, may not yield the correct results.

The autocorrelation, reported as r, is the strength of the association among the residuals. The r has the same meaning as the coefficient of correlation. That is, val- ues close to �1.00 or 1.00 indicate a strong association, and values near 0 indi- cate no association. Instead of directly conducting a hypothesis test on r, we use the Durbin-Watson statistic.

The Durbin-Watson statistic, identified by the letter d, is computed by first deter- mining the residuals for each observation. That is, et � (Yt � t). Next, we compute d using the following relationship.

Ŷ

DURBIN-WATSON STATISTIC [16–4]d � a n

t�2 (et � et�1)

a n

t�1 (et)

To determine the numerator of formula (16–4), we lag each of the residuals one period and then square the difference between consecutive residuals. This may also be called finding the differences. This accounts for summing the observations from 2, rather than from 1, up to n. In the denominator, we square the residuals and sum over all n observations.

The value of the Durbin-Watson statistic can range from 0 to 4. The value of d is 2.00 when there is no autocorrelation among the residuals. When the value of d gets close to 0, this indicates positive autocorrelation. Values beyond 2 indicate neg- ative autocorrelation. Negative autocorrelation seldom exists in practice. To occur, successive residuals would tend to be large, but would have opposite signs.

To conduct a test for autocorrelation, the null and alternate hypotheses are:

H0: No residual correlation (� � 0) H1: Positive residual correlation (� 0)

Recall from the previous chapter that r refers to the sample correlation and that � is the correlation coefficient in the population. The critical values for d are reported in Appen- dix B.10. To determine the critical value, we need � (the significance level), n (the sample size), and k (the number of independent variables). The decision rule for the Durbin-Watson test is altered from what we are used to. As usual, there is a range of values where the null hypothesis is rejected and a range where it is not rejected. How- ever, there is also a range of values where d is inconclusive. That is, in the inconclusive range the null hypothesis is neither rejected nor not rejected. To state this more formally:

• Values less than dl cause the rejection of the null hypothesis. • Values greater then du will result in the null hypothesis not being rejected. • Values of d between dl and du yield inconclusive results.

The subscript l refers to the lower limit of d and the subscript u the upper limit. How do we interpret the various decisions for the test for residual correlation?

If the null hypothesis is not rejected, we conclude that autocorrelation is not present. The residuals are not correlated, there is no autocorrelation present, and the regression assumption has been met. There will not be any problem with the esti- mated value of the standard error of estimate. If the null hypothesis is rejected, then we conclude that autocorrelation is present.

The usual remedy for autocorrelation is to include another predictor variable that captures time order. For example, we might use the square root of Y instead of Y. This transformation will result in a change in the distribution of the residuals. If the result falls in the inconclusive range, more sophisticated tests are needed, or conservatively, we treat the conclusion as rejecting the null hypothesis.

An example will show the details of the Durbin-Watson test and how the results are interpreted.

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Time Series and Forecasting 633

Example

Solution The first step is to determine the regression equation.

Banner Rocker Company manufactures and markets rocking chairs. The com- pany developed a special rocker for senior citizens, which it advertises exten- sively on TV. Banner’s market for the spe- cial chair is the Carolinas, Florida, and Arizona where there are many senior cit- izens and retired people. The president of Banner Rocker is studying the associa- tion between his advertising expense (X ) and the number of rockers sold over the last 20 months (Y ). He collected the fol-

lowing data. He would like to create a model to forecast sales, based on the amount spent on advertising, but is concerned that, because he gathered these data over consecutive months, there might be problems with autocorrelation.

Advertising Advertising Month Sales (000) ($ millions) Month Sales (000) ($ millions)

1 153 $5.5 11 169 $6.3 2 156 5.5 12 176 5.9 3 153 5.3 13 176 6.1 4 147 5.5 14 179 6.2 5 159 5.4 15 184 6.2 6 160 5.3 16 181 6.5 7 147 5.5 17 192 6.7 8 147 5.7 18 205 6.9 9 152 5.9 19 215 6.5

10 160 6.2 20 209 6.4

Determine the regression equation. Is advertising a good predictor of sales? If the owner were to increase the amount spent on advertising by $1,000,000, how many additional chairs can he expect to sell? Investigate the possibility of autocorrelation.

Regression Analysis: Chairs (000) versus Advertising ($mil)

The regression equation is Chairs (000) � �43.8 � 36.0 Advertising ($mil)

Predictor Coef SE Coef T P Constant �43.80 34.44 �1.27 0.220 Advertising ($mil) 35.950 5.746 6.26 0.000

S � 12.3474 R-Sq � 68.5% R-Sq(adj) � 66.8%

Analysis of Variance

Source DF SS MS F P Regression 1 5967.7 5967.7 39.14 0.000 Residual Error 18 2744.3 152.5

The coefficient of determination is 68.5 percent. So we know there is a strong pos- itive association between the variables. We conclude that, as we increase the amount spent on advertising, we can expect to sell more chairs. Of course this is what we had hoped.

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634 Chapter 16

How many more chairs can we expect to sell if we increase advertising by $1,000,000? We must be careful with the units of the data. Sales are in thousands of chairs and advertising expense is in millions of dollars. The regression equa- tion is:

This equation indicates that an increase of 1 in X will result in an increase of 35.95 in Y. So an increase of $1,000,000 in advertising will increase sales by 35,950 chairs. To put it another way, it will cost $27.82 in additional advertising expense per chair sold, found by $1,000,000�35,950.

What about the potential problem with autocorrelation? Many software pack- ages, such as Minitab, will calculate the value of the Durbin-Watson test and output the results. To understand the nature of the test and to see the details of formula (16–4), we use an Excel spreadsheet.

Ŷ � �43.80 � 35.950X

To investigate the possible autocorrelation, we need to determine the residuals for each observation. We find the fitted values—that is, the —for each of the 20 months. This information is shown in the fourth column, column D. Next we find the residual, which is the difference between the actual value and the fitted values. So for the first month:

The residual, reported in column E, is slightly different due to rounding in the soft- ware. Notice in particular the string of five negative residuals in rows 9 through 13. In column F, we lag the residuals one period. In column G, we find the difference

e1 � Y1 � Ŷ1 � 153 � 153.925 � �0.925

Ŷ � �43.80 � 35.950X � �43.80 � 35.950(5.5) � 153.925

Ŷ

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Time Series and Forecasting 635

between the current residual and the residual in the previous and square this dif- ference. Using the values from the software:

The other values in column G are found the same way. The values in column H are the squares of those in column E.

To find the value of d, we need the sums of columns G and H. These sums are noted in yellow in the spreadsheet.

Now to answer the question as to whether there is significant autocorrelation. The null and the alternate hypotheses are stated as follows.

H0: No residual correlation H1: Positive residual correlation

The critical value of d is found in Appendix B.10, a portion of which is shown below. There is one independent variable, so k � 1, the level of significance is 0.05, and the sample size is 20. We move to the .05 table, the columns where k � 1, and the row of 20. The reported values are dl � 1.20 and du � 1.41. The null hypothesis is rejected if d 1.20 and not rejected if d 1.41. No conclusion is reached if d is between 1.20 and 1.41.

d � a n

t�2 (et � et�1)

a n

t�1 (et)

� 2338.583 2744.269

� 0.8522

(e1) 2 � (�0.9237)2 � 0.8531

(et � et�1) 2 � (e2 � e2�1)

2 � [2.0763 � (�0.9237)]2 � (3.0000)2 � 9.0000

n k

uddl

dudl 15 1.08 1.36 0.95 1.54 16 1.10 1.37 0.98 1.54 17 1.13 1.38 1.02 1.54 18 1.16 1.39 1.05 1.53 19 1.18 1.40 1.08 1.53 20 1.20 1.41 1.10 1.54 21 1.22 1.42 1.13 1.54 22 1.24 1.43 1.15 1.54 23

25 1.29 1.45 1.21 1.55

1.26 1.44 1.17 1.54 24 1.27 1.45 1.19 1.55

Positive autocorrelation Undetermined

InconclusiveReject Ho

0.85

HoAccept

1.20

1.41

autocorrelation No

Because the computed value of d is 0.8522, which is less than the dl, we reject the null hypothesis and accept the alternate hypothesis. We conclude that the residuals are autocorrelated. We have violated one of the regression assumptions. What do we do? The presence of autocorrelation usually means that the regres- sion model has not been correctly specified. It is likely we need to add one or more independent variables that have some time-ordered effects on the depen- dent variable. The simplest independent variable to add is one that represents the time periods.

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636 Chapter 16

Exercises 15. Recall Exercise 9 from Chapter 14 and the regression equation to predict job performance.

See page 544. a. Plot the residuals in the order in which the data are presented. b. Test for autocorrelation at the .05 significance level.

16. Consider the data in Exercise 10 from Chapter 14 and the regression equation to pre- dict commissions earned. See page 545. a. Plot the residuals in the order in which the data are presented. b. Test for autocorrelation at the .01 significance level.

Chapter Summary I. A time series is a collection of data over a period of time.

A. The trend is the long-run direction of the time series. B. The cyclical component is the fluctuation above and below the long-term trend line

over a longer period of time. C. The seasonal variation is the pattern in a time series within a year. These patterns tend

to repeat themselves from year to year for most businesses. D. The irregular variation is divided into two components.

1. The episodic variations are unpredictable, but they can usually be identified. A flood is an example.

2. The residual variations are random in nature. II. A moving average is used to smooth the trend in a time series.

III. The linear trend equation is where a is the Y-intercept, b is the slope of the line, and t is the coded time. A. We use least squares to determine the trend equation. B. If the trend is not linear, but rather the increases tend to be a constant percent, the Y

values are converted to logarithms, and a least squares equation is determined using the logarithms.

IV. A seasonal factor can be estimated using the ratio-to-moving-average method. A. The six-step procedure yields a seasonal index for each period.

1. Seasonal factors are usually computed on a monthly or a quarterly basis. 2. The seasonal factor is used to adjust forecasts, taking into account the effects of

the season. V. The Durbin-Watson statistic [16–4] is used to test for autocorrelation.

[16–4]

Chapter Exercises 17. Refer to the following diagram.

d � a n

t�2 (et � et�1)

a n

t�1 (et)

Ŷ � a � bt,

20,000

18,000

16,000

14,000

12,000

10,000

0 1990 1995 2000 2005 2010

Pr od

uc tio

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Time Series and Forecasting 637

a. Estimate the linear trend equation for the production series by drawing a straight line through the data.

b. What is the average annual decrease in production? c. Based on the trend equation, what is the forecast for the year 2014?

18. Refer to the following diagram. a. Estimate the linear trend equation for the personal income series. b. What is the average annual increase in personal income?

18,000 16,000 14,000 12,000 10,000 8,000 6,000 4,000 2,000

0 1995 2000 2005 2010P

er so

na l i

nc om

e (in

d ol

la rs

)

19. The asset turnovers, excluding cash and short-term investments, for RNC Company from 2000 to 2010 are:

2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010

1.11 1.28 1.17 1.10 1.06 1.14 1.24 1.33 1.38 1.50 1.65

a. Plot the data. b. Determine the least squares trend equation. c. Calculate the points on the trend line for 2003 and 2008, and plot the line on the graph. d. Estimate the asset turnover for 2015. e. How much did the asset turnover increase per year, on the average, from 2000 to 2010?

20. The sales, in billions of dollars, of Keller Overhead Door Inc. for 2005 to 2010 are:

a. Plot the data. b. Determine the least squares trend equation. c. Use the trend equation to calculate the points for 2007 and 2010. Plot them on the

graph and draw the regression line. d. Estimate the net sales for 2013. e. By how much have sales increased (or decreased) per year on the average during the

period? 21. The number of employees, in thousands, of Keller Overhead Door Inc. for the years 2005

to 2010 are:

a. Plot the data. b. Determine the least squares trend equation. c. Use the trend equation to calculate the points for 2007 and 2010. Plot them on the

graph and draw the regression line. d. Estimate the number of employees in 2013. e. By how much has the number of employees increased (or decreased) per year, on the

average, during the period?

Year Sales Year Sales

2005 7.45 2008 7.94 2006 7.83 2009 7.76 2007 8.07 2010 7.90

Year Employees Year Employees

2005 45.6 2008 39.3 2006 42.2 2009 34.0 2007 41.1 2010 30.0

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638 Chapter 16

Year Price Year Price Year Price Year Price

1990 12.9135 1995 27.7538 2000 49.5625 2005 59.85 1991 16.8250 1996 29.0581 2001 48.68 2006 62.00 1992 20.6125 1997 36.0155 2002 42.22 2007 77.51 1993 20.3024 1998 40.6111 2003 46.62 2008 54.77 1994 18.3160 1999 35.0230 2004 52.20 2009 60.80

a. Plot the data. b. Determine the least squares trend equation. c. Calculate the points for the years 1995 and 2000. d. Estimate the selling price in 2011. Does this seem like a reasonable estimate based

on the historical data? e. By how much has the stock price increased or decreased (per year) on average dur-

ing the period? 23. If plotted, the following sales series would appear curvilinear. This indicates that sales

are increasing at a somewhat constant annual rate (percent). To fit the sales, therefore, a logarithmic equation should be used.

Sales Sales Year ($ millions) Year ($ millions)

2000 8.0 2006 39.4 2001 10.4 2007 50.5 2002 13.5 2008 65.0 2003 17.6 2009 84.1 2004 22.8 2010 109.0 2005 29.3

22. Listed below is the selling price for a share of PepsiCo Inc. at the close of each year.

a. Determine the logarithmic equation. b. Determine the coordinates of the points on the logarithmic straight line for 1997 and 2006. c. By what percent did sales increase per year, on the average, during the period from

2000 to 2008? d. Based on the equation, what are the estimated sales for 2009?

24. Reported below are the amounts spent on advertising ($ millions) by a large firm from 2000 to 2010.

Year Amount Year Amount

2000 88.1 2006 132.6 2001 94.7 2007 141.9 2002 102.1 2008 150.9 2003 109.8 2009 157.9 2004 118.1 2010 162.6 2005 125.6

a. Determine the logarithmic trend equation. b. Estimate the advertising expenses for 2013. c. By what percent per year did advertising expense increase during the period?

25. Listed below is the selling price for a share of Oracle Inc. stock at the close of the year.

Year Price Year Price Year Price Year Price

1990 0.1944 1995 3.1389 2000 29.0625 2005 12.21 1991 0.3580 1996 4.6388 2001 13.81 2006 19.11 1992 0.7006 1997 3.7188 2002 10.80 2007 20.23 1993 1.4197 1998 7.1875 2003 13.23 2008 17.73 1994 2.1790 1999 28.0156 2004 13.72 2009 24.53

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Time Series and Forecasting 639

a. Plot the data. b. Determine the least squares trend equation. Use both the actual stock price and the

logarithm of the price. Which seems to yield a more accurate forecast? c. Calculate the points for the years 1993 and 1998. d. Estimate the selling price in 2012. Does this seem like a reasonable estimate based

on the historical data? e. By how much has the stock price increased or decreased (per year) on average dur-

ing the period? Use your best answer from part (b). 26. The production of Reliable Manufacturing Company for 2009 and part of 2010 follows.

2009 2010 2009 2010 Production Production Production Production

Month (thousands) (thousands) Month (thousands) (thousands)

January 6 7 July 3 4 February 7 9 August 5 March 12 14 September 14 April 8 9 October 6 May 4 5 November 7 June 3 4 December 6

a. Using the ratio-to-moving-average method, determine the specific seasonals for July, August, and September 2009.

b. Assume that the specific seasonal indexes in the following table are correct. Insert in the table the specific seasonals you computed in part (a) for July, August, and September 2009, and determine the 12 typical seasonal indexes.

c. Interpret the typical seasonal index. 27. The sales of Andre’s Boutique for 2009 and part of 2010 are:

Year Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec.

2009 ? ? ? 92.1 106.5 92.9 2010 88.9 102.9 178.9 118.2 60.1 43.1 44.0 74.0 200.9 90.0 101.9 90.9 2011 87.6 103.7 170.2 125.9 59.4 48.6 44.2 77.2 196.5 89.6 113.2 80.6 2012 79.8 105.6 165.8 124.7 62.1 41.7 48.2 72.1 203.6 80.2 103.0 94.2 2013 89.0 112.1 182.9 115.1 57.6 56.9

2009 Sales 2010 Sales 2009 Sales 2010 Sales Month (thousands) (thousands) Month (thousands) (thousands)

January 78 65 July 81 65 February 72 60 August 85 61 March 80 72 September 90 75 April 110 97 October 98 May 92 86 November 115 June 86 72 December 130

a. Using the ratio-to-moving-average method, determine the specific seasonals for July, August, September, and October 2009.

b. Assume that the specific seasonals in the following table are correct. Insert in the table the specific seasonals you computed in part (a) for July, August, September, and October 2009, and determine the 12 typical seasonal indexes.

Year Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec.

2009 ? ? ? ? 123.6 150.9 2010 83.9 77.6 86.1 118.7 99.7 92.0 87.0 91.4 97.3 105.4 124.9 140.1 2011 86.7 72.9 86.2 121.3 96.6 92.0 85.5 93.6 98.2 103.2 126.1 141.7 2012 85.6 65.8 89.2 125.6 99.6 94.4 88.9 90.2 100.2 102.7 121.6 139.6 2013 77.3 81.2 85.8 115.7 100.3 89.7

c. Interpret the typical seasonal index.

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640 Chapter 16

28. The quarterly production of pine lumber, in millions of board feet, by Northwest Lumber since 2006 is:

Quarter

Year Winter Spring Summer Fall

2006 7.8 10.2 14.7 9.3 2007 6.9 11.6 17.5 9.3 2008 8.9 9.7 15.3 10.1 2009 10.7 12.4 16.8 10.7 2010 9.2 13.6 17.1 10.3

a. Determine the typical seasonal pattern for the production data using the ratio-to- moving-average method.

b. Interpret the pattern. c. Deseasonalize the data and determine the linear trend equation. d. Project the seasonally adjusted production for the four quarters of 2011.

29. Work Gloves Corp. is reviewing its quarterly sales of Toughie, the most durable glove it produces. The numbers of pairs produced (in thousands) by quarter are:

a. Using the ratio-to-moving-average method, determine the four typical quarterly indexes. b. Interpret the typical seasonal pattern.

30. Sales of roof material, by quarter, since 2004 by Carolina Home Construction Inc. are shown below (in $000).

Quarter

I II III IV Year Jan.–Mar. Apr.–June July–Sept. Oct.–Dec.

2005 142 312 488 208 2006 146 318 512 212 2007 160 330 602 187 2008 158 338 572 176 2009 162 380 563 200 2010 162 362 587 205

Quarter

Year I II III IV

2004 210 180 60 246 2005 214 216 82 230 2006 246 228 91 280 2007 258 250 113 298 2008 279 267 116 304 2009 302 290 114 310 2010 321 291 120 320

a. Determine the typical seasonal patterns for sales using the ratio-to-moving-average method.

b. Deseasonalize the data and determine the trend equation. c. Project the sales for 2011, and then seasonally adjust each quarter.

31. Blueberry Farms Golf and Fish Club of Hilton Head, South Carolina, wants to find monthly seasonal indexes for package play, nonpackage play, and total play. The pack- age play refers to golfers who visit the area as part of a golf package. Typically, the greens fees, cart fees, lodging, maid service, and meals are included as part of a golfing package.

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Time Series and Forecasting 641

The course earns a certain percentage of this total. The nonpackage play includes play by local residents and visitors to the area who wish to play golf. The following data, beginning with July 2007, report the package and nonpackage play by month, as well as the total amount, in thousands of dollars.

Using statistical software: a. Develop a seasonal index for each month for the package sales. What do you note

about the various months? b. Develop a seasonal index for each month for the nonpackage sales. What do you

note about the various months? c. Develop a seasonal index for each month for the total sales. What do you note about

the various months? d. Compare the indexes for package sales, nonpackage sales, and total sales. Are the

busiest months the same? 32. The following is the number of retirees receiving benefits from the State Teachers Retire-

ment System of Ohio from 1991 until 2009.

Year Month Package Local Total

2007 July $ 18.36 $43.44 $ 61.80 August 28.62 56.76 85.38 September 101.34 34.44 135.78 October 182.70 38.40 221.10 November 54.72 44.88 99.60 December 36.36 12.24 48.60

2008 January 25.20 9.36 34.56 February 67.50 25.80 93.30 March 179.37 34.44 213.81 April 267.66 34.32 301.98 May 179.73 40.80 220.53 June 63.18 40.80 103.98 July 16.20 77.88 94.08 August 23.04 76.20 99.24 September 102.33 42.96 145.29 October 224.37 51.36 275.73 November 65.16 25.56 90.72 December 22.14 15.96 38.10

Year Month Package Local Total

2009 January 30.60 9.48 40.08 February 63.54 30.96 94.50 March 167.67 47.64 215.31 April 299.97 59.40 359.37 May 173.61 40.56 214.17 June 64.98 63.96 128.94 July 25.56 67.20 92.76 August 31.14 52.20 83.34 September 81.09 37.44 118.53 October 213.66 62.52 276.18 November 96.30 35.04 131.34 December 16.20 33.24 49.44

2010 January 26.46 15.96 42.42 February 72.27 35.28 107.55 March 131.67 46.44 178.11 April 293.40 67.56 360.96 May 158.94 59.40 218.34 June 79.38 60.60 139.98

a. Plot the data. b. Determine the least squares trend equation. Use a linear equation. c. Calculate the points for the years 1993 and 1998. d. Estimate the number of retirees that will be receiving benefits in 2012. Does this seem

like a reasonable estimate based on the historical data? e. By how much has the number of retirees increased or decreased (per year) on average

during the period? 33. Ray Anderson, owner of Anderson Ski Lodge in upstate New York, is interested in fore-

casting the number of visitors for the upcoming year. The following data are available, by quarter, since 2004. Develop a seasonal index for each quarter. How many visitors would you expect for each quarter of 2010, if Ray projects that there will be a 10 per- cent increase from the total number of visitors in 2011? Determine the trend equation,

Year Service Year Service Year Service Year Service

1991 58,436 1996 70,448 2001 83,918 2006 99,248 1992 59,994 1997 72,601 2002 86,666 2007 102,771 1993 61,515 1998 75,482 2003 89,257 2008 106,099 1994 63,182 1999 78,341 2004 92,574 2009 109,031 1995 67,989 2000 81,111 2005 95,843

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642 Chapter 16

Year Quarter Visitors Year Quarter Visitors

2004 I 86 2008 I 188 II 62 II 172 III 28 III 128 IV 94 IV 198

2005 I 106 2009 I 208 II 82 II 202 III 48 III 154 IV 114 IV 220

2006 I 140 2010 I 246 II 120 II 240 III 82 III 190 IV 154 IV 252

2007 I 162 II 140 III 100 IV 174

34. The enrollment in the College of Business at Midwestern University by quarter since 2006 is:

Quarter

Year Winter Spring Summer Fall

2006 2,033 1,871 714 2,318 2007 2,174 2,069 840 2,413 2008 2,370 2,254 927 2,704 2009 2,625 2,478 1,136 3,001 2010 2,803 2,668 — —

Using the ratio-to-moving-average method: a. Determine the four quarterly indexes. b. Interpret the quarterly pattern of enrollment. Does the seasonal variation surprise you? c. Compute the trend equation, and forecast the 2011 enrollment by quarter.

35. The Jamie Farr Kroger Classic is an LPGA (women’s professional golf) tournament played in Toledo, Ohio, each year. Listed below are the total purse and the prize for the winner for the 22 years from 1988 through 2009. Develop a trend equation for both variables. Which variable is increasing at a faster rate? Project both the amount of the purse and the prize for the winner in 2011. Find the ratio of the winner’s prize to the total purse. What do you find? Which variable can we estimate more accurately, the size of the purse or the winner’s prize?

project the number of visitors for 2011, and seasonally adjust the forecast. Which fore- cast would you choose?

Year Purse Prize Year Purse Prize

1988 $275,000 $ 41,250 1999 $ 800,000 $120,000 1989 275,000 41,250 2000 1,000,000 150,000 1990 325,000 48,750 2001 1,000,000 150,000 1991 350,000 52,500 2002 1,000,000 150,000 1992 400,000 60,000 2003 1,000,000 150,000 1993 450,000 67,500 2004 1,200,000 180,000 1994 500,000 75,000 2005 1,200,000 180,000 1995 500,000 75,000 2006 1,200,000 180,000 1996 575,000 86,250 2007 1,300,000 195,000 1997 700,000 105,000 2008 1,300,000 195,000 1998 800,000 120,000 2009 1,400,000 210,000

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36. Go to the Bureau of Labor Statistics website, www.bls.gov, and click on the Consumer Price Index option, select Consumer Price Index—All Urban Consumers (Current Series), select U.S. All items, 1982–84 � 100, and click Retrieve data at the bottom. Ask for the yearly output for the last 10 to 20 years. Develop a regression equation for the annual Consumer Price Index for the selected period. Use both the linear and the log approach. Which do you think is best?

37. Develop a trend line for a large or well-known company, such as GM, General Electric, or Microsoft, for the last 10 years. You could go to the company website. Most compa- nies have a section called “Financial Information” or similar. Go to that location and look for sales over the last 10 years. If you do not know the website of the company, go to the financial section of Yahoo or USA Today, where there is a location for “symbol lookup.” Type in the company name, which should then give you the symbol. Look up the company via the symbol, and you should find the information. The symbol for GM is just GM, the symbol for General Electric is GE. Comment on the trend line of the com- pany you selected over the period. Is the trend increasing or decreasing? Does the trend follow a linear or log equation?

38. Select one of the major economic indicators, such as the Dow Jones Industrial Average, Nasdaq, or the S&P 500. Develop a trend line for the index over the last 10 years by using the value of the index at the end of the year, or for the last 30 days by selecting the closing value of the index for the last 30 days. You can locate this information in many places. For example, go to http://finance.yahoo.com, click on Nasdaq on the left hand, select Historical Prices, and a period of time, perhaps the last 30 days, and you will find the information. You should be able to download it directly to Excel to create your trend equation. Comment on the trend line you created. Is it increasing or decreas- ing? Does the trend line follow a linear or log equation?

Data Set Exercise 39. Refer to the Baseball 2009 data, which include information on the 2009 Major League

Baseball season. The data include the mean player salary since 1989. Plot the informa- tion and develop a linear trend equation. Write a brief report on your findings.

Software Commands 1. The MegaStat commands for creating the seasonal

indexes on pages 625 and 626 are: a. Enter the coded time period and the value of the

time series in two columns. You may also want to include information on the years and quarters.

b. Select MegaStat, Time Series/Forecasting, and Deseasonalization, and hit Enter.

c. Input the range of the data, indicate the data are in the first quarter, and click OK.

Time Series and Forecasting 643

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644 Chapter 16

Chapter 16 Answers to Self-Review

16–1 16–3 a.

b. About 156.7 percent. The antilog of 0.40945 is 2.567. Subtracting 1 yields 1.567.

c. About 454.5, found by � 0.20081 � .40945(6) � 2.65751. The antilog of 2.65751 is 454.5.

16–4 a. The following values are from a software package. Due to rounding, your figures might be slightly different.

Ŷ

a � 0.20081

b � 0.40945

Three-Year Three-Year Production Moving Moving

Year (thousands) Total Average

2005 2 — — 2006 6 12 4 2007 4 15 5 2008 5 12 4 2009 3 18 6 2010 10 — —

0 2005 2006 2007 2008 2009 2010

Original data Moving average

Nu m

be r p

ro du

ce d

(0 00

)

0 2002 2003 2004 2005 2006 2007 2008 2009 Years 1 2 3 4 5 6 7 8 Codes

X = 1 Y = 4.5833

X = 8 Y = 12.9168

Pr od

uc tio

n (0

00 )

16–2 a.

b. � a � bt � 3.3928 � 1.1905t (in thousands) c. For 2002:

for 2009:

d. For 2012, t � 11, so

or 16,488 king-size rockers.

Ŷ � 3.3928 � 1.1905(11) � 16.4883

Ŷ � 3.3928 � 1.1905(8) � 12.9168

Ŷ � 3.3928 � 1.1905(1) � 4.5833

Ŷ

Year Y log Y t

2006 2.13 0.3284 1 2007 18.10 1.2577 2 2008 39.80 1.5999 3 2009 81.40 1.9106 4 2010 112.00 2.0492 5

Winter Spring Summer Fall

Mean 119.35 81.66 125.31 74.24 Typical seasonal 119.18 81.55 125.13 74.13

The correction factor is 0.9986. b. Total sales at Teton Village for the winter season

are typically 19.18 percent above the annual average.

16–5 The forecast value for January of the sixth year is 34.9, found by

Seasonally adjusting the forecast, 34.9(120)�100 � 41.88. For February, � 4.4 � 0.5(62) � 35.4. Then, (35.4)95�100 � 33.63.

Ŷ

Ŷ � 4.4 � 0.5(61) � 34.9

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A Review of Chapters 15 and 16 645

A Review of Chapters 15 and 16 Chapter 15 presents index numbers. An index number describes the relative change in value from one period, called the base period, to another called the given period. It is actually a percent, but the percent sign is usually omitted. Indexes are used to compare the change in unlike series over time. For example, a company might wish to compare the change in sales with the change in the num- ber of sales representatives employed over the same period of time. A direct comparison is not meaningful because the units for one set of data are dollars and the other people. Index numbers also facilitate the comparison of very large values, where the amount of change in the actual val- ues is very large and therefore difficult to interpret.

There are two types of price indexes. In an unweighted price index, the quantities are not con- sidered. To form an unweighted index, we divide the base period value into the current period (also called the given period) and report the percent change. So if sales were $12,000,000 in 2004 and $18,600,000 in 2010, the simple unweighted price index for 2010 is:

We conclude there is a 55 percent increase in the sales during the six-year period. In a weighted price index, quantities are considered. The most widely used weighted index is

the Laspeyres Price Index. It uses the base period quantities as weights to compare changes in prices. It is computed by multiplying the base period quantities by the base period price for each product considered and summing the result. This result is the denominator of the fraction. The numerator of the fraction is the product of the base period quantities and the current price. For example, an appliance store sold 50 computers at $1,000 and 200 DVDs at $150 each in year 2004. In 2010, the same store sold 60 computers at $1,200 and 230 DVDs at $175. The Laspeyres Price Index is:

Notice the same base period quantities are used as weights in both the numerator and the denom- inator. The index indicates there has been an 18.75 percent increase in the value of sales during the six-year period.

The most widely used and reported index is the Consumer Price Index (CPI). The CPI is a Laspeyres type index. It is reported monthly by the U.S. Department of Labor and is often used to report the rate of inflation in the prices of goods and services in the United States. The current base period is 1982–84.

In Chapter 16, we studied time series and forecasting. A time series is a collection of data over a period of time. The earnings per share of General Electric common stock over the last ten years is an example of a time series. There are four components to a time series: the trend, cyclic effects, seasonal effects, and irregular effects.

Trend is the long-term direction of the time series. It can be either increasing or decreasing. The cyclical component is the fluctuation above and below the trend line over a period of sev-

eral years. Economic cycles are examples of the cyclical component. Most businesses shift between relative expansion and reduction periods over a cycle of several years.

Seasonal variation is the recurring pattern of the time series within a year. The consumption of many products and services is seasonal. Beach homes along the Gulf Coast are seldom rented dur- ing the winter and ski lodges in Wyoming are not used in the summer months. Hence, we say the rental of beach front properties and ski lodges are seasonal.

The irregular component includes any unpredictable events. In other words, the irregular com- ponent includes events that cannot be forecast. There are two types of irregular components. Episodic variations are unpredictable, but can usually be identified. The Nashville flooding in the sum- mer of 2010 is an example. The residual variation is random in nature and not predicted or identified.

The linear trend for a time series is given by the equation � a � bt, where is the esti- mated trend value, a is the intercept with the Y axis, b is the slope of the trend line (the rate of change), and t refers to the coded values for the time periods. We use the least squares method described in Chapter 13 to determine the trend line. Autocorrelation is often a problem when using the trend equation. Autocorrelation means that successive values of the time series are correlated.

ŶŶ

P � ©pt q0 ©p0q0

(100) � $1,200 � 50 � $175 � 200 $1,000 � 50 � $150 � 200

(100) � $95,000 $80,000

(100) � 118.75

P � pt p0

(100) � $18,600,000 $12,000,000

(100) � 155.0

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646 Chapter 16

Glossary

Chapter 15 Consumer Price Index An index reported monthly by the U.S. Department of Labor. It describes the change in a market basket of goods and services from the base period of 1982–84 to the present. Simple index The value in the given period divided by the value in the base period. The result is usually multiplied by 100 and reported as a percent. Weighted index The prices in the base period and the given period are multiplied by quantities (weights).

Chapter 16 Autocorrelation Successive residuals in a time series are correlated.

Cyclical variation The rise and fall of a time series over periods longer than one year. Episodic variation It is variation that is random in na- ture, but a cause can be identified. Irregular variation Variation in a times series that is ran- dom in nature and does not regularly repeat itself. Residual variation It is variation that is random in nature and cannot be identified or predicted. Seasonal variation Patterns of change in a time series within a year. These patterns of change repeat themselves each year. Secular trend The smoothed long-term direction of a time series.

Problems

1. Listed below is the consolidated revenue (in $ billions) for General Electric for the period from 2005 to 2009.

Consolidated Year Revenues ($ billions)

2005 148 2006 151 2007 172 2008 182 2009 157

a. Determine the index for 2009, using 2005 as the base period. b. Use the period 2005 to 2007 as the base period and find the index for 2009. c. With 2005 as the base year, use the least squares method to find the trend equation.

What is the estimated consolidated revenue for 2012? What is the rate of increase per year?

2. The table below shows the unemployment rate and the available workforce for three counties in northwest Pennsylvania for June 2007 and May 2010.

June 2007 May 2010

Labor Unemployed Labor Unemployed County Force % Force %

Erie 141,500 4.8 141,800 10.0 Warren 22,700 4.7 21,300 8.5 McKean 22,200 4.9 21,900 10.8

a. In June 2007, the national unemployment rate was 4.6%. For June 2007, compute a simple average unemployment index for the region using the national unemployment rate as the base. Interpret the simple average index.

b. In May 2010, the national unemployment rate was 9.7%. For May 2010, compute a sim- ple average unemployment index for the region using the national unemployment rate as the base. Interpret the simple average index.

c. Use the data for this region of northwest Pennsylvania to create a weighted unemploy- ment index using the Laspeyres method. Use the June 2007 data as the base period. Interpret the index.

3. Based on five years of monthly data (the period from January 2006 to December 2010), the trend equation for a small company is � 3.5 � 0.7t. The seasonal index for January is 120 and for June it is 90. What is the seasonally adjusted sales forecast for January 2011 and June 2011?

Ŷ

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A Review of Chapters 15 and 16 647

2006 2010

Price Quantity Price Quantity

Driver $250.00 5 $275.00 6 Putter 60.00 12 75.00 10 Irons 700.00 3 750.00 4

Year Sales

2006 $130,000 2007 145,000 2008 120,000 2009 170,000 2010 190,000

a. Find the simple index for each year using 2006 as the base year. b. Find the simple index for each year using 2006–2007 as the base year.

2. Listed below are the price and quantity of several golf items purchased by members of the men’s golf league at Indigo Creek Golf and Tennis Club.

a. Determine the simple aggregate price index, with 2006 as the base period. b. Determine a Laspeyres price index. c. Determine the Paasche price index. d. Determine a value index.

3. The monthly linear trend equation for the Hoopes ABC Beverage Store is:

� 5.50 � 1.25t

The equation is based on four years of monthly data and is reported in thousands of dollars. The index for January is 105.0 and for February it is 98.3. Determine the seasonally adjusted forecast for January and February of the fifth year.

Ŷ

Practice Test

Part 1—Objective 1. To compute an index, the base period is always in the . (numerator, denominator, can be in either, always 100)

1. 2. A number that measures the relative change from one period to another is called a/an .

2. 3. In a weighted index, both the price and the are considered. 3. 4. In a Laspeyres index, the quantities are used in both the numerator and denominator. (base period, given pe-

riod, oldest, newest—pick one) 4. 5. The current base period for the Consumer Price Index is . 5. 6. The long-term direction of a time series is called the . 6. 7. One method used to smooth the trend in a time series is a . 7. 8. When successive residuals are correlated, this condition is called . 8. 9. Irregular variation in a time series that is random in nature is called . 9.

10. In a three-year moving average, the weights given to each period are . (the same, oldest year has most weight, oldest year has the least weight) 10.

Part 2—Problems 1. Listed below are the sales at Roberta’s Ice Cream Stand for the last five years.

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17 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Conduct a test of hypothesis comparing an observed set of frequencies to an expected distribution.

LO2 List and explain the characteristics of the chi-square distribution.

LO3 Compute a goodness- of-fit test for unequal expected frequencies.

LO4 Conduct a test of hypothesis to verify that data grouped into a frequency distribution are a sample from a normal population.

LO5 Use graphical and statistical methods to determine whether a set of sample data is from a normal population.

LO6 Perform a chi-square test for independence on a contingency table.

Nonparametric Methods: Goodness-of-Fit Tests

For many years, TV executives used the guideline that 30 percent of

the audience were watching each of the traditional big three prime-time

networks and 10 percent were watching cable stations on a weekday

night. A random sample of 500 viewers in the Tampa–St. Petersburg,

Florida, area last Monday night showed that 165 homes were tuned in

to the ABC affiliate, 140 to the CBS affiliate, 125 to the NBC affiliate,

and the remainder were viewing a cable station. At the .05

significance level, can we conclude that the guideline is still

reasonable? (See Exercise 12 and LO2.)

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Nonparametric Methods: Goodness-of-Fit Tests 649

17.1 Introduction Chapters 9 through 12 discuss data of interval or ratio scale, such as weights of steel ingots, incomes of minorities, and years of employment. We conducted hypoth- esis tests about a single population mean, two population means, and three or more population means. For these tests, we assume the populations follow the normal probability distribution. However, there are tests available in which no assumption regarding the shape of the population is necessary. These tests are referred to as nonparametric. This means the assumption of a normal population is not necessary.

There are also tests exclusively for data of nominal scale of measurement. Recall from Chapter 1 that nominal data is the “lowest” or most primitive. For this type of measurement, data are classified into categories where there is no natural order. Examples include gender of Congressional representatives, state of birth of students, or brand of peanut butter purchased. In this chapter, we introduce a new test sta- tistic, the chi-square statistic.

17.2 Goodness-of-Fit Test: Equal Expected Frequencies The goodness-of-fit test is one of the most commonly used statistical tests. It is par- ticularly useful because it requires only the nominal level of measurement. So we are able to conduct a test of hypothesis on data that has been classified into groups. Our first illustration of this test involves the case when the expected cell frequencies are equal. As the full name implies, the purpose of the goodness-of-fit test is to compare an observed distribution to an expected distribution. An example will describe the hypothesis-testing situation.

Example Bubba’s Fish and Pasta is a chain of restaurants located along the Gulf Coast of Florida. Bubba, the owner, is considering adding steak to his menu. Before doing so, he decides to hire Magnolia Research, LLC, to conduct a survey of adults as to their favorite meal when eating out. Magnolia selected a sample 120 adults and asked each to indicate their favorite meal when dining out. The results are reported below.

Is it reasonable to conclude there is no preference among the four entrées?

Solution If there is no difference in the popularity of the four entrées, we would expect the observed frequencies to be equal—or nearly equal. To put it another way, we would expect as many adults to indicate they preferred chicken as fish. Thus, any dis- crepancy in the observed and expected frequencies is attributed to sampling error or chance.

What is the level of measurement in this problem? Notice that when a person is selected, we can only classify the selected adult as to the entrée preferred. We do not get a reading or a measurement of any kind. The “measurement” or “classification” is based on the selected entrée. In addition, there is no natural order to the favorite entrée.

Favorite Entrée Frequency

Chicken 32 Fish 24 Meat 35 Pasta 29

Total 120

TABLE 17–1 Favorite Entrée as Selected by a Sample of 120 Adults

LO1 Conduct a test of hypothesis comparing an observed set of frequencies to an expected distribution.

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650 Chapter 17

No one entrée is assumed better than another. Therefore, the nominal scale is appropriate.

If the entrées are equally popular, we would expect 30 adults to select each meal. Why is this so? If there are 120 adults in the sample and four categories, we expect that one-fourth of those surveyed would select each entrée. So 30, found by 120/4, is the expected frequency for each category or cell, assuming there is no preference for any of the entrées. This information is summarized in Table 17–2. An examination of the data indicates meat is the entrée selected most frequently (35 out of 120) and fish is selected least frequently (24 out of 120). Is the difference in the number of times each entrée is selected due to chance,

or should we conclude that the entrées are not equally preferred?

with k � 1 degrees of freedom, where:

k is the number of categories. is an observed frequency in a particular category. is an expected frequency in a particular category.fe

To investigate the issue, we use the five-step hypothesis-testing procedure.

Step 1: State the null hypothesis and the alternate hypothesis. The null hypothesis, , is that there is no difference between the set of observed frequencies and the set of expected frequencies. In other words, any dif- ference between the two sets of frequencies is attributed to sampling error. The alternate hypothesis, , is that there is a difference between the observed and expected sets of frequencies. If the null hypothesis is rejected and the alternate hypothesis is accepted, we conclude the pref- erences are not equally distributed among the four categories (cells).

: There is no difference in the proportion of adults selecting each entrée.

: There is a difference in the proportion of adults selecting each entrée.

Step 2: Select the level of significance. We selected the .05 significance level. The probability is .05 that a true null hypothesis is rejected.

Step 3: Select the test statistic. The test statistic follows the chi-square distrib- ution, designated by �2.

Frequency Frequency Favorite Meal Observed, Expected,

Chicken 32 30 Fish 24 30 Meat 35 30 Pasta 29 30

Total 120 120

fefo

TABLE 17–2 Observed and Expected Frequency for Survey of 120 Adults

CHI-SQUARE TEST STATISTIC �2 � © c (fo � fe)

fe d [17–1]

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Nonparametric Methods: Goodness-of-Fit Tests 651

We will examine the characteristics of the chi-square distribution in more detail shortly.

Step 4: Formulate the decision rule. Recall that the decision rule in hypothesis testing is the value that separates the region where we do not reject from the region where is rejected. This number is called the critical value. As we will soon see, the chi-square distribution is really a family of distributions. Each distribution has a slightly different shape, depending on the number of degrees of freedom. The number of degrees of freedom is k � 1, where k is the number of categories. In this particular problem, there are four categories, the four meal entrées. Because there are four cate- gories, there is k � 1 � 4 � 1 � 3 degrees of freedom. As noted, a category is called a cell, and there are four cells. The critical value for 3 degrees of freedom and the .05 level of significance is found in Appendix B.3. A por- tion of that table is shown in Table 17–3. The critical value is 7.815, found by locating 3 degrees of freedom in the left margin and then moving hori- zontally (to the right) and reading the critical value in the .05 column.

H0 H0

The decision rule is to reject the null hypothesis if the computed value of chi-square is greater than 7.815. If it is less than or equal to 7.815, we fail to reject the null hypothesis. Chart 17–1 shows the decision rule.

The decision rule indicates that if there are large differences between the observed and expected frequencies, resulting in a computed �2 of more than 7.815, the null hypothesis should be rejected. However, if the differ- ences between and are small, the computed �2 value will be 7.815 or less, and the null hypothesis should not be rejected. The reasoning is that such small differences between the observed and expected frequencies are probably due to chance. Remember, the 120 observations are a sam- ple of the population.

Step 5: Compute the value of chi-square and make a decision. Of the 120 adults in the sample, 32 indicated their favorite entrée was chicken. The counts were reported in Table 17–1. The calculations for chi-square

fefo

TABLE 17–3 A Portion of the Chi-Square Table

Right-Tail AreaDegrees of Freedom

df .10 .05 .02 .01

1 2.706 3.841 5.412 6.635 2 4.605 5.991 7.824 9.210 3 6.251 7.815 9.837 11.345 4 7.779 9.488 11.668 13.277 5 9.236 11.070 13.388 15.086

Pr ob

ab ili

Do not reject

7.815 Scale of �2 Critical value

Region of rejection

.05

CHART 17–1 Chi-Square Probability Distribution for 3 Degrees of Freedom, Showing the Region of Rejection, .05 Level of Significance

Statistics in Action

For many years, researchers and sta- tisticians believed that all variables were normally dis- tributed. In fact, it was generally as- sumed to be a uni- versal law. However, Karl Pearson ob- served that experi- mental data were not always normally dis- tributed but there was no way to prove his observations were correct. To solve this problem, Pearson discovered the chi- square statistic that basically compares an observed fre- quency distribution with an assumed nor- mal distribution. His discovery proved that all variables were not normally distributed.

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follow. (Note again that the expected frequencies are the same for each cell.)

Column 1: Determine the differences between each and . That is, . The sum of these differences is zero.

Column 2: Square the difference between each observed and expected frequency, that is, .

Column 3: Divide the result for each observation by the expected frequency, that is, Finally, sum these values. The result is the value of �2, which is 2.20.

(fo � fe) 2�fe.

(fo � fe) 2

fo � fe fefo

The computed �2 of 2.20 is not in the rejection region. It is less than the critical value of 7.815. The decision, therefore, is to not reject the null hypothesis. We conclude that the differences between the observed and the expected frequencies could be due to chance. That means there is no preference among the four entrées.

We can use software to compute the value of chi-square. The output of MegaStat follows. The steps are shown in the Software Commands section at the end of the chapter. The computed value of chi-square is 2.20, the same value obtained in our earlier calculations. Also note the p-value is .5319, much larger than .05.

Favorite Entrée

Chicken 32 30 2 4 0.133 Fish 24 30 �6 36 1.200 Meat 35 30 5 25 0.833 Pasta 29 30 �1 1 0.033

Total 120 120 0 2.200

(fo � fe) 2�fe(fo � fe)2(fo � fe)fefo

Goodness-of-Fit Test

observed expected O � E (O � E)2/E % of chisq 32 30.000 2.000 0.133 6.06 24 30.000 �6.000 1.200 54.55 35 30.000 5.000 0.833 37.88 29 30.000 �1.000 0.033 1.52

120 120.000 0.000 2.200 100.00

2.20 chi-square 3 df

.5319 p-value

The chi-square distribution, which is used as the test statistic in this chapter, has the following characteristics.

1. Chi-square values are never negative. This is because the difference between and is squared, that is, ( � )2.

2. There is a family of chi-square distributions. There is a chi-square distribu- tion for 1 degree of freedom, another for 2 degrees of freedom, another for 3 degrees of freedom, and so on. In this type of problem, the number of degrees of freedom is determined by k � 1, where k is the number of categories. There- fore, the shape of the chi-square distribution does not depend on the size of the sample, but on the number of categories used. For example, if 200 employees of an airline were classified into one of three categories—flight personnel, ground

fefofefo LO2 List and explain the characteristics of the chi-square distribution.

652 Chapter 17

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support, and administrative personnel—there would be k � 1 � 3 � 1 � 2 degrees of freedom.

3. The chi-square distribution is positively skewed. However, as the number of degrees of freedom increases, the distribution begins to approximate the normal probability distribution. Chart 17–2 shows the distributions for selected degrees of freedom. Notice that for 10 degrees of freedom the curve is approaching a normal distribution.

0 1 2 3 4 5 6 7 8 9 10

.40

.30

.20

.10

.00

Pr ob

ab ili

Chi-square values (�2)

df = 1

df = 3

df = 5 df = 10

11 12 13 14 15 16 17 18 19

Shape of �2 distribution approaches normal distribution as df becomes larger

CHART 17–2 Chi-Square Distributions for Selected Degrees of Freedom

Self-Review 17–1 The human resources director at Georgetown Paper Inc. is concerned about absen- teeism among hourly workers. She decides to sample the company records to determine whether absenteeism is distributed evenly throughout the six-day work week. The hypotheses are:

H0: Absenteeism is evenly distributed throughout the work week. H1: Absenteeism is not evenly distributed throughout the work week.

The sample results are:

Number Absent Number Absent

Monday 12 Thursday 10 Tuesday 9 Friday 9 Wednesday 11 Saturday 9

Nonparametric Methods: Goodness-of-Fit Tests 653

(a) What are the numbers 12, 9, 11, 10, 9, and 9 called? (b) How many categories (cells) are there? (c) What is the expected frequency for each day? (d) How many degrees of freedom are there? (e) What is the chi-square critical value at the 1 percent significance level? (f) Compute the �2 test statistic. (g) What is the decision regarding the null hypothesis? (h) Specifically, what does this indicate to the human resources director?

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Day Rounds

Monday 124 Tuesday 74 Wednesday 104 Thursday 98 Friday 120

Opinion Number of Buyers Opinion Number of Buyers

Outstanding 47 Good 39 Excellent 45 Fair 35 Very good 40 Undesirable 34

654 Chapter 17

Exercises 1. In a particular chi-square goodness-of-fit test, there are four categories and 200 obser-

vations. Use the .05 significance level. a. How many degrees of freedom are there? b. What is the critical value of chi-square?

2. In a particular chi-square goodness-of-fit test, there are six categories and 500 observa- tions. Use the .01 significance level. a. How many degrees of freedom are there? b. What is the critical value of chi-square?

3. The null hypothesis and the alternate are:

: The frequencies are equal. : The frequencies are not equal.

a. State the decision rule, using the .05 significance level. b. Compute the value of chi-square. c. What is your decision regarding ?

4. The null hypothesis and the alternate are:

: The frequencies are equal. : The frequencies are not equal.

a. State the decision rule, using the .05 significance level. b. Compute the value of chi-square. c. What is your decision regarding ?

5. A six-sided die is rolled 30 times and the numbers 1 through 6 appear as shown in the following frequency distribution. At the .10 significance level, can we conclude that the die is fair?

H1 H0

Category

Current 320 300 1.33 Late 120 150 6.00 Uncollectible 60 50 2.00

500 500 9.33

(fo � fe ) 2

fefefo

Reject H0. The accounts receivable data does not reflect the national average.

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18 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Define a nonparametric test and know when one is applied.

LO2 Conduct the sign test for dependent samples using the binomial and standard normal distributions as the test statistics.

LO3 Conduct a test of hypothesis for dependent samples using the Wilcoxon signed-rank test.

LO4 Conduct and interpret the Wilcoxon rank-sum test for independent samples.

LO5 Conduct and interpret the Kruskal-Wallis test for several independent samples.

LO6 Compute and interpret Spearman’s coefficient of rank correlation.

LO7 Conduct a test of hypothesis to determine whether the correlation among the ranks in the population is different from zero.

Nonparametric Methods: Analysis of Ranked Data

Assembly workers at Coastal Computers Inc. assemble one or two

subassemblies and insert them in a frame. Executives at CC think that the

employees would have more pride in their work if they assembled all

components and tested the completed computer. A sample of 25

employees is selected to test the idea. Twenty liked assembling the entire

unit and testing it. At the .05 level, can we conclude the employees

preferred assembling the entire unit? (See Exercise 8 and LO2.)

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Nonparametric Methods: Analysis of Ranked Data 681

18.1 Introduction In Chapter 17, we introduced tests of hypothesis for nominal-scale variables. Recall from Chapter 1 that nominal level of measurement implies the data can only be clas- sified into categories, and there is no particular order to the categories. The purpose of these tests is to determine whether an observed set of frequencies, , is signifi- cantly different from a corresponding set of expected frequencies, . Likewise, if you are interested in the relationship between two characteristics—such as the age of an individual and his or her music preference—you would tally the data into a contingency table and use the chi-square distribution as the test statistic. For both these types of problems, no assumptions need to be made about the shape of the population. We do not have to assume, for example, that the population of interest follows the nor- mal distribution, as we did with the tests of hypotheses in Chapters 10 through 12.

This chapter continues our discussion of hypothesis tests designed especially for nonparametric data. For these tests, we do not need to assume anything about the shape of the population distribution. Sometimes, we use the term distribution- free tests. These tests do require that the responses can be ranked or ordered. So the responses must be measured with an ordinal, interval, or ratio scale. An exam- ple of an ordinal scale is executive title. Corporate executives can be ranked as assistant vice president, vice president, senior vice president, and president. A vice president is ranked higher than an assistant vice president, a senior vice president is ranked higher than a vice president, and so on.

In this chapter, we consider five distribution-free tests and the Spearman coef- ficient of rank correlation. The tests are: the sign test, the median test, the Wilcoxon signed-rank test, the Wilcoxon rank-sum test, and the Kruskal-Wallis analysis of vari- ance by ranks.

18.2 The Sign Test The sign test is based on the sign of a difference between two related observa- tions. We usually designate a plus sign for a positive difference and a minus sign for a negative difference. For example, a dietitian wishes to see if a person’s cho- lesterol level will decrease if the diet is supplemented by a certain mineral. She selects a sample of 20 production workers over the age of 40 and measures the workers’ cholesterol level. After the 20 subjects take the mineral for six weeks, they are tested again. If the cholesterol level has dropped, a plus sign is recorded. If it has increased, a negative sign is recorded. If there is no change, a zero is recorded (and that person is dropped from the study). For a sign test, we are not concerned with the magnitude of the difference, only the direction of the difference.

The sign test has many applications. One is for “before/ after” experiments. To illustrate, an evaluation is to be made on a new tune-up program for automobiles. We record the number of miles traveled per gallon of gasoline before the tune-up and again after the tune-up. If the tune-up is not effective—that is, it had no effect on performance— then about half of the automobiles tested would show an increase in miles per gallon and the other half a decrease. A “�” sign is assigned to an increase, a “�” sign to a decrease.

A product-preference experiment illustrates another use of the sign test. Taster’s Choice markets two kinds of cof- fee in a 4-ounce jar: decaffeinated and regular. Its market research department wants to determine whether coffee drinkers prefer decaffeinated or regular coffee. Coffee drinkers are given two small, unmarked cups of coffee, and

fe fo

LO2 Conduct the sign test for dependent samples using the binomial and standard normal distributions as the test statistics.

LO1 Define a nonparametric test and know when it is applied.

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682 Chapter 18

each is asked his or her preference. Preference for decaffeinated could be coded “�” and preference for regular “�.” In a sense, the data are ordinal level because the coffee drinkers give their preferred coffee the higher rank; they rank the other kind below it. Here again, if the population of consumers do not have a preference, we would expect half of the sample of coffee drinkers to prefer decaffeinated and the other half regular coffee.

We can best show the application of the sign test by an example. We will use a “before/after” experiment.

Example The director of information systems at Samuelson Chemicals recommended that an in-plant training program be instituted for certain managers. The objective is to improve the computer knowledge base in the Payroll, Accounting, and Production Planning Departments.

A sample of 15 managers is randomly selected from the three departments. Each manager is rated on their computer knowledge by a panel of experts. They were rated as being either outstanding, excellent, good, fair, or poor. (See Table 18–1.) After the three-month training program, the same panel of experts rated each manager again. The two ratings (before and after) are shown along with the sign of the difference. A “�” sign indicates improvement, and a “�” sign indicates that the manager’s competence using databases had declined after the training program.

Solution

We are interested in whether the in-plant training program increased the knowl- edge base of the managers. That is, are the managers more knowledgeable after the training program than before?

TABLE 18–1 Competence Before and After the Training Program

Name Before After Sign of Difference

T. J. Bowers Good Outstanding � Sue Jenkins Fair Excellent � James Brown Excellent Good � Tad Jackson Poor Good � Andy Love Excellent Excellent 0 Sarah Truett Good Outstanding � Antonia Aillo Poor Fair � Jean Unger Excellent Outstanding � Coy Farmer Good Poor � Troy Archer Poor Good � V. A. Jones Good Outstanding � Juan Guillen Fair Excellent � Candy Fry Good Fair � Arthur Seiple Good Outstanding � Sandy Gumpp Poor Good �

Dropped from analysis

We will use the five-step hypothesis-testing procedure.

Step 1: State the null hypothesis and the alternate hypothesis.

: � � .50 There has been no change in the computer knowl- edge base of the managers as a result of the training program.

: � � .50 There has been an increase in the computer knowl- edge base of the managers as a result of the training program.

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Nonparametric Methods: Analysis of Ranked Data 683

The symbol � refers to the proportion in the population with a partic- ular characteristic. If we do not reject the null hypothesis, it will indicate the training program has produced no change in the knowledge base, or that knowledge actually decreased. If we reject the null hypothesis, it will indicate that the knowledge of the managers has increased as a result of the training program.

The test statistic follows the binomial probability distribution. It is appropriate because the sign test meets all the binomial assumptions, namely:

1. There are only two outcomes: a “success” and a “failure.” A manager either increased his or her knowledge (a success) or did not.

2. For each trial, the probability of success is assumed to be .50. Thus, the probability of a success is the same for all trials (managers in this case).

3. The total number of trials is fixed (15 in this experiment). 4. Each trial is independent. This means, for example, that Arthur

Seiple’s performance in the three-month course is unrelated to Sandy Gumpp’s performance.

Step 2: Select a level of significance. We chose the .10 level. Step 3: Decide on the test statistic. It is the number of plus signs resulting from

the experiment. Step 4: Formulate a decision rule. Fifteen managers were enrolled in the

training course, but Andy Love showed no increase or decrease in knowledge. (See Table 18–1.) He was, therefore, eliminated from the study because he could not be assigned to either group, so n � 14. From the binomial probability distribution table in Appendix B.9, for an n of 14 and a probability of .50, we copied the binomial probability distribution in Table 18–2. The number of successes is in column 1, the probabilities of success in column 2, and the cumulative probabilities in column 3. To arrive at the cumulative probabilities, we add the prob- abilities of success in column 2 from the bottom. For illustration, to get the cumulative probability of 11 or more successes, we add .000 � .001 � .006 � .022 � .029.

This is a one-tailed test because the alternate hypothesis gives a di- rection. The inequality (�) points to the right. Thus, the region of rejection is in the upper tail. If the inequality sign pointed toward the left tail (�), the region of rejection would be in the lower tail. If that were the case, we would add the probabilities in column 2 down to get the cumulative prob- abilities in column 3.

Recall that we selected the .10 level of significance. To arrive at the decision rule for this problem, we go to the cumulative probabilities in Table 18–2, column 3. We read up from the bottom until we come to the cumulative probability nearest to but not exceeding the level of signifi- cance (.10). That cumulative probability is .090. The number of successes (plus signs) corresponding to .090 in column 1 is 10. Therefore, the deci- sion rule is: If the number of pluses in the sample is 10 or more, the null hypothesis is rejected and the alternate hypothesis accepted.

To repeat: We add the probabilities up from the bottom because the direction of the inequality (�) is toward the right, indicating that the region of rejection is in the upper tail. If the number of plus signs in the sample is 10 or more, we reject the null hypothesis; otherwise, we do not reject

. The region of rejection is portrayed in Chart 18–1. What procedure is followed for a two-tailed test? We combine (add)

the probabilities of success in the two tails until we come as close to the

Statistics in Action

A recent study of un- dergraduate students at the University of Michigan revealed the students with the worst attendance records also tended to earn the lowest grades. Does that surprise you? Stu- dents who were absent less than 10 percent of the time tended to earn a B or better. The same study also found that students who sat in the front of the class earned higher grades than those who sat in the back.

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desired level of significance ( ) as possible without exceeding it. In this example, is .10. The probability of 3 or fewer successes is .029, found by .000 � .001 � .006 � .022. The probability of 11 or more successes is also .029. Adding the two probabilities gives .058. This is the closest we can come to .10 without exceeding it. Had we included the probabil- ities of 4 and 10 successes, the total would be .180, which exceeds .10. Hence, the decision rule for a two-tailed test would be to reject the null hypothesis if there are 3 or fewer plus signs, or 11 or more plus signs.

Step 5: Make a decision regarding the null hypothesis. Eleven out of the 14 managers in the training course increased their computer knowledge. The number 11 is in the rejection region, which starts at 10, so is re- jected. We conclude that the three-month training course was effective. It increased the computer knowledge of the managers.

684 Chapter 18

CHART 18–1 Binomial Distribution, n � 14, � � .50

TABLE 18–2 Binomial Probability Distribution for n � 14, � � .50

(1) (2) (3) Number of Probability of Cumulative Successes Success Probability

0 .000 1.000 1 .001 .999 2 .006 .998 3 .022 .992 4 .061 .970 5 .122 .909 6 .183 .787 7 .209 .604 8 .183 .395 9 .122 .212

10 .061 .090 11 .022 .029 .000 � .001 � 12 .006 .007 .006 � .022 13 .001 Add up .001 14 .000 .000

.20

.15

.10

.05

0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Pr ob

ab ili

ty o

f s uc

ce ss

Number of successes (number of + signs)

Region of rejection

10 or more +

H0: � < 0.50 H1: � > 0.50 n = 14 α = 0.10

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Nonparametric Methods: Analysis of Ranked Data 685

It should be noted again that if the alter- nate hypothesis does not give a direction— for example, : � � .50 and : � 50— the test of hypothesis is two-tailed. In such cases, there are two rejection regions—one in the lower tail and one in the upper tail. If � .10 and the test is two-tailed, the area in each tail is .05 ( �2 � .10�2 � .05). Self- Review 18–1 illustrates this.

H1H0

Self-Review 18–1 Recall the Taster’s Choice example described on page 681, involving a consumer test to determine the preference for decaffeinated versus regular coffee. Use the .10 significance level. The null and alternate hypotheses are:

(a) Is this a one-tailed or a two-tailed test of hypothesis? (b) Show the decision rule in a chart. (c) Letting consumer preference for decaffeinated coffee be a “�” and preference for reg-

ular coffee a “�,” it was found that two customers preferred decaffeinated. What is your decision? Explain.

H1: � .50 H0: � � .50 n � 12

Exercises 1. The following hypothesis-testing situation is given: : � � .50 and : � � .50. The sig-

nificance level is .10, and the sample size is 12. a. What is the decision rule? b. There were nine successes. What is your decision regarding the null hypothesis? Explain.

2. The following hypothesis-testing situation is given: : � � .50 and : � .50. The sig- nificance level is .05, and the sample size is 9. a. What is the decision rule? b. There were five successes. What is your decision regarding the null hypothesis?

3. Calorie Watchers has low-calorie breakfasts, lunches, and dinners. If you join the club, you receive two packaged meals a day. CW claims that you can eat anything you want for the third meal and still lose at least five pounds the first month. Members of the club are weighed before commencing the program and again at the end of the first month. The experiences of a random sample of 11 enrollees are:

H1H0

Name Weight Change Name Weight Change

Foster Lost Hercher Lost Taoka Lost Camder Lost Lange Gained Hinckle Lost Rousos Lost Hinkley Lost Stephens No change Justin Lost Cantrell Lost

We are interested in whether there has been a weight loss as a result of the Calorie Watchers program. a. State and . b. Using the .05 level of significance, what is the decision rule? c. What is your conclusion about the Calorie Watchers program?

H1H0

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686 Chapter 18

4. Many new stockbrokers resist giving presentations to bankers and certain other groups. Sensing this lack of self-confidence, management arranged to have a confidence- building seminar for a sample of new stockbrokers and enlisted Career Boosters for a three-week course. Before the first session, Career Boosters measured the level of confidence of each participant. It was measured again after the three-week seminar. The before and after levels of self-confidence for the 14 in the course are shown below. Self-confidence was classified as being either negative, low, high, or very high.

Before After Before After Stockbroker Seminar Seminar Stockbroker Seminar Seminar

J. M. Martin Negative Low F. M. Orphey Low Very high T. D. Jagger Negative Negative C. C. Ford Low High A. D. Hammer Low High A. R. Utz Negative Low T. A. Jones, Jr. Very high Low M. R. Murphy Low High B. G. Dingh Low High P. A. Lopez Negative Low D. A. Skeen Low High B. K. Pierre Low High C. B. Simmer Negative High N. S. Walker Low Very high

The purpose of this study is to find whether Career Boosters was effective in raising the self-confidence of the new stockbrokers. That is, was the level of self-confidence higher after the seminar than before it? Use the .05 significance level. a. State the null and alternate hypotheses. b. Using the .05 level of significance, state the decision rule—either in words or in chart form. c. Draw conclusions about the seminar offered by Career Boosters.

Using the Normal Approximation to the Binomial If the number of observations in the sample is larger than 10, the normal distribu- tion can be used to approximate the binomial. Recall that in Section 6.5 of Chap- ter 6 on page 197, we computed the mean of the binomial distribution from � � n� and the standard deviation from . In this case, � � .50, so the � � 1n�(1 � �) equations reduce to � � .50n and , respectively.

The test statistic z is

SIGN TEST, n � 10 [18–1]

If the number of pluses or minuses is more than n�2, we use the following form as the test statistic:

SIGN TEST, n � 10, [18–2]

� SIGNS MORE THAN n�2

If the number of pluses or minuses is less than n�2, the test statistic z is

SIGN TEST, n � 10, [18–3]

� SIGNS LESS THAN n�2

In the preceding formulas, X is the number of plus (or minus) signs. The value �.50 or �.50 is the continuity correction factor, discussed in Section 7.5 of Chapter 7. Briefly, it is applied when a continuous distribution such as the normal distribution (which we are using) is used to approximate a discrete distribution (the binomial).

The following example illustrates the details of the sign test when n is greater than 10.

z � (X � .50) � �

� �

(X � .50) � .50n .501n

z � (X � .50) � �

� �

(X � .50) � .50n .501n

z � (X .50) � �

�

� � .501n

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Self-Review 18–2 The human resources department at Ford Motor Company began blood pressure screen- ing and education for the 100 employees in the paint department at the first of the year. As a follow-up in July, the same 100 employees were again screened for blood pressure and 80 showed a reduction. Can we conclude the screening was effective in reducing blood pressure readings? (a) State the null hypothesis and the alternate hypothesis. (b) What is the decision rule for a significance level of .05? (c) Compute the value of the test statistic. (d) What is your decision regarding the null hypothesis? (e) Interpret your decision.

Example The market research department of Cola Inc. has the assignment of testing a new soft drink. There are two versions of the drink under consideration. One version is sweet and the other bitter. The market research department selects a random sam- ple of 64 consumers. Each consumer will taste both the sweet cola (labeled A) and the bitter one (labeled B) and indicate a preference. Conduct a test of hypothesis to determine if there is a difference in the preference for the sweet and bitter tastes. Use the .05 significance level.

Solution Step 1: State the null and alternate hypotheses. : � � .50 There is no preference. : � .50 There is a preference.

Step 2: Select a level of significance. It is .05, which is stated in the problem. Step 3: Select the test statistic. It is z, given in formula (18–1).

where � � .50n and . Step 4: Formulate the decision rule. Referring to Appendix B.1, Areas under

the Normal Curve, for a two-tailed test (because states that � .50) and the .05 significance level, the critical values are �1.96 and �1.96. Recall from Chapter 10 that for a two-tailed test we split the rejection probability in half and place one half in each tail. That is, �2 � .05�2 � .025. Continuing, .5000 � .0250 � .4750. Searching for .4750 in the body of the table and reading the z value in the left margin gives 1.96, the crit- ical value. Therefore, do not reject if the computed z value is between �1.96 and �1.96. Otherwise, reject and accept .

Step 5: Compute z, compare the computed value with the critical value, and make a decision regarding Preference for cola A was given a “�” sign and preference for B a “�” sign. Out of the 64 in the sample, 42 pre- ferred the sweet taste, which is cola A. Therefore, there are 42 pluses. Since 42 is more than n�2 � 64�2 � 32, we use formula (18–2) for z:

The computed z of 2.38 is beyond the critical value of 1.96. Therefore, the null hypothesis of no difference is rejected at the .05 significance level. There is evidence of a difference in consumer preference. That is, we conclude consumers prefer one cola over another.

The p-value is the probability of finding a z value larger than 2.38 or smaller than �2.38. From Appendix B.1, the probability of finding a z value greater than 2.38 is .5000 � .4913 � .0087. Thus, the two-tailed p-value is .0174. So the probability of obtaining a sample statistic this extreme when the null hypothesis is true is less than 2 percent.

z � (X � .50) � .50n

.501n �

(42 � .50) � .50(64) .50164

� 2.38

H0.

H1H0 H0

� � .501n

z � (X .50) � �

�

H1 H0

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688 Chapter 18

Exercises 5. A sample of 45 overweight men participated in an exercise program. At the conclusion

of the program, 32 had lost weight. At the .05 significance level, can we conclude the program is effective? a. State the null hypothesis and the alternate hypothesis. b. State the decision rule. c. Compute the value of the test statistic. d. What is your decision regarding the null hypothesis?

6. A sample of 60 college students was given a special training program designed to improve their time management skills. One month after completing the course, the students were con- tacted and asked whether the skills learned in the program were effective. A total of 42 responded yes. At the .05 significance level, can we conclude the program is effective? a. State the null hypothesis and the alternate hypothesis. b. State the decision rule. c. Compute the value of the test statistic. d. What is your decision regarding the null hypothesis?

7. Pierre’s Restaurant announced that on Thursday night the menu would consist of unusual gourmet items, such as squid, rabbit, snails from Scotland, and dandelion greens. As part of a larger survey, a sample of 81 regular customers was asked whether they pre- ferred the regular menu or the gourmet menu. Forty-three preferred the gourmet menu. At the .02 level, can we conclude the customers preferred the gourmet menu?

8. Assembly workers at Coastal Computers Inc. assemble just one or two subassemblies and insert them in a frame. The executives at CC think that the employees would have more pride in their work if they assembled all of the subassemblies and tested the com- plete computer. A sample of 25 employees was selected to experiment with the idea. After a training program, each was asked his or her preference. Twenty liked assembling the entire unit and testing it. At the .05 level, can we conclude the employees preferred assembling the entire unit? Explain the steps you used to arrive at your decision.

Testing a Hypothesis about a Median Most of the tests of hypothesis we have conducted so far involved the population mean or a proportion. The sign test is one of the few tests that can be used to test the value of a median. Recall from Section 3.6 in Chapter 3 that the median is the value above which half of the observations lie and below which the other half lie. For hourly wages of $7, $9, $11, and $18, the median is $10. Half of the wages are above $10 an hour and the other half below $10.

To conduct a test of hypothesis, a value above the median is assigned a plus sign, and a value below the median is assigned a minus sign. If a value is the same as the median, it is dropped from further analysis.

Example A study several years ago by the consumer research department of Superior Gro- cers found the median weekly amount spent on grocery items by young married couples was $123. The CEO would like to repeat the research to determine whether the median amount spent has changed. The department’s new sample information showed that, in a random sample of 102 young adult married couples, 60 spent more than $123 last week on grocery items, 40 spent less, and 2 spent exactly $123. At the .10 significance level, is it reasonable to conclude that the median amount spent is not equal to $123?

Solution If the population median is $123, then we expect about half the sampled couples to have spent more than $123 last week and about half less than $123. After dis- carding the two customers who spent exactly $123, we would expect 50 to be above the median and 50 to be below the median. Is this difference attributable to chance, or is the median some value other than $123? The statistical test for the median will help answer that question.

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The null and the alternate hypotheses are:

: Median � $123 : Median $123

This is a two-tailed test because the alternate hypothesis does not indicate a direc- tion. That is, we are not interested in whether the median is less than or greater than $123, only that it is different from $123. The test statistic meets the binomial assumptions. That is:

1. An observation is either larger or smaller than the proposed median, so there are only two possible outcomes.

2. The probability of a success remains constant at .50. That is, � � .50. 3. The couples selected as part of the sample represent independent trials. 4. We count the number of successes in a fixed number of trials. In this case, we

consider 100 couples and count the number who spend more than $123 per week on grocery items.

The usable sample size is 100 and � is .50, so n� � 100(.50) � 50 and n(1 � �) � 100(1 � .50) � 50, which are both larger than 5, so we use the normal dis- tribution to approximate the binomial. That is, we actually use the standard normal distribution as the test statistic. The significance level is .10, so �2 � .10�2 � .05 of the area is in each tail of a normal distribution. From Appendix B.1, which shows the areas under a normal curve, the critical values are �1.65 and 1.65. The deci- sion rule is to reject if z is less than �1.65 or greater than 1.65.

We use formula (18–2) for z because 60 is greater than n�2 or (100�2 � 50).

The null hypothesis is rejected because the computed value of 1.90 is greater than the critical value of 1.65. The sample evidence indicates that the median amount spent per week on grocery items by young couples is not $123. The p-value is .0574 found by 2(.5000 � .4713). The p-value is smaller than the significance level of .10 for this test.

z � (X � .50) � .50n

.501n �

(60 � .5) � .50(100) .501100

� 1.90

H1 H0

Exercises 9. The median salary for a chiropractor in the United States is $81,500 per year, according

to the U.S. Department of Labor. A group of recent graduates believe this amount is too low. In a random sample of 205 chiropractors who recently graduated, 170 began with a salary of more than $81,500 and five earned a salary of exactly $81,500. a. State the null and alternate hypotheses. b. State the decision rule. Use the .05 significance level. c. Do the necessary computations and interpret the results.

10. Central Airlines claims that the median price of a round-trip ticket from Chicago to Jackson Hole, Wyoming, is $503. This claim is being challenged by the Association of Travel Agents, who believe the median price is less than $503. A random sample of 400 round-trip tickets

Self-Review 18–3 After receiving the results regarding the weekly amount spent on grocery items for young couples from the consumer research department, the CEO of Superior Grocers wondered whether the same was true for senior citizen couples. In this case, the CEO wants the consumer research department to investigate whether the median weekly amount spent per week by senior citizens is greater than $123. A sample of 64 senior citizen couples revealed 42 spent more than $123 per week on grocery items. Use the .05 significance level.

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690 Chapter 18

from Chicago to Jackson Hole revealed 160 tickets were below $503. None of the tickets were exactly $503. Let � .05. a. State the null and alternate hypotheses. b. What is your decision regarding ? Interpret.

18.3 Wilcoxon Signed-Rank Test for Dependent Samples The paired t test (page 392), described in Chapter 11, has two requirements. First, the samples must be dependent. Recall that dependent samples are characterized by a measurement, some type of intervention, and then another measurement. For example, a large company began a “wellness” program at the start of the year. Twenty workers were enrolled in the weight reduction portion of the program. To begin, all participants were weighed. Next they dieted, did the exercise, and so forth in an attempt to lose weight. At the end of the program, which lasted six months, all participants were weighed again. The difference in their weight between the start and the end of the program is the variable of interest. Note that there is a mea- surement, an intervention, and then another measurement.

The second requirement for the paired t test is that the distribution of the differ- ences follow the normal probability distribution. In the company wellness example,

this would require that the differences in the weights of the population of partici- pants follow the normal probability distri- bution. In that case, this assumption is reasonable. However, there are instances when we want to study the differences between dependent observations where we cannot assume that the distribution of the differences approximates a normal distribution. Frequently, we encounter a problem with the normality assumption when the level of measurement in the samples is ordinal, rather than interval or ratio. For example, suppose there are 10 surgical patients on 3 East today. The

nursing supervisor asks Nurse Benner and Nurse Jurris to rate each of the 10 patients on a scale of 1 to 10, according to the difficulty of patient care. The distribution of the differences in the ratings probably would not approximate the normal distribution, and, therefore, the paired t test would not be appropriate.

In 1945, Frank Wilcoxon developed a nonparametric test, based on the differ- ences in dependent samples, where the normality assumption is not required. This test is called the Wilcoxon signed-rank test. The following example details its application.

LO3 Conduct a test of hypothesis for dependent samples using the Wilcoxon signed-rank test.

Example Fricker’s is a family restaurant chain located primarily in the southeastern part of the United States. It offers a full dinner menu, but its specialty is chicken. Recently, Bernie Frick, the owner and founder, developed a new spicy flavor for the batter in which the chicken is cooked. Before replacing the current flavor, he wants to con- duct some tests to be sure that patrons will like the spicy flavor better.

To begin, Bernie selects a random sample of 15 customers. Each sampled cus- tomer is given a small piece of the current chicken and asked to rate its overall taste on a scale of 1 to 20. A value near 20 indicates the participant liked the flavor, whereas a score near 0 indicates they did not like the flavor. Next, the same 15 participants

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Solution The samples are dependent or related. That is, the participants are asked to rate both flavors of chicken. Thus, if we compute the difference between the rating for the spicy flavor and the current flavor, the resulting value shows the amount the par- ticipants favor one flavor over the other. If we choose to subtract the current flavor score from the spicy flavor score, a positive result is the “amount” the participant favors the spicy flavor. Negative difference scores indicate the participant favored the current flavor. Because of the somewhat subjective nature of the scores, we are not sure the distribution of the differences follows the normal distribution. We decide to use the nonparametric Wilcoxon signed-rank test.

As usual, we will use the five-step hypothesis-testing procedure. The null hypothesis is that there is no difference in the rating of the chicken flavors by the participants. That is, as many participants in the study rated the spicy flavor higher as rated the regular flavor higher. The alternate hypothesis is that the ratings are higher for the spicy flavor. More formally:

: There is no difference in the ratings of the two flavors. : The spicy ratings are higher.

This is a one-tailed test. Why? Because Bernie Frick, the owner of Fricker’s, will want to change his chicken flavor only if the sample participants show that the population of customers like the new flavor better. The significance level is .05, as stated above.

The steps to conduct the Wilcoxon signed-rank test are as follows.

1. Compute the difference between the spicy flavor score and the current flavor score for each participant. For example, Arquette’s spicy flavor score was 14 and current flavor score was 12, so the amount of the difference is 2. For Jones, the difference is �8, found by 8 � 16, and for Fish it is 4, found by 6 � 2. The differences for all participants are shown in column 4 of Table 18–3.

2. Only the positive and negative differences are considered further. That is, if the difference in flavor scores is 0, that participant is dropped from further analysis and the number in the sample reduced. From Table 18–3, Hall, the sixth partici- pant, scored both the spicy and the current flavor a 16. Hence, Hall is dropped from the study and the usable sample size reduced from 15 to 14.

3. Determine the absolute differences for the values computed in column 4. Recall that in an absolute difference we ignore the sign of the difference. The absolute differences are shown in column 5.

4. Next, rank the absolute differences from smallest to largest. Arquette, the first participant, scored the spicy chicken a 14 and the current a 12. The difference of 2 in the two taste scores is the smallest absolute difference, so it is given a ranking of 1. The next largest difference is 3, given by Miller, so it is given a

H1 H0

are given a sample of the new chicken with the spicier flavor and again asked to rate its taste on a scale of 1 to 20. The results are reported below. Is it reasonable to con- clude that the spicy flavor is preferred? Use the .05 significance level.

Spicy Flavor Current Flavor Spicy Flavor Current Flavor Participant Score Score Participant Score Score

Arquette 14 12 Garcia 19 10 Jones 8 16 Sundar 18 10 Fish 6 2 Miller 16 13 Wagner 18 4 Peterson 18 2 Badenhop 20 12 Boggart 4 13 Hall 16 16 Hein 7 14 Fowler 14 5 Whitten 16 4 Virost 6 16

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rank of 2. The other differences are ranked in a similar manner. There are three participants who rated the difference in the flavor as 8. That is, Jones, Badenhop, and Sundar each had a difference of 8 between their rating of the spicy flavor and the current flavor. To resolve this issue, we average the ranks involved and report the average rank for each. This situation involves the ranks 5, 6, and 7, so all three participants are assigned the rank of 6. The same situation occurs for those participants with a difference of 9. The ranks involved are 8, 9, and 10, so those participants are assigned a rank of 9.

5. Each assigned rank in column 6 is then given the same sign as the original dif- ference, and the results are reported in column 7. For example, the second par- ticipant has a difference score of �8 and a rank of 6. This value is located in the section of column 7.

6. The and columns are totaled. The sum of the positive ranks is 75 and the sum of the negative ranks is 30. The smaller of the two rank sums is used as the test statistic and referred to as T.

The critical values for the Wilcoxon signed-rank test are located in Appendix B.7. A portion of that table is shown on the following page. The row is used for one- tailed tests and the 2 row for two-tailed tests. In this case, we want to show that customers like the spicy taste better, which is a one-tailed test, so we select the row. We chose the .05 significance level, so move to the right to the column headed .05. Go down that column to the row where n is 14. (Recall that one person in the study rated the chicken flavors the same and was dropped from the study, making the usable sample size 14.) The value at the intersection is 25, so the critical value is 25. The decision rule is to reject the null hypothesis if the smaller of the rank sums is 25 or less. The value obtained from Appendix B.7 is the largest value in the rejection region. To put it another way, our decision rule is to reject if the smaller of the two rank sums is 25 or less. In this case, the smaller rank sum is 30, so the decision is not to reject the null hypothesis. We cannot conclude there is a differ- ence in the flavor ratings between the current and the spicy. Mr. Frick has not shown that customers prefer the new flavor. Likely, he would stay with the current flavor of chicken and not change to the spicier flavor.

R�R� R�

TABLE 18–3 Flavor Rating for Current and Spicy Flavors

(1) (2) (3) (4) (5) (6) (7) Spicy Current Difference Absolute Signed Rank

Participant Score Score in Score Difference Rank R � R�

Arquette 14 12 2 2 1 1 Jones 8 16 �8 8 6 6 Fish 6 2 4 4 3 3 Wagner 18 4 14 14 13 13 Badenhop 20 12 8 8 6 6 Hall 16 16 * * * * Fowler 14 5 9 9 9 9 Virost 6 16 �10 10 11 11 Garcia 19 10 9 9 9 9 Sundar 18 10 8 8 6 6 Miller 16 13 3 3 2 2 Peterson 18 2 16 16 14 14 Boggart 4 13 �9 9 9 9 Hein 7 14 �7 7 4 4 Whitten 16 4 12 12 12 12

Total 75 30

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2� .15 .10 .05 .04 .03 .02 .01 n � .075 .05 .025 .02 .015 .01 .005

4 0 5 1 0 6 2 2 0 0 7 4 3 2 1 0 0 8 7 5 3 3 2 1 0 9 9 8 5 5 4 3 1

10 12 10 8 7 6 5 3 11 16 13 10 9 8 7 5 12 19 17 13 12 11 9 7 13 24 21 17 16 14 12 9 14 28 25 21 19 18 15 12 15 33 30 25 23 21 19 15

Exercises 11. An industrial psychologist selected a random sample of seven young urban professional

couples who own their homes. The size of their home (square feet) is compared with that of their parents. At the .05 significance level, can we conclude that the professional cou- ples live in larger homes than their parents?

Self-Review 18–4 The assembly area of Gotrac Products was recently redesigned. Installing a new lighting system and purchasing new workbenches were two features of the redesign. The produc- tion supervisor would like to know if the changes resulted in an improvement in worker productivity. To investigate, she selected a sample of 11 workers and determined the production rate before and after the changes. The sample information is reported below.

(a) How many usable pairs are there? That is, what is n? (b) Use the Wilcoxon signed-rank test to determine whether the new procedures actually

increased production. Use the .05 level and a one-tailed test. (c) What assumption are you making about the distribution of the differences in production

before and after redesign?

Production Production Production Production Operator Before After Operator Before After

S. M. 17 18 U. Z. 10 22 D. J. 21 23 Y. U. 20 19 M. D. 25 22 U. T. 17 20 B. B. 15 25 Y. H. 24 30 M. F. 10 28 Y. Y. 23 26 A. A. 16 16

Couple Name Professional Parent Couple Name Professional Parent

Gordon 1,725 1,175 Kuhlman 1,290 1,360 Sharkey 1,310 1,120 Welch 1,880 1,750 Uselding 1,670 1,420 Anderson 1,530 1,440 Bell 1,520 1,640

12. Toyota USA is studying the effect of regular versus high-octane gasoline on the fuel economy of its new high-performance, 3.5-liter, V6 engine. Ten executives are selected and asked to maintain records on the number of miles traveled per gallon of gas. The results are:

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Using the Wilcoxon signed-rank test, determine whether the suggested changes are worthwhile. a. State the null hypothesis. b. You decide on the alternate hypothesis. c. You decide on the level of significance. d. State the decision rule. e. Compute T and arrive at a decision. f. What did you assume about the distribution of the differences?

Miles per Gallon Miles per Gallon

Executive Regular High-Octane Executive Regular High-Octane

Bowers 25 28 Rau 38 40 Demars 33 31 Greolke 29 29 Grasser 31 35 Burns 42 37 DeToto 45 44 Snow 41 44 Kleg 42 47 Lawless 30 44

Production Production

Employee Old System Mump Method Employee Old System Mump Method

A 60 64 I 87 84 B 40 52 J 80 80 C 59 58 K 56 57 D 30 37 L 21 21 E 70 71 M 99 108 F 78 83 N 50 56 G 43 46 O 56 62 H 40 52

Past Production after Past Production after Production Installing, Lighting, Production Installing, Lighting,

Employee Record Music, etc. Employee Record Music, etc.

JD 23 33 WWJ 21 25 SB 26 26 OP 25 22 MD 24 30 CD 21 23 RCF 17 25 PA 16 17 MF 20 19 RRT 20 15 UHH 24 22 AT 17 9 IB 30 29 QQ 23 30

At the .05 significance level, is there a difference in the number of miles traveled per gallon between regular and high-octane gasoline?

13. A new assembly-line procedure to increase production has been suggested by Mr. Mump. To test whether the new procedure is superior to the old procedure, a random sample of 15 assembly line workers was selected. The number of units produced in an hour under the old procedure is determined, then the new Mump procedure was introduced. After an appro- priate break-in period, their production was measured again. The results were:

At the .05 significance level, can we conclude the production is greater using the Mump method? a. State the null and alternate hypotheses. b. State the decision rule. c. Arrive at a decision regarding the null hypothesis.

14. It has been suggested that daily production of a subassembly would be increased if better lighting were installed and background music and free coffee and doughnuts were provided during the day. Management agreed to try the scheme for a limited time. The number of subassemblies produced per week by a sample of employees follows.

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Example Dan Thompson, the president of CEO Airlines, recently noted an increase in the number of no-shows for flights out of Atlanta. He is particularly interested in deter- mining whether there are more no-shows for flights that originate from Atlanta com- pared with flights leaving Chicago. A sample of nine flights from Atlanta and eight from Chicago are reported in Table 18–4. At the .05 significance level, can we con- clude that there are more no-shows for the flights originating in Atlanta?

Solution If the populations of no-shows followed the normal probability distribution and had equal variances, the two-sample t test, discussed in Section 11.4 in Chapter 11, would be appropriate. In this case, Mr. Thompson believes these two conditions cannot be met. Therefore, a nonparametric test, the Wilcoxon rank-sum test, is appropriate.

If the number of no-shows is the same for Atlanta and Chicago, then we expect the sum of the ranks for the two distributions to be about the same. Or to put it another

TABLE 18–4 Number of No-Shows for Scheduled Flights

Atlanta Chicago Atlanta Chicago

11 13 20 9 15 14 24 17 10 10 22 21 18 8 25 11 16

18.4 Wilcoxon Rank-Sum Test for Independent Samples One test specifically designed to determine whether two independent samples came from equivalent populations is the Wilcoxon rank-sum test. This test is an alternative to the two-sample t test described on page 383 in Chapter 11. Recall that the t test requires that the two populations follow the normal distribution and have equal popu- lation variances. These conditions are not required for the Wilcoxon rank-sum test.

The Wilcoxon rank-sum test is based on the sum of ranks. The data are ranked as if the observations were part of a single sample. If the null hypothesis is true, then the ranks will be about evenly distributed between the two samples, and the sum of the ranks for the two samples will be about the same. That is, the low, medium, and high ranks should be about equally divided between the two sam- ples. If the alternate hypothesis is true, one of the samples will have more of the lower ranks and, thus, a smaller rank sum. The other sample will have more of the higher ranks and, therefore, a larger rank sum. If each of the samples contains at least eight observations, the standard normal distribution is used as the test statis- tic. The formula is:

WILCOXON RANK-SUM TEST [18–4]

where: is the number of observations from the first population. is the number of observations from the second population.

W is the sum of the ranks from the first population. n2 n1

z � W �

n1(n1 � n2 � 1) 2

n1n2(n1 � n2 � 1) 12

Test is based on the sum of the ranks

LO4 Conduct and interpret the Wilcoxon rank-sum test for independent samples.

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way, the average rank of the two groups will be about the same. If the number of no-shows is not the same, we expect the average of ranks to be quite different.

Mr. Thompson believes there are more no-shows for Atlanta flights. Thus, a one- tailed test is appropriate, with the rejection region located in the upper tail. The null and alternate hypotheses are:

: The population distribution of no-shows is the same or less for Atlanta and Chicago.

: The population distribution of no-shows is larger for Atlanta than for Chicago.

The test statistic follows the standard normal distribution. At the .05 significance level, we find from Appendix B.1 the critical value of z is 1.65. The null hypothesis is rejected if the computed value of z is greater than 1.65.

The alternate hypothesis is that there are more no-shows in Atlanta, which means that distribution is located to the right of the Chicago distribution. The details of rank assignment are shown in Table 18–5. We rank the observations from both samples as if they were a single group. The Chicago flight with only 8 no-shows had the fewest, so it is assigned a rank of 1. The Chicago flight with 9 no-shows is ranked 2, and so on. The Atlanta flight with 25 no-shows is the highest, so it is assigned the largest rank, 17. There are also two instances of tied ranks. There are Atlanta and Chicago flights that each have 10 no-shows. There are also two Atlanta flights with 11 no-shows. How do we handle these ties? The solution is to aver- age the ranks involved and assign the average rank to both flights. In the case involving 10 no-shows, the ranks involved are 3 and 4. The mean of these ranks is 3.5, so a rank of 3.5 is assigned to both the Atlanta and the Chicago flights with 10 no-shows.

The sum of the ranks for the Atlanta flights is 96.5. This is the value of W in formula 18-4. From Table 18–5, there are nine flights originating in Atlanta and eight in Chicago, so � 9 and � 8. Computing z from formula (18–4) gives:

Because the computed z value (1.49) is less than 1.65, the null hypothesis is not rejected. The evidence does not show a difference in the distributions of the num- ber of no-shows. That is, it appears that the number of no-shows is the same in Atlanta as in Chicago. The p-value is .0681, found by determining the area to the right of 1.49 (.5000 � .4319), indicates the same result.

z � W �

n1(n1 � n2 � 1) 2

n1n2(n1 � n2 � 1) 12

�

96.5 � 9(9 � 8 � 1)

9(8)(9 � 8 � 1) 12

� 1.49

n2n1

TABLE 18–5 Ranked Number of No-Shows for Scheduled Flights

Atlanta Chicago

No-Shows Rank No-Shows Rank

11 5.5 13 7 15 9 14 8 10 3.5 10 3.5 18 12 8 1 11 5.5 16 10 20 13 9 2 24 16 17 11 22 15 21 14 25 17

96.5 56.5

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MegaStat software can produce the same results. The MegaStat p-value is .0742, which is close to the value we calculated. The difference is due to rounding in the statistical software and a correction for ties.

In using the Wilcoxon rank-sum test, you may number the two populations in either order. However, once you have made a choice, W must be the sum of the ranks identified as population 1. If, in the no-show example, the population of Chicago was identified as number 1, the direction of the alternate hypothe- sis would be changed. The value of z would be the same but have the opposite sign.

: The population distribution of no-shows is the same or larger for Chicago than for Atlanta.

: The population distribution of no-shows is smaller for Chicago than for Atlanta.

The computed value of z is �1.49, found by:

Our conclusion is the same as described earlier. There is no difference in the typical number of no-shows for Chicago and Atlanta.

z � W �

n1(n1 � n2 � 1) 2

n1n2(n1 � n2 � 1) 12

�

56.5 � 8(8 � 9 � 1)

8(9)(8 � 9 � 1) 12

� �1.49

Self-Review 18–5 The research director for Top Flite wants to know whether there is a difference in the distribution of the distances traveled by two of the company’s golf balls. Eight of its XL-5000 brand and eight of its D2 brand balls were hit by an automatic fairway metal. The distances (in yards) were as follows:

Do not assume the distributions of the distances traveled follow the normal probability distribution. At the .05 significance level, is there a difference between the two distributions?

XL-5000: 252, 263, 279, 273, 271, 265, 257, 280 D2: 262, 242, 256, 260, 258, 243, 239, 265

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698 Chapter 18

Exercises 15. The following observations were randomly selected from populations that were not nec-

essarily normally distributed. Use the .05 significance level, a two-tailed test, and the Wilcoxon rank-sum test to determine whether there is a difference between the two populations.

Population A: 38, 45, 56, 57, 61, 69, 70, 79 Population B: 26, 31, 35, 42, 51, 52, 57, 62

Population A: 12, 14, 15, 19, 23, 29, 33, 40, 51 Population B: 13, 16, 19, 21, 22, 33, 35, 43

Campus 28, 16, 42, 29, 31, 22, 50, 42, 23, 25 Online 26, 42, 65, 38, 29, 32, 59, 42, 27, 41, 46, 18

15 years, fixed rate 41, 36, 42, 39, 36, 48, 49, 38 30 years, fixed rate 21, 27, 36, 20, 19, 21, 39, 24, 22

16. The following observations were randomly selected from populations that are not nec- essarily normally distributed. Use the .05 significance level, a two-tailed test, and the Wilcoxon rank-sum test to determine whether there is a difference between the two populations.

17. Tucson State University offers two MBA programs. In the first program, the students meet two nights per week at the university’s main campus in downtown Tucson. In the sec- ond program, students only communicate online with the instructor. The director of the MBA experience at Tucson wishes to compare the number of hours studied last week by the two groups of students. A sample of 10 on-campus students and 12 online stu- dents revealed the following information.

Do not assume the two distributions of study times, which are reported in hours, follow a normal distribution. At the .05 significance level, can we conclude the online students study more?

18. In recent times, with mortgage rates at low levels, financial institutions have had to pro- vide more customer convenience. One of the innovations offered by Coastal National Bank and Trust is online entry of mortgage applications. Listed below is the time, in min- utes, to complete the application process for customers applying for a 15-year fixed-rate and a 30-year fixed-rate mortgage.

At the .05 significance level, is it reasonable to conclude that it takes less time for those customers applying for the 30-year fixed-rate mortgage? Do not assume the distribution times follow a normal distribution for either group.

18.5 Kruskal-Wallis Test: Analysis of Variance by Ranks The analysis of variance (ANOVA) procedure discussed in Chapter 12 tests the hypothesis that several population means were equal. The data were interval or ratio level. Also, it was assumed the populations followed the normal probability distrib- ution and their standard deviations were equal. What if the data are ordinal scale and/or the populations do not follow a normal distribution? W. H. Kruskal and W. A. Wallis reported a nonparametric test in 1952 requiring only ordinal-level (ranked)

LO5 Conduct and interpret the Kruskal- Wallis test for several independent samples.

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Nonparametric Methods: Analysis of Ranked Data 699

data. No assumptions about the shape of the populations are required. The test is referred to as the Kruskal-Wallis one-way analysis of variance by ranks.

For the Kruskal-Wallis test to be applied, the samples selected from the popula- tions must be independent. For example, if samples from three groups—executives, staff, and supervisors—are to be selected and interviewed, the responses of one group (say, the executives) must in no way influence the responses of the others.

To compute the Kruskal-Wallis test statistic, (1) all the samples are combined, (2) the combined values are ordered from low to high, and (3) the ordered values are replaced by ranks, starting with 1 for the smallest value. An example will clarify the details of the procedure.

Test requires independent samples but the populations do not have to be normal.

Example The Hospital Systems of the Carolinas operate three hospitals in the Greater Charlotte area: St. Luke’s Memorial on the west side of the city, Swedish Med- ical Center to the south, and Piedmont Hospital on the east side of town. The director of administration is concerned about the waiting time of patients with non-life-threatening athletic-type injuries that arrive during weekday evenings at the three hospitals. Specifically, is there a difference in the waiting times at the three hospitals?

Solution To investigate, the director selected random samples of patients at the three loca- tions and determined the time, in minutes, between entering the particular facility and when treatment was completed. The times in minutes are reported in Table 18–6.

From Table 18–6, we observe that the shortest waiting time is 35 minutes for the fifth sampled patient at Piedmont Hospital. The longest waiting time is 107 minutes by the seventh patient at the Swedish Medical Center.

Likely the first thought for comparing the waiting times is to determine whether there is a difference in the mean waiting time at the three hospitals, that is, use the one-way ANOVA described in Section 12.5. However, as we described in Section 12.4, there are three requirements for that test:

1. The samples are from independent populations. 2. The population variances must be equal. 3. The samples are from normal populations.

In this instance, the samples are from independent populations, the three differ- ent hospitals. But suppose we do not want to assume equal variance in the waiting times at the three hospitals or that these waiting times follow a normal probability dis- tribution. Failing these two criteria means we do not meet the ANOVA requirements, and so we cannot use the ANOVA techniques. Instead, we turn to the Kruskal-Wallis test, where these assumptions are not required.

St. Luke’s Memorial Swedish Medical Center Piedmont Hospital

56 103 42 39 87 38 48 51 89 38 95 75 73 68 35 60 42 61 62 107

TABLE 18–6 Waiting Times for Emergency Room Patients at Hospital Systems of the Carolinas

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700 Chapter 18

The first step in conducting the Kruskal-Wallis test is to state the null and the alternate hypotheses.

: The population distributions of waiting times are the same for the three hospitals.

: The population distributions are not all the same for the three hospitals.

The director of administration selected the .05 significance level. The test statistic used for the Kruskal-Wallis test is designated H. Its formula is:

with k � 1 degrees of freedom (k is the number of populations), where:

are the sums of the ranks of samples 1, 2, . . . , k, respectively.

, , . . . , are the sizes of samples 1, 2, . . . , k, respectively. n is the combined number of observations for all samples.

The distribution of the sample H statistic is very close to the chi-square distri- bution with k � 1 degrees of freedom. We prefer that each sample include at least five observations. We use chi-square to formulate the decision rule. In this example, there are three populations—a population of waiting times for patients at St. Luke’s Memorial, another for patients at the Swedish Medical Center, and a third for Pied- mont Hospital patients. Thus, there are k � 1, or 3 � 1 � 2 degrees of freedom. Refer to the chi-square table of critical values in Appendix B.3. The critical value for 2 degrees of freedom and the .05 level of significance is 5.991. So our decision rule is: Do not reject the null hypothesis if the computed value of the test statistic H is less than or equal to 5.991. If the computed value of H is greater than 5.991, reject the null hypothesis and accept the alternate hypothesis.

The next step is to determine the value of the test statistic. We replace the waiting times at the three hospitals with the corresponding ranks. Considering the wait- ing times as a single population, the Piedmont patient with a waiting time of 35 minutes waited the shortest time and hence is given the lowest rank of 1. There are two patients that waited 38 minutes, one at St. Luke’s and one at Piedmont. To resolve this tie, each patient is given a rank of 2.5, found by (2 � 3)�2. This process is continued for all waiting times. The longest waiting time is 107 minutes, and that Swedish Medical Center patient is given a rank of 21. The scores, the ranks, and the sum of the ranks for each of the three hospitals are given in Table 18–7.

nkn2n1

©R1, ©R2, . . . , ©Rk

KRUSKAL- WALLIS TEST

[18–5]H � 12

n(n � 1) c (©R1)2

n1 �

(©R2)2

n2 � . . . �

(©Rk)2

nk d � 3(n � 1)

St. Luke’s Memorial Swedish Medical Center Piedmont Hospital

Waiting Rank of Waiting Rank of Waiting Rank of Time Waiting Time Time Waiting Time Time Waiting Time

56 9.0 103 20.0 42 5.5 39 4.0 87 16.0 38 2.5 48 7.0 51 8.0 89 17.5 38 2.5 95 19.0 75 15.0 73 14.0 68 13.0 35 1.0 60 10.0 42 5.5 61 11.0 62 12.0 107 21.0

89 17.5

©R1 � 58.5 ©R2 � 120.0 ©R3 � 52.5

TABLE 18–7 Waiting Times, Ranks, and Sums of Ranks for Hospital Systems of the Carolinas

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Nonparametric Methods: Analysis of Ranked Data 701

Solving for H gives

Because the computed value of H(5.38) is less than the critical value of 5.991, we do not reject the null hypothesis. There is not enough evidence to conclude that there is a difference among the waiting times at the three hospitals.

The Kruskal-Wallis procedure can be done using Minitab software. Output for the example regarding the hospital waiting time follows. The computed value of H is 5.38 and the p-value reported on the output is .068. This agrees with our earlier calculations.

� 12

21(21 � 1) c 58.52

7 �

1202

8 �

52.52

6 d � 3(21 � 1) � 5.38

H � 12

n(n � 1) c (©R1)2

n1 �

(©R2)2

n2 �

(©R3)2

n3 d � 3(n � 1)

Recall from Chapter 12 that, for the analysis of variance technique to apply, we assume: (1) the populations are normally distributed, (2) these populations have equal standard deviations, and (3) the samples are selected from indepen- dent populations. If these assumptions are met in the hospital waiting time exam- ple, we use the F distribution as the test statistic. If these assumptions cannot be met, we apply the distribution-free test by Kruskal-Wallis. To highlight the sim- ilarities between the two approaches, we will solve the hospital waiting time example using the ANOVA technique.

To begin, we state the null and the alternate hypotheses for the three hospitals.

: � � : The treatment means are not all the same.

For the .05 significance level, with k � 1 � 3 � 1 � 2 degrees of freedom in the numerator and n � k � 21 � 3 � 18 degrees of freedom in the denominator, the crit- ical value of F is 3.55. The decision rule is to reject the null hypothesis if the computed value of F is greater than 3.55. The output using Excel follows.

H1 �3�2�1H0

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702 Chapter 18

From the previous output, the computed value of F is 3.822 and the p-value is .041. Our decision is to reject the null hypothesis and accept the alternate hypothesis. Using the one-way ANOVA test, we conclude that the treatment means are not the same. That is, the mean waiting times at the three Hospital Systems of the Carolinas hospitals are different.

We have contradictory conclusions on the same data. How can this happen? If we compare the results using p-values, the answers are similar. For the Kruskal-Wallis test, the p-value was .057, which is only slightly more than the significance level of .05, but our decision was not to reject the null hypothesis. The p-value using ANOVA is .041, which is not far beyond the critical value into the rejection region. So, to sum- marize, we just missed rejecting with the Kruskal-Wallis test, and we were just in the rejection region using ANOVA. The difference in the p-values is .016. Thus, the results are actually quite close in terms of the p-values.

Exercises 19. Under what conditions should the Kruskal-Wallis test be used instead of analysis of

variance? 20. Under what conditions should the Kruskal-Wallis test be used instead of the Wilcoxon

rank-sum test? 21. The following sample data were obtained from three populations that did not follow a

normal distribution.

Self-Review 18–6 The regional bank manager of Statewide Financial Bank is interested in the turnover rate of personal checking accounts at four of the branches. (Turnover rate is the speed at which the money in an account is deposited and withdrawn. An extremely active account may have a rate of 300; if only one or two checks were written, the rate could be about 30.) The turnover rates of the samples selected from the four branch banks are shown in the table below. Using the .01 level and the Kruskal-Wallis test, determine whether there is a difference in the turnover rates of the personal checking accounts among the four branches.

West Great Englewood Side Northern Sylvania

Branch Branch Branch Branch

208 91 302 99 307 62 103 116 199 86 319 189 142 91 340 103

91 80 180 100 296 131

Similar results for one-way ANOVA and Kruskal-Wallis.

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Nonparametric Methods: Analysis of Ranked Data 703

a. State the null hypothesis. b. Using the .05 level of risk, state the decision rule. c. Compute the value of the test statistic. d. What is your decision on the null hypothesis?

22. The following sample data were obtained from three populations where the variances were not equal, and you wish to compare the populations.

Sample 1 Sample 2 Sample 3

50 48 39 54 49 41 59 49 44 59 52 47 65 56 51

Sample 1 Sample 2 Sample 3

21 15 38 29 17 40 35 22 44 45 27 51 56 31 53 71

a. State the null hypothesis. b. Using the .01 level of risk, state the decision rule. c. Compute the value of the test statistic. d. What is your decision on the null hypothesis?

23. Davis Outboard Motors Inc. recently developed an epoxy painting process for corrosion protection on exhaust components. Bill Davis, the owner, wishes to determine whether the lengths of life for the paint are equal for three different conditions: saltwater, fresh- water without weeds, and freshwater with a heavy concentration of weeds. Accelerated- life tests were conducted in the laboratory, and the number of hours the paint lasted before peeling was recorded.

Freshwater Saltwater Freshwater with Weeds

167.3 160.6 182.7 189.6 177.6 165.4 177.2 185.3 172.9 169.4 168.6 169.2 180.3 176.6 174.7

Use the Kruskal-Wallis test and the .01 level to determine whether the lasting quality of the paint is the same for the three water conditions.

24. The National Turkey Association wants to experiment with the effects of three different feed mixtures on weight gain in poults. Because no experience exists regarding the three mixtures, no assumptions regarding the population distribution of weights exists. To study the effects of the three mixtures, five poults were given feed A, six were given feed B, and five were given feed C over a three-week time period. Test at the .05 level the hypoth- esis that there is no effect of feed mixture on weight.

Weight (in pounds)

Feed Mixture A Feed Mixture B Feed Mixture C

11.2 12.6 11.3 12.1 10.8 11.9 10.9 11.3 12.4 11.3 11.0 10.6 12.0 12.0 12.0

10.7

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Example Lorrenger Plastics Inc. recruits management trainees at colleges and universities throughout the United States. Each trainee is given a score by the recruiter during the on-campus interview. This score is an expression of future potential and may range from 0 to 200, with the higher score indicating more potential. If the appli- cant is hired by Lorrenger, he or she then enters an in-plant training program. At the completion of this program, the recruit is given another composite score. This score, which is based on tests and the opinions of group leaders and in-plant training offi- cers, can range from 0 to 100. Again, a higher score indicates more potential. The on-campus scores and the in-plant training scores are given in Table 18–8.

704 Chapter 18

18.6 Rank-Order Correlation In Chapter 13, we discussed r, the sample coefficient of correlation. Recall that it mea- sures the association between two interval- or ratio-scaled variables. For example, the coefficient of correlation reports the association between the salary of executives and their years of experience, or the association between the number of miles a ship- ment had to travel and the number of days it took to arrive at its destination.

Charles Spearman, a British statistician, introduced a measure of correlation for ordinal-level data. This measure allows us to describe the relationship between sets of ranked data. For example, two staff members in the Office of Research at the University of the Valley are asked to rank 10 faculty research proposals for funding purposes. We want to study the association between the ratings of the two staff members. That is, do the two staff members rate the same proposals as the most worthy and the least worthy of funding? Spearman’s coefficient of rank correlation, denoted , provides a measure of the association.

The coefficient of rank correlation is computed by the following formula. rs

SPEARMAN’S COEFFICIENT OF RANK CORRELATION

[18–6]rs � 1 � 6©d 2

n(n2 � 1)

where: d is the difference between the ranks for each pair. n is the number of paired observations.

Like the coefficient of correlation, the coefficient of rank correlation can assume any value from �1.00 up to 1.00. A value of �1.00 indicates perfect negative cor- relation and a value of 1.00 perfect positive correlation among the ranks. A rank correlation of 0 indicates that there is no association among the ranks. Rank cor- relations of �.84 and .80 both indicate a strong association, but the former indi- cates an inverse relationship between the ranks and the latter a direct relationship.

Graduate On-Campus Score In-Plant Score

Spina, Sal 83 45 Gordon, Ray 106 45 Althoff, Roberta 92 45 Alvear, Ginny 48 36 Wallace, Ann 127 68 Lyons, George 113 83 Harbin, Joe 118 88

(continued )

TABLE 18–8 On-Campus Scores and In-Plant Scores for a Sample of Recent College Graduates Hired at Lorrenger Plastics Inc.

LO6 Compute and interpret Spearman‘s coefficient of rank correlation.

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Determine the association between the on-campus and in-plant scores. Do those recruits that earn high scores during the on-campus interview also earn high scores during their in-plant training?

Nonparametric Methods: Analysis of Ranked Data 705

Graduate On-Campus Score In-Plant Score

Davison, Jack 78 61 Brydon, Tom 83 66 Bobko, Jack 193 94 Koppel, Marty 101 56 Nyland, Patricia 123 91

Solution In Section 4.6, we investigated the association between two variables using a scatter diagram. This is a good place to begin. Below is a scatter diagram of the associa- tion between the on-campus and in-plant scores. Clearly there is a direct or positive association between the two scores. However, note the plot for Jack Bobko, the third graduate from the bottom. His on-campus score of 193 is 66 points higher than Ann Wallace, the next highest score. Bobko’s plot is a potential outlier from the others and may be distorting the association between the two variables.

The usual measure of association is the coefficient of correlation, described in Section 13.3 on page 468. This measure of association requires both variables to be interval scale. In this case, the scores are interval scale, but the fact that the one score is so much larger, an outlier, is a problem. Because that point seems so dif- ferent from the others, statisticians often suggest the rank of the scores be used in place of the actual scores. Spearman’s Coefficient of Rank Correlation uses the ranks of the scores rather than the actual scores. That is, it correlates the ranks rather than the actual scores and reduces the impact of Bobko’s on-campus score being so much larger than the others.

To calculate the coefficient of rank correlation, we first rank the scores from low to high. We start with the on-campus scores. The lowest score awarded by the on- campus recruiter is a 48 to Ginny Alvear, so she is ranked 1. The next lowest is 78 to Jack Davison, so he is ranked 2. There are two on-campus scores of 83. The tie is resolved by giving each a rank of 3.5, which is the average of ranks 3 and 4. The highest on-campus score is Jack Bobko’s 193; he is given the highest rank of 12.

30 0 50 100 150 200 250

100

On-Campus Score

Scatter Diagram of On-Campus and In-Plant Scores for

Lorrenger Plastics Recruits

In -P

la nt

S co

reStatistics in Action

Manatees are large mammals that like to float just below the water’s surface. Be- cause they float just below the surface, they are in danger from powerboat pro- pellers. A study of the correlation be- tween the number of powerboat registra- tions in coastal Florida counties and the number of acci- dental manatee deaths revealed a strong positive corre- lation. As a result, Florida created re- gions where power- boats are prohibited, so that manatees could thrive.

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706 Chapter 18

Testing the Significance of rs In Section 13.4 of Chapter 13, we tested the significance of Pearson’s r. For ranked data, the question also arises whether the correlation in the population is actually zero. For instance, there were only 12 graduates sampled in the preceding Example. The rank correlation coefficient of .785 indicates a rather strong direct or positive relationship. Is it possible that the correlation of .785 is due to chance and that the correlation among the ranks in the population is really 0? We can conduct a test of hypothesis to answer that question.

For a sample of 10 or more, the significance of rs is determined by computing t using the following formula. The sampling distribution of rs follows the t distribu- tion with n � 2 degrees of freedom.

Testing whether correlation in the population is zero

HYPOTHESES TEST, RANK CORRELATION [18–7]t � rs A

n � 2 1 � r s

The null and the alternate hypotheses are:

: The rank correlation in the population is zero. : There is a positive association among the ranks.H1

The same procedure is followed for the in-plant scores. Again, Ginny Alvear had the lowest score of 36, so her in-plant ranking is 1. There are three in-plant scores of 45. The mean of the three tied ranks is 3, found by (2 � 3 � 4)�3 � 3, so each of these recruits is awarded an in-plant ranking of 3. This is illustrated along with the necessary calculations for rs in Table 18–9.

The coefficient of rank correlation is .785, found by using formula (18–6).

The value of .785 indicates a strong positive association between the scores of the on-campus recruiter and the scores of the in-plant training staff. The graduates who received high scores from the on-campus recruiter also tended to be the ones who received high scores from the training staff. It would be reasonable to conclude that there is association between the two sets of scores.

rs � 1 � 6©d2

n(n2 � 1) � 1 �

6(61.50) 12(122 � 1)

� 1 � .215 � .785

Scores Rank Difference between

On-Campus In-Plant On-Campus In-Plant Graduate Score Score Rank Rank Ranks (d ) d 2

Spina, Sal 83 45 3.5 3 0.5 0.25 Gordon, Ray 106 45 7 3 4.0 16.00 Althoff, Roberta 92 45 5 3 2.0 4.00 Alvear, Ginny 48 36 1 1 0.0 0.00 Wallace, Ann 127 68 11 8 3.0 9.00 Lyons, George 113 83 8 9 �1.0 1.00 Harbin, Joe 118 88 9 10 �1.0 1.00 Davison, Jack 78 61 2 6 �4.0 16.00 Brydon, Tom 83 66 3.5 7 �3.5 12.25 Bobko, Jack 193 94 12 12 0.0 0.00 Koppel, Marty 101 56 6 5 1.0 1.00 Nyland, Patricia 123 91 10 11 �1.0 1.00

0 61.50

TABLE 18–9 Calculations for the Coefficient of Rank Correlation ( )rs

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Nonparametric Methods: Analysis of Ranked Data 707

The decision rule is to reject if the computed value of t is greater than 1.812 (from Appendix B.2, .05 significance level, one-tailed test, and 10 degrees of freedom, found by n � 2 � 12 � 2 � 10).

The computed value of t is 4.007:

The null hypothesis is rejected because the computed t of 4.007 is greater than 1.812. The alternate hypothesis is accepted. There is evidence of a positive correlation between the ranks given by the on-campus recruiters and the ranks assigned during training.

t � rs A

n � 2 1 � r s

2 � .785A 12 � 2

1 � .7852 � 4.007

Exercises 25. Do husbands and wives like the same TV shows? A recent study by Nielsen Media

Research asked a young married couple to rank shows from best to worst. A rank of 1 indicates the best liked show and a rank of 14 the least liked show. The results for one married couple follow.

Program Male Rating Female Rating

60 Minutes 4 5 CSI—New York 6 4 Bones 7 8 SportsCenter 2 7 Late Show with David Letterman 12 11 NBC Nightly News 8 6 Law and Order: Los Angeles 5 3 Miami Medical 3 9 Survivor 13 2 Office 14 10 American Idol 1 1 Grey’s Anatomy 9 13 House 10 12 Criminal Minds 11 14

a. Draw a scatter diagram. Place the male rankings on the horizontal axis and the female rankings on the vertical axis.

b. Compute the coefficient of rank correlation between the male and female rankings. c. At the .05 significance level, is it reasonable to conclude there is a positive associa-

tion between the two rankings?

Self-Review 18–7 A sample of individuals applying for factory positions at Davis Enterprises revealed the following scores on an eye perception test (X ) and a mechanical aptitude test (Y ):

(a) Compute the coefficient of rank correlation. (b) At the .05 significance level, can we conclude that the correlation in the population is

different from 0?

Eye Mechanical Eye Mechanical Subject Perception Aptitude Subject Perception Aptitude

001 805 23 006 810 28 002 777 62 007 805 30 003 820 60 008 840 42 004 682 40 009 777 55 005 777 70 010 820 51

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708 Chapter 18

26. Far West University offers both day and evening classes in business administration. One question in a survey of students inquires how they perceive the prestige associated with certain careers. A day student was asked to rank the careers from 1 to 8, with 1 having the most prestige and 8 the least prestige. An evening student was asked to do the same. The results follow.

Ranking Ranking Ranking Ranking by Day by Evening by Day by Evening

Career Student Student Career Student Student

Accountant 6 3 Statistician 1 7 Computer programmer 7 2 Marketing researcher 4 8 Branch bank manager 2 6 Stock analyst 3 5 Hospital administrator 5 4 Production manager 8 1

Annual Ranking in Annual Ranking in Sales Training Sales Training

Representative ($ thousands) Program Representative ($ thousands) Program

Kitchen 319 3 Arden 300 10 Bond 150 9 Crane 280 5 Gross 175 6 Arthur 200 2 Arbuckle 460 1 Keene 190 7 Greene 348 4 Knopf 300 8

Rank, by Assistant Coach Rank, by Assistant Coach

Player Jean Cann John Cannelli Player Jean Cann John Cannelli

Cora Jean Seiple 7 5 Candy Jenkins 3 1 Bette Jones 2 4 Rita Rosinski 5 7 Jeannie Black 10 10 Anita Lockes 4 2 Norma Tidwell 1 3 Brenda Towne 8 9 Kathy Marchal 6 6 Denise Ober 9 8

Find Spearman’s coefficient of rank correlation. 27. New representatives for Clark Sprocket and Chain Inc. attended a brief training pro-

gram before being assigned to a regional sales office. At the end of the training program, the representatives were ranked with respect to future sales potential by the vice presi- dent of sales. At the end of the first sales year, their rankings were paired with their first year sales:

a. Compute and interpret the coefficient of rank correlation between first-year sales and rank after the training program.

b. At the .05 significance level, can we conclude that there is a positive association between first-year sales dollars and ranking in the training program?

28. Suppose Texas A & M University—Commerce has five scholarships available for the women’s basketball team. The head coach provided the two assistant coaches with the names of 10 high school players with potential to play at the university. Each assis- tant coach attended three games and then ranked the 10 players with respect to poten- tial. To explain, the first coach ranked Norma Tidwell as the best player among the 10 scouted and Jeannie Black the worst.

a. Determine Spearman’s rank correlation coefficient. b. At the .05 significance level, can we conclude there is a positive association between

the ranks?

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Nonparametric Methods: Analysis of Ranked Data 709

Chapter Summary I. The sign test is based on the sign difference between two related observations.

A. No assumptions need to be made about the shape of the two populations. B. It is based on paired or dependent samples. C. For small samples, find the number of � or � signs and refer to the binomial distri-

bution for the critical value. D. For a sample of 10 or more, use the standard normal distribution and the following

formula.

[18–2] [18–3]

II. The median test is used to test a hypothesis about a population median. A. Find � and � for a binomial distribution. B. The z distribution is used as the test statistic. C. The value of z is computed from the following formula, where X is the number of obser-

vations above or below the median.

[18–1]

III. The Wilcoxon signed-rank test is a nonparametric test where the normality assumption is not required. A. Data must be at least ordinal scale, and the samples must be dependent. B. The steps to conduct the test are:

1. Rank absolute differences between the related observations. 2. Apply the sign of the differences to the ranks. 3. Sum negative ranks and positive ranks. 4. The smaller of the two sums is the computed T value. 5. Refer to Appendix B.7 for the critical value, and make a decision regarding H0.

IV. The Wilcoxon rank-sum test is used to test whether two independent samples came from equal populations. A. No assumption about the shape of either population is required. B. The data must be at least ordinal scale. C. Each sample must contain at least eight observations. D. To determine the value of the test statistic W, the sample observations are ranked from

low to high as if they were from a single group. E. The sum of ranks for each of the two samples is determined. F. W is used to compute z, where W is the sum of the ranks for population 1.

[18–4]

G. The standard normal distribution, found in Appendix B.1, is the test statistic. V. The Kruskal-Wallis one-way ANOVA by ranks is used to test whether several populations

are the same. A. No assumption regarding the shape of any of the populations is required. B. The samples must be independent and at least ordinal scale. C. The sample observations are ranked from smallest to largest as though they were a

single group. D. The test statistic follows the chi-square distribution, provided there are at least five obser-

vations in each sample. E. The value of the test statistic is computed from the following:

[18–5]

VI. Spearman’s coefficient of rank correlation is a measure of the association between two ordinal-scale variables.

H � 12

n(n � 1) c (©R1)2

n1 �

(©R2)2

n2 � . . . �

(©Rk )2

nk d � 3(n � 1)

z � W �

n1(n1 � n2 � 1) 2

n1n2(n1 � n2 � 1) 12

z � (X .50) � �

�

z � (X .50) � .50n

.501n

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710 Chapter 18

A. It can range from �1 up to 1. 1. A value of 0 indicates there is no association between the variables. 2. A value of �1 indicates perfect negative correlation, and 1 indicates perfect positive

correlation. B. The value of is computed from the following formula.

[18–6]

C. Provided the sample size is at least 10, we can conduct a test of hypothesis using the following formula:

[18–7]

1. The test statistic follows the t distribution. 2. There are n � 2 degrees of freedom.

Pronunciation Key SYMBOL MEANING PRONUNCIATION

Spearman’s coefficient of r sub s rank correlation

Chapter Exercises 29. The vice president of programming at NBC is finalizing the prime-time schedule for the

fall. She has decided to include a hospital drama but is unsure which of two possibili- ties to select. She has a pilot called “The Surgeon” and another called “Critical Care.” To help her make a final decision, a sample of 20 viewers from throughout the United States was asked to watch the two pilots and indicate which show they prefer. The results were that 12 liked The Surgeon, 7 liked Critical Care, and one had no preference. Is there a preference for one of the two shows? Use the .10 significance level.

30. IBM Inc. is going to award a contract for fine-line pens to be used nationally in its offices. Two suppliers, Bic and Pilot, have submitted bids. To determine the preference of office employees, brokers, and others, a personal preference test is to be conducted using a randomly selected sample of 20 employees. The .05 level of significance is to be used. a. If the alternate hypothesis states that Bic is preferred over Pilot, is the sign test to be

conducted as a one-tailed or a two-tailed test? Explain. b. As each of the sample members told the researchers his or her preference, a “�” was

recorded if it was Bic and a “�” if it was the Pilot fine-line pen. A count of the pluses revealed that 12 employees preferred Bic, 5 preferred Pilot, and 3 were undecided. What is n?

c. What is the decision rule in words? d. What conclusion did you reach regarding pen preference? Explain.

31. Cornwall and Hudson, a large retail department store, wants to handle just one brand of high-quality CD player. The list has been narrowed to two brands: Sony and Panasonic. To help make a decision, a panel of 16 audio experts met. A passage using Sony components (labeled A) was played. Then the same passage was played using Panasonic components (labeled B). A “�” in the following table indicates an individual’s preference for the Sony components, a “�” indicates preference for Panasonic, and a 0 signifies no preference.

t � rs B

n � 2 1 � r 2s

rs � 1 � 6©d 2

n(n2 � 1)

Expert

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

� � � � � � � 0 � � � � � � � �

Conduct a test of hypothesis at the .10 significance level to determine whether there is a difference in preference between the two brands.

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Nonparametric Methods: Analysis of Ranked Data 711

32. The Greater Jacksonville, Florida, Real Estate Association claims that the median rental for three-bedroom condominiums is more than $1,200 a month. A random sample of 149 units showed 5 rented for exactly $1,200 a month, and 75 rented for more than $1,200. At the .05 level, can we conclude that the median rental is more than $1,200? a. State and . b. Give the decision rule. c. Do the necessary calculations, and arrive at a decision.

33. The Citrus Council of America wants to determine whether consumers prefer plain orange juice or juice with some orange pulp in it. A random sample of 212 consumers was selected. Each member of the sample tasted a small, unlabeled cup of one kind and then tasted the other kind. Twelve consumers said they had no preference, 40 preferred plain juice, and the remainder liked the juice with pulp better. Test at the .05 level that the pref- erences for plain juice and for orange juice with pulp are equal.

34. The objective of a community research project is to determine whether women are more community conscious before marriage or after five years of marriage. A test designed to mea- sure community consciousness was administered to a sample of women before marriage, and the same test was given to them five years after marriage. The test scores are:

H1H0

Before After Before After Name Marriage Marriage Name Marriage Marriage

Beth 110 114 Carol 186 196 Jean 157 159 Lisa 116 116 Sue 121 120 Sandy 160 140 Cathy 96 103 Petra 149 142 Mary 130 139

Test at the .05 level. is: There is no difference in community consciousness before and after marriage. is: There is a difference.

35. Is there a difference in the annual divorce rates in predominantly rural counties among three geographic regions, namely, the Southwest, the Southeast, and the Northwest? Test at the .05 level. Annual divorce rates per 1,000 population for randomly selected counties are:

H1 H0

Southwest: 5.9, 6.2, 7.9, 8.6, 4.6 Southeast: 5.0, 6.4, 7.3, 6.2, 8.1, 5.1 Northwest: 6.7, 6.2, 4.9, 8.0, 5.5

36. The production manager of MPS Audio Systems Inc. is concerned about the idle time of workers. In particular, he would like to know if there is a difference in the idle minutes for workers on the day shift and the evening shift. The information to the left is the num- ber of idle minutes yesterday for the five day-shift workers and the six evening-shift work- ers. Use the .05 significance level.

37. Drs. Trythall and Kerns are studying the mobility of executives in selected industries. Their research measures mobility using a score based on the number of times an executive has moved, changed companies, or changed jobs within a company over the last 10 years. The highest number of points is awarded for moving and changing companies, the fewest for changing jobs within a company and not moving. The distribution of scores does not follow the normal probability distribution. Develop an appropriate test to determine if there is a difference in the mobility scores in the four industries. Use the .05 significance level.

Day Evening Shift Shift

92 96 103 114 116 80

81 82 89 88

Chemical Retail Internet Space

4 3 62 30 17 12 40 38 8 40 81 46

20 17 96 40 16 31 76 21

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712 Chapter 18

38. A series of questions on sports and world events was asked of a randomly selected group of young adult naturalized citizens. The results were translated into a “knowledge” score. The scores were:

Citizen Sports World Events Citizen Sports World Events

J. C. McCarthy 47 49 L. M. Zaugg 87 75 A. N. Baker 12 10 J. B. Simon 59 86 B. B. Beebe 62 76 J. Goulden 40 61 L. D. Gaucet 81 92 A. A. Fernandez 87 18 C. A. Jones 90 86 A. M. Carbo 16 75 J. N. Lopez 35 42 A. O. Smithy 50 51 A. F. Nissen 61 61 J. J. Pascal 60 61

a. Determine the degree of association between how the citizens ranked with respect to knowledge of sports and how they ranked on world events.

b. At the .05 significance level, is the rank correlation in the population greater than zero? 39. Early in the basketball season, 12 college teams appeared to be outstanding. A panel of

sportswriters and a panel of college basketball coaches were asked to rank the 12 teams. Their composite rankings were as follows.

Team Coaches Sportswriters Team Coaches Sportswriters

Duke 1 1 Syracuse 7 10 UNLV 2 5 Georgetown 8 11 Indiana 3 4 Villanova 9 7 North Carolina 4 6 LSU 10 12 Louisville 5 3 St. Johns 11 8 Ohio State 6 2 Michigan 12 9

Determine the correlation between the rankings of the coaches and the sportswriters. At the .05 significance level, can we conclude there is a positive correlation between the rankings?

40. Professor Bert Forman believes the students who complete his examinations in the short- est time receive the highest grades and those who take the longest to complete them receive the lowest grades. To verify his suspicion, he assigns a rank to the order of finish and then grades the examinations. The results are shown below:

Order of Score Order of Score Student Completion (50 possible) Student Completion (50 possible)

Gromney 1 48 Smythe 7 39 Bates 2 48 Arquette 8 30 MacDonald 3 43 Govito 9 37 Sosa 4 49 Gankowski 10 35 Harris 5 50 Bonfigilo 11 36 Cribb 6 47 Matsui 12 33

Convert the test scores to a rank and find the coefficient of rank correlation. At the .05 significance level, can Professor Forman conclude there is a positive association between the order of finish and the test scores?

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Nonparametric Methods: Analysis of Ranked Data 713

Data Set Exercises 41. Refer to the Real Estate data, which report information on homes sold in Goodyear,

Arizona. a. Use an appropriate nonparametric test to determine whether there is a difference in

the typical selling price of the homes in the several townships. Assume the selling prices are not normally distributed. Use the .05 significance level.

b. Combine the homes with 6 or more bedrooms into one group and determine whether there is a difference according to the number of bedrooms in the typical selling prices of the homes. Use the .05 significance level and assume the distribution of selling prices is not normally distributed.

c. Assume that the distribution of the distance from the center of the city is positively skewed. That is, the normality assumption is not reasonable. Compare the distribu- tion of the distance from the center of the city of the homes that have a pool with those that do not have a pool. Can we conclude there is a difference in the distribu- tions? Use the .05 significance level.

42. Refer to the Baseball 2009 data, which report information on the 2009 Major League Baseball season. a. Rank the teams by the number of wins and their total team salary. Compute the coef-

ficient of rank correlation between the two variables. At the .01 significance level, can you conclude that it is greater than zero?

b. Assume that the distributions of team salaries for the American League and National League do not follow the normal distribution. Conduct a test of hypothesis to see whether there is a difference in the two distributions.

c. Rank the 30 teams by attendance and by team salary. Determine the coefficient of rank correlation between these two variables. At the .05 significance level, is it rea- sonable to conclude the ranks of these two variables are related?

43. Refer to the data on the school buses in the Buena School District. a. Assume the distribution of the maintenance cost for the three bus manufacturers does

not follow a normal distribution. Conduct a test of hypothesis at the .05 significance level to determine whether the distributions differ.

b. Assume the distribution of the maintenance cost for the bus capacities does not fol- low a normal distribution. Conduct a test of hypothesis at the .05 significance level to determine whether the distributions differ.

c. Assume the distribution of the maintenance cost for the bus types, diesel or gasoline, does not follow a normal distribution. Conduct a test of hypothesis at the .05 signif- icance level to determine whether the distributions differ.

1. The MegaStat for Excel commands necessary for the Wilcoxon rank-sum test on page 697 are: a. Enter the number of no-shows for Atlanta

in column A and for Chicago in column B. b. Select MegaStat, Nonparametric Tests,

and Wilcoxon-Mann/Whitney Test, then hit Enter.

c. For Group 1, use the data on Atlanta flights (B3:B11) and for Group 2 use the data on Chicago flights (D3:D10). Click on Correct for ties and one-tailed, and less than as the Alternative, then click on OK.

Software Commands

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714 Chapter 18

2. The Minitab commands for the Kruskal-Wallis test on page 701 are: a. Enter the scores in column 1 and a code corre-

sponding to their group in column 2. Name the variable in C1 Scores and the variable in C2 Group.

b. From the menu bar, select Stat, Nonparametric, and Kruskal-Wallis, and hit Enter.

c. Select the variables Scores as the Response variable and Group as the Factor.

3. The Excel commands for the one-way ANOVA on page 702 are: a. Enter the names Manufacturing, Finance, and

Trade in the first row and the data in the columns under them.

b. Select the Data tab on the top of the menu. Then, on the far right, select Data Analysis. Select ANOVA: Single Factor, then click OK.

c. In the dialog box, the Input Range is A1:C9, click on Labels in First Row, and enter E1 as the Output Range, then click OK.

Chapter 18 Answers to Self-Review

18–1 a. Two-tailed because does not state a direction. b.

.25

.20

.15

.10

.05

0 1 2 3 4 5 6 7 8 9 10 11 12

Pr ob

ab ili

ty o

f s uc

ce ss

Number of successes

Region of rejection

not exceeding .050, which is half the level of significance. The decision rule is to reject if the number of plus signs is 2 or less or 10 or more.

c. Reject ; accept . There is a preference. 18–2 a. : � � 0.50, : � � 0.50.

b. Reject if z � 1.65. c. Since 80 is more than n�2 � 100�2 � 50,

we use:

d. is rejected. e. The screening was effective.

18–3 : The median � $123, : The median is more than $123. The decision rule is to reject if z � 1.65.

Reject , because 2.38 is larger than 1.65. The median amount spent is more than $123.

z � (42 � .50) � 32

.50164 �

9.5 4

� 2.38

H0 H1H0

z � (80 � .50) � .50(100)

.501100 �

29.5 5

� 5.9

H0 H1H0

H1H0

Adding down, .000 � .003 � .016 � .019. This is the largest cumulative probability up to but

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Nonparametric Methods: Analysis of Ranked Data 715

Abso- lute

Differ- Differ- Before After ence ence Rank R � R�

17 18 �1 1 1.5 1.5 21 23 �2 2 3.0 3.0 25 22 3 3 5.0 5.0 15 25 �10 10 8.0 8.0 10 28 �18 18 10.0 10.0 16 16 — — — — — 10 22 �12 12 9.0 9.0 20 19 1 1 1.5 1.5 17 20 �3 3 5.0 5.0 24 30 �6 6 7.0 7.0 23 26 �3 3 5.0 5.0

48.5 6.5

XL-5000 D2

Distance Rank Distance Rank

252 4 262 9 263 10 242 2 279 15 256 5 273 14 260 8 271 13 258 7 265 11.5 243 3 257 6 239 1 280 16 265 11.5

Total 89.5 46.5

Reject ; accept . There is evidence of a difference in the distances traveled by the two golf balls.

18–6

H1H018–4 a. n � 10 (because there was no change for A. A.) b.

: Production is the same. : Production has increased.

The sum of the positive signed ranks is 6.5; the negative sum is 48.5. From Appendix B.7, one- tailed test, n � 10, the critical value is 10. Since 6.5 is less than 10, reject the null hypothesis and accept the alternate. New procedures did increase production.

c. No assumption regarding the shape of the distribution is necessary.

18–5 : There is no difference in the distances traveled by the XL-5000 and by the D2.

: There is a difference in the distances traveled by the XL-5000 and by the D2.

Do not reject if the computed z is between 1.96 and �1.96 (from Appendix B.1); otherwise, reject and accept . � 8, the number of observations in the first sample.

n1H1 H0

H1 H0

Ranks

Englewood West Side Great Northern Sylvania

17 5 19 7 20 1 9.5 11 16 3 21 15 13 5 22 9.5

5 2 14 8 18 12

Rank

X Y X Y d d 2

805 23 5.5 1 4.5 20.25 777 62 3.0 9 �6.0 36.00 820 60 8.5 8 0.5 0.25 682 40 1.0 4 �3.0 9.00 777 70 3.0 10 �7.0 49.00 810 28 7.0 2 5.0 25.00 805 30 5.5 3 2.5 6.25 840 42 10.0 5 5.0 25.00 777 55 3.0 7 �4.0 16.00 820 51 8.5 6 2.5 6.25

0 193.00

: The population distributions are identical. : The population distributions are not identical.

The critical value of chi-square for k � 1 � 4 � 1 � 3 degrees of freedom is 11.345. Since the computed value of 13.635 is greater than 11.345, the null hypothesis is rejected. We conclude that the turnover rates are not the same.

18–7 a.

� 13.635

� 3(22 � 1)

H � 12

22(22 � 1) c (89)2

6 �

(16)2

5 �

(85.5)2

5 �

(62.5)2

6 d

H1 H0

n1 � 6 n2 � 5 n3 � 5 n4 � 6

©R1 � 89 ©R2 � 16 ©R3 � 85.5 ©R4 � 62.5

b. : � � 0; : � 0. Reject if t � �2.306 or t � 2.306.

is not rejected. We have not shown a relationship between the two tests. H0

t � �.170 A

10 � 2 1 � ( �0.170)2

� �0.488

H0H1H0

rs � 1 � 6(193) 10(99)

� �.170

W � 89.5

� 21.5 9.52

� 2.26

z � 89.5 �

8(8 � 8 � 1) 2

(8)(8)(8 � 8 � 1) 12

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716 Chapter 18

A Review of Chapters 17 and 18 In Chapters 17 and 18, we describe statistical methods to study data that is either the nominal or the ordinal scale of measurement. These methods are nonparametric or distribution-free statistics. They do not require assumptions regarding the shape of the population. Recall, for example, in Chapter 12 when investigating the means of several populations, we assume the populations follow the normal probability distribution.

In Chapter 17, we describe the chi-square distribution. We use this distribution to compare the observed set of frequencies in a random sample with the corresponding set of expected frequen- cies in the population. The level of measurement is the nominal scale. Recall that when data are measured at the nominal level, the observations can only be classified according to some label, name, or characteristic.

In Chapter 17, we also explore the relationship between two variables in a contingency table. That is, we observe two characteristics of each sampled individual or object. For example, is there a rela- tionship between the quality of the product (acceptable or unacceptable) and the shift on which it was manufactured (day, afternoon, or night)? The chi-square distribution is used as the test statistic.

In Chapter 18, we describe five nonparametric tests of hypothesis and the coefficient of rank correlation. Each of these tests requires at least the ordinal scale of measurement. That is, we are able to rank, or order, the variables of interest.

The sign test for dependent samples is based on the sign of the difference between related observations. The binomial distribution is the test statistic. In cases where the sample is greater than 10, the normal approximation to the binomial probability distribution serves as the test statistic.

The first step when using the median test is to count the number of observations above (or below) the proposed median. Next, we use the standard normal distribution to determine if this num- ber is reasonable or too large to have occurred by chance.

The Wilcoxon signed-rank test requires dependent samples. It is an extension of the sign test in that it makes use of both the direction and the magnitude of the difference between related val- ues. It has its own sampling distribution, which is reported in Appendix B.7.

The Wilcoxon ranked-sum test assumes independent populations, but does not require the pop- ulations to follow the normal probability distribution. It is an alternative to the t test for independent samples described in Chapter 11. When there are at least eight observations in each sample, the test statistic is the standard normal distribution.

The Kruskal-Wallis test is an extension of the Wilcoxon ranked-sum test in that it handles more than two populations. It is an alternative to the one-way ANOVA method described in Chapter 12. It does not require the populations to follow the normal probability distribution or that the popula- tions have equal standard deviations.

The statistic, Spearman’s coefficient of rank correlation, is a special case of the Pearson coef- ficient of correlation, described in Chapter 13. It is based on the correlation between the ranks of related observations. It may range from �1.00 to 1.00, with 0 indicating no association between the ranks.

Glossary

Chapter 17 Chi-square distribution A distribution with these char- acteristics: (1) Its value can only be positive. (2) There is a family of chi-square distributions, a different one for each different degree of freedom. (3) The distributions are positively skewed, but as the number of degrees of free- dom increases, the distribution approaches the normal distribution. Chi-square goodness-of-fit test A test with the objec- tive of determining how well an observed set of frequen- cies fits an expected set of frequencies. It is concerned with one nominal-scale variable, such as the color of a car. Contingency table If two characteristics, such as gen- der and highest degree earned for a sample of stock bro- kers, are cross-classified into a table, the result is called a contingency table. The chi-square test statistic is used to investigate whether the two characteristics are related.

Nominal level of measurement The “lowest” level of measurement. Such data can only be classified into cate- gories, and there is no particular order for the categories. For example, it makes no difference whether the cate- gories “male” and “female” are listed in that order, or fe- male first and male second. The categories are mutually exclusive—meaning, in this illustration, that a person can- not be a male and a female at the same time. Nonparametric or distribution-free tests Hypothesis tests involving nominal- and ordinal-level data. No as- sumptions need be made about the shape of the popula- tion distribution; that is, we do not assume the population is normally distributed.

Chapter 18 Kruskal-Wallis one-way analysis of variance by ranks A test used when the assumptions for parametric analysis

Lin01803_ch18_680-719.qxd 11/11/10 11:23 AM Page 716

of variance (ANOVA) cannot be met. Its purpose is to test whether several populations are equal. The data must be at least ordinal scale. Sign test A test used for dependent samples. The sign test is used to find whether there is a brand preference for two products or to determine whether performance after an experiment is greater than before the experiment. Also, the sign test is used to test a hypothesis about the median. Spearman’s coefficient of rank correlation A measure of the association between the ranks of two variables. It can range from �1.00 to 1.00. A value of �1.00 indicates a perfect negative association among the ranks and a value of 1.00 means a perfect positive association among the ranks. A value of 0 indicates no association among the ranks.

Wilcoxon rank-sum test A nonparametric test requiring independent samples. The data must be at least ordinal level. That is, the data must be capable of being ranked. The test is used when the assumptions for the parametric Student t test cannot be met. The objective of the test is to find whether two independent samples can be considered as coming from the same population. Wilcoxon signed-rank test A nonparametric test re- quiring at least ordinal-level data and dependent sam- ples. Its purpose is to find whether there is any difference between two sets of paired (related) observations. It is used if the assumptions required for the paired t test can- not be met.

Problems

1. The owner of Beach Front Snow Cones Inc. believes the median number of snow cones sold per day between Memorial Day and Labor Day is 60. Below is a sample of 20 days. Is it reasonable to conclude that the median is actually greater than 60? Use the .05 sig- nificance level.

A Review of Chapters 17 and 18 717

2. The manufacturer of children’s raincoats wants to know if there is a preference among children for any specific color. The information below is the color preference for a sample of 50 children between the ages 6 and 10. Use the .05 significance level to investigate.

3. Is there a difference (in feet) of the length of suspension bridges in the northeast, south- east, and far west parts of the United States? Conduct an appropriate test of hypothe- sis on the following data. Do not assume the bridge lengths follow a normal probability distribution. Use the .05 significance level.

65 70 65 64 66 54 68 61 62 67 65 50 64 55 74 57 67 72 66 65

Color Frequency

Blue 17 Red 8 Green 12 Yellow 13

Northeast Southeast Far West

3,645 3,502 3,547 3,727 3,645 3,636 3,772 3,718 3,659 3,837 3,746 3,673 3,873 3,758 3,728 3,882 3,845 3,736 3,894 3,940 3,788

4,070 3,802 4,081

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718 Chapter 18

Cases

A. Century National Bank Is there a relationship between the location of the branch bank and whether the customer has a debit card? Based on the information available, develop a table that shows the relationship between these two variables. At the .05 significance level, can we conclude there is a relationship between the branch location and whether the customer uses a debit card?

B. Thomas Testing Labs John Thomas, the owner of Thomas Testing, has for some time done contract work for insurance companies regarding drunk driving. To improve his research capabil- ities, he recently purchased the Rupple Driving Simula- tor. This device will allow a subject to take a “road test” and provide a score indicating the number of driving errors committed during the test drive. Higher scores indicate more driving errors. Driving errors would include: not coming to a complete stop at a stop sign, not using turning signals, not exercising caution on wet or snowy pavement, and so on. During the road test, problems appear at random and not all problems appear in each road test. These are major advantages to the Rupple Dri- ving Simulator because subjects do not gain any advan- tage by taking the test several times.

With the new driving simulator, Mr. Thomas would like to study in detail the problem of drunk driving. He begins by selecting a random sample of 25 drivers. He asks each of the selected individuals to take the test drive on the Rupple Driving Simulator. The number of errors for each driver is recorded. Next, he has each of the individuals in the group drink three 16-ounce cans of beer in a 60-minute period and return to the Rupple Driving Simulator for another test drive. The number of driving errors after drinking the beer is also shown. The research

question is: Does alcohol impair the driver’s ability and, therefore, increase the number of driving errors?

Mr. Thomas believes the distribution of scores on the test drive does not follow a normal distribution and, therefore, a nonparametric test should be used. Because the observations are paired, he decides to use both the sign test and the Wilcoxon signed-rank test.

Driving Errors Driving Errors

Without With Without With Subject Alcohol Alcohol Subject Alcohol Alcohol

1 75 89 14 72 106 2 78 83 15 83 89 3 89 80 16 99 89 4 100 90 17 75 77 5 85 84 18 58 78 6 70 68 19 93 108 7 64 84 20 69 69 8 79 104 21 86 84 9 83 81 22 97 86

10 82 88 23 65 92 11 83 93 24 96 97 12 84 92 25 85 94 13 80 103

a. Compare the results using these two procedures. Conduct an appropriate test of hypothesis to deter- mine if alcohol is related to driving errors.

b. Write a report that summarizes your findings.

Practice Test

Part 1––Objective 1. The level of measurement is required for the chi-square goodness-of-fit test. 1. 2. Which of the following is not a characteristic of the chi-square distribution? (positively skewed, based on degrees

of freedom, cannot be negative, at least 30 observations) 2. 3. In a contingency table, how many traits are considered for each variable? 3. 4. In a contingency table, there are four rows and three columns; hence, there are degrees of freedom.

4. 5. In a goodness-of-fit test, the critical value of chi-square is based on . (sample size, number of cate-

gories, number of variables, or none of these.) 5. 6. In a sign test, are the samples dependent or independent? 6. 7. In a sign test of eight paired observations, the test statistic is the distribution. (binomial, z, t, or chi-

square.) 7. 8. What is the major difference between the Kruskal-Wallis test and the Wilcoxon rank-sum test? (one is based

on dependent samples and the other independent samples, or one is for comparing two independent samples and the other two or more independent samples) 8.

9. Under what conditions can the coefficient of rank correlation be less than �1.00? 9.

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A Review of Chapters 17 and 18 719

10. The Kruskal-Wallis test is used in the place of ANOVA when which two of the following criteria are not met? (normal population, equal standard deviations, more than 12 items in the sample, the populations are independent)

10.

Part 2––Problems Use the standard five-step hypothesis testing procedure. 1. A recent census report indicated that 65 percent of families have two parents present, 20 percent have only

a mother present, 10 percent have only a father present, and 5 percent have no parent present. A random sample of 200 children from a large rural school district revealed the following:

Two Parents Mother Only Father Only No Parent Total

120 40 30 10 200

8 A.M. 10 A.M. 1:30 P.M.

68 59 67 84 59 69 75 63 75 78 62 76 70 78 79 77 76 83 88 80 86 71 86

Mystery Romance Self-Help Total

Men 250 100 190 540 Women 130 170 200 500

Is there sufficient evidence to conclude that the proportion of families by type of parent present in this particular school district differs from those reported in the recent census?

2. A book publisher wants to investigate the type of books selected for recreational reading by men and women. A random sample provided the following information. At the .05 significance level, should we conclude that gender is related or unrelated to type of book selected?

3. An instructor has three sections of basic statistics. Listed below are the grades on the first exam for each sec- tion. Assume that the distributions do not follow the normal probability distribution. At the .05 significance level, is there a difference in the distributions of scores?

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19 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Explain the purpose of quality control in production and service operations.

LO2 Discuss the two causes of process variation.

LO3 Use a Pareto chart to identify sources of variation.

LO4 Construct and interpret a fishbone diagram.

LO5 Compare an attribute versus a variable measure of quality.

LO6 Compute upper and lower control limits for mean and range charts.

LO7 Compare in-control and out-of-control quality control charts.

LO8 Construct and interpret percent defective and c -bar charts.

LO9 Explain the process of acceptance sampling.

LO10 Describe an operating characteristic curve for a sam- pling plan.

Statistical Process Control and Quality Management

A bicycle manufacturer randomly selects 10 frames each day and tests

for defects. The number of defective frames found over the last

14 days is 3, 2, 1, 3, 2, 2, 8, 2, 0, 3, 5, 2, 0, and 4. Construct a control

chart for this process and comment on whether the process is “in

control.” (See Exercise 11 and LO8.)

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19.1 Introduction Throughout this text, we present many applications of hypothesis testing. In Chap- ter 10, we describe methods for testing a hypothesis regarding a single population value. In Chapter 11, we describe methods for testing a hypothesis about two pop- ulations. In this chapter, we present another, somewhat different application of hypothesis testing, called statistical process control or SPC.

Statistical process control is a collection of strategies, techniques, and actions taken by an organization to ensure it is producing a quality product or providing a quality service. SPC begins at the product planning stage, when we specify the attributes of the product or service. It continues through the production stage. Each attribute throughout the process contributes to the overall quality of the product. To effectively use quality control, measurable attributes and specifications are devel- oped against which the actual attributes of the product or service are compared.

19.2 A Brief History of Quality Control Prior to the 1900s, U.S. industry was largely characterized by small shops making relatively simple products, such as candles or furniture. In these small shops, the individual worker was generally a craftsman who was completely responsible for the quality of the work. The worker could ensure the quality through the personal selec- tion of the materials, skillful manufacturing, and selective fitting and adjustment.

In the early 1900s, factories sprang up, where people with limited training were formed into large assembly lines. Products became much more complex. The individ- ual worker no longer had complete control over the quality of the product. A semi- professional staff, usually called the Inspection Department, became responsible for the quality of the product. The quality responsibility was usually fulfilled by a 100 per- cent inspection of all the important characteristics. If there were any discrepancies noted, these problems were handled by the manufacturing department supervisor. In essence, quality was attained by “inspecting the quality into the product.”

During the 1920s, Dr. Walter A. Shewhart, of Bell Telephone Laboratories, devel- oped the concepts of statistical quality control. He introduced the concept of “con- trolling” the quality of a product as it was being manufactured, rather than inspecting the quality into the product after it was manufactured. For the purpose of control- ling quality, Shewhart developed charting techniques for controlling in-process man- ufacturing operations. In addition, he introduced the concept of statistical sample inspection to estimate the quality of a product as it was being manufactured. This replaced the old method of inspecting each part after it was completed in the pro- duction operation.

Statistical quality control really came into its own during World War II. The need for mass-produced war-related items, such as bomb sights, accurate radar, and other electronic equipment, at the lowest possible cost hastened the use of statis- tical sampling and quality control charts. Since World War II, these statistical tech- niques have been refined and sharpened. The use of computers has also widened the use of these techniques.

World War II virtually destroyed the Japanese production capability. Rather than retool their old production methods, the Japanese enlisted the aid of the late Dr. W. Edwards Deming, of the United States Department of Agriculture, to help them develop an overall plan. In a series of seminars with Japanese planners, he stressed a philosophy that is known today as Deming’s 14 points. These 14 points are listed on the following page. He emphasized that quality originates from improving the process, not from inspection, and that quality is determined by the customers. The manufacturer must be able, via market research, to anticipate the needs of cus- tomers. Senior management has the responsibility for long-term improvement. Another of his points, and one that the Japanese strongly endorsed, is that every

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member of the company must contribute to long-term improvement. To achieve this improvement, ongoing education and training are necessary.

Deming had some ideas that did not mesh with contemporary management philosophies in the United States. Two areas where Deming’s ideas differed from U.S. management philosophy were with production quotas and merit ratings. He claimed these two practices, which are both common in the United States, are not productive and should be eliminated. He also pointed out that U.S. managers are mostly interested in good news. Good news, however, does not provide an oppor- tunity for improvement. On the other hand, bad news opens the door for new prod- ucts and allows for company improvement.

Listed below, in a condensed form, are Dr. Deming’s 14 points. He was adamant that the 14 points needed to be adopted as a package in order to be successful. The underlying theme is cooperation, teamwork, and the belief that workers want to do their jobs in a quality fashion.

DEMING’S 14 POINTS 1. Create constancy of purpose for the continual improvement of products

and service to society. 2. Adopt a philosophy that we can no longer live with commonly accepted

levels of delays, mistakes, defective materials, and defective workmanship. 3. Eliminate the need for mass inspection as the way to achieve quality.

Instead achieve quality by building the product correctly in the first place. 4. End the practice of awarding business solely on the basis of price. Instead,

require meaningful measures of quality along with the price. 5. Improve constantly and forever every process for planning, production, and

service. 6. Institute modern methods of training on the job for all employees, including

managers. This will lead to better utilization of each employee. 7. Adopt and institute leadership aimed at helping people do a better job. 8. Encourage effective two-way communication and other means to drive out

fear throughout the organization so that everyone may work more effectively and more productively for the company.

9. Break down barriers between departments and staff areas. 10. Eliminate the use of slogans, posters, and exhortations demanding zero

defects and new levels of productivity without providing methods. 11. Eliminate work standards that prescribe quotas for the workforce and numeri-

cal goals for people in management. Substitute aids and helpful leadership in order to achieve continual improvement in quality and productivity.

12. Remove the barriers that rob hourly workers and the people in management of their right to pride of workmanship.

13. Institute a vigorous program of education and encourage self-improvement for everyone. What an organization needs is good people and people who are improving with education. Advancement to a competitive position will have its roots in knowledge.

14. Define clearly management’s permanent commitment to ever-improving quality and productivity to implement all of these principles.

Deming’s 14 points did not ignore statistical quality control, which is often abbreviated as SQC. The objective of statistical quality control is to monitor pro- duction through many stages of manufacturing. We use the tools of statistical qual- ity control, such as X-bar and R charts, to monitor the quality of many processes and services. Control charts allow us to identify when a process or service is “out

LO1 Explain the purpose of quality control in production and service operations.

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of control,” that is, when the point in time is reached where an excessive number of defective units are being produced.

Interest in quality has accelerated dramatically in the United States since the late 1980s. Turn on the television and watch the commercials sponsored by Ford, Nissan, and GM to verify the emphasis on quality control on the assembly line. It is now one of the “in” topics in all facets of business. V. Daniel Hunt, a noted American Quality Control consultant, reports that in the United States 20 to 25 percent of the cost of produc- tion is currently spent finding and correcting mistakes. And, he added, the additional cost incurred in repairing or replacing faulty products in the field drives the total cost of poor quality to nearly 30 percent. In Japan, he indicated, this cost is about 3 percent!

In recent years, companies have been motivated to improve quality by the challenge of being recognized for their quality achievements. The Malcolm Baldrige National Quality Award,

established in 1988, is awarded annually to U.S. companies that demonstrate excel- lence in quality achievement and management. The award categories include man- ufacturing, service, small business, health care, and education. Past winners include Xerox, IBM, the University of Wisconsin–Stout, Ritz-Carlton Hotel Corporation, Federal Express, and Cadillac. The 2009 winners were:

• Honeywell Federal Manufacturing & Technologies, LLC (FM&T) was the winner in the manufacturing category. They are one of the nation’s most diverse low- volume, high-reliability production facilities, serving government agencies, national laboratories, universities, and U.S. industry. The company’s overall customer satisfaction rate was measured at or above 95 percent for the past four years, which compares favorably to similar companies, whose levels range from 78 per- cent to 85 percent for the same period.

• AtlantiCare was a 2009 winner in the health care category. The organization is a nonprofit health system in southeastern New Jersey, delivering acute and chronic care, preventive and at-risk services, and health information. Among other achievements, survey responses from 2007 to 2009 show that customer satisfaction results were above the 90th percentile national benchmark, includ- ing those for the surgery center, the spine institute, urgent care, and clinical labs.

• MidwayUSA was the award recipient in the small business category. The firm is a family-owned, catalog/Internet-based retail merchant that offers shooting, reloading, gunsmithing, and hunting products. Retail customers represent 90 per- cent of the firm's total business, with dealers and international customers making up the remaining 10 percent. More than 95,000 different products from more than 700 different vendors are distributed. By focusing their processes on serving the customer, the company improved their overall cus- tomer satisfaction rating from 91 percent in 2007 and 2008 to 93 percent in 2009.

• Based in St. Joseph, Missouri, Heartland Health was a 2009 recipient in the health care category. Heartland Health is a not-for-profit, community-based, integrated health system serving the residents of northwest Missouri, northeast Kansas, southeast Nebraska, and southwest Iowa. With more than 3,200 care- givers (employees, volunteers, and health care practitioners), Heartland Health is the region’s largest health system. Heartland Health uses Six Sigma methods for continuous improvement. Improvements in reducing errors, inspections, tests, and audits resulted in cost savings of $8 million in 2005 to more than $25 million in 2009.

Statistics in Action

Does excellence in quality management lead to higher finan- cial performance? Recent research compared the finan- cial performance of companies that re- ceived the Baldrige National Quality Award to similar companies who did not receive the award. The research showed that the com- panies receiving the award had an average of 39 percent higher operating income and 26 percent higher sales, and were 1.22 percent lower in their cost per sales dollar.

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• The Veterans Affairs Cooperative Studies Program (VACSP) Clinical Research Pharmacy Coordinating Center (the Center) was the winner in the nonprofit category. The Center manufactures, packages, stores, labels, distributes, and tracks clinical trial materials (drugs and devices) and monitors patient safety. A significant accomplishment of the company is customer retention. Seventy- five percent of the Center’s customer relationships exceed 10 years.

You can obtain more information on these and other winners by visiting the website http://www.quality.nist.gov.

Six Sigma Many service, manufacturing, and nonprofit organizations are committed to improv- ing the quality of their services and products. “Six Sigma” is a name given to an organization-wide program designed to improve quality and performance through- out an organization. The focus of the program is to reduce the variation in any process used to produce and deliver services and products to customers. Six Sigma programs apply to production processes as well as accounting and other organiza- tional support processes. The ultimate outcomes of a Six Sigma program are to reduce the costs of defects and errors, increase customer satisfaction and sales of products and services, and to increase profits.

Six Sigma gets its name from the normal distribution. The term sigma means standard deviation, and “plus or minus” three standard deviations gives a total range of six standard deviations. So Six Sigma means that a process should not gener- ate more than 3.4 defects per million for any product or service. Many companies strive for even fewer defects.

To attain this goal, a Six Sigma program trains every organization member in processes to identify sources of process variation that significantly affect quality. The process includes identifying and defining problems, collecting and analyzing data to investigate and become knowledgeable about the problem, making process improvements to reduce process variation, and implementing procedures for improv- ing the process.

Six Sigma uses many statistical techniques to collect and analyze the data needed to reduce process variation. The following are included in this text: his- tograms, analysis of variation, chi-square test of independence, regression, and correlation.

General Electric, Motorola, and AlliedSignal (now a part of Honeywell) are large companies that have used Six Sigma methods and achieved significant quality improvement and cost savings. Even cities like Fort Wayne, Indiana, have used six sigma techniques to improve their operations. The city is reported to have saved $10 million since 2000 and improved customer service at the same time. For exam- ple, the city reduced missed trash pickups by 50 percent and cut the response time to repair potholes from 21 to 3 hours. You can learn more about Six Sigma ideas, methods, and training at www.6sigma.us.

19.3 Causes of Variation No two products are exactly the same. There is always some variation. The weight of each McDonald’s Quarter Pounder is not exactly 0.25 pounds. Some will weigh more than 0.25 pounds, others less. The standard time for the TARTA (Toledo Area Regional Transit Authority) bus run from downtown Toledo, Ohio, to Perrysburg is 25 minutes. However, each run does not take exactly 25 minutes. Some runs take longer. Other times the TARTA driver must wait in Perrysburg before returning to Toledo. In some cases, there is a reason for the bus being late, an accident on the expressway or a snowstorm, for example. In other cases, the driver may not “hit”

724 Chapter 19

LO2 Discuss the two causes of process variation.

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ASSIGNABLE VARIATION Variation that is not random. It can be eliminated or reduced by investigating the problem and finding the cause.

There are several reasons why we should be concerned with variation. Two are stated below.

1. It will change the shape, dispersion, and central location of the distribution of the product characteristic being measured.

2. Assignable variation is usually correctable, whereas chance variation usually can- not be corrected or stabilized economically.

19.4 Diagnostic Charts There are a variety of diagnostic techniques available to investigate quality problems. Two of the more prominent of these techniques are Pareto charts and fishbone diagrams.

Pareto Charts Pareto analysis is a technique for tallying the number and type of defects that happen within a product or service. The chart is named after a nineteenth-century Italian scientist, Vilfredo Pareto. He noted that most of the “activity” in a process is caused by relatively few of the “factors.” His concept, often called the 80–20 rule, is that 80 percent of the activity is caused by 20 percent of the factors. By concen- trating on 20 percent of the factors, managers can attack 80 percent of the prob- lem. For example, Emily’s Family Restaurant, located at the junction of Interstates 75 and 70, is investigating “customer complaints.” The five complaints heard most frequently are: discourteous service, cold food, long wait for seating, few menu choices, and unruly young children. Suppose discourteous service was mentioned most frequently and cold food second. These two factors total more than 85 per- cent of the complaints and hence are the two that should be addressed first because this will yield the largest reduction in complaints.

To develop a Pareto chart, we begin by tallying the type of defects. Next, we rank the defects in terms of frequency of occurrence from largest to smallest. Finally, we produce a vertical bar chart, with the height of the bars corresponding to the frequency of each defect. The following example illustrates these ideas.

the green lights or the traffic is unusually heavy and slow for no apparent reason. There are two general sources of variation in a process—chance and assignable.

CHANCE VARIATION Variation that is random in nature. This type of variation cannot be completely eliminated unless there is a major change in the techniques, technologies, methods, equipment, or materials used in the process.

Internal machine friction, slight variations in material or process conditions (such as the temperature of the mold being used to make glass bottles), atmospheric con- ditions (such as temperature, humidity, and the dust content of the air), and vibra- tions transmitted to a machine from a passing forklift are a few examples of sources of chance variation.

If the hole drilled in a piece of steel is too large due to a dull drill, the drill may be sharpened or a new drill inserted. An operator who continually sets up the machine incorrectly can be replaced or retrained. If the roll of steel to be used in the process does not have the correct tensile strength, it can be rejected. These are examples of assignable variation.

LO3 Use a Pareto chart to identify sources of variation.

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Example The city manager of Grove City, Utah, is concerned with water usage, particularly in single family homes. She would like to develop a plan to reduce the water usage in Grove City. To investigate, she selects a sample of 100 homes and determines the typical daily water usage for various purposes. These sample results are as follows.

Solution A Pareto chart is useful for identifying the major areas of water usage and focus- ing on those areas where the greatest reduction can be achieved. The first step is to convert each of the activities to a percent and then to order them from largest to smallest. The total water usage per day is 339.3 gallons, found by totaling the gallons used in the eight activities. The activity with the largest use is watering lawns. It accounts for 143.7 gallons of water per day, or 42.4 percent of the amount of water used. The next largest category is personal bathing, which accounts for 31.4 percent of the water used. These two activities account for 73.8 percent of the water usage.

What is the area of greatest usage? Where should she concentrate her efforts to reduce the water usage?

Reasons for Water Usage Gallons per Day

Laundering 24.9 Watering lawn 143.7 Personal bathing 106.7 Cooking 5.1 Swimming pool 28.3 Dishwashing 12.3 Car washing 10.4 Drinking 7.9

Reasons for Water Usage Gallons per Day Percent

Laundering 24.9 7.3 Watering lawn 143.7 42.4 Personal bathing 106.7 31.4 Cooking 5.1 1.5 Swimming pool usage 28.3 8.3 Dishwashing 12.3 3.6 Car washing 10.4 3.1 Drinking 7.9 2.3

Total 339.3 100.0

To draw the Pareto chart, we begin by scaling the number of gallons used on the left vertical axis and the corresponding percent on the right vertical axis. Next we draw a vertical bar with the height of the bar corresponding to the activity with the largest number of occurrences. In the Grove City example, we draw a vertical bar for the activity watering lawns to a height of 143.7 gallons. (We call this the count.) We continue this procedure for the other activities, as shown in the Minitab output in Chart 19–1.

Below the chart, we list the activities, their frequency of occurrence, and the per- cent of the time each activity occurs. In the last row, we list the cumulative percent- age. This cumulative row will allow us to quickly determine which set of activities account for most of the activity. These cumulative percents are plotted above the

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vertical bars. In the Grove City example, the activities of watering lawn, personal bathing, and pools account for 82.1 percent of the water usage. The city manager can attain the greatest gain by looking to reduce the water usage in these three areas.

CHART 19–1 Pareto Chart for Water Usage in Grove City, Utah

Fishbone Diagrams Another diagnostic chart is a cause-and-effect diagram or a fishbone diagram. It is called a cause-and-effect diagram to emphasize the relationship between an effect and a set of possible causes that produce the particular effect. This diagram is useful to help organize ideas and to identify relationships. It is a tool that encourages open brain- storming for ideas. By identifying these relationships, we can determine factors that are the cause of variability in our process. The name fishbone comes from the manner in which the various causes and effects are organized on the diagram. The effect is usu- ally a particular problem, or perhaps a goal, and it is shown on the right-hand side of the diagram. The major causes are listed on the left-hand side of the diagram.

The usual approach to a fishbone diagram is to consider four problem areas, namely, methods, materials, equipment, and personnel. The problem, or the effect, is the head of the fish. See Chart 19–2.

Materials

Personnel

Methods

Problem or

Effect

Equipment

CHART 19–2 Fishbone Diagram

LO4 Construct and interpret a fishbone diagram.

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Materials

Personnel

Methods

Complaints of

Cold Food

Food at correct starting temperature

Packaging insulates enough

Food heated to correct temperature

Food placed under heating lights

Thermostat working properly

Heating lights at correct height

Employees operating equipment correctly

Servers deliver food quickly

Equipment

Under each of the possible causes are subcauses that are identified and inves- tigated. The subcauses are factors that may be producing the particular effect. Infor- mation is gathered about the problem and used to fill in the fishbone diagram. Each of the subcauses is investigated and those that are not important eliminated, until the real cause of the problem is identified.

Chart 19–3 illustrates the details of a fishbone diagram. Suppose a family restau- rant, such as those found along an interstate highway, has recently been experiencing complaints from customers that the food being served is cold. Notice each of the subcauses are listed as assumptions. Each of these subcauses must be investi- gated to find the real problem regarding the cold food. In a fishbone diagram, there is no weighting of the subcauses.

CHART 19–3 Fishbone Diagram for a Restaurant Investigation of Cold Food Complaints

Self-Review 19–1 The Rouse Home, located on the south side of Chicago, is a mental health facility. Recently there have been complaints regarding conditions at the home. The administrator would like to use a Pareto chart to investigate. When a patient or patient’s relative has a complaint, they are asked to complete a complaint form. Listed below is a summary of the complaint forms received during the last 12 months.

Develop a Pareto chart. What complaints would you suggest the administrator work on first to achieve the most significant improvement?

Complaint Number Complaint Number

Nothing to do 45 Dirty conditions 63 Poor care by staff 71 Poor quality of food 84 Medication error 2 Lack of respect by staff 35

Exercises 1. Tom Sharkey is the owner of Sharkey Chevy, Buick, GMC, Isuzu. At the start of the

year, Tom instituted a customer opinion program to find ways to improve service. The day after the service is performed, Tom’s administrative assistant calls the customer to find out whether the service was performed satisfactorily and how the service might be improved. Listed below is a summary of the complaints for the first six months.

728 Chapter 19

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Complaint Frequency Complaint Frequency

Problem not corrected 38 Price too high 23 Error on invoice 8 Wait too long for service 10 Unfriendly atmosphere 12

2. Out of 110 diesel engines tested, a rework and repair facility found 9 had leaky water pumps, 15 had faulty cylinders, 4 had ignition problems, 52 had oil leaks, and 30 had cracked blocks. Draw a Pareto chart to identify the key problem in the engines.

19.5 Purpose and Types of Quality Control Charts Control charts identify when assignable causes of variation or changes have entered the process. For example, Wheeling Company makes vinyl-coated aluminum replacement windows for older homes. The vinyl coating must have a thickness between certain limits. If the coating becomes too thick, it will cause the windows to jam. On the other hand, if the coating becomes too thin, the window will not seal properly. The mechanism that determines how much coating is put on each window becomes worn and begins making the coating too thick. Thus, a change has occurred in the process. Control charts are useful for detecting the change in process con- ditions. It is important to know when changes have entered the process, so that the cause may be identified and corrected before a large number of unacceptable items are produced.

Control charts are similar to the scoreboard in a baseball game. By looking at the scoreboard, the fans, coaches, and players can tell which team is winning the game. However, the scoreboard can do nothing to win or lose the game. Control charts provide a similar function. These charts indicate to the workers, group leaders, quality control engineers, production supervisor, and management whether the production of the part or service is “in control” or “out of control.” If the production is “out of control,” the control chart will not fix the situation; it is just a piece of paper with numbers and dots on it. Instead, the person responsible must adjust the machine manufac- turing the part or do what is necessary to return production to “in control.”

There are two types of control charts. A variable control chart portrays measurements, such as the amount of cola in a two-liter bot- tle or the outside diameter of a piece of pipe. A variable control chart requires the interval or the ratio scale of measurement. An attribute control chart classifies a product or service as either acceptable or unacceptable. It is based on the nominal scale of measurement. The Marines stationed at Camp Lejeune are asked to rate the meals served as acceptable or unacceptable; bank loans are either repaid or they are defaulted.

Control Charts for Variables To develop control charts for variables, we rely on the sampling theory discussed in connection with the central limit theorem in Chapter 8. Suppose a sample of five

Develop a Pareto chart. What complaints would you suggest that Tom work on to improve the quality of service?

LO5 Compare an attribute versus a variable measure of quality.

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pieces is selected each hour from the production process and the mean of each sample computed. The sample means are and so on. The mean of these sample means is denoted as . We use k to indicate the number of sample means. The overall or grand mean is found by:

GRAND MEAN [19–1]

The standard error of the distribution of the sample means is designated by . It is found by:

STANDARD ERROR OF THE MEAN [19–2]

These relationships allow limits to be established around the sample means to show how much variation can be expected for a given sample size. These expected limits are called the upper control limit (UCL) and the lower control limit (LCL). An example will illustrate the use of control limits and how the limits are determined.

sX � s 1n

X � © of the sample means

Number of sample means �

©X k

X X1, X2, X3,

Example Statistical Software Inc. offers a toll-free number where customers can call with problems involving the use of their products from 7 A.M. until 11 P.M. daily. It is impossible to have every call answered immediately by a technical representative, but it is important customers do not wait too long for a person to come on the line. Customers become upset when they hear the message “Your call is important to us. The next available representative will be with you shortly” too many times. To understand its process, Statistical Software decides to develop a control chart describing the total time from when a call is received until the representative answers the call and resolves the issue raised by the caller. Yesterday, for the 16 hours of operation, five calls were sampled each hour. This information is reported below, in minutes, until the issue was resolved.

Sample Number

Time 1 2 3 4 5

A.M. 7 8 9 15 4 11 8 7 10 7 6 8 9 11 12 10 9 10

10 12 8 6 9 12 11 11 10 6 14 11

P.M. 12 7 7 10 4 11 1 10 7 4 10 10 2 8 11 11 7 7 3 8 11 8 14 12 4 12 9 12 17 11 5 7 7 9 17 13 6 9 9 4 4 11 7 10 12 12 12 12 8 8 11 9 6 8 9 10 13 9 4 9

10 9 11 8 5 11

LO6 Compute upper and lower control limits for mean and range charts.

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Solution

Based on this information, develop a control chart for the mean duration of the call. Does there appear to be a trend in the calling times? Is there any period in which it appears that customers wait longer than others?

A mean chart has two limits, an upper control limit (UCL) and a lower control limit (LCL). These upper and lower control limits are computed by:

CONTROL LIMITS FOR [19–3] THE MEAN

where s is an estimate of the standard deviation of the population, �. Notice that in the calculation of the upper and lower control limits the number 3 appears. It represents the 99.74 percent confidence limits. The limits are often called the 3-sigma limits. However, other levels of confidence (such as 90 or 95 percent) can be used.

This application was developed before computers were widely available, and computing standard deviations was difficult. Rather than calculate the standard devi- ation from each sample as a measure of variation, it is easier to use the range. For fixed sized samples, there is a constant relationship between the range and the stan- dard deviation, so we can use the following formulas to determine the 99.74 per- cent control limits for the mean. It can be demonstrated that the term ) from formula (19–3) is equivalent to in the following formula.

CONTROL LIMITS FOR THE MEAN

[19–4]

where: is a constant used in computing the upper and the lower control limits. It is based on the average range, . The factors for various sample sizes are available in Appendix B.8. (Note: n in this table refers to the number in the sample.) A portion of Appendix B.8 is shown below. To locate the

factor for this problem, find the sample size for n in the left margin. It is 5. Then move horizontally to the column, and read the factor. It is 0.577.

A2 A2

R A2

UCL � X � A2R LCL � X � A2R

A2R 3(s�1n

UCL � X � 3 s 1n

and LCL � X � 3 s 1n

n d2 2 1.880 1.128 0 3.267 3 1.023 1.693 0 2.575 4 0.729 2.059 0 2.282 5 0.577 2.326 0 2.115 6 0.483 2.534 0 2.004

D4D3A˛2

is the mean of the sample means, computed by where k is the num- ber of samples selected. In this problem, a sample of five observations is taken each hour for 16 hours, so k � 16.

©X�k,X

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The centerline for the chart is . It is 9.413 minutes, found by 150.6�16. The mean of the ranges ( ) is 6.375 minutes, found by 102�16. Thus, the upper control limit of the X bar chart is:

The lower control limit of the X bar chart is:

UCL, and LCL, and the sample means are portrayed in Chart 19–4. The mean, is 9.413 minutes, the upper control limit is located at 13.091 minutes, and the lower control limit is located at 5.735. There is some variation in the duration of the calls, but all sample means are within the control limits. Thus, based on 16 samples of five calls, we conclude that 99.74 percent of the time the mean length of a sample of five calls will be between 5.735 minutes and 13.091 minutes.

X,X,

LCL � X � A2R � 9.413 � 0.577(6.375) � 5.735

UCL � X � A2R � 9.413 � 0.577(6.375) � 13.091

R X

TABLE 19–1 Duration of 16 Samples of Five Help Sessions

Time 1 2 3 4 5 Mean Range

A.M. 7 8 9 15 4 11 9.4 11 8 7 10 7 6 8 7.6 4 9 11 12 10 9 10 10.4 3

10 12 8 6 9 12 9.4 6 11 11 10 6 14 11 10.4 8

P.M. 12 7 7 10 4 11 7.8 7 1 10 7 4 10 10 8.2 6 2 8 11 11 7 7 8.8 4 3 8 11 8 14 12 10.6 6 4 12 9 12 17 11 12.2 8 5 7 7 9 17 13 10.6 10 6 9 9 4 4 11 7.4 7 7 10 12 12 12 12 11.6 2 8 8 11 9 6 8 8.4 5 9 10 13 9 4 9 9.0 9

10 9 11 8 5 11 8.8 6

Total 150.6 102

CHART 19–4 Control Chart for Mean Length of Customer Calls to Statistical Software Inc.

7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 A.M. P.M.

Time

Ca ll

le ng

th (i

n m

in ut

es )

UCL

LCL

13 12 11 10 9 8 7 6

is the mean of the ranges of the sample. It is R�k. Remember the range is the difference between the largest and the smallest value in each sample. It describes the variability occurring in that particular sample. (See Table 19–1.)

©R

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Statistical Process Control and Quality Management 733

Example The length of time customers of Statistical Software Inc. waited from the time their call was answered until a technical representative answered their question or solved their problem is recorded in Table 19–1. Develop a control chart for the range. Does it appear that there is any time when there is too much variation in the operation?

Solution The first step is to find the mean of the sample ranges. The range for the five calls sampled in the 7 A.M. hour is 11 minutes. The longest call selected from that hour was 15 minutes and the shortest 4 minutes; the difference in the lengths is 11 minutes. In the 8 A.M. hour, the range is 4 minutes. The total of the 16 ranges is 102 minutes, so the average range is 6.375 minutes, found by . Referring to Appendix B.8 or the partial table on page 731, D3 and D4 are 0 and 2.115, respectively. The lower and upper control limits are 0 and 13.483.

The range chart with the 16 sample ranges plotted is shown in Chart 19–5. This chart shows all the ranges are well within the control limits. Hence, we conclude the variation in the time to service the customers’ calls is within normal limits, that is, “in control.” Of course, we should be determining the control limits based on one set of data and then applying them to evaluate future data, not the data we already know.

LCL � D3R � 0(6.375) � 0

UCL � D4R � 2.115(6.375) � 13.483

R � 102�16

Because the statistical theory is based on the normality of large samples, con- trol charts should be based on a stable process, that is, a fairly large sample, taken over a long period of time. One rule of thumb is to design the chart after at least 25 samples have been selected.

Range Charts In addition to the central location in a sample, we must also monitor the amount of variation from sample to sample. A range chart shows the variation in the sample ranges. If the points representing the ranges fall between the upper and the lower limits, it is concluded that the operation is in control. According to chance, about 997 times out of 1,000 the range of the samples will fall within the limits. If the range should fall above the limits, we conclude that an assignable cause affected the oper- ation and an adjustment to the process is needed. Why are we not as concerned about the lower control limit of the range? For small samples, the lower limit is often zero. Actually, for any sample of six or less, the lower control limit is 0. If the range is zero, then logically all the parts are the same and there is not a problem with the variability of the operation.

The upper and lower control limits of the range chart are determined from the following equations.

CONTROL CHART FOR RANGES [19–5]

The values for and , which reflect the usual three (sigma) limits for various sample sizes, are found in Appendix B.8 or in the table on page 731.

�D4D3

UCL � D4R LCL � D3R

Statistics in Action

Control charts were used to help convict a person who bribed jai alai players to lose. and R charts showed unusual bet- ting patterns and that some contestants did not win as much as expected when they made certain bets. A quality control expert was able to identify times when assign- able variation stopped, and prose- cutors were able to tie those times to the arrest of the suspect.

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734 Chapter 19

19.6 In-Control and Out-of-Control Situations Three illustrations of in-control and out-of-control processes follow.

1. The mean chart and the range chart together indicate that the process is in control. Note the sample means and sample ranges are clustered close to the centerlines. Some are above and some below the centerlines, indicating the process is quite stable. That is, there is no visible tendency for the means and ranges to move toward the out-of-control areas.

Everything OK

UCL

LCL

Mean Chart

7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 A.M. P.M.

Time

Ra ng

e of

c al

ls (in

m in

ut es

)

UCL14 12 10 8 6 4 2

CHART 19–5 Control Chart for Ranges of Length of Customer Calls to Statistical Software Inc.

Minitab will draw a control chart for the mean and the range. Following is the output for the Statistical Software example. The data are in Table 19–1. The minor differences in the control limits are due to rounding.

LO7 Compare in-control and out-of- control quality control charts.

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Statistical Process Control and Quality Management 735

UCL

LCL

Mean Chart

UCL

LCL

Range Chart

Mean out of control

Considerable variation in ranges

UCL

LCL

Range Chart

UCL

LCL

Mean Chart

UCL

LCL

Range Chart

2. The sample means are in control, but the ranges of the last two samples are out of control. This indicates there is considerable variation in the samples. Some sample ranges are large; others are small. An adjustment in the process is probably necessary.

3. The mean is in control for the first samples, but there is an upward trend toward the UCL. The last two sample means are out of control. An adjustment in the process is indicated.

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736 Chapter 19

The preceding chart for the mean is an example of a control chart that offers some additional information. Note the direction of the last five observations of the mean. They are all above and increasing, and, in fact, the last two observations are out of control. The fact that the sample means were increasing for six consecutive obser- vations is very improbable and another indication that the process is out of control.

Self-Review 19–2 The manager of River City McDonald’s randomly selects four customers each hour. For these selected customers, she determines the time, in minutes, between order entry and order delivery. The results are shown below.

(a) Compute the mean wait, the mean range, and determine the control limits for the mean and the range and chart them.

(b) Are the measurements within the control limits? Interpret the chart.

Sample Times

Time 1 2 3 4

9 A.M. 1 4 5 2 10 A.M. 2 3 2 1 11 A.M. 1 7 3 5

Exercises 3. Describe the difference between assignable variation and chance variation. 4. Describe the difference between an attribute control chart and a variable control chart. 5. Samples of size n � 4 are selected from a production line.

a. What is the value of the factor used to determine the upper and lower control lim- its for the mean?

b. What are the values of the and factors used to determine the lower and upper control limits for the range?

6. Samples of size 5 are selected from a manufacturing process. The mean of the sample ranges is .50. What is the estimate of the standard deviation of the population?

7. A new industrial oven has just been installed at Piatt Bakery. To develop experience re- garding the oven temperature, an inspector reads the temperature at four different places inside the oven each half hour. The first reading, taken at 8:00 A.M., was 340 degrees Fahrenheit. (Only the last two digits are given in the following table to make the computa- tions easier.)

D4D3

Reading

Time 1 2 3 4

8:00 A.M. 40 50 55 39 8:30 A.M. 44 42 38 38 9:00 A.M. 41 45 47 43 9:30 A.M. 39 39 41 41

10:00 A.M. 37 42 46 41 10:30 A.M. 39 40 39 40

a. On the basis of this initial experience, determine the control limits for the mean tem- perature. Determine the grand mean. Plot the experience on a chart.

b. Interpret the chart. Does there seem to be a time when the temperature is out of control? 8. Refer to Exercise 7.

a. On the basis of this initial experience, determine the control limits for the range. Plot the experience on a chart.

b. Does there seem to be a time when there is too much variation in the temperature?

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Statistical Process Control and Quality Management 737

19.7 Attribute Control Charts Often the data we collect are the result of counting rather than measuring. That is, we observe the presence or absence of some attribute. For example, the screw top on a bottle of shampoo either fits onto the bottle and does not leak (an “acceptable” condition) or does not seal and a leak results (an “unacceptable” condition), or a bank makes a loan to a customer and the loan is either repaid or it is not repaid. In other cases, we are interested in the number of defects in a sample. British Airways might count the number of its flights arriving late per day at Gatwick Airport in London. In this section, we discuss two types of attribute charts: the p-chart (percent defective) and the c-bar chart (number of defectives).

Percent Defective Charts If the item recorded is the fraction of unacceptable parts made in a larger batch of parts, the appropriate control chart is the percent defective chart. This chart is based on the binomial distribution, discussed in Chapter 6, and proportions, discussed in Chapter 9. The centerline is at p, the mean proportion defective. The p replaces the of the variable control chart. The mean proportion defective is found by:

MEAN PROPORTION DEFECTIVE [19–6]

The variation in the sample proportion is described by the standard error of a pro- portion. It is found by:

STANDARD ERROR OF THE SAMPLE PROPORTION [19–7]

Hence, the upper control limit (UCL) and the lower control limit (LCL) are com- puted as the mean percent defective plus or minus three times the standard error of the percents (proportions). The formula for the control limits is:

CONTROL LIMITS FOR PROPORTIONS [19–8]

An example will show the details of the calculations and the conclusions.

LCL, UCL � p � 3 B

p(1 � p) n

sp � B

p(1 � p) n

p � Total number defective

Total number of items sampled

Example Jersey Glass Company Inc. produces small hand mirrors. Jersey Glass runs a day and evening shift each weekday. The quality assurance department (QA) mon- itors the quality of the mirrors twice during the day shift and twice during the evening shift. QA selects and carefully inspects a random sample of 50 mirrors once every four hours. Each mirror is classified as either acceptable or unaccept- able. Finally, QA counts the number of mirrors in the sample that do not conform to quality specifications. Listed next are the results of these checks over the last 10 business days.

LO8 Construct and interpret percent defective and c-bar charts.

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Solution

738 Chapter 19

Number Number Date Sampled Defects Date Sampled Defects

10-Oct 50 1 17-Oct 50 7 50 0 50 9 50 9 50 0 50 9 50 8

11-Oct 50 4 18-Oct 50 6 50 4 50 9 50 5 50 6 50 3 50 1

12-Oct 50 9 19-Oct 50 4 50 3 50 5 50 10 50 2 50 2 50 5

13-Oct 50 2 20-Oct 50 0 50 4 50 0 50 9 50 4 50 4 50 7

14-Oct 50 6 21-Oct 50 5 50 9 50 1 50 2 50 9 50 4 50 9

Construct a percent defective chart for this process. What are the upper and lower control limits? Interpret the results. Does it appear the process is out of control dur- ing the period?

Number Fraction Number Fraction Date Sampled Defects Defective Date Sampled Defects Defective

10-Oct 50 1 0.02 17-Oct 50 7 0.14 50 0 0.00 50 9 0.18 50 9 0.18 50 0 0.00 50 9 0.18 50 8 0.16

11-Oct 50 4 0.08 18-Oct 50 6 0.12 50 4 0.08 50 9 0.18 50 5 0.10 50 6 0.12 50 3 0.06 50 1 0.02

12-Oct 50 9 0.18 19-Oct 50 4 0.08 50 3 0.06 50 5 0.10 50 10 0.20 50 2 0.04 50 2 0.04 50 5 0.10

13-Oct 50 2 0.04 20-Oct 50 0 0.00 50 4 0.08 50 0 0.00 50 9 0.18 50 4 0.08 50 4 0.08 50 7 0.14

14-Oct 50 6 0.12 21-Oct 50 5 0.10 50 9 0.18 50 1 0.02 50 2 0.04 50 9 0.18 50 4 0.08 50 9 0.18

Total 2000 196

The first step is to determine the overall proportion defective. We use formula (19–6).

So we estimate that .098 of the mirrors produced during the period do not meet specifications.

p � Total number defective

Total number of items sampled �

196 2,000

� .098

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Statistical Process Control and Quality Management 739

The upper and lower control limits are computed by using formula (19–8)

From the above calculations, the upper control limit is .2241, found by .098 � .1261. The lower control limit is 0. Why? The lower limit by the formula is .098 � .1261 � �0.0281. However, a negative proportion defective is not possible, so the smallest value is 0. We set the control limits at 0 and 0.2241. Any sample outside these lim- its indicates the quality level of the process has changed.

This information is summarized in Chart 19–6, which is output from the Minitab system.

LCL, UCL � p � 3 B

p(1 � p) n

� .098 � 3 B

.098(1 � .098) 50

� .098 � .1261

CHART 19–6 Percent Defective Chart for Mirrors at Jersey Glass

Number Fraction Number Fraction Date Sampled Defects Defective Date Sampled Defects Defective

24-Oct 50 1 0.02 27-Oct 50 2 0.04 50 13 0.26 50 1 0.02 50 10 0.20 50 7 0.14 50 7 0.14 50 12 0.24

25-Oct 50 4 0.08 28-Oct 50 5 0.10 50 5 0.10 50 5 0.10 50 6 0.12 50 10 0.20 50 10 0.20 50 9 0.18

26-Oct 50 6 0.12 50 1 0.02 50 8 0.16 50 4 0.08

After establishing the limits, the process is monitored for the next week—five days, two shifts per day—with two quality checks per shift. The results are shown below.

The process was out of control on two occasions, on October 24 when the num- ber of defects was 13 and again on October 27 when the number of defects was 12. QA should report this information to the production department for the appro- priate action. The Minitab output follows.

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740 Chapter 19

c-Bar Charts The c-bar chart plots the number of defects or failures per unit. It is based on the Poisson distribution discussed in Chapter 6. The number of bags mishandled on a flight by Southwest Airlines might be monitored by a c-bar chart. The “unit” under consideration is the flight. On most flights, there are no bags mishandled. On oth- ers, there may be only one, on others two, and so on. The Internal Revenue Service might count and develop a control chart for the number of errors in arithmetic per tax return. Most returns will not have any errors, some returns will have a single error, others will have two, and so on. We let be the mean number of defects per unit. Thus, is the mean number of bags mishandled by Southwest Airlines per flight or the mean number of arithmetic errors per tax return. Recall from Chapter 6 that the standard deviation of a Poisson distribution is the square root of the mean. Thus, we can determine the 3-sigma, or 99.74 percent, limits on a c-bar chart by:

CONTROL LIMITS FOR THE NUMBER OF DEFECTS PER UNIT

[19–9]LCL, UCL � c � 32c

c c

Example The publisher of the Oak Harbor Daily Telegraph is concerned about the number of misspelled words in the daily newspaper. It does not print a paper on Saturday or Sunday. In an effort to control the problem and promote the need for correct spelling, a control chart will be used. The number of misspelled words found in the final edition of the paper for the last 10 days is: 5, 6, 3, 0, 4, 5, 1, 2, 7, and 4. Determine the appropriate control limits and interpret the chart. Were there any days during the period that the number of misspelled words was out of control?

Solution During the 10-day period, there were a total of 37 misspelled words. The mean num- ber of misspelled words per edition is 3.7. The number of misspelled words per edi- tion follows the Poisson probability distribution. The standard deviation is the square root of the mean.

To find the upper control limit, we use formula (19–9). The lower control limit is zero.

UCL � c � 32c � 3.7 � 323.7 � 3.7 � 5.77 � 9.47

c � ©X n

� 5 � 6 � . . . � 4

10 �

37 10

� 3.7 s � 2c � 23.7 � 1.924

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Statistical Process Control and Quality Management 741

Self-Review 19–3 Auto-Lite Company manufactures car batteries. At the end of each shift, the quality assur- ance department selects a sample of batteries and tests them. The number of defective batteries found over the last 12 shifts is 2, 1, 0, 2, 1, 1, 7, 1, 1, 2, 6, and 1. Construct a control chart for the process and comment on whether the process is in control.

Exercises 9. Below is a percent defective chart for a manufacturing process.

The computed lower control limit would be 3.7 � 3(1.924) � �2.07. However, the num- ber of misspelled words cannot be less than 0, so we use 0 as the lower limit. The lower control limit is 0 and the upper limit is 9.47. When we compare each of the data points to the value of 9.47, we see they are all less than the upper control limit; the number of misspelled words is “in control.” Of course, newspapers are going to strive to eliminate all misspelled words, but control charting techniques offer a means of tracking daily results and determining whether there has been a change. For example, if a new proofreader was hired, her work could be compared with others. These results are summarized in Chart 19–7, which is output from the Minitab system.

CHART 19–7 c-Bar Chart for Number of Misspelled Words per Edition of the Oak Harbor Daily Telegraph

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742 Chapter 19

a. What is the mean percent defective? What are the upper and lower control limits? b. Are there any sample observations that indicate the process is out of control? Which

sample numbers are they? c. Does there seem to be any trend in the process? That is, does the process seem to

be getting better, getting worse, or staying the same? 10. Inter State Moving and Storage Company is setting up a control chart to monitor the pro-

portion of residential moves that result in written complaints due to late delivery, lost items, or damaged items. A sample of 50 moves is selected for each of the last 12 months. The number of written complaints in each sample is 8, 7, 4, 8, 2, 7, 11, 6, 7, 6, 8, and 12.

a. Design a percent defective chart. Insert the mean percent defective, UCL, and LCL. b. Plot the proportion of written complaints in the last 12 months. c. Interpret the chart. Does it appear that the number of complaints is out of control for

any of the months? 11. A bicycle manufacturer randomly selects 10 frames each day and tests for defects. The

number of defective frames found over the last 14 days is 3, 2, 1, 3, 2, 2, 8, 2, 0, 3, 5, 2, 0, and 4. Construct a control chart for this process and comment on whether the process is “in control.”

12. Scott Paper tests its toilet paper by subjecting 15 rolls to a wet stress test to see whether and how often the paper tears during the test. Following are the number of defectives found over the last 15 days: 2, 3, 1, 2, 2, 1, 3, 2, 2, 1, 2, 2, 1, 0, and 0. Construct a control chart for the process and comment on whether the process is “in control.”

13. Sam’s Supermarkets tests its checkout clerks by randomly examining the printout receipts for scanning errors. The following numbers are the number of errors on each receipt for October 27: 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0. Construct a control chart for this process and comment on whether the process is “in control.”

14. Dave Christi runs a car wash chain with outlets scattered throughout Chicago. He is con- cerned that some local managers are giving away free washes to their friends. He decides to collect data on the number of “voided” sales receipts. Of course, some of them are legitimate voids. Would the following data indicate a reasonable number of voids at his facilities: 3, 8, 3, 4, 6, 5, 0, 1, 2, 4? Construct a control chart for this process and com- ment on whether the process is “in control.”

19.8 Acceptance Sampling The previous section was concerned with maintaining the quality of the product as it is being produced. In many business situ- ations, we are also concerned with the qual- ity of the incoming finished product. What do the following cases have in common?

• Sims Software Inc. purchases DVDs from DVD International. The normal pur- chase order is for 100,000 DVDs, pack- aged in lots of 1,000. Todd Sims,

president, does not expect each DVD to be perfect. In fact, he has agreed to accept lots of 1,000 with up to 10 percent defective. He would like to develop a plan to inspect incoming lots, to ensure that the quality standard is met. The purpose of the inspection procedure is to separate the acceptable from the unacceptable lots.

• Zenith Electric purchases magnetron tubes from Bono Electronics for use in its new microwave oven. The tubes are shipped to Zenith in lots of 10,000. Zenith allows the incoming lots to contain up to 5 percent defective tubes. It would like to develop a sampling plan to determine which lots meet the criterion and which do not.

Statistics in Action

It was reported during the late 1980s that a Canadian firm or- dered some parts from a Japanese company with instructions that there should be “no more than three de- fective parts per one thousand.” When the parts arrived, there was a note attached that said, “Your three defective parts are wrapped separately in the upper left com- partment of the ship- ment.” This is a far cry from the days when “Made in Japan” meant cheap but not reliable.

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Statistical Process Control and Quality Management 743

• General Motors purchases windshields from many suppliers. GM insists that the windshields be in lots of 1,000, and is willing to accept 50 or fewer defects in each lot, that is, 5 percent defective. They would like to develop a sampling procedure to verify that incoming shipments meet the criterion.

The common thread in these cases is a need to verify that an incoming prod- uct meets the stipulated requirements. The situation can be likened to a screen door, which allows the warm summer air to enter the room while keeping the bugs out. Acceptance sampling lets the lots of acceptable quality into the manufacturing area and screens out lots that are not acceptable.

Of course, the situation in modern business is more complex. The buyer wants protection against accepting lots that are below the quality standard. The best pro- tection against inferior quality is 100 percent inspection. Unfortunately, the cost of 100 percent inspection is often prohibitive. Another problem with checking each item is that the test may be destructive. If all lightbulbs were tested until burning out before they were shipped, there would be none left to sell. Also, 100 percent inspec- tion may not lead to the identification of all defects, because boredom might cause a loss of perception on the part of the inspectors. Thus, complete inspection is rarely employed in practical situations.

The usual procedure is to screen the quality of incoming parts by using a sta- tistical sampling plan. According to this plan, a sample of n units is randomly selected from the lots of N units (the population). This is called acceptance sam- pling. The inspection will determine the number of defects in the sample. This num- ber is compared with a predetermined number called the critical number or the acceptance number. The acceptance number is usually designated c. If the num- ber of defects in the sample of size n is less than or equal to c, the lot is accepted. If the number of defects exceeds c, the lot is rejected and returned to the supplier, or perhaps submitted to 100 percent inspection.

Acceptance sampling is a decision-making process. There are two possible decisions: accept or reject the lot. In addition, there are two situations under which the decision is made: the lot is good or the lot is bad. These are the states of nature. If the lot is good and the sample inspection reveals the lot to be good, or if the lot is bad and the sample inspection indicates it is bad, then a correct deci- sion is made. However, there are two other possibilities. The lot may actually con- tain more defects than it should, but it is accepted. This is called consumer’s risk. Similarly, the lot may be within the agreed-upon limits, but it is rejected dur- ing the sample inspection. This is called the producer’s risk. The following sum- mary table for acceptance decisions shows these possibilities. Notice how this discussion is very similar to the ideas of Type I and Type II errors starting on page 359 in Section 10.10.

Consumer’s risk

Producer’s risk

States of Nature

Decision Good Lot Bad Lot

Accept lot Correct Consumer’s risk Reject lot Producer’s risk Correct

To evaluate a sampling plan and determine that it is fair to both the producer and the consumer, the usual procedure is to develop an operating characteristic curve, or an OC curve as it is usually called. An OC curve reports the percent defec- tive along the horizontal axis and the probability of accepting that percent defective along the vertical axis. A smooth curve is usually drawn connecting all the possible levels of quality. The binomial distribution is used to develop the probabilities for an OC curve.

Acceptance sampling

Acceptance number

LO9 Explain the process of acceptance sampling.

LO10 Describe an operating characteristic curve for a sampling plan.

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744 Chapter 19

Example Sims Software, as mentioned earlier, purchases DVDs from DVD International. The DVDs are packaged in lots of 1,000 each. Todd Sims, president of Sims Software, has agreed to accept lots with 10 percent or fewer defective DVDs. Todd has directed his inspection department to select a random sample of 20 DVDs and examine them carefully. He will accept the lot if it has two or fewer defectives in the sample. Develop an OC curve for this inspection plan. What is the probability of accepting a lot that is 10 percent defective?

Solution This type of sampling is called attribute sampling because the sampled item, a DVD in this case, is classified as acceptable or unacceptable. No “reading” or “mea- surement” is obtained on the DVD. Let � represent the actual proportion defective in the population.

The lot is good if � � .10. The lot is bad if � .10.

Let X be the number of defects in the sample. The decision rule is:

Accept the lot if X � 2. Reject the lot if X 3.

Here the acceptable lot is one with 10 percent or fewer defective DVDs. If the lot is acceptable when it has exactly 10 percent defectives, it would be even more acceptable if it contained fewer than 10 percent defectives. Hence, it is the usual practice to work with the upper limit of the percent of defectives.

The binomial distribution is used to compute the various values on the OC curve. Recall that for us to use the binomial there are four requirements:

1. There are only two possible outcomes. Here the DVD is either acceptable or unacceptable.

2. There is a fixed number of trials. In this instance, the number of trials is the sam- ple size of 20.

3. There is a constant probability of success. A success is finding a defective DVD. The probability of success is assumed to be .10.

4. The trials are independent. The probability of obtaining a defective DVD on the third one selected is not related to the likelihood of finding a defect on the fourth DVD selected.

Appendix B.9 gives various binomial probabilities. However, the tables in Appen- dix B.9 go up to only 15, that is, n � 15. For this problem, n � 20, so we will use Excel to compute the various binomial probabilities. The following Excel output shows the binomial probabilities for n � 20 when � is equal to .05, .10, .15, .20, .25, and .30.

We need to convert the terms used in Chapter 6 to acceptance sampling vocab- ulary. We let � refer to the probability of finding a defect, c the number of defects allowed, and n the number of items sampled. In this case, we will allow up to two defects, so c � 2. This means that we will allow 0, 1, or 2 of the 20 items sampled to be defective and still accept the incoming shipment of DVDs.

To begin, we determine the probability of accepting a lot that is 5 percent defec- tive. This means that � � .05, c � 2, and n � 20. From the Excel output, the like- lihood of selecting a sample of 20 items from a shipment that contained 5 percent defective and finding exactly 0 defects is .358. The likelihood of finding exactly 1 defect is .377, and finding 2 is .189. Hence, the likelihood of 2 or fewer defects is .924, found by .358 � .377 � .189. This result is usually written in shorthand nota- tion as follows (recall that bar “|” means “given that”).

P(x � 2 ƒ � � .05 and n � 20) � .358 � .377 � .189 � .924

Attribute sampling

Decision rule

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Statistical Process Control and Quality Management 745

Self-Review 19–4 Compute the probability of accepting a lot of DVDs that is actually 30 percent defective, using the sampling plan for Sims Software.

Continuing, the likelihood of accepting a lot that is actually 10 percent defective is .677. That is:

The complete OC curve in Chart 19–8 shows the smoothed curve for all values of � between 0 and about 30 percent. There is no need to show values larger than 30 percent because their probability is very close to 0. The likelihood of accepting lots of selected quality levels is shown in table form on the right-hand side of Chart 19–8. With the OC curve, the management of Sims Software will be able to quickly evaluate the probabilities of various quality levels.

P(x � 2 ƒ � � .10 and n � 20) � .122 � .270 � .285 � .677

CHART 19–8 OC Curve for Sampling Plan (n � 20, c � 2)

1.00

.80

.60

.40

.20

Lot percent defective

Probability of accepting lot

0 5

10 20 30

1.000 .924 .677 .207 .036

Probability of accepting a lot

that is 10% defective is .677

Pr ob

ab ili

ty of

a cc

ep tin

g lo

0 5 10 15 20 25 30 Incoming lot percent defective

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746 Chapter 19

Exercises 15. Determine the probability of accepting lots that are 10 percent, 20 percent, 30 percent,

and 40 percent defective using a sample of size 12 and an acceptance number of 2. 16. Determine the probability of accepting lots that are 10 percent, 20 percent, 30 percent,

and 40 percent defective using a sample of size 14 and an acceptance number of 3. 17. Warren Electric manufactures fuses for many customers. To ensure the quality of the outgo-

ing product, it tests 10 fuses each hour. If no more than one fuse is defective, it packages the fuses and prepares them for shipment. Develop an OC curve for this sampling plan. Compute the probabilities of accepting lots that are 10 percent, 20 percent, 30 percent, and 40 percent defective. Draw the OC curve for this sampling plan using the four quality levels.

18. Grills Video Products purchases LCDs from Mira Electronics. According to his sampling plan, Art Grills, owner of Grills Video, will accept a shipment of LCDs if 3 or fewer are defective in a sample of 25. Develop an OC curve for these percents defective: 10 percent, 20 per- cent, 30 percent, and 40 percent. You will need a statistical software package.

Chapter Summary I. The objective of statistical quality control is to monitor the quality of the product or ser-

vice as it is being developed. II. A Pareto chart is a technique for tallying the number and type of defects that happen

within a product or service. A. This chart was named after an Italian scientist, Vilfredo Pareto. B. The concept of the chart is that 80 percent of the activity is caused by 20 percent of

the factors. III. A fishbone diagram emphasizes the relationship between a possible problem cause that

will produce the particular effect. A. It is also called a cause-and-effect diagram. B. The usual approach is to consider four problem areas: methods, materials, equipment,

and personnel. IV. The purpose of a control chart is to monitor graphically the quality of a product or service.

A. There are two types of control charts. 1. A variable control chart is the result of a measurement. 2. An attribute chart shows whether the product or service is acceptable or not

acceptable. B. There are two sources of variation in the quality of a product or service.

1. Chance variation is random in nature and cannot be controlled or eliminated. 2. Assignable variation is not due to random causes and can be eliminated.

C. Four control charts were considered in this chapter. 1. A mean chart shows the mean of a variable, and a range chart shows the range of

the variable. a. The upper and lower control limits are set at plus or minus 3 standard errors from

the mean. b. The formulas for the upper and lower control limits for the mean are:

[19–4]

c. The formulas for the upper and lower control limits for the range are:

[19–5]

2. A percent defective chart is an attribute chart that shows the proportion of the prod- uct or service that does not conform to the standard. a. The mean percent defective is found by

[19–6]

b. The control limits for the proportion defective are determined from the equation

[19–8]LCL, UCL � p � 3 B

p(1 � p) n

p � Total number defective

Total number of items sampled

UCL � D4R LCL � D3R

UCL � X � A2R LCL � X � A2R

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Statistical Process Control and Quality Management 747

3. A c-bar chart refers to the number of defects per unit. a. It is based on the Poisson distribution. b. The mean number of defects per unit is . c. The control limits are determined from the following equation.

[19–9]

V. Acceptance sampling is a method to determine whether an incoming lot of a product meets specified standards. A. It is based on random sampling techniques. B. A random sample of n units is selected from a population of N units. C. c is the maximum number of defective units that may be found in the sample of n and

the lot is still considered acceptable. D. An OC (operating characteristic) curve is developed using the binomial probability dis-

tribution to determine the probability of accepting lots of various quality levels.

Pronunciation Key SYMBOL MEANING PRONUNCIATION

Mean of the sample means X double bar

Standard error of the mean s sub X bar

Constant used to determine the upper A sub 2 and lower control limit for the mean

Mean of the sample ranges R bar

Constant used to determine the D sub 4 upper control limit for the range

Mean number of defects per unit c bar

Chapter Exercises 19. The production supervisor at Westburg Electric Inc. noted an increase in the number of

electric motors rejected at the time of final inspection. Of the last 200 motors rejected, 80 of the defects were due to poor wiring, 60 contained a short in the coil, 50 involved a defective plug, and 10 involved other defects. Develop a Pareto chart to show the major problem areas.

20. The manufacturer of running shoes conducted a study on its newly developed jogging shoe. Listed below are the type and frequency of the nonconformities and failures found. Develop a Pareto chart to show the major problem areas.

s ˛X

LCL, UCL � c � 31c

Type of Nonconformity Frequency Type of Nonconformity Frequency

Sole separation 34 Lace breakage 14 Heel separation 98 Eyelet failure 10 Sole penetration 62 Other 16

21. At Rumsey’s Old Fashion Roast Beef, cola drinks are filled by an automatic machine whose operation is based on the weight of the drink. When the process is in control, the machine fills each cup so that the grand mean is 10.0 ounces and the mean range is 0.25 for samples of 5. a. Determine the upper and lower control limits for the process for both the mean and

the range. b. The manager of the I-280 store tested five soft drinks served last hour and found that

the mean was 10.16 ounces and the range was 0.35 ounces. Is the process in con- trol? Should other action be taken?

22. A new machine has just been installed to cut and rough-shape large slugs. The slugs are then transferred to a precision grinder. One of the critical measurements is the outside diameter. The quality control inspector randomly selected five slugs each half-hour, mea- sured the outside diameter, and recorded the results. The measurements (in millimeters) for the period 8:00 A.M. to 10:30 A.M. follow.

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748 Chapter 19

a. Determine the control limits for the mean and the range. b. Plot the control limits for the mean outside diameter and the range. c. Are there any points on the mean or the range chart that are out of control? Com-

ment on the chart. 23. Long Last Tire Company, as part of its inspection process, tests its tires for tread wear

under simulated road conditions. Twenty samples of three tires each were selected from different shifts over the last month of operation. The tread wear is reported below in hun- dredths of an inch.

Outside Diameter (millimeters)

Time 1 2 3 4 5

8:00 87.1 87.3 87.9 87.0 87.0 8:30 86.9 88.5 87.6 87.5 87.4 9:00 87.5 88.4 86.9 87.6 88.2 9:30 86.0 88.0 87.2 87.6 87.1

10:00 87.1 87.1 87.1 87.1 87.1 10:30 88.0 86.2 87.4 87.3 87.8

a. Determine the control limits for the mean and the range. b. Plot the control limits for the mean tread wear and the range. c. Are there any points on the mean or the range chart that are “out of control”? Com-

ment on the chart. 24. Charter National Bank has a staff of loan officers located in its branch offices through-

out the Southwest. Robert Kerns, vice president of consumer lending, would like some information on the typical amount of loans and the range in the amount of the loans. A staff analyst of the vice president selected a sample of 10 loan officers and from each officer selected a sample of five loans he or she made last month. The data are reported below. Develop a control chart for the mean and the range. Do any of the officers appear to be “out of control”? Comment on your findings.

Sample Tread Wear Sample Tread Wear

1 44 41 19 11 11 33 34 2 39 31 21 12 51 34 39 3 38 16 25 13 30 16 30 4 20 33 26 14 22 21 35 5 34 33 36 15 11 28 38 6 28 23 39 16 49 25 36 7 40 15 34 17 20 31 33 8 36 36 34 18 26 18 36 9 32 29 30 19 26 47 26

10 29 38 34 20 34 29 32

Loan Amount ($000) Loan Amount ($000)

Officer 1 2 3 4 5 Officer 1 2 3 4 5

Weinraub 59 74 53 48 65 Bowyer 66 80 54 68 52 Visser 42 51 70 47 67 Kuhlman 74 43 45 65 49 Moore 52 42 53 87 85 Ludwig 75 53 68 50 31 Brunner 36 70 62 44 79 Longnecker 42 65 70 41 52 Wolf 34 59 39 78 61 Simonetti 43 38 10 19 47

25. The producer of a candy bar, called the “A Rod” Bar, reports on the package that the calorie content is 420 per 2-ounce bar. A sample of five bars from each of the last 10 days is sent for a chemical analysis of the calorie content. The results are shown next. Does it appear that there are any days where the calorie count is out of control? Develop an appropriate control chart and analyze your findings.

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Statistical Process Control and Quality Management 749

26. Early Morning Delivery Service guarantees delivery of small packages by 10:30 A.M. Of course, some of the packages are not delivered by 10:30 A.M. For a sample of 200 pack- ages delivered each of the last 15 working days, the following number of packages were delivered after the deadline: 9, 14, 2, 13, 9, 5, 9, 3, 4, 3, 4, 3, 3, 8, and 4. a. Determine the mean proportion of packages delivered after 10:30 A.M. b. Determine the control limits for the proportion of packages delivered after 10:30 A.M.

Were any of the sampled days out of control? c. If 10 packages out of 200 in the sample were delivered after 10:30 A.M. today, is this

sample within the control limits? 27. An automatic machine produces 5.0-millimeter bolts at a high rate of speed. A quality

control program has been initiated to control the number of defectives. The quality con- trol inspector selects 50 bolts at random and determines how many are defective. The number of defectives in the first 10 samples is 3, 5, 0, 4, 1, 2, 6, 5, 7, and 7.

a. Design a percent defective chart. Insert the mean percent defective, UCL, and LCL. b. Plot the percent defective for the first 10 samples on the chart. c. Interpret the chart.

28. Steele Breakfast Foods Inc. produces a popular brand of raisin bran cereal. The pack- age indicates it contains 25.0 ounces of cereal. To ensure the product quality, the Steele inspection department makes hourly checks on the production process. As a part of the hourly check, four boxes are selected and their contents weighed. The results are reported below.

Calorie Count Calorie Count

Sample 1 2 3 4 5 Sample 1 2 3 4 5

1 426 406 418 431 432 6 427 417 408 418 422 2 421 422 415 412 411 7 422 417 426 435 426 3 425 420 406 409 414 8 419 417 412 415 417 4 424 419 402 400 417 9 417 432 417 416 422 5 421 408 423 410 421 10 420 422 421 415 422

Sample Weights Sample Weights

1 26.1 24.4 25.6 25.2 14 23.1 23.3 24.4 24.7 2 25.2 25.9 25.1 24.8 15 24.6 25.1 24.0 25.3 3 25.6 24.5 25.7 25.1 16 24.4 24.4 22.8 23.4 4 25.5 26.8 25.1 25.0 17 25.1 24.1 23.9 26.2 5 25.2 25.2 26.3 25.7 18 24.5 24.5 26.0 26.2 6 26.6 24.1 25.5 24.0 19 25.3 27.5 24.3 25.5 7 27.6 26.0 24.9 25.3 20 24.6 25.3 25.5 24.3 8 24.5 23.1 23.9 24.7 21 24.9 24.4 25.4 24.8 9 24.1 25.0 23.5 24.9 22 25.7 24.6 26.8 26.9

10 25.8 25.7 24.3 27.3 23 24.8 24.3 25.0 27.2 11 22.5 23.0 23.7 24.0 24 25.4 25.9 26.6 24.8 12 24.5 24.8 23.2 24.2 25 26.2 23.5 23.7 25.0 13 24.4 24.5 25.9 25.5

Develop an appropriate control chart. What are the limits? Is the process out of control at any time?

29. An investor believes there is a 50–50 chance that a stock will increase on a particular day. To investigate this idea, for 30 consecutive trading days the investor selects a random sample of 50 stocks and counts the number that increase. The number of stocks in the sample that increased is reported below.

14 12 13 17 10 18 10 13 13 14 13 10 12 11 9 13 14 11 12 11 15 13 10 16 10 11 12 15 13 10

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750 Chapter 19

Develop a percent defective chart and write a brief report summarizing your findings. Based on these sample results, is it reasonable that the odds are 50–50 that a stock will increase? What percent of the stocks would need to increase in a day for the process to be “out of control”?

30. Lahey Motors specializes in selling cars to buyers with a poor credit history. Listed below is the number of cars that were repossessed from Lahey customers because they did not meet the payment obligations over each of the last 36 months.

Develop a c-bar chart for the number repossessed. Were there any months when the number was out of control? Write a brief report summarizing your findings.

31. A process engineer is considering two sampling plans. In the first, a sample of 10 will be selected and the lot accepted if 3 or fewer are found defective. In the second, the sam- ple size is 20 and the acceptance number is 5. Develop an OC curve for each. Compare the probability of acceptance for lots that are 5, 10, 20, and 30 percent defective. Which of the plans would you recommend if you were the supplier?

32. Christina Sanders is a member of the women’s basketball team at Windy City College. Last season, she made 55 percent of her free throw attempts. In an effort to improve this statistic, she attended a summer camp devoted to foul-shooting techniques. The next 20 days she shot 100 foul shots a day. She carefully recorded the number of attempts made each day. The results are reported below.

6 5 8 20 11 10 9 3 9 9 15 12 4 11 9 9 6 18 6 8

9 7 13 7 11 8 11 13 6 14 13 5 5 8 10 11

To interpret, the first day she made 55 out of 100, or 55 percent. The last day she made 67 out of 100 or 67 percent. a. Develop a control chart for the proportion of shots made. Over the 20 days of prac-

tice, what percent of attempts did she make? What are the upper and lower control limits for the proportion of shots made?

b. Is there any trend in her proportion made? Does she seem to be improving, staying the same, or getting worse?

c. Find the percent of attempts made for the last five days of practice. Use the hypothesis testing procedure, formula (10–4), to determine if there is an improvement from 55 percent.

33. Eric’s Cookie House sells chocolate chip cookies in shopping malls. Of concern is the num- ber of chocolate chips in each cookie. Eric, the owner and president, would like to estab- lish a control chart for the number of chocolate chips per cookie. He selects a sample of 15 cookies from today’s production and counts the number of chocolate chips in each. The results are as follows: 6, 8, 20, 12, 20, 19, 11, 23, 12, 14, 15, 16, 12, 13, and 12. a. Determine the centerline and the control limits. b. Develop a control chart and plot the number of chocolate chips per cookie. c. Interpret the chart. Does it appear that the number of chocolate chips is out of con-

trol in any of the cookies sampled? 34. The number of “near misses” recorded for the last 20 months at Lima International Air-

port is 3, 2, 3, 2, 2, 3, 5, 1, 2, 2, 4, 4, 2, 6, 3, 5, 2, 5, 1, and 3. Develop an appropriate control chart. Determine the mean number of misses per month and the limits on the number of misses per month. Are there any months where the number of near misses is out of control?

35. The following number of robberies was reported during the last 10 days to the robbery divi- sion of the Metro City Police: 10, 8, 8, 7, 8, 5, 8, 5, 4, and 7. Develop an appropriate con- trol chart. Determine the mean number of robberies reported per day and determine the con- trol limits. Are there any days when the number of robberies reported is out of control?

36. Swiss Watches, Ltd. purchases watch stems for their watches in lots of 10,000. Their sampling plan calls for checking 20 stems, and if 3 or fewer stems are defective, the lot is accepted. a. Based on the sampling plan, what is the probability that a lot of 40 percent defective

will be accepted? b. Design an OC curve for incoming lots that have zero, 10 percent, 20 percent, 30 per-

cent, and 40 percent defective stems.

55 61 52 59 67 57 61 59 69 58 57 66 63 63 63 65 63 68 64 67

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Statistical Process Control and Quality Management 751

37. Automatic Screen Door Manufacturing Company purchases door latches from a number of vendors. The purchasing department is responsible for inspecting the incoming latches. Automatic purchases 10,000 door latches per month and inspects 20 latches selected at random. Develop an OC curve for the sampling plan if three latches can be defective and the incoming lot is still accepted.

38. At the beginning of each football season, Team Sports, the local sporting goods store, purchases 5,000 footballs. A sample of 25 balls is selected, and they are inflated, tested, and then deflated. If more than two balls are found defective, the lot of 5,000 is returned to the manufacturer. Develop an OC curve for this sampling plan. a. What are the probabilities of accepting lots that are 10 percent, 20 percent, and 30 per-

cent defective? b. Estimate the probability of accepting a lot that is 15 percent defective. c. John Brennen, owner of Team Sports, would like the probability of accepting a lot that

is 5 percent defective to be more than 90 percent. Does this appear to be the case with this sampling plan?

1. The Minitab commands for the Pareto chart on page 727 are: a. Enter the reasons for water usage in column C1

and the gallons used in C2. Give the columns appropriate names.

b. Click on Stat, Quality Tools, Pareto Chart, and then hit Enter.

c. Select Chart defects table, indicate the location of the labels and frequencies, click on Options and type a chart title, and click OK.

Software Commands

2. The Minitab commands for the X-bar and R charts on page 734 are: a. Enter the information in Table 19–1 or from the

CD. The file name is Table 19-1. b. Click on Stat, Control Charts, Variables Charts

for Subgroups, Xbar-R, and hit Enter. c. Select All observations for a chart are in one

column. Then in the box below, select the vari- able Minutes.

3. The Minitab commands for the percent defective chart on page 739 are: a. Enter the data on the number of defects from

page 738. b. Click on Stat, Control Charts, Attribute Charts,

P, and hit Enter. c. Under Variables, select Defects, then enter 50 for

Subgroup sizes. Click on Labels, type in the title, and click OK twice.

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752 Chapter 19

4. The Minitab commands for the c-bar chart on page 741 are: a. Enter the data on the number of misspelled

words from page 740. b. Click on Stat, Control Charts, Attribute Charts,

C, and hit Enter. c. Select the Variable indicating the number of mis-

spelled words, then click on Labels and type the title in the space provided, and click OK twice.

Chapter 19 Answers to Self-Review

19–1

Po or

fo od

Po or

c ar

Di rty

Count 84 71 63 45 35 2 Percent 28 24 21 15 12 0 Cum. % 28 52 73 88 100 100

100

Co un

Percent

to d

o No

th in

No re

sp ec

M ed

. e rr

300 250 200 150 100 50

Seventy-three percent of the complaints involve poor food, poor care, or dirty conditions. These are the factors the administrator should address.

19–2 a.

Sample Times

1 2 3 4 Total Average Range

1 4 5 2 12 3 4 2 3 2 1 8 2 2 1 7 3 5 16 4 6

9 12

UCL � 5.916 LCL � 0.084

� 3 � 0.729(4)

UCL and LCL � X � A2R

X � 9 3

� 3 R � 12 3

� 4

UCL � D4R � 2.282(4) � 9.128

LCL � D3 R � 0(4) � 0

5.916

.084

Bar X Chart UCL

LCL

Time (A.M.)

9 10 11

9.128

Range Chart UCL

LCL 9 10 11

Time (A.M.)

b. Yes. Both the mean chart and the range chart indicate that the process is in control.

19–3

Because LCL is a negative value, we set LCL � 0. The shift with seven defects is out of control.

19–4 P(X � � .30 and n � 20) � .0362 ƒ �

LCL � 2.083 � 312.083 � �2.247

UCL � 2.083 � 312.083 � 6.413

c � 25 12

� 2.083

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20 Learning Objectives When you have completed this chapter, you will be able to:

LO1 Identify and apply the three components of a decision.

LO2 Compute and interpret the expected values for a payoff table.

LO3 Explain and interpret opportunity loss.

LO4 Describe three strategies for decision making.

LO5 Compute and describe the expected value of perfect information.

LO6 Organize possible out- comes into a decision tree and interpret the result.

An Introduction to Decision Theory

Blackbeard’s Phantom Fireworks is considering introducing two new

bottle rockets. The company can add both to the current line, neither,

or just one of the two. The success of these products depends on

consumers’ reactions. These reactions can be summarized as good,

fair, or poor. The company’s revenues are estimated in the payoff table

in Exercise 11. Compute the expected monetary value for each

decision. (See Exercise 11a and LO2.)

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20-2 Chapter 20

20.1 Introduction A branch of statistics called statistical decision theory that uses probability has been developed since the early 1950s. As the name implies, the focus is on the process of making decisions and explicitly includes the payoffs that may result. In contrast, classical statistics focuses on estimating a parameter, such as the popu- lation mean, constructing a confidence interval, or conducting a hypothesis test. Classical statistics does not address the financial consequences.

Statistical decision theory is concerned with determining which decision, from a set of possible alternatives, is optimal for a particular set of conditions. Consider the following examples of decision-theory problems.

• Ford Motor Company must decide whether to purchase assembled door locks for the 2010 Ford F-150 truck or to manufacture and assemble the door locks at

its Sandusky, Ohio, plant. If sales of the F-150 truck continue to increase, it will be more profitable to manufac- ture and assemble the parts. If sales level off or decline, it will be more profitable to purchase the door locks assembled. Should it make or buy the door locks?

• Banana Republic developed a new line of summer rain jackets that are very popular in the cold-weather regions of the country. It would like to

purchase commercial television time during the upcoming NCAA basketball final. If both teams that play in the game are from warm parts of the country, it estimates that only a small proportion of the viewers will be interested in the jackets. However, a matchup between two teams who come from cold climates would reach a large proportion of viewers who wear jackets. Should it purchase commercial television time?

• General Electric is considering three options regarding the prices of refrigera- tors for next year. GE could (1) raise the prices 5 percent, (2) raise the prices 2.5 percent, or (3) leave the prices as they are. The final decision will be based on sales estimates and on GE’s knowledge of what other refrigerator manufac- turers might do.

In each of these cases, the decision is characterized by several alternative courses of action and several factors not under the control of the decision maker. For example, Banana Republic has no control over which teams reach the NCAA basketball final. These cases characterize the nature of decision making. Possible decision alternatives can be listed, possible future events determined, and even probabilities established, but the decisions are made in the face of uncertainty.

20.2 Elements of a Decision There are three components to any decision: (1) the choices available, or alterna- tives; (2) the states of nature, which are not under the control of the decision maker; and (3) the payoffs. These concepts will be explained in the following paragraphs.

The alternatives, or acts, are the choices available to the decision maker. Ford can decide to manufacture and assemble the door locks in Sandusky, or it can decide to purchase them. To simplify our presentation, we assume the decision maker can select from a rather small number of outcomes. With the help of com- puters, however, the decision alternatives can be expanded to a large number of possibilities.

LO1 Identify and apply the three components of a decision.

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An Introduction to Decision Theory 20-3

The states of nature are the uncontrollable future events. The state of nature that actually happens is outside the control of the decision maker. Ford does not know whether demand will remain high for the F-150. Banana Republic cannot determine whether warm-weather or cold-weather teams will play in the NCAA basket- ball final.

A payoff is needed to compare each combination of decision alternative and state of nature. Ford may estimate that if it assembles door locks at its Sandusky plant and the demand for F-150 trucks is low, the payoff will be $40,000. Conversely, if it purchases the door locks assembled and the demand is high, the payoff is esti- mated to be $22,000.

The main elements of the decision under conditions of uncertainty are identi- fied schematically:

Act

Event Uncertainty regarding future demand. State of nature (future demand) unknown. Decision maker has no control over state of nature.�

� Profit.Breakeven. Loss.

Two or more courses of action open to decision maker. Decision maker must evaluate alternatives. Decision maker selects a course of action based on certain criteria. Depending on the set of circumstances, these criteria may be quantitative, psychological, sociological, and so on.

�

Explanatory Statements

Outcome Payoff

Consequence

In many cases, we can make better decisions if we establish probabilities for the states of nature. These probabilities may be based on historical data or sub- jective estimates. Ford may estimate the probability of continued high demand as .70. GE may estimate the probability to be .25 that Amana and other manufacturers will raise the prices of their refrigerators.

20.3 A Case Involving Decision Making under Conditions of Uncertainty At the outset, it should be emphasized that this case description includes only the fun- damental concepts found in decision making. The purpose of examining the case is to explain the logical procedure followed. The first step is to set up a payoff table.

Payoff Table Bob Hill, a small investor, has $1,100 to invest. He has studied several common stocks and narrowed his choices to three, namely, Kayser Chemicals, Rim Homes, and Texas Electronics. He estimated that, if his $1,100 were invested in Kayser Chemicals and a strong bull market developed by the end of the year (that is, stock prices increased drastically), the value of his Kayser stock would more than double, to $2,400. How- ever, if there was a bear market (i.e., stock prices declined), the value of his Kayser stock could conceivably drop to $1,000 by the end of the year. His predictions regard- ing the value of his $1,100 investment for the three stocks for a bull market and for a bear market are shown in Table 20–1. This table is a payoff table.

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20-4 Chapter 20

The various choices are called the decision alternatives or the acts. There are three in this situation. Let A1 be the purchase of Kayser Chemicals, A2 the purchase of Rim Homes, and A3 the purchase of Texas Electronics. Whether the market turns out to be bear or bull is not under the control of Bob Hill. These uncontrolled future events are the states of nature. Let the bull market be represented by S1 and the bear market by S2.

Expected Payoff If the payoff table was the only information available, the investor might take a con- servative action and buy Texas Electronics to be assured of at least $1,150 at the end of the year (a slight profit). A speculative venture, however, might be to buy Kayser Chemicals, with the possibility of more than doubling the $1,100 investment.

Any decision regarding the purchase of one of the three common stocks made solely on the information in the payoff table would ignore the valuable historical records kept by Moody’s, Value Line, and other investment services relative to stock price movements over a long period. A study of these records, for example, revealed that during the past 10 years stock market prices increased six times and declined only four times. According to this information, the probability of a market rise is .60 and the probability of a market decline is .40.

Assuming these historical frequencies are reliable, the payoff table and the prob- ability estimates (.60 and .40) can be combined to arrive at the expected payoff of buying each of the three stocks. Expected payoff is also called expected mone- tary value, shortened to EMV. It can also be described as the mean payoff. The calculations needed to arrive at the expected payoff for the act of purchasing Kayser Chemicals are shown in Table 20–2.

Bull Market, Bear Market, Purchase S1 S2 Kayser Chemicals (A1) $2,400 $1,000 Rim Homes (A2) 2,200 1,100 Texas Electronics (A3) 1,900 1,150

TABLE 20–1 Payoff Table for Three Common Stocks under Two Market Conditions

Probability of Expected State of Nature Payoff State of Nature Value

Market rise, S1 $2,400 .60 $1,440 Market decline, S2 1,000 .40 400

$1,840

TABLE 20–2 Expected Payoff for the Act of Buying Kayser Chemicals, EMV (A1)

To explain one expected monetary value calculation, note that if the investor had purchased Kayser Chemicals and the market prices declined, the value of the stock would be only $1,000 at the end of the year (from Table 20–1). Past experi- ence, however, revealed that this event (a market decline) occurred only 40 percent of the time. In the long run, therefore, a market decline would contribute $400 to the total expected payoff from the stock, found by $1,000 � .40. Adding the $400 to the $1,440 expected under rising market conditions gives $1,840, the “expected” payoff in the long run.

These calculations are summarized as follows.

LO2 Compute and interpret the expected values for a payoff table.

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An Introduction to Decision Theory 20-5

where: EMV(Ai) refers to the expected monetary value of decision alternative i. There

may be many decisions possible. We will let 1 refer to the first deci- sion, 2 to the second, and so on. The lowercase letter i represents the entire set of decisions.

P(Sj) refers to the probability of the states of nature. There can be an unlim- ited number, so we will let j represent this possible outcome.

V(Ai, Sj) refers to the value of the payoffs. Note that each payoff is the result of a combination of a decision alternative and a state of nature.

EMV(A1), the expected monetary value for the decision alternative of purchasing Kayser Chemicals stock, is computed by:

Purchasing Kayser Chemicals stock is only one possible choice. The expected payoffs for the acts of buying Kayser Chemicals, Rim Homes, and Texas Electron- ics are given in Table 20–3.

� .60($2,400) � .40($1,000) � $1,840

EMV(A1) � [P(S1) � V(A1, S1)] � [P(S2) � V(A1, S2)]

Expected Purchase Payoff

Kayser Chemicals $1,840 Rim Homes 1,760 Texas Electronics 1,600

TABLE 20–3 Expected Payoffs for Three Stocks

An analysis of the expected payoffs in Table 20–3 indicates that purchasing Kayser Chemicals would yield the greatest expected profit. This outcome is based on (1) the investor’s estimated future value of the stocks and (2) historical experi- ence with respect to the rise and decline of stock prices. It should be emphasized that although purchasing Kayser stock represents the best action under the expected-value criterion, the investor still might decide to buy Texas Electronics stock in order to minimize the risk of losing some of the $1,100 investment.

Self-Review 20–1 Verify the conclusion, shown in Table 20–3, that the expected payoff for the act of pur- chasing Rim Homes stock is $1,760.

Exercises 1. The following payoff table was developed. Let P(S1) � .30, P(S2) � .50, and P(S3) � .20.

Compute the expected monetary value for each of the alternatives. What decision would you recommend?

State of Nature

Alternative S1 S2 S3 A1 $50 $70 $100 A2 90 40 80 A3 70 60 90

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2. Wilhelms Cola Company plans to market a new lime-flavored cola this summer. The deci- sion is whether to package the cola in returnable or in nonreturnable bottles. Currently, the state legislature is considering eliminating nonreturnable bottles. Tybo Wilhelms, pres- ident of Wilhelms Cola Company, has discussed the problem with his state representa- tive and established the probability to be .70 that nonreturnable bottles will be eliminated. The following table shows the estimated monthly profits (in thousands of dollars) if the lime cola is bottled in returnable versus nonreturnable bottles. Of course, if the law is passed and the decision is to bottle the cola in nonreturnable bottles, all profits would be from out-of-state sales. Compute the expected profit for both bottling decisions. Which decision do you recommend?

Law Is Law Is Not Passed ($000), Passed ($000),

Alternative S1 S2 Returnable bottle 80 40 Nonreturnable bottle 25 60

Opportunity Loss Another method to analyze a decision regarding which common stock to purchase is to determine the profit that might be lost because the state of nature (the market behav- ior) was not known at the time the investor bought the stock. This potential loss is called opportunity loss or regret. To illustrate, suppose the investor had purchased the common stock of Rim Homes, and a bull market developed. Further, suppose the value of his Rim Homes stock increased from $1,100 to $2,200, as anticipated. But had the investor bought Kayser Chemicals stock and market values increased, the value of his Kayser stock would be $2,400 (from Table 20–1). Thus, the investor missed making an extra profit of $200 by buying Rim Homes instead of Kayser Chemicals. To put it another way, the $200 represents the opportunity loss for not knowing the cor- rect state of nature. If market prices did increase, the investor would have regretted buying Rim Homes. However, had the investor bought Kayser Chemicals and market prices increased, he would have had no regret, that is, no opportunity loss.

The opportunity losses corresponding to this example are given in Table 20–4. Each amount is the outcome (opportunity loss) of a particular combination of acts and a state of nature, that is, stock purchase and market reaction.

Notice that the stock of Kayser Chemicals would be a good investment choice in a rising (bull) market, Texas Electronics would be the best buy in a declining (bear) market, and Rim Homes is somewhat of a compromise.

Opportunity Loss

Purchase Market Rise Market Decline

Kayser Chemicals $ 0 $150 Rim Homes 200 50 Texas Electronics 500 0

TABLE 20–4 Opportunity Losses for Various Combinations of Stock Purchase and Market Movement

Self-Review 20–2 Refer to Table 20–4. Verify that the opportunity loss for: (a) Rim Homes, given a market decline, is $50. (b) Texas Electronics, given a market rise, is $500.

LO3 Explain and interpret opportunity loss.

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Exercises 3. Refer to Exercise 1. Develop an opportunity loss table. Determine the opportunity loss

for each decision. 4. Refer to Exercise 2, involving Wilhelms Cola Company. Develop an opportunity loss table,

and determine the opportunity loss for each decision.

Expected Opportunity Loss The opportunity losses in Table 20–4 again ignore the historical experience of mar- ket movements. Recall that the probability of a market rise is .60 and that of a market decline .40. These probabilities and the opportunity losses can be com- bined to determine the expected opportunity loss. These calculations are shown in Table 20–5 for the decision to purchase Rim Homes. The expected opportunity loss is $140.

Interpreting, the expected opportunity loss of $140 means that, in the long run, the investor would lose the opportunity to make an additional profit of $140 if he decided to buy Rim Homes stock. This expected loss would be incurred because the investor was unable to accurately predict the trend of the stock market. In a bull market, he could earn an additional $200 by purchasing the common stock of Kayser Chemicals, but in a bear market an investor could earn an additional $50 by buying Texas Electronics stock. When weighted by the probability of the event, the expected opportunity loss is $140.

Probability Expected Opportunity of State of Opportunity

State of Nature Loss Nature Loss

Market rise, S1 $200 .60 $120 Market decline, S2 50 .40 20

$140

TABLE 20–5 Expected Opportunity Loss for the Act of Buying Rim Homes Stock

These calculations are summarized as follows:

EXPECTED OPPORTUNITY LOSS [20–2]

where: EOL(Ai) refers to the expected opportunity loss for a particular decision alternative. P(Sj) refers to the probability associated with the states of nature j. R(Ai, Sj) refers to the regret or loss for a particular combination of a state of

nature and a decision alternative.

EOL(A2), the regret, or expected opportunity loss, for selecting Rim Homes, is com- puted as follows:

The expected opportunity losses for the three decision alternatives are given in Table 20–6. The lowest expected opportunity loss is $60, meaning that the investor would experience the least regret on average if he purchased Kayser Chemicals.

� .60($200) � .40($50) � $140

EOL(A2) � [P(S1) � R(A2, S1)] � [P(S2) � R(A2, S2)]

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Incidentally, note that the decision to purchase Kayser Chemicals stock because it offers the lowest expected opportunity loss reinforces the decision made previ- ously, that Kayser stock would ultimately result in the highest expected payoff ($1,840). These two approaches (lowest expected opportunity loss and highest expected payoff) will always lead to the same decision concerning which course of action to follow.

Expected Opportunity

Purchase Loss

Kayser Chemicals $ 60 Rim Homes 140 Texas Electronics 300

TABLE 20–6 Expected Opportunity Losses for the Three Stocks

Exercises 5. Refer to Exercises 1 and 3. Compute the expected opportunity losses. 6. Refer to Exercises 2 and 4. Compute the expected opportunity losses.

20.4 Maximin, Maximax, and Minimax Regret Strategies Several financial advisors consider the purchase of Kayser Chemicals stock too risky. They note that the payoff might not be $1,840, but only $1,000 (from Table 20–1). Arguing that the stock market is too unpredictable, they urge the investor to take a more conservative position and buy Texas Electronics. This is called a max- imin strategy: it maximizes the minimum gain. On the basis of the payoff table (Table 20–1), they reason that the investor would be assured of at least a $1,150 return, that is, a small profit. Those who subscribe to this somewhat pessimistic strategy are sometimes called maximiners.

At the other extreme are the optimistic maximaxers, who would select the stock that maximizes the maximum gain. If their maximax strategy was followed, the investor would purchase Kayser Chemicals stock. These optimists stress that there is a possibility of selling the stock in the future for $2,400 instead of only $1,150, as advocated by the maximiners.

Another strategy is the minimax regret strategy. Advisors advocating this approach would scan the opportunity losses in Table 20–4 and select the stock that minimizes the maximum regret. In this example, it would be Kayser Chemicals stock, with a maximum opportunity loss of $150. Recall that you wish to avoid opportu- nity losses! The maximum regrets were $200 for Rim Homes and $500 for Texas Electronics.

Self-Review 20–3 Referring to Table 20–6, verify that the expected opportunity loss for the act of purchasing Texas Electronics is $300.

Minimax strategy

Maximin strategy

Maximax strategy

LO4 Describe three strategies for decision making.

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20.5 Value of Perfect Information Before deciding on a stock, the investor might want to consider ways of predicting the movement of the stock market. If he knew precisely what the market would do, he could maximize profit by always purchasing the correct stock. The question is: What is this advance information worth? The dollar value of this information is called the expected value of perfect information, written EVPI. In this example, it would mean that Bob Hill knew beforehand whether the stock market would rise or decline in the near future.

An acquaintance who is an analyst with a large brokerage firm said that he would be willing to supply Bob with information that he might find valuable in predicting market rises and declines. Of course, there would be a fee, as yet undetermined, for this information, regardless of whether the investor used it. What is the maximum amount that Bob should pay for this special service? $10? $100? $500?

The value of the information from the analyst is, in essence, the expected value of perfect information, because the investor would then be assured of buying the most profitable stock.

How much is “perfect” information worth?

VALUE OF PERFECT INFORMATION The difference between the maximum payoff under conditions of certainty and the maximum payoff under uncertainty.

In this example, it is the difference between the maximum value of the stock at the end of the year under conditions of certainty and the value associated with the opti- mum decision using the expected-value criterion.

From a practical standpoint, the maximum expected value under conditions of certainty means that the investor would buy Kayser Chemicals if a market rise were predicted and Texas Electronics if a market decline were imminent. The expected payoff under conditions of certainty is $1,900. (See Table 20–7.)

Probability of State of Expected

State of Nature Decision Payoff Nature Payoff

Market rise, S1 Buy Kayser $2,400 .60 $1,440 Market decline, S2 Buy Texas Electronics 1,150 .40 460

$1,900

TABLE 20–7 Calculations for the Expected Payoff under Conditions of Certainty

Recall that if the actual behavior of the stock market was unknown (conditions of uncertainty), the stock to buy would be Kayser Chemicals; its expected value at the end of the period was computed to be $1,840 (from Table 20–3). The value of perfect information is, therefore, $60, found by:

$1,900 Expected value of stock purchased under conditions of certainty �1,840 Expected value of purchase (Kayser) under conditions of uncertainty

$ 60 Expected value of perfect information

In general, the expected value of perfect information is computed as follows:

[20–3]

It would be worth up to $60 for the information the stock analyst might supply. In essence, the analyst would be “guaranteeing” a selling price on average of

� Expected value under conditions of uncertainty EVPI � Expected value under conditions of certaintyEXPECTED VALUE OF

PERFECT INFORMATION

LO5 Compute and describe the expected value of perfect information.

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$1,900, and if the analyst asked $40 for the information, the investor would be assured of a $1,860 payoff, found by $1,900 � $40. Thus, it would be worthwhile for the investor to agree to this fee ($40) because the expected outcome ($1,860) would be greater than the expected value under conditions of uncertainty ($1,840). However, if his acquaintance wanted a fee of $100 for the service, the investor would realize only $1,800 on average, found by $1,900 � $100. Logically, the service would not be worth $100, because the investor could expect $1,840 on average without agreeing to this financial arrangement. Notice that the expected value of perfect information ($60) is the same as the minimum of the expected regrets (Table 20–6). That is not an accident.

The output for the investment example using the Excel system is shown above. The expected payoff and the expected opportunity loss are the same as reported in Table 20–3 and Table 20–6, respectively. We used the Excel Formula Bar (the fx key) to find the expected values. In a larger problem, this would be helpful. The cal- culations in the preceding investment example were kept at a minimum to emphasize the new terms and the decision-making procedures. When the number of decision alternatives and the number of states of nature become large, a computer package or spreadsheet is recommended.

20.6 Sensitivity Analysis In the foregoing stock selection situation, the set of probabilities applied to the pay- off values was derived from historical experience with similar market conditions. Objections may be voiced, however, that future market behavior may be different from past experiences. Despite these differences, the rankings of the decision alternatives are frequently not highly sensitive to changes within a plausible range. As an exam- ple, suppose the investor’s brother believes that instead of a 60 percent chance of a market rise and a 40 percent chance of a decline, the reverse is true—that is, there is a .40 probability that the stock market will rise and a .60 probability of a decline. Further, the investor’s cousin thinks the probability of a market rise is .50 and that of a decline is .50. A comparison of the original expected payoffs (left column), the expected payoffs for the set of probabilities suggested by the investor’s brother (cen- ter column), and those cited by the cousin (right column) is shown in Table 20–8. The decision is the same in all three cases—purchase Kayser Chemicals.

Expected payoffs are not highly sensitive.

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A comparison of the three sets of expected payoffs in Table 20–8 reveals the best alternative would still be to purchase Kayser Chemicals. As might be expected, there are some differences in the expected future values for each of the three stocks.

If there are drastic changes in the assigned probabilities, the expected values and the optimal decision may change. As an example, suppose the prognostication for a market rise was .20 and for a market decline .80. The expected payoffs would be as shown in Table 20–9. In the long run, the best alternative would be to buy Rim Homes stock. Thus, sensitivity analysis lets you see how accurate the proba- bility estimates need to be in order to feel comfortable with your choice.

Expected Payoffs

Historical Experience Brother’s Estimate Cousin’s Estimate (probability of .60 (probability of .40 (probability of .50

Purchase rise, .40 decline) rise, .60 decline) rise, .50 decline)

Kayser Chemicals $1,840 $1,560 $1,700 Rim Homes 1,760 1,540 1,650 Texas Electronics 1,600 1,450 1,525

TABLE 20–8 Expected Payoffs for Three Sets of Probabilities

Self-Review 20–4 Referring to Table 20–8, verify that: (a) The expected payoff for Texas Electronics for the brother’s set of probabilities is $1,450. (b) The expected payoff for Kayser Chemicals for the cousin’s set of probabilities is $1,700.

Expected Purchase Payoff

Kayser Chemicals $1,280 Rim Homes 1,320 Texas Electronics 1,300

TABLE 20–9 Expected Values for Purchasing the Three Stocks

Self-Review 20–5 Is there any choice of probabilities for which the best alternative would be to purchase Texas Electronics stock? (Hint: This can be arrived at algebraically or by using a trial-and- error method. Try a somewhat extreme probability for a market rise.)

Exercises 7. Refer to Exercises 1, 3, and 5. Compute the expected value of perfect information. 8. Refer to Exercises 2, 4, and 6. Compute the expected value of perfect information. 9. Refer to Exercise 1. Revise the probabilities as follows: P(S1) � .50, P(S2) � .20, and

P(S3) � .30. Does this change the decision? 10. Refer to Exercise 2. Reverse the probabilities; that is, let P(S1) � .30 and P(S2) � .70. Does

this alter your decision?

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20.7 Decision Trees An analytic tool introduced in Chapter 5 that is also useful for studying a decision situation is a decision tree. It is a picture of all the possible courses of action and the consequent possible outcomes. A box is used to indicate the point at which a decision must be made, and the branches going out from the box indicate the alter- natives under consideration. Referring to Chart 20–1, on the left is the box with three branches radiating from it, representing the acts of purchasing Kayser Chemicals, Rim Homes, or Texas Electronics.

Decision tree: A picture of all possible outcomes

Decision tree shows Kayser Chemicals is the best buy

CHART 20–1 Decision Tree for the Investor’s Decision

$2,400

$1,000

$2,200

$1,100

$1,900

$1,150

Market rise (.6

0) $1,840

1 Market drop (.40)

Market rise (.6

0) $1,760

Market drop (.40)

Market rise (.6

0) $1,600

Market drop (.40)

Buy Ka

yse r

Buy Texas

Buy Rim 2

$1,840

The three nodes, or circles, numbered 1, 2, and 3, represent the expected pay- off of each of the three stocks. The branches going out to the right of the nodes show the chance events (market rise or decline) and their corresponding probabilities in parentheses. The numbers at the extreme ends of the branches are the estimated future values of ending the decision process at those points. This is sometimes called the conditional payoff to denote that the payoff depends on a particular choice of action and a particular chance outcome. Thus, if the investor purchased Rim Homes stock and the market rose, the conditional value of the stock would be $2,200.

After the decision tree has been constructed, the best decision strategy can be found by what is termed backward induction. For example, suppose the investor is considering the act of purchasing Texas Electronics. Starting at the lower right in Chart 20–1 with the anticipated payoff given a market rise ($1,900) versus a market decline ($1,150) and going backward (moving left), the appropriate probabilities are applied to give the expected payoff of $1,600 [found by .60($1,900) � .40($1,150)]. The investor would mark the expected value of $1,600 above circled node 3 as shown in Chart 20–1. Similarly, the investor would determine the expected values for Rim Homes and Kayser Chemicals.

Assuming the investor wants to maximize the expected value of his stock pur- chase, $1,840 would be preferred over $1,760 or $1,600. Continuing to the left toward the box, the investor would draw a double bar across branches represent- ing the two alternatives he rejected (numbers 2 and 3, representing Rim Homes and Texas Electronics). The unmarked branch that leads to the box is clearly the best action to follow, namely, buy Kayser Chemicals stock.

The expected value under conditions of certainty can also be portrayed via a decision tree analysis (see Chart 20–2). Recall that under conditions of certainty the

LO6 Organize possible outcomes into a decision tree and interpret the result.

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investor would know before the stock is purchased whether the stock market would rise or decline. Hence, he would purchase Kayser Chemicals in a rising market and Texas Electronics in a falling market, and the expected payoff would be $1,900. Again, backward induction would be used to arrive at the expected payoff of $1,900.

CHART 20–2 Decision Tree Given Perfect Information

Mark et ris

e (.60 )

Market drop (.40)

$1,900

$1,100

$1,000

$1,150

Buy K ayser

$1,150 Buy Texas

Buy Rim

$2,200

$2,400

$1,900

Buy K ayser

$2,400

Buy Texas

Buy Rim

The monetary difference based on perfect information in Chart 20–2 and the de- cision based on imperfect information in Chart 20–1 is $60, found by $1,900 � $1,840. Recall that the $60 is the expected value of perfect information.

Decision tree analysis provides an alternative way to perform the calculations presented earlier in the chapter. Some managers find these graphic sketches help them in following the decision logic.

If perfect information is available: Buy Kayser in rising market; buy Texas in declining market

Chapter Summary I. Statistical decision theory is concerned with making decisions from a set of alternatives.

A. The various courses of action are called the acts or alternatives. B. The uncontrollable future events are called the states of nature. Probabilities are usu-

ally assigned to the states of nature. C. The consequence of a particular decision alternative and state of nature is called the

payoff. D. All possible combinations of decision alternatives and states of nature result in a pay-

off table. II. There are several criteria for selecting the best decision alternative.

A. In the expected monetary value (EMV) criterion, the expected value for each decision alter- native is computed, and the optimal (largest if profits, smallest if costs) is selected.

B. An opportunity loss table can be developed. 1. An opportunity loss table is constructed by taking the difference between the opti-

mal decision for each state of nature and the other decision alternatives. 2. The difference between the optimal decision and any other decision is the

opportunity loss or regret due to making a decision other than the optimum. 3. The expected opportunity loss (EOL) is similar to the expected monetary value. The

opportunity loss is combined with the probabilities of the various states of nature for each decision alternative to determine the expected opportunity loss.

C. The strategy of maximizing the minimum gain is referred to as maximin. D. The strategy of maximizing the maximum gain is called maximax. E. The strategy that minimizes the maximum regret is designated minimax regret.

III. The expected value of perfect information (EVPI) is the difference between the best expected payoff under certainty and the best expected payoff under uncertainty.

IV. Sensitivity analysis examines the effects of various probabilities for the states of nature on the expected values.

V. Decision trees are useful for structuring the various alternatives. They present a picture of the various courses of action and the possible states of nature.

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Chapter Exercises 11. Blackbeard’s Phantom Fireworks is considering introducing two new bottle rockets. The

company can add both to the current line, neither, or just one of the two. The success of these products depends on consumers’ reactions. These reactions can be summarized as good, P(S1) � .30; fair, P(S2) � .50; or poor, P(S3) � .20. The company’s revenues, in thousands of dollars, are estimated in the following payoff table.

State of Nature

Decision S1 S2 S3 Neither 0 0 0 Product 1 only 125 65 30 Product 2 only 105 60 30 Both 220 110 40

a. Compute the expected monetary value for each decision. b. What decision would you recommend? c. Develop an opportunity loss table. d. Compute the expected opportunity loss for each decision. e. Compute the expected value of perfect information.

12. A financial executive for Fidelity Investments lives in Boston but frequently must travel to New York. She can go to New York by car, train, or plane. The cost for a plane ticket from Boston to New York is $200, and it is estimated that the trip takes 30 min- utes in good weather and 45 minutes in bad weather. The cost for a train ticket is $100, and the trip takes an hour in good weather and two hours in bad weather. The cost to drive her own car from Boston to New York is $40, and this trip takes three hours in good weather and four in bad weather. The executive places a value of $60 per hour on her time. The weather forecast is for a 60 percent chance of bad weather tomorrow. What decision would you recommend? (Hint: Set up a payoff table, and remember that you want to minimize costs.) What is the expected value of perfect information?

13. Thomas Manufacturing Company has $100,000 available to invest. John Thomas, the president and CEO of the company, would like to either expand his production, invest the money in stocks, or purchase a certificate of deposit from the bank. Of course, the unknown is whether the economy will continue at a high level or there will be a reces- sion. He estimates the likelihood of a recession at .20. Whether there is a recession or not, the certificate of deposit will result in a gain of 6 percent. If there is a recession, he predicts a 10 percent loss if he expands his production and a 5 percent loss if he invests in stocks. If there is not a recession, an expansion of production will result in a 15 per- cent gain, and stock investment will produce a 12 percent gain. a. What decision should he make if he uses the maximin strategy? b. What decision should John Thomas make if the maximax strategy is used? c. What decision would be made if he uses the expected monetary value criterion? d. What is the expected value of perfect information?

14. The quality assurance department at Malcomb Products must either inspect each part in a lot or not inspect any of the parts. That is, there are two decision alterna- tives: inspect all the parts or inspect none of the parts. The proportion of parts defec- tive in the lot, Sj, is known from historical data to assume the following probability distribution.

State of Nature, Probability,

Sj P(Sj)

.02 .70

.04 .20

.06 .10

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For the decision not to inspect any parts, the cost of quality is C � NSjK. For inspecting all the items in the lot, it is C � Nk, where:

N � 20 (lot size) K � $18.00 (the cost of finding a defect) k � $0.50 (the cost of sampling one item)

a. Develop a payoff table. b. What decision should be made if the expected value criterion is used? c. What is the expected value of perfect information?

15. Dude Ranches Incorporated was founded on the idea that many families in the eastern and southern areas of the United States do not have a sufficient amount of vacation time to drive to the dude ranches in the Southwest and Rocky Mountain areas for their vaca- tions. Various surveys indicated, however, that there was a considerable interest in this type of family vacation, which includes horseback riding, cattle drives, swimming, fish- ing, and the like. Dude Ranches Incorporated bought a large farm near several eastern cities and constructed a lake, a swimming pool, and other facilities. However, to build a number of family cottages on the ranch would have required a considerable investment. Further, the owners reasoned that most of this investment would be lost should the ranch– farm complex be a financial failure. Instead, they decided to enter into an agreement with Mobile Homes Manufacturing Company to supply a very attractive authentic ranch-type mobile home. Mobile Homes agreed to deliver a mobile home on Saturday for $300 a week. Mobile Homes must know early Saturday morning how many mobile homes Dude Ranches Incorporated wants for the forthcoming week. It has other customers to supply and can only deliver the homes on Saturday. This presents a problem. Dude Ranches will have some reservations by Saturday, but indications are that many families do not make them. Instead, they prefer to examine the facilities before making a decision. An analysis of the various costs involved indicated that $350 a week should be charged for a ranch home, including all privileges. The basic problem is how many mobile ranch homes to order from Mobile Homes each week. Should Dude Ranches Incorporated order 10 (considered the minimum), 11, 12, 13, or 14 (considered the maximum)?

Any decision made solely on the information in the payoff table would ignore, how- ever, the valuable experience that Dude Ranches Incorporated has acquired in the past four years (about 200 weeks) actually operating a dude ranch in the Southwest. Its records showed that it always had nine advance reservations. Also, it never had a demand for 15 or more cottages. The occupancy of 10, 11, 12, 13, or 14 ranch cottages, in part, represented families who drove in and inspected the facilities before renting. A frequency distribution showing the number of weeks in which 10, 11, . . . , 14 ranch cot- tages were rented during the 200-week period is found in the following table.

Number of Cottages Number of Rented Weeks

10 26 11 50 12 60 13 44 14 20

200

a. Construct a payoff table. b. Determine the expected payoffs, and arrive at a decision. c. Set up an opportunity loss table. d. Compute the expected opportunity losses, and arrive at a decision. e. Determine the expected value of perfect information.

16. The proprietor of the newly built White Mountain Ski and Swim Lodge has been consid- ering purchasing or leasing several snowmobiles for the use of guests. The owner found that other financial obligations made it impossible to purchase the machines. Snowmo- biles Incorporated (SI) will lease a machine for $20 a week, including any needed main- tenance. According to SI, the usual rental charge to the guests of the lodge is $25 a week. Gasoline and oil are extra. Snowmobiles Incorporated only leases a machine for the full season. The proprietor of Ski and Swim, knowing that leasing an excessive number

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of snowmobiles might cause a net loss for the lodge, investigated the records of other resort owners. The combined experience at several other lodges was found to be:

Number of Snowmobiles Number of Demanded by Guests Weeks

7 10 8 25 9 45

10 20

a. Design a payoff table. b. Compute the expected profits for leasing 7, 8, 9, and 10 snowmobiles based on the

cost of leasing of $20, the rental charge of $25, and the experience of other lodges. c. Which alternative is the most profitable? d. Design an opportunity loss table. e. Find the expected opportunity losses for leasing 7, 8, 9, and 10 snowmobiles. f. Which act would give the least expected opportunity loss? g. Determine the expected value of perfect information. h. Suggest a course of action to the proprietor of the Ski and Swim Lodge. Include in

your explanation the various figures, such as expected profit. 17. Casual Furniture World has had numerous inquiries regarding the availability of furniture

and equipment that could be rented for large outdoor summer parties. This includes such items as folding chairs and tables, a deluxe grill, propane gas, and lights. No rental equip- ment of this nature is available locally, and the management of the furniture store is con- sidering forming a subsidiary to handle rentals.

An investigation revealed that most people interested in renting wanted a complete group of party essentials (about 12 chairs, four tables, a deluxe grill, a bottle of propane gas, tongs, etc.). Management decided not to buy a large number of complete sets because of the financial risk involved. That is, if the demand for the rental groups was not as large as anticipated, a large financial loss might be incurred. Further, outright pur- chase would mean that the equipment would have to be stored during the off-season.

It was then discovered that a firm in Boston leased a complete party set for $560 for the summer season. This amounts to about $5 a day. In the promotional literature from the Boston firm, a rental fee of $15 was suggested. For each set rented, a profit of $10 would thus be earned. It was then decided to lease from the Boston firm, at least for the first season.

The Boston firm suggested that, based on the combined experience of similar rental firms in other cities, either 41, 42, 43, 44, 45, or 46 complete sets be leased for the sea- son. Based on this suggestion, management must now decide on the most profitable number of complete sets to lease for the season.

The leasing firm in Boston also made available some additional information gathered from several rental firms similar to the newly formed subsidiary. Note in the following table (which is based on the experience of the other rental firms) that for 360 days of the total of 6,000 days’ experience—or about 6 percent of the days—these rental firms rented out 41 complete party sets. On 10 percent of the days during a typical summer, they rented 42 complete sets, and so on.

Number of Number of Number of Number of Sets Rented Days Sets Rented Days

40 0 44 2,400 41 360 45 1,500 42 600 46 300 43 840 47 0

a. Construct a payoff table. (As a check figure, for the act of having 41 complete sets available and the event of renting 41, the payoff is $410.)

b. The expected daily profit for leasing 43 complete sets from the Boston firm is $426.70; for 45 sets, $431.70; and for 46 sets, $427.45. Organize these expected daily profits into a table, and complete the table by finding the expected daily profit for leasing 41, 42, and 44 sets from the Boston firm.

c. On the basis of the expected daily profit, what is the most profitable action to take? d. The expected opportunity loss for leasing 43 party sets from the Boston firm is $11.60;

for 45 sets, $6.60; for 46 sets, $10.85. Organize these into an expected opportunity

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loss table, and complete the table by computing the expected opportunity loss for 41, 42, and 44.

e. According to the expected opportunity loss table, what is the most profitable course of action to take? Does this agree with your decision for part (c)?

f. Determine the expected value of perfect information. Explain what it indicates in this problem.

18. Tim Waltzer owns and operates Waltzer’s Wrecks, a discount car rental agency near Cleveland Hopkins International Airport. He rents a wreck for $20 a day. He has an arrangement with Landrum Leasing to purchase used cars at $6,000 each. His cars receive only needed maintenance and, as a result, are worth only $2,000 at the end of the year of operation. Tim has decided to sell all his wrecks every year and purchase a complete set of wrecks from Landrum Leasing.

His clerk-accountant provided him with a probability distribution with respect to the number of cars rented per day.

Numbers of Cars Rented per Day

20 21 22 23

Probability .10 .20 .50 .20

Tim is an avid golfer and tennis player. He is either on the golf course on weekends or playing tennis indoors. Thus, his car rental agency is only open weekdays. Also, he closes for two weeks during the summer and goes on a golfing tour.

The clerk-accountant estimated that it cost $1.50 per car rental for minimal mainte- nance and cleaning. a. How many cars should he purchase to maximize profit? b. What is the expected value of perfect information?

19. You sign up for a cell phone plan and are presented with this chart showing how your plan “automatically adjusts” to the minutes you use each month. For example: If you select Option 1 and you use 700 minutes the first month, you’ll only pay $79.99. If your usage then goes down to 200 minutes the second month, you’ll only pay $29.99. You guess your monthly usage will be 100, 300, 500, or 700 anytime minutes. Assume the proba- bilities for each event are the same.

Option 1— Starting at $29.99 per Month

Anytime Minutes Cost

0–200 $29.99 201–700 $5 for each 50 minutes Above 700 Additional anytime minutes only 10¢ each

Option 2 — Starting at $34.99 per Month

Anytime Minutes Cost

0–400 $34.99 401–900 $5 for each 50 minutes Above 900 Additional anytime minutes only 10¢ each

Option 3 — Starting at $59.99 per Month

Anytime Minutes Cost

0–1,000 $59.99 1,001–1,500 $5 for each 50 minutes Above 1,500 Additional anytime minutes only 10¢ each

a. Create a payoff (cost) table for this decision. b. Using the expected monetary value principle, which decision would you suggest? c. Using the optimistic (maximax cost) approach, which decision would you suggest? d. Using the pessimistic (maximin cost) strategy, which decision would you suggest? e. Work out an opportunity loss table for this decision. f. Using the minimax regret strategy, which choice would you suggest? g. What is the expected value of perfect information?

20. You’re about to drive to New York. If your car’s engine is out of tune, your gas cost will increase by $100. Having the engine tested will cost $20. If it’s out of tune, repairs will cost $60. Before testing, the probability is 30 percent that the engine is out of tune. What should you do?

Lin01803_ch20_001-018.qxd 11/11/10 11:26 AM Page 20-17

20-18 Chapter 20

Chapter 20 Answers to Self-Review

20–1 20–4 a. Probability Expected

Event Payoff of Event Value

Market rise $2,200 .60 $1,320 Market decline 1,100 .40 440

$1,760

20–2 a. Suppose the investor purchased Rim Homes stock, and the value of the stock in a bear market dropped to $1,100 as anticipated (Table 20–1). Instead, had the investor purchased Texas Electronics and the market declined, the value of the Texas Electronics stock would be $1,150. The difference of $50, found by $1,150 � $1,100, represents the investor’s regret for buying Rim Homes stock.

b. Suppose the investor purchased Texas Electronics stock, and then a bull market developed. The stock rose to $1,900, as anticipated (Table 20–1). However, had the investor bought Kayser Chemicals stock and the market value increased to $2,400 as anticipated, the difference of $500 represents the extra profit the investor could have made by purchasing Kayser Chemicals stock.

20–3 Expected

Probability Opportunity Event Payoff of Event Value

Market rise $500 .60 $300 Market decline 0 .40 0

$300

20–5 For probabilities of a market rise (or decline) down to .333, Kayser Chemicals stock would provide the largest expected profit. For probabilities .333 to .143, Rim Homes would be the best buy. For .143 and below, Texas Electronics would give the largest expected profit. Algebraic solutions:

Kayser: 2,400p � (1 � p)1,000 Rim: 2,200p � (1 � p)1,100

1,400p � 1,000 � 1,100p � 1,100 p � .333

Rim: 2,200p � (1 � p)1,100 Texas: 1,900p � (1 � p)1,150

1,100p � 1,100 � 750p � 1,150 p � .143

Probability Expected Event Payoff of Event Value

Market rise $1,900 .40 $ 760 Market decline 1,150 .60 690

$1,450

Probability Expected Event Payoff of Event Value

Market rise $2,400 .50 $1,200 Market decline 1,000 .50 500

$1,700

Lin01803_ch20_001-018.qxd 11/11/10 11:26 AM Page 20-18

Appendixes

APPENDIX A: DATA SETS

A.1 Data Set 1—Goodyear, Arizona, Real Estate Sales Data

A.2 Data Set 2—Baseball Statistics, 2009 Season

A.3 Data Set 3—Buena School District Bus Data

A.4 Applewood Auto Group

A.5 Banking Data Set—Century National Bank Case

APPENDIX B: TABLES

B.1 Areas under the Normal Curve

B.2 Student’s t Distribution

B.3 Critical Values of Chi-Square

B.4 Critical Values of the F Distribution

B.5 Poisson Distribution

B.6 Table of Random Numbers

B.7 Wilcoxon T Values

B.8 Factors for Control Charts

B.9 Binomial Probability Distribution

B.10 Critical Values for the Durbin–Watson d Statistic

APPENDIX C:

Answers to Odd-Numbered Chapter Exercises and Review Exercises

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Appendix A: Data Sets A.1 Data Set 1—Goodyear, Arizona, Real Estate Sales Data

Variables x1 � Selling price in $000 x2 � Number of bedrooms x3 � Size of the home in square feet x4 � Pool (1 � yes, 0 � no) x5 � Distance from the center of the city in miles x6 � Township x7 � Garage attached (1 � yes, 0 � no) x8 � Number of bathrooms 105 homes sold

x1 x2 x3 x4 x5 x6 x7 x8 263.1 4 2,300 0 17 5 1 2.0 182.4 4 2,100 1 19 4 0 2.0 242.1 3 2,300 1 12 3 0 2.0 213.6 2 2,200 1 16 2 0 2.5 139.9 2 2,100 1 28 1 0 1.5

245.4 2 2,100 0 12 1 1 2.0 327.2 6 2,500 1 15 3 1 2.0 271.8 2 2,100 1 9 2 1 2.5 221.1 3 2,300 0 18 1 0 1.5 266.6 4 2,400 1 13 4 1 2.0

292.4 4 2,100 1 14 3 1 2.0 209.0 2 1,700 1 8 4 1 1.5 270.8 6 2,500 1 7 4 1 2.0 246.1 4 2,100 1 18 3 1 2.0 194.4 2 2,300 1 11 3 0 2.0

281.3 3 2,100 1 16 2 1 2.0 172.7 4 2,200 0 16 3 0 2.0 207.5 5 2,300 0 21 4 0 2.5 198.9 3 2,200 0 10 4 1 2.0 209.3 6 1,900 0 15 4 1 2.0

252.3 4 2,600 1 8 4 1 2.0 192.9 4 1,900 0 14 2 1 2.5 209.3 5 2,100 1 20 5 0 1.5 345.3 8 2,600 1 9 4 1 2.0 326.3 6 2,100 1 11 5 1 3.0

173.1 2 2,200 0 21 5 1 1.5 187.0 2 1,900 1 26 4 0 2.0 257.2 2 2,100 1 9 4 1 2.0 233.0 3 2,200 1 14 3 1 1.5 180.4 2 2,000 1 11 5 0 2.0

234.0 2 1,700 1 19 3 1 2.0 207.1 2 2,000 1 11 5 1 2.0 247.7 5 2,400 1 16 2 1 2.0 166.2 3 2,000 0 16 2 1 2.0 177.1 2 1,900 1 10 5 1 2.0

754

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755

Appendix A A.1 Data Set 1—Goodyear, Arizona, Real Estate Sales Data (continued )

x1 x2 x3 x4 x5 x6 x7 x8 182.7 4 2,000 0 14 4 0 2.5 216.0 4 2,300 1 19 2 0 2.0 312.1 6 2,600 1 7 5 1 2.5 199.8 3 2,100 1 19 3 1 2.0 273.2 5 2,200 1 16 2 1 3.0

206.0 3 2,100 0 9 3 0 1.5 232.2 3 1,900 0 16 1 1 1.5 198.3 4 2,100 0 19 1 1 1.5 205.1 3 2,000 0 20 4 0 2.0 175.6 4 2,300 0 24 4 1 2.0

307.8 3 2,400 0 21 2 1 3.0 269.2 5 2,200 1 8 5 1 3.0 224.8 3 2,200 1 17 1 1 2.5 171.6 3 2,000 0 16 4 0 2.0 216.8 3 2,200 1 15 1 1 2.0

192.6 6 2,200 0 14 1 0 2.0 236.4 5 2,200 1 20 3 1 2.0 172.4 3 2,200 1 23 3 0 2.0 251.4 3 1,900 1 12 2 1 2.0 246.0 6 2,300 1 7 3 1 3.0

147.4 6 1,700 0 12 1 0 2.0 176.0 4 2,200 1 15 1 1 2.0 228.4 3 2,300 1 17 5 1 1.5 166.5 3 1,600 0 19 3 0 2.5 189.4 4 2,200 1 24 1 1 2.0

312.1 7 2,400 1 13 3 1 3.0 289.8 6 2,000 1 21 3 1 3.0 269.9 5 2,200 0 11 4 1 2.5 154.3 2 2,000 1 13 2 0 2.0 222.1 2 2,100 1 9 5 1 2.0

209.7 5 2,200 0 13 2 1 2.0 190.9 3 2,200 0 18 3 1 2.0 254.3 4 2,500 0 15 3 1 2.0 207.5 3 2,100 0 10 2 0 2.0 209.7 4 2,200 0 19 2 1 2.0

294.0 2 2,100 1 13 2 1 2.5 176.3 2 2,000 0 17 3 0 2.0 294.3 7 2,400 1 8 4 1 2.0 224.0 3 1,900 0 6 1 1 2.0 125.0 2 1,900 1 18 4 0 1.5

236.8 4 2,600 0 17 5 1 2.0 164.1 4 2,300 1 19 4 0 2.0 217.8 3 2,500 1 12 3 0 2.0 192.2 2 2,400 1 16 2 0 2.5 125.9 2 2,400 1 28 1 0 1.5

(continued )

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Appendix A A.1 Data Set 1—Goodyear, Arizona, Real Estate Sales Data (concluded)

x1 x2 x3 x4 x5 x6 x7 x8 220.9 2 2,300 0 12 1 1 2.0 294.5 6 2,700 1 15 3 1 2.0 244.6 2 2,300 1 9 2 1 2.5 199.0 3 2,500 0 18 1 0 1.5 240.0 4 2,600 1 13 4 1 2.0

263.2 4 2,300 1 14 3 1 2.0 188.1 2 1,900 1 8 4 1 1.5 243.7 6 2,700 1 7 4 1 2.0 221.5 4 2,300 1 18 3 1 2.0 175.0 2 2,500 1 11 3 0 2.0

253.2 3 2,300 1 16 2 1 2.0 155.4 4 2,400 0 16 3 0 2.0 186.7 5 2,500 0 21 4 0 2.5 179.0 3 2,400 0 10 4 1 2.0 188.3 6 2,100 0 15 4 1 2.0

227.1 4 2,900 1 8 4 1 2.0 173.6 4 2,100 0 14 2 1 2.5 188.3 5 2,300 1 20 5 0 1.5 310.8 8 2,900 1 9 4 1 2.0 293.7 6 2,400 1 11 5 1 3.0

179.0 3 2,400 1 8 4 1 2.0 188.3 6 2,100 0 14 2 1 2.5 227.1 4 2,900 1 20 5 0 1.5 173.6 4 2,100 1 9 4 1 2.0 188.3 5 2,300 1 11 5 1 3.0

756

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757

Appendix A A.2 Data Set 2—Baseball Statistics, 2009 Season

Variables x1 � Team x2 � League (American � 1, National � 0) x3 � Built (year stadium was built) x4 � Size (stadium capacity) x5 � Salary (total 2009 team salary, $ millions) x6 � Wins x7 � Attendance (total for team in millions) x8 � BA (team batting average) x9 � ERA (team earned run average)

x10 � HR (team home runs) x11 � Errors (team errors) x12 � SB (team stolen bases) x13 � Year x14 � Average player salary ($)

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759

Appendix A A.3 Data Set 3—Buena School District Bus Data

Variables x1 � Bus number x2 � Maintenance cost ($) x3 � Age x4 � Miles x5 � Bus type (diesel or gasoline) x6 � Bus manufacturer (Bluebird, Keiser, Thompson) x7 � Passengers

Bus Number, Maintenance Cost, Age, Miles, Bus Type, Bus Manufacturer, Passengers, x1 x2 x3 x4 x5 x6 x7

135 329 7 853 Diesel Bluebird 55 120 503 10 883 Diesel Keiser 42 200 505 10 822 Diesel Bluebird 55

40 466 10 865 Gasoline Bluebird 55 427 359 7 751 Gasoline Keiser 55

759 546 8 870 Diesel Keiser 55 10 427 5 780 Gasoline Keiser 14

880 474 9 857 Gasoline Keiser 55 481 382 3 818 Gasoline Keiser 6 387 422 8 869 Gasoline Bluebird 55

326 433 9 848 Diesel Bluebird 55 861 474 10 845 Gasoline Bluebird 55 122 558 10 885 Gasoline Bluebird 55 156 561 12 838 Diesel Thompson 55 887 357 8 760 Diesel Bluebird 6

686 329 3 741 Diesel Bluebird 55 490 497 10 859 Gasoline Bluebird 55 370 459 8 826 Gasoline Keiser 55 464 355 3 806 Gasoline Bluebird 55 875 489 9 858 Diesel Bluebird 55

883 436 2 785 Gasoline Bluebird 55 57 455 7 828 Diesel Bluebird 55

482 514 11 980 Gasoline Bluebird 55 704 503 8 857 Diesel Bluebird 55 989 380 9 803 Diesel Keiser 55

731 432 6 819 Diesel Bluebird 42 75 478 6 821 Diesel Bluebird 55

162 406 3 798 Gasoline Keiser 55 732 471 9 815 Diesel Keiser 42 751 444 2 757 Diesel Keiser 14

600 493 10 1008 Diesel Bluebird 55 948 452 9 831 Diesel Keiser 42 358 461 6 849 Diesel Bluebird 55 833 496 8 839 Diesel Thompson 55 692 469 8 812 Diesel Bluebird 55

(continued )

Lin01803_appa_753-763.qxd 11/12/10 7:06 AM Page 759

Appendix A A.3 Data Set 3—Buena School District Bus Data (concluded)

760

Bus Number, Maintenance Cost, Age, Miles, Bus Type, Bus Manufacturer, Passengers, x1 x2 x3 x4 x5 x6 x7 61 442 9 809 Diesel Keiser 55 9 414 4 864 Gasoline Keiser 55

314 459 11 859 Diesel Thompson 6 396 457 2 815 Diesel Thompson 55 365 462 6 799 Diesel Keiser 55

398 570 9 844 Diesel Thompson 14 43 439 9 832 Gasoline Bluebird 55

500 369 5 842 Gasoline Bluebird 55 279 390 2 792 Diesel Bluebird 55 693 469 9 775 Gasoline Keiser 55

884 381 9 882 Diesel Bluebird 55 977 501 7 874 Diesel Bluebird 55

38 432 6 837 Gasoline Keiser 14 725 392 5 774 Diesel Bluebird 55 982 441 1 823 Diesel Bluebird 55

724 448 8 790 Diesel Keiser 42 603 468 4 800 Diesel Keiser 14 168 467 7 827 Gasoline Thompson 55

45 478 6 830 Diesel Keiser 55 754 515 14 895 Diesel Keiser 14

39 411 6 804 Gasoline Bluebird 55 671 504 8 866 Gasoline Thompson 55 418 504 9 842 Diesel Bluebird 55 984 392 8 851 Diesel Bluebird 55 953 423 10 835 Diesel Bluebird 55

507 410 7 866 Diesel Bluebird 55 540 529 4 846 Gasoline Bluebird 55 695 477 2 802 Diesel Bluebird 55 193 540 11 847 Diesel Thompson 55 321 450 6 856 Diesel Bluebird 6

918 390 5 799 Diesel Bluebird 55 101 424 4 827 Diesel Bluebird 55 714 433 7 817 Diesel Bluebird 42 678 428 7 842 Diesel Keiser 55 768 494 7 815 Diesel Bluebird 42

29 396 6 784 Gasoline Bluebird 55 554 458 4 817 Diesel Bluebird 14 767 493 6 816 Diesel Keiser 55 699 475 9 816 Gasoline Bluebird 55 954 476 10 827 Diesel Bluebird 42

705 403 4 806 Diesel Keiser 42 660 337 6 819 Gasoline Bluebird 55 520 492 10 836 Diesel Bluebird 55 814 426 4 757 Diesel Bluebird 55 353 449 4 817 Gasoline Keiser 55

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Appendix A A.4 Data Set 4—Applewood Auto Group

x1 � Age—the age of the buyer at the time of the purchase x2 � Profit—the amount earned by the dealership on the sale of each vehicle x3 � Location—the dealership where the vehicle was purchased x4 � Vehicle type—SUV, sedan, compact, hybrid, or truck x5 � Previous—the number of vehicles previously purchased at any of the four

Applewood dealerships by the customer

761

(continued )

Age Profit Location Vehicle-Type Previous x1 x2 x3 x4 x5 21 $1,387 Tionesta Sedan 0 23 1,754 Sheffield SUV 1 24 1,817 Sheffield Hybrid 1 25 1,040 Sheffield Compact 0 26 1,273 Kane Sedan 1 27 1,529 Sheffield Sedan 1 27 3,082 Kane Truck 0 28 1,951 Kane SUV 1 28 2,692 Tionesta Compact 0 29 1,206 Sheffield Sedan 0 29 1,342 Kane Sedan 2 30 443 Kane Sedan 3 30 754 Olean Sedan 2 30 1,621 Sheffield Truck 1 31 870 Tionesta Sedan 1 31 1,174 Kane Truck 0 31 1,412 Sheffield Sedan 1 31 1,809 Tionesta Sedan 1 31 2,415 Kane Sedan 0 32 1,546 Sheffield Truck 3 32 2,148 Tionesta SUV 2 32 2,207 Sheffield Compact 0 32 2,252 Tionesta SUV 0 33 1,428 Kane SUV 2 33 1,889 Olean SUV 1 34 1,166 Olean Sedan 1 34 1,320 Tionesta Sedan 1 34 2,265 Olean Sedan 0 35 1,323 Olean Sedan 2 35 1,761 Kane Sedan 1 35 1,919 Tionesta SUV 1 36 2,357 Kane SUV 2 36 2,866 Kane Sedan 1 37 732 Olean SUV 1 37 1,464 Olean Sedan 3 37 1,626 Tionesta Compact 4 37 1,761 Olean SUV 1 37 1,915 Tionesta SUV 2 37 2,119 Kane Hybrid 1 38 1,766 Sheffield SUV 0 38 2,201 Sheffield Truck 2 39 996 Kane Compact 2 39 2,813 Tionesta SUV 0 40 323 Kane Sedan 0 40 352 Sheffield Compact 0 40 482 Olean Sedan 1 40 1,144 Tionesta Truck 0

Age Profit Location Vehicle-Type Previous x1 x2 x3 x4 x5 40 $1,485 Sheffield Compact 0 40 1,509 Kane SUV 2 40 1,638 Sheffield Sedan 0 40 1,961 Sheffield Sedan 1 40 2,127 Olean Truck 0 40 2,430 Tionesta Sedan 1 41 1,704 Sheffield Sedan 1 41 1,876 Kane Sedan 2 41 2,010 Tionesta Sedan 1 41 2,165 Tionesta SUV 0 41 2,231 Tionesta SUV 2 41 2,389 Kane Truck 1 42 335 Olean SUV 1 42 963 Kane Sedan 0 42 1,298 Tionesta Sedan 1 42 1,410 Kane SUV 2 42 1,553 Tionesta Compact 0 42 1,648 Olean SUV 0 42 2,071 Kane SUV 0 42 2,116 Kane Compact 2 43 1,500 Tionesta Sedan 0 43 1,549 Kane SUV 2 43 2,348 Tionesta Sedan 0 43 2,498 Tionesta SUV 1 44 294 Kane SUV 1 44 1,115 Kane Truck 0 44 1,124 Tionesta Compact 2 44 1,532 Tionesta SUV 3 44 1,688 Kane Sedan 4 44 1,822 Kane SUV 0 44 1,897 Sheffield Compact 0 44 2,445 Kane SUV 0 44 2,886 Olean SUV 1 45 820 Kane Compact 1 45 1,266 Olean Sedan 0 45 1,741 Olean Compact 2 45 1,772 Olean Compact 1 45 1,932 Tionesta Sedan 1 45 2,350 Sheffield Compact 0 45 2,422 Kane Sedan 1 45 2,446 Olean Compact 1 46 369 Olean Sedan 1 46 978 Kane Sedan 1 46 1,238 Sheffield Compact 1 46 1,818 Kane SUV 0 46 1,824 Olean Truck 0 46 1,907 Olean Sedan 0

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762

Appendix A A.4 Data Set 4—Applewood Auto Group (concluded)

Age Profit Location Vehicle-Type Previous x1 x2 x3 x4 x5 46 $1,938 Kane Sedan 0 46 1,940 Kane Truck 3 46 2,197 Sheffield Sedan 1 46 2,646 Tionesta Sedan 2 47 1,461 Kane Sedan 0 47 1,731 Tionesta Compact 0 47 2,230 Tionesta Sedan 1 47 2,341 Sheffield SUV 1 47 3,292 Olean Sedan 2 48 1,108 Sheffield Sedan 1 48 1,295 Sheffield SUV 1 48 1,344 Sheffield SUV 0 48 1,906 Kane Sedan 1 48 1,952 Tionesta Compact 1 48 2,070 Kane SUV 1 48 2,454 Kane Sedan 1 49 1,606 Olean Compact 0 49 1,680 Kane SUV 3 49 1,827 Tionesta Truck 3 49 1,915 Tionesta SUV 1 49 2,084 Tionesta Sedan 0 49 2,639 Sheffield SUV 0 50 842 Kane SUV 0 50 1,963 Sheffield Sedan 1 50 2,059 Sheffield Sedan 1 50 2,338 Tionesta SUV 0 50 3,043 Kane Sedan 0 51 1,059 Kane SUV 1 51 1,674 Sheffield Sedan 1 51 1,807 Tionesta Sedan 1 51 2,056 Sheffield Hybrid 0 51 2,236 Tionesta SUV 2 51 2,928 Kane SUV 0 52 1,269 Tionesta Sedan 1 52 1,717 Sheffield SUV 3 52 1,797 Kane Sedan 1 52 1,955 Olean Hybrid 2 52 2,199 Tionesta SUV 0 52 2,482 Olean Compact 0 52 2,701 Sheffield SUV 0 52 3,210 Olean Truck 4 53 377 Olean SUV 1 53 1,220 Olean Sedan 0

Age Profit Location Vehicle-Type Previous x1 x2 x3 x4 x5 53 $1,401 Tionesta SUV 2 53 2,175 Olean Sedan 1 54 1,118 Sheffield Compact 1 54 2,584 Olean Compact 2 54 2,666 Tionesta Truck 0 54 2,991 Tionesta SUV 0 55 934 Sheffield Truck 1 55 2,063 Kane SUV 1 55 2,083 Sheffield Sedan 1 55 2,856 Olean Hybrid 1 55 2,989 Tionesta Compact 1 56 910 Sheffield SUV 0 56 1,536 Kane SUV 0 56 1,957 Sheffield SUV 1 56 2,240 Olean Sedan 0 56 2,695 Kane Sedan 2 57 1,325 Olean Sedan 1 57 2,250 Sheffield Sedan 2 57 2,279 Sheffield Hybrid 1 57 2,626 Sheffield Sedan 2 58 1,501 Sheffield Hybrid 1 58 1,752 Kane Sedan 3 58 2,058 Kane SUV 1 58 2,370 Tionesta Compact 0 58 2,637 Sheffield SUV 1 59 1,426 Sheffield Sedan 0 59 2,944 Olean SUV 2 60 2,147 Olean Compact 2 61 1,973 Kane SUV 3 61 2,502 Olean Sedan 0 62 783 Sheffield Hybrid 1 62 1,538 Olean Truck 1 63 2,339 Olean Compact 1 64 2,700 Kane Truck 0 65 2,222 Kane Truck 1 65 2,597 Sheffield Truck 0 65 2,742 Tionesta SUV 2 68 1,837 Sheffield Sedan 1 69 2,842 Kane SUV 0 70 2,434 Olean Sedan 4 72 1,640 Olean Sedan 1 72 1,821 Tionesta SUV 1 73 2,487 Olean Compact 4

Lin01803_appa_753-763.qxd 10/22/10 8:49 AM Page 762

x1 x2 x3 x4 x5 x6 1,756 13 4 0 1 2

748 9 2 1 0 1 1,501 10 1 0 0 1 1,831 10 4 0 1 3 1,622 14 6 0 1 4

1,886 17 3 0 1 1 740 6 3 0 0 3

1,593 10 8 1 0 1 1,169 6 4 0 0 4 2,125 18 6 0 0 2

1,554 12 6 1 0 3 1,474 12 7 1 0 1 1,913 6 5 0 0 1 1,218 10 3 1 0 1 1,006 12 4 0 0 1

2,215 20 3 1 0 4 137 7 2 0 0 3 167 5 4 0 0 4 343 7 2 0 0 1

2,557 20 7 1 0 4

2,276 15 4 1 0 3 1,494 11 2 0 1 1 2,144 17 3 0 0 3 1,995 10 7 0 0 2 1,053 8 4 1 0 3

1,526 8 4 0 1 2 1,120 8 6 1 0 3 1,838 7 5 1 1 3 1,746 11 2 0 0 2 1,616 10 4 1 1 2

x1 x2 x3 x4 x5 x6 1,958 6 2 1 0 2

634 2 7 1 0 4 580 4 1 0 0 1

1,320 4 5 1 0 1 1,675 6 7 1 0 2

789 8 4 0 0 4 1,735 12 7 0 1 3 1,784 11 5 0 0 1 1,326 16 8 0 0 3 2,051 14 4 1 0 4

1,044 7 5 1 0 1 1,885 10 6 1 1 2 1,790 11 4 0 1 3

765 4 3 0 0 4 1,645 6 9 0 1 4

32 2 0 0 0 3 1,266 11 7 0 0 4

890 7 1 0 1 1 2,204 14 5 0 0 2 2,409 16 8 0 0 2

1,338 14 4 1 0 2 2,076 12 5 1 0 2 1,708 13 3 1 0 1 2,138 18 5 0 1 4 2,375 12 4 0 0 2

1,455 9 5 1 1 3 1,487 8 4 1 0 4 1,125 6 4 1 0 2 1,989 12 3 0 1 2 2,156 14 5 1 0 2

Appendix A A.5 Banking Data Set—Century National Bank Case (Review Sections)

x1 � Account balance in $ x2 � Number of ATM transactions in the month x3 � Number of other bank services used x4 � Has a debit card (1 � yes, 0 � no) x5 � Receives interest on the account (1 � yes, 0 � no) x6 � City where banking is done 60 Accounts

763

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Appendix B: Tables B.1 Areas under the Normal Curve

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141 0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517 0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879

0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224 0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549 0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852 0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133 0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389

1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621 1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830 1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015 1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319

1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441 1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633 1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706 1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767

2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952 2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964 2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974 2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981 2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986

3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990

Example: If z = 1.96, then P(0 to z ) = 0.4750.

z 0 1.96

0.4750

764

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765

Appendix B B.2 Student’s t Distribution

0 0

1 2� ��

1 2

Confidence interval t–t 0

Left-tailed test –t t

Right-tailed test t

Two-tailed test t–t

Confidence Intervals, c 80% 90% 95% 98% 99% 99.9%

Level of Significance for One-Tailed Test, � df 0.10 0.05 0.025 0.01 0.005 0.0005

Level of Significance for Two-Tailed Test, � 0.20 0.10 0.05 0.02 0.01 0.001

1 3.078 6.314 12.706 31.821 63.657 636.619 2 1.886 2.920 4.303 6.965 9.925 31.599 3 1.638 2.353 3.182 4.541 5.841 12.924 4 1.533 2.132 2.776 3.747 4.604 8.610 5 1.476 2.015 2.571 3.365 4.032 6.869

6 1.440 1.943 2.447 3.143 3.707 5.959 7 1.415 1.895 2.365 2.998 3.499 5.408 8 1.397 1.860 2.306 2.896 3.355 5.041 9 1.383 1.833 2.262 2.821 3.250 4.781

10 1.372 1.812 2.228 2.764 3.169 4.587

11 1.363 1.796 2.201 2.718 3.106 4.437 12 1.356 1.782 2.179 2.681 3.055 4.318 13 1.350 1.771 2.160 2.650 3.012 4.221 14 1.345 1.761 2.145 2.624 2.977 4.140 15 1.341 1.753 2.131 2.602 2.947 4.073

16 1.337 1.746 2.120 2.583 2.921 4.015 17 1.333 1.740 2.110 2.567 2.898 3.965 18 1.330 1.734 2.101 2.552 2.878 3.922 19 1.328 1.729 2.093 2.539 2.861 3.883 20 1.325 1.725 2.086 2.528 2.845 3.850

21 1.323 1.721 2.080 2.518 2.831 3.819 22 1.321 1.717 2.074 2.508 2.819 3.792 23 1.319 1.714 2.069 2.500 2.807 3.768 24 1.318 1.711 2.064 2.492 2.797 3.745 25 1.316 1.708 2.060 2.485 2.787 3.725

26 1.315 1.706 2.056 2.479 2.779 3.707 27 1.314 1.703 2.052 2.473 2.771 3.690 28 1.313 1.701 2.048 2.467 2.763 3.674 29 1.311 1.699 2.045 2.462 2.756 3.659 30 1.310 1.697 2.042 2.457 2.750 3.646

31 1.309 1.696 2.040 2.453 2.744 3.633 32 1.309 1.694 2.037 2.449 2.738 3.622 33 1.308 1.692 2.035 2.445 2.733 3.611 34 1.307 1.691 2.032 2.441 2.728 3.601 35 1.306 1.690 2.030 2.438 2.724 3.591

Confidence Intervals, c 80% 90% 95% 98% 99% 99.9%

Level of Significance for One-Tailed Test, � df 0.10 0.05 0.025 0.01 0.005 0.0005

Level of Significance for Two-Tailed Test, � 0.20 0.10 0.05 0.02 0.01 0.001

36 1.306 1.688 2.028 2.434 2.719 3.582 37 1.305 1.687 2.026 2.431 2.715 3.574 38 1.304 1.686 2.024 2.429 2.712 3.566 39 1.304 1.685 2.023 2.426 2.708 3.558 40 1.303 1.684 2.021 2.423 2.704 3.551

41 1.303 1.683 2.020 2.421 2.701 3.544 42 1.302 1.682 2.018 2.418 2.698 3.538 43 1.302 1.681 2.017 2.416 2.695 3.532 44 1.301 1.680 2.015 2.414 2.692 3.526 45 1.301 1.679 2.014 2.412 2.690 3.520

46 1.300 1.679 2.013 2.410 2.687 3.515 47 1.300 1.678 2.012 2.408 2.685 3.510 48 1.299 1.677 2.011 2.407 2.682 3.505 49 1.299 1.677 2.010 2.405 2.680 3.500 50 1.299 1.676 2.009 2.403 2.678 3.496

51 1.298 1.675 2.008 2.402 2.676 3.492 52 1.298 1.675 2.007 2.400 2.674 3.488 53 1.298 1.674 2.006 2.399 2.672 3.484 54 1.297 1.674 2.005 2.397 2.670 3.480 55 1.297 1.673 2.004 2.396 2.668 3.476

56 1.297 1.673 2.003 2.395 2.667 3.473 57 1.297 1.672 2.002 2.394 2.665 3.470 58 1.296 1.672 2.002 2.392 2.663 3.466 59 1.296 1.671 2.001 2.391 2.662 3.463 60 1.296 1.671 2.000 2.390 2.660 3.460

61 1.296 1.670 2.000 2.389 2.659 3.457 62 1.295 1.670 1.999 2.388 2.657 3.454 63 1.295 1.669 1.998 2.387 2.656 3.452 64 1.295 1.669 1.998 2.386 2.655 3.449 65 1.295 1.669 1.997 2.385 2.654 3.447

66 1.295 1.668 1.997 2.384 2.652 3.444 67 1.294 1.668 1.996 2.383 2.651 3.442 68 1.294 1.668 1.995 2.382 2.650 3.439 69 1.294 1.667 1.995 2.382 2.649 3.437 70 1.294 1.667 1.994 2.381 2.648 3.435

(continued )

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Appendix B B.2 Student’s t Distribution (concluded )

766

Confidence Intervals, c 80% 90% 95% 98% 99% 99.9%

Level of Significance for One-Tailed Test, � df 0.10 0.05 0.025 0.01 0.005 0.0005

Level of Significance for Two-Tailed Test, � 0.20 0.10 0.05 0.02 0.01 0.001

71 1.294 1.667 1.994 2.380 2.647 3.433 72 1.293 1.666 1.993 2.379 2.646 3.431 73 1.293 1.666 1.993 2.379 2.645 3.429 74 1.293 1.666 1.993 2.378 2.644 3.427 75 1.293 1.665 1.992 2.377 2.643 3.425

76 1.293 1.665 1.992 2.376 2.642 3.423 77 1.293 1.665 1.991 2.376 2.641 3.421 78 1.292 1.665 1.991 2.375 2.640 3.420 79 1.292 1.664 1.990 2.374 2.640 3.418 80 1.292 1.664 1.990 2.374 2.639 3.416

81 1.292 1.664 1.990 2.373 2.638 3.415 82 1.292 1.664 1.989 2.373 2.637 3.413 83 1.292 1.663 1.989 2.372 2.636 3.412 84 1.292 1.663 1.989 2.372 2.636 3.410 85 1.292 1.663 1.988 2.371 2.635 3.409

86 1.291 1.663 1.988 2.370 2.634 3.407 87 1.291 1.663 1.988 2.370 2.634 3.406 88 1.291 1.662 1.987 2.369 2.633 3.405

Confidence Intervals, c 80% 90% 95% 98% 99% 99.9%

Level of Significance for One-Tailed Test, � df 0.10 0.05 0.025 0.01 0.005 0.0005

Level of Significance for Two-Tailed Test, � 0.20 0.10 0.05 0.02 0.01 0.001

89 1.291 1.662 1.987 2.369 2.632 3.403 90 1.291 1.662 1.987 2.368 2.632 3.402

91 1.291 1.662 1.986 2.368 2.631 3.401 92 1.291 1.662 1.986 2.368 2.630 3.399 93 1.291 1.661 1.986 2.367 2.630 3.398 94 1.291 1.661 1.986 2.367 2.629 3.397 95 1.291 1.661 1.985 2.366 2.629 3.396

96 1.290 1.661 1.985 2.366 2.628 3.395 97 1.290 1.661 1.985 2.365 2.627 3.394 98 1.290 1.661 1.984 2.365 2.627 3.393 99 1.290 1.660 1.984 2.365 2.626 3.392

100 1.290 1.660 1.984 2.364 2.626 3.390

120 1.289 1.658 1.980 2.358 2.617 3.373 140 1.288 1.656 1.977 2.353 2.611 3.361 160 1.287 1.654 1.975 2.350 2.607 3.352 180 1.286 1.653 1.973 2.347 2.603 3.345 200 1.286 1.653 1.972 2.345 2.601 3.340

� 1.282 1.645 1.960 2.326 2.576 3.291

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767

Appendix B B.3 Critical Values of Chi-Square

This table contains the values of that correspond to a specific right-tail area and specific number of degrees of freedom.

�20

Example: With 17 df and a .02 area in the upper tail, �2 � 30.995

�2

Right-Tail AreaDegrees of Freedom,

df 0.10 0.05 0.02 0.01

1 2.706 3.841 5.412 6.635 2 4.605 5.991 7.824 9.210 3 6.251 7.815 9.837 11.345 4 7.779 9.488 11.668 13.277 5 9.236 11.070 13.388 15.086

6 10.645 12.592 15.033 16.812 7 12.017 14.067 16.622 18.475 8 13.362 15.507 18.168 20.090 9 14.684 16.919 19.679 21.666

10 15.987 18.307 21.161 23.209

11 17.275 19.675 22.618 24.725 12 18.549 21.026 24.054 26.217 13 19.812 22.362 25.472 27.688 14 21.064 23.685 26.873 29.141 15 22.307 24.996 28.259 30.578

16 23.542 26.296 29.633 32.000 17 24.769 27.587 30.995 33.409 18 25.989 28.869 32.346 34.805 19 27.204 30.144 33.687 36.191 20 28.412 31.410 35.020 37.566

21 29.615 32.671 36.343 38.932 22 30.813 33.924 37.659 40.289 23 32.007 35.172 38.968 41.638 24 33.196 36.415 40.270 42.980 25 34.382 37.652 41.566 44.314

26 35.563 38.885 42.856 45.642 27 36.741 40.113 44.140 46.963 28 37.916 41.337 45.419 48.278 29 39.087 42.557 46.693 49.588 30 40.256 43.773 47.962 50.892

Lin01803_appb_764-781.qxd 10/22/10 8:52 AM Page 767

Appendix B B.4 Critical Values of the F Distribution at a 5 Percent Level of Significance

768

.05

Degrees of Freedom for the Numerator

1 2 3 4 5 6 7 8 9 10 12 15 20 24 30 40

1 161 200 216 225 230 234 237 239 241 242 244 246 248 249 250 251 2 18.5 19.0 19.2 19.2 19.3 19.3 19.4 19.4 19.4 19.4 19.4 19.4 19.4 19.5 19.5 19.5 3 10.1 9.55 9.28 9.12 9.01 8.94 8.89 8.85 8.81 8.79 8.74 8.70 8.66 8.64 8.62 8.59 4 7.71 6.94 6.59 6.39 6.26 6.16 6.09 6.04 6.00 5.96 5.91 5.86 5.80 5.77 5.75 5.72 5 6.61 5.79 5.41 5.19 5.05 4.95 4.88 4.82 4.77 4.74 4.68 4.62 4.56 4.53 4.50 4.46

6 5.99 5.14 4.76 4.53 4.39 4.28 4.21 4.15 4.10 4.06 4.00 3.94 3.87 3.84 3.81 3.77 7 5.59 4.74 4.35 4.12 3.97 3.87 3.79 3.73 3.68 3.64 3.57 3.51 3.44 3.41 3.38 3.34 8 5.32 4.46 4.07 3.84 3.69 3.58 3.50 3.44 3.39 3.35 3.28 3.22 3.15 3.12 3.08 3.04 9 5.12 4.26 3.86 3.63 3.48 3.37 3.29 3.23 3.18 3.14 3.07 3.01 2.94 2.90 2.86 2.83

10 4.96 4.10 3.71 3.48 3.33 3.22 3.14 3.07 3.02 2.98 2.91 2.85 2.77 2.74 2.70 2.66

11 4.84 3.98 3.59 3.36 3.20 3.09 3.01 2.95 2.90 2.85 2.79 2.72 2.65 2.61 2.57 2.53 12 4.75 3.89 3.49 3.26 3.11 3.00 2.91 2.85 2.80 2.75 2.69 2.62 2.54 2.51 2.47 2.43 13 4.67 3.81 3.41 3.18 3.03 2.92 2.83 2.77 2.71 2.67 2.60 2.53 2.46 2.42 2.38 2.34 14 4.60 3.74 3.34 3.11 2.96 2.85 2.76 2.70 2.65 2.60 2.53 2.46 2.39 2.35 2.31 2.27 15 4.54 3.68 3.29 3.06 2.90 2.79 2.71 2.64 2.59 2.54 2.48 2.40 2.33 2.29 2.25 2.20

16 4.49 3.63 3.24 3.01 2.85 2.74 2.66 2.59 2.54 2.49 2.42 2.35 2.28 2.24 2.19 2.15 17 4.45 3.59 3.20 2.96 2.81 2.70 2.61 2.55 2.49 2.45 2.38 2.31 2.23 2.19 2.15 2.10 18 4.41 3.55 3.16 2.93 2.77 2.66 2.58 2.51 2.46 2.41 2.34 2.27 2.19 2.15 2.11 2.06 19 4.38 3.52 3.13 2.90 2.74 2.63 2.54 2.48 2.42 2.38 2.31 2.23 2.16 2.11 2.07 2.03 20 4.35 3.49 3.10 2.87 2.71 2.60 2.51 2.45 2.39 2.35 2.28 2.20 2.12 2.08 2.04 1.99

21 4.32 3.47 3.07 2.84 2.68 2.57 2.49 2.42 2.37 2.32 2.25 2.18 2.10 2.05 2.01 1.96 22 4.30 3.44 3.05 2.82 2.66 2.55 2.46 2.40 2.34 2.30 2.23 2.15 2.07 2.03 1.98 1.94 23 4.28 3.42 3.03 2.80 2.64 2.53 2.44 2.37 2.32 2.27 2.20 2.13 2.05 2.01 1.96 1.91 24 4.26 3.40 3.01 2.78 2.62 2.51 2.42 2.36 2.30 2.25 2.18 2.11 2.03 1.98 1.94 1.89 25 4.24 3.39 2.99 2.76 2.60 2.49 2.40 2.34 2.28 2.24 2.16 2.09 2.01 1.96 1.92 1.87

30 4.17 3.32 2.92 2.69 2.53 2.42 2.33 2.27 2.21 2.16 2.09 2.01 1.93 1.89 1.84 1.79 40 4.08 3.23 2.84 2.61 2.45 2.34 2.25 2.18 2.12 2.08 2.00 1.92 1.84 1.79 1.74 1.69 60 4.00 3.15 2.76 2.53 2.37 2.25 2.17 2.10 2.04 1.99 1.92 1.84 1.75 1.70 1.65 1.59

120 3.92 3.07 2.68 2.45 2.29 2.18 2.09 2.02 1.96 1.91 1.83 1.75 1.66 1.61 1.55 1.50 � 3.84 3.00 2.60 2.37 2.21 2.10 2.01 1.94 1.88 1.83 1.75 1.67 1.57 1.52 1.46 1.39

De gr

ee s

of F

re ed

om fo

r t he

D en

om in

at or

Lin01803_appb_764-781.qxd 10/22/10 8:52 AM Page 768

769

Appendix B B.4 Critical Values of the F Distribution at a 1 Percent Level of Significance (concluded )

.01

Degrees of Freedom for the Numerator

1 2 3 4 5 6 7 8 9 10 12 15 20 24 30 40

1 4052 5000 5403 5625 5764 5859 5928 5981 6022 6056 6106 6157 6209 6235 6261 6287 2 98.5 99.0 99.2 99.2 99.3 99.3 99.4 99.4 99.4 99.4 99.4 99.4 99.4 99.5 99.5 99.5 3 34.1 30.8 29.5 28.7 28.2 27.9 27.7 27.5 27.3 27.2 27.1 26.9 26.7 26.6 26.5 26.4 4 21.2 18.0 16.7 16.0 15.5 15.2 15.0 14.8 14.7 14.5 14.4 14.2 14.0 13.9 13.8 13.7 5 16.3 13.3 12.1 11.4 11.0 10.7 10.5 10.3 10.2 10.1 9.89 9.72 9.55 9.47 9.38 9.29

6 13.7 10.9 9.78 9.15 8.75 8.47 8.26 8.10 7.98 7.87 7.72 7.56 7.40 7.31 7.23 7.14 7 12.2 9.55 8.45 7.85 7.46 7.19 6.99 6.84 6.72 6.62 6.47 6.31 6.16 6.07 5.99 5.91 8 11.3 8.65 7.59 7.01 6.63 6.37 6.18 6.03 5.91 5.81 5.67 5.52 5.36 5.28 5.20 5.12 9 10.6 8.02 6.99 6.42 6.06 5.80 5.61 5.47 5.35 5.26 5.11 4.96 4.81 4.73 4.65 4.57

10 10.0 7.56 6.55 5.99 5.64 5.39 5.20 5.06 4.94 4.85 4.71 4.56 4.41 4.33 4.25 4.17

11 9.65 7.21 6.22 5.67 5.32 5.07 4.89 4.74 4.63 4.54 4.40 4.25 4.10 4.02 3.94 3.86 12 9.33 6.93 5.95 5.41 5.06 4.82 4.64 4.50 4.39 4.30 4.16 4.01 3.86 3.78 3.70 3.62 13 9.07 6.70 5.74 5.21 4.86 4.62 4.44 4.30 4.19 4.10 3.96 3.82 3.66 3.59 3.51 3.43 14 8.86 6.51 5.56 5.04 4.69 4.46 4.28 4.14 4.03 3.94 3.80 3.66 3.51 3.43 3.35 3.27 15 8.68 6.36 5.42 4.89 4.56 4.32 4.14 4.00 3.89 3.80 3.67 3.52 3.37 3.29 3.21 3.13

16 8.53 6.23 5.29 4.77 4.44 4.20 4.03 3.89 3.78 3.69 3.55 3.41 3.26 3.18 3.10 3.02 17 8.40 6.11 5.18 4.67 4.34 4.10 3.93 3.79 3.68 3.59 3.46 3.31 3.16 3.08 3.00 2.92 18 8.29 6.01 5.09 4.58 4.25 4.01 3.84 3.71 3.60 3.51 3.37 3.23 3.08 3.00 2.92 2.84 19 8.18 5.93 5.01 4.50 4.17 3.94 3.77 3.63 3.52 3.43 3.30 3.15 3.00 2.92 2.84 2.76 20 8.10 5.85 4.94 4.43 4.10 3.87 3.70 3.56 3.46 3.37 3.23 3.09 2.94 2.86 2.78 2.69

21 8.02 5.78 4.87 4.37 4.04 3.81 3.64 3.51 3.40 3.31 3.17 3.03 2.88 2.80 2.72 2.64 22 7.95 5.72 4.82 4.31 3.99 3.76 3.59 3.45 3.35 3.26 3.12 2.98 2.83 2.75 2.67 2.58 23 7.88 5.66 4.76 4.26 3.94 3.71 3.54 3.41 3.30 3.21 3.07 2.93 2.78 2.70 2.62 2.54 24 7.82 5.61 4.72 4.22 3.90 3.67 3.50 3.36 3.26 3.17 3.03 2.89 2.74 2.66 2.58 2.49 25 7.77 5.57 4.68 4.18 3.85 3.63 3.46 3.32 3.22 3.13 2.99 2.85 2.70 2.62 2.54 2.45

30 7.56 5.39 4.51 4.02 3.70 3.47 3.30 3.17 3.07 2.98 2.84 2.70 2.55 2.47 2.39 2.30 40 7.31 5.18 4.31 3.83 3.51 3.29 3.12 2.99 2.89 2.80 2.66 2.52 2.37 2.29 2.20 2.11 60 7.08 4.98 4.13 3.65 3.34 3.12 2.95 2.82 2.72 2.63 2.50 2.35 2.20 2.12 2.03 1.94

120 6.85 4.79 3.95 3.48 3.17 2.96 2.79 2.66 2.56 2.47 2.34 2.19 2.03 1.95 1.86 1.76 � 6.63 4.61 3.78 3.32 3.02 2.80 2.64 2.51 2.41 2.32 2.18 2.04 1.88 1.79 1.70 1.59

De gr

ee s

of F

re ed

om fo

r t he

D en

om in

at or

Lin01803_appb_764-781.qxd 10/22/10 8:52 AM Page 769

Appendix B B.5 Poisson Distribution

�

x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

5 0.0000 0.0000 0.0000 0.0001 0.0002 0.0004 0.0007 0.0012 0.0020 6 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0002 0.0003 7 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

�

x 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0

0 0.3679 0.1353 0.0498 0.0183 0.0067 0.0025 0.0009 0.0003 0.0001 1 0.3679 0.2707 0.1494 0.0733 0.0337 0.0149 0.0064 0.0027 0.0011 2 0.1839 0.2707 0.2240 0.1465 0.0842 0.0446 0.0223 0.0107 0.0050 3 0.0613 0.1804 0.2240 0.1954 0.1404 0.0892 0.0521 0.0286 0.0150 4 0.0153 0.0902 0.1680 0.1954 0.1755 0.1339 0.0912 0.0573 0.0337

5 0.0031 0.0361 0.1008 0.1563 0.1755 0.1606 0.1277 0.0916 0.0607 6 0.0005 0.0120 0.0504 0.1042 0.1462 0.1606 0.1490 0.1221 0.0911 7 0.0001 0.0034 0.0216 0.0595 0.1044 0.1377 0.1490 0.1396 0.1171 8 0.0000 0.0009 0.0081 0.0298 0.0653 0.1033 0.1304 0.1396 0.1318 9 0.0000 0.0002 0.0027 0.0132 0.0363 0.0688 0.1014 0.1241 0.1318

10 0.0000 0.0000 0.0008 0.0053 0.0181 0.0413 0.0710 0.0993 0.1186 11 0.0000 0.0000 0.0002 0.0019 0.0082 0.0225 0.0452 0.0722 0.0970 12 0.0000 0.0000 0.0001 0.0006 0.0034 0.0113 0.0263 0.0481 0.0728 13 0.0000 0.0000 0.0000 0.0002 0.0013 0.0052 0.0142 0.0296 0.0504 14 0.0000 0.0000 0.0000 0.0001 0.0005 0.0022 0.0071 0.0169 0.0324

15 0.0000 0.0000 0.0000 0.0000 0.0002 0.0009 0.0033 0.0090 0.0194 16 0.0000 0.0000 0.0000 0.0000 0.0000 0.0003 0.0014 0.0045 0.0109 17 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0006 0.0021 0.0058 18 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0002 0.0009 0.0029 19 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0004 0.0014

20 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0002 0.0006 21 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0003 22 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001

770

Lin01803_appb_764-781.qxd 10/22/10 8:52 AM Page 770

771

Appendix B B.6 Table of Random Numbers

02711 08182 75997 79866 58095 83319 80295 79741 74599 84379 94873 90935 31684 63952 09865 14491 99518 93394 34691 14985 54921 78680 06635 98689 17306 25170 65928 87709 30533 89736 77640 97636 37397 93379 56454 59818 45827 74164 71666 46977 61545 00835 93251 87203 36759 49197 85967 01704 19634 21898

17147 19519 22497 16857 42426 84822 92598 49186 88247 39967 13748 04742 92460 85801 53444 65626 58710 55406 17173 69776 87455 14813 50373 28037 91182 32786 65261 11173 34376 36408 08999 57409 91185 10200 61411 23392 47797 56377 71635 08601 78804 81333 53809 32471 46034 36306 22498 19239 85428 55721

82173 26921 28472 98958 07960 66124 89731 95069 18625 92405 97594 25168 89178 68190 05043 17407 48201 83917 11413 72920 73881 67176 93504 42636 38233 16154 96451 57925 29667 30859 46071 22912 90326 42453 88108 72064 58601 32357 90610 32921 44492 19686 12495 93135 95185 77799 52441 88272 22024 80631

31864 72170 37722 55794 14636 05148 54505 50113 21119 25228 51574 90692 43339 65689 76539 27909 05467 21727 51141 72949 35350 76132 92925 92124 92634 35681 43690 89136 35599 84138 46943 36502 01172 46045 46991 33804 80006 35542 61056 75666 22665 87226 33304 57975 03985 21566 65796 72915 81466 89205

39437 97957 11838 10433 21564 51570 73558 27495 34533 57808 77082 47784 40098 97962 89845 28392 78187 06112 08169 11261 24544 25649 43370 28007 06779 72402 62632 53956 24709 06978 27503 15558 37738 24849 70722 71859 83736 06016 94397 12529 24590 24545 06435 52758 45685 90151 46516 49644 92686 84870

48155 86226 40359 28723 15364 69125 12609 57171 86857 31702 20226 53752 90648 24362 83314 00014 19207 69413 97016 86290 70178 73444 38790 53626 93780 18629 68766 24371 74639 30782 10169 41465 51935 05711 09799 79077 88159 33437 68519 03040 81084 03701 28598 70013 63794 53169 97054 60303 23259 96196

69202 20777 21727 81511 51887 16175 53746 46516 70339 62727 80561 95787 89426 93325 86412 57479 54194 52153 19197 81877 08199 26703 95128 48599 09333 12584 24374 31232 61782 44032 98883 28220 39358 53720 80161 83371 15181 11131 12219 55920 84568 69286 76054 21615 80883 36797 82845 39139 90900 18172

04269 35173 95745 53893 86022 77722 52498 84193 22448 22571 10538 13124 36099 13140 37706 44562 57179 44693 67877 01549 77843 24955 25900 63843 95029 93859 93634 20205 66294 41218 12034 94636 49455 76362 83532 31062 69903 91186 65768 55949 10524 72829 47641 93315 80875 28090 97728 52560 34937 79548

68935 76632 46984 61772 92786 22651 07086 89754 44143 97687 89450 65665 29190 43709 11172 34481 95977 47535 25658 73898 90696 20451 24211 97310 60446 73530 62865 96574 13829 72226 49006 32047 93086 00112 20470 17136 28255 86328 07293 38809 74591 87025 52368 59416 34417 70557 86746 55809 53628 12000

06315 17012 77103 00968 07235 10728 42189 33292 51487 64443 62386 09184 62092 46617 99419 64230 95034 85481 07857 42510 86848 82122 04028 36959 87827 12813 08627 80699 13345 51695 65643 69480 46598 04501 40403 91408 32343 48130 49303 90689 11084 46534 78957 77353 39578 77868 22970 84349 09184 70603

Lin01803_appb_764-781.qxd 10/22/10 8:52 AM Page 771

2�

.15 .10 .05 .04 .03 .02 .01

�

n .075 .050 .025 .020 .015 .010 .005

4 0 5 1 0 6 2 2 0 0 7 4 3 2 1 0 0 8 7 5 3 3 2 1 0

9 9 8 5 5 4 3 1 10 12 10 8 7 6 5 3 11 16 13 10 9 8 7 5 12 19 17 13 12 11 9 7 13 24 21 17 16 14 12 9

14 28 25 21 19 18 15 12 15 33 30 25 23 21 19 15 16 39 35 29 28 26 23 19 17 45 41 34 33 30 27 23 18 51 47 40 38 35 32 27

19 58 53 46 43 41 37 32 20 65 60 52 50 47 43 37 21 73 67 58 56 53 49 42 22 81 75 65 63 59 55 48 23 89 83 73 70 66 62 54

24 98 91 81 78 74 69 61 25 108 100 89 86 82 76 68 26 118 110 98 94 90 84 75 27 128 119 107 103 99 92 83 28 138 130 116 112 108 101 91

29 150 140 126 122 117 110 100 30 161 151 137 132 127 120 109 31 173 163 147 143 137 130 118 32 186 175 159 154 148 140 128 33 199 187 170 165 159 151 138

34 212 200 182 177 171 162 148 35 226 213 195 189 182 173 159 40 302 286 264 257 249 238 220 50 487 466 434 425 413 397 373 60 718 690 648 636 620 600 567

70 995 960 907 891 872 846 805 80 1,318 1,276 1,211 1,192 1,168 1,136 1,086 90 1,688 1,638 1,560 1,537 1,509 1,471 1,410

100 2,105 2,045 1,955 1,928 1,894 1,850 1,779

772

Appendix B B.7 Wilcoxon T Values

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773

Appendix B B.8 Factors for Control Charts

Chart for Chart for Averages Ranges

Factors for Factors for Factors for Control Limits Central Line Control Limits

A2 d2 D3 D4 2 1.880 1.128 0 3.267 3 1.023 1.693 0 2.575 4 .729 2.059 0 2.282 5 .577 2.326 0 2.115

6 .483 2.534 0 2.004 7 .419 2.704 .076 1.924 8 .373 2.847 .136 1.864 9 .337 2.970 .184 1.816

10 .308 3.078 .223 1.777

11 .285 3.173 .256 1.744 12 .266 3.258 .284 1.716 13 .249 3.336 .308 1.692 14 .235 3.407 .329 1.671 15 .223 3.472 .348 1.652

SOURCE: Adapted from American Society for Testing and Materials, Manual on Quality Control of Materials, 1951, Table B2, p. 115. For a more detailed table and explanation, see J. Duncan Acheson, Quality Control and Industrial Statistics, 3d ed. (Homewood, Ill.: Richard D. Irwin, 1974), Table M, p. 927.

Number of Items in Sample,

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Appendix B B.9 Binomial Probability Distribution

n � 1

Probability

x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

0 0.950 0.900 0.800 0.700 0.600 0.500 0.400 0.300 0.200 0.100 0.050 1 0.050 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 0.950

n � 2

Probability

x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

0 0.903 0.810 0.640 0.490 0.360 0.250 0.160 0.090 0.040 0.010 0.003 1 0.095 0.180 0.320 0.420 0.480 0.500 0.480 0.420 0.320 0.180 0.095 2 0.003 0.010 0.040 0.090 0.160 0.250 0.360 0.490 0.640 0.810 0.903

n � 3

Probability

x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

0 0.857 0.729 0.512 0.343 0.216 0.125 0.064 0.027 0.008 0.001 0.000 1 0.135 0.243 0.384 0.441 0.432 0.375 0.288 0.189 0.096 0.027 0.007 2 0.007 0.027 0.096 0.189 0.288 0.375 0.432 0.441 0.384 0.243 0.135 3 0.000 0.001 0.008 0.027 0.064 0.125 0.216 0.343 0.512 0.729 0.857

n � 4

Probability

x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

0 0.815 0.656 0.410 0.240 0.130 0.063 0.026 0.008 0.002 0.000 0.000 1 0.171 0.292 0.410 0.412 0.346 0.250 0.154 0.076 0.026 0.004 0.000 2 0.014 0.049 0.154 0.265 0.346 0.375 0.346 0.265 0.154 0.049 0.014 3 0.000 0.004 0.026 0.076 0.154 0.250 0.346 0.412 0.410 0.292 0.171 4 0.000 0.000 0.002 0.008 0.026 0.063 0.130 0.240 0.410 0.656 0.815

n � 5

Probability

x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

0 0.774 0.590 0.328 0.168 0.078 0.031 0.010 0.002 0.000 0.000 0.000 1 0.204 0.328 0.410 0.360 0.259 0.156 0.077 0.028 0.006 0.000 0.000 2 0.021 0.073 0.205 0.309 0.346 0.313 0.230 0.132 0.051 0.008 0.001 3 0.001 0.008 0.051 0.132 0.230 0.313 0.346 0.309 0.205 0.073 0.021 4 0.000 0.000 0.006 0.028 0.077 0.156 0.259 0.360 0.410 0.328 0.204 5 0.000 0.000 0.000 0.002 0.010 0.031 0.078 0.168 0.328 0.590 0.774

774

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775

Appendix B B.9 Binomial Probability Distribution (continued )

n � 6

Probability

x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

0 0.735 0.531 0.262 0.118 0.047 0.016 0.004 0.001 0.000 0.000 0.000 1 0.232 0.354 0.393 0.303 0.187 0.094 0.037 0.010 0.002 0.000 0.000 2 0.031 0.098 0.246 0.324 0.311 0.234 0.138 0.060 0.015 0.001 0.000 3 0.002 0.015 0.082 0.185 0.276 0.313 0.276 0.185 0.082 0.015 0.002 4 0.000 0.001 0.015 0.060 0.138 0.234 0.311 0.324 0.246 0.098 0.031

5 0.000 0.000 0.002 0.010 0.037 0.094 0.187 0.303 0.393 0.354 0.232 6 0.000 0.000 0.000 0.001 0.004 0.016 0.047 0.118 0.262 0.531 0.735

n � 7

Probability

x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

0 0.698 0.478 0.210 0.082 0.028 0.008 0.002 0.000 0.000 0.000 0.000 1 0.257 0.372 0.367 0.247 0.131 0.055 0.017 0.004 0.000 0.000 0.000 2 0.041 0.124 0.275 0.318 0.261 0.164 0.077 0.025 0.004 0.000 0.000 3 0.004 0.023 0.115 0.227 0.290 0.273 0.194 0.097 0.029 0.003 0.000 4 0.000 0.003 0.029 0.097 0.194 0.273 0.290 0.227 0.115 0.023 0.004

5 0.000 0.000 0.004 0.025 0.077 0.164 0.261 0.318 0.275 0.124 0.041 6 0.000 0.000 0.000 0.004 0.017 0.055 0.131 0.247 0.367 0.372 0.257 7 0.000 0.000 0.000 0.000 0.002 0.008 0.028 0.082 0.210 0.478 0.698

n � 8

Probability

x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

0 0.663 0.430 0.168 0.058 0.017 0.004 0.001 0.000 0.000 0.000 0.000 1 0.279 0.383 0.336 0.198 0.090 0.031 0.008 0.001 0.000 0.000 0.000 2 0.051 0.149 0.294 0.296 0.209 0.109 0.041 0.010 0.001 0.000 0.000 3 0.005 0.033 0.147 0.254 0.279 0.219 0.124 0.047 0.009 0.000 0.000 4 0.000 0.005 0.046 0.136 0.232 0.273 0.232 0.136 0.046 0.005 0.000

5 0.000 0.000 0.009 0.047 0.124 0.219 0.279 0.254 0.147 0.033 0.005 6 0.000 0.000 0.001 0.010 0.041 0.109 0.209 0.296 0.294 0.149 0.051 7 0.000 0.000 0.000 0.001 0.008 0.031 0.090 0.198 0.336 0.383 0.279 8 0.000 0.000 0.000 0.000 0.001 0.004 0.017 0.058 0.168 0.430 0.663

(continued )

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Appendix B B.9 Binomial Probability Distribution (continued )

n � 9

Probability

x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

0 0.630 0.387 0.134 0.040 0.010 0.002 0.000 0.000 0.000 0.000 0.000 1 0.299 0.387 0.302 0.156 0.060 0.018 0.004 0.000 0.000 0.000 0.000 2 0.063 0.172 0.302 0.267 0.161 0.070 0.021 0.004 0.000 0.000 0.000 3 0.008 0.045 0.176 0.267 0.251 0.164 0.074 0.021 0.003 0.000 0.000 4 0.001 0.007 0.066 0.172 0.251 0.246 0.167 0.074 0.017 0.001 0.000

5 0.000 0.001 0.017 0.074 0.167 0.246 0.251 0.172 0.066 0.007 0.001 6 0.000 0.000 0.003 0.021 0.074 0.164 0.251 0.267 0.176 0.045 0.008 7 0.000 0.000 0.000 0.004 0.021 0.070 0.161 0.267 0.302 0.172 0.063 8 0.000 0.000 0.000 0.000 0.004 0.018 0.060 0.156 0.302 0.387 0.299 9 0.000 0.000 0.000 0.000 0.000 0.002 0.010 0.040 0.134 0.387 0.630

n � 10

Probability

x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

0 0.599 0.349 0.107 0.028 0.006 0.001 0.000 0.000 0.000 0.000 0.000 1 0.315 0.387 0.268 0.121 0.040 0.010 0.002 0.000 0.000 0.000 0.000 2 0.075 0.194 0.302 0.233 0.121 0.044 0.011 0.001 0.000 0.000 0.000 3 0.010 0.057 0.201 0.267 0.215 0.117 0.042 0.009 0.001 0.000 0.000 4 0.001 0.011 0.088 0.200 0.251 0.205 0.111 0.037 0.006 0.000 0.000

5 0.000 0.001 0.026 0.103 0.201 0.246 0.201 0.103 0.026 0.001 0.000 6 0.000 0.000 0.006 0.037 0.111 0.205 0.251 0.200 0.088 0.011 0.001 7 0.000 0.000 0.001 0.009 0.042 0.117 0.215 0.267 0.201 0.057 0.010 8 0.000 0.000 0.000 0.001 0.011 0.044 0.121 0.233 0.302 0.194 0.075 9 0.000 0.000 0.000 0.000 0.002 0.010 0.040 0.121 0.268 0.387 0.315

10 0.000 0.000 0.000 0.000 0.000 0.001 0.006 0.028 0.107 0.349 0.599

n � 11

Probability

x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

0 0.569 0.314 0.086 0.020 0.004 0.000 0.000 0.000 0.000 0.000 0.000 1 0.329 0.384 0.236 0.093 0.027 0.005 0.001 0.000 0.000 0.000 0.000 2 0.087 0.213 0.295 0.200 0.089 0.027 0.005 0.001 0.000 0.000 0.000 3 0.014 0.071 0.221 0.257 0.177 0.081 0.023 0.004 0.000 0.000 0.000 4 0.001 0.016 0.111 0.220 0.236 0.161 0.070 0.017 0.002 0.000 0.000

5 0.000 0.002 0.039 0.132 0.221 0.226 0.147 0.057 0.010 0.000 0.000 6 0.000 0.000 0.010 0.057 0.147 0.226 0.221 0.132 0.039 0.002 0.000 7 0.000 0.000 0.002 0.017 0.070 0.161 0.236 0.220 0.111 0.016 0.001 8 0.000 0.000 0.000 0.004 0.023 0.081 0.177 0.257 0.221 0.071 0.014 9 0.000 0.000 0.000 0.001 0.005 0.027 0.089 0.200 0.295 0.213 0.087

10 0.000 0.000 0.000 0.000 0.001 0.005 0.027 0.093 0.236 0.384 0.329 11 0.000 0.000 0.000 0.000 0.000 0.000 0.004 0.020 0.086 0.314 0.569

776

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777

Appendix B B.9 Binomial Probability Distribution (continued )

n � 12

Probability

x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

0 0.540 0.282 0.069 0.014 0.002 0.000 0.000 0.000 0.000 0.000 0.000 1 0.341 0.377 0.206 0.071 0.017 0.003 0.000 0.000 0.000 0.000 0.000 2 0.099 0.230 0.283 0.168 0.064 0.016 0.002 0.000 0.000 0.000 0.000 3 0.017 0.085 0.236 0.240 0.142 0.054 0.012 0.001 0.000 0.000 0.000 4 0.002 0.021 0.133 0.231 0.213 0.121 0.042 0.008 0.001 0.000 0.000

5 0.000 0.004 0.053 0.158 0.227 0.193 0.101 0.029 0.003 0.000 0.000 6 0.000 0.000 0.016 0.079 0.177 0.226 0.177 0.079 0.016 0.000 0.000 7 0.000 0.000 0.003 0.029 0.101 0.193 0.227 0.158 0.053 0.004 0.000 8 0.000 0.000 0.001 0.008 0.042 0.121 0.213 0.231 0.133 0.021 0.002 9 0.000 0.000 0.000 0.001 0.012 0.054 0.142 0.240 0.236 0.085 0.017

10 0.000 0.000 0.000 0.000 0.002 0.016 0.064 0.168 0.283 0.230 0.099 11 0.000 0.000 0.000 0.000 0.000 0.003 0.017 0.071 0.206 0.377 0.341 12 0.000 0.000 0.000 0.000 0.000 0.000 0.002 0.014 0.069 0.282 0.540

n � 13

Probability

x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

0 0.513 0.254 0.055 0.010 0.001 0.000 0.000 0.000 0.000 0.000 0.000 1 0.351 0.367 0.179 0.054 0.011 0.002 0.000 0.000 0.000 0.000 0.000 2 0.111 0.245 0.268 0.139 0.045 0.010 0.001 0.000 0.000 0.000 0.000 3 0.021 0.100 0.246 0.218 0.111 0.035 0.006 0.001 0.000 0.000 0.000 4 0.003 0.028 0.154 0.234 0.184 0.087 0.024 0.003 0.000 0.000 0.000

5 0.000 0.006 0.069 0.180 0.221 0.157 0.066 0.014 0.001 0.000 0.000 6 0.000 0.001 0.023 0.103 0.197 0.209 0.131 0.044 0.006 0.000 0.000 7 0.000 0.000 0.006 0.044 0.131 0.209 0.197 0.103 0.023 0.001 0.000 8 0.000 0.000 0.001 0.014 0.066 0.157 0.221 0.180 0.069 0.006 0.000 9 0.000 0.000 0.000 0.003 0.024 0.087 0.184 0.234 0.154 0.028 0.003

10 0.000 0.000 0.000 0.001 0.006 0.035 0.111 0.218 0.246 0.100 0.021 11 0.000 0.000 0.000 0.000 0.001 0.010 0.045 0.139 0.268 0.245 0.111 12 0.000 0.000 0.000 0.000 0.000 0.002 0.011 0.054 0.179 0.367 0.351 13 0.000 0.000 0.000 0.000 0.000 0.000 0.001 0.010 0.055 0.254 0.513

(continued )

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Appendix B B.9 Binomial Probability Distribution (concluded )

n � 14

Probability

x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

0 0.488 0.229 0.044 0.007 0.001 0.000 0.000 0.000 0.000 0.000 0.000 1 0.359 0.356 0.154 0.041 0.007 0.001 0.000 0.000 0.000 0.000 0.000 2 0.123 0.257 0.250 0.113 0.032 0.006 0.001 0.000 0.000 0.000 0.000 3 0.026 0.114 0.250 0.194 0.085 0.022 0.003 0.000 0.000 0.000 0.000 4 0.004 0.035 0.172 0.229 0.155 0.061 0.014 0.001 0.000 0.000 0.000

5 0.000 0.008 0.086 0.196 0.207 0.122 0.041 0.007 0.000 0.000 0.000 6 0.000 0.001 0.032 0.126 0.207 0.183 0.092 0.023 0.002 0.000 0.000 7 0.000 0.000 0.009 0.062 0.157 0.209 0.157 0.062 0.009 0.000 0.000 8 0.000 0.000 0.002 0.023 0.092 0.183 0.207 0.126 0.032 0.001 0.000 9 0.000 0.000 0.000 0.007 0.041 0.122 0.207 0.196 0.086 0.008 0.000

10 0.000 0.000 0.000 0.001 0.014 0.061 0.155 0.229 0.172 0.035 0.004 11 0.000 0.000 0.000 0.000 0.003 0.022 0.085 0.194 0.250 0.114 0.026 12 0.000 0.000 0.000 0.000 0.001 0.006 0.032 0.113 0.250 0.257 0.123 13 0.000 0.000 0.000 0.000 0.000 0.001 0.007 0.041 0.154 0.356 0.359 14 0.000 0.000 0.000 0.000 0.000 0.000 0.001 0.007 0.044 0.229 0.488

n � 15

Probability

x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

0 0.463 0.206 0.035 0.005 0.000 0.000 0.000 0.000 0.000 0.000 0.000 1 0.366 0.343 0.132 0.031 0.005 0.000 0.000 0.000 0.000 0.000 0.000 2 0.135 0.267 0.231 0.092 0.022 0.003 0.000 0.000 0.000 0.000 0.000 3 0.031 0.129 0.250 0.170 0.063 0.014 0.002 0.000 0.000 0.000 0.000 4 0.005 0.043 0.188 0.219 0.127 0.042 0.007 0.001 0.000 0.000 0.000

5 0.001 0.010 0.103 0.206 0.186 0.092 0.024 0.003 0.000 0.000 0.000 6 0.000 0.002 0.043 0.147 0.207 0.153 0.061 0.012 0.001 0.000 0.000 7 0.000 0.000 0.014 0.081 0.177 0.196 0.118 0.035 0.003 0.000 0.000 8 0.000 0.000 0.003 0.035 0.118 0.196 0.177 0.081 0.014 0.000 0.000 9 0.000 0.000 0.001 0.012 0.061 0.153 0.207 0.147 0.043 0.002 0.000

10 0.000 0.000 0.000 0.003 0.024 0.092 0.186 0.206 0.103 0.010 0.001 11 0.000 0.000 0.000 0.001 0.007 0.042 0.127 0.219 0.188 0.043 0.005 12 0.000 0.000 0.000 0.000 0.002 0.014 0.063 0.170 0.250 0.129 0.031 13 0.000 0.000 0.000 0.000 0.000 0.003 0.022 0.092 0.231 0.267 0.135 14 0.000 0.000 0.000 0.000 0.000 0.000 0.005 0.031 0.132 0.343 0.366

15 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.005 0.035 0.206 0.463

778

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779

Appendix B B.10A Critical Values for the Durbin–Watson d Statistic (� � .05)

k � 1 k � 2 k � 3 k � 4 k � 5 n dL,.05 dU,.05 dL,.05 dU,.05 dL,.05 dU,.05 dL,.05 dU,.05 dL,.05 dU,.05 15 1.08 1.36 0.95 1.54 0.82 1.75 0.69 1.97 0.56 2.21 16 1.10 1.37 0.98 1.54 0.86 1.73 0.74 1.93 0.62 2.15 17 1.13 1.38 1.02 1.54 0.90 1.71 0.78 1.90 0.67 2.10 18 1.16 1.39 1.05 1.53 0.93 1.69 0.82 1.87 0.71 2.06 19 1.18 1.40 1.08 1.53 0.97 1.68 0.86 1.85 0.75 2.02 20 1.20 1.41 1.10 1.54 1.00 1.68 0.90 1.83 0.79 1.99 21 1.22 1.42 1.13 1.54 1.03 1.67 0.93 1.81 0.83 1.96 22 1.24 1.43 1.15 1.54 1.05 1.66 0.96 1.80 0.86 1.94 23 1.26 1.44 1.17 1.54 1.08 1.66 0.99 1.79 0.90 1.92 24 1.27 1.45 1.19 1.55 1.10 1.66 1.01 1.78 0.93 1.90 25 1.29 1.45 1.21 1.55 1.12 1.66 1.04 1.77 0.95 1.89 26 1.30 1.46 1.22 1.55 1.14 1.65 1.06 1.76 0.98 1.88 27 1.32 1.47 1.24 1.56 1.16 1.65 1.08 1.76 1.01 1.86 28 1.33 1.48 1.26 1.56 1.18 1.65 1.10 1.75 1.03 1.85 29 1.34 1.48 1.27 1.56 1.20 1.65 1.12 1.74 1.05 1.84 30 1.35 1.49 1.28 1.57 1.21 1.65 1.14 1.74 1.07 1.83 31 1.36 1.50 1.30 1.57 1.23 1.65 1.16 1.74 1.09 1.83 32 1.37 1.50 1.31 1.57 1.24 1.65 1.18 1.73 1.11 1.82 33 1.38 1.51 1.32 1.58 1.26 1.65 1.19 1.73 1.13 1.81 34 1.39 1.51 1.33 1.58 1.27 1.65 1.21 1.73 1.15 1.81 35 1.40 1.52 1.34 1.58 1.28 1.65 1.22 1.73 1.16 1.80 36 1.41 1.52 1.35 1.59 1.29 1.65 1.24 1.73 1.18 1.80 37 1.42 1.53 1.36 1.59 1.31 1.66 1.25 1.72 1.19 1.80 38 1.43 1.54 1.37 1.59 1.32 1.66 1.26 1.72 1.21 1.79 39 1.43 1.54 1.38 1.60 1.33 1.66 1.27 1.72 1.22 1.79 40 1.44 1.54 1.39 1.60 1.34 1.66 1.29 1.72 1.23 1.79 45 1.48 1.57 1.43 1.62 1.38 1.67 1.34 1.72 1.29 1.78 50 1.50 1.59 1.46 1.63 1.42 1.67 1.38 1.72 1.34 1.77 55 1.53 1.60 1.49 1.64 1.45 1.68 1.41 1.72 1.38 1.77 60 1.55 1.62 1.51 1.65 1.48 1.69 1.44 1.73 1.41 1.77 65 1.57 1.63 1.54 1.66 1.50 1.70 1.47 1.73 1.44 1.77 70 1.58 1.64 1.55 1.67 1.52 1.70 1.49 1.74 1.46 1.77 75 1.60 1.65 1.57 1.68 1.54 1.71 1.51 1.74 1.49 1.77 80 1.61 1.66 1.59 1.69 1.56 1.72 1.53 1.74 1.51 1.77 85 1.62 1.67 1.60 1.70 1.57 1.72 1.55 1.75 1.52 1.77 90 1.63 1.68 1.61 1.70 1.59 1.73 1.57 1.75 1.54 1.78 95 1.64 1.69 1.62 1.71 1.60 1.73 1.58 1.75 1.56 1.78

100 1.65 1.69 1.63 1.72 1.61 1.74 1.59 1.76 1.57 1.78

SOURCE: J. Durbin and G. S. Watson, “Testing for Serial Correlation in Least Squares Regression, II,” Biometrika 30 (1951), pp. 159–78. Reproduced by permission of the Biometrika Trustees.

Lin01803_appb_764-781.qxd 10/22/10 8:52 AM Page 779

Appendix B B.10B Critical Values for the Durbin–Watson d Statistic (� � .025)

k � 1 k � 2 k � 3 k � 4 k � 5 n dL,.025 dU,.025 dL,.025 dU,.025 dL,.025 dU,.025 dL,.025 dU,.025 dL,.025 dU,.025 15 0.95 1.23 0.83 1.40 0.71 1.61 0.59 1.84 0.48 2.09 16 0.98 1.24 0.86 1.40 0.75 1.59 0.64 1.80 0.53 2.03 17 1.01 1.25 0.90 1.40 0.79 1.58 0.68 1.77 0.57 1.98 18 1.03 1.26 0.93 1.40 0.82 1.56 0.72 1.74 0.62 1.93 19 1.06 1.28 0.96 1.41 0.86 1.55 0.76 1.72 0.66 1.90 20 1.08 1.28 0.99 1.41 0.89 1.55 0.79 1.70 0.70 1.87 21 1.10 1.30 1.01 1.41 0.92 1.54 0.83 1.69 0.73 1.84 22 1.12 1.31 1.04 1.42 0.95 1.54 0.86 1.68 0.77 1.82 23 1.14 1.32 1.06 1.42 0.97 1.54 0.89 1.67 0.80 1.80 24 1.16 1.33 1.08 1.43 1.00 1.54 0.91 1.66 0.83 1.79 25 1.18 1.34 1.10 1.43 1.02 1.54 0.94 1.65 0.86 1.77 26 1.19 1.35 1.12 1.44 1.04 1.54 0.96 1.65 0.88 1.76 27 1.21 1.36 1.13 1.44 1.06 1.54 0.99 1.64 0.91 1.75 28 1.22 1.37 1.15 1.45 1.08 1.54 1.01 1.64 0.93 1.74 29 1.24 1.38 1.17 1.45 1.10 1.54 1.03 1.63 0.96 1.73 30 1.25 1.38 1.18 1.46 1.12 1.54 1.05 1.63 0.98 1.73 31 1.26 1.39 1.20 1.47 1.13 1.55 1.07 1.63 1.00 1.72 32 1.27 1.40 1.21 1.47 1.15 1.55 1.08 1.63 1.02 1.71 33 1.28 1.41 1.22 1.48 1.16 1.55 1.10 1.63 1.04 1.71 34 1.29 1.41 1.24 1.48 1.17 1.55 1.12 1.63 1.06 1.70 35 1.30 1.42 1.25 1.48 1.19 1.55 1.13 1.63 1.07 1.70 36 1.31 1.43 1.26 1.49 1.20 1.56 1.15 1.63 1.09 1.70 37 1.32 1.43 1.27 1.49 1.21 1.56 1.16 1.62 1.10 1.70 38 1.33 1.44 1.28 1.50 1.23 1.56 1.17 1.62 1.12 1.70 39 1.34 1.44 1.29 1.50 1.24 1.56 1.19 1.63 1.13 1.69 40 1.35 1.45 1.30 1.51 1.25 1.57 1.20 1.63 1.15 1.69 45 1.39 1.48 1.34 1.53 1.30 1.58 1.25 1.63 1.21 1.69 50 1.42 1.50 1.38 1.54 1.34 1.59 1.30 1.64 1.26 1.69 55 1.45 1.52 1.41 1.56 1.37 1.60 1.33 1.64 1.30 1.69 60 1.47 1.54 1.44 1.57 1.40 1.61 1.37 1.65 1.33 1.69 65 1.49 1.55 1.46 1.59 1.43 1.62 1.40 1.66 1.36 1.69 70 1.51 1.57 1.48 1.60 1.45 1.63 1.42 1.66 1.39 1.70 75 1.53 1.58 1.50 1.61 1.47 1.64 1.45 1.67 1.42 1.70 80 1.54 1.59 1.52 1.62 1.49 1.65 1.47 1.67 1.44 1.70 85 1.56 1.60 1.53 1.63 1.51 1.65 1.49 1.68 1.46 1.71 90 1.57 1.61 1.55 1.64 1.53 1.66 1.50 1.69 1.48 1.71 95 1.58 1.62 1.56 1.65 1.54 1.67 1.52 1.69 1.50 1.71

100 1.59 1.63 1.57 1.65 1.55 1.67 1.53 1.70 1.51 1.72

SOURCE: J. Durbin and G. S. Watson, “Testing for Serial Correlation in Least Squares Regression, II,” Biometrika 30 (1951), pp. 159–78. Reproduced by permission of the Biometrika Trustees.

780

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781

Appendix B B.10C Critical Values for the Durbin–Watson d Statistic (� � .01)

k � 1 k � 2 k � 3 k � 4 k � 5 n dL,.01 dU,.01 dL,.01 dU,.01 dL,.01 dU,.01 dL,.01 dU,.01 dL,.01 dU,.01 15 0.81 1.07 0.70 1.25 0.59 1.46 0.49 1.70 0.39 1.96 16 0.84 1.09 0.74 1.25 0.63 1.44 0.53 1.66 0.44 1.90 17 0.87 1.10 0.77 1.25 0.67 1.43 0.57 1.63 0.48 1.85 18 0.90 1.12 0.80 1.26 0.71 1.42 0.61 1.60 0.52 1.80 19 0.93 1.13 0.83 1.26 0.74 1.41 0.65 1.58 0.56 1.77 20 0.95 1.15 0.86 1.27 0.77 1.41 0.68 1.57 0.60 1.74 21 0.97 1.16 0.89 1.27 0.80 1.41 0.72 1.55 0.63 1.71 22 1.00 1.17 0.91 1.28 0.83 1.40 0.75 1.54 0.66 1.69 23 1.02 1.19 0.94 1.29 0.86 1.40 0.77 1.53 0.70 1.67 24 1.04 1.20 0.96 1.30 0.88 1.41 0.80 1.53 0.72 1.66 25 1.05 1.21 0.98 1.30 0.90 1.41 0.83 1.52 0.75 1.65 26 1.07 1.22 1.00 1.31 0.93 1.41 0.85 1.52 0.78 1.64 27 1.09 1.23 1.02 1.32 0.95 1.41 0.88 1.51 0.81 1.63 28 1.10 1.24 1.04 1.32 0.97 1.41 0.90 1.51 0.83 1.62 29 1.12 1.25 1.05 1.33 0.99 1.42 0.92 1.51 0.85 1.61 30 1.13 1.26 1.07 1.34 1.01 1.42 0.94 1.51 0.88 1.61 31 1.15 1.27 1.08 1.34 1.02 1.42 0.96 1.51 0.90 1.60 32 1.16 1.28 1.10 1.35 1.04 1.43 0.98 1.51 0.92 1.60 33 1.17 1.29 1.11 1.36 1.05 1.43 1.00 1.51 0.94 1.59 34 1.18 1.30 1.13 1.36 1.07 1.43 1.01 1.51 0.95 1.59 35 1.19 1.31 1.14 1.37 1.08 1.44 1.03 1.51 0.97 1.59 36 1.21 1.32 1.15 1.38 1.10 1.44 1.04 1.51 0.99 1.59 37 1.22 1.32 1.16 1.38 1.11 1.45 1.06 1.51 1.00 1.59 38 1.23 1.33 1.18 1.39 1.12 1.45 1.07 1.52 1.02 1.58 39 1.24 1.34 1.19 1.39 1.14 1.45 1.09 1.52 1.03 1.58 40 1.25 1.34 1.20 1.40 1.15 1.46 1.10 1.52 1.05 1.58 45 1.29 1.38 1.24 1.42 1.20 1.48 1.16 1.53 1.11 1.58 50 1.32 1.40 1.28 1.45 1.24 1.49 1.20 1.54 1.16 1.59 55 1.36 1.43 1.32 1.47 1.28 1.51 1.25 1.55 1.21 1.59 60 1.38 1.45 1.35 1.48 1.32 1.52 1.28 1.56 1.25 1.60 65 1.41 1.47 1.38 1.50 1.35 1.53 1.31 1.57 1.28 1.61 70 1.43 1.49 1.40 1.52 1.37 1.55 1.34 1.58 1.31 1.61 75 1.45 1.50 1.42 1.53 1.39 1.56 1.37 1.59 1.34 1.62 80 1.47 1.52 1.44 1.54 1.42 1.57 1.39 1.60 1.36 1.62 85 1,48 1.53 1.46 1.55 1.43 1.58 1.41 1.60 1.39 1.63 90 1.50 1.54 1.47 1.56 1.45 1.59 1.43 1.61 1.41 1.64 95 1.51 1.55 1.49 1.57 1.47 1.60 1.45 1.62 1.42 1.64

100 1.52 1.56 1.50 1.58 1.48 1.60 1.46 1.63 1.44 1.65

SOURCE: J. Durbin and G. S. Watson, “Testing for Serial Correlation in Least Squares Regression, II,” Biometrika 30 (1951), pp. 159–78. Reproduced by permission of the Biometrika Trustees.

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CHAPTER 1 1. a. Interval

b. Ratio c. Nominal d. Nominal e. Ordinal f. Ratio

3. Answers will vary. 5. Qualitative data is not numerical, whereas quantitative data

is numerical. Examples will vary by student. 7. A discrete variable may assume only certain values. A

continuous variable may assume an infinite number of values within a given range. The number of traffic citations issued each day during February in Garden City Beach, South Carolina, is a discrete variable. The weight of commercial trucks passing the weigh station at milepost 195 on Interstate 95 in North Carolina is a continuous variable.

9. a. Ordinal b. Ratio c. The newer system provides information on the distance

between exits. 11. If you were using this store as typical of all Barnes & Noble

stores, then it would be sample data. However, if you were considering it as the only store of interest, then the data would be population data.

13.

782

Appendix C: Answers Answers to Odd-Numbered Chapter Exercises

Discrete Variable Continuous Variable

Qualitative b. Gender d. Soft drink preference

Quantitative f. SAT scores a. Salary g. Student rank in class c. Sales volume of h. Rating of a finance MP3 players

professor e. Temperature i. Number of home

computers

15. According to the sample information, 120�300 or 40% would accept a job transfer.

17. a. Total sales increased by 106,041, found by 1,255,337 � 1,149,296, which is 9.2 percent.

b. Market shares are:

Ford has gained 3.7% and Toyota lost 3.9% of their market shares.

c. Percent changes are:

Ford and Nissan had increases of more than 20 percent. General Motors and Hyundai had increases of more than 10 percent. Meanwhile, Toyota had a decrease of over 10 percent.

19. Earnings increase each year over the previous year until a large peak in 2008. Then there is a rather large drop in 2009.

21. a. League is a qualitative variable; the others are quantitative.

b. League is a nominal-level variable; the others are ratio- level variables.

CHAPTER 2 1. 25 percent market share. 3.

Season Frequency Relative Frequency

Winter 100 .10 Spring 300 .30 Summer 400 .40 Fall 200 .20

1,000 1.00

5. a. A frequency table.

Color Frequency Relative Frequency

Bright White 130 0.10 Metallic Black 104 0.08 Magnetic Lime 325 0.25 Tangerine Orange 455 0.35 Fusion Red 286 0.22

Total 1300 1.00

2010 2009

General Motors 22.9% 22.0% Ford Motor 19.9% 16.2% Chrysler 11.3% 12.7% Toyota 15.8% 19.7% American Honda 11.8% 12.4% Nissan NA 10.6% 9.4% Hyundai 5.1% 4.8% Mazda 2.6% 2.8%

General Motors increase of 13.7% Ford Motor increase of 34.3% Chrysler decrease of 3.2% Toyota decrease of 12.4% American Honda increase of 3.9% Nissan NA increase of 22.8% Hyundai increase of 17.0% Mazda increase of 2.9%

Discrete Continuous

Nominal b. Gender

Ordinal d. Soft drink preference g. Student rank in class h. Rating of a finance

professor

Interval f. SAT scores e. Temperature

Ratio i. Number of home a. Salary computers c. Sales volume of

MP3 players

Lin01803_appc_782-828.qxd 11/18/10 12:51 PM Page 782

15. a. Histogram b. 100 c. 5 d. 28 e. 0.28 f. 12.5

g. 13 17. a. 50

b. 1.5 thousand miles, or 1,500 miles. c.

783

Tangerine Orange 35.0%

Magnetic Lime

25.0%

Metallic Black 8.0%

Bright White 10.0%

Fusion Red 22.0%

Pie Chart of Color Frequencies

d. 350,000 orange, 250,000 lime, 220,000 red, 100,000 white, and 80,000 black, found by multiplying relative frequency by 1,000,000 production.

7. 25 � 32, 26 � 64, therefore, 6 classes 9. 27 � 128, 28 � 256, suggests 8 classes

Class intervals of 40, 45, or 50 all

would be acceptable.

11. a. 24 � 16 Suggests 5 classes.

b. Use interval of 1.5.

c. 24 d.

i � 31 � 25

5 � 1.2

i � $567 � $235

8 � 41

Units f Relative Frequency

24.0 up to 25.5 2 0.125 25.5 up to 27.0 4 0.250 27.0 up to 28.5 8 0.500 28.5 up to 30.0 0 0.000 30.0 up to 31.5 2 0.125

Total 16 1.000

Number of Visits f

0 up to 3 9 3 up to 6 21 6 up to 9 13 9 up to 12 4

12 up to 15 3 15 up to 18 1

Total 51

Number of Percent Visits of Total

0 up to 3 17.65 3 up to 6 41.18 6 up to 9 25.49 9 up to 12 7.84

12 up to 15 5.88 15 up to 18 1.96

Total 100.00

e. The largest concentration is in the 27.0 up to 28.5 class (8). 13. a.

b. The largest group of shoppers (21) shop at the BiLo Supermarket 3, 4, or 5 times during a month period. Some customers visit the store only 1 time during the month, but others shop as many as 15 times.

Frequent Flier Miles f CF

0 up to 3 5 5 3 up to 6 12 17 6 up to 9 23 40 9 up to 12 8 48

12 up to 15 2 50

0 3 6 9 12 15

25 20 15 10 5

Nu m

be r o

f e m

pl oy

ee s

Frequent flier miles

–1.5

25 20 15 10 5 0Nu

m be

r o f e

m pl

oy ee

Frequent flier miles 1.5 4.5 7.5 10.5 13.5 16.5

f. For the 50 employees, about half traveled between 6,000 and 9,000 miles. Five employees traveled less than 3,000 miles, and 2 traveled more than 12,000 miles.

19. a. 40 b. 5 c. 11 or 12 d. About $18/hr e. About $9/hr f. About 75%

21. a. 5 b.

d. X � 1.5, Y � 5 e.

Fr eq

ue nc

Chart of Frequency vs. Color

500

400

300

200

100

0 Metallic Black

Color

Bright White

Magnetic Lime

Tangerine Orange

Fusion Red

Lin01803_appc_782-828.qxd 11/18/10 12:51 PM Page 783

784

Planned Activities

No Planned Activities

Not Sure

No Answer

d. A pie chart would be better because it clearly shows that nearly half of the customers prefer no planned activities.

27. 26 � 64 and 27 � 128, suggest 7 classes 29. a. 5, because 24 � 16 � 25 and 25 � 32 � 25

b. Use interval of 7.

c. 15 d.

i � 48 � 16

5 � 6.4

31. a. 25 � 32, 26 � 64, 6 classes recommended.

b. use an interval of 2.

c. 0 d.

i � 10 � 1

6 � 1.5,

$ 70 up to $105 4 105 up to 140 17 140 up to 175 14 175 up to 210 2 210 up to 245 6 245 up to 280 1

Depreciation

Insurance

Repairs

Interest

Fuel

0 200

Ex pe

nd itu

re s

400 Amount

600 800 1,000

Wages 73%

Other 3%Pensions

3%IRA 8%

Social Security

Dividends 11%

Class Frequency

15 up to 22 3 22 up to 29 8 29 up to 36 7 36 up to 43 5 43 up to 50 2

|| |||| |||| || |||| ||| |||

Class Frequency

0 up to 2 1 2 up to 4 5 4 up to 6 12 6 up to 8 17 8 up to 10 8

Class Frequency

0 up to 200 19 200 up to 400 1 400 up to 600 4 600 up to 800 1 800 up to 1000 2

e. The distribution is fairly symmetric or bell-shaped with a large peak in the middle of the two classes of 4 up to 8.

33.

This distribution is positively skewed with a large “tail” to the right or positive values. Notice that the top 7 tunes account for 4,342 plays out of a total of 5,968 or about 73 percent of all plays.

35. a. 56 c. 55 b. 10 (found by 60 � 50) d. 17

37. a. $30.50, found by ($265 � $82)�6. b. $35 c.

d. The purchases range from a low of about $70 to a high of about $280. The concentration is in the $105 up to $140 and $140 up to $175 classes.

39.

41.

c. 50

Frequent flier miles

Fr eq

ue nc

ie s

0 3 6 9 12 15

50%

20%

40%

60%

80%

d. About 8.7 thousand miles 23. a. A qualitative variable uses either the nominal or ordinal

scale of measurement. It is usually the result of counts. Quantitative variables are either discrete or continuous. There is a natural order to the results for a quantitative variable. Quantitative variables can use either the interval or ratio scale of measurement.

b. Both types of variables can be used for samples and populations.

25. a. Frequency table b.

0 20 40 60 80

100 120 140 160

No answerNot sureNo planned activities

Planned activities

e. It is fairly symmetric, with most of the values between 22 and 36.

Lin01803_appc_782-828.qxd 11/18/10 12:51 PM Page 784

785

By far the largest part of income in South Carolina is wages. Almost three-fourths of the adjusted gross income comes from wages. Dividends and IRAs each contribute roughly another 10%.

43. a. Since 26 � 64 � 70 � 128 � 27, 7 classes are recom- mended. The interval should be at least (1,002.2 � 3.3)�7 � 142.7. Use 150 as a convenient value.

0 75 225 375 525 675 825 975

Value

Fr eq

ue nc

45. a. Pie chart b. 215, found by 0.43 � 500 c. Seventy-eight percent are in either a house of worship

(43%) or outdoors (35%). 47. a.

SC Income Percent Cumulative

Wages 73 73 Dividends 11 84 IRA 8 92 Pensions 3 95 Social Security 2 97 Other 3 100

Selling Price f CF

110 up to 145 3 3 145 up to 180 19 22 180 up to 215 31 53 215 up to 250 25 78 250 up to 285 14 92 285 up to 320 10 102 320 up to 355 3 105

51. Use interval of 35.i � 345.3 � 125.0

7 � 31.47

a. Most homes (53%) are in the 180 up to 250 range. b. The largest value is near 355; the smallest, near 110. c.

Nu m

be r o

f H om

145 180110 215 250 285 320 355

120 100

80 60 40 20

1.2 1.0 0.8 0.6 0.4 0.2 0

Selling Price

Percent

About 42 homes sold for less than 200. About 55% of the homes sold for less than 220. So 45% sold for more. Less than 1% of the homes sold for less than 125.

100

120

140

Orange GreenBlueYellow RedBrown

Color Frequency

Brown 130 Yellow 98 Red 96 Blue 52 Orange 35 Green 33

444

Pe tro

leu m

pro du

cts

Pa sse

ng er

ca rs

Ca r p

art s a

ac ce

ss ori

Alu mi

nu m

Lu mb

Am ou

nt ($

bi lli

on )

Product

Chart of Amount ($billion)

b. 0.33, found by (63.7 � 36.6)/303.4 c. 0.77, found by (63.7 � 36.6)/130.2

49. Co un

0 1 2 3 4

Chart of Township

Township

Townships 3 and 4 have more sales than the average and Townships 1 and 5 have somewhat less than the average.

Lin01803_appc_782-828.qxd 11/18/10 12:51 PM Page 785

786

State Mean Median Range

California 33.10 34.0 32 Iowa 24.50 25.0 19

CHAPTER 3 1. � 5.4, found by 27�5 3. a. , found by 28�4

b. (5 � 7) � (9 � 7) � (4 � 7) � (10 � 7) � 0 5. , found by 43.74�3. 7. a. 15.4, found by 154�10

b. Population parameter, since it includes all the sales at Midtown Ford

9. a. $54.55, found by $1,091�20 b. A sample statistic—assuming that the power company

serves more than 20 customers

11. so

13. $22.91, found by

15. $17.75, found by ($400 � $750 � $2,400)�200 17. a. No mode

b. The given value would be the mode. c. 3 and 4 bimodal

19. a. Mean � 3.25 b. Median � 5 c. Mode � 5

21. a. Median � 2.9 b. Mode � 2.9

23.

Median � 58, Mode � 58 Any of the three measures would be satisfactory.

25. a.

b. Median � 7.45. There are several modes: 6.5, 7.3, 7.8, and 8.7

c. ,

Median � 8.7 About 1 percentage point higher in Winter

27. 12.8 percent increase, found by

29. 12.28 percent increase, found by

31. 2.47%, found by

33. 33.5%, found by

35. a. 7, found by 10 � 3. b. 6, found by 30�5. c. 2.4, found by 12�5. d. The difference between the highest number sold (10)

and the smallest number sold (3) is 7. On average, the number of HDTVs sold deviates by 2.4 from the mean of 6.

37. a. 30, found by 54 � 24. b. 38, found by 380�10. c. 7.2, found by 72�10. d. The difference of 54 and 24 is 30. On average, the

number of minutes required to install a door deviates 7.2 minutes from the mean of 38 minutes.

39.

262,700,000 340,213

� 1

214.5 172.2

� 1

2 5 (1.094)(1.138)(1.117)(1.119)(1.147) � 1.1228

2 5 (1.08)(1.12)(1.14)(1.26)(1.05) � 1.128

X � 33.8

4 � 8.45

X � 90.4 12

� 7.53

X � 647 11

� 58.82

300($20) � 400($25) � 400($23) 300 � 400 � 400

©X n

X � 14.58

X � 7.0

Class f

730 up to 770 5 770 up to 810 17 810 up to 850 37 850 up to 890 18 890 up to 930 1 930 up to 970 0 970 up to 1010 2

53. Since 26 � 64 � 80 � 128 � 27, use 7 classes. The interval should be at least (1008 � 741)/7 � 38.14 miles. Use 40. The resulting frequency distribution is:

a. The typical amount driven is 830 miles. The range is from 730 up to 1010 miles.

b. The distribution is “bell shaped” around 830. However, there are two outliers up around 1000 miles.

725 0

Cu m

. F re

Cumulative Frequency of Miles Driven per Month

Miles 775 825 875 925 975 1025 1075

Forty percent of the buses were driven fewer than 820 miles. Fifty-nine buses were driven less than 850 miles.

The first chart shows that about two-thirds of the buses are diesel. The second diagram shows that nearly three fourths of the buses have 55 seats.

The mean and median ratings were higher, but there was also more variation in California.

Pie Chart of Bus Type

Diesel 66.3%

Gasoline 33.8%

Pie Chart of Seats

55 Passenger 75.0%

42 Passenger 11.3%

14 Passenger 8.8%

6 Passenger 5.0%

Lin01803_appc_782-828.qxd 11/18/10 12:51 PM Page 786

41. a. 5 b. 4.4, found by

43. a. $2.77 b. 1.26, found by

45. a. Range: 7.3, found by 11.6 � 4.3. Arithmetic mean: 6.94, found by 34.7�5. Variance: 6.5944, found by 32.972�5. Standard deviation: 2.568, found by .

b. Dennis has a higher mean return (11.76 � 6.94). However, Dennis has greater spread in its returns on equity (16.89 � 6.59).

47. a.

b. s � 2.3452 49. a.

b. s � 9.0921

51. a.

53. About 69%, found by 1 � 1�(1.8)2

55. a. About 95% b. 47.5%, 2.5%

57. Because the exact values in a frequency distribution are not known, the midpoint is used for every member of that class.

59.

s � 2123.66 � 11.12

� 1,112.9

9 � 123.66

s2 � (101 � 95.1)2 � · · · � (88 � 95.1)2

10 � 1

X � 951 10

� 95.1

s2 � 744

10 � 1 � 82.667

s2 � (28 � 38)2 � · · · � (42 � 38)2

10 � 1 � 82.667

X � 38

s2 � (7 � 4)2 � · · · � (3 � 4)2

5 � 1 �

22 5 � 1

� 5.5

X � 4

16.5944

(2.68 � 2.77)2 � (1.03 � 2.77)2 � (2.26 � 2.77)2

� (4.30 � 2.77)2 � (3.58 � 2.77)2

(8 � 5)2 � (3 � 5)2 � (7 � 5)2 � (3 � 5)2 � (4 � 5)2

787

Class f M fM (M � ) f (M � )2

20 up to 30 7 25 175 �22.29 3,477.909 30 up to 40 12 35 420 �12.29 1,812.529 40 up to 50 21 45 945 �2.29 110.126 50 up to 60 18 55 990 7.71 1,069.994 60 up to 70 12 65 780 17.71 3,763.729

70 3,310 10,234.287

61.

s � B

10,234.287 70 � 1

� 12.18

X � 3,310

70 � 47.29

Number of Clients f M fM (M � ) f (M � )2

20 up to 30 1 25 25 �19.8 392.04 30 up to 40 15 35 525 �9.8 1,440.60 40 up to 50 22 45 990 0.2 0.88 50 up to 60 8 55 440 10.2 832.32 60 up to 70 4 65 260 20.2 1,632.16

50 2,240 4,298.00

63. a. Mean � 5, found by (6 � 4 � 3 � 7 � 5)�5. Median is 5, found by rearranging the values and selecting the middle value.

b. Population, because all partners were included c.

65.

Median � 37.50

67. The mean is 35.675, found by 1427/40. The median is 36, found by sorting the data and averaging the 20th and 21st observations.

69.

71.

73. GM � B

12 6,286,800 5,164,900

� 1 � 1.0094 � 1.0 � .0094

Xw � [15,300(4.5) � 10,400(3.0) � 150,600(10.2)]

176,300 � 9.28

Xw � $5.00(270) � $6.50(300) � $8.00(100)

270 � 300 � 100 � $6.12

X � 545 16

� 34.06

Class M f fM M � (M � )2 f(M � )2

3.5 up to 6.5 5 10 50 �4 16 160 6.5 up to 9.5 8 6 48 �1 1 6 9.5 up to 12.5 11 9 99 2 4 36

12.5 up to 15.5 14 4 56 5 25 100 15.5 up to 18.5 17 1 17 8 64 64

270 366

XXX

75. a. 55, found by 72 � 17 b. 14.4, found by 144�10, where c. 17.6245

77. a. This is a population, because it includes all the public universities in Ohio.

b. The mean is 22,163. c. The median is 18,989. d. The range is 57,271. e. The standard deviation is 14,156.

79. a. There were 13 flights, so all items are considered.

Median � 195 c. Range � 301 � 7 � 294

81. a. The mean is $717.20, found by $17,930�25. The median is $717.00 and there are two modes, $710 and $722.

b. The range is $90, found by $771 � $681, and the standard deviation is $24.87, found by the square root of 14,850�24.

c. From $667.46 up to $766.94, found by $717.20 2($24.87).

83. a. The mean 0.8654 is found by 17.309�2. The median is 0.86 and the mode is 0.792.

b. The range is 0.269, found by 1.025 � 0.756 and the standard deviation is 0.0853, found by the square root of 0.138167�19.

c. From 0.6948 up to 1.036, found by 0.8654 2(0.0853).

85. a.

b. Range � 18 � 4 � 14

c. 25 � 32, so suggest 5 classes

Use i � 3i � 18 � 4

5 � 2.8

s � B

368.7 30 � 1

� 3.57

X � 273 30

� 9.1, Median � 9

s � B

133,846 13

� 101.47

� 2,259

13 � 173.77

X � 43.2

s � B

4,298 50 � 1

� 9.37

X � 2,240

50 � 44.8

Lin01803_appc_782-828.qxd 11/18/10 12:51 PM Page 787

788

7 8 9 10 X

Scatter Diagram of Y versus X

11 12 13

19. a. The mean is 30.8, found by 154�5. The median is 31.0, and the standard deviation is 3.96, found by

b. �0.15, found by

3(30.8 � 31.0) 3.96

s � B

62.8 4

� 3.96

Salary

36 1.313131 2.264250504 26 �1.212121 �1.780894343 33 0.555556 0.171467764 28 �0.707071 �0.353499282 31 0.050505 0.000128826

0.301453469

a (X � X )

s b

a (X � X )

s b

There were a total of 16 calls studied. The number of calls ranged from 5 to 52. Seven of the 16 subscribers made between 30 and 39 calls.

11. Median � 53, found by (11 � 1)1 2 � 6th value in from lowest Q1 � 49, found by (11 � 1)1 2 � 3rd value in from lowest Q3 � 55, found by (11 � 1)1 2 � 9th value in from lowest

13. a. Q1 � 33.25, Q3 � 50.25 b. D2 � 27.8, D8 � 52.6 c. P67 � 47

15. a. 350 b. Q1 � 175, Q3 � 930 c. 930 � 175 � 755 d. Less than 0, or more than about 2,060 e. There are no outliers. f. The distribution is positively skewed.

17.

3 4

1 4

1 2

--------- -----------------------

-----+---------+---------+---------+---------+---------+--- +-------------------------+

+-------------------------+ ++ +++

14.0 21.0 28.0 35.0 42.0 49.0

0.125, found by [5�(4 � 3)] � 0.301 21. a. The mean is 21.93, found by 328.9�15. The median is

15.8, and the standard deviation is 21.18, found by

b. 0.868, found by [3(21.93 � 15.8)] c. 2.444, found by [15�(14 � 13)] � 29.658

23.

s � B

6283 14

� 21.18

There is a positive relationship between the variables. 25. a. Both variables are nominal scale.

b. Contingency table c. Men are about twice as likely to order a dessert. From

the table, 32 percent of the men ordered dessert, but only 15 percent of the women.

27. a. Dot plot b. 15 c. 5

29. Stem-and-leaf N � 23 3 3 222 3 3 5 3 77 5 3

10 4 00000 11 4 2 11 4 (6) 4 666666 6 4 6 5 6 5 222222

31. a.

Median � 83.7 � 85.6

2 � 84.65

L50 � (20 � 1) 50

100 � 10.50

The mean and standard deviation from grouped data are estimates of the mean and standard deviations of the actual values.

87. a. 1. The mean team salary is $88,510,000 and the median is $80,350,000. Since the distribution is skewed, the median value of $80,350,000 is more typical.

2. The range is $164,700,000; found by 201,500,000 � 36,800,000. The standard deviation is $33,900,000. About 95 percent of the team salaries are between $20,710,000 and $156,310,000; found by $88,510,000 plus or minus 2($33,900,000).

b. 9.65% per year, found by

CHAPTER 4 1. In a histogram, observations are grouped so their individual

identity is lost. With a dot plot, the identity of each observation is maintained.

3. a. Dot plot b. 15 c. 1, 7 d. 2 and 3

5. a. 620 to 629 b. 5 c. 621, 623, 623, 627, 629

7. a. 25 b. One c. 38,106 d. 60, 61, 63, 63, 65, 65, 69 e. No values f. 9 g. 9 h. 76 i. 16

3,240,000 512,930

� 1

s � B

366 30 � 1

� 3.552

x � 270 30

� 9.0

Stem Leaves

0 5 1 28 2 3 0024789 4 12366 5 2

The distribution is somewhat positively skewed. Note that the dashed line above 35 is longer than below 18.

Lin01803_appc_782-828.qxd 11/18/10 12:51 PM Page 788

------ ---------------------- ---------------------

---------------------

------+---------+---------+---------+---------+---------+

+I I

32.5 35.0 37.5 40.0 42.5 45.0

------------- ------------------- -----------------------

-----------------------

------+---------+---------+---------+---------+---------+

+I I

24.5 28.0 31.5 35.0 38.5 42.0

------------- --------------- -------------------------

-------------------------

--------+---------+---------+---------+---------+--------C20

+I I

64.0 72.0 80.0 88.0 96.0

There is a direct association between them. The first observation is larger on both scales.

c. � 95.579

P83 � 93.3 � .43(98.6 � 93.3) L83 � 21(.83) � 17.43 P26 � 66.6 � .46(72.9 � 66.6) � 69.498 L26 � 21(.26) � 5.46 Q3 � 87.1 � .75(90.2 � 87.1) � 89.425 L75 � 21(.75) � 15.75 Q1 � 66.6 � .25(72.9 � 66.6) � 68.175 L25 � (21)(.25) � 5.25

789

33. a. Q1 � 26.25, Q3 � 35.75, Median � 31.50

c. The median time for public transportation is about 6 minutes less. There is more variation in public transportation. The difference between Q1 and Q3 is 9.5 minutes for public transportation and 5.5 minutes for private transportation.

35. The distribution is positively skewed. The first quartile is about $20 and the third quartile is about $90. There is one outlier located at $255. The median is about $50.

37. a.

b. Q1 � 33.25, Q3 � 38.75, Median � 37.50

Box Plot of Price

Price

* * *

0 10000 20000 30000 40000 50000

Median is 3733. First quartile is 1478. Third quartile is 6141. So prices over 13,135.5, found by 6141 � 1.5 (6141 � 1478), are outliers. There are three (13,925, 20,413, and 44,312).

10000

20000

30000

40000

50000

0 1 2 3 4 5 Size

Scatterplot of Price versus Size

Pr ic

Box Plot of Size

Size 0 1 2 3 4 5

* * *

Shape\ Ultra Cut Average Good Ideal Premium Ideal All

Emerald 0 0 1 0 0 1 Marquise 0 2 0 1 0 3 Oval 0 0 0 1 0 1 Princess 1 0 2 2 0 5 Round 1 3 3 13 3 23

Total 2 5 6 17 3 33

15.0 17.5 20.0 22.5 25.0 Age

Ac ci

de nt

Scatterplot of Accidents versus Age

27.5 30.0 32.5

As age increases, the number of accidents decreases. 43. a. 139,340,000

b. 5.4% unemployed, found by (7523�139,340)100 c. Men � 5.64%

Women � 5.12% 45. a.

The majority of the diamonds are round (23). Premium cut is most common (17). The Round Premium combination occurs most often (13).

39.

41.

sk � 0.065 or sk � 3(7.7143 � 8.0) 3.9036

� �0.22

Median is 0.84. First quartile is 0.515. Third quartile is 1.12. So sizes over 2.0275, found by 1.12 � 1.5 (1.12 � 0.515), are outliers. There are three (2.03, 2.35, and 5.03).

0 20 40 60 80 100

** **

Age

Boxplot of Age

There are five outliers. There is a group of three around 40 years (Angels, Athletics, and Dodgers) and a group of two close to one hundred years old (Cubs and Red Sox).

Lin01803_appc_782-828.qxd 11/18/10 12:51 PM Page 789

790

Person Outcome 1 2

1 A A 2 A F 3 F A 4 F F

The first quartile is $66,650,000 and the third is $105,500,000. The distribution is positively skewed, with the New York Yankees a definite outlier.

CHAPTER 5 1.

9. a. Answers will vary. Here are some possibilities: 123, 124, 125, 999

b. (1�10)3

c. Classical 11. P(A or B) � P(A) � P(B) � .30 � .20 � .50

P(neither) � 1 � .50 � .50. 13. a. 102�200 � .51

b. .49, found by 61�200 � 37�200 � .305 � .185. Special rule of addition.

15. P(above C) � .25 � .50 � .75 17. P(A or B) � P(A) � P(B) � P(A and B)

� .20 � .30 � .15 � .35 19. When two events are mutually exclusive, it means that if

one occurs, the other event cannot occur. Therefore, the probability of their joint occurrence is zero.

21. a. P(P and F) � 0.20 b. P(P and D) � 0.30 c. No d. Joint probability e.

23. P(A and B) � P(A) � P(B A) � .40 � .30 � .12 25. .90, found by (.80 � .60) � .5.

.10, found by (1 � .90). 27. a. P(A1) � 3�10 � .30

b. P(B1 A2) � 1�3 � .33 c. P(B2 and A3) � 1�10 � .10

29. a. A contingency table b. .27, found by 300�500 � 135�300 c. The tree diagram would appear as:

ƒ � 1 � .10 � .90

P(P or D or F ) � 1 � P(P and D and F )

50 75 100 125 150 175 200

Salary

Boxplot of Salary

50 75 150 Salary

Scatter Diagram of Wins vs. Salary

100

W in

200175100 125

Higher salaries lead to more wins. d.

The distribution is fairly uniform between 59 and 103.

Wins

Dotplot of Wins

60 66 72 78 84 90 96 102

Conditional Probabilities

Joint Probabilities

50/ 500

Bel ow

Av era

Above Average 300/500

150/500 Average

16/50

12/50 22/50

45/15 0

60/150

45/150

93/30 0

72/300

135/300

Fair

Good

Excel Fair

Good

Excel Fair

Good

Excel

(50/500) (16/50) = .032

(50/500) (12/50) = .024

(50/500) (22/50) = .044

(150/500) (45/150) = .090

(150/500) (60/150) = .120

(150/500) (45/150) = .090

(300/500) (93/300) = .186

(300/500) (72/300) = .144

(300/500) (135/300) = .270

Total 1.000

3. a. .176, found by b. Empirical

5. a. Empirical b. Classical c. Classical d. Empirical, based on seismological data

7. a. The survey of 40 people about environmental issues b. 26 or more respond yes, for example. c. 10�40 � .25 d. Empirical e. The events are not equally likely, but they are mutually

exclusive.

6 34

31. Probability the first presentation wins � 3�5 � .60 Probability the second presentation wins � (2�5)(3�4) � .30 Probability the third presentation wins � (2�5)(1�4)(3�3) � .10

33.

35.

� (.70)(.50)

[(.70)(.50)] � [(.30)(.90)] � .5645

P(night ƒ win) � P(night)P(win ƒ night) P(night)P(win ƒ night) � P(day)P(win ƒ day)

� .60 � .05

(.60 � .05) � (.40 � .10) � .4286

P(A1 ƒ B1) � P(A1) � P(B1 ƒ A1)

P(A1) � P(B1 ƒ A1) � P(A2) � P(B1 ƒ A2)

Lin01803_appc_782-828.qxd 11/18/10 12:51 PM Page 790

791

37.

39. a. 78,960,960 b. 840, found by (7)(6)(5)(4). That is 7!�3! c. 10, found by 5!�3!2!

41. 210, found by (10)(9)(8)(7)�(4)(3)(2) 43. 120, found by 5! 45. 10,897,286,400, found by

15P10 � (15)(14)(13)(12)(11)(10)(9)(8)(7)(6) 47. a. Asking teenagers to compare their reactions to a newly

developed soft drink. b. Answers will vary. One possibility is more than half of

the respondents like it. 49. Subjective 51. a. 4/9, found by

b. because 53. a. .8145, found by (.95)4

b. Special rule of multiplication c. P(A and B and C and D) � P(A) � P(B) � P(C) � P(D)

55. a. .08, found by .80 � .10 b. No; 90% of females attended college, 78% of males c.

(3/4) # (2/3) � 0.5.3/4, (2/3) # (2/3).

� (.30)(.20)

(.30)(.20) � (.30)(.90) � (.40)(.60) � .1053

� P(cash or check) P( 7 $50 ƒ cash or check) P(cash or check) P( 7 $50 ƒ cash or check)

� P(credit) P( 7 $50 ƒ credit) � P(debit) P( 7 $50 ƒ debit)

P(cash or check ƒ 7 $50)

Female

Total 1.000

.80

.20

Male

College Attended .80 � .90 = .720.90

.10

.78

.22

attended .80 � .10 = .080

Attended .20 � .78 = .156

Not

JointSex

attended .20 � .22 = .044

d. Yes, because all the possible outcomes are shown on the tree diagram.

57. a. 0.57, found by 57�100 b. 0.97, found by (57�100) � (40�100) c. Yes, because an employee cannot be both. d. 0.03, found by 1 � 0.97

59. a. 1�2, found by (2�3)(3�4) b. 1�12, found by (1�3)(1�4) c. 11�12, found by 1 � 1�12

61. a. 0.9039, found by (0.98)5

b. 0.0961, found by 1 � 0.9039 63. a. 0.0333, found by (4�10)(3�9)(2�8)

b. 0.1667, found by (6�10)(5�9)(4�8) c. 0.8333, found by 1 � 0.1667 d. Dependent

65. a. 0.3818, found by (9�12)(8�11)(7�10) b. 0.6182, found by 1 � 0.3818

67. a. P(S) P(R S) � .60(.85) � 0.51 b. P(S) P(PR S) � .60(1 � .85) � 0.09

69. a.

b. P(defective/not perfect) � .031 .143

� .217

� 112

1,000 �

31 1,000

� .143

P(not perfect) � P(bad sector) � P(defective) ƒ#

ƒ#

Supplier Joint Revised

Tyson .00600 .1159 Fuji .01200 .2319 Kirkpatricks .01750 .3382 Parts .01625 .3140

.05175 1.0000

71.

73. a. P(P or D) � (1�50)(9�10) � (49�50)(1�10) � 0.116 b. P(No) � (49�50)(9�10) � 0.882 c. P(No on 3) � (0.882)3 � 0.686 d. P(at least one prize) � 1 � 0.686 � 0.314

75. Yes, 256 is found by 28. 77. .9744, found by 1 � (.40)4

79. a. .185, found by (.15)(.95) � (.05)(.85) b. .0075, found by (.15)(.05)

81. a. P(F and �60) � .25, found by solving with the general rule of multiplication:

b. 0 c. .3333, found by 1�3

83. 264 � 456,976 85. 1�3, 628,800 87. a.

b. P(Tyson ƒ defective) � .20(.03) .20(.03) � .30(.04)

� .25(.07) � .25(.065)

� .1159

� .05175 P(D) � .20(.03) � .30(.04) � .25(.07) � .25(.065)

P(F ) # P(760 ƒ F) � (.5)(.5)

� .0294

P(poor ƒ profit) � .10(.20) .10(.20) � .60(.80) � .30(.60)

89. 0.512, found by (0.8)3

91. .525, found by 1 � (.78)3

93. a. Winning Attendance

Season Low Moderate High Total

No 9 3 2 14 Yes 2 7 7 16

Total 11 10 9 30

Losing Winning Season Season Total

New 8 8 16 Old 6 8 14

Total 14 16 30

1. 0.5333, found by 16�30 2. 0.6000, found by 16�30 � 9�30 � 7�30 � 18�30 3. 0.7778, found by 7�9 4. 0.0667, found by 2�30

1. 0.53330, found by 16�30 2. 0.2667, found by 8�30 3. 0.8000, found by 16�30 � 16�30 � 8�30

Lin01803_appc_782-828.qxd 11/18/10 12:52 PM Page 791

792

CHAPTER 6 1. Mean � 1.3, variance � .81, found by:

3. Mean � 14.5, variance � 27.25, found by:

5. a. � 27.25 � (15 � 14.5)2(.2) � (20 � 14.5)2(.4)

�2 � (5 � 14.5)2(.1) � (10 � 14.5)2(.3)

� 5(.1) � 10(.3) � 15(.2) � 20(.4) � 14.5

� .81 � (2 � 1.3)2(.3) � (3 � 1.3)2(.1)

�2 � (0 � 1.3)2(.2) � (1 � 1.3)2(.4)

� 0(.20) � 1(.40) � 2(.30) � 3(.10) � 1.3

a. � �xP(x) � 21 b. �2 � �(x � )2P(x) � 259

9. a.

11. a.

13. a. .2668, found by

b. .1715, found by

c. .0404, found by

15. a. .2824, found by

b. .3765, found by

c. .2301, found by

d. � 1.2, found by 12(.10)

� � 1.0392, found by 21.08

P(2) � 12!

(12 � 2)!2! (.10)2(.9)10

P(1) � 12!

(12 � 1)!1! (.10)1(.9)11

P(0) � 12!

(12 � 0)!0! (.10)0(.9)12

P(0) � 9!

(9 � 0)!0! (.3)0(.7)9

P(4) � 9!

(9 � 4)!4! (.3)4(.7)5

P(2) � 9!

(9 � 2)!2! (.3)2(.7)7

� � 20.72 � .8485 �2 � 0.72 � 1.8

X P(X)

0 .064 1 .288 2 .432 3 .216

P(3) � 4!

3!(4 � 3)! (.25)3(.75)4�3 � .0469

P(2) � 4!

2!(4 � 2)! (.25)2(.75)4�2 � .2109

� � 2259 � 16.093

Calls, (x � �)2

x Frequency P(x) xP(x) P(x)

0 8 .16 0 .4624 1 10 .20 .20 .0980 2 22 .44 .88 .0396 3 9 .18 .54 .3042 4 1 .02 .08 .1058

50 1.70 1.0100

b. Discrete distribution, because only certain outcomes are possible.

c. � �x P(x) � 1.70 d.

7. � � 21.01 � 1.005

Amount P(x) xP(x) (x � �)2 P(x)

10 .50 5 60.50 25 .40 10 6.40 50 .08 4 67.28

100 .02 2 124.82

21 259.00

17. a. 0.1858, found by

b. 0.1416, found by

c. 3.45, found by (0.23)(15) 19. a. 0.296, found by using Appendix B.9 with n of 8, � of

0.30, and x of 2 b. P(x � 2) � 0.058 � 0.198 � 0.296 � 0.552 c. 0.448, found by P(x � 3) � 1 � P(x � 2) � 1 � 0.552

21. a. 0.387, found from Appendix B.9 with n of 9, � of 0.90, and x of 9

b. P(X � 5) � 0.001 c. 0.992, found by 1 � 0.008 d. 0.947, found by 1 � 0.053

23. a. � 10.5, found by 15(0.7) and 1.7748

b. 0.2061, found by

c. 0.4247, found by 0.2061 � 0.2186 d. 0.5154, found by

0.2186 � 0.1700 � 0.0916 � 0.0305 � 0.0047

25.

27.

29.

31. a. .6703 b. .3297

33. a. .0613 b. .0803

35. � 6

37. A random variable is a quantitative or qualitative outcome that results from a chance experiment. A probability distribution also includes the likelihood of each possible outcome.

39. � $1,000(.25) � $2,000(.60) � $5,000(.15) � $2,200 �2 � (1,000 � 2,200)2 .25 � ($2,000 � $2,200)2 .60 �

(5,000 � 2,200)2 .15 � 1,560,000

41. � 12(.25) � � 15(.1) � 13.2

43. a. � 10(.35) � 3.5 b. P (X � 4) � 10C4 (.35)

4 (.65)6 � 210(.0150) (.0754) � .2375 c. P (X 4) � 10Cx (.35)

X (.65)10�X

� .2375 � .1536 � � .0000 � .4862 45. a. 6, found by 0.4 � 15

b. 0.0245, found by

c. 0.0338, found by 0.0245 � 0.0074 � 0.0016 � 0.0003 � 0.0000

d. 0.0093, found by 0.0338 � 0.0245 47. a. � 20(0.075) � 1.5

b. 0.2103, found by

c. 0.7897, found by 1 � 0.2103

49. a. 0.1311, found by

b. 2.4, found by (0.15)(16) c. 0.2100, found by

1 � 0.0743 � 0.2097 � 0.2775 � 0.2285

51. P(2) � [6C2][4C2]

[10C4] �

(15)(6) 210

� 0.4286

16! 4!12!

(0.15)4(0.85)12

20! 0!20!

(0.075)0(0.925)20 � � 220(0.075)(0.925) � 1.1779

15! 10!5!

(0.4)10(0.6)5

· · · �

� � 20.86 � .927

�2 � (12 � 13.2)2.25 � · · · � (15 � 13.2)2.10 � 0.86 · · ·

� .7149 � 1 � (.0025 � .0149 � .0446 � .0892 � .1339)P(X � 5)

P(2) � [9C3][6C2]

[15C5] �

84(15) 3003

� .4196

P(0) � [7C2][3C0]

[10C2] �

21(1) 45

� .4667

P(2) � [6C2][4C1]

10C3 �

15(4) 120

� .50

15! 10!5!

(0.7)10(0.3)5

� � 215(0.7)(0.3) �

15! 5!10!

(0.23)5(0.77)10

15! 2!13!

(0.23)2(0.77)13

Lin01803_appc_782-828.qxd 11/18/10 12:52 PM Page 792

793

53. a. 0 0.0002 1 0.0019 2 0.0116 3 0.0418 4 0.1020 5 0.1768 6 0.2234

7 0.2075 8 0.1405 9 0.0676

10 0.0220 11 0.0043 12 0.0004

b. � 12(0.52) � 6.24

c. 0.1768 d. 0.3343, found by 0.0002 � 0.0019

� 0.0116 � 0.0418 � 0.1020 � 0.1768

55. a.

57.

59. a. .0498 b. .7746, found by (1 � .0498)5

61. � 4.0, from Appendix B.5 a. .0183 b. .1954 c. .6289 d. .5665

63. a. 0.1733, found by

b. 0.0450, found by

c. 0.9550, found by 1 � 0.0450

65.

67. Let � n� � 155(1�3,709) � 0.042

Very unlikely!

69. a. � n� � 15(.67) � 10.05

b. P(8) �15C8(.67) 8(.33)7 � 6435(.0406)(.000426) � .1114

c. P(x � 8) � .1114 � .1759 � · · · � .0025 � .9163 71. The mean number of home runs per game is 2.0749, found

by 5042�(15 � 162).

c. P(X �� 4) � 0.1566, found by 1 � (0.1257 � 0.2605 � 0.2703 � 0.1869)

CHAPTER 7 1. a. b � 10, a � 6

d. Area � 1

(10 � 6) # (10 � 6)

1 � 1

� � B

(10 � 6)2

12 � 1.1547

� 6 � 10

2 � 8

P(2) � 2.07492 e�2.0749

2! � 0.2703

P(0) � 2.07490 e�2.0749

0! � 0.1257

� � 2n�(1 � �) � 215(.67)(.33) � 1.8211

P(5) � 0.0425 e�0.042

5! � 0.000000001

P(0) � (.407)0 e� .407

0! � 0.6656

P(2) � (.407)2e� .407

2! � 0.0551

� n� � 23a 2

113 b � .407

(3.1)0 e�3.1

(3.1)4 e�3.1

P(X � 0) � [8C4][4C0]

[12C4] �

70 495

� .141

P(X � 1) � 1 � P(0) � 1 � .2917 � .7083

P(0) � [7C3][3C0]

[10C3] �

(35)(1) 120

� .2917

P(1) � [7C2][3C1]

[10C3] �

(21)(3) 120

� .5250

� � 212(0.52)(0.48) � 1.7307

3. a. 0.30, found by (30 � 27)�(30 � 20) b. 0.40, found by (24 � 20)�(30 � 20)

5. a. a � 0.5, b � 3.00

d. 0, found by

7. The actual shape of a normal distribution depends on its mean and standard deviation. Thus, there is a normal distribution, and an accompanying normal curve, for a mean of 7 and a standard deviation of 2. There is another normal curve for a mean of $25,000 and a standard deviation of $1,742, and so on.

9. a. 490 and 510, found by 500 1(10) b. 480 and 520, found by 500 2(10) c. 470 and 530, found by 500 3(10)

11.

Adjusting for their industries, Rob is well below average and Rachel well above.

13. a. 1.25, found by

b. 0.3944, found in Appendix B.1

c. 0.3085, found by

Find 0.1915 in Appendix B.1 for z � �0.5, then 0.5000 � 0.1915 � 0.3085

15. a. 0.3413, found by ,

then find 0.3413 in Appendix B.1 for z � 1 b. 0.1587, found by 0.5000 � 0.3413 � 0.1587

c. 0.3336, found by

Find 0.1664 in Appendix B.1, for z � �0.43, then 0.5000 � 0.1664 � 0.3336

17. a. 0.8276: First find z � �1.5, found by (44 � 50)�4 and z � 1.25 � (55 � 50)�4. The area between �1.5 and 0 is 0.4332 and the area between 0 and 1.25 is 0.3944, both from Appendix B.1. Then adding the two areas we find that 0.4332 � 0.3944 � 0.8276.

b. 0.1056, found by 0.5000 � .3944, where z � 1.25 c. 0.2029: Recall that the area for z � 1.25 is 0.3944, and

the area for z � 0.5, found by (52 � 50)�4, is 0.1915. Then subtract 0.3944 � 0.1915 and find 0.2029.

19. a. 0.3264, found by 0.5000 � 0.1736, where z � 0.45, found by [(3000 � 2708)/650].

b. 0.2152; the z-value for $3,500 is 1.22, found by [(3500 � 2708)�650], and the corresponding area is 0.3888, which leads to 0.3888 � 0.1736 � 0.2152.

c. 0.5143; the z-value for $2,500 is � 0.32, found by [(2,500 � 2,708)�650], and the corresponding area is 0.1255, which leads to 0.1255 � 0.3888 � 0.5143.

z � $19.00 � $20.50

$3.50 � �0.43

z � $24 � $20.50

$3.50 � 1.00

z � 18 � 20

2.5 � �0.5

z � 25 � 20

4.0 � 1.25

ZRachel � $50,000 � $35,000

$8,000 � 1.875

ZRob � $50,000 � $60,000

$5,000 � �2

P(x 7 1.5) � 1

(3.0 � 0.5) # 3.0 � 1.5

1 �

1.5 2.5

� 0.6

1 (3.0 � 0.5)

(1.0 � 1.0)

P(x 6 1) � 1

(3.0 � 0.5) # 1 � .5

1 �

.5 2.5

� 0.2

� � B

(3.00 � .50)2

12 � .72

� 0.5 � 3.00

2 � 1.75

P(7 � x � 9) � 1

(10 � 6) # (9 � 7)

1 �

2 4

� .50

P(X 7 7) � 1

(10 � 6) # 10 � 7

1 �

3 4

� .75

Lin01803_appc_782-828.qxd 11/18/10 12:52 PM Page 793

21. a. 0.0764, found by z � (20 � 15)�3.5 � 1.43, then 0.5000 � 0.4236 � 0.0764

b. 0.9236, found by 0.5000 � 0.4236, where z � 1.43 c. 0.1185, found by z � (12 � 15)�3.5 � �0.86.

The area under the curve is 0.3051, then z � (10 � 15)�3.5 � �1.43. The area is 0.4236. Finally, 0.4236 � 0.3051 � 0.1185.

23. X � 56.60, found by adding 0.5000 (the area left of the mean) and then finding a z value that forces 45 percent of the data to fall inside the curve. Solving for X: 1.65 � (X � 50)�4 � 56.60.

25. $1,630, found by $2,100 � 1.88($250) 27. a. 214.8 hours: Find a z value where 0.4900 of area is

between 0 and z. That value is z � 2.33. Then solve for X: 2.33 � (X � 195)�8.5, so X � 214.8 hours.

b. 270.2 hours: Find a z value where 0.4900 of area is between 0 and (�z). That value is z � �2.33. Then solve for X: �2.33 � (X � 290)�8.5, so X � 270.2 hours.

29. 41.7 percent, found by 12 � 1.65(18) 31. a. � n� � 50(0.25) � 12.5

794

� � 29.375 � 3.0619

�2 � n� (1 � �) � 12.5(1 � 0.25) � 9.375

b. 0.2578, found by (14.5 � 12.5)�3.0619 � 0.65. The area is 0.2422. Then 0.5000 � 0.2422 � 0.2578.

c. 0.2578, found by (10.5 � 12.5)�3.0619 � �0.65. The area is 0.2422. Then 0.5000 � 0.2422 � 0.2578.

33. a. � � 25.208 � 2.2821

� n� � 80(0.07) � 5.6

0.3483, found from z � (6.5 � 5.6)�2.2821 � 0.39 with the corresponding area of 0.1517, then 0.5000 � 0.1517 � 0.3483

b. 0.5160, found from z � (5.5 � 5.6)�2.2821 � � 0.04 with the corresponding area of 0.0160, then 0.5000 � 0.0160 � 0.5160

c. .1677, found by .5160 � 0.3483. 35. a. Yes. (1) There are two mutually exclusive outcomes:

overweight and not overweight. (2) It is the result of counting the number of successes (overweight members). (3) Each trial is independent. (4) The probability of 0.30 remains the same for each trial.

b. 0.0084, found by

The area under the curve for 2.39 is 0.4916. Then 0.5000 � 0.4916 � 0.0084.

c. 0.8461, found by

The area between 139.5 and 150 is 0.3461. Adding 0.3461 � 0.5000 � 0.8461.

37. a. 0.3935, found by 1 � e[(�1�60) (30)]

b. 0.1353, found by e[(�1�60) (120)]

c. 0.1859, found by e[(�1�60) (45)] � e[(�1�60) (75)]

d. 41.59 seconds, found by �60 ln(0.5) 39. a. 0.5654, found by 1 � e[(�1�18) (15)]

and 0.2212, found by 1 � e[(�1�60) (15)]

b. 0.0013, found by e[(�1�18) (120)]

and 0.1353, found by e[(�1�60) (120)]

c. 0.1821, found by e[(�1�18) (30)] � e[(�1�18) (90)]

and 0.3834, found by e[(�1�60) (30)] � e[(�1�60) (90)]

d. 4 minutes, found by �18 ln(0.8) and 13.4 minutes, found by �60 ln(0.8)

41. a.

b. � � B

(12.05 � 11.96)2

12 � .0260

� 11.96 � 12.05

2 � 12.005

z � 139.5 � 150

10.24695 � �1.02

z � X �

� �

174.5 � 150 10.24695

� 2.39

� � 2105 � 10.24695

�2 � 500(.30)(.70) � 105 � 500(0.30) � 150

e. All cans have more than 11.00 ounces, so the probability is 100%.

43. a.

45. a. �0.4 for net sales, found by (170 � 180)�25. 2.92 for employees, found by (1,850 � 1,500)�120.

b. Net sales are 0.4 standard deviations below the mean. Employees is 2.92 standard deviations above the mean.

c. 65.54 percent of the aluminum fabricators have greater net sales compared with Clarion, found by 0.1554 � 0.5000. Only 0.18 percent have more employees than Clarion, found by 0.5000 � 0.4982.

47. a. 0.5000, because

b. 0.2514, found by 0.5000 � 0.2486 c. 0.6374, found by 0.2486 � 0.3888 d. 0.3450, found by 0.3888 � 0.0438

49. a. 0.3015, found by 0.5000 � 0.1985 b. 0.2579, found by 0.4564 � 0.1985 c. 0.0011, found by 0.5000 � 0.4989 d. 1,818, found by 1,280 � 1.28(420)

51. a. 90.82%: First find z � 1.33, found by (40 � 34)�4.5. The area between 0 and 1.33 is 0.4082. Then add 0.5000 and 0.4082 and find 0.9082 or 90.82%.

b. 78.23%: First find z � �0.78 found by (25 � 29)�5.1. The area between 0 and (�0.78) is 0.2823. Then add 0.5000 and 0.2823 and find 0.7823 or 78.23%.

c. 44.5 hours/week for women: Find a z value where 0.4900 of the area is between 0 and z. That value is 2.33. Then solve for X: 2.33 � (X � 34)�4.5, so X � 44.5 hours/week. 40.9 hours/week for men: 2.33 � (X � 29)�5.1, so X � 40.9 hours/week.

53. About 4,099 units, found by solving for X. 1.65 � (X � 4,000)�60

55. a. 15.39%, found by (8 � 10.3)�2.25 � �1.02, then 0.5000 � 0.3461 � 0.1539.

b. 17.31%, found by: z � (12 � 10.3)�2.25 � 0.76. Area is 0.2764. z � (14 � 10.3)�2.25 � 1.64. Area is 0.4495. The area between 12 and 14 is 0.1731, found by 0.4495 � 0.2764.

c. Yes, but it is rather remote. Reasoning: On 99.73% of the days, returns are between 3.55 and 17.05, found by 10.3 3(2.25). Thus, the chance of less than 3.55 returns is rather remote.

57. a. 0.9678, found by:

Then (31.5 � 38.4)�3.72 � �1.85, for which the area is 0.4678. Then 0.5000 � 0.4678 � 0.9678.

b. 0.0853, found by (43.5 � 38.4)�3.72 � 1.37, for which the area is 0.4147. Then 0.5000 � 0.4147 � .0853.

c. 0.8084, found by 0.4441 � 0.3643 d. 0.0348, found by 0.4495 � 0.4147

� � 213.824 � 3.72

�2 � 60(0.64)(0.36) � 13.824 � 60(0.64) � 38.4

z � 30 � 490

90 � �5.11

P(X 7 5) � 1

(10 � 4) # a10 � 5

1 b �

5 6

� .83

P(X 6 6) � 1

(10 � 4) # a6 � 4

1 b �

2 6

� .33

� � B

(10 � 4)2

12 � 1.732

� 4 � 10

2 � 7

� .07 .09

� .78

P(X 7 11.98) � 1

(12.05 � 11.96) a12.05 � 11.98

1 b

P(X 6 12) � 1

(12.05 � 11.96) 12.00 �11.96

1 �

.04

.09 � .44

Lin01803_appc_782-828.qxd 11/18/10 12:52 PM Page 794

795

59. 0.0968, found by:

The area is 0.4032. Then, for 25 or more, 0.5000 � 0.4032 � 0.0968.

61. a. b. c. z � (30 � 28.5)�10 � 0.15,

then 0.5000 � 0.0596 � 0.5596 63. a. 21.19 percent found by z � (9.00 � 9.20)�0.25 � �0.80,

so 0.5000 � 0.2881 � 0.2119 b. Increase the mean. z � (9.00 � 9.25)�0.25 � �1.00,

P � 0.5000 � 0.3413 � 0.1587. Reduce the standard deviation. � � (9.00 � 9.20)� 0.15 � �1.33; P � 0.5000 � 0.4082 � 0.0918. Reducing the standard deviation is better because a smaller percent of the hams will be below the limit.

65. a. z � (60 � 52)�5 � 1.60, so 0.5000 � 0.4452 � 0.0548 b. Let z � 0.67, so 0.67 � (X � 52)�5 and X � 55.35, set

mileage at 55,350 c. z � (45 � 52)�5 � �1.40, so 0.5000 � 0.4192 � 0.0808

67. and

69.

71. a. 0.4262, found by 1 � e[(�1�27) (15)]

b. 0.1084, found by e[(�1�27) (60)]

c. 0.1403, found by e[(�1�27) (30)] � e[(�1�27) (45)]

d. 2.84 secs, found by �27 ln(0.9) 73. a. 0.2835, found by 1 � e[(�1�300,000) (100,000)]

b. 0.1889, found by e[(�1�300,000) (500,000)]

c. 0.2020, found by e[(�1�300,000) (200,000)] � e[(�1�300,000) (350,000)]

d. Both the mean and standard deviation are 300,000 hours.

75. a. 0.0655, found by 0.5000 � 0.4345 with z � (3500 � 2448)�698 � 1.51; leads to 2.0 teams, found by 30(0.0655). Three teams actually had attendance of more than 3.5 million, so the estimate is fairly accurate.

b. 0.8729, found by 0.5000 � 0.3729, with z � (50 � 88.51)�33.90 � �1.14; leads to 26.2 teams, found by 30(0.8729). Twenty-seven teams actually had salaries of more than $50 million, so the estimate is accurate.

CHAPTER 8 1. a. 303 Louisiana, 5155 S. Main, 3501 Monroe,

2652 W. Central b. Answers will vary. c. 630 Dixie Hwy, 835 S. McCord Rd, 4624 Woodville Rd d. Answers will vary.

3. a. Bob Schmidt Chevrolet Great Lakes Ford Nissan Grogan Towne Chrysler Southside Lincoln Mercury Rouen Chrysler Jeep Eagle

b. Answers will vary. c. Yark Automotive

Thayer Chevrolet Toyota Franklin Park Lincoln Mercury Mathews Ford Oregon Inc. Valiton Chrysler

P(z 7 1.60) � .05000 � 0.4452 � 0.0548 z � (29.5 � 22.5)�4.37 � 1.60 � 150(0.15) � 22.5 � � 2150(0.15)(0.85) � 4.37 � 462,718

470 � �

� 0.25 500 �

� � 1.28 � � 29,126

1.65 � (45 � )�10 � 28.5 1.65 � (45 � )�5 � 36.75

z � (24.5 � 20)�3.46 � 1.30. � � 212 � 3.46

�2 � 50(0.40)(0.60) � 12 � 50(0.40) � 20

5. a.

c. More dispersion with population data compared to the sample means. The sample means vary from 12 to 15, whereas the population varies from 12 to 16.

7. a.

c. The dispersion of the population is greater than that of the sample means. The sample means vary from 12.66 to 16.33, whereas the population varies from 12 to 20.

9. a. 20, found by 6C3 b.

They are equal. d.

� 2.67, found by (3 � 6 � 3 � 3 � 0 � 1)�6.

X � 2.67, found by 53.33

20 .

Sample Cases Sum Mean

Ruud, Wu, Sass 3, 6, 3 12 4.00 Ruud, Sass, Flores 3, 3, 3 9 3.00

o o o o Sass, Flores, Schueller 3, 3, 1 7 2.33

� (12 � 12 � 14 � 15 � 20)�5 � 14.6

X � (12.66 � · · · � 15.33 � 15.33)

10 � 14.6

Sample Values Sum Mean

1 12, 12, 14 38 12.66 2 12, 12, 15 39 13.00 3 12, 12, 20 44 14.66 4 14, 15, 20 49 16.33 5 12, 14, 15 41 13.66 6 12, 14, 15 41 13.66 7 12, 15, 20 47 15.66 8 12, 15, 20 47 15.66 9 12, 14, 20 46 15.33

10 12, 14, 20 46 15.33

� (12 � 12 � 14 � 16)�4 � 13.5 X � (12 � 13 � 14 � 13 � 14 � 15)�6 � 13.5

Sample Values Sum Mean

1 12, 12 24 12 2 12, 14 26 13 3 12, 16 28 14 4 12, 14 26 13 5 12, 16 28 14 6 14, 16 30 15

.50

.40

.30

.20

.10

Pr ob

ab ili

Number of cases

Population

0 2 µ 4

Lin01803_appc_782-828.qxd 11/18/10 12:52 PM Page 795

11. a.

� 0 � 1 � · · · � 9

10 � 4.5

.05

0 1 2 3 4 5 6 7 8 9

796

The population has more dispersion than the sample means. The sample means vary from 1.33 to 4.0. The population varies from 0 to 6.

.50

.40

.30

.20

.10

Pr ob

ab ili

Sample means number of cases

Distribution of Sample Means

1.33 2.33 3.33 3.0 4.02.01

Sample Mean Number of Means Probability

1.33 3 .1500 2.00 3 .1500 2.33 4 .2000 3.00 4 .2000 3.33 3 .1500 4.00 3 .1500

20 1.0000

Sample Sum

1 11 2.2 2 31 6.2 3 21 4.2 4 24 4.8 5 21 4.2

Sample Sum

6 20 4.0 7 23 4.6 8 29 5.8 9 35 7.0

10 27 5.4

Fr eq

ue nc

Values 2.0 3 4 5 6 7 8

The mean of the 10 sample means is 4.84, which is close to the population mean of 4.5. The sample means range from 2.2 to 7.0, whereas the population values range from 0 to 9. From the above graph, the sample means tend to cluster between 4 and 5.

13. a.–c. Answers will vary depending on the coins in your possession.

15. a.

c. P � .6147, found by 0.3413 � 0.2734

17. or virtually certain

19. a. Formal Man, Summit Stationers, Bootleggers, Leather Ltd, Petries

b. Answers may vary. c. Elder-Beerman, Frederick’s of Hollywood, Summit

Stationers, Lion Store, Leather Ltd., Things Remembered, County Seat, Coach House Gifts, Regis Hairstylists

21. a.

z � 1,950 � 2,200

250�150 � �7.07 P � 1,

P � .1587, found by .5000 � .3413

z � 56 � 60 12�19

� �1.00

P � .2266, found by .5000 � .2734

z � 63 � 60 12�19

� 0.75

Sample Value Sum Mean

1 79, 64 143 71.5 2 79, 84 163 81.5

o o o o 15 92, 77 169 84.5

1,195.0

b. Mean of sample means is (1.0 � 1.5 � 2.0 � � 3.0)�9 � 18�9 � 2.0. The population mean is (1 � 2 � 3)�3 � 6�3 � 2. They are the same value.

c. Variance of sample means is (1.0 � 0.25 � 0.0 � �p

Deviation from Square of Samples Mean Mean Deviation

1, 1 1.0 �1.0 1.0 1, 2 1.5 �0.5 0.25 1, 3 2.0 0.0 0.0 2, 1 1.5 �0.5 0.25 2, 2 2.0 0.0 0.0 2, 3 2.5 0.5 0.25 3, 1 2.0 0.0 0.0 3, 2 2.5 0.5 0.25 3, 3 3.0 1.0 1.0

3.0)�9 � 18�9 � 2.0. The population mean is (1 � 1.0)�9 � 1�3. Variance of the population values is (1 � 0 � 1)�3 � 2�3. The variance of the population is twice as large as that of the sample means.

d. Sample means follow a triangular shape peaking at 2. The population is uniform between 1 and 3.

23. Larger samples provide narrower estimates of a population mean. So the company with 200 sampled customers can provide more precise estimates. In addition, they are selected consumers who are familiar with laptop computers and may be better able to evaluate the new computer.

25. a. We selected 60, 104, 75, 72, and 48. Answers will vary. b. We selected the third observation. So the sample

consists of 75, 72, 68, 82, 48. Answers will vary. c. Number the first 20 motels from 00 to 19. Randomly

select three numbers. Then number the last five numbers 20 to 24. Randomly select two numbers from that group.

27. a. 15, found by 6C2 b.

Lin01803_appc_782-828.qxd 11/18/10 12:52 PM Page 796

797

c. found by 1,195�15.

found by 478�6. They are equal.

d. No. The student is not graded on all available information. He/she is as likely to get a lower grade based on the sample as a higher grade.

29. a. 10, found by 5C2 b.

� 79.67,

X � 79.67,

Number of Number of Shutdowns Mean Shutdowns Mean

4, 3 3.5 3, 3 3.0 4, 5 4.5 3, 2 2.5 4, 3 3.5 5, 3 4.0 4, 2 3.0 5, 2 3.5 3, 5 4.0 3, 2 2.5

Sample Mean Frequency Probability

2.5 2 .20 3.0 2 .20 3.5 3 .30 4.0 2 .20 4.5 1 .10

10 1.00

The two means are equal. d. The population values are relatively uniform in shape.

The distribution of sample means tends toward normality.

31. a. The distribution will be normal.

e. 0.4922, found by 0.3186 � 0.1736

33.

35.

37. Between 5,954 and 6,046, found by

39.

41. a. Alaska, Connecticut, Georgia, Kansas, Nebraska, South Carolina, Virginia, Utah

b. Arizona, Florida, Iowa, Massachusetts, Nebraska, North Carolina, Rhode Island, Vermont

43. a.

or virtually never

P � 0.5000 � 0.4864 � 0.0136

z � 500 � 510 14.28�110

� �2.21,

P � 0.4864 � 0.5000 � 0.9864

z � 500 � 510 14.28�110

� �2.21,

z � 600 � 510 14.28�110

� 19.9, P � 0.00,

P � 0.0375, found by 0.5000 � 0.4625

z � 900 � 947 205�160

� �1.78

6,000 1.96 (150�140)

P � 0.8238, found by 0.5000 � 0.3238

z � 25.1 � 24.8

2.5�160 � 0.93

P � 0.9826, found by 0.5000 � 0.4826

z � $335 � $350

$45�140 � �2.11

P � 0.6736, found by 0.5000 � 0.1736

z � 34.5 � 35 5.5�125

� �0.45

P � 0.1814, found by 0.5000 � 0.3186

z � 36 � 35 5.5�125

� 0.91

� ˛ X �

5.5 125

� 1.1

� (4 � 3 � 5 � 3 � 2)�5 � 3.4 X � (3.5 � 4.5 � · · · � 2.5)�10 � 3.4

45. a.

47. Mean 2009 attendance is 2.448 million. Likelihood of a sample mean this large or larger is 0.0606, found by 0.5000 � 0.4394. The z value is 1.55.

CHAPTER 9 1. 51.314 and 58.686, found by 3. a. 1.581, found by

b. The population is normally distributed and the population variance is known.

c. 16.901 and 23.099, found by 20 3.099 5. a. $20. It is our best estimate of the population mean.

b. $18.60 and $21.40, found by . About 95 percent of the intervals similarly constructed will include the population mean.

7. a. 8.60 gallons. b. 7.83 and 9.37, found by c. If 100 such intervals were determined, the population

mean would be included in about 99 intervals. 9. a. 2.201

b. 1.729 c. 3.499

11. a. The population mean is unknown, but the best estimate is 20, the sample mean.

b. Use the t distribution since the standard deviation is unknown. However, assume the population is normally distributed.

c. 2.093 d. Between 19.06 and 20.94, found by

e. Neither value is reasonable, because they are not inside the interval.

13. Between 95.39 and 101.81, found by

15. a. 0.8, found by 80�100 b. Between 0.72 and 0.88, found by

c. We are reasonably sure the population proportion is between 72 and 88 percent.

17. a. 0.625, found by 250�400 b. Between 0.563 and 0.687, found by

c. We are reasonably sure the population proportion is between 56 and 69 percent.

19. 33.41 and 36.59, found by

21. 1.683 and 2.037, found by

23. 97, found by

25. 196, found by n � 0.15(0.85)a 1.96 0.05

b 2

� 195.9216

n � a 1.96 � 10

2 b

� 96.04

1.86 2.680a 0.5 150

b B

400 � 50 400 � 1

35 2.030a 5 136

b B

300 � 36 300 � 1

0.625 2.58a B

0.625(1 � 0.625) 400

0.8 1.96a B

0.8(1 � 0.8) 100

98.6 1.833(5.54�110)

20 2.093(2�120)

8.60 2.58(2.30�160)

$20 1.96($5�149)

�x � 25�1250 55 2.58(10�149)

.0162 found by .5000 � .4838 P � .3577 � .3577 � .7154

z � 6.75 � 6.5 2.1�181

� 1.07, z � 6.25 � 6.5 2.1�181

� �1.07,

P � .4838 � .4838 � .9676

z � 7.0 � 6.5 2.1�181

� 2.14, z � 6.0 � 6.5 2.1�181

� �2.14,

� X � 2.1 181

� 0.23

Lin01803_appc_782-828.qxd 11/18/10 12:53 PM Page 797

27. 554, found by

29. a. 577, found by

b. 601, found by

31. 6.13 years to 6.87 years, found by 33. a. Between $313.41 and 332.59, found by

b. $350 is not reasonable, because it is outside of the confidence interval.

35. a. The population mean is unknown. b. Between 7.50 and 9.14, found by

c. 10 is not reasonable because it is outside the confidence interval.

37. a. 65.49 up to 71.71 hours, found by b. The value suggested by the NCAA is included in the

confidence interval. Therefore, it is reasonable. c. Changing the confidence interval to 95 would reduce

the width of the interval. The value of 2.680 would change to 2.010.

39. 61, found by 41. Between $13,734 up to $15,028, found by

15,000 is reasonable because it is inside the confidence interval.

43. a. $62.583, found by $751�12 b. Between $60.54 and $64.63, found by

c. $60 is not reasonable, because it is outside the confidence interval.

45. a. 89.4667, found by 1,342�15 b. Between 84.99 and 93.94, found by

c. Yes, because even the lower limit of the confidence interval is above 80.

47. The confidence interval is between 0.011 and 0.059, found

by . It would not be

reasonable to conclude that fewer than 5 percent of the employees are now failing the test, because 0.05 is inside the confidence interval.

49. $52.51 and $55.49, found by

51. 369, found by n � 0.60(1 � 0.60)(1.96�0.05)2

53. 97, found by [(1.96 � 500)�100]2

55. a. Between 7,849 and 8,151, found by

b. 554, found by

57. a. Between 75.44 and 80.56, found by

b. 221, found by

59. a. 30, found by 180�

b. $355.10 and $476.90, found by

c. About 1,245, found by

61. a. 708.13, rounded up to 709, found by 0.21(1 � 0.21)(1.96�0.03)2

b. 1,068, found by 0.50(0.50)(1.96�0.03)2

a 1.96(180)

10 b

$416 � 2.030a $180 136

136

n � a (1.65)(9)

1.0 b

2 78 � 2.010(9�150)

n � a (1.96)(300)

25 b

2 8,000 � 2.756(300�130)

$54.00 � 2.032 $4.50 135

(500 � 35) 500 � 1

0.035 � 2.58a B

0.035(1 � 0.035) 400

89.4667 � 2.145(8.08�115)

62.583 � 1.796(3.94�112)

14,381 � 1.711(1,892�125).

1.96(16�1n) � 4

68.6 � 2.680(8.2�150)

8.32 � 1.685(3.07�140)

323 � 2.426a 25 140

6.5 � 1.989(1.7�185)

n � 0.50(0.50)a 1.96 0.04

b 2

� 600.25

n � 0.60(0.40)a 1.96 0.04

b 2

� 576.24

n � a 1.96 � 3

0.25 b

� 553.19

798

63. Between 0.573 and 0.653, found by

. Yes, because even the

lower limit of the confidence interval is above 0.500.

65. a. Between 0.156 and 0.184, found by

b. Yes, because 18 percent is inside the confidence interval. c. 21,682; found by 0.17(1 � 0.17)[1.96�0.005]2

67. Between 12.69 and 14.11, found by 13.4 � 1.96 (6.8� ) 69. a. For selling price: 211.99 up to 230.22, found by

b. For distance: 13.685 up to 15.572, found by

c. For garage: 0.5867 up to 0.7657, found by

d. Answers may vary.

71. a. Between $438.34 and 462.24, found by

b. Between 820.72 and 839.50, found by

c. Answers will vary.

CHAPTER 10 1. a. Two-tailed

b. Reject H0 when z does not fall in the region between �1.96 and 1.96.

c. �1.2, found by d. Fail to reject H0. e. p � .2302, found by 2(.5000 � .3849). A 23.02%

chance of finding a z value this large when H0 is true. 3. a. One-tailed

b. Reject H0 when z � 1.65. c. 1.2, found by d. Fail to reject H0 at the .05 significance level e. p � .1151, found by .5000 � .3849. An 11.51 percent

chance of finding a z value this large or larger. 5. a.

b. Reject H0 if z � �1.96 or z � 1.96. c. �0.69, found by:

d. Do not reject H0. e. p � .4902, found by 2(.5000 � .2549). Crosset’s

experience is not different from that claimed by the manufacturer. If H0 is true, the probability of finding a value more extreme than this is .4902.

7. a. H0: � 6.8 H1: � � 6.8 b. Reject H0 if z � �1.65

d. H0 is rejected. e. p � 0. The mean number of DVDs watched is less than

6.8 per month. If H0 is true, there is virtually no chance of getting a statistic this small.

9. a. Reject H0 when t � 1.833.

c. Reject H0. The mean is greater than 10.

t � 12 � 10 (3�110)

� 2.108

z � 6.2 � 6.8 0.5�136

� �7.2

z � 59,500 � 60,000

(5,000�148) � �0.69

H0: � � 60,000 H1: � 60,000

z � (21 � 20)�(5�136) � 1.2

z � (49 � 50)�(5�136) � �1.2

830.11 � 1.99 a 42.19 180

450.29 � 1.99 a 53.69 180

(1.96) B

0.6762(1 � 0.6762) 105

� 0.6762 � 0.0895

0.6762 � 14.629 � (1.983)(4.874�1105) � 14.629 � 0.943

221.1 � (1.983)(47.11�1105) � 221.1 � 9.12

1352

0.17 � 1.96 B

(0.17)(1 � 0.17) 2700

.613 � 2.58a B

0.613(1 � 0.613) 1,000

Lin01803_appc_782-828.qxd 11/18/10 1:35 PM Page 798

799

11. 1

Reject H0 if t � 1.703.

Reject H0 and conclude that the mean number of calls is greater than 40 per week.

13.

Reject H0 if t � 1.833.

Reject H0 and conclude that the mean income in Wilmington is greater than $40,000.

15. a. Reject H0 if t � �3.747.

c. Do not reject H0. We cannot conclude the population mean is less than 20.

d. Between .05 and .10, about .065 17.

Reject H0 if t � 2.821.

Reject H0 and conclude that the drug has increased the amount of urine. The p-value is between 0.01 and 0.005. There is a slight probability (between one chance in 100 and one chance in 200) this rise could have arisen by chance.

19.

Reject H0 if t � 1.796.

Reject H0 and conclude that the mean number of text messages is greater than 50. The p-value is less than 0.05. There is a slight probability (less than one chance in 20) this could happen by chance.

21. a. H0 is rejected if z � 1.65. b. 1.09, found by c. H0 is not rejected.

23. a. b. H0 is rejected if z � 2.33. c. 1.62, found by d. H0 is not rejected. We cannot conclude that the proportion

of men driving on the Ohio Turnpike is larger than 0.52. 25. a.

b. H0 is rejected if z � �1.28. c. �2.67, found by d. H0 is rejected. Fewer than 90 percent of the customers

receive their orders in less than 10 minutes. 27. 1.05, found by . Then

0.5000 � 0.3531 � 0.1469, which is the probability of a Type II error.

29. Reject H0 if z � �1.65 or z � 1.65.

Reject H0. We can conclude that the mean salary is not $45,000. p-value 0.0672, found by 2(0.5000 � 0.4664).

z � 45,500 � 45,000

$3,000�1120 � 1.83

H0: � $45,000 H1: � $45,000

z � (9,992 � 9,880)�(400�1100)

z � (0.82 � 0.90)�1(0.90 � 0.10)�100

H0: � � 0.90 H1: � 6 0.90

z � (.5667 � .52)�1(0.52 � 0.48)�300

H0: � � 0.52 H1: � 7 0.52

z � (0.75 � 0.70)�1(0.70 � 0.30)�100

t � 82.5 � 50 59.5�112

� 1.89

H0: � 50 H1: 7 50

t � 1.6 � 1.4

0.216�110 � 2.93

H0: � 1.4 H1: 7 1.4

t � 17 � 20

(3.536�15) � �1.90

X � 17 and s � B

50 5 � 1

� 3.536

t � 50,000 � 40,000

10,000�110 � 3.16

H0: � 40,000 H1: 7 40,000

t � 42 � 40

(2.1�128) � 5.040

H0: � 40 H1: 7 40 31. Reject H0 if z � �1.65.

Reject H0. The mean weight loss is less than 10 pounds. p-value � 0.5000 � 0.4943 � 0.0057

33.

Assuming a 5% significance level, reject H0 if t � �1.677.

Do not reject H0. West Virginia students are not sleeping less than 6 hours. p-value is between .05 and .10.

35. Reject H0 if t � �1.711

We fail to reject H0 and conclude that the mean number of residents is not necessarily less than 3.13.

37. Reject H0 if t � 2.821.

� 15.66 s � 1.544

Reject H0. The average rate is greater than 14 percent. 39. Assume a normal population.

Reject H0 if t � �2.201 or t � 2.201.

Do not reject H0. Cannot show a difference between senior citizens and the national average. p-value is about 0.09.

41. Assume a normal population. Reject H0 if t � �2.718.

Do not reject H0. The p-value is greater than 0.05. 43.

Reject H0 if t � �2.110 or t � 2.110.

Reject H0. The mean gain or loss does not equal 0. The p-value is less than 0.01, but greater than 0.001.

45. Assume a normal population. Reject H0 if t � 1.761.

Reject H0. The mean number with the scanner is greater than 100. p-value is 0.001.

t � 109.4 � 100 9.9628�115

� 3.654

s � B

1,389.6 15 � 1

� 9.9628

X � 1,641

15 � 109.4

H0: � 100 H1: 7 100

t � �0.2322 � 0 0.3120�118

� �3.158

X � �0.2322 s � 0.3120

H0: � 0 H1: � 0

t � 5.1667 � 6.5 3.1575�112

� �1.463

X � 5.1667 s � 3.1575

H0: � 6.5 H1: 6 6.5

t � 3.425 � 3.1 .6077�112

� 1.853

s � B

4.0625 12 � 1

� .6077

X � 41.1 12

� 3.425

H0: � 3.1 H1: � 3.1

t � 15.66 � 14.00

1.544�110 � 3.400

H0: � 14 H1: 7 14

t � 2.86 � 3.13 1.20�125

� �1.13

H0: � 3.13 H1: 6 3.13

t � 6.8 � 7.0 0.9�150

� �1.57

H0: � 7.0 H1: 6 7.0

z � 9.0 � 10.0 2.8�150

� �2.53

H0:̌ � 10 H1:̌ 6 10

Lin01803_appc_782-828.qxd 11/18/10 12:53 PM Page 799

47. Reject H0 if t � 3.250 or t � �3.250.

Fail to reject H0. 49. a. This is a binomial situation with both the mean number of

successes and failures equal to 21.5, found by 0.5 � 43. b. c.

H0: � � 0.50 H1: � � 0.50

t � 1.3 � 1.5 0.9�110

� �0.703

H0: � 1.5 H1: � 1.5

800

0.4

0.3

0.2

0.1

0.0

De ns

ity

–2.576

0.005 0.005

0 2.576 z value

Distribution Plot Normal, Mean = 0, StDev = 1

Reject H0 if z is not between �2.576 and 2.576.

d. We fail to reject the

null hypothesis. These data do not prove the coin flip is biased.

e. The p-value is 0.0220, found by 2 � (0.5000 � 0.4890). A value this extreme will happen about once out of fifty times with a fair coin.

51. H0 is rejected if z � 2.33.

H0 is rejected. Ms. Dennis is correct. More than 60 percent of the accounts are more than three months old.

53. H0 is rejected if z � 1.65.

H0 is rejected. We conclude that there has been an increase in the proportion of people wanting to go to Europe.

55. H0 is rejected if z � 2.33

H0 is rejected. More than 20 percent of the owners move during a particular year. p-value � 0.5000 � 0.4977 � 0.0023.

57. Reject H0 if z is greater than 2.326.

We fail to reject the null hypothesis. These data do not show that college students are more likely to skip breakfast.

z � (16�30) � 0.40

1[0.40 (1 � 0.40)�30] � 1.49

H0: � � 0.40 H1: � 7 0.40

z � (56�200) � 0.20 1(0.20 � 0.80)�200

� 2.83

H0: � � 0.20 H1: � 7 0.20

z � 0.480 � 0.44

1(0.44 � 0.56)�1.000 � 2.55

H0: � � 0.44 H1: � 7 0.44

z � .70 � .60

.60(.40) 200

� 2.89

H0: � � 0.60 H1: � 7 0.60

z � a

29 43 b � 0.50

20.50(1 � 0.50)�43 � 2.29

59. .

These data do not prove there is a reduced fatality rate. 61. a.

So the limits are 8.725 and 9.275. b.

P(z � �1.05) � 0.5000 � 0.3531 � 0.8531 c.

P(z � �0.15) � 0.5000 � 0.0596 � 0.4404

63.

Let n � 33 65.

Reject H0 if t � �1.714.

Do not reject the null hypothesis. The time is not less. 67. a.

Reject H0 if t is not between �2.045 and 2.045.

Do not reject the null.

The mean salary could be $80.0 million. b.

Reject the null. The mean attendance was more than 2,000,000.

CHAPTER 11 1. a. Two-tailed test

b. Reject if z � �2.05 or z � 2.05

d. Reject H0 e. found by 2(.5000 �.4952)

3. Step 1 Step 2 The .05 significance level was chosen. Step 3 Reject H0 if Step 4 �0.94, found by:

Step 5 Fail to reject Babies using the Gibbs brand did not gain less weight. p-value � .1736, found by .5000 �.3264.

5. If , reject H0.

Reject the null. It is reasonable to conclude that those who had a Caesarean section are shorter. The p-value is virtually zero. That much of a difference could almost never be due to sampling error.

z � 61.4 � 60.6

(1.2)2

45 �

(1.1)2

� 3.187

z 7 1.65 H0: 1 � 2 H1: 1 7 2

H0.

z � 7.6 � 8.1

(2.3)2

40 �

(2.9)2

� �0.94

z 6 �1.65.

H1: 1 6 2H0: 1 � 2 p-value � .0096,

z � 102 � 99

40 �

� 2.59

t � 2,448,000 � 2,000,000

698,000�130 � 3.51

Reject H0 if t is 7 1.699. H0: � 2,000,000 H1: 7 2,000,000

t � 83.51 � 80 33.90�130

� 1.38

H0: � 80 H1: � 80

t � 7.5 � 8 3.2�124

� �0.77

H0: � 8 H1: 6 8

50 � 2.33 10 1n

� 55 � .525 10 1n

n � (5.71)2 � 32.6

z � (9.275 � 9.300)�(1�136) � �0.15

z � (8.725 � 8.900)�(1�136) � �1.05

9.00 1.65(1�136) � 9.00 0.275

z � 0.0006 � 0.0008

0.0008 (0.9992) 10,000

� �0.707 H0 is not rejected

H0 is rejected if z 6 �1.645 H0: � � 0.0008 H1: � 6 0.0008

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7. a. H0 is rejected if z � 1.65.

b. 0.64, found by

c. 1.61, found by

d. is not rejected. 9. a.

b. is rejected if or .

d. found by

e. is rejected. The proportion infested is not the same in the two fields.

11. is rejected if .

is not rejected. We cannot conclude that a larger proportion of Democrats favor lowering the standards.

13. a. Reject if or

d. Do not reject . e. p-value is greater than 0.10 and less than 0.20.

15. Reject if t is not between �2.807 and 2.807.

Do not reject There is no difference in the mean salaries. 17.

Reject if .

Reject The mean daily expenses are greater for the sales staff. The p-value is between 0.05 and 0.10.

19. a.

� 12.96 S 12df

df � a

25 15

� 225 12 b

a 25 15 b

15 � 1 �

a 225 12 b

12 � 1

� 416.84

0.1984 � 31.9602

H0.

t � 142.5 � 130.3

A 203.82a

1 6

� 1 7 b

� 1.536

s2p � (6 � 1)(12.2)2 � (7 � 1)(15.8)2

6 � 7 � 2 � 203.82

t 7 1.363H0 df � 6 � 7 � 2 � 11

�s 7 �aH1:�s � �aH0: H0.

t � 7240 � 9188

B 88,584,000a

1 12

� 1

13 b

� �0.517

s2p � (12 � 1)(8,242)2 � (13 � 1)(10,369)2

12 � 13 � 2 � 88,584,000

�1 � �2 H1: �1 � �2 df � 12 � 13 � 2 � 23H0:

t � 23 � 26

A 19.9375a

1 10

� 1 8 b

� �1.416

s2p � (10 � 1)(4)2 � (8 � 1)(5)2

10 � 8 � 2 � 19.9375

df � 10 � 8 � 2 � 16 t 6 �2.120.t 7 2.120H0

p-value � 0.3015.

z � 0.21 � 0.20

(0.2044)(0.7956) 800

� (0.2044)(0.7956)

1,000

� 0.52

pc � 168 � 200

800 � 1,000 � 0.2044

z 7 2.05H0 d 7 rH1: d � rH0:

z � 0.06 � 0.10

1[(0.08 0.92)�400] � [(0.08 0.92)�400]

�2.09,

pc � 24 � 40

400 � 400 � 0.08

z 7 1.96z 6 �1.96H0 1 � 2H1: 1 � 2H0:

z � 0.70 � 0.60

1[(0.64 0.36)�100] � [(0.64 0.36)�150]

pc � 70 � 90

100 � 150

b. Reject if or .

d. Fail to reject the null hypothesis.

21. a.

b. Reject if .

d. Reject the null hypothesis. The mean adoption cost from Russia is greater than the mean adoption cost from China.

23. a. Reject if .

d. Reject There are more defective parts produced on the day shift.

e. p-value is less than 0.005, but greater than 0.0005. 25.

Reject if

Reject The incentive plan resulted in an increase in daily income. The p-value is about .025.

27. Reject if df � 35 � 40 � 2, or .

Do not reject There is no difference in the number of times men and women buy take-out dinner in a month. The p-value is between .05 and .10.

29. Reject if or .

is not rejected. There is no difference in the H0.

z � 4.77 � 5.02

(1.05)2

40 �

(1.23)2

� �1.04

z 7 1.96z 6 �1.96H0 �1 � �2H1:�1 � �2H0:

H0.

t � 24.51 � 22.69

A 17.3079a

1 35

� 1

40 b

� 1.890

sp 2 �

(35 � 1)(4.48)2 � (40 � 1)(3.86)2

35 � 40 � 2 � 17.3079

t 7 2.645t 6 �2.645H0 �M � �WH1:�M � �WH0:

H0.

t � 25.917

40.791�112 � 2.20

t 7 1.796H0 sd � 40.791 d � 25.917

�d 7 0H1:�d � 0H0:

H0.

t � 3.00

0.816�14 � 7.35

sd � A

2 3

� 0.816d � 12 4

� 3.00

t 7 2.353H0

t � 12,840 � 11,045

2,387,025 18

� 697,225

� 4.276

t 7 1.706H0 �Russia 7 �ChinaH1:�Russia � �ChinaH0:

df � a

697,225 16

� 2,387,025

18 b

a 697,225

16 b

16 � 1 �

a 2,387,025

18 b

18 � 1

� 26.7 S 26df

t � 50 � 46

25 15

� 225 12

� 0.8852

t 6 �2.179t 7 2.179H0 H0: �1 � �2 H1: �1 � �2

mean number of calls. p-value � 2(0.5000 � 0.3508) � 0.2984.

31. Reject if

Reject H0. The mean income is larger for Plan B. The p-value � .5000 � .4798 � .0202. The skewness does not matter because of the sample sizes.

t � $61,000 � $57,000

($7,100)2

30 �

($9,200)2

� $4000.00 $1948.42

� 2.05

t 7 1.668H0 �B 7 �AH1:�B � �AH0:

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802

33. Reject if .

is not rejected. There is no difference in the proportions that found relief with the new and the old drugs.

35. If reject

Reject the null hypothesis. We can conclude the proportion of men who believe the division is fair is greater.

37. is rejected if z � 1.65.

H0 is not rejected. There has not been an increase in the proportion calling conditions “good.” The p-value is 0.1446, found by 0.5000 � 0.3554. The increase in the percentages will happen by chance in one out of every seven cases.

39. is rejected if z is not between �1.96 and 1.96.

is rejected. There is a difference in the replies of the sexes.

41. a.

b. Reject if or .

d. Reject the null hypothesis. There is a difference. 43. Assume equal population standard deviations.

Reject if or .

p-value � 0.10. Do not reject There is no difference in the mean number of hamburgers sold at the two locations.

H0.

t � 83.55 � 78.8

A 161.2969a

1 10

� 1

12 b

� 0.874

s2p � (10 � 1)(10.5)2 � (12 � 1)(14.25)2

10 � 12 � 2 � 161.2969

t 7 2.086t 6 �2.086H0 n � sH1: n � sH0:

t � 1.65 � 2.20

0.3136 12

� 0.0900

� �3.00

t 6 �2.120t 7 2.120H0 a � wH1: a � wH0:

� 0.0011

0.000062 � 0.0000051 � 16.37 S 16df

df � a

0.3136 12

� 0.0900

12 b

a 0.3136

12 b

12 � 1 �

a 0.0900

12 b

12 � 1

z �

100 300

� 36 200

(0.272) (0.728) 300

� (0.272) (0.728)

200

� 3.775

pc � 100 � 36

300 � 200 � .272

�1 � �2 H1: �1 � �2H0:

z � 0.091 � 0.085

(0.088) (0.912) 5000

� (0.088) (0.912)

5000

� 1.059

pc � .091 � .085

2 � .088

H0�1 � �2 H1: �1 7 �2H0:

z � .6600 � .60625

.63(.37) 1,500

� .63(.37) 1,600

� 3.10

pc � 990 � 970

1,500 � 1,600 � 0.63

H0.z 7 2.33, �1 7 �2H1:�1 � �2H0:

z � 0.90 � 0.87

0.882(0.118) 200

� 0.882(0.118)

300

� 1.019

pc � 180 � 261 200 � 300

� 0.882

z 7 1.65H0 �1 7 �2H1:�1 � �2H0: 45. Assume equal population standard deviations.

Reject if or .

Do not reject There is no difference in the mean amount purchased.

47. Assume equal population standard deviations. Reject if .

Reject The mean number of transactions by the young adults is more than for the senior citizens.

49. Reject if .

is not rejected. There is no difference in the mean number sold at the regular price and the mean number sold at the reduced price.

51. Reject if .

Do not reject There is no difference in the mean number of absences. The p-value is greater than 0.05 but less than .10.

53. Reject H0 if t � �2.024 or t � 2.204.

Reject the null hypothesis. The population means are different.

55. Reject if .

Reject H0. The mean is lower. 57.

Reject if or .

Do not reject There is no difference in the mean number of cars in the two lots.

H0.

t � 86.24 � 92.04

A 566.335a

1 25

� 1

28 b

� �0.886

s2p � (25 � 1)(23.43)2 � (28 � 1)(24.12)2

25 � 28 � 2 � 566.335

sR � 24.12XR � 92.04, sO � 23.43XO � 86.24,

t 7 2.008t 6 � 2.008H0 df � 25 � 28 � 2 � 51

O � RH1: O � R,H0:

t � 3.11

2.91�18 � 3.02

sd � 2.91d � 3.11 t 7 1.895H0

d 7 0H1: d � 0H0:

t � 150 � 180

A 1157.89a

1 15

� 1 25 b

� �2.699

sp 2 �

(15 � 1)(40)2 � (25 � 1)(30)2

15 � 25 � 2 � 1157.89

1 � 2H1: 1 � 2H0:

H0.

t � 1.75

2.9155�18 � 1.698

sd � 2.9155d � 1.75 t 7 1.895H0 d 7 0H1: d � 0H0:

t � 125.125 � 117.714

A 305.708a

1 8

� 1 7 b

� 0.819

s2p � (8 � 1)(15.094)2 � (7 � 1)(19.914)2

8 � 7 � 2 � 305.708

s2 � 19.914X2 � 117.714 s1 � 15.094X1 � 125.125

t 7 2.650H0 1 7 2H1: 1 � 2H0:

H0.

t � 10.375 � 5.636

A 5.672a

1 8

� 1

11 b

� 4.28

s2p � (8 � 1)(2.2638)2 � (11 � 1)(2.4606)2

8 � 11 � 2 � 5.672

t 7 2.567H0 1 7 2H1: 1 � 2H0:

H0.

t � 15.87 � 18.29

A 6.06a

1 10

� 1 14 b

� �2.374

s2p � (10 � 1)(2.33)2 � (14 � 1)(2.55)2

10 � 14 � 2 � 6.06

t 6 �2.819t 7 2.819H0 1 � 2H1: 1 � 2H0:

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59.

Round down degrees of freedom.

Starting in the first five rows (in contrast to the last four) lowers the odds.

61. a. 1 � without pool

Reject if or .

Reject There is a difference in mean selling price for homes with and without a pool.

b. 2 � with garage 1 � without attached garage

H0.

t � 202.8 � 231.5

A 2,041.05a

1 38

� 1

67 b

� �3.12

s2p � (38 � 1)(33.7)2 � (67 � 1)(50.46)2

38 � 67 � 2 � 2,041.05

n2 � 67s2 � 50.46X2 � 231.5 n1 � 38s1 � 33.7X1 � 202.8

t 6 �2.000t 7 2.000H0 1 � 2H1: 1 � 2H0:

2 � with pool

t � 39.4 � 187.5

(53.2)2

15 �

(48.3)2

� �7.57 Reject H0.

Reject H0 if t 6 �2.492.

df � 24, found by a

53.22

15 �

48.32

12 b

a 53.22

15 b

14 �

a 48.32

12 b

� 24.546

F 6 BH1: F � BH0:

Do not reject There may be no difference in the mean maintenance cost for the two types of buses.

CHAPTER 12 1. 9.01, from Appendix B.4 3. Reject if where degrees of freedom in the

numerator are 7 and 5 in the denominator. Computed found by:

Do not reject There is no difference in the variations of the two populations.

5. Reject where (3.10 is about halfway between 3.14 and 3.07.) Computed found by:

Do not reject There is no difference in the variations of the two populations.

7. a. Treatment means are not all the same.

b. Reject if . c & d.

F 7 4.26H0

H1: 1 � 2 � 3;H0:

H0.

F � (12)2

(10)2 � 1.44

F � 1.44, F 7 3.10.H0

�21 � � 2 2H1:�

2 1 � �

2 2H0:

H0.

F � s21 s22

� (10)2

(7)2 � 2.04

F � 2.04,

F 7 10.5,H0

H0.

t � 454.8 � 441.5

B 2878a

1 53

� 1

27 b

� 1.05

Source SS df MS F

Treatment 62.17 2 31.08 21.94 Error 12.75 9 1.42

Total 74.92 11

e. Reject The treatment means are not all the same. 9. Treatment means are not all the

same. Reject if F 7 4.26.H0 H1: 1 � 2 � 3;H0:

H0.

Source SS df MS F

Treatment 276.50 2 138.25 14.18 Error 87.75 9 9.75

Reject The treatment means are not all the same. 11. a. Not all means are the same.

b. Reject if . c. d.

SS total � 116.67.SSE � 9.47,SST � 107.20, F 7 4.26H0

H1: 1 � 2 � 3;H0: H0.

Source SS df MS F

Treatment 107.20 2 53.600 50.96 Error 9.47 9 1.052

Total 116.67 11

e. Since is rejected. At least one of the means differs.

Yes, we can conclude that treatments 1 and 2 have different means.

13. Not all means are equal. is rejected if F 7 3.71.H0

H1: 1 � 2 � 3 � 4;H0:

� [5.777, 9.157] � 7.467 1.69 � (9.667 � 2.20) 2.262 11.052(1�3 � 1�5)

(X1 � X2) t1MSE(1�n1 � 1�n2)

H050.96 7 4.26.

Source SS df MS F

Treatment 32.33 3 10.77 2.36 Error 45.67 10 4.567

Total 78.00 13

H0: 1 � 2 Reject if or .

Reject There is a difference in mean selling price for homes with and without an attached garage.

c. Reject if or .

Reject There is a difference in mean selling price for homes in Township 1 and Township 2.

d. If z is not between and 1.96, reject H0.

Reject the null hypothesis. There is a difference. 63.

If t is not between �1.991 and 1.991, reject H0.

s2p � (53 � 1) (52.9)2 � (27 � 1)(55.1)2

53 � 27 � 2 � 2878

1 � 2H1: 1 � 2H0:

z � 0.462 � 0.811

10.64 � 0.36�52 � 0.64 � 0.36�53 � �3.73

pc � 24 � 43 52 � 53

� 0.64

�1.96 �1 � �2H1:�1 � �2H0:

H0.

t � 196.91 � 227.45

A 1,667.43a

1 15

� 1

20 b

� �2.19

s2p � (15 � 1)(35.78)2 � (20 � 1)(44.19)2

15 � 20 � 2 � 1,667.43

n2 � 20s2 � 44.19X2 � 227.45 n1 � 15s1 � 35.78X1 � 196.91

t 6 �2.036t 7 2.036H0 1 � 2H1: 1 � 2H0:

H0.

t � 185.45 � 238.18

A 1,620.07a

1 34

� 1

71 b

� �6.28

s2p � (34 � 1)(28.00)2 � (71 � 1)(44.88)2

103 � 1,620.07

s2 � 44.88X2 � 238.18 s1 � 28.00X1 � 185.45 df � 34 � 71 � 2 � 103� � 0.05

t 6 �2.000t 7 2.000H0 1 � 2H1:

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804

Because 2.36 is less than 3.71, is not rejected. There is no difference in the mean number of weeks.

15. a. Not all treatment means are equal. b. Reject if F � 18.5. c. Not all block means are equal.

H0 is rejected if d.

e. SSE � 289.50 � 204.1667 � 76 � 9.3333 2(39.5 � 36.5)2 � 8 � 50 � 18 � 76 SSB � 2(38.5 � 36.5)2 � 2(31.5 � 36.5)2 � SST � 289.5 � 85.3333 � 204.1667

� 85.3333 SSE � (46 � 42.3333)2 � . . . � (35 � 30.6667)2 SS total � (46.0 � 36.5)2 � . . . � (35� 36.5)2 � 289.5

F 7 19.0. H1: 1 � 2 � 3;H0:

H1: 1 � 2;H0:

Source SS df MS F

Treatment 204.167 1 204.167 43.75 Blocks 76.000 2 38.000 8.14 Error 9.333 2 4.667

Total 289.5000 5

f. so reject There is a difference in the treatments. so do not reject for blocks. There is no difference among blocks.

17. For treatment: For blocks:

Not all means equal H1: Not all means equal Reject if . Reject if .F 7 3.84F 7 4.46 H1:

1 � 2 � 3 � 4 � 5H0: 1 � 2 � 3H0:

H08.14 6 19.0, H0.43.75 7 18.5,

Source SS df MS F

Treatment 62.53 2 31.2650 5.75 Blocks 33.73 4 8.4325 1.55 Error 43.47 8 5.4338

Total 139.73

Source SS df MS F P

Size 156.333 2 78.1667 1.98 0.180 Weight 98.000 1 98.000 2.48 0.141 Interaction 36.333 2 18.1667 0.46 0.642 Error 473.333 12 39.444

Total 764.000 17

There is a difference in shifts, but not by employee. 19.

a. Since the p-value (0.18) is greater than 0.05, there is no difference in the Size means.

b. The p-value for Weight (0.141) is also greater than 0.05. Thus, there is no difference in those means.

c. There is no significant interaction because the p-value (0.642) is greater than 0.05.

21. a.

d-320 j-1000

M ea

Machine

30 uv-57

Interaction Plot (data means) for Sales

Position Inside Outside

Two-way ANOVA: Sales versus Position, Machine Source df SS MS F P

Position 1 104.167 104.167 9.12 0.007 Machine 2 16.333 8.167 0.72 0.502 Interaction 2 457.333 228.667 20.03 0.000 Error 18 205.500 11.417

Total 23 783.333

One-way ANOVA: D-320 Sales versus Position

Source df SS MS F P

Position 1 364.50 364.50 40.88 0.001 Error 6 53.50 8.92

Total 7 418.00

One-way ANOVA: J-1000 Sales versus Position

Source df SS MS F P

Position 1 84.5 84.5 5.83 0.052 Error 6 87.0 14.5

Total 7 171.5

One-way ANOVA: UV-57 Sales versus Position

Source df SS MS F P

Position 1 112.5 112.5 10.38 0.018 Error 6 65.0 10.8

Total 7 177.5

The position and the interaction of position and machine effects are significant. The effect of machine on sales is not significant. c.

Recommendations using the statistical results and mean sales plotted in part (a): Position the D-320 machine outside. Statistically, the position of the J-1000 does not matter. Position the UV-57 machine inside.

23. df1 � 21 � 1 � 20; df2 � 18 � 1 � 17. H0 is rejected if F � 3.16.

Reject H0. There is more variation in the selling price of oceanfront homes.

25. Sharkey: White:

dfs � 7 � 1 � 6;

dfw � 8 � 1 � 7. Reject H0 if F � 8.26.

Cannot reject H0. There is no difference in the variation of the monthly sales.

27. a. H0: 1 � 2 � 3 � 4 H1: Treatment means are not all equal.

b. � � .05 Reject H0 if F � 3.10. c.

F � (22.95)2

(14.79)2 � 2.41

�2w 7 �2s.H1:�2w � �2s;H0: sw � 22.95n � 8 ss � 14.79n � 7

F � (45,600)2

(21,330)2 � 4.57

�21 7 �22.H1:�21 � �22;H0:

Yes, there appears to be an interaction effect. Sales are different based on machine position, either in the inside or outside position.

Source SS df MS F

Treatment 50 4 � 1 � 3 50�3 1.67 Error 200 24 � 4 � 20 10

Total 250 24 � 1 � 23

d. Do not reject H0.

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H0 is rejected. There is a difference in the treatment means. 31. H0: 1 � 2 � 3 � 4; H1: Not all means are equal.

H0 is rejected if F � 3.10.

29. H0: 1 � 2 � 3; H1: Not all treatment means are equal. H0 is rejected if F � 3.89.

Source SS df MS F

Treatment 63.33 2 31.667 13.38 Error 28.40 12 2.367

Total 91.73 14

Source SS df MS F

Treatment 219.43 1 219.43 23.10 Error 114.00 12 9.5

Total 333.43 13

Source SS df MS F

Treatment 25.0 3 8.3333 5.88 Blocks 21.5 2 10.75 7.59 Error 8.5 6 1.4167

Total 55.0 11

Source SS df MS F

Factor 87.79 3 29.26 9.12 Error 64.17 20 3.21

Total 151.96 23

Because the computed F of 9.12 � 3.10, the null hypothesis of no difference is rejected at the .05 level.

33. a. H0: 1 � 2; H1: 1 � 2. Critical value of F � 4.75.

Then t2 � F. That is (�4.806)2 � 23.10. c. H0 is rejected. There is a difference in the mean scores.

35. The null hypothesis is rejected because the F statistic (8.26) is greater than the critical value (5.61) at the .01 significance level. The p-value (.0019) is also less than the significance level. The mean gasoline mileages are not the same.

37. H0: 1 � 2 � 3 � 4. H1: At least one mean is different. Reject H0 if F � 2.7395. Since 2.72 is less than 2.7395, H0 is not rejected. You can also see this conclusion from the p-value of 0.051, which is greater than 0.05. There is no difference in the means for the different types of first-class mail.

39. For color, the critical value of F is 4.76; for size, it is 5.14.

t � 19 � 27

A 9.5a

1 6

� 1 8 b

� �4.806

H0s for both treatment and blocks (color and size) are rejected. At least one mean differs for color and at least one mean differs for size.

41. a. Critical value of F is 3.49. Computed F is .668. Do not reject H0.

b. Critical value of F is 3.26. Computed F value is 100.204. Reject H0 for block means.

There is a difference in homes but not assessors. 43. For gasoline:

H0: 1 � 2 � 3; H1: Mean mileage is not the same. Reject H0 if F � 3.89.

Source SS df MS F

Treatment 0.03478 5 0.00696 3.86 Error 0.10439 58 0.0018

Total 0.13917 63

ANOVA Table

Source SS df MS F

Gasoline 44.095 2 22.048 26.71 Autos 77.238 6 12.873 15.60 Error 9.905 12 0.825

Total 131.238 20

45. H0: 1 � 2 � 3 � 4 � 5 � 6; H1: The treatment means are not equal. Reject H0 if F � 2.37.

There is a difference in both autos and gasoline.

H0 is rejected. There is a difference in the mean weight of the colors.

47. a.

Private Sector

Public

1250

1200

1150

1100

1050

1000

M ea

Interaction Plot (data means) for Wage

Gender Men Women

b. Two-way ANOVA: Wage versus Gender, Sector

Source DF SS MS F P Gender 1 44086 44086 11.44 0.004 Sector 1 156468 156468 40.61 0.000 Interaction 1 14851 14851 3.85 0.067 Error 16 61640 3853 Total 19 277046

There is no interaction effect of gender and sector on wages. However, there are significant differences in mean wages based on gender and significant differences in mean wages based on sector.

For automobile: H0: 1 � 2 � . . . � 7; H1: Mean mileage is not the same. Reject H0 if F � 3.00.

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c. One-way ANOVA: Wage versus Sector Source DF SS MS F P Sector 1 156468 156468 23.36 0.000 Error 18 120578 6699 Total 19 277046

s � 81.85 R-Sq � 56.48% R-Sq(adj) � 54.06%

Individual 95% CIs For Mean Based on Pooled StDev

Level N Mean StDev --+---------+---------+---------+------- Private 10 1175.2 95.9 (------*------) Public 10 998.3 64.8 (------*------)

--+---------+---------+---------+------- 960 1040 1120 1200

One-way ANOVA: Wage versus Gender Source DF SS MS F P Gender 1 44086 44086 3.41 0.081 Error 18 232960 12942 Total 19 277046

s � 113.8 R-Sq � 15.91% R-Sq(adj) � 11.24%

Individual 95% CIs For Mean Based on Pooled StDev

Level N Mean StDev --+---------+---------+---------+------- Men 10 1133.7 137.9 (----------*----------) Women 10 1039.8 82.9 (----------*---------)

--+---------+---------+---------+------- 980 1050 1120 1190

d. The statistical results show that only sector, private or public, has a significant effect on the wages of accountants.

49. a.

Reject H0 if F � 2.05 (estimated). df1 � 67 � 1 � 66; df2 � 38 � 1 � 37

Reject H0. There is a difference in the variance of the two selling prices.

b. H0: H1:

Reject H0 if F � 2.21 (estimated).

Reject H0. There is a difference in the variance of the two selling prices.

F � (44.88)2

(28.00)2 � 2.57

�2g � � 2 ng.�

2 g � �

2 ng;

F � (50.57)2

(33.71)2 � 2.25

�2np � � 2 p.H1:�

2 np � �

2 pH0:

Source SS df MS F

Township 13,263 4 3,316 1.52 Error 217,505 100 2,175

Total 230,768 104

Source SS df MS F

Treatment 28,996 2 14,498 5.62 Error 198,696 77 2,580

Total 227,692 79

H0: 1 � 2 � 3 � 4 � 5; H1: Not all treatment means are equal. Reject H0 if F � 2.46. Do not reject H0. There is no difference in the mean selling prices in the five townships.

51. a. H0: 1 � 2 � 3 H1: Not all treatment means are equal. Reject H0 if F � 4.89.

b. H0: 1 � 2 � 3 H1: Not all treatment means are equal. Reject H0 if F � 3.12.

Reject H0. The mean maintenance costs are different.

Source SS df MS F

Treatment 5,095 2 2,547 1.45 Error 135,513 77 1,760

Total 140,608 79

Do not reject H0. The mean miles traveled are not different.

This reduces to �64.94 38.68, so the difference is between �103.62 and �26.26. To put it another way, Bluebird is less costly than Thompson by an amount between $26.26 and $103.62.

CHAPTER 13 1.

3. a. Sales. b.

Number of advertisements 1 2 3 4 5 6

Sa le

s ($

00 0)

r � 10.6

(5 � 1)(2.7019)(1.3038) � 0.7522

(441.81 � 506.75) 1.991 B

2580a 1

47 �

1 8 b

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d. There is a strong positive association between the variables.

5. a. Police is the independent variable, and crime is the dependent variable.

d. Strong inverse relationship. As the number of police increases, the crime decreases.

7. Reject H0 if t � 1.812.

Do not reject . 9. Reject H0 if

Reject There is a positive correlation between gallons sold and the pump price.

11. Reject H0 if .

Reject H0. There is a positive correlation between the number of passengers and plane weight.

13. a.

b. 6.3081, found by 15. a. sy � 2.011

b. Ŷ � 1.333 � .667(6) � 5.335

a � 7.4 � .677(9.1) � 1.333

b � .904a 2.011 2.726

b � 0.667

r � 44.6

(10 � 1)(2.726)(2.011) � .904

a � 5.8 � 0.3630(5.6) � 3.7671

b � 0.7522a 1.3038 2.7019

b � 0.3630

Ŷ � 3.7778 � 0.3630X

t � 0.667215 � 2 11 � 0.6672

� 3.228

t 7 2.650 H1: � 7 0H0: � � 0

H0.

t � .78220 � 2 11 � (.78)2

� 5.288

df � 18.t 7 2.552.H0: � � 0; H1: � 7 0. H0

t � .32212 � 2 11 � (.32)2

� 1.068

r � �231.75

(8 � 1)(5.8737)(6.4462) � �0.8744

18.0

12.0

6.0

Police 12.0 16.0 20.0 24.0 28.0

Cr im

es r �

36 (5 � 1)(1.5811)(6.1237)

� 0.9295

d. ($ millions)

19. a.

b. 10.1957, found by c. For each policeman added, crime goes down by

almost one. 21.

Reject H0 if .

Reject H0 and conclude the slope is less than zero. 23.

Reject H0 if t not between �2.228 and 2.228

Reject H0 and conclude the slope is different from zero.

25. The standard error of estimate is 3.379, found by .

The coefficient of determination is 0.76, found by (�0.874)2. Seventy-six percent of the variation in crimes can be explained by the variation in police.

27. The standard error of estimate is 0.913, found by .

The coefficient of determination is 0.82, found by 29.733�36.4. Eighty-two percent of the variation in kilowatt hours can be explained by the variation in the number of rooms.

29. a.

31. a.

b. � [2.751, 9.865] 6.308 (3.182)(.993)21 � 1�5 � .0671 � [4.675, 7.941] � 6.308 1.633

6.308 (3.182)(.993) B

.2 � (7 � 5.6)2

29.2

sy # x � B

500 13

� 6.2017

r � 2.6667 � .8165

r2 � 1000 1500

� .6667

6.667 10 � 2

68.4877 8 � 2

t � 0.08/0.03 � 2.667

df � n � 2 � 12 � 2 � 10H1: � � 0H0: � � 0

t � �0.96�0.22 � �4.364

t 6 �1.943 df � n � 2 � 8 � 2 � 6H1: � 6 0H0: � � 0

29.3877 � 0.9596(20)

a � 95 8

� (�0.9596)a 146

8 b � 29.3877

b � �.8744a 6.4462 5.8737

b � �0.9596

Ŷ � 1.8507 � 0.0836(50.0) � 6.0307

a � 64.1 12

� 0.0836a 501.10

12 b � 1.8507

b � .6734a 3.248

26.170 b � 0.0836

r � 629.64

(12 � 1)(26.17)(3.248) � .6734

Ea rn

in gs

Sales 0 10 20 30 40 50 60 70 80 90

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33. a. 4.2939, 6.3721 b. 2.9854, 7.6806

35. The correlation between the two variables is 0.298. By squaring X, the correlation increases to .998.

37. Reject H0 if

Reject There is a positive correlation between passengers and weight of luggage.

39. Reject H0 if

Do not reject There is not a positive correlation between engine size and performance. p-value is greater than .05, but less than .10.

41. a. The total number of cars sold decreases as the percent market share decreases. The relationship is inverse, so as one increases, the other decreases.

The value of r indicates a fairly strong negative association between the variables.

c. Reject H0 if .

Reject H0. There is a negative correlation. d. 77.6 percent, found by (�0.881)2, of the variation in

market share is accounted for by variation in cars sold.

43. a. b. c. Reject if

H0 is rejected. There is a positive association between family size and the amount spent on food.

45. a.

There is an inverse relationship between the variables. As the months owned increase, the number of hours exercised decreases.

b. c. Reject

Reject H0. There is a negative association between months owned and hours exercised.

47. a. Median age and population are directly related.

c. The slope of 0.272 indicates that for each increase of 1 million in the population, the median age increases on average by 0.272 years.

r � 11.93418

(10 � 1)(2.207)(1.330) � 0.452

t � �0.827210 � 2 11 � (�0.827)2

� �4.16

H0 if t 6 �2.896.H0: � � 0; H1: � 6 0. r � �0.827

Months 2 4 6 8 10 12

Ho ur

t � 0.589210 � 2 11 � (.589)2

� 2.062

t 7 1.860.H0H0: � � 0; H1: � 7 0. r 2 � (0.589)2 � 0.3469 r � 0.589

t � �0.881212 � 2 11 � (�0.881)2

� �5.89

t 6 �2.764 H1: � 6 0H0: � � 0

r � �305.19

(12 � 1)(3.849)(8.185) � �0.881

H0.

t � .47212 � 2 11 � (.47)2

� 1.684

t 7 2.764.H0: � � 0; H1: � 7 0.

H0.

t � .94225 � 2 11 � (.94)2

� 13.213

t 7 1.714.H0: � � 0; H1: � 7 0.

d. The median age is 32.08 years, found by 31.4 � 0.272(2.5). e. The p-value (0.190) for the population variable is greater

than, say 0.05. A test for significance of that coefficient would fail to be rejected. In other words, it is possible the population coefficient is zero.

f. Reject H0 if t is not between �2.306 and 2.306.

df � 8 Do not reject H0.

There may be no relationship between age and population.

49. a. b.

d. Nearly 50 percent of the variation in the amount of the bid is explained by the number of bidders.

51. a.

There appears to be a relationship between the two variables. As the distance increases, so does the shipping time.

b. Reject H0 if

H0 is rejected. There is a positive association between shipping distance and shipping time.

c. Nearly half of the variation in shipping time is explained by shipping distance.

d. 53. a.

The regression equation is: Dividend. For each additional dollar of dividend, the price increases by $2.41.

b. Thus, 65.8 percent of the

variation in price is explained by the dividend. c.

At the 5 percent level, reject H0 when

Thus, H0 is rejected. The population correlation is positive. 55. a. 35

c. r 2 � 13,548,662,082 14,531,349,474

� 0.932

sy # x � 229,778,406 � 5,456.96

t � 0.811230 � 2 11 � (0.811)2

� 7.34

t 7 1.701. H1: � 7 0H0: � � 0r � 1.658 � 0.811

r 2 � 5,057.6 7,682.7

� 0.658

Price � 26.8 � 2.41 � a � 26.8 b � 2.41 sy # x � 1.987

r 2 � 0.479.

t � 0.692220 � 2 11 � (0.692)2

� 4.067

t 7 1.734.H0: � � 0; H1: � 7 0. r � 0.692

Da ys

Distance 600 700 800 900

r 2 � 0.499. � [5.4835, 10.4543] � 7.9689 2.4854

7.9689 (2.160)(1.114) B

1 � 1

15 �

(7 � 7.1333)2

73.7333

Ŷ � 11.2358 � 0.4667(7.0) � 7.9689 a � 11.2358b � �0.4667,

t � 0.452210 � 2 11 � (0.452)2

� 1.433

H1: p 0H0: � � 0

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d. e. reject if .

Reject H0. There is a direct relationship between size of the house and its market value.

57. a. The regression equation is Price � �773 � 1,408 Speed. b. The second laptop (1.6, 922) with a residual of �557.60,

is priced $557.60 below the predicted price. That is a noticeable “bargain.”

c. The correlation of Speed and Price is 0.835. Reject if .

Reject It is reasonable to say the population correlation is positive.

59. a. Reject if

b. an additional cup increases the dog’s weight by almost 23 pounds.

c. Dog number 4 is an overeater. 61. The correlation of Box Office and Adjusted Budget is 0.027.

At the 5% level, reject if .

Do not reject H0. The population correlation is not necessarily positive. “Big budget” movies do not always lead to large box office returns.

63. a. There does seem to be a direct relationship between the variables.

t � 0.027250 � 2

11 � (0.027)2 � 0.187

t 7 1.677H0 H1: � 7 0H0: � � 0

Ŷ � �29.7 � 22.93X;

t � .987218 � 2 11 � (.987)2

� 24.564

t 7 1.746.H0H0: � � 0, H1: � 7 0.r � .987,

H0.

t � 0.835212 � 2 11 � (0.835)2

� 4.799

t 7 1.8125H0H0: � � 0 H1: � 7 0

t � .966235 � 2 11 � (.966)2

� 21.46

t 7 1.692H0H0: � � 0, H1: � 7 0; r � 20.932 � 0.966 For ERA:

At the 5% level, reject if .

Reject The ERA correlation is negative.

CHAPTER 14 1. a. Multiple regression equation

b. The Y-intercept c. � 64,100 � 0.394(796,000) � 9.6(6,940)

� 11,600(6.0) � $374,748 3. a. 497.736, found by

� 16.24 � 0.017(18) � 0.0028(26,500) � 42(3) � 0.0012(156,000) � 0.19(141) � 26.8(2.5)

b. Two more social activities. Income added only 28 to the index; social activities added 53.6.

5. a.

95% of the residuals will be between 6.136, found by 2(3.068)

The independent variables explain 11.8% of the variation.

7. a. � 84.998 � 2.391X1 � 0.4086X2 b. 90.0674, found by � 84.998 � 2.391(4) � 0.4086(11) c. n � 65 and k � 2 d. H0: H1: Not all are 0

Reject H0 if F � 3.15. F � 4.14, reject H0. Not all net regression coefficients equal zero.

e. For X1 For X2 H0: �1 � 0 H0: �2 � 0 H1: �1 � 0 H1: �2 � 0 t � 1.99 t � �2.38 Reject H0 if t � 2.0 or t � �2.0. Delete variable 1 and keep 2.

f. The regression analysis should be repeated with only X2 as the independent variable.

9. a. The regression equation is: Performance � 29.3 � 5.22 Aptitude � 22.1 Union

�’s�1 � �2 � 0

Ŷ Ŷ

� 1 � 9.414

10.3375 � 1 � .911 � .089

R2adj � 1 �

SSE n � (k � 1)

SS total n � 1

� 1 �

583.693 65 � (2 � 1)

661.6 65 � 1

R2 � SSR

SS total �

77.907 661.6

� .118

� 29.414 � 3.068

sY # 12 � B

SSE n � (k � 1)

� B

583.693 65 � (2 � 1)

Ŷ

H0.

t � �0.635 230 � 2

21 � (�0.635)2 � �4.35

t 6 �1.701H0 H1: � 6 0H0: � � 0

Predictor Coef SE Coef T P Constant 29.28 12.77 2.29 0.041 Aptitude 5.222 1.702 3.07 0.010 Union 22.135 8.852 2.50 0.028

S � 16.9166 R-Sq � 53.3% R-Sq (adj) � 45.5%

Analysis of Variance Source DF SS MS F P Regression 2 3919.3 1959.6 6.85 0.010 Residual Error 12 3434.0 286.2 Total 14 7353.3

100

50 75 100 125 150 Salary

Fitted Line Plot Wins = 67.12 + 0.1568 Salary

W in

175 200

b. 82.8, found by 67.12 � 0.1568 � 100 c. 0.78, found by 0.1568(5) d. df � n � 2 � 30 � 2 � 28H1: � 7 0H0: � � 0

b. These variables are effective in predicting performance. They explain 53.3 percent of the variation in performance. In particular, union membership increases the typical performance by 22.1.

Reject if t � 1.701 t � 0.1568�0.0564 � 2.78 Reject and conclude the slope is positive.

e. 0.216 or 21.6%, found by 819/3792 f. The correlation between wins and batting average

is 0.467. The correlation between wins and ERA is �0.635. ERA is the stronger. For batting average: At the 5% level, reject if .

Reject The batting average correlation is positive.H0.

t � 0.467 230 � 2

21 � (0.467)2 � 2.795

t 7 1.701H0 H1: � 7 0H0: � � 0

H0 H0

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The price decreases 54.2 as each additional bidder participates. Meanwhile the price increases 16.3 as the painting gets older. While one would expect older paintings to be worth more, it is unexpected that the price goes down as more bidders participate!

b. The regression equation is Price � 3,972 � 185 Bidders � 6.35 Age � 1.46 X1X2

Predictor Coef SE Coef T P Constant 3971.7 850.2 4.67 0.000 Bidders �185.0 114.9 �1.61 0.122 Age 6.353 9.455 0.67 0.509 X1X2 1.462 1.277 1.15 0.265

The t value corresponding to the interaction term is 1.15. This is not significant. So we conclude there is no interaction.

c. In the stepwise procedure, the number of bidders enters the equation first. Then the interaction term enters. The variable age would not be included as it is not significant. Response is Price on 3 predictors, with N � 25.

Step 1 2 Constant 4,507 4,540

Bidders �57 �256 T-Value �3.53 �5.59 P-Value 0.002 0.000

X1X2 2.25 T-Value 4.49 P-Value 0.000

S 295 218 R-Sq 35.11 66.14 R-Sq(adj) 32.29 63.06

13. a. n � 40 b. 4

e. H0: �1 � �2 � �3 � �4 � 0 H1: Not all the �s equal zero.

sy # 1234 � 2500�35 � 3.7796

R2 � 750

1250 � .60

Predictor Coef SE Coef T P Constant 3080.1 343.9 8.96 0.000 Bidders �54.19 12.28 �4.41 0.000 Age 16.289 3.784 4.30 0.000

H0 is rejected if F � 2.65.

H0 is rejected. At least one �i does not equal zero. 15. a. n � 26

b. R2 � 100/140 � .7143 c. 1.4142, found by d. H0: �1 � �2 � �3 � �4 � �5 � 0

H1: Not all the �s are 0. H0 is rejected if F � 2.71. Computed F � 10.0. Reject H0. At least one regression coefficient is not zero.

e. H0 is rejected in each case if t � �2.086 or t � 2.086. X1 and X5 should be dropped.

17. a. $28,000

c. 9.199, found by d. H0 is rejected if F � 2.97 (approximately)

H0 is rejected. At least one regression coefficient is not zero.

e. If computed t is to the left of �2.056 or to the right of 2.056, the null hypothesis in each of these cases is rejected. Computed t for X2 and X3 exceed the critical value. Thus, “population” and “advertising expenses” should be retained and “number of competitors,” X1, dropped.

19. a. The strongest correlation is between GPA and legal. No problem with multicollinearity.

c. H0 is rejected if F � 5.41.

At least one coefficient is not zero. d. Any H0 is rejected if t � �2.571 or t � 2.571. It appears

that only GPA is significant. Verbal and math could be eliminated.

R2 has only been reduced .0303. f. The residuals appear slightly skewed (positive), but

acceptable. g. There does not seem to be a problem with the plot.

21. a. The correlation of Screen and Price is 0.893. So there does appear to be a linear relationship between the two.

b. Price is the “dependent” variable. c. The regression equation is Price � �2484 � 101

Screen. For each inch increase in screen size, the price increases $101 on average.

d. Using “dummy” indicator variables for Sharp and Sony, the regression equation is Price � � 2308 � 94.1 Screen � 15 Manufacturer Sharp � 381 Manufacturer Sony. Sharp can obtain on average $15 more than Samsung and Sony can collect an additional benefit of $381 more than Samsung.

e. Here is some of the output.

R2 � 4.2061 5.0631

� .8307

F � 1.4532 0.1407

� 10.328

R2 � 4.3595 5.0631

� .8610

Computed F � 1,016.67 84.62

� 12.01

284.62

R2 � SSR

SS total � 3,050 5,250

� .5809

F � 750�4 500�35

� 13.125

c. Reject H0 if t � �2.179 or t � 2.179. Since 2.50 is greater than 2.179, we reject the null hypothesis and conclude that union membership is significant and should be included.

d. When you consider the interaction variable, the regression equation is Performance � 38.7 � 3.80 Aptitude �0.1 Union � 3.61 X1X2

H0: �2 � 0 H1: �2 � 0

Predictor Coef SE Coef T P Constant 38.69 15.62 2.48 0.031 Aptitude 3.802 2.179 1.74 0.109 Union �0.10 23.14 �0.00 0.997 X1X2 3.610 3.473 1.04 0.321

The t value corresponding to the interaction term is 1.04. This is not significant. So we conclude there is no interaction between aptitude and union membership when predicting job performance.

11. a. The regression equation is Price � 3080 � 54.2 Bidders � 16.3 Age

Predictor Coef SE Coef T P Constant –2308.2 492.0 -4.69 0.000 Screen 94.12 10.83 8.69 0.000 Manufacturer_Sharp 15.1 171.6 0.09 0.931 Manufacturer_Sony 381.4 168.8 2.26 0.036

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The p-value for Sharp is relatively large. A test of their coefficient would not be rejected. That means they may not have any real advantage over Samsung. On the other hand, the p-value for the Sony coefficient is quite small. That indicates that it did not happen by chance and there is some real advantage to Sony over Samsung.

f. A histogram of the residuals indicates they follow a normal distribution.

g. The residual variation may be increasing for larger fitted values.

23. a.

Sales seem to fall with the number of competitors and rise with the number of accounts and potential.

b. Pearson correlations

The number of accounts and the market potential are moderately correlated.

c. The regression equation is: Sales � 178 � 1.81 Advertising � 3.32 Accounts � 21.2 Competitors � 0.325 Potential

300

2 4 6 8 30 40 50 60 70

200

100

300

200

100

0 4 6 8 10 12 5 10 15 20

CompetitorsS al

Potential

Advertising Accounts Scatter Diagram of Sales vs. Advertising, Accounts, Competitors, Potential

Sales Advertising Accounts Competitors

Advertising 0.159

Accounts 0.783 0.173

Competitors �0.833 �0.038 �0.324

Potential 0.407 �0.071 0.468 �0.202

Predictor Coef SE Coef T P Constant 178.32 12.96 13.76 0.000 Advertising 1.807 1.081 1.67 0.109 Accounts 3.3178 0.1629 20.37 0.000 Competitors �21.1850 0.7879 �26.89 0.000 Potential 0.3245 0.4678 0.69 0.495

S � 9.60441 R-Sq � 98.9% R-Sq(adj) � 98.7%

The computed F value is quite large. So we can reject the null hypothesis that all of the regression coefficients are zero. We conclude that some of the independent variables are effective in explaining sales.

d. Market potential and advertising have large p-values (0.495 and 0.109, respectively). You would probably drop them.

e. If you omit potential, the regression equation is: Sales � 180 � 1.68 Advertising � 3.37 Accounts � 21.2 Competitors

Analysis of Variance

Source DF SS MS F P Regression 4 176777 44194 479.10 0.000 Residual Error 21 1937 92 Total 25 178714

Now advertising is not significant. That would also lead you to cut out the advertising variable and report that the polished regression equation is: Sales � 187 � 3.41 Accounts � 21.2 Competitors

Predictor Coef SE Coef T P Constant 179.84 12.62 14.25 0.000 Advertising 1.677 1.052 1.59 0.125 Accounts 3.3694 0.1432 23.52 0.000 Competitors �21.2165 0.7773 �27.30 0.000

Predictor Coef SE Coef T P Constant 186.69 12.26 15.23 0.000 Accounts 3.4081 0.1458 23.37 0.000 Competitors �21.1930 0.8028 �26.40 0.000

Residual

Fr eq

ue nc

–20 –10 0 10 20

1 2 3 4 5 6 7 8 9

Histogram of the Residuals (response is Sales)

The histogram looks to be normal. There are no problems shown in this plot.

g. The variance inflation factor for both variables is 1.1. They are less than 10. There are no troubles as this value indicates the independent variables are not strongly correlated with each other.

25. The computer output is:

Predictor Coef StDev t-ratio p Constant 651.9 345.3 1.89 0.071 Service 13.422 5.125 2.62 0.015 Age �6.710 6.349 �1.06 0.301 Gender 205.65 90.27 2.28 0.032 Job �33.45 89.55 �0.37 0.712

Analysis of Variance SOURCE DF SS MS F p Regression 4 1066830 266708 4.77 0.005 Error 25 1398651 55946 Total 29 2465481

a. � 651.9 � 13.422X1 � 6.710X2 � 205.65X3 � 33.45X4 b. R2 � .433, which is somewhat low for this type of study. c. H0: �1 � �2 � �3 � �4 � 0; H1: not all �s equal zero.

Reject H0 if F � 2.76.

H0 is rejected. Not all the �i’s equal 0. d. Using the .05 significance level, reject the hypothesis

that the regression coefficient is 0 if t � �2.060 or t � 2.060. Service and gender should remain in the analyses; age and job should be dropped.

e. Following is the computer output using the independent variables service and gender.

F � 1,066,830�4

1,398,651�25 � 4.77

Ŷ

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812

Predictor Coef StDev t-ratio p Constant 784.2 316.8 2.48 0.020 Service 9.021 3.106 2.90 0.007 Gender 224.41 87.35 2.57 0.016

Analysis of Variance SOURCE DF SS MS F p Regression 2 998779 499389 9.19 0.001 Error 27 1466703 54322 Total 29 2465481

A man earns $224 more per month than a woman. The difference between technical and clerical jobs is not significant.

27. a. � 29.913 � 5.324X1 � 1.449X2 b. EPS is (t � �3.26, p-value � .005). Yield is not (t � 0.81,

p-value � .431). c. An increase of 1 in EPS results in a decline of 5.324 in P�E. d. Stock number 2 is undervalued. e. Below is a residual plot. It does not appear to follow the

normal distribution.

Ŷ

g. The correlation between yield and EPS is not a problem. No problem with multicollinearity.

P�E EPS EPS �0.602 Yield .054 .162

29. a. The regression equation is Sales (000) � 1.02 � 0.0829 Informercials

Predictor Coef SE Coef T P Constant 1.0188 0.3105 3.28 0.006 Informericals 0.08291 0.01680 4.94 0.000

Analysis of Variance Source DF SS MS F P Regression 1 2.3214 2.3214 24.36 0.000 Residual Error 13 1.2386 0.0953 Total 14 3.5600

The global test demonstrates there is a relationship between sales and the number of infomercials.

−0.4 −0.2 0.0 0.2 0.4 RESI1

Histogram of RESI1

Fr eq

ue nc

0.0

0.5

1.0

1.5

2.0

2.5

3.0

The residuals appear to follow the normal distribution. 31. a. The regression equation is

Auction Price � �118929 � 1.63 Loan � 2.1 Monthly Payment � 50 Payments Made

f. There does not seem to be a problem with the plot of the residuals versus the fitted values.

Analysis of Variance

Source DF SS MS F P

Regression 3 5966725061 1988908354 39.83 0.000

Residual

Error 16 798944439 49934027

Total 19 6765669500

Predictor Coef SE Coef T P Constant �118929 19734 �6.03 0.000 Loan 1.6268 0.1809 8.99 0.000 Monthly Payment 2.06 14.95 0.14 0.892

Payments Made 50.3 134.9 0.37 0.714

The computed F is 39.83. It is much larger than the critical value 3.24. The p value is also quite small. Thus, the null hypothesis that all the regression coefficients are zero can be rejected. At least one of the multiple regression coefficients is different from zero.

The null hypothesis is that the coefficient is zero in the individual test. It would be rejected if t is less than �2.120 or more than 2.120. In this case, the t value for the loan variable is larger than the critical value. Thus, it should not be removed. However, the monthly payment and payments made variables would likely be removed.

c. The revised regression equation is: Auction Price � �119893 � 1.67 Loan

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813

Price Bedrooms Size Pool Distance Garage

Bedrooms 0.467 Size 0.371 0.383 Pool 0.294 0.005 0.201 Distance �0.347 �0.153 �0.117 �0.139 Garage 0.526 0.234 0.083 0.114 �0.359 Baths 0.382 0.329 0.024 0.055 �0.195 0.221

Maintenance Age Miles

Age 0.465 Miles 0.450 0.522 GasolineIndicato �0.118 �0.068 0.025

Predictor Coef SE Coef T P Constant 38.71 39.02 .99 .324 Bedrooms 7.118 2.551 2.79 0.006 Size 0.03800 0.01468 2.59 0.011 Pool 18.321 6.999 2.62 0.010 Distance �0.9295 0.7279 �1.28 0.205 Garage 35.810 7.638 4.69 0.000 Baths 23.315 9.025 2.58 0.011

S � 33.21 R-Sq � 53.2% R-Sq (adj) � 50.3%

Analysis of Variance SOURCE DF SS MS F P Regression 6 122676 20446 18.54 0.000 Residual Error 98 108092 1103 Total 104 230768

a. Each additional bedroom adds about $7,000 to the selling price, each additional square foot adds $38, a pool adds $18,300 to the value, an attached garage increases the value by $35,800, and each mile the home is from the center of the city reduces the selling price by $929.

b. The R-square value is 0.532. c. The correlation matrix is as follows:

35. a. The regression equation is

Maintenance � 102 � 5.94 Age � 0.374 Miles � 11.8 GasolineIndicator

Each additional year of age adds $5.94 to upkeep cost. Every extra mile adds $0.374 to maintenance total. Gasoline buses are cheaper to maintain than diesel by $11.80 per year.

b. The coefficient of determination is 0.286, found by 65135/227692. Twenty-nine percent of the variation in maintenance cost is explained by these variables.

c. The correlation matrix is:

Predictor Coef SE Coef T P Constant 17.01 35.24 .48 .630 Bedrooms 7.169 2.559 2.80 0.006 Size 0.03919 0.01470 �2.67 0.009 Pool 19.110 6.994 2.73 0.007 Garage 38.847 7.281 5.34 0.000 Baths 24.624 8.995 2.74 0.007

S � 33.32 R-Sq � 52.4% R-Sq(adj) � 50.0%

Analysis of Variance SOURCE DF SS MS F P Regression 5 120877 24175 21.78 0.000 Residual Error 99 109890 1110 Total 104 230768

The independent variable garage has the strongest correlation with price. Distance is inversely related, as expected, and there does not seem to be a problem with correlation among the independent variables.

d. The results of the global test suggest that some of the independent variables have net regression coefficients different from zero.

e. We can delete distance. f. The new regression output follows.

33. The computer output is as follows:

In reviewing the p-values for the various regression coefficients, all are less than .05. We leave all the independent variables.

g. & h. Analysis of the residuals, not shown, indicates the normality assumption is reasonable. In addition, there is no pattern to the plots of the residuals and the fitted values of Y.

Age and Miles both have moderately strong correlations with maintenance cost. The highest correlation among the independent variables is 0.522 between Age and Miles. That is smaller than 0.70 so multicollinearity may not be a problem.

Analysis of Variance Source DF SS MS F P Regression 3 65135 21712 10.15 0.000 Residual Error 76 162558 2139 Total 79 227692

Predictor Coef SE Coef T P Constant 102.3 112.9 0.91 0.368 Age 5.939 2.227 2.67 0.009 Miles 0.3740 0.1450 2.58 0.012 GasolineIndicator �11.80 10.99 �1.07 0.286

The p-value is zero. Reject the null hypothesis of all coefficients being zero and say at least one is important.

The p-value of the gasoline indicator is bigger than 0.10. Consider deleting it.

f. The condensed regression equation is

Maintenance � 106 � 6.17 Age � 0.363 Miles

The normality conjecture appears realistic.

Residuals

Pe rc

en t

0 –100–140 –60 20–20

60 100

Histogram

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814

138.9

–46.3

–92.6

–138.9

–185.2

0.0

46.3

92.6

350300 400 450 500 Predicted

Residuals by Predicted

Re si

du al

550

This plot appears to be random and to have a constant variance.

CHAPTER 15 1. 114.6, found by ($19,989�$17,446)(100)

123.1, found by ($21,468�$17,446)(100) 124.3, found by ($21,685�$17,446)(100) 91.3, found by ($15,922�$17,446)(100) 105.3, found by ($18,375�$17,446)(100) 314.2, found by ($54,818�$17,446)(100)

3. 2003: 115.2, found by (581.9�505.2)�(100) 2004: 98.2, found by (496.1�505.2)�(100) 2005: 90.4, found by (456.6�505.2)(100) 2006: 85.8, found by (433.3�505.2)(100)

5. a.

b. P � 14.52 9.16

(100) � 158.52

Pa � 2.49 1.79

(100) � 139.11Pc � 4.19 1.59

(100) � 263.52

Ps � 4.49 3.29

(100) � 136.47Pt � 3.35 2.49

(100) � 134.54

d. P � $3.35(6) � 4.49(5) � 4.19(3) � 2.49(4) $2.49(6) � 3.29(5) � 1.59(3) � 1.79(4)

(100) � 150.2

P � $3.35(6) � 4.49(4) � 4.19(2) � 2.49(3) $2.49(6) � 3.29(4) � 1.59(2) � 1.79(3)

(100) � 147.1

7. a.

� 349.06

V � $5.95(214) � 9.80(489) � 6.00(203) � 3.29(106) $1.52(200) � 2.10(565) � 1.48(291) � 3.05(87)

(100)

P � 2102.92(103.32) � 103.12 (100) � 103.32

P � 0.10(20,000) � 0.03(130,000) � 0.15(42,000) � 0.10(65,000) 0.07(20,000) � 0.04(130,000) � 0.15(42,000) � 0.08(65,000)

(100) � 102.92

P � 0.10(17,000) � 0.03(125,000) � 0.15(40,000) � 0.10(62,000) 0.07(17,000) � 0.04(125,000) � 0.15(40,000) � 0.08(62,000)

P � 0.38 0.34

(100) � 111.8

PH � 0.10 0.08

(100) � 125.0PS � 0.15 0.15

(100) � 100

PC � 0.03 0.04

(100) � 75.0PW � 0.10 0.07

(100) � 142.9

l � 2(147.1)(150.2) � 148.64

11. a.

Index is 126.3. b. Business activity increased 26.3 percent from

2000 to 2005. 13.

“Real” salary increased 15.

$42,053 � $19,800 � $22,253. X � ($89,673)�2.1324 � $42,053

(0.15) � 1.263.

l � 6.8 5.3

(0.20) � 362.26 265.88

(0.40) � 125.0 109.6

(0.25) � 622,864 529,917

Year Tinora Tinora Index National Index

1995 $28,650 100.0 100 2004 $33,972 118.6 122.5 2009 $37,382 130.5 136.9

The Tinora teachers received smaller increases than the national average.

17. The index (2000 � 100) for selected years is:

Year 2001 2002 2003 2004 2005 2006 2007 2008 2009

Index 105.4 116.8 139.9 165.1 186.7 198.6 241.7 265.2 261.5

Year 2001 2002 2003 2004 2005 2006 2007 2008 2009

Index 114.5 129.7 146.0 160.4 163.9 172.0 187.4 186.6 178.4

Year 2001 2002 2003 2004 2005 2006 2007 2008 2009

Index 100.9 107.3 109.6 108.9 114.6 121.1 118.1 117.6 114.5

The domestic sales almost doubled between 2000 and 2007 and then firmed up.

19. The index (2000 � 100) for selected years is:

International sales grew by about 160% between 2000 and 2009.

21. The index (2000 � 100) for selected years is:

Year 2005 2006 2007 2008 2009

Index 113.4 117.2 125.4 132.1 136.6

23. The index (2004 � 100) for selected years is:

The number of employees increased almost 15 percent between 2000 and 2009.

Revenue increased about 37 percent over the period. 25. The index (2004 � 100) for selected years is:

Year 2005 2006 2007 2008 2009

Index 94.5 97.2 98.2 100.6 99.4

The number of employees decreased one percent between 2004 and 2009.

27.

29.

31. I � 2179.37(178.23) � 178.80

P � $2.00(18) � 1.88(5) � 2.89(70) � 3.99(27) $0.81(18) � 0.84(5) � 1.44(70) � 2.91(27)

(100) � 179.37

Ppo � 3.99 2.91

(100) � 137.11Pmi � 2.89 1.44

(100) � 200.69

Psh � 1.88 0.84

(100) � 223.81Pma � 2.00 0.81

(100) � 246.91

33.

PW � 1.00 0.85

(100) � 117.65

PS � 0.90 1.20

(100) � 75.0PR � 0.60 0.50

(100) � 120

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815

35.

37.

39.

41.

43.

45.

47.

49.

51.

The economy is up 23.05 percent from 1996 to 2009.

53. February:

March:

55. For 1995: $1,876,466, found by $2,400,000�1.279 For 2009: $2,028,986, found by $3,500,000�1.725

CHAPTER 16 1. The weighted moving averages are: 31,584.8, 33,088.9,

34,205.4, 34,899.8, 35,155.0, 34,887.1 3. The regression equation is:

For 2010, t � 12 and 5.

7. a.

b. 28.95%, found by c.

Antilog of 0.8300459 � 6.76 Ŷ � �0.0531997 � 0.1104057 (8) � 0.8300459 Ŷ � �0.0531997 � 0.1104057t for 2010, t � 8

1.28945 � 1.0

a � 1.390087

5 � 0.1104057a

15 5 b � �0.0531997

� 1.104057

10 � 0.1104057

b � 5.274318 � (1.390087) (15)�5

55 � (15)2�5

� 1.30 � 0.90 (7) � 7.6Ŷ � 1.30 � 0.90tŶ

� 8842 � 88.1273(12) � 7784.47Ŷ � 8842 � 88.1273tŶ

� 93.5

l � 100 c 6.4 8.0

(0.40) � 21 20

(0.35) � 297 300

(0.25) d

� 99.50

l � 100 c 6.8 8.0

(0.40) � 23 20

(0.35) � 303 300

(0.25) d

� 1,501 1,214

(0.30) d � 123.05

l � 100 c 1,971.0 1,159.0

(0.20) � 91 87

(0.10) � 114.7 110.6

(0.40)

P � 2(703.56)(686.58) � 695.02

(100) � 703.56

P � 0.76(1,000) � 2.50(5,000) � 26(60,000) � 490(500)

0.287(1,000) � 0.17(5,000) � 3.18(60,000) � 133(500)

PP � 490 133

(100) � 368.42PP � 26.00 3.18

(100) � 817.61

PN � 2.50 0.17

(100) � 1,470.59PA � 0.76

0.287 (100) � 264.8

P � 2(89.79)(91.25) � 90.52

� 89.79

P � 0.05(2,000) � 0.12(200) � 0.18(400) � 0.15(100) 0.06(2,000) � 0.10(200) � 0.20(400) � 0.15(100)

(100)

PE � .015 0.15

(100) � 100PP � 0.18 0.20

(100) � 90

PC � 0.12 0.10

(100) � 120PC � 0.05 0.06

(100) � 83.33

P � 2(106.87)(106.04) � 106.45

P � 0.60(320) � 0.90(110) � 1.00(230) 0.50(320) � 1.20(110) � 0.85(230)

(100) � 106.87

11. Estimated Pairs Seasonal Quarterly Forecast

t (millions) Index (millions)

21 40.05 110.0 44.055 22 41.80 120.0 50.160 23 43.55 80.0 34.840 24 45.30 90.0 40.770

Average SI Seasonal Quarter Component Index

1 0.6859 0.6911 2 1.6557 1.6682 3 1.1616 1.1704 4 0.4732 0.4768

13. The following are the sales estimates.

15. a. The ordered residuals are: 2.61, 2.83, 15.50, 17.17, 6.39, 7.72, 3.81, 7.25,

8.03, and �0.75.�1.08, �16.86,�0.41,�3.72,

�48.50,

Estimate Index Seasonally Adjusted

10.080 0.6911 6.966 10.458 1.6682 17.446 10.837 1.1704 12.684 11.215 0.4768 5.343

Ŷ � 5.1658 � .37805t.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Index

–50

–40

–30

–20

–10

RE SI

Time Series Plot of RESI1

b. There are 2 independent variables (k) and the sample size (n) is 15. For a significance level of 0.05, the upper value is 1.54. Since the computed value of the Durbin- Watson statistic is 2.48, which is above the upper limit, the null hypothesis is not rejected. There is no autocorrelation among these residuals.

17. a. assuming the line starts at 18,000 in 1990 and goes down to 10,000 in 2010.

b. 400 c. 8,000, found by

19. a.

b. using for 2000 c. For 2003, and for 2004, d. For 2015, e. Each asset turned over 0.044 times.

21. a.

b. c. For 2003, For 2005, Ŷ � 34.2255.Ŷ � 40.1913.

Ŷ � 49.140 � 2.9829t

25 01 0302 04 05 06

Ŷ � 1.70999 Ŷ � 1.40136Ŷ � 1.18091,

t � 1Ŷ � 1.00455 � 0.04409t,

0 00 01 02 03 04 05 06 07 08 09 10

1.6 1.8

1.4 1.2 1.0 0.8 0.6 0.4 0.2

18,000 � 400(25)

Ŷ � 18,000 � 400t,

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816

d. For 2009, e. The number of employees decreases at a rate of 2,983

per year. 23. a. Log

b. Log found by 0.790231 � 0.113669(0), antilog is 6.169. Log found by 0.790231 � 0.113669(9), antilog is 65.051.

c. 29.92, which is the antilog of .113669 minus 1 d. Log antilog is 142.65.

25. a. Ŷ � 2.154258,

Ŷ � 1.813252,

Ŷ � 0.790231, Ŷ � 0.790231 � .113669t

Ŷ � 22.2939

b. The equations are and/or log The equation using the logarithm appears better because R2 is larger.

c. log , antilog is 1.4355. log , antilog is 1.8767.

d. log , antilog is 7.0463. It is reasonable if the price rises at the historical rate!

e. The annual rate of increase is 9.91 percent, found by the antilog of 0.0945 minus 1.

27. a. July 87.5; August 92.9; September 99.3; October 109.1 b.

c. April, November, and December are periods of high sales, while February’s sales are lowest.

Note: The solution to Exercises 29 to 33 may vary due to rounding and the particular software package used. 29. a.

Seasonal Index by Quarter

Average SI Seasonal Quarter Component Index

1 0.5014 0.5027 2 1.0909 1.0936 3 1.7709 1.7753 4 0.6354 0.6370

Correction � 1,200�1,206.2 � 0.99486

Month Total Mean Corrected

July 348.9 87.225 86.777 Aug. 368.1 92.025 91.552 Sept. 395.0 98.750 98.242 Oct. 420.4 105.100 104.560 Nov. 496.2 124.050 123.412 Dec. 572.3 143.075 142.340 Jan. 333.5 83.375 82.946 Feb. 297.5 74.375 73.993 March 347.3 86.825 86.379 April 481.3 120.325 119.707 May 396.2 99.050 98.541 June 368.1 92.025 91.552

1,206.200

� �0.221 � 0.0945(23) � 1.9525Ŷ

� �0.221 � 0.0945(9) � 0.6295Ŷ � �0.221 � 0.0945(4) � 0.157Ŷ

� �0.221 � 0.0945t.Ŷ � �1.35 � 1.20 tŶ

1990

30 25 20 15 10 5 0

Pr ic

Year 1995 2000 2005 2010

Oracle Selling Price

b. Production is the largest in the third quarter. It is 77.5 percent above the average quarter. The second quarter is also above average. The first and fourth quarters are well below average, with the first quarter at about 50 percent of a typical quarter.

31. a. The seasonal indices for package play are shown below. Recall that period 1 is actually July, because the data begin with July.

Period Index Period Index

1 0.19792 7 0.26874 2 0.25663 8 0.63189 3 0.87840 9 1.67943 4 2.10481 10 2.73547 5 0.77747 11 1.67903 6 0.18388 12 0.60633

Period Index Period Index

1 1.73270 7 0.23673 2 1.53389 8 0.69732 3 0.94145 9 1.00695 4 1.29183 10 1.13226 5 0.66928 11 0.98282 6 0.52991 12 1.24486

Period Index Period Index

1 0.63371 7 0.25908 2 0.61870 8 0.65069 3 0.89655 9 1.49028 4 1.86415 10 2.28041 5 0.74353 11 1.48235 6 0.29180 12 0.78876

Notice the 4th period (October) and the 10th period (April) are more than twice the average.

b. The seasonal indices for nonpackage play are:

These indices are more constant. Notice the very low values in the 6th (December) and 7th (January) periods.

c. The seasonal indices for total play are:

The regression equation is: Ŷ � 43.611 � 7.21153t

Period Visitors Index Forecast

29 252.86 1.2053 304.77 30 260.07 1.0212 265.58 31 267.29 0.6301 168.42 32 274.50 1.1457 314.50

In 2010, there were 928 visitors. A 10 percent increase in 2011 means there will be 1021 visitors. The quarterly estimates are visitors per quarter.1,021�4 � 255.25

These indices show both the peaks in October (4th period) and April (10th period) and the valleys in December (6th period) and January (7th period).

d. Package play is relatively highest in April. Nonpackage play is relatively high in July. Since 70% of total play comes from package play, total play is very similar to package play.

33. Seasonal Index by Quarter

Average SI Seasonal Quarter Component Index

1 1.1962 1.2053 2 1.0135 1.0212 3 0.6253 0.6301 4 1.1371 1.1457

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The regression approach is probably superior, because the trend is considered.

35. Purse regression equation is Purse � 134,740 � 57,651 � t. Prize regression equation is Prize � 20,211 � 8648 � t. Notice that both the slope and the intercept of the second equation are 15 percent of the corresponding part of the first equation. The prize is always 15% of the purse. The projected purse for 2011 is $1.52 million, found by 134,740 � 57,651 � (24). The fitted prize is $227,755.

37. Answers will vary. 39. With 1988 as the base year, the regression equation is:

Salary increased at a rate of $138,682 per year over the period.

� 316,683 � 138,682t.Ŷ

Period Visitors Index Forecast

Winter 255.25 1.2053 307.65 Spring 255.25 1.0212 260.66 Summer 255.25 0.6301 160.83 Fall 255.25 1.1457 292.44

500,000

1,000,000

2,000,000

2,500,000

$3,500,000

3,000,000

1,500,000

1985 1990 1995 2000 20102005

Salary Salary = 138682 year – 3E+08

CHAPTER 17 1. a. 3

b. 7.815 3. a. Reject if

c. Reject The proportions are not equal. 5. The outcomes are the same; The outcomes are not

the same. Reject if

Do not reject Cannot reject that outcomes are the same.

7. There is no difference in the proportions. There is a difference in the proportions.

Reject if

Do not reject There is no difference in the proportions. 9. a. Reject if .

c. Do not reject . 11. Proportions are as stated; Proportions are not as

stated. Reject if

Reject The proportions are not as stated.H0.

�2 � (50 � 25)2

25 � . . . � (160 � 275)

275 � 115.22

�2 7 11.345.H0 H1:H0:

�2 � (30 � 24)2

24 �

(20 � 24)2

24 �

(10 � 12)2

12 � 2.50

�2 7 9.210H0 H0.

�2 � (47 � 40)2

40 � . . . � (34 � 40)

40 � 3.400

�2 7 15.086.H0 H1: H0:

H0H0.

�2 � (3 � 5)2

5 � . . . � (7 � 5)

5 � 7.60

�2 7 9.236H0 H1:H0:

H0.

�2 � (10 � 20)2

20 �

(20 � 20)2

20 �

(30 � 20)2

20 � 10.0

�2 7 5.991H0

Number of Clients z-values Area Found by fe Under 30 Under �1.58 0.0571 0.5000 � 0.4429 2.855 30 up to 40 �1.58 up to �0.51 0.2479 0.4429 � 0.1950 12.395 40 up to 50 �0.51 up to 0.55 0.4038 0.1950 � 0.2088 20.19 50 up to 60 0.55 up to 1.62 0.2386 0.4474 � 0.2088 11.93 60 or more 1.62 or more 0.0526 0.5000 � 0.4474 2.63

Number of Clients Area fe fo fe � fo (fo � fe)2 [(fo � fe)2]�fe Under 40 0.3050 15.25 16 �0.75 0.5625 0.0369 40 up to 50 0.4038 20.19 22 �1.81 3.2761 0.1623 50 or more 0.2912 14.56 12 2.56 6.5536 0.4501

Total 1.0000 50.00 50 0 0.6493

13. The population of clients follows a normal distribution. The population of clients does not follow a normal

distribution. Reject the null if chi-square is greater than 5.991.

H1: H0:

The first and last class both have expected frequencies smaller than 5. They are combined with adjacent classes.

Since 0.6493 is not greater than 5.991, we fail to reject the null hypothesis. These data could be from a normal distribution.

15. The p-value of 0.746 is greater than 0.05 and the plotted values are close to the line. Thus it is reasonable to say the readings are normally distributed.

17. There is no relationship between community size and section read. There is a relationship. Reject if

Do not reject There is no relationship between community size and section read.

19. No relationship between error rates and item type. There is a relationship between error rates and

item type. Reject if

Do not reject There is not a relationship between error rates and item type.

21. Distribution is not as given above.

Reject if �2 7 4.605.H0df � 2. H1:

�r � �e � 0.25�s � 0.50,H0:

H0.

�2 � (20 � 14.1)2

14.1 � . . . � (225 � 225.25)

225.25 � 8.033

�2 7 9.21.H0 H1: H0:

H0.

�2 � (170 � 157.50)2

157.50 � . . . � (88 � 83.62)

83.62 � 7.340

�2 7 9.488. H0H1:

H0:

Turn fo fe fo � fe (fo � fe)2�fe Straight 112 100 12 1.44 Right 48 50 �2 0.08 Left 40 50 �10 2.00

Total 200 200 3.52

TV Station fo fe fo � fe (fo � fe)2 (fo � fe)2�fe WNAE 53 50 3 9 0.18 WRRN 64 50 14 196 3.92 WSPD 33 50 �17 289 5.78

150 150 0 9.88

is not rejected. The proportions are as given in the null hypothesis.

23. There is no preference with respect to TV stations. There is a preference with respect to TV stations.

is rejected if �2 7 5.991.H0df � 3 � 1 � 2. H1: H0:

is rejected. There is a preference for TV stations.H0

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Region fo fe fo � fe (fo � fe)2�fe Northeast 68 84 �16 3.0476 Midwest 104 96 8 0.6667 South 155 140 15 1.6071 West 73 80 �7 0.6125

Total 400 400 0 5.9339

25. The distribution is not as given.

Reject if �2 7 11.345.H0 H1:

�w � 0.20�s � 0.35,�m � 0.24,�n � 0.21,H0:

Color fo fe (fo � fe)2�fe Blue 12 16.56 1.256 Brown 14 8.64 3.325 Yellow 13 10.80 0.448 Red 14 8.64 3.325 Orange 7 16.56 5.519 Green 12 10.80 0.133

Total 72 14.006

Do not reject The color distribution agrees with the manufacturer’s information.

39. a. Salary and winning are not related. Salary and winning are related.

Reject if .�2 7 3.84H0 H1: H0:

H0.

Do not reject . The digits are evenly distributed. 29.

fo fe fo – fe (fo – fe)2 (fo � fe)2�fe 44 28 16 256 9.143 32 28 4 16 0.571 23 28 �5 25 0.893 27 28 �1 1 0.036 23 28 �5 25 0.893 24 28 �4 16 0.571 31 28 3 9 0.321 27 28 �1 1 0.036 28 28 0 0 0.000 21 28 �7 49 1.750

14.214

Salary

Winning Lower Half Top Half Total

No 9 5 14 Yes 6 10 16

Total 15 15

is not rejected. The distribution of order destinations reflects the population.

27. The proportions are the same. The proportions are not the same.

Reject if . �2 7 16.919H0 H1: H0:

Do not reject H0. Conclude that salary and winning may not be related.

�2 � (9 � 7)2

7 �

(5 � 7)2

7 �

(6 � 8)2

8 �

(10 � 8)2

8 � 2.14

Hourly Wage f M fM M � x (M � x)2 f(M � x)2

$5.50 up to 6.50 20 6 120 �2.222 4.938 98.8 6.50 up to 7.50 24 7 168 �1.222 1.494 35.9 7.50 up to 8.50 130 8 1040 �0.222 0.049 6.4 8.50 up to 9.50 68 9 612 0.778 0.605 41.1 9.50 up to 10.50 28 10 280 1.778 3.161 88.5

Total 270 2220 270.7

Wage z-values Area Found by fe fo fe � fo (fo � fe)2 [(fo � fe)2]�fe Under Under 0.5000 � $6.50 �1.72 0.0427 0.4573 11.529 20 �8.471 71.7578 6.2241

6.50 up �1.72 up 0.4573 � to 7.50 to �0.72 0.1931 0.2642 52.137 24 28.137 791.6908 15.1848

7.50 up �0.72 up 0.2642 � to 8.50 to 0.28 0.3745 0.1103 101.115 130 �28.885 834.3432 8.2514

8.50 up 0.28 up 0.3980 � to 9.50 to 1.27 0.2877 0.1103 77.679 68 9.679 93.6830 1.2060

9.50 or 1.27 or 0.5000 � more more 0.1020 0.3980 27.54 28 �0.46 0.2116 0.0077

Total 1.0000 270 270 0 30.874

The sample mean is 8.222, found by 2220�270. The sample standard deviation is 1.003, found as the square root of 270.7�269.

The population of wages follows a normal distribution. The population of hourly wages does not follow a

normal distribution. Reject the null if chi-square is greater than 7.779.

H1: H0:

Since 30.874 is greater than 7.779, we reject the null hypothesis not from a normal distribution.

31.

The p-value (0.097) is greater than 0.05. Do not reject the null hypothesis. The data could be normally distributed.

33. Gender and attitude toward the deficit are not related. Gender and attitude toward the deficit are related.

Reject if

Since you reject A person’s position on the deficit is influenced by his or her gender.

35. Whether a claim is filed and age are not related. Whether a claim is filed and age are related.

Reject if

Reject Age is related to whether a claim is filed. 37. .

The proportions are not as given. Reject if .�2 7 15.086H0 H1:�BR � �R � .12�Y � �G � .15,�BL � �O � .23,H0:

H0.

�2 � (170 � 203.33)2

203.33 � . . . � (24 � 35.67)

35.67 � 53.639

�2 7 7.815.H0 H1: H0:

H0.43.578 7 5.991,

� (114 � 143.95) 2

143.95 �

(25 � 43.47)2

43.47 � 43.578

� (68 � 49.53) 2

49.53 �

(305 � 256.59)2

256.59

�2 � (244 � 292.41)2

292.41 �

(194 � 164.05)2

164.05

�2 7 5.991.H0 H1: H0:

99 Mean 22163 StDev 14734 N 13 AD 0.594 P-Value 0.097

0 20000 40000 60000 Enrollment

Probability Plot of Enrollment Normal

Pe rc

en t

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5 10

30 40

99 Mean 2.448 StDev 0.6983 N 30 AD 0.602 P-Value 0.108

50 60 70 80

95 90

1.0 1.5 2.0 2.5 3.0 4.03.5 4.5 Attendance

Probability Plot of Attendance Normal

Pe rc

en t

99 Mean 88.51 StDev 33.90 N 30 AD 0.774 P-Value 0.039

0 50 100 150 200 Salary

Probability Plot of Salary Normal

Pe rc

en t

The salary p-value is 0.039, which is less than 0.05. Reject the null hypothesis. Salaries are not normally distributed. However, the attendance p-value is 0.108, which is greater than 0.05. Do not reject the null hypothesis. Attendance could be normally distributed.

CHAPTER 18 1. a. If the number of pluses (successes) in the sample is 9 or

more, reject b. Reject because the cumulative probability associated

with nine or more successes (.073) does not exceed the significance level (.10).

3. a. b. is rejected if there are nine or more plus signs. A “�”

represents a loss. c. Reject It is an effective program, because there were

9 people who lost weight. 5. a. (There is no change in weight.)

(There is a loss of weight.) b. Reject if .

d. Reject The weight loss program is effective. 7. is rejected if

Because do not reject No preference. 9. a.

b. is rejected if .

is rejected. The median income is greater than $81,500.H0

z � 170 � .50 � 100

7.07 � 9.83

z 7 1.65H0 Median 7 $81,500H1:Median � $81,500;H0:

H0..44 6 2.05,

z � 42.5 � 40.5

4.5 � .44

z 7 2.05.H0� 7 .50.H1:� � .50,H0: H0.

z � (32 � .50) � .50(45)

.50145 � 2.68

z 7 1.65H0 � 7 .50H1: � � .50H0:

H0.

n � 10� 7 .50;H1:� � .50;H0:

H0 H0.

b. 11.

Sums: So (the smaller of the two sums). From Appendix B.7, .05 level, one-tailed test, the critical value is 3. Since the T of , do not reject (one-tailed test). There is no difference in square footage. Professional couples do not live in larger homes.

13. a. The production is the same for the two systems. Production using the Mump method is greater.

b. is rejected if c. The calculations for the first three employees are:

The sum of the negative ranks is 6.5. Since 6.5 is less than 21, is rejected. Production using the Mump method is greater.

15. The distributions are the same. The distributions are not the same. Reject if or z 7 1.96.z 6 �1.96H0

H1:H0:

Employee Old Mump d Rank R � R �

A 60 64 4 6 6 B 40 52 12 12.5 12.5 C 59 58 �1 2 2

n � 13.T � 21,H0 H1: H0:

H04 7 3 n � 7,

T � 4�24.�4,

Couple Difference Rank

1 550 7 2 190 5 3 250 6 4 �120 3 5 �70 1 6 130 4 7 90 2

Campus Online

Age Rank Age Rank

26 6 28 8 42 16.5 16 1 65 22 42 16.5 38 13 29 9.5 29 9.5 31 11 32 12 22 3 59 21 50 20 42 16.5 42 16.5 27 7 23 4 41 14 25 5 46 19 94.5 18 2

158.5

is not rejected. There is no difference in the two populations.

17. : The distributions are the same. H1: The distribution of Campus is to the right. Reject if z � 1.65.H0

z � 86.5 �

8(8 � 8 � 1) 2

8(8)(8 � 8 � 1) 12

� 1.943

A B Score Rank Score Rank

38 4 26 1 45 6 31 2 56 9 35 3 57 10.5 42 5 61 12 51 7 69 14 52 8 70 15 57 10.5 79 16 62 13

86.5 49.5

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820

is not rejected. There is no difference in the distributions. 19. ANOVA requires that we have two or more populations, the

data are interval- or ratio-level, the populations are normally distributed, and the population standard deviations are equal. Kruskal-Wallis requires only ordinal-level data, and no assumptions are made regarding the shape of the populations.

21. a. The three population distributions are equal. Not all of the distributions are the same.

b. Reject if . c.

H 7 5.991H0

H1:H0:

z � 158.5 �

12(12 � 10 � 1) 2

12(10)(12 � 10 � 1) 12

� 1.35

Rank Rank Rank

8 5 1 11 6.5 2 14.5 6.5 3 14.5 10 4 16 12 9

64 13 19 53

d. Reject because 8.98 � 5.991. The three distributions are not equal.

23. The distributions of the lengths of life are the same. The distributions of the lengths of life are not the same.

is rejected if H � 9.210.H0

H1: H0:

� 59.98 � 51 � 8.98

H � 12

16(16 � 1) c (64)2

5 �

(53)2

6 �

(19)2

5 d � 3(16 � 1)

Salt Fresh Others

Hours Rank Hours Rank Hours Rank

167.3 3 160.6 1 182.7 13 189.6 15 177.6 11 165.4 2 177.2 10 185.3 14 172.9 7 169.4 6 168.6 4 169.2 5 180.3 12 176.6 9 174.7 8

46 39 35

is not rejected. There is no difference in the three distributions.

25. a.

H � 12

15(16) c (46)2

5 �

(39)2

5 �

(35)2

5 d � 3(16) � 0.62

0 0 2 4 6 8 10 12 14

Male

Fe m

al e

Scatter Diagram of Female versus Male

Representative Sales Rank Training Rank d d 2

1 319 3 3 0 0 2 150 10 9 1 1 3 175 9 6 3 9 4 460 1 1 0 0 5 348 2 4 �2 4 6 300 4.5 10 �5.5 30.25 7 280 6 5 1 1 8 200 7 2 5 25 9 190 8 7 1 1

10 300 4.5 8 �3.5 12.25

83.50

A moderate positive correlation. b. No correlation among the ranks. A positive

correlation among the ranks. Reject if t � 1.860.

is not rejected. The correlation in population among the ranks could be 0.

29. � � .50; � � .50; Use a software package to develop the binomial probability distribution for n � 19 and � � .50.

is rejected if there are either 5 or fewer “�” signs, or 14 or more. The total of 12 “�” signs falls in the acceptance region. is not rejected. There is no preference between the two shows.

31. � � .50 � � .50 is rejected if there are 12 or more or 3 or fewer plus

signs. Because there are only 8 plus signs, is not rejected. There is no preference with respect to the two brands of components.

33. � � .50; � � .50. Reject if z � 1.96 or z � �1.96.

Reject There is a difference in the preference for the two types of orange juice.

H0:

z � 159.5 � 100

7.071 � 8.415

H0H1:H0:

H0 H0

H1:H0:

t � 0.494 A

10 � 2 1 � (0.494)2

� 1.607

H0 H1:H0:

rs � 1 � 6(83.5)

10(102 � 1) � 0.494

b. Male Female d d 2

4 5 �1 1 6 4 2 4 7 8 �1 1 2 7 �5 25

12 11 1 1 8 6 2 4 5 3 2 4 3 9 �6 36

13 2 11 121 14 10 4 16

1 1 0 0 9 13 �4 16

10 12 �2 4 11 14 �3 9

Total 242

c. No correlation among the ranks. A positive correlation among the ranks.

Reject if t � 1.782.

is rejected. We conclude the correlation in population among the ranks is positive. Husbands and wives generally like the same shows.

27.

t � 0.47 A

14 � 2 1 � (0.47)2

� 1.84

H1: H0:

rs � 1 � 6(242)

14(142 � 1) � 0.47

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35. Rates are the same; The rates are not the same. is rejected if H � 5.991. H � .082. Do not reject

37. The populations are the same. The populations differ. Reject if H � 7.815. H � 14.30. Reject .

39.

There is no correlation between the rankings of the coaches and of the sportswriters.

There is a positive correlation between the rankings of the coaches and of the sportswriters. Reject if t � 1.812.

is rejected. There is a positive correlation between the sportswriters and the coaches.

41. a. There is no difference in the distributions of the selling prices in the five townships. There is a difference in the distributions of the selling prices of the five townships. is rejected if H is greater than 9.488. The computed value of H is 4.70, so the null hypothesis is not rejected. The sample data do not suggest a difference in the distributions of selling prices.

b. There is no difference in the distributions of the selling prices depending on the number of bedrooms.

There is a difference in the distributions of the selling prices depending on the number of bedrooms. is rejected if H is greater than 9.448. The computed value of H is 16.34, so the null hypothesis is rejected. The sample data indicate there is a difference in the distributions of selling prices based on the number of bedrooms. Note: Combine 6 or more into a single group.

c. There is no difference in the distributions of the distance from the center of the city depending on whether the home had a pool or not. There is a difference in the distributions of the distances from the center of the city depending on whether the home has a pool or not. is rejected if H is greater than 3.84. The computed value of H is 3.37, so the null hypothesis is not rejected. The sample data do not suggest a difference in the distributions of the distances.

43. a. The distributions of the maintenance costs are the same for all manufacturers.

The distributions of the costs are not the same. is rejected if H � 5.991.

is rejected. There is a difference in the maintenance cost for the three bus manufacturers.

b. The distributions of the maintenance costs are the same for bus capacities.

The distributions of the costs are not the same

is rejected if H � 7.815.

� 3(81) � 2.74 is not rejected. There is no difference in the

maintenance cost for the four different capacities. c. The distributions are the same.

The distributions are different. Reject H0 if z � �1.96 or z � 1.96.

We fail to reject H0. The distributions could be the same.

W � 2252 �

53(53 � 27 � 1) 2

(53) (27) (53 � 27 � 1) 12

� 1.07

H1: H0:

H � 12

80(81) c

(96.5)2

4 �

(332.5)2

7 �

(388.5)2

9 �

(2422.5)2

60 d

H1:

H0:

H � 12

80(81) c

(1765)2

47 �

(972)2

25 �

(503)2

8 d � 3(81) � 8.29

H1:

H0:

H1:

H0:

H0 H1:

H0:

H1: H0:

t � 0.727 A

12 � 2 1 � (.727)2

� 3.348

H1:

H0:

rs � 1 � 6(78)

12(122 � 1) � 0.727

H0H0 H1:H0:

H0.H0

H1:H0:

0.5 0.4 0.3 0.2 0.1 0

Problem not corrected

Unfriendly Invoice error

Price high Long wait

CHAPTER 19 1.

Count 38 23 12 10 8 Percent 42 25 13 11 9 Cum % 42 67 80 91 100

, Arithmetic R,

Time Means Range

8:00 A.M. 46 16 8:30 A.M. 40.5 6 9:00 A.M. 44 6 9:30 A.M. 40 2

10:00 A.M. 41.5 9 10:30 A.M. 39.5 1

251.5 40

UCL

LCL 8 8:30 9 9:30 10 10:30

46.78

41.92

37.06

About 67 percent of the complaints concern the problem not being corrected and the price being too high.

3. Chance variation is random in nature; because the cause is a variety of factors, it cannot be entirely eliminated. Assignable variation is not random; it is usually due to a specific cause and can be eliminated.

5. a. The A2 factor is 0.729. b. The value for D3 is 0, and for D4 it is 2.282.

7. a.

UCL � 41.92 � 0.729(6.67) � 46.78

LCL � 41.92 � 0.729(6.67) � 37.06 b. Interpreting, the mean reading was 341.92 degrees

Fahrenheit. If the oven continues operating as evidenced by the first six hourly readings, about 99.7 percent of the mean readings will lie between 337.06 degrees and 346.78 degrees.

9. a. The fraction defective is 0.0507. The upper control limit is 0.0801 and the lower control limit is 0.0213.

b. Yes, the 7th and 9th samples indicate the process is out of control.

c. The process appears to stay the same.

11.

The control limits are 0 and 7.5. The process is out of control on the seventh day.

2.64 3 12.64

c � 37 14

� 2.64

R � 40 6

� 6.67� 251.5

6 � 41.92X

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Percent Probability of Defective Accepting Lot

10 .889 20 .558 30 .253 40 .083

1.00

.80

.60

.40

.20

0 5 10 15 20 25 30 35 40

Pr ob

ab ili

ty o

f a cc

ep ta

nc e

(30, .149)

(10, .736)

17.

P(X � 1 ƒ n � 10, � � .40) � .046 P(X � 1 ƒ n � 10, � � .30) � .149 P(X � 1 ƒ n � 10, � � .20) � .375 P(X � 1 ƒ n � 10, � � .10) � .736

19.

Co un

Percent

100

150

200

100

Reason Count

Percent Cum %

Poor wiring 80

40.0 40.0

Short coil 60

30.0 70.0

Defective plug 50

25.0 95.0

Other 10

5.0 100.0

21. a.

b. The mean is 10.16, which is above the upper control limit and is out of control. There is too much cola in the soft drinks. The process is in control for variation; an adjustment is needed.

23. a.

UCL � 2.575(15.6) � 40.17 LCL � 30.5665 � (1.023)(15.6) � 14.61 UCL � 30.5665 � (1.023)(15.6) � 46.53

R � 312 20

� 15.6

� 611.3333

20 � 30.57X

LCL � 0(0.25) � 0 UCL � 2.115(0.25) � 0.52875

� 9.85575 LCL � 10.0 � 0.577(0.25) � 10.0 � 0.14425

� 10.14425 UCL � 10.0 � 0.577(0.25) � 10.0 � 0.14425

0 Subgroup

M ea

10 20

UCL = 46.53

X = 30.57

LCL = 14.61

Ra ng

UCL = 40.16

R = 15.60

LCL = 0.000

c. The points all seem to be within the control limits. No adjustments are necessary.

25.

All the points are in control for both the mean and the range.

27. a.

LCL � 0.08 � 0.115 � 0 UCL � 0.08 � 0.115 � 0.195

3 B

0.08(0.92) 50

� 0.115

p � 40

10(50) � 0.08

UCL � 2.115(16.2) � 34.26 LCL � 418.3 � (0.577)(16.2) � 408.95 UCL � 418.3 � (0.577)(16.2) � 427.65

R � 162 10

� 16.2

� 4,183

10 � 418.3X

b. 0.2

0.15

0.1

0.05

0 1 2 3 4 5

Samples

Pe rc

en t D

ef ec

tiv e

6 7 8 9 10

0.5

0.4

0.3

0.2

0.1

0.0

Pr op

or tio

10 20 Sample Number

P Chart for C1

UCL = 0.4337

P = 0.25

LCL = 0.06629

c. There are no points that exceed the limits. 29.

13.

The control limits are from 0 to 2.760, so there are no receipts out of control.

15.

0.545 3 10.545 � 0.545 2.215

c � 6

11 � 0.545

Lin01803_appc_782-828.qxd 11/18/10 1:05 PM Page 822

823

0.1 C1

0.2 0.3

1.0

0.9

0.8

0.7

0.6

0.5

0.4

Pl an

These sample results indicate that the odds are much less than 50-50 for an increase. The percent of stocks that increase is “in control” around 0.25, or 25%. The control limits are 0.06629 and 0.4337.

31.

P(X � 5 ƒ n � 20, � � 0.30) � 0.417 P(X � 5 ƒ n � 20, � � 0.20) � 0.805 P(X � 5 ƒ n � 20, � � 0.10) � 0.989 P(X � 5 ƒ n � 20, � � 0.05) � 0.999 P(X � 3 ƒ n � 10, � � 0.30) � 0.649 P(X � 3 ƒ n � 10, � � 0.20) � 0.878 P(X � 3 ƒ n � 10, � � 0.10) � 0.987 P(X � 3 ƒ n � 10, � � 0.05) � 0.999

The solid line is the operating characteristic curve for the first plan, and the dashed line, the second. The supplier would prefer the first because the probability of acceptance is higher (above). However, if he is really sure of his quality, the second plan seems higher at the very low range of defect percentages and might be preferred.

33. a.

b. LCL � 14.2 � 11.3 � 2.9 UCL � 14.2 � 11.3 � 25.5

3114.2 � 11.30c � 213 15

� 14.2;

0 1 2 3 4 5 6 7 8 9 10

Day

Ro bb

er ie

37.

P(X � 3 ƒ n � 20, � � .30) � .108 P(X � 3 ƒ n � 20, � � .20) � .412 P(X � 3 ƒ n � 20, � � .10) � .867

1.0

0.8

0.6

0.4

0.2

0.0 0 5 10 15 20 25 30 35 40

Pr ob

ab ili

ty o

f a cc

ep tin

g lo

Incoming lot percent defective

26 24 22 20 18 16 14 12 10 8 6 4 2 0

Sample

Ch oc

ol at

e Ch

ip s

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

c. All the points are in control.

35.

LCL � 7.0 � 317 � 0 UCL � 7.0 � 317 � 14.9

c � 70 10

� 7.0

Lin01803_appc_782-828.qxd 11/18/10 1:06 PM Page 823

Appendix C Answers to Odd-Numbered Review Exercises

5. a. Box plot. b. Median is 48, the first quartile is 24, and the third

quartile is 84. c. Positively skewed with the long tail to the right. d. You cannot determine the number of observations.

REVIEW OF CHAPTERS 5–7 PROBLEMS

1. a. .035 b. .018 c. .648

3. a. .0401 b. .6147 c. 7,440

5. a.

b. About 550 c.

REVIEW OF CHAPTERS 8 AND 9 PROBLEMS

1. z � � 0.59, .5000 � .2224 � .2776

3. 160 2.426 , 152.33 up to 167.67

5. 985.5 2.571 , 864.27 up to 1,106.73

7. 240 2.131 , 221.35 up to 258.65

Because 250 is in the interval, the evidence does not indicate an increase in production.

9. n � � 150

11. n � .08(.92) � 999

13. n � .4(.6) � 1,448

REVIEW OF CHAPTERS 10–12 PROBLEMS

1. H0: � 36; H1: � 36. Reject H0 if t � �1.683.

t � � �3.60

Reject H0. The mean height is less than 36 inches. 3. H0: � 20, H1: � 20. Reject H0 if t � 1.860.

t � � 0.485

H0 is not rejected. The mean amount of unproductive time is not more than 20 minutes.

5. H0: d � 0; H1: d � 0. Reject H0 if t � 1.883.

H0 is not rejected. There is no difference in the life of the paints.

d � 0.4 sd � 6.11 t � 0.4

6.11�110 � 0.21

21 � 20 6.185�19

35.5 � 36.0 0.9�142

¢2.33 0.03 ≤2

¢2.33 0.02 ≤2

B1.96(25) 4

R 2

35 116

115.5 16

20 140

8.8 � 8.6 2.0�135

� 1.833

� � 1.18 � 1.10

REVIEW OF CHAPTERS 1–4 PROBLEMS

1. a. Mean is 147.9. Median is 148.5. Standard deviation is 69.24.

b. The first quartile is 106. The third quartile is 186.25. c.

824

0 50 100 200150 250 300

There are no outliers. The distribution is symmetric. The whiskers and the boxes are about equal on the two sides.

d. 26 � 64, use 6 classes; i � 299 � 14

6 � 47.5, use i � 6.

Amount Frequency

$ 0 up to $ 50 3 50 up to 100 8

100 up to 150 15 150 up to 200 13 200 up to 250 7 250 up to 300 7

Total 50

Amounts Frequency

$24,000 up to 30,000 8 30,000 up to 36,000 22 36,000 up to 42,000 15 42,000 up to 48,000 4 48,000 up to 54,000 1 54,000 up to 60,000 1

Total 51

e. Answers will vary but include all of the above information. 3. a. Mean is $35,768. Median is $34,405. Standard deviation

is $5,992. b. The first quartile is $32,030. The third quartile is 38,994. c.

30000 35000 40000 5000045000 55000 60000

There are two outliers above $50,000. The distribution is positively skewed. The whiskers and the boxes on the right are much larger than the ones on the left. d.

e. Answers will vary but include all of the above information.

Lin01803_appc_782-828.qxd 11/18/10 1:06 PM Page 824

825

7. For Social Status H0: The mean self-rated social status of the employees is the same. H1: The mean self-rated social status of the employees is not the same. Reject H0 if F � 4.26. For Educational Background H0: The mean scores for school type is the same. H1: The mean scores for school type is not the same. Reject H0 if F � 4.26. For Interaction H0: There is no interaction between social status and school type. H1: There is interaction between social status and school type. Reject H0 if F � 3.63.

3. a. 0.9261 b. 2.0469, found by c. �1 � �2 � �3 � �4 � 0

Not all coefficients are zero.

Reject if F � 2.87, computed F � 62.697, found by 162.70�4.19.

d. Could delete X2 because t-ratio (1.29) is less than the critical t value of 2.086. Otherwise, reject for X1, X3, and X4 because all of those t-ratios are greater than 2.086.

REVIEW OF CHAPTERS 15 AND 16 PROBLEMS

1. a. 106.1, found by (157�148(100) b. 100.0, found by 157�157(100) c. 147.3 � 4.9t and 186.5, found by 147.3 � 4.9(8)

REVIEW OF CHAPTERS 17 AND 18 PROBLEMS

is rejected if There are 16 observations greater than 60.

Reject The median sales per day is greater than 60. 3. The population lengths are the same.

The population lengths are not the same. is rejected if

Do not reject The population lengths are the same.H0.

� 78.451 � 75 � 3.451

H � 12

24(24 � 1) c (104.5)2

7 �

(125.5)2

9 �

(70)2

8 d � 3(24 � 1)

H is 7 5.991.H0 H1: H0:

H0.

z � 15.5 � 10.0

2.2361 � 2.46

z 7 1.65.H0 � � 120(.5)(.5) � 2.2361 � 20(.5) � 10

Median 7 60H1: Median � 60H0:

Ŷ � [3.5 � 0.7(66)]0.90 � (49.7)(0.90) � 44.73 Ŷ � [3.5 � 0.7(61)]1.20 � [46.2][1.20] � 55.44

H1: H0:

183.8�20

Two-way ANOVA: Sales versus Social, School

Source df SS MS F P

Social 2 84.000 42.0000 8.49 0.008 School 2 22.333 11.1667 2.26 0.160 Interaction 4 337.667 84.4167 17.07 0.000 Error 9 44.500 4.9444

Total 17 488.500

There is a difference between the mean sales for social status but not for schools. There is an interaction between social status and schools.

REVIEW OF CHAPTERS 13 AND 14 PROBLEMS

1. a. Profit b. � a � b1 X1 � b2 X2 � b3 X3 � b4 X4 c. $163,200 d. About 86 percent of the variation in net profit is

explained by the four variables. e. About 68 percent of the net profits would be within

$3,000 of the estimates; about 95% would be within 2($3,000), or $6,000, of the estimates; and virtually all would be within 3($3,000), or $9,000, of the estimates.

Ŷ

Lin01803_appc_782-828.qxd 11/18/10 1:06 PM Page 825

PRACTICE TEST (AFTER CHAPTER 4) Part 1

1. statistics 2. descriptive statistics 3. population 4. quantitative and qualitative 5. discrete 6. nominal 7. nominal 8. zero 9. seven

10. 50 11. variance 12. never 13. median

PROBLEMS

1. 2. a. 30 thousands of dollars

b. 105 c. 52 d. 0.19, found by 20/105 e. 165 f. 120 and 330

3. a. 70 b. 71.5 c. 67.8 d. 28 e. 9.34

4. $44.20, found by [(200)$36 � (300)$40 � (500)$50]�1,000 5. a. pie chart

b. 11.1 c. three times d. 65%

PRACTICE TEST (AFTER CHAPTER 7) Part 1

1. never 2. experiment 3. event 4. joint 5. a. permutation

b. combination 6. one 7. three or more outcomes 8. infinite 9. one

10. 0.2764 11. 0.0475 12. independent 13. mutually exclusive 14. only two outcomes 15. bell-shaped

2 3 (1.18)(1.04)(1.02) � 1.0777 or 7.77%

PROBLEMS 1. a. 0.0526, found by (5�20)(4�19)

b. 0.4474, found by 1� (15�20)(14�19) 2. a. 0.2097, found by 16(.15)(.85)15

b. 0.9257, found by 1 � (.85)16

3. 720, found by 6 � 5 � 4 � 3 � 2 4. a. 2.2, found by .2(1) � .5(2) � .2(3) � .1(4)

b. 0.76, found by .2(1.44) � .5(0.04) � .2(0.64) � .1(3.24) 5. a. 0.1808. The z value for $2,000 is 0.47, found by

(2,000 � 1,600)�850. b. 0.4747, found by 0.2939 � 0.1808 c. 0.0301, found by 0.5000 � 0.4699

6. a. contingency table b. 0.625, found by 50�80 c. 0.75, found by 60�80 d. 0.40, found by 20�50 e. 0.125, found by 10�80

7. a. 0.0498, found by

b. 0.2240, found by

c. 0.1847, found by 1 � [0.0498 � 0.1494 � 0.2240 � 0.2240 � 0.1680]

d. .0025

PRACTICE TEST (AFTER CHAPTER 9) PART 1

1. random sample 2. sampling error 3. standard error 4. become smaller 5. point estimate 6. confidence interval 7. population size 8. proportion 9. positively skewed

10. 0.5

PART 2 1. 0.0351, found by 0.5000 � 0.4649. The corresponding

2. a. The population mean is unknown. b. 9.3 years, which is the sample mean c. 0.3922, found by d. The confidence interval is from 8.63 up to 9.97, found

by 9.3 1.708

3. 2675, found by .27(1 � .27)

4. The confidence interval is from 0.5459 up to 0.7341, found

by .64 1.96 B

.64(1 � .64) 100

a 2.33 .02 b

a 2 126

2/126

z � 11 � 12.2 2.3/112

� � 1.81

33 e�3

30e�3

826

Appendix C Solutions to Practice Tests

Lin01803_appc_782-828.qxd 11/18/10 1:06 PM Page 826

827

PRACTICE TEST (AFTER CHAPTER 12) PART 1

1. null hypothesis 2. significance level 3. five 4. standard deviation 5. normality 6. test statistic 7. split evenly between the two tails 8. range from negative infinity to positive infinity 9. independent

10. three and 20

PART 2

1. H0: � 90 H1: � 90 If t � 2.567, reject H0.

Do not reject the null. The mean time in the park could be 90 minutes.

2. H0: 1 � 2 H1: 1 � 2

If t � � 2.064 or t � 2.064, then reject H0.

Reject the null hypothesis. There is a difference in the mean miles traveled.

3. a. three, because there are 2 df between groups. b. 21, found by the total degrees of freedom plus 1. c. If the significance level is 0.05, the critical value is 3.55. d. H0: 1 � 2 � 3 H1: Treatment means are not all the

same. e. At a 5 percent significance level, the null hypothesis is

rejected. f. At a 5 percent significance level, we can conclude the

treatment means differ.

PRACTICE TEST (AFTER CHAPTER 14) PART 1

1. vertical 2. interval 3. zero 4. �0.77 5. never 6. 7 7. decrease of .5 8. �0.9 9. zero

10. unlimited 11. linear 12. residual 13. two 14. correlation matrix 15. normal distribution

PART 2 1. a. 30

b. The regression equation is � 90.619X �0.9401. If X is zero, the line crosses the vertical axis at �0.9401. As the independent variable increases by one unit, the dependent variable increases by 90.619 units.

c. 905.2499

Ŷ

t � 837 � 797

A1220.83¢ 114 � 1

12 ≤

� 40.0

13.7455 � 2.910

sp 2 �

(14 � 1)(30)2 � (12 � 1)(40)2

14 � 12 � 2 � 1220.83

df � 14 � 12 � 2 � 24

t � 96 � 90 12/118

� 2.12

d. 0.3412, found by 129.7275�380.1667. Thirty-four percent of the variation in the dependent variable is explained by the independent variable.

e. 0.5842, found by H0: p � 0 H1: p � 0 Using a significance level of 0.01, reject H0 if t � 2.467.

Reject H0. There is a negative correlation between the variables.

2. a. 30 b. 4 c. 0.5974, found by 227.0928�380.1667 d. H0: �1 � �2 � �3 � �4 � 0 H1 : Not all �s are 0.

Reject H0 if F � 4.18 (using a 1 percent level of significance). Since the computed value of F is 9.27, reject H0. Not all of the regression coefficients are zero.

e. Reject H0 if t � 2.787 or t � �2.787 (using a 1 percent level of significance). Drop variable 2 initially and then rerun. Perhaps you will delete variable(s) 1 or 4 also.

PRACTICE TEST (AFTER CHAPTER 16) PART 1

1. denominator 2. index 3. quantity 4. base period 5. 1982–1984 6. trend 7. moving average 8. autocorrelation 9. residual

10. same

PART 2 1. a. 111.54, found by (145,000�130,000) � 100 for 2006

92.31, found by (120,000�130,000) � 100 for 2007 130.77, found by (170,000�130,000) � 100 for 2008 146.15, found by (190,000�130,000) � 100 for 2009

b. 87.27, found by (120,000�137,500) � 100 for 2007 126.64, found by (170,000�137,500) � 100 for 2008 138.18, found by (190,000�137,500) � 100 for 2009

2. a. 108.91, found by (1100�1010) � 100 b. 111.18, found by (4525�4070) � 100 c. 110.20, found by (5400�4900) � 100 d. 110.69, found by the square root of (111.18) � (110.20)

3. For January of the fifth year, the seasonally adjusted forecast is 70.0875, found by 1.05 � [5.50 � 1.25(49)]. For February of the fifth year, the seasonally adjusted forecast is 66.844, found by 0.983 � [5.50 � 1.25(50)].

PRACTICE TEST (AFTER CHAPTER 18) PART 1

1. nominal 2. at least 30 observations 3. two 4. 6 5. number of categories 6. dependent 7. binomial 8. comparing two or more independent samples 9. never

10. normal populations, equal standard deviations

t � 0.5842230 � 2

21 � (0.5842)2 � 3.81

10.3412

Lin01803_appc_782-828.qxd 11/18/10 1:07 PM Page 827

PART 2 1. H0: The proportions are as stated. H1: The proportions

are not as stated. Using a significance level of 0.05, reject H0 if � 7.815.

Do not reject H0. Proportions could be as declared. 2. H0: No relationship between gender and book type.

H1: There is a relationship between gender and book type. Using a significance level of 0.01, reject H0 if

Reject H0. There is a relationship between gender and book type.

3. H0: The distributions are the same. H1: The distributions are not the same. H0 is rejected if H � 5.99.

�2 � (250 � 197.3)2

197.3 � . . . �

(200 � 187.5)2

187.5 � 54.84

�2 7 9.21.

� (30 � 20)2

20 �

(10 � 10)2

10 � 5.769

�2 � (120 � 130)2

130 �

(40 � 40)2

�2

828

H0 is not rejected. There is no difference in the three distributions.

H � 12

24(25) c 1072

8 �

562

7 �

1372

9 d � 3(25) � 4.29

8:00 A.M. Ranks 10:00 A.M. Ranks 1:30 P.M. Ranks

68 6 59 1.5 67 5 84 20 59 1.5 69 7 75 10.5 63 4 75 10.5 78 15.5 62 3 76 12.5 70 8 78 15.5 79 17 77 14 76 12.5 83 19 88 24 80 18 86 21.5 71 9 86 21.5

87 23

Sums 107 56 137 Count 8 7 9

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829

Photo Credits

Chapter 1 Page 1: Courtesy Barnes & Noble; Page 2: John A. Rizzo / Getty Images; Page 5: Image Source / PictureQuest; Page 9: Rachel Epstein / The Image Works; Page 11: Royalty Free / Corbis

Chapter 2 Page 21: Courtesy Merrill Lynch; Cover photo by Kara Phelps; Page 22: Justin Sullivan / Getty Images; Page 23: Photodisc / Getty Images

Chapter 3 Page 57: Andy Lyons / Getty Images; Page 58: Digital Vision / Getty Images; Page 60: Bloomberg via Getty Images; Page 77: Spencer Grant / Photoedit

Chapter 4 Page 102: Randy Faris / Corbis; Page 108: Somos / Veer / Getty Images; Page 112: Ryan McVay / Getty Images; Page 124: Steve Mason / Getty Images

Chapter 5 Page 144: Karin Slade / Getty Images; Page 145: Robert Galbraith / Reuters / Landov; Page 153: Teri Stratford; Page 156: Tony Arruga / Corbis; Page 168: Courtesy Intel Corporation

Chapter 6 Page 186: JGI / Jamie Grill / Getty Images; Page 192: ThinkStock / Jupiter Images; Page 195: Kent Gilbert / AP Photo; Page 205: Howard Berman / Getty Images

Chapter 7 Page 222: Ilene MacDonald / Alamy; Page 223: C. Sherburne / PhotoLink / Getty Images; Page 239: JupiterImages / Getty Images; Page 247: Zumawireworldphotostwo / Newscom

Chapter 15 Page 573: Steve Cole / Photodisc / Getty Images; Page 574: Digital Vision / Punchstock; Page 589: Image Ideas Inc. / Picture Quest

Chapter 16 Page 604: Bob Levey / Getty Images; Page 605: Flying Colours, Ltd / Photodisc / Getty Images; Page 612: © First Light / Alamy; Page 621: © Daniel Belenguer / Alamy; Page 633: Arthur Tilley / Getty Images

Chapter 17 Page 648: Najiah Feanny / Corbis; Page 650: © Ian Dagnall / Alamy; Page 667: Scott Olson / Getty Images; Page 660: Steve Mason / Getty Images

Chapter 18 Page 680: ITAR-TASS / Landov; Page 681: Photo courtesy of Nestle; Page 685: © Corbis; Page 690: Ryan McVay / Getty Images

Chapter 19 Page 720: Jerry Lampen / Reuters / Landov; Page 723: Courtesy of the National Institute of Standards and Technology, Office of Quality Programs, Gaithersburg, MD; Page 729: © Kevpix / Alamy; Page 742: Comstock / Getty Images

Chapter 20 (on the website: www.mhhe.com/lind15e) Page P20-1: Mark Horn / Getty Images; Page P20-2: Gary C. Knapp / AP Photo

Chapter 8 Page 265: JB Reed / Landov; Page 267: David Epperson / Getty Images

Chapter 9 Page 297: Jack Hollingsworth / Photodisc / Getty Images; Page 299: © Corbis All Rights Reserved; Page 301: Del Monte Corporation; Page 311: PhotoLink / Getty Images; Page 313: Rich Pedroncelli / AP Photo

Chapter 10 Page 333: Photo Source Hawaii / Alamy; Page 334: Russell Illig / Getty Images; Page 337: Jim Stern / Bloomberg via Getty Images; Page 342: Robert Nicholas / Getty Images; Page 344: Gene J. Puskar / AP Photo

Chapter 11 Page 371: Charles O’Rear / Corbis; Page 372: Joe Raedle / Getty Images; Page 375: NCR Corporation; Page 379: Mick Broughton / Alamy; Page 392: Photodisc / Getty Images

Chapter 12 Page 410: George Nikitin / AP Photo; Page 412: The McGraw-Hill Companies, Inc. / John Flournoy, photographer; Page 413: Daniel Acker / Bloomberg News / Getty Images; Page 430: John A. Rizzo / Getty Images

Chapter 13 Page 461: © Twentieth Century-Fox Film Corporation / Photofest NYC; Page 462: Friend Giving Samples by Sue R. Day; Page 476: Thinkstock / Superstock; Page 495: Matt Slocum / AP Photo

Chapter 14 Page 512: Keith Brofsky / Getty Images

Lin01803_pc_829-830.qxd 11/23/10 5:27 PM Page 829

Lin01803_pc_829-830.qxd 11/10/10 5:07 PM Page 830

831

INDEX

A. C. Nielsen Company, 287 ABC, 275, 648 Acceptance number, 743 Acceptance sampling, 742–745 Addition rules

general, 155–157 special, 153–155

Adjusted coefficient of determination, 522

AlliedSignal, 724 Alpha (�), 337 Alternate hypothesis, 336 American Automobile Association

(AAA), 159 American Coffee Producers

Association, 162 American Management Association,

302 American Restaurant Association,

290 Analysis of ranked data; see Ranked-

data analysis Analysis of variance (ANOVA); see

also F distributions ANOVA tables, 421

linear regression, 488–489 multiple regression, 519–520

assumptions, 416 compared to Kruskal-Wallis test,

701–702 differences in treatment means,

426–428 importance, 416–417 test procedure, 418–424 two-way, 430–433, 435–440 use of, 411

Anderson-Darling test of normality, 663–664

Applewood Auto Group data set, 761–762

Arithmetic mean, 58, 61, 88–89 Arm and Hammer Company,

286–287

Asbury Automotive Group, 22 Assignable variation, 725 Asymptotic distributions, 227; see

also Normal probability distributions

AtlantiCare, 723 Attribute control charts, 729,

737–741 Attribute sampling, 744 Attributes; see Qualitative variables Autocorrelation, 537, 631–635 AutoNation, 22 Average percent increase over

time, 73 Averages, 58

Backward elimination, 544 Baldrige National Quality Award,

723–724 Banking data set, 763 Bar charts, 24–25, 26–27 Base periods, 576, 577–578 Bayes, Thomas, 167 Bayes’ Theorem, 167–170 Bell Telephone Laboratories, 721 Bell-shaped distributions, 227; see

also Normal probability distributions

Best Buy Inc., 298 Best subset regression, 530, 544 Beta (�), 337, 359–362 Beta coefficient, of stock, 479–480,

617 Bethlehem Steel, 145 Bias, 268, 330 Bimodal distributions, 67, 120 Binomial probability distributions

characteristics, 195–196 compared to hypergeometric

distributions, 206 constructing, 196–197 cumulative, 202–203 definition, 195

Lin01803_ndx_831-850.qxd 11/10/10 5:11 PM Page 831

832 Index

Binomial probability distributions—Cont.

formula, 196 mean, 197–198 normal approximation to, 242–245,

686–687 Poisson distributions, 210–211 sampling with replacement, 204 shapes, 200–201 software examples, 199, 200 tables, 198, 774–778 variance, 197–198

Bivariate data, 124 Blocking variables, 431–432 BLS; see Bureau of Labor Statistics BMW, 22, 298 Boeing Inc., 3 Box plots, 116–118 British Airways, 737 Bureau of Labor Statistics (BLS), 575,

588, 589, 592 Burger King, 313 Busch Gardens, 155–156 Bush, George W., 162 Business cycles, 606–607

Cadillac, 723 Carey, Anne R., 10 Carli, G. R., 577 Categories; see Nominal level data Causation, correlation and, 469 Cause-and-effect diagrams,

727–728 c-bar charts, 740–741 CBS, 266, 314, 648 Cedar Fair, 612–614 Cells, 651 Central limit theorem, 279–280,

284–285 Central location, measures of; see

Measures of location Chance variation, 725 Charts, 6; see also Graphical displays

bar, 24–25, 26–27 control; see Control charts pie, 25–27

Chebyshev, P. L., 85 Chebyshev’s theorem, 85–86 Chevron, 4 Chi-square distribution, 651, 652–653

Chi-square test contingency table analysis, 667–670 goodness-of-fit test

equal expected frequencies, 649–652

normal distributions, 659–662 unequal expected frequencies,

655–656 limitations, 657–658

Chi-square test statistic, 650, 651 computing, 651–652 critical values, 651, 767

Chrysler, 22 Churchill Downs, 57 Circuit City, 335 Class frequencies, 24–25 Class intervals, 31, 32, 33–34 Class midpoints, 33–34, 88 Class widths, 31 Classical probability, 148–149 Cluster sampling, 271 Coefficient of correlation, 465–470

characteristics, 466 computing, 468 definition, 466 derivation, 466–468 formula, 468 interpretation, 468–469 relationship to coefficient of

determination and standard error of estimate, 488–490

strength of relationship, 465 testing significance of, 472–475

Coefficient of determination, 487–488 adjusted, 522 from ANOVA table, 489 formula, 489 multiple, 521–522 relationship to correlation

coefficient and standard error of estimate, 488–490

Coefficient of multiple determination, 521–522

Coefficient of skewness, 120–122 Colgate-Palmolive Co., 5 Collectively exhaustive events, 149 Combination formula, 174–175 Combinations, 174 Complement rule, 154 Computer applications; see Software Conditional probability, 160

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Index 833

Confidence intervals computer simulation, 304–305 computing, 302–303, 310–311 definition, 298 for difference in treatment

means, 427 in linear regression, 492–494 90 percent, 302 95 percent, 300–302 99 percent, 300, 302 for population mean, 299, 310

with known standard deviation, 300–303

with unknown standard deviation, 306–308

for proportion, 313–316 Confidence levels, 317–319 Consumer Price Index (CPI), 588

base periods, 592 compilation of, 592 components, 577, 592 functions, 592 history, 592 publication of, 574 special uses of, 592–595 specific indexes, 592

Consumer Satisfaction Index, 587 Consumer’s risk, 743 Contingency table analysis, 667–670 Contingency tables, 126–127,

162–164, 668 Continuity correction factor, 242–245 Continuous probability distributions

area within, 224 examples, 190–191 exponential, 246–250 F; see F distributions normal; see Normal probability

distributions t; see t distribution uniform, 223–226

Continuous random variables, 190 Continuous variables, 9 Control charts

attribute, 729, 737–741 c-bar charts, 740–741 factors, 773 in-control and out-of-control

processes, 734–736 percent defective ( p) chart,

737–740

range charts, 733–734 use of, 235, 722–723, 729, 733 variable, 729–733

Control limits for defects per unit, 740 lower, 730, 731 for proportions, 737 upper, 730, 731

Cooper Tire and Rubber Company, 7 Correction factors

for adjusting quarterly means, 624–625

continuity, 242–245 finite-population (FPC), 320–321

Correlation analysis, 463–464 Correlation coefficient; see

Coefficient of correlation Correlation matrix, 546–547 Costco, 2 Cost-of-living adjustments, 595 Cost-of-living index; see Consumer

Price Index Counting principles

combination formula, 174–175 multiplication formula, 171–172 permutation formula, 172–174

CPI; see Consumer Price Index Critical number, 743 Critical values, 339

chi-square test statistic, 651, 767 Durbin-Watson d statistic, 779–781 f distribution, 768–769 Wilcoxon signed-rank test,

692–693, 772 Cumulative binomial probability

distributions, 202–203 Cumulative frequency distributions,

42–44 Cumulative frequency polygons,

42–44 Curvilinear relationships, 495–497 Cyclical variations, 606–607

Data; see also Variables bivariate, 124 collection, 5–6 interval level, 11–12, 61 measurement levels, 9–13 nominal level, 10, 24–27, 313 ordinal level, 11

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834 Index

Data; see also Variables—Cont. ratio level, 12–13, 61 raw, 30 transforming, 495–497 univariate, 124

Data sets Applewood Auto Group, 761–762 banking, 763 Major League Baseball, 757–758 real estate, 754–756 school district bus data, 759–760

Deciles, 111 Decision rules, 338–339 Defects; see Quality control Deflators, 594 Degrees of freedom (df), 310, 348,

388 Delta Airlines, 208–209 Deming, W. Edwards, 721–722 Deming’s 14 points, 721–722 Dependent events, 160–161 Dependent samples

independent samples vs., 375–377 two-sample tests of, 392–395, 690 Wilcoxon signed-rank test,

690–693 Dependent variables, 464 Descriptive statistics, 6 Deseasonalized data, 627–630 Deviation, mean, 76–78; see also

Standard deviation Discrete probability distributions

binomial; see Binomial probability distributions

definition, 190 hypergeometric, 204–206 mean, 191–193 Poisson, 207–209 standard deviation, 191–193 variance, 191

Discrete random variables, 190 Discrete variables, 9 Disney World, 155–156 Dispersion, 58

measures of; see Measures of dispersion

reasons for studying, 74 Distribution-free tests, 681 Distributions; see Frequency

distributions; Probability distributions

DJIA; see Dow Jones Industrial Average

Dole Pineapple, 333 Dollar, purchasing power of, 594–595 Dot plots, 103–104, 108 Dow Jones Industrial Average (DJIA),

574, 589, 596–597 Dummy variables, 537–539 Durbin-Watson statistic, 631–635,

779–781

Empirical probability, 149–150 Empirical Rule, 86–87, 231–232 Enron, 14 Environmental Protection Agency

(EPA), 4, 298 Errors; see Sampling error; Standard

error; Type I error; Type II error Ethics

issues in statistics, 14 reporting results, 92

Events collectively exhaustive, 149 definition, 147 dependent, 160–161 independent, 159 joint, 156 mutually exclusive, 148, 153, 154

Excel, 14–15 ANOVA tables, 423–424 area under normal curve, 234–235 binomial probability distribution,

198, 200, 203 coefficient of skewness, 120 combinations, 175 confidence intervals, 311–312 correlation coefficient, 468 Durbin-Watson statistic, 634 global test, 525 histograms, 38 hypergeometric distribution,

206–207 index numbers, 576 Laspeyres price index, 582 mean, median, and mode, 69 moving averages, 610 multiple regression, 514–517 normal probability plots, 534 Paasche’s price index, 583 paired t test, 396–397

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Index 835

pie charts, 26 quartiles, 113–114 random sampling, 268–269 regression analysis, 480, 484, 487 scatter diagrams, 125 standard deviation, 84 test of variances, 415 time series, 619, 630 two-sample t test, 386, 395 two-way ANOVA, 433–434, 439

Expected frequencies, 669 Expected values, 191 Experiments

definition, 146 random variables, 189 two-factor, 433

Exponential distributions, 246–250 ExxonMobil, 4

F distributions characteristics, 411–412 comparing two variances, 412–415 comparison of population means,

416–418 critical values, 768–769 global test, 524–526 test statistics, 412–413, 421 use of, 412

Facebook, 65 Factors, 436; see also Treatments Federal Express, 723 Federal Reserve, 3, 574 Federalist, The, 31 Finite populations, 204, 320 Finite-population correction (FPC)

factor, 320–321 Fishbone diagrams, 727–728 Fisher, Irving, 584 Fisher, Ronald A., 266, 411 Fisher’s ideal index, 584 Forbes, 4 Ford Motor Company, 22, 667, 723 Forecasting; see also Time series

with deseasonalized data, 628–630 errors in, 629 long-term, 605

Forward selection method, 544 FPC; see Finite-population correction

factor Frequency distributions, 6

class frequencies, 24–25 class intervals, 31, 32, 33–34 class widths, 31 classes, 31–32 constructing, 29–33 cumulative, 42–44 definition, 29 graphical presentations, 36

cumulative frequency polygons, 42–44

frequency polygons, 38–40 histograms, 36–38, 39

relative, 34–35 skewed, 70–71, 119–121 software example, 34 symmetric, 69–70, 86

Frequency polygons, 38–40 cumulative, 42–44

Frequency tables, 23–24 Frito-Lay, 4–5

Gates, William, 4 General Electric, 724 General Foods Corporation, 340 General Motors, 4, 22, 356, 378,

723, 743 General rule of addition, 155–157 General rule of multiplication,

160–161 Geometric mean, 72–73 Global test, 524–526 Goodness-of-fit test; see also Chi-

square test equal expected frequencies,

649–652 normal distributions, 659–662 unequal expected frequencies,

655–656 Gosset, William, 307 Gould, Stephen Jay, 120 Grand mean, 730 Graphical displays; see also Charts

box plots, 116–118 cumulative frequency polygons,

42–44 dot plots, 103–104, 108 of frequency distributions, 36 frequency polygons, 38–40 histograms, 36–38, 39 normal probability plots, 534

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836 Index

Graphical displays; see also Charts—Cont.

quality control charts fishbone diagrams, 727–728 Pareto charts, 725–727

residual plots, 532–533 scatter diagrams, 124–125,

463, 532 of statistical information, 4–5 stem-and-leaf displays, 105–108 tree diagrams, 164–165 Venn diagrams, 154

Graunt, John, 10 Guinness Brewery, 307 Gwynn, Tony, 88

Hamilton, Alexander, 31 Hammond Iron Works, 74 Harris International, 266 Heartland Health, 723 Hendrick Auto Group, 22 Histograms, 36–38, 39 Home Depot, 605–606 Homeland Security, Department of,

11 Homoscedasticity, 534 Honeywell Federal Manufacturing &

Technologies, 723 Hunt, V. Daniel, 723 Hypergeometric probability

distributions, 204–206 Hypotheses

alternate, 336 definition, 334–335 null, 336

Hypothesis testing; see also Analysis of variance

definition, 335 five-step procedure, 335–340 goodness-of-fit test, 649–652 for median, 688–689 nonparametric tests; see Chi-square

test; Ranked-data analysis one-sample; see One-sample

hypothesis tests p-values, 345–346, 354–355 two-sample; see Two-sample

hypothesis tests Type I and Type II error, 337–338

Hyundai, 22

IBM, 723 Inclusive events, 157 Income, real, 593 Independent events, 159 Independent observations, 537 Independent samples, 372–373; see

also Two-sample hypothesis tests

dependent samples vs., 375–377 Kruskal-Wallis test, 698–702 Wilcoxon rank-sum test,

695–697 Independent variables, 464; see also

Interaction multicollinearity, 534–536 qualitative, 537–539 selecting, 525–530, 544

Index numbers constructing, 577–578 definition, 574 development of, 577 examples, 575–576

Indexes; see also Consumer Price Index

base periods, 576, 577–578 as deflators, 594 importance, 574 purpose, 577 seasonal, 621–626 shifting bases of, 595–597 simple, 577 special-purpose, 587–588 unweighted, 579–581 value, 585–586 weighted, 581–584

Inductive statistics; see Inferential statistics

Inferential statistics, 6–7, 145 Interaction

hypothesis tests for, 437–440 in multiple regression, 540–542 two-way ANOVA with, 435–440

Interaction plots, 436–437 Intercept

in multiple regression, 513–514 of regression line, 478

Internal Revenue Service, 31, 740

Interval level data, 11–12, 61 Irregular variation, 608

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Index 837

J. D. Power & Associates, 587 Jay, John, 31 Johnson & Johnson, 59 Joint events, 156 Joint probability, 156

Kellogg Company, 2 Kennedy, John F., 105 Kia, 22 Kruskal, W. H., 698 Kruskal-Wallis one-way analysis of

variance by ranks, 698–702 Kutner, Michael H., 532

Labor, Department of, 574, 592 Landon, Alfred, 270, 372 Laplace, Pierre-Simon, 167 Laspeyres, Etienne, 581 Laspeyres price index, 581–582, 583 Law of large numbers, 149 LCL; see Lower control limit Least squares method, in

forecasting, 616–617 Least squares principle, 476–477 Leaves, 105 Level of significance, 337 Li, William, 532 Linear regression

assumptions, 490–492 confidence intervals, 492–494 drawing line, 479–480 least squares principle, 476–477 prediction intervals, 492, 493–494 standard error of estimate,

486–487, 488–490 testing significance of slope,

483–485 transforming data, 495–497

Linear trend equation, 615–616 Literary Digest poll, 372–373 Lockheed, 462 Log trend equation, 619–620 Long-term forecasting; see

Forecasting Lorrange Plastics, 7 Lotteries, 656 Lower control limit (LCL), 730, 731

Madison, James, 31 Madoff, Bernie, 14 Major League Baseball data set,

757–758 Malcolm Baldrige National Quality

Award, 723–724 Margin of error, 314, 317 Martin Marietta, 462 Mauer, Joe, 87–88 McDonald’s, 724 Mean

arithmetic, 58, 61, 88–89 of binomial probability distribution,

197–198 difference between two, 374 difference from median,

120–121 of discrete probability distribution,

191–193 Empirical Rule, 86–87, 231–232 geometric, 72–73 of grouped data, 88–89 median, mode, and, 66–67, 70 of normal distribution, 227–228 of Poisson distribution, 208 population; see Population mean sample; see Sample mean standard error of, 285, 730 treatment, 423, 426–428 of uniform distribution, 224 weighted, 63

Mean deviation, 76–78 Mean proportion defective, 737 Mean square, 423 Mean square error (MSE), 423,

426–427 Mean square for treatments

(MST), 423 Measurement levels,

9–10, 13 interval, 11–12, 61 nominal, 10 ordinal, 11 ratio, 12–13, 61

Measures of dispersion, 58 mean deviation, 76–78 range, 75–76 standard deviation; see Standard

deviation variance; see Variance

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838 Index

Measures of location, 58 average, 58 mean; see Mean median; see Median mode, 65–66, 67 software example, 69

Measures of position, 111 deciles, 111 percentiles, 111, 113–114 quartiles, 111–112

Median, 64–65, 111 difference from mean, 120–121 hypothesis tests for, 688–689 mean, mode, and, 67, 70

MegaStat, 15 binomial probability

distributions, 199 chi-square test statistic, 652 frequency distributions, 34 quartiles, 113 seasonal indexes, 622–623,

625–626 Wilcoxon rank-sum test, 697

Merrill Lynch, 5, 21, 61 Microsoft Corporation, 4, 14–15, 605 MidwayUSA, 723 Minitab, 14, 15

ANOVA tables, 424 box plots, 117 c-bar charts, 741 chi-square test, 670 coefficient of skewness, 120, 122 confidence intervals, 310–311,

493–494 control charts, 734 correlation analysis, 473–474 correlation matrix, 536 dot plots, 104 Kruskal-Wallis test, 701 multiple regression, 514–517, 528 normal probability plots, 534 one-sample hypothesis tests,

353–354 Pareto charts, 726–727 percent defective charts, 739 Poisson probability distribution,

209 prediction intervals, 493–494 quartiles, 113, 114 regression coefficients, testing,

528–529

residuals histogram, 534 standard deviation, 84 stem-and-leaf displays, 107 stepwise regression, 543 time series data, 616–617 two-sample t test, 390 two-sample test of proportions,

381 Minnesota Twins, 87–88 Mode, 65–66, 67, 70 Morton Thiokol, 462–463 Motorola Inc., 724 Moving-average method

in time series, 608–611 weighted, 611–614

MSE; see Mean square error MST; see Mean square for treatments Multicollinearity, 534–536 Multiple regression

ANOVA tables, 519–520 assumptions, 531–532 autocorrelation, 537 coefficient of multiple

determination, 521–522 evaluating regression equation

with correlation matrix, 546–547 global test, 524–526 individual regression coefficients,

525–530 with scatter diagrams, 532 selecting variables, 525–530, 544

example, 546–551 general equation, 513 homoscedasticity, 534 inferences about population

parameters, 523–530 interaction, 540–542 intercept, 513–514 linear relationships, 531–533 multicollinearity, 534–536 multiple standard error of estimate,

520–521 qualitative independent variables,

537–539 regression coefficients, 513–514,

525–530 residuals analysis, 533–534 stepwise, 530, 542–544

Multiple standard error of estimate, 520–521

Multiplication formula, 171–172

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Index 839

Multiplication rules general, 160–161 special, 159–160

Mutually exclusive events, 148, 153, 154

Nachtscheim, Chris J., 532 NASA, 150 NASDAQ, 22, 574, 596–597 National Collegiate Athletic

Association (NCAA), 167 National Science Foundation, 605 NBC, 648 NCAA; see National Collegiate

Athletic Association Negatively skewed distributions,

70–71, 119 Neter, John, 532 New York Stock Exchange, 22, 589 New York Stock Exchange Index, 589 Nightingale, Florence, 38 NIKKEI 225, 574 90 percent confidence intervals, 302 95 percent confidence intervals,

300–302 99 percent confidence intervals, 300,

302 Nissan, 22, 723 Nixon, Richard, 105 Nominal level data, 10; see also

Chi-square graphical displays, 24–27 proportions, 313

Nonlinear relationships, 495–497 Nonlinear trends, 618–620 Nonparametric methods; see Chi-

square test; Ranked-data analysis

Normal approximation to binomial distribution, 242–245, 686–687

Normal probability distributions area between values, 234 area under curve, 229, 230,

233–236, 764 characteristics, 227–228 combining two areas, 237–238 confirmation tests, 662–664 formula, 227 goodness-of-fit test, 659–662 mean, 227–228

percentages of observations, 249 of residuals, 534 standard; see Standard normal

distribution standard deviation, 227–228

Normal probability plots, 534 Normal Rule, 86–87 Null hypothesis, 336 Numeric data; see Quantitative

variables

Objective probability, 148 OC curve; see Operating

characteristic curve Ohio State Lottery, 25–26 O’Neal, Shaquille, 244 One-sample hypothesis tests

for population mean with known standard deviation,

341–345 software solution, 353–354 with unknown standard

deviation, 348–352 One-tailed tests of significance,

340–341, 345 Operating characteristic (OC) curve,

743 Ordinal level data, 11; see also

Ranked-data analysis Outcomes

counting, 171 definition, 146–147

Outliers, 118

Paasche’s price index, 582–583 Paired samples, 392–395 Paired t test, 293, 690 Palmer, Chad, 10 Parameters, population, 59, 274 Pareto, Vilfredo, 725 Pareto charts, 725–727 Pearson, Karl, 120, 465, 651 Pearson product-moment correlation

coefficient; see Coefficient of correlation

Pearson’s coefficient of skewness, 120 Pearson’s r; see Coefficient of

correlation Penske Auto Group, 22

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840 Index

Percent defective (p) chart, 737–740 Percentiles, 111, 113–114 Permutation formula, 172–174 Permutations, 173 Pie charts, 25–27 Point estimates, 298–299 Poisson probability distributions, 207–209

binomial probability estimation, 210–211 characteristics, 208–209 definition, 207 exponential distribution and, 247 formula, 208 mean, 208 tables, 209, 770 variance, 208

Pooled proportion, 379 Pooled variance, 383 Population mean, 58–59

confidence intervals for, 299, 310 with known standard deviation, 300–303 with unknown standard deviation, 306–308

hypothesis tests for comparing three or more, 416–418 with equal population standard deviations,

383–386 with known standard deviation, 341–345 one-tailed test, 345 with unequal population standard

deviations, 388–390 with unknown standard deviation, 348–352,

383–390 point estimates, 298–299 sample size for estimating, 317–318 two-tailed test for, 341–344

Population proportion, 314 hypothesis tests for, 356–358 sample size for estimating, 318–319

Population standard deviation, 82, 300–303, 306–308

Population variance, 80–81, 412–415 Populations

definition, 7 finite, 204, 320 inferences in multiple regression, 523–530 parameters, 59, 274 relationship to samples, 8 strata, 270–271

Positively skewed distributions, 70, 119–120 Posterior probability, 167–168 PPI; see Producer Price Index Prediction intervals, 492, 493–494

Prior probability, 167 Probability

approaches, 151 Bayes’ Theorem, 167–170 classical, 148–149 conditional, 160 counting principles

combination formula, 174–175 multiplication formula, 171–172 permutation formula, 172–174

definition, 146 empirical, 149–150 events, 147 experiments, 146 joint, 156 objective, 148 outcomes, 146–147 posterior, 167–168 prior, 167 special rule of multiplication, 159–160 subjective, 150–151

Probability distributions binomial; see Binomial probability

distributions characteristics, 187 continuous; see Continuous probability

distributions definition, 187 discrete; see Discrete probability distributions F distributions; see F distributions generating, 187–188 hypergeometric, 204–206 normal; see Normal probability distributions Poisson, 207–209 uniform, 223–226

Probability rules complement rule, 154 general rule of addition, 155–157 general rule of multiplication, 160–161 special rule of addition, 153–155

Probability theory, 145 Processes; see Quality control Producer Price Index (PPI), 574, 589, 594 Producer’s risk, 743 Proportions

confidence intervals for, 313–316 control limits, 737 definition, 314 hypothesis tests for

one-sample, 356–358 two-sample, 378–381

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Index 841

pooled, 379 population, 314, 318–319 sample, 314

Pseudo-random numbers, 266 Purchasing power of dollar, 594–595 P-values, 345–346, 354–355, 473

Qualitative variables; see also Nominal level data

definition, 8 in multiple regression, 537–539

Quality control acceptance sampling, 742–745 attribute sampling, 744 Baldrige National Quality Award,

723–724 causes of variation, 724–725 control charts; see Control charts diagnostic charts, 725–728 fishbone diagrams, 727–728 history, 721–722 Pareto charts, 725–727 six sigma, 724 statistical (SQC), 721, 722–723 statistical process control, 721

Quantitative variables continuous, 9 definition, 9 discrete, 9

Quartiles, 111–112, 116

RAND Corporation, 266 Random numbers

finding, 266 in lotteries, 656 pseudo-, 266 tables, 268, 771

Random samples; see Sampling Random variables

continuous, 190 definition, 189 discrete, 190

Random variation, 419 Range, 75–76 Range charts, 733–734 Ranked-data analysis

Kruskal-Wallis test, 698–702 rank-order correlation, 704–707

sign test; see Sign tests Spearman’s coefficient of rank

correlation, 704–706 Wilcoxon rank-sum test,

695–697 Wilcoxon signed-rank test,

690–693 Rank-order correlation, 704–707 Ratio level data, 12–13, 61 Ratio-to-moving-average method,

622–626 Raw data, 30 Real estate data set, 754–756 Real income, 593 Regression analysis, 462, 476; see

also Linear regression; Multiple regression

Regression coefficients, 513–514, 525–530

Regression equation, 476, 478 Regression line, 479–480 Relative class frequencies, 24–25 Relative frequencies, 149–150 Relative frequency distributions,

34–35 Residual plots, 532–533 Residuals

calculating, 480–481 correlated, 631–635 variation in, 533–534

Ritz-Carlton Hotel Corporation, 723 Rockwell International, 462 Roosevelt, Franklin D., 270, 372 Royal Viking, 222 Rules of probability; see Probability

rules

Sample mean, 60 sampling distribution of,

275–277 central limit theorem,

279–280, 284 standard deviation, 285 use of, 286–288

z values, 287 Sample proportion, 314

standard error of, 737 Sample standard deviation, 83–84 Sample statistics, 60, 274 Sample variance, 83

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842 Index

Samples definition, 7 dependent, 392–397, 690 independent; see Independent

samples paired, 392–395 relationship to population, 8 sizes, 316–317 use of, 7–8

Sampling acceptance, 742–745 attribute, 744 cluster, 271 reasons for, 7–8, 266–267 with replacement, 204 without replacement, 204 simple random, 267–268 stratified random, 270–271 systematic random, 270

Sampling distribution of sample mean, 275–277

central limit theorem, 279–280, 284 standard deviation, 285 use of, 286–288

Sampling error, 274–275 Scatter diagrams, 124–125, 463, 532 School district bus data set, 759–760 Seasonal indexes, 621–626 Seasonal variation, 607–608, 621 Seasonally adjusted data, 627–630 Secular trends, 605–606 Serial correlation; see Autocorrelation Shewhart, Walter A., 721 Sign tests, 681–685

hypothesis tests for median, 688–689

using normal approximation to binomial, 686–687

Significance, statistical, 346 Significance level, 337 Simple aggregate index, 580–581 Simple average of price indexes,

579–580 Simple indexes, 577 Simple random samples, 267–268 Six sigma, 724 Skewed distributions, 70–71, 119–120 Skewness

coefficient of, 120–122 Pearson’s coefficient of, 120 software coefficient of, 120–121

Slope, of regression line, 478, 483–485 Smith Barney, 111–112, 113–114 Smucker’s, 60 Software, statistical, 14–16; see also

Excel; MegaStat; Minitab Software coefficient of skewness,

120–121 Southwest Airlines, 740 SPC; see Statistical process control Spearman, Charles, 704 Spearman’s coefficient of rank

correlation, 704–706 Special rule of addition, 153–155 Special rule of multiplication, 159–160 Spread; see Dispersion Spurious correlations, 469 SQC; see Statistical quality control SSB; see Sum of squares due to

blocks SSE; see Sum of squares error SSI; see Sum of squares interaction SST; see Sum of squares treatment Standard & Poor’s 500 Index,

479–480, 574, 590, 617 Standard deviation

Chebyshev’s theorem, 85–86 definition, 80 of discrete probability distribution,

191–193 Empirical Rule, 86–87, 231–232 of grouped data, 88, 89–90 of normal distribution, 227–228 population, 82, 300–303, 306–308 sample, 83–84 software example, 84 of uniform distribution, 224 use of, 85

Standard error finite-population correction factor,

320–321 of mean, 285, 730 of sample proportion, 737

Standard error of estimate definition, 486 formula, 487, 490 multiple, 520–521 relationship to coefficients of

correlation and determination, 488–490

Standard normal distribution, 229–231 applications of, 231, 233–236

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Index 843

computing probabilities, 230 probabilities table, 230, 764

Standard normal values, 229–230

Standardizing, 120–121 Starbucks, 77–78 State Farm Insurance, 7 Statistic

definition, 60 test, 338

Statistical inference; see Inferential statistics

Statistical process control (SPC), 721 Statistical quality control (SQC), 721,

722–723 Statistical significance, 346 Statistics

computer applications, 14–16 definition, 4, 5 descriptive, 6 history, 10, 307 inferential, 6–7, 145 misleading, 14 reasons for studying, 2–4

Stem-and-leaf displays, 105–108 Stems, 105 Stepwise regression, 530, 542–544 Stock indexes; see Dow Jones

Industrial Average; NASDAQ; Standard & Poor’s 500 Index

Strata, 270–271 Stratified random samples, 270–271 Student’s t distribution, 307, 527,

765–766 Subjective probability, 150–151 Sum of squared residuals, 480–481 Sum of squares due to blocks

(SSB), 432 Sum of squares error (SSE), 421, 489

with interaction, 437, 438 two-way, 432–433

Sum of squares interaction (SSI), 438

Sum of squares total (SS total), 421 Sum of squares treatment (SST),

421, 423 Sutter Home Winery, 267 Symmetric distributions, 69–70, 86,

119; see also Normal probability distributions

Systematic random samples, 270

t distribution characteristics, 307 confidence interval for population

mean, 307–308 development of, 307 Student’s, 307, 527, 765–766 use of, 308–309

t tests for coefficient of correlation, 472–475 paired, 293

Taster’s Choice, 681–682 Test statistic, 338 Time series

cyclical variations, 606–607 definition, 605 deseasonalized data, 627–630 Durbin-Watson statistic, 631–635 irregular variations, 608 least squares method, 616–617 linear trend equation, 615–616 moving-average method, 608–611 nonlinear trends, 618–620 seasonal indexes, 621–626 seasonal variation, 607–608, 621 secular trends, 605–606 weighted moving average, 611–614

Tippett, L., 266 Total variation, 418

in Y, 488–489 Transformations, 495–497 Treatment means, 423, 426–428 Treatment variation, 418–419 Treatments, 417 Tree diagrams, 164–165 Tukey, John W., 105 Two-factor experiments, 433 Two-sample hypothesis tests

dependent samples, 392–395 independent samples, 372–376

dependent samples vs., 395–397 standard deviations equal,

383–386 standard deviations known,

383–390 standard deviations unequal,

388–390 paired t test, 293 for proportion, 378–381

Two-tailed tests of significance, 341–344

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844 Index

Two-way analysis of variance, 430–433 with interaction, 435–440

Tyco, 14 Type I error, 337 Type II error, 337–338, 359–362

UCL; see Upper control limit Unexplained variation, 489 Ungrouped data, 30 Uniform probability distributions,

223–226 United States Postal Service, 74 Univariate data, 124 University of Michigan, 683 University of Michigan Institute for

Social Research, 515 University of Wisconsin-Stout, 723 Unweighted indexes, 579–581 Upper control limit (UCL), 730, 731 USA Today, 3, 10

Value indexes, 585–586 Variable control charts, 729–733 Variables

blocking, 431–432 dependent, 464 dummy, 537–539 independent, 464

multicollinearity, 534–536 qualitative, 537–539 selecting, 525–530

measurement levels, 9–13 qualitative, 8, 537–539 quantitative, 9 random, 189–190 relationship between two, 124–127 types, 8–9

Variance; see also Analysis of variance (ANOVA)

of binomial probability distribution, 197–198

definition, 79–80 of discrete probability distribution,

191 of distribution of differences,

373–374 Kruskal-Wallis test, 698–702 of Poisson distribution, 208 pooled, 383 population, 80–81

sample, 83 Variance inflation factor (VIF), 535–536 Variation; see also Dispersion

assignable, 725 causes, 724–725 chance, 725 explained, 488–489 irregular, 608 random, 419 seasonal; see Seasonal variation total, 418, 488–489 treatment, 418–419 unexplained, 489

Venn, J., 154 Venn diagrams, 154 Veterans Affairs Cooperative Studies

Program, 724 VIF; see Variance inflation factor Volvo, 22

Wallis, W. A., 698 Walmart, 4 Weighted indexes

Fisher’s ideal index, 584 Laspeyres price index, 581–582, 583 Paasche’s price index, 582–583

Weighted mean, 63 Weighted moving average, 611–614 Wells, H. G., 2 Wendy’s, 63 Wilcoxon, Frank, 690 Wilcoxon rank-sum test, 695–697 Wilcoxon signed-rank test, 690–693

critical values, 692–693, 772 Williams, Ted, 88 World War II, 208, 339, 721

Xerox, 723

Yates, F., 266 Y-intercept, 478

z distribution as test statistic, 338 use of, 308–309

z values (scores), 229, 230, 287 Zogby International, 266

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CHAPTER 3 • Population mean

� � [3–1]

• Sample mean, raw data

[3–2]

• Weighted mean

[3–3]

• Geometric mean

[3–4]

• Geometric mean rate of increase

[3–5]

• Range

Range � Largest value � Smallest value [3–6]

• Mean deviation

[3–7]

• Population variance

[3–8]

• Population standard deviation

[3–9]

• Sample variance

[3–10]

• Sample standard deviation

[3–11]

• Sample mean, grouped data

[3–12]

• Sample standard deviation, grouped data

[3–13]

CHAPTER 4 • Location of a percentile

[4–1]

• Pearson’s coefficient of skewness

[4–2]

• Software coefficient of skewness

[4–3]

CHAPTER 5 • Special rule of addition

P( A or B) � P( A) � P(B) [5–2]

• Complement rule

P( A) � 1 � P(~A) [5–3]

• General rule of addition

P( A or B) � P( A) � P(B) � P( A and B) [5–4]

• Special rule of multiplication

P( A and B) � P( A)P(B) [5–5]

• General rule of multiplication

P( A and B) � P( A)P(B� A) [5–6]

• Bayes’ Theorem

[5–7]

• Multiplication formula

Total arrangements � ( m)(n) [5–8]

• Number of permutations

[5–9]

• Number of combinations

[5–10]

CHAPTER 6 • Mean of a probability distribution

� � �[xP(x)] [6–1]

• Variance of a probability distribution

�2 � �[(x � �)2P(x)] [6–2]

• Binomial probability distribution

P(x) � nCx � x(1 � �)n � x [6–3]

• Mean of a binomial distribution

� � n� [6–4]

• Variance of a binomial distribution

�2 � n�(1 � �) [6–5]

• Hypergeometric probability distribution

[6–6]

• Poisson probability distribution

[6–7]

• Mean of a Poisson distribution

� � n� [6–8]

P(x) � �xe��

P(x) � (SCx)(N�SCn�x)

NCn

nCr � n!

r!(n � r)!

n Pr � n!

(n � r)!

P( A1�B) � P( A1)P(B�A1)

P( A1)P(B�A1) � P( A2)P(B�A2)

sk � n

(n � 1)( n � 2) ca a

X � X s

b 3

sk � 3( X � Median)

Lp � (n � 1) P

100

s � B g f(M � X )2

n � 1

X � g fM

s � B g (X � X )2

n � 1

s2 � �(X � X

_ )2

n � 1

� � ��(X � �)2N

� 2 � �(X � �)2

MD � ��X � X

_ �

GM � n B

Value at end of period Value at start of period

� 1.0

GM � 1n (X1)(X2)(X3) # # # (Xn)

X _

w � w1X1 � w2X2 � � � � � wn Xn

w1 � w2 � � � � � wn

X _

� �X n

�X N

KEY FORMULAS Lind, Marchal, and Wathen • Statistical Techniques in Business and Economics, 15th edition CHAPTER 7 • Mean of a uniform distribution

[7–1]

• Standard deviation of a uniform distribution

[7–2]

• Uniform probability distribution

[7–3]

if a x b and 0 elsewhere

• Normal probability distribution

[7–4]

• Standard normal value

[7–5]

• Exponential distribution

[7–6]

• Finding a probability using the exponential distribution

[7–7]

CHAPTER 8 • Standard error of mean

[8–1]

• z-value, � and � known

[8–2]

CHAPTER 9 • Confidence interval for �, with � known

[9–1]

• Confidence interval for �, � unknown

[9–2]

• Sample proportion

[9–3]

• Confidence interval for proportion

p z [9–4]

• Sample size for estimating mean

[9–5]

• Sample size for a proportion

[9–6]n � �(1 � �)a z E b

n � a z� E b

�p(1 � p) n

p � X n

X _

t s

�n

X z �

z � X � � �/1n

� X � �

�

�n

P ( Arrival time 6 x ) � 1 � e�l x

P( x) � le�lx

z � X � �

�

P(x) � 1

��2� e� B(X��)2

2�2 R

P(x) � 1

b � a

� � � (b � a)2 12

� � a � b

CHAPTER 10 • Testing a mean, � known

[10–1]

• Testing a mean, � unknown

[10–2]

• Test of hypothesis, one proportion

[10–3]

• Type II error

[10–4]

CHAPTER 11 • Variance of the distribution of difference in means

[11–1]

• Two-sample test of means, known �

[11–2]

• Pooled proportion

[11–4]

• Two-sample test of proportions

[11–3]

• Pooled variance

[11–5]

• Two-sample test of means, unknown but equal �

[11–6]

• Two-sample tests of means, unknown and unequal �s

[11–7]

• Degrees of freedom for unequal variance test

[11–8]

• Paired t test

[11–9]

CHAPTER 12 • Test for comparing two variances

[12–1]F � s21 s22

t � d _

sd ��n

df � [(s21/n1) � (s22/n2)]2

(s21/n1)2

n1 � 1 �

(s22/n 2)2

n2 � 1

t � X1 � X2

s21 n1

� s21 n2

t � X _

1 � X _

�s2p� 1n1 � 1n2

s2p � (n1 � 1) s21 � (n2 � 1) s22

n1 � n2 � 2

z � p1 � p2

�pc(1 � pc) n1

� pc(1 � pc)

pc � X1 � X2 n1 � n2

z � X1 � X2

�21 n1

� �22 n 2

�x 2

1�x2 � �21 n1

� �2

z � X _

c � �1

�/�n

z � p � �

��(1 � �)n

t � X � � s/1n

z � X _

� �

�/�n

Lin01803_keyform.qxd 10/28/10 1:36 PM Page 2

• Real income

Real income � (100) [15–8]

• Using an index as a deflator

[15–9]

• Purchasing power

[15–10]

CHAPTER 16 • Linear trend

� a � bt [16–1]

• Log trend equation

log � log a � log b(t ) [16–2]

• Correction factor for adjusting quarterly means

[16–3]

• Durbin-Watson statistic

[16–4]

CHAPTER 17 • Chi-square test statistic

�2 � [17–1]

• Expected frequency

fe � [17–2]

CHAPTER 18 • Sign test, n � 10

[18–1]

• Wilcoxon rank-sum test

[18–4]

• Kruskal-Wallis test

H �

� 3(n � 1) [18–5]

• Spearman coefficient of rank correlation

rs � 1 � [18–6]

• Hypothesis test, rank correlation

t � rs [18–7]

CHAPTER 19 • Grand mean

X �

� [19–1]

• Control limits, mean

UCL � X �

� A2R �

LCL � X �

� A2R �

[19–4]

• Control limits, range

UCL � D4R �

LCL � D3R �

[19–5]

• Mean proportion defective

p � [19–6]

• Control limits, proportion

UCL and LCL � p 3 [19–8]

• Control limits, c-bar chart

UCL and LCL � 3 [19–9]

CHAPTER 20 (ON THE WEBSITE: www.mhhe.com/lind15e) • Expected monetary value

EMV(Ai ) � �[P(Sj ) · V(Ai , Sj )] [20–1]

• Expected opportunity loss

EOL(Ai ) � �[P(Sj) · R(Ai, Sj )] [20–2]

• Expected value of perfect information

EVPI � Expected value under conditions of certainty

� Expected value of optimal decision under

conditions of uncertainty [20–3]

c�� c�

�p(1 � p) n

Sum of the number defective Total number of items sampled

� X k

� n � 2 1 � r 2s

6�d 2

n(n2 � 1)

12 n(n � 1)

(�R1)2n1 � (�R2)2

n 2 � · · · �

(�Rk)2

nk �

z � W �

n1(n1 � n2 � 1) 2

�n1n2(n1 � n2 � 1) 12

z � (X .50) � �

�

(Row total)(Column total) Grand total

a c (fo � fe)

fe d

d � a n

t�2 (et � et�1)

a n

t�1 et

Correction factor � 4.00

Total of four means

Ŷ

Purchasing power � $1 CPI

(100)

Deflated sales � Actual sales

Index (100)

Money income CPI

• Sum of squares, total

SS total � [12–2]

• Sum of squares, error

SSE � [12–3]

• Sum of squares, treatments

SST � SS total � SSE [12–4]

• Confidence interval for differences in treatment means

[12–5]

• Sum of squares, blocks

SSB � k [12–6]

• Sum of squares, two-way ANOVA

SSE � SS total � SST � SSB [12–7]

• Sum of squares, interaction

[12–8]

• Sum of squares error, with interaction

SSE � SS total � SS factor A � SS factor B � SSI [12–9]

CHAPTER 13 • Correlation coefficient

[13–1]

• Test for significant correlation

[13–2]

• Linear regression equation

� a � bX [13–3]

• Slope of the regression line

[13–4]

• Intercept of the regression line

[13–5]

• Test for a zero slope

[13–6]

• Standard error of estimate

sy · x � [13–7]

• Coefficient of determination

[13–8]

• Confidence interval

t(sy · x) [13–9]

• Prediction interval

t(sy · x) [13–10]

CHAPTER 14 • Multiple regression equation

� a � b1X1 � b2X2 � · · · � bk Xk [14–1]

• Multiple standard error of estimate

sY · 123 · · · k � [14–2]

• Coefficient of multiple determination

R2 � [14–3]

• Adjusted coefficient of determination

[14–4]

• Global test of hypothesis

[14–5]

• Testing for a particular regression coefficient

[14–6]

• Variance inflation factor

[14–7]

CHAPTER 15 • Simple index

P � (100) [15–1]

• Simple average of price relatives

P � [15–2]

• Simple aggregate index

P � (100) [15–3]

• Laspeyres’ price index

P � (100) [15–4]

• Paasche’s price index

P � (100) [15–5]

• Fisher’s ideal index [15–6]

• Value index

V � (100) [15–7] �ptqt �p0q0

�(Laspeyres’ price index)(Paasche’s price index)

�ptqt �p0qt

�ptq0 �p0q0

�pt �p0

�Pi n

pt p0

VIF � 1

1 � R2j

t � bi � 0

sbi

F � SSR /k

SSE /[n � (k � 1)]

R2adj � 1 �

SSE n � (k � 1)

SS total n � 1

SSR SS total

g(Y � Ŷ )2

n � (k � 1)

Ŷ

�1 � 1 n

� (X � X

_ )2

�(X � X _

)2 Ŷ

�1 n

� (X � X

_ )2

�(X � X _

)2 Ŷ

r2 � SSR

SS Total � 1 � SSE

SS Total

�(Y � Ŷ )2

n � 2

t � b � 0

a � Y � bX

b � r sy sx

Ŷ

t � r �n � 2 �1 � r 2

r � g (X � X )(Y � Y )

(n � 1) sx sy

g (Xb � XG)2

(X _

1 � X _

2) t �MSE� 1n1 � 1n2

g (X � Xc)2

g (X � XG)2

Lin01803_keyform.qxd 10/28/10 1:36 PM Page 3

ISBN: 0073401803 Author: Lind, Marchal, Wathen Title: Statistical Techniques in Business & Economics

Fifteenth edition

Back endsheets Color: 4 Pages: 5

Student’s t Distribution

0 0

1 2� ��

1 2

Confidence interval t–t 0

Left-tailed test –t t

Right-tailed test t

Two-tailed test t–t

Confidence Intervals, c 80% 90% 95% 98% 99% 99.9%

Level of Significance for One-Tailed Test, � 0.10 0.05 0.025 0.01 0.005 0.0005