CLA 1 Paper - Data Analysis & Business Intelligence
Statistical Techniques in Business & Economics
LIND
MARCHAL
WATHEN
Seventeenth Edition
Statistical Techniques in
BUSINESS & ECONOMICS
The McGraw-Hill/Irwin Series in Operations and Decision Sciences
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Statistical Techniques in
BUSINESS & ECONOMICS
S E V E N 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
STATISTICAL TECHNIQUES IN BUSINESS & ECONOMICS, SEVENTEENTH EDITION Published by McGraw-Hill Education, 2 Penn Plaza, New York, NY 10121. Copyright © 2018 by McGraw-Hill Education. All rights reserved. Printed in the United States of America. Previous editions © 2015, 2012, and 2010. 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 McGraw- Hill Education, 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.
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Library of Congress Cataloging-in-Publication Data
Names: Lind, Douglas A., author. | Marchal, William G., author. | Wathen, Samuel Adam. author. Title: Statistical techniques in business & economics/Douglas A. Lind, Coastal Carolina University and The University of Toledo, William G. Marchal, The University of Toledo, Samuel A. Wathen, Coastal Carolina University. Other titles: Statistical techniques in business and economics Description: Seventeenth Edition. | Dubuque, IA : McGraw-Hill Education, [2017] | Revised edition of the authors’ Statistical techniques in business & economics, [2015] Identifiers: LCCN 2016054310| ISBN 9781259666360 (alk. paper) | ISBN 1259666360 (alk. paper) Subjects: LCSH: Social sciences—Statistical methods. | Economics—Statistical methods. | Commercial statistics. Classification: LCC HA29 .M268 2017 | DDC 519.5—dc23 LC record available at https://lccn.loc.gov/2016054310
The Internet addresses listed in the text were accurate at the time of publication. The inclusion of a website does not indicate an endorsement by the authors or McGraw-Hill Education, and McGraw-Hill Education does not guarantee the accuracy of the information presented at these sites.
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DEDICATION
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, Jake, and Brady), and Mark and Sarah (Jared, Drew, and Nate).
Douglas A. Lind
To Oscar Sambath Marchal, Julian Irving Horowitz, Cecilia Marchal Nicholson and Andrea.
William G. Marchal
To my wonderful family: Barb, Hannah, and Isaac.
Samuel A. Wathen
vi
Over the years, we received many compliments on this text and understand that it’s a favorite among students. We accept that as the highest 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 descriptive and infer- ential statistics. To illustrate the application of statistics, we use many examples and exercises that focus on business applications, but also relate to the current world of the college 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 successful in a basic statistics course. This step-by-step approach enhances performance, accel- erates preparedness, and significantly improves motivation. 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, locating relevant business data was difficult. That has changed! Today, locating data is not a problem. The number of items you purchase at the grocery store is automatically recorded at the checkout counter. Phone companies track the time of our calls, the length of calls, and the identity of the person called. Credit card companies maintain information on the number, time and date, and amount of our purchases. Medical devices automati- cally monitor our heart rate, blood pressure, and temperature 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 in real time.
Today, the practice of data analytics is widely applied to “big data.” The practice of data analytics requires skills and knowledge in several areas. Computer skills are needed to process large volumes of information. Analytical skills are needed to evaluate, summarize, organize, and analyze the information. Critical thinking skills are needed to interpret and communicate the results of processing the information.
Our text supports the development of basic data analytical skills. In this edition, we added a new section at the end of each chapter called Data Analytics. As you work through the text, this section provides the instructor and student with opportu- nities to apply statistical knowledge and statistical software to explore several busi- ness environments. Interpretation of the analytical results is an integral part of these exercises.
A variety of statistical software is available to complement our text. Microsoft Excel includes an add-in with many statistical analyses. Megastat is an add-in available for Microsoft Excel. Minitab and JMP are stand-alone statistical software available to down- load for either PC or MAC computers. In our text, Microsoft Excel, Minitab, and Megastat are used to illustrate statistical software analyses. When a software application is pre- sented, the software commands for the application are available in Appendix C. 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 longer necessary to dwell on calculations. We have replaced many of the calculation examples with interpre- tative ones, to assist the student in understanding and interpreting the statistical results. In addition, we 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 con- cepts, along with supporting interesting and relevant examples.
A N O T E F R O M T H E A U T H O R S
vii
WHAT’S NEW IN THE SEVENTEENTH EDITION? We have made many changes to examples and exercises throughout the text. The sec- tion on “Enhancements” to our text details them. The major change to the text is in response to user interest in the area of data analytics. Our approach is to provide in- structors and students with the opportunity to combine statistical knowledge, computer and statistical software skills, and interpretative and critical thinking skills. A set of new and revised exercises is included at the end of chapters 1 through 18 in a section titled “Data Analytics.”
In these sections, exercises refer to three data sets. The North Valley Real Estate sales data set lists 105 homes currently on the market. The Lincolnville School District bus data lists information on 80 buses in the school district’s bus fleet. The authors de- signed these data so that students will be able to use statistical software to explore the data and find realistic relationships in the variables. The Baseball Statistics for the 2016 season is updated from the previous edition.
The intent of the exercises is to provide the basis of a continuing case analysis. We suggest that instructors select one of the data sets and assign the corresponding exer- cises as each chapter is completed. Instructor feedback regarding student performance is important. Students should retain a copy of each chapter’s results and interpretations to develop a portfolio of discoveries and findings. These will be helpful as students progress through the course and use new statistical techniques to further explore the data. The ideal ending for these continuing data analytics exercises is a comprehensive report based on the analytical findings.
We know that working with a statistics class to develop a very basic competence in data analytics is challenging. Instructors will be teaching statistics. In addition, instruc- tors will be faced with choosing statistical software and supporting students in develop- ing or enhancing their computer skills. Finally, instructors will need to assess student performance based on assignments that include both statistical and written compo- nents. Using a mentoring approach may be helpful.
We hope that you and your students find this new feature interesting and engaging.
HOW ARE CHAPTERS ORGANIZED TO ENGAGE STUDENTS AND PROMOTE LEARNING?
Chapter Learning Objectives Each chapter begins with a set of learning objectives designed to pro- vide focus for the chapter and motivate student learning. These objectives, lo- cated in the margins next to the topic, indicate what the student should be able to do after completing each sec- tion in 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.
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO2-1 Summarize qualitative variables with frequency and relative frequency tables.
LO2-2 Display a frequency table using a bar or pie chart.
LO2-3 Summarize quantitative variables with frequency and relative frequency distributions.
LO2-4 Display a frequency distribution using a histogram or frequency polygon.
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 LO2-3.)
Describing Data: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS,
AND GRAPHIC PRESENTATION2
Source: © rido/123RF
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Introduction to the Topic Each chapter starts with a review of the important concepts of the previ- ous chapter and provides a link to the material in the current chapter. This step-by-step approach increases com- prehension by providing continuity across the concepts.
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 19
INTRODUCTION The United States automobile retailing industry is highly competitive. It is dominated by megadealerships that own and operate 50 or more franchises, employ over 10,000 people, and generate several billion dollars in annual sales. Many of the top dealerships
are publicly owned with shares traded on the New York Stock Exchange or NASDAQ. In 2014, the largest megadealership was AutoNation (ticker symbol AN), followed by Penske Auto Group (PAG), Group 1 Automotive, Inc. (ticker symbol GPI), and the privately owned Van Tuyl Group.
These large corporations use statistics and analytics to summarize and analyze data and information to support their decisions. As an ex- ample, we will look at the Applewood Auto group. It owns four dealer- ships and sells a wide range of vehicles. These include the 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 Kane 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 Apple- wood dealerships? What type of vehicle did they purchase?
The Applewood Auto Group operates four dealerships:
• Tionesta Ford Lincoln sells Ford and Lincoln cars and trucks. • Olean Automotive Inc. has the Nissan franchise as well as the General Motors
brands of Chevrolet, Cadillac, and GMC Trucks. • Sheffield Motors Inc. sells Buick, GMC trucks, Hyundai, and Kia. • Kane 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 them 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 to the left. The variables collected include:
• Age—the age of the buyer at the time of the purchase. • Profit—the amount earned by the dealership on the sale of each
vehicle. • 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 (www.mhhe .com/lind17e) and in Appendix A.4 at the end of the text.
Source: © Justin Sullivan/Getty Images
CONSTRUCTING FREQUENCY TABLES Recall from Chapter 1 that techniques used to describe a set of data are called descrip- tive statistics. Descriptive statistics organize data to show the general pattern of the data, to identify where values tend to concentrate, and to expose extreme or unusual data values. The first technique we discuss is a frequency table.
LO2-1 Summarize qualitative variables with frequency and relative frequency tables.
FREQUENCY TABLE A grouping of qualitative data into mutually exclusive and collectively exhaustive classes showing the number of observations in each class.
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Example/Solution After important concepts are introduced, a solved example is given. This example provides a how-to illustration and shows a relevant business application that helps students answer the question, “How can I apply this concept?”
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 95
INTRODUCTION Chapter 2 began our study of descriptive statistics. In order to transform raw or un- grouped data into a meaningful form, we organize the data into a frequency distribution. We present the frequency distribution in graphic form as a histogram or a frequency polygon. This allows us to visualize where the data tend 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, median, and mode. 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, variance, and standard deviation. These measures of dispersion allow us to de- scribe 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. Examples include the number of hours a student studied and the points earned on an examination; if a sampled product meets quality specifications and the shift on which it is manufactured; or the amount of electric- ity used in a month by a homeowner and the mean daily high temperature in the region for the month. These charts and graphs provide useful insights as we use business analytics to enhance our understanding of data.
DOT PLOTS Recall for the Applewood Auto Group data, we summarized the profit earned on the 180 vehicles sold with a frequency distribution using eight classes. When we orga- nized 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 individual observation. To develop a dot plot, we display a dot for each observation 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 individually, 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 histo- grams tend to be most useful for large data sets. An example will show how to con- struct and interpret dot plots.
LO4-1 Construct and interpret a dot plot.
E X A M P L E
The service departments at Tionesta Ford Lincoln and Sheffield Motors Inc., two of the four Applewood Auto Group dealerships, were both open 24 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
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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Self-Reviews Self-Reviews are interspersed throughout each chapter and follow Example/Solution sec- tions. They help students mon- itor their progress and provide immediate reinforcement for that particular technique. An- swers are in Appendix E.
106 CHAPTER 4
calculate quartiles. Excel 2013 and Excel 2016 offer both methods. The Excel function, Quartile.exc, will result in the same answer as Equation 4–1. The Excel function, Quar- tile.inc, will result in the Excel Method answers.
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:
7.69 7.72 7.8 7.86 7.90 7.94 7.97 8.06 8.09
(a) What is the median weight? (b) Determine the weights corresponding to the first and third quartiles.
S E L F - R E V I E W 4–2
11. Determine the median and the first and third quartiles in the following data.
46 47 49 49 51 53 54 54 55 55 59
12. Determine the median and the first and third quartiles in the following data.
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
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 im- portant. 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.
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 re- port 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.
E X E R C I S E S
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viii
ix
Statistics in Action Statistics in Action articles are scattered through- out the text, usually about two per chapter. They provide unique, interesting applications and his- torical insights in the field of statistics.
144 CHAPTER 5
The General Rule of Addition The outcomes of an experiment may not be mutually exclusive. For example, 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 probabilities is 1.10. We know, however, that this probability cannot be greater than 1. The explanation is that many tour- ists 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 World and the probability he or she visited Busch Gardens, and (2) subtract the proba- bility of visiting both. Thus:
P(Disney or Busch) = P(Disney) + P(Busch) − P(both Disney and Busch) = .60 + .50 − .30 = .80
When two events both occur, the probability is called a joint probability. The prob- ability (.30) that a tourist visits both attractions is an example of a joint probability.
© Rostislav Glinsky/Shutterstock.com
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.
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?
S E L F - R E V I E W 5–3
STATISTICS IN ACTION
If you wish to get some attention at the next gath- ering 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 necessarily 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 people 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 birthday. Hint: To compute this, find the probability everyone was born on a different day and use the complement rule. Try this in your class.
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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. They also appear in the Glossary at the end of the book.
A SURVEY OF PROBABILITY CONCEPTS 145
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.
So the general rule of addition, which is used to compute the probability of two events that are not mutually exclusive, is:
GENERAL RULE OF ADDITION P(A or B) = P(A) + P(B) − P(A and B) [5–4]
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 sometimes 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 difference 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. Other- wise, we must account for the joint probability and use the general rule of addition.
E X A M P L E
What is the probability that a card chosen at random from a standard deck of cards will be either a king or a heart?
S O L U T I O N
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
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Formulas Formulas that are used for the first time are boxed and numbered for reference. In addi- tion, a formula card is bound into the back of the text that lists all the key formulas.
A SURVEY OF PROBABILITY CONCEPTS 147
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 probability of passing both is .50. What is the probability of passing at least one?
21. The aquarium at Sea Critters Depot contains 140 fish. Eighty of these fish are green swordtails (44 female and 36 male) and 60 are orange swordtails (36 female and 24 males). A fish is randomly captured from the aquarium:
a. What is the probability the selected fish is a green swordtail? b. What is the probability the selected fish is male? c. What is the probability the selected fish is a male green swordtail? d. What is the probability the selected fish is either a male or a green swordtail?
22. A National Park Service survey of visitors to the Rocky Mountain region revealed that 50% visit Yellowstone Park, 40% visit the Tetons, and 35% 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 TO CALCULATE PROBABILITY In this section, we discuss the rules for computing the likelihood that two events both happen, or their joint probability. For example, 16% of the 2016 tax returns were pre- pared by H&R Block and 75% of those returns showed a refund. What is the likelihood a person’s tax form was prepared by H&R Block and the person received a refund? 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 multipli- cation: 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 probabil- ity of the occurrence of the other event.
LO5-4 Calculate probabilities using the rules of multiplication.
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 differ- ent 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 P(A and B) = P(A)P(B) [5–5]
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Exercises Exercises are included after sec- tions within the chapter and at the end of the chapter. Section exercises cover the material stud- ied in the section. Many exercises have data files available to import into statistical software. They are indicated with the FILE icon. Answers to the odd-numbered exercises are in Appendix D.
DESCRIBING DATA: NUMERICAL MEASURES 79
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 computed to be $7.51. Suppose these employees are located in Georgia. If the standard 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 em- ployees 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 out of every four, or 75%, of the values 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 values, or 88.9%, will lie between plus three standard deviations and minus three standard deviations of the mean. At least 24 of 25 values, or 96%, will lie between plus and minus five standard deviations of the mean.
Chebyshev’s theorem states:
LO3-5 Explain and apply Chebyshev’s theorem and the Empirical Rule.
STATISTICS IN ACTION
Most colleges report the “average class size.” This information 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
(continued)
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 value greater than 1.
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. 49. Dave’s Automatic Door, referred to in Exercise 37, installs automatic garage
door openers. 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 Exer- cise 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
E X E R C I S E S
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Computer Output The text includes many software examples, using Excel, MegaStat®, and Minitab. The software results are illustrated in the chapters. Instructions for a particular software example are in Appendix C.
64 CHAPTER 3
E X A M P L E
Table 2–4 on page 26 shows the profit on the sales of 180 vehicles at Applewood Auto Group. Determine the mean and the median selling price.
S O L U T I O N
The mean, median, and modal amounts of profit are reported in the following output (highlighted in the screen shot). (Reminder: The instructions to create the output appear in the Software Commands in Appendix C.) There are 180 vehicles in the study, so using a calculator would be tedious and prone to error.
Software Solution We can use a statistical software package to find many measures of location.
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? 22. Big Orange Trucking is designing an information system for use in “in-cab”
communications. 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 direction, 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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HOW DOES THIS TEXT REINFORCE STUDENT LEARNING?
x
BY CHAPTER
Chapter Summary Each chapter contains a brief summary of the chapter material, including vocab- ulary, definitions, and critical formulas.
202 CHAPTER 6
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.
In summary, the Poisson distribution is a family of discrete distributions. All that is needed to construct a Poisson probability distribution is the mean number of defects, errors, or other random variable, designated as μ.
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?
S E L F - R E V I E W 6–6
31. In a Poisson distribution μ = 0.4. a. What is the probability that x = 0? b. What is the probability that x > 0?
32. In a Poisson distribution μ = 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 three loans will be defaulted? b. What is the probability that at least three 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% 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 3 days each year for weather emergencies. What is the probability that schools in Los Angeles County will close for 4 days next year?
E X E R C I S E S
C H A P T E R S U M M A R Y
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:
μ = Σ[xP(x)] (6–1) B. The variance is equal to:
σ2 = Σ[(x − μ)2P(x)] (6–2)
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Pronunciation Key This section lists the mathematical symbol, its meaning, and how to pronounce it. We believe this will help the student retain the meaning of the symbol and generally en- hance course communications.
168 CHAPTER 5
P R O N U N C I A T I O N K E Y
SYMBOL MEANING PRONUNCIATION
P(A) 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(A | B) Probability of A given B has happened P of A given B
nPr Permutation of n items selected r at a time Pnr
nCr Combination of n items selected r at a time Cnr
C H A P T E R E X E R C I S E S
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 in- clude 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 racehorse. After a successful Kentucky Derby, Corn on the Cob is a heavy favorite at 2 to 1 odds to win the Preakness Stakes. a. If he is a 2 to 1 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?
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Chapter Exercises Generally, the end-of-chapter exercises are the most challenging and integrate the chapter concepts. The answers and worked-out solutions for all odd- numbered exercises are in Appendix D at the end of the text. Many exercises are noted with a data file icon in the margin. For these exercises, there are data files in Excel format located on the text’s website, www.mhhe.com/Lind17e. These files help students use statistical software to solve the exercises.
348 CHAPTER 10
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. 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 α.
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 β. 2. The likelihood of a Type II error must be calculated comparing the hypothesized
distribution to an alternate distribution based on sample results.
P R O N U N C I A T I O N K E Y
SYMBOL MEANING PRONUNCIATION
H0 Null hypothesis H sub zero
H1 Alternate hypothesis H sub one
α/2 Two-tailed significance level Alpha divided by 2 xc Limit of the sample mean x bar sub c
μ0 Assumed population mean mu sub zero
C H A P T E R E X E R C I S E S
25. 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 reasonable to conclude that the mean income is not equal to $45,000? Deter- mine the p-value.
26. 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 the packaging process is 3 pounds per bag. At the end of each day, Jeff Rutter, the production manager, weighs 10 bags and computes the mean weight of the sample. Below are the weights of 10 bags from today’s production.
45.6 47.7 47.6 46.3 46.2 47.4 49.2 55.8 47.5 48.5
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.
27. A new weight-watching company, Weight Reducers International, advertises that those who join will lose an average of 10 pounds after the first two weeks. The standard devi- ation is 2.8 pounds. A random sample of 50 people who joined the weight reduction program revealed a mean loss of 9 pounds. At the .05 level of significance, can we conclude that those joining Weight Reducers will lose less than 10 pounds? Determine the p-value.
28. 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 ounce. The quality-con- trol department took a random sample of 50 cans and found that the arithmetic mean weight was 16.05 ounces. At the 5% level of significance, can we conclude that the mean weight is greater than 16 ounces? Determine the p-value.
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Data Analytics The goal of the Data Analytics sec- tions is to develop analytical skills. The exercises present a real world context with supporting data. The data sets are printed in Appendix A and available to download from the text’s website www.mhhe.com/Lind17e. Statistical software is required to analyze the data and respond to the exercises. Each data set is used to explore questions and dis- cover findings that relate to a real world context. For each business context, a story is uncovered as students progress from chapters one to seventeen.
244 CHAPTER 7
68. In establishing warranties on HDTVs, 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 deviation of 3.34 months. Where should the warranty limits be set so that only 10% 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% 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 manu- facturer’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 probability 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 distribu- tion with a mean of 27 seconds. What is the probability his “boot” will require: a. Less than 15 seconds? b. More than 60 seconds? c. Between 30 and 45 seconds? d. What is the point below which only 10% 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: a. Will visit an emergency room within the next 6 months? b. Will not visit the ER over the next 6 years? c. Will visit an ER 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?
D A T A A N A L Y T I C S
(The data for these exercises are available at the text website: www.mhhe.com/lind17e.)
74. Refer to the North Valley Real Estate data, which report information on homes sold during the last year. a. The mean selling price (in $ thousands) of the homes was computed earlier to be $357.0,
with a standard deviation of $160.7. Use the normal distribution to estimate the percent- age of homes selling for more than $500.000. Compare this to the actual results. Is price normally distributed? Try another test. If price is normally distributed, how many homes should have a price greater than the mean? Compare this to the actual number of homes. Construct a frequency distribution of price. What do you observe?
b. The mean days on the market is 30 with a standard deviation of 10 days. Use the normal distribution to estimate the number of homes on the market more than 24 days. Compare this to the actual results. Try another test. If days on the market is normally distributed, how many homes should be on the market more than the mean number of days? Compare this to the actual number of homes. Does the normal
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Software Commands Software examples using Excel, Mega- Stat®, and Minitab are included through- out the text. The explanations of the computer input commands are placed at the end of the text in Appendix C.
780
11–2. The Minitab commands for the two-sample t-test on page 368 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 col- umns, select C1 Store for the First column and C2 Name of the Second, click the box next to Assume equal variances, and click OK.
11–3. The Excel commands for the paired t-test on page 373 are: a. Enter the data into columns B and C (or any other two col-
umns) 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 Variable 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.
CHAPTER 12 12–1. The Excel commands for the test of variances on page 391 are: a. Enter the data for U.S. 25 in column A and for I-75 in col-
umn 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.
12–2. The Excel commands for the one-way ANOVA on page 400 are: a. Key in data into four columns labeled Northern, WTA, Po-
cono, 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 0.05, and finally select Output Range as F1 and click OK.
c. In the dialog box, indicate that the range of Variable 1 is from A1 to A6 and Variable 2 from B1 to B7, the Hypothe- sized Mean Difference is 0, click Labels, Alpha is 0.05, and the Output Range is D1. Click OK.
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xi
Answers to Self-Review The worked-out solutions to the Self-Reviews are pro- vided at the end of the text in Appendix E.
11
16–7 a. 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
rs = 1 − 6(193)
10(99) = −.170
b. H0: ρ = 0; H1: ρ ≠ 0. Reject H0 if t < −2.306 or t > 2.306.
t = −.170√ 10 − 2
1 − (−0.170)2 = −0.488
H0 is not rejected. We have not shown a relationship between the two tests.
CHAPTER 17 17–1 1.
Country Amount Index (Based=US) China 822.7 932.8 Japan 110.7 125.5 United States 88.2 100.0 India 86.5 98.1 Russia 71.5 81.1
China Produced 832.8% more steel than the US
2. a.
Year Average Hourly Earnings Index (1995 = Base) 1995 11.65 100.0 2000 14.02 120.3 2005 16.13 138.5 2013 19.97 171.4 2016 21.37 183.4
2016 Average wage Increased 83.4% from 1995
b.
Year Average Hourly Earnings Index (1995 – 2000 = Base) 1995 11.65 90.8 2000 14.02 109.2 2005 16.13 125.7 2013 19.97 155.6 2016 21.37 166.5
2016 Average wage Increased 86.5% from the average of 1995, 2000
17–2 1. a. P1 = ($85/$75)(100) = 113.3 P2 = ($45/$40)(100) = 112.5 P = (113.3 + 112.5)/2 = 112.9 b. P = ($130/$115)(100) = 113.0
c. P = $85(500) + $45(1,200) $75(500) + $40(1,200)
(100)
= $96,500 85,500
(100) = 112.9
d. P = $85(520) + $45(1,300) $75(520) + $40(1,300)
(100)
= $102,700
$91,000 (100) = 112.9
e. P = √(112.9) (112.9) = 112.9
17–3 a. P = $4(9,000) + $5(200) + $8(5,000) $3(10,000) + $1(600) + $10(3,000)
(100)
= $77,000 60,600
(100) = 127.1
b. The value of sales went up 27.1% from 2001 to 2017
17–4 a. For 2011
Item Weight
Cotton ($0.25/$0.20)(100)(.10) = 12.50 Autos (1,200/1,000)(100)(.30) = 36.00 Money turnover (90/80)(100)(.60) = 67.50 Total 116.00
For 2016
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 Total 108.25
b. Business activity increased 16% from 2004 to 2009. It increased 8.25% from 2004 to 2014.
17–5 In terms of the base period, Jon’s salary was $14,637 in 2000 and $17,944 in 2016. This indicates that take-home pay in- creased at a faster rate than the rate of prices paid for food, transportation, etc.
17–6 $0.42, round by ($1.00/238.132)(100). The purchasing power has declined by $0.58.
17–7 Year IPI PPI
2007 111.07 92.9 2008 107.12 100.2 2009 94.80 95.3 2010 100.00 100.0 2011 102.93 107.8 2012 105.80 110.1 2013 107.83 110.5 2014 110.98 111.5 2015 111.32 105.8
The Industrial Production index (IPI) increased 11.32% from 2010 to 2015. The Producer Price Index (PPI) increases 5.8%.
CHAPTER 18 18–1
Year Number Produced Moving Average
2011 2 2012 6 4 2013 4 5 2014 5 4 2015 3 6 2016 10
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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 Section Review is included. Much like a review before an exam, these include a brief overview of the chap- ters and problems for review.
126 A REVIEW OF CHAPTERS 1–4
D A T A A N A L Y T I C S
44. Refer to the North Valley real estate data recorded on homes sold during the last year. Prepare a report on the selling prices of the homes based on the answers to the following questions. a. Compute the minimum, maximum, median, and the first and the third quartiles of
price. Create a box plot. Comment on the distribution of home prices. b. Develop a scatter diagram with price on the vertical axis and the size of the home on
the horizontal. Is there a relationship between these variables? Is the relationship direct or indirect?
c. For homes without a pool, develop a scatter diagram with price on the vertical axis and the size of the home on the horizontal. Do the same for homes with a pool. How do the relationships between price and size for homes without a pool and homes with a pool compare?
45. Refer to the Baseball 2016 data that report information on the 30 Major League Baseball teams for the 2016 season. a. In the data set, the year opened, is the first year of operation for that stadium. For
each team, use this variable to create a new variable, stadium age, by subtracting the value of the variable, year opened, from the current year. Develop a box plot with the new variable, age. Are there any outliers? If so, which of the stadiums are outliers?
b. Using the variable, salary, create a box plot. Are there any outliers? Compute the quartiles using formula (4–1). Write a brief summary of your analysis.
c. Draw a scatter diagram with the variable, wins, on the vertical axis and salary on the horizontal axis. What are your conclusions?
d. Using the variable, wins, draw a dot plot. What can you conclude from this plot? 46. Refer to the Lincolnville School District bus data.
a. Referring to the maintenance cost variable, develop a box plot. What are the mini- mum, first quartile, median, third quartile, and maximum values? Are there any outliers?
b. Using the median maintenance cost, develop a contingency table with bus manufac- turer as one variable and whether the maintenance cost was above or below the median as the other variable. What are your conclusions?
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 distribution as a histogram or a frequency polygon. Chapter 3 began by describing measures of location, 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, 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 skew- ness, 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 of variation calculated, and the information presented in graphical form.
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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.
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?
A REVIEW OF CHAPTERS 1–4 129
C A S E S
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 writ- ten 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 custom- ers. 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? To better understand the customers, Mr. Selig asked Ms. Wendy Lamberg, director of planning, to select a sam- ple of customers and prepare a report. To begin, 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 (automatic teller machine) transac- tions in the last month; (2) the number of other bank ser- vices (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 customer’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 ac- count 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.
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 deviation 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 es- sentially the same job but at different pay. To investi- gate, he collected the information below. Suppose you are a student intern in the Accounting Department and have been given the task to write a report summarizing the situation.
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
To kick off the project, Mr. Cory St. Julian held a meeting with his staff and you were invited. At this meeting, it was suggested that you calculate several measures of
Lin66360_ch04_094-131.indd 129 1/10/17 7:41 PM
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.
130 A REVIEW OF CHAPTERS 1–4
location, create charts or draw graphs such as a cumula- tive 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 represen- tatives. The company sells sporting goods to two major
markets. There are 40 sales representatives who call di- rectly on large-volume customers, such as the athletic de- partments 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.
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
P R A C T I C E T E S T
There is a practice test at the end of each review section. The tests are in two parts. The first part contains several objec- tive 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 decisions is called . 1. 2. Methods of organizing, summarizing, and presenting data in an informative way are
called . 2. 3. The entire set of individuals or objects of interest or the measurements obtained from all
individuals or objects of interest are 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 are 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 the 2k method suggest 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,
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.
Part 2—Problems 1. The Russell 2000 index of stock prices increased by the following amounts over the last 3 years.
18% 4% 2%
What is the geometric mean increase for the 3 years?
Lin66360_ch04_094-131.indd 130 1/10/17 7:41 PM
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ADDITIONAL RESOURCES
xiv
xv
ACKNOWLEDGMENTS
Stefan Ruediger Arizona State University Anthony Clark St. Louis Community College Umair Khalil West Virginia University Leonie Stone SUNY Geneseo
Golnaz Taghvatalab Central Michigan University John Yarber Northeast Mississippi Community College John Beyers University of Maryland
Mohammad Kazemi University of North Carolina Charlotte Anna Terzyan Loyola Marymount University Lee O. Cannell El Paso Community College
This edition of Statistical Techniques in Business and Economics is the product of many people: students, colleagues, reviewers, and the staff at McGraw-Hill Education. We thank them all. We wish to express our sincere gratitude to the reviewers:
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. Shelly Moore, College of Western Idaho, and John Arcaro, Lakeland Community College, accuracy checked the Connect exercises. Ed Pappanastos, Troy University, built new data sets and revised Smartbook. Rene Ordonez, Southern Oregon University, built the Connect guided examples. Wendy Bailey, Tory University, prepared the test bank. Stephanie Campbell, Mineral Area College, prepared the Powerpoint decks. Vickie Fry, Westmoreland County Community College, provided countless hours of digital accuracy checking and support.
We also wish to thank the staff at McGraw-Hill. This includes Dolly Womack, Senior Brand Man- ager; Michele Janicek, Product Developer Coordinator; Camille Corum and Ryan McAndrews, Product Developers; Harvey Yep and Bruce Gin, Content Project Managers; and others we do not know per- sonally, but who have made valuable contributions.
xvi CONTENTS
xvi
ENHANCEMENTS TO STATISTICAL TECHNIQUES IN BUSINESS & ECONOMICS, 17E
MAJOR CHANGES MADE TO INDIVIDUAL CHAPTERS:
CHAPTER 1 What Is Statistics? • Revised Self-Review 1-2.
• New Section describing Business Analytics and its integration with the text.
• Updated exercises 2, 3, 17, and 19.
• New Data Analytics section with new data and questions.
CHAPTER 2 Describing Data: Frequency Tables, Frequency Distributions, and Graphic Presentation • Revised chapter introduction.
• Added more explanation about cumulative relative frequency distributions.
• Updated exercises 47 and 48 using real data.
• New Data Analytics section with new data and questions.
CHAPTER 3 Describing Data: Numerical Measures • Updated Self-Review 3-2.
• Updated Exercises 16, 18, 73, 77, and 82.
• New Data Analytics section with new data and questions.
CHAPTER 4 Describing Data: Displaying and Exploring Data • Updated exercise 22 with 2016 New York Yankee player
salaries.
• New Data Analytics section with new data and questions.
CHAPTER 5 A Survey of Probability Concepts • Revised the Example/Solution in the section on Bayes
Theorem.
• Updated exercises 45 and 58 using real data.
• New Data Analytics section with new data and questions.
CHAPTER 6 Discrete Probability Distributions • Expanded discussion of random variables.
• Revised the Example/Solution in the section on Poisson distribution.
• Updated exercises 18, 58, and 68.
• New Data Analytics section with new data and questions.
CHAPTER 7 Continuous Probability Distributions • Revised Self-Review 7-1.
• Revised the Example/Solutions using Uber as the context.
• Updated exercises 19, 22, 28, 36, 47, and 64.
• New Data Analytics section with new data and questions.
CHAPTER 8 Sampling Methods and the Central Limit Theorem • New Data Analytics section with new data and questions.
CHAPTER 9 Estimation and Confidence Intervals • New Self-Review 9-3 problem description.
• Updated exercises 5, 6, 12, 14, 23, 24, 33, 41, 43, and 61.
• New Data Analytics section with new data and questions.
CHAPTER 10 One-Sample Tests of Hypothesis • Revised the Example/Solutions using an airport, cell phone
parking lot as the context.
• Revised the section on Type II error to include an additional example.
• New Type II error exercises, 23 and 24.
• Updated exercises 19, 31, 32, and 43.
• New Data Analytics section with new data and questions.
CHAPTER 11 Two-Sample Tests of Hypothesis • Updated exercises 5, 9, 12, 26, 27, 30, 32, 34, 40, 42,
and 46.
• New Data Analytics section with new data and questions.
CHAPTER 12 Analysis of Variance • Revised Self-Reviews 12-1 and 12-3.
• Updated exercises 10, 21, 24, 33, 38, 42, and 44.
• New Data Analytics section with new data and questions.
CHAPTER 13 Correlation and Linear Regression • Added new conceptual formula, to relate the standard error
to the regression ANOVA table.
• Updated exercises 36, 41, 42, 43, and 57.
• New Data Analytics section with new data and questions.
CHAPTER 14 Multiple Regression Analysis • Updated exercises 19, 21, 23, 24, and 25.
• New Data Analytics section with new data and questions.
CHAPTER 15 Nonparametric Methods: Nominal Level Hypothesis Tests • Updated the context of Manelli Perfume Company Example/
Solution.
• Revised the “Hypothesis Test of Unequal Expected Frequen- cies” Example/Solution.
• Updated exercises 3, 31, 42, 46, and 61.
• New Data Analytics section with new data and questions.
xvii
CHAPTER 16 Nonparametric Methods: Analysis of Ordinal Data • Revised the “Sign Test” Example/Solution.
• Revised the “Testing a Hypothesis About a Median” Example/ Solution.
• Revised the “Wilcoxon Rank-Sum Test for Independent Popu- lations” Example/Solution.
• Revised Self-Reviews 16-3 and 16-6.
• Updated exercise 25.
• New Data Analytics section with new data and questions.
CHAPTER 17 Index Numbers • Revised Self-Reviews 17-1, 17-2, 17-3, 17-4, 17-5, 17-6, 17-7.
• Updated dates, illustrations, and examples.
• New Data Analytics section with new data and questions.
CHAPTER 18 Time Series and Forecasting • Updated dates, illustrations, and examples.
• New Data Analytics section with new data and questions.
CHAPTER 19 Statistical Process Control and Quality Management • Updated 2016 Malcolm Baldridge National Quality Award
winners.
• Updated exercises 13, 22, and 25.
xix
B R I E F C O N T E N T S
1 What is Statistics? 1 2 Describing Data: Frequency Tables, Frequency Distributions,
and Graphic Presentation 18
3 Describing Data: Numerical Measures 51 4 Describing Data: Displaying and Exploring Data 94 Review Section
5 A Survey of Probability Concepts 132 6 Discrete Probability Distributions 175 7 Continuous Probability Distributions 209 Review Section 8 Sampling Methods and the Central Limit Theorem 250 9 Estimation and Confidence Intervals 282 Review Section 10 One-Sample Tests of Hypothesis 318 11 Two-Sample Tests of Hypothesis 353 12 Analysis of Variance 386 Review Section 13 Correlation and Linear Regression 436 14 Multiple Regression Analysis 488 Review Section 15 Nonparametric Methods:
Nominal Level Hypothesis Tests 545
16 Nonparametric Methods: Analysis of Ordinal Data 582 Review Section
17 Index Numbers 621 18 Time Series and Forecasting 653 Review Section 19 Statistical Process Control and Quality Management 697 20 An Introduction to Decision Theory 728
Appendixes: Data Sets, Tables, Software Commands, Answers 745
Glossary 847
Index 851
xx
C O N T E N T S
1 What is Statistics? 1 Introduction 2
Why Study Statistics? 2
What is Meant by Statistics? 3
Types of Statistics 4
Descriptive Statistics 4 Inferential Statistics 5
Types of Variables 6
Levels of Measurement 7
Nominal-Level Data 7 Ordinal-Level Data 8 Interval-Level Data 9 Ratio-Level Data 10
EXERCISES 11
Ethics and Statistics 12
Basic Business Analytics 12
Chapter Summary 13
Chapter Exercises 14
Data Analytics 17
2 Describing Data: FREQUENCY TABLES, FREQUENCY
DISTRIBUTIONS, AND GRAPHIC PRESENTATION 18 Introduction 19
Constructing Frequency Tables 19
Relative Class Frequencies 20
Graphic Presentation of Qualitative Data 21
EXERCISES 25
Constructing Frequency Distributions 26
Relative Frequency Distribution 30
EXERCISES 31
Graphic Presentation of a Distribution 32
Histogram 32 Frequency Polygon 35
EXERCISES 37
Cumulative Distributions 38
EXERCISES 41
Chapter Summary 42
Chapter Exercises 43
Data Analytics 49
3 Describing Data: NUMERICAL MEASURES 51
Introduction 52
Measures of Location 52
The Population Mean 53 The Sample Mean 54 Properties of the Arithmetic Mean 55
EXERCISES 56
The Median 57 The Mode 59
EXERCISES 61
The Relative Positions of the Mean, Median, and Mode 62
EXERCISES 63
Software Solution 64
The Weighted Mean 65
EXERCISES 66
The Geometric Mean 66
EXERCISES 68
Why Study Dispersion? 69
Range 70 Variance 71
EXERCISES 73
Population Variance 74 Population Standard Deviation 76
EXERCISES 76
Sample Variance and Standard Deviation 77 Software Solution 78
EXERCISES 79
Interpretation and Uses of the Standard Deviation 79
Chebyshev’s Theorem 79 The Empirical Rule 80
A Note from the Authors vi
CONTENTS xxi
EXERCISES 81
The Mean and Standard Deviation of Grouped Data 82
Arithmetic Mean of Grouped Data 82 Standard Deviation of Grouped Data 83
EXERCISES 85
Ethics and Reporting Results 86
Chapter Summary 86
Pronunciation Key 88
Chapter Exercises 88
Data Analytics 92
4 Describing Data: DISPLAYING AND EXPLORING DATA 94
Introduction 95
Dot Plots 95
Stem-and-Leaf Displays 96
EXERCISES 101
Measures of Position 103
Quartiles, Deciles, and Percentiles 103
EXERCISES 106
Box Plots 107
EXERCISES 109
Skewness 110
EXERCISES 113
Describing the Relationship between Two Variables 114
Contingency Tables 116
EXERCISES 118
Chapter Summary 119
Pronunciation Key 120
Chapter Exercises 120
Data Analytics 126
Problems 127
Cases 129
Practice Test 130
5 A Survey of Probability Concepts 132 Introduction 133
What is a Probability? 134
Approaches to Assigning Probabilities 136
Classical Probability 136 Empirical Probability 137 Subjective Probability 139
EXERCISES 140
Rules of Addition for Computing Probabilities 141
Special Rule of Addition 141 Complement Rule 143 The General Rule of Addition 144
EXERCISES 146
Rules of Multiplication to Calculate Probability 147
Special Rule of Multiplication 147 General Rule of Multiplication 148
Contingency Tables 150
Tree Diagrams 153
EXERCISES 155
Bayes’ Theorem 157
EXERCISES 161
Principles of Counting 161
The Multiplication Formula 161 The Permutation Formula 163 The Combination Formula 164
EXERCISES 166
Chapter Summary 167
Pronunciation Key 168
Chapter Exercises 168
Data Analytics 173
6 Discrete Probability Distributions 175 Introduction 176
What is a Probability Distribution? 176
Random Variables 178
Discrete Random Variable 179 Continuous Random Variable 179
The Mean, Variance, and Standard Deviation of a Discrete Probability Distribution 180
Mean 180 Variance and Standard Deviation 180
EXERCISES 182
Binomial Probability Distribution 184
How Is a Binomial Probability Computed? 185 Binomial Probability Tables 187
EXERCISES 190
Cumulative Binomial Probability Distributions 191
EXERCISES 193
Hypergeometric Probability Distribution 193
xxii CONTENTS
EXERCISES 197
Poisson Probability Distribution 197
EXERCISES 202
Chapter Summary 202
Chapter Exercises 203
Data Analytics 208
7 Continuous Probability Distributions 209 Introduction 210
The Family of Uniform Probability Distributions 210
EXERCISES 213
The Family of Normal Probability Distributions 214
The Standard Normal Probability Distribution 217
Applications of the Standard Normal Distribution 218 The Empirical Rule 218
EXERCISES 220
Finding Areas under the Normal Curve 221
EXERCISES 224
EXERCISES 226
EXERCISES 229
The Normal Approximation to the Binomial 229
Continuity Correction Factor 230 How to Apply the Correction Factor 232
EXERCISES 233
The Family of Exponential Distributions 234
EXERCISES 238
Chapter Summary 239
Chapter Exercises 240
Data Analytics 244
Problems 246
Cases 247
Practice Test 248
8 Sampling Methods and the Central Limit Theorem 250 Introduction 251
Sampling Methods 251
Reasons to Sample 251 Simple Random Sampling 252 Systematic Random Sampling 255 Stratified Random Sampling 255 Cluster Sampling 256
EXERCISES 257
Sampling “Error” 259
Sampling Distribution of the Sample Mean 261
EXERCISES 264
The Central Limit Theorem 265
EXERCISES 271
Using the Sampling Distribution of the Sample Mean 273
EXERCISES 275
Chapter Summary 275
Pronunciation Key 276
Chapter Exercises 276
Data Analytics 281
9 Estimation and Confidence Intervals 282 Introduction 283
Point Estimate for a Population Mean 283
Confidence Intervals for a Population Mean 284
Population Standard Deviation, Known σ 284 A Computer Simulation 289
EXERCISES 291
Population Standard Deviation, σ Unknown 292 EXERCISES 299
A Confidence Interval for a Population Proportion 300
EXERCISES 303
Choosing an Appropriate Sample Size 303
Sample Size to Estimate a Population Mean 304 Sample Size to Estimate a Population Proportion 305
EXERCISES 307
Finite-Population Correction Factor 307
EXERCISES 309
Chapter Summary 310
Chapter Exercises 311
Data Analytics 315
Problems 316
Cases 317
Practice Test 317
10 One-Sample Tests of Hypothesis 318 Introduction 319
What is Hypothesis Testing? 319
CONTENTS xxiii
Six-Step Procedure for Testing a Hypothesis 320
Step 1: State the Null Hypothesis (H0) and the Alternate Hypothesis (H1) 320 Step 2: Select a Level of Significance 321 Step 3: Select the Test Statistic 323 Step 4: Formulate the Decision Rule 323 Step 5: Make a Decision 324 Step 6: Interpret the Result 324
One-Tailed and Two-Tailed Hypothesis Tests 325
Hypothesis Testing for a Population Mean: Known Population Standard Deviation 327
A Two-Tailed Test 327 A One-Tailed Test 330
p-Value in Hypothesis Testing 331
EXERCISES 333
Hypothesis Testing for a Population Mean: Population Standard Deviation Unknown 334
EXERCISES 339
A Statistical Software Solution 340
EXERCISES 342
Type II Error 343
EXERCISES 346
Chapter Summary 347
Pronunciation Key 348
Chapter Exercises 348
Data Analytics 352
11 Two-Sample Tests of Hypothesis 353 Introduction 354
Two-Sample Tests of Hypothesis: Independent Samples 354
EXERCISES 359
Comparing Population Means with Unknown Population Standard Deviations 360
Two-Sample Pooled Test 360
EXERCISES 364
Unequal Population Standard Deviations 366
EXERCISES 369
Two-Sample Tests of Hypothesis: Dependent Samples 370
Comparing Dependent and Independent Samples 373
EXERCISES 375
Chapter Summary 377
Pronunciation Key 378
Chapter Exercises 378
Data Analytics 385
12 Analysis of Variance 386 Introduction 387
Comparing Two Population Variances 387
The F Distribution 387 Testing a Hypothesis of Equal Population Variances 388
EXERCISES 391
ANOVA: Analysis of Variance 392
ANOVA Assumptions 392 The ANOVA Test 394
EXERCISES 401
Inferences about Pairs of Treatment Means 402
EXERCISES 404
Two-Way Analysis of Variance 406
EXERCISES 411
Two-Way ANOVA with Interaction 412
Interaction Plots 412 Testing for Interaction 413 Hypothesis Tests for Interaction 415
EXERCISES 417
Chapter Summary 418
Pronunciation Key 420
Chapter Exercises 420
Data Analytics 429
Problems 431
Cases 433
Practice Test 434
13 Correlation and Linear Regression 436 Introduction 437
What is Correlation Analysis? 437
The Correlation Coefficient 440
EXERCISES 445
Testing the Significance of the Correlation Coefficient 447
EXERCISES 450
Regression Analysis 451
Least Squares Principle 451 Drawing the Regression Line 454
EXERCISES 457
Testing the Significance of the Slope 459
xxiv CONTENTS
EXERCISES 461
Evaluating a Regression Equation’s Ability to Predict 462
The Standard Error of Estimate 462 The Coefficient of Determination 463
EXERCISES 464
Relationships among the Correlation Coefficient, the Coefficient of Determination, and the Standard Error of Estimate 464
EXERCISES 466
Interval Estimates of Prediction 467
Assumptions Underlying Linear Regression 467 Constructing Confidence and Prediction Intervals 468
EXERCISES 471
Transforming Data 471
EXERCISES 474
Chapter Summary 475
Pronunciation Key 477
Chapter Exercises 477
Data Analytics 487
14 Multiple Regression Analysis 488 Introduction 489
Multiple Regression Analysis 489
EXERCISES 493
Evaluating a Multiple Regression Equation 495
The ANOVA Table 495 Multiple Standard Error of Estimate 496 Coefficient of Multiple Determination 497 Adjusted Coefficient of Determination 498
EXERCISES 499
Inferences in Multiple Linear Regression 499
Global Test: Testing the Multiple Regression Model 500 Evaluating Individual Regression Coefficients 502
EXERCISES 505
Evaluating the Assumptions of Multiple Regression 506
Linear Relationship 507 Variation in Residuals Same for Large and Small ŷ Values 508 Distribution of Residuals 509 Multicollinearity 509 Independent Observations 511
Qualitative Independent Variables 512
Regression Models with Interaction 515
Stepwise Regression 517
EXERCISES 519
Review of Multiple Regression 521
Chapter Summary 527
Pronunciation Key 528
Chapter Exercises 529
Data Analytics 539
Problems 541
Cases 542
Practice Test 543
15 Nonparametric Methods: NOMINAL LEVEL HYPOTHESIS TESTS 545
Introduction 546
Test a Hypothesis of a Population Proportion 546
EXERCISES 549
Two-Sample Tests about Proportions 550
EXERCISES 554
Goodness-of-Fit Tests: Comparing Observed and Expected Frequency Distributions 555
Hypothesis Test of Equal Expected Frequencies 555
EXERCISES 560
Hypothesis Test of Unequal Expected Frequencies 562
Limitations of Chi-Square 563
EXERCISES 565
Testing the Hypothesis That a Distribution is Normal 566
EXERCISES 569
Contingency Table Analysis 570
EXERCISES 573
Chapter Summary 574
Pronunciation Key 575
Chapter Exercises 576
Data Analytics 581
16 Nonparametric Methods: ANALYSIS OF ORDINAL DATA 582
Introduction 583
The Sign Test 583
CONTENTS xxv
EXERCISES 587
Using the Normal Approximation to the Binomial 588
EXERCISES 590
Testing a Hypothesis About a Median 590
EXERCISES 592
Wilcoxon Signed-Rank Test for Dependent Populations 592
EXERCISES 596
Wilcoxon Rank-Sum Test for Independent Populations 597
EXERCISES 601
Kruskal-Wallis Test: Analysis of Variance by Ranks 601
EXERCISES 605
Rank-Order Correlation 607
Testing the Significance of rs 609
EXERCISES 610
Chapter Summary 612
Pronunciation Key 613
Chapter Exercises 613
Data Analytics 616
Problems 618
Cases 619
Practice Test 619
17 Index Numbers 621 Introduction 622
Simple Index Numbers 622
Why Convert Data to Indexes? 625 Construction of Index Numbers 625
EXERCISES 627
Unweighted Indexes 628
Simple Average of the Price Indexes 628 Simple Aggregate Index 629
Weighted Indexes 629
Laspeyres Price Index 629 Paasche Price Index 631 Fisher’s Ideal Index 632
EXERCISES 633
Value Index 634
EXERCISES 635
Special-Purpose Indexes 636
Consumer Price Index 637 Producer Price Index 638 Dow Jones Industrial Average (DJIA) 638
EXERCISES 640
Consumer Price Index 640
Special Uses of the Consumer Price Index 641 Shifting the Base 644
EXERCISES 646
Chapter Summary 647
Chapter Exercises 648
Data Analytics 652
18 Time Series and Forecasting 653 Introduction 654
Components of a Time Series 654
Secular Trend 654 Cyclical Variation 655 Seasonal Variation 656 Irregular Variation 656
A Moving Average 657
Weighted Moving Average 660
EXERCISES 663
Linear Trend 663
Least Squares Method 665
EXERCISES 667
Nonlinear Trends 668
EXERCISES 669
Seasonal Variation 670
Determining a Seasonal Index 671
EXERCISES 676
Deseasonalizing Data 677
Using Deseasonalized Data to Forecast 678
EXERCISES 680
The Durbin-Watson Statistic 680
EXERCISES 686
Chapter Summary 686
Chapter Exercises 686
Data Analytics 693
Problems 695
Practice Test 696
19 Statistical Process Control and Quality Management 697 Introduction 698
A Brief History of Quality Control 698
Six Sigma 700
xxvi CONTENTS
Sources of Variation 701
Diagnostic Charts 702
Pareto Charts 702 Fishbone Diagrams 704
EXERCISES 705
Purpose and Types of Quality Control Charts 705
Control Charts for Variables 706 Range Charts 709
In-Control and Out-of-Control Situations 711
EXERCISES 712
Attribute Control Charts 713
p-Charts 713 c-Bar Charts 716
EXERCISES 718
Acceptance Sampling 719
EXERCISES 722
Chapter Summary 722
Pronunciation Key 723
Chapter Exercises 724
20 An Introduction to Decision Theory 728 Introduction 729
Elements of a Decision 729
Decision Making Under Conditions of Uncertainty 730
Payoff Table 730 Expected Payoff 731
EXERCISES 732
Opportunity Loss 733
EXERCISES 734
Expected Opportunity Loss 734
EXERCISES 735
Maximin, Maximax, and Minimax Regret Strategies 735
Value of Perfect Information 736
Sensitivity Analysis 737
EXERCISES 738
Decision Trees 739
Chapter Summary 740
Chapter Exercises 741
APPENDIXES 745
Appendix A: Data Sets 746
Appendix B: Tables 756
Appendix C: Software Commands 774
Appendix D: Answers to Odd-Numbered Chapter Exercises 785
Review Exercises 829
Solutions to Practice Tests 831
Appendix E: Answers to Self-Review 834
Glossary 847
Index 851
What is Statistics? 1
BEST BUY sells Fitbit wearable technology products that track a person’s physical activity and sleep quality. The Fitbit technology collects daily information on a person’s number of steps so that a person can track calories consumed. The information can be synced with a cell phone and displayed with a Fitbit app. Assume you know the daily number of Fitbit Flex 2 units sold last month at the Best Buy store in Collegeville, Pennsylvania. Describe a situation where the number of units sold is considered a sample. Illustrate a second situation where the number of units sold is considered a population. (See Exercise 11 and LO1-3.)
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO1-1 Explain why knowledge of statistics is important.
LO1-2 Define statistics and provide an example of how statistics is applied.
LO1-3 Differentiate between descriptive and inferential statistics.
LO1-4 Classify variables as qualitative or quantitative, and discrete or continuous.
LO1-5 Distinguish between nominal, ordinal, interval, and ratio levels of measurement.
LO1-6 List the values associated with the practice of statistics.
© Kelvin Wong/Shutterstock.com
2 CHAPTER 1
INTRODUCTION Suppose you work for a large company and your supervisor asks you to decide if a new version of a smartphone should be produced and sold. You start by thinking about the product’s innovations and new features. Then, you stop and realize the consequences of the decision. The product will need to make a profit so the pricing and the costs of production and distribution are all very important. The decision to introduce the product is based on many alternatives. So how will you know? Where do you start?
Without a long experience in the industry, beginning to develop an intelligence that will make you an expert is essential. You select three other people to work with and meet with them. The conversation focuses on what you need to know and what information and data you need. In your meeting, many questions are asked. How many competitors are already in the market? How are smartphones priced? What design features do com- petitors’ products have? What features does the market require? What do customers want in a smartphone? What do customers like about the existing products? The answers will be based on business intelligence consisting of data and information collected through customer surveys, engineering analysis, and market research. In the end, your presentation to support your decision regarding the introduction of a new smartphone is based on the statistics that you use to summarize and organize your data, the statistics that you use to compare the new product to existing products, and the statistics to esti- mate future sales, costs, and revenues. The statistics will be the focus of the conversa- tion that you will have with your supervisor about this very important decision.
As a decision maker, you will need to acquire and analyze data to support your decisions. The purpose of this text is to develop your knowledge of basic statistical techniques and methods and how to apply them to develop the business and personal intelligence that will help you make decisions.
WHY STUDY STATISTICS? If you look through your university catalogue, you will find that statistics is required for many college programs. As you investigate a future career in accounting, economics,
human resources, finance, business analytics, or other business area, you also will discover that statistics is required as part of these college pro- grams. So why is statistics a requirement in so many disciplines?
A major driver of the requirement for statistics knowledge is the tech- nologies available for capturing data. Examples include the technology that Google uses to track how Internet users access websites. As people use Google to search the Internet, Google records every search and then uses these data to sort and prioritize the results for future Internet searches. One recent estimate indicates that Google processes 20,000 terabytes of information per day. Big-box retailers like Target, Walmart, Kroger, and others scan every purchase and use the data to manage the distribution of products, to make decisions about marketing and sales, and to track daily and even hourly sales. Police departments collect and use data to provide city residents with maps that communicate informa- tion about crimes committed and their location. Every organization is col- lecting and using data to develop knowledge and intelligence that will help people make informed decisions, and to track the implementation of their decisions. The graphic to the left shows the amount of data gener- ated every minute (www.domo.com). A good working knowledge of sta- tistics is useful for summarizing and organizing data to provide information that is useful and supportive of decision making. Statistics is used to make valid comparisons and to predict the outcomes of decisions.
In summary, there are at least three reasons for studying statistics: (1) data are collected everywhere and require statistical knowledge to
LO1-1 Explain why knowledge of statistics is important.
© Gregor Schuster/Getty Images RF
WHAT IS STATISTICS? 3
make the information useful, (2) statistical techniques are used to make professional and personal decisions, and (3) no matter what your career, you will need a knowl- edge of statistics to understand the world and to be conversant in your career. An understanding of statistics and statistical method will help you make more effective personal and professional decisions.
WHAT IS MEANT BY STATISTICS? This question can be rephrased in two, subtly different ways: what are statistics and what is statistics? To answer the first question, a statistic is a number used to communi- cate a piece of information. Examples of statistics are:
• The inflation rate is 2%. • Your grade point average is 3.5. • The price of a new Tesla Model S sedan is $79,570.
Each of these statistics is a numerical fact and communicates a very limited piece of in- formation that is not very useful by itself. However, if we recognize that each of these statistics is part of a larger discussion, then the question “what is statistics” is applicable. Statistics is the set of knowledge and skills used to organize, summarize, and analyze data. The results of statistical analysis will start interesting conversations in the search for knowledge and intelligence that will help us make decisions. For example:
• The inflation rate for the calendar year was 0.7%. By applying statistics we could compare this year’s inflation rate to the past observations of inflation. Is it higher, lower, or about the same? Is there a trend of increasing or decreasing inflation? Is there a relationship between interest rates and government bonds?
• Your grade point average (GPA) is 3.5. By collecting data and applying statistics, you can determine the required GPA to be admitted to the Master of Business Administration program at the University of Chicago, Harvard, or the University of Michigan. You can determine the likelihood that you would be admitted to a partic- ular program. You may be interested in interviewing for a management position with Procter & Gamble. What GPA does Procter & Gamble require for college grad- uates with a bachelor’s degree? Is there a range of acceptable GPAs?
• You are budgeting for a new car. You would like to own an electric car with a small carbon footprint. The price for the Tesla Model S Sedan is $79,570. By collecting additional data and applying statistics, you can analyze the alternatives. For exam- ple, another choice is a hybrid car that runs on both gas and electricity such as a 2015 Toyota Prius. It can be purchased for about $28,659. Another hybrid, the Chevrolet Volt, costs $33,995. What are the differences in the cars’ specifications? What additional information can be collected and summarized so that you can make a good purchase decision?
Another example of using statistics to provide information to evaluate decisions is the distribution and market share of Frito-Lay products. Data are collected on each of the Frito-Lay product lines. These data include the market share and the pounds of product sold. Statistics is used to present this information in a bar chart in Chart 1–1. It clearly shows Frito-Lay’s dominance in the potato, corn, and tortilla chip markets. It also shows the absolute measure of pounds of each product line consumed in the United States.
These examples show that statistics is more than the presentation of numerical in- formation. Statistics is about collecting and processing information to create a conversa- tion, to stimulate additional questions, and to provide a basis for making decisions. Specifically, we define statistics as:
LO1-2 Define statistics and provide an example of how statistics is applied.
STATISTICS IN ACTION
A feature of our textbook is called Statistics in Action. Read each one carefully to get an appreciation of the wide application of statis- tics in management, economics, nursing, law enforcement, sports, and other disciplines. • In 2015, Forbes pub-
lished a list of the rich- est Americans. William Gates, founder of Microsoft Corporation, is the richest. His net worth is estimated at $76.0 billion. (www .forbes.com)
• In 2015, the four largest privately owned American companies, ranked by revenue, were Cargill, Koch Industries, Dell, and Albertsons. (www .forbes.com)
• In the United States, a typical high school grad- uate earns $668 per week, a typical college graduate with a bache- lor’s degree earns $1,101 per week, and a typical college graduate with a master’s degree earns $1,326 per week. (www.bls.gov/emp/ ep_chart_001.htm)
STATISTICS The science of collecting, organizing, presenting, analyzing, and interpreting data to assist in making more effective decisions.
4 CHAPTER 1
In this book, you will learn the basic techniques and applications of statistics that you can use to support your decisions, both personal and professional. To start, we will differentiate between descriptive and inferential statistics.
TYPES OF STATISTICS When we use statistics to generate information for decision making from data, we use either descriptive statistics or inferential statistics. Their application depends on the questions asked and the type of data available.
Descriptive Statistics 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, descriptive statistics can be used to organize data into a meaningful form. We define descriptive statistics as:
LO1-3 Differentiate between descriptive and inferential statistics.
DESCRIPTIVE STATISTICS Methods of organizing, summarizing, and presenting data in an informative way.
The following are examples that apply descriptive statistics to summarize a large amount of data and provide information that is easy to understand.
• There are a total of 46,837 miles of interstate highways in the United States. The interstate system represents only 1% of the nation’s total roads but carries more than 20% 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 mile in length. Alaska does not have any interstate highways, Texas has the most inter- state miles at 3,232, and New York has the most interstate routes with 28.
• The average person spent $133.91 on traditional Valentine’s Day merchandise in 2014. This is an increase of $2.94 from 2013. As in previous years, men spent more than twice the amount women spent on the holiday. The average man spent $108.38 to impress the people in his life while women only spent $48.41.
Statistical methods and techniques to generate descriptive statistics are presented in Chapters 2 and 4. These include organizing and summarizing data with frequency distributions and presenting frequency distributions with charts and graphs. In addition, statistical measures to summarize the characteristics of a distribution are discussed in Chapter 3.
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
WHAT IS STATISTICS? 5
Inferential Statistics Sometimes we must make decisions based on a limited set of data. For example, we would like to know the operating characteristics, such as fuel efficiency measured by miles per gallon, of sport utility vehicles (SUVs) currently in use. If we spent a lot of time, money, and effort, all the owners of SUVs could be surveyed. In this case, our goal would be to survey the population of SUV owners.
POPULATION The entire set of individuals or objects of interest or the measurements obtained from all individuals or objects of interest.
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.
However, based on inferential statistics, we can survey a limited number of SUV owners and collect a sample from the population.
Samples often are used to obtain reliable estimates of population parameters. (Sam- pling is discussed in Chapter 8.) In the process, we make trade-offs between the time, money, and effort to collect the data and the error of estimating a population parameter. The process of sampling SUVs is illustrated in the following graphic. In this example, we would like to know the mean or average SUV fuel efficiency. To estimate the mean of the population, six SUVs are sampled and the mean of their MPG is calculated.
Population All items
Sample Items selected
from the population
So, the sample of six SUVs represents evidence from the population that we use to reach an inference or conclusion about the average MPG for all SUVs. The process of sampling from a population with the objective of estimating properties of a population is called inferential statistics.
STATISTICS IN ACTION
Where did statistics get its start? In 1662 John Graunt published an article called “Natural and Political Obser- vations Made upon Bills of Mortality.” The author’s “observations” were the re- sult of a study and analysis of a weekly church publica- tion called “Bill of Mortality,” which listed births, christen- ings, and deaths and their causes. Graunt realized that the Bills of Mortality repre- sented only a fraction of all births and deaths in London. However, he used the data to reach broad conclusions or inferences about the im- pact of disease, such as the plague, on the general population. His logic is an example of statistical inference. His analysis and interpretation of the data are thought to mark the start of statistics.
6 CHAPTER 1
Inferential statistics is widely applied to learn something about a population in busi- ness, agriculture, politics, and government, as shown in the following examples:
• Television networks constantly monitor the popularity of their programs by hiring Nielsen and other organizations to sample the preferences of TV viewers. For example, 9.0% of a sample of households with TVs watched The Big Bang Theory during the week of November 2, 2015 (www.nielsen.com). These program ratings are used to make decisions about advertising rates and whether to continue or cancel a program.
• In 2015, a sample of U.S. Internal Revenue Service tax preparation volunteers were tested with three standard tax returns. The sample indicated that tax returns were completed with a 49% accuracy rate. In other words there were errors on about half of the returns. In this example, the statistics are used to make decisions about how to improve the accuracy rate by correcting the most common errors and improving the training of volunteers.
A feature of our text is self-review problems. There are a number of them inter- spersed throughout each chapter. The first self-review follows. Each self-review tests your comprehension of preceding material. The answer and method of solution are given in Appendix E. You can find the answer to the following self-review in 1–1 in Appendix E. We recommend that you solve each one and then check your answer.
The answers are in Appendix E.
The Atlanta-based advertising firm Brandon and Associates asked a sample of 1,960 con- sumers 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) Is this an example of descriptive statistics or inferential statistics? Explain. (b) What could Brandon and Associates report to Boston Market regarding acceptance of
the chicken dinner in the population?
TYPES OF VARIABLES There are two basic types of variables: (1) qualitative and (2) quantitative (see Chart 1–2). When an object or individual is observed and recorded as a nonnumeric characteristic, it is a qualitative variable or an attribute. Examples of qualitative variables are gender, bev- erage preference, type of vehicle owned, state of birth, and eye color. When a variable is qualitative, we usually count the number of observations for each category and determine
LO1-4 Classify variables as qualitative or quantitative, and discrete or continuous.
S E L F - R E V I E W 1–1
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
WHAT IS STATISTICS? 7
what percent are in each category. For example, if we observe the variable eye color, what percent of the population has blue eyes and what percent has brown eyes? If the variable is type of vehicle, what percent of the total number of cars sold last month were SUVs? Qualitative variables are often summarized in charts and bar graphs (Chapter 2).
When a variable can be reported numerically, it is called a quantitative variable. Examples of quantitative variables are the balance in your checking account, the num- ber of gigabytes of data used on your cell phone plan last month, the life of a car battery (such as 42 months), and the number of people employed by a company.
Quantitative variables are either discrete or continuous. Discrete variables can as- sume only certain values, and there are “gaps” between the values. Examples of dis- crete variables are the number of bedrooms in a house (1, 2, 3, 4, etc.), the number 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 students in each section. Notice that a home can have 3 or 4 bedrooms, but it cannot have 3.56 bedrooms. Thus, there is a “gap” between possible values. Typically, discrete variables are counted.
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 ounces of raisins in a box of raisin bran cereal 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.
LEVELS OF MEASUREMENT Data can be classified according to levels of measurement. The level of measurement determines how data should be summarized and presented. It also will indicate the type of statistical analysis that can be performed. Here are two examples of the relationship between measurement and how we apply statistics. There are six colors of candies in a bag of M&Ms. Suppose we assign brown a value of 1, yellow 2, blue 3, orange 4, green
5, and red 6. What kind of variable is the color of an M&M? It is a qualita- tive variable. Suppose someone summarizes M&M color by adding the assigned color values, divides the sum by the number of M&Ms, and re- ports that the mean color is 3.56. How do we interpret this statistic? You are correct in concluding that it has no meaning as a measure of M&M color. As a qualitative variable, we can only report the count and per- centage of each color in a bag of M&Ms. As a second example, in a high school track meet there are eight competitors in the 400-meter run. We report the order of finish and that the mean finish is 4.5. What does the mean finish tell us? Nothing! In both of these instances, we have not used the appropriate statistics for the level of measurement.
There are four levels of measurement: nominal, ordinal, interval, and ratio. The low- est, or the most primitive, measurement is the nominal level. The highest is the ratio level of measurement.
Nominal-Level Data For the nominal level of measurement, observations of a qualitative variable are mea- sured and recorded as labels or names. The labels or names can only be classified and counted. There is no particular order to the labels.
LO1-5 Distinguish between nominal, ordinal, interval, and ratio levels of measurement.
© Ron Buskirk/Alamy Stock Photo
NOMINAL LEVEL OF MEASUREMENT Data recorded at the nominal level of measurement is represented as labels or names. They have no order. They can only be classified and counted.
8 CHAPTER 1
The classification of the six colors of M&M milk chocolate candies is an example of the nominal level of measurement. We simply classify the candies by color. There is no natural order. That is, we could report the brown candies first, the orange first, or any of the other colors first. Recording the variable 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 data measured at the nominal level, we are limited to counting the number in each category of the variable. Often, we convert these counts to percentages. For example, a random sample of M&M candies reports the following percentages for each color:
Color Percent in a bag
Blue 24% Green 20% Orange 16% Yellow 14% Red 13% Brown 13%
To process the data for a variable measured at the nominal level, we often numer- ically code the labels or names. For example, if we are interested in measuring the home state for students at East Carolina University, we would assign a student’s home state of Alabama a code of 1, Alaska a code of 2, Arizona a 3, and so on. Using this procedure with an alphabetical listing of states, Wisconsin is coded 49 and Wyoming 50. Realize that the number assigned to each state is still a label or name. The reason we assign numerical codes is to facilitate counting the number of students from each state with statistical software. Note that assigning numbers to the states does not give us license to manipulate the codes as numerical information. Specifically, in this exam- ple, 1 + 2 = 3 corresponds to Alabama + Alaska = Arizona. Clearly, the nominal level of measurement does not permit any mathematical operation that has any valid interpretation.
Ordinal-Level Data The next higher level of measurement is the ordinal level. For this level of measure- ment a qualitative variable or attribute is either ranked or rated on a relative scale.
ORDINAL LEVEL OF MEASUREMENT Data recorded at the ordinal level of measurement is based on a relative ranking or rating of items based on a defined attribute or qualitative variable. Variables based on this level of measurement are only ranked or counted.
For example, many businesses make decisions about where to locate their facil- ities; in other words, where is the best place for their business? Business Facilities (www.businessfacilities.com) publishes a list of the top 10 states for the “best business climate.” The 2016 rankings are shown to the left. They are based on the evaluation of many different factors, including the cost of labor, business tax climate, quality of life, transportation infrastructure, educated workforce, and economic growth potential.
This is an example of an ordinal scale because the states are ranked in order of best to worst business climate. That is, we know the relative order of the states based
Best Business Climate
1. Florida 2. Utah 3. Texas 4. Georgia 5. Indiana 6. Tennessee 7. Nebraska 8. North Carolina 9. Virginia 10. Washington
WHAT IS STATISTICS? 9
on the attribute. For example, in 2016 Florida had the best business climate and Utah was second. Indiana was fifth, and that was better than Tennessee but not as good as Georgia. Notice we cannot say that Floridaʼs business climate is five times better than Indianaʼs business climate because the magnitude of the differences between the states is not known. To put it another way, we do not know if the magnitude of the differ- ence between Louisiana and Utah is the same as between Texas and Georgia.
Another example of the ordinal level measure is based on a scale that measures an attribute. This type of scale is used when students rate instructors on a variety of attri- butes. One attribute may be: “Overall, how do you rate the quality of instruction in this class?” A student’s response is recorded on a relative scale of inferior, poor, good, ex- cellent, and superior. An important characteristic of using a relative measurement scale is that we cannot distinguish the magnitude of the differences between groups. We do not know if the difference between “Superior” and “Good” is the same as the difference between “Poor” and “Inferior.”
Table 1–1 lists the frequencies of 60 student ratings of instructional quality for Pro- fessor James Brunner in an Introduction to Finance course. The data are summarized based on the order of the scale used to rate the instructor. That is, they are summarized by the number of students who indicated a rating of superior (6), good (26), and so on. We also can convert the frequencies to percentages. About 43.3% (26/60) of the stu- dents rated the instructor as good.
TABLE 1–1 Rating of a Finance Professor
Rating Frequency Percentage
Superior 6 10.0% Good 26 43.3% Average 16 26.7% Poor 9 15.0% Inferior 3 5.0%
Interval-Level Data The interval level of measurement is the next highest level. It includes all the character- istics of the ordinal level, but, in addition, the difference or interval between values is meaningful.
INTERVAL LEVEL OF MEASUREMENT For data recorded at the interval level of measurement, the interval or the distance between values is meaningful. The interval level of measurement is based on a scale with a known unit of measurement.
The Fahrenheit temperature scale is an example of the interval level of measurement. Suppose the high temperatures on three consecutive winter days in Boston are 28, 31, and 20 degrees Fahrenheit. These temperatures can be easily ranked, but we can also determine the interval or distance between temperatures. This is possible because 1 de- gree Fahrenheit represents a constant unit of measurement. That is, the distance between 10 and 15 degrees Fahrenheit is 5 degrees, and is the same as the 5-degree distance between 50 and 55 degrees Fahrenheit. It is also important to note that 0 is just a point on the scale. It does not represent the absence of the condition. The measurement of zero degrees Fahrenheit does not represent the absence of heat or cold. But by our own measurement scale, it is cold! A major limitation of a variable measured at the interval level is that we cannot make statements similar to 20 degrees Fahrenheit is twice as warm as 10 degrees Fahrenheit.
10 CHAPTER 1
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. woman’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 that as the size changes by two units (say from size 10 to size 12 or from size 24 to size 26), each of the mea- surements 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. More- over, 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 a size 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.
Ratio-Level Data Almost all quantitative variables are 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 and the ratio between two numbers are both meaningful.
RATIO LEVEL OF MEASUREMENT Data recorded at the ratio level of measurement are based on a scale with a known unit of measurement and a meaningful interpretation of zero on the scale.
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 also a good illustration. If you have zero dollars, then you have no money, and a wage of $50 per hour is two times the wage of $25 per hour. Weight also is measured at the ratio level of measurement. If a scale is correctly calibrated, then it will read 0 when nothing is on the scale. Further, something that weighs 1 pound is half as heavy as something that weighs 2 pounds.
Table 1–2 illustrates the ratio scale of measurement for the variable, annual income for four father-and-son combinations. Observe that the senior Lahey earns twice as much as his son. In the Rho family, the son makes twice as much as the father.
Name Father Son
Lahey $80,000 $ 40,000 Nale 90,000 30,000 Rho 60,000 120,000 Steele 75,000 130,000
TABLE 1–2 Father–Son Income Combinations
WHAT IS STATISTICS? 11
Chart 1–3 summarizes the major characteristics of the various levels of measure- ment. The level of measurement will determine the type of statistical methods that can be used to analyze a variable. Statistical methods to analyze variables measured on a nominal level are discussed in Chapter 15; methods for ordinal-level variables are dis- cussed in Chapter 16. Statistical methods to analyze variables measured on an interval or ratio level are presented in Chapters 9 through 14.
Levels of Measurement
RatioNominal Ordinal Interval
Meaningful 0 point and ratio between values
Data may only be classi�ed
Data are ranked Meaningful difference between values
• Jersey numbers of football players • Make of car
• Your rank in class • Team standings in the Southeastern Conference
• Temperature • Dress size
• Number of patients seen • Number of sales calls made • Distance to class
CHART 1–3 Summary and Examples of the Characteristics for Levels of Measurement
(a) The mean age of people who listen to talk radio is 42.1 years. What level of measure- ment is used to assess the variable age?
(b) In a survey of luxury-car owners, 8% of the U.S. population owned luxury cars. In California and Georgia, 14% of people owned luxury cars. Two variables are included in this information. What are they and how are they measured?
S E L F - R E V I E W 1–2
The answers to the odd-numbered exercises are in Appendix D.
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 student’s state of birth. e. A student’s academic class—that is, freshman, sophomore, junior, or senior. f. Number of hours students study per week.
2. Slate is a daily magazine on the Web. Its business activities can be described by a number of variables. What is the level of measurement for each of the following variables?
a. The number of hits on their website on Saturday between 8:00 am and 9:00 am. b. The departments, such as food and drink, politics, foreign policy, sports, etc. c. The number of weekly hits on the Sam’s Club ad. d. The number of years each employee has been employed with Slate.
3. On the Web, go to your favorite news source and find examples of each type of variable. Write a brief memo that lists the variables and describes them in terms of qualitative or quantitative, discrete or continuous, and the measurement level.
E X E R C I S E S
12 CHAPTER 1
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 mis- representation of business and financial information. 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 informa- tion was reported, the companies were worth much less than advertised. The result was many investors lost all or nearly all of the money they had invested.
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 practice 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 mislead 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” when analyzing and reporting findings and results.
As you progress through this text, we will highlight ethical issues in the collection, 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 informa- tion. 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 bias in a delib- erate attempt to mislead or misrepresent.
BASIC BUSINESS ANALYTICS A knowledge of statistics is necessary to support the increasing need for companies and organizations to apply business analytics. Business analytics is used to process and analyze data and information to support a story or narrative of a company’s business, such as “what makes us profitable,” “how will our customers respond to a change in marketing”? In addition to statistics, an ability to use computer software to summarize, organize, analyze, and present the findings of statistical analysis is essential. In this text, we will be using very elementary applications of business analytics using common and available computer software. Throughout our text, we will use Microsoft Excel and, oc- casionally, Minitab. Universities and colleges usually offer access to Microsoft Excel. Your computer already may be packaged with Microsoft Excel. If not, the Microsoft Office package with Excel often is sold at a reduced academic price through your uni- versity or college. In this text, we use Excel for the majority of the applications. We also use an Excel “Add-in” called MegaStat. If your instructor requires this package, it is avail- able at www.mhhe.com/megastat. This add-in gives Excel the capability to produce additional statistical reports. Occasionally, we use Minitab to illustrate an application. See www.minitab.com for further information. Minitab also offers discounted academic pricing. The 2016 version of Microsoft Excel supports the analyses in our text. However,
LO1-6 List the values associated with the practice of statistics.
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. People on welfare in Cook County (Chicago), Illinois. d. The 30 stocks that make up the Dow Jones Industrial Average.
WHAT IS STATISTICS? 13
earlier versions of Excel for Apple Mac computers do not have the necessary add-in. If you do not have Excel 2016 and are using an Apple Mac computer with Excel, you can download the free, trial version of Stat Plus at www.analystsoft.com. It is a statistical software package that will integrate with Excel for Mac computers.
The following example shows the application of Excel to perform a statistical summary. It refers to sales information from the Applewood Auto Group, a multi-location car sales and service company. The Applewood information has sales information for 180 vehicle sales. Each sale is described by several variables: the age of the buyer, whether the buyer is a re- peat customer, the location of the dealership for the sale, the type of vehicle sold, and the profit for the sale. The following shows Excel’s summary of statistics for the variable profit. The summary of profit shows the mean profit per vehicle was $1,843.17, the median profit was slightly more at $1,882.50, and profit ranged from $294 to $3,292.
Throughout the text, we will motivate the use of computer software to summarize, describe, and present information and data. The applications of Excel are supported by instructions so that you can learn how to apply Excel to do statistical analysis. The in- structions are presented in Appendix C of this text. These data and other data sets and files are available on the text’s student website, www.mhhe.com/lind17e.
C H A P T E R S U M M A R Y
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.
14 CHAPTER 1
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 be-
tween 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 constant 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.
C H A P T E R E X E R C I S E S
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 situations, would you collect information using a sample or a popula-
tion? Why? a. Statistics 201 is a course taught at a university. Professor Rauch 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 profit as a percent- age of revenue for the #1-ranked corporation in the Fortune 500 for each of the last 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.
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. Usually, the advertisers assume relatively short, popular songs 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 border 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 level of measurement for each of these three variables?
11. Best Buy sells Fitbit wearable technology products that track a person’s activity. For ex- ample, the Fitbit technology collects daily information on a person’s number of steps so that a person can track calories consumed. The information can be synced with a cell phone and displayed with a Fitbit app. Assume you know the daily number of Fitbit Flex
WHAT IS STATISTICS? 15
2 units sold last month at the Best Buy store in Collegeville, Pennsylvania. Describe a situation where the number of units sold is considered a sample. Illustrate a second sit- uation where the number of units sold is considered a population.
12. Using the concepts of sample and population, 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 video screens
Discrete Variable Continuous Variable
Qualitative
Quantitative a. Salary
Discrete Continuous
Nominal
Ordinal
Interval
Ratio a. Salary
14. Using data from such publications as the Statistical Abstract of the United States, Forbes, or any news source, give examples of variables measured with nominal, ordinal, interval, and ratio scales.
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 U.S. 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 Home Entertainment, Inc., recently began a “no-hassles” return policy. A sample of 500 customers who 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 Wall Street Journal’s website, www.wsj.com, reported the number of cars and light-duty trucks sold through October of 2014 and October of 2015. The top six- teen manufacturers are listed here. Sales data often is reported in this way to compare current sales to last year’s sales.
16 CHAPTER 1
a. Using computer software, compare the October 2015 sales to the October 2014 sales for each manufacturer by computing the difference. Make a list of the manufac- turers that increased sales compared to 2014; make a list of manufacturers that de- creased sales.
b. Using computer software, compare 2014 sales to 2015 sales for each manufacturer by computing the percentage change in sales. Make a list of the manufacturers in order of increasing percentage changes. Which manufacturers are in the top five in percentage change? Which manufacturers are in the bottom five in percentage change?
c. Using computer software, first sort the data using the 2015 year-to-date sales. Then, design a bar graph to illustrate the 2014 and 2015 year-to-date sales for the top 12 manufacturers. Also, design a bar graph to illustrate the percentage change for the top 12 manufacturers. Compare these two graphs and prepare brief written comments.
18. The following chart depicts the average amounts spent by consumers on holiday gifts.
Write a brief report summarizing the amounts spent during the holidays. Be sure to in- clude the total amount spent and the percent spent by each group.
19. The following chart depicts the earnings in billions of dollars for ExxonMobil for the pe- riod 2003 until 2014. Write a brief report discussing the earnings at ExxonMobil during
Year-to-Date Sales
Through October Through October Manufacturer 2015 2014
General Motors Corp. 2,562,840 2,434,707 Ford Motor Company 2,178,587 2,065,612 Toyota Motor Sales USA Inc. 2,071,446 1,975,368 Chrysler 1,814,268 1,687,313 American Honda Motor Co Inc. 1,320,217 1,281,777 Nissan North America Inc. 1,238,535 1,166,389 Hyundai Motor America 638,195 607,539 Kia Motors America Inc. 526,024 489,711 Subaru of America Inc. 480,331 418,497 Volkswagen of America Inc. 294,602 301,187 Mercedes-Benz 301,915 281,728 BMW of North America Inc. 279,395 267,193 Mazda Motor of America Inc. 267,158 259,751 Audi of America Inc. 165,103 146,133 Mitsubishi Motors N A, Inc. 80,683 64,564 Volvo 53,803 47,823
WHAT IS STATISTICS? 17
the period. Was one year higher than the others? Did the earnings increase, decrease, or stay the same over the period?
Year Earnings ($ billions)
A B C D E F G H
2003 14.5
2004 16.7
2005 24.3
2006 26.2
2007 26.5
2008 45.2
2009 19.3
2010 30.5
2011 41.1
2012 44.9
2013 32.6
2014
1 2
3
4
5
6
7
8
9
10
11
12
13 32.5
50
40
D o
lla rs
(b ill
io ns
)
30
20
10
20 03
20 04
20 05
20 06
20 07
20 08
20 09
20 10
20 11
20 12
20 13
20 14
0
Year
ExxonMobile Annual Earnings
D A T A A N A L Y T I C S
20. Refer to the North Valley Real Estate data, which report information on homes sold in the area last year. Consider the following variables: selling price, number of bed- rooms, township, and mortgage type. a. Which of the variables are qualitative and which are quantitative? b. How is each variable measured? Determine the level of measurement for each of the
variables. 21. Refer to the Baseball 2016 data, which report information on the 30 Major League
Baseball teams for the 2016 season. Consider the following variables: number of wins, payroll, 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 Lincolnville School District bus data, which report information on the school district’s bus fleet. a. Which of the variables are qualitative and which are quantitative? b. Determine the level of measurement for each variable.
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO2-1 Summarize qualitative variables with frequency and relative frequency tables.
LO2-2 Display a frequency table using a bar or pie chart.
LO2-3 Summarize quantitative variables with frequency and relative frequency distributions.
LO2-4 Display a frequency distribution using a histogram or frequency polygon.
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 LO2-3.)
Describing Data: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS,
AND GRAPHIC PRESENTATION2
© rido/123RF
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 19
INTRODUCTION The United States automobile retailing industry is highly competitive. It is dominated by megadealerships that own and operate 50 or more franchises, employ over 10,000 people, and generate several billion dollars in annual sales. Many of the top dealerships
are publicly owned with shares traded on the New York Stock Exchange or NASDAQ. In 2014, the largest megadealership was AutoNation (ticker symbol AN), followed by Penske Auto Group (PAG), Group 1 Automotive, Inc. (ticker symbol GPI), and the privately owned Van Tuyl Group.
These large corporations use statistics and analytics to summarize and analyze data and information to support their decisions. As an ex- ample, we will look at the Applewood Auto group. It owns four dealer- ships and sells a wide range of vehicles. These include the 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 Kane 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 Apple- wood dealerships? What type of vehicle did they purchase?
The Applewood Auto Group operates four dealerships:
• Tionesta Ford Lincoln sells Ford and Lincoln cars and trucks. • Olean Automotive Inc. has the Nissan franchise as well as the General Motors
brands of Chevrolet, Cadillac, and GMC Trucks. • Sheffield Motors Inc. sells Buick, GMC trucks, Hyundai, and Kia. • Kane 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 them 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 to the left. The variables collected include:
• Age—the age of the buyer at the time of the purchase. • Profit—the amount earned by the dealership on the sale of each
vehicle. • 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 (www.mhhe .com/lind17e) and in Appendix A.4 at the end of the text.
© Justin Sullivan/Getty Images
CONSTRUCTING FREQUENCY TABLES Recall from Chapter 1 that techniques used to describe a set of data are called descrip- tive statistics. Descriptive statistics organize data to show the general pattern of the data, to identify where values tend to concentrate, and to expose extreme or unusual data values. The first technique we discuss is a frequency table.
LO2-1 Summarize qualitative variables with frequency and relative frequency tables.
FREQUENCY TABLE A grouping of qualitative data into mutually exclusive and collectively exhaustive classes showing the number of observations in each class.
20 CHAPTER 2
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. Examples of qualitative data include political affiliation (Republican, Demo- crat, Independent, or other), state of birth (Alabama, . . . , Wyoming), and method of payment for a purchase at Barnes & Noble (cash, digital wallet, 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 num- ber of credit hours they are registered for this 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 pre- vious purchases are quantitative variables.
Suppose Ms. Ball wants to summarize last month’s sales by location. The first step is to sort the vehicles sold last month according to their location and then tally, or count, the number sold at each location of the four locations: Tionesta, Olean, Sheffield, or Kane. The four locations are used to develop a frequency table with four mutually exclusive (distinctive) classes. Mutually exclu- sive classes means that a particular vehicle can be assigned to only one class. In addition, the frequency table must be collectively exhaustive. That is every vehi- cle sold last month is accounted for in the table. If every vehicle is included in the frequency table, the table will be collectively exhaustive and the total number of vehicles will be 180. How do we obtain these counts? Excel provides a tool called a Pivot Table that will quickly and accurately establish the four classes and do the counting. The Excel results follow in Table 2–1. The table shows a total of 180 vehicles and, of the 180 vehicles, 52 were sold at Kane Motors. © Steve Cole/Getty Images RF
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
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 relationship between a class frequency and the total number of observations. In the vehicle sales ex- ample, we may want to know the percentage of total cars sold at each of the four locations. To convert a frequency table to a relative frequency table, each of the class frequencies is divided by the total number of observations. Again, this is easily accomplished using Excel. 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.
TABLE 2–2 Relative Frequency Table of Vehicles Sold by Location Last Month at Applewood Auto Group
Location Number of Cars Relative Frequency Found by
Kane 52 .289 52/180 Olean 40 .222 40/180 Sheffield 45 .250 45/180 Tionesta 43 .239 43/180
Total 180 1.000
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 21
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 in- terest 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.
LO2-2 Display a frequency table using a bar or pie chart.
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.
PIE CHART A chart that shows the proportion or percentage that each class represents of the total number of frequencies.
We use the Applewood Auto Group data as an example (Chart 2–1). The variables of interest are the location where the vehicle was sold and the number of vehicles sold at each location. We label the horizontal axis with the four locations and scale the verti- cal axis with the number sold. The variable location is of nominal scale, so the order of the locations on the horizontal axis does not matter. In Chart 2–1, the locations are listed alphabetically. The locations could also be in order of decreasing or increasing frequencies.
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.
Nu m
be r o
f V eh
ic le
s So
ld
50
40
30
20
10
0 Kane Olean
Location
Shef�eld Tionesta
CHART 2–1 Number of Vehicles Sold by Location
Another useful type of chart for depicting qualitative information is a pie chart.
We explain the details of constructing a pie chart using the information in Table 2–3, which shows the frequency and percent of cars sold by the Applewood Auto Group for each vehicle type.
22 CHAPTER 2
The first step to develop a pie chart is to mark the percentages 0, 5, 10, 15, and so on evenly around the circumference of a circle (see Chart 2–2). To plot the 40% of total sales represented by sedans, draw a line from the center of the circle to 0 and another line from the center of the circle to 40%. The area in this “slice” represents the number of sedans sold as a percentage of the total sales. Next, add the SUV’s percentage of total sales, 30%, to the sedan’s percentage of total sales, 40%. The result is 70%. Draw a line from the center of the circle to 70%, so the area between 40 and 70 shows the sales of SUVs as a percentage of total sales. Continuing, add the 15% of total sales for compact vehicles, which gives us a total of 85%. Draw a line from the center of the circle to 85, so the “slice” between 70% and 85% represents the number of compact vehicles sold as a percentage of the total sales. The remaining 10% for truck sales and 5% for hybrid sales are added to the chart using the same method.
Vehicle Type Number Sold Percent Sold
Sedan 72 40 SUV 54 30 Compact 27 15 Truck 18 10 Hybrid 9 5
Total 180 100
TABLE 2–3 Vehicle Sales by Type at Applewood Auto Group
25%
50%
70%
85%
95% 0%
40%
75%
Hybrid
Truck
Sedan
SUV
Compact
CHART 2–2 Pie Chart of Vehicles by Type
Because each slice of the pie represents the relative frequency of each vehicle type as a percentage of the total sales, we can easily compare them:
• The largest percentage of sales is for sedans. • Sedans and SUVs together account for 70% of vehicle sales. • Hybrids account for 5% of vehicle sales, in spite of being on the market for only a
few years.
We can use Excel software to quickly count the number of cars for each vehicle type and create the frequency table, bar chart, and pie chart shown in the following summary. The Excel tool is called a Pivot Table. The instructions to produce these de- scriptive statistics and charts are given in Appendix C.
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 23
Pie and bar charts both serve to illustrate frequency and relative frequency ta- bles. When is a pie chart preferred to a bar chart? In most cases, pie charts are used to show and compare the relative differences in the percentage of observations for each value or class of a qualitative variable. Bar charts are preferred when the goal is to compare the number or frequency of observations for each value or class of a qualitative variable. The following Example/Solution shows another application of bar and pie charts.
E X A M P L E
SkiLodges.com is test marketing its new website and is interested in how easy its website design is to navigate. It randomly selected 200 regular Internet users and asked them to perform a search task on the website. Each person was asked to rate the relative ease of navigation as poor, good, excellent, or awesome. The re- sults are shown in the following table:
Awesome 102 Excellent 58 Good 30 Poor 10
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.
S O L U T I O N
The data are measured on an ordinal scale. That is, the scale is ranked in relative ease of navigation when moving from “awesome” to “poor.” 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 relative frequency and the horizontal scale shows the values of the ease-of- navigation variable.
24 CHAPTER 2
A pie chart can also be used to graph these data. The pie chart emphasizes that more than half of the respondents rate the relative ease of using the website awesome.
Re la
tiv e
Fr eq
ue nc
y %
60
50
40
30
20
10
0 PoorGoodExcellentAwesome
Ease of Navigation of SkiLodges.com website
Ease of Navigation
Beverage Number
Cola-Plus 40 Coca-Cola 25 Pepsi 20 Lemon-Lime 15
Total 100
The answers are in Appendix E.
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 prefer most. The results are shown in the following table:
S E L F - R E V I E W 2–1
Poor 5%
Ease of Navigation of SkiLodges.com website
Good 15%
Awesome 51% Excellent
29%
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 25
(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.
The answers to the odd-numbered exercises are at the end of the book in Appendix D.
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 con- sumer 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. One hundred liked winter best, 300 liked spring, 400 liked summer, and 200 liked fall. Develop a frequency table and a relative frequency table to summarize this information.
4. Two thousand frequent business travelers are asked which midwestern city they prefer: Indianapolis, Saint Louis, Chicago, or Milwaukee. One hundred liked India- napolis best, 450 liked Saint Louis, 1,300 liked Chicago, and the remainder pre- ferred Milwaukee. Develop a frequency table and a relative frequency table to summarize this information.
5. Wellstone Inc. produces and markets replacement covers for cell phones in five different colors: bright white, metallic black, magnetic lime, tangerine orange, and fusion red. To estimate the demand for each color, the company 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:
E X E R C I S E S
Bright white 130 Metallic black 104 Magnetic lime 325 Tangerine orange 455 Fusion red 286
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 com- panies were:
Fourth-Quarter Sales Company ($ 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
The consultant wants to include a chart in his report comparing the sales of the six companies. Use a bar chart to compare the fourth-quarter sales of these corpora- tions and write a brief report summarizing the bar chart.
26 CHAPTER 2
CONSTRUCTING FREQUENCY DISTRIBUTIONS 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 two qualitative variables: the location of the sale and the type of vehicle sold. We created frequency and relative frequency tables and depicted the results in bar and pie charts.
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 previ- ous purchases. Suppose Ms. Ball wants to summarize last month’s sales by profit earned for each vehicle. We can describe profit using a frequency distribution.
LO2-3 Summarize quantitative variables with frequency and relative frequency distributions.
FREQUENCY DISTRIBUTION A grouping of quantitative data into mutually exclusive and collectively exhaustive classes showing the number of observations in each class.
How do we develop a frequency distribution? The following example shows the steps to construct a frequency distribution. Remember, our goal is to construct tables, charts, and graphs that will quickly summarize the data by showing the location, extreme values, and shape of the data’s distribution.
TABLE 2–4 Profit on Vehicles Sold Last Month by the Applewood Auto Group Maximum
Minimum
$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 1,760 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 1,762 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
S O L U T I O N
To begin, we need the profits for each of the 180 vehicle sales listed in Table 2–4. This information is called raw or ungrouped data because it is simply a listing
E X A M P L E
Ms. Kathryn Ball of the Applewood Auto Group wants to summarize the quantitative variable profit with a frequency distribution and display the distribution with charts and graphs. With this information, Ms. Ball can easily answer the following ques- tions: What is the typical profit on each sale? What is the largest or maximum profit on any sale? What is the smallest or minimum profit on any sale? Around what value do the profits tend to cluster?
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 27
of the individual, observed profits. It is possible to search the list and find the smallest or minimum profit ($294) and the largest or maximum profit ($3,292), but that is about all. It is difficult to determine a typical profit or to visualize where the profits tend to cluster. The raw data are more easily interpreted if we summarize the data with a frequency distribution. The steps to create this frequency distribu- tion follow.
Step 1: Decide on the number of classes. 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 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. Generally, the class interval is the same for all classes. The classes all taken together must cover at least the distance from the minimum value in the data up to the max- imum value. Expressing these words in a formula:
i ≥ Maximum Value − Minimum Value
k where i is the class interval, and k is the number of classes.
For the Applewood Auto Group, the minimum value is $294 and the maximum value is $3,292. If we need 8 classes, the interval should be:
i ≥ Maximum Value − Minimum Value
k =
$3,292 − $294 8
= $374.75
In practice, this interval size is usually rounded up to some conve- nient number, such as a multiple of 10 or 100. The value of $400 is a reasonable choice.
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. 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 always round the class interval up to get a conve- nient class size, we cover a larger than necessary range. For ex- ample, using 8 classes with an interval 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 range (Maximum − Minimum), it is natural to put ap- proximately equal amounts of the excess in each of the two tails. Of course, we also should select convenient class limits. A guide- line 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 these data.
28 CHAPTER 2
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
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 ||||
Step 4: Tally the vehicle profit into the classes and determine the number of observations in each class. 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 similar manner. When all the profits are tallied, the table would appear as:
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. There- fore, the class frequency in the first class is 8 and the class frequency in the second class is 11. There are a total of 180 observations in the entire set of data. So the sum of all the frequencies should be equal to 180. The results of the frequency distribution are in Table 2–5.
Now that we have organized the data into a frequency distribution (see Table 2–5), we can summarize the profits of the vehicles for the Applewood Auto Group. Observe the following:
1. The profits from vehicle sales range between $200 and $3,400. 2. The vehicle profits are classified using a class interval of $400. The class inter-
val is determined by subtracting consecutive lower or upper class limits. For
TABLE 2–5 Frequency Distribution of Profit for Vehicles Sold Last Month at 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
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 29
example, the lower limit of the first class is $200, and the lower limit of the second class is $600. The difference is the class interval of $400.
3. The profits are concentrated between $1,000 and $3,000. The profit on 157 vehicles, or 87%, was within this range.
4. For each class, we can determine the typical profit or class midpoint. It is half- way between the lower or upper limits of two consecutive classes. It is com- puted by adding the lower or upper limits of consecutive classes and dividing by 2. Referring to Table 2–5, 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 midpoint best represents, or is typical of, the profits of the vehi- cles in that class. Applewood sold 8 vehicles with a typical profit of $400.
5. The largest concentration, or highest frequency, of vehicles sold is in the $1,800 up to $2,200 class. There are 45 vehicles in this class. The class midpoint is $2,000. So we say that the typical profit in the class with the highest frequency is $2,000.
By presenting this information to Ms. Ball, we give her a clear picture of the distribu- tion 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 minimum profit for any vehicle sold is $294 or that the maximum profit was $3,292. However, the lower limit of the first class and the upper limit of the last class convey essen- tially the same meaning. Likely, Ms. Ball will make the same judgment if she knows the smallest profit is about $200 that she will if she knows the exact profit is $292. The advantages of summarizing the 180 profits into a more understandable and organized form more than offset this disadvantage.
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
TABLE 2–6 Adjusted Gross Income for Individuals Filing Income Tax Returns
When we summarize raw data with frequency distributions, equal class intervals are pre- ferred. However, in certain situations unequal class intervals may be necessary to avoid a large number of classes with very small frequencies. Such is the case in Table 2–6. The U.S. Internal Revenue Service uses unequal-sized class intervals for adjusted gross income on individual tax returns to summarize the number of individual tax returns. If we use our method to find equal class intervals, the 2k rule results in 25 classes, and
STATISTICS IN ACTION
In 1788, James Madison, John Jay, and Alexander Hamilton anonymously published a series of essays entitled The Federalist. These Federalist papers were an attempt to convince the people 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 re- mained contested. Through the use of statistical analysis, and particularly studying the frequency distributions of various words, we can now conclude that James Madison is the likely author of the 12 papers. In fact, the statistical evidence that Madison is the author is overwhelming.
30 CHAPTER 2
a class interval of $400,000, assuming $0 and $10,000,000 as the minimum and maximum values for adjusted gross income. Using equal class intervals, the first 13 classes in Table 2–6 would be combined into one class of about 99.9% of all tax returns and 24 classes for the 0.1% of the returns with an adjusted gross income above $400,000. Using equal class inter- vals does not provide a good understanding of the raw data. In this case, good judgment in the use of unequal class intervals, as demonstrated in Table 2–6, is required to show the distribution of the number of tax returns filed, especially for incomes under $500,000.
In the first quarter of last year, the 11 members of the sales staff at Master Chemical Company earned the following commissions:
$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 distribu-
tion. What is the largest concentration of commissions earned? What is the smallest, and the largest? What is the typical amount earned?
Relative Frequency Distribution It may be desirable, as we did earlier with qualitative data, to convert class frequencies 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 percentage of the vehi- cle profits are in the $1,000 up to $1,400 class. To convert a frequency distribution to a relative frequency distribution, each of the class frequencies is divided by the total num- ber of observations. From the distribution of vehicle profits, Table 2–5, the relative fre- quency for the $1,000 up to $1,400 class is 0.128, found by dividing 23 by 180. That is, profit on 12.8% of the vehicles sold is between $1,000 and $1,400. The relative fre- quencies for the remaining classes are shown in Table 2–7.
S E L F - R E V I E W 2–2
TABLE 2–7 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
There are many software packages that perform statistical calculations. Throughout this text, we will show the output from Microsoft Excel, MegaStat (a Microsoft Excel add-in), and Minitab (a statistical software package). Because Excel is most readily available, it is used most frequently.
Within the earlier Graphic Presentation of Qualitative Data section, we used the Pivot Table tool in Excel to create a frequency table. To create the table to the left, we use the same Excel tool to
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 31
compute frequency and relative frequency distributions for the profit variable in the Applewood Auto Group data. The necessary steps are given in the Software Commands section in Appendix C.
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. Listed below is the sorted distance of each of the 73 home runs.
S E L F - R E V I E W 2–3
(a) For this data, show that seven classes would be used to create a frequency distribution using the 2k rule.
(b) Show that a class interval of 30 would summarize the data in seven classes. (c) Construct frequency and relative frequency distributions for the data with
seven classes and a class interval of 30. Start the first class with a lower limit of 300.
(d) How many home runs traveled a distance of 360 up to 390 feet? (e) What percentage of the home runs traveled a distance of 360 up to 390 feet? (f) What percentage of the home runs traveled a distance of 390 feet or more?
7. A set of data consists of 38 observations. How many classes would you recom- mend for the frequency distribution?
8. A set of data consists of 45 observations between $0 and $29. What size would you recommend 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 minimum value is 42 and the maximum value 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.
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
frequency distribution. e. Comment on the shape of the distribution.
E X E R C I S E S This icon indicates that the data are available at the text website: www.mhhe.com/ Lind17e. 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
320 320 347 350 360 360 360 361 365 370 370 375 375 375 375 380 380 380 380 380 380 390 390 391 394 396 400 400 400 400 405 410 410 410 410 410 410 410 410 410 410 410 411 415 415 416 417 417 420 420 420 420 420 420 420 420 429 430 430 430 430 430 435 435 436 440 440 440 440 440 450 480 488
32 CHAPTER 2
The data are 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 number of oil changes into a frequency distribution. e. Comment on the shape of the frequency distribution. Also determine the relative
frequency distribution.
13. The manager of the BiLo Supermarket in Mt. Pleasant, Rhode Island, gathered the following information on the number of times a customer visits the store during a month. The responses of 51 customers were:
65 98 55 62 79 59 51 90 72 56 70 62 66 80 94 79 63 73 71 85
12. The Quick Change Oil Company has a number of outlets in the metropolitan Seat- tle area. The daily number of oil changes at the Oak Street outlet in the past 20 days are:
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, organize 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 of money spent per day on food and drink by families who visit the amuse- ment park. 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.
GRAPHIC PRESENTATION OF A DISTRIBUTION Sales managers, stock analysts, hospital administrators, and other busy executives of- ten 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 histogram, the fre- quency 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
LO2-4 Display a distribution using a histogram or frequency polygon.
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 33
horizontal axis and the class frequencies on the vertical axis. The class frequencies are represented by the heights of the bars. However, there is one important differ- ence based on the nature of the data. Quantitative data are usually measured 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 continu- ous nature of the data.
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.
E X A M P L E
Below is the frequency distribution of the profits on vehicle sales last month at the Applewood Auto Group.
Construct a histogram. What observations can you reach based on the information presented in the histogram?
S O L U T I O N
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.
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
200 600 1,000 1,400
32
24
16
8 8
11
23
Nu m
be r o
f V eh
ic le
s (c
la ss
fr eq
ue nc
y)
Pro�t $
CHART 2–3 Construction of a Histogram
34 CHAPTER 2
From Chart 2–3 we note the profit on eight vehicles was $200 up to $600. There- fore, the height of the column for that class is 8. There are 11 vehicle 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 profit, plotted on the horizontal axis, is a continuous variable. In a bar chart, the scale of measurement is usually nominal and the vertical bars are separated. This is an important distinction be- tween the histogram and the bar chart.
We can make the following statements using Chart 2–4. They are the same as the observations based on Table 2–5.
1. The profits from vehicle sales range between $200 and $3,400. 2. The vehicle profits are classified using a class interval of $400. The class inter-
val is determined by subtracting consecutive lower or upper class limits. For example, the lower limit of the first class is $200, and the lower limit of the second class is $600. The difference is the class interval or $400.
3. The profits are concentrated between $1,000 and $3,000. The profit on 157 vehicles, or 87%, was within this range.
4. For each class, we can determine the typical profit or class midpoint. It is halfway between the lower or upper limits of two consecutive classes. It is computed by adding the lower or upper limits of consecutive classes and dividing by 2. Refer- ring to Chart 2–4, 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 best represents, or is typical of, the profits of the vehicles in that class. Applewood sold 8 vehicles with a typical profit of $400.
5. The largest concentration, or highest frequency of vehicles sold, is in the $1,800 up to $2,200 class. There are 45 vehicles in this class. The class midpoint is $2,000. So we say that the typical profit in the class with the highest frequency is $2,000.
Thus, the histogram provides an easily interpreted visual representation of a frequency 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 use the relative frequencies of Table 2–7, the result is 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. The Excel commands to create Chart 2–4 are given in Appendix C.
20 0–
60 0
60 0–
1,0 00
1,0 00
–1 ,40
0
1,4 00
–1 ,80
0
1,8 00
–2 ,20
0
2,2 00
–2 ,60
0
2,6 00
–3 ,00
0
3,0 00
–3 ,40
0
10
0
30
20
Pro�t
11
23
38
45
32
19
4 8
40
v
Fr eq
ue nc
y
CHART 2–4 Histogram of the Profit on 180 Vehicles Sold at the Applewood Auto Group
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 35
Frequency Polygon A frequency polygon also shows the shape of a distribution and is similar to a histo- gram. It consists of line segments connecting the points formed by the intersections of the class midpoints and the class frequencies. The construction of a frequency polygon is illustrated in Chart 2–5. We use the profits from the cars sold last month at the Apple- wood 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 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.
STATISTICS IN ACTION
Florence Nightingale is known as the founder of the nursing profession. However, she also saved many lives by using statisti- cal analysis. When she encountered an unsanitary condition or an undersup- plied hospital, she improved the conditions and then used statistical data to document the improve- ment. Thus, she was able to convince others of the need for medical reform, particularly in the area of sanitation. She developed original graphs to demon- strate that, during the Crimean War, more soldiers died from unsanitary condi- tions than were killed in combat.
Fr eq
ue nc
y
8
24
40
48
16
4000
Pro�t $
32
800 1,200 1,600 2,000 2,400 2,800 3,200 3,600
CHART 2–5 Frequency Polygon of Profit on 180 Vehicles Sold at Applewood Auto Group
As noted previously, the $200 up to $600 class is represented by the midpoint $400. To construct a frequency polygon, move horizontally on the graph to the mid- point, $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 midpoint ($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 rectangular bar representing
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
36 CHAPTER 2
8
24
40
48
56
16
4000
Pro�t $
32
Fr eq
ue nc
y
800 1,200 1,600 2,000 2,400 2,800 3,200 3,600
Fowler Motors Applewood
CHART 2–6 Distribution of Profit at Applewood Auto Group and Fowler Motors
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 frequency distributions. Sup- pose 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 Applewood and about $2,400 for Fowler.
• There is less variation or dispersion in the profits at Fowler Motors than at Apple- wood. 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: $3,600.
The total number of cars sold at the two dealerships is about the same, so a direct comparison is possible. If the difference in the total number of cars sold is large, then converting the frequencies to relative frequencies and then plotting the two distribu- tions would allow a clearer comparison.
The annual imports of a selected group of electronic suppliers are shown in the following frequency distribution.
S E L F - R E V I E W 2–4
Imports ($ millions) Number of Suppliers
2 up to 5 6 5 up to 8 13 8 up to 11 20 11 up to 14 10 14 up to 17 1
(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).
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 37
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 fol- lowing chart shows the number of packages shipped per day for the last 100 days. For example, the first class shows that there were 5 days when the number of pack- ages shipped was 0 up to 5.
Fr eq
ue nc
y Number of Packages
10
0 5 10 15 20 25 30 35
20
30
13
28 23
18
10 35
a. What is this chart called? b. What is the total number of packages shipped? c. What is the class interval? d. What is the number of packages shipped in the 10 up to 15 class? e. What is the relative frequency of packages shipped in 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.
0
10
20
30
2 4 6 8 10 12
Fr eq
ue nc
y
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. What is the class interval? d. 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.
E X E R C I S E S
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
38 CHAPTER 2
Cumulative Distributions Consider once again the distribution of the profits on vehicles sold by the Applewood Auto Group. Suppose we were interested in the number of vehicles that sold for a profit of less than $1,400. These values can be approximated by developing a cumulative frequency distribution and portraying it graphically in a cumulative frequency polygon. Or, suppose we were interested in the profit earned on the lowest-selling 40% of the ve- hicles. These values can be approximated by developing a cumulative relative frequency distribution and portraying it graphically in a cumulative relative frequency polygon.
a. How many employees were studied? b. What is the midpoint of the first class? c. Construct a histogram. 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. 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.
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.
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
E X A M P L E
The frequency distribution of the profits earned at Applewood Auto Group is repeated from Table 2–5.
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
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 39
Construct a cumulative frequency polygon to answer the following question: sixty of the vehicles earned a profit of less than what amount? Construct a cumulative relative frequency polygon to answer this question: seventy-five percent of the vehicles sold earned a profit of less than what amount?
S O L U T I O N
As the names imply, a cumulative frequency distribution and a cumulative fre- quency polygon require cumulative frequencies. To construct a cumulative fre- quency distribution, refer to the preceding table and note that there were eight vehicles in which 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 2–8.)
TABLE 2–8 Cumulative Frequency Distribution for Profit on Vehicles Sold Last Month at Applewood Auto Group
Profit Cumulative Frequency Found by
Less than $ 600 8 8 Less than 1,000 19 8 + 11 Less than 1,400 42 8 + 11 + 23 Less than 1,800 80 8 + 11 + 23 + 38 Less than 2,200 125 8 + 11 + 23 + 38 + 45 Less than 2,600 157 8 + 11 + 23 + 38 + 45 + 32 Less than 3,000 176 8 + 11 + 23 + 38 + 45 + 32 + 19 Less than 3,400 180 8 + 11 + 23 + 38 + 45 + 32 + 19 + 4
TABLE 2–9 Cumulative Relative Frequency Distribution for Profit on Vehicles Sold Last Month at Applewood Auto Group
Profit Cumulative Frequency Cumulative Relative Frequency
Less than $ 600 8 8/180 = 0.044 = 4.4% Less than $ 1,000 19 19/180 = 0.106 = 10.6% Less than $ 1,400 42 42/180 = 0.233 = 23.3% Less than $ 1,800 80 80/180 = 0.444 = 44.4% Less than $2,200 125 125/180 = 0.694 = 69.4% Less than $2,600 157 157/180 = 0.872 = 87.2% Less than $3,000 176 176/180 = 0.978 = 97.8% Less than $3,400 180 180/180 = 1.000 = 100%
To construct a cumulative relative frequency distribution, we divide the cumulative frequencies by the total number of observations, 180. As shown in Table 2-9, the cumulative relative frequency of the fourth class is 80/180 = 44%. This means that 44% of the vehicles sold for less than $1,800.
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 right in terms of cumulative relative frequencies. In the Applewood Auto Group,
40 CHAPTER 2
the vertical axis on the left is labeled from 0 to 180 and on the right from 0 to 100%. Note, as an example, that 50% on the right axis should be opposite 90 vehicles on the left axis.
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 Chart 2–7.
We should point out that the shape of the distribution is the same if we use cumulative relative frequencies instead of the cumulative frequencies. The only difference is that the vertical axis is scaled in percentages. In the following charts, a percentage scale is added to the right side of the graphs to help answer ques- tions about cumulative relative frequencies.
200 600 1,000 1,400 1,800 2,200 2,600 3,000 3,400
Nu m
be r o
f V eh
ic le
s So
ld
Pe rc
en t o
f V eh
ic le
s So
ld Pro�t $
100
75
50
25
0
20
40
60
80
100
120
140
160
180
CHART 2–7 Cumulative Frequency Polygon for Profit on Vehicles Sold Last Month at Applewood Auto Group
Using Chart 2–7 to find the amount of profit on 75% of the cars sold, draw a hori- zontal line from the 75% mark on the right-hand vertical axis over to the polygon, 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% of the vehicles sold earned a profit of $2,300 or less for the Applewood group.
To find the highest profit earned on 60 of the 180 vehicles, we use Chart 2–7 to locate the value of 60 on the left-hand vertical axis. Next, we draw a horizontal line from the value of 60 to the polygon and then drop down to the X-axis and read the profit. It is about $1,600, so we estimate that 60 of the vehicles sold for a profit of less than $1,600. We can also make estimates of the percentage of vehicles that sold for less than a particular amount. To explain, suppose we want to estimate the percentage of vehicles that sold for a profit of less than $2,000. We begin by locat- ing the value of $2,000 on the X-axis, move vertically to the polygon, and then horizontally to the vertical axis on the right. The value is about 56%, so we conclude 56% of the vehicles sold for a profit of less than $2,000.
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 41
A sample of the hourly wages of 15 employees at Home Depot in Brunswick, Georgia, was organized into the following table.
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
(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 less
than $11 per hour?
S E L F - R E V I E W 2–5
19. The following cumulative frequency and the cumulative relative frequency polygon for the distribution of hourly wages of a sample of certified welders in the Atlanta, Georgia, area is shown in the graph.
Fr eq
ue nc
y
Hourly Wage
Pe rc
en t
0 5 10 15 20 25 30
100
75
50
25
40
30
20
10
a. How many welders were studied? b. What is the class interval? c. About how many welders earn less than $10.00 per hour? d. About 75% 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 cumulative frequency and the cumulative relative frequency polygon for a dis- tribution of selling prices ($000) of houses sold in the Billings, Montana, area is shown in the graph.
Fr eq
ue nc
y
Pe rc
en t
200
150
100
50
100
75
50
25
Selling Price ($000)
500 100 150 200 250 350300
E X E R C I S E S
42 CHAPTER 2
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% 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 Inc. 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 relative frequencies, about 75% of the employees
accumulated how many miles or less? 22. The frequency distribution of order lead time of the retailer from Exercise 18 is
repeated below.
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
a. How many orders were filled in less than 10 days? In less than 15 days? b. Convert the frequency distribution to cumulative frequency and cumulative rela-
tive frequency distributions. c. Develop a cumulative frequency polygon. d. About 60% of the orders were filled in less than how many days?
C H A P T E R S U M M A R Y
I. A frequency table is a grouping of qualitative data into mutually exclusive and collectively exhaustive classes showing 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
observations. V. A frequency distribution is a grouping of data into mutually exclusive and collectively ex-
haustive 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 and determine the frequency in each class.
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 43
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 several methods for graphically portraying a frequency distribution.
A. A histogram portrays the frequencies in the form of a rectangle or bar for each class. The height of the rectangles is proportional to the class frequencies.
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 graph of a cumulative frequency distribution shows the number of observations less than a given value.
D. A graph of a cumulative relative frequency distribution shows the percent of observa- tions less than a given value.
C H A P T E R E X E R C I S E S
23. Describe the similarities and differences of qualitative and quantitative variables. Be sure to include the following: 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 between a frequency table and a frequency distribution. Be sure to include which requires qualitative data and which requires quan- titative 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:
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,
metropolitan area. To maintain customer loyalty, one of Speedy Swift’s performance objectives 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 (pack- age delivered within 5 minutes of the promised time), late (package delivered more than 5 minutes past the promised time), or lost (package never delivered). Speedy Swift’s objective is to deliver 99% of all packages either early or on-time. 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
44 CHAPTER 2
a. What kind of variable is delivery performance? What scale is used to measure 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. Write a memo reporting the results of the analyses. Include your tables and graphs with
written descriptions of what they show. Conclude with a general statement of last month’s delivery performance as it relates to Speedy Swift’s performance objectives.
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 25 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 45 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
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.
31. A social scientist is studying the use of iPods by college students. A sample of 45 students revealed they played the following number of songs yesterday.
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 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 shape 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 36 stocks.
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
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?
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 45
d. Using your responses to parts (a), (b), and (c), create a frequency distribution. e. Describe the shape of the frequency distribution.
33. You are exploring the music in your iTunes library. The total play counts over the past year for the 27 songs on your “smart playlist” are shown below. Make a frequency distribu- tion 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?
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
34. The monthly issues of the Journal of Finance are available on the Internet. The table below shows the number of times an issue was downloaded over the last 33 months. Suppose that you wish to summarize the number of downloads with a frequency distribution.
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
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. Describe the shape of the frequency distribution.
35. The following histogram shows the scores on the first exam for a statistics class.
50 60 70 80 90 100
25 20 15 10
5 0
Score
Fr eq
ue nc
y
3
14
21
12
6
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.
100
75
50
25
250 200 150 100 50
0 50 100 150 Selling Price ($000)
200 250 300 350
Fr eq
ue nc
y
Pe rc
en t
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% of the houses sold for less than what amount? e. One hundred seventy-five of the homes sold for less than what amount?
46 CHAPTER 2
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 44 cash register receipts revealed these initial purchases:
$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
a. Arrive at a suggested class interval. b. Organize the data into a frequency distribution using a lower limit of $70. c. Interpret your findings.
38. The numbers of outstanding shares for 24 publicly traded companies are listed in the following table.
Number of Outstanding Shares Company (millions)
Southwest Airlines 738 FirstEnergy 418 Harley Davidson 226 Entergy 178 Chevron 1,957 Pacific Gas and Electric 430 DuPont 932 Westinghouse 22 Eversource 314 Facebook 1,067 Google, Inc. 64 Apple 941
Number of Outstanding Shares Company (millions)
Costco 436 Home Depot 1,495 DTE Energy 172 Dow Chemical 1,199 Eastman Kodak 272 American Electric Power 485 ITT Corporation 93 Ameren 243 Virginia Electric and Power 575 Public Service Electric & Gas 506 Consumers Energy 265 Starbucks 744
a. Using the number of outstanding shares, summarize the companies with a frequency distribution.
b. Display the frequency distribution with a frequency polygon. c. Create a cumulative frequency distribution of the outstanding shares. d. Display the cumulative frequency distribution with a cumulative frequency polygon. e. Based on the cumulative relative frequency distribution, 75% of the companies have
less than “what number” of outstanding shares? f. Write a brief analysis of this group of companies based on your statistical summaries
of “number of outstanding shares.” 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.
Expenditure Item Amount
Fuel $ 603 Interest on car loan 279 Repairs 930 Insurance and license 646 Depreciation 492
Total $2,950
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 47
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
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 “pre-
ferred customer.” Estimate the percentage of preferred customers. d. The bank is also considering a service charge to the lowest 10% of the ending bal-
ances. 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% in dividends, interest, and capital gains; 8% in IRAs and taxable pensions; 3% in business income pensions; 2% in Social Security; and the remaining 3% from other sources. Develop a pie chart depicting the breakdown of adjusted gross income. Write a paragraph summa- rizing 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
a. Organize the data into a frequency distribution. How many classes would you sug- gest? What value would you suggest for a class interval?
b. Draw a histogram. Describe 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 cli- ents in the 40 up to 50 years old age group. Listed following is the value of all the in- vestments in thousands of dollars for the 70 participants in the study.
$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 sug- gest? What value would you suggest for a class interval?
b. Draw a histogram. Financial experts suggest that this age group of people have at least five times their salary saved. As a benchmark, assume an investment portfolio of $500,000 would support retirement in 10–15 years. In writing, summarize your results.
48 CHAPTER 2
44. A total of 5.9% of the prime-time viewing audience watched shows on ABC, 7.6% watched shows on CBS, 5.5% on Fox, 6.0% on NBC, 2.0% on Warner Brothers, and 2.2% on UPN. A total of 70.8% of the audience watched shows on other cable net- works, such as CNN and ESPN. You can find the latest information on TV viewing from the following website: http://www.nielsen.com/us/en/top10s.html/. Develop a pie chart or a bar chart to depict this information. Write a paragraph summarizing your findings.
45. Refer to the following chart:
Contact for Job Placement at Wake Forest University
Networking and
Connections 70%
On-Campus Recruiting
10%
Job Posting Websites
20%
a. What is the name given to this type of chart? b. Suppose that 1,000 graduates will start a new job shortly after graduation. Estimate
the number of graduates whose first contact for employment occurred through net- working and other connections.
c. Would it be reasonable to conclude that about 90% of job placements were made through networking, connections, and job posting websites? Cite evidence.
46. The following chart depicts the annual revenues, by type of tax, for the state of Georgia.
Sales 44.54%Income
43.34%
Other 0.9%
License 2.9%
Corporate 8.31%
Annual Revenue State of Georgia
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.
DESCRIBING DATA: FREQUENCY TABLES, FREQUENCY DISTRIBUTIONS, AND GRAPHIC PRESENTATION 49
47. In 2014, the United States exported a total of $376 billion worth of products to Canada. The five largest categories were:
Product Amount
Vehicles $63.3 Machinery 59.7 Electrical machinery 36.6 Mineral fuel and oil 24.8 Plastic 17.0
a. Use a software package to develop a bar chart. b. What percentage of the United States’ total exports to Canada is represented by the
two categories “Machinery” and “Electrical Machinery”? c. What percentage of the top five exported products do “Machinery” and “Electrical
Machinery” represent? 48. In the United States, the industrial revolution of the early 20th century changed
farming by making it more efficient. 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. An outcome of making farming more efficient is the reduction of the number of farms from over 6 million in 1920 to about 2.2 million farms today. Listed below is the number of farms, in thousands, for each of the 50 states. Summarize the data and write a paragraph that describes your findings.
50 12 5 28 59 19 35 22 80 5 8 48 3 75 25 77 46 68 10 69 77 25 13 20 35 6 52 61 36 38 88 1 75 246 59 50 44 98 74 2 32 42 7 31 28 9 8 44 25 37
49. One of the most popular candies in the United States is M&M’s produced by the Mars Company. In the beginning M&M’s were all brown. Now they are produced in red, green, blue, orange, brown, and yellow. 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:
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. Based on cumulative relative frequencies, how busy were the highest 80% of the days?
D A T A A N A L Y T I C S
51. Refer to the North Valley Real Estate data that reports information on homes sold during the last year. For the variable price, select an appropriate class interval and orga- nize the selling prices into a frequency distribution. Write a brief report summarizing your findings. Be sure to answer the following questions in your report. a. Around what values of price do the data tend to cluster? b. Based on the frequency distribution, what is the typical selling price in the first class?
What is the typical selling price in the last class?
50 CHAPTER 2
c. Draw a cumulative relative frequency distribution. Using this distribution, fifty percent of the homes sold for what price or less? Estimate the lower price of the top ten percent of homes sold. About what percent of the homes sold for less than $300,000?
d. Refer to the variable bedrooms. Draw a bar chart showing the number of homes sold with 2, 3, 4 or more bedrooms. Write a description of the distribution.
52. Refer to the Baseball 2016 data that report information on the 30 Major League Baseball teams for the 2016 season. Create a frequency distribution for the Team Salary variable and answer the following questions. a. What is the typical salary for a team? What is the range of the salaries? b. Comment on the shape of the distribution. Does it appear that any of the teams have
a salary that is out of line with the others? c. Draw a cumulative relative frequency distribution of team salary. Using this distribu-
tion, forty percent of the teams have a salary of less than what amount? About how many teams have a total salary of more than $220 million?
53. Refer to the Lincolnville School District bus data. Select the variable referring to the number of miles traveled since the last maintenance, 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 relative frequency distribution. Forty percent of the buses
were driven fewer than how many miles? How many buses were driven less than 10,500 miles?
d. Refer to the variables regarding the bus manufacturer and the bus capacity. Draw a pie chart of each variable and write a description of your results.
Describing Data: NUMERICAL MEASURES 3
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO3-1 Compute and interpret the mean, the median, and the mode.
LO3-2 Compute a weighted mean.
LO3-3 Compute and interpret the geometric mean.
LO3-4 Compute and interpret the range, variance, and standard deviation.
LO3-5 Explain and apply Chebyshev’s theorem and the Empirical Rule.
LO3-6 Compute the mean and standard deviation of grouped data.
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 LO3-1.)
© Andy Lyons/Getty Images
52 CHAPTER 3
INTRODUCTION Chapter 2 began our study of descriptive statistics. To summarize raw data into a mean- ingful form, we organized qualitative data into a frequency table and portrayed the re- sults 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 vari- ous websites, in the newspaper, and in other jour- nals. 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% 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 $8.31, according to the National Association of Theater 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 addition 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 distribu- tions of the two salaries are the same. However, we need to examine the disper- sion 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 execu- tives in the Internet firms range from $70,000 to $90,000, but salaries for the mar- keting 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 execu- tives. To describe the dispersion, we will consider the range, the variance, and the standard deviation.
MEASURES OF LOCATION 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 median, the mode, the weighted mean, 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.
LO3-1 Compute and interpret the mean, the median, and the mode.
© Andersen Ross/Getty Images RF
STATISTICS IN ACTION
Did you ever meet the “average” American man? Well, his name is Robert (that is the nominal level of measurement) and he is 31 years old (that is the ratio level), is 69.5 inches tall (again the ratio level of measurement), weighs 172 pounds, wears a size 9½ 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 American 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. Idol- ized in the 1950s, Marilyn Monroe would be consid- ered overweight by today’s standards. She fluctuated between a size 14 and a size 18 dress, and was a healthy and attractive woman.
DESCRIBING DATA: NUMERICAL MEASURES 53
The Population Mean Many studies involve all the values in a population. For example, there are 12 sales as- sociates employed at the Reynolds Road Carpet Outlet. The mean amount of commis- sion they earned last month was $1,345. This is 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 $95.47. • 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 num- ber of values in the population. To find the population mean, we use the following formula.
Population mean = Sum of all the values in the population
Number of values in the population
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 mathe- matics. The mean of the population using mathematical symbols is:
POPULATION MEAN μ = Σx N
(3–1)
where: μ 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. Σ is the Greek capital letter “sigma” and indicates the operation of adding. Σx 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.
PARAMETER A characteristic of a population.
E X A M P L E
There are 42 exits on I-75 through the state of Kentucky. Listed below are the distances between exits (in miles).
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
54 CHAPTER 3
Why is this information a population? What is the mean number of miles between exits?
S O L U T I O N
This is a population because we are considering all the exits on I-75 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):
μ = Σx N
= 11 + 4 + 10 + … + 1
42 =
192 42
= 4.57
How do we interpret the value of 4.57? It is the typical number of miles between exits. Because we considered all the exits on I-75 in Kentucky, this value is a popu- lation parameter.
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 orange marmalade in the jar labeled as containing 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 sample is determined, and that value is used to estimate the amount 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:
Sample mean = Sum of all the values in the sample
Number of values in the 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 sam- ple is:
© Bloomberg/Getty Images
SAMPLE MEAN x = Σx n
(3–2)
where: x represents the sample mean. It is read “x bar.” n is the number of values in the sample. x represents any particular value. Σ is the Greek capital letter “sigma” and indicates the operation of adding. Σx 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 orange marmalade is 11.5 ounces, this is an example of a statistic.
DESCRIBING DATA: NUMERICAL MEASURES 55
Properties of the Arithmetic Mean The arithmetic mean is a widely used measure of location. It has several important properties:
1. To compute a mean, the data must be measured at the interval or ratio level. 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 a measure of location 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:
Σ (x − x) = 0
As an example, the mean of 3, 8, and 4 is 5. Then:
Σ(x − x ) = (3 − 5) + (8 − 5) + (4 − 5)
= −2 + 3 − 1
= 0
Thus, we can consider the mean as a balance point for a set of data. To illustrate, 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
STATISTIC A characteristic of a sample.
E X A M P L E
Verizon is studying the number of monthly 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 last month?
S O L U T I O N
Using formula (3–2), the sample mean is:
Sample mean = Sum of all values in the sample Number of values in the sample
x = Σx n
= 90 + 77 + … + 83
12 =
1,170 12
= 97.5
The arithmetic mean number of minutes used last month by the sample of cell phone users is 97.5 minutes.
56 CHAPTER 3
board is balanced perfectly! The deviations below the mean (−3) are equal to the devi- ations above the mean (+3). Shown schematically:
21
22
+3
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 sample of financial planners 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 financial planner has an income in the $60,000 to $63,000 range. One income ($1.2 million) is unduly affecting the mean.
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. The six students in Computer Science 411 are a population. Their final 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?
S E L F - R E V I E W 3–1
The answers to the odd-numbered exercises are in Appendix D.
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 Σ (x − x) = 0. 4. a. Compute the mean of the following sample values: 1.3, 7.0, 3.6, 4.1, 5.0.
b. Show that Σ (x − x) = 0. 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?
E X E R C I S E S
DESCRIBING DATA: NUMERICAL MEASURES 57
The Median We have stressed that, for data containing one or two very large or very small values, the arithmetic mean may not be representative. The center for such data is better de- scribed by a measure of location called the median.
To illustrate the need for a measure of location other than the arithmetic mean, sup- pose 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 up- ward, causing it to be an unrepresentative average. It does seem that a price around $70,000 is a more typical or representative average, 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 minimum to the maximum values.
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. A mail-order company counted the number of incoming calls per day to the compa- ny’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 custom- ers 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. A Human Resources manager at Metal Technologies studied the overtime hours of welders. A sample of 15 welders showed the following number of overtime hours worked 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 reve- nue 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 products. The company is growing rapidly and dedicates only 1 day of sales train- ing 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 3 months. Com- ment on using the current mean sales per month as a sales goal for new employ- ees. Why do new employees leave the company?
58 CHAPTER 3
The median price of the units available is $70,000. To determine this, we order the prices from the minimum value ($60,000) to the maximum value ($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 from Prices Ordered from Minimum to Maximum Maximum to Minimum
$ 60,000 $275,000 65,000 80,000 70,000 ← Median → 70,000 80,000 65,000 275,000 60,000
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 are an odd number of observations (five). How is the median determined for an even number of observations? As before, the observa- tions 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.
E X A M P L E
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.
3 5 7 5 9 1 3 9 17 10
Find the median number of hours.
S O L U T I O N
Note that the number of adults sampled is even (10). The first step, as before, is to order the hours using Facebook from the minimum value to the maximum value. Then identify the two middle times. The arithmetic mean of the two middle observations gives us the median hours. Arranging the values from minimum to maximum:
1 3 3 5 5 7 9 9 10 17
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 adult 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.
DESCRIBING DATA: NUMERICAL MEASURES 59
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.
The Mode The mode is another measure of location.
MODE The value of the observation that appears most frequently.
The mode is especially useful in summarizing nominal-level data. As an example 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 con- sumers prefer. The largest number of respondents favored Lamoure, as evidenced by the highest bar. Thus, Lamoure is the mode.
N um
be r
of R
es po
ns es
Bath oil
Amor Lamoure Soothing
300
200
100
0
400
Smell Nice Far Out
Mode
CHART 3–1 Number of Respondents Favoring Various Bath Oils
E X A M P L E
Recall the data regarding the distance in miles between exits on I-75 in Kentucky. The information is repeated below.
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
What is the modal distance?
S O L U T I O N
The first step is to organize the distances into a frequency table. This will help us determine the distance that occurs most frequently.
60 CHAPTER 3
In summary, we can determine the mode for all levels of data—nominal, ordinal, in- terval, 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 because every value occurs once: $19, $21, $23, $20, and $18. 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 question the use of two modes to represent the location of this set of age data.
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
The distance that occurs most often is 1 mile. This happens eight times—that is, there are eight exits that are 1 mile apart. So the modal distance between exits is 1 mile.
Which of the three measures of location (mean, median, or mode) best rep- resents the central location of these data? Is the mode the best measure of location to represent the Kentucky data? No. The mode assumes only the nominal scale of measurement and the variable miles is measured using the ratio scale. We calcu- lated the mean to be 4.57 miles. See page 54. Is the mean the best measure of location to represent these 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 representative measure of the distance between exits.
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 United States over the last 10 years are 22, 20,
21, 15, 5, 11, 19, 19, 15, and 11. (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?
S E L F - R E V I E W 3–2
DESCRIBING DATA: NUMERICAL MEASURES 61
13. 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 14–16, determine the (a) mean, (b) median, and (c) mode.
14. 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
15. 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 6 5 −1 11
16. The following are the ages of the 10 people in the Java Coffee Shop at the Southwyck Shopping Mall at 10 a.m.
21 41 20 23 24 33 37 42 23 29
17. Several indicators of long-term economic growth in the United States and their annual percent change 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?
18. Sally Reynolds sells real estate along the coastal area of Northern California. Below are her total annual commissions between 2005 and 2015. Find the mean, median, and mode of the commissions she earned for the 11 years.
Year Amount (thousands)
2005 292.16 2006 233.80 2007 206.97 2008 202.67 2009 164.69 2010 206.53 2011 237.51 2012 225.57 2013 255.33 2014 248.14 2015 269.11
19. The accounting firm of Rowatti and Koppel specializes in income tax returns for self-employed professionals, such as physicians, dentists, architects, and law- yers. 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
E X E R C I S E S
62 CHAPTER 3
The Relative Positions of the Mean, Median, and Mode Refer to the histogram in Chart 3–2. It is a symmetric distribution, which is also mound- shaped. This distribution has the same shape on either side of the center. If the histo- gram 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 30 years in Chart 3–2. We should point out that there are symmetric distributions that are not mound-shaped.
Age
Fr eq
ue nc
y
Mean = 30 Median = 30 Mode = 30
CHART 3–2 A Symmetric Distribution
The number of years corresponding to the highest point of the curve is the mode (30 years). Because the distribution is symmetrical, the median corresponds to the point where the distribution is cut in half (30 years). Also, because the arithmetic mean is the balance point of a distribution (as shown in the Properties of the Arithmetic Mean sec- tion on page 56), and the distribution is symmetric, the arithmetic mean is 30. 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, such as the distribution of weekly income in Chart 3–3, 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 smallest of the three measures.
If the distribution is highly skewed, the mean would not be a good measure to use. The median and mode would be more representative.
Find the mean, median, and mode for the number of returns prepared by each accountant. If you could report only one, which measure of location would you recommend reporting?
20. 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 scores for the 10 applicants. What would you report to the president? Does it seem that the applicants are better than the general population?
DESCRIBING DATA: NUMERICAL MEASURES 63
Conversely, if a distribution is negatively skewed, such as the distribution of tensile strength in Chart 3–4, 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, the mean should not be used to represent the data.
Mode = 25
Fr eq
ue nc
y
Median = 29 Mean = 60
Weekly Income
CHART 3–3 A Positively Skewed Distribution
CHART 3–4 A Negatively Skewed Distribution
Mean = 45
Fr eq
ue nc
y
Median = 76 Mode = 80
Tensile Strength
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.
S E L F - R E V I E W 3–3
21. 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
E X E R C I S E S
64 CHAPTER 3
E X A M P L E
Table 2–4 on page 26 shows the profit on the sales of 180 vehicles at Applewood Auto Group. Determine the mean and the median selling price.
S O L U T I O N
The mean, median, and modal amounts of profit are reported in the following output (highlighted in the screen shot). (Reminder: The instructions to create the output appear in the Software Commands in Appendix C.) There are 180 vehicles in the study, so using a calculator would be tedious and prone to error.
Software Solution We can use a statistical software package to find many measures of location.
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? 22. Big Orange Trucking is designing an information system for use in “in-cab”
communications. 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 direction, 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
DESCRIBING DATA: NUMERICAL MEASURES 65
THE WEIGHTED MEAN The weighted mean is a convenient way to compute the arithmetic mean 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 $1.84, $2.07, and $2.40, 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).
x = $1.84 + $1.84 + $1.84 + $2.07 + $2.07 + $2.07 + $2.07 + $2.40 + $2.40 + $2.40
10
x = $21.00
10 = $2.10
The mean selling price of the last 10 drinks is $2.10. 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 occurs. We will refer to the weighted mean as xW . This is read “x bar sub w.”
xw = 3($1.84) + 4($2.07) + 3($2.40)
10 =
$21.00 10
= $2.10
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:
LO3-2 Compute a weighted mean.
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 measures 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. Man- agement at Applewood might use this value for revenue projections. For example, if the dealership could increase the number of vehicles sold in a month from 180 to 200, this would result in an additional estimated $37,000 of revenue, found by 20($1,850).
WEIGHTED MEAN xw = w1x1 + w2x2 + w3x3 + … + wnxn
w1 + w2 + w3 + … + wn (3–3)
This may be shortened to:
xw = Σ (wx)
Σw Note that the denominator of a weighted mean is always the sum of the weights.
E X A M P L E
The Carter Construction Company pays its hourly employees $16.50, $19.00, or $25.00 per hour. There are 26 hourly employees, 14 of whom 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?
66 CHAPTER 3
S O L U T I O N
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
xw = 14($16.50) + 10($19.00) + 2($25.00)
14 + 10 + 2 =
$471.00 26
= $18.1154
The weighted mean hourly wage is rounded to $18.12.
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.
S E L F - R E V I E W 3–4
THE GEOMETRIC MEAN The geometric mean is useful in finding the average change of percentages, ratios, in- dexes, or growth rates over time. It has a wide application in business and economics 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:
LO3-3 Compute and interpret the geometric mean.
GEOMETRIC MEAN GM = √n (x1) (x2) … (xn) (3–4)
The geometric mean will always be less than or equal to (never more than) the arithme- tic mean. Also, all the data values must be positive.
As an example of the geometric mean, suppose you receive a 5% increase in salary this year and a 15% increase next year. The average annual percent increase is 9.886%,
23. 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?
24. 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 hard- cover. What was the weighted mean price of a book?
25. 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?
26. 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?
E X E R C I S E S
DESCRIBING DATA: NUMERICAL MEASURES 67
not 10.0%. Why is this so? We begin by calculating the geometric mean. Recall, for ex- ample, that a 5% increase in salary is 105%. We will write it as 1.05.
GM = √(1.05) (1.15) = 1.09886
This can be verified by assuming that your monthly earning was $3,000 to start and you received two increases of 5% and 15%.
Raise 1 = $3,000(.05) = $150.00
Raise 2 = $3,150(.15) = 472.50 Total $622.50
Your total salary increase is $622.50. This is equivalent to:
$3,000.00(.09886) = $296.59
$3,296.58(.09886) = 325.91 $622.50
The following example shows the geometric mean of several percentages.
E X A M P L E
The return on investment earned by Atkins Construction Company for four succes- sive years was 30%, 20%, −40%, and 200%. What is the geometric mean rate of return on investment?
S O L U T I O N
The number 1.3 represents the 30% return on investment, which is the “original” investment of 1.0 plus the “return” of 0.3. The number 0.6 represents the loss of 40%, which is the original investment of 1.0 less the loss of 0.4. This calculation 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%, found by
GM = √n (x1) (x2) … (xn) = √ 4 (1.3) (1.2) (0.6) (3.0) = √4 2.808 = 1.294
The geometric mean is the fourth root of 2.808. So, the average rate of return (com- pound annual growth rate) is 29.4%.
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!
A second application of the geometric mean is to find an average percentage change over a period of time. For example, if you earned $45,000 in 2004 and $100,000 in 2016, what is your annual rate of increase over the period? It is 6.88%. The rate of increase is determined from the following formula.
RATE OF INCREASE OVER TIME GM = √ n Value at end of period
Value at start of period − 1 (3–5)
In formula 3-5 above, n is the number of periods. An example will show the details of finding the average annual percent increase.
68 CHAPTER 3
E X A M P L E
During the decade of the 1990s, and into the 2000s, Las Vegas, Nevada, was one of the fastest-growing cities in the United States. The population increased from 258,295 in 1990 to 613,599 in 2014. This is an increase of 355,304 people, or a 137.56% increase over the period. The population has more than doubled. What is the average annual percent increase?
S O L U T I O N
There are 24 years between 1990 and 2014, so n = 24. Then the geometric mean formula (3–5) as applied to this problem is:
GM = √ n Value at end of period
Value at start of period − 1.0 = √
24 613,599 258,295
− 1.0 = 1.0367 − 1.0 = .0367
To summarize, the steps to compute the geometric mean are:
1. Divide the value at the end of the period by the value at the beginning of the period.
2. Find the nth root of the ratio, where n is the number of periods. 3. Subtract one.
The value of .0367 indicates that the average annual growth over the period was 3.67%. To put it another way, the population of Las Vegas increased at a rate of 3.67% per year from 1990 to 2014.
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 1996 to 120,520 in 2016.
Find the geometric mean annual percent increase.
S E L F - R E V I E W 3–5
27. Compute the geometric mean of the following monthly percent increases: 8, 12, 14, 26, and 5.
28. Compute the geometric mean of the following weekly 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
30. In 2001, a total of 40,244,000 taxpayers in the United States filed their individual tax returns electronically. By the year 2015, the number increased to 128,653,000. 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 Statis- tics. 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 2015, it increased to 236.525. 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 million. By 2014, revenue had increased to $5,817.0 million. What was the geometric mean annual increase for the period?
E X E R C I S E S
DESCRIBING DATA: NUMERICAL MEASURES 69
WHY STUDY DISPERSION? A measure of location, such as the mean, median, or mode, only describes the center 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 aver- aged 3 feet in depth, would you want to wade across on foot without additional informa- tion? 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 foot to 5.5 feet? Your decision 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 representative 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.
LO3-4 Compute and interpret the range, variance, and standard deviation.
STATISTICS IN ACTION
The U.S. Postal Service has tried to become more “user friendly” in the last several years. A recent 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 common complaint. The level of consistency is mea- sured by the standard devi- ation of the delivery times.
0
0
10
Years
20
Em pl
oy ee
s
10 20
CHART 3–5 Histogram of Years of Employment at Hammond Iron Works Inc.
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
33. In 2000, there were 720,000 cell phone subscribers worldwide. By 2015, the num- ber of cell phone subscribers increased to 752,000,000. 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 2002–03 and 2015–16. What is the geometric mean annual increase for the period for the two types of colleges? Compare the rates of increase.
Type of College 2002–03 2015–16
Public $ 4,960 $23,893 Private 18,056 32,405
70 CHAPTER 3
the two means, you might conclude that the distributions of the hourly outputs are iden- tical. Production records for 9 hours at the two plants, however, reveal that this conclu- sion 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.
We will consider several measures of dispersion. The range is based on the maxi- mum and minimum values in the data set; that is, only two values are considered. The variance and the standard deviation use all the values in a data set and are based on deviations from the arithmetic mean.
Range The simplest measure of dispersion is the range. It is the difference between the maxi- mum and minimum values in a data set. In the form of an equation:
48 49 50 51 52 _ X
48 49 50 51 52 _ X
53 54 55 56 57 58 59 604746454443424140
Baton Rouge
Tucson
Hourly Production
CHART 3–6 Hourly Production of Computer Monitors at the Baton Rouge and Tucson Plants
RANGE Range = Maximum value − Minimum value (3–6)
The range is widely used in production management and control applications be- cause it is very easy to calculate and understand.
E X A M P L E
Refer to Chart 3–6 above. Find the range in the number of computer monitors pro- duced per hour for the Baton Rouge and the Tucson plants. Interpret the two ranges.
S O L U T I O N
The range of the hourly production of computer monitors at the Baton Rouge plant is 4, found by the difference between the maximum hourly production of 52 and
DESCRIBING DATA: NUMERICAL MEASURES 71
Variance A limitation of the range is that it is based on only two values, the maximum and the minimum; it does not take into consideration all of the values. The variance does. It measures the mean amount by which the values in a population, or sample, vary from their mean. In terms of a definition:
the minimum of 48. The range in the hourly production for the Tucson plant is 20 computer monitors, found by 60 − 40. We therefore conclude that (1) there is less dispersion 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 com- puter 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.
VARIANCE The arithmetic mean of the squared deviations from the mean.
The following example illustrates how the variance is used to measure dispersion.
E X A M P L E
The chart below shows the number of cappuccinos sold at the Starbucks in the Orange County airport and the Ontario, California, airport between 4 and 5 p.m. for a sample of 5 days last month.
Determine the mean, median, range, and variance for each location. Comment on the similarities and differences in these measures.
S O L U T I O N
The mean, median, and range for each of the airport locations are reported as part of an Excel spreadsheet.
© Sorbis/Shutterstock.com
72 CHAPTER 3
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 variance. First, for Orange County:
Variance = Σ(x − μ)2
N =
(−302) + (−102) + 02 + 102 + 302
5 =
2,000 5
= 400
The variance is 400. That is, the average squared deviation from the mean is 400. The following shows the detail of determining the variance for the number of
cappuccinos sold at the Ontario Airport.
Variance = Σ(x − μ)2
N =
(−302) + (−52) + 02 + 52 + 302
5 =
1,850 5
= 370
So the mean, median, and range of the cappuccinos sold are the same at the two airports, but the variances are different. The variance at Orange County is 400, but it is 370 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 suggest the two distributions are the same. The range for both locations is also the same, 60. However, recall that
DESCRIBING DATA: NUMERICAL MEASURES 73
the range provides limited information about the dispersion because it is based on only two values, the minimum and maximum.
The variances are not the same for the two airports. The variance 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 variance for Orange County of 400 to the variance for Ontario of 370. Based on the variance, we conclude that the dispersion for the sales distribution of the Ontario Starbucks is more concentrated—that is, nearer the mean of 50—than for the Orange County location.
The variance has an important advantage over the range. It uses all the values in the computation. Recall that the range uses only the highest and the lowest values.
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 variance of the weights.
S E L F - R E V I E W 3–6
For Exercises 35–38, calculate the (a) range, (b) arithmetic mean, (c) variance, and (d) interpret the statistics.
35. During last weekend’s sale, there were five customer service representatives on duty at the Electronic Super Store. The numbers of HDTVs these representatives sold were 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.
37. Dave’s Automatic Door installs automatic garage door openers. The following list indicates the number of minutes needed to install 10 door openers: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42.
38. All eight companies in the aerospace industry were 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%.
39. Ten young adults living in California rated the taste of a newly developed su- shi pizza topped with tuna, rice, and kelp on a scale of 1 to 50, with 1 indicating 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 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 for sushi pizza in the two markets. 40. The personnel files of all eight employees at the Pawnee location of Acme
Carpet Cleaners Inc. revealed that during the last 6-month period they lost the fol- lowing number of days due to illness:
2 0 6 3 10 4 1 2
E X E R C I S E S
74 CHAPTER 3
Population Variance In the previous example, we developed the concept of variance as a measure of disper- sion. Similar to the mean, we can calculate the variance of a population or the variance of a sample. The formula to compute the population variance is:
POPULATION VARIANCE σ2 = Σ(x − μ)2
N (3–7)
where:
σ2 is the population variance (σ is the lowercase Greek letter sigma). It is read as “sigma squared.”
x is the value of a particular observation in the population. μ is the arithmetic mean of the population. N is the number of observations in the population.
The process for computing the variance is implied by the formula.
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 the population variance is the mean of the squared difference 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 vari- ance 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 maximum and minimum values. We over- come the issue where Σ(x − μ) = 0 by squaring the differences. Squaring the differences will always result in nonnegative values. The following is another example that illus- trates the calculation and interpretation of the variance.
E X A M P L E
The number of traffic citations issued last year by month in Beaufort County, South Carolina, is reported below.
Citations by Month
January February March April May June July August September October November December 19 17 22 18 28 34 45 39 38 44 34 10
Determine the population variance.
All 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 resources, compare the two locations. What would you recommend?
DESCRIBING DATA: NUMERICAL MEASURES 75
S O L U T I O N
Because we are studying all the citations for a year, the data comprise a population. To determine the population variance, we use formula (3–7). The table below de- tails the calculations.
Citations Month (x) x − μ (x − μ)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.
μ = Σx N
= 19 + 17 + … + 10
12 =
348 12
= 29
2. Next we find the difference between each observation and the mean. This is shown in the third column of the table. Recall on page 55 in this chapter, the Verizon example showed that the sum of the differences between each value and the mean is 0. This principle is repeated here. 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 for each month. That is shown in the fourth column of the table. All the squared differences will be positive. Note that squaring a negative value, or multiplying a negative value by itself, always results in a positive value.
4. The squared differences are totaled. The total of the fourth column is 1,488. That is the term Σ(x − μ)2.
5. Finally, we divide the squared differences by N, the number of observations in the population.
σ2 = Σ(x − σ)2
N =
1,488 12
= 124
So, the population variance for the number of citations is 124.
Like the range, the variance can be used to compare the 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 num- ber of citations issued in Marlboro County, South Carolina, is 342.9, we conclude that (1) there is less dispersion in the distribution of the number 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 mea- sure of location than the mean number of citations in Marlboro County.
76 CHAPTER 3
Population Standard Deviation When we compute the variance, it is important to understand the unit of measure and what happens when the differences in the numerator are squared. That is, in the previ- ous example, the number of monthly citations is the variable. When we calculate the variance, the unit of measure for the variance is citations squared. Using “squared cita- tions” as a unit of measure is cumbersome.
There is a way out of this difficulty. By taking the square root of the population vari- ance, we can transform it to the same unit of measurement used for the original 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 σ = √ Σ(x − μ)2
N (3–8)
The Philadelphia office of PricewaterhouseCoopers 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.
S E L F - R E V I E W 3–7
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 population 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 5-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. Com- pare the means and dispersions in the two firms.
E X E R C I S E S
DESCRIBING DATA: NUMERICAL MEASURES 77
Sample Variance and Standard Deviation The formula for the population mean is μ = Σx/N. We just changed the symbols for the sample mean; that is, x = Σx/n. Unfortunately, the conversion from the population vari- ance 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 s2 = Σ(x − x )2
n − 1 (3–9)
where:
s2 is the sample variance. x is the value of each observation in the sample. x 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 x is used to estimate μ, it tends to underestimate the population variance, σ2. 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 parame- ters like σ2, (n − 1) is preferred to n in defining the sample variance. We will also use this convention when computing the sample standard deviation.
E X A M P L E
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?
S O L U T I O N
The sample variance is computed by using formula (3–9).
x = Σx n
= $85
5 = $17
Hourly Wage (x) x − x (x − x )2
$12 −$5 25 20 3 9 16 −1 1 18 1 1 19 2 4
$85 0 40
s2 = Σ(x − x )2
n − 1 =
40 5 − 1
= 10 in dollars squared
78 CHAPTER 3
The sample standard deviation is used as an estimator of the population standard deviation. As noted previously, the population standard deviation is the square root of the population variance. Likewise, the sample standard deviation is the square root of the sample variance. The sample standard deviation is determined by:
SAMPLE STANDARD DEVIATION s = √ Σ(x − x )2
n − 1 (3–10)
E X A M P L E
The sample variance in the previous example involving hourly wages was com- puted to be 10. What is the sample standard deviation?
S O L U T I O N
The sample standard deviation is $3.16, found by √10. Note again that the sample 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.
Software Solution On page 64, we used Excel to determine the mean, median, and mode of profit for the Applewood Auto Group data. You also will note that it lists the sample variance and sample standard deviation. Excel, like most other statistical software, assumes the data are from a sample.
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? Compute the sample standard deviation.
S E L F - R E V I E W 3–8
DESCRIBING DATA: NUMERICAL MEASURES 79
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 computed to be $7.51. Suppose these employees are located in Georgia. If the standard 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 em- ployees 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 out of every four, or 75%, of the values 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 values, or 88.9%, will lie between plus three standard deviations and minus three standard deviations of the mean. At least 24 of 25 values, or 96%, will lie between plus and minus five standard deviations of the mean.
Chebyshev’s theorem states:
LO3-5 Explain and apply Chebyshev’s theorem and the Empirical Rule.
STATISTICS IN ACTION
Most colleges report the “average class size.” This information 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
(continued)
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 value greater than 1.
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. 49. Dave’s Automatic Door, referred to in Exercise 37, installs automatic garage
door openers. 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 Exer- cise 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
E X E R C I S E S
80 CHAPTER 3
it found the mean from a list of the class sizes of each student, it was 147. Why the disparity? Because there are few 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 student by reducing the number of students in each class. That is, cut out the large freshman lecture classes.
(continued from p. 79)
EMPIRICAL RULE For a symmetrical, bell-shaped frequency distribution, approximately 68% of the observations will lie within plus and minus one standard deviation of the mean; about 95% of the observations will lie within plus and minus two standard deviations of the mean; and practically all (99.7%) will lie within plus and minus three standard deviations of the mean.
E X A M P L E
Dupree Paint Company employees contribute a mean of $51.54 to the company’s profit-sharing plan every two weeks. The standard deviation of biweekly contributions 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, that is between $25.26 and $77.83?
S O L U T I O N
About 92%, found by
1 − 1
k2 = 1 −
1
(3.5)2 = 1 −
1 12.25
= 0.92
The Empirical Rule Chebyshev’s theorem applies to 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.
These relationships are portrayed graphically in Chart 3–7 for a bell-shaped distribution 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 Percentage of Observations
Applying the Empirical Rule, if a distribution is symmetrical and bell-shaped, practically all of the observations lie between the mean plus and minus three standard deviations. Thus, if x = 100 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.
DESCRIBING DATA: NUMERICAL MEASURES 81
Conversely, if we know that the range is 60 and the distribution is bell-shaped, we can approximate the standard deviation by dividing the range by 6. For this illustration: range ÷ 6 = 60 ÷ 6 = 10, the standard deviation.
E X A M P L E
A sample of the rental rates at University Park Apartments approximates a symmet- rical, bell-shaped distribution. The sample mean is $500; the standard deviation is $20. Using the Empirical Rule, answer these questions:
1. About 68% of the monthly rentals are between what two amounts? 2. About 95% of the monthly rentals are between what two amounts? 3. Almost all of the monthly rentals are between what two amounts?
S O L U T I O N
1. About 68% are between $480 and $520, found by x ± 1s = $500 ± 1($20). 2. About 95% are between $460 and $540, found by x ± 2s = $500 ± 2($20). 3. Almost all (99.7%) are between $440 and $560, found by x ± 3s = $500 ± 3($20).
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 inch. (a) If we do not know the shape of the distribution of outside pipe diameters, 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% of the observations will be between what two values?
S E L F - R E V I E W 3–9
53. According to Chebyshev’s theorem, at least what percent of any set of observa- tions will be within 1.8 standard deviations of the mean?
54. The mean income of a group of sample observations is $500; the standard devia- tion is $40. According to Chebyshev’s theorem, at least what percent of the in- comes 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 x − 2s and x + 2s? b. Between x and x + 2s ? Above x + 2s?
56. The following graph portrays the distribution of the number of spicy chicken sand- wiches sold at a nearby Wendy’s for the last 141 days. The mean number of sand- wiches sold per day is 91.9 and the standard deviation is 4.67.
10090 Sales
If we use the Empirical Rule, sales will be between what two values on 68% of the days? Sales will be between what two values on 95% of the days?
E X E R C I S E S
82 CHAPTER 3
THE MEAN AND STANDARD DEVIATION OF GROUPED DATA In most instances, measures of location, such as the mean, and measures of dispersion, 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 distribution and wish to estimate the mean or standard deviation. In the following discussion, we show how we can esti- mate the mean and standard deviation from data organized into a frequency distribu- tion. We should stress that a mean or a standard deviation from grouped data is an estimate of the corresponding actual values.
Arithmetic Mean of Grouped Data 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 midpoint of the class. The mean of a sample of data organized in a frequency distribution is computed by:
LO3-6 Compute the mean and standard deviation of grouped data.
STATISTICS IN ACTION
During the 2016 Major League Baseball season, DJ LeMahieu of the Colorado Rockies had the highest batting average at .348. 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 bat- ting average has remained constant at about .260 for more than 100 years, but the standard deviation declined from .049 to .031. This indicates less disper- sion in the batting averages today and helps explain the lack of any .400 hitters in recent times.
ARITHMETIC MEAN OF GROUPED DATA x = ΣfM
n (3–11)
where:
x is 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. Σfm is the sum of these products. n is the total number of frequencies.
E X A M P L E
The computations for the arithmetic mean of data grouped into a frequency distribution will be shown based on the Applewood Auto Group profit data. Recall in Chapter 2, in Table 2–7 on page 30, 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
S O L U T I O N
The mean vehicle selling price can be estimated from data grouped into a fre- quency distribution. To find the estimated mean, assume the midpoint of each class is representative of the data values in that class. Recall that the midpoint of a class
DESCRIBING DATA: NUMERICAL MEASURES 83
Standard Deviation of Grouped Data To calculate the standard deviation of data grouped into a frequency distribution, we need to adjust formula (3–10) slightly. We weight each of the squared differences by the number of frequencies in each class. The formula is:
is halfway between the lower class limits of two consecutive classes. To find the midpoint 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 summarized in Table 3–1.
TABLE 3–1 Profit on 180 Vehicles Sold Last Month at Applewood Auto Group
Solving for the arithmetic mean using formula (3–11), we get:
x = ΣfM
n =
$333,200 180
= $1,851.11
We conclude that the mean profit per vehicle is about $1,851.
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
STANDARD DEVIATION, GROUPED DATA s = √ Σf(M − x )2
n − 1 (3–12)
where:
s is 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. x is the sample mean.
E X A M P L E
Refer to the frequency distribution for the Applewood Auto Group profit data re- ported in Table 3–1. Compute the standard deviation of the vehicle selling prices.
S O L U T I O N
Following the same practice used earlier for computing the mean of data grouped into a frequency distribution, f is the class frequency, M the class midpoint, and n the number of observations.
84 CHAPTER 3
Profit Frequency (f ) Midpoint (M) fM (M − x ) (M −x )2 f(M − x )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
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 sec- ond 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 f(M − x )2. The total is 76,169,920. To find the standard devi- ation, we insert these values in formula (3–12).
s = √ Σf(M − x )2
n − 1 = √
76,169,780 180 − 1
= 652.33
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 64 is $1,843.17 and the standard deviation is $643.63. The respective values estimated from data grouped into a frequency distribution are $1,851.11 and $652.33. The difference in the means is $7.94, or about 0.4%. The standard deviations differ by $8.70, or 1.4%. Based on the percentage difference, the estimates are very close to the actual values.
The net incomes of a sample of twenty container shipping companies were organized into the following table:
Net Income ($ millions) Number of Companies
2 up to 6 1 6 up to 10 4 10 up to 14 10 14 up to 18 3 18 up to 22 2
(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?
S E L F - R E V I E W 3–10
DESCRIBING DATA: NUMERICAL MEASURES 85
57. When we compute the mean of a frequency distribution, why do we refer to this as an estimated mean?
58. Estimate the mean and the standard deviation of the following frequency distribu- tion showing the number of times students eat at campus dining places in a month.
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. Estimate the mean and the standard deviation of the following frequency dis- tribution showing the ages of the first 60 people in line on Black Friday at a retail store.
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 distribution on the age of Internet users. Estimate 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
61. The IRS was interested in the number of individual tax forms prepared by small accounting firms. The IRS randomly sampled 50 public accounting firms with 10 or fewer employees in the Dallas–Fort Worth area. The following frequency ta- ble 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
E X E R C I S E S
86 CHAPTER 3
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 sta- tistics, it is also important to understand statistics so that you can be an intelligent con- sumer of information.
In this chapter, we learned how to compute numerical descriptive statistics. Specifi- cally, 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 suppose that the client also asks what the median sales price is, and the median is $60,000. Why was the developer only reporting the mean price? This information is extremely important to a person’s decision making when buying a home. Knowing the advantages and disadvantages of the mean, median, and mode is important as we report statistics and as we use statistical information to make decisions.
We also learned how to compute measures of dispersion: range, variance, and standard deviation. Each of these statistics also has advantages and disadvantages. Remember that the range provides information about the overall spread of a distribu- tion. However, it does not provide any information about how the data are clustered or concentrated around the center of the distribution. As we learn more about statistics, we need to remember that when we use statistics we must maintain an independent and principled point of view. Any statistical report requires objective and honest com- munication of the results.
C H A P T E R S U M M A R Y
I. A measure of location is a value used to describe the central tendency of a set of data. A. The arithmetic mean is the most widely reported measure of location.
1. It is calculated by adding the values of the observations and dividing by the total number of observations. a. The formula for the population mean of ungrouped or raw data is
μ = Σx N
(3–1)
b. The formula for the sample mean is
x = Σx n
(3–2)
62. Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 60 man- ufacturing companies located in the Southwest. Estimate the mean and the stan- dard deviation of advertising expenses.
Advertising Expenditure Number of ($ 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
DESCRIBING DATA: NUMERICAL MEASURES 87
c. The formula for the sample mean of data in a frequency distribution is
x = ΣfM
n (3–11)
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 median is the value in the middle of a set of ordered data. 1. To find the median, sort the observations from minimum to maximum 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.
C. 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.
D. The weighted mean is found by multiplying each observation by its corresponding weight. 1. The formula for determining the weighted mean is
xw = w1x1 + w2 x2 + w3 x3 + … + wn xn
w1 + w2 + w3 + … + wn (3–3)
E. The geometric mean is the nth root of the product of n positive values. 1. The formula for the geometric mean is
GM = √n (x1) (x2) (x3) … (xn) (3–4) 2. The geometric mean is also used to find the rate of change from one period to another.
GM = √ n Value at end of period
Value at beginning of period − 1 (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 maximum and minimum values in a set of data. 1. The formula for the range is
Range = Maximum value − Minimum 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 variance is the mean of the squared deviations from the arithmetic mean. 1. The formula for the population variance is
σ2 = Σ(x − μ)2
N (3–7)
2. The formula for the sample variance is
s2 = Σ(x − x )2
n − 1 (3–9)
3. The major characteristics of the variance are: a. All observations are used in the calculation. b. The units are somewhat difficult to work with; they are the original units squared.
C. 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.
88 CHAPTER 3
2. The formula for the sample standard deviation is
s = √ Σ(x − x )2
n − 1 (3–10)
3. The formula for the standard deviation of grouped data is
s = √ Σf(M − x )2
n − 1 (3–12)
III. We use the standard deviation to describe a frequency distribution by applying Chebyshev’s theorem or the Empirical Rule. 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% of the values will be within one standard deviation of the mean, 95% within two, and virtually all within three.
P R O N U N C I A T I O N K E Y
SYMBOL MEANING PRONUNCIATION
μ Population mean mu Σ Operation of adding sigma Σx Adding a group of values sigma x x Sample mean x bar
xw Weighted mean x bar sub w
GM Geometric mean G M
ΣfM Adding the product of the frequencies and the class midpoints sigma f M
σ2 Population variance sigma squared σ Population standard deviation sigma
C H A P T E R E X E R C I S E S
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 Σ(x − μ) = 0.
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 Σ(x − x ) = 0.
65. A sample of households that subscribe to United Bell Phone Company for landline phone service revealed the following number of calls received per household last week. Determine the mean and the median number of calls received.
52 43 30 38 30 42 12 46 39 37 34 46 32 18 41 5
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 number 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.
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
DESCRIBING DATA: NUMERICAL MEASURES 89
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 me- dian of a typical wash cycle.
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
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 appoint- ments she made per hour? What is the median number of appointments per hour? Write a brief report summarizing the findings.
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
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. Yesterday, 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 semester he took courses in statistics and accounting, 3 hours each, and earned an A in both. He earned a B in a 5-hour history course and a B in a 2-hour history of jazz course. In addi- tion, he took a 1-hour course dealing with the rules of basketball so he could get his li- cense 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 central tendency did you calculate? What method did you use?
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 Director of Economic Development. He must present a report to several companies that are consid- ering locating in northwest Ohio. What would be an appropriate unemployment rate to show for the entire region?
County Percent Unemployed Size of Workforce
Wood 4.5 15,300 Ottawa 3.0 10,400 Lucas 10.2 150,600
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.
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
a. What is the arithmetic mean glucose reading? b. What is the median glucose reading? c. What is the modal glucose reading?
73. The first Super Bowl was played in 1967. The cost for a 30-second commercial was $42,000. The cost of a 30-second commercial for Super Bowl 50 was $4.6 million. What was the geometric mean rate of increase for the 50 year period?
90 CHAPTER 3
74. A recent article suggested that, if you earn $25,000 a year today and the inflation rate continues at 3% 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%. 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 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 standard deviation.
77. The enrollments of the 13 public universities in the state of Ohio are listed below.
College Enrollment
University of Akron 26,106 Bowling Green State University 18,864 Central State University 1,718 University of Cincinnati 44,354 Cleveland State University 17,194 Kent State University 41,444 Miami University 23,902 Ohio State University 62,278 Ohio University 36,493 Shawnee State University 4,230 University of Toledo 20,595 Wright State University 17,460 Youngstown State University 12,512
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 medi- cal 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.
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
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.
9 195 241 301 216 260 7 244 192 147 10 295 142
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.
DESCRIBING DATA: NUMERICAL MEASURES 91
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 restau- rant 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.
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
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 enter- tainment expenses last year:
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
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 that includes about 95% of the
observations. 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 pay- off on a $2 bet.
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 2011 Animal Kingdom 2.75 2.034 43.80 2012 I’ll Have Another 1.5 2.03050 32.60 2013 Orb 2.5 2.04817 12.80 2014 California Chrome 1.75 2.0610 7.00 2015 American Pharaoh 1.00 2.05033 7.80
92 CHAPTER 3
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 winning time and payoff
on a $2 bet. 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.
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
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 distri-
bution. 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.
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
a. Estimate the mean cost. b. Estimate the standard deviation. c. Use the Empirical Rule to estimate the proportion of costs within two standard devia-
tions 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.
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
D A T A A N A L Y T I C S
86. Refer to the North Valley Real Estate data and prepare a report on the sales prices of the homes. Be sure to answer the following questions in your report. a. Around what values of price do the data tend to cluster? What is the mean sales
price? What is the median sales price? Is one measure more representative of the typical sales prices than the others?
b. What is the range of sales prices? What is the standard deviation? About 95% of the sales prices are between what two values? Is the standard deviation a useful statistic for describing the dispersion of sales price?
c. Repeat (a) and (b) using FICO score.
DESCRIBING DATA: NUMERICAL MEASURES 93
87. Refer to the Baseball 2016 data, which report information on the 30 Major League Baseball teams for the 2016 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 representa- tive of the typical team salary than the others?
2. What is the range of the team salaries? What is the standard deviation? About 95% of the salaries are between what two values?
b. Refer to the information on the average salary for each year. In 2000 the average player salary was $1.99 million. By 2016 the average player salary had increased to $4.40 million. What was the rate of increase over the period?
88. Refer to the Lincolnville School District bus data. Prepare a report on the mainte- nance 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
maintenance cost last month? What is the median cost? Is one measure more repre- sentative of the typical cost than the others?
b. What is the range of maintenance costs? What is the standard deviation? About 95% of the maintenance costs are between what two values?
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO4-1 Construct and interpret a dot plot.
LO4-2 Construct and describe a stem-and-leaf display.
LO4-3 Identify and compute measures of position.
LO4-4 Construct and analyze a box plot.
LO4-5 Compute and interpret the coefficient of skewness.
LO4-6 Create and interpret a scatter diagram.
LO4-7 Develop and explain a contingency table.
MCGIVERN JEWELERS recently posted an advertisement on a social media site 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 LO4-4.)
Describing Data: DISPLAYING AND EXPLORING DATA4
© Denis Vrublevski/Shutterstock.com
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 95
INTRODUCTION Chapter 2 began our study of descriptive statistics. In order to transform raw or un- grouped data into a meaningful form, we organize the data into a frequency distribution. We present the frequency distribution in graphic form as a histogram or a frequency polygon. This allows us to visualize where the data tend 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, median, and mode. 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, variance, and standard deviation. These measures of dispersion allow us to de- scribe 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. Examples include the number of hours a student studied and the points earned on an examination; if a sampled product meets quality specifications and the shift on which it is manufactured; or the amount of electric- ity used in a month by a homeowner and the mean daily high temperature in the region for the month. These charts and graphs provide useful insights as we use business analytics to enhance our understanding of data.
DOT PLOTS Recall for the Applewood Auto Group data, we summarized the profit earned on the 180 vehicles sold with a frequency distribution using eight classes. When we orga- nized 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 individual observation. To develop a dot plot, we display a dot for each observation 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 individually, 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 histo- grams tend to be most useful for large data sets. An example will show how to con- struct and interpret dot plots.
LO4-1 Construct and interpret a dot plot.
E X A M P L E
The service departments at Tionesta Ford Lincoln and Sheffield Motors Inc., two of the four Applewood Auto Group dealerships, were both open 24 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
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
96 CHAPTER 4
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
S O L U T I O N
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 each dealership over the last 24 working days.
The dot plots, shown in the center of the output, graphically illustrate the distribu- tions for each dealership. The plots show the difference in the location and dis- persion 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 Sheffield serviced a mean of 35.83 vehicles 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 Shef- field (4.96 vehicles per day) than at Tionesta (4.112 cars per day).
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 organizing the data into a frequency distribution is we get a quick visual picture of the shape of the
LO4-2 Construct and describe a stem-and-leaf display.
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 97
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 Theater of the Republic in Erie, Pennsylvania, books live theater and musical performances. The the- ater’s capacity is 160 seats. Last year, among the forty-five performances, there were eight different plays and twelve different bands. The following frequency distribution shows that between eighty up to ninety people attended two of the forty-five perfor- mances; there were seven performances where ninety up to one hundred people at- tended. However, is the attendance within this class clustered about 90, spread evenly throughout the class, or clustered near 99? We cannot tell.
Attendance 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 used to display quantitative information in a condensed form and provide more information than the frequency distribution is the stem-and-leaf display. An advantage of the stem-and-leaf display over a frequency distribution is 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 construc- tion of a stem-and-leaf display using the number people attending each perfor- mance, 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:
9 ∣ 6 4 3 4 5 6 7
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:
9 ∣ 3 4 4 5 6 6 7
With the stem-and-leaf display, we can quickly observe that 94 people attended two performances and the number attending 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.
98 CHAPTER 4
The following example explains the details of developing a stem-and-leaf display.
E X A M P L E
Listed in Table 4–1 is the number of people attending each of the 45 performances at the Theater of the Republic last year. Organize the data into a stem-and-leaf display. Around what values does attendance tend to cluster? What is the smallest attendance? The largest attendance?
S O L U T I O N
From the data in Table 4–1, we note that the smallest attendance is 88. So we will make the first stem value 8. The largest attendance 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 has 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, the chart is as follows.
Stem Leaf
8 8 9 6 3 10 11 12 13 14 15
Organizing all the data, the stem-and-leaf chart looks as follows.
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
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
TABLE 4–1 Number of People Attending Each of the 45 Performances at the Theater of the Republic
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
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 99
The final table would appear as follows, where we have sorted all of the leaf values.
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
You can draw several conclusions from the stem-and-leaf display. First, the mini- mum number of people attending is 88 and the maximum is 156. There were two per- formances with less than 90 people attending, and three performances with 150 or more. You can observe, for example, that for the three performances with more than 150 people attending, the actual attendances were 155, 155, and 156. The concentra- tion of attendance is between 110 and 130. There were fifteen performances with at- tendance between 110 and 119 and eight performances between 120 and 129. We can also tell that within the 120 to 129 group the actual attendances were spread evenly throughout the class. That is, 120 people attended two performances, 124 peo- ple attended one performance, 125 people attended three performances, and 127 peo- ple attended two performances.
We also can generate this information on the Minitab software system. We have named the variable Attendance. The Minitab output is below. You can find the Minitab commands that will produce this output in Appendix C.
The Minitab solution provides some additional information regarding cumulative 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 there are 9 observations that have occurred before the value of 100. The number 15 indicates that 15 observations 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 there are nine values in this group. In this case, we describe the middle value as the value below which half of the observations oc- cur. 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 represent the “more than” cumulative totals. There are 21 observations of 120 or more, 13 of 130 or more, and so on.
100 CHAPTER 4
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 num- ber 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, partic- ularly on a smaller set of data.
© Somos/Veer/Getty Images RF
1. The number of employees at each of the 142 Home Depot stores in the Southeast region is shown in the following dot plot.
100 10484 88 92 Number of employees
9680
(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
S E L F - R E V I E W 4–1
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 101
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?
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.
6 72 3 4 51
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 a big-box retail store for the last 26 days.
199 144
a. What are the maximum and the minimum numbers 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 shows the number of units produced per day in a factory.
Stem Leaf 3 8 4 5 6 6 0133559 7 0236778 8 59 9 00156 10 36
a. How many days were studied? b. How many observations are in the first class?
E X E R C I S E S
102 CHAPTER 4
c. What are the minimum value and the maximum value? 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 prescriptions filled per day at the pharmacy on the corner of Fourth and Main Streets.
Stem Leaf 12 689 13 123 14 6889 15 589 16 35 17 24568 18 268 19 13456 20 034679 21 2239 22 789 23 00179 24 8 25 13 26 27 0
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 prescriptions filled? g. On how many days were 220 or more prescriptions filled? h. What is the middle value? i. How many days did the number of filled prescriptions range between 170 and 210?
9. A survey of the number of phone calls made by a sample of 16 Verizon sub- scribers 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?
52 43 30 38 30 42 12 46 39 37 34 46 32 18 41 5
10. Aloha Banking Co. is studying ATM use in suburban Honolulu. Yesterday, for a sample of 30 ATM's, the bank counted the number of times each machine was used. The data is presented in the table. Develop a stem-and-leaf chart to summa- rize the data. What were 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
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 103
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 the minimum to the maximum. In Chapter 3, we called the middle value of a set of data arranged from the minimum to the maximum the median. That is, 50% of the observations are larger than the median and 50% 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, usu- ally labeled Q1, is the value below which 25% of the observations occur, and the third quartile, usually labeled Q3, is the value below which 75% of the observations occur.
Similarly, deciles divide a set of observations into 10 equal parts and percentiles into 100 equal parts. So if you found that your GPA was in the 8th decile at your univer- sity, you could conclude that 80% of the students had a GPA lower than yours and 20% had a higher GPA. If your GPA was in the 92nd percentile, then 92% of students had a GPA less than your GPA and only 8% of students had a GPA greater than your GPA. Per- centile 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 percen- tile. So if we want to find the 92nd percentile we would use L92, and if we wanted the median, the 50th percentile, then L50. For a number of observations, n, the location of the Pth percentile, can be found using the formula:
LO4-3 Identify and compute measures of position.
LOCATION OF A PERCENTILE Lp = (n + 1) P
100 [4–1]
An example will help to explain further.
E X A M P L E
Morgan Stanley is an investment company with offices located throughout the United States. Listed below are the commissions earned last month by a sample of 15 brokers at the Morgan Stanley office in Oakland, California.
$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.
S O L U T I O N
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
104 CHAPTER 4
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 formula indicated we should find the fourth ordered value. What if there were 20 observations in the sample, that is n = 20, and we wanted to locate the first quartile? From the loca- tion formula (4–1):
L25 = (n + 1) P
100 = (20 + 1)
25 100
= 5.25
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 the minimum to the maximum: 43, 61, 75, 91, 101, and 104. The first quartile is located at
L25 = (n + 1) P
100 = (6 + 1)
25 100
= 1.75
The position formula tells us that the first quartile is located between the first and the second values and it is .75 of the distance between the first and the second 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 distance between the first and second values, so .75(18) = 13.5. To complete the procedure, we add 13.5 to the first value, 43, and report that the first quartile is 56.5.
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.
L23 = (n + 1) P
100 = (80 + 1)
23 100
= 18.63
The median value is the observation in the center and is the same as the 50th percen- tile, so P equals 50. So the median 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 com- missions more than $2,038 and half earned less than $2,038. The result using
formula (4–1) to find the median is the same as the method presented in Chapter 3.
Recall the definition of a quartile. Quartiles divide a set of observations into four equal parts. Hence 25% of the observations 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:
L25 = (n + 1) P
100 = (15 + 1)
25 100
= 4
and to locate the third quartile, n = 15 and P = 75:
L75 = (n + 1) P
100 = (15 + 1)
75 100
= 12
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.
© Ramin Talaie/Getty Images
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 105
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.
Statistical software is very helpful when describing and summarizing data. Excel, Minitab, and MegaStat, a statistical analysis Excel add-in, all provide summary statistics that include quartiles. For example, the Minitab summary of the Morgan Stanley com- mission data, shown below, includes the first and third quartiles, and other statistics. Based on the reported quartiles, 25% of the commissions earned were less than $1,721 and 75% were less than $2,205. These are the same values we calculated using formula (4–1).
There are ways other than formula (4–1) to lo- cate quartile values. For example, another method 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. We will call this the Excel Method. In the Morgan Stanley data, this method would place the first quartile at position 4.5 (.25 × 15 + .75) and the third quartile at position 11.5 (.75 × 15 + .25). The first quartile would be interpolated as 0.5, or one-half the difference between the fourth- and the fifth-ranked values. Based on this method, the first quartile is $1739.5, found by ($1,721 + 0.5[$1,758 − $1,721]). The third quar- tile, at position 11.5, would be $2,151, or one-half the distance between the eleventh- and the
twelfth-ranked values, found by ($2,097 + 0.5[$2,205 − $2,097]). Excel, as shown in the Morgan Stanley and Applewood examples, can compute quartiles using either of the two methods. Please note the text uses formula (4–1) to calculate quartiles.
Is the difference between the two methods important? No. Usually it is just a nui- sance. In general, both methods calculate values that will support the statement that ap- proximately 25% of the values are less than the value of the first quartile, and approximately 75% of the data values are less than the value of the third quartile. When the sample is
large, the difference in the results from the two methods is small. For example, in the Applewood Auto Group data there are 180 vehicles. The quartiles computed using both methods are shown to the left. Based on the variable profit, 45 of the 180 values (25%) are less than both values of the first quartile, and 135 of the 180 values (75%) are less than both values of the third quartile.
When using Excel, be careful to understand the method used to
STATISTICS IN ACTION
John W. Tukey (1915–2000) received a PhD in mathe- matics from Princeton in 1939. However, when he joined the Fire Control Re- search Office during World War II, his interest in ab- stract mathematics shifted to applied statistics. He de- veloped effective numerical and graphical methods for studying patterns in data. Among the graphics he developed 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 victory for Richard Nixon in the presi- dential election won by John F. Kennedy.
Morgan Stanley Commissions
1460 Equation 4-1 2047 1471
Quartile 1 Quartile 3
1721 2205
Alternate Method Quartile 1 Quartile 3
1739.5 2151
2054 1637 2097 1721 2205 1758 2287 1787 2311 1940 2406 2038
Pro�tAge Applewood
Equation 4-1 Quartile 1 Quartile 3
1415.5 2275.5
Alternate Method Quartile 1 Quartile 3
1422.5 2268.5
$1,387 $1,754 $1,817 $1,040 $1,273 $1,529 $3,082 $1,951 $2,692 $1,342
21 23 24 25 26 27 27 28 28 29
106 CHAPTER 4
calculate quartiles. Excel 2013 and Excel 2016 offer both methods. The Excel function, Quartile.exc, will result in the same answer as Equation 4–1. The Excel function, Quar- tile.inc, will result in the Excel Method answers.
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:
7.69 7.72 7.8 7.86 7.90 7.94 7.97 8.06 8.09
(a) What is the median weight? (b) Determine the weights corresponding to the first and third quartiles.
S E L F - R E V I E W 4–2
11. Determine the median and the first and third quartiles in the following data.
46 47 49 49 51 53 54 54 55 55 59
12. Determine the median and the first and third quartiles in the following data.
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
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 im- portant. 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.
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 re- port 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.
E X E R C I S E S
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 107
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.
LO4-4 Construct and analyze a box plot.
E X A M P L E
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:
Minimum value = 13 minutes
Q1 = 15 minutes
Median = 18 minutes
Q3 = 22 minutes
Maximum value = 30 minutes
Develop a box plot for the delivery times. What conclusions can you make about the delivery times?
S O L U T I O N
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
Q3
Minimum value
Maximum value
Minutes
The box plot also shows the interquartile range of delivery times between Q1 and Q3. The interquartile range is 7 minutes and indicates that 50% of the deliveries are between 15 and 22 minutes.
The box plot also reveals that the distribution of delivery times is positively skewed. In Chapter 3, we defined skewness as the lack of symmetry 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,
108 CHAPTER 4
the 25% of the data larger than the third quartile is more spread out than the 25% 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 de- livery times between 18 minutes and 22 minutes.
E X A M P L E
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?
S O L U T I O N
Minitab was used to develop the following chart and summary statistics.
The median age of the purchaser is 46 years, 25% of the purchasers are less than 40 years of age, and 25% are more than 52.75 years of age. Based on the sum- mary information and the box plot, we conclude:
• Fifty percent of the purchasers are between the ages of 40 and 52.75 years. • The distribution of ages is fairly symmetric. There are two reasons for this con-
clusion. 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.
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 inter- quartile range smaller than Q1 or larger than Q3. In this example, an outlier would be a value larger than 71.875 years, found by:
Outlier > Q3 + 1.5(Q3 − Q1) = 52.75 + 1.5(52.75 − 40) = 71.875
An outlier would also be a value less than 20.875 years.
Outlier < Q1 − 1.5(Q3 − Q1) = 40 − 1.5(52.75 − 40) = 20.875
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 109
The following box plot shows the assets in millions of dollars for credit unions in Seattle, Washington.
0 10 20 30 40 50 60 70 80 90 100
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?
S E L F - R E V I E W 4–3
From the box plot, we conclude 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 limitations of the software and space available. It is a good idea to check the actual data. In this in- stance, there are three purchasers 72 years old or older; two are 72 and one is 73.
15. The box plot below shows the amount spent for books and supplies per year by students at four-year public colleges.
0 350 700 1,050 1,400 $1,750
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? 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 tuition per credit hour at four-year public colleges.
*
0 300 600 900 1,200 $1,500
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 2016 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?
E X E R C I S E S
110 CHAPTER 4
SKEWNESS In Chapter 3, we described measures of central location for a distribution of data by re- porting the mean, median, and mode. We also described measures that show the amount of spread or variation in a distribution, such as the range and the standard deviation.
Another characteristic of a distribution is the shape. There are four shapes com- monly observed: symmetric, positively skewed, negatively skewed, and bimodal. In a symmetric distribution the mean and median are equal and the data values are evenly spread around these values. The shape of the distribution below the mean and median is a mirror image of distribution above the mean and median. A distribution of values is skewed to the right or positively skewed if there is a single peak, but the values extend much farther 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 observations extend farther to the left, in the negative direction, than to the right. In a negatively skewed distribution, the mean is smaller than the median. Positively skewed 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 distribu- tion 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.
LO4-5 Compute and interpret the coefficient of skewness.
M ed
ia n
M ea
n
45
Fr eq
ue nc
y
Fr eq
ue nc
y
Fr eq
ue nc
y
Fr eq
ue nc
y
Years
Ages
Symmetric
Monthly Salaries
Positively Skewed
$3,000 $4,000
M ed
ia n
M ea
n
Median Mean
Test Scores
Negatively Skewed
75 80 Score
Mean
Outside Diameter
Bimodal
.98 1.04 Inches$
CHART 4–1 Shapes of Frequency Polygons
There are several formulas in the statistical literature used to calculate skewness. The simplest, developed by Professor Karl Pearson (1857–1936), is based on the differ- ence between the mean and the median.
18. A sample of 28 time shares in the Orlando, Florida, area revealed the follow- ing 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.
$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
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 111
Using this relationship, the 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 there is no skewness present.
In this text, we present output from Minitab and Excel. Both of these software pack- ages compute a value for the coefficient of skewness based on the cubed deviations from the mean. The formula is:
SOFTWARE COEFFICIENT OF SKEWNESS
sk = n
(n − 1) (n − 2)[ ∑(
x − x s )
3
] [4–3]
Formula (4–3) offers an insight into skewness. The right-hand side of the formula is the difference between each value and the mean, divided by the standard deviation. That is the portion (x − x )/s of the formula. This idea is called standardizing. We will discuss the idea of standardizing a value in more detail in Chapter 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 of the standard deviation. If this difference is positive, the particular value is larger than the mean; if the value is nega- tive, 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 (3–10)] we squared the difference between each value and the mean, so that the result was all nonnegative values.
If the set of data values under consideration is symmetric, when we cube the stan- dardized 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. If there are several small values clearly separate from the others, the sum of the cubed differences will be negative.
An example will illustrate the idea of skewness.
PEARSON’S COEFFICIENT OF SKEWNESS sk = 3(x − Median)
s [4–2]
STATISTICS IN ACTION
The late Stephen Jay Gould (1941–2002) was a profes- sor of zoology and professor of geology at Harvard University. In 1982, he was diagnosed with cancer and had an expected survival time of 8 months. However, never to be discouraged, his research showed that the distribution of survival time is dramatically skewed to the right and showed that not only do 50% of similar cancer patients survive more than 8 months, but that the survival time could be years rather than months! In fact, Dr. Gould lived an- other 20 years. Based on his experience, he wrote a widely published essay titled “The Median Is Not the Message.”
E X A M P L E
Following are the earnings per share for a sample of 15 software companies for the year 2016. The earnings per share are arranged from smallest to largest.
Compute the mean, median, and standard deviation. Find the coefficient of skewness using Pearson’s estimate and the software methods. What is your conclusion regarding the shape of the distribution?
S O L U T I O N
These are sample data, so we use formula (3–2) to determine the mean
x = Σx 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
112 CHAPTER 4
The median is the middle value in a set of data, arranged from smallest to largest. In this case, there is an odd-number of observations, so the middle value is the median. It is $3.18.
We use formula (3–10) on page 78 to determine the sample standard deviation.
s = √ Σ(x − x )2
n − 1 = √
($0.09 − $4.95)2 + … + ($16.40 − $4.95)2
15 − 1 = $5.22
Pearson’s coefficient of skewness is 1.017, found by
sk = 3(x − Median)
s =
3($4.95 − $3.18) $5.22
= 1.017
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. We have 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 earnings per share of $0.09, are:
( x − x
s ) 3
= ( 0.09 − 4.95
5.22 ) 3
= (−0.9310)3 = −0.8070
When we sum the 15 cubed values, the result is 11.8274. That is, the term Σ[(x − x )/s]3 = 11.8274. To find the coefficient of skewness, we use formula (4–3), with n = 15.
sk = n
(n − 1) (n − 2) ∑(
x − x s )
3
= 15
(15 − 1) (15 − 2) (11.8274) = 0.975
We conclude that the earnings per share values are somewhat positively skewed. The following Minitab summary reports the descriptive measures, such as
TABLE 4–2 Calculation of the Coefficient of Skewness
Earnings per Share (x − x )
s (
x − x s )
3
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
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 113
A sample of five data entry clerks employed in the Horry County Tax Office revised the fol- lowing 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?
S E L F - R E V I E W 4–4
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.
36.0 26.0 33.0 28.0 31.0
20. Listed below are the salaries, in $000, for a sample of 15 chief financial offi- cers in the electronics industry.
$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
E X E R C I S E S
the mean, median, and standard deviation of the earnings per share data. Also in- cluded are the coefficient of skewness and a histogram with a bell-shaped curve superimposed.
114 CHAPTER 4
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 histogram in Chapter 2 to summarize the profit on vehicles sold by the Applewood Auto Group. Earlier in
this chapter, we used dot plots and stem-and-leaf displays 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 ana- lysts frequently wish to understand the relationship between two variables. Here are some examples:
• Tybo and Associates is a law firm that advertises ex- tensively on local TV. The partners are considering increasing their advertising budget. Before doing so, they would like to know the relationship be- tween 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?
LO4-6 Create and interpret a scatter diagram.
© Steve Mason/Getty Images RF
21. Listed below are the commissions earned ($000) last year by the 15 sales representatives at Furniture Patch Inc.
$ 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
22. Listed below are the salaries for the 2016 New York Yankees Major League Baseball team.
Player Salary Player Salary
CC Sabathia $25,000,000 Dustin Ackley $3,200,000 Mark Teixeira 23,125,000 Martin Prado 3,000,000 Masahiro Tanaka 22,000,000 Didi Gregorius 2,425,000 Jacoby Ellsbury 21,142,857 Aaron Hicks 574,000 Alex Rodriguez 21,000,000 Austin Romine 556,000 Brian McCann 17,000,000 Chasen Shreve 533,400 Carlos Beltran 15,000,000 Greg Bird 525,300 Brett Gardner 13,500,000 Luis Severino 521,300 Chase Headley 13,000,000 Bryan Mitchell 516,650 Aroldis Chapman 11,325,000 Kirby Yates 511,900 Andrew Miller 9,000,000 Mason Williams 509,700 Starlin Castro 7,857,143 Ronald Torreyes 508,600 Nathan Eovaldi 5,600,000 John Barbato 507,500 Michael Pineda 4,300,000 Dellin Betances 507,500 Ivan Nova 4,100,000 Luis Cessa 507,500
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 115
• 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 relation- ship between the height of fathers and the height of their sons. That is, do tall fathers tend to have tall children? Would you expect LeBron James, the 6′8″, 250 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 example 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 indirect rela- tionship 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 de- picts 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.
$24,000 21,000 18,000 15,000 12,000A
uc tio
n pr
ic e
10,000 30,000 50,000 Odometer
Auction Price versus Odometer $10,000
8,000 6,000 4,000 2,000
0
Co st
(a nn
ua l)
0 1 2 3 4 5 6 Age (years)
Age of Buses and Maintenance Cost Height versus Salary
125 120 115 110 105 100
95 90S
al ar
y ($
00 0)
54 55 56 57 58 59 60 61 62 63 Height (inches)
CHART 4–2 Three Examples of Scatter Diagrams.
E X A M P L E
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?
116 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 in- creases, 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 younger drivers and less for older drivers.
• 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 positions and not in a position to issue as many citations. But in any event, as age increases, the num- ber of citations decreases.
CONTINGENCY TABLES 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.
LO4-7 Develop and explain a contingency table.
S O L U T I O N
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 assume profit depends on the age of the purchaser. As people age, they earn more income and purchase more expensive cars which, in turn, produce higher profits. We use Excel to develop the scatter diagram. The Excel commands are in Appendix C.
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 variables more extensively, even calculating several numerical measures to ex- press the relationship between variables.
0 10 20 30 40 Age (Years)
Profit and Age of Buyer at Applewood Auto Group Pr
ofi t p
er V
eh ic
le ($
)
50 60 70 80 $0
$500
$1,000
$1,500
$2,000
$2,500
$3,000
$3,500
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 117
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 (freshman, sophomore, junior, or senior).
• A product is classified as acceptable or unacceptable and by the shift (day, after- noon, or night) on which it is manufactured.
• A voter in a school bond referendum is classified as to party affiliation (Democrat, 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.
E X A M P L E
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?
S O L U T I O N
In a contingency table, both variables only need to be nominal or ordinal. In this example, the variable dealership is a nominal variable and the variable profit is a ratio variable. To convert profit to an ordinal variable, we classify the variable profit into two categories, those cases where the profit earned is more than the median and those cases where it is less. On page 64, 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
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%, 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% for Olean, 42% for Sheffield, and 60% for Tionesta.
We will return to the study of contingency tables in Chapter 5 during the study of probability and in Chapter 15 during the study of nonparametric methods of analysis.
118 CHAPTER 4
The rock group Blue String Beans is touring the United States. The following chart shows the relationship between concert seating capacity and revenue in $000 for a sample of concerts.
5800 6300 6800 Seating Capacity
8
7
6
5
4
3
2
Am ou
nt ($
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?
S E L F - R E V I E W 4–5
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
24. Silver Springs Moving and Storage Inc. is studying the relationship between the number 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.
1 2 3 Rooms
40
30
20
10
0
Ho ur
s
54
E X E R C I S E S
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 119
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 be- tween the gender of a guest and whether the guest orders dessert. To investigate the relationship, the manager collected the following information on 200 recent customers.
Gender
Dessert Ordered Male Female Total
Yes 32 15 47 No 68 85 153
Total 100 100 200
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
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?
C H A P T E R S U M M A R Y
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% are less than the second quartile, and 75% 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 10 equal parts and percentiles into 100 equal parts.
120 CHAPTER 4
IV. A box plot is a graphic display of a set of data. A. A box is drawn enclosing the regions between the first quartile and the third quartile.
1. A line is drawn inside the box at the median value. 2. Dotted line segments are drawn from the third quartile to the largest value to
show the highest 25% of the values and from the first quartile to the smallest value to show the lowest 25% of the values.
B. A box plot is based on five statistics: the maximum and minimum values, the first and third quartiles, and the median.
V. 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:
sk = 3(x − Median)
s [4–2]
2. The coefficient of skewness computed by statistical software is:
sk = n
(n − 1) (n − 2)[ ∑(
x − x s )
3
] [4–3]
VI. 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 un-
der 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. VII. A contingency table is used to classify nominal-scale observations according to two
characteristics.
P R O N U N C I A T I O N K E Y
SYMBOL MEANING PRONUNCIATION
Lp Location of percentile L sub p
Q1 First quartile Q sub 1
Q3 Third quartile Q sub 3
C H A P T E R E X E R C I S E S
27. A sample of students attending Southeast Florida University is asked the number of so- cial activities in which they participated last week. The chart below was prepared from the sample data.
41 2 Activities
30
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, Moncks Corner, 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.
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 121
5020 30 Patients
4010
Location
Georgetown
Moncks Corner
Aynor
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. Below is the number of customers who visited Smith’s True-Value hardware store in Bellville, Ohio, over the last twenty-three days. Make a stem-and-leaf display of this variable.
46 52 46 40 42 46 40 37 46 40 52 32 37 32 52 40 32 52 40 52 46 46 52
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 de- scription of your findings.
Company Name Year Founded Employees
AES Corp. 1981 30,000 American Capital Ltd. 1986 484 AvalonBay Communities Inc. 1978 1,767 Capital One Financial Corp. 1995 31,800 Constellation Energy Group Inc. 1816 9,736 Coventry Health Care Inc. 1986 10,250 Danaher Corp. 1984 45,000 Dominion Resources Inc. 1909 17,500 Fannie Mae 1938 6,450 Freddie Mac 1970 5,533 Gannett Co. 1906 49,675 General Dynamics Corp. 1952 81,000 Genworth Financial Inc. 2004 7,200 Harman International Industries Inc. 1980 11,246 Host Hotels & Resorts Inc. 1927 229 Legg Mason 1899 3,800 Lockheed Martin Corp. 1995 140,000 Marriott International Inc. 1927 151,000 MedImmune LLC 1988 2,516 NII Holdings Inc. 1996 7,748 Norfolk Southern Corp. 1982 30,594 Pepco Holdings Inc. 1896 5,057 Sallie Mae 1972 11,456 T. Rowe Price Group Inc. 1937 4,605 The Washington Post Co. 1877 17,100
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
122 CHAPTER 4
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.
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 25 senior citizens and 30 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, an on-line banking company, 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 employees to use public transportation. Below is a listing of the time to get to work this morning according to whether the em- ployee 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
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 pub- lished. Be sure to include information on the values of the first and third quartiles,
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 123
the median, and whether there is any skewness. If there are any outliers, estimate their value.
Number of Newspapers
****
0 20 40 60 80 100
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 in- voices. 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.
Invoice Amount
*
0 50 100 150 200 250
36. The American Society of PeriAnesthesia Nurses (ASPAN; www.aspan.org) is a national organization serving nurses practicing in ambulatory surgery, preanesthesia, and postanesthesia care. The organization consists of the 40 components listed below.
State/Region Membership
Alabama 95 Arizona 399 Maryland, Delaware, DC 531 Connecticut 239 Florida 631 Georgia 384 Hawaii 73 Illinois 562 Indiana 270 Iowa 117 Kentucky 197 Louisiana 258 Michigan 411 Massachusetts 480 Maine 97 Minnesota, Dakotas 289 Missouri, Kansas 282 Mississippi 90 Nebraska 115 North Carolina 542 Nevada 106
State/Region Membership
New Jersey, Bermuda 517 Alaska, Idaho, Montana, Oregon, Washington 708 New York 891 Ohio 708 Oklahoma 171 Arkansas 68 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
Use statistical software to answer the following questions. a. Find the mean, median, and standard deviation of the number of members per
component.
124 CHAPTER 4
b. Find the coefficient of skewness, using the software. What do you conclude about the shape of the distribution of component size?
c. Compute the first and third quartiles using formula (4–1). 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 posted an advertisement on a social media site reporting the shape, size, price, and cut grade for 33 of its diamonds currently in stock. The information is re- ported below.
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 outliers? What is the median price? What are the values of the first and the third quartiles?
b. Develop a box plot of the variable size and comment on the result. Are there any outliers? What is the median price? What are the values of the first and the third quartiles?
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 associ- ation 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?
38. Listed below is the amount of commissions earned last month for the eight mem- bers of the sales staff at Best Electronics. Calculate the coefficient of skewness using both methods. Hint: Use of a spreadsheet will expedite the calculations.
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 expe- dite the calculations.
3 12 13 7 8 3 8
DESCRIBING DATA: DISPLAYING AND EXPLORING DATA 125
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, copier, 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 re- lationship between the variables?
34 39 44 Months
$130
120
110
100
90
80
M on
th ly
M ai
nt en
an ce
C os
t 49
41. An auto insurance company reported the following information regarding the age of a driver 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
42. Wendy’s offers eight different condiments (mustard, catsup, onion, mayonnaise, pickle, lettuce, tomato, and relish) on hamburgers. A store manager collected the following in- formation 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 number of condiments?
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
43. Here is a table showing the number of employed and unemployed workers 20 years or older by gender in the United States.
Number of Workers (000)
Gender Employed Unemployed
Men 70,415 4,209 Women 61,402 3,314
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.
126 A REVIEW OF CHAPTERS 1–4
D A T A A N A L Y T I C S
44. Refer to the North Valley real estate data recorded on homes sold during the last year. Prepare a report on the selling prices of the homes based on the answers to the following questions. a. Compute the minimum, maximum, median, and the first and the third quartiles of
price. Create a box plot. Comment on the distribution of home prices. b. Develop a scatter diagram with price on the vertical axis and the size of the home on
the horizontal. Is there a relationship between these variables? Is the relationship direct or indirect?
c. For homes without a pool, develop a scatter diagram with price on the vertical axis and the size of the home on the horizontal. Do the same for homes with a pool. How do the relationships between price and size for homes without a pool and homes with a pool compare?
45. Refer to the Baseball 2016 data that report information on the 30 Major League Baseball teams for the 2016 season. a. In the data set, the year opened, is the first year of operation for that stadium. For
each team, use this variable to create a new variable, stadium age, by subtracting the value of the variable, year opened, from the current year. Develop a box plot with the new variable, age. Are there any outliers? If so, which of the stadiums are outliers?
b. Using the variable, salary, create a box plot. Are there any outliers? Compute the quartiles using formula (4–1). Write a brief summary of your analysis.
c. Draw a scatter diagram with the variable, wins, on the vertical axis and salary on the horizontal axis. What are your conclusions?
d. Using the variable, wins, draw a dot plot. What can you conclude from this plot? 46. Refer to the Lincolnville School District bus data.
a. Referring to the maintenance cost variable, develop a box plot. What are the mini- mum, first quartile, median, third quartile, and maximum values? Are there any outliers?
b. Using the median maintenance cost, develop a contingency table with bus manufac- turer as one variable and whether the maintenance cost was above or below the median as the other variable. What are your conclusions?
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 distribution as a histogram or a frequency polygon. Chapter 3 began by describing measures of location, 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, 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 skew- ness, 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 of variation calculated, and the information presented in graphical form.
A REVIEW OF CHAPTERS 1–4 127
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 following 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 distri-
bution 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 45 U.S. presidents and their age as they began their terms in office.
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 23 B. Harrison 55
Number Name Age
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 45 Trump 70
Use a statistical software package such as Excel or Minitab to help answer the following 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 distri-
bution 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.
P R O B L E M S
1. The duration in minutes of a sample of 50 power outages last year in the state of South Carolina is listed below.
128 A REVIEW OF CHAPTERS 1–4
3. Listed below is the 2014 median household income for the 50 states and the District of Columbia. https://www.census.gov/hhes/www/income/data/historical/ household/
State Amount
Alabama 42,278 Alaska 67,629 Arizona 49,254 Arkansas 44,922 California 60,487 Colorado 60,940 Connecticut 70,161 Delaware 57,522 D.C. 68,277 Florida 46,140 Georgia 49,555 Hawaii 71,223 Idaho 53,438 Illinois 54,916 Indiana 48,060 Iowa 57,810 Kansas 53,444 Kentucky 42,786 Louisiana 42,406 Maine 51,710 Maryland 76,165 Massachusetts 63,151 Michigan 52,005 Minnesota 67,244 Mississippi 35,521 Missouri 56,630
State Amount
Montana 51,102 Nebraska 56,870 Nevada 49,875 New Hampshire 73,397 New Jersey 65,243 New Mexico 46,686 New York 54,310 North Carolina 46,784 North Dakota 60,730 Ohio 49,644 Oklahoma 47,199 Oregon 58,875 Pennsylvania 55,173 Rhode Island 58,633 South Carolina 44,929 South Dakota 53,053 Tennessee 43,716 Texas 53,875 Utah 63,383 Vermont 60,708 Virginia 66,155 Washington 59,068 West Virginia 39,552 Wisconsin 58,080 Wyoming 55,690
Use a statistical software package such as Excel or Minitab to help answer the following 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 distri-
bution 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 information. 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?
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
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?
A REVIEW OF CHAPTERS 1–4 129
C A S E S
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 writ- ten 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 custom- ers. 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? To better understand the customers, Mr. Selig asked Ms. Wendy Lamberg, director of planning, to select a sam- ple of customers and prepare a report. To begin, 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 (automatic teller machine) transac- tions in the last month; (2) the number of other bank ser- vices (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 customer’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 ac- count 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.
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 deviation 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 es- sentially the same job but at different pay. To investi- gate, he collected the information below. Suppose you are a student intern in the Accounting Department and have been given the task to write a report summarizing the situation.
Yearly Salary ($000) Women Men
Less than 30 2 0 30 up to 40 3 1 40 up to 50 17 4 50 up to 60 17 24 60 up to 70 8 21 70 up to 80 3 7 80 or more 0 3
To kick off the project, Mr. Cory St. Julian held a meeting with his staff and you were invited. At this meeting, it was suggested that you calculate several measures of
130 A REVIEW OF CHAPTERS 1–4
location, create charts or draw graphs such as a cumula- tive frequency distribution, and determine the quartiles for both men and women. Develop the charts and write the report summarizing the yearly salaries of employees at Wildcat Plumbing Supply. Does it appear that there are pay differences based on gender?
C. Kimble Products: Is There a Difference In the Commissions?
At the January national sales meeting, the CEO of Kimble Products was questioned extensively regarding the com- pany policy for paying commissions to its sales represen- tatives. The company sells sporting goods to two major
markets. There are 40 sales representatives who call di- rectly on large-volume customers, such as the athletic de- partments at major colleges and universities and professional sports franchises. There are 30 sales repre- sentatives who represent the company to retail stores lo- cated in shopping malls and large discounters such as Kmart and Target. Upon his return to corporate headquarters, the CEO asked the sales manager for a report comparing the com- missions earned last year by the two parts of the sales team. The information is reported below. Write a brief re- port. Would you conclude that there is a difference? Be sure to include information in the report on both the cen- tral tendency and dispersion of the two groups.
Commissions Earned by Sales Representatives Calling on Large Retailers ($)
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
P R A C T I C E T E S T
There is a practice test at the end of each review section. The tests are in two parts. The first part contains several objec- tive 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 decisions is called . 1. 2. Methods of organizing, summarizing, and presenting data in an informative way are
called . 2. 3. The entire set of individuals or objects of interest or the measurements obtained from all
individuals or objects of interest are 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 are 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 the 2k method suggest 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,
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.
Part 2—Problems 1. The Russell 2000 index of stock prices increased by the following amounts over the last 3 years.
18% 4% 2%
What is the geometric mean increase for the 3 years?
2. The information below refers to the selling prices ($000) of homes sold in Warren, Pennsylvania, during 2016.
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 2016? 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 2013, 300 shares at $40 each in September 2015, and 500 shares at $50 each in January 2016. What is the investor’s weighted mean price per share?
5. During the 50th Super Bowl, 30 million pounds of snack food were eaten. The chart below depicts this information.
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?
A REVIEW OF CHAPTERS 1–4 131
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO5-1 Define the terms probability, experiment, event, and outcome.
LO5-2 Assign probabilities using a classical, empirical, or subjective approach.
LO5-3 Calculate probabilities using the rules of addition.
LO5-4 Calculate probabilities using the rules of multiplication.
LO5-5 Compute probabilities using a contingency table.
LO5-6 Calculate probabilities using Bayes’ theorem.
LO5-7 Determine the number of outcomes using principles of counting.
RECENT SURVEYS indicate 60% of 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 terra-cotta 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 LO5-3.)
A Survey of Probability Concepts5
© Karin Slade/Getty Image
A SURVEY OF PROBABILITY CONCEPTS 133
INTRODUCTION The emphasis in Chapters 2, 3, and 4 is on descriptive statistics. In Chapter 2, we orga- nize the profits on 180 vehicles sold by the Applewood Auto Group into a frequency 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 mea- sures of location and dispersion to locate a typical profit on vehicle sales and to exam- ine 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 deviation. In Chapter 4, we de- velop charts and graphs, such as a scatter diagram or a dot plot, 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 infer- ence or inferential statistics.
Seldom does a decision maker have complete information to make a decision. For example:
• Toys and Things, a toy and puzzle manufac- turer, 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. To investigate, the president of Toys and Things decided to hire a market research firm. The firm selected a sample of 800 consumers from the population and asked each respon- dent 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 U.S. Steel mill must assure management that the quarter-inch wire being produced has an acceptable tensile strength. Clearly, not all the wire produced can be tested for tensile strength because testing re- quires the wire to be stretched until it breaks—thus destroying it. So a random sam- ple 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 sample taken from that population. (The populations for the preceding illustrations are all con- sumers who like sports trivia games, all the quarter-inch steel wire produced, all televi- sion 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 theory, often referred to as the science of uncertainty. Probability theory allows the decision maker 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 addi- tion and multiplication rules.
© Ballda/Shutterstock.com
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 lottery 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 2 × 1.7/100,000,000 = 0.000000034. 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/PowerballOdds.html
134 CHAPTER 5
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% chance of rain for Super Bowl Sunday. Based on a survey of consumers who tested a newly developed toothpaste with a banana flavor, the probability is .03 that, if marketed, it will be a financial success. (This means that the chance of the banana-flavor tooth- paste being accepted by the public is rather remote.) What is a probability? In general, it is a numerical value that describes the chance that something will happen.
LO5-1 Define the terms probability, experiment, event, and outcome.
PROBABILITY A value between zero and one, inclusive, describing the relative possibility (chance or likelihood) an event will occur.
A probability is frequently expressed as a decimal, such as .70, .27, or .50, or a percent such as 70%, 27% or 50%. It also may be reported as a fraction such as 7/10, 27/100, or 1/2. It can assume any number from 0 to 1, inclusive. Expressed as a per- centage, the range is between 0% and 100%, inclusive. If a company has only five sales regions, and each region’s name or number is written on a slip of paper and the slips put in a hat, the probability of selecting one of the five regions is 1. The probability of select- ing from the hat a slip of paper that reads “Pittsburgh Steelers” is 0. Thus, the probability of 1 represents something that is certain to happen, and the probability of 0 represents something 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 likely it will happen. The relationship is shown in the following diagram along with a few of our personal beliefs. You might, however, se- lect 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
Sometimes, the likelihood of an event is expressed using the term odds. To explain, someone says the odds are “five to two” that an event will occur. This means that in a total of seven trials (5 + 2), the event will occur five times and not occur two times. Using odds, we can compute the probability that the event occurs as 5/(5 + 2) or 5/7. So, if the odds in favor of an event are x to y, the probability of the event is x/(x + y).
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 spe- cific meanings.
EXPERIMENT A process that leads to the occurrence of one and only one of several possible results.
A SURVEY OF PROBABILITY CONCEPTS 135
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 proba- bility, an experiment has two or more possible results, and it is uncertain which will occur.
OUTCOME A particular result of an experiment.
EVENT A collection of one or more outcomes of an experiment.
For example, the tossing of a coin is an experiment. You are unsure of the outcome. When a coin is tossed, one particular outcome is a “head.” The alternative outcome is a “tail.” Similarly, asking 500 college students if they would travel more than 100 miles to attend a Mumford and Sons concert is an experiment. In this experiment, one possible outcome is that 273 students indicate they would travel more than 100 miles to attend the concert. Another outcome is that 317 students would attend the concert. Still an- other outcome is that 423 students indicate they would attend the concert. When one or more of the experiment’s outcomes are observed, we call this an event.
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 number 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 is 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
136 CHAPTER 5
APPROACHES TO ASSIGNING PROBABILITIES In this section, we describe three ways to assign a probability to an event: classical, empirical, and subjective. The classical and empirical methods are objective and are based on information and data. The subjective method is based on a person’s belief or estimate of an event’s likelihood.
Classical Probability Classical probability is based on the assumption that the outcomes of an experiment are equally likely. Using the classical viewpoint, the probability of an event happening is com- puted by dividing the number of favorable outcomes by the number of possible outcomes:
LO5-2 Assign probabilities using a classical, empirical, or subjective approach.
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 of the 80 players testing the new game 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.
S E L F - R E V I E W 5–1
Probability of an event =
Number of favorable outcomes Total number of possible outcomes
[5–1]CLASSICAL PROBABILITY
E X A M P L E
Consider an experiment of rolling a six-sided die. What is the probability of the event “an even number of spots appear face up”?
S O L U T I O N
The possible outcomes are:
a one-spot
a two-spot
a three-spot
a four-spot
a five-spot
a six-spot
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 = 3 6
← ←
Number of favorable outcomes
Total number of possible outcomes = .5
The mutually exclusive concept appeared earlier in our study of frequency distri- butions 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.
A SURVEY OF PROBABILITY CONCEPTS 137
MUTUALLY EXCLUSIVE The occurrence of one event means that none of the other events can occur at the same time.
COLLECTIVELY EXHAUSTIVE At least one of the events must occur when an experiment is conducted.
EMPIRICAL PROBABILITY The probability of an event happening is the fraction of the time similar events happened in the past.
LAW OF LARGE NUMBERS Over a large number of trials, the empirical probability of an event will approach its true probability.
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 manufac- tured part is acceptable or unacceptable. The part cannot be both acceptable and unac- ceptable 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 mutually exclusive.
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 out- come will be either even or odd. So the set is collectively exhaustive.
If the set of events is collectively exhaustive and the events are mutually exclusive, the sum of the probabilities is 1. Historically, the classical approach to probability 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 probability 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 outcomes; the roll of a die has six possible outcomes. We can logically arrive at the probability 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 lottery 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 order 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.
The formula to determine an empirical probability is:
Empirical probability = Number of times the event occurs
Total number of observations 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.
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 probability for
138 CHAPTER 5
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 seven different experiments 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.
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 ap- proaches 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 empirical approach to assigning a probability, we estimate the likelihood a student at Scandia will earn an A is .15.
• Stephen Curry of the Golden State Warriors made 363 out of 400 free throw attempts during the 2015–16 NBA season. Based on the empirical approach to probability, the likelihood of him making his next free throw attempt is .908.
Life insurance companies rely on past data to determine the acceptability of an appli- cant 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 likelihood a 20-year-old female will die within the next year is .00105.
The empirical concept is illustrated with the following example.
E X A M P L E
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?
S O L U T I O N
We use letters or numbers to simplify the equations. P stands for probability and A represents the event of a successful mission. In this case, P(A) stands for the probability a future mission is successfully completed.
Probability of a successful flight = Number of successful flights
Total number of flights
P(A) = 111 113
= .98
We can use this as an estimate of probability. In other words, based on past experience, the probability is .98 that a future space shuttle mission will be safely completed.
A SURVEY OF PROBABILITY CONCEPTS 139
Subjective Probability If there is little or no experience or information on which to base a probability, it is esti- mated subjectively. Essentially, this means an individual evaluates the available opin- ions and information and then estimates or assigns the probability. This probability is called a subjective probability.
SUBJECTIVE CONCEPT OF PROBABILITY The likelihood (probability) of a particular event happening that is assigned by an individual based on whatever information is available.
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 are involved in an automobile accident during the next 12 months.
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 al- ways assigns a likelihood to an event that has not yet occurred. There is, of course, considerable latitude in the degree of uncertainty that surrounds this probability, based primarily on the knowledge possessed by the individual concerning the underlying pro- cess. 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 acceptance 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% 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 prob- ability value to an event of interest, and a difference exists only in the predictor’s confi- dence in the precision of the estimate. However, regardless of the viewpoint, the same laws of probability (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
140 CHAPTER 5
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 probability a particular child chosen at random will have a parent who is divorced? Which approach did you use?
3. What is the probability you will save one million dollars by the time you retire? Which approach to probability did you use to answer this question?
S E L F - R E V I E W 5–2
1. Some people are in favor of reducing federal taxes to increase consumer spending and others 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 experiment 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
From the 34 students, suppose you randomly select a student. 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 must hire a new president. The Board of Directors prepares a 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 probability 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 randomly chosen as the spokesperson?
c. You purchase a ticket for the Lotto Canada lottery. Over 5 million tickets were sold. 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 all possible outcomes. b. What probability concept would be used to assign probabilities to the outcomes?
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?
E X E R C I S E S
A SURVEY OF PROBABILITY CONCEPTS 141
RULES OF ADDITION FOR COMPUTING PROBABILITIES There are two rules of addition, the special rule of addition and the general rule of addi- tion. We begin with the special rule of addition.
Special Rule of Addition When we use the special rule of addition, the events must be mutually exclusive. Recall that mutually exclusive means that when one event occurs, 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 probabili- ties. This rule is expressed in the following formula:
LO5-3 Calculate probabilities using the rules of addition.
8. A sample of 2,000 licensed drivers revealed the following number of speed- ing 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. 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 num- ber (PIN) for use at ATMs.
a. Think of this as an experiment and list four possible outcomes. b. What is the probability that a customer will pick 259 as their 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. Which concept of probability did you use in (a)?
SPECIAL RULE OF ADDITION P(A or B) = P(A) + P(B) [5–2]
For three mutually exclusive events designated A, B, and C, the rule is written:
P(A or B or C) = P(A) + P(B) + P(C)
An example will show the details.
142 CHAPTER 5
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 dia- gram, 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 that is drawn inside the rectangle proportional to the probability of the event. The following Venn diagram represents the mutually exclusive concept. There is no overlap- ping of events, meaning that the events are mutually exclusive. In the following Venn diagram, assume the events A, B, and C are about equally likely.
E X A M P L E
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 over- weight. A check of 4,000 packages filled in the past month revealed:
Number of Probability of Weight Event Packages Occurrence
Underweight A 100 .025 ← 100
4,000Satisfactory B 3,600 .900 Overweight C 300 .075
4,000 1.000
What is the probability that a particular package will be either underweight or overweight?
S O L U T I O N
The outcome “underweight” is the event A. The outcome “overweight” is the event C. Applying the special rule of addition:
P(A or C) = P(A) + P(C) = .025 + .075 = .10
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.
© Ian Dagnall/Alamy Stock Photo
Event A
Event B
Event C
A SURVEY OF PROBABILITY CONCEPTS 143
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(∼A) and read “not A,” must logi- cally equal 1. This is written:
P(A) + P(∼A) = 1
This can be revised to read:
COMPLEMENT RULE P(A) = 1 − P(∼A) [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 happening by determining the probability of it not happening and subtracting the result from 1. Notice that the events A and ∼A are mutually exclusive and collectively exhaustive. Therefore, the probabilities of A and ∼A sum to 1. A Venn diagram illustrating the complement rule is shown as:
Event A
, A
E X A M P L E
Referring to the previous example/solution, the probability a bag of mixed vegeta- bles is underweight is .025 and the probability 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.
S O L U T I O N
The probability the bag is unsatisfactory equals the probability the bag is over- weight 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 .025
not (A or C) .90
C .075
144 CHAPTER 5
The General Rule of Addition The outcomes of an experiment may not be mutually exclusive. For example, 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 probabilities is 1.10. We know, however, that this probability cannot be greater than 1. The explanation is that many tour- ists 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 World and the probability he or she visited Busch Gardens, and (2) subtract the proba- bility of visiting both. Thus:
P(Disney or Busch) = P(Disney) + P(Busch) − P(both Disney and Busch) = .60 + .50 − .30 = .80
When two events both occur, the probability is called a joint probability. The prob- ability (.30) that a tourist visits both attractions is an example of a joint probability.
© Rostislav Glinsky/Shutterstock.com
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.
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?
S E L F - R E V I E W 5–3
STATISTICS IN ACTION
If you wish to get some attention at the next gath- ering 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 necessarily 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 people 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 birthday. Hint: To compute this, find the probability everyone was born on a different day and use the complement rule. Try this in your class.
A SURVEY OF PROBABILITY CONCEPTS 145
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.
So the general rule of addition, which is used to compute the probability of two events that are not mutually exclusive, is:
GENERAL RULE OF ADDITION P(A or B) = P(A) + P(B) − P(A and B) [5–4]
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 sometimes 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 difference 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. Other- wise, we must account for the joint probability and use the general rule of addition.
E X A M P L E
What is the probability that a card chosen at random from a standard deck of cards will be either a king or a heart?
S O L U T I O N
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
146 CHAPTER 5
From formula (5–4):
P(A or B) = P(A) + P(B) − P(A and B) = 4/52 + 13/52 − 1/52 = 16/52, or .3077
A Venn diagram portrays these outcomes, which are not mutually exclusive.
Kings
Hearts
Both
A B A
and
B
Routine physical examinations are conducted annually as part of a health service program for General Concrete Inc. employees. It was discovered that 8% of the employees need corrective shoes, 15% need major dental work, and 3% 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.
S E L F - R E V I E W 5–4
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 in- come 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% chance this company will earn a profit, a 30% chance it will break even, and a 20% 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?
E X E R C I S E S
A SURVEY OF PROBABILITY CONCEPTS 147
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 probability of passing both is .50. What is the probability of passing at least one?
21. The aquarium at Sea Critters Depot contains 140 fish. Eighty of these fish are green swordtails (44 female and 36 male) and 60 are orange swordtails (36 female and 24 males). A fish is randomly captured from the aquarium:
a. What is the probability the selected fish is a green swordtail? b. What is the probability the selected fish is male? c. What is the probability the selected fish is a male green swordtail? d. What is the probability the selected fish is either a male or a green swordtail?
22. A National Park Service survey of visitors to the Rocky Mountain region revealed that 50% visit Yellowstone Park, 40% visit the Tetons, and 35% 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 TO CALCULATE PROBABILITY In this section, we discuss the rules for computing the likelihood that two events both happen, or their joint probability. For example, 16% of the 2016 tax returns were pre- pared by H&R Block and 75% of those returns showed a refund. What is the likelihood a person’s tax form was prepared by H&R Block and the person received a refund? 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 multipli- cation: 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 probabil- ity of the occurrence of the other event.
LO5-4 Calculate probabilities using the rules of multiplication.
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 differ- ent 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 P(A and B) = P(A)P(B) [5–5]
148 CHAPTER 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)
STATISTICS IN ACTION
In 2000 George W. Bush won the U.S. presidency by the slimmest of margins. Many election stories resulted, some involving voting irregularities, others raising interesting election questions. In a local Michigan election, there was a tie between two candidates for an elected position. To break the tie, the candi- dates drew a slip of paper from a box that contained two slips of paper, one marked “Winner” and the other unmarked. To deter- mine 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, because the two events are indepen- dent. Winning the coin flip did not alter the probability of either candidate drawing the winning slip of paper.
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?
S E L F - R E V I E W 5–5
General Rule of Multiplication If two events are not independent, they are referred to as dependent. To illustrate depen- dency, 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.)
E X A M P L E
A survey by the American Automobile Association (AAA) revealed 60% of its mem- bers made airline reservations last year. Two members are selected at random. What is the probability both made airline reservations last year?
S O L U T I O N
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. The probability that the second member selected made a reservation is also .60, so P(R2) = .60. Because 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:
P(R1 and R2) = P(R1)P(R2) = (.60)(.60) = .36
All possible outcomes can be shown as follows. R means a reservation is made, and ∼R means no reservation is 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 (special addition rule) making a reservation is .48 (.24 + .24). You can also observe that the outcomes are mutually exclusive and collectively exhaustive. Therefore, the proba- bilities sum to 1.00.
Outcomes Joint Probability
R1 R2 (.60)(.60) = .36 R1 ∼R2 (.60)(.40) = .24 ∼R1 R2 (.40)(.60) = .24 ∼R1 ∼R2 (.40)(.40) = .16 Total 1.00
A SURVEY OF PROBABILITY CONCEPTS 149
The fraction 2/9 (or 3/9) is called a conditional probability because its value is conditional on (dependent on) whether a diet or regular soda was the first selection from the cooler.
CONDITIONAL PROBABILITY The probability of a particular event occurring, given that another event has occurred.
In the general rule of multiplication, the conditional probability is required to compute the joint probability of two events that are not independent. For two events, A and B, that are not independent, the conditional probability is represented as P(B | A), and ex- pressed as the probability of B given A. Or the probability of B is conditional on the oc- currence and effect of event A. Symbolically, the general rule of multiplication for two events that are not independent is:
GENERAL RULE OF MULTIPLICATION P(A and B) = P(A)P(B | A) [5–6]
We can extend the general rule of multiplication to more than two events. For three events A, B, and C, the formula is:
P(A and B and C) = P(A)P(B | A)P(C | A and B)
In the case of the golf shirt example, the probability of selecting three white shirts with- out replacement is:
P(W1 and W2 and W3) = P(W1)P(W2 | W1)P(W3 | W1 and W2) = ( 9 12)(
8 11)(
7 10) = .38
So the likelihood of selecting three shirts without replacement and all being white is .38.
E X A M P L E
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 launder and return the used shirts to the closet. What is the likelihood both shirts selected are white?
S O L U T I O N
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 | W1) = 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).
P(W1 and W2) = P(W1)P(W2 | W1) = ( 9 12)(
8 11) = .55
So the likelihood of selecting two shirts and finding them both to be white is .55.
150 CHAPTER 5
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.
S E L F - R E V I E W 5–6
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 on page 116 in Chapter 4. To review, we refer to a two-way table as a contingency table.
LO5-5 Compute probabilities using a contingency table.
CONTINGENCY TABLE A table used to classify sample observations according to two or more identifiable categories or classes.
A contingency table is a cross-tabulation that simultaneously summarizes two variables of interest and their relationship. The level of measurement can be nominal. Below are several examples.
• One hundred fifty adults were asked their gender and the number of Facebook accounts they used. The following table summarizes the results.
Gender
Facebook Accounts 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.
The following example shows how the rules of addition and multiplication are used when we employ contingency tables.
A SURVEY OF PROBABILITY CONCEPTS 151
E X A M P L E
Last month, the National Association of Theater Managers conducted a survey of 500 randomly selected adults. The survey asked respondents their age and the number of times they saw a movie in a theater. The results are summarized in Table 5–1.
TABLE 5–1 Number of Movies Attended per Month by Age
Age
Less than 30 30 up to 60 60 or Older Movies per Month B1 B2 B3 Total
0 A1 15 50 10 75 1 or 2 A2 25 100 75 200 3, 4, or 5 A3 55 60 60 175 6 or more A4 5 15 30 50
Total 100 225 175 500
The association is interested in understanding the probabilities that an adult will see a movie in a theater, especially for adults 60 and older. This information is useful for making decisions regarding discounts on tickets and concessions for seniors. Determine the probability of:
1. Selecting an adult who attended 6 or more movies per month. 2. Selecting an adult who attended 2 or fewer movies per month. 3. Selecting an adult who attended 6 or more movies per month or is 60 years of
age or older. 4. Selecting an adult who attended 6 or more movies per month given the person
is 60 years of age or older. 5. Selecting an adult who attended 6 or more movies per month and is 60 years
of age or older.
Determine the independence of:
6. Number of movies per month attended and the age of the adult.
S O L U T I O N
Table 5–1 is called a contingency table. In a contingency table, an individual or an object is classified according to two criteria. In this example, a sampled adult is classi- fied by age and by the number of movies attended per month. The rules of addition [formulas (5–2) and (5–4)] and the rules of multiplication [formulas (5–5) and (5–6)] allow us to answer the various probability questions based on the contingency table.
1. To find the probability that a randomly selected adult attended 6 or more mov- ies per month, focus on the row labeled “6 or more” (also labeled A4) in Table 5–1. The table shows that 50 of the total of 500 adults are in this class. Using the empirical approach, the probability is computed:
P(6 or more) = P(A4) = 50
500 = .10
This probability indicates 10% of the 500 adults attend 6 or more movies per month.
2. To determine the probability of randomly selecting an adult who went to 2 or fewer movies per month, two outcomes must be combined: attending 0 mov- ies per month and attending 1 or 2 movies per month. These two outcomes are mutually exclusive. That is, a person can only be classified as attending 0
152 CHAPTER 5
movies per month, or 1 or 2 movies per month, not both. Because the two outcomes are mutually exclusive, we use the special rule of addition [formula (5–2)] by adding the probabilities of attending no movies and attending 1 or 2 movies:
P [(attending 0) or (attending 1 or 2)] = P(A1) + P(A2) = ( 75
500 +
200 500) = .55
So 55% of the adults in the sample attended 2 or fewer movies a month. 3. To determine the probability of randomly selecting an adult who went to “6 or
more” movies per month or whose age is “60 or older,” we again use the rules of addition. However, in this case the outcomes are not mutually exclu- sive. Why is this? Because a person can attend more than 6 movies per month, be 60 or older, or be both. So the two groups are not mutually exclu- sive because it is possible that a person would be counted in both groups. To determine this probability, the general rule of addition [formula (5–4)] is used.
P [(6 or more) or (60 or older)] = P(A4) + P(B3) − P(A4 and B3)
= ( 50
500 +
175 500
− 30
500) = .39
So 39% of the adults are either 60 or older, attend 6 or more movies per month, or both.
4. To determine the probability of selecting a person who attends 6 or more mov- ies per month given that the person is 60 or older, focus only on the column labeled B3 in Table 5–1. That is, we are only interested in the 175 adults who are 60 or older. Of these 175 adults, 30 attended 6 or more movies. Using the general rule of multiplication [formula (5–6)]:
P[(6 or more) given (60 or older)] = P(A4 | B3) = 30 175
= .17
Of the 500 adults, 17% of adults who are 60 or older attend 6 or more movies per month. This is called a conditional probability because the probability is based on the “condition” of being the age of 60 or older. Recall that in part (1), 10% of all adults attend 6 or more movies per month; here we see that 17% of adults who are 60 or older attend movies. This is valuable information for the- ater managers regarding the characteristics of their customers.
5. The probability a person attended 6 or more movies and is 60 or older is based on two conditions and they must both happen. That is, the two outcomes “6 or more movies” (A4) and “60 or older” (B3) must occur jointly. To find this joint probability we use the special rule of multiplication [formula (5–6)].
P[(6 or more) and (60 or older)] = P(A4 and B3) = P(A4)P(B3| A4)
To compute the joint probability, first compute the simple probability of the first outcome, A4, randomly selecting a person who attends 6 or more movies. To find the probability, refer to row A4 in Table 5–1. There are 50 of 500 adults that attended 6 or more movies. So P(A4) = 50/500.
Next, compute the conditional probability P(B3 | A4). This is the probability of selecting an adult who is 60 or older given that the person attended 6 or more movies. The conditional probability is:
P[(60 or older) given (60 or more)] = P(B3| A4) = 30/50
Using these two probabilities, the joint probability that an adult attends 6 or more movies and is 60 or older is:
P[(6 or more) and (60 or older)] = P(A4 and B3) = P(A4)P(B3| A4) = (50/500)(30/50) = .06
A SURVEY OF PROBABILITY CONCEPTS 153
Based on the sample information from Table 5–1, the probability that an adult is both over 60 and attended 6 or more movies is 6%. It is important to know that the 6% is relative to all 500 adults.
Is there another way to determine this joint probability without using the special rule of multiplication formula? Yes. Look directly at the cell where row A4, attends 6 or more movies, and column B3, 60 or older, intersect. There are 30 adults in this cell that meet both criteria, so P(A4 and B3) = 30/500 = .06. This is the same as computed with the formula.
6. Are the events independent? We can answer this question with the help of the results in part 4. In part 4 we found the probability of selecting an adult who was 60 or older given that the adult attended 6 or more movies was .17. If age is not a factor in movie attendance then we would expect the probability of a person who is 30 or less that attended 6 or more movies to also be 17%. That is, the two conditional probabilities would be the same. The probability that an adult attends 6 or more movies per month given the adult is less than 30 years old is:
P[(6 or more) given (less than 30)] = 5
100 = .05
Because these two probabilities are not the same, the number of movies at- tended and age are not independent. To put it another way, for the 500 adults, age is related to the number of movies attended. In Chapter 15, we investigate this concept of independence in greater detail.
Refer to Table 5–1 on page 151 to find the following probabilities. (a) What is the probability of selecting an adult that is 30 up to 60 years old? (b) What is the probability of selecting an adult who is under 60 years of age? (c) What is the probability of selecting an adult who is less than 30 years old or attended
no movies? (d) What is the probability of selecting an adult who is less than 30 years old and went to
no movies?
S E L F - R E V I E W 5–7
Tree Diagrams A tree diagram is a visual that is helpful in organizing and calculating probabilities for problems similar to the previous example/solution. This type of problem involves sev- eral stages and each stage is illustrated with a branch of the tree. The branches of a tree diagram are labeled with probabilities. We will use the information in Table 5–1 to show the construction of a tree diagram.
1. We begin the construction by drawing a box with the variable, age, on the left to represent the root of the tree (see Chart 5–2).
2. There are three main branches going out from the root. The upper branch rep- resents the outcome that an adult is less than 30 years old. The branch is labeled with the probability, P(B1) = 100/500. The next branch represents the outcome that adults are 30 up to 60 years old. This branch is labeled with the probability P(B2) = 225/500. The remaining branch is labeled P(B3) = 175/500.
3. Four branches “grow” out of each of the four main branches. These branches rep- resent the four categories of movies attended per month—0; 1 or 2; 3, 4, or 5; and 6 or more. The upper branches of the tree represent the conditional probabilities that an adult did not attend any movies given they are less than 30 years old. These are written P(A1 | B1), P(A2 | B1), P(A3 | B1), and P(A4 | B1) where A1 refers to attending no movies; A2 attending 1 or 2 movies per month; A3 attending 3, 4, or 5 movies
154 CHAPTER 5
Age 30 up to 60years old
30 years old or younger
60 years old or older
175 500 5 .35
15 100 5.15
25 100 5.25
55 100 5.55
5 100 5.05
50 225 5.22
100 225 5.44
60 225 5.27
15 225 5.07
10 175 5.06
75 175 5.43
60 175 5.34
30 175 5.17
100 500 5 .20
225 500 5 .45
0 movies
1 or 2 movies
3, 4, or 5 movies
6 or more movies
0 movies
1 or 2 movies
3, 4, or 5 movies
6 or more movies
0 movies
1 or 2 movies
3, 4, or 5 movies
6 or more movies
100 500
15 100 5 .033
100 500
25 100 5 .053
5 .11100500 55
1003
5 .01100500 5
1003
5 .10225500 50
2253
5 .20225500 100 2253
5 .12225500 60
2253
5 .03225500 15
2253
5 .02175500 10
1753
5 .15175500 75
1753
5 .12175500 60
1753
5 .06175500 30
1753
CHART 5–2 Tree Diagram Showing Age and Number of Movies Attended
per month; and A4 attending 6 or more movies per month. For the upper branch of the tree, these probabilities are 15/100, 25/100, 55/100, and 5/100. We write the conditional probabilities in a similar fashion on the other branches.
4. Finally we determine the various joint probabilities. For the top branches, the events are an adult attends no movies per month and is 30 years old or younger; an adult attends 1 or 2 movies and is 30 years old or younger; an adult attends 3, 4, or 5 movies per month and is 30 years old or younger; and an adult attends 6 or more movies per month and is 30 years old or younger. These joint probabilities are shown on the right side of Chart 5–2. To explain, the joint probability that a ran- domly selected adult is less than 30 years old and attends 0 movies per month is:
P(B1 and A1) = P(B1)P(A1| B1) = ( 100 500)(
15 100) = .03
The tree diagram summarizes all the probabilities based on the contingency table in Table 5–1. For example, the conditional probabilities show that the 60-and-older group has the highest percentage, 17%, attending 6 or movies per month. The
A SURVEY OF PROBABILITY CONCEPTS 155
30-to-60-year-old group has the highest percentage, 22%, of seeing no movies per month. Based on the joint probabilities, 20% of the adults sampled attend 1 or 2 movies per month and are 30 up to 60 years of age. As you can see, there are many observations that we can make based on the information presented in the tree diagram.
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
What is the probability of selecting a shopper who:
(a) Visited a Sears store often? (b) Visited a Sears store in an enclosed mall? (c) Visited a Sears store in an enclosed mall or visited a Sears store often? (d) Visited a Sears store often, given that the shopper went to a Sears store in an
enclosed mall?
In addition:
(e) Are the number of visits and the enclosed mall variables independent? (f) What is the probability of selecting a shopper who visited a Sears store often and it
was in an enclosed mall? (g) Draw a tree diagram and determine the various joint probabilities.
S E L F - R E V I E W 5–8
23. Suppose P(A) = .40 and P(B | A) = .30. What is the joint probability of A and B? 24. Suppose P(X1) = .75 and P(Y2 | X1) = .40. What is the joint probability of X1 and Y2? 25. A local bank reports that 80% of its customers maintain a checking account, 60%
have a savings account, and 50% have both. If a customer is chosen at random, 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 prob- ability 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 probabil- ity neither truck is available?
27. Refer to the following table.
First Event
Second Event A1 A2 A3 Total
B1 2 1 3 6 B2 1 2 1 4
Total 3 3 4 10
E X E R C I S E S
156 CHAPTER 5
a. Determine P(A1). b. Determine P(B1 | A2). c. Determine P(B2 and A3).
28. Three defective electric toothbrushes were accidentally shipped to a drugstore by Cleanbrush 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 also is 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 aver-
age 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) re- maining the same value. List the possible outcomes of this experiment. Estimate the probability at least two of the stocks increase in value.
31. A survey of 545 college students asked: What is your favorite winter sport? And, what type of college do you attend? The results are summarized below:
Favorite Winter Sport
College Type Snowboarding Skiing Ice Skating Total
Junior College 68 41 46 155 Four-Year College 84 56 70 210 Graduate School 59 74 47 180 Total 211 171 163 545
Using these 545 students as the sample, a student from this study is randomly selected.
a. What is the probability of selecting a student whose favorite sport is skiing? b. What is the probability of selecting a junior-college student? c. If the student selected is a four-year-college student, what is the probability that
the student prefers ice skating? d. If the student selected prefers snowboarding, what is the probability that the
student is in junior college? e. If a graduate student is selected, what is the probability that the student prefers
skiing or ice skating? 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 was born on Saturday?
A SURVEY OF PROBABILITY CONCEPTS 157
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 at- tempted 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.” The formula for Bayes’ theorem is:
LO5-6 Calculate probabilities using Bayes’ theorem.
BAYES’ THEOREM P(Ai ∣ B) = P(Ai)P(B ∣ Ai)
P(A1)P(B ∣ A1) + P(A2)P(B ∣ A2) [5–7]
Assume in formula (5–7) that the events A1 and A2 are mutually exclusive and collectively exhaustive, and Ai refers to either event A1 or A2. Hence A1 and A2 are in this case com- plements. The meaning of the symbols used is illustrated by the following example.
Suppose 5% of the population of Umen, a fictional Third World country, have a dis- ease that is peculiar to that country. We will let A1 refer to the event “has the disease” and A2 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 P(A1) = .05. This probability, P(A1) = P(has the disease) = .05, is called the prior probability. It is given this name because the probability is assigned before any empiri- cal data are obtained.
STATISTICS IN ACTION
A recent study by the National Collegiate Athletic Association (NCAA) reported that of 150,000 senior boys playing on their high school basketball team, 64 would make a professional team. To put it another way, the odds of a high school senior basketball player making a professional team are 1 in 2,344. From the same study: 1. The odds of a high
school senior basket- ball player playing some college basket- ball 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 basketball as a senior in college, the odds of making a professional team are about 1 in 37.5.
PRIOR PROBABILITY The initial probability based on the present level of information.
POSTERIOR PROBABILITY A revised probability based on additional information.
The prior probability a person is not afflicted with the disease is therefore .95, or P(A2) = .95, found by 1 − .05.
There is a diagnostic technique to detect the disease, but it is not very accurate. Let B denote the event “test shows the disease is present.” Assume that historical 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 probability definitions devel- oped earlier in this chapter, this statement is written as:
P(B | A1) = .90
Assume the probability is .15 that for a person who actually does not have the disease the test will indicate the disease is present.
P(B | A2) = .15
Let’s randomly select a person from Umen and perform the test. The test results in- dicate the disease is present. What is the probability the person actually has the disease? In symbolic form, we want to know P(A1 | B), which is interpreted as: P(has the disease | the test results are positive). The probability P(A1 | B) is called a posterior probability.
With the help of Bayes’ theorem, formula (5–7), we can determine the posterior probability.
P(A1 ∣ B) = P(A1)P(B ∣ A1)
P(A1)P(B ∣ A1) + P(A2)P(B ∣ A2)
= (.05) (.90)
(.05) (.90) + (.95) (.15) =
.0450 .1875
= .24
158 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 ex- haustive events, A1 and A2. If there are n such events, A1, A2, . . ., An, Bayes’ theorem, formula (5–7), becomes
P(Ai ∣ B) = P(Ai)P(B ∣ Ai)
P(A1)P(B ∣ A1) + P(A2)P(B ∣ A2) + … + P(An)P(B ∣ An)
With the preceding notation, the calculations for the Umen problem are summarized in the following table.
Prior Conditional Joint Posterior Event, Probability, Probability, Probability, Probability,
Ai P(Ai ) P(B ∣ Ai ) P(Ai and B) P(Ai ∣ B) Disease, A1 .05 .90 .0450 .0450/.1875 = .24 No disease, A2 .95 .15 .1425 .1425/.1875 = .76 P(B) = .1875 1.00
Another illustration of Bayes’ theorem follows.
E X A M P L E
A manufacturer of cell phones purchases a microchip, called the LS-24, from three suppliers: Hall Electronics, Schuller Sales, and Crawford Components. Forty-five percent of the LS-24 chips are purchased from Hall Electronics, 30% from Schuller Sales, and the remaining 25% from Crawford Compo- nents. The manufacturer has extensive histories on the three suppliers and knows that 3% of the LS-24 chips from Hall Elec- tronics are defective, 6% of chips from Schuller Sales are de- fective, and 4% of the chips purchased from Crawford Components are defective.
When the LS-24 chips arrive from the three suppliers, they are placed directly in a bin and not inspected or otherwise identified by supplier. A worker selects a chip for installation and finds it defective. What is the probability that it was manu- factured by Schuller Sales?
S O L U T I O N
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.
A1 The LS-24 was purchased from Hall Electronics. A2 The LS-24 was purchased from Schuller Sales. A3 The LS-24 was purchased from Crawford Components.
© McGraw-Hill Education/ Marker Dierker, photographer
A SURVEY OF PROBABILITY CONCEPTS 159
• The prior probabilities are:
P(A1) = .45 The probability the LS-24 was manufactured by Hall Electronics.
P(A2) = .30 The probability the LS-24 was manufactured by Schuller Sales.
P(A3) = .25 The probability the LS-24 was manufactured by Crawford Components.
• The additional information can be either:
B1 The LS-24 is defective, or B2 The LS-24 is not defective.
• The following conditional probabilities are given.
P(B1 | A1) = .03 The probability that an LS-24 chip produced by Hall Elec- tronics is defective.
P(B1 | A2) = .06 The probability that an LS-24 chip produced by Schuller Sales is defective.
P(B1 | A3) = .04 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 supplier, 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 P(A2 | B1).
Look at Schuller’s quality record. It is the worst of the three suppliers. They produce 30 percent of the product, but 6% are defective. Now that we have found a defec- tive LS-24 chip, we suspect that P(A2 | B1) is greater than the 30% of P(A2). That is, we expect the revised probability to be greater than .30. But how much greater? Bayes’ theorem can give us the answer. As a first step, consider the tree diagram in Chart 5–3.
B1 = Defective
B2 = Good
B1 = Defective
B2 = Good
B1 = Defective
B2 = Good
P (B1| A1) = .03
P (B2| A1) = .97
P (B1| A2) = .06
P (B2|A2) = .94
P (B1|A3) = .04
P (B2|A3) = .96
A2 = Schuller P (A2) = .30
A1 = Hall P (A1) = .45
A3 = Crawford P (A3) = .25
Joint probability Conditional probability
Prior probability
P (A1 and B1) = P (A1) P (B1| A1) = (.45) (.03) = .0135
P (A1 and B2) = P (A1) P (B2|A1) = (.45) (.97) = .4365
P (A2 and B1) = P (A2) P (B1|A2) = (.30) (.06) = .0180
P (A2 and B2) = P (A2) P (B2|A2) = (.30) (.94) = .2820
P (A3 and B1) = P (A3) P (B1|A3) = (.25) (.04) = .0100
P (A3 and B2) = P (A3) P (B2|A3) = (.25) (.96) = .2400
Total 1.000
CHART 5–3 Tree Diagram of the Cell Phone Manufacturing Problem
160 CHAPTER 5
REVISION OF CHART 5-3 The events are dependent, so the prior probability in the first branch is multiplied by the conditional probability in the second branch to obtain the joint probability. 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 supplier to the determination of whether the chip was defective.
What we need to do is reverse the time process. That is, instead of mov- ing from left to right in Chart 5–3, we need to move from right to left. We have a defective 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 probabil- ities as relative frequencies out of 10,000 cases. For example, the likelihood of a defective LS-24 chip that was produced by Hall Electronics is .0135. So of 10,000 cases, we would expect to find 135 defective chips produced by Hall Electronics. We observe that in 415 of 10,000 cases the LS-24 chip selected for assembly is defective, found by 135 + 180 + 100. Of these 415 defective chips, 180 were produced by Schuller Sales. Thus, the probability that the defective LS-24 chip was purchased from Schuller Sales is 180/415 = .4337. We have now determined the revised probability of P(A2 | B1). Before we found the defective chip, the likelihood that it was purchased from Schuller Sales was .30. This likeli- hood has been increased to .4337. What have we accomplished by using Bayes’ Theorem? Once we found the defective part, we conclude that it is much more likely it is a product of Schuller Sales. The increase in the probability is rather dramatic moving from .30 to .4337.
This information is summarized in the following table.
Prior Conditional Joint Posterior Event, Probability, Probability, Probability, Probability,
Ai P(Ai ) P(B1 ∣ Ai ) P(Ai and B1) P(Ai ∣ B1) Hall .45 .03 .0135 .0135/.0415 = .3235 Schuller .30 .06 .0180 .0180/.0415 = .4337 Crawford .25 .04 .0100 .0100/.0415 = .2410 P(B1) = .0415 1.0000
The probability the defective LS-24 chip came from Schuller Sales can be formally found by using Bayes’ theorem. We compute P(A2 | B1), where A2 refers to Schuller Sales and B1 to the fact that the selected LS-24 chip was defective.
P(A2 ∣ B1) = P(A2)P(B1 ∣ A2)
P(A1)P(B1 ∣ A1) + P(A2)P(B1 ∣ A2) + P(A3) (B1 ∣ A3)
= (.30) (.06)
(.45) (.03) + (.30) (.06) + (.25) (.04) =
.0180 .04850
= .4337
This is the same result obtained from Chart 5–3 and from the conditional probabil- ity table.
Refer to the preceding example and solution.
(a) Design a formula to find the probability the part selected came from Crawford Compo- nents, given that it was a good chip.
(b) Compute the probability using Bayes’ theorem.
S E L F - R E V I E W 5–9
A SURVEY OF PROBABILITY CONCEPTS 161
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 describe three formulas: the multiplication formula (not to be confused with the multiplication rule described earlier in the chapter), the permutation formula, and the combination formula.
The Multiplication Formula We begin with the multiplication formula.
LO5-7 Determine the number of outcomes using principles of counting.
MULTIPLICATION FORMULA If there are m ways of doing one thing and n ways of doing another thing, there are m x n ways of doing both.
In terms of a formula:
MULTIPLICATION FORMULA Total number of arrangements = (m)(n) [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)
33. P(A1) = .60, P(A2) = .40, P(B1 | A1) = .05, and P(B1 | A2) = .10. Use Bayes’ theorem to determine P(A1 | B1).
34. P(A1) = .20, P(A2) = .40, P(A3) = .40, P(B1 | A1) = .25, P(B1 | A2) = .05, and P(B1 | A3) = .10. Use Bayes’ theorem to determine P(A3 | B1).
35. The Ludlow Wildcats baseball team, a minor league team in the Cleveland Indians or- ganization, plays 70% of their games at night and 30% during the day. The team wins 50% of their night games and 90% of their day games. According to today’s news- paper, 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% of the students will complete the assigned problems. She has also determined that among those who do their assignments, 90% will pass the course. Among those students who do not do their homework, 60% will pass. Mike Fishbaugh took statis- tics 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% of their sales are cash, 30% are paid with a credit card, and 40% with a debit card. Twenty percent of the cash purchases, 90% of the credit card purchases, and 60% of the debit card purchases are for more than $50. Ms. Tina Stevens just purchased a new dress that cost $120. What is the probability that she paid cash?
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% of the garages with open doors will have something stolen, but only 1% of those closed will have something stolen. If a garage is robbed, what is the probability the doors were left open?
E X E R C I S E S
162 CHAPTER 5
E X A M P L E
An automobile dealer wants to advertise that for $29,999 you can buy a convert- ible, a two-door sedan, or a four-door model with your choice of either wire wheel covers or solid wheel covers. Based on the number of models and wheel covers, how many different vehicles can the dealer offer?
S O L U T I O N
Of course, the dealer could determine the total number of different cars by pictur- ing and counting them. There are six.
Convertible with wire wheel covers
Convertible with solid wheel covers
Four-door with wire wheel covers
Two-door with solid wheel covers
Four-door with solid wheel covers
Two-door with wire wheel covers
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
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 (m)(n) = (8)(6) = 48 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 exam- ple, 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 interior floor plans, the multiplication formula would be used to find how many different arrangements were possible. There are 12 possibilities.
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?
S E L F - R E V I E W 5–10
A SURVEY OF PROBABILITY CONCEPTS 163
2. Pioneer manufactures three models of Wifi Internet radios, two MP3 docking stations, four different sets of speakers, and three CD carousel changers. 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 The multiplication formula is applied to find the number of possible arrangements for two or more groups. In contrast, we use the permutation formula to find the number of possible arrangements when there is a single group of objects. Illustrations of this type of problem are:
• Three electronic parts, a transistor, an LED, and a synthesizer, are assembled into a plug-in component for a HDTV. 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?
One order for the first illustration might be the transistor first, the LED 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:
where: n is the total number of objects. r is the number of objects selected.
Before we solve the two problems illustrated, the permutations and combinations (to be discussed shortly) use a notation called n factorial. It is written n! and means the product of n (n − 1 )(n − 2)(n − 3) ⋅ ⋅ ⋅ (1). For instance, 5! = 5 · 4 · 3 · 2 · 1 = 120.
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 Pro Scientific calculator has the following key:
10x
LOG
x !
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.
6!3! 4!
= 6 · 5 · 4 · 3 · 2 · 1 (3 · 2 · 1)
4 · 3 · 2 · 1 = 180
By definition, zero factorial, written 0!, is 1. That is, 0! = 1.
PERMUTATION FORMULA n Pr = n!
(n − r)! [5–9]
164 CHAPTER 5
The Combination Formula If the order of the selected objects is not important, any selection is called a combination. Logically, the number of combinations is always less than the number of permutations. The formula to count the number of r object combinations from a set of n objects is:
E X A M P L E
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?
S O L U T I O N
There are three electronic parts to be assembled, so n = 3. Because all three are to be inserted into the plug-in component, r = 3. Solving using formula (5–9) gives:
n Pr = n!
(n − r)! =
3! (3 − 3)!
= 3! 0!
= 3! 1
= 6
We can check the number of permutations arrived at by using the permutation for- mula. 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:
(3)(2)(1) = 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
In the previous example, we selected and arranged all the objects, that is n = r. 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.
E X A M P L E
The Fast Media Company is producing a one-minute video advertisement. In the pro- duction process, eight different video segments were made. To make the one-minute ad, they can only select three of the eight segments. How many different ways can the eight video segments be arranged in the three spaces available in the ad?
S O L U T I O N
There are eight possibilities for the first available space in the ad, seven for the second space (one has been used up), and six for the third space. Thus:
(8)(7)(6) = 336,
that is, there are a total of 336 different possible arrangements. This could also be found by using formula (5–9). If n = 8 video segments and r = 3 spaces available, the formula leads to
n Pr = n!
(n − r)! =
8! (8 − 3)!
= 8! 5!
= (8) (7) (6)5!
5! = 336
COMBINATION FORMULA nCr = n!
r!(n − r)! [5–10]
A SURVEY OF PROBABILITY CONCEPTS 165
For example, if executives Able, Baker, and Chauncy are to be chosen as a committee to negotiate a merger, there is only one possible combination of these three; the com- mittee 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
E X A M P L E
The Grand 16 movie theater uses teams of three employees to work the conces- sion stand each evening. There are seven employees available to work each eve- ning. How many different teams can be scheduled to staff the concession stand?
S O L U T I O N
According to formula (5–10), there are 35 combinations, found by
7C3 = n!
r!(n − r)! =
7! 3!(7 − 3)!
= 7!
3!4! = 35
The seven employees taken three at a time would create the possibility of 35 differ- ent teams.
When the number of permutations or combinations is large, the calculations are tedious. Computer software and handheld calculators have “functions” to compute these num- bers. The Excel output for the selection of three video segments for the eight available at the Fast Media Company is shown below. There are a total of 336 arrangements.
Below is the output for the number of teams at the Grand 16 movie theater. Three em- ployees are chosen from seven possible employees.
166 CHAPTER 5
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 permitted. 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 preceding example/solution involving the Grand 16 movie theater, there were 35 possible teams of three taken from seven employees.
(a) Use formula (5–10) to show this is true. (b) The manager of the theater wants to plan for staffing the concession stand with
teams of five employees on the weekends to serve the larger crowds. From the seven employees, how many teams of five employees are possible?
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?
S E L F - R E V I E W 5–11
39. Solve the following: a. 40!/35! b. 7P4 c. 5C2
40. Solve the following: a. 20!/17! b. 9P3 c. 7C2
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 ex- change. 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 dif- ferent 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. Sam Snead’s restaurant in Conway, South Carolina, offers an early bird special from 4–6 pm each week day evening. If each patron selects a Starter Selection (4 options), an Entrée (8 options), and a Dessert (3 options), how many different meals are possible?
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.
E X E R C I S E S
C H A P T E R S U M M A R Y
I. A probability is a value between 0 and 1 inclusive that represents the likelihood a partic- ular 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 probability that any of two or more events can occur. A. The special rule of addition is used when events are mutually exclusive.
P(A or B) = P(A) + P(B) [5–2] B. The general rule of addition is used when the events are not mutually exclusive.
P(A or B) = P(A) + P(B) − P(A and B) [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.
P(A) = 1 − P(~A) [5–3] VI. The rules of multiplication are applied when two or more events occur simultaneously.
A. The special rule of multiplication refers to events that are independent.
P(A and B) = P(A)P(B) [5–5] B. The general rule of multiplication refers to events that are not independent.
P(A and B) = P(A)P(B | A) [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 an-
other 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:
P(A1 ∣ B) = P(A1)P(B ∣ A1)
P(A1)P(B ∣ A1) + P(A2)P(B ∣ A2) [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 = (m)(n) [5–8] B. A permutation is an arrangement in which the order of the objects selected from a
specific pool of objects is important.
n Pr = n!
(n − r)! [5–9]
C. A combination is an arrangement where the order of the objects selected from a specific pool of objects is not important.
n Cr = n!
r!(n − r)! [5–10]
A SURVEY OF PROBABILITY CONCEPTS 167
168 CHAPTER 5
P R O N U N C I A T I O N K E Y
SYMBOL MEANING PRONUNCIATION
P(A) 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(A | B) Probability of A given B has happened P of A given B
nPr Permutation of n items selected r at a time Pnr
nCr Combination of n items selected r at a time Cnr
C H A P T E R E X E R C I S E S
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 in- clude 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 racehorse. After a successful Kentucky Derby, Corn on the Cob is a heavy favorite at 2 to 1 odds to win the Preakness Stakes. a. If he is a 2 to 1 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?
A SURVEY OF PROBABILITY CONCEPTS 169
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% 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 ar-
rived at the answer to part (a). 54. Refer to the following picture.
B
,B
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 P(B) + P(∼B) equal? 55. In a management trainee program at Claremont Enterprises, 80% of the trainees are
female and 20% male. Ninety percent of the females attended college, and 78% 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 randomly selected a Delta flight on four different days. 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
hourly 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 an hourly worker? b. What is the probability the selected employee is either an hourly worker or a
supervisor? c. Refer to part (b). Are these events mutually exclusive? d. What is the probability the selected employee is neither an hourly worker nor a
supervisor? 58. DJ LeMahieu of the Colorado Rockies had the highest batting average in the 2016 Major
League Baseball season. His average was .348. So assume the probability of getting a hit is .348 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?
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59. Four women’s college basketball teams are participating in a single-elimination holiday basketball tournament. If one team is favored in its semifinal match by odds of 2 to 1 and another squad is favored in its contest 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. Mor- tality 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 has 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 has 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?
64. There are 20 families living in the Willbrook Farms Development. Of these families, 10 prepared 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 three 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 one member of the committee is a woman?
66. A recent survey reported in BloombergBusinessweek dealt with the salaries of CEOs at large corporations and whether company shareholders made money or lost money.
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
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 two 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
A SURVEY OF PROBABILITY CONCEPTS 171
staff estimates that 60% of Althoff and Roll’s potential market read the newspaper. It is further estimated that 85% 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 presidents to attend games with him, and claims he selects the person to attend at ran- dom. 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 format 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. Assume the two events are independent. 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% of the time the prospects were rated as good, 30% of the time fair, and 10% of the time poor. Of those rated good, 80% made a profit the first year; of those rated fair, 60% made a profit the first year; and of those rated poor, 20% 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 Super-T shirts. One of the boxes was selected at random, and a shirt was chosen at random 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 probability of winning a free soft drink is 0.10, and the probability of winning a free large pizza is 0.02. 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 num-
ber 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.
172 CHAPTER 5
76. Recent surveys indicate 60% of 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 with its magnificent terra-cotta soldiers, horses, and chariots, which lay bur- ied 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 smoking. If 60% 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 standardized 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% of the time and a defective steering mechanism 5% of the time. Let’s assume (and hope) that these problems occur inde- pendently. 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 purchased 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 com- pany 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 three cans could be selected? b. What is the probability that the contaminated can is selected for testing?
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 in- dependent. What is the probability the system works under these conditions? The prob- ability 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% of the tubes, Fuji Importers 30%, Kirkpatricks 25%, and Parts Inc. 25%. Tyson Wholesale tends to have the best quality, as only 3% of its tubes arrive defective. Fuji Importers’ tubes are 4% defective, Kirkpatricks’ 7%, and Parts Inc.’s are 6.5% defective. a. What is the overall percent defective?
A SURVEY OF PROBABILITY CONCEPTS 173
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%, 50%, and 20%, 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 probability a “poor” driver will have an accident is .10. The company sells Mr. Brophy an insurance 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% chance each specific leg of the trip is on time, what is the probability all three flights arrive on time?
90. The probability a D-Link 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 light emitting diode (LED) displays are manufactured by Sam- sung. What is the probability that in a collection of three independent LED HDTV pur- chases, at least one is a Samsung?
D A T A A N A L Y T I C S
92. Refer to the North Valley Real Estate data, which report information on homes sold 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 has a pool. 2. The home is in Township 1 or has a pool. 3. Given that it is in Township 3, that it has a pool. 4. 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 at-
tached 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 2016 data, which reports information on the 30 Major League Baseball teams for the 2016 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 million, 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 20 years old versus one that is 20 years old or more.
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.
174 CHAPTER 5
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 follow- ing probabilities:
1. Selecting a team with a stadium that is at least 20 years old. 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 Lincolnville school bus data. Set up a variable that divides the age of the buses into three groups: new (less than 5 years old), medium (5 but less than 10 years), and old (10 or more years). The median maintenance cost is $4,179. Based on this value, create a variable for those less than or equal to the median (low maintenance) and those more than the median (high maintenance cost). Finally, develop a table to show the relationship between maintenance cost and age of the bus. a. What percentage of the buses are less than five years old? b. What percentage of the buses less than five years old have low maintenance costs? c. What percentage of the buses ten or more years old have high maintenance costs? 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?
Discrete Probability Distributions 6
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO6-1 Identify the characteristics of a probability distribution.
LO6-2 Distinguish between discrete and continuous random variables.
LO6-3 Compute the mean, variance, and standard deviation of a discrete probability distribution.
LO6-4 Explain the assumptions of the binomial distribution and apply it to calculate probabilities.
LO6-5 Explain the assumptions of the hypergeometric distribution and apply it to calculate probabilities.
LO6-6 Explain the assumptions of the Poisson distribution and apply it to calculate probabilities.
RECENT STATISTICS SUGGEST that 15% 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 LO6-4.)
© JGI/Jamie Grill/Getty Images
176 CHAPTER 6
INTRODUCTION Chapters 2 through 4 are devoted to descriptive statistics. We describe raw data by or- ganizing the data into a frequency distribution and portraying the distribution in tables, graphs, and charts. Also, we compute a measure of location—such as the arithmetic mean, median, or mode—to locate a typical value near the center of the distribution. 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 could 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 Chapter 5, we state that a probability is a value between 0 and 1 inclusive, and we examine how probabilities can be combined using rules of addition and multiplication.
This chapter begins the study of probability distributions. A probability distribu- tion is like a relative frequency distribution. However, instead of describing the past, it is used to provide estimates of the likelihood of future events. Probability distributions can be described by measures of location and dispersion so we show how to com- pute a distribution’s mean, variance, and standard deviation. We also discuss three frequently occurring discrete probability distributions: the binomial, hypergeometric, and Poisson.
WHAT IS A PROBABILITY DISTRIBUTION? A probability distribution defines or describes the likelihoods for a range of possible future outcomes. For example, Spalding Golf Products, Inc. assembles golf clubs with three components: a club head, a shaft, and a grip. From experience five percent of the shafts received from their Asian supplier are defective. As part of Spalding’s statistical process control they inspect twenty shafts from each arriving shipment. From experi- ence, we know that the probability of a defective shaft is five percent. Therefore, in a sample of twenty shafts, we would expect one shaft to be defective and the other nine- teen shafts to be acceptable. But, by using a probability distribution we can completely describe the range of possible outcomes. For example, we would know the probability that none of the twenty shafts are defective, or that two, or three, or four, or continuing up to twenty shafts in the sample are defective. Given the small probability of a defec- tive shaft, the probability distribution would show that there is a very small probability of four or more defective shafts.
LO6-1 Identify the characteristics of a probability distribution.
PROBABILITY DISTRIBUTION A listing of all the outcomes of an experiment and the probability associated with each outcome.
The important characteristics of a probability distribution are:
CHARACTERISTICS OF A PROBABILITY DISTRIBUTION
1. The probability of a particular outcome is between 0 and 1 inclusive. 2. The outcomes are mutually exclusive. 3. The list of outcomes is exhaustive. So the sum of the probabilities of the out-
comes is equal to 1.
How can we generate a probability distribution? The following example will explain.
DISCRETE PROBABILITY DISTRIBUTIONS 177
E X A M P L E
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 num- ber of heads?
S O L U T I O N
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 shown in the following table.
Possible Coin Toss 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
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.
TABLE 6–1 Probability Distribution for the Events of Zero, One, Two, and Three Heads Showing Face Up on Three Tosses of a Coin
Number of Probability Heads, of Outcome, x P(x)
0 1 8
= .125
1 3 8
= .375
2 3 8
= .375
3 1 8
= .125
Total 8 8
= 1.000
178 CHAPTER 6
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?
S E L F - R E V I E W 6–1
RANDOM VARIABLES In any experiment of chance, the outcomes occur randomly. So it is often called a ran- dom 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 measured with quantitative variables (such as dollars, weight, or number of children), and other experi- mental outcomes are measured with qualitative variables (such as color or religious preference). A few examples will further illustrate what is meant by a random variable. • 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. • The hourly wage of a sample of 50 plumbers in Jacksonville, FL. The hourly wage is
the random variable. • The number of defective lightbulbs produced in an hour at the Cleveland Electric
Company, Inc. • The grade level (Freshman, Sophomore, Junior, or Senior) of the members of the
St. James High School Varsity girls’ basketball team. The grade level is the random variable and notice that it is a qualitative variable.
• The number of participants in the 2016 New York City Marathon. • The daily number of drivers charged with driving under the influence of alcohol in
Brazoria County, Texas, last month.
A random variable is defined as follows:
LO6-2 Distinguish between discrete and continuous random variables.
RANDOM VARIABLE A variable measured or observed as the result of an experiment. By chance, the variable can have different values.
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, .125 + .375 + .375 + .125 = 1.00.
Pr ob
ab ili
ty
Number of Heads
0 1 2 3 0
3 8
1 2
1 4
1 8
P (x )
CHART 6–1 Graphical Presentation of the Number of Heads Resulting from Three Tosses of a Coin and the Corresponding Probability
DISCRETE PROBABILITY DISTRIBUTIONS 179
In Chapter 5 we defined the terms experiment, outcome, and event. Consider the example we just described regarding the experiment of tossing a fair coin three times. In this case the random variable is the number of heads that appear in the three tosses. There are eight possible outcomes to this experiment. These outcomes are shown in the following diagram.
TTT TTH THT HTT
THH HTH HHT
HHH
Possible outcomes for three coin tosses
The event {one head} occurs and the random variable x 5 1.
So, one possible outcome is that a tail appears on each toss: TTT. This single out- come would describe the event of zero heads appearing in three tosses. Another pos- sible outcome is a head followed by two tails: HTT. If we wish to determine the event of exactly one head appearing in the three tosses, we must consider the three possible outcomes: TTH, THT, and HTT. These three outcomes describe the event of exactly one head appearing in three tosses.
In this experiment, the random variable is the number of heads in three tosses. The random variable can have four different values, 0, 1, 2, or 3. The outcomes of the exper- iment are unknown. But, using probability, we can compute the probability of a single head in three tosses as 3/8 or 0.375. As shown in Chapter 5, the probability of each value of the random variable can be computed to create a probability distribution for the random variable, number of heads in three tosses of a coin.
There are two types of random variables: discrete or continuous.
Discrete Random Variable A discrete random variable can assume only a certain number of separated values. For example, the Bank of the Carolinas counts the number of credit cards carried for a group of customers. The data are summarized with the following relative frequency table.
Number of Credit Cards Relative Frequency
0 .03 1 .10 2 .18 3 .21 4 or more .48 Total 1.00
In this frequency table, the number of cards carried is the discrete random variable.
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 val- ues. To be a discrete random variable, these values must be separated—that is, have distance between them. As an example, a department store offers coupons with dis- counts of 10%, 15%, and 25%. In terms of probability, we could compute the probability that a customer would use a 10% coupon versus a 15% or 25% coupon.
180 CHAPTER 6
Continuous Random Variable On the other hand, a continuous random variable can assume an infinite number of values within a given range. It is measured on a continuous interval or ratio scale. Examples include: • 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 and is measured on a continuous scale of time.
• The annual snowfall in Minneapolis, Minnesota. The random variable is the amount of snow, measured on a continuous scale.
As with discrete random variables, the likelihood of a continuous random variable can be summarized with a probability distribution. For example, with a probability distribution for the flight time between Atlanta and Los Angeles, we could say that there is a probability of 0.90 that the flight will be less than 4.5 hours. This also implies that there is a probability of 0.10 that the flight will be more than 4.5 hours. With a probability of snowfall in Minneap- olis, we could say that there is probability of 0.25 that the annual snowfall will exceed 48 inches. This also implies that there is a probability of 0.75 that annual snowfall will be less than 48 inches. Notice that these examples refer to a continuous range of values.
THE MEAN, VARIANCE, AND STANDARD DEVIATION OF A DISCRETE PROBABILITY DISTRIBUTION In Chapter 3, we discussed measures of location and variation for a frequency distribu- tion. 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 distribution by the lowercase Greek letter mu (μ) and the standard deviation by the lowercase Greek letter sigma (σ).
Mean The mean is a typical value used to represent the central location of a probability distri- bution. It also is the long-run average value of the random variable. The mean of a prob- ability 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 corresponding probabilities of occurrence.
The mean of a discrete probability distribution is computed by the formula:
LO6-3 Compute the mean, variance, and standard deviation of a probability distribution.
MEAN OF A PROBABILITY DISTRIBUTION μ = Σ[xP(x)] (6–1)
VARIANCE OF A PROBABILITY DISTRIBUTION σ2 = Σ[(x − μ)2P(x)] (6–2)
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 The mean is a typical value used to summarize a discrete probability distribution. How- ever, it does not describe the amount of spread (variation) in a distribution. The variance does this. The formula for the variance of a probability distribution is:
The computational steps are:
1. Subtract the mean from each value of the random variable, and square this difference. 2. Multiply each squared difference by its probability. 3. Sum the resulting products to arrive at the variance.
DISCRETE PROBABILITY DISTRIBUTIONS 181
The standard deviation, σ, is found by taking the positive square root of σ2; that is, σ = √σ2
An example will help explain the details of the calculation and interpretation of the mean and standard deviation of a probability distribution.
E X A M P L E
John Ragsdale sells new cars for Pelican Ford. John usually sells the largest number of cars on Saturday. He has developed the following probability distribution for the number of cars he expects to sell on a particular Saturday.
Number of Probability, Cars Sold, x P(x)
0 .1 1 .2 2 .3 3 .3 4 .1
1.0
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?
S O L U T I O N
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):
μ = Σ[xP(x)] = 0(.1) + 1(.2) + 2(.3) + 3(.3) + 4(.1) = 2.1
These calculations are summarized in the following table.
Number of Cars Sold, Probability, x P(x) x · P(x)
0 .1 0.0 1 .2 0.2 2 .3 0.6 3 .3 0.9 4 .1 0.4
1.0 μ = 2.1
How do we interpret a mean of 2.1? This value indicates that, over a large num- ber of Saturdays, John Ragsdale expects to sell a mean of 2.1 cars a day. Of
© Thinkstock/JupiterImages RF
182 CHAPTER 6
course, it is not possible for him to sell exactly 2.1 cars on any particular Satur- day. However, the expected value can be used to predict the arithmetic mean number of cars sold on Saturdays in the long run. For example, 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. The following table illustrates the steps to calculate the variance using formula (6–2). The first two columns repeat the probability distribution. In column three, the mean is subtracted from each value of the random variable. In column four, the differences from column three are squared. In the fifth column, each squared difference in column four is multiplied by the corresponding probabil- ity. The variance is the sum of the values in column five.
Number of Cars Sold, Probability, x P(x) (x − μ) (x − μ)2 (x − μ)2P(x) 0 .1 0 − 2.1 4.41 0.441 1 .2 1 − 2.1 1.21 0.242 2 .3 2 − 2.1 0.01 0.003 3 .3 3 − 2.1 0.81 0.243 4 .1 4 − 2.1 3.61 0.361 σ2 = 1.290
Recall that the standard deviation, σ, is the positive square root of the variance. In this example, √σ2 = √1.290 = 1.136 cars. How do we apply 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).
The Pizza Palace offers three sizes of cola. The smallest size sells for $1.99, the medium for $2.49, and the large for $2.89. Thirty percent of the drinks sold are small, 50% are medium, and 20% are large. Create a probability distribution for the random variable price and an- swer the following questions. (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?
S E L F - R E V I E W 6–2
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
E X E R C I S E S
DISCRETE PROBABILITY DISTRIBUTIONS 183
3. Compute the mean and variance of the following probability distribution.
x P(x)
5 .1 10 .3 15 .2 20 .4
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 volunteer ambulance service of Walterboro, South Carolina, for the last 50 days. To explain, there were 22 days on which there were two emergency calls, and 9 days on which there were three emergency calls.
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 experi- ence. 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 purchases of more than $50 to their Belk credit card will be given a spe- cial 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 percent 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
184 CHAPTER 6
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? 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.
Number of Hours Frequency Amount Charged
1 20 $ 3 2 38 6 3 53 9 4 45 12 5 40 14 6 13 16 7 5 18 8 36 20
250
a. Convert the information on the number of hours parked to a probability distribu- tion. 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.
BINOMIAL PROBABILITY DISTRIBUTION The binomial probability distribution is a widely occurring discrete probability distribu- tion. To describe experimental outcomes with a binomial distribution, there are four re- quirements. The first requirement is there are only two possible outcomes on a particular experimental trial. For example, on a test, a true/false question is either answered cor- rectly or incorrectly. In a resort, a housekeeping supervisor reviews an employee’s work and evaluates it as acceptable or unacceptable. A key characteristic of the two out- comes is that they must be mutually exclusive. This means that the answer to a true/ false question must be either correct or incorrect but cannot be both correct and incor- rect at the same time. Another example is the outcome of a sales call. Either a customer purchases or does not purchase the product, but the sale cannot result in both out- comes. Frequently, we refer to the two possible outcomes of a binomial experiment as a “success” and a “failure.” However, this distinction does not imply that one outcome is good and the other is bad, only that there are two mutually exclusive outcomes.
The second binomial requirement is that the random variable is the number of suc- cesses for a fixed and known number of trials. For example, we flip a coin five times and count the number of times a head appears in the five flips, we randomly select 10 em- ployees and count the number who are older than 50 years of age, or we randomly se- lect 20 boxes of Kellogg’s Raisin Bran and count the number that weigh more than the amount indicated on the package. In each example, we count the number of successes from the fixed number of trials.
A third requirement is that we know the probability of a success and it is the same for each trial. Three examples are:
• For a test with 10 true/false questions, we know there are 10 trials and the proba- bility of correctly guessing the answer for any of the 10 trials is 0.5. Or, for a test with 20 multiple-choice questions with four options and only one correct answer, we know that there are 20 trials and the probability of randomly guessing the correct answer for each of the 20 trials is 0.25.
LO6-4 Explain the assumptions of the binomial distribution and apply it to calculate probabilities.
DISCRETE PROBABILITY DISTRIBUTIONS 185
• Bones Albaugh is a Division I college basketball player who makes 70% of his foul shots. If he has five opportuni- ties in tonight’s game, the likelihood he will be successful on each of the five attempts is 0.70.
• In a recent poll, 18% of adults indicated a Snickers bar was their favorite candy bar. We select a sample of 15 adults and ask each for his or her favorite candy bar. The likelihood a Snickers bar is the answer for each adult is 0.18.
The final requirement of a binomial probability distribution is that each trial is indepen- dent of any other trial. Independent means 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 gender of the other two.
• Suppose 20% of the patients served in the emergency room at Waccamaw Hospital do not have insurance. If the second patient served on the afternoon 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.
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 is the number of successes in a fixed number of trials. 3. The probability of success is 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.
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 the Hannah Landscaping Company plants 10 Norfolk pine trees today knowing that 90% of these trees survive, we can compute the binomial probability that exactly 8 trees survive. In this case the number of trials is the 10 trees, the probability of success is .90, and the number of successes is eight. In fact, we can compute a binomial probability for any number of successes from 0 to 10 surviving trees.
A binomial probability is computed by the formula:
BINOMIAL PROBABILITY FORMULA P(x) = nCxπx(1 − π)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. π is the probability of a success on each trial.
We use the Greek letter π (pi) to denote a binomial population parameter. Do not con- fuse it with the mathematical constant 3.1416.
© David Madison/Digital Vision/Getty Images
186 CHAPTER 6
E X A M P L E
There are five flights daily from Pittsburgh via American Airlines 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?
S O L U T I O N
We can use formula (6–3). The probability that a particular flight is late is .20, so let π = .20. There are five flights, so n = 5, and X, the random variable, refers to the number of successes. In this case, a “success” is a flight that arrives late. The ran- dom variable, x, can be equal to 0 late flights in the five trials, 1 late flight in the five trials, or 2, 3, 4, or 5. The probability for no late arrivals, x = 0, is,
P(0) = nCx(π)x(1 − π)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
P(1) = nCx(π)x(1 − π)n−x
= 5C1(.20)1(1 − .20)5−1 = (5)(.20)(.4096) = .4096
The entire binomial probability distribution with π = .20 and n = 5 is shown in the following bar chart. We observe that the probability of exactly three late flights is .0512 and from the bar chart that the distribution of the number of late arrivals is positively skewed.
Pr ob
ab ili
ty
.30
.35
.40
.45
.25
.20
.15
.10
.05
.00 0
0.3277
1
0.4096
2
0.2048
3
0.0512
4
0.0064
5
0.0003
Probability Distribution for the Number of Late Flights
Number of Late Flights
DISCRETE PROBABILITY DISTRIBUTIONS 187
The mean (μ) and the variance (σ2) of a binomial distribution are computed in a “shortcut” fashion by:
MEAN OF A BINOMIAL DISTRIBUTION μ = nπ (6–4)
VARIANCE OF A BINOMIAL DISTRIBUTION σ2 = nπ(1 − π) (6–5)
For the example regarding the number of late flights, recall that π = .20 and n = 5. Hence: μ = nπ = (5)(.20) = 1.0
σ2 = nπ(1 − π) = 5(.20)(1 − .20) = .7997
The mean of 1.0 and the variance of .7997 can be verified from formulas (6–1) and (6–2). The probability distribution from the bar chart shown earlier and the details of the calculations are shown below.
Number of Late Flights, x P(x) xP(x) x − μ (x − μ)2 (x − μ)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 μ = 1.0000 σ2 = 0.7997
Binomial Probability Tables Formula (6–3) can be used to build a binomial probability distribution for any value of n and π. However, for a larger n, the calculations take more time. For convenience, the tables in Appendix B.1 show the result of using the formula for various values of n and π. Table 6–2 shows part of Appendix B.1 for n = 6 and various values of π.
TABLE 6–2 Binomial Probabilities for n = 6 and Selected Values of π
n = 6 Probability
x\π .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
E X A M P L E
In the Southwest, 5% of all cell phone calls are dropped. What is the probability that out of six randomly selected calls, none was dropped? Exactly one? Exactly two? Exactly three? Exactly four? Exactly five? Exactly six out of six?
188 CHAPTER 6
S O L U T I O N
The binomial conditions are met: (a) there are only two possible outcomes (a particular call is either dropped or not dropped), (b) there are 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 on the previous page for the probability of exactly zero dropped calls. Go down the left margin to an x of 0. Now move horizontally to the column headed by a π of .05 to find the probability. It is .735. The values in Table 6–2 are rounded to three decimal places.
The probability of exactly one dropped call in a sample of six calls is .232. The complete binomial probability distribution for n = 6 and π = .05 is:
Number of Number of Dropped Probability of Dropped Probability of Calls, Occurrence, Calls, Occurrence, x P(x) x P(x)
0 .735 4 .000 1 .232 5 .000 2 .031 6 .000 3 .002
Of course, there is a slight chance of getting exactly five dropped calls out of six random selections. It is .00000178, found by inserting the appropriate values in the binomial formula:
P(5) = 6C5(.50)5(.95)1 = (6)(.05)5(.95) = .00000178
For six out of the six, the exact probability is .000000016. Thus, the probability is very small that five or six calls will be dropped in six trials.
We can compute the mean or expected value of the distribution of the num- ber defective:
μ = nπ = (6)(.05) = 0.30
σ2 = nπ(1 − π) = 6(.05)(.95) = 0.285
Ninety-five 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 trans- fer. 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 Excel to verify your answers to parts (b) and (c).
S E L F - R E V I E W 6–3
Appendix B.1 is limited. It gives probabilities for n values from 1 to 15 and π values of .05, .10, . . . , .90, and .95. A software program can generate the probabilities for a specified number of successes, given n and π. The Excel output on the next page shows the probability when n = 40 and π = .09. Note that the number of successes stops at 15 because the probabilities for 16 to 40 are very close to 0. The instructions are detailed in the Software Commands in Appendix C.
DISCRETE PROBABILITY DISTRIBUTIONS 189
Several additional points should be made regarding the binomial probability distribution.
1. If n remains the same but π increases from .05 to .95, the shape of the distribution changes. Look at Table 6–3 and Chart 6–2. The distribution for a π of .05 is posi- tively skewed. As π approaches .50, the distribution becomes symmetrical. As π goes beyond .50 and moves toward .95, the probability distribution becomes negatively skewed. Table 6–3 highlights probabilities for n = 10 and a π of .05, .10, .20, .50, and .70. The graphs of these probability distributions are shown in Chart 6–2.
2. If π, 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 situa- tion where π remains constant at .10 but n increases from 7 to 40.
P (x )
.60
.50
.40
.30
.20
.10
.00
p = .05 n = 10
p = .10 n = 10
p = .20 n = 10
p = .50 n = 10
p = .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
x Successes
x Successes
x Successes
x Successes
CHART 6–2 Graphing the Binomial Probability Distribution for a π of .05, .10, .20, .50, and .70, and an n of 10
TABLE 6–3 Probability of 0, 1, 2, . . . Successes for a π of .05, .10, .20, .50, and .70, and an n of 10
x \π .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
190 CHAPTER 6
11
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 π of .10 and an n of 7, 12, 20, and 40
9. In a binomial situation, n = 4 and π = .25. Determine the probabilities of the follow- ing events using the binomial formula.
a. x = 2 b. x = 3
10. In a binomial situation, n = 5 and π = .40. Determine the probabilities of the follow- ing events using the binomial formula.
a. x = 1 b. x = 2
11. Assume a binomial distribution where n = 3 and π = .60. a. Refer to Appendix B.1, 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
definitions given in formulas (6–1) and (6–2). 12. Assume a binomial distribution where n = 5 and π = .30.
a. Refer to Appendix B.1 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
definitions given in formulas (6–1) and (6–2). 13. An American Society of Investors survey found 30% 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 has used a discount broker?
14. The U.S. Postal Service reports 95% of first-class mail within the same city is delivered within 2 days of the time of mailing. Six letters are randomly sent to differ- ent locations.
a. What is the probability that all six arrive within 2 days? b. What is the probability that exactly five arrive within 2 days? c. Find the mean number of letters that will arrive within 2 days. d. Compute the variance and standard deviation of the number that will arrive
within 2 days.
E X E R C I S E S
DISCRETE PROBABILITY DISTRIBUTIONS 191
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 the Chapter 2, Cumulative Distribution section on page 38. The following example will illustrate.
E X A M P L E
A study by the Illinois Department of Transportation concluded that 76.2% of front seat occupants used seat belts. That is, both occupants of the front seat were using their seat belts. Suppose we decide to compare that information with current us- age. 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?
S O L U T I O N
This situation meets the binomial requirements.
15. Industry standards suggest that 10% of new vehicles require warranty ser- vice 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? 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% of these contacts. During the next 2 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 2-hour period.
17. A recent survey by the American Accounting Association revealed 23% 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 41% of American households use a cell phone exclusively for their telephone service. In a sample of eight households,
a. Find the probability that no household uses a cell phone as their exclusive tele- phone service.
b. Find the probability that exactly 5 households exclusively use a cell phone for telephone service.
c. Find the mean number of households exclusively using cell phones.
192 CHAPTER 6
• 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%.
• 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 π = .762.
P(x = 7) = 12C7(.762)7(1 − .762)12−7 = 792(.149171)(.000764) = .0902
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%.
To find the probability that the occupants in seven 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 141.
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 vehi- cles 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 probability that occupants of all 12 sample vehicles are wearing seat belts. The probability of each of these outcomes is then totaled.
P(x ≥ 7) = P(x = 7) + P(x = 8) + P(x = 9) + P(x = 10) + P(x = 11) + P(x = 12) = .0902 + .1805 + .2569 + .2467 + .1436 + .0383 = .9562
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 Software Commands in Appendix C.
DISCRETE PROBABILITY DISTRIBUTIONS 193
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% of the registered voters in a precinct are
LO6-5 Explain the assumptions of the hypergeometric distribution and apply it to calculate probabilities.
A recent study revealed that 40% of women in the San Diego metropolitan area who work full time also volunteer in the community. Suppose we randomly select eight women in the San Diego area. (a) What are the values for n and π? (b) What is the probability exactly three of the women volunteer in the community? (c) What is the probability at least one of the women volunteers in the community?
S E L F - R E V I E W 6–4
19. In a binomial distribution, n = 8 and π = .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).
20. In a binomial distribution, n = 12 and π = .60. Find the following probabilities. a. x = 5. b. x ≤ 5. c. x ≥ 6.
21. In a recent study, 90% 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% of the production 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 import- ant. GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 70% 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 standard 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% of the cars approaching an individual toll booth in New Jersey 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.
E X E R C I S E S
194 CHAPTER 6
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 sampling is done with replacement, mean- ing 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 of a success will change for each observation. For example, if the population 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 selection 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 med- ical school, and the eighteen Dakota 4x4 Crew Cabs at Helfman Dodge Chrysler Jeep in Houston, Texas.
Recall that one of the criteria for the binomial distribution is that the probabil- ity 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 selected from a finite population without replacement and (2) if the size of the sample n is more than 5% of the size of the population N, then the hypergeo- metric 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 P(x) = (SCx) (N−SCn−x)
NCn [6–6]
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 mutually 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.
DISCRETE PROBABILITY DISTRIBUTIONS 195
E X A M P L E
PlayTime Toys Inc. employs 50 people in the Assembly Department. Forty of the employ- ees belong to a union and 10 do not. Five employees are selected at random to form a committee to meet with management re- garding shift starting times. What is the probability that four of the five selected for the committee belong to a union?
S O L U T I O N
The population in this case is the 50 Assembly Department employees. An employee can be selected for the committee only once. Hence, the sampling is done without replacement. Thus, the probability of selecting a union employee, for example, changes from one trial to the next. The hypergeometric 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.
We wish to find the probability 4 of the 5 committee members belong to a union. Inserting these values into formula (6–6):
P(4) = (40C4) (50−40C5−4)
50C5 =
(91,390) (10) 2,118,760
= .431
Thus, the probability of selecting 5 assembly workers at random from the 50 work- ers and finding 4 of the 5 are union members is .431.
© Howard Berman/Getty Images
Table 6–4 shows the hypergeometric probabilities of finding 0, 1, 2, 3, 4, and 5 union members on the committee.
Union Members Probability
0 .000 1 .004 2 .044 3 .210 4 .431 5 .311 1.000
TABLE 6–4 Hypergeometric Probabilities (n = 5, N = 50, and S = 40) for the Number of Union Members on the Committee
Table 6–5 shows a comparison of the results using the binomial distribution and the hypergeometric distribution. Because 40 of the 50 Assembly Department employees belong to the union, we let π = .80 for the binomial distribution. The bi- nomial probabilities for Table 6–5 come from the binomial distribution with n = 5 and π = .80.
196 CHAPTER 6
As Table 6–5 shows, when the binomial requirement of a constant probability of success cannot be met, the hypergeometric distribution should be used. There are clear differences between the probabilities.
However, 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 distribution can be used to closely ap- proximate the hypergeometric distribution when n < .05N. In other words, the binomial will closely approximate the hypergeometric distribution if the sample is less than 5% of the population. For example, if the population, N, is 150, the number of successes in the population, S, is 120, and the sample size, n, is five, then the rule of thumb is true. That is, 5 < 0.05(150), or 5 < 7.5. The sample size is less than 5% of the population. In the following table, hypergeometric and binomial probability distributions are compared for this situation. The probabilities are very close.
Number of Union Hypergeometric Binomial Probability Members on Committee Probability, P(x) (n = 5 and π = .80) 0 .000 .000 1 .004 .006 2 .044 .051 3 .210 .205 4 .431 .410 5 .311 .328 1.000 1.000
TABLE 6–5 Hypergeometric and Binomial Probabilities for PlayTime Toys Inc. Assembly Department
Hypergeometric Binomial Probability (n = 5 x Probability, P(x) and π = .80 = (120/150) 0 .000 .000 1 .006 .006 2 .049 .051 3 .206 .205 4 .417 .410 5 .322 .328 1.000 1.000
TABLE 6–6 A Comparison of Hypergeometric and Binomial Probabilities When the Sample Size is Less than 0.05(n)
Horwege Discount Brokers plans to hire five 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 eight men and four women among the approved appli- cants. What is the probability that three of the five hired are men?
S E L F - R E V I E W 6–5
A hypergeometric distribution can be created using Excel. See the output for Table 6–5 on the left. The necessary steps are given in Appendix C in the back of the text.
DISCRETE PROBABILITY DISTRIBUTIONS 197
POISSON PROBABILITY DISTRIBUTION The Poisson probability distribution describes the number of times some event occurs during a specified interval. Examples of an interval may be time, distance, area, or volume.
The distribution is based on two assumptions. The first assumption is that the prob- ability 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 in- tervals. This distribution is a limiting form of the binomial distribution when the probabil- ity 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, π, of a particular event’s happening is quite small. The Poisson distribution is a discrete probability distribution because it is formed by counting.
LO6-6 Explain the assumptions of the Poisson distribution and apply it to calculate probabilities.
E X E R C I S E S 25. A CD contains 10 songs; 6 are classical and 4 are rock and roll. In a sample of three
songs, what is the probability that exactly two are classical? Assume the samples are drawn without replacement.
26. A population consists of 15 items, 10 of which are acceptable. In a sample of four items, what is the probability that exactly three are acceptable? Assume the sam- ples are drawn without replacement.
27. The Riverton Branch of the National Bank of Wyoming has 10 real estate loans over $1,000,000. Of these 10 loans, 3 are “underwater.” A loan is underwater if the amount of the loan is greater than the value of the property. The chief loan officer decided to randomly select two of these loans to determine if they met all banking standards. What is the probability that neither of the selected loans is underwater?
28. The Computer Systems Department has eight faculty, six of whom are tenured. Dr. Vonder, the chairman, wants to establish a committee of three 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
question, use the complement rule.) 29. Keith’s Florists has 15 delivery trucks, used mainly to deliver flowers and flower ar-
rangements in the Greenville, South Carolina, area. Of these 15 trucks, 6 have brake problems. A sample of five trucks is randomly selected. What is the probabil- ity that two 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 four or five of the six
winning numbers. 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.
198 CHAPTER 6
The 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.
STATISTICS IN ACTION
Near the end of World War II, the Germans developed rocket bombs, which were fired at the city of London. The Allied military command didn’t know whether these bombs were fired at random or whether they had an aiming device. To investi- gate, the city of London was divided into 586 square regions. The distribution of hits in each square was recorded as follows:
Hits 0 1 2 3 4 5 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 regions were hit four times. Using the Poisson distribution, with a mean of 0.93 hits per re- gion, 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 bombs were falling at random. The Germans had not developed a bomb with an aiming device.
This probability distribution has many applications. It is used as a model to describe the distribution 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 is described mathematically by the formula:
POISSON DISTRIBUTION P(x) = μxe−μ
x! (6–7)
MEAN OF A POISSON DISTRIBUTION μ = nπ (6–8)
where: μ (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, μ, is found by nπ, where n is the total number of trials and π the probability of success.
The variance of the Poisson is equal to its mean. If, for example, the probability 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 μ = nπ = 10,000(.0003) = 3.0.
Recall that for a binomial distribution there are a fixed number of trials. For example, 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, how- ever, can assume an infinite number of values—that is, 0, 1, 2, 3, 4, 5, . . . However, the probabilities become very small after the first few occurrences (successes).
E X A M P L E
Budget Airlines is a seasonal airline that operates flights from Myrtle Beach, South Carolina, to various cities in the northeast. The destinations include Boston, Pittsburgh, Buffalo, and both LaGuardia and JFK airports in New York City. Recently Budget has been concerned about the number of lost bags. Ann Poston from the Analytics Depart- ment was asked to study the issue. She randomly selected a sample of 500 flights and found that a total of twenty bags were lost on the sampled flights.
DISCRETE PROBABILITY DISTRIBUTIONS 199
Part of Appendix B.2 is repeated as Table 6–7. For certain values of µ, the mean of the Poisson distribution, we can read the probability directly from the table. Turning to another example, NewYork-LA Trucking Company finds the mean number of break- downs on the New York to Los Angeles route is 0.30. From Table 6–7 we can locate the probability of no breakdowns on a particular run. First find the column headed “0.30” then read down that column to the row labeled “0”. The value at the intersec- tion is .7408, so this value is the probability of no breakdowns on a particular run. The probability of one breakdown is .2222.
Show that this situation follows the Poisson distribution. What is the mean number of bags lost per flight? What is the likelihood that no bags are lost on a flight? What is the probability at least one bag is lost?
S O L U T I O N
To begin, let’s confirm that the Budget Airlines situation follows a Poisson Distribu- tion. Refer to the highlighted box labeled Poisson Probability Experiment in this section. We count the number of bags lost on a particular flight. On most flights there were no bags lost, on a few flights one was lost, and perhaps in very rare cir- cumstances more than one bag was lost. The continuum or interval is a particular flight. Each flight is assumed to be independent of any other flight.
Based on the sample information we can estimate the mean number of bags lost per flight. There were 20 bags lost in 500 flights so the mean number of bags lost per flight is .04, found by 20/500. Hence μ = .04.
We use formula (6–7) to find the probability of any number of lost bags. In this case x, the number of lost bags is 0.
P(0) = μxe−μ
x! =
.040e−0.04
0! = .9608
The probability of exactly one lost bag is:
P(1) = μxe−μ
x! =
.040e−0.04
1! = .0384
The probability of one or more lost bags is:
1 − P(0) = 1 − μxe−μ
x! = 1 −
.040e−0.04
0! = 1 − .9608 = .0392
These probabilities can also be found using Excel. The commands to compute Poisson probabilities are in Appendix C.
200 CHAPTER 6
Earlier in this section, we mentioned that the Poisson probability distribution is a limiting form of the binomial. That is, we could estimate a binomial probability using the Poisson. In the following example, we use the Poisson distribution to estimate a binomial proba- bility when n, the number of trials, is large and π, the probability of a success, small.
TABLE 6–7 Poisson Table for Various Values of μ (from Appendix B.2)
E X A M P L E
Coastal Insurance Company underwrites insurance for beachfront properties along the Virginia, North and South Carolina, and Georgia coasts. It uses the esti- mate 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 mortgage period?
S O L U T I O N
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:
μ = nπ = 30(.05) = 1.5 where:
n is the number of years, 30 in this case. π is the probability a hurricane meeting the strength criteria comes ashore. μ 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.
P(x ≥ 1) = 1 − P(x = 0) = 1 − μ0e−1.5
0! = 1 − .2231 = .7769
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 mortgage 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%.
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.
μ
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
DISCRETE PROBABILITY DISTRIBUTIONS 201
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 183.
• There are only two possible outcomes: a hurricane hits the St. Simons area or it does not.
• There are a fixed number of trials, in this case 30 years. • There is a constant probability of success; that is, the probability of a hurricane hit-
ting 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:
P(x ≥ 1) = 1 − P(x = 0) = 1 − [30C0(.05)0(.95)30] = 1 − [(1)(1)(.2146)] = .7854
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 bino- mial is the more “technically correct” solution. The Poisson can be thought of as an ap- proximation for the binomial, when n, the number of trials is large, and π, 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 difference in the answers. In fact, as n gets larger and π smaller, the difference between the two distributions gets smaller.
The Poisson probability distribution is always positively skewed and the random variable has no specific upper limit. In the lost bags example/solution, the Poisson distribution, with μ = 0.04, is highly skewed. As μ becomes larger, the Poisson distri- bution becomes more symmetrical. For example, Chart 6–4 shows the distributions of
.50
.60
.40
.30
.20
.10
.00
Pr ob
ab ili
ty o
f O cc
ur re
nc e
Number of Occurrences
Transmission Services
Muffler Replacements
Oil changes 110 1 2 3 4 5 6 7 8 9100 1 2 3 4 5 60 1 2 3 4
P(x)
m = 0.7
m = 2.0
m = 6.0
CHART 6–4 Poisson Probability Distributions for Means of 0.7, 2.0, and 6.0
202 CHAPTER 6
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.
In summary, the Poisson distribution is a family of discrete distributions. All that is needed to construct a Poisson probability distribution is the mean number of defects, errors, or other random variable, designated as μ.
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?
S E L F - R E V I E W 6–6
31. In a Poisson distribution μ = 0.4. a. What is the probability that x = 0? b. What is the probability that x > 0?
32. In a Poisson distribution μ = 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 three loans will be defaulted? b. What is the probability that at least three 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% 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 3 days each year for weather emergencies. What is the probability that schools in Los Angeles County will close for 4 days next year?
E X E R C I S E S
C H A P T E R S U M M A R Y
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:
μ = Σ[xP(x)] (6–1) B. The variance is equal to:
σ2 = Σ[(x − μ)2P(x)] (6–2)
DISCRETE PROBABILITY DISTRIBUTIONS 203
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. 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:
P(x) = nCxπx(1 − π)n−x (6–3)
F. The mean is computed as:
μ = nπ (6–4)
G. The variance is
σ2 = nπ(1 − π) (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:
P(x) = (SCx) (N−SCn−x) (NCn)
(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 π is small. E. A Poisson probability is determined from the following equation:
P(x) = μxe−μ
x! (6–7)
F. The mean and the variance are:
μ = nπ σ2 = nπ (6–8)
C H A P T E R E X E R C I S E S
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
probabilities of these values are .25, .60, and .15, respectively. Determine the mean and variance of the investment’s dollar value.
204 CHAPTER 6
40. The following notice appeared in the golf shop at a Myrtle Beach, South Carolina, golf course.
Blackmoor Golf Club Members
The golf shop is holding a raffle to win a TaylorMade M1 10.5° Regular Flex Driver ($300 value).
Tickets are $5.00 each. Only 80 tickets will be sold.
Please see the golf shop to get your tickets!
John Underpar buys a ticket. a. What are Mr. Underpar’s possible monetary outcomes? b. What are the probabilities of the possible outcomes? c. Summarize Mr. Underpar’s “experiment” as a probability distribution. d. What is the mean or expected value of the probability distribution? Explain your result. e. If all 80 tickets are sold, what is the expected return to the Club?
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. Com- pute 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 probabili- ties 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 corresponding 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 study, 35% of people surveyed indicated chocolate was their favorite flavor of ice cream. Suppose we select a sample of 10 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 Americans. A Spanish-speaking person is accused of killing a non-Spanish-speaking Amer- ican 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.
DISCRETE PROBABILITY DISTRIBUTIONS 205
45. An auditor for Health Maintenance Services of Georgia reports 40% of policyholders 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? 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 final-
ized, 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 stockholders approve?
47. A federal study reported that 7.5% of the U.S. workforce has a drug problem. A drug enforcement official for the state of Indiana wished to investigate this statement. 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% 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% of those who visit a retail site on the internet to 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 acceptance sampling. Acceptance sampling is a statistical method used to monitor the quality of purchased parts and components. To ensure the quality of incoming parts, a purchaser or manufacturer normally samples 20 parts and allows one defect. a. What is the likelihood of accepting a lot that is 1% defective? b. If the quality of the incoming lot was actually 2%, what is the likelihood of accepting it? c. If the quality of the incoming lot was actually 5%, what is the likelihood of accepting it?
51. Unilever Inc. recently developed a new body wash with a scent of ginger. Their research indicates that 30% of men like the new scent. To further investigate, Unilever’s market- ing research group randomly selected 15 men and asked them if they liked the scent. What is the probability that six or more men like the ginger scent in the body wash?
52. Dr. Richmond, a psychologist, is studying the daytime television viewing habits of col- lege students. She believes 45% of college students watch soap operas during the af- ternoon. 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 students who 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% 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.
206 CHAPTER 6
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?
54. The Internal Revenue Service is studying the category of charitable contributions. 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 contri- butions of more than $1,000. 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 one member of the committee lives in northern Kentucky?
56. Topten is a leading source on energy-efficient products. Their list of the top seven vehi- cles in terms of fuel efficiency for 2017 includes three Hondas. a. Determine the probability distribution for the number of Hondas in a sample of two
cars chosen from the top seven. b. What is the likelihood that in the sample of two at least one Honda is included?
57. The position of chief of police in the city of Corry, Pennsylvania, is vacant. A search com- mittee of Corry residents is charged with the responsibility of recommending a new chief to the city council. There are 12 applicants, 4 of whom are either female or mem- bers of a minority. The search committee decides to interview all 12 of the applicants. 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 popula- tion. 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 38,802,500 Yes 2 Texas 26,956,958 Yes 3 Florida 19,893,297 Yes 4 New York 19,746,227 Yes 5 Illinois 12,880,580 No 6 Pennsylvania 12,787,209 No 7 Ohio 11,594,163 No 8 Georgia 10,097,343 Yes 9 North Carolina 9,943,964 Yes 10 Michigan 9,909,877 No 11 New Jersey 8,938,175 Yes 12 Virginia 8,326,289 Yes 13 Washington 7,061,530 Yes 14 Massachusetts 6,745,408 Yes 15 Arizona 6,731,484 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 has any coastline? b. Exactly one of the selected states has a coastline? c. At least one of the selected states has a coastline?
DISCRETE PROBABILITY DISTRIBUTIONS 207
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 5 consecutive days at least one Lexus is sold?
60. Suppose 1.5% of the antennas on new Nokia cell phones are defective. For a random 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 custom- ers waiting in line. What is the probability that you visit Safeway today during this pe- riod 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 non-work-related e-mails per hour. As- sume the arrival of these e-mails is approximated by the Poisson distribution. a. What is the probability Linda Lahey, company president, received exactly one non-
work-related e-mail between 4 p.m. and 5 p.m. yesterday? b. What is the probability she received five or more non-work-related e-mails during the
same period? c. What is the probability she did not receive any non-work-related e-mails 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. Recent difficult economic times have caused an increase in the foreclosure rate of home mortgages. Statistics from the Penn Bank and Trust Company show their monthly foreclo- sure rate is now 1 loan out of every 136 loans. Last month the bank approved 300 loans. a. How many foreclosures would you expect the bank to have last month? b. What is the probability of exactly two foreclosures? c. What is the probability of at least one foreclosure?
65. The National Aeronautics and Space Administration (NASA) has experienced two disasters. 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 mis- sions. 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. Sup- pose 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 partic- ipating in the second round that day. Estimate the probability that four golfers would score a hole in one on the sixth hole.
68. According to sales information in the first quarter of 2016, 2.7% of new vehicles sold in the United States were hybrids. This is down from 3.3% for the same period a year ear- lier. An analyst’s review of the data indicates that the reasons for the sales decline in- clude the low price of gasoline and the higher price of a hybrid compared to similar vehicles. Let’s assume these statistics remain the same for 2017. That is, 2.7 percent of
208 CHAPTER 6
new car sales are hybrids in the first quarter of 2017. For a sample of 40 vehicles sold in the Richmond, Virginia area: a. How many vehicles would you expect to be hybrid? b. Use the Poisson distribution to find the probability that five of the sales were hybrid
vehicles. c. Use the binomial distribution to find the probability that five of the sales were hybrid
vehicles. 69. A recent CBS News survey reported that 67% 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 eight adults would indicate the Treasury should
continue making pennies? c. What is the likelihood at least eight adults would indicate the Treasury should con-
tinue making pennies?
D A T A A N A L Y T I C S
70. Refer to the North Valley Real Estate data, which report information on homes sold in the 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 2016 data. Compute the mean number of home runs per
game. To do this, first find the mean number of home runs per team for 2016. 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.
Continuous Probability Distributions 7
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO7-1 Describe the uniform probability distribution and use it to calculate probabilities.
LO7-2 Describe the characteristics of a normal probability distribution.
LO7-3 Describe the standard normal probability distribution and use it to calculate probabilities.
LO7-4 Approximate the binomial probability distribution using the standard normal probability distribution to calculate probabilities.
LO7-5 Describe the exponential probability distribution and use it to calculate probabilities.
CRUISE SHIPS of the Royal Viking line report that 80% 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 LO7-4.)
© Ilene MacDonald/Alamy Stock Photo
210 CHAPTER 7
INTRODUCTION Chapter 6 began our study of probability distributions. We consider three discrete prob- ability 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 2014. The number still operating in 2017 can be 0, 1, 2, . . . , 10. There cannot be 3.7, 12, or −7 still operating in 2017. In this example, only certain outcomes are possi- ble and these outcomes are represented by clearly separated values. In addition, the result is usually found by counting the number of successes. We count the number of the businesses in the study that are still in operation in 2017.
We continue our study of probability distributions by examining continuous proba- bility 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. As an example, at Dave’s Inlet Fish Shack flounder is the featured, fresh-fish menu item. The distribution of the amount of flounder sold per day has a mean of 10.0 pounds per day and a standard deviation of 3.0 pounds per day. This distribution is continuous because Dave, the owner, “measures” the amount of flounder sold each day. It is important to realize that a contin- uous random variable has an infinite number of values within a particular range. So, for a continuous random variable, probability is for a range of values. The probability for a specific value of a continuous random variable is 0.
This chapter shows how to use three continuous probability distributions: the uni- form probability distribution, the normal probability distribution, and the exponential probability distribution.
THE FAMILY OF UNIFORM PROBABILITY DISTRIBUTIONS The uniform probability distribution is the simplest distribution for a continuous ran- dom variable. This distribution is rectangular in shape and is completely defined by its minimum and maximum values. Here are some examples that follow a uniform distribution.
• The sales of gasoline at the Kwik Fill in Medina, New York, fol- low a uniform distribution that varies between 2,000 and 5,000 gallons per day. The random variable is the number of gallons sold per day and is continuous within the interval between 2,000 gallons and 5,000 gallons.
• Volunteers at the Grand Strand Public Library prepare federal 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 rectangular 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.
The mean of a uniform distribution is located in the middle of the interval between the minimum and maximum values. It is computed as:
LO7-1 Describe the uniform probability distribution and use it to calculate probabilities.
© C. Sherburne/PhotoLink/Getty Imagess
MEAN OF THE UNIFORM DISTRIBUTION μ = a + b
2 (7–1)
CONTINUOUS PROBABILITY DISTRIBUTIONS 211
The standard deviation describes the dispersion of a distribution. In the uniform distribu- tion, the standard deviation is also related to the interval between the maximum and minimum values.
a
P (x )
1 b 2 a
b
CHART 7–1 A Continuous Uniform Distribution
UNIFORM DISTRIBUTION P(x) = 1
b − a if a ≤ x ≤ b and 0 elsewhere (7–3)
The equation for the uniform probability distribution is:
As we described in Chapter 6, probability distributions are useful for making prob- ability statements concerning the values of a random variable. For distributions describ- ing 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 multiplying its length by its height. For the uniform distribution, the height of the rectangle 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 distribu- tion 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 probability distribution is equal to 1.00. In general
Area = (height) (base) = 1
(b − a) (b − a) = 1.00
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:
Area = (height) (base) = 1
(15 − 10) (15 − 10) = 1.00
The following example illustrates the features of a uniform distribution and how we use it to calculate probabilities.
E X A M P L E
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
σ = √ (b − a)2
12 (7–2)
STANDARD DEVIATION OF THE UNIFORM DISTRIBUTION
212 CHAPTER 7
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?
S O L U T I O N
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 distribution is 30 minutes.
2. The times students must wait for the bus are uniform over the interval from 0 minutes to 30 minutes, so in this case a is 0 and b is 30.
Area = (height) (base) = 1
(30 − 0) (30 − 0) = 1.00
3. To find the mean, we use formula (7–1).
μ = a + b
2 =
0 + 30 2
= 15
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).
σ = √ (b − a)2
12 = √
(30 − 0)2
12 = 8.66
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 < wait time < 30) = (height) (base) = 1
(30 − 0) (5) = .1667
0
.060
.0333
0 10 40
Length of Wait (minutes)
Pr ob
ab ili
ty
3020
CHART 7–2 Uniform Probability Distribution of Student Waiting Times
CONTINUOUS PROBABILITY DISTRIBUTIONS 213
Microwave ovens only last so long. The life-time of a microwave oven 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 microwave oven lasts between 10 and 14 years? (e) What is the probability a microwave oven will last less than 9 years?
S E L F - R E V I E W 7–1
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.
E X E R C I S E S
So the probability a student waits between 25 and 30 minutes is .1667. This conclusion is illustrated by the following graph.
0
P (x )
.0333
10 20 25
Area 5 .1667
m 5 15 30
5. The area within the distribution for the interval 10 to 20 represents the probability.
P(10 < wait time < 20) = (height) (base) = 1
(30 − 0) (10) = .3333
We can illustrate this probability as follows.
0
P (x )
.0333
10 20
Area 5 .3333
m 5 15 30
214 CHAPTER 7
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.
LO7-2 Describe the characteristics of a normal probability distribution.
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?
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 service may call an 800 number for technical support. It takes the technician between 30 seconds and 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% of the problem-solving times. What are
the end points of these two times?
NORMAL PROBABILITY DISTRIBUTION P(x) = 1
σ√2π e−[
(x−μ)2
2σ 2 ] (7–4)
However, do not be bothered by how complex this formula looks. You are already familiar with many of the values. The symbols μ and σ refer to the mean and the stan- dard deviation, as usual. The Greek symbol π is a constant and its value is approximately 22/7 or 3.1416. The letter e is also a constant. It is the base of the natural log system and is approximately 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 calculations using formula (7–4). Instead you will use a table, given in Appendix B.3, to find various probabilities. These probabilities can also be calculated using Excel functions as well as other statistical software.
CONTINUOUS PROBABILITY DISTRIBUTIONS 215
The normal probability distribution has the following characteristics:
• It is bell-shaped and has a single peak at the center of the distribution. The arithme- tic 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 normal 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 shapes of the curves will be mirror images. Also, the area of each half is 0.5.
• It falls off smoothly in either direction from the central value. That is, the distribution 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 indefinitely in both directions.
• The location of a normal distribution is determined by the mean, μ. The dispersion or spread of the distribution is determined by the standard deviation, σ.
These characteristics are shown graphically in Chart 7–3.
STATISTICS IN ACTION
Many variables are approxi mately, normally distributed, such as IQ scores, life ex pectancies, and adult height. This implies that nearly all observations oc cur within 3 standard devia tions of the mean. On the other hand, observations that occur beyond 3 stan dard 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 standard 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). LeBron James, a professional bas ketball player with the Cleveland Cavaliers, is 80 inches, or 6 feet 8 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 per son like LeBron James. As another example, the driver’s seat in most ve hicles is set to comfortably fit a person who is at least 159 cm (62.5 inches) tall. The distribution of heights of adult women is approxi mately a normal distribu tion with a mean of 161.5 cm and a standard deviation of 6.3 cm. Thus about 35% of adult women will not fit comfortably in the driver’s seat.
Normal curve is symmetrical Two halves identical
Mean, median, and mode are
equal
Theoretically, curve extends to – `
Theoretically, curve extends to + `
Tail Tail
CHART 7–3 Characteristics of a Normal Distribution
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 are compared. In the Camden plant, the mean is 20 years and the standard deviation is 3.1 years. There is another normal probability distribution for the length of service in the Dunkirk plant, where μ = 20 years and σ = 3.9 years. In the Elmira plant, μ = 20 years and σ = 5.0 years. Note that the means are the same but the standard deviations are different. As the standard deviation gets smaller, the distribution becomes more narrow and “peaked.”
CHART 7–4 Normal Probability Distributions with Equal Means but Different Standard Deviations
0 4 7 10 13 16 19 22 25 28 m 5 20 years of service
s 5 3.1 years, Camden plant
s 5 3.9 years, Dunkirk plant
s 5 5.0 years, Elmira plant
37 403431
216 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.
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.
Sugar Yummies
s = 1.6 grams
Alphabet Gems
s = 1.6 grams
Weight Droppers
s = 1.6 grams
m 283
grams
m 301
grams
m 321
grams
CHART 7–5 Normal Probability Distributions Having Different Means but Equal Standard Deviations
m 2,000
psi
s 5 41 psi s 5 52 psi
s 5 26 psi
m 2,107
psi
m 2,186
psi
CHART 7–6 Normal Probability Distributions with Different Means and Standard Deviations
In Chapter 6, recall that discrete probability distributions show the specific likeli- hood a discrete value will occur. For example, on page 186 the binomial distribution is used to calculate the probability that none of the five flights arriving at the Bradford Pennsylvania Regional Airport will be late.
With a continuous probability distribution, areas below the curve define probabili- ties. The total area under the normal curve is 1.0. This accounts for all possible out- comes. 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 normally distrib- uted 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 also can 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.
CONTINUOUS PROBABILITY DISTRIBUTIONS 217
THE STANDARD NORMAL PROBABILITY DISTRIBUTION The number of normal distributions is unlimited, each having a different mean (μ), stan- dard deviation (σ), or both. While it is possible to provide a limited number of probability tables for discrete distributions such as the binomial and the Poisson, providing tables for the infinite number of normal distributions is impractial. 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 prob- ability distribution by subtracting the mean from each observation and dividing this dif- ference by the standard deviation. The results are called z values or z scores.
LO7-3 Describe the standard normal probability distribution and use it to calculate probabilities.
z VALUE The signed distance between a selected value, designated x, and the mean, μ, divided by the standard deviation, σ.
So, a z value is the distance from the mean, measured in units of the standard devi- ation. The formula for this conversion is:
STANDARD NORMAL VALUE z = x − μ
σ (7–5)
where:
x is the value of any particular observation or measurement. μ is the mean of the distribution. σ is the standard deviation of the distribution.
As we noted in the preceding definition, a z value expresses the distance or differ- ence 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. Therefore, the z distribution has all the characteristics of any normal probability distribution. These char- acteristics are listed on page 215 in the Family of Normal Probability Distributions sec- tion. The table in Appendix B.3 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 combination of many hereditary and environ mental factors, each having about the same amount of weight or influence on the skills. Thus, much like a binomial distribution with a large number of trials, many skills and attributes follow the normal distribution. For example, the SAT Reasoning Test is the most widely used standardized test for college admissions in the United States. Scores are based on a normal dis tribution with a mean of 1,500 and a standard deviation of 300.
TABLE 7–1 Areas under the Normal Curve
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 . . .
218 CHAPTER 7
Applications of the Standard Normal Distribution The standard normal distribution is very useful for determining probabilities for any normally distributed random variable. The basic procedure is to find the z value for a particular value of the random variable based on the mean and standard deviation of its distribution. Then, using the z value, we can use the standard normal distribution to find various probabilities. The following example/solution describes the details of the application.
E X A M P L E
In recent years a new type of taxi service has evolved in more than 300 cities world- wide, where the customer is connected directly with a driver via a smartphone. The idea was first developed by Uber Technologies, which is headquartered in San Francisco, California. It uses the Uber mobile app, which allows customers with a smartphone to submit a trip request which is then routed to a Uber driver who picks up the customer and takes the customer to the desired location. No cash is in- volved, the payment for the transaction is handled via a digital payment.
Suppose the weekly income of Uber drivers follows the normal probability distribu- tion with a mean of $1,000 and a standard deviation of $100. What is the z value of in- come for a driver who earns $1,100 per week? For a driver who earns $900 per week?
S O L U T I O N
Using formula (7–5), the z values corresponding to the two x values ($1,100 and $900) are:
For x = $1,100: For x = $900:
z = x − μ
σ z =
x − μ σ
= $1,100 − $1,000
$100 =
$900 − $1,000 $100
= 1.00 = −1.00
The z of 1.00 indicates that a weekly income of $1,100 is one standard deviation above the mean, and a z of −1.00 shows that a $900 income is one standard devi- ation below the mean. Note that both incomes ($1,100 and $900) are the same distance ($100) from the mean.
A recent national survey concluded that the typical person consumes 48 ounces of water per day. Assume daily water consumption follows a normal probability distribution with a standard deviation of 12.8 ounces. (a) What is the z value for a person who consumes 64 ounces of water per day? Based on
this z value, how does this person compare to the national average? (b) What is the z value for a person who consumes 32 ounces of water per day? Based on
this z value, how does this person compare to the national average?
S E L F - R E V I E W 7–2
The Empirical Rule The Empirical Rule is introduced on page 80 of Chapter 3. It states that if a random vari- able is normally distributed, then:
1. Approximately 68% of the observations will lie within plus and minus one standard deviation of the mean.
CONTINUOUS PROBABILITY DISTRIBUTIONS 219
2. About 95% of the observations will lie within plus and minus two standard devia- tions of the mean.
3. Practically all, or 99.7% of the observations, will lie within plus and minus three stan- dard deviations of the mean.
Now, knowing how to apply the standard normal probability distribution, we can verify the Empirical Rule. For example, one standard deviation from the mean is the same as a z value of 1.00. When we refer to the standard normal probability table, a z value of 1.00 corresponds to a probability of 0.3413. So what percent of the observations will lie within plus and minus one standard deviation of the mean? We multiply (2)(0.3413), which equals 0.6826, or approximately 68% of the observations are within plus and minus one standard deviation of the mean.
The Empirical Rule is summarized in the following graph.
converts to
– 3 – 2 – 1 1 2 30
m – 3s m – 2s m – 1s m + 3sm + 2sm + 1s
68%
95%
Practically all
Scale of z
Scale of xm
Transforming measurements to standard normal deviates changes the scale. The conversions are also shown in the graph. For example, μ + 1σ is converted to a z value of 1.00. Likewise, μ − 2σ 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, μ.
E X A M P L E
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% of the batteries failed between what two values? 2. About 95% of the batteries failed between what two values? 3. Virtually all of the batteries failed between what two values?
S O L U T I O N
We can use the Empirical Rule to answer these questions.
1. About 68% of the batteries will fail between 17.8 and 20.2 hours, found by 19.0 ± 1(1.2) hours.
2. About 95% 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.
220 CHAPTER 7
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% of the incomes lie between what two amounts? (b) About 95% 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?
S E L F - R E V I E W 7–3
7. Explain what is meant by this statement: “There is not just one normal probability distribution 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% of the observations lie between what two values? b. About 95% 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’ ex- perience 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 deviation 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 re- pair 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 la- bor costs follows the normal probability distribution. Compute z values for each and comment on your findings.
E X E R C I S E S
This information is summarized on the following chart.
m m – 3s m – 2s m – 1s m + 3s m + 2sm + 1s 15.4 16.6 17.8 20.2 21.4 22.619.0
68% 95%
Practically all
Scale of hours
CONTINUOUS PROBABILITY DISTRIBUTIONS 221
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/solution will illustrate the details.
E X A M P L E
In the first example/solution described on page 218 in this section, we reported that the weekly income of Uber drivers followed the normal distribution with a mean of $1,000 and a standard deviation of $100. That is, μ = $1,000 and σ = $100. What is the likeli- hood of selecting a driver whose weekly income is between $1,000 and $1,100?
S O L U T I O N
We have already converted $1,100 to a z value of 1.00 using formula (7–5). To repeat:
z = x − μ
σ =
$1,100 − $1,000 $100
= 1.00
The probability associated with a z of 1.00 is available in Appendix B.3. A portion of Appendix B.3 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% of Uber drivers earn between $1,000 and $1,100 weekly, or the likelihood of selecting a driver and finding his or her income is between $1,000 and $1,100 is .3413.
This information is summarized in the following diagram.
Scale of z1.0
$1,000 Scale of dollars
0
$1,100
.3413
222 CHAPTER 7
In the example/solution just completed, we are interested in the probability be- tween the mean and a given value. Let’s change the question. Instead of wanting to know the probability of selecting a random driver who earned between $1,000 and $1,100, suppose we wanted the probability of selecting a driver 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 driver 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 driver earning less than $1,000 is .5000. Finally, we add the two probabil- ities, so .3413 + .5000 = .8413. About 84% of Uber drivers earn less than $1,100 per week. See the following diagram.
Scale of z1.0
$1,000 Scale of dollars
0
$1,100
.3413.5000
Excel will calculate this probability. The necessary commands are in the Software Commands in Appendix C. The answer is .8413, the same as we calculated.
STATISTICS IN ACTION
Many processes, such as filling soda bottles and can ning fruit, are normally dis tributed. Manufacturers must guard against both over and underfilling. If they put too much in the can or bottle, they are giv ing away their product. If they put too little in, the customer may feel cheated and the government may question the label descrip tion. “Control charts,” with limits drawn three standard deviations above and be low the mean, are routinely used to monitor this type of production process.
E X A M P L E
Refer to the first example/solution discussed on page 218 in this section regarding the weekly income of Uber drivers. The distribution of weekly incomes follows the normal probability distribution, with a mean of $1,000 and a standard deviation of $100. What is the probability of selecting a driver whose income is:
1. Between $790 and $1,000? 2. Less than $790?
CONTINUOUS PROBABILITY DISTRIBUTIONS 223
The temperature of coffee sold at the Coffee Bean Cafe follows the normal probability distri- bution, 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?
S E L F - R E V I E W 7–4
S O L U T I O N
We begin by finding the z value corresponding to a weekly income of $790. From formula (7–5):
z = x − μ
s =
$790 − $1,000 $100
= −2.10
See Appendix B.3. 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 normal curve corresponding to a z value of 2.10 is .4821. However, because the normal 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 driver earning between $790 and $1,000 is .4821. In probability nota- tion, we write P($790 < weekly income < $1,000) = .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. Be- cause 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.
So we conclude that 48.21% of the Uber drivers have weekly incomes be- tween $790 and $1,000. Further, we can anticipate that 1.79% earn less than $790 per week. This information is summarized in the following diagram.
.5000
Scale of z0 Scale of dollars$1,000
–2.10 $790
.0179
.4821
224 CHAPTER 7
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.
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?
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 dis- tribution. 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 devia- tion 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?
E X E R C I S E S
E X A M P L E
Continuing the example/solution first discussed on page 218 using the weekly in- come of Uber drivers, weekly income follows 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?
S O L U T I O N
The problem can be divided into two parts. For the area between $840 and the mean of $1,000:
z = $840 − $1,000
$100 =
−$160 $100
= −1.60
For the area between the mean of $1,000 and $1,200:
z = $1,200 − $1,000
$100 =
$200 $100
= 2.00
The area under the curve for a z of −1.60 is .4452 (from Appendix B.3). 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% of the drivers have weekly incomes be- tween $840 and $1,200. This is shown in a diagram:
CONTINUOUS PROBABILITY DISTRIBUTIONS 225
Another application of the normal distribution involves determining the area be- tween values on the same side of the mean.
Scale of z2.00
.4772.4452
21.6 Scale of dollars$1,200$1,000$840
What is this probability?
E X A M P L E
Returning to the weekly income distribution of Uber drivers (μ = $1,000, σ = $100), what is the area under the normal curve between $1,150 and $1,250?
S O L U T I O N
The situation is again separated into two parts, and formula (7–5) is used. First, we find the z value associated with a weekly income of $1,250:
z = $1,250 − $1,000
$100 = 2.50
Next we find the z value for a weekly income of $1,150:
z = $1,150 − $1,000
$100 = 1.50
From Appendix B.3, the area associated with a z value of 2.50 is .4938. So the probability of a weekly income 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 in- come between $1,000 and $1,150 is .4332. The probability of a weekly income 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 prob- ability of a weekly income between $1,150 and $1,250 is .0606. In probability no- tation, we write P($1,150 < weekly income < $1,250) = .4938 − .4332 = .0606.
Scale of incomes Scale of z
$1,250 2.50
$1,000 0
.0606
$1,150 1.50
.4332
226 CHAPTER 7
To summarize, there are four situations for finding the area under the standard nor- mal 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.
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?
S E L F - R E V I E W 7–5
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 of 75.0 or less. c. Compute the probability of a value between 55.0 and 70.0.
19. Suppose the Internal Revenue Service reported that the mean tax refund for the year 2016 was $2,800. Assume the standard deviation is $450 and that the amounts refunded follow a normal probability distribution.
a. What percent of the refunds are more than $3,100? b. What percent of the refunds are more than $3,100 but less than $3,500? c. What percent of the refunds are more than $2,250 but less than $3,500?
20. The distribution of the number of viewers for the American Idol television show follows a normal distribution with a mean of 29 million and a standard deviation of 5 million. What is the probability 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 lis- teners 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 for:
a. More than 20 minutes? b. 20 minutes or less? c. Between 10 and 12 minutes?
22. Among the thirty largest U.S. cities, the mean one-way commute time to work is 25.8 minutes. https://deepblue.lib.umich.edu/bitstream/handle/2027.42/112057/103196. pdf?sequence=1&isAllowed=y. The longest one-way travel time is in New York City, where the mean time is 39.7 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 50 minutes?
E X E R C I S E S
CONTINUOUS PROBABILITY DISTRIBUTIONS 227
The previous example/solutions require finding the percent of the observations located between two observations or the percent of the observations above, or be- low, a particular observation x. A further application of the normal distribution involves finding the value of the observation x when the percent above or below the observa- tion is given.
E X A M P L E
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 standard deviation of 2,050 miles and that the distribution of miles follows the normal probability distribu- tion. Layton wants to set the minimum guar- anteed mileage so that no more than 4% of the tires will have to be replaced. What min- imum guaranteed mileage should Layton announce?
S O L U T I O N
The facets of this case are shown in the fol- lowing diagram, where x represents the minimum guaranteed mileage.
Scale of milesm 67,900
.5000
4% or .0400
x ?
Tire replaced if the mileage is less than this value
.4600
Inserting these values in formula (7–5) for z gives:
z = x − μ
σ =
x − 67,900 2,050
There are two unknowns in this equation, z and x. To find x, we first find z, and then solve for x. Recall from the characteristics of a normal curve that the area to the left of μ is .5000. The area between μ and x is .4600, found by .5000 − .0400. Now refer to Appendix B.3. 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
© JupiterImages/Getty Images
228 CHAPTER 7
Knowing that the distance between μ and x is −1.75σ or z = −1.75, we can now solve for x (the minimum guaranteed mileage):
z = x − 67,900
2,050
−1.75 = x − 67,900
2,050 −1.75(2,050) = x − 67,900 x = 67,900 − 1.75(2,050) = 64,312
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% of the tires will be replaced under this plan.
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
TABLE 7–2 Selected Areas under the Normal Curve
Excel will also find the mileage value. See the following output. The necessary com- mands are given in the Software Commands in Appendix C.
An analysis of the final test scores for Introduction to Business reveals the scores fol- low the normal probability distribution. The mean of the distribution is 75 and the stan- dard deviation is 8. The professor wants to award an A to students whose score is in the highest 10%. What is the dividing point for those students who earn an A and those earning a B?
S E L F - R E V I E W 7–6
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 7–2.
CONTINUOUS PROBABILITY DISTRIBUTIONS 229
THE NORMAL APPROXIMATION TO THE BINOMIAL Chapter 6 describes the binomial probability distribution, which is a discrete distribu- tion. The table of binomial probabilities in Appendix B.1 goes successively from an n of 1 to an n of 15. If a problem involved taking a sample of 60, generating a binomial dis- tribution for that large a number would be very time-consuming. A more efficient ap- proach 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 π = .50 from
LO7-4 Approximate the binomial probability distribution using the standard normal probability distribution to calculate probabilities.
23. A normal distribution has a mean of 50 and a standard deviation of 4. Determine the value below which 95% 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% of the values will occur.
25. Assume that the 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% of the airplanes?
26. The SAT Reasoning Test is perhaps the most widely used standardized test for col- lege 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% 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 4 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 devia- tion of 8.5 hours.
a. If Dick is in the top 1% in terms of listening time, how many hours did he listen last year?
b. Assume that the distribution of times 4 years earlier also follows the normal probability distribution with a standard deviation of 8.5 hours. How many hours did the 1% who listen to the least music actually listen?
28. For the most recent year available, the mean annual cost to attend a private univer- sity in the United States was $42,224. Assume the distribution of annual costs fol- lows 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 he or she will make an investment. A research report says that annual re- turns from a specific class of common equities are distributed according to a normal distribution with a mean of 12% and a standard deviation of 18%. 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?
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 poten- tial customers as to how long they can expect a cartridge to last. How many pages should the manufacturer advertise for each cartridge if it wants to be correct 99% of the time?
E X E R C I S E S
230 CHAPTER 7
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.
When can we use the normal approximation to the binomial? The normal probability distribution is a good approximation to the binomial probability distribution when nπ and n(1 − π) 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 binomial 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 number of trials.
3. The probability of a success, π, 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% of its new customers return for another meal. For a week 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 customers 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 re- mains at .70 for all 80 customers.
Therefore, we could use the binomial formula (6–3) described on page 185.
P(x) = nCx (π)x (1 − π)n−x
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:
P(x = 60) = 80C60 (.70)60 (1 − .70)20 = .063
Next we find the probability that exactly 61 customers return. It is:
P(x = 61) = 80C61 (.70)61 (1 − .70)19 = .048
n = 1
P (x
)
.40
x
.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
0
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 π = .50
CONTINUOUS PROBABILITY DISTRIBUTIONS 231
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 preceding problem in this manner is tedious. We can also use statistical software packages to find the various probabilities. Listed below are the binomial probabilities for n = 80, π = .70, and x, the number of customers returning, 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 probabilities are 0.000.
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
We can find the probability of 60 or more returning by summing .063 + .048 + . . . + .001, which is .197. However, a look at the plot below shows the similarity of this dis- tribution to a normal distribution. All we need do is “smooth out” the discrete probabili- ties into a continuous distribution. Furthermore, 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.
.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
ty
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 is made because a continuous dis- tribution (the normal distribution) is being used to approximate a discrete distribution (the binomial distribution).
232 CHAPTER 7
How to Apply the Correction Factor Only four cases may arise. These cases are:
1. For the probability at least x occur, use the area above (x − .5). 2. For the probability that more than x occur, use the area above (x + .5). 3. For the probability that x or fewer occur, use the area below (x + .5). 4. For the probability that fewer than x occur, 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 value 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:
μ = nπ = 80(.70) = 56 σ2 = nπ(1 − π) = 80(.70) (1 − .70) = 16.8 σ = √16.8 = 4.10
z = x − μ
σ =
59.5 − 56 4.10
= 0.85
Step 2: Determine the area under the normal curve between a μ 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.3 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.
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:
Scale of x Scale of z
.5000
59.5 .85
Probability is .1977 that 60 or more out of
80 will return to Santoni’s
.3023
.1977
56 0
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.
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 customers returning. The result compares favorably with that computed on page 230 using the binomial distribution. The probability using the binomial distribution is .197, whereas the probability using the normal approximation is .1977.
CONTINUOUS PROBABILITY DISTRIBUTIONS 233
A study by Great Southern Home Insurance revealed that none of the stolen goods were recovered by the homeowners in 80% 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?
S E L F - R E V I E W 7–7
31. Assume a binomial probability distribution with n = 50 and π = .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 π = .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% 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?
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% 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%
of its new members are 15 pounds overweight. A membership drive in a metropol- itan area resulted in 500 new members.
a. It has been suggested that the normal approximation to the binomial be used to determine the probability that 175 or more of the new members are 15 pounds overweight. Does this problem qualify as a binomial problem? Explain.
b. What is the probability that 175 or more of the new members are 15 pounds overweight?
c. What is the probability that 140 or more new members are 15 pounds overweight? 36. The website, herecomestheguide.com, suggested that couples planning their wed-
ding should expect eighty percent 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 200 invitations.
a. How many guests would you expect to accept the invitation? b. What is the standard deviation? c. What is the probability 150 or more will accept the invitation? d. What is the probability exactly 150 will accept the invitation?
E X E R C I S E S
234 CHAPTER 7
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 exponential 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 al- ways positive. The exponential distribution usually describes situations such as:
• The service time for customers at the information desk of the Dallas Public Library. • The time between “hits” on a website. • The lifetime of a kitchen appliance. • 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 distribution is described by only one parameter, which we will identify as λ (pronounced “lambda”). λ 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 λ from 1/3 to 1 to 2. Observe that as we decrease λ, the shape of the distribution is “less skewed.”
2.5
2
1.5
x
1
0.5
0 0 1 2 3 4
Three Exponential Distributions
l 5 0.33 l 5 1.0 l 5 2.0
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, μ. We described the Poisson distribution starting in Chapter 6 on page 197. It too is a positively skewed distribution. To ex- plain the relationship between the Poisson and the exponential distributions, suppose customers arrive at a family restaurant during the dinner hour at a rate of six per hour. The Poisson distribution would have a mean of six. For a time interval of one hour, we can use the Poisson distribution to find the probability that one, or two, or ten customers arrive. But suppose instead of studying the number of customers arriving in an hour, we wish to study the time between their arrivals. The time between arrivals is a continuous distri- bution because time is measured as a continuous random variable. If cus-
tomers arrive at a rate of six per hour, then logically 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 ar- rive at a certain rate per hour, which we call μ, then we can expect the mean time be- tween arrivals to be 1/μ. The rate parameter λ is equal to 1/μ. So in our restaurant arrival example, the mean time between customer arrivals is λ = 1/6 of an hour.
The graph of the exponential distribution starts at the value of λ when the random 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 λ as rate parameter. As we described with the Poisson distribution on page 197,
LO7-5 Describe the exponential probability distribution and use it to calculate probabilities.
© Robert Cicchetti/Shutterstock.com
CONTINUOUS PROBABILITY DISTRIBUTIONS 235
e is a mathematical constant equal to 2.71828. It is the base for the natural logarithm system. It is a pleasant surprise that both the mean and the standard deviation of the exponential probability distribution are equal to 1/λ.
EXPONENTIAL DISTRIBUTION P(x) = λe−λx (7–6)
FINDING A PROBABILITY USING THE EXPONENTIAL DISTRIBUTION
P(Arrival time < x) = 1 − e−λx (7–7)
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 distribution be- tween two specified values give the probability the random variable is in that interval. A table, such as Appendix B.3 for the normal distribution, is not necessary for the expo- nential distribution. The area under the exponential density function is found by a for- mula and the necessary calculations can be accomplished with a handheld calculator with an ex key. Most statistical software packages will also calculate exponential proba- bilities by inputting the rate parameter, λ, only. The probability of obtaining an arrival value less than a particular value of x is:
E X A M P L E
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. Find the probability the next order arrives in more than 40 seconds.
S O L U T I O N
To begin, we determine the rate parameter λ, which in this case is 1/20. To find the probability, we insert 1/20 for λ and 5 for x in formula (7–7).
P( Arrival time < 5) = 1 − e − 1
20 (5)
= 1 − e−0.25 = 1 − .7788 = .2212
So we conclude there is a 22% chance the next order will arrive in less than 5 seconds. The region is identified as the 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, l 5 1/20
The preceding computations addressed the area in the left-tail area of the expo- nential distribution with λ = 1/20 and the area between 0 and 5—that is, the area that is below 5 seconds. What if you are interested in the right-tail area? It is found
236 CHAPTER 7
using the complement rule. See formula (5–3) in Chapter 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.
P(Arrival < 40) = 1 − e − 1
20 (40)
= 1 − .1353 = .8647
2. Find the probability an order is received in more than 40 seconds.
P(Arrival > 40) = 1 − P(Arrival < 40) = 1 − .8647 = .1353
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%.
In the preceding example/solution, when we apply the exponential probability dis- tribution to compute the probability that the arrival time is greater than 40 seconds, you probably observed that there is some redundancy. In general, if we wish to find the likelihood of a time greater than some value x, such as 40, the complement rule is ap- plied as follows:
P(Arrival > x) = 1 − P(Arrival < x) = 1 − (1 − e−λx) = e−λx
In other words, when we subtract formula (7–7) from 1 to find the area in the right tail, the result is e−λx. Thus, the probability that more than 40 seconds go by before the next order arrives is computed without the aid of the complement rule as follows:
P(Arrival > 40) = e − 1
20 (40)
= .1353
The result is shown in the following graph.
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, l 5 1/20
What if you wish to determine the probability that it will take more than 5 seconds 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.
In symbols, you can write this as:
P( 5 ≤ x ≤ 40) = P(Arrival ≤ 40) − P(Arrival ≤ 5)
= (1 − e − 1
20 (40)
) − (1 − e − 1
20 (5)
) = .8647 − .2212 = .6435
We conclude that about 64% of the time, the time between orders will be between 5 seconds and 40 seconds.
CONTINUOUS PROBABILITY DISTRIBUTIONS 237
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, l 5 1/20
Previous examples require finding the percentage of the observations located be- tween two values or the percentage of the observations above or below a particular 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 exam- ple/solution illustrates this situation.
E X A M P L E
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 distribu- tion with a mean of 4,000 hours. Compton wants a warranty period such that only 5% of the power supply units fail during that period. What value should they set for the warranty period?
S O L U T I O N
Note that 4,000 hours is a mean and not a rate. Therefore, we must compute λ as 1/4,000, or 0.00025 failure per hour. A diagram of the situation is shown below, where x represents the minimum guaranteed lifetime.
0 2000 4000 6000 8000 10000 12000 0
0.00005
0.0001
0.00015
0.0002
0.00025
0.0003 Exponential, l 5 0.00025
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.
P (Arrival time < x) = 1 − e(−λ x )
= 1 − e − 1
4,000 (x)
= .05
238 CHAPTER 7
Next, we solve this equation for x. So, we subtract 1 from both sides of the equa- tion and multiply by −1 to simplify the signs. The result is:
.95 = e − 1
4,000 (x)
Next, we take the natural log of both sides and solve for x:
ln (.95 ) = − 1
4,000 x
−(.051293294) = − 1
4,000 x
x = 205.17 In this case, x = 205.17. Hence, Compton can set the warranty period at 205 hours and expect about 5% of the power supply units to be returned.
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% of the runs are longer than this time.)
S E L F - R E V I E W 7–8
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 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% of the TV sets last more than what length of time?
39. The Bureau of Labor Statistics’ American Time Use Survey, www.bls.gov/data, 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 follow an exponential distribution, find the pro- portion of each group that spends:
a. Less than 15 minutes per day using a computer for leisure. b. More than 2 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?
E X E R C I S E S
CONTINUOUS PROBABILITY DISTRIBUTIONS 239
C H A P T E R S U M M A R Y
I. The uniform distribution is a continuous probability distribution with the following characteristics. 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 described by the following equation for the region from a to b:
P(x) = 1
b − a (7–3)
E. The mean and standard deviation of a uniform distribution are computed as follows:
μ = (a + b)
2 (7–1)
σ = √ (b − a)2
12 (7–2)
II. The normal probability distribution is a continuous distribution with the following characteristics. 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 standard deviation changes.
2. The normal probability distribution is described by the following formula:
P(x) = 1
σ√2π e−[
(x − μ)2
2σ2 ] (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.
z = x − μ
σ (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. nπ and n(1 − π) must both be at least 5.
1. n is the number of observations. 2. π is the probability of a success.
B. The four conditions for a binomial probability distribution are: 1. There are only two possible outcomes. 2. π (pi) 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 num-
ber of trials. C. The mean and variance of a binomial distribution are computed as follows:
μ = nπ σ2 = nπ(1 − π)
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.
240 CHAPTER 7
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 probabilities are computed using the formula:
P( x ) = λe−λ x (7–6)
C. It is nonnegative, is positively skewed, declines steadily to the right, and is asymptotic.
D. The area under the curve is given by the formula
P(Arrival time < x ) = 1 − e−λ x (7–7)
E. Both the mean and standard deviation are:
μ = 1/λ σ2 = 1/λ
C H A P T E R E X E R C I S E S
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 remaining 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% discount on the purchase. The time required for the credit ap- plication 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 eleva- tor 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 Fabricators 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.
CONTINUOUS PROBABILITY DISTRIBUTIONS 241
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 be-
tween 32 hours and 34 hours to erect? b. What percent of the garages take between 29 hours and 34 hours to erect? c. What percent of the garages take 28.7 hours or less to erect? d. Of the garages, 5% take how many hours or more to erect?
47. In 2015 The United States Department of Agriculture issued a report (http://www.cnpp. usda.gov/sites/default/files/CostofFoodMar2015.pdf) indicating a family of four spent an av- erage of about $890 per month on food. Assume the distribution of food expenditures for a family of four follows the normal distribution, with a standard deviation of $90 per month. a. What percent of the families spend more than $430 but less than $890 per month on
food? b. What percent of the families spend less than $830 per month on food? c. What percent spend between $830 and $1,000 per month on food? d. What percent spend between $900 and $1,000 per month on food?
48. A study of long-distance phone calls made from General Electric Corporate Headquar- ters in Fairfield, Connecticut, 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 minute. a. What is the probability that calls last between 4.2 and 5 minutes? b. What is the probability that calls last more than 5 minutes? c. What is the probability that calls last between 5 and 6 minutes? d. What is the probability that 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% 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 is the probability that annual dental expenses are more than $1,500? b. What is the probability that annual dental expenses are between $1,500 and
$2,000? c. Estimate the probability that an employee had no annual dental expenses. d. What was the cost for the 10% 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% of the representa- tives. 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 number of hours of TV view- ing per week is higher among adult women than adult 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 1% of women who watch the most TV per week watch?
Find the comparable value for the men.
242 CHAPTER 7
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 deviation 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% of production based on past experience. Past records indicate weekly production follows the normal distribu- tion. The mean of this distribution is 4,000 units per week and the standard deviation is 60 units per week. If the bonus is paid on the upper 5% of production, 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 daily number of customers
returning items follows the normal distribution. The mean number of customers return- ing items is 10.3 per day and the standard deviation is 2.25 per day. a. For any day, what is the probability that eight or fewer customers returned items? b. For any day, what is the probability that the number of customers returning items is
between 12 and 14? c. Is there any chance of a day with no customer returns?
56. A recent news report indicated that 20% of all employees steal from their company each year. If a company employs 50 people, what is the probability that: a. Fewer than five employees steal? b. More than five employees steal? c. Exactly five 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% 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% of those taking the quantitative methods portion of the CPA ex- amination 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% of high-speed chases in- volving automobiles result in a minor or major accident. If 50 high-speed chases occur in a year, 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% 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% of the flights are cleared in 45 minutes, so 5% 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?
CONTINUOUS PROBABILITY DISTRIBUTIONS 243
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. As- suming 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? 63. The weights of canned hams processed at Henline Ham Company follow the normal
distribution, with a mean of 9.20 pounds and a standard deviation of 0.25 pound. 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 pound to 0.15. Which change would you recommend?
64. A recent Gallup study (http://www.gallup.com/poll/175286/hour-workweek-actually- longer- seven-hours.aspx) found the typical American works an average of 46.7 hour per week. The study did not report the shape of the distribution of hours worked or the standard deviation. It did however indicate that 40% of the workers worked less than 40 hours a week and that 18 percent worked more than 60 hours. a. If we assume that the distribution of hours worked is normally distributed, and know-
ing 40% of the workers worked less than 40 hours, find the standard deviation of the distribution.
b. If we assume that the distribution of hours worked is normally distributed and 18% of the workers worked more than 60 hours, find the standard deviation of the distribution.
c. Compare the standard deviations computed in parts a and b. Is the assumption that the distribution of hours worked is approximately normal reasonable? Why?
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% 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 is the probability that the end-of-day trading price is over $45.00? Estimate the
number of days in a year when the trading price finished above $45.00. b. What percent of the days was the price between $38.00 and $40.00? c. What is the minimum share price for the top 15% of end-of-day trading prices?
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% of the time sales are more than 500,000. What are the mean and the standard deviation?
244 CHAPTER 7
68. In establishing warranties on HDTVs, 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 deviation of 3.34 months. Where should the warranty limits be set so that only 10% 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% 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 manu- facturer’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 probability 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 distribu- tion with a mean of 27 seconds. What is the probability his “boot” will require: a. Less than 15 seconds? b. More than 60 seconds? c. Between 30 and 45 seconds? d. What is the point below which only 10% 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: a. Will visit an emergency room within the next 6 months? b. Will not visit the ER over the next 6 years? c. Will visit an ER 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?
D A T A A N A L Y T I C S
(The data for these exercises are available at the text website: www.mhhe.com/lind17e.)
74. Refer to the North Valley Real Estate data, which report information on homes sold during the last year. a. The mean selling price (in $ thousands) of the homes was computed earlier to be $357.0,
with a standard deviation of $160.7. Use the normal distribution to estimate the percent- age of homes selling for more than $500.000. Compare this to the actual results. Is price normally distributed? Try another test. If price is normally distributed, how many homes should have a price greater than the mean? Compare this to the actual number of homes. Construct a frequency distribution of price. What do you observe?
b. The mean days on the market is 30 with a standard deviation of 10 days. Use the normal distribution to estimate the number of homes on the market more than 24 days. Compare this to the actual results. Try another test. If days on the market is normally distributed, how many homes should be on the market more than the mean number of days? Compare this to the actual number of homes. Does the normal
CONTINUOUS PROBABILITY DISTRIBUTIONS 245
distribution yield a good approximation of the actual results? Create a frequency dis- tribution of days on the market. What do you observe?
75. Refer to the Baseball 2016 data, which report information on the 30 Major League Base- ball teams for the 2016 season. a. The mean attendance per team for the season was 2.439 million, with a standard
deviation of 0.618 million. Use the normal distribution to estimate the number of teams with attendance of more than 3.5 million. Compare that estimate with the ac- tual number. Comment on the accuracy of your estimate.
b. The mean team salary was $121 million, with a standard deviation of $40.0 million. Use the normal distribution to estimate the number of teams with a team salary of more than $100 million. Compare that estimate with the actual number. Comment on the accuracy of the estimate.
76. Refer to the Lincolnville School District bus data. a. Refer to the maintenance cost variable. The mean maintenance cost for last year is
$4,552 with a standard deviation of $2332. Estimate the number of buses with a maintenance cost of more than $6,000. Compare that with the actual number. Create a frequency distribution of maintenance cost. Is the distribution normally distributed?
b. Refer to the variable on the number of miles driven since the last maintenance. The mean is 11,121 and the standard deviation is 617 miles. Estimate the number of buses traveling more than 11,500 miles since the last maintenance. Compare that number with the actual value. Create a frequency distribution of miles since mainte- nance cost. Is the distribution normally distributed?
A REVIEW OF CHAPTERS 5–7 The chapters in this section consider methods of dealing with uncertainty. In Chapter 5, we describe the concept of prob- ability. A probability is a value between 0 and 1 that expresses the likelihood a particular event will occur. We also looked at methods to calculate probabilities using rules of addition and multiplication; presented principles of counting, including permutations and combinations; and described situations for using Bayes’ theorem.
Chapter 6 describes discrete probability distributions. Discrete probability distributions list all possible outcomes of an exper- iment and the probability associated with each outcome. We describe three discrete probability distributions: the binomial distribution, the hypergeometric distribution, and the Poisson distribution. The requirements for the binomial distribution are there are only two possible outcomes for each trial, there is a constant probability of success, there are a fixed number of trials, and the trials are independent. The binomial distribution lists the probabilities for the number of successes in a fixed number of trials. The hypergeometric distribution is similar to the binomial, but the probability of success is not constant, so the trials are not independent. The Poisson distribution is characterized by a small probability of success in a large number of trials. It has the following characteristics: the random variable is the number of times some event occurs in a fixed interval, the probability of a success is proportional to the size of the interval, and the intervals are independent and do not overlap.
Chapter 7 describes three continuous probability distributions: the uniform distribution, the normal distribution, and the exponential distribution. The uniform probability distribution is rectangular in shape and is defined by minimum and maxi- mum values. The mean and the median of a uniform probability distribution are equal, and it does not have a mode.
A normal probability distribution is the most widely used and widely reported distribution. Its major characteristics are that it is bell-shaped and symmetrical, completely described by its mean and standard deviation, and asymptotic, that is, it falls smoothly in each direction from its peak but never touches the horizontal axis. There is a family of normal probability distributions—each with its own mean and standard deviation. There are an unlimited number of normal probability distributions.
To find the probabilities for any normal probability distribution, we convert a normal distribution to a standard normal prob- ability distribution by computing z values. A z value is the distance between x and the mean in units of the standard devi- ation. 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 computed using standard normal probability tables (see Appendix B.3).
The exponential probability distribution describes the time between events in a sequence. These events occur inde- pendently at a constant rate per unit of time or length. The exponential probability distribution is positively skewed, with λ as the “rate” parameter. The mean and standard deviation are equal and are the reciprocal of λ.
246 CHAPTER 7
P R O B L E M S
1. Proactine, a new medicine for acne, is claimed by the manufacturer to be 80% 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. 12 or more people will show significant improvement?
2. Customers at the Bank of Commerce of Idaho Falls, Idaho, default at a rate of .005 on small home-improvement loans. 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 three 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 ExxonMobil oil production platform against weather losses for one year. The president of Daniel-James estimates 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 ExxonMobil 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 Grand Junction, Colorado, 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 estimate?
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 number of children among families that have children?
6. The following table shows a breakdown of the 114th U.S. Congress by party affiliation. (There are two independent senators included in the count of Democratic senators. There is one vacant House seat.)
Party
Democrats Republicans Total
House 188 246 434 Senate 46 54 100 Total 234 300 534
CONTINUOUS PROBABILITY DISTRIBUTIONS 247
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?
C A S E S
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 customers. Compare the actual distribution with the theo- retical distribution. Cite some specific examples and com- ment 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 ap- pear that account balances are related to the branch? Cite some examples and comment on your findings.
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 con- cerned that five may not be enough to make a good deci- sion. They suggest that you should check 10 signatures and reject the page if 3 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 dis- trict, 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. Calcu- late 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.
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 13
C. Geoff Applies Data Analytics 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 successful calls per hour for the staff. These data ap- pear next 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 different kinds of probability distributions (bino- mial, 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 probationary employee should be accepted as having reached full production status? This is important be- cause it means a pay raise for the employee, and there have been some probationary employees in the past who have quit because 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
248 CHAPTER 7
The score is the sum of the points on the six items. For example, Tracy 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 $450 (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.
Score 30 40 50 60 70 80 90 Probability .70 .78 .85 .90 .94 .95 .96
Tracy’s score of 44 would translate into a probability of being profitable of approximately .81. In other words, 81% of customers like Tracy 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 $450 $650 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.
Descriptive statistics:
N MEAN MEDIAN STANDARD DEVIATION 35 3.229 3.000 1.682 MIN MAX 1ST QUARTILE 3RD QUARTILE 0.0 8.0 2.0 5.0
Analyze the distribution of sales calls. Which distribution do you think Geoff should use for his analysis? Support your recommendation with your analysis. What standard should be used to determine if an employee has reached “full production” status? Explain your recommendation.
D. CNP Bank Card Before banks issue a credit card, they usually rate or score the customer in terms of his or her projected proba- bility of being a profitable customer. A typical scoring ta- ble appears 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.)
P R A C T I C E T E S T
Part 1—Objective 1. Under what conditions will a probability be greater than 1 or 100%? 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 important. 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.
CONTINUOUS PROBABILITY DISTRIBUTIONS 249
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% 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, uniform, 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 are businesses, and 3 are for charitable organizations. If he selects the two returns at random, what is the probability: a. Both are businesses? b. At least one is a business?
2. The IRS reports that 15% of returns where the adjusted gross income is more than $1,000,000 will be subject to a computer audit. For the year 2017, 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?
3. Fred works in a tax office with five other CPAs. There are five parking spots beside the office. In how many different 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 2017. 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 2017, 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.
Dependents
Refund 1 2 3 or more Total
Yes 20 20 10 50 No 10 20 0 30 Total 30 40 10 80
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 that 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?
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO8-1 Explain why populations are sampled and describe four methods to sample a population.
LO8-2 Define sampling error.
LO8-3 Demonstrate the construction of a sampling distribution of the sample mean.
LO8-4 Recite the central limit theorem and define the mean and standard error of the sampling distribution of the sample mean.
LO8-5 Apply the central limit theorem to calculate probabilities.
THE NIKE annual report says that the average American buys 6.5 pairs of sports shoes per year. Suppose a sample of 81 customers is surveyed and the population standard deviation of sports shoes purchased per year is 2.1 What is the standard error of the mean in this experiment? (See Exercise 45 and LO8-4.)
Sampling Methods and the Central Limit Theorem
8
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SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 251
INTRODUCTION Chapters 2 through 4 emphasize techniques to describe data. To illustrate these techniques, 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 measures of location and dispersion. Such measures as the mean and the standard deviation describe the typical profit and the spread in the profits. In these chapters, the emphasis is on describing the distribution of the data. That is, we describe something that has already happened.
In Chapter 5, we begin to lay the foundation for statistical inference with the study of probability. Recall that in statistical inference our goal is to determine something about a population based only on the sample. The population is the entire group of in- dividuals or objects under consideration, and the sample is a part or subset of that pop- ulation. Chapter 6 extends the probability concepts by describing three discrete probability distributions: the binomial, the hypergeometric, and the Poisson. Chapter 7 describes three continuous probability distributions: the uniform, normal, and exponen- tial. Probability distributions encompass all possible 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 process of selecting items from a population so we can use this information to make judgments or inferences about the pop- ulation. We begin this chapter by discussing methods of selecting a sample from a popula- tion. Next, we construct a distribution of the sample mean to understand 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.
SAMPLING METHODS In Chapter 1, we said the purpose of inferential statistics is to find something about a pop- ulation 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 population. In this section, we discuss the 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. Public opin- ion polls and consumer testing organizations, such as Harris Interactive Inc., CBS News Polls, and Zogby Analytics, usually contact fewer than 2,000 of the nearly 60 million families in the United States. One consumer panel–type organization charges $40,000 to mail samples and tabulate responses to test a product (such as breakfast cereal, cat food, or perfume). The same product test using all 60 million families would be too expensive to be worthwhile.
3. The physical impossibility of checking all items in the population. Some popu- lations 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 populations of fish, birds, snakes, deer, and the like are large and are constantly moving, being
LO8-1 Explain why populations are sampled and describe four methods to sample a population.
252 CHAPTER 8
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 se- lected at random, tracking fish catches, or netting fish at predeter- mined 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 Department selects a sample from the current production. Each piece is stretched until it breaks and the breaking point (usually measured 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 ad- ditional accuracy of a 100% sample—that is, studying the entire population—is essen- tial 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 inclusion of all grocery stores in the United States would significantly affect the index because 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 sampling method is a simple random sampling.
© David Epperson/Getty Images
SIMPLE RANDOM SAMPLE A sample selected so that each item or person in the population has the same chance of being included.
To illustrate the selection process for a simple random sample, suppose the popula- tion of interest is the 750 Major League Baseball players on the active rosters of the 30 teams at the end of the 2017 season. The president of the players’ union wishes to form a committee of 10 players to study the issue of concussions. One way of ensuring that every player in the population has the same chance of being chosen to serve on the Con- cussion Committee is to write each name of the 750 players on a slip of paper and place all the slips of paper in a box. After the slips of paper have been thoroughly mixed, the first selection is made by drawing a slip of paper from the box identifying the first player. The slip of paper is not returned to the box. This process is repeated nine more times to form the committee. (Note that the probability of each selection does increase slightly because the slip is not replaced. However, the differences are very small because the population is 750. The probability of each selection is about 0.0013, rounded to four decimal places.)
Of course, the process of writing all the players’ names on a slip of paper is very time-consuming. A more convenient method of selecting a random sample is to use a table of random numbers such as the one in Appendix B.4. In this case the union president would prepare a list of all 750 players and number each of the players from 1 to 750 with a computer application. Using a table of random numbers, we would randomly pick a starting place in the table, and then select 10 three-digit numbers between 001 and 750. A computer can also generate random numbers. These numbers would correspond with the 10 players in the list that will be asked to participate on the committee. As the name simple random sampling implies, the probability of selecting any number between 001 and 750 is the same. Thus, the probability of selecting the player assigned the number 131 is
SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 253
the same as the probability of selecting player 722 or player 382. Using random numbers to select players for the committee removes any bias from the selection process.
The following example shows how to select random numbers using a portion of a random number table illustrated below. First, we choose a starting point in the table. One way of selecting the starting point is to close your eyes and point at a number in the table. Any starting point will do. Another way is to randomly pick a column and row. Suppose the time is 3:04. Using the hour, three o’clock, pick the third column and then, using the min- utes, four, move down to the fourth row of numbers. The number is 03759. Because there are only 750 players, we will use the first three digits of a five-digit random number. Thus, 037 is the number of the first player to be a member of the sample. To continue selecting players, we could move in any direction. Suppose we move right. The first three digits of the number to the right of 03759 are 447. Player number 447 is the second player se- lected to be on the committee. The next three-digit number to the right is 961. You skip 961 as well as the next number 784 because there are only 750 players. The third player selected is number 189. We continue this process until we have 10 players.
STATISTICS IN ACTION
To insure that an unbiased, representative sample is selected from a population, lists of random numbers are needed. In 1927, L. Tippett published the first book of random numbers. In 1938, R. A. Fisher and F. Yates published 15,000 random digits generated using two decks of cards. In 1955, RAND Corporation published a million random digits, generated by the random frequency pulses of an electronic roulette wheel. Since then, com- puter programs have been developed for generating digits that are “almost” random and hence are called pseudo-random. The question of whether a com- puter program can be used to generate numbers that are truly random remains a debatable issue.
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 point player player
Statistical packages such as Minitab and spreadsheet packages such as Excel have software that will select a simple random sample. The following example/solution uses Excel to select a random sample from a list of the data.
E X A M P L E
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. For each day of June 2017, the number of rooms rented is listed. Use Excel to select a sample of five nights during the month of June.
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
S O L U T I O N
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 rooms were rented. The information is reported in column D of the Excel
254 CHAPTER 8
The following roster lists the students enrolled in an introductory course in business statis- tics. Three students will be randomly selected and asked questions about 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 are in the sample? (b) Now use the table of random digits, Appendix B.4, to select your own sample. (c) What would you do if you encountered the number 59 in the table of random digits?
S E L F - R E V I E W 8–1
STAT 264 BUSINESS STATISTICS 9:00 AM - 9:50 AM MW; 118 CARLSON HALL; PROFESSOR LIND
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 in Appendix C. The Excel system performs the sampling with replacement. This means it is possi- ble for the same day to appear more than once in a sample.
SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 255
Systematic Random Sampling The simple random sampling procedure is awkward in some research situations. For exam- ple, Stood’s Grocery Market needs to sample their customers to study the length of time customers spend in the store. Simple random sampling is not an effective method. Practi- cally, we do not have a list of customers, so assigning random numbers to customers is im- possible. Instead, we can use systematic random sampling to select a representative sample. Using this method for Stood’s Grocery Market, we decide to select 100 customers over 4 days, Monday through Thursday. We will select 25 customers a day and begin the sampling at different times each day: 8 a.m., 11 a.m., 4 p.m., and 7 p.m. We write the 4 times and 4 days on slips of paper and put them in two hats—one hat for the days and the other hat for the times. We select one slip from each hat. This ensures that the time of day is ran- domly assigned for each day. Suppose we selected 4 p.m. for the starting time on Monday. Next we select a random number between 1 and 10; it is 6. Our selection process begins on Monday at 4 p.m. by selecting the sixth customer to enter the store. Then, we select every 10th (16th, 26th, 36th) customer until we reach the goal of 25 customers. For each of these sampled customers, we measure the length of time the customer spends in the store.
SYSTEMATIC RANDOM SAMPLE A random starting point is selected, and then every kth member of the population is selected.
STRATIFIED RANDOM SAMPLE A population is divided into subgroups, called strata, and a sample is randomly selected from each stratum.
STATISTICS IN ACTION
Random and unbiased sampling methods are ex- tremely important to make valid statistical inferences. In 1936, the Literary Di- gest conducted a straw vote to predict the out- come of the presidential race between Franklin Roosevelt and Alfred Landon. Ten million ballots in the form of returnable postcards were sent to ad- dresses taken from Literary Digest subscribers, tele- phone directories and au- tomobile registrations. In 1936 not many people could afford a telephone or an automobile. Thus, the population that was sam- pled did not represent the population of voters. A sec- ond problem was with the non-responses. 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 represented a cross-section of all the vot- ers. On Election Day, Roos- evelt won with 61% of the vote. Landon had 39%. In the mid-1930s people who had telephones and drove automobiles clearly did not represent American voters!
Simple random sampling is used in the selection of the days, the times, and the starting point. But the systematic procedure is used to select the actual customer.
Before using systematic random sampling, we should carefully observe the physi- cal order of the population. When the physical order is related to the population charac- teristic, then systematic random sampling should not be used because the sample could be biased. For example, if we wanted to audit the invoices in a file drawer that were ordered in increasing dollar amounts, systematic random sampling would not guarantee an unbiased 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 called strata. For example, college students can be grouped as full time or part time; as male or female; or as freshman, sophomore, junior, or senior. Usually the strata are formed based on members’ shared attributes or characteristics. A random sample from each stratum is taken in a number proportional to the stratum’s size when compared to the population. Once the strata are defined, we 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 companies in the United States. The objective of the study is to determine whether firms with high returns on equity (a measure of profitability) spend more on advertising than firms with low returns on equity. To make sure the sample 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 is used, observe that firms in the 3rd and 4th strata have a high chance of selection (probabil- ity of 0.87) while firms in the other strata have a small 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 each of strata 1 and
256 CHAPTER 8
5 is represented in the sample. Let’s say that 50 firms are selected for intensive study. Then based on probability, 1 firm, or (0.02)(50), should be randomly selected from stratum 1. We would randomly select 5, or (0.10)(50), firms from stratum 2. 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 some cases, of more accurately reflecting the characteristics of the population than does simple random or systematic random sampling.
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.
TABLE 8–1 Number Selected for a Proportional Stratified Random Sample
Profitability Number of Relative Number Stratum (return on equity) Firms Frequency Sampled
1 30% and over 8 0.02 1* 2 20 up to 30% 35 0.10 5* 3 10 up to 20% 189 0.54 27 4 0 up to 10% 115 0.33 16 5 Deficit 5 0.01 1
Total 352 1.00 50
*0.02 of 50 = 1, 0.10 of 50 = 5, etc.
CLUSTER SAMPLING 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 the greater Chicago, Illinois, metropolitan area about state and federal environmental protection policies. Selecting a random sample of residents in this region and personally contacting each one would be time-consuming and very expensive. Instead, you could employ cluster sampling by subdi- viding the region into small units, perhaps by counties. These are often called primary units.
There are 12 counties in the greater Chicago metropolitan area. Suppose you ran- domly select 3 counties. The 3 chosen are La Porte, Cook, and Kenosha (see Chart 8–1 below). Next, you select a random sample of the residents in each of these counties and interview them. This is also referred to as sampling through an intermediate unit. In this case, the intermediate unit is the county. (Note that this is a combination of cluster sam- pling and simple random sampling.)
Lake Michigan
La Porte
Po rte
r Lake
Cook
Will
Gr un
dy
Ke nd
all
Kane
McHenry Lake
Du Pa
ge
Kenosha
CHART 8–1 The Counties of the Greater Chicago, Illinois, Metropolitan Area
SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 257
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.
Refer to Self-Review 8–1 and the class roster on page 254. 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?
S E L F - R E V I E W 8–2
1. The following is a list of 24 Marco’s Pizza stores in Lucas County. The stores are iden- tified by numbering them 00 through 23. Also noted is whether the store is corpo- rate-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
00 2607 Starr Av C 01 309 W Alexis Rd C 02 2652 W Central Av C 03 630 Dixie Hwy M 04 3510 Dorr St C 05 5055 Glendale Av C 06 3382 Lagrange St M 07 2525 W Laskey Rd C 08 303 Louisiana Av C 09 149 Main St C 10 835 S McCord Rd M 11 3501 Monroe St M
ID No. Address Type
12 2040 Ottawa River Rd C 13 2116 N Reynolds Rd C 14 3678 Rugby Dr C 15 1419 South Av C 16 1234 W Sylvania Av C 17 4624 Woodville Rd M 18 5155 S Main M 19 106 E Airport Hwy C 20 6725 W Central M 21 4252 Monroe C 22 2036 Woodville Rd C 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. Using systematic random sampling, every seventh location is selected starting
with the third store in the list. Which locations will be included in the sample? d. Using stratified random sampling, select three locations. Two should be corpo-
rate-owned and one should be manager-owned. 2. The following is a list of 29 hospitals in the Cincinnati, Ohio, and Northern Kentucky
region. Each hospital is assigned a number, 00 through 28. The hospitals are clas- sified by type, either 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.
E X E R C I S E S
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
ID Number Name Address Type
04 Mercy Hospital– 100 Riverfront Plaza M/S Hamilton Hamilton, Ohio 45011 05 Middletown Regional 105 McKnight Drive M/S 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
258 CHAPTER 8
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. c. Using systematic random sampling, every fifth location is selected starting with
the second hospital in the list. Which hospitals will be included in the sample? d. Using stratified random sampling, select five hospitals. Four should be medical
and surgical hospitals and one should be a specialty hospital. Select an appro- priate 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. The members are identified by numbering them 00 through 34.
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?
ID Number Name Address Type
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 10 Christ Hospital 2139 Auburn Avenue M/S Cincinnati, Ohio 45219 11 Deaconess Hospital 311 Straight Street M/S 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
ID Number Name Address Type
19 St. Luke’s Hospital 7380 Turfway Drive M/S West Florence, Kentucky 41075 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 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
ID Number Dealer
00 Dave White Acura 01 Autofair Nissan 02 Autofair Toyota-Suzuki 03 George Ball’s Buick GMC
Truck 04 York Automotive Group 05 Bob Schmidt Chevrolet 06 Bowling Green Lincoln Mercury
Jeep Eagle 07 Brondes Ford 08 Brown Honda 09 Brown Mazda 10 Charlie’s Dodge
ID Number Dealer
11 Thayer Chevrolet/Toyota 12 Spurgeon Chevrolet Motor
Sales, Inc. 13 Dunn Chevrolet 14 Don Scott Chevrolet 15 Dave White Chevrolet Co. 16 Dick Wilson Infinity 17 Doyle Buick 18 Franklin Park Lincoln Mercury 19 Genoa Motors 20 Great Lakes Ford Nissan 21 Grogan Towne Chrysler 22 Hatfield Motor Sales
ID Number Dealer
23 Kistler Ford, Inc. 24 Lexus of Toledo 25 Mathews Ford Oregon, Inc. 26 Northtown Chevrolet 27 Quality Ford Sales, Inc. 28 Rouen Chrysler Jeep Eagle 29 Saturn of Toledo 30 Ed Schmidt Jeep Eagle 31 Southside Lincoln Mercury 32 Valiton Chrysler 33 Vin Divers 34 Whitman Ford
SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 259
SAMPLING “ERROR” In the previous section, we discussed sampling methods that are used to select a sam- ple that is an 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 standard deviation would be exactly equal to the population standard deviation. We can therefore expect a differ- ence between a sample statistic and its corresponding population parameter. This dif- ference is called sampling error.
LO8-2 Define sampling error.
SAMPLING ERROR The difference between a sample statistic and its corresponding population parameter.
The following example/solution clarifies the idea of sampling error.
E X A M P L E
Refer to the example/solution on page 253, where we studied the number of rooms rented at the Foxtrot Inn bed and breakfast in Tryon, North Carolina. The population is the number of rooms rented each of the 30 days in June 2017. Find the mean of
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. Using systematic random sampling, every fifth dealer is selected starting with
the third dealer in the list. Which dealers are included in the sample?
b. Use the table of random numbers to select your own sample of five dealers. c. Using systematic random sampling, every seventh dealer is selected starting
with the fourth dealer in the list. Which dealers are included in the sample? 4. Listed next are the 27 Nationwide Insurance agents in the El Paso, Texas metropol-
itan area. The agents are numbered 00 through 26. We would like to estimate the mean number of years employed with Nationwide.
ID Number Agent
00 Bly Scott 3332 W Laskey Rd 01 Coyle Mike 5432 W Central Av 02 Denker Brett 7445 Airport Hwy 03 Denker Rollie 7445 Airport Hwy 04 Farley Ron 1837 W Alexis Rd 05 George Mark 7247 W Central Av 06 Gibellato Carlo 6616 Monroe St 07 Glemser Cathy 5602 Woodville Rd 08 Green Mike 4149 Holland Sylvania Rd 09 Harris Ev 2026 Albon Rd
ID Number Agent
10 Heini Bernie 7110 W Centra 11 Hinckley Dave 14 N Holland Sylvania Rd 12 Joehlin Bob 3358 Navarre Av 13 Keisser David 3030 W Sylvania Av 14 Keisser Keith 5902 Sylvania Av 15 Lawrence Grant 342 W Dussel Dr 16 Miller Ken 2427 Woodville Rd 17 O’Donnell Jim 7247 W Central Av 18 Priest Harvey 5113 N Summit St 19 Riker Craig 2621 N Reynolds Rd
ID Number Agent
20 Schwab Dave 572 W Dussel Dr 21 Seibert John H 201 S Main 22 Smithers Bob 229 Superior St 23 Smithers Jerry 229 Superior St 24 Wright Steve 105 S Third St 25 Wood Tom 112 Louisiana Av 26 Yoder Scott 6 Willoughby Av
260 CHAPTER 8
the population. Select three random samples of 5 days. Calculate the mean rooms rented for each sample and compare it to the population mean. What is the sam- pling error in each case?
S O L U T I O N
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 μ.
μ = Σx N
= 0 + 2 + 3 + . . . + 3
30 =
94 30
= 3.13
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 is 3.80 rooms, which we desig- nate as x1. The bar over the x reminds us that it is a sample mean and the sub- script 1 indicates it is the mean of the first sample.
x1 = Σx n
= 4 + 7 + 4 + 3 + 1
5 =
19 5
= 3.80
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 (x1 − μ) = 3.80 − 3.13 = 0.67. The second random sample of 5 days from the pop- ulation 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 3.40, found by
x2 = Σx n
= 3 + 3 + 2 + 3 + 6
5 = 3.40
The sampling error is (x2 − μ) = 3.4 − 3.13 = 0.27. In the third random sample, the mean was 1.80 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, indicating that the sample mean overestimated the population mean; other times they are negative values, indicating the sample mean was less than the popula- tion 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) on page 164. Each of the 142,506 dif- ferent 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 summed the sampling errors for all 142,506 samples, the result would equal zero. This is true because the sample mean is an unbiased estimator of the popula- tion mean.
SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 261
SAMPLING DISTRIBUTION OF THE SAMPLE MEAN In the previous section, we defined sampling error and presented the results when we compared a sample statistic, such as the sample mean, to the population mean. To put it another way, when we use the sample mean to estimate the population mean, how can we determine how accurate the estimate is? How does:
• A quality-assurance supervisor decide if a machine is filling 20-ounce bottles with 20 ounces of cola based only on a sample of 10 filled bottles?
• FiveThirtyEight.com or Gallup make accurate statements about the demographics of voters in a presidential race based on relatively small samples from a voting pop- ulation 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/solution varied from one sample to the
next. The mean of the first sample of 5 days was 3.80 rooms, and the second sample mean 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.
LO8-3 Demonstrate the construction of a sampling distribution of the sample mean.
SAMPLING DISTRIBUTION OF THE SAMPLE MEAN A probability distribution of all possible sample means of a given sample size.
The following example/solution illustrates the construction of a sampling distribu- tion of the sample mean. We have intentionally used a small population to highlight the relationship between the population mean and the various sample means.
E X A M P L E
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
Employee Hourly Earnings Employee Hourly Earnings
Joe $14 Jan 14 Sam 14 Art 16 Sue 16 Ted 18 Bob 16
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?
S O L U T I O N
Here are the solutions to the questions.
1. The population is small so it is easy to calculate the population mean. It is $15.43, found by:
μ = Σx N
= $14 + $14 + $16 + $16 + $14 + $16 + $18
7 = $15.43
262 CHAPTER 8
Hourly Hourly Sample Employees Earnings Sum Mean Sample Employees Earnings Sum Mean
1 Joe, Sam $14, $14 $28 $14 12 Sue, Bob 16,16 32 16 2 Joe, Sue 14, 16 30 15 13 Sue, Jan 16,14 30 15 3 Joe, Bob 14, 16 30 15 14 Sue, Art 16,16 32 16 4 Joe, Jan 14, 14 28 14 15 Sue, Ted 16,18 34 17 5 Joe, Art 14, 16 30 15 16 Bob, Jan 16,14 30 15 6 Joe, Ted 14, 18 32 16 17 Bob, Art 16,16 32 16 7 Sam, Sue 14, 16 30 15 18 Bob, Ted 16,18 34 17 8 Sam, Bob 14, 16 30 15 19 Jan, Art 14,16 30 15 9 Sam, Jan 14,14 28 14 20 Jan, Ted 14,18 32 16 10 Sam, Art 14,16 30 15 21 Art, Ted 16,18 34 17 11 Sam, Ted 14,18 32 16
TABLE 8–3 Sample Means for All Possible Samples of 2 Employees
TABLE 8–4 Sampling Distribution of the Sample Mean for n = 2
Sample Mean Number of Means Probability
$14 3 .1429 15 9 .4285 16 6 .2857 17 3 .1429 21 1.0000
We identify the population mean with the Greek letter μ. Recall from earlier chapters, Greek letters are used to represent population parameters.
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) on page 164.
NCn = N!
n!(N − n)! =
7! 2!(7 − 2)!
= 21
where N = 7 is the number of items in the population and n = 2 is the number of items in the sample.
The 21 sample means from all possible samples of 2 that can be drawn from the population of 7 employees are shown in Table 8–3. These 21 sample means are used to construct a probability distribution. This is called the sam- pling distribution of the sample mean, and it is summarized in Table 8–4.
3. Using the data in Table 8–3, the mean of the sampling distribution of the sample mean is obtained by summing the various sample means and dividing the sum by the number of samples. The mean of all the sample means is usually written μx. The μ reminds us that it is a population value because we have considered all possible samples of two employees from the population of seven employees. The subscript x indicates that it is the sampling distribution of the sample mean.
μx = Sum of all sample means Total number of samples
= $14 + $15 + $15 + … + $16 + $17
21
= $324
21 = $15.43
SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 263
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 sampling distribu- tion 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 the sample mean is narrower than the
population distribution. 3. The sampling distribution of the sample mean tends to become bell-shaped and to
approximate the normal probability distribution.
Given a bell-shaped or normal probability distribution, we will be able to apply con- cepts 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 sample size as it re- lates to the sampling distribution of the sample mean.
The years of service of the five executives employed by Standard Chemicals are:
Name Years
Mr. Snow 20 Ms. Tolson 22 Mr. Kraft 26 Ms. Irwin 24 Mr. Jones 28
S E L F - R E V I E W 8–3
4. Refer to Chart 8–2. It shows the population distribution based on the data in Table 8–2 and the distribution of the sample mean based on the data in Table 8–4. These observations can be made:
a. The mean of the distribution of the sample mean ($15.43) is equal to the mean of the population: μ = μx.
b. The spread in the distribution of the sample mean is less than the spread in the population values. The sample means range from $14 to $17 while the popula- tion values vary from $14 up to $18. If we continue to increase the sample size, 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 approxi- mate the normal probability distribution.
Population distribution
.40
.30
.20
14
.10
.40
.30
.20
.10
Hourly earnings16 18m m
Distribution of sample mean
14 Sample mean of hourly earnings
16 1815 17 x x
Pr ob
ab ili
ty
Pr ob
ab ili
ty
CHART 8–2 Distributions of Population Values and Sample Means
264 CHAPTER 8
(a) Using the combination formula, how many samples of size 2 are possible? (b) List all possible samples of two 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)?
1
0 20 22 24 26 28
Years of service
Fr eq
ue nc
y
(g) Is the distribution of the sample mean computed in part (c) starting to show some tendency toward being bell-shaped?
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 is the number of cases each partner actually tried in court last month.
Partner Number of Cases
Ruud 3 Wu 6 Sass 3 Flores 3 Wilhelms 0 Schueller 1
a. How many different samples of size 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 8–2, compare the dispersion in the population with
that of the sample means.
E X E R C I S E S
SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 265
THE CENTRAL LIMIT THEOREM In this section, we examine the central limit theorem. Its application to the sampling distri- bution of the sample mean, introduced in the previous section, allows us to use the normal probability distribution to create confidence intervals for the population 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 probability distribution. The approximation is more accurate for large samples than for small samples. This is one of the most useful conclu- sions in statistics. We can reason 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 population distributions.
LO8-4 Recite the central limit theorem and define the mean and standard error of the sampling distribution of the sample mean.
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.
To further illustrate the central limit theorem, if the population follows a normal proba- bility distribution, then for any sample size the sampling distribution of the sample mean will also be normal. If the population distribution is symmetrical (but not normal), you will see the normal shape of the distribution 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 8–3 for various population shapes. Observe the convergence to a normal distribution regardless of the shape of the population distribution.
The idea that a distribution of sample means will converge to normality when the population is not normal is illustrated in Charts 8–4, 8–5, and 8–6. We will discuss this example in more detail shortly, but Chart 8–4 is a graph of a discrete probability distri- bution 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 from the population portrayed in Chart 8–4 and compute the mean of each sample. These results are shown in Chart 8–5. 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 n = 5 from a population that is positively skewed and found the distribution
10. There are five sales associates at Mid-Motors Ford. The five associates and the number of cars they sold last week are:
Sales Associate 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 the sample mean with that of
the population. d. On a chart similar to Chart 8–2, compare the dispersion in sample means with
that of the population.
266 CHAPTER 8
of sample means is different from the shape of the population. As we take larger sam- ples, that is, n = 20 instead of n = 5, we will find the distribution of the sample mean will approach the normal distribution. Chart 8–6 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 follow- ing example/solution will underscore this condition.
n 5 2
n 5 6
n 5 30
n 5 2
n 5 6
n 5 30
n 5 2
n 5 6
n 5 30
n 5 2
n 5 6
n 5 30
x x x x
Populations
Sampling Distributions
x _
x _
x _
x _
x _
x _
x _
x _
x _
x _
x _
x _
CHART 8–3 Results of the Central Limit Theorem for Several Populations
E X A M P L E
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 representative 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
SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 267
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 service of all employees (the population) compare with the shape of the sampling distribution of the mean? The years of service (rounded to the nearest year) of the 40 employees 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
CHART 8–4 Years of Service for Spence Sprockets Inc. Employees
S O L U T I O N
Chart 8–4 shows a histogram for the frequency distribution of the years of ser- vice for the population of 40 current employees. This distribution is positively skewed. Why? Because the business has grown in recent years, the distribution shows that 29 of the 40 employees have been with the company less than 6 years. Also, there are 11 employees who have worked at Spence Sprockets for more than 6 years. In particular, four employees have been with the company 12 years or more (count the frequencies above 12). So there is a long tail in the distribution of service years to the right, that is, the distribution is positively skewed.
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 years of service for those on the committee?
To begin, Ed writes the years 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 and randomly selects five slips of paper. The years of service for these five employees are 1, 9, 0, 19, and 14 years. Thus, the mean years of service for these five sampled employees is 8.60 years. How does that com- pare with the population mean? At this point, Ed does not know the population mean, but the number of employees in the population is only 40, so he decides to calculate the mean years of service for all his employees. It is 4.8 years,
268 CHAPTER 8
found by adding the years of service for all the employees and dividing the total by 40.
μ = 11 + 4 + 18 + … + 2 + 3
40 = 4.80
The difference between a sample mean (x̄) and the population mean (μ) is called sampling error. In other words, the difference of 3.80 years between the sample mean of 8.60 and the population mean of 4.80 is the sampling error. It is due to chance. Thus, if Ed selected these five employees to constitute the com- mittee, their mean years 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 years 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 and computing the mean for each sample is shown in Table 8–5 and Chart 8–5. 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 the distribution of these sample means. The population of the years of service for employees (Chart 8–4) is positively skewed, but the distribution of these 25 sample means does not reflect the same positive skew. There is also a differ- ence 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 8–5 Twenty-Five Random Samples of Five Employees
Sample Data
Sample Obs 1 Obs 2 Obs 3 Obs 4 Obs 5 Sum Mean
A 1 9 0 19 14 43 8.6 B 7 4 4 1 3 19 3.8 C 8 19 8 2 1 38 7.6 D 4 18 2 0 11 35 7.0 E 4 2 4 7 18 35 7.0 F 1 2 0 3 2 8 1.6 G 2 3 2 0 2 9 1.8 H 11 2 9 2 4 28 5.6 I 9 0 4 2 7 22 4.4 J 1 1 1 11 1 15 3.0 K 2 0 0 10 2 14 2.8 L 0 2 3 2 16 23 4.6 M 2 3 1 1 1 8 1.6 N 3 7 3 4 3 20 4.0 O 1 2 3 1 4 11 2.2 P 19 0 1 3 8 31 6.2 Q 5 1 7 14 9 36 7.2 R 5 4 2 3 4 18 3.6 S 14 5 2 2 5 28 5.6 T 2 1 1 4 7 15 3.0 U 3 7 1 2 1 14 2.8 V 0 1 5 1 2 9 1.8 W 0 3 19 4 2 28 5.6 X 4 2 3 4 0 13 2.6 Y 1 1 2 3 2 9 1.8
SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 269
Now let’s change the example by increasing the size of each sample from 5 employees to 20. Table 8–6 reports the result of selecting 25 samples of 20 em- ployees each and computing their sample means. These sample means are shown graphically in Chart 8–6. Compare the shape of this distribution to the population (Chart 8–4) and to the distribution of sample means where the sample is n = 5 (Chart 8–5). You should observe two important features:
1. The shape of the distribution of the sample mean is different from that of the population. In Chart 8–4, the distribution of all employees is positively skewed. However, as we select random samples from this population, the shape of the
Sample Data
Sample Obs 1 Obs 2 Obs 3 – Obs 19 Obs 20 Sum Mean
A 3 8 3 – 4 16 79 3.95 B 2 3 8 – 3 1 65 3.25 C 14 5 0 – 19 8 119 5.95 D 9 2 1 – 1 3 87 4.35 E 18 1 2 – 3 14 107 5.35 F 10 4 4 – 2 1 80 4.00 G 5 7 11 – 2 4 131 6.55 H 3 0 2 – 16 5 85 4.25 I 0 0 18 – 2 3 80 4.00 J 2 7 2 – 3 2 81 4.05 K 7 4 5 – 1 2 84 4.20 L 0 3 10 – 0 4 81 4.05 M 4 1 2 – 1 2 88 4.40 N 3 16 1 – 11 1 95 4.75 O 2 19 2 – 2 2 102 5.10 P 2 18 16 – 4 3 100 5.00 Q 3 2 3 – 3 1 102 5.10 R 2 3 1 – 0 2 73 3.65 S 2 14 19 – 0 7 142 7.10 T 0 1 3 – 2 0 61 3.05 U 1 0 1 – 9 3 65 3.25 V 1 9 4 – 2 11 137 6.85 W 8 1 9 – 8 7 107 5.35 X 4 2 0 – 2 5 86 4.30 Y 1 2 1 – 1 18 101 5.05
TABLE 8–6 Twenty-Five Random Samples of 20 Employees
CHART 8–5 Histogram of Mean Years of Service for 25 Samples of Five Employees
270 CHAPTER 8
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 the sample mean than in the population distribution. In the population, the years 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 8–6 Histogram of Mean Years 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.
μx = 3.95 + 3.25 + … + 4.30 + 5.05
25 = 4.676
We use the symbol μx 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 × bar.” We observe that the mean of the sample means, 4.676 years, is very close to the population mean of 4.80.
What should we conclude from this example? The central limit theorem indicates 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 sampled or selected, the stronger the convergence. The Spence Sprockets Inc. example shows how the central limit theorem works. We began with a positively skewed population (Chart 8–4). Next, we selected 25 random sam- ples of 5 observations, computed the mean of each sample, and finally organized these 25 sample means into a histogram (Chart 8–5). We observe that the shape of the sampling distribution of the sample mean is very different from that of the popula- tion. The population distribution is positively skewed compared to the nearly normal shape of the sampling distribution of the sample mean.
To further illustrate the effects of the central limit theorem, we increased the num- ber of observations in each sample from 5 to 20. We selected 25 samples of 20 obser- vations each and calculated the mean of each sample. Finally, we organized these sample means into a histogram (Chart 8–6). The shape of the histogram in Chart 8–6 is clearly moving toward the normal probability distribution.
If you go back to Chapter 6 where several binomial distributions with a “success” pro- portion of .10 are shown in Chart 6–3 on page 190, 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 8–6 also shows the convergence to normality as n increases. This
SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 271
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, defined on page 265, does not say anything about the dispersion of the sampling distribution of the sample mean or about the compar- ison 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 distri- bution is exactly equal to the population mean (i.e., μx = μ ), and if the standard devi- ation in the population is σ, the standard deviation of the sample means is σ∕√n where n is the number of observations in each sample. We refer to σ∕√n as the stan- dard error of the mean. Its longer name is actually the standard deviation of the sampling distribution of the sample mean.
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 popula- tion mean if we are able to select all possible samples of the same size from a given population. That is:
μ = μx Even if we do not select all samples, we can expect the mean of the distribution 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 σ, the standard devi- ation of the distribution of sample means is σ∕√n. Note that when we increase the size of the sample, the standard error of the mean decreases.
STANDARD ERROR OF THE MEAN σ x = σ
√n (8–1)
Refer to the Spence Sprockets Inc. data on page 267. Select 10 random samples of five employees each. Use the methods described earlier in the chapter and the Table of Random Numbers (Appendix B.4) 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 8–4. What is the mean of your 10 sample means?
S E L F - R E V I E W 8–4
11. Appendix B.4 is a table of random numbers that are uniformly distributed. Hence, each digit from 0 to 9 has the same likelihood of occurrence.
a. Draw a graph showing the population distribution of random numbers. What is the population mean?
b. Following are the first 10 rows of five digits from the table of random numbers in Appendix B.4. 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
E X E R C I S E S
272 CHAPTER 8
Chart 8–4. Compare the mean of the sampling distribution of the sample mean 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 throughout the United States and Canada. The number of units sold last month by each representative is listed below. Assume these sales figures to be the popula- tion 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.4 to determine the items to be included in the sample.
d. Compare the mean of the sampling distribution of the sample mean to the pop- ulation 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 backward to record the ages (in years) of the coins. For example, if the current year is 2017, then a coin with 2015 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 ran- domly selected phone numbers. For example, if a phone number is 555-9704, re- cord a 4.
a. Draw a histogram or other graph of this population distribution. Using the uni- form distribution, 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 sam- ple means.
c. Compare the shapes of the two histograms.
SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 273
USING THE SAMPLING DISTRIBUTION OF THE SAMPLE MEAN The previous discussion is important because most business decisions are made on the basis of sample information. Here are some examples.
1. Arm & Hammer Company wants to ensure that its laundry detergent actually con- tains 100 fluid ounces, as indicated on the label. Historical summaries from the fill- ing process indicate the mean amount per container is 100 fluid ounces and the standard deviation is 2 fluid ounces. At 10 a.m., a quality technician measures 40 containers and finds the mean amount per container is 99.8 fluid ounces. Should the technician shut down the filling operation?
2. A. C. Nielsen Company provides information to organizations advertising on televi- sion. Prior research indicates that adult Americans watch an average of 6.0 hours per day of television. The standard deviation is 1.5 hours. What is the probability that we could randomly select a sample of 50 adults and find that they watch an average of 6.5 hours or more 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 of an adult is 160 pounds and the standard deviation is 15 pounds. However, the distri- bution of weights does not follow the normal probability distribution. It is positively skewed. For a sample of 30 adults, what is the likelihood that their mean weight is 170 pounds or more?
We can answer the questions in each of these situations using the ideas discussed in the previous section. In each case, we have a population with information about its mean and standard deviation. Using this information and sample size, we can determine the distribu- tion of sample means and compute the probability that a sample mean will fall within a certain range. The sampling distribution will be normally distributed under two conditions:
1. When the samples are taken from populations known to follow the normal distribu- tion. In this case, the size of the sample is not a factor.
2. When the shape of the population distribution is not known, sample size is import- ant. In general, the sampling distribution will be normally distributed as the sample size approaches infinity. In practice, a sampling distribution will be close to a normal distribution with samples of at least 30 observations.
We use formula (7–5) from the previous chapter to convert any normal distribution to the standard normal distribution. Using formula (7–5) to compute z values, we can use the standard normal table, Appendix B.3, to find the probability that an observation is within a specific range. The formula for finding a z value is:
z = x − μ
σ In this formula, x is the value of the random variable, μ is the population mean, and σ is the population standard deviation.
However, when we sample from populations, we are interested in the distribution of X , 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 σ∕√n in the denominator rather than σ. Therefore, to find the likelihood of a sample mean within a specified range, we first use the following formula to find the corresponding z value. Then we use Appendix B.3 or statistical software to determine the probability.
LO8-5 Apply the central limit theorem to calculate probabilities.
z = x − μ σ∕√n
(8–2)FINDING THE z VALUE OF x − WHEN THE
POPULATION STANDARD DEVIATION IS KNOWN
274 CHAPTER 8
The following example/solution will show the application.
E X A M P L E
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 critical 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. Records maintained by the Quality Assurance Depart- ment 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 ounce. At 8 a.m. today the quality technician randomly selected 16 bottles from the filling line. The mean amount of cola contained 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 ounce unusual?
S O L U T I O N
We 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 (μ) ounces and a population standard deviation of 0.4 (σ) ounce and find the sample mean to be 31.38 (x ) or more. We use formula (8–2) to find the value of z.
z = x − μ σ∕√n
= 31.38 − 31.20
0.4∕√16 = 1.80
The numerator of this equation, x − μ = 31.38 − 31.20 = .18, is the sampling error. The denominator, σ∕√n = 0.4∕√16 = 0.1, is the standard error of the sampling 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.3, locate the probability corresponding to a z value of 1.80. It is .4641. The likeli- hood 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% 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 ounce 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 8–7.
CHART 8–7 Sampling Distribution of the Mean Amount of Cola in a Jumbo Bottle
31.20 31.38 Ounces (x– )
0 1.80 z value
.4641
.0359
SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 275
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?
S E L F - R E V I E W 8–5
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. In a certain section of Southern California, the distribution of monthly rent for a one-bedroom apartment has a mean of $2,200 and a standard deviation of $250. The distribution of the monthly rent does not follow the normal distribution. In fact, it is positively skewed. What is the probability of selecting a sample of 50 one- bedroom apartments 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 distribution and the standard deviation is 80 minutes. A consumer watch- dog 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?
E X E R C I S E S
C H A P T E R S U M M A R Y
I. The characteristics of the F distribution are: A. There are many reasons for sampling a population. B. The results of a sample may adequately estimate the value of the population parame-
ter, thus saving time and money. C. It may be too time-consuming to contact all members of the population. D. It may be impossible to check or locate all the members of the population. E. The cost of studying all the items in the population may be prohibitive. F. 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. III. The sampling error is the difference between a population parameter and a sample
statistic.
276 CHAPTER 8
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:
σ X =
σ √n
(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. If the population is not normally distributed, the sampling distribution of the sample mean will approach a normal distribution when the sample size is at least 30. Assume the population stan- dard deviation is known. To determine the probability that a sample mean falls in a particular region, use the following formula.
z = x − μ σ∕√n
(8–2)
P R O N U N C I A T I O N K E Y
SYMBOL MEANING PRONUNCIATION
μX Mean of the sampling distribution mu sub x bar of the sample mean
σ X Population standard error of sigma sub x bar the sample mean
C H A P T E R E X E R C I S E S
19. The 25 retail stores located in the North Towne Square Mall numbered 00 through 24 are:
00 Elder-Beerman 01 Sears 02 Deb Shop 03 Frederick’s of Hollywood 04 Petries 05 Easy Dreams 06 Summit Stationers 07 E. B. Brown Opticians 08 Kay-Bee Toy & Hobby
09 Lion Store 10 Bootleggers 11 Formal Man 12 Leather Ltd. 13 Barnes and Noble 14 Pat’s Hallmark 15 Things Remembered 16 Pearle Vision Express 17 Dollar Tree
18 County Seat 19 Kid Mart 20 Lerner 21 Coach House Gifts 22 Spencer Gifts 23 CPI Photo Finish 24 Regis Hairstylists
a. If the following random numbers are selected, which retail stores should be con- tacted 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.4. c. A systematic sampling procedure will be used. The first store will be selected and
then every third store. Which stores will be in the sample? 20. The Medical Assurance Company is investigating the cost of a routine office visit to fam-
ily-practice physicians in the Rochester, New York, area. The following is a list of 39 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).
SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 277
a. The random numbers obtained from Appendix B.4 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.4.
c. Using systematic random sampling, every fifth physician is selected starting with the fourth physician in the list. Which physicians will be contacted?
d. Select a sample that includes two physicians in solo practice (S), two in partnership (P), and one in group practice (G). Explain your procedure.
21. A population consists of the following three values: 1, 2, and 3. a. Sampling with replacement, list all possible samples of size 2 and compute the mean
of every sample. b. Find the means of the distribution of the sample mean and the population mean.
Compare 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. Based on all student records at Camford University, students spend 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 Incorpo- rated (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 eComputers, an economy-priced computer, recently completed the design for a new laptop model. eComputer’s top management would like some assistance in pricing the new laptop. Two market research firms were contacted and asked to prepare a pricing strategy. Marketing-Gets-Results tested the new eComput- ers laptop with 50 randomly selected consumers who indicated they plan to purchase a laptop within the next year. The second marketing research firm, called Market- ing-Reaps-Profits, test-marketed the new eComputers laptop with 200 current laptop owners. Which of the marketing research companies’ test results will be more useful? Discuss why.
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
278 CHAPTER 8
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? b. What happens to the distribution of the sample means if the sample size is
increased? c. When using 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
a. Using a table of random numbers (Appendix B.4), 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 will be in- terviewed about airport facilities, service, and so on. To select the sample, 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.4. b. The sample of 10 could have been chosen using a systematic sample. Choose the
first number using Appendix B.4, and then list the numbers to be interviewed. c. Evaluate the two methods by giving the advantages and possible disadvantages. d. 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 re- ceived the following exam scores (percent correct): 79, 64, 84, 82, 92, and 77. To com- pute your final course grade, the instructor decided to randomly select two exam scores, compute their mean, and use this score to determine your final course grade. a. Compute the population mean. b. How many different samples of two test grades are possible? c. List all possible samples of size 2 and compute the mean of each. d. Compute the mean of the sample means and compare it to the population mean. e. 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 tell-
ers made the following number of errors each: 2, 3, 5, 3, and 5. a. How many different samples of two 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.
Technician Shutdowns Technician Shutdowns
Taylor 4 Rousche 3 Hurley 3 Huang 2 Gupta 5
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. The
following table lists the number of refrigerators sold by each representative last month.
SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 279
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 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. Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal probability distribution with a mean of 35.0 hours and a standard deviation of 5.5 hours. As a part of its quality assurance program, Power +, Inc. 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 samples will have a mean useful life greater than 34.5
hours? e. What proportion of the samples will have a mean useful life between 34.5 and 36.0
hours? 32. Majesty Video Production Inc. wants the mean length of its advertisements to be 30
seconds. Assume the distribution of ad length follows the normal distribution with a population standard deviation of 2 seconds. Suppose we select a sample of 16 ads produced by Majesty. a. What can we say about the shape of the distribution of the sample mean time? b. What is the standard error of the mean time? c. What percent of the sample means will be greater than 31.25 seconds? d. What percent of the sample means will be greater than 28.25 seconds? e. What percent of the sample means will be greater than 28.25 but less than 31.25
seconds? 33. Recent studies indicate that the typical 50-year-old woman spends $350 per year for
personal-care products. The distribution of the amounts spent follows a normal distribu- tion 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 $165,000. 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 $167,000? d. What is the likelihood of selecting a sample with a mean of more than $155,000? e. Find the likelihood of selecting a sample with a mean of more than $155,000 but
less than $167,000. 35. In the United States, the mean age of men when they marry for the first time follows the
normal distribution with a mean of 29 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 when they were first married is less than 29.3. 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?
280 CHAPTER 8
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% 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% of the sample means occur?
39. The mean performance score on a physical fitness test for Division I student-athletes is 947 with a standard deviation of 205. If you select a random sample of 60 of these stu- dents, what is the probability the mean is below 900?
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 8–2, 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
0 Alabama 1 Alaska 2 Arizona 3 Arkansas 4 California 5 Colorado 6 Connecticut 7 Delaware 8 Florida 9 Georgia 10 Hawaii 11 Idaho 12 Illinois 13 Indiana 14 Iowa 15 Kansas 16 Kentucky 17 Louisiana 18 Maine 19 Maryland 20 Massachusetts 21 Michigan 22 Minnesota 23 Mississippi 24 Missouri
Number State
25 Montana 26 Nebraska 27 Nevada 28 New Hampshire 29 New Jersey 30 New Mexico 31 New York 32 North Carolina 33 North Dakota 34 Ohio 35 Oklahoma 36 Oregon 37 Pennsylvania 38 Rhode Island 39 South Carolina 40 South Dakota 41 Tennessee 42 Texas 43 Utah 44 Vermont 45 Virginia 46 Washington 47 West Virginia 48 Wisconsin 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. Select a systematic sample of every sixth item using the digit 02 as the starting point. Which states are included?
42. Human Resource Consulting (HRC) surveyed a random sample of 60 Twin Cities construc- tion 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.
SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM 281
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 hacking attacks experienced by members of the Information Systems Security Association is 510 per year with a standard deviation of 14.28 attacks. The number of attacks per year is normally distributed. Suppose noth- ing 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 be-
tween 500 and 600. c. What is the possibility they will experience an average of less than 500 attacks over
the next 10 years? 44. An economist uses the price of a gallon of milk as a measure of inflation. She finds that
the average price is $3.82 per gallon and the population standard deviation is $0.33. You decide to sample 40 convenience stores, collect their prices for a gallon of milk, and compute the mean price for the sample. a. What is the standard error of the mean in this experiment? b. What is the probability that the sample mean is between $3.78 and $3.86? 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 $3.92?
45. Nike's annual report says that the average American buys 6.5 pairs of sports shoes per year. Suppose a sample of 81 customers is surveyed and the population standard devi- ation of sports shoes purchased per year is 2.1. 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 pair? d. What is the likelihood the sample mean is greater than 7 pairs?
D A T A A N A L Y T I C S
46. Refer to the North Valley Real Estate data, which report information on the homes sold last year. Assume the 105 homes is a population. Compute the population mean and the standard deviation of price. Select a sample of 10 homes. Compute the mean. Determine the likelihood of a sample mean price this high or higher.
47. Refer to the Baseball 2016 data, which report information on the 30 Major League Baseball teams for the 2016 season. Over the last decade, the mean attendance per team followed a normal distribution with a mean of 2.45 million per team and a standard deviation of .71 million. Compute the mean attendance per team for the 2016 season. Determine the likelihood of a sample mean attendance this large or larger from the population.
48. Refer to the Lincolnville School District bus data. Information provided by manufac- turers of school buses suggests the mean maintenance cost per year is $4,400 per bus with a standard deviation of $1,000. Compute the mean maintenance cost for the Lin- colnville buses. Does the Lincolnville data seem to be in line with that reported by the manufacturer? Specifically, what is the probability of Lincolnville’s mean annual mainte- nance cost, or greater, given the manufacturer’s data?
Estimation and Confidence Intervals
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO9-1 Compute and interpret a point estimate of a population mean.
LO9-2 Compute and interpret a confidence interval for a population mean.
LO9-3 Compute and interpret a confidence interval for a population proportion.
LO9-4 Calculate the required sample size to estimate a population proportion or population mean.
LO9-5 Adjust a confidence interval for finite populations.
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 meal per week. Construct a 99% confidence interval for the population mean. (See Exercise 36 and LO9-2.)
© Jack Hollingsworth/Photodisc/Getty Images
9
ESTIMATION AND CONFIDENCE INTERVALS 283
INTRODUCTION The previous chapter began our discussion of sampling. We introduced both the rea- sons 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 sampling 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, is known. In most business situations, such information is not available. In fact, one purpose of sampling is 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 study- ing point estimates. A point estimate is a single value (point) computed from sample information and used to estimate a population value. For example, we may be inter- ested in the number of hours worked by consultants employed by Boston Consulting Group. Using simple random sampling, we select 50 consultants and ask each of them how many hours they worked last week. The sample’s mean is a point estimate of the unknown population mean. A more informative approach is to present a range of values where 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.
POINT ESTIMATE FOR A POPULATION MEAN A point estimate is a single statistic used to estimate a population parameter. Suppose Best Buy Inc. wants to estimate the mean age of people who purchase LCD HDTV tele- visions. They select a random sample of 75 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.
POINT ESTIMATE The statistic, computed from sample information, that estimates a population parameter.
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 feasible to contact each tourist. Therefore, 500 tourists are randomly selected as they depart the country and asked in detail about their spending while visiting Barbados. The mean amount spent by the sample of 500 tourists is an estimate of the unknown popula- tion parameter. That is, we let the sample mean serve as a point estimate of the population mean.
LO9-1 Compute and interpret a point estimate of a population mean.
STATISTICS IN ACTION
On all new cars, a fuel economy estimate is promi- nently displayed on the window sticker as required by the Environmental Pro- tection Agency (EPA). Often, fuel economy is a factor in a consumer’s choice of a new car because of fuel costs or environmental con- cerns. The fuel estimates for a 2016 BMW 328i Sedan (4-cylinder, auto- matic) are 23 miles per gallon (mpg) in the city and 35 on the highway. The EPA recognizes that actual fuel economy may differ from the estimates by noting, “No test can simulate all possible combinations of conditions and climate, driver behavior, and car care habits. Actual mileage depends on how, when, and where the vehicle is driven. The EPA has found that the mpg obtained by most drivers will be within a few mpg of the estimates.”
284 CHAPTER 9
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 re- vealed 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 conclude that the population mean is still 45 days and that the difference between the sam-
ple mean (46.7 days) and the proposed population mean (45 days) is sampling error? In other words, is the sample mean significantly differ- ent from the population mean? 3. Recent medical studies indicate that exercise is an important part of
a person’s overall health. The director of human resources at OCF, a large glass manufacturer, wants an estimate 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. This value is a point estimate of the unknown population mean.
The sample mean, x , is not the only point estimate of a popula- tion parameter. For example, p, a sample proportion, is a point esti- mate of π, the population proportion; and s, the sample standard deviation, is a point estimate of σ, the population standard deviation.
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 describe how confident we are that the population parameter is in the interval. We might say, for in- stance, that we are 90% confident 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 for a population mean, we will consider two situations:
• We use sample data to estimate μ with x and the population standard deviation (σ) is known.
• We use sample data to estimate μ with x and the population standard deviation is unknown. In this case, we substitute the sample standard deviation (s) for the pop- ulation standard deviation (σ).
There are important distinctions in the assumptions between these two situations. We first consider the case where σ is known.
Population Standard Deviation, Known σ A confidence interval is computed using two statistics: the sample mean, x , and the standard deviation. From previous chapters, you know that the standard deviation is an important statistic because it measures the dispersion, or variation, of a population or
LO9-2 Compute and interpret a confidence interval for a population mean.
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ESTIMATION AND CONFIDENCE INTERVALS 285
sampling distribution. In computing a confidence interval, the standard deviation is used to compute the limits 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, σ. Typically, we know the population standard deviation in situations where we have a long history of collected data. Examples are data from monitoring processes that fill soda bottles or cereal boxes, and the results of the SAT Reason- ing Test (for college admission). Knowing σ allows us to simplify the development of a confidence interval 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, x , of sample size n from a population. The population standard deviation, σ, 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 μ and a standard deviation σ∕√n. Also recall that this value is called the standard error.
The results of the central limit theorem allow us to make the following general con- fidence interval statements using z-statistics:
1. Ninety-five percent of all confidence intervals computed from random samples se- lected from a population will contain the population mean. These intervals are com- puted using a z-statistic equal to 1.96.
2. Ninety percent of all confidence intervals computed from random samples selected from a population will contain the population mean. These confidence intervals are computed using a z-statistic equal to 1.65.
These confidence interval statements provide examples of levels of confidence and are called a 95% confidence interval and a 90% confidence interval. The 95% and 90% are the levels of confidence and refer to the percentage of similarly constructed intervals that would include the parameter being estimated—in this case, μ, the popu- lation mean.
How are the values of 1.96 and 1.65 obtained? First, let’s look for the z value for a 95% confidence interval. The following diagram and Table 9–1 will help explain. Table 9–1 is a reproduction of the standard normal table in Appendix B. However, many rows and columns have been eliminated to allow us to better focus on particular rows and columns.
1. First, we divide the confidence level in half, so .9500/2 = .4750. 2. Next, we find the value .4750 in the body of Table 9–1. Note that .4750 is located
in the table at the intersection of a row and a column. 3. Locate the corresponding row value in the left margin, which is 1.9, and the column
value in the top margin, which is .06. Adding the row and column values gives us a z value of 1.96.
4. Thus, the probability of finding a z value between 0 and 1.96 is .4750. 5. Likewise, because the normal distribution is symmetric, the probability of finding a
z value between −1.96 and 0 is also .4750. 6. When we add these two probabilities, the probability that a z value is between
−1.96 and 1.96 is .9500.
For the 90% level of confidence, we follow the same steps. First, one-half of the desired confidence interval is .4500. A search of Table 9–1 does not reveal this exact value. However, it is between two values, .4495 and .4505. As in step three, we locate each value in the table. The first, .4495, corresponds to a z value of 1.64 and the sec- ond, .4505, corresponds to a z value of 1.65. To be conservative, we will select the larger of the two z values, 1.65, and the exact level of confidence is 90.1%, or 2(0.4505). Next, the probability of finding a z value between −1.65 and 0 is .4505, and the proba- bility that a z value is between −1.65 and 1.65 is .9010.
286 CHAPTER 9
How do we determine a 95% confidence interval? The width of the interval is deter- mined by two factors: (1) the level of confidence, as described in the previous section, and (2) the size of the standard error of the mean. To find the standard error of the mean, recall from the previous chapter [see formula (8–1) on page 271] that 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:
σ x =
σ √n
where:
σ x is the symbol for the standard error of the mean. We use a Greek letter because
it is a population value, and the subscript x reminds us that it refers to a sam- pling distribution of the sample means.
σ 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 standard deviation of the population. The larger the population standard deviation, σ, the larger σ∕√n. If the population is homogeneous, resulting in a small population standard devi- ation, 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, indicating that there is less vari- ability in the sample means.
We can summarize the calculation for a 95% confidence interval using the following formula:
x ± 1.96 σ
√n
TABLE 9–1 The Standard Normal Table for Selected Values
z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07
⫶ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884
z value
.025
1.96–1.96
.025 .4750.4750
ESTIMATION AND CONFIDENCE INTERVALS 287
Similarly, a 90.1% confidence interval is computed as follows:
x ± 1.65 σ
√n
The values 1.96 and 1.65 are z values corresponding to the 95% and the 90.1% con- fidence intervals, respectively. However, we are not restricted to these values. We can select any confidence level between 0 and 100% and find the corresponding value for z. In general, a confidence interval for the population mean when the population follows the normal distribution and the population standard deviation is known is computed by:
x ± z σ
√n (9–1)CONFIDENCE INTERVAL FOR A
POPULATION MEAN WITH σ KNOWN
To explain these ideas, consider the following example. Del Monte Foods distributes diced peaches in 4.5-ounce plastic cups. To ensure that each cup contains at least the required amount, Del Monte sets the filling operation to dispense 4.51 ounces of peaches and gel in each cup. Of course, not every cup will contain exactly 4.51 ounces of peaches and gel. Some cups will have more and others less. From historical data, Del Monte knows that 0.04 ounce is the standard deviation of the filling process and that the amount, in ounces, follows the normal probability distribution. The quality control technician selects a sample of 64 cups at the start of each shift, measures the amount in each cup, computes the mean fill amount, and then develops a 95% confidence interval for the population mean. Using the confidence interval, is the process filling
the cups to the desired amount? This morning’s sample of 64 cups had a sample mean of 4.507 ounces. Based on this information, the 95% confidence interval is:
x ± 1.96 σ
√n = 4.507 ± 1.96
0.04 √64
= 4.507 ± 0.0098
The 95% confidence interval estimates that the population mean is between 4.4972 ounces and 4.5168 ounces of peaches and gel. Recall that the process is set to fill each cup with 4.51 ounces. Because the desired fill amount of 4.51 ounces is in this interval, we conclude that the filling process is achieving the desired results. In other words, it is reasonable to conclude that the sample mean of 4.507 could have come from a population distribution with a mean of 4.51 ounces.
In this example, we observe that the population mean of 4.51 ounces is in the con- fidence interval. But this is not always the case. If we selected 100 samples of 64 cups from the population, calculated the sample mean, and developed a confidence interval based on each sample, we would expect to find the population mean in about 95 of the 100 intervals. Or, in contrast, about five of the intervals would not contain the population mean. From Chapter 8, this is called sampling error. The following example details re- peated sampling from a population.
Courtesy Del Monte Corporation
E X A M P L E
The American Management Association (AMA) is studying the income of store man- agers in the retail industry. A random sample of 49 managers reveals a sample mean of $45,420. The standard deviation of this population is $2,050. The associ- ation 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?
288 CHAPTER 9
mm – 1.96 σ n
x1
x2
x3
x4
x5
x6
Sample 1 of size 49. It includes the population mean. Sample 2 of size 49. It includes the population mean. Sample 3 of size 49. It includes the population mean. Sample 4 of size 49. It includes the population mean.
Sample 6 of size 49. It includes the population mean.
Sample 5 of size 49. It does not include the population mean.
Population mean
m + 1.96 σ n
Scale of x
S O L U T I O N
Generally, distributions of salary and income are positively skewed because a few individuals earn considerably more than others, thus skewing the distribution in the positive direction. Fortunately, the central limit theorem states that the sampling distribution of the mean becomes a normal distribution as sample size increases. In this instance, a sample of 49 store managers is large enough that we can assume that the sampling 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 AMA decides to use the 95% level of confidence. To determine the corresponding confidence interval, we use formula (9–1):
x ± z σ
√n = $45,420 ± 1.96
$2,050 √49
= $45,420 ± $574
The confidence interval limits are $44,846 and $45,994 determined by sub- tracting $574 and adding $574 to the sample mean. The degree or level of confidence is 95% and the confidence interval is from $44,846 to $45,994. The value, $574, is called the margin of error.
3. How do we interpret these results? Suppose we select many samples of 49 store managers, perhaps several hundred. For each sample, we compute the mean and then construct a 95% confidence interval, such as we did in the previ- ous section. We could expect about 95% of these confidence intervals to contain the population mean. About 5% of the intervals would not contain the population mean annual income, which is μ. 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), deter- mining a 95% confidence interval for the population mean. Note that not all inter- vals include the population mean. Both the endpoints of the fifth sample are less than the population mean. We attribute this to sampling error, and it is the risk we assume when we select the level of confidence.
ESTIMATION AND CONFIDENCE INTERVALS 289
A Computer Simulation With statistical software, we can create random samples of a desired sample size, n, from a population. For each sample of n observations with corresponding numerical values, we can calculate the sample mean. With the sample mean, population standard deviation, and confidence level, we can determine the confidence interval for each sam- ple. Then, using all samples and the confidence intervals, we can find the frequency that the population mean is included in the confidence intervals. The following example does just that.
E X A M P L E
From many years in the automobile leasing business, Town Bank knows that the mean distance driven on an automobile with a four-year lease is 50,000 miles and the standard deviation is 5,000 miles. These are population values. Suppose Town Bank would like to experiment with the idea of sampling to estimate the population mean of 50,000 miles. Town Bank decides to choose a sample size of 30 observa- tions and a 95% confidence interval to estimate the population mean. Based on the experiment, we want to count the number of confidence intervals that include the population mean of 50,000. We expect about 95%, or 57 of the 60 intervals, will include the population mean. To make the calculations easier to understand, we’ll conduct the study in thousands of miles, instead of miles.
S O L U T I O N
Using statistical software, 60 random samples of 30 observations, n = 30, are gen- erated and the sample means for each sample computed. Then, using the n of 30 and a standard error of 0.913 (σ∕√n = 5∕√30), a 95% confidence interval is com- puted for each sample. The results of the experiment are shown next.
Sample Observations Sample 95% Confidence Limits
Sample 1 2 3 4 5 – – – 26 27 28 29 30 Mean Lower Limit Upper Limit
1 56 47 47 48 58 – – – 55 62 48 61 57 51.6 49.811 53.389 2 55 51 52 40 53 – – – 47 54 55 55 45 50.77 48.981 52.559 3 42 46 48 46 41 – – – 50 52 50 47 45 48.63 46.841 50.419 4 52 49 55 47 49 – – – 46 56 49 43 50 49.9 48.111 51.689 5 48 50 53 48 45 – – – 46 51 61 49 47 49.03 47.241 50.819 6 49 44 47 46 48 – – – 51 44 51 52 43 47.73 45.941 49.519 7 50 53 39 50 46 – – – 55 47 43 50 57 50.2 48.411 51.989 8 47 51 49 58 44 – – – 49 57 54 48 48 51.17 49.381 52.959 9 51 44 47 56 45 – – – 45 51 49 49 52 50.33 48.541 52.119 10 45 44 52 52 56 – – – 52 51 52 50 48 50 48.211 51.789 11 43 52 54 46 54 – – – 43 46 49 52 52 51.2 49.411 52.989 12 57 53 48 42 55 – – – 49 44 46 46 48 49.8 48.011 51.589 13 53 39 47 51 53 – – – 42 44 44 55 58 49.6 47.811 51.389 14 56 55 45 43 57 – – – 48 51 52 55 47 49.03 47.241 50.819 15 49 50 39 45 44 – – – 49 43 44 51 51 49.37 47.581 51.159 16 46 44 55 53 55 – – – 44 53 53 43 44 50.13 48.341 51.919 17 64 52 55 55 43 – – – 58 46 52 58 55 52.47 50.681 54.259 18 57 51 60 40 53 – – – 50 51 53 46 52 50.1 48.311 51.889 19 50 49 51 57 45 – – – 53 52 40 45 52 49.6 47.811 51.389 20 45 46 53 57 49 – – – 49 43 43 53 48 49.47 47.681 51.259 21 52 45 51 52 45 – – – 43 49 49 58 53 50.43 48.641 52.219 22 48 48 52 49 40 – – – 50 47 54 51 45 47.53 45.741 49.319
(continued)
290 CHAPTER 9
23 48 50 50 53 44 – – – 48 57 52 44 39 49.1 47.311 50.889 24 51 51 40 54 52 – – – 54 45 50 57 48 50.13 48.341 51.919 25 48 63 41 52 41 – – – 48 50 48 44 53 49.33 47.541 51.119 26 47 45 48 59 49 – – – 44 47 49 55 42 49.63 47.841 51.419 27 52 45 60 51 52 – – – 52 50 54 46 52 49.4 47.611 51.189 28 46 48 46 57 51 – – – 51 50 51 41 52 49.33 47.541 51.119 29 46 48 45 42 48 – – – 49 43 59 46 50 48.27 46.481 50.059 30 55 48 47 48 48 – – – 47 59 54 51 42 50.53 48.741 52.319 31 58 49 56 46 46 – – – 44 51 47 51 46 50.77 48.981 52.559 32 53 54 52 58 55 – – – 53 52 45 44 51 50 48.211 51.789 33 50 57 56 51 51 – – – 58 47 50 56 46 49.7 47.911 51.489 34 61 48 49 53 54 – – – 46 46 56 45 54 50.03 48.241 51.819 35 43 42 43 46 49 – – – 49 49 56 51 45 49.43 47.641 51.219 36 39 48 48 51 44 – – – 54 52 47 50 52 50.07 48.281 51.859 37 48 43 57 42 54 – – – 52 50 59 50 52 50.17 48.381 51.959 38 55 43 49 57 45 – – – 41 51 51 52 52 49.5 47.711 51.289 39 47 49 58 54 54 – – – 50 56 51 56 58 50.37 48.581 52.159 40 47 56 41 50 54 – – – 46 56 61 61 45 51.6 49.811 53.389 41 48 47 42 47 62 – – – 44 47 49 55 43 49.43 47.641 51.219 42 46 49 43 36 52 – – – 45 51 46 51 43 47.67 45.881 49.459 43 44 48 49 48 51 – – – 47 52 51 48 49 49.63 47.841 51.419 44 45 52 54 54 49 – – – 49 45 53 50 52 49.07 47.281 50.859 45 54 46 54 45 48 – – – 55 38 56 50 62 49.53 47.741 51.319 46 48 50 49 52 51 – – – 53 57 58 46 50 49.9 48.111 51.689 47 54 55 46 55 50 – – – 56 54 50 55 51 50.5 48.711 52.289 48 45 47 47 63 44 – – – 45 53 42 53 50 50.1 48.311 51.889 49 47 47 48 54 56 – – – 50 48 54 49 51 49.93 48.141 51.719 50 45 61 51 45 54 – – – 55 52 47 45 53 51.03 49.241 52.819 51 49 62 43 49 48 – – – 49 58 42 58 52 51.07 49.281 52.859 52 54 52 62 43 54 – – – 51 57 49 58 55 50.17 48.381 51.959 53 46 50 59 56 46 – – – 50 51 52 54 53 50.47 48.681 52.259 54 52 50 48 48 58 – – – 58 52 43 61 54 51.77 49.981 53.559 55 45 44 46 56 46 – – – 43 45 63 48 56 49.37 47.581 51.159 56 60 50 56 51 43 – – – 45 43 49 59 54 50.37 48.581 52.159 57 59 56 43 47 52 – – – 49 54 50 50 57 49.53 47.741 51.319 58 52 55 48 51 40 – – – 53 51 51 52 47 49.77 47.981 51.559 59 53 50 44 53 52 – – – 47 50 55 46 51 50.07 48.281 51.859 60 55 54 50 52 43 – – – 57 50 48 47 53 52.07 50.281 53.859
To explain, in the first row, the statistical software computed 30 random obser- vations from a population distribution with a mean of 50 and a standard deviation of 5. To conserve space, only observations 1 through 5 and 26 through 30 are listed. The first sample’s mean is computed and listed as 51.6. In the next columns, the upper and lower limits of the 95% confidence interval for the first sample are shown. The confidence interval calculation for the first sample follows:
x ± 1.96 σ
√n = 51.6 ± 1.96
5 √30
= 51.6 ± 1.789
This calculation is repeated for all samples. The results of the experiment show that 93.33%, or 56 of the sixty confidence intervals include the population mean of 50. 93.33% is close to the estimate that 95%, or 57, of the intervals will include the population mean. Using the complement, we expected 5%, or three, of the intervals would not include the population mean. The experiment resulted in 6.67%, or four, of the 60 intervals that did not include the population mean. The particular intervals, 6, 17, 22, and 42, are highlighted in yellow. This is another example of sampling
ESTIMATION AND CONFIDENCE INTERVALS 291
error, or the possibility that a particular random sample may not be a good repre- sentation of the population. In each of these four samples, the mean of the sample is either much less or much more than the population mean. Because of random sampling, the mean of the sample is not a good estimate of the population mean, and the confidence interval based on the sample’s mean does not include the pop- ulation mean.
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 also are available. The Marketing Department of Bun-and-Run Inc. reports that the distribution of daily sales for their restaurants follows the normal distribution and that the population standard deviation is $3,000. A sample of 40 franchises showed the mean daily sales to be $20,000. (a) What is the population mean of daily sales for Bun-and-Run franchises? (b) What is the best estimate of the population mean? What is this value called? (c) Develop a 95% confidence interval for the population mean of daily sales. (d) Interpret the confidence interval.
S E L F - R E V I E W 9–1
1. A sample of 49 observations is taken from a normal population with a standard deviation of 10. The sample mean is 55. Determine the 99% confidence interval for the population mean.
2. A sample of 81 observations is taken from a normal population with a standard deviation of 5. The sample mean is 40. Determine the 95% confidence interval for the population mean.
3. A sample of 250 observations is selected from a normal population with a popula- tion standard deviation of 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% confidence interval. c. Determine the 95% confidence interval for the population mean.
4. Suppose you know σ and you want an 85% confidence level. What value would you use as z in formula (9–1)?
5. A research firm conducted a survey to determine the mean amount Americans spend on coffee during a week. They found the distribution of weekly spending followed the normal distribution with a population standard deviation of $5. A sam- ple of 49 Americans revealed that x = $20.
a. What is the point estimate of the population mean? Explain what it indicates. b. Using the 95% level of confidence, determine the confidence interval for μ. Ex-
plain what it indicates. 6. Refer to the previous exercise. Instead of 49, suppose that 64 Americans were sur-
veyed about their weekly expenditures on coffee. Assume the sample mean re- mained the same.
a. What is the 95% confidence interval estimate of μ? b. Explain why this confidence interval is narrower than the one determined in the
previous 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% confidence interval for the population mean. c. Interpret the meaning of part (b).
E X E R C I S E S
292 CHAPTER 9
Population Standard Deviation, σ Unknown In the previous section, we assumed the population standard deviation was known. In the case involving Del Monte 4.5-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 sampling situations the popula- tion standard deviation (σ) is not known. Here are some examples where we wish to esti- mate 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 students, 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 deviation (σ) required in formula (9–1).
• The Dean of Students wants to estimate the distance the typical commuter student 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 cam- pus. From the sample data, she calculates the mean travel distance, that is, x. 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 estimate 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, the director can estimate the mean amount. However, to develop a confidence interval using formula (9–1), the population standard deviation is neces- sary. 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 σ, the population standard deviation. But in doing so, we cannot use formula (9–1). Because we do not know σ, 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 charac- teristics 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:
t = x − μ s∕√n
where s is an estimate of σ. He noticed differences between estimating σ based on s, especially when s was calculated from a very small sample. The t distribution 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 normal distribution. This is because the standard deviation of the t distribution is larger than that of the standard normal distribution.
The following characteristics of the t distribution are based on the assumption that the population of interest is normal, or nearly normal.
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% confidence interval for the mean num- ber of misspelled words in the population of student essays.
STATISTICS IN ACTION
The t distribution was cre- ated by William Gosset who 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 em- ployees 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 paper, he first described the properties of the t distribu- tion and used it to monitor the brewing process so that the beer met Guinness’ quality standards.
ESTIMATION AND CONFIDENCE INTERVALS 293
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 corresponding z value. Chart 9–2 shows the values of z for a 95% level of confidence and of t for the
0
z distribution
t distribution
CHART 9–1 The Standard Normal Distribution and Student’s t Distribution
Distribution of z
.025
1.96 Scale of z
Distribution of t
.025
.025 .95
.95 .025
2.776
-1.96 0
-2.776 0 Scale of t
n = 5
CHART 9–2 Values of z and t for the 95% Level of Confidence
• 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 distributions
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 nor- mal distribution (see Chart 9–1). As the sample size increases, however, the t distri- bution approaches the standard normal distribution because the errors in using s to estimate σ decrease with larger samples.
294 CHAPTER 9
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 confi- dence the t distribution is flatter or more spread out than the standard normal distribu- tion. Note that the 95% confidence interval using a t statistic will be wider compared to an interval using a z statistic.
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, σ UNKNOWN
x ± t s
√n (9–2)
To determine a confidence interval for the population mean with an unknown pop- ulation standard deviation, we:
1. Assume the sampled population is either normal or approximately normal. This as- sumption may be questionable for small sample sizes, and becomes more valid with larger sample sizes.
2. Estimate the population standard deviation (σ) with the sample standard devia- tion (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 σ, 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.
Use the t distribution Use the z distribution
Is the population standard
deviation known?
Assume the population is
normal
YesNo
CHART 9–3 Determining When to Use the z Distribution or the t Distribution
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.
E X A M P L E
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 inch of tread remain- ing with a standard deviation of 0.09 inch. Construct a 95% confidence interval
ESTIMATION AND CONFIDENCE INTERVALS 295
for the population mean. Would it be reasonable for the manufacturer to con- clude that after 50,000 miles the population mean amount of tread remaining is 0.30 inch?
S O L U T I O N
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 inch. We use formula (9–2):
x ± t s
√n
From the information given, x = 0.32, s = 0.09, and n = 10. To find the value of t, we use Appendix B.5, a portion of which is reproduced in Table 9–2. Appendix B.5 is also reproduced on the inside back cover of the text. The first step for locating t is to move across the columns identified for “Confidence Intervals” to the level of confidence requested. In this case, we want the 95% 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 samples, 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.
TABLE 9–2 A Portion of the t Distribution
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
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
296 CHAPTER 9
Here is another example to clarify the use of confidence intervals. Suppose an article in your local newspaper reported that the mean time to sell a residential prop- erty 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 de- velop a 95% 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 proposed 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 state- ment by the local newspaper may not be correct. To put it another way, it seems un- reasonable 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 interpreting a confidence interval. We used Minitab to perform the calculations.
the standard deviation of the sample because one number (the arithmetic mean) is known. For a 95% 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).
x ± t s
√n = 0.32 ± 2.262
0.09 √10
= 0.32 ± 0.64
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% confidence interval with each sample’s mean and the standard deviation, we expect 190 of the intervals 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 reasonably sure (95% confident) that the mean remaining tread depth is between 0.256 and 0.384 inch. Be- cause the value of 0.30 is in this interval, it is possible that the mean of the population is 0.30.
E X A M P L E
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% confidence interval. Interpret the result. Would it be reasonable to conclude that the population mean is $50? What about $60?
S O L U T I O N
The mall manager assumes that the population of the amounts spent follows the normal distribution. This is a reasonable assumption in this case. Additionally, the
ESTIMATION AND CONFIDENCE INTERVALS 297
© McGraw-Hill Education/Andrew Resek, photographer
confidence interval technique is quite powerful and tends to commit any errors on the conservative side if the population is not normal. We should not make the normality assumption when the population is se- verely skewed or when the distribution has “thick tails.” In Chapter 16, we present methods for han- dling this issue if we cannot make the normality as- sumption. 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 re- sults 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 $49.348, 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 avail- able from Appendix B.5. 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% con- fidence level. The value at this intersection is 2.093. We substitute these values into formula (9–2) to find the confidence interval.
x ± t s
√n = $49.348 ± 2.093
$9.012 √20
= $49.348 ± $4.218
The endpoints of the confidence interval are $45.130 and $53.566. It is rea- sonable 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 reason- able that the population mean could be $50. The value of $60 is not in the confi- dence interval. Hence, we conclude that the population mean is unlikely to be $60.
The calculations to construct a confidence interval are also available in Excel. The output follows. Note that the sample mean ($49.348) and the sample standard devia- tion ($9.012) are the same as those in the Minitab calculations. In the Excel output, the last line 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
√n = 2.093
$9.012 √20
= $4.218
298 CHAPTER 9
Before doing the confidence interval exercises, we would like to point out a useful characteristic of the t distribution that will allow us to use the t table to quickly find both z and t values. Earlier in this section on page 293, we detailed the characteristics of the t distribution. The last point indicated that as we increase the sample size the t distribu- tion approaches the z distribution. In fact, when we reach an infinitely large sample, the t distribution is exactly equal to the z distribution.
To explain, Table 9–3 is a portion of Appendix B.5, with the degrees of freedom between 4 and 99 omitted. To find the appropriate z value for a 95% confidence interval,
TABLE 9–3 Student’s t Distribution
Confidence Interval
80% 90% 95% 98% 99% 99.9%
Level of Significance for One-Tailed Test, α 0.1 0.05 0.025 0.01 0.005 0.0005
Level of Significance for Two-Tailed Test, α 0.2 0.1 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 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 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
df (degrees of freedom)
ESTIMATION AND CONFIDENCE INTERVALS 299
we begin by going to the confidence interval section and selecting the column headed “95%.” Move down that column to the last row, which is labeled “∞,” or infinite degrees of freedom. The value reported is 1.960, the same value that we found using the stan- dard normal distribution in Appendix B.3. This confirms the convergence of the t dis- tribution to the z distribution.
What does this mean for us? Instead of searching in the body of the z table, we can go to the last row of the t table and find the appropriate value to build a confi- dence interval. An additional benefit is that the values have three decimal places. So, using this table for a 90% confidence interval, go down the column headed “90%” and see the value 1.645, which is a more precise z value that can be used for the 90% confidence level. Other z values for 98% and 99% confidence intervals are also available with three decimals. Note that we will use the t table, which is summa- rized in Table 9–3, to find the z values with three decimals for all following exercises and problems.
Dottie Kleman is the “Cookie Lady.” She bakes and sells cookies at 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% confidence interval for the population mean. Assume that the popula-
tion distribution is normal. (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?
S E L F - R E V I E W 9–2
9. Use Appendix B.5 to locate the value of t under the following conditions. a. The sample size is 12 and the level of confidence is 95%. b. The sample size is 20 and the level of confidence is 90%. c. The sample size is 8 and the level of confidence is 99%.
10. Use Appendix B.5 to locate the value of t under the following conditions. a. The sample size is 15 and the level of confidence is 95%. b. The sample size is 24 and the level of confidence is 98%. c. The sample size is 12 and the level of confidence is 90%.
11. The owner of Britten’s Egg Farm wants to estimate the mean number of eggs pro- duced per chicken. A sample of 20 chickens shows they produced 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% confidence interval, what is the value of t? d. Develop the 95% confidence interval for the population mean. e. Would it be reasonable to conclude that the population mean is 21 eggs? What
about 25 eggs? 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 45 gallons
E X E R C I S E S
300 CHAPTER 9
A CONFIDENCE INTERVAL FOR A POPULATION 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 Technical Institute reports that 80% of its graduates enter the job market in a position related to their field of study.
• A company representative claims that 45% of Burger King sales are made at the drive-through window.
• A survey of homes in the Chicago area indicated that 85% of the new con- struction had central air conditioning.
• A recent survey of married men between the ages of 35 and 50 found that 63% felt that both partners should earn a living.
These examples illustrate the nominal scale of measurement when the outcome is limited to two values. In these cases, an observation is classified into one of two 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 win- dow. We can talk about the groups in terms of proportions.
PROPORTION The fraction, ratio, or percent indicating the part of the sample or the population having a particular trait of interest.
LO9-3 Compute and interpret a confidence interval for a population proportion.
© Vytautas Kielaitis/Shutterstock.com
with a standard deviation of 20 gallons. Assume the population distribution is normal.
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% confidence interval, what is the value of t? d. Develop the 90% confidence interval for the population mean. e. Would it be reasonable to conclude that the population mean is 48 gallons?
13. Merrill Lynch Securities and Health Care Retirement Inc. are two large employ- ers in downtown 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% confidence interval for the population mean. Interpret the result. 14. The Columbus, Ohio 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.
14 24 24 19 24 7 31 20 26 23 23 28 16 15 21
Develop a 98% confidence interval for the population mean. Interpret the result.
ESTIMATION AND CONFIDENCE INTERVALS 301
As an example of a proportion, a recent survey indicated that 92 out of 100 people surveyed favored the continued use of daylight savings time in the summer. The sam- ple proportion is 92/100, or .92, or 92%. If we let p represent the sample proportion, x the number of “successes,” and n the number of items sampled, we can determine a sample proportion as follows.
SAMPLE PROPORTION p = x n
(9–3)
The population proportion is identified by π. Therefore, π refers to the percent of successes in the population. Recall from Chapter 6 that π is the proportion of “suc- cesses” 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 two requirements:
1. The binomial conditions, discussed in Chapter 6, have been met. These conditions are: a. The sample data are the number of successes in n trials. 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 nπ and n(1 − π) should both be greater than or equal to 5. This 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 interval 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 π, which is not known.
To develop a confidence interval for a population proportion, we use:
p ± z√ p(1 − p)
n (9–4)
CONFIDENCE INTERVAL FOR A POPULATION PROPORTION
An example will help to explain the details of determining a confidence interval and interpreting the result.
E X A M P L E
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 sam- ple of 2,000 current BBA members reveals 1,600 plan to vote for the merger pro- posal. What is the estimate of the population proportion? Develop a 95% confidence interval for the population proportion. Basing your decision on this sample information, can you conclude that the necessary proportion of BBA mem- bers favor the merger? Why?
302 CHAPTER 9
S O L U T I O N
First, calculate the sample proportion from formula (9–3). It is .80, found by
p = x n
= 1,600 2,000
= .80
Thus, we estimate that 80% of the population favor the merger proposal. We deter- mine the 95% confidence interval using formula (9–4). The z value corresponding to the 95% level of confidence is 1.96.
p ± z √ p(1 − p)
n = .80 ± 1.96 √
.80(1 − .80) 2,000
= .80 ± .018
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 only values greater than 75% of the union membership.
To review the interpretation of the confidence interval: If the poll was conducted 100 times with 100 different samples, we expect the confidence intervals constructed from 95 of the samples would contain the true population proportion. In addition, the interpretation 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% of those in the sample voted for Mr. Obermeyer. Based on formula (9–3):
p = x n
= 275 500
= .55
Now, to be assured of election, he must earn more than 50% 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 percent in the pop- ulation that will ultimately vote for the candidate. So the question is: Could we take a sample of 500 voters from a population where 50% or less of the voters support Mr. Obermeyer and find that 55% of the sample support him? To put it another way, could the sampling error, which is p − π = .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 the lower endpoint is greater than .50, 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 he is not assured of a majority and we cannot conclude he will be elected. In this case, using the 95% significance level and formula (9–4):
p ± z √ p(1 − p)
n = .55 ± 1.96 √
.55(1 − .55) 500
= .55 ± .044
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% 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 polling organi- zations, television networks, and surveys of public opinion on election night.
STATISTICS IN ACTION
The results of many surveys include confidence intervals. For example, a recent sur- vey of 800 TV viewers in Toledo, Ohio, found 44% watched the evening news on the local CBS affiliate. The article also reported a margin of error of 3.4%. The margin of error is actu- ally the amount that is added and subtracted from the point estimate to find the endpoints of a confidence interval. For a 95% level of confidence, the margin of error is:
z √ p(1 − p)
n
= 1.96 √ .44(1 − .44)
800 = 0.034
The estimate of the pro- portion of all TV viewers in Toledo, Ohio who watch the local news on CBS is between (.44 − .034) and (.44 + .034) or 40.6% and 47.4%.
ESTIMATION AND CONFIDENCE INTERVALS 303
A market research consultant was hired to estimate the proportion of homemakers who associate the brand name of a laundry detergent with the container’s shape and color. The consultant randomly selected 1,400 homemakers. From the sample, 420 were able to identify the brand by name based only on the shape and color of the container. (a) Estimate the value of the population proportion. (b) Develop a 99% confidence interval for the population proportion. (c) Interpret your findings.
S E L F - R E V I E W 9–3
15. The owner of the West End Kwick Fill Gas Station wishes to determine the propor- tion of customers who pay at the pump using a credit card or debit card. He surveys 100 customers and finds that 80 paid at the pump.
a. Estimate the value of the population proportion. b. Develop a 95% confidence interval for the population proportion. c. Interpret your findings.
16. Ms. Maria Wilson is considering running for mayor 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.
1. Estimate the value of the population proportion. 2. Develop a 99% confidence interval for the population proportion. 3. Interpret your findings.
17. The Fox TV network is considering replacing one of its prime-time crime investiga- tion shows with a new family-oriented comedy show. Before a final decision is made, network executives designed an experiment to estimate the proportion of their viewers who would prefer the comedy show over the crime investigation show. A random sample of 400 viewers was selected and asked to watch the new comedy show and the crime investigation show. After viewing the shows, 250 indi- cated they would watch the new comedy show and suggested it replace the crime investigation show.
a. Estimate the value of the population proportion of people who would prefer the comedy show.
b. Develop a 99% confidence interval for the population proportion of people who would prefer the comedy show.
c. Interpret your findings. 18. Schadek Silkscreen Printing Inc. purchases plastic cups and imprints them with
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 and inspected them for defects. He found 15 to be defective.
a. What is the estimated proportion defective in the population? b. Develop a 95% confidence interval for the proportion defective. c. Zack has an agreement with his supplier that if 10% or more of the cups are de-
fective, he can return the order. Should he return this lot? Explain your decision.
E X E R C I S E S
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%. 3. The variation or dispersion of the population being studied.
LO9-4 Calculate the required sample size to estimate a population proportion or population mean.
304 CHAPTER 9
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 determine the endpoints of the confidence interval. For example, in a study of wages, we may decide that we want to estimate the mean wage of the population with a margin 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 3.5%. 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 be- tween 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 result in a wider confidence interval.
The second choice is the level of confidence. In working with confidence intervals, we logically choose relatively high levels of confidence such as 95% and 99%. To com- pute the sample size, we need the z-statistic that corresponds to the chosen level of confidence. The 95% level of confidence corresponds to a z value of 1.96, and a 90% level of confidence corresponds to a z value of 1.645 (using the t table). 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 deviation. If the population is widely dispersed, a large sample is required to get a good estimate. On the other hand, if the population is concentrated (homogeneous), the required sample size to get a good estimate will be smaller. Often, we do not know the population standard devia- tion. Here are three suggestions to estimate 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 question- naire, we use it on a small sample of students. From this small sample, 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 stan- dard deviation from another study. Suppose we want to estimate the number 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, assuming that the distribu- tion 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 ATM transactions per month made by college stu- dents. She believes that the distribution of ATM transactions follows the normal distribution. The minimum and maximum of ATM transactions per month are 2 and 50, so the range is 48, found by (50 − 2). Then the estimated value of the popula- tion standard deviation would be eight ATM transactions per month, 48/6.
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 margin of error used to calculate the endpoints of confidence intervals to estimate a population mean! See formula 9–1.
E = z σ
√n
ESTIMATION AND CONFIDENCE INTERVALS 305
Solving this equation for n yields the following result.
n = ( zσ E )
2
(9–5) SAMPLE SIZE FOR ESTIMATING THE POPULATION MEAN
where:
n is the size of the sample. z is the standard normal z-value corresponding to the desired level of confidence. σ 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.
Sample Size to Estimate a Population Proportion To determine the sample size to estimate a population proportion, the same three vari- ables 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.
E X A M P L E
A student in public administration wants to estimate the mean monthly earnings of city council members in large cities. She can tolerate a margin of error of $100 in estimating the mean. She would also prefer to report the interval estimate with a 95% 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?
S O L U T I O N
The maximum allowable error, E, is $100. The value of z for a 95% level of confi- dence 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:
n = ( zσ E )
2
= ( (1.96) ($1,000)
$100 ) 2
= (19.6)2 = 384.16
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%, this will require a larger sample. Using the t table with infinite degrees of freedom, the z value for a 99% level of confidence is 2.576.
n = ( zσ E )
2
= ( (2.576) ($1,000)
$100 ) 2
= (25.76)2 = 663.58
We recommend a sample of 664. Observe how much the change in the confidence level changed the size of the sample. An increase from the 95% to the 99% level of confidence resulted in an increase of 279 observations, or 72% [(664/385) × 100]. This would greatly increase the cost of the study, in terms of both time and money. Hence, the level of confidence should be considered carefully.
306 CHAPTER 9
For the binomial distribution, the margin of error is:
E = z √ π(1 − π)
n
Solving this equation for n yields the following equation
n = π(1 − π)( z E)
2
(9–6) SAMPLE SIZE FOR THE POPULATION PROPORTION
where:
n is the size of the sample. z is the standard normal z-value corresponding to the desired level of confidence. π is the population proportion. E is the maximum allowable error.
As before, the z-value is associated with our choice of confidence level. We also decide the margin of error, E. However, the population variance of the binomial distribu- tion is represented by π(1 − π). To estimate the population variance, we need a value of the population proportion. If a reliable value cannot be determined with a pilot study or found in a comparable study, then a value of .50 can be used for π. Note that π (1 − π) has the largest value using 0.50 and, therefore, without a good estimate of the popula- tion proportion, using 0.50 as an estimate of π overstates the sample size. Using a larger sample size will not hurt the estimate of the population proportion.
A university’s office of research wants to estimate the arithmetic mean grade point av- erage (GPA) of all graduating seniors during the past 10 years. GPAs range between 2.0 and 4.0. The estimate of the population mean GPA should be within plus or minus .05 of the population mean. Based on prior experience, the population standard devia- tion is 0.279. Using a 99% level of confidence, how many student records need to be selected?
S E L F - R E V I E W 9–4
E X A M P L E
The student in the previous example also wants to estimate the proportion of cities that have private refuse collectors. The student wants to estimate the pop- ulation proportion with a margin of error of .10, prefers a level of confidence of 90%, and has no estimate for the population proportion. What is the required sample size?
S O L U T I O N
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 1.645, us- ing the t table with infinite degrees of freedom. Because no estimate of the population proportion is available, we use .50. The suggested number of obser- vations is
n = (.5) (1 − .5)( 1.645
.10 ) 2
= 67.65
The student needs a random sample of 68 cities.
ESTIMATION AND CONFIDENCE INTERVALS 307
19. A population’s standard deviation is 10. We want to estimate the population mean within 2, with a 95% level of confidence. How large a sample is required?
20. We want to estimate the population mean within 5, with a 99% 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 should be within plus or minus .05, with a 95% 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 should be within plus or minus .10, with a 99% level of confidence. The best estimate of the population proportion is .45. How large a sample is required?
23. A large on-demand, video streaming company is designing a large-scale survey to determine the mean amount of time corporate executives watch on-demand televi- sion. A small pilot survey of 10 executives indicated that the mean time per week is 12 hours, with a standard deviation of 3 hours. The estimate of the mean viewing time should be within one-quarter hour. The 95% level of confidence is to be used. How many executives 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. Each box of carrots should weigh 20.4 pounds. The processor knows that the standard deviation of box weight is 0.5 pound. The processor wants to know if the current packing process meets the 20.4 weight standard. How many boxes must the pro- cessor sample to be 95% confident that the estimate of the population mean is within 0.2 pound?
25. Suppose the U.S. president wants to estimate the proportion of the population that supports 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% level of confidence. The president’s political advisors found a similar survey from two years ago that reported that 60% of people supported health care revisions.
a. How large of a sample is required? b. How large of a sample would be necessary if no estimate were available for the
proportion supporting current policy? 26. Past surveys reveal that 30% 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. How many tourists should be randomly selected to estimate the population
proportion with a 90% confidence level and a 1% margin of error. 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.
E X E R C I S E S
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 sam- ple proportions.
A population that has a fixed upper bound is finite. For example, there are 7,640 students enrolled at Eastern Illinois University, there are 40 employees at Spence Sprockets, there were 917 Jeep Wranglers assembled 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 students regis- tered for your statistics class. It can also be very large, such as all senior citizens living in Florida.
LO9-5 Adjust a confidence interval for finite populations.
308 CHAPTER 9
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 proportion, then we need to adjust the standard error of the proportion in formula (9–4).
This adjustment is called the finite-population correction factor. It is often short- ened to FPC and is:
FPC = √ N − n N − 1
Why is it necessary to apply a factor, and what is its effect? Logically, if the sample is a substantial percentage of the population, the estimate of the population parameter is more precise. Note the effect of the term (N − n)/(N − 1). Suppose the population is 1,000 and the sample 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 standard error reduces the standard error by about 5% (1 − .9492 = .0508). This reduction in the size of the standard error yields a smaller range of values in estimating 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%. Table 9–4 shows the effects of various sample sizes.
TABLE 9–4 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 popula- tion and the population standard deviation was unknown, we would adjust formula (9–2) as follows:
x ± t s
√n(√ N − n N − 1 )
We would make a similar adjustment to formula (9–4) in the case of a proportion. The following example summarizes the steps to find a confidence interval for the
mean.
E X A M P L E
There are 250 families residing 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.
1. What is the population mean? What is the best estimate of the population mean?
2. Develop a 90% confidence interval for the population mean. What are the end- points of the confidence interval?
3. Using the confidence interval, explain why the population mean could be $445. Could the population mean be $425? Why?
ESTIMATION AND CONFIDENCE INTERVALS 309
S O L U T I O N
First, note the population is finite. That is, there is a limit to the number of people residing 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.
x ± t s
√n(√ N − n N − 1 )
In this case, we know x = 450, s = 75, N = 250, and n = 40. We do not know the population standard deviation, so we use the t distribution. To find the ap- propriate value of t, we use Appendix B.5, and move across the top row to the column headed 90%. The degrees of freedom are df = n − 1 = 40 − 1 = 39, so we move to the cell where the df row of 39 intersects with the column headed 90%. The value is 1.685. Inserting these values in the formula:
x ± t s
√n(√ N − n N − 1 )
= $450 ± 1.685 $75 √40(√
250 − 40 250 − 1 ) = $450 ± $19.98
√.8434 = $450 ± $18.35
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 confidence interval and $425 is not within the confidence interval.
The same study of church contributions in Scandia revealed that 15 of the 40 families sam- pled attend church regularly. Construct the 95% confidence interval for the proportion of families attending church regularly.
S E L F - R E V I E W 9–5
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% confidence inter- val 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% confidence inter- val 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 num- ber of soft drinks consumed per person was 1.86, with a standard deviation of 0.50. Develop a 99% confidence interval for the mean number of soft drinks con- sumed 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. Con- struct the 95% confidence interval for the proportion of all welders who graduated from a registered welding course.
E X E R C I S E S
310 CHAPTER 9
C H A P T E R S U M M A R Y
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 devia-
tion, 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
x ± z σ
√n (9–1)
b. To determine the confidence limits when the population standard deviation is unknown, we use the t distribution. The formula is
x ± t s
√n (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 popu- lation that has a particular characteristic. A. A sample proportion, p, is found by x, the number of successes, divided by n, the num-
ber of observations. B. We construct a confidence interval for a sample proportion from the following
formula.
p ± z √ p(1 − p)
n (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
n = ( zσ E )
2
(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 π to calculate the variation in the population. 4. The formula to determine the sample size for a proportion is
n = π(1 − π)( z E)
2
(9–6)
VI. For a finite population, the standard error is adjusted by the factor: √ N − n N − 1
ESTIMATION AND CONFIDENCE INTERVALS 311
C H A P T E R E X E R C I S E S
31. A random sample of 85 group leaders, supervisors, and similar personnel at General Motors revealed that, on average, they spent 6.5 years in a particular job before being promoted. The standard deviation of the sample was 1.7 years. Construct a 95% confi- dence interval.
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 always 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 pound. a. What is the estimated population mean? b. Determine a 95% 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 to lease a one-bedroom apartment. The mean cost per month for a random sample of 40 apartments currently available for lease was $884. The standard deviation of the sample was $50. a. Develop a 98% confidence interval for the population mean. b. Would it be reasonable to conclude that the population mean is $950 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% confidence interval for the population 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 for the sample is 8.32 years and the standard devia- tion is 3.07 years. a. What is the mean of the population? b. Develop a 90% 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 meal per week. Construct a 99% confidence interval for the population mean.
37. The National Collegiate Athletic Association (NCAA) reported that college football assis- tant coaches spend a mean of 70 hours per week on coaching and recruiting during the season. A random sample of 50 assistant coaches showed the sample mean to be 68.6 hours, with a standard deviation of 8.2 hours. a. Using the sample data, construct a 99% confidence interval for the population
mean. b. Does the 99% confidence interval include the value suggested by the NCAA? Inter-
pret this result. c. Suppose you decided to switch from a 99% to a 95% confidence interval. Without
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% 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.
312 CHAPTER 9
39. A student conducted a study and reported that the 95% 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 size 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 surveyed a random sample of 20 dealers. The data revealed a mean amount of profit per car sold was $290, with a standard deviation of $125. Develop a 95% confidence interval for the population mean of profit per car.
41. A study of 25 graduates of four-year public colleges revealed the mean amount owed by a student in student loans was $55,051. The standard deviation of the sample was $7,568. Construct a 90% confidence interval for the population mean. Is it reasonable to conclude that the mean of the population is actually $55,000? Explain why or why not.
42. An important factor in selling a residential property is the number of times real estate agents show a home. A sample of 15 homes recently sold in the Buffalo, New York, area revealed the mean number of times a home was shown was 24 and the standard deviation of the sample was 5 people. Develop a 98% confidence interval for the population mean.
43. In 2003, the Accreditation Council for Graduate Medical Education (ACGME) imple- mented new rules limiting work hours for all residents. A key component of these rules is that residents should work no more than 80 hours per week. The following is the number of weekly hours worked in 2017 by a sample of residents at the Tidelands Medical Center.
84 86 84 86 79 82 87 81 84 78 74 86
a. What is the point estimate of the population mean for the number of weekly hours worked at the Tidelands Medical Center?
b. Develop a 90% confidence interval for the population mean. c. Is the Tidelands Medical Center within the ACGME guideline? Why?
44. PrintTech, Inc. is introducing a new line of ink-jet printers and would like to promote 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% 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 sample 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% confidence level for the population mean. c. According to Dr. Benner’s test, is it reasonable to conclude that the mean stress level
of Internet executives is 80? Explain. 46. Pharmaceutical companies promote their prescription drugs using television advertis-
ing. In a survey of 80 randomly sampled television viewers, 10 indicated that they asked their physician about using a prescription drug they saw advertised on TV. Develop a 95% confidence interval for the proportion of viewers who discussed a drug seen on TV with their physician. Is it reasonable to conclude that 25% of the viewers discuss an ad- vertised drug with their physician?
47. HighTech, Inc. randomly tests its employees about company policies. Last year in the 400 random tests conducted, 14 employees failed the test. Develop a 99% confidence interval for the proportion of applicants that fail the test. Would it be reasonable to con- clude that 5% of the employees cannot pass the company policy test? Explain.
ESTIMATION AND CONFIDENCE INTERVALS 313
48. During a national debate on changes to health care, a cable news service performs an opinion poll of 500 small-business owners. It shows that 65% of small-business owners do not approve of the changes. Develop a 95% 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% confidence interval for the proportion of voters in the county who plan to vote for Ms. Miller. From this sample information, is it reasonable to con- clude that Ms. Miller will receive a majority of the votes?
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.
A total of 560 respondents selected the first statement, indicating they thought the president was doing a good job. a. Construct a 95% 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 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% 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% confidence interval for the proportion of checking account customers who have a Visa card with the bank.
53. It is estimated that 60% of U.S. households subscribe to cable TV. You would like to ver- ify this statement for your class in mass communications. If you want your estimate to be within 5 percentage points, with a 95% level of confidence, how many households should you sample?
54. You need to estimate the mean number of travel days per year for 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 salespeople should you sample? Use the 90% confidence level.
55. You want to estimate the mean family income in a rural area of central Indiana. The question is, how many families should be sampled? In a pilot sample of 10 families, the standard deviation of the sample was $500. The sponsor of the survey wants you to use the 95% 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 premi- ums 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% confidence interval for the popula-
tion mean yearly premium. b. How large a sample is needed to find the population mean within $250 at 99%
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 equivalent air pressure on 30 randomly chosen flights. The study revealed a mean equivalent air pressure of 8,000 feet with a standard deviation of 300 feet. a. Develop a 99% confidence interval for the population mean equivalent air
pressure. b. How large a sample is needed to find the population mean within 25 feet at 95%
confidence?
314 CHAPTER 9
58. A survey of 25 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $65.00 per hour. The sample stan- dard 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% 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% confidence? 59. Based on a sample of 50 U.S. citizens, the American Film Institute found that a typical
American spent 78 hours watching movies last year. The standard deviation of this sam- ple was 9 hours. a. Develop a 95% confidence interval for the population mean number of hours spent
watching movies last year. b. How large a sample should be used to be 90% 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% 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?
61. A survey of 36 randomly selected iPhone owners showed that the purchase price has a mean of $650 with a sample standard deviation of $24. a. Compute the standard error of the sample mean. b. Compute the 95% 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% confidence level and a margin of error of 2%. A pilot survey reveals that 5 of the 50 sampled hold two or more jobs. How many in the workforce should be interviewed to meet your requirements?
63. A study conducted several years ago reported that 21 percent of public accountants changed companies within 3 years. The American Institute of CPA’s would like to update the study. They would like to estimate the population proportion of public accountants who changed companies within 3 years with a margin of error of 3% and a 95% level of confidence. 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 selected a random
sample of 36 customers and reviewed the value of their accounts. The mean was $32,000 with a sample standard deviation of $8,200. What is a 90% confidence interval for the mean account value of the population of customers?
65. The National Weight Control Registry tries to mine secrets of success from people who lost at least 30 pounds and kept it off for at least a year. It reports that out of 2,700 reg- istrants, 459 were on a low-carbohydrate diet (less than 90 grams a day). a. Develop a 95% confidence interval for the proportion of people on a low-carbohy-
drate diet. b. Is it possible that the population percentage is 18%? c. How large a sample is needed to estimate the proportion within 0.5%?
66. Near the time of an election, a cable news service performs an opinion poll of 1,000 prob- able voters. It shows that the Republican contender has an advantage of 52% to 48%. a. Develop a 95% 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% confidence interval for the mean time Wired subscribers spend on the Internet.
68. The Tennessee Tourism Institute (TTI) plans to sample information center visitors entering the state to learn the fraction of visitors who plan to camp in the state. Current estimates are that 35% of visitors are campers. How many visitors would you sample to estimate the population proportion of campers with a 95% confidence level and an allowable error of 2%?
ESTIMATION AND CONFIDENCE INTERVALS 315
D A T A A N A L Y T I C S
69. Refer to the North Valley Real Estate data, which reports information on homes sold in the area during the last year. Select a random sample of twenty homes. a. Based on your random sample of twenty homes, develop a 95% confidence interval
for the mean selling price of the homes. b. Based on your random sample of twenty homes, develop a 95% confidence interval
for the mean days on the market. c. Based on your random sample of twenty homes, develop a 95% confidence interval
for the proportion of homes with a pool. d. Suppose that North Valley Real Estate employs several agents. Each agent will be
randomly assigned twenty homes to sell. The agents are highly motivated to sell homes based on the commissions they earn. They are also concerned about the twenty homes they are assigned to sell. Using the confidence intervals you created, write a general memo informing the agents about the characteristics of the homes they may be assigned to sell.
e. What would you do if your confidence intervals did not include the mean of all 105 homes? How could this happen?
70. Refer to the Baseball 2016 data, which report information on the 30 Major League Baseball teams for the 2016 season. Assume the 2016 data represents a sample. a. Develop a 95% confidence interval for the mean number of home runs per team. b. Develop a 95% confidence interval for the mean batting average by each team. c. Develop a 95% confidence interval for the mean earned run average (ERA) for each team.
71. Refer to the Lincolnville School District bus data. a. Develop a 95% confidence interval for the mean bus maintenance cost. b. Develop a 95% confidence interval for the mean bus odometer miles. c. Write a business memo to the state transportation official to report your results.
A REVIEW OF CHAPTERS 8–9 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. For example, to contact all U.S. bank officers and record their annual incomes would be too expensive and time-consuming, . 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. Therefore, to estimate a popu- lation parameter, we select a sample from the population. A sample is a part of the population. Care must be taken to ensure that every member of our population has a chance of being selected; otherwise, the conclusions might be biased. A number of probability-type sampling methods 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 corresponding population parame- ter. 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 parameter 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 population 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 sample, we decrease the variation in the sampling distribution.
The central limit theorem is the foundation of statistical inference. It states that, if the population from which we select the sam- ples 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 single 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 prob- ably 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 standard 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.
316 CHAPTER 9
P R O B L E M S
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 normal 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 customers 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% confidence interval for the mean number of customers per day.
4. The manager of the local Hamburger Express wishes to estimate the mean time custom- ers 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 minute. Develop a 90% confidence interval for the mean waiting time.
5. Defiance Tool and Die has 293 sales offices throughout the world. The VP of Sales is studying the usage of its copy machines. A random sample of six of the sales offices revealed the following number of copies made in each selected office last week.
826 931 1,126 918 1,011 1,101
Develop a 95% confidence interval for the mean number of copies per week. 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 five 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 inter- val 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. Con- struct a confidence interval for the mean number of widgets produced per day. Does it seem reasonable to conclude that the mean daily widget production has changed? Jus- tify 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% 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% 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% confidence interval for the pro- portion of residents who believe the tax rate is too high. Does it seem reasonable to conclude that 50% of the voters believe that 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 Internet. How large of a sample of purchasers is necessary for the estimate to be
ESTIMATION AND CONFIDENCE INTERVALS 317
within 2 percentage points with a 98% level of confidence? Current thinking is that about 8% 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% of those sampled smoked. Develop a 98% confidence interval for the proportion of adults who currently smoke. Does it seem reasonable to conclude that the proportion of adults who smoke has changed?
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% regularly play, but the auditor would like some current information. How large a sample is necessary for the estimate to be within 3 percentage points, with a 98% level of confidence?
C A S E S
Century National Bank Refer to the description of Century National Bank at the end of the Review of Chapters 1–4 on page 129. 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% confi- dence interval for the proportion of customers using these cards. On the basis of the confidence interval, is it reason- able to conclude that more than half of the customers use a debit card? Write a brief report interpreting the results.
P R A C T I C E T E S T
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 does 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, South Carolina, 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% confidence interval for the population mean.
3. A recent federal report indicated that 27% of children ages 2 to 5 ate a vegetable at least five times a week. How large a sample is needed to estimate the true population proportion within 2% with a 98% level of confidence? 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% confi- dence interval for the population proportion.
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO10-1 Explain the process of testing a hypothesis.
LO10-2 Apply the six-step procedure for testing a hypothesis.
LO10-3 Distinguish between a one-tailed and a two-tailed test of hypothesis.
LO10-4 Conduct a test of a hypothesis about a population mean.
LO10-5 Compute and interpret a p-value.
LO10-6 Use a t statistic to test a hypothesis.
LO10-7 Compute the probability of a Type II error.
© Franco Salmoiraghi/Photo Resource Hawaii/Alamy Stock Photo
One-Sample Tests of Hypothesis10
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 ounce. 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% level of significance, can we conclude that the mean weight is greater than 16 ounces? Determine the p-value. (See Exercise 26 and LO10-4.)
ONE-SAMPLE TESTS OF HYPOTHESIS 319
INTRODUCTION Chapter 8 began our study sampling and statistical inference. We described how we could select a random sample to estimate the value of a population parameter. For ex- ample, we selected a sample of five employees at Spence Sprockets, 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 confidence interval. A confidence interval is a range of values within which we expect the popula- tion 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 auto- mobiles passing milepost 150 on the West Virginia Turnpike is 68 miles per hour.
• The mean number of miles driven by those leasing a Chevy Trail- Blazer for 3 years is 32,000 miles.
• The mean time an American family lives in a particular single-family dwelling is 11.8 years.
• In 2016, the mean start- ing salary for a graduate from a four-year business program is $51,541.
• According to the Kelley Blue Book (www.kbb.com), a 2017 Ford Edge averages 21 miles per gallon in the city.
• The mean cost to remodel a kitchen is $20,000.
This chapter and several of the following chapters cover statistical hypothesis test- ing. We begin by defining what we mean by a statistical hypothesis and statistical hy- pothesis testing. Next, we outline the steps in statistical hypothesis testing. Then we conduct tests of hypothesis for means. In the last section of the chapter, we describe possible errors due to sampling in hypothesis testing.
WHAT IS HYPOTHESIS TESTING? The terms hypothesis testing and testing a hypothesis are used interchangeably. Hypoth- esis testing starts with a statement, or assumption, about a population parameter—such as the population mean. This statement is referred to as a hypothesis.
HYPOTHESIS A statement about a population parameter subject to verification.
A hypothesis might be that the mean monthly commission of sales associates in retail electronics stores, such as hhgregg, is $2,000. We cannot contact all hhgregg sales associates to determine that the mean is $2,000. The cost of locating and inter- viewing every hhgregg electronics sales associate in the United States would be exor- bitant. To test the validity of the hypothesis (μ = $2,000), we must select a sample from the population of all hhgregg electronics sales associates, calculate sample statistics,
LO10-1 Explain the process of testing a hypothesis.
© Russell Ilig/Getty Images
320 CHAPTER 10
and based on certain decision rules reject or fail to reject the hypothesis. A sample mean of $1,000 per month is much less than $2,000 per month and we would most likely reject the hypothesis. However, suppose the sample mean is $1,995. Can we at- tribute the $5 difference between $1,995 and $2,000 to sampling error? Or is this dif- ference of $5 statistically significant?
HYPOTHESIS TESTING A procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement.
SIX-STEP PROCEDURE FOR TESTING A HYPOTHESIS There is a six-step procedure that systematizes hypothesis testing; when we get to step 6 we are ready to interpret the results of the test based on the decision to reject or not reject the hypothesis. However, hypothesis 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 man- ner 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.
LO10-2 Apply the six-step procedure for testing a hypothesis.
Step 1 Step 2 Step 3 Step 4 Step 5 Step 6
Take a sample, arrive at decision
Formulate a decision
rule
Select a level of
Identify the test
statistic
State null and alternate
hypotheses
Interpret the
result
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 different from 60,000. The null hypothesis would be written H0: μ = 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 pa- rameter would have to be known. To actually determine it, we would have to test, sur- vey, or count every item in the population. This is usually not feasible. The alternative is to take a sample from the population.
Often, the null hypothesis begins by stating, “There is no significant difference between . . .” or “The mean impact strength of the glass is not significantly different
ONE-SAMPLE TESTS OF HYPOTHESIS 321
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 im- pact 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 param- eter (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:
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.
LEVEL OF SIGNIFICANCE The probability of rejecting the null hypothesis when it is true.
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. commercial 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: μ = 15. The alternate hypothesis is that the statement is not true, that is, H1: μ ≠ 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 hypothesis. 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 suggest the null hypoth- esis is untrue.
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.
The level of significance is designated α, the Greek letter alpha. It is also sometimes 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% level), the .01 level, the .10 level, or any other level between 0 and 1. Traditionally, the .05 level is selected for consumer re- search projects, .01 for quality assurance, and .10 for political polling. You, the re- searcher, must decide on the level of significance before formulating a decision rule and collecting sample data.
322 CHAPTER 10
To illustrate how it is possible to reject a true hypothesis, suppose a firm manufacturing 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 speci- fies that the computer manufacturer’s quality-assurance department will randomly sample all incoming shipments of circuit boards. If more than 6% of the boards sampled are substandard, the shipment will be rejected. The null hypothesis is that the incoming shipment of boards meets the quality standards of the contract and contains 6% or less defective boards. The alternate hypothesis is that more than 6% of the boards are defective.
A shipment of 4,000 circuit boards was received from Allied Electronics, and the quality assurance department selected a random sample of 50 circuit boards for testing. Of the 50 circuit boards
sampled, 4 boards, or 8%, were substandard. The shipment was rejected because it exceeded the maximum of 6% substandard printed circuit boards. If the shipment was actually substandard, then the decision to return the boards to the supplier was correct.
However, because of sampling error, there is a small probability of an incorrect de- cision. Suppose there were only 40, or 4%, defective boards in the shipment (well under the 6% threshold) and 4 of these 40 were randomly selected in the sample of 50. The sample evidence indicates that the percentage of defective boards is 8% (4 out of 50 is 8%) so we reject the shipment. But, in fact, of the 4,000 boards, there are only 40 defec- tive units. The true defect rate is 1.00%. In this instance our sample evidence estimates 8% defective but there is only 1% defective in the population. Based on the sample evi- dence, an incorrect decision was made. In terms of hypothesis testing, we rejected the null hypothesis when we should have failed to reject the null hypothesis. By rejecting a true null hypothesis, we committed a Type I error. The probability of committing a Type I error is represented by the Greek letter alpha (α).
© Jim Stern/Bloomberg via Getty Images
TYPE I ERROR Rejecting the null hypothesis, H0, when it is true.
TYPE II ERROR Not rejecting the null hypothesis when it is false.
The other possible error in hypothesis testing is called Type II error. The probability of committing a Type II error is designated by the Greek letter beta (β).
The firm manufacturing personal computers would commit a Type II error if, un- known to the manufacturer, an incoming shipment of printed circuit boards from Allied Electronics contained 15% substandard boards, yet the shipment was accepted. How could this happen? A random sample of 50 boards could have 2 (4%) substandard boards, and 48 good boards. According to the stated procedure, because the sample contained less than 6% substandard boards, the decision is to accept the shipment. This is a Type II error. While this event is extremely unlikely, it is possible based on the process of randomly sampling from a population. In a later section, we show how to calculate the probability of a Type II error.
In retrospect, the researcher cannot study every item or individual in the popula- tion. Thus, there is a possibility of two types of error—a Type I error, wherein the null hypothesis is rejected when it should not be rejected, 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, α, and beta, β. Alpha (α) is the probability of making a Type I error, and beta (β) is the probability of making a Type II error. The following table summarizes the decisions the researcher could make and the possible consequences.
ONE-SAMPLE TESTS OF HYPOTHESIS 323
Step 3: Select the Test Statistic There are many test statistics. In this chapter, we use both z and t as the test statistics. In later chapters, we will use such test statistics as F and χ2, called chi-square.
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
TEST STATISTIC A value, determined from sample information, used to determine whether to reject the null hypothesis.
In hypothesis testing for the mean (μ) when σ is known, the test statistic z is com- puted by:
TESTING A MEAN, σ KNOWN z = x − μ σ∕√n
(10–1)
The z value is based on the sampling distribution of x , which follows the normal dis- tribution with a mean (μ
x ) equal to μ and a standard deviation σ x, which is equal to σ∕√n. We can thus determine whether the difference between x and μ is statistically significant by finding the number of standard deviations x is from μ, using formula (10–1).
Step 4: Formulate the Decision Rule A decision rule is a statement of the specific conditions under which the null hypothesis is rejected and the conditions under which it is not rejected. The region or area of rejec- tion defines the location of all those values that are so large or so small that the proba- bility 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 con- ducted later in the chapter.
Scale of z0 1.645 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
324 CHAPTER 10
Note in the chart that:
• The area where the null hypothesis is not rejected is to the left of 1.645. We will explain how to get the 1.645 value shortly.
• The area of rejection is to the right of 1.645. • 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 1.645 separates the regions where the null hypothesis is rejected and
where it is not rejected. • The value 1.645 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 step in hypothesis testing is to compute the value of the test statistic, compare its value to the critical value, and make 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 1.645. We reject the null hypothesis, reasoning that it is highly improbable that a computed z value this large is due to sampling error (chance).
Had the computed value been 1.645 or less, say 0.71, the null hypothesis is not rejected. It is reasoned that such a small computed value could be attributed to chance, that is, sampling error. As we have emphasized, only one of two decisions is possible in hypothesis testing—either reject or do not reject the null hypothesis.
However, because the decision is based on a sample, it is always possible to make either of two decision errors. It is possible to make a Type I error when the null hypoth- esis is rejected when it should not be rejected. Or it is also possible to make a Type II error when the null hypothesis is not rejected and it should have been rejected. Fortu- nately, we select the probability of making a Type I error, α (alpha), and we can compute the probabilities associated with a Type II error, β (beta).
Step 6: Interpret the Result The final step in the hypothesis testing procedure is to interpret the results. The process does not end with the value of a sample statistic or the decision to reject or not reject the null hypothesis. What can we say or report based on the results of the statistical test? Here are two examples:
• An investigative reporter for a Colorado newspaper reports that the mean monthly income of convenience stores in the state is $130,000. You decide to conduct a test of hypothesis to verify the report. The null hypothesis and the alternate hypothesis are:
H0: μ = $130,000
H1: μ ≠ $130,000
A sample of convenience stores provides a sample mean and standard deviation, and you compute a z statistic. The results of the hypothesis test result in a decision to not reject the null hypothesis. How do you interpret the result? Be cautious with
STATISTICS IN ACTION
During World War II, allied military planners needed estimates of the number of German tanks. The infor- mation provided by tradi- tional spying methods was not reliable, but statistical methods proved to be valu- able. For example, espio- nage and reconnaissance led analysts to estimate that 1,550 tanks were pro- duced during June 1941. However, using the serial numbers of captured tanks and statistical analysis, mil- itary planners estimated that only 244 tanks were produced. The actual num- ber produced, as deter- mined from German production records, was 271. The estimate using statistical 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.
ONE-SAMPLE TESTS OF HYPOTHESIS 325
your interpretation because by not rejecting the null hypothesis, you did not prove the null hypothesis to be true. Based on the sample data, the difference between the sample mean and hypothesized population mean was not large enough to re- ject the null hypothesis.
• In a recent speech to students, the dean of the College of Business reported that the mean credit card debt for college students is $3,000. You decide to conduct a test of the dean’s statement or hypothesis to investigate the statement’s truth. The null hypothesis and the alternate hypothesis are:
H0: μ = $3,000
H1: μ ≠ $3,000
A random sample of college students provides a sample mean and standard devia- tion, and you compute a z statistic. The hypothesis test results in a decision to reject the null hypothesis. How do you interpret the result? The sample evidence does not support the dean’s statement. Based on the sample data, the mean amount of stu- dent credit card debt is different from $3,000. You have disproved the null hypothesis with a stated probability of a Type I error, α. That is, there is a small probability that the decision to reject the null hypothesis was an error due to random sampling.
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, α. 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 information. 6. Interpret the results of the test.
Before actually conducting a test of hypothesis, we describe the difference between a one-tailed and a two-tailed hypothesis test.
ONE-TAILED AND TWO-TAILED HYPOTHESIS TESTS Refer to Chart 10–1. It shows a one-tailed test. It is called a one-tailed test because the rejection region is only in one tail of the curve. In this case, it is in the right, or upper, tail of the curve. To illustrate, suppose that the packaging department 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: μ ≤ 453. This is read, “the population mean (μ) is equal to or less than 453.” The alternate hypothesis is, therefore, H1: μ > 453. This is read, “μ 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 observe that the null hypothesis includes the equal sign. That is, H0: μ ≤ 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 automobile manufacturers, large automobile leasing companies, and other organizations that pur- chase 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 ship- ment 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 conclude
LO10-3 Distinguish between a one-tailed and a two- tailed test of hypothesis.
STATISTICS IN ACTION
LASIK is a 15-minute surgi- cal procedure that uses a laser to reshape an eye’s cornea with the goal of im- proving eyesight. Research shows that about 5% of all surgeries involve complica- tions such as glare, corneal haze, overcorrection or un- dercorrection 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 hypothesis that the surgery will improve eyesight. The sample data of LASIK sur- gery shows that 5% of all cases result in complica- tions. The 5% represents a Type I error rate. When a person decides to have the surgery, he or she expects to reject the null hypothesis. In 5% of future cases, this expectation will not be met. (Source: American Academy of Ophthalmology Journal, Vol. 16, no. 43.)
326 CHAPTER 10
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 the tires is not less than 60,000 miles. This statement appears in the null hypothesis. The null and alternate hypotheses in this case are written H0: μ ≥ 60,000 and H1: μ < 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 the tire wear 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 direc- tion, such as:
H0: The mean income of female stockbrokers is less than or equal to $65,000 per year. H1: The mean income of female 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 female stockbrokers is $65,000 per year. H1: The mean income of female 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% area of rejection is divided equally into the two tails of the sampling distribution (2.5% 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.
CHART 10–2 Sampling Distribution for the Statistic z, Left-Tailed Test, .05 Level of Significance
Scale of z0–1.645 Critical value
Do not reject H0
Region of rejection
CHART 10–3 Regions of Nonrejection and Rejection for a Two-Tailed Test, .05 Level of Significance
Scale of z0–1.96
Do not reject H0
Critical value
Region of rejection
.025
1.96 Critical value
Region of rejection
.025
.95
ONE-SAMPLE TESTS OF HYPOTHESIS 327
HYPOTHESIS TESTING FOR A POPULATION MEAN: KNOWN POPULATION STANDARD DEVIATION
A Two-Tailed Test An example will show the details of the six-step hypothesis testing procedure. We also wish to use a two-tailed test. That is, we are not concerned whether the sample 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 historical information about the population and in fact know its standard deviation.
LO10-4 Conduct a test of a hypothesis about a population mean.
E X A M P L E
Jamestown Steel Company manufac- tures and assembles desks and other office equipment 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 expan- sion, new production methods have been introduced and new employees hired. The vice president of manufac- turing 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?
S O L U T I O N
In this example, we know two important pieces of information: (1) the population of weekly production follows the normal distribution and (2) the standard devia- tion of this normal distribution is 16 desks per week. So it is appropriate to use the z statistic. 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 hy- pothesis 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:
H0: μ = 200
H1: μ ≠ 200
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 production is greater than 200 or less than 200. The vice president wants only to find out whether the production rate is dif- ferent from 200.
© Robert Nicholas/Getty Images
328 CHAPTER 10
Before moving to Step 2, we wish to emphasize two points.
• The null hypothesis has the equal sign. Why? Because the value we are testing is always in the null hypothesis. Logically, the alter- nate hypothesis never contains the equal sign.
• Both the null hypothesis and the alternate hypothesis contain Greek letters—in this case μ, which is the symbol for the population mean. Tests of hypothesis always refer to population parameters, never to sample statistics. To put it another way, you will never see the sym- bol x as part of the null hypothesis or the alternate hypothesis.
Step 2: Select the level of significance. In the example description, the signif- icance level selected is .01. This is α, 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 standard units (z values) permits their use not only in this problem but also in other hypothesis-testing problems. Formula (10–1) for z is re- peated next with the various letters identified.
Sample mean Population mean
Sample sizeStandard deviation of population
z = x _
– m
s n
Step 4: Formulate the decision rule. We formulate the decision rule by first de- termining the critical values of z. Because this is a two-tailed test, half of .01, or .005, is placed in each tail. The area where H0 is not rejected, lo- cated between the two tails, is therefore .99. Using the Student’s t Distri- bution table in Appendix B.5, move to the top margin called “Level of Significance for Two-Tailed Tests, α,” select the column with α = .01, and move to the last row, which is labeled ∞, or infinite degrees of freedom. The z value in this cell is 2.576. For your convenience, Appendix B.5, Student’s t Distribution, is repeated in the inside back cover. All the facets of this problem are shown in the diagram in Chart 10–4.
CHART 10–4 Decision Rule for the .01 Significance Level
Scale of z0
.01 ___ 2
a_ 2 5
a_ 2 55 .005
.01 ___ 2 = .005
–2.576 2.576
.5000.5000
.4950.4950
H0 not rejectedRegion of rejection
Region of rejection
H0: m 5 200 H1: m Þ 200
Critical value Critical value
ONE-SAMPLE TESTS OF HYPOTHESIS 329
Did we prove that the assembly rate is still 200 per week? Not really. We failed 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. For example, in the U.S. judi- cial system, a person is presumed innocent until proven guilty. The trial starts with a null hypothesis that the individual is innocent. If the individual is acquitted, the trial did not provide enough evidence to reject the presumption of innocence and conclude that the individual was not innocent or guilty as charged. That is what we do in statisti- cal hypothesis testing when we do not reject the null hypothesis. The correct interpre- tation is that, based on the evidence or sample information, we have failed to disprove the null hypothesis.
We selected the significance level, .01 in this case, before setting up the decision rule and sampling the population. This is the appropriate strategy. The significance level should be set by the investigator, but it should be determined before gathering the sam- ple 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
The decision rule is: if the computed value of z is not between −2.576 and 2.576, reject the null hypothesis. If z falls between −2.576 and 2.576, do not reject the null hypothesis.
Step 5: Make a decision. Take a sample from the population (weekly pro- duction), compute a test statistic, 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):
z = x − μ σ∕√n
= 203.5 − 200
16∕√50 = 1.547
Because 1.547 is between −2.576 and 2.576, we decide not to reject H0.
Step 6: Interpret the result. We did not reject the null hypothesis, so we have failed to show that the population mean has changed from 200 per week. To put it another way, the difference between the population mean of 200 per week and the sample mean of 203.5 could simply be due to chance. What should we tell the vice presi- dent? The sample information fails to indicate that the new produc- tion methods resulted in a change in the 200-desks-per-week production rate.
z scale
Reject H0 Reject H0
Do not reject H0
Computed value of z
2.5761.5470–2.576
330 CHAPTER 10
week to a z value. Then we compared the computed value of the test statistic (1.547) to that of the critical values (−2.576 and 2.576). Because the computed value of the test statistic was in the region where the null hypothesis was not rejected, we concluded that the population mean could be 200. To use the confidence interval approach, on the other hand, we would develop a confidence interval, based on formula (9–1). See page 287. The interval would be from 197.671 to 209.329, found by 203.5 ± 2.576(16/√50). 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 hypothe- sized 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 pro- posed population value occurring in the confidence interval.
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 dis- pensing 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.
S E L F - R E V I E W 10–1
© Kevin Lorenzi/Bloomberg/Getty Images
A One-Tailed Test In the previous example/solution, we emphasized that we were concerned only with reporting 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 methods, 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 differently. The biggest difference occurs in the alternate hypothe- sis. Before, we stated the alternate hypothesis as “different from”; now we want to state it as “greater than.” In symbols:
A two-tailed test: A one-tailed test:
H0: μ = 200 H0: μ ≤ 200
H1: μ ≠ 200 H1: μ > 200
ONE-SAMPLE TESTS OF HYPOTHESIS 331
The critical values for a one-tailed test are different from a two-tailed test at the same significance level. In the previous example/solution, 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.
Scale of z0
.005 Region of rejection
H0 is not rejected
.99
.005 Region of rejection
0
H0 is not rejected
.99
.01 Region of rejection
One-tailed testTwo-tailed test
2.326 Critical value
2.576 Critical value
–2.576 Critical value
H0: m 5 200 H1: m Þ 200
H0: m # 200 H1: m . 200
CHART 10–5 Rejection Regions for Two-Tailed and One-Tailed Tests, α = .01
For the one-tailed test, the critical value of z is 2.326. Using the Student’s t Distribu- tion table in Appendix B.5, move to the top heading called “Level of Significance for One-Tailed Tests, α,” select the column with α = .01, and move to the last row, which is labeled ∞, or infinite degrees of freedom. The z value in this cell is 2.326.
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 or not reject the null hypothesis. So, for example, if the critical value is 1.96 and the computed value of the test statistic is 2.19, the decision is to reject the null hypothesis.
In recent years, spurred by the availability of computer software, additional informa- tion is often reported on the strength of the rejection. That is, how confident are we in rejecting the null hypothesis? This approach reports the 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.
LO10-5 Compute and interpret a p-value.
STATISTICS IN ACTION
There is a difference be- tween statistically significant and practically 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 2 years lost 1 pound. Do you think many people would be interested in tak- ing the pill to lose 1 pound? The results of using the new pill were statistically significant but not practi- cally significant.
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.
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 find the p-value? To calculate p-values, we will need to use the z table (Appendix B.3) and, to use this table, we will round z test statistics to two decimals. To illustrate how to compute a p-value, we will use the example where we tested the null hypothesis that the mean number of desks produced per week at Fredonia was 200.
332 CHAPTER 10
We did not reject the null hypothesis because the computed z test statistic of 1.547 fell in the region between −2.576 and 2.576. We agreed not to reject the null hypothesis if the z test statistic fell in this region. Rounding 1.547 to 1.55 and using the z table, 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 x greater than 203.5 if μ = 200 is .0606. To compute 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. Notice for the two-tailed hypothesis test, the p-value is represented by areas in both tails of the distribution. Then the p-value can easily be compared with the significance level. The same decision rule is used as in the one- sided test.
Rejection region Rejection region
p-value +
1.55
.0606.0606
Scale of z2.576–2.576 –1.55 0
.01 ___ 2
a 2 5 5 .005
.01 ___ 2
a 2 5 5 .005
A p-value is a way to express the likelihood that H0 is false. But how do we inter- pret a p-value? We have already said that if the p-value is less than the significance level, then we reject H0; if it is greater than the significance level, then we do not re- ject 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.
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.
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)?
S E L F - R E V I E W 10–2
ONE-SAMPLE TESTS OF HYPOTHESIS 333
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.
H0: μ = 50 H1: μ ≠ 50
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 .01 significance level.
H0: μ ≤ 10 H1: μ > 10
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.
H0: μ ≤ 20 H1: μ > 20
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 hypothesis using the .025 significance level.
H0: μ ≥ 220 H1: μ < 220
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 devi- ation 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 experi- ence different from that claimed by the manufacturer at the .05 significance level?
6. The waiting time for customers at MacBurger Restaurants follows a normal dis- tribution with a population 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 movies per month with a population standard deviation of 1.8. The distribution of number of movies watched per month follows the normal distribu- tion. A random sample of 36 college students revealed that the mean number of movies watched last month was 6.2. At the .05 significance level, can we conclude that college students watch fewer movies 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 $9.95. 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?
E X E R C I S E S
334 CHAPTER 10
E X A M P L E
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 and recorded the cost to process each claim. The sample information is reported below.
At the .01 significance level, is it reasonable to conclude that the mean cost to process a claim is now less than $60?
HYPOTHESIS TESTING FOR A POPULATION MEAN: POPULATION STANDARD DEVIATION UNKNOWN In the preceding example, we knew σ, the population standard deviation, and that the population followed the normal distribution. In most cases, however, the population standard deviation is unknown. Thus, σ must be based on prior studies or estimated by the sample standard deviation, s. The population standard deviation in the following example is not known, so the sample standard deviation is used to estimate σ.
To find the value of the test statistic, we use the t distribution and revise formula (10–1) as follows:
LO10-6 Use a t statistic to test a hypothesis.
TESTING A MEAN, σ UNKNOWN t = x − μ s∕√n
(10–2)
with n − 1 degrees of freedom, where:
x is the sample mean. μ 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 292–294 in Chapter 9. We summarized this prob- lem in Chart 9–3 on page 294. Under these conditions, the correct statistical proce- dure 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/solution shows the details.
$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
ONE-SAMPLE TESTS OF HYPOTHESIS 335
S O L U T I O N
We will use the six-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:
H0: μ ≥ $60
H1: μ < $60 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 distri- bution. 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 histogram in the center of the following Minitab output. Observe the normal distribution superimposed on the frequency distribution.
We do not know the standard deviation of the population. So we substitute the sample standard deviation. The value of the test statis- tic is computed by formula (10–2):
t = x − μ s∕√n
A B C D E F G H I J K
1 Descriptive Statistics: Cost
2 Statistics
3 Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
4 Cost 26 0 56.423 1.969 10.041 38.000 48.750 56.500 63.000 78.000 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
Fr eq
ue nc
y
0 40 50 60 70 80
1
2
3
4
5
Histogram (with Normal Curve) of Cost
Cost
336 CHAPTER 10
Step 4: Formulate the decision rule. The critical values of t are given in Appendix B.5, a portion of which is shown in Table 10–1. Appendix B.5 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 num- ber 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 sample is 26, and we sampled 1 population, so there are 26 – 1 = 25 degrees of freedom.
TABLE 10–1 A Portion of the t Distribution Table
Confidence Intervals
80% 90% 95% 98% 99% 99.9%
Level of Significance for One-Tailed Test, α
df 0.10 0.05 0.025 0.01 0.005 0.0005
Level of Significance for Two-Tailed Test, α
0.20 0.10 0.05 0.02 0.01 0.001
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 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
ONE-SAMPLE TESTS OF HYPOTHESIS 337
To find the critical value, first locate the row with the appropriate de- grees 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 significance 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 re- gion 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. From the Minitab output, the mean cost per claim for the sample of 26 observations is $56.423. The standard deviation of this sample is $10.041. We insert these values in formula (10–2) and compute the value of t:
t = x − μ s∕√n
= $56.423 − $60
$10.041∕√26 = −1.816
Because –1.816 lies in the region to the right of the critical value of –2.485 (see Chart 10–6), the null hypothesis is not rejected at the .01 significance level.
CHART 10–6 Rejection Region, t Distribution, .01 Significance Level
0–1.816 Computed value of t
Region of rejection
–2.485 Critical value
Scale of t
a 5 .01
H0: m $ $60 H1: m , $60
df 5 26 2 1 5 25
Step 6: Interpret the result. We have not disproved the null hypothesis. The sample of claims could have been selected from a population with a mean cost of $60 per claim. To put it another way, the difference of $3.577 ($56.423 – $60.00) between the sample mean and the population mean could be due to sampling error. The test results do not allow the claims department manager to conclude that the cost- cutting measures have been effective.
E X A M P L E
The Myrtle Beach International Airport provides a cell phone parking lot where peo- ple can wait for a message to pick up arriving passengers. To decide if the cell phone lot has enough parking places, the manager of airport parking needs to
In the previous example, the mean and the standard deviation were computed using Minitab. The following example/solution shows the details when the sample mean and sample standard deviation are calculated from sample data.
338 CHAPTER 10
S O L U T I O N
We begin by stating the null hypothesis and the alternate hypothesis. In this case, the question is whether the population mean could be more than 15 minutes. So this is a one-tailed test. We state the two hypotheses as follows:
H0: μ ≤ 15
H1: μ > 15
There are 11 degrees of freedom, found by n − 1 = 12 − 1 = 11. The critical t value is 1.796, found by referring to Appendix B.5 for a one-tailed test, using α = .05 with 11 degrees of freedom. The decision rule is: Reject the null hypothesis if the computed t is greater than 1.796. This information is summarized in Chart 10–7.
0 1.796 Scale of t
Rejection region
a = .05
CHART 10–7 Rejection Region, One-Tailed Test, Student’s t Distribution, α = .05
We calculate the sample mean using formula (3–2) and the sample standard devi- ation using formula (3–8). The sample mean is 23 minutes, and the sample standard deviation is 9.835 minutes. The details of the calculations are shown in Table 10–2.
TABLE 10–2 Calculations of Sample Mean and Standard Deviation Parking Times
Customer x, Minutes (x − x )2
Chmura 30 49 Will 24 1 x =
Σx n
= 276 12
= 23 Crompton 28 25 Craver 22 1 Cao 14 81 s = √
Σ(x − x )2
n − 1 = √
1064 12 − 1
= 9.835 Nowlin 2 441 Esposito 39 256 Colvard 23 0 Hoefle 23 0 Lawler 28 25 Trask 12 121 Grullon 31 64 Total 276 1064
At the .05 significance level, is it reasonable to conclude that the mean time in the lot is more than 15 minutes?
know if the mean time in the lot is more than 15 minutes. A sample of 12 recent customers showed they were in the lot the following lengths of time, in minutes.
30 24 28 22 14 2 39 23 23 28 12 31
ONE-SAMPLE TESTS OF HYPOTHESIS 339
Now we are ready to compute the value of t, using formula (10–2).
t = x − μ s∕√n
= 23 − 15
9.835∕√12 = 2.818
The null hypothesis that the population mean is less than or equal to 15 min- utes is rejected because the computed t value of 2.818 lies in the area to the right of 1.796. We conclude that the time customers spend in the lot is more than 15 min- utes. This result indicates that the airport may need to add more parking places.
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.
S E L F - R E V I E W 10–3
9. Given the following hypotheses:
H0: μ ≤ 10 H1: μ > 10
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 signifi- cance 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 hypotheses:
H0: μ = 400 H1: μ ≠ 400
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 signifi- cance 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 text- book publishing 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 sam- ple 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 a mean time of 42.3 minutes 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?
E X E R C I S E S
340 CHAPTER 10
A Statistical Software Solution The Minitab statistical software system, used in earlier chapters and the previous sec- tion, provides an efficient method for conducting a one-sample test of hypothesis for a population mean. The steps to generate the following output are shown in Appendix C.
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 probability of a t value as extreme or more extreme than the computed t value, given that the null hypothesis is true. Using the Minitab analysis from the previous cell phone parking lot example, the p-value of .008 is the likelihood of a t value of 2.82 or larger, given a pop- ulation mean of 15. 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.
13. The mean income per person in the United States is $50,000, and the distribution of incomes follows a normal distribution. A random sample of 10 residents of Wilm- ington, Delaware, had a mean of $60,000 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 han- dle than paper ticketing. 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 random sample of 20 airports and col- lected information on the number of complaints the airport had with e-tickets for the month of March. The information is reported below.
14 14 16 12 12 14 13 16 15 14 12 15 15 14 13 13 12 13 10 13
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.
ONE-SAMPLE TESTS OF HYPOTHESIS 341
To explain further, refer to the diagram below. The p-value of .008 is the brown shaded area and the significance level is the total amber and brown shaded area. Because the p-value of .008 is less than the significance level of .05, the null hypothesis is rejected. Had the p-value been larger—say, .06, .19, or .57—than the significance level, the null hypothesis would not be rejected.
0 1.796 2.818 Scale of t
Rejection region
a = .05
In the preceding example, the alternate hypothesis was one-sided, and the upper (right) tail of the t distribution contained the rejection region. The p-value is the area to the right of 2.818 for a t distribution with 11 degrees of freedom.
What if we were conducting a two-sided test, so that the rejection region is in both the upper and the lower tails? That is, in the cell phone parking lot example, if H1 were stated as μ ≠ 15, we would have reported the p-value as the area to the right of 2.818 plus the value to the left of −2.818. Both of these values are .008, so the p-value is .008 + .008 = .016.
How can we estimate a p-value without a computer? To illustrate, recall that, in the example/solution regarding the length of time at the cell phone parking lot, we rejected the null hypothesis that μ ≤ 15 and accepted the alternate hypothesis that μ > 15. The significance level was .05, so logically the p-value is less than .05. To estimate the p-value more accurately, go to Appendix B.5 and find the row with 11 degrees of freedom. The computed t value of 2.818 is between 2.718 and 3.106. (A portion of Appendix B.5 is reproduced as Table 10–3.) The one-tailed significance level corre- sponding to 2.718 is .01, and for 3.106 it is .005. Therefore, the p-value is between .005 and .01. 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 .01.”
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, α
df 0.10 0.05 .0025 0.01 0.005 0.0005
Level of Significance for Two-Tailed Test, α
0.20 0.10 0.05 0.02 0.01 0.001
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 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
342 CHAPTER 10
A machine is set to fill a small bottle with 9.0 grams of medicine. A sample of eight bottles revealed the following amounts (grams) in each bottle.
9.2 8.7 8.9 8.6 8.8 8.5 8.7 9.0
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.
S E L F - R E V I E W 10–4
15. Given the following hypotheses:
H0: μ ≥ 20 H1: μ > 20
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 hypotheses:
H0: μ = 100 H1: μ ≠ 100
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 distribution 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 consumption in liters:
1.5 1.6 1.5 1.4 1.9 1.4 1.3 1.9 1.8 1.7
At the .01 significance level, can we conclude that water consumption has in- creased? 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, Hold- longer was added to the chlorine to find whether it would increase the shelf life. A sample of nine jugs of chlorine had these shelf lives (in hours):
2,159 2,170 2,180 2,179 2,160 2,167 2,171 2,181 2,185
At the .025 level, has Holdlonger increased the shelf life of the chlorine? Estimate the p-value.
E X E R C I S E S
ONE-SAMPLE TESTS OF HYPOTHESIS 343
TYPE II ERROR Recall that the level of significance, identified by the symbol α, 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.
In a hypothesis-testing situation there is also the possibility that a null hypothesis is not rejected when it is actually false. This is called a Type II error. The probability of a Type II error is identified by the Greek letter beta (β). In contrast to selecting a value for α in the hypothesis testing procedure, the value of β is calculated after the hypothesis testing procedure is finished. The following example illustrates the details of determin- ing the value of β.
LO10-7 Compute the probability of a Type II error.
19. A Washington, D.C., “think tank” announces the typical teenager sent 67 text messages per day in 2017. To update that estimate, you phone a sample of 12 teenagers and ask them how many text messages they sent the previous day. Their responses were:
51 175 47 49 44 54 145 203 21 59 42 100
At the .05 level, can you conclude that the mean number is greater than 67? Esti- mate 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 evaluate its effectiveness. The number of in-depth surveys conducted during a week by a random sample of 15 agents are:
53 57 50 55 58 54 60 52 59 62 60 60 51 59 56
At the .05 level of significance, can we conclude that the mean number of inter- views conducted by the agents is more than 53 per week? Estimate the p-value.
E X A M P L E
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, σ, is 400 psi. To monitor the quality of the cotter pins, samples of 100 pins are randomly selected and tested for their strength. In our hypothesis testing procedure the hypotheses are:
H0: μ = 10,000
H1: μ ≠ 10,000
To determine if a shipment of steel bars meets the quality standard, Western Wire Products set up a rule for the quality-control inspector to follow: “Take a sam- ple of 100 steel bars. Test each of the bars for tensile strength. Using a .05 signifi- cance level, accept the shipment if the sample mean ( x ) strength falls between 9,922 psi and 10,078 psi.” These values are the critical values for the hypothesis test. If the sample mean is more than 10,078 or less than 9,922, the hypothesis is rejected and we conclude that the shipment does not meet the quality standard.
Refer to Chart 10–8, Graph A. Given that the population mean is 10,000 psi, designated μ0, with a standard deviation of 400, the distribution shows the regions where the hypothesis is rejected and where it is not rejected, that is, whether the shipment meets the quality standard for tensile strength.
344 CHAPTER 10
Suppose that the result of testing 100 bars results in a sample mean of 9,900 psi. Clearly, based on Graph A, the shipment does not meet the quality stan- dard and is rejected with a .05 probability of a Type I error, a small chance of reject- ing the shipment in error. To calculate the Type II error, we assume that the sample mean of 9,900 psi is the true population mean. Compare Graph A and Graph B. Graph A represents the company’s distribution of tensile strength centered on 10,000 psi. Graph B, based on the sample data, suggests that the distribution is centered on 9,900 psi.
Now, let’s use Graph B to determine the probability of a Type II error, β. From the quality standards we know that 9,922 psi is used to reject the null hypothesis. Any sample mean greater than 9,922 and less than 10,078 is accepted. If the dis- tribution is really centered on 9,900 psi, it is possible to find sample means more than 9,922, and we would fail to reject the null hypothesis, μ = 10,000. This is the area in Graph B labeled “Probability of β” where the null hypothesis would not be rejected. The graph indicates that the probability of a Type II error is .2912. We can also calculate the power of the test as (1 − β). The power is the probability of not making a Type II error, rejecting the null hypothesis correctly. Here, the power of the test is .7088.
S O L U T I O N
The probability of committing a Type II error, is represented by the shaded area in Chart 10–8, Graph B. It is computed by determining the area under the normal curve that lies above 9,922 pounds. The calculation of the areas under the nor- mal curve was discussed in Chapter 7. Reviewing briefly, first determine the prob- ability of the sample mean falling between 9,900 and 9,922. Then this probability is subtracted from .5000 (which represents all the area to the right of the mean of 9,900) to arrive at the probability of making a Type II error.
0
9,922
Reject lot
Reject lot
10,07810,000 –1.96 x– 1.96 x–
Graph A
Graph B
psi
psi9,900 9,922
.2912.2088.5000
1
xc
CHART 10–8 Charts Showing Type I and Type II Errors
ONE-SAMPLE TESTS OF HYPOTHESIS 345
The number of standard errors (z value) between the mean of the incoming lot (9,900), designated by μ1, and xC, representing the critical value for 9,922, is com- puted by:
With n = 100 and σ = 400, the value of z is 0.55:
z = xc − μ1 σ∕√n
= 9,922 − 9,900
400∕√100 =
22 40
= 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, based on the company’s standard of 10,000 psi, when the sample suggests that the population mean is 9,900 psi and the shipment should be rejected.
Let’s use another example, illustrated in Chart 10–9. Suppose the testing of a sample of 100 bars results in a mean of 10,120 psi. Based on the company’s stan- dards, represented in Graph A, the shipment should be rejected. The sample mean of 10,120 is more than the critical value of 10,078. Based on the sample mean, Graph C shows the distribution of sample means based on the distribution cen- tered on the sample mean, 10,120.
If the distribution is centered on the sample mean of 10,120, there is a probability that a sample mean could be less than the upper critical value of 10,078. In Graph C, this is labeled, β. To find the probability, we calculate the z value of 10,078 in Graph C:
z = xc − μ1 σ∕√n
= 10,078 − 10,120
400∕√100 = −1.05
TYPE II ERROR z = xc − μ1 σ∕√n
(10–3)
CHART 10–9 Type I and Type II Errors (Another Example)
0
9,922
Rejection region
Rejection region
10,07810,000 –1.96 s– 1.96 sm
m
xx –
2 = .025a
2 = .025a
10,120 10,078
1
Probability of making a Type II
error Probability of not making a Type II error
b 1 2b
Region A
Region C
psi
psi
xc
0.1469 0.3531 0.5000
346 CHAPTER 10
The probability that z is less than −1.05 is .1469, found by .5000 − .3531. Therefore, β, or the probability of a Type II error, is .1469. The power of the test is 1.000 − .1469 or .8531. Caution: If the difference between μ0 and μ1 is relatively small, the probability of a Type II error may occur in both tails. This eventuality is not considered here.
Using the methods illustrated by Charts 10–8 Region B and 10–9 Region C, the probability of a Type II error can be determined for any sample mean used as an estimate of μ1.
To review, in hypothesis testing, we select the probability of making a Type I error, α. Naturally, we pick small probabilities, typically less than 0.10. The probabil- ity of a Type II error depends on the sample result. The larger the difference be- tween the hypothesized mean and the sample mean, the smaller the probability of a Type II error, failing to reject the null hypothesis when it should have been rejected.
For this example/solution, Table 10–4 shows the probabilities of a Type II error and the power of the test for selected values of μ1. The right column gives the prob- ability of not making a Type II error, which is also known as the power of a test.
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 10–9, Region C, represents the probability of falsely accepting 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?
S E L F - R E V I E W 10–5
21. Refer to Table 10–4 and the example just completed. With n = 100, σ = 400, xc = 9,922 and μ1 = 9,880, verify that the probability of a Type II error is .1469.
22. Refer to Table 10–4 and the example just completed. With n = 100, σ = 400, xc = 10,078, and μ1 = 10,100, verify that the probability of a Type II error is .2912.
23. The management of KSmall Industries is considering a new method of assembling a computer. The current assembling method requires a mean time of 60 minutes
E X E R C I S E S
TABLE 10–4 Probabilities of a Type II Error for μ0 = 10,000 Pounds and Selected Alternative Means, .05 Level of Significance
Selected Probability of Power of a Mean (μ1) Type II Error (β) Test (1 − β) 9,820 .0054 .9946 9,880 .1469 .8531 9,900 .2912 .7088 9,940 .6736 .3264 10,000 — * — 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 μ1 = μ0.
ONE-SAMPLE TESTS OF HYPOTHESIS 347
with a standard deviation of 2.7 minutes. Using the new method, the mean assem- bly time for a random sample of 24 computers was 58 minutes. Using the .10 level of significance, can we conclude that the assembly time using the new method is faster? What is the probability of a Type II error?
24. A recent national survey found that parents read an average (mean) of 10 books per month to their children under five years old. The population standard deviation is 5. The distribution of books read per month follows the normal distribution. A random sample of 25 households revealed that the mean number of books read last month was 12. At the .01 significance level, can we conclude that parents read more than the average number of books to their children? What is the probability of a Type II error?
C H A P T E R S U M M A R Y
I. The objective of hypothesis testing is to verify the validity of a statement about a popula- tion 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 or probability of rejecting a true null hypothesis.
2. The most frequently used probabilities used as significance levels are .01, .05, and .10. As a probability, any value between 0 and 1.00 is possible, but we prefer small probabilities of making a Type I error.
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 popula- tion 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, and make a decision regard- ing the null hypothesis.
F. Interpret the results of your decision. 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, σ, is known, the test statistic is the standard nor- mal distribution and is determined from:
z = x − μ σ∕√n
(10–1)
B. If the population standard deviation is not known, s is substituted for σ. The test statis- tic is the t distribution, and its value is determined from:
t = x − μ s∕√n
(10–2)
348 CHAPTER 10
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. 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 α.
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 β. 2. The likelihood of a Type II error must be calculated comparing the hypothesized
distribution to an alternate distribution based on sample results.
P R O N U N C I A T I O N K E Y
SYMBOL MEANING PRONUNCIATION
H0 Null hypothesis H sub zero
H1 Alternate hypothesis H sub one
α/2 Two-tailed significance level Alpha divided by 2 xc Limit of the sample mean x bar sub c
μ0 Assumed population mean mu sub zero
C H A P T E R E X E R C I S E S
25. 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 reasonable to conclude that the mean income is not equal to $45,000? Deter- mine the p-value.
26. 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 the packaging process is 3 pounds per bag. At the end of each day, Jeff Rutter, the production manager, weighs 10 bags and computes the mean weight of the sample. Below are the weights of 10 bags from today’s production.
45.6 47.7 47.6 46.3 46.2 47.4 49.2 55.8 47.5 48.5
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.
27. A new weight-watching company, Weight Reducers International, advertises that those who join will lose an average of 10 pounds after the first two weeks. The standard devi- ation is 2.8 pounds. A random sample of 50 people who joined the weight reduction program revealed a mean loss of 9 pounds. At the .05 level of significance, can we conclude that those joining Weight Reducers will lose less than 10 pounds? Determine the p-value.
28. 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 ounce. The quality-con- trol department took a random sample of 50 cans and found that the arithmetic mean weight was 16.05 ounces. At the 5% level of significance, can we conclude that the mean weight is greater than 16 ounces? Determine the p-value.
ONE-SAMPLE TESTS OF HYPOTHESIS 349
29. According to a recent survey, Americans get a mean of 7 hours of sleep per night. A random sample of 50 students at West Virginia University revealed the mean length of time slept last night was 6 hours and 48 minutes (6.8 hours). The standard deviation of the sample was 0.9 hour. At the 5% level of significance, is it reasonable to conclude that students at West Virginia sleep less than the typical American? Compute the p-value.
30. A statewide real estate sales agency, Farm Associates, specializes in selling farm prop- erty 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 recently sold farms revealed a mean selling time of 94 days, with a standard deviation of 22 days. At the .10 significance level, has there been an increase in selling time?
31. 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 num- ber 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?
32. 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 significance level. Determine the p-value and explain its meaning.
33. A recent survey by nerdwallet.com indicated Americans paid a mean of $6,658 interest on credit card debt in 2017. A sample of 12 households with children revealed the follow amounts. At the .05 significance level is it reasonable to conclude that these households paid more interest?
7077 5744 6753 7381 7625 6636 7164 7348 8060 5848 9275 7052
34. A recent article in The Wall Street Journal reported that the home equity loan rate is now less than 4%. A sample of eight small banks in the Midwest revealed the follow- ing home equity loan rates (in percent):
3.6 4.1 5.3 3.6 4.9 4.6 5.0 4.4
At the .01 significance level, can we conclude that the home equity loan rate for small banks is less than 4%? Estimate the p-value.
35. A recent study revealed the typical American coffee drinker consumes an average of 3.1 cups per day. A sample of 12 senior citizens revealed they consumed the follow- ing amounts of coffee, reported in cups, yesterday.
3.1 3.3 3.5 2.6 2.6 4.3 4.4 3.8 3.1 4.1 3.1 3.2
At the .05 significance level, do these sample data suggest there is a difference be- tween the national average and the sample mean from senior citizens?
36. The postanesthesia care area (recovery room) at St. Luke’s Hospital in Maumee, Ohio, was recently enlarged. The hope was that the change would increase the mean number of patients served per day to more than 25. A random sample of 15 days re- vealed the following numbers of patients.
25 27 25 26 25 28 28 27 24 26 25 29 25 27 24
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.
350 CHAPTER 10
37. www.golfsmith.com receives an average of 6.5 returns per day from online shop- pers. For a sample of 12 days, it received the following numbers of returns.
0 4 3 4 9 4 5 9 1 6 7 10
At the .01 significance level, can we conclude the mean number of returns is less than 6.5? 38. 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 sample 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 min- utes is reported at 2.40 hours.)
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
Can we conclude that the mean time for a game is less than 3.50 hours? Use the .05 significance level.
39. 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.
−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
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.
40. Listed below is the annual rate of return (reported in percent) for a sample of 12 taxable mutual funds.
4.63 4.15 4.76 4.70 4.65 4.52 4.70 5.06 4.42 4.51 4.24 4.52
Using the .05 significance level, is it reasonable to conclude that the mean rate of return is more than 4.50%?
41. Many grocery stores and large retailers such as Kroger and Walmart 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 number of customers using the service for a sample of 15 days at a Walmart location.
120 108 120 114 118 91 118 92 104 104 112 97 118 108 117
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.
42. For a recent year, the mean fare to fly from Charlotte, North Carolina, to Chicago, Illinois, on a discount ticket was $267. A random sample of 13 round-trip discount fares on this route last month shows:
$321 $286 $290 $330 $310 $250 $270 $280 $299 $265 $291 $275 $281
At the .01 significance level, can we conclude that the mean fare has increased? What is the p-value?
43. The publisher of Celebrity Living claims that the mean sales for personality magazines that feature people such as Megan Fox or Jennifer Lawrence 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. Do these data contradict the publisher’s claim? Use the 0.01 significance level.
ONE-SAMPLE TESTS OF HYPOTHESIS 351
44. 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.
45. The number of “destination weddings” has skyrocketed in recent years. For example, 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 $30,000. Listed below is a total cost in $000 for a sample of 8 Caribbean weddings.
29.7 29.4 31.7 29.0 29.1 30.5 29.1 29.8
At the .05 significance level, is it reasonable to conclude the mean wedding cost is less than $30,000 as advertised?
46. The American Water Works Association reports that the per capita water use in a sin- gle-family home is 69 gallons per day. Legacy Ranch is a relatively new housing devel- opment. The builders installed more efficient water fixtures, such as low-flush toilets, and subsequently conducted a survey of the residences. Thirty-six owners responded, and the sample mean water use per day was 64 gallons with a standard deviation 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?
47. A cola-dispensing machine is set to dispense 9.00 ounces of cola per cup, with a standard deviation of 1.00 ounce. The manufacturer of the machine would like to set the control limit in such a way that, for samples of 36, 5% of the sample means will be greater than the upper control limit, and 5% of the sample means will be less than the lower control limit. a. At what value should the control limit be set? b. If the population mean shifts to 8.6, what is the probability of detecting the change? c. If the population mean shifts to 9.6, what is the probability of detecting the change?
48. 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 hour at the mall, with a standard deviation of 0.10 hour. 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 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 hour. a. Develop a hypothesis test to determine if the mean time spent in the mall changed.
Use the .10 significance level. b. Suppose the mean shopping time actually increased from 0.75 hour to 0.79 hours.
What is the probability of making a Type II error? c. When Brunner and Swanson reported the information in part (b) to the mall owners,
the owners believed that the probability of making a Type II error was too high. How could this probability be reduced?
49. The following null and alternate hypotheses are given.
H0: μ ≤ 50 H1: μ > 50
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?
50. An insurance company, based on past experience, estimates the mean damage for a natural disaster in its area is $5,000. After introducing several plans to prevent loss, it randomly 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.
51. A national grocer’s magazine reports the typical shopper spends 8 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.
352 CHAPTER 10
D A T A A N A L Y T I C S
52. The North Valley Real Estate data reports information on the homes sold last year. a. Adam Marty recently joined North Valley Real Estate and was assigned twenty homes
to market and show. When he was hired, North Valley assured him that the twenty homes would be fairly assigned to him. When he reviewed the selling prices of his assigned homes, he thought that the prices were much below the average of $357,000. Adam was able to find the data of how the other agents in the firm were assigned to the homes. Use statistical inference to analyze the “fairness” that homes were assigned to the agents.
53. Refer to the Baseball 2016 data, which report information on the 30 Major League Baseball teams for the 2016 season. a. Conduct a test of hypothesis to determine whether the mean salary of the teams was
different from $100.0 million. Use the .05 significance level. b. Using a 5% significance level, conduct a test of hypothesis to determine whether the
mean attendance was more than 2,000,000 per team. 54. Refer to the Lincolnville School District bus data.
a. Select the variable for the number of miles traveled last month. Conduct a hypothesis test to determine whether the mean miles traveled last month equals 10,000. Use the .01 significance level. Find the p-value and explain what it means.
b. A study of school bus fleets reports that the average per bus maintenance cost is $4,000 per year. Using the maintenance cost variable, conduct a hypothesis test to determine whether the mean maintenance cost for Lincolnville’s bus fleet is more than $4,000 at the .05 significance level. Determine the p-value and report the results.
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO11-1 Test a hypothesis that two independent population means are equal, assuming that the population standard deviations are known and equal.
LO11-2 Test a hypothesis that two independent population means are equal, with unknown population standard deviations.
LO11-3 Test a hypothesis about the mean population difference between paired or dependent observations.
LO11-4 Explain the difference between dependent and independent samples.
Two-Sample Tests of Hypothesis 11
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 conclude that babies using the Gibbs brand gained less weight? (See Exercise 3 and LO11-1.)
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354 CHAPTER 11
INTRODUCTION Chapter 10 began our study of hypothesis testing. We described the nature of hypoth- esis 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 random sample from a population and conducted a test of whether the proposed population value was rea- sonable. Recall in Chapter 10 that we selected a sample of the number of desks assem- bled per week at Jamestown Steel Company to determine whether there was a change
in the production rate. Similarly, we sam- pled the cost to process insurance claims to determine if cost-cutting measures resulted in a mean less than the current $60 per claim. In both cases, we compared the re- sults of a single sample statistic to a popula- tion parameter.
In this chapter, we expand the idea of hypothesis testing to two populations. That is, we select random samples from two dif- ferent populations to determine whether the population means 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. At Grabit Software, Inc., do customer service employees receive more calls for assis- tance during the morning or afternoon?
3. In the fast-food industry, is there a difference in the mean number of days absent be- tween young workers (under 21 years of age) and older workers (more than 60 years of age)?
4. 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.
© John Lund/Drew Kelly/Blend Images LLC RF
TWO-SAMPLE TESTS OF HYPOTHESIS: INDEPENDENT SAMPLES A city planner in Tampa, Florida wishes to know whether there is a difference in the mean hourly wage rate of plumbers and electricians in central Florida. A financial accountant wishes to know whether the mean rate of return for domestic, U.S., 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, domestic, U.S., mutual funds are one population and global mutual funds, the other.
To investigate the question in each of these cases, 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 plumb- ers and the electricians, we would expect the difference between the two sample means to be zero. But what if our sample results yield a difference other than 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.
Return to the results of Chapter 8. Recall that we showed that a distribution of sam- ple means would tend to approximate the normal distribution. We need to again assume that a distribution of sample means will follow the normal distribution. It can be shown
LO11-1 Test a hypothesis that two independent population means are equal, assuming that the population standard deviations are known and equal.
TWO-SAMPLE TESTS OF HYPOTHESIS 355
mathematically that the distribution of the differences 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 difference 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 differ- ence between the two sample means to be zero. If there is a difference 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 dis- tribution of the differences in the sample means. Because of the Central Limit Theorem in Chapter 8, we know that the distribution of the sample 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 differences will also follow the normal distri- bution. 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 differences 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
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 Mean Hourly Earnings of 20 Random Samples of Plumbers and Electricians and the Differences between the Means
356 CHAPTER 11
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 be- tween the sample means is $1.04. This process was repeated 19 more times. Observe that in 17 of the 20 cases, the differences are positive because the mean of the plumb- ers is larger than the mean of the electricians. In two cases, the differences are negative because the mean of the electricians is larger than the mean of the plumbers. In one case, the means are equal.
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 independent populations, as in this case, the distribution of the differences has a variance (standard deviation squared) equal to the sum of the two individual variances. This means that we can add the variances of the two sampling distributions. To put it another way, the vari- ance of the difference in sample means (x1 − x2) is equal to the sum of the variance for the plumbers and the variance for the electricians.
σ2 x1−x2 =
σ21 n1
+ σ22 n2
(11–1) VARIANCE OF THE DISTRIBUTION OF DIFFERENCES IN MEANS
The term σ2 x1−x2 looks complex but need not be difficult to interpret. The σ2 portion
reminds us that it is a variance, and the subscript x1 − x2 that it is a distribution of differ- ences 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 or “standard error” of the distribution of differences. Finally, we standardize the distribution of the differences. The result is the following equation.
z = x1 − x2
√ σ21 n1
+ σ 2 2
n2
(11–2)TWO-SAMPLE TEST OF MEANS—KNOWN σ
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 are unrelated, that is, independent. • The standard deviations for both populations are known.
The following example shows the details of the test of hypothesis for two popula- tion means and shows how to interpret the results.
E X A M P L E
Customers at the FoodTown Super- market have a choice when paying for their groceries. 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 FoodTown employee scans each item and puts it on a short conveyor, where another employee puts it in a bag and then into the gro- cery cart. In the Fast Lane procedure, © Teschner/Agencja Fotograficzna Caro/Alamy Stock Photo
TWO-SAMPLE TESTS OF HYPOTHESIS 357
the customer scans each item, bags it, and places the bags in the cart him- or her- self. The Fast Lane procedure is designed to reduce the time a customer spends in the checkout line.
The Fast Lane facility was recently installed at the Byrne Road FoodTown loca- tion. 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 all his or her bags are in the cart. Hence the time includes both waiting in line and checking out. What is the p-value?
Population Customer Type Sample Size Sample Mean Standard Deviation
Standard 50 5.50 minutes 0.40 minute Fast Lane 100 5.30 minutes 0.30 minute
S O L U T I O N
We use the six-step hypothesis-testing procedure to investigate the question.
Step 1: State the null hypothesis and the alternate hypothesis. The null hy- pothesis is that the mean standard checkout time is less than or equal to the mean Fast Lane checkout time. In other words, the difference of 0.20 minute 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 μS refer to the mean checkout time for the population of standard customers and μF the mean checkout time for the Fast Lane customers. The null and alter- native hypotheses are:
H0: μS ≤ μF H1: μS > μF
Step 2: Select the level of significance. The significance level is the probabil- ity that we reject the null hypothesis when it is actually true. This like- lihood is determined prior to selecting the sample or performing any calculations. The .05 and .01 significance levels are the most com- mon, 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 normal 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 populations 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 significance, 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 indi- cate that the mean checkout time is longer for those using the stan- dard 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, go to Student’s t distribution
358 CHAPTER 11
(Appendix B.5). In the table headings, find the row labeled “Level of Significance for One-Tailed Test” and select the column for an alpha of .01. Go to the bottom row with infinite degrees of freedom. The z critical value is 2.326. So the decision rule is to reject the null hypothesis if the value of the test statistic exceeds 2.326. Chart 11–1 depicts the decision rule.
Step 5: Make the decision regarding H0. FoodTown randomly selected 50 customers using the standard checkout and computed a sample mean checkout time of 5.5 minutes, and 100 customers using the Fast Lane checkout and computed a sample mean checkout time of 5.3 minutes. We assume that the population standard deviations for the two methods is known. We use formula (11-2) to compute the value of the test statistic.
z = xS − xF
√ σ 2S nS
+ σ 2F nF
= 5.5 − 5.3
√ 0.402
50 +
0.302
100
= 0.2
0.064031 = 3.123
The computed value of 3.123 is larger than the critical value of 2.326. Our decision is to reject the null hypothesis and accept the alternate hypothesis.
Step 6: Interpret the result. The difference of .20 minute between the mean checkout times is too large to have occurred by chance. 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 3.123. From Appendix B.3, we cannot find the probability associated with 3.123. 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 .5000 − .4990. We conclude that there is very little likelihood that the null hypothesis is true! The checkout time is less using the fast lane.
CHART 11–1 Decision Rule for One-Tailed Test at .01 Significance Level
.5000 .4900
H0: mS # mF H1: mS . mF
2.326 Critical value
Scale of z
Rejection region
.01
0
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 customers. For ex- ample, Mr. Smith’s checkout time does not affect any other customer’s checkout time.
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 num- ber 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.
TWO-SAMPLE TESTS OF HYPOTHESIS 359
2. Both populations follow the normal distribution. In the FoodTown example, the popu- lation of times in both the standard checkout line and the Fast Lane follow normal distributions.
3. Both population standard deviations are known. In the FoodTown example, the popu- lation standard deviation of the Fast Lane times was 0.30 minute. The population stan- dard deviation of the standard checkout times was 0.40 minute.
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.
S E L F - R E V I E W 11–1
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 signifi- cance level.
H0: μ1 = μ2 H1: μ1 ≠ μ2
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?
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.
H0: μ1 ≤ μ2 H1: μ1 > μ2
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 six-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 re- vealed 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
E X E R C I S E S
360 CHAPTER 11
COMPARING POPULATION MEANS WITH UNKNOWN POPULATION STANDARD DEVIATIONS In the previous section, we used the standard normal distribution and z as the test sta- tistic to test a hypothesis that two population means from independent populations were equal. The hypothesis tests presumed that the populations were normally distrib- uted and that we knew the population standard deviations. However, in most cases, we do not know the population standard deviations. 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 (σ). See formula (10–2) on page 334.
Two-Sample Pooled Test In this section, we describe 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
LO11-2 Test a hypothesis that two independent population means are equal, with unknown population standard deviations.
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 conclude 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 deviations for the Cincinnati and Pittsburgh em- ployees 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. Do married and unmarried women spend the same amount of time per week using Facebook? A random sample of 45 married women who use Facebook spent an average of 3.0 hours per week on this social media website. A random sample of 39 unmarried women who regularly use Facebook spent an average of 3.4 hours per week. Assume that the weekly Facebook time for married women has a popu- lation standard deviation of 1.2 hours, and the population standard deviation for unmarried, regular Facebook users is 1.1 hours per week. Using the .05 signifi- cance level, do married and unmarried women differ in the amount of time per week spent on Facebook? Find the p-value and interpret the result.
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 unionized seem to offer higher wages. She decided to investigate and gathered the following information.
Sample Population Group Sample Size Mean Wage Standard Deviation
Union 40 $20.75 $2.25 Nonunion 45 $19.80 $1.90
Would it be reasonable for her to conclude that union nurses earn more? Use the .02 significance level. What is the p-value?
TWO-SAMPLE TESTS OF HYPOTHESIS 361
investigating the difference in sample means. There are two major differences in this test and the previous test described in this chapter.
1. We assume the sampled populations have equal but unknown standard deviations. 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 formula (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 de- grees of freedom that each sample provides. Why do we need to pool the sample stan- dard deviations? Because we assume that the two populations 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.
where: s21 is the variance (standard deviation squared) of the first sample. s22 is the variance of the second sample.
The value of t is computed from the following equation.
POOLED VARIANCE s2p = (n1 − 1)s21 + (n2 − 1)s22
n1 + n2 − 2 (11–3)
t = x1 − x2
√s 2 p (
1 n1
+ 1
n2)
(11–4)TWO-SAMPLE TEST OF MEANS— UNKNOWN σ′S
where: x1 is the mean of the first sample. x2 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. s2p is the pooled estimate of the population variance.
The number of degrees of freedom in the test is the total number of items sampled mi- nus 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 are approximately normally distributed. 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.
E X A M P L E
Owens Lawn Care Inc. manufactures and assembles lawnmowers that are shipped to dealers throughout the United States and Canada. Two different procedures have been proposed for mounting the engine on the frame of the lawnmower. The question is: Is there a difference in the mean time to mount the engines on the
362 CHAPTER 11
frames of the lawnmowers? The first procedure was developed by longtime Owens employee Herb Welles (designated as procedure W), and the other procedure was developed by Owens Vice President of Engineering William Atkins (designated as procedure A). To evaluate the two methods, we conduct a time and motion study. A sample of five employees is 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.
Welles Atkins (minutes) (minutes)
2 3 4 7 9 5 3 8 2 4 3
S O L U T I O N
Following the six 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.
H0: μW = μA H1: μW ≠ μA
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 nW + nA − 2. Five assem- blers 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 Appendix B.5 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 11–2. We do not reject the null hypothesis if the computed value of t falls between −1.833 and 1.833.
0
Rejection region
.05
Rejection region
.05
–1.833 Critical value
Scale of t
Do not reject H0
1.833 Critical value
H0: mW 5 mA H1: mW Þ mA
CHART 11–2 Regions of Rejection, Two-Tailed Test, df = 9, and .10 Significance Level
TWO-SAMPLE TESTS OF HYPOTHESIS 363
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 (3–9). See the details below.
Welles Method Atkins Method
xW (xW − xW )2 xA (xA − xA)2
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
xW = ΣxW nW
= 20 5
= 4 xA = ΣxA nA
= 30 6
= 5
sW = √ Σ(xW − xW)2
nW − 1 = √
34 5 − 1
= 2.9155 sA = √ Σ(xA − xA)2
nA − 1 = √
22 6 − 1
= 2.0976
Step 2: Pool the sample variances. We use formula (11–3) to pool the sam- ple variances (standard deviations squared).
s2p = (nW − 1)s2W + (nA − 1)s2A
nW + nA − 2 =
(5 − 1) (2.9155)2 + (6 − 1) (2.0976)2
5 + 6 − 2 = 6.2222
Step 3: Determine the value of t. The mean mounting time for the Welles method is 4.00 minutes, found by xW = 20∕5. The mean mounting time for the Atkins method is 5.00 minutes, found by xA = 30∕6. We use formula (11–4) to calculate the value of t.
t = xW − xA
√s 2 p (
1 nW
+ 1
nA)
= 4.00 − 5.00
√6.2222 ( 1 5
+ 1 6)
= −0.662
The decision is not to reject the null hypothesis because −0.662 falls in the region between −1.833 and 1.833. Our conclusion is that the sample data failed to show a difference between the mean assembly times of the two methods.
We also can estimate the p-value using Appendix B.5. Locate the row with 9 degrees of freedom, and use the two-tailed test column. Find the t value, without regard to the sign, that is closest to our computed value of 0.662. It is 1.383, corre- sponding to a significance level of .20. Thus, even had we used the 20% signifi- cance level, we would not have rejected the null hypothesis of equal means. We can report that the p-value is greater than .20.
Excel has a procedure called “t-Test: Two Sample Assuming Equal Variances” that will perform the calculations of formulas (11–3) and (11–4) as well as find the sample means and sample variances. The details of the procedure are provided in Appendix C. 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.
364 CHAPTER 11
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?
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 re- vealed the following number of defects.
S E L F - R E V I E W 11–2
Day 5 8 7 6 9 7 Afternoon 8 10 7 11 9 12 14 9
For Exercises 7 and 8: (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.
7. The null and alternate hypotheses are:
H0: μ1 = μ2 H1: μ1 ≠ μ2
A random sample of 10 observations from one population revealed a sample mean of 23 and a sample standard 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 pop- ulation means?
8. The null and alternate hypotheses are:
H0: μ1 = μ2 H1: μ1 ≠ μ2
E X E R C I S E S
TWO-SAMPLE TESTS OF HYPOTHESIS 365
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 obser- vations 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 six-step hypothesis testing procedure for the following exercises.
9. Listed below are the 25 players on the opening-day roster of the 2016 New York Yankees Major League Baseball team, their salaries, and fielding positions.
Player Position Salary (US$)
C.C. Sabathia Starting Pitcher $25,000,000 Mark Teixeira First Base $23,125,000 Masahiro Tanaka Starting Pitcher $22,000,000 Jacoby Ellsbury Center Field $21,142,857 Alex Rodriguez Designated Hitter $21,000,000 Brian McCann Catcher $17,000,000 Carlos Beltran Right Field $15,000,000 Brett Gardner Left Field $13,500,000 Chase Headley Third Base $13,000,000 Andrew Miller Relief Pitcher $ 9,000,000 Starlin Castro Second Base $ 7,857,142 Nathan Eovaldi Starting Pitcher $ 5,600,000 Michael Pineda Starting Pitcher $ 4,300,000 Ivan Nova Relief Pitcher $ 4,100,000 Dustin Ackley Left Field $ 3,200,000 Didi Gregorius Shortstop $ 2,425,000 Aaron Hicks Center Field $ 574,000 Austin Romine Catcher $ 556,000 Chasen Shreve Relief Pitcher $ 533,400 Luis Severino Starting Pitcher $ 521,300 Kirby Yates Relief Pitcher $ 511,900 Ronald Torreyes Second Base $ 508,600 Johnny Barbato Relief Pitcher $ 507,500 Dellin Betances Relief Pitcher $ 507,500 Luis Cessa Relief Pitcher $ 507,500
Sort the players into two groups, all pitchers (relief and starting) and position play- ers (all others). Assume equal population standard deviations for the pitchers and the position players. Test the hypothesis that mean salaries of pitchers and position players are equal using the .01 significance level.
10. A recent study compared the time spent together by single- and dual-earner cou- ples. According to the records kept by the wives during the study, the mean amount of time spent together watching television among the single-earner cou- ples was 61 minutes per 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 deviation 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.
366 CHAPTER 11
11. 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 col- lected the following sample information.
Sales ($) 131 135 146 165 136 142 Audit ($) 130 102 129 143 149 120 139
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?
12. 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 collected the following information on the amounts earned last week by a sample of school teachers and a sample of nurses.
School Teachers ($) 1,095 1,076 1,077 1,125 1,034 1,059 1,052 1,070 1,079 1,080 1.092 1,082 Nurses ($) 1,091 1,140 1,071 1,021 1,100 1,109 1,075 1,079
Is it reasonable to conclude that the mean weekly salary of nurses is higher? Use the .01 significance level. What is the p-value?
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 standard deviations, but we assumed they were equal. In many cases, this is a reasonable as- sumption, but what if it is not? In the next chapter, we present a formal method to test the assumption of equal variances. If the variances are not equal, we describe a test of hypothesis that does not require either the equal variance or the normality assumption in Chapter 16.
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 for- mula. 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:
t = x1 − x2
√ s21 n1
+ s22 n2
(11–5)TEST STATISTIC FOR NO DIFFERENCE IN MEANS, UNEQUAL VARIANCES
df = [(s21 ∕n1) + (s
2 2∕n2)]
2
(s21 ∕n1) 2
n1 − 1 +
(s22∕n2) 2
n2 − 1
(11–6)DEGREES OF FREEDOM FOR UNEQUAL VARIANCE TEST
The degrees of freedom statistic is found by:
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.
TWO-SAMPLE TESTS OF HYPOTHESIS 367
E X A M P L E
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 2 minutes, and then evaluate the amount of liquid the paper has taken up from the vat. A random sample of nine 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.
S O L U T I O N
To begin, let’s assume that the amounts of liquid absorbed follow the normal probability 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 distribution as the test statistic. The assumption of equal population stan- dard deviations 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 ab- sorption 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 ob- serve the difference in the variation in the following dot plot provided by Minitab. The software commands to create a Minitab dot plot are given in Appendix C, Chapter 4, 4-1.
So we decide to use the t distribution and assume that the population standard deviations are not the same.
In the six-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.
H0: μ1 = μ2 H0: μ1 ≠ μ2
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 (11–6). To do so, we need to find the sample standard deviations. We can use statistical software to quickly find these results.
368 CHAPTER 11
The respective sample sizes are n1 = 9 and n2 = 12 and the respective standard deviations are 3.321 ml and 1.621 ml.
Variable n Mean Standard Deviation
Store 9 6.444 3.321 Name 12 9.417 1.621
Inserting this information into formula (11–6):
df = [(s21 ∕n1) + (s22∕n2)]
2
(s21 ∕n1)2
n1 − 1 +
(s22∕n2)2
n2 − 1
= [(3.3212∕9) + (1.6212∕12)]2
(3.3212∕9)2
9 − 1 +
(1.6212∕12)2
12 − 1
= 1.44442
.1877 + .0044 = 10.86
The usual practice is to round down to the integer, so we use 10 degrees of free- dom. From Appendix B.5 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 (11–5). Recall that the mean amount of absorption for the store paper towels is 6.444 ml and 9.417 ml for the brand.
t = x1 − x2
√ s21 n1
+ s22 n2
= 6.444 − 9.417
√ 3.3212
9 +
1.6212
12
= −2.474
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.
For this analysis there are many calculations. Statistical software often provides an option to compare two population means with different standard deviations. The Minitab output for this example follows.
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 his or her own or was contacted by a telemarketer. The company obtained the following sample information regarding end-of-the-month balances for the two groups.
S E L F - R E V I E W 11–3
TWO-SAMPLE TESTS OF HYPOTHESIS 369
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 Sample Size Mean Standard Deviation
Applied 10 $1,568 $356 Contacted 8 1,967 857
For exercises 13 and 14, assume the sample populations do not have equal standard deviations 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 decision about the null hypothesis.
13. The null and alternate hypotheses are:
H0: μ1 = μ2 H1: μ1 ≠ μ2
A random sample of 15 items from the first population showed a mean of 50 and a standard deviation of 5. A sample of 12 items for the second population showed a mean of 46 and a standard deviation of 15.
14. The null and alternate hypotheses are:
H0: μ1 ≤ μ2 H1: μ1 > μ2
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.
15. A recent survey compared the costs of adoption through public and private agen- cies. For a sample of 16 adoptions through a public agency, the mean cost was $21,045, with a standard deviation of $835. For a sample of 18 adoptions through a private agency, the mean cost was $22,840, with a standard deviation of $1,545. Can we conclude the mean cost is larger for adopting children through a private agency? Use the .05 significance level.
16. Suppose you are an expert on the fashion industry and wish to gather infor- mation 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 15 Claiborne models:
E X E R C I S E S
$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
$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 12 Klein models.
Is it reasonable to conclude that Claiborne models earn more? Use the .05 signifi- cance level and assume the population standard deviations are not the same.
370 CHAPTER 11
TWO-SAMPLE TESTS OF HYPOTHESIS: DEPENDENT SAMPLES In the Owens Lawn Care example/solution on page 361, we tested the difference between the means from two independent populations. 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 assem- bly 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 Savings 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 important that these two firms be similar in their appraisal values. To review the consistency of the two appraisal firms, Nickel Savings randomly selects 10 homes and has both Schadek Appraisals and Bowyer Real Estate appraise the values 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 re- ferred 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 larger appraisal values, then the mean of the distribution of the differences will not be 0.
We will use the symbol μd to indicate the population mean of the distribution of dif- ferences. We assume the distribution of the population of differences is approximately normally distributed. The test statistic follows the t distribution and we calculate its value from the following formula:
LO11-3 Test a hypothesis about the mean population difference between paired or dependent observations.
© Photodisc/Getty Images
PAIRED t TEST t = d
sd ∕√n (11–7)
There are n − 1 degrees of freedom and
d is the mean of the difference between the paired or related observations. sd is the standard deviation of the differences between the paired or related
observations. n is the number of paired observations.
The standard deviation of the differences is computed by the familiar formula for the standard deviation [see formula (3–9)], except d is substituted for x. The formula is:
sd = √ Σ (d − d)2
n − 1
The following example illustrates this test.
TWO-SAMPLE TESTS OF HYPOTHESIS 371
E X A M P L E
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
A 235 228 B 210 205 C 231 219 D 242 240 E 205 198 F 230 223 G 231 227 H 210 215 I 225 222 J 249 245
At the .05 significance level, can we conclude there is a difference between the firms’ mean appraised home values?
S O L U T I O N
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 firms’ appraised values. We are not interested in show- ing 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 between the two firms’ appraised values. The null and alternate hypotheses are:
H0: μd = 0 H1: μd ≠ 0
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.5 and 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 in- tersection 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.
Home Schadek Bowyer Difference, d (d − d) (d − d)2
A 235 228 7 2.4 5.76 B 210 205 5 0.4 0.16 C 231 219 12 7.4 54.76 D 242 240 2 −2.6 6.76 E 205 198 7 2.4 5.76 F 230 223 7 2.4 5.76 G 231 227 4 −0.6 0.36 H 210 215 −5 −9.6 92.16 I 225 222 3 −1.6 2.56 J 249 245 4 −0.6 0.36 46 0 174.40
372 CHAPTER 11
d = Σd n
= 46 10
= 4.60
sd = √ Σ (d − d )2
n − 1 = √
174.4 10 − 1
= 4.402
Using formula (11–7), the value of the test statistic is 3.305, found by
t = d
sd ∕√n =
4.6 4.402∕√10
= 4.6
1.3920 = 3.305
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 between the firms’ mean appraised home values. 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.5 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.
TABLE 11–2 A Portion of the t Distribution from Appendix B.5
Excel’s statistical analysis software has a procedure called “t-Test: Paired Two- Sample for Means” that will perform the calculations of formula (11–7). The output from this procedure is given below.
The computed value of t is 3.305, and the two-tailed p-value is .009. Be- cause the p-value is less than .05, we reject the hypothesis that the mean of the distribution 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.
TWO-SAMPLE TESTS OF HYPOTHESIS 373
COMPARING DEPENDENT AND INDEPENDENT SAMPLES Beginning students are often confused by the difference between tests for independent samples [formula (11–4)] and tests for dependent samples [formula (11–7)]. 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 obser- vations. 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 sooth- ing 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 pro- duction 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 summer between the junior and senior year, they participate in the course that gives 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 measurement (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. The previous example/solution regarding Nickel Savings illus- trates dependent samples. A property is selected and both firms appraise 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.
LO11-4 Explain the difference between dependent and independent samples.
374 CHAPTER 11
Why do we prefer dependent samples to independent samples? By using dependent samples, we are able to reduce the variation in the sampling distribution. To illustrate, we will use the Nickel Savings and Loan example/solution just completed. Suppose we assume that we have two independent samples of real estate property for appraisal and conduct the following test of hypothesis, using formula (11–4). The null and alternate hypotheses are:
H0: μ1 = μ2 H1: μ1 ≠ μ2
There are now two independent samples of 10 each. So the number of degrees of freedom is 10 + 10 − 2 = 18. From Appendix B.5, for the .05 significance level, H0 is rejected if t is less than −2.101 or greater than 2.101.
We use Excel to find the means and standard deviations of the two independent samples as shown in the Chapter 3 section of Appendix C. The Excel instructions to find the pooled variance and the value of the “t Stat” are in the Chapter 11 section in Appendix C. These values are highlighted in yellow.
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 (11–3) is
s2p = (n1 − 1)s21 + (n2 − 1)s22
n1 + n2 − 2 =
(10 − 1) (14.452) + (10 − 1) (14.29)2
10 + 10 − 2 = 206.50
From formula (11–4), t is 0.716.
t = x1 − x2
√s 2 p (
1 n1
+ 1
n2)
= 226.8 − 222.2
√206.50 ( 1
10 +
1 10)
= 4.6
6.4265 = 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 numerator 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 372 in the previous section). 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
TWO-SAMPLE TESTS OF HYPOTHESIS 375
statistical decisions. The denominator measures the standard error of the statistic. When the samples are not paired, two kinds of variation are present: differences be- tween the two appraisal firms and the difference in the value of the real estate. Proper- ties 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 inter- ested in the difference between the two appraisal firms.
In sum, when we can pair or match observations that measure differences for a common variable, a hypothesis test based on dependent samples is more sensitive to detecting a significant difference than a hypothesis test based on independent sam- ples. In the case of comparing the property valuations by Schadek Appraisals and Bowyer Real Estate, the hypothesis test based on dependent samples eliminates the variation between the values of the properties and focuses only on the comparisons in the two appraisals for each property. There is a bit of bad news here. In the dependent samples 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 observations are paired. However, in most cases, this is a small price to pay for a better test.
Advertisements by Core 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 stu- dents lost weight?
S E L F - R E V I E W 11–4
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
(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?
17. The null and alternate hypotheses are:
H0: μd ≤ 0 H1: μd > 0
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.
E X E R C I S E S
Day
1 2 3 4
Day shift 10 12 15 19 Afternoon shift 8 9 12 15
376 CHAPTER 11
At the .05 significance level, can we conclude there are more defects produced on the day shift?
18. The null and alternate hypotheses are:
H0: μd = 0 H1: μd ≠ 0
The following paired observations show the number of traffic citations given for speeding by Officer Dhondt and Officer Meredith of the South Carolina Highway Patrol for the last five months.
Number of Citations Issued
May June July August September
Officer Dhondt 30 22 25 19 26 Officer Meredith 26 19 20 15 19
At the .05 significance level, is there a difference in the mean number of citations given by the two officers?
Note: Use the six-step hypothesis testing procedure to solve the following exercises.
19. The management of Discount Furniture, a chain of discount furniture stores in the Northeast, designed an incentive plan for salespeople. To evaluate this inno- vative plan, 12 salespeople were selected at random, and their weekly incomes before and after the plan were recorded.
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
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.
20. The federal government recently granted funds for a special program de- signed 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 program? Use the .01 significance level. Estimate the p-value.
TWO-SAMPLE TESTS OF HYPOTHESIS 377
C H A P T E R S U M M A R Y
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. The two populations follow normal distributions. 2. The samples are from independent populations. 3. The formula to compute the value of z is
z = x1 − x2
√ σ21 n1
+ σ22 n2
(11–2)
II. The test statistic to compare two means is the t distribution if the population standard deviations are not known. A. Both populations are approximately normally distributed. 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:
s2p = (n1 − 1)s21 + (n2 − 1)s22
n1 + n2 − 2 (11–3)
2. The value of t is computed from the following formula:
t = x1 − x2
√s 2 p (
1 n1
+ 1
n2)
(11–4)
3. The degrees of freedom for the test are n1 + n2 − 2. III. 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.
df = [(s21 ∕n1) + (s22∕n2)]2
(s21 ∕n1)2
n1 − 1 +
(s22∕n2)2
n2 − 1
(11–6)
B. The value of the test statistic is computed from the following formula.
t = x1 − x2
√ s21 n1
+ s22 n2
(11–5)
IV. For dependent samples, we assume the population distribution of the paired differences 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:
t = d
sd ∕√n (11–7)
378 CHAPTER 11
P R O N U N C I A T I O N K E Y
SYMBOL MEANING PRONUNCIATION
s2p Pooled sample variance s squared sub p
x1 Mean of the first sample x bar sub 1
x2 Mean of the second sample x bar sub 2
d Mean of the difference between d bar dependent observations
sd Standard deviation of the difference s sub d between dependent observations
C H A P T E R E X E R C I S E S
21. A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distributions follow the normal probability distribution and the population standard deviations are equal. The information is sum- marized below.
Statistic Men Women
Sample mean 24.51 22.69 Sample standard deviation 4.48 3.86 Sample size 35 40
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?
22. 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. The mean number of units produced by a sample of 54 day-shift workers was 345. The mean number of units produced by a sample of 60 night-shift workers was 351. Assume the population standard deviation of the number of units produced on the day shift is 21 and 28 on the night shift. Using the .05 significance level, is the number of units pro- duced on the night shift larger?
23. 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 calls they make per day. 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. Assume the population standard deviation for Larry Clark is 1.05 calls per day and 1.23 calls per day for George Murnen. 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?
24. 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. Com- pute the p-value.
25. A cell phone company offers two plans to its subscribers. At the time new subscrib- ers sign up, they are asked to provide some demographic information. The mean
TWO-SAMPLE TESTS OF HYPOTHESIS 379
yearly income for a sample of 40 subscribers to Plan A is $57,000 with a standard deviation of $9,200. For a sample of 30 subscribers to Plan B, the mean income is $61,000 with a standard deviation of $7,100. At the .05 significance level, is it rea- sonable to conclude the mean income of those selecting Plan B is larger? What is the p-value?
26. A computer manufacturer offers technical support that is available 24 hours a day, 7 days a week. Timely resolution of these calls is important to the company’s image. For 35 calls that were related to software, technicians resolved the issues in a mean time of 18 minutes with a standard deviation of 4.2 minutes. For 45 calls related to hardware, technicians resolved the problems in a mean time of 15.5 minutes with a standard devi- ation of 3.9 minutes. At the .05 significance level, does it take longer to resolve software issues? What is the p-value?
27. Music streaming services are the most popular way to listen to music. Data gathered over the last 12 months show Apple Music 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 Spotify was used by an average of 2.2 million families with a sample stan- dard deviation of 0.30 million. Assume the population standard deviations are not the same. Using a significance level of .05, test the hypothesis of no difference in the mean number of households picking either service.
28. Businesses such as General Mills, Kellogg’s, and Betty Crocker regularly use cou- pons to build brand allegiance and stimulate sales. Marketers believe that the users of paper coupons are different from the users of e-coupons accessed through the Internet. One survey recorded the age of each person who redeemed a coupon along with the type of coupon (either paper or electronic). The sample of 25 tradi- tional paper-coupon clippers had a mean age of 39.5 with a standard deviation of 4.8. The sample of 35 e-coupon users had a mean age of 33.6 years with a standard deviation of 10.9. Assume the population standard deviations are not the same. Us- ing a significance level of .01, test the hypothesis of no difference in the mean ages of the two groups of coupon clients.
29. 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?
30. Educational Technology, Inc. sells software to provide guided homework problems for a statistics course. They would like to know if students who use the software score better on exams. A sample of students who used the software had the following exam scores: 86, 78, 66, 83, 84, 81, 84, 109, 65, and 102. Students who did not use the soft- ware had the following exam scores: 91, 71, 75, 76, 87, 79, 73, 76, 79, 78, 87, 90, 76, and 72. Assume the population standard deviations are not the same. At the .10 signifi- cance level, can we conclude that there is a difference in the mean exam scores for the two groups of students?
31. 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 differ- ently, but both store managers claim their layout maximizes the amounts customers will purchase on impulse. A sample of 10 customers at the Peach Street store re- vealed they spent the following amounts on impulse purchases: $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 on impulse purchases: $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 signifi- cance level, is there a difference in the mean amounts purchased on impulse at the two stores?
32. Grand Strand Family Medical Center treats minor medical emergencies for visitors to the Myrtle Beach area. There are two facilities, one in the Little River Area and the other in Murrells Inlet. The Quality Assurance Department wishes to compare the mean
380 CHAPTER 11
waiting time for patients at the two locations. Samples of the waiting times for each loca- tion, reported in minutes, follow:
Location Waiting Time
Little River 31 28 29 22 29 18 32 25 29 26 Murrells Inlet 22 23 26 27 26 25 30 29 23 23 27 22
Assume the population standard deviations are not the same. At the .05 significance level, is there a difference in the mean waiting time?
33. 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?
Under 25 10 10 11 15 7 11 10 9 Over 60 4 8 7 7 4 5 1 7 4 10 5
34. Two of the teams competing in the America’s Cup race are Team Oracle U.S.A. and Land Rover BAR. They race their boats over a part of the course several times. Below are a sample of times in minutes for each boat. Assume the population standard devia- tions are not the same. At the .05 significance level, can we conclude that there is a difference in their mean times?
Boat Time (minutes)
Land Rover BAR 12.9 12.5 11.0 13.3 11.2 11.4 11.6 12.3 14.2 11.3 Team Oracle 14.1 14.1 14.2 17.4 15.8 16.7 16.1 13.3 13.4 13.6 10.8 19.0
35. The manufacturer of an MP3 player wanted to know whether a 10% 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 regular and reduced prices at the randomly se- lected outlets. At the .01 significance level, can the manufacturer conclude that the price reduction resulted in an increase in sales?
Regular price 138 121 88 115 141 125 96 Reduced price 128 134 152 135 114 106 112 120
36. 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 proposed 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:
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
TWO-SAMPLE TESTS OF HYPOTHESIS 381
At the .01 significance level, is it reasonable to conclude that the modification reduced the number of traffic accidents?
37. Lester Hollar is vice president for human resources for a large manufacturing com- pany. In recent years, he has noticed an increase in absenteeism that he thinks is re- lated 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 evaluate the program, he selected a random sample of eight participants and found the number of days each was absent in the six months before the exercise program began and in the six months following the exercise program. Below are the results. At the .05 significance level, can he conclude that the number of absences has declined? Estimate the p-value.
Employee Before After
Bauman 6 5 Briggs 6 2 Dottellis 7 1 Lee 7 3 Perralt 4 3 Rielly 3 6 Steinmetz 5 3 Stoltz 6 7
38. 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 data for the sample of families and their two insurance quotes are reported below. At the .10 significance level, can we conclude that there is a difference in the amounts quoted?
Midstates Gecko 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
39. Fairfield Homes is developing two parcels near Pigeon Fork, Tennessee. In order to test different advertising approaches, it uses different media to reach potential
382 CHAPTER 11
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?
40. A candy company taste-tested two chocolate bars, one with almonds and one without almonds. A panel of testers rated the bars on a scale of 0 to 5, with 5 indicating the highest taste rating. Assume the population standard deviations are equal. At the .05 significance level, do the ratings show a difference between chocolate bars with or with- out almonds?
With Almonds Without Almonds
3 0 1 4 2 4 3 3 1 4 1 2
41. 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 sample of eight operating rooms in the greater Seattle area during the last year. The following table reports the contamination levels before and after the use of the soap for each operating room.
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
At the .05 significance level, can we conclude the contamination measurements are lower after use of the new soap?
42. The following data on annual rates of return were collected from eleven randomly selected stocks listed on the New York Stock Exchange (“the big board”) and twelve randomly selected stocks listed on NASDAQ. Assume the population standard devia- tions are the same. At the .10 significance level, can we conclude that the annual rates of return are higher on the big board?
NYSE NASDAQ
15.0 8.8 10.7 6.0 20.2 14.4 18.6 19.1 19.1 17.6 8.7 17.8 17.8 15.9 13.8 17.9 22.7 21.6 14.0 6.0 26.1 11.9 23.4
TWO-SAMPLE TESTS OF HYPOTHESIS 383
43. The city of Laguna Beach operates two public parking lots. The Ocean Drive parking lot can accommodate up to 125 cars and the Rio Rancho parking lot can accommodate up to 130 cars. City planners are considering increasing the size of the lots and changing the fee structure. To begin, the Planning Office would like some information on the number of cars in the lots at various times of the day. A junior plan- ner officer is assigned the task of visiting the two lots at random times of the day and evening and counting the number 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.
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.
44. The amount of income spent on housing is an important component of the cost of living. The total costs of housing for homeowners might include mortgage pay- ments, property taxes, and utility costs (water, heat, electricity). An economist se- lected a sample of 20 homeowners in New England and then calculated these total housing costs as a percent of monthly income, 5 years ago and now. The informa- tion is reported below. Is it reasonable to conclude the percent is less now than 5 years ago?
Homeowner Five Years Ago Now Homeowner Five Years Ago Now
Holt 17% 10% Lozier 35% 32% Pierse 20 39 Cieslinski 16 32 Merenick 29 37 Rowatti 23 21 Lanoue 43 27 Koppel 33 12 Fagan 36 12 Rumsey 44 40 Bobko 43 41 McGinnis 44 42 Kippert 45 24 Pierce 28 22 San Roman 19 26 Roll 29 19 Kurimsky 49 28 Lang 39 35 Davison 49 26 Miller 22 12
45. The CVS Pharmacy located on US 17 in Murrells Inlet has been one of the busiest pharmaceutical retail stores in South Carolina for many years. To try and capture more business in the area, CVS top management opened another store about 6 miles west on SC 707. After a few months, CVS management decided to compare the business vol- ume at the two stores. One way to measure business volume is to count the number of cars in the store parking lots on random days and times. The results of the survey from the last 3 months of the year are reported below. To explain, the first observation was on October 2 at 20:52 military time (8:52 p.m.). At that time there were four cars in the US 17 lot and nine cars in the SC 707 lot. At the .05 significance level, is it reasonable to
384 CHAPTER 11
46. A goal of financial literacy for children is to learn how to manage money wisely. One question is: How much money do children have to manage? A recent study by Schnur Educational Research Associates randomly sampled 15 children between 8 and 10 years old and 18 children between 11 and 14 years old and recorded their monthly allowance. Is it reasonable to conclude that the mean allowance re- ceived by children between 11 and 14 years is more than the allowance received by children between 8 and 10 years? Use the .01 significance level. What is the p-value?
Vehicle Count
Date Time US 17 SC 707
Oct 2 20:52 4 9 Oct 11 19:30 5 7 Oct 15 22:08 9 12 Oct 19 11:42 4 5 Oct 25 15:32 10 8 Oct 26 11:02 9 15 Nov 3 11:22 13 7 Nov 5 19:09 20 3 Nov 8 15:10 15 14 Nov 9 13:18 15 11 Nov 15 22:38 13 11 Nov 17 18:46 16 12 Nov 21 15:44 17 8 Nov 22 15:34 15 3 Nov 27 21:42 20 6 Nov 29 9:57 17 13 Nov 30 17:58 5 9 Dec 3 19:54 7 13 Dec 15 18:20 11 6 Dec 16 18:25 14 15 Dec 17 11:08 8 8 Dec 22 21:20 10 3 Dec 24 15:21 4 6 Dec 25 20:21 7 9 Dec 30 1 4:25 19 4
8–10 Years 11–14 Years 8–10 Years 11–14 Years
26 49 26 41 33 44 25 38 30 42 27 44 26 38 29 39 34 39 34 50 26 41 32 49 27 39 41 27 38 42 30 38 30
conclude that, based on vehicle counts, the US 17 store has more business volume than the SC 707 store?
TWO-SAMPLE TESTS OF HYPOTHESIS 385
D A T A A N A L Y T I C S
47. The North Valley Real Estate data reports information on the homes sold 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
selling price of homes with an attached garage and homes without an attached garage?
c. At the .05 significance level, can we conclude that there is a difference in the mean selling price of homes that are in default on the mortgage?
48. Refer to the Baseball 2016 data, which report information on the 30 Major League Baseball teams for the 2016 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? At the .05 significance level, is there a difference in the mean attendance for the two groups?
49. Refer to the Lincolnville School District bus data. Is there a difference in the mean maintenance cost for the diesel versus the gasoline buses? Use the .05 significance level.
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO12-1 Apply the F distribution to test a hypothesis that two population variances are equal.
LO12-2 Use ANOVA to test a hypothesis that three or more population means are equal.
LO12-3 Use confidence intervals to test and interpret differences between pairs of population means.
LO12-4 Use a blocking variable in a two-way ANOVA to test a hypothesis that three or more population means are equal.
LO12-5 Perform a two-way ANOVA with interaction and describe the results.
ONE VARIABLE THAT GOOGLE uses to rank pages on the Internet is page speed, the time it takes for a web page to load into your browser. A source for women’s clothing is redesigning their page to improve the images that show its products and to reduce its load time. The new page is clearly faster, but initial tests indicate there is more variation in the time to load. A sample of 16 different load times showed that the standard deviation of the load time was 22 hundredths of a second for the new page and 12 hundredths of a second for the current page. At the .05 significance level, can we conclude that there is more variation in the load time of the new page? (See Exercise 24 and LO12-1.)
Analysis of Variance12
© Alexander Hassenstein/Getty Images
ANALYSIS OF VARIANCE 387
INTRODUCTION In this chapter, we continue our discussion of hypothesis testing. Recall that in Chapters 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 distribution (the stan- dard 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. In this chapter, we expand our idea of hypothesis tests. We describe a test for variances and then a test that simultaneously compares several population means to determine if they are equal.
COMPARING TWO POPULATION VARIANCES In Chapter 11, we tested hypotheses about equal population means. The tests differed based on our assumptions regarding whether the population standard deviations or variances were equal or unequal. In this chapter, the assumption about equal popula- tion variances is also important. In this section, we present a way to statistically test this assumption. The test is based on the F distribution.
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 popu- lation means is called analysis of variance (ANOVA). In both of these situations, the popu- lations 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 determined by two parameters: the degrees of freedom in the numerator and the degrees of freedom in the denominator. The shape of the distribution is illustrated by the fol- lowing graph. There is one F distribution for the combination of 29 degrees of free- dom in the numerator (df ) and 28 degrees of freedom in the denominator. There is another F distribution for 19 degrees of freedom in the numerator 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.
LO12-1 Apply the F distribution to test a hypothesis that two population variances are equal.
0 1 2 3 4 5
Re la
tiv e
fre qu
en cy
df � (29, 28)
df � (19, 6)
df � (6, 6)
F
388 CHAPTER 12
2. The F distribution is continuous. This means that the value of F can assume an infinite number of values between zero and positive infinity.
3. The F statistic cannot be negative. The smallest value F can assume is 0. 4. The F distribution 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 numer- ator and denominator, the distribution approaches a normal distribution.
5. The F distribution is asymptotic. As the values of F increase, the distribution ap- proaches the horizontal axis but never touches it. This is similar to the behavior of the normal probability distribution, described in Chapter 7.
Testing a Hypothesis of Equal Population Variances The first application of the F distribution that we describe occurs when we test the hy- pothesis that the variance of one normal population equals the variance of another normal population. The following examples will show the use of the test:
• A health services corporation manages two hospitals in Knoxville, Tennessee: St. Mary’s North and St. Mary’s South. In each hospital, the mean waiting time in the Emergency Department is 42 minutes. The hospital administrator believes that the St. Mary’s North Emergency Department has more variation in waiting time than St. Mary’s South.
• 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 sam- ple of 10 technology and 10 utility stocks shows the same mean rate of return, but there is likely more variation in the technology stocks.
• An on-line newspaper found that men and women spend about the same amount of time per day accessing news apps. However, the same report indicated the times of men had nearly twice as much variation compared to the times of women.
The F distribution is also used to test the assumption that the variances of two nor- mal populations are equal. Recall that in the previous chapter the t test to investigate whether the means of two independent populations differed assumes that the variances of the two normal populations are the same. See this list of assumptions on page 361. The F distribution is used to test the assumption that the variances are equal.
To compare two population variances, we first state the null hypothesis. The null hypothesis is that the variance of one normal population, σ21 , equals the variance of an- other normal population, σ22. The alternate hypothesis is that the variances differ. In this instance, the null hypothesis and the alternate hypothesis are:
H0: σ21 = σ22 H1: σ21 ≠ σ22 To conduct the test, we select a random sample of observations, n1, from one popula- tion and a random sample of observations, n2, from the second population. The test statistic is defined as follows.
© McGraw-Hill Education/ John Flournoy, photographer John Flournoy
F = s21 s22
(12–1)TEST STATISTIC FOR COMPARING TWO VARIANCES
ANALYSIS OF VARIANCE 389
The terms s1 2 and s22 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 free- dom. 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 (α/2) and then referring to the appropri- ate degrees of freedom in Appendix B.6. An example will illustrate.
S O L U T I O N
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 (3–9) to compute the sample standard deviations. To obtain the sample variances, we square the standard deviations.
U.S. ROUTE 25
x = Σx n
= 408
7 = 58.29 s = √
Σ(x − x )2
n − 1 = √
485.43 7 − 1
= 8.9947
INTERSTATE 75
x = Σx n
= 472
8 = 59.00 s = √
Σ(x − x )2
n − 1 = √
134 8 − 1
= 4.3753
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 high- way. However, the I-75 route is several miles longer. It is important that the service
E X A M P L E
Lammers Limos offers limousine service from Government Center in downtown 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 air- port using each route and then compare the results. He collected the following sample data, which is reported in minutes. Using the .10 significance level, is there a difference in the variation in the driving times for the two routes?
© Daniel Acker/Bloomberg/Getty Images RF
U.S. Route 25 Interstate 75
52 59 67 60 56 61 45 51 70 56 54 63 64 57 65
390 CHAPTER 12
offered be both timely and consistent, so he decides to conduct a statistical test to determine whether there really is a difference in the variation of the two routes.
We use the six-step hypothesis test procedure.
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. For this example/solution, the sub- script 1 indicates information for U.S. 25; the subscript 2 indicates in- formation for I-75.
H0: σ21 = σ22 H1: σ21 ≠ σ22
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.6, a portion of which is
reproduced as Table 12–1. Because we are conducting a two-tailed test, the tabled significance level is .05, found by α/2 = .10/2 = .05. There are n1 − 1 = 7 − 1 = 6 degrees of freedom in the numerator and n2 − 1 = 8 − 1 = 7 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.6) for the .05 significance level to 6 de- grees of freedom in the numerator. Then move down that column to the critical value opposite 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.
Degrees Degrees of Freedom for Numerator 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
TABLE 12–1 Critical Values of the F Distribution, α = .05
Step 5: Next we compute the ratio of the two sample variances, determine the value of the test statistic, and make a decision regarding the null hypothesis. Note that formula (12–1) refers to the sample variances, but we calculated the sample standard deviations. We need to square the standard deviations to determine the variances.
F = s21 s22
= (8.9947)2
(4.3753)2 = 4.23
The decision is to reject the null hypothesis because the computed F value (4.23) is larger than the critical value (3.87).
Step 6: We conclude there is a difference in the variation in the time to travel the two routes. Mr. Lammers will want to consider this in his scheduling.
ANALYSIS OF VARIANCE 391
The usual practice is to determine the F ratio by putting the larger of the two sam- ple 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 ex- tensive F tables.
A logical question arises: Is it possible to conduct one-tailed tests? For example, suppose in the previous example we suspected that the variance of the times using the U.S. 25 route, σ 21, is larger than the variance of the times along the I-75 route, σ22. We would state the null and the alternate hypothesis as
H0: σ21 ≤ σ22 H1: σ21 > σ22 The test statistic is computed as s1
2∕s22. Notice that we labeled the population with the suspected large variance as population 1. So s1
2 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.6 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 has a procedure to perform a test of variances. Below is the out- put. The computed value of F is the same as that determined by using formula (12–1). The result of the one-tail hypothesis test is to reject the null hypothesis. The F of 4.23 is greater than the critical value of 3.87. Also, the p-value is less than 0.05. We conclude the variance of travel times on U.S. 25 is greater than the variance of travel times on I-75.
Steele Electric Products Inc. assembles cell phones. For the last 10 days, Mark Nagy completed a mean of 39 phones per day, with a standard deviation of 2 per day. Debbie Richmond com- pleted a mean of 38.5 phones per day, with a standard deviation of 1.5 per day. At the .05 sig- nificance level, can we conclude that there is more variation in Mark’s daily production?
S E L F - R E V I E W 12–1
1. What is the critical F value when the sample size for the numerator is six and the sample size for the denominator is four? Use a two-tailed test and the .10 significance level.
2. What is the critical F value when the sample size for the numerator is four and the sample size for the denominator is seven? Use a one-tailed test and the .01 signifi- cance level.
3. The following hypotheses are given.
H0: σ21 = σ22 H1: σ21 ≠ σ22
E X E R C I S E S
392 CHAPTER 12
ANOVA: ANALYSIS OF VARIANCE The F distribution is used to perform a wide variety of hypothesis tests. For example, when testing the equality of three or more population means, the Analysis of Variance (ANOVA) technique is used and the F statistic is used as the test statistic.
ANOVA Assumptions The ANOVA to test the equality of three or more population means requires that three assumptions are true:
1. The populations follow the normal distribution. 2. The populations have equal standard deviations (σ). 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 differences 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 training program, we administer a test to measure understanding of firefighting techniques to the four groups. 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. For each t test, suppose we choose an α = .05. Therefore, the probability of
LO12-2 Use ANOVA to test a hypothesis that three or more population means are equal.
A random sample of eight observations from the first population resulted in a stan- dard deviation of 10. A random sample of six observations from the second popu- lation resulted in a standard deviation of 7. At the .02 significance level, is there a difference in the variation of the two populations?
4. The following hypotheses are given.
H0: σ21 ≤ σ22 H1: σ21 > σ22
A random sample of five observations from the first population resulted in a stan- dard 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 discov- ered that the mean listening time for a sample of 10 men was 35 minutes per day. The standard deviation was 10 minutes per day. The mean listening time for a sam- ple of 12 women 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 differ- ence 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% with a standard deviation of 3.9%. The mean rate of return on a sample of 8 utility stocks was 10.9% with a standard deviation of 3.5%. At the .05 significance level, can we conclude that there is more variation in the oil stocks?
ANALYSIS OF VARIANCE 393
a Type I error, rejecting the null when it is true, is .05. The complement is the probability of .95 that we do not reject the null when it is true. Because we conduct six separate (independent) tests, the probability that all six tests result in correct decisions is:
P(All correct) = (.95)(.95)(.95)(.95)(.95)(.95) = .735
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 1 − .735 = .265. To summarize, if we conduct six independent tests using the t distribution, the likelihood of rejecting a true null hypothesis because of sampling error is an unsatisfac- tory .265. The ANOVA technique allows us to compare population means simultane- ously at a selected significance level. It avoids the buildup of Type I error associated with testing many hypotheses.
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 identify 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 illustration will clarify the term treatment and demonstrate an application of ANOVA.
E X A M P L E
Joyce Kuhlman manages a regional financial center. She wishes to compare the productivity, as measured by the number of customers served, among three em- ployees. Four days are randomly selected and the number of customers served by each employee is recorded. The results are:
Wolfe White Korosa
55 66 47 54 76 51 59 67 46 56 71 48
S O L U T I O N
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.
Em pl
oy ee
Customers served
White
Korosa
Wolfe
µ1 µ2 µ3
CHART 12–1 Case Where Treatment Means Are Different
394 CHAPTER 12
Suppose there is no difference in the treatment means. This would indicate that the population means are the same. This is shown in Chart 12–2. Note again that the populations follow the normal distribution and the variation in each of the popula- tions 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
The ANOVA Test How does the ANOVA test work? Recall that we want to determine whether the various 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 392 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 ad- vantage 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 conclude 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 example/solution in the previous section. The manager wants to deter- mine 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 (55 + 54 + … + 48)/12. 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.
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 (55 − 58)2 + (54 − 58)2 + … + (48 − 58)2.
Next, break this total variation into two components: variation due to the treatment variation and random variation.
ANALYSIS OF VARIANCE 395
The variation due to treatments is also called variation between treatment means. In this example, we first square the difference between each treatment mean and the overall mean. The mean for Wolfe is 56 customers, found by (55 + 54 + 59 + 56)/4. The other means are 70 and 48, respectively. Then, each of the squared differences is mul- tiplied by the number of observations in each treatment. In this case, the value is 4. Last, these values are summed together. This term is 992. The sum of the squares due to the treatments is:
4(56 − 58)2 + 4(70 − 58)2 + 4(48 − 58)2 = 992
If there is considerable variation among the treatment means compared to the overall mean, it is logical that this term will be a large value. If the treatment means are similar, this value will be small. The smallest possible value would be zero. This would occur when all the treatment means are the same. In this case, all the treatment means would also equal the overall mean.
The other source of variation is referred to as random variation, or the error component.
TREATMENT VARIATION The sum of the squared differences between each treatment mean and the grand or overall mean.
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 each treatment or employee. This is also called the variation within the treatments. The error variation is 90.
(55 − 56)2 + (54 − 56)2 + … + (48 − 48)2 = 90
We determine the test statistic, which is the ratio of the two estimates of the popu- lation variance, from the following equation.
F =
Estimate of the population variance based on the differences between the treatment means
Estimate of the population variance based on the variation within the treatments
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 (3–9)], 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, 90/(12 − 3). Hence, our second estimate of the population variance is 90/9. This is actually a generalization of formula (11–4), we pooled the sample variances from two populations.
The last step is to take the ratio of these two estimates.
F = 992∕2 90∕9
= 49.6
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.
396 CHAPTER 12
E X A M P L E
Recently airlines cut services, such as meals and snacks during flights, and started charging for checked luggage. A group of four carriers hired Brunner Marketing Research Inc. to survey passengers regarding their level of satisfaction with a re- cent 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 possible 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.
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
S O L U T I O N
We will use the six-step hypothesis-testing procedure.
Step 1: State the null hypothesis and the alternate hypothesis. The null hy- pothesis is that the mean scores are the same for the four airlines.
H0: μN = μW = μP = μB The alternate hypothesis is that the mean scores are not all the same
for the four airlines.
H1: 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 difference in the mean scores for the four airlines. If H0 is re- jected, we conclude 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.6. 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 equal the number of treatments, designated as k, minus 1.
ANALYSIS OF VARIANCE 397
The degrees of freedom in the denominator are 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 numerator = k − 1 = 4 − 1 = 3
Degrees of freedom in the denominator = n − k = 22 − 4 = 18
Refer to Appendix B.6 and the .01 significance level. Move horizon- tally across the top of the page to 3 degrees of freedom in the numer- ator. 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 H0 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 to- tal 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:
SS total = Σ (x − xG)2 (12–2)
where: x is each sample observation. xG 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:
SSE = Σ (x − xc)2 (12–3) where:
xc is the sample mean for treatment c.
The SSE is calculated:
SSE = Σ (x − xN)2 + Σ (x − xW)2 + Σ (x − xP)2 + Σ (x − xB)2
The detailed calculations of SS total and SSE for this example follow. To determine 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
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
398 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 (x − xG) = 94 − 75.64 = 18.36. For the last passenger, (x − xG) = 65 − 75.64 = −10.64. The calcula- tions for all other passengers follow.
Northern WTA Pocono Branson
18.36 −0.64 −5.64 −7.64 14.36 −7.64 −2.64 −5.64 9.36 1.36 0.36 −3.64 4.36 7.36 2.36 −10.64 12.36 4.36 −1.64 −7.64 −10.64 −10.64
Then square each of these differences and sum all the values. Thus, for the first passenger:
(x − xG)2 = (94 − 75.64)2 = (18.36)2 = 337.09
Finally, sum all the squared differences as formula (12–2) directs. Our SS total value is 1,485.10.
To compute the term SSE, find the deviation between each observa- tion and its treatment mean. In the example, the mean of the first treatment (that is, the passengers on Northern Airlines) is 87.25, found by xN = 349∕4. The subscript N refers to Northern Airlines.
ANALYSIS OF VARIANCE 399
The first passenger rated Northern a 94, so (x − xN) = (94 − 87.25) = 6.75. The first passenger in the WTA group responded with a total score of 75, so (x − xW) = (75 − 78.20) = −3.2. The detail for all the passengers follows.
Each of these values is squared and then summed for all 22 observa- tions. The four column totals can also be summed to find SSE. The values are shown in the following table.
So the SSE value is 594.41. That is, Σ (x − xc)2 = 594.41. Finally, we determine SST, the sum of the squares due to the
treatments, by subtraction.
SST = SS total − SSE (12–4)
For this example:
SST = SS total − SSE = 1,485.10 − 594.41 = 890.69.
To find the computed value of F, work your way across the ANOVA table. The degrees of freedom for the numerator and the denomina- tor are the same as in step 4 on page 396 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 treatments is SST di- vided 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 fashion. 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 com- pute the value of F as follows.
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
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
Have you ever waited in line for a telephone and it seemed like the person us- ing the phone talked on and on? There is evidence that people actually talk longer on public tele- phones when someone is waiting. In a recent survey, researchers measured the length of time that 56 shoppers in a mall spent on the phone (1) when they were alone, (2) when a per- son was using the adjacent phone, and (3) when a per- son was using an adjacent 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.
STATISTICS IN ACTION
400 CHAPTER 12
The computed value of F is 8.99, which is greater than the critical value of 5.09, so the null hypothesis is rejected.
Step 6: Interpret the result. We conclude the population means are not all equal. At this point, the results of the ANOVA only show that at least one pair of mean satisfaction scores are not the same among the four airlines. We cannot statistically show which airlines differ in satisfaction or which air- lines have the highest or lowest satisfaction scores. The techniques for determining how the airlines differ are presented in the next section.
The calculations in the previous example/solution are tedious if the number of ob- servations in each treatment is large. Many statistical software packages will perform the calculations and output the results. In the following illustration, Excel is used to cal- culate the descriptive statistics and ANOVA for the previous example/solution involving airlines and passenger ratings. There are some slight differences between the output and the previous calculations. These differences are due to rounding.
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 proba- bility 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 freedom in the denomi- nator. So when we reject the null hypothesis in this instance, there is a very small likeli- hood of committing a Type I error!
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.
S E L F - R E V I E W 12–2
Near Bread Near Beer With Cleaners
18 12 26 14 18 28 19 10 30 17 16 32
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?
ANALYSIS OF VARIANCE 401
7. The following are four observations collected from each of three treatments. Test the hypothesis that the treatment means are equal. Use the .05 significance level.
a. State the null 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 are six observations collected from treatment 1, four observations collected from treatment 2, and five observations collected from treatment 3. Test the hypothesis at the .05 significance level that the treatment means are equal.
E X E R C I S E S
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 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.
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 partic- ular importance is the income in the area surrounding the proposed mall. A ran- dom 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 six-step hypothesis test- ing 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 per week senior executives by type of industry spend at their desktop computers. The manager selected a sample of five executives from each of three
402 CHAPTER 12
INFERENCES ABOUT PAIRS OF TREATMENT MEANS Suppose we carry out the ANOVA procedure, make the decision to reject the null hy- pothesis, and 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 this analysis.
Recall in the previous example/solution regarding airline passenger ratings, we concluded that there was a difference in the treatment means. That is, the null hypothe- sis was rejected and the alternate hypothesis accepted. The conclusion is that at least one of the airline’s mean level of satisfaction is different for the others. Now, the ques- tion is which of the four airlines 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 example on page 400, 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 conclusion that there is a significant differ- ence 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 mean square error, or MSE, and is determined by SSE/(n − k). A confidence interval for the difference between two populations is found by:
LO12-3 Use confidence intervals to test and interpret differences between pairs of population means.
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
32 28 30 30 28 28 30 26 26 32 28 28 30 30 30
(x1 − x2) ± t√MSE ( 1 n1
+ 1
n2) (12–5)
CONFIDENCE INTERVAL FOR THE DIFFERENCE IN TREATMENT MEANS
where: x1 is the mean of the first sample. x2 is the mean of the second sample. t is obtained from Appendix B.5. The degrees of freedom are equal to n − k. 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 x1 − x2 = 5 and
ANALYSIS OF VARIANCE 403
t√MSE ( 1 n1
+ 1
n2) = 12—the confidence interval would range from −7.00 up to 17.00.
To put it in symbols:
(x1 − x2) ± t√MSE ( 1 n1
+ 1
n2) = 5.00 ± 12.00 = −7.00 up to 17.00
Note that zero is in this interval. Therefore, we conclude that there is no significant dif- ference 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 x1 − x2 = −0.35 and
t√MSE ( 1 n1
+ 1
n2) = 0.25, 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 dif- ference between the mean scores of passengers on Northern and Branson. With a 95% level of confidence, the endpoints of the confidence interval are 10.457 and 26.043.
(xN − xB) ± t√MSE ( 1
nN +
1 nB)
= (87.25 − 69.00) ± 2.101√33.023 ( 1 4
+ 1 6)
= 18.25 ± 7.793 where:
xN is 87.25. xB is 69.00. t is 2.101: from Appendix B.5 with (n − k) = 22 − 4 = 18 degrees of freedom. MSE is 33.023: from the ANOVA table with SSE/(n − k) = 594.4/18. nN is 4. nB is 6.
The 95% confidence interval ranges from 10.457 up to 26.043. Both endpoints are positive; hence, we can conclude these treatment means differ significantly. That is, passengers on Northern Airlines rated service significantly different from those on Branson Airlines.
The confidence intervals for the differences between each pair of means can be obtained directly using statistical software. The following confidence intervals were computed using the one-way ANOVA in Minitab. Statistical software, such as Minitab, offers a variety of methods to control Type I error when making multiple comparisons. The following analysis used Fisher’s method to compare means.
404 CHAPTER 12
The output shows the confidence intervals for the difference between each pair of treatment means. The first row shows the confidence interval that compares WTA and Northern. It shows a confidence interval that does not include zero. It also shows the p-value for a hypothesis test that the means of WTA and Northern are equal. The hypothesis is rejected because a p-value of 0.031 is less than an assumed α of 0.05. Both results indicate that the WTA and Northern means are significantly different. Reviewing the entire table, only two pairs of means are not significantly different: Pocono and WTA, and Branson and Pocono. All other confi- dence intervals do not include zero and have p-values less than 0.05. Therefore, all other pairs of means are significantly different.
The graphic illustrates the results of the confidence interval analysis. Each confi- dence interval is represented by its endpoints and treatment mean. Note that a differ- ence of zero is illustrated with the vertical dotted line. Two of the intervals include zero, Pocono and WTA, and Branson and Pocono. The others do not include zero so the means are significantly different. The following pairs of means are different: WTA and Northern, Pocono and Northern, Branson and Northern, and Branson and WTA.
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 treatment means are equal is rejected should any analysis of the individual treatment means be attempted.
The following data are the semester tuition charges ($000) for a sample of five private col- leges in the Northeast region of the United States, four in the Southeast region, and five in the West region. At the .05 significance level, can we conclude there is a difference in the mean tuition rates for the various regions?
S E L F - R E V I E W 12–3
(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% confidence interval for that difference.
Northeast Southeast West ($000) ($000) ($000)
40 38 37 41 39 38 42 40 36 40 38 37 42 36
11. The following are three observations collected from treatment 1, five observa- tions collected from treatment 2, and four observations collected from treatment 3. Test the hypothesis that the treatment means are equal at the .05 significance level.
E X E R C I S E S
Treatment 1 Treatment 2 Treatment 3
8 3 3 11 2 4 10 1 5 3 4 2
ANALYSIS OF VARIANCE 405
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 H0 is rejected, can we conclude that treatment 2 and treatment 3 differ? Use
the 95% 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 in- formation is submitted to Minitab with the following results:
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 return on equity for three types of stock: utility, retail, and banking stocks. The fol- lowing output is obtained:
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 H0 is rejected, can we conclude that treatment 1 and treatment 2 differ? Use
the 95% level of confidence. 12. The following are six observations collected from treatment 1, ten observa-
tions collected from treatment 2, and eight observations collected from treat- ment 3. Test the hypothesis that the treatment means are equal at the .05 significance level.
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
406 CHAPTER 12
TWO-WAY ANALYSIS OF VARIANCE In the example/solution in the previous section, we divided the total variation in passen- ger ratings of the airlines into two categories: 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 passen- ger ratings example, there may be other causes of variation. These factors might
LO12-4 Use a blocking variable in a two-way ANOVA to test a hypothesis that three or more population means are equal.
a. Using the .05 level of significance, is there a difference in the mean return on equity among the three types of stock?
b. Can the analyst conclude there is a difference between the mean return on equity for utility and retail stocks? For utility and banking stocks? For banking and retail stocks? Explain.
ANALYSIS OF VARIANCE 407
include, for example, the season 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 variance. That is, if we can reduce the denominator of the F statistic (reducing the error vari- ance 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 reduction in the error variance.
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?
S O L U T I O N
To begin, we conduct a test of hypothesis using a one-way ANOVA. That is, we consider only the four routes. Under this condition, differences in travel times are due to either treatment or random variation. In this example/solution, the subscripts correspond to the treatments or routes: 1 for U.S. 6, 2 for West End, 3 for Hickory Street, and 4 for Route 59. The null hypothesis and the alternate hypothesis for comparing the mean travel time along the four routes are:
H0: μ1 = μ2 = μ3 = μ4 H1: Not all treatment means are the same.
There are four routes, so the numerator degrees of freedom is (k − 1) = (4 − 1) = 3. There are 20 observations, so the degrees of freedom in the denominator is
E X A M P L E
WARTA, the Warren Area Regional Transit Authority, is expanding bus ser- vice 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. Be- cause there will be many different driv- ers, 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.
© John A. Rizzo/Getty Images RF
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
408 CHAPTER 12
(n − k) = (20 − 4) = 16. From Appendix B.6, at the .05 significance level, the critical value of F is 3.24. The decision rule is to reject the null hypothesis if the computed F test statistic’s value is greater than 3.24.
We use Excel to perform the calculations and output the results. The com- puted value of F is 2.483, so we decide to not reject the null hypothesis. We conclude there is no difference in the mean travel time along the four routes. There is no reason to conclude that any one of the routes is faster than any other.
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 differ- ences 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.
In this example, we only considered the variation due to the treatments (routes) and took all the remaining variation to be random. If we include the effect or variance of the drivers, this would allow us to reduce the SSE term, and the computed values of the F statistics would be larger.
In this case, we let the drivers be the blocking variable. To include the variance due to the drivers, 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 following formula.
BLOCKING VARIABLE A second treatment variable that when included in the ANOVA analysis will have the effect of reducing the SSE term.
SSB = kΣ (xb − xG)2 (12–6) where:
k is the number of treatments. b is the number of blocks. xb is the sample mean of block b. xG is the overall or grand mean.
From the calculations on the next page, 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.
ANALYSIS OF VARIANCE 409
Substituting this information into formula (12–6) we determine SSB, the sum of squares due to the drivers (the blocking variable), is 119.7.
SSB = kΣ (xb − xG)2
= 4(19.5 − 22.8)2 + 4(21.0 − 22.8)2 + 4(22.5 − 22.8)2
+ 4(24.75 − 22.8)2 + 4(26.25 − 22.8)2
= 119.7
The SSE term is found by subtraction.
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 19.50 Snaverly 16 23 23 22 84 21.00 Ormson 21 21 26 22 90 22.50 Zollaco 23 22 29 25 99 24.75 Filbeck 25 24 28 28 105 26.25
SUM OF SQUARES ERROR, TWO-WAY SSE = SS total − SST − SSB (12–7)
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 calcu- lated as before, and SSB is found from formula (12–6). The values for the various com- ponents of the ANOVA table are computed as follows.
SSE is found by formula (12–7).
SSE = SS total − SST − SSB = 229.2 − 72.8 − 119.7 = 36.7
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
(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 driv- ers in this example) was only to reduce the error variation, we should not conduct 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 blocking variable. On the other hand, we may wish to give the blocks the same status as the treatments and con- duct a hypothesis test. 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
410 CHAPTER 12
the travel time for the different drivers, so we will conduct the hypothesis test of equal block means. The subscripts are the first letter of each driver’s name. The two sets of hypotheses are:
1. H0: The treatment means are equal (μ1 = μ2 = μ3 = μ4). H1: At least one treatment mean is different. 2. H0: The block means are equal (μD = μS = μO = μZ = μF). H1: At least one block mean is different.
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.
F = MST MSE
= 24.27 3.06
= 7.93
The null hypothesis is rejected and we conclude that at least one of the route’s mean travel times is different from the other routes. WARTA will want to conduct some tests to determine which treatment means differ.
Next, we test to find whether the travel times for the various drivers are equal. The degrees of freedom in the numerator for blocks are (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.
F = MSB MSE
= 29.93 3.06
= 9.78
The null hypothesis about the block means is rejected, and we conclude that at least one driver’s mean travel time is different from the other drivers. Thus, WARTA manage- ment can conclude, based on the sample results, that there is a difference in the mean travel times of drivers.
The Excel spreadsheet has a two-factor ANOVA procedure. The output for the WARTA example just completed follows. This output also includes the p-values. The p-value for the null hypothesis 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.
ANALYSIS OF VARIANCE 411
Vive Shampoo sells three shampoos, one each for dry, normal, and oily hair. Sales, in mil- lions of dollars, for the past 5 months are given in the following table. Using the .05 signifi- cance level, test whether the mean sales differ for the three types of shampoo or by month.
S E L F - R E V I E W 12–4
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
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 regarding the two sets of hypotheses and interpret the results.
15. The following data was collected for a two-factor ANOVA with two treatments and three blocks.
16. The following data was collected for a two-factor ANOVA with three treatments and three blocks.
17. Chapin Manufacturing Company operates 24 hours a day, 5 days a week. The workers 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?
E X E R C I S E S
Treatment
Block 1 2
A 46 31 B 37 26 C 44 35
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
412 CHAPTER 12
TWO-WAY ANOVA WITH INTERACTION In the previous section, we studied the separate or independent effects of two vari- ables, or factors, on a response variable, travel time. In the example/solution, the two factors were the bus routes and the drivers and the response was travel time. The anal- ysis shows two significant results. First, the mean travel times between routes, averaged over all drivers, are different. Second, the mean travel times between the five drivers, averaged over all the routes, are different. What could explain these differences? The differences between routes may simply be related to differences in the distance of the routes. We really didn’t study the distance of the various routes. Perhaps the differences are explained by how fast, on average, the drivers drive regardless of the route.
There is another effect that may influence travel time that we have not considered. This is called the interaction effect between route and driver on travel time. That is, differences in travel time may depend on both the driver and the route. For example, it is possible that one of the drivers is especially good driving one of the routes. Perhaps one driver knows how to effectively time the traffic lights or how to avoid heavily con- gested intersections for one or more of the routes. In this case, differences in mean travel time may depend on the combined effect of driver and route. The results of the interaction of driver and route can provide interesting information.
LO12-5 Perform a two-way ANOVA with interaction and describe the results.
18. There are three hospitals in the Tulsa, Oklahoma, area. The following data show the number of outpatient surgeries performed on Monday, Tuesday, Wednesday, Thursday, and Friday 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?
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
INTERACTION The effect of one factor on a response variable differs depending on the value of another factor.
Interaction Plots An everyday illustration of interaction is the effect of diet and exercise on body weight. It is generally agreed that body weight (the response variable) can be affected by two factors, diet and exercise. However, research shows there is also a combined or interac- tion effect of diet and exercise on weight loss. That is, the amount of weight loss will be different and depend on diet AND whether people exercise.
The following graph, or interaction plot, illustrates the interaction of diet and exer- cise. First, for people who do not exercise, the mean weight losses for Diet 1 and Diet 2 are plotted. These are connected with the blue line. Clearly, there is a difference in weight loss for the two diets. Second, for people who did exercise, the mean weight losses for Diet 1 and Diet 2 are plotted. These are connected with the green line. Again, there is a clear difference in weight loss between Diet 1 and Diet 2 for people who ex- ercise. The plot also shows an interaction effect between diet and exercise on weight
ANALYSIS OF VARIANCE 413
loss. Notice the two lines are not parallel to each other. For Diet 1, the mean weight loss is more when people also exercise. For Diet 2, the mean weight loss is also more when people also exercise, but the weight loss is much greater than for Diet 1. So, what is the effect of diet and exercise on weight loss? It depends on the combined, or interaction, effects of diet and exercise.
0.00
5.00
10.00
15.00
20.00
25.00
30.00
Diet 1 No exercise
Plot of Interaction
W ei
gh t L
os s
Diet 2 Exercise
0.00
5.00
10.00
15.00
20.00
25.00
30.00
Diet 1 No exercise
Plot of No Interaction
W ei
gh t L
os s
Diet 2 Exercise
How would the interaction plot look if there were no interaction? The following graph shows an analysis of diet and exercise with no interaction.
In this case the lines are parallel. Comparing the means, the effect of exercise on weight loss for Diet 1 and Diet 2 is the same. The estimated weight loss is about two pounds. In addition, the effect of diet is the same whether people exercise or not. It is about 13 pounds.
Testing for Interaction To test for an interaction effect, we use a two-way ANOVA with interaction. To illustrate, we return to the previous WARTA example/solution. Restating the issue facing WARTA management: They want to expand bus service from downtown Warren to Starbrick. So far they have concluded, based on statistical analysis, that there is a difference in the mean travel time along the four proposed routes and a difference in the mean travel times of the five drivers. But it is possible that the combination, or the interaction be- tween routes and drivers, has a significant effect on mean travel time.
In this analysis, we call the two variables, route and driver, factors. We refer to the variable, travel time, as the response variable. To test for interaction, the sample data must be replicated for each route. In this case, each driver drives each route three times so there are three observed times for each route/driver combination. This information is summarized in the following Excel spreadsheet.
414 CHAPTER 12
The graph below shows the interaction plot using the information in the above table. The vertical axis is the travel time in minutes. The four routes are labeled on the horizontal axis and each line plots the mean travel times for each driver for all four routes. For ex- ample, the green line reports average travel times for Deans for each of the four routes.
To evaluate interaction effects, a useful first step is to plot the means for each driver/ route combination. For the driver/route combination Driver Deans using Route 6, the mean is 18 minutes, found by (18 + 15 + 21)/3. For the driver/route combination Driver Filbeck using Route 59, the mean is 28 minutes, found by (28 + 30 + 26)/3. In a similar manner we calculate the means for the other cells and summarize the results in the following table.
US 6 Deans
Tr av
el T
im e
15.0
20.0
25.0
30.0
Hickory StWest End
Interaction Plot
Route 59 Snaverly Ormson Zollaco Filbeck
ANALYSIS OF VARIANCE 415
From the graph, what observations can be made about the interaction of driver and route on travel time? Most importantly, the lines are not parallel. Because the five lines are clearly not parallel, there is an interaction effect of driver and route on travel time; that is, travel time depends on the combined effect of driver and route.
Note the differences in travel times. For the U.S. 6 route, Snaverly has the lowest or fastest mean travel time. Deans has the lowest mean travel time for the West End and Hickory Street routes. Zollaco has the slowest average time for the Hickory Street route. There are many other observations that lead to the general observation that travel time is related to the combined effects of driver AND route. The critical question is whether the observed interactions are significant or the differences are due to chance.
Hypothesis Tests for Interaction The next step is to conduct statistical tests to further investigate the possible interaction effects. In summary, our study of travel times has several questions:
• Is there an interaction effect of routes and drivers on mean travel times? • Are the mean travel times for drivers the same? • Are the mean travel times for the routes the same?
Of the three questions, we are most interested in the test for interactions. We formalize these ideas into three sets of hypotheses:
1. H0: There is no interaction between drivers and routes. H1: There is interaction between drivers and routes. 2. H0: The driver means are equal. H1: At least one driver travel time mean is different. 3. H0: The route means are equal. H1: At least one route travel time mean is different.
We test each of these hypotheses as we did in the previous section using the F distribution. The tests are summarized with the following ANOVA table. It is similar to the two-way ANOVA in the previous section with the addition of the Interaction source of variation. In addition, we refer to the driver effect as Factor A and the route effect as Factor B. Each of these hypotheses is tested using the familiar F statistic.
Source of Variation Sum of Squares df Mean Square F
Factor A (driver) SSA k − 1 MSA = SSA/(k − 1) MSA/MSE Factor B (route) SSB b − 1 MSB = SSA/(b − 1) MSB/MSE Interaction SSI (k − 1)(b − 1) MSI = SSI/[(k − 1)(b − 1)] MSI/MSE Error SSE n − kb MSE = SSE/(n − kb) Total n − 1
To test the hypotheses for a two-way ANOVA with interaction, we use the ANOVA: Two-Factor with Replication in the Data Analysis add-in for Excel. The details of using Excel are summarized in Appendix C. The following ANOVA table shows the results of the analysis. We use the p-values to test each hypothesis. Using the .05 significance level, the null hypotheses are rejected if the computed p-value is less than .05.
416 CHAPTER 12
Reviewing the results of the ANOVA, the p-value for the interaction effect of .0456 is less than our significance level of .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.
A significant interaction effect provides important information about the combined effects of the variables. If interaction is present, then a test of differences in the factor means using a one-way ANOVA for each level of the other factor is the next step. 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 route by testing the hypothesis H0: Driver travel times are equal. The results follow.
The results of the one-way ANOVA show there are significant differences in the mean travel times among the drivers for every route, except Route 59 with a p-value of 0.06. A review of the interaction plot may reveal some of the differences. For ex- ample, for the West End route, the graph suggests that Deans has the best mean travel time. Further statistical analysis would test pairs of mean travel times to deter- mine the significant differences between driver travel times for each route that has a significant p-value.
ANALYSIS OF VARIANCE 417
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
Factor A means? How do you know? (b) How many levels does Factor B have? Is there a significant difference among the
Factor 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?
See the following ANOVA table.
S E L F - R E V I E W 12–5
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
19. Consider the following sample data for a two-factor ANOVA analysis. There are two levels (heavy and light) of factor A (weight), and three levels (small, medium, and large) of factor B (size). For each combination of size and weight, there are three observations.
Compute an ANOVA with statistical software, and 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?
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 differ- ence in Factor A means, Factor B means, or the interaction means. Use the .05 significance level. (Hint: estimate the values from the F table.)
E X E R C I S E S
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
ANOVA
Source SS df MS F
Factor A 75 Factor B 25 Interaction 300 Error 600
Total 1000
418 CHAPTER 12
21. A vending machine company sells its packaged foods in a variety of different machines. The company is considering three types of new vending machines. Man- agement wants to know if the different machines affect sales. These vending ma- chines are designated as J-1000, D-320, and UV-57. Management also wants to know if the position of the machines indoors or outdoors affects sales. Each of six similar locations was randomly assigned a machine and position combination. The data below are the number of purchases over four days.
a. Draw the interaction graph. Based on your observations, is there an interaction ef- fect? Based on the graph, describe the interaction effect of machine and position.
b. Compute an ANOVA with statistical software, and 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 techniques. What do you conclude?
22. A large company is organized into three functional areas: manufacturing, mar- keting, 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. Compute an ANOVA with statistical software, and 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 salary for each area using statistical tech- niques. What do you recommend to the distributor?
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
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
C H A P T E R S U M M A R Y
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 change, 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:
F = s21 s22
(12–1)
ANALYSIS OF VARIANCE 419
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 populations 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:
SS total = Σ (x − xG)2 (12–2)
2. The formula for SSE, the sum of squares error, is:
SSE = Σ (x − xc)2 (12–3)
3. The formula for the SST, the sum of squares treatment, is found by subtraction.
SST = SS total − SSE (12–4)
4. This information is summarized in the following ANOVA table and the value of F is determined.
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.
(x1 − x2) ± t√MSE ( 1 n1
+ 1
n2) (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:
SSB = kΣ (xb − xG)2 (12–6)
C. The SSE term, or sum of squares error, is found from the following equation.
SSE = SS total − SST − SSB (12–7)
D. The F statistics for the treatment variable and the blocking variable are determined in the following 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
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
VI. In a two-way ANOVA with repeated observations, we consider two treatment variables and the possible interaction between the variables. The complete ANOVA table including interactions is:
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
420 CHAPTER 12
P R O N U N C I A T I O N K E Y
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
C H A P T E R E X E R C I S E S
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. One variable that Google uses to rank pages on the Internet is page speed, the time it takes for a web page to load into your browser. A source for women’s clothing is rede- signing their page to improve the images that show its products and to reduce its load time. The new page is clearly faster, but initial tests indicate there is more variation in the time to load. A sample of 16 different load times showed that the standard deviation of the load time was 22 hundredths of a second for the new page and 12 hundredths of a second for the current page. At the .05 significance level, can we conclude that there is more variation in the load time of the new page?
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 are the numbers of new cars sold at Sharkey in the last 7 months and for the last 8 months at Dave White. Do you agree with Mr. Sharkey? Use the .01 significance level.
Sharkey 98 78 54 57 68 64 70 Dave White 75 81 81 30 82 46 58 101
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. Create the ANOVA table. What is the value of F? d. What is your decision regarding the null hypothesis?
27. In an ANOVA table, the MSE is 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. Create 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.
Sum of Mean Source Squares df Square F
Treatment 2 Error 20 Total 500 11
ANALYSIS OF VARIANCE 421
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
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 re- sults are shown below. Use the .05 significance level.
Discount Variety Department
$12 $15 $19 13 17 17 14 14 16 12 18 20 15 17 19
30. Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minutes on three different airlines. The results are:
a. Use the .05 significance level and the six-step hypothesis-testing process to check if there is a difference in the mean flight times among the three airlines.
b. Develop a 95% 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 determine whether there is a difference in the mean number of crimes committed among the four districts. He examined the records from six randomly selected days and recorded the number of crimes. At the .05 significance level, can Chief North conclude that there is a difference in the mean number of crimes among the four districts?
Goust Jet Red Cloudtran
51 50 52 51 53 55 52 52 60 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
422 CHAPTER 12
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
a. Using analysis of variance techniques, test H0 that the two mean test scores are equal; α = .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 repair cars. To check if there is any difference in repair times, customers are randomly selected from each repair shop and their repair times in days are recorded. The output from a statisti- cal software package is:
Traditional Lecture Distance
37 50 35 46 41 49 40 44 35 41 34 42 45 43
Summary
Groups Sample Size 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
33. When only two treatments are involved, ANOVA and the Student’s t test (Chapter 11) result in the same conclusions. Also, for computed test statistics, t2 = F. To demonstrate this relationship, use the following example. Fourteen randomly selected students en- rolled in a history course were divided into two groups, one consisting of 6 students who took the course in the normal lecture format. The other group of 8 students took the course in a distance format. At the end of the course, each group was examined with a 50-item test. The following is a list of the number correct for each of the two groups.
ANALYSIS OF VARIANCE 423
Is there evidence to suggest a difference in the mean repair 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 miles per gallon of the three car sizes. What do you conclude? Use the .01 significance level.
Summary
Groups Sample Size 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
Additional results are shown below.
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 as- sembly line and the number of components produced during these periods for each line is recorded. The output from a statistical software package is:
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
Summary
Groups Sample Size 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
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
424 CHAPTER 12
Using the ANOVA results, compare the average delivery times of the four different types of mail.
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% confidence interval for the difference in the means between Line B and Line C.
37. The postal service sorts mail as Priority Mail Express, Priority Mail, First-Class Mail, or Standard Mail. Over a period of 3 weeks, 18 of each type were mailed from the Network Distribution Center in Atlanta, Georgia, to Des Moines, Iowa. The total delivery time in days was recorded. Minitab was used to perform the ANOVA. The results follow:
ANALYSIS OF VARIANCE 425
38. To prevent spam from entering your email inbox, you use a filter. You are interested in knowing if the number of spam emails differs by day of the week. The number of spam emails by day of week is counted and recorded. Minitab is used to perform the data analysis. Here are the results:
Using the ANOVA results, compare the average number of spam emails for each day of the week.
426 CHAPTER 12
a. Is there a difference in the treatment means? b. Is there a difference in the block means?
42. A task requires the completion of four activities. A teacher would like to know if differences in the sequence of the four activities results in different task completion times. The teacher selects three students and demonstrates the activities in random
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
39. Shank’s Inc., a nationwide advertising firm, wants to know whether the size of an advertisement and the color of the advertisement make a difference in the response of magazine 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).
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 6 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?
41. The city of Tucson, Arizona, employs people to assess the value of homes for the purpose of calculating 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 dollars. Can we conclude that there is a difference in the assessors? Use the .05 significance level.
ANALYSIS OF VARIANCE 427
Using the .05 level of significance: a. Is there a difference in the task completion times among the different sequences? b. Is there a difference in task completion times between the students?
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 dif- ferent 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?
order to the students. Then each student completes the task with each of the activity sequences. The completion times are recorded. The following table shows the minutes for each student to complete each task.
Time (minutes)
Sequence Allen Carla Henry
A 22.4 20.8 21.5 B 17.0 19.4 20.7 C 19.2 20.2 21.2 D 20.3 18.6 20.4
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
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. Each of three supermarket chains in the Denver area claims to have the lowest overall prices. As part of an investigative study on supermarket advertising, a local television station conducted a study by randomly selecting nine grocery items. Then, on the same day, an intern was sent to each of the three stores to purchase the nine items. From the receipts, the following data was recorded. At the .05 significance level, is there a difference in the mean price for the nine items between the three supermarkets?
428 CHAPTER 12
45. Listed below are the weights (in grams) of a sample of M&M’s Plain candies, classi- fied 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.
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
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 randomly selected hours from each station, the sum of squared differences between each observation and the mean for its respective radio station, Σ(x − xC)2, are:
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
Hard rock station: 126.29 Classical station: 233.34
Country/western station: 166.79 Easy listening station: 77.57
The total sum of squares for the data is: SS total = 1,099.61. a. Determine SSE. b. Determine SST. c. Complete an ANOVA table. d. At the .05 significance level, is there a difference in the treatment means? e. If the mean for the hard rock station is 51.32 and the mean for the country/western
station is 50.85, determine if there is a difference using the .05 significance level. 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. Random samples of five men and five women were selected in each group.
ANALYSIS OF VARIANCE 429
a. Draw an interaction plot of men and women means by sector. b. Compute an ANOVA with statistical software and, using the .05 significance level,
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
differences in factor means. d. Interpret the results in a brief report.
48. Robert Altoff is vice president of engineering for a manufacturer of household washing machines. As part of a new product development project, he wishes to deter- mine the optimal length of time for the washing cycle. Included in the project is a study of the relationship between the detergent used (four brands) and the length of the washing cycle (18, 20, 22, or 24 minutes). 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
a. Draw an interaction plot of the detergent means by cycle time. b. Compute the ANOVA with statistical software and, using the .05 significance level,
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
differences in factor means. d. Interpret the results in a brief report.
D A T A A N A L Y T I C S
49. The North Valley Real Estate data reports information on the homes sold 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? 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?
d. Adam Marty recently joined North Valley Real Estate and was assigned twenty homes to market and show. When he was hired, North Valley assured him that the twenty homes would be fairly assigned to him. When he reviewed the selling prices of his assigned homes, he thought that the prices were much below the average of $357,000. Adam was able to find the data of the homes assigned to agents in the firm. Use statistical inference to compare the mean price of homes assigned to him to the mean price of homes assigned to the other agents. What do the results indicate? How is your analysis defining fairness?
e. Home buyers finance the purchase of their home with a mortgage. In this data, the mortgages are either a fixed rate mortgage paid over 30 years, or an adjustable rate mortgage. The adjustable rate mortgage provides a lower introductory interest rate for the first five years of occupancy. Then, in the fifth year, the rate is adjusted to the current rate plus an additional percent. Usually, the adjusted rate is higher than
430 CHAPTER 12
the “introductory” rate. With this information, we may predict that the average years of occupancy would be different for home owners based on the type of mortgage and whether they defaulted on the mortgage. Use the data to evaluate this prediction.
50. Refer to the Baseball 2016 data, which report information on the 30 Major League Baseball teams for the 2016 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?
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 .05 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 .05 significance level.
51. Refer to the Lincolnville School District bus data. a. Conduct a test of hypothesis to reveal whether the mean maintenance cost is equal
for each of the bus manufacturers. Use the .01 significance level. b. Conduct a test of hypothesis to determine whether the mean miles traveled since the
last maintenance is equal for each bus manufacturer. Use the .05 significance level.
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 an assumption about a characteristic of a population. 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 hypothesis. 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 pop- ulation parameter under study equaled a particular value. For example, we wish to investigate whether the mean tenure 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 avail- able and when it was not available.
In Chapter 11, we extended the idea of hypothesis testing to whether two independent random samples came from pop- ulations 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 facilities. The alternate hy- pothesis 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 test statistic. 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 distribution and we assume that the distribution of differences follows the normal distribution. One typical paired sample problem calls for recording an individual’s blood pressure before administering medication and then again afterward in order to evaluate the effec- tiveness of the medication. We also considered the case of testing two population proportions. For example, the produc- tion manager 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 inde- pendent normal populations have the same mean. This is accomplished by comparing the variances of the random sam- ples selected from these populations. We apply the usual hypothesis-testing procedure, but we use the F distribution as the test statistic. Often the calculations are tedious, so a statistical software package is recommended.
ANALYSIS OF VARIANCE 431
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 simulta- neous 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.
An additional feature of the two-factor ANOVA is 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 is easily extended to include a test for interactions.
P R O B L E M S
For problems 1–6, state: (a) the null and the alternate hypotheses, (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 sample of 12 balls was dropped from the platform and the mean height of the bounce was 35.5 inches, with a standard deviation of 0.9 inch. At the .05 significance level, can the supervisor conclude that the mean bounce height is less than 36 inches?
2. It was hypothesized that road construction workers, on the average, spend 20 min- utes of each hour not engaged in productive work. 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 pe- riod (exclusive 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 unproduc- tive time is greater than 20 minutes?
3. Stiktite, Inc. plans a test of the mean holding power of two glues designed for plastic. 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 size 12 10 Sample mean 250 252 Sample standard deviation 5 8
At the .01 significance level, is there a difference between the mean holding power of Epox and that of Holdtite?
4. 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 includ- ing the additive. The same procedure was followed for a total of 10 pieces. Then each
432 CHAPTER 12
piece was subjected to brilliant light. The data, the number of hours each piece lasted before it faded beyond a certain point, follow:
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
Using the .05 significance level, determine whether the additive is effective in pro- longing the life of the paint.
5. A Buffalo, New York, cola distributor is featuring a 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 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:
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.
6. 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 analysis of variance (including the possibility of interactions) on the data and de- scribe 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
Major/Year 2014 2015 2016
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
7. A school supervisor is reviewing initial wages of former students (in $000). Samples were taken over 3 years for four different majors (accounting, administration, finance, and mar- keting). For each combination of major and year, three former students were sampled.
ANALYSIS OF VARIANCE 433
a. Here is an interaction plot of the information. What does it reveal?
Accounting 2014
M ea
n W
ag e
60.0
65.0
70.0
75.0
2015
Interaction Plot
2016 Administration Finance Marketing
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.
C A S E S
A. Century National Bank Refer to the description of Century National Bank on page 129 at the end of the Review of Chapters 1–4. With many other options available, customers no lon- ger let their money sit in a checking account. For many years the mean checking balance has been $1,600. Do 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 cus- tomer was 8; now he believes it has increased to more than 10. In fact, the advertising agency that prepares
TV commercials for Century would like to use this on the new commercial being designed. Is there sufficient evidence to conclude that the mean number of transac- tions per customer 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 ac- count balances among the four branches. If there are dif- ferences, between which branches do these differences occur?
d. If proper, test the other hypotheses at the .05 significance level. If it is not appropri- ate, describe why you should not do the tests.
434 CHAPTER 12
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 differently 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 to have enough nurses so that incoming patients need- ing service can be handled promptly. It is stressful for pa- tients 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 dif- ferences 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.
Date Day Patients
9-29-16 Monday 38 9-30-16 Tuesday 28 10-1-16 Wednesday 28 10-2-16 Thursday 30 10-3-16 Friday 35 10-6-16 Monday 35 10-7-16 Tuesday 25 10-8-16 Wednesday 22 10-9-16 Thursday 21 10-10-16 Friday 32 10-13-16 Monday 37 10-14-16 Tuesday 29 10-15-16 Wednesday 27 10-16-16 Thursday 28 10-17-16 Friday 35 10-20-16 Monday 37 10-21-16 Tuesday 26 10-22-16 Wednesday 28 10-23-16 Thursday 23 10-24-16 Friday 33
P R A C T I C E T E S T
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. 3. Assuming the null hypothesis is true, the likelihood of finding a value of the
test statistic at least as extreme as the one found in the sample is called the . 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, if the population standard deviations
are not known, what must we assume 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 −∞ to ∞, 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 freedom 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 hypothesis, 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 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 taxicab companies operating in the Grand Strand area? The Sun News, the local paper, is investigating and obtained the following sample information. At the
ANALYSIS OF VARIANCE 435
.05 significance level, is it reasonable to conclude there is a difference in the mean miles traveled? Assume equal population variances.
Horse and Buggy Variable Yellow Cab Cab Company
Sample size 14 12 Mean miles 837 797 Standard deviation 30 40
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
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. Why can we conclude the treatment means differ?
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO13-1 Explain the purpose of correlation analysis.
LO13-2 Calculate a correlation coefficient to test and interpret the relationship between two variables.
LO13-3 Apply regression analysis to estimate the linear relationship between two variables.
LO13-4 Evaluate the significance of the slope of the regression equation.
LO13-5 Evaluate a regression equation’s ability to predict using the standard estimate of the error and the coefficient of determination.
LO13-6 Calculate and interpret confidence and prediction intervals.
LO13-7 Use a log function to transform a nonlinear relationship.
© Ingram Publishing/Superstock RF
Correlation and Linear Regression13
TRAVELAIR.COM samples domestic airline flights to explore the relationship between airfare and distance. The service would like to know if there is a correlation between airfare and flight distance. If there is a correlation, what percentage of the variation in airfare is accounted for by distance? How much does each additional mile add to the fare? (See Exercise 61 and LO13-2, LO13-3, and LO13-5.)
CORRELATION AND LINEAR REGRESSION 437
INTRODUCTION Chapters 2 through 4 presented descriptive statistics. We organized 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 Chapters 8 through 12, we studied statistical inference, where we collected a sample to estimate a population parameter such as the popula- tion mean or population proportion. In addition, we used the sample data to test a hypothesis about a population mean or a population proportion, the difference be- tween 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 relationships between two interval- or ratio-level variables. In all business fields, identifying and studying relation- ships between variables 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 example, 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 reduction 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 price and demand.
As another familiar example, recall 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 page 116. In that graph, 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 be- tween 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 mean- ing and purpose of correlation analysis. We continue by developing an equation 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.
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 rela- tionship between the variables. The next step is usually to calculate the correlation co- efficient. It provides a quantitative measure of the strength of the relationship between
LO13-1 Explain the purpose of correlation analysis.
STATISTICS IN ACTION
The space shuttle Challenger exploded on January 28, 1986. An investigation of the cause examined four contractors: Rockwell International for the shuttle and engines, Lockheed Martin for ground support, Martin Marietta for the external fuel tanks, and Morton Thiokol for the solid fuel booster rockets. After sev- eral months, the investiga- tion blamed the explosion on defective O-rings pro- duced by Morton Thiokol. A study of the contractor’s stock prices showed an interesting happenstance. On the day of the Chal- lenger explosion, Morton Thiokol stock was down 11.86% and the stock of the other three lost only 2 to 3%. Can we conclude that financial markets predicted the outcome of the investigation?
438 CHAPTER 13
two variables. As an example, the sales manager of North American Copier Sales, 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 15 representatives and determines, for each representative, the number of sales calls made and the number of copiers sold. This information is reported in Table 13–1.
By reviewing the data, we observe that there does seem to be some relationship between the number of sales calls and the number of units sold. That is, the salespeo- ple who made the most sales calls sold the most units. However, the relationship 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.
TABLE 13–1 Number of Sales Calls and Copiers Sold for 15 Salespeople
Sales Representative Sales Calls Copiers Sold
Brian Virost 96 41 Carlos Ramirez 40 41 Carol Saia 104 51 Greg Fish 128 60 Jeff Hall 164 61 Mark Reynolds 76 29 Meryl Rumsey 72 39 Mike Kiel 80 50 Ray Snarsky 36 28 Rich Niles 84 43 Ron Broderick 180 70 Sal Spina 132 56 Soni Jones 120 45 Susan Welch 44 31 Tom Keller 84 30
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 vari- ables. The usual first step is to plot the data in a scatter diagram. An example will show how a scatter diagram is used.
E X A M P L E
North American Copier Sales 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 representa- tives 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 15 sales representatives and determines 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
CORRELATION AND LINEAR REGRESSION 439
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.
S O L U T I O N
Based on the information in Table 13–1, Ms. Bancer suspects there is a relationship between the number of sales calls made in a month and the number of copiers sold. Ron Broderick sold the most copiers last month and made 180 sales calls. On the other hand, Ray Snarsky, Carlos Ramirez, and Susan Welch made the fewest calls: 36, 40, and 44. They also 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 estimating or predicting the dependent variable. For example, we would like to predict the expected number of copiers sold if a salesperson makes 100 sales calls. In the randomly selected sam- ple data, the independent variable—sales calls—is 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 particular value of the independent variable. The dependent variable is random. That is, for a given value of the indepen- dent variable, there are many possible outcomes for the dependent variable.
It is common practice to scale the dependent variable (copiers sold) on the vertical or Y-axis and the independent variable (number of sales calls) on the hori- zontal or X-axis. To develop the scatter diagram of the North American Copier Sales information, we begin with the first sales representative, Brian Virost. Brian made 96 sales calls last month and sold 41 copiers, so x = 96 and y = 41. To plot this point, move along the horizontal axis to x = 96, then go vertically to y = 41 and place a dot at the intersection. This process is continued until all the paired data are plotted, as shown in Chart 13–1.
CHART 13–1 Scatter Diagram Showing Sales Calls and Copiers Sold
0 0
10 20 30 40 50 60 70 80
20 40 60 80 100
Sales Calls (x )
Co pi
er s
So ld
(y )
120 140 160 180 200
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 na- tional sales manager, 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 straight line. In the following section, you will measure the strength and direction of this relationship between two variables by determining the correlation coefficient.
440 CHAPTER 13
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 −1.00 to +1.00 inclusive. A cor- relation coefficient of −1.00 or +1.00 indicates perfect correlation. For example, a cor- relation coefficient for the preceding example computed to be +1.00 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 −1.00 would reveal 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 variables were linear and perfect is shown in Chart 13–2.
LO13-2 Calculate a correlation coefficient to test and interpret the relationship between two variables.
X
Y
X
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
If there is absolutely no relationship between the two sets of variables, Pear- son’s r is zero. A correlation coefficient r close to 0 (say, .08) shows that the linear relationship is quite weak. The same conclusion is drawn if r = −.08. Coefficients of −.91 and +.91 have equal strength; both indicate very strong correlation between the two variables. Thus, the strength of the correlation does not depend on the di- rection (either − or +).
Scatter diagrams for r = 0, a weak r (say, −.23), and a strong r (say, +.87) 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 representing 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.
Strong Positive Correlation Between Hours Studied and Score
0 5 Hours
Sc or
e
10 15 60 65 70 75 80 85 90 95
100
Weak Negative Correlation Between Price and Quantity
200 250 Price
Qu an
tit y
300 350
600 550 500 450 400 350 300 25040
–1 0 1 2 Number of Children
No Correlation Between Income and Number of Children
In co
m e
($ 00
0)
3 4 5 6
45 50 55 60 65 70 75 80 85
CHART 13–3 Scatter Diagrams Depicting Zero, Weak, and Strong Correlation
CORRELATION AND LINEAR REGRESSION 441
The following drawing summarizes the strength and direction of the correlation coefficient.
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 −1 up to and including +1. 4. A value near 0 indicates there is little linear relationship between the variables. 5. A value near 1 indicates a direct or positive linear relationship between the variables. 6. A value near −1 indicates an inverse or negative linear relationship between the
variables.
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
The characteristics of the correlation coefficient are summarized below.
How is the value of the correlation coefficient determined? We will use the North American Copier Sales in Table 13–1 as an example. It is replicated in Table 13–2 for your convenience.
Sales Representative Sales Calls Copiers Sold
Brian Virost 96 41 Carlos Ramirez 40 41 Carol Saia 104 51 Greg Fish 128 60 Jeff Hall 164 61 Mark Reynolds 76 29 Meryl Rumsey 72 39 Mike Kiel 80 50 Ray Snarsky 36 28 Rich Niles 84 43 Ron Broderick 180 70 Sal Spina 132 56 Soni Jones 120 45 Susan Welch 44 31 Tom Keller 84 30
Total 1440 675
TABLE 13–2 Number of Sales Calls and Copiers Sold for 15 Salespeople
442 CHAPTER 13
We begin 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 96 calls (x = Σx/n = 1440/15 = 96) and a horizontal line at 45 copiers (y = Σy/n = 675/15 = 45). 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 (96, 45).
0 0
10
20
30
40
50
60
70
80
20 40 60 80 100 Sales Calls (x )
Co pi
er s
So ld
(y )
120 140 160 180 200
Quadrant IV Quadrant I
Quadrant III Quadrant II
y = 45
x = 96
CHART 13–4 Computation of the Correlation Coefficient
TABLE 13–3 Deviations from the Mean and Their Products
Sales Representative Sales Calls (x) Copiers Sold (y) x − x x − y (x − x ) ( y − y )
Brian Virost 96 41 0 −4 0 Carlos Ramirez 40 41 −56 −4 224 Carol Saia 104 51 8 6 48 Greg Fish 128 60 32 15 480 Jeff Hall 164 61 68 16 1,088 Mark Reynolds 76 29 −20 −16 320 Meryl Rumsey 72 39 −24 −6 144 Mike Kiel 80 50 −16 5 −80 Ray Snarsky 36 28 −60 −17 1,020 Rich Niles 84 43 −12 −2 24 Ron Broderick 180 70 84 25 2,100 Sal Spina 132 56 36 11 396 Soni Jones 120 45 24 0 0 Susan Welch 44 31 −52 −14 728 Tom Keller 84 30 −12 −15 180 Totals 1440 675 0 0 6,672
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 third person on the list in Table 13–2, Carol Saia, made 104 sales calls and sold 51 copiers. These values are above their respective means, so this point is located in Quadrant I, which is in the up- per-right quadrant. She made 8 more calls than the mean number of sales calls and sold 6 more than the mean number sold. Tom Keller, the last name on the list in Table 13–2, made 84 sales calls and sold 30 copiers. Both of these values are less than their respective means, hence this point is in the lower-left quadrant. Tom made 12 fewer sales calls and sold 15 fewer copiers than the respective means. The deviations from the mean number of sales calls and the mean number of copiers sold are summarized in Table 13–3 for the
CORRELATION AND LINEAR REGRESSION 443
15 sales representatives. The sum of the products of the deviations from the respective means is 6672. That is, the term Σ(x − x )(y − y ) = 6672.
In both the upper-right and the lower-left quadrants, the product of (x − x ) (y − y ) is positive because both of the factors have the same sign. In our example, this happens for all sales representatives except Mike Kiel. Mike made 80 sales calls (which is less than the mean) but sold 50 machines (which is more than the mean). 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, Quadrants II and IV. Now (x − x ) and (y − y ) will have opposite signs, so their product is negative. The resulting correlation coefficient 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 (x − x ) (y − y ) offset the positive products, so the sum is near zero. This leads to a cor- relation coefficient near zero. So, the term Σ (x − x ) (y − y ) drives the strength as well as the sign of the relationship between the two variables.
The correlation coefficient is also unaffected by the units of the two variables. For example, if we had used hundreds of copiers sold instead of the number sold, the cor- relation coefficient would be the same. The correlation coefficient is independent of the scale used if we divide the term Σ (x − x ) (y − y ) by the sample standard deviations. It is also made independent of the sample size and bounded by the values +1.00 and −1.00 if we divide by (n − 1).
This reasoning leads to the following formula:
To compute the correlation coefficient, we use the standard deviations of the sam- ple of 15 sales calls and 15 copiers sold. We could use formula (3–9) to calculate the sample standard deviations or we could use a statistical software package. For the specific Excel and Minitab commands, see the Software Commands in Appendix C. The following is the Excel output. The standard deviation of the number of sales calls is 42.76 and of the number of copiers sold 12.89.
CORRELATION COEFFICIENT r = Σ (x − x ) ( y − y)
(n − 1)sx sy [13–1]
We now insert these values into formula (13–1) to determine the correlation coefficient:
r = Σ (x − x )( y − y)
(n − 1)sx sy =
6672 (15 − 1)(42.76)(12.89)
= 0.865
444 CHAPTER 13
How do we interpret a correlation of 0.865? First, it is positive, so we conclude there is a direct relationship between the number of sales calls and the number of copi- ers sold. This confirms our reasoning based on the scatter diagram, Chart 13–4. The value of 0.865 is fairly close to 1.00, so we conclude that the association is strong.
We must be careful with the interpretation. The correlation of 0.865 indicates a strong positive linear 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 statistically related.
If there is a strong relationship (say, .97) between two variables, we are tempted to assume that an increase or decrease in one variable causes a change in the other vari- able. For example, historically, the consumption of Georgia peanuts and the consump- tion 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 in- creased proportionately. Further, as the population of donkeys has decreased, there has been an increase in the number of doctoral degrees granted. Relationships such as these are called spurious correlations. What we can conclude when we find two vari- ables 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.
E X A M P L E
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, for whom 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?
S O L U T I O N
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.
CHART 13–5 Scatter Diagram of Profit versus Age for the Applewood Auto Group Data
3500
3000
2500
2000
1500
1000
500
0 0 20 40 60 80
Age
Scatter Plot of Pro t vs. Age
Pr o
t
The scatter diagram suggests that a positive relationship does exist between age and profit; however, that relationship does not appear strong.
CORRELATION AND LINEAR REGRESSION 445
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.
A B C
1 Age Profit
2 Age 1
3 Profit 0.262 1
For these 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 correlation coefficient is: r = 0.262. It is much closer to zero than one. Therefore, the relationship between the two variables is weak. We would observe that the relationship between the age of a buyer and the profit of their purchase is not very strong.
For Applewood Auto Group, the data does not support a business decision to cre- ate an advertising campaign to attract older buyers.
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 In- ternet, 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.
S E L F - R E V I E W 13–1
(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.
Advertising Expense Sales Revenue Month ($ million) ($ million)
July 2 7 August 1 3 September 3 8 October 4 10
1. The following sample of observations were randomly selected.
E X E R C I S E S
x 4 5 3 6 10 y 4 6 5 7 7
Determine the correlation coefficient and interpret the relationship between x and y.
446 CHAPTER 13
2. The following sample of observations were randomly selected.
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
x 5 3 6 3 4 4 6 8 y 13 15 7 12 13 11 9 5
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 information 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 advertisement was aired and digital camera sales. The pairings are:
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 assem- ble the subassemblies. They produced 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 production 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 be-
tween 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 of police and the number of crimes reported. The chief gath- ered the following sample information.
CORRELATION AND LINEAR REGRESSION 447
Testing the Significance of the Correlation Coefficient Recall that the sales manager of North American Copier Sales found the correlation be- tween the number of sales calls and the number of copiers sold was 0.865. This indicated a strong positive association between the two variables. However, only 15 salespeople were sampled. Could it be that the correlation in the population is actually 0? This would mean the correlation of 0.865 was due to chance, or sampling error. The population in this example is all the salespeople employed by the firm.
Resolving this dilemma requires a test to answer the question: Could there be zero correlation in the population from which the sample was selected? To put it an- other 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 ρ represent the correlation in the population. It is pronounced “rho.”
We will continue with the illustration involving sales calls and copiers sold. We em- ploy the same hypothesis testing steps described in Chapter 10. The null hypothesis and the alternate hypothesis are:
H0: ρ = 0 (The correlation in the population is zero.) H1: ρ ≠ 0 (The correlation in the population is different from zero.)
This is a two-tailed test. The null hypothesis can be rejected with either large or small sample values of the correlation coefficient.
a. Which variable is the dependent variable and which is the independent variable? Hint: Which of the following makes better sense: Cities with more police have fewer crimes, or cities with fewer crimes have more police? Explain your choice.
b. Draw a scatter diagram. c. Determine the correlation coefficient. d. Interpret the correlation coefficient. Does it surprise you that the correlation co-
efficient is negative? 6. The owner of Maumee Ford-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.
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?
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
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
448 CHAPTER 13
The formula for t is:
t TEST FOR THE CORRELATION COEFFICIENT
t = r√n − 2 √1 − r 2
with n − 2 degrees of freedom [13–2]
Using the .05 level of significance, the decision rule states that if the computed t falls in the area between plus 2.160 and minus 2.160, the null hypothesis is not rejected. To locate the critical value of 2.160, refer to Appendix B.5 for df = n − 2 = 15 − 2 = 13. See Chart 13–6.
Applying formula (13–2) to the example regarding the number of sales calls and units sold:
t = r√n − 2 √1 − r 2
= .865√15 − 2
√1 − .8652 = 6.216
The computed t is in the rejection region. Thus, H0 is rejected at the .05 significance level. Hence we conclude the correlation in the population is not zero. This 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 computed, when H0 is true. To determine the p-value, go to the t distribution in Appendix B.5 and find the row for 13 degrees of freedom. The value of the test statistic is 6.216, so in the row for 13 degrees of freedom and a two-tailed test, find the value closest to 6.216. For a two-tailed test at the 0.001 significance level, the critical value is 4.221. Because 6.216 is greater than 4.221, we conclude that the p-value is less than 0.001.
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 correla- tion in the population between the two variables is 0. The Minitab output follows.
–2.160 0 2.160 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 13 df
CORRELATION AND LINEAR REGRESSION 449
E X A M P L E
In the Applewood Auto Group example on page 444, we found that the correla- tion coefficient between the profit on the sale of a vehicle by the Applewood Auto Group and the age of the person that purchased the vehicle was 0.262. The sign of the correlation coefficient was positive, so we concluded there was a direct relationship between the two variables. However, because the value of the correlation coefficient was small—that is, near zero—we concluded that an advertising campaign directed toward the older buyers was not warranted. We can test our conclusion by conducting a hypothesis test that the correlation coef- ficient is greater than zero using the .05 significance level.
S O L U T I O N
To test the hypothesis, we need to clarify the sample and population issues. Let’s assume that the data collected on the 180 vehicles 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 ρ is the correlation 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 or less than zero. The alternate hypothesis is that there is positive correlation between the two variables.
H0: ρ ≤ 0 (The correlation in the population is negative or equal to zero.) H1: ρ > 0 (The correlation in the population is positive.)
This is a one-tailed test because we are interested in confirming a positive associa- tion between the variables. The test statistic follows the t distribution with n − 2 degrees of freedom, so the degrees of freedom are 180 − 2 = 178. However, the value for 178 degrees of freedom is not in Appendix B.5. 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.
t = r √n − 2 √1 − r 2
= 0.262√180 − 2
√1 − 0.2622 = 3.622
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.
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 re- lationship between age and profit exists. How can this be? We must be very careful about the application of the hypothesis test results. The hypothesis test shows a statistically significant result. However, this result does not necessarily support a practical decision to start a new marketing and promotion campaign to older pur- chasers. In fact, the relatively low correlation coefficient is an indication that the outcome of a new marketing and promotion campaign to older potential purchas- ers is, at best, uncertain.
450 CHAPTER 13
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?
S E L F - R E V I E W 13–2
7. The following hypotheses are given.
H0: ρ ≤ 0 H1: ρ > 0
A random sample of 12 paired observations indicated a correlation of .32. Can we conclude that the correlation in the population is greater than zero? Use the .05 significance level.
8. The following hypotheses are given.
H0: ρ ≥ 0 H1: ρ < 0
A random sample of 15 paired observations has a correlation of −.46. Can we con- clude that the correlation in the population is less than zero? Use the .05 signifi- cance 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 passengers on a particular flight and the cost of the flight. It seems logical that more passengers 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 number 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 significance level.
12. The Student Government Association at Middle Carolina University wanted to demonstrate the relationship between the number of beers a student drinks and his or her 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 they consumed their assigned number of beers, a member of the local sheriff’s office measured their blood alco- hol content. The sample information is reported below.
E X E R C I S E S
Student Beers BAC Student Beers BAC
Charles 6 0.10 Jaime 3 0.07 Ellis 7 0.09 Shannon 3 0.05 Harriet 7 0.09 Nellie 7 0.08 Marlene 4 0.10 Jeanne 1 0.04 Tara 5 0.10 Michele 4 0.07 Kerry 3 0.07 Seth 2 0.06 Vera 3 0.10 Gilberto 7 0.12 Pat 6 0.12 Lillian 2 0.05 Marjorie 6 0.09 Becky 1 0.02
CORRELATION AND LINEAR REGRESSION 451
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 coef- ficient. Regression analysis is another method to examine a linear relationship between two variables. This analysis uses the basic concepts of correlation but pro- vides much more information by expressing the linear relationship between two vari- ables in the form of an equation. 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 number of units sold for a sample of 15 sales representatives employed by North American Copier Sales. Chart 13–1 portrayed this information in a scatter diagram. Recall that we tested the significance of the correlation coefficient (r = 0.865) and concluded that a significant relationship exists between the two vari- ables. Now we want to develop a linear equation that expresses the relationship between the number of sales calls, the in- dependent 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.
LO13-3 Apply regression analysis to estimate the linear relationship between two variables.
© Image Source/Getty Images RF
Use a statistical software package to answer the following questions. a. Develop a scatter diagram for the number of beers consumed and BAC. Com-
ment 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
relationship in the population between the number of beers consumed and the BAC? What is the p-value?
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 company’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 regression 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 example, if a stock has a beta of 1.5, then when the S&P index in- creases 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 de- crease 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.
REGRESSION EQUATION An equation that expresses the linear relationship between two variables.
Least Squares Principle In regression analysis, our objective is to use the data to position a line that best rep- resents 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. 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. How- ever, each would result in a different estimate of units sold for a particular number of sales calls.
452 CHAPTER 13
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.
0 0 50 100
Sales Calls (x )
Co pi
er s
So ld
(y )
150 200
10 20 30 40 50 60 70 80
CHART 13–7 Sales Calls and Copiers Sold for 15 Sales Representatives
0 0 50 100
Sales Calls (x )
Co pi
er s
So ld
(y )
150 200
10 20 30 40 50 60 70 80
Line A
CHART 13–8 Four Lines Superimposed on the Scatter Diagram
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 follow. 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 (x = 3, y = 8) deviates by 2 from the line, found by 10 − 8. The deviation squared is 4. The squared deviation for the plot x = 4, y = 18 is 16. The squared deviation for the plot x = 5, y = 16 is 4. The sum of the squared deviations is 24, found by 4 + 16 + 4.
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, it is 132. Both sums are greater than the sum for the line in Chart 13–9, found by using the least squares method.
CORRELATION AND LINEAR REGRESSION 453
The equation of a line has the form
26
22
18
14
10
6 2 3 4 5 6
Ac hi
ev em
en t s
co re
Years of service with company
6
2
2
26
22
18
14
10
6
4
2
2
2 3 4 5 6
Ac hi
ev em
en t s
co re
Years of service with company
CHART 13–9 The Least Squares Line CHART 13–10 Line Drawn with a Straight Edge
26
22
18
14
10
6 2 3 4 5 6
Ac hi
ev em
en t s
co re
Years of service with company
8
2 8
CHART 13–11 Different Line Drawn with a Straight Edge
ŷ = a + bx [13–3]GENERAL FORM OF LINEAR REGRESSION EQUATION
where: ŷ , 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 x = 0. 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 ŷ 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 equation that best fits the data.
The formulas for a and b are:
SLOPE OF THE REGRESSION LINE b = r ( sy sx)
[13–4]
Y-INTERCEPT a = y − bx [13–5]
where: r is the correlation coefficient. sy is the standard deviation of y (the dependent variable). sx is the standard deviation of x (the independent variable).
where: y is the mean of y (the dependent variable). x is the mean of x (the independent variable).
454 CHAPTER 13
E X A M P L E
Recall the example involving North American Copier Sales. The sales manager gathered information on the number of sales calls made and the number of copiers sold for a random sample of 15 sales representatives. As a part of her presentation at the upcoming sales meeting, Ms. Bancer, the sales manager, would like to offer specific information about the relationship between the num- ber of sales calls and the number of copiers sold. Use the least squares method to determine a linear equation to express the relationship between the two variables. What is the expected number of copiers sold by a representative who made 100 calls?
S O L U T I O N
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. In the previous section on page 443, we determined the correlation coefficient r (0.865). In the Excel output on page 443, we determined the standard deviation of the independent variable x (42.76) and the standard deviation of the dependent variable y (12.89). The values are inserted in formula (13–4).
b = r ( sy sx)
= .865 ( 12.89 42.76) = 0.2608
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 worksheet on page 443. From formula (13–5):
a = y − bx = 45 − .2608(96) = 19.9632
Thus, the regression equation is
ŷ = 19.9632 + 0.2608x.
So if a salesperson makes 100 calls, he or she can expect to sell 46.0432 copiers, found by
ŷ = 19.9632 + 0.2608x = 19.9632 + 0.2608(100) = 46.0432
The b value of .2608 indicates that for each additional sales call, the sales repre- sentative can expect to increase the number of copiers sold by about 0.2608. To put it another way, 20 additional sales calls in a month will result in about five more copiers being sold, found by 0.2608 (20) = 5.216.
The a value of 19.9632 is the point where the equation crosses the Y-axis. A literal translation is that if no sales calls are made, that is x = 0, 19.9632 copiers will be sold. Note that x = 0 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 36 to 180, so estimates should be limited to that range.
Drawing the Regression Line The least squares equation ŷ = 19.9632 + 0.2608x can be drawn on the scatter dia- gram. The fifth sales representative in the sample is Jeff Hall. He made 164 calls. His estimated number of copiers sold is ŷ = 19.9632 + 0.2608(164) = 62.7344. The plot x = 164 and ŷ = 62.7344 is located by moving to 164 on the X-axis and then going vertically to 62.7344. The other points on the regression equation can be determined
CORRELATION AND LINEAR REGRESSION 455
The least squares regression line has some interesting and unique features. First, it will always pass through the point (x , y ). 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 96, found by x = 1440/15. The mean number of copiers sold is 45.0, found by y = 675/15. If we let x = 96 and then use the regression equation to find the estimated value for ŷ the result is:
ŷ = 19.9632 + 0.2608(96) = 45
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 repre- sented by the two means. In this case, the regression equation will pass through the point x = 96 and y = 45.
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 Σ( y − ŷ )2 is smaller for the least squares regression equation than for any other equation. We use the Excel system to demonstrate this result in the following printout.
0 20
30
40
50
60
70
80
(x = 40, y = 30.3952)
(x = 120, y = 51.2592)
(164, 62.7344)
50 100
Sales Calls (x )
Co pi
er s
So ld
(y )
150 200
CHART 13–12 The Line of Regression Drawn on the Scatter Diagram
Sales Representative Sales Calls (x) Copiers Sold (y) Estimated Sales (ŷ)
Brian Virost 96 41 45.0000 Carlos Ramirez 40 41 30.3952 Carol Saia 104 51 47.0864 Greg Fish 128 60 53.3456 Jeff Hall 164 61 62.7344 Mark Reynolds 76 29 39.7840 Meryl Rumsey 72 39 38.7408 Mike Kiel 80 50 40.8272 Ray Snarsky 36 28 29.3520 Rich Niles 84 43 41.8704 Ron Broderick 180 70 66.9072 Sal Spina 132 56 54.3888 Soni Jones 120 45 51.2592 Susan Welch 44 31 31.4384 Tom Keller 84 30 41.8704
by substituting a particular value of x into the regression equation and calculating ŷ . All the points are connected to give the line. See Chart 13–12.
456 CHAPTER 13
In columns A, B, and C in the Excel spreadsheet above, we duplicated the sample infor- mation on sales and copiers sold from Table 13–1. In column D, we provide the esti- mated sales values, the ŷ values, as calculated above.
In column E, we calculate the residuals, or the error values. This is the difference be- tween the actual values and the predicted values. That is, column E is (y − ŷ ). For Soni Jones,
ŷ = 19, 9632 + 0.2608(120) = 51.2592
Her actual value is 45. So the residual, or error of estimate, is
(y − ŷ ) = (45 − 51.2592) = −6.2592
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 587.111.
Σ(y − ŷ )2 = 16.0000 + 112.4618 + ⋅ ⋅ ⋅ + 140.9064 = 587.1108
This is the sum of the squared differences or the least squares value. There is no other line through these 15 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 squared differences for these equations. In column G, we use the equation y* = 18 + 0.275x 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, Brian Virost,
y* = 18 + 0.275(96) = 44.4
(y − y*)2 = (41 − 44.4)2 = 11.56
This procedure is continued for the other 14 sales representatives and the squared residuals totaled. The result is 597.8. This is a larger value (597.8 is more than 587.1108) 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 y** = 20 + 0.225x. Again, this equation is similar to the least squares equation. The details for Brian Virost are:
y** = 20 + 0.225x = 20 + 0.225(96) = 41.6
( y − y**)2 = (41 − 41.6)2 = 0.36
This procedure is continued for the other 14 sales representatives and the residuals totaled. The result is 793, which is also larger than the least squares values.
A B C D E F G H I J 1 Sales Rep Sales Calls (x) Copiers Sold (y) Estimated Sales (y – ŷ ) (y – ŷ )2 y* (y – y* )2 y** (y – y** )2
2 Brian Virost 96 41 45.0000 –4.0000 16.0000 44.4000 11.5600 41.6000 0.3600 3 Carlos Ramirez 40 41 30.3952 10.6048 112.4618 29.0000 144.0000 29.0000 144.0000 4 Carol Saia 104 51 47.0864 3.9136 15.3163 46.6000 19.3600 43.4000 57.7600 5 Greg Fish 128 60 53.3456 6.6544 44.2810 53.2000 46.2400 48.8000 125.4400 6 Jeff Hall 164 61 62.7344 –1.7344 3.0081 63.1000 4.4100 56.9000 16.8100 7 Mark Reynolds 76 29 39.7840 –10.7840 116.2947 38.9000 98.0100 37.1000 65.6100 8 Meryl Rumsey 72 39 38.7408 0.2592 0.0672 37.8000 1.4400 36.2000 7.8400 9 Mike Kiel 80 50 40.8272 9.1728 84.1403 40.0000 100.0000 38.0000 144.0000
10 Ray Snarsky 36 28 29.3520 –1.3520 1.8279 27.9000 0.0100 28.1000 0.0100 11 Rich Niles 84 43 41.8704 1.1296 1.2760 41.1000 3.6100 38.9000 16.8100 12 Ron Broderick 180 70 66.9072 3.0928 9.5654 67.5000 6.2500 60.5000 90.2500 13 Sal Spina 132 56 54.3888 1.6112 2.5960 54.3000 2.8900 49.7000 39.6900 14 Soni Jones 120 45 51.2592 –6.2592 39.1776 51.0000 36.0000 47.0000 4.0000 15 Susan Welch 44 31 31.4384 –0.4384 0.1922 30.1000 0.8100 29.9000 1.2100 16 Tom Keller 84 30 41.8704 –11.8704 140.9064 41.1000 123.2100 38.9000 79.2100 17 Total 0.0000 587.1108 597.8000 793.0000
CORRELATION AND LINEAR REGRESSION 457
What have we shown with the example? The sum of the squared residuals [Σ(y − ŷ )2] 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.
Refer to Self-Review 13–1, where the owner of Haverty’s Furniture Company was study- ing the relationship between sales and the amount spent on advertising. The advertising expense and sales revenue, both in millions of dollars, for the last 4 months are re- peated below.
S E L F - R E V I E W 13–3
(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
13. The following sample of observations was 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 of observations was randomly selected.
x: 5 3 6 3 4 4 6 8 y: 13 15 7 12 13 11 9 5
a. Determine the regression equation. b. Determine the value of ŷ 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.
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
a. Determine the regression equation. b. Determine the number of kilowatt-hours, in thousands, for a six-room house.
E X E R C I S E S
458 CHAPTER 13
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 $50.0 million in sales, estimate the earnings.
18. We are studying mutual bond funds for the purpose of investing in several funds. For this particular study, we want to focus on the assets of a fund and its five- year performance. The question is: Can the five-year rate of return be estimated based on the assets of the fund? Nine mutual funds were selected at random, and their assets and rates of return are shown below.
Sales Earnings Sales Earnings Company ($ 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
Assets Return Assets Return Fund ($ millions) (%) Fund ($ millions) (%)
AARP High Quality Bond $622.2 10.8 MFS Bond A $494.5 11.6 Babson Bond L 160.4 11.3 Nichols Income 158.3 9.5 Compass Capital Fixed Income 275.7 11.4 T. Rowe Price Short-term 681.0 8.2 Galaxy Bond Retail 433.2 9.1 Thompson Income B 241.3 6.8 Keystone Custodian B-1 437.9 9.2
16. Mr. James McWhinney, president of Daniel-James Financial Services, believes there is a relationship between the number of client contacts and the dollar amount of sales. To document this assertion, Mr. McWhinney gathered the following sample information. The x column indicates the number of client contacts last month and the y column shows the value of sales ($ thousands) last month for each client sampled.
a. Determine the regression equation. b. Determine the estimated sales if 40 contacts are made.
17. A recent article in Bloomberg Businessweek listed the “Best Small Compa- nies.” 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.
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
CORRELATION AND LINEAR REGRESSION 459
a. Draw a scatter diagram. b. Compute the correlation coefficient. c. Write a brief report of your findings for parts (a) and (b). 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. Assume the dependent variable is number of crimes.
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.
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 princi- ple. 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 hypoth- esis to see if the slope of the regression line is different from zero. Why is this import- ant? If we can show that the slope of the line in the population is different 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 variable. If we cannot demonstrate that this slope is different from zero, then we conclude 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 depen- dent variable as a predictor, rather than use the regression equation.
Following from the hypothesis-testing procedure in Chapter 10, the null and alter- native hypotheses are:
H0 : β = 0
H1 : β ≠ 0
We use β (the Greek letter beta) to represent the population slope for the regression equation. This is consistent with our policy to identify population parameters by Greek letters. We assumed the information regarding North American Copier Sales, Table 13–2, is a sample. Be careful here. Remember, this is a single sample of salespeople, but when we selected a particular salesperson we identified two pieces of information: how many customers they called on and how many copiers they sold.
We identified the slope value as b. So b is our computed slope based on a sample and is an estimate of the population’s slope, identified as β. The null hypothesis is that the slope of the regression equation in the population is zero. If this is the case, the re- gression line is horizontal and there is no relationship between the independent vari- able, 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 al- ternate hypothesis accepted, this indicates that the slope of the regression line for the population is not equal to zero. To put it another way, a significant relationship exists between the two variables. Knowing the value of the independent variable allows us to estimate the value of the dependent variable.
LO13-4 Evaluate the significance of the slope of the regression equation.
460 CHAPTER 13
Before we test the hypothesis, we use statistical software to determine the needed regression statistics. We continue to use the North American Copier Sales data from Table 13–2 and use Excel to perform the necessary calculations. The following spread- sheet 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 .865, which is the correlation coefficient we calculated using formula (13–1).
2. Next is an ANOVA table. This is a useful table for summarizing regression informa- tion. 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 0.2606, and the intercept, which is 19.98. (Note that these values for the slope and the intercept are slightly different from those computed in the Example/Solution on page 454. 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 standard 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 intercept values. The standard error of the slope coefficient is 0.0420.
To test the null hypothesis, we use the t-distribution with (n − 2) and the following formula.
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:
H0 : β ≤ 0
H1 : β > 0
Notice that we have a one-tailed test. If we do not reject the null hypothesis, we con- clude 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
t = b − 0
sb with n − 2 degrees of freedom [13–6]TEST FOR THE SLOPE
CORRELATION AND LINEAR REGRESSION 461
variable. In our case, this means that knowing the number of sales calls made by a rep- resentative 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 salesperson, we can predict or forecast their 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 13 degrees of freedom, found by n − 2 = 15 − 2. We use the .05 significance level. From Appendix B.5, the critical value is 1.771. Our decision rule is to reject the null hypothesis if the value computed from formula (13–6) is greater than 1.771. We apply formula (13–6) to find t.
t = b − 0
sb =
0.2606 − 0 0.042
= 6.205
The computed value of 6.205 exceeds our critical value of 1.771, 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, number of sales calls, is useful in esti- mating copier sales.
The table also provides us information on the p-value of this test. This cell is high- lighted 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 reported in exponential notation and is equal to 0.0000319, 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 448, 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 = 6.205. Actually, when comparing the results of simple linear regression and correlation analysis, the two tests are equivalent and will always yield exactly the same values of t and the same p-values.
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 ŷ = 1.5 + 2.2x for a sam- ple of 5 months. Conduct a test of hypothesis to show there is a positive relationship be- tween advertising and sales. From statistical software, the standard error of the regression coefficient is 0.42. Use the .05 significance level.
S E L F - R E V I E W 13–4
21. Refer to Exercise 5. The regression equation is ŷ = 29.29 − 0.96x, the sample size is 8, and the standard error of the slope is 0.22. Use the .05 significance level. Can we conclude that the slope of the regression line is less than zero?
22. Refer to Exercise 6. The regression equation is ŷ = 11.18 − 0.49x, 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 ŷ = 1.85 + .08x, the sample size is 12, and the standard error of the slope is 0.03. Use the .05 significance level. Can we conclude that the slope of the regression line is different from zero?
24. Refer to Exercise 18. The regression equation is ŷ = 9.9198 − 0.00039x, the sample size is 9, and the standard error of the slope is 0.0032. Use the .05 signifi- cance level. Can we conclude that the slope of the regression line is less than zero?
E X E R C I S E S
462 CHAPTER 13
EVALUATING A REGRESSION EQUATION’S ABILITY TO PREDICT
The Standard Error of Estimate The results of the regression analysis for North American Copier Sales show a signifi- cant 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 = 19.9632 + 0.2608 (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 84, then we can predict the number of copiers sold. It is 41.8704, found by 19.9632 + 0.2608(84). However, the data show two sales representatives with 84 sales calls and 30 and 43 copiers sold. So, is the regression equation a good predictor of “Number of copiers sold”?
Perfect prediction, which is finding the exact outcome, is practically impossible in almost all disciplines including economics and business. For example:
• A large electronics firm, with production facilities throughout the United States, has a stock option plan for employees. Suppose there is a relationship between the number of years employed and the number of shares owned. This relationship is likely because, as number of years of service increases, the number of shares an employee earns also increases. If we observe all employees with 20 years of ser- vice, they would most likely own different numbers of shares.
• A real estate developer in the southwest United States studied the relationship between the income of buyers and the size, in square feet, of the home they pur- chased. The developer’s analysis shows that as the income of a purchaser in- creases, the size of the home purchased will also increase. However, all buyers with an income of $70,000 will not purchase a home of exactly the same size.
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 measure is called the standard error of estimate. The standard error of estimate is symbolized by sy⋅x. The subscript, y ⋅ x, 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.
LO13-5 Evaluate a regression equation’s ability to predict using the standard estimate of the error and the coefficient of determination.
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.
The standard error of estimate is found using formula (13–7).
sy · x = √ Σ ( y − ŷ )2
n − 2 [13–7]STANDARD ERROR OF ESTIMATE
The calculation of the standard error of estimate requires the sum of the squared differ- ences between each observed value of y and the predicted value of y, which is identi- fied as ŷ in the numerator. This calculation is illustrated in the following spreadsheet. See the highlighted cell in the lower right corner.
CORRELATION AND LINEAR REGRESSION 463
The calculation of the standard error of estimate is:
sy · x = √ Σ ( y − ŷ )2
n − 2 = √
587.1108 15 − 2
= 6.720
The standard error of estimate can be calculated using statistical software such as Excel. It is included in Excel’s regression analysis on page 460 and highlighted in yellow. Its value is 6.720.
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 pre- cise estimate of y.
The Coefficient of Determination Using the standard error of estimate provides a relative measure of a regression equa- tion’s ability to predict. We will use it to provide more specific information about a pre- diction 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 predict. It is called the coefficient of determination, or R-square.
The coefficient of determination is easy to compute. It is the correlation coefficient squared. Therefore, the term R-square is also used. With the North American Copier Sales data, the correlation coefficient for the relationship between the number of copiers sold and the number of sales calls is 0.865. If we compute (0.865)2, the coefficient of determination is 0.748. See the blue (Multiple R) and green (R-square) highlighted cells in the spreadsheet on page 460. To better interpret the coefficient of determination, convert it to a percentage. Hence, we say that 74.8% of the variation in the number of copiers sold is explained, or accounted for, by the variation in the number of sales calls.
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.
A B C D E F
1 Sales Rep Sales Calls (x) Copiers Sold (y) Estimated Sales (y – ŷ ) (y – ŷ )2
2 Brian Virost 96 41 45.0000 –4.0000 16.0000 3 Carlos Ramirez 40 41 30.3952 10.6048 112.4618 4 Carol Saia 104 51 47.0864 3.9136 15.3163 5 Greg Fish 128 60 53.3456 6.6544 44.2810 6 Jeff Hall 164 61 62.7344 –1.7344 3.0081 7 Mark Reynolds 76 29 39.7840 –10.7840 116.2947 8 Meryl Rumsey 72 39 38.7408 0.2592 0.0672 9 Mike Kiel 80 50 40.8272 9.1728 84.1403 10 Ray Snarsky 36 28 29.3520 –1.3520 1.8279 11 Rich Niles 84 43 41.8704 1.1296 1.2760 12 Ron Broderick 180 70 66.9072 3.0928 9.5654 13 Sal Spina 132 56 54.3888 1.6112 2.5960 14 Soni Jones 120 45 51.2592 –6.2592 39.1776 15 Susan Welch 44 31 31.4384 –0.4384 0.1922 16 Tom Keller 84 30 41.8704 –11.8704 140.9064 17 Total 0.0000 587.1108
464 CHAPTER 13
How well can the regression equation predict number of copiers sold with number of sales calls made? If it were possible to make perfect predictions, the coefficient of determination would be 100%. That would mean that the independent variable, number of sales calls, explains or accounts for all the variation in the number of copiers sold. A coefficient of determination of 100% 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 dia- gram. Our analysis shows that only 74.8% of the variation in copiers sold is explained by the number of sales calls. Clearly, these data do not form a perfect line. Instead, the data are scattered around the best-fitting, least squares regression line, and there will be error in the predictions. In the next section, the standard error of estimate is used to provide more specific information regarding the error associated with using the regres- sion equation to make predictions.
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.
S E L F - R E V I E W 13–5
(You may wish to use a statistical software package such as Excel, Minitab, or Megastat to assist in your calculations.)
25. Refer to Exercise 5. Determine the standard error of estimate and the coefficient of determination. Interpret the coefficient of determination.
26. Refer to Exercise 6. Determine the standard error of estimate and the coefficient of determination. 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.
E X E R C I S E S
Relationships among the Correlation Coefficient, the Coefficient of Determination, and the Standard Error of Estimate In formula 13-7 shown on page 462, 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
Σ( y − ŷ )2
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 correlation 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.
CORRELATION AND LINEAR REGRESSION 465
We also noted that the square of the correlation coefficient is the coefficient of de- termination. The coefficient of determination measures the percentage of the variation 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 highlighted portion of the spreadsheet below. This table is similar to the analysis of variance table developed in Chapter 12. In that chapter, the total varia- tion was divided into two components: variation due to the treatments and variation due to random error. The concept is similar in regression analysis. The total variation is divided into two components: (1) variation explained by the regression (explained by the indepen- dent variable) and (2) the error, or residual. This is the unexplained variation. These three sources of variance (total, regression, and residual) 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 indepen- dent variable. The number of degrees of freedom associated 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 are 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.
r2 = SSR
SS Total = 1 −
SSE SS Total
[13–8]COEFFICIENT OF DETERMINATION
Using the values from the ANOVA table, the coefficient of determination is 1738.89/ 2326.0 = 0.748. Therefore, the more variation of the dependent variable (SS Total) ex- plained by the independent variable (SSR), the higher the coefficient of determination.
The ANOVA sum of squares are:
Regression Sum of Squares = SSR = Σ (ŷ − y )2 = 1738.89
Residual or Error Sum of Squares = SSE = Σ ( y − ŷ)2 = 587.11
Total Sum of Squares = SS Total = Σ ( y − y )2 = 2326.0
Recall that the coefficient of determination is defined as the percentage of the total vari- ation (SS Total) explained by the regression equation (SSR). Using the ANOVA table, the reported value of R-square can be validated.
A B C D E F G H I J 1 Sales Representive Sales Calls (x) Copiers Sold (y) SUMMARY OUTPUT 2 Brian Virost 96 41 Regression Statistics 3 Carlos Ramirez 40 41 Multiple R 0.865 4 Carol Saia 104 51 R Square 0.748 5 Greg Fish 128 60 Adjusted R Square 0.728 6 Jeff Hall 164 61 Standard Error 6.720 7 Mark Reynolds 76 29 Observations 15 8 Meryl Rumsey 72 39 9 Mike Kiel 80 50 ANOVA 10 Ray Snarsky 36 28 df SS MS F Significance F 11 Rich Niles 84 43 Regression 1 1738.890 1738.890 38.503 0.000 12 Ron Broderick 180 70 Residual 13 587.110 45.162 13 Sal Spina 132 56 Total 14 2326 14 Sani Jones 120 45 15 Susan Welch 44 31
Coefficients Standard
Error t Stat P-value
16 Tom Keller 84 30 Intercept 19.980 4.390 4.552 0.001 17 Sales Calls (x) 0.261 0.042 6.205 0.000
466 CHAPTER 13
We can also express the coefficient of determination in terms of the error or resid- ual variation:
r2 = 1 − SSE
SS Total = 1 −
587.11 2326.0
= 1 − 0.252 = 0.748
As illustrated in formula 13–8, 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, 25.2% of the total variation in the dependent variable is error or residual variation.
The final observation that relates the correlation coefficient, the coefficient of de- termination, and the standard error of estimate is to show the relationship between the standard error of estimate and SSE. By substituting [SSE Residual or Error Sum of Squares = SSE = Σ( y − ŷ )2] into the formula for the standard error of estimate, we find:
sy · x = √ SSE
n − 2 [13–9]
STANDARD ERROR OF ESTIMATE
sy · x = √Residual mean square [13–10]STANDARD ERROR OF THE ESTIMATE
Note that sy ⋅x can also be computed using the residual mean square from the ANOVA table.
In sum, regression analysis provides two statistics to evaluate the predictive ability of a regression equation: the standard error of the estimate and the coefficient of determina- tion. When reporting the results of a regression analysis, the findings must be clearly explained, especially when using the results to make predictions of the dependent vari- able. The report must always include a statement regarding the coefficient of determi- nation 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.
29. Given the following ANOVA table:
Source DF SS MS F
Regression 1 1000.0 1000.0 26.00 Error 13 500.0 38.46
Total 14 1500.0
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%. 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.
E X E R C I S E S
CORRELATION AND LINEAR REGRESSION 467
INTERVAL ESTIMATES OF PREDICTION The standard error of estimate and the coefficient of determination are two statistics 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 lin- ear regression should be reviewed. Chart 13–13 illustrates these assumptions.
LO13-6 Calculate and interpret confidence and prediction intervals.
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 be- tween height and salary. For each additional inch of height, a man can expect to earn an additional $250 per year. So a man 6′6″ tall receives a $3,000 “stature” bonus over his 5′6″ coun- terpart. Being overweight or underweight is also re- lated to earnings, particu- larly among women. A study of young women showed the heaviest 10% earned about 6% less than their lighter counterparts.
CHART 13–13 Regression Assumptions Shown Graphically
x
y
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.
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. 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 esti- mate (sy ·x).
4. The y values are statistically independent. This means that in selecting a sample, 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% of the observations, the mean plus or minus two standard deviations will encompass 95% of the observations, and
468 CHAPTER 13
the mean plus or minus three standard deviations will encompass virtually all of the observations. The same relationship exists between the predicted values ŷ and the standard error of estimate (sy ⋅x).
1. ŷ ± sy · x will include the middle 68% of the observations. 2. ŷ ± 2sy · x will include the middle 95% of the observations. 3. ŷ ± 3sy · x will include virtually all the observations.
We can now relate these assumptions to North American Copier Sales, where we studied the relationship between the number of sales calls and the number of copi- ers sold. If we drew a parallel line 6.72 units above the regression line and another 6.72 units below the regression line, about 68% of the points would fall between the two lines. Similarly, a line 13.44 [2sy · x = 2 (6.72)] units above the regression line and another 13.44 units below the regression line should include about 95% of the data values.
As a rough check, refer to column E in the Excel spreadsheet appearing on page 456. Four of the 15 deviations exceed one standard error of estimate. That is, the deviations of Carlos Ramirez, Mark Reynolds, Mike Keil, and Tom Keller all ex- ceed 6.72 (one standard error). All values are less than 13.44 units away from the regression line. In short, 11 of the 15 deviations are within one standard error and all are within two standard errors. That is a fairly 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 Ŷ. Each distribution of Y has a standard deviation equal to the regression analysis’s standard error of estimate.
The first interval estimate is called a confidence interval. This is used when the re- gression equation is used to predict the mean value of Y for a given value of x. For ex- ample, 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 ± tsy · x √ 1 n
+ (x − x)2
Σ (x − x )2 [13–11]CONFIDENCE INTERVAL FOR
THE MEAN OF Y, GIVEN X
ŷ ± tsy · x √1 + 1 n
+ (x − x )2
Σ (x − x )2 [13–12]PREDICTION INTERVAL
FOR Y, GIVEN X
The second interval estimate is called a prediction interval. This is used when the regression equation is used to predict an individual y for a given value of x. For exam- ple, we would estimate the salary of a particular retail executive who has 20 years of experience. To calculate a prediction interval, formula 13-11 is modified by adding a 1 under the radical. To determine the prediction interval for an estimate of an individual for a given x, the formula is:
CORRELATION AND LINEAR REGRESSION 469
E X A M P L E
We return to the North American Copier Sales illustration. Determine a 95% confi- dence interval for all sales representatives who make 50 calls, and determine a pre- diction interval for Sheila Baker, a West Coast sales representative who made 50 calls.
S O L U T I O N
We use formula (13–11) to determine a confidence level. Table 13–4 includes the necessary totals and a repeat of the information of Table 13–2.
The first step is to determine the number of copiers we expect a sales representa- tive to sell if he or she makes 50 calls. It is 33.0032, found by
ŷ = 19.9632 + 0.2608x = 19.9632 + 0.2608 (50) = 33.0032
To find the t value, we need to first know the number of degrees of freedom. In this case, the degrees of freedom are n − 2 = 15 − 2 = 13. We set the confidence level at 95%. To find the value of t, move down the left-hand column of Appendix B.5 to 13 degrees of freedom, then move across to the column with the 95% level of con- fidence. The value of t is 2.160.
In the previous section, we calculated the standard error of estimate to be 6.720. We let x = 50, and from Table 13–4, the mean number of sales calls is 96.0 (1440/15) and Σ(x − x )2 = 25, 600 Inserting these values in formula (13–11), we can determine the confidence interval.
Confidence Interval = ŷ ± tsy · x √ 1 n
+ (x − x )2
Σ (x − x )2
= 33.0032 ± 2.160(6.720)√ 1
15 +
(50 − 96)2
25,600 = 33.0032 ± 5.6090
TABLE 13–4 Determining Confidence and Prediction Intervals
Sales Copiers Sales Representative Calls (x) Sold (y) (x − x ) (x − x )2
Brian Virost 96 41 0 0 Carlos Ramirez 40 41 −56 3,136 Carol Saia 104 51 8 64 Greg Fish 128 60 32 1,024 Jeff Hall 164 61 68 4,624 Mark Reynolds 76 29 −20 400 Meryl Rumsey 72 39 −24 576 Mike Kiel 80 50 −16 256 Ray Snarsky 36 28 −60 3,600 Rich Niles 84 43 −12 144 Ron Broderick 180 70 84 7,056 Sal Spina 132 56 36 1,296 Sani Jones 120 45 24 576 Susan Welch 44 31 −52 2,704 Tom Keller 84 30 −12 144 Total 1440 675 0 25,600
470 CHAPTER 13
Thus, the 95% confidence interval for all sales representatives who make 50 calls is from 27.3942 up to 38.6122. To interpret, let’s round the values. If a sales repre- sentative makes 50 calls, he or she can expect to sell 33 copiers. It is likely the sales will range from 27.4 to 38.6 copiers.
Suppose we want to estimate the number of copiers sold by Sheila Baker, who made 50 sales calls. Using formula (13–12), the 95% prediction interval is deter- mined as follows:
Prediction Interval = ŷ ± tsy · x √1 + 1 n
+ (x − x )2
Σ (x − x )2
= 33.0032 ± 2.160(6.720)√1 + 1
15 +
(50 − 96)2
25,600 = 33.0032 ± 15.5612
Thus, the interval is from 17.442 up to 48.5644 copiers. We conclude that the num- ber of office machines sold will be between about 17.4 and 48.6 for a particular sales representative, such as Sheila Baker, who makes 50 calls. This interval is quite large. It is much larger than the confidence interval for all sales representa- tives who made 50 calls. It is logical, however, that there should be more variation in the sales estimate for an individual than for a group.
The following Minitab graph shows the relationship between the least squares re- gression line (in the center), the confidence interval (shown in crimson), and the predic- tion interval (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 (96) in either direction, the confidence inter- val and prediction interval bands widen. This is caused by the numerator of the right- hand term under the radical in formulas (13–11) and (13–12). 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.
CORRELATION AND LINEAR REGRESSION 471
The regression equation was computed to be ŷ = 1.5 + 2.2x and the standard error 0.9487. Both variables are reported in millions of dollars. Determine the 90% confidence interval for the typical month in which $3 million was spent on advertising.
Refer to the sample data in Self-Review 13–1, where the owner of Haverty’s Furniture was studying the relationship between sales and the amount spent on advertising. The adver- tising expense and sales revenue, both in millions of dollars, for the last 4 months are re- peated below.
S E L F - R E V I E W 13–6
Advertising Expense Sales Revenue Month ($ million) ($ million)
July 2 7 August 1 3 September 3 8 October 4 10
TRANSFORMING DATA Regression analysis describes the relationship between two variables. A requirement is that this relationship be linear. The same is true of the correlation coefficient. It mea- sures the strength of a linear relationship between two variables. But what if the rela- tionship is not linear? The remedy is to rescale one or both of the variables so the new relationship is linear. For example, instead of using the actual values of the dependent variable, y, we would create a new dependent variable by computing the log to the
LO13-7 Use a log function to transform a nonlinear relationship.
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 par-
ticular six-room home. 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 salesperson who makes 40 contacts.
E X E R C I S E S
We wish to emphasize again the distinction between a confidence interval and a prediction interval. A confidence interval refers to the mean of all cases for a given value of x and is computed by formula (13–11). A prediction interval refers to a particu- lar, single case for a given value of x and is computed using formula (13–12). The pre- diction interval will always be wider because of the extra 1 under the radical in the second equation.
472 CHAPTER 13
E X A M P L E
GroceryLand Supermarkets is a regional grocery chain with over 300 stores located in the midwestern United States. The corporate director of marketing for GroceryLand wishes to study the effect of price on the weekly sales of two-liter bottles of their private-brand diet cola. The objectives of the study are:
1. To determine whether there is a relationship between selling price and weekly sales. Is this relationship direct or indirect? Is it strong or weak?
2. To determine the effect of price increases or decreases on sales. Can we effec- tively forecast sales based on the price?
S O L U T I O N
To begin the project, the marketing director meets with the vice president of sales and other company staff members. They decide that it would be reason- able to price the two-liter bottle of their private-brand diet cola from $0.50 up to $2.00. To collect the data needed to analyze the relationship between price and sales, the marketing director selects a random sample of 20 stores and then randomly assigns a selling price for the two-liter bottle of diet cola be- tween $0.50 and $2.00 to each selected store. The director contacts each of the 20 store managers included in the study to tell them the selling price and ask them to report the sales for the product at the end of the week. The results are reported below. For example, store number A-17 sold 181 two-liter bottles of diet cola at $0.50 each.
GroceryLand Sales and Price Data GroceryLand Sales and Price Data
Store Number Price Sales
A-17 0.50 181 A-121 1.35 33 A-227 0.79 91 A-135 1.71 13 A-6 1.38 34 A-282 1.22 47 A-172 1.03 73 A-296 1.84 11 A-143 1.73 15 A-66 1.62 20
Store Number Price Sales
A-30 0.76 91 A-127 1.79 13 A-266 1.57 22 A-117 1.27 34 A-132 0.96 74 A-120 0.52 164 A-272 0.64 129 A-120 1.05 55 A-194 0.72 107 A-105 0.75 119
base 10 of y, Log(y). This calculation is called a transformation. Other common transfor- mations include taking the square root, taking the reciprocal, or squaring one or both of the variables.
Thus, two variables could be closely related, but their relationship is not linear. Be cautious when you are interpreting the correlation coefficient or a regression equation. These statistics may indicate there is no linear relationship, but there could be a rela- tionship of some other nonlinear or curvilinear form. The following example explains the details.
To examine the relationship between Price and Sales, we use regression analysis setting Price as the independent variable and Sales as the dependent
CORRELATION AND LINEAR REGRESSION 473
variable. The analysis will provide important information about the relationship between the variables. The analysis is summarized in the following Minitab output.
From the output, we can make these conclusions:
1. The relationship between the two variables is inverse or indirect. As the Price of the cola increases, the Sales of the product decreases. Given basic eco- nomic theory of price and demand, this is expected.
2. There is a strong relationship between the two variables. The coefficient of determination is 88.9%. So 88.9% of the variation in Sales is accounted for by the variation in Price. From the coefficient of determination, we can compute the correlation coefficient as the square root of the coefficient of determina- tion. The correlation coefficient is the square root of 0.889, or 0.943. The sign of the correlation coefficient is negative because sales are inversely related to price. Therefore, the correlation coefficient is −0.943.
3. Before continuing our summary of conclusions, we should look carefully at the scatter diagram and the plot of the regression line. The assumption of a linear relationship is tenuous. If the relationship is linear, the data points should be distributed both above and below the line over the entire range of the independent variable. However, for the highest and lowest prices, the data points are above the regression line. For the selling prices in the mid- dle, most of the data points are below the regression line. So the linear re- gression equation does not effectively describe the relationship between Price and Sales. A transformation of the data is needed to create a linear relationship.
By transforming one of the variables, we may be able to change the nonlinear relationship between the variables to a linear relationship. Of the possible choices, the director of marketing decides to transform the dependent variable, Sales, by taking the logarithm to the base 10 of each Sales value. Note the new variable, Log-Sales, in the following analysis. Now, the regression analysis uses Log-Sales as
474 CHAPTER 13
the dependent variable and Price as the independent variable. This analysis is re- ported below.
What can we conclude from the regression analysis using the transformation of the dependent variable Sales?
1. By transforming the dependent variable, Sales, we increase the coefficient of determination from 0.889 to 0.989. So Price now explains nearly all of the vari- ation in Log-Sales.
2. Compare this result with the scatter diagram before we transformed the de- pendent variable. The transformed data seem to fit the linear relationship re- quirement much better. Observe that the data points are both above and below the regression line over the range of Price.
3. The regression equation is ŷ = 2.685 − 0.8738x. The sign of the slope value is negative, confirming the inverse association between the variables. We can use the new equation to estimate sales and study the effect of changes in price. For example, if we decided to sell the two-liter bottle of diet cola for $1.25, the predicted Log-Sales is:
ŷ = 2.685 − 0.8738x = 2.685 − 0.8738 (1.25) = 1.593
Remember that the regression equation now predicts the log, base10, of Sales. Therefore, we must undo the transformation by taking the antilog of 1.593, which is 101.593, or 39.174. So, if we price the two-liter diet cola product at $1.25, the predicted weekly sales are 39 bottles. If we increase the price to $2.00, the regression equation would predict a value of .9374. Taking the anti- log, 10.9374, the predicted sales decrease to 8.658, or, rounding, 9 two-liter bot- tles per week. Clearly, as price increases, sales decrease. This relationship will be very helpful to GroceryLand when making pricing decisions for this product.
35. Given the following sample of five observations, develop a scatter diagram, using x as the independent variable and y as the dependent variable, and compute the correlation coefficient. Does the relationship between the variables appear to be linear? Try squaring the x variable and then develop a scatter diagram and de- termine the correlation coefficient. Summarize your analysis.
E X E R C I S E S
x −8 −16 12 2 18 y 58 247 153 3 341
CORRELATION AND LINEAR REGRESSION 475
a. Using Score as the independent variable and Prize as the dependent variable, develop a scatter diagram. Does the relationship appear to be linear? Does it seem reasonable that as Score increases the Prize decreases?
b. What percentage of the variation in the dependent variable, Prize, is accounted for by the independent variable, Score?
c. Calculate a new variable, Log-Prize, computing the log to the base 10 of Prize. Draw a scatter diagram with Log-Prize as the dependent variable and Score as the independent variable.
d. Develop a regression equation and compute the coefficient of determination using Log-Prize as the dependent variable.
e. Compare the coefficient of determination in parts (b) and (d). What do you conclude?
f. Write out the regression equation developed in part (d). If a player shot an even par score of 288 for the four rounds, how much would you expect that player to earn?
36. Every April, The Masters—one of the most prestigious golf tournaments on the PGA golf tour—is played in Augusta, Georgia. In 2016, 55 players received prize money. The 2016 winner, Danny Willett, earned a prize of $1,800,000. Jordan Spieth and Lee Westwood tied for second place, earning $880,000. Two amateur players finished “in the money” but they could not accept the prize money. They are not included in the data. The data are briefly summarized below. Each player has three corresponding variables: finishing position, score, and prize (in dollars). We want to study the relationship between score and prize.
Position Player Score Prize
1 Danny Willett 283 $1,800,000 2(tied) Jordan Spieth 286 $880,000 2(tied) Lee Westwood 286 $880,000 4(tied) Paul Casey 287 $413,333 4(tied) J.B. Holmes 287 $413,333 4(tied) Dustin Johnson 287 $413,333 .. .. .. .. .. .. .. .. .. .. .. .. 52(tied) Keegan Bradley 301 $24,900 52(tied) Larry Mize 301 $24,900 54 Hunter Mahan 302 $24,000 55(tied) Kevin Na 303 $23,400 55(tied) Cameron Smith 303 $23,400 57 Thongchai Jaidee 307 $23,000
C H A P T E R S U M M A R Y
I. A scatter diagram is a graphic tool used 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.
476 CHAPTER 13
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 negative sign means there is an inverse relationship.
F. It is designated by the letter r and found by the following equation:
r = Σ (x − x ) ( y − y)
(n − 1)sxsy [13–1]
G. To test a hypothesis that a population correlation is different from 0, we use the fol- lowing statistic:
t = r√n − 2 √1 − r 2
with n − 2 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 ŷ = a + bx A. ŷ 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 ŷ when x = 0. 2. a is computed using the following equation.
a = y − bx [13–5]
C. b is the slope of the fitted line. 1. It shows the amount of change in ŷ 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.
b = r ( sy sx)
[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 be-
tween 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:
t = b − 0
sb [13–6]
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.
sy · x = √ Σ ( y − ŷ)2
n − 2 [13–7]
CORRELATION AND LINEAR REGRESSION 477
VII. The coefficient of determination is the proportion of the variation of a dependent vari- able 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.
r 2 = SSR
SS Total = 1 −
SSE SS Total
[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.
ŷ ± tsy · x √ 1 n
+ (x − x )2
Σ (x − x )2 [13–11]
2. The width of the interval is affected by the level of confidence, the size of the standard error of estimate, and the size of the sample, as well as the value of the independent 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.
ŷ ± tsy · x √1 + 1 n
+ (x − x )2
Σ (x − x )2 [13–12]
2. The difference between formulas (13–11) and (13–12) 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 in- dependent variable.
P R O N U N C I A T I O N K E Y
SYMBOL MEANING PRONUNCIATION
Σxy Sum of the products of x and y Sum x y ρ Correlation coefficient in the population Rho ŷ Estimated value of Y y hat
sy · x Standard error of estimate s sub y dot x
r2 Coefficient of determination r square
C H A P T E R E X E R C I S E S
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?
478 CHAPTER 13
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 revenue 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 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
number of rooms occupied? 41. The table below shows the number of cars (in millions) sold in the United States for
various years and the percent of those cars manufactured by GM.
Day Revenue Occupied Day Revenue 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 1985 15.4 40.1 1955 7.8 50.4 1990 13.5 36.0 1960 7.3 44.0 1995 15.5 31.7 1965 10.3 49.9 2000 17.4 28.6 1970 10.1 39.5 2005 16.9 26.9 1975 10.8 43.1 2010 11.6 19.1 1980 11.5 44.0 2015 17.5 17.6
Use a statistical software package to answer the following questions. a. Is the number of cars sold directly or indirectly related to GM’s percentage of the
market? 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
variables? 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 40 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 −0.363. At the .05 significance level, can we conclude that there is a negative associa- tion between the two variables.
CORRELATION AND LINEAR REGRESSION 479
43. For each of the 32 National Football League teams, the numbers of points scored and allowed during the 2016 season are shown below.
PTS PTS TEAM Conference Scored Allowed
Baltimore AFC 343 321 Buffalo AFC 399 378 Cincinnati AFC 325 315 Cleveland AFC 264 452 Denver AFC 333 297 Houston AFC 279 328 Indianapolis AFC 411 392 Jacksonville AFC 318 400 Kansas City AFC 389 311 Miami AFC 363 380 New England AFC 441 250 NY Jets AFC 275 409 Oakland AFC 416 385 Pittsburgh AFC 399 327 San Diego AFC 410 423 Tennessee AFC 381 378
PTS PTS TEAM Conference Scored Allowed
Arizona NFC 418 362 Atlanta NFC 540 406 Carolina NFC 369 402 Chicago NFC 279 399 Dallas NFC 421 306 Detroit NFC 346 358 Green Bay NFC 432 388 Los Angeles NFC 224 394 Minnesota NFC 327 307 NY Giants NFC 469 454 New Orleans NFC 310 284 Philadelphia NFC 367 331 San Francisco NFC 309 480 Seattle NFC 354 292 Tampa Bay NFC 354 369 Washington NFC 396 383
Store Size (thousands of Sales square feet) (millions $)
3.7 9.18 2.0 4.58 5.0 8.22 0.7 1.45 2.6 6.51 2.9 2.82 5.2 10.45 5.9 9.94 3.0 4.43 2.4 4.75 2.4 7.30 0.5 3.33 5.0 6.76
Store Size (thousands of Sales square feet) (millions $)
0.4 0.55 4.2 7.56 3.1 2.23 2.6 4.49 5.2 9.90 3.3 8.93 3.2 7.60 4.9 3.71 5.5 5.47 2.9 8.22 2.2 7.17 2.3 4.35
Assuming these are sample data, answer the following questions. You may use statisti- cal software to assist you. a. What is the correlation coefficient between these variables? Are you surprised the
association is negative? Interpret your results. b. Find the coefficient of determination. What does it say about the relationship? c. At the .05 significance level, can you conclude there is a negative association be-
tween “points scored” and “points allowed”? d. At the .05 significance level, can you conclude there is a negative association be-
tween “points scored” and “points allowed” for each conference? 44. The Cotton Mill is an upscale chain of women’s clothing stores, located primarily in
the southwest United States. Due to recent success, The Cotton Mill’s top management is planning to expand by locating new stores in other regions of the country. The direc- tor of planning has been asked to study the relationship between yearly sales and the store size. As part of the study, the director selects a sample of 25 stores and deter- mines the size of the store in square feet and the sales for last year. The sample data follow. The use of statistical software is suggested.
480 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:
ŷ = 15 − 5x
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 investigate the relationship, data on population and median age in 10 large cities were collected.
a. Draw a scatter diagram. Use store size as the independent variable. Does there ap- pear to be a relationship between the two variables. Is it positive or negative?
b. Determine the correlation coefficient and the coefficient of determination. Is the rela- tionship strong or weak? Why?
c. At the .05 significance level, can we conclude there is a significant positive correlation? 45. The manufacturer of Cardio Glide exercise equipment wants to study the relation-
ship between the number of months since the glide was purchased and the time, in hours, the equipment was used last week.
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 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 these 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?
CORRELATION AND LINEAR REGRESSION 481
a. Plot these 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?
48. Emily Smith decides to buy a fuel-efficient used car. Here are several vehicles she is considering, with the estimated cost to purchase and the age of the vehicle.
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 Cruz $2,475 8 Mazda3 $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
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 particular interest is whether the number of bidders increases or decreases the amount of the winning bid.
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% prediction interval for the winning bid. d. Determine the coefficient of determination. Interpret its value.
482 CHAPTER 13
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 ran- dom sample of 20 shipments made last month. Shipping distance is the independent variable and shipping time is the dependent variable. The results are as follows:
50. Mr. William Profit is studying companies going public for the first time. He is particu- larly 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. 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 cor- relation 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 oper- ating committee of the board of directors. As a part of your proposal, you need to include information 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.
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
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
CORRELATION AND LINEAR REGRESSION 483
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
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.
Household Amount Spent Monthly Income
1 $ 555 $4,388 2 489 4,558
39 1,206 9,862 40 1,145 9,883
. . .
. . .
. . .
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
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
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 num-
ber of construction work-zone fatalities and the number of unemployed people in a state. The regression equation is Fatalities = 12.7 + 0.000114 (Unemp). Some additional output is:
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 associa-
tion 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.
484 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 be-
tween 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 comput- ers along with their corresponding processor speeds in gigahertz.
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 signifi- cance 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.
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 Grumman 10.8 6.0 Rockwell Collins 27.3 8.7 United Technologies 20.1 3.2
Computer Speed Price Computer Speed Price
1 2.0 1008.50 7 2.0 1098.50 2 1.6 461.00 8 1.6 693.50 3 1.6 532.00 9 2.0 1057.00 4 1.8 971.00 10 1.6 1001.00 5 2.0 1068.50 11 1.0 468.50 6 1.2 506.00 12 1.4 434.50
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
CORRELATION AND LINEAR REGRESSION 485
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.
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.
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 be-
tween the distance from the fire station and the amount of damage. Use the .01 sig- nificance level and a two-tailed test.
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
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
486 CHAPTER 13
61. TravelAir.com samples domestic airline flights to explore the relationship between airfare and distance. The service would like to know if there is a correlation between airfare and flight distance. If there is a correlation, what percentage of the variation in airfare is accounted for by distance? How much does each additional mile add to the fare? The data follow.
Origin Destination Distance Fare
Detroit, MI Myrtle Beach, SC 636 $109 Baltimore, MD Sacramento, CA 2,395 252 Las Vegas, NV Philadelphia, PA 2,176 221 Sacramento, CA Seattle, WA 605 151 Atlanta, GA Orlando, FL 403 138 Boston, MA Miami, FL 1,258 209 Chicago, IL Covington, KY 264 254 Columbus, OH Minneapolis, MN 627 259 Fort Lauderdale, FL Los Angeles, CA 2,342 215 Chicago, IL Indianapolis, IN 177 128 Philadelphia, PA San Francisco, CA 2,521 348 Houston, TX Raleigh/Durham, NC 1,050 224 Houston, TX Midland/Odessa, TX 441 175 Cleveland, OH Dallas/Ft.Worth, TX 1,021 256 Baltimore, MD Columbus, OH 336 121 Boston, MA Covington, KY 752 252 Kansas City, MO San Diego, CA 1,333 206 Milwaukee, WI Phoenix, AZ 1,460 167 Portland, OR Washington, DC 2,350 308 Phoenix, AZ San Jose, CA 621 152 Baltimore, MD St. Louis, MO 737 175 Houston, TX Orlando, FL 853 191 Houston, TX Seattle, WA 1,894 231 Burbank, CA New York, NY 2,465 251 Atlanta, GA San Diego, CA 1,891 291 Minneapolis, MN New York, NY 1,028 260 Atlanta, GA West Palm Beach, FL 545 123 Kansas City, MO Seattle, WA 1,489 211 Baltimore, MD Portland, ME 452 139 New Orleans, LA Washington, DC 969 243
a. Draw a scatter diagram with Distance as the independent variable and Fare as the dependent variable. Is the relationship direct or indirect?
b. Compute the correlation coefficient. At the .05 significance level, is it reasonable to conclude that the correlation coefficient is greater than zero?
c. What percentage of the variation in Fare is accounted for by Distance of a flight? d. Determine the regression equation. How much does each additional mile add to the
fare? Estimate the fare for a 1,500-mile flight. e. A traveler is planning to fly from Atlanta to London Heathrow. The distance is 4,218
miles. She wants to use the regression equation to estimate the fare. Explain why it would not be a good idea to estimate the fare for this international flight with the re- gression equation.
D A T A A N A L Y T I C S
62. The North Valley Real Estate data reports information on homes on the market. 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% confidence interval for all 2,200 square foot homes and the 95% prediction interval for the selling price of a home with 2,200 square feet.
CORRELATION AND LINEAR REGRESSION 487
b. Let days-on-the-market be the dependent variable and price be the independent variable. Determine the regression equation. Estimate the days-on-the-market of a home that is priced at $300,000. Determine the 95% confidence interval of days-on- the-market for homes with a mean price of $300,000, and the 95% prediction inter- val of days-on-the-market for a home priced at $300,000.
c. Can you conclude that the independent variables “days on the market” and “selling price” are positively correlated? Are the size of the home and the selling price posi- tively correlated? Use the .05 significance level. Report the p-value of the test. Sum- marize your results in a brief report.
63. Refer to the Baseball 2016 data, which reports information on the 2016 Major League Baseball season. Let attendance be the dependent variable and total team sal- ary 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. What is the expected attendance for a team with a salary of $100.0 million? c. If the owners pay an additional $30 million, how many more people could they ex-
pect to attend? 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 attendance is accounted for by salary? f. Determine the correlation between attendance and team batting average and be-
tween attendance and team ERA. Which is stronger? Conduct an appropriate test of hypothesis for each set of variables.
64. Refer to the Lincolnville School bus data. Develop a regression equation that ex- presses the relationship between age of the bus and maintenance cost. The age of the bus is the independent variable. a. Draw a scatter diagram. What does this diagram suggest as to the relationship be-
tween 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.
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO14-1 Use multiple regression analysis to describe and interpret a relationship between several independent variables and a dependent variable.
LO14-2 Evaluate how well a multiple regression equation fits the data.
LO14-3 Test hypotheses about the relationships inferred by a multiple regression model.
LO14-4 Evaluate the assumptions of multiple regression.
LO14-5 Use and interpret a qualitative, dummy variable in multiple regression.
LO14-6 Include and interpret an interaction effect in multiple regression analysis.
LO14-7 Apply stepwise regression to develop a multiple regression model.
LO14-8 Apply multiple regression techniques to develop a linear model.
© Image Source/Getty Images RF
Multiple Regression Analysis14
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 within the Review of Multiple Regression section.)
MULTIPLE REGRESSION ANALYSIS 489
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 (−.88 or .78, for example) indicates a very strong linear relationship, whereas a value near 0 (−.12 or .18, for example) indicates that the relationship is weak. Next we developed a proce- dure to determine a linear equation to express the relationship between the two vari- ables. We referred to this as a regression line. This line describes the relationship 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 x1, x2, . . . , and so on) that help us better explain or predict the depen- dent variable (y). Almost all of the ideas we saw in simple linear correlation and regression extend to this more general situation. However, the additional independent variables do lead to some new considerations. Multiple regression analysis can be used either as a descriptive or as an inferential technique.
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.
LO14-1 Use multiple regression analysis to describe and interpret a relationship between several independent variables and a dependent variable.
where: a is the intercept, the value of ŷ when all the X’s are zero. bj is the amount by which ŷ changes when that particular xj 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 to identify the independent variable, age.
In Chapter 13, the regression analysis described and tested the relationship between a dependent variable, ŷ and a single independent variable, x. The relationship between ŷ and x was graphically portrayed by a line. When there are two independent variables, the regression equation is
ŷ = a + b1x1 + b2x2 Because there are two independent variables, this relationship is graphically portrayed 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 ŷ on the plane. If a multiple regression analysis in- cludes 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 coeffi- cients, suppose the selling price of a home is directly related to the number of rooms and inversely related to its age. We let x1 refer to the number of rooms, x2 to the age of the home in years, and y to the selling price of the home in thousands of dollars ($000).
ŷ = a + b1x1 + b2 x2 + b3 x3 + . . . + bk xk (14–1) GENERAL MULTIPLE REGRESSION EQUATION
490 CHAPTER 14
Suppose the regression equation, calculated using statistical software, is:
ŷ = 21.2 + 18.7x1 − 0.25x2 The intercept value of 21.2 indicates the regression equation (plane) intersects the y-axis at 21.2. This happens when both the number of rooms and the age of the home are zero. We could say that $21,200 is the average value of a property without a house.
The first regression coefficient, 18.7, indicates that for each increase of one room in the size of a home, the selling price will increase by $18.7 thousand ($18,700), re- gardless of the age of the home. The second regression coefficient, −0.25, indicates that for each increase of one year in age, the selling price will decrease by $.25 thou- sand ($250), regardless of the number of rooms. As an example, a seven-room home that is 30 years old is expected to sell for $144,600.
ŷ = 21.2 + 18.7x1 − 0.25x2 = 21.2 + 18.7(7) − 0.25(30) = 144.6
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 Σ(y − ŷ)2 is minimized. The calculations are very te- dious, so they are usually performed by a statistical software package.
In the following example, we show a multiple regression analysis using three inde- pendent variables employing Excel that produces a standard set of statistics and re- ports. Statistical software such as Minitab, MegaStat, and others provide advanced regression analysis techniques.
x1
Y
x2
Estimated point ( y )
Observed point ( y )
Plane formed through the sample points 5 a 1 b1 x1 1 b2 x2
^̂
ŷ
CHART 14–1 Regression Plane with 10 Sample Points
E X A M P L E
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 re- search department at Salsberry has been asked to develop some guidelines re- garding heating costs for single-family homes. Three variables are thought to relate to the heating costs: (1) the mean daily outside temperature, (2) the num- ber 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.
MULTIPLE REGRESSION ANALYSIS 491
The data in Table 14–1 are available in Excel worksheet format at the textbook website, www.mhhe.com/Lind17e. The basic instructions for using Excel for these data are in the Software Commands in Appendix C.
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?
S O L U T I O N
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 x1. • The number of inches of insulation in the attic, represented by x2. • The age in years of the furnace, represented by x3.
Given these definitions, the general form of the multiple regression equation fol- lows. The value ŷ is used to estimate the value of y.
ŷ = a + b1x1 + b2 x2 + b3 x3 Now that we have defined the regression equation, we are ready to use Excel to compute all the statistics needed for the analysis. The output from Excel is shown on the following page.
To use the regression equation to predict the January heating cost, we need to know the values of the regression coefficients: b1, b2, and b3. These are highlighted in the software reports. The software uses the variable names or labels associated with
TABLE 14–1 Factors in January Heating Cost for a Sample of 20 Homes
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
492 CHAPTER 14
each independent variable. The regression equation intercept, a, is labeled “inter- cept” in the Excel output.
In this case, the estimated regression equation is:
ŷ = 427.194 − 4.583x1 − 14.831x2 + 6.101x3 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 (x1), there are 5 inches of insulation in the attic (x2), and the furnace is 10 years old (x3). By substituting the values for the independent variables:
ŷ = 427.194 − 4.583(30) − 14.831(5) + 6.101(10) = 276.56
The estimated January heating cost is $276.56. 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 surprising. As the outside temperature increases, the cost to heat the home de- creases. The numeric value of the regression coefficient provides more information. If the outside temperature increases by 1 degree and the other two independent variables remain constant, we can estimate a decrease of $4.583 in monthly heat- ing 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 ex- pect 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.
STATISTICS IN ACTION
Many studies indicate a woman will earn about 70% of what a man would for the same work. Re- searchers at the University of Michigan Institute for Social Research found that about one-third of the dif- ference can be explained by such social factors as differences in education, seniority, and work inter- ruptions. The remaining two-thirds is not explained by these social factors.
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
S E L F - R E V I E W 14–1
MULTIPLE REGRESSION ANALYSIS 493
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 the SkyWheel, a landmark in Myrtle Beach. x4 the number of servers employed. x5 the number of years the current owner operated the restaurant.
The following is part of the output obtained using statistical software.
(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 the SkyWheel, has 20 servers, and has been op- erated by the current owner for 5 years?
(b) Interpret the values of b2 and b3 in the multiple regression equation.
1. The director of marketing at Reeves Wholesale Products is studying monthly sales. Three independent variables were selected as estimators of sales: regional popula- tion, 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 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%? 2. Thompson Photo Works purchased several new, highly sophisticated processing
machines. The production department needed some guidance with respect to qualifications 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 esti- mate 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:
E X E R C I S E S
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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The equation is:
ŷ = 11.6 + 0.4x1 + 0.286x2 + 0.112x3 + 0.002x4
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
6 years and scored 280 on the mechanical aptitude test. Carl’s prior on-the-job performance rating is 97, and he is 35 years old. Estimate Carl’s performance on the new machine.
3. A consulting group was hired by the Human Resources Department at General Mills, Inc. to survey company employees regarding their degree of satisfaction 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 mar- riage (x1), annual income (x2), number of children living (x3), value of all assets (x4), status of health in the form of an index (x5), and the average number of social activ- ities per week—such as bowling and dancing (x6). Suppose the multiple regression equation is:
ŷ = 16.24 + 0.017x1 + 0.0028x2 + 42x3 + 0.0012x4 + 0.19x5 + 26.8x6
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 find- ings are:
On the basis of the sample results, the regression equation is:
ŷ = 62.65 − 1.86x1 − 0.52x2
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 b1 and b2 negative? Is this logical?
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
MULTIPLE REGRESSION ANALYSIS 495
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 eval- uate how well the equation fits the data.
The ANOVA Table As in Chapter 13, the statistical analysis of a multiple regression equation is summa- rized 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 category. 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 number of degrees of freedom associated with the error term is equal to the total degrees of free- dom, n − 1, minus the regression degrees of freedom, k. So, the residual or error degrees of freedom is (n − 1) − k, and is the same as n − (k + 1).
LO14-2 Evaluate how well a multiple regression equation fits the data.
In the ANOVA table, the column headed, “SS”, lists the sum of squares for each source of variation: regression, residual or error, and total. The sum of squares is the amount of variation attributable to each source.
The total variation of the dependent variable, y, is summarized in “SS total”. You should 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:
Total Sum of Squares = SS total = Σ(y − y )2
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 be- tween the estimated or predicted values, ŷ and the overall mean of y. The regression sum of squares is found by:
Regression Sum of Squares = SSR = Σ(ŷ − y )2
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, ŷ. Notice that this difference is the error of estimating or predicting the dependent variable with the multiple regression equation. It is calculated as:
Residual or Error Sum of Squares = SSE = Σ(y− ŷ )2
We will use the ANOVA table information from the previous example to evaluate the regression equation to estimate January heating costs.
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
496 CHAPTER 14
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 row 2 in the Excel spreadsheet on the previous page. The actual heating cost for the first observation, y, is $250; the out- side temperature, x1, is 35 degrees; the depth of insulation, x2, is 3 inches; and the age of the furnace, x3, is 6 years. Using the regression equation developed in the previous section, the estimated heating cost for this home is:
ŷ = 427.194 − 4.583x1 − 14.831x2 + 6.101x3 = 427.194 − 4.583(35) − 14.831(3) + 6.101(6) = 258.90
So we would estimate that a home with a mean January outside temperature of 35 de- grees, 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 ac- tual value and the estimated value—is y − ŷ = 250 − 258.90 = −8.90. 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.
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:
Sy ·123…k = √ Σ ( y − ŷ)2
n − (k + 1) = √
SSE n − (k + 1)
(14–2)MULTIPLE STANDARD ERROR OF ESTIMATE
where: y is the actual observation. ŷ 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. SSE is the Residual Sum of Squares from an ANOVA table.
There is more information in the ANOVA table that can be used to compute the multiple standard error of estimate. The column headed “MS” reports the mean squares for the regression and residual variation. These values are calculated as the sum of squares divided by the corresponding degrees of freedom. The multiple standard error
MULTIPLE REGRESSION ANALYSIS 497
of estimate is equal to the square root of the residual MS, which is also called the mean square error or the MSE.
sy·123...k = √MSE = √2605.995 = $51.05
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% of the residuals will be within ±$51.05 and about 95% within ±2(51.05), or ±$102.10. As before with similar measures of dispersion, such as the standard error of estimate in Chapter 13, a smaller multiple standard error indicates a better 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 de- pendent variable explained, or accounted for, by the independent variable. In the multi- ple regression case, we extend this definition as follows.
COEFFICIENT OF MULTIPLE DETERMINATION The percent of variation in the dependent variable, y, explained by the set of independent variables, x1, x2, x3, … xk.
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 R2 because it is calculated as the square of a correlation coefficient.
2. It can range from 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 R2 is a value between 0 and 1, it is easy to interpret, compare, and understand.
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.
R2 = SSR
SS total (14–3)COEFFICIENT OF MULTIPLE
DETERMINATION
We can use the regression and the total sum of squares from the ANOVA table high- lighted in the Excel output appearing earlier in this section and compute the coefficient of determination.
R2 = SSR
SS total =
171,220.473 212,915.750
= .804
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% of the variation in heating cost. To put it another way, 19.6% of the varia- tion is due to other sources, such as random error or variables not included in the analysis. Using the ANOVA table, 19.6% is the error sum of squares divided by the
498 CHAPTER 14
total sum of squares. Knowing that the SSR + SSE = SS total, the following relation- ship is true.
1 − R2 = 1 − SSR
SS total =
SSE SS total
= 41,695.277 212,915.750
= .196
Adjusted Coefficient of Determination The coefficient of determination tends to increase as more independent variables are added to the multiple regression model. Each new independent variable causes the predictions to be more accurate. That, in turn, makes SSE smaller and SSR larger. Hence, R2 increases only because the total number of independent variables increases 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 variables has on the coefficient of multiple determination, statistical software packages use an adjusted coefficient of multiple determination.
The error and total sum of squares are divided by their degrees of freedom. Notice es- pecially the degrees of freedom for the error sum of squares include k, the number of independent variables. For the cost of heating example, the adjusted coefficient of de- termination is:
R2adj = 1 −
41,695.277 20 − (3 + 1) 212,915.750
20 − 1
= 1 − 2,605.955 11,206.092
= 1 − .233 = .767
If we compare the R2 (0.80) to the adjusted R2 (0.767), the difference in this case is small.
R2adj = 1 −
SSE n − (k + 1)
SS total n − 1
(14–4)ADJUSTED COEFFICIENT OF DETERMINATION
Refer to Self-Review 14–1 on the subject of restaurants in Myrtle Beach. The ANOVA por- tion of the regression output is presented below.
S E L F - R E V I E W 14–2
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% 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.
MULTIPLE REGRESSION ANALYSIS 499
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 variables. How- ever, 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 population re- gression equation that relates the dependent variable to the k independent variables. This is sometimes called a model of the relationship. In symbols we write:
Y = α + β1X1 + β2X2 + … + βk Xk
This equation is analogous to formula (14–1) except the coefficients are now re- ported 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 com- puted values of a and bi are sample statistics. These sample statistics are point estimates of the corresponding population parameters α and βi. For example, the sample regression coefficient b2 is a point estimate of the population parameter β2. The sampling 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 parameter values to be estimated. Thus, by using the properties of the sampling distributions of these statistics, inferences about the population pa- rameters are possible.
LO14-3 Test hypotheses about the relationships inferred by a multiple regression model.
5. Consider the ANOVA table that follows.
a. Determine the standard error of estimate. About 95% 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% 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.
E X E R C I S E S
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
500 CHAPTER 14
Global Test: Testing the Multiple Regression Model We can test the ability of the independent variables X1, X2, . . . , Xk to explain the behav- ior of the dependent variable Y. To put this in question form: Can the dependent vari- able be estimated without relying on the independent variables? The test used is referred to as the global test. Basically, it investigates whether it is possible that all the independent variables have zero regression coefficients.
To relate this question to the heating cost example, we will test whether the three independent variables (amount of insulation in the attic, mean daily outside temperature, and age of furnace) effectively estimate home heating costs. In testing the hypothesis, we first state the null hypothesis and the alternate hypothesis in terms of the three pop- ulation parameters, β1, β2, and β3 . Recall that b1, b2, and b3 are sample regression coeffi- cients and are not used in the hypothesis statements. In the null hypothesis, we test whether the regression coefficients in the population are all zero. The null hypothesis is:
H0: β1 = β2 = β3 = 0
The alternate hypothesis is:
H1: Not all the βi’s are 0.
If the hypothesis test fails to reject the null hypothesis, it implies the regression coeffi- cients are all zero and, logically, are of no value in estimating the dependent variable (heating cost). Should that be the case, we would have to search for some other inde- pendent 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 signifi- cance. 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 change, 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 denom- inator, 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 horizon- tal 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 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.
Using the values from the ANOVA table on page 492, the F-statistic is
F = SSR∕k
SSE∕[n − (k + 1)] =
171,220.473∕3 41,695.277∕[20 − (3 + 1)]
= 21.90
Remember that the F-statistic tests the basic null hypothesis that two variances or, in this case, two mean squares are equal. 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
GLOBAL TEST F = SSR∕k
SSE∕[n − (k + 1)] (14–5)
MULTIPLE REGRESSION ANALYSIS 501
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 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) calculating a p-value based on the test statistic and comparing the p-value to the significance level. The critical value method using the F-statistic 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 vari- ables. The middle 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.
The critical value of F is found in Appendix B.6. Using the table for the .05 significance 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 H0 is not rejected and the region where H0 is rejected are shown in the following diagram.
3.24 Scale of F
F distribution df = (3, 16)
Region of rejection
(.05 level) Region
where H0 is not
rejected
Continuing with the global test, the decision rule is: Do not reject the null hypothesis, H0, 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 H0 and accept the alternate hypothesis, H1.
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 varia- tion in the dependent variable (heating cost). We expected this decision. Logically, the
502 CHAPTER 14
outside temperature, the amount of insulation, or the age of the furnace has a great bearing 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 statistical 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.
Evaluating Individual Regression Coefficients So far we have shown that at least one, but not necessarily all, of the regression coeffi- cients is not equal to zero and thus useful for predictions. The next step is to test the independent variables individually to determine which regression coefficients may be 0 and which are not.
Why is it important to know if any of the βi’s equal 0? If a β could equal 0, it implies that this particular independent variable is of no value in explaining any variation in the dependent value. If there are coefficients for which H0 cannot be rejected, we may want to eliminate them from the regression equation.
Our strategy is to use three sets of hypotheses: one for temperature, one for insu- lation, and one for age of the furnace.
For temperature: For insulation: For furnace age:
H0: β1 = 0 H0: β2 = 0 H0: β3 = 0 H1: β1 ≠ 0 H1: β2 ≠ 0 H1: β3 ≠ 0
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 n − (k + 1) degrees of freedom.
The number of sample observations is n. There are 20 homes in the study, so n = 20. The number of independent variables is k, which is 3. Thus, there are n − (k + 1) = 20 − (3 + 1) = 16 degrees of freedom.
The critical value for t is in Appendix B.5. For a two-tailed test with 16 degrees of freedom using the .05 significance level, H0 is rejected if t is less than −2.120 or greater than 2.120.
Refer to the Excel output earlier in this section. The column highlighted in orange, headed Coefficients, shows the values for the multiple regression equation:
ŷ = 427.194 − 4.583x1 − 14.831x2 + 6.101x3 Interpreting the term −4.583x1 in the equation: For each degree increase in tempera- ture, we predict that heating cost will decrease $4.58, holding the insulation and age of the furnace 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 col- umn labeled “Standard Error” estimates the variability for each of these regression coeffi- cients. The sampling distributions of the coefficients follow the t distribution with n − (k + 1) degrees of freedom. Hence, we are able to test the independent variables individually to determine whether the regression coefficients differ from zero. The formula is:
t = bi − 0
sbi (14–6)TESTING INDIVIDUAL
REGRESSION COEFFICIENTS
MULTIPLE REGRESSION ANALYSIS 503
The bi refers to any one of the regression coefficients, and sbi refers to the standard deviation of that distribution of the regression coefficient. We include 0 in the equation because the null hypothesis is βi = 0.
To illustrate this formula, refer to the test of the regression coefficient for the inde- pendent variable temperature. From the output earlier in this section, the regression coefficient for temperature is −4.583. The standard deviation of the sampling distribu- tion of the regression coefficient for the independent variable temperature is 0.772. In- serting these values in formula (14–6):
t = b1 − 0
sb1 =
−4.583 − 0 0.772
= −5.937
The computed value of t is −5.937 for temperature (the small difference 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 temperature and insulation variables are not zero. The computed t for the age of the furnace is 1.521, so we con- clude that the coefficient could equal 0. The independent variable age of the furnace is not a significant predictor of heating cost. The results of these hypothesis tests indi- cate that the analysis should focus on temperature and insulation as predictors of heating cost.
We can also use p-values to test the individual regression coefficients. Again, these are commonly reported in statistical software output. The computed value of t for tem- perature 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 temperature is not equal to zero and should be included in the equation to predict heating costs. For insulation, the value of t 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 fur- nace” 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. For the age of the furnace variable, we failed to reject the null hypothesis that the regression coef- ficient was zero. It is clear that we should drop that variable and rerun the regres- sion equation. Below is the Excel output where heating cost is the dependent variable and outside temperature and amount of insulation are the independent variables.
504 CHAPTER 14
Summarizing the results from this new output:
1. The new regression equation is:
ŷ = 490.286 − 5.150x1 − 14.718x2 Notice that the regression coefficients for outside temperature (x1) and amount of
insulation (x2) are similar to but not exactly the same as when we included the inde- pendent variable age of the furnace. Compare the above equation to that in the Excel output earlier in this section. Both of the regression coefficients are negative as in the earlier equation.
2. The details of the global test are as follows:
H0 : β1 = β2 = 0
H1 : Not all of the βi’s = 0
The F distribution is the test statistic and there are k = 2 degrees of freedom in the numerator and n − (k + 1) = 20 − (2 + 1) = 17 degrees of freedom in the denomi- nator. Using the .05 significance level and Appendix B.6, the decision rule is to re- ject H0 if F is greater than 3.59. We compute the value of F as follows:
F = SSR∕k
SSE∕ [n − (k + 1)] =
165,194.521∕2 47,721.229∕[20 − (2 + 1)]
= 29.424
Because the computed value of F (29.424) 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 (29.424) has a p-value (0.000), which is clearly less than 0.05. Therefore, we reject the null hypothesis and accept the alter- nate. We conclude that at least one of the regression coefficients is different from 0.
3. The next step is to conduct a test of the regression coefficients individually. We want to determine 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
H0: β1 = 0 H0: β2 = 0
H1: β1 ≠ 0 H1: β2 ≠ 0
The test statistic is the t distribution with n − (k + 1) = 20 − (2 + 1) = 17 degrees of freedom. Using the .05 significance level and Appendix B.5, the decision rule is to reject H0 if the computed value of t is less than −2.110 or greater than 2.110.
Outside Temperature Insulation
t = b1 − 0
sb1 =
−5.150 − 0 0.702
= −7.337 t = b2 − 0
sb2 =
−14.718 − 0 4.934
= −2.983
In both tests, we reject H0 and accept H1. We conclude that each of the regression co- efficients 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 ex- plain, suppose we develop a multiple regression equation based on five independent variables. We conduct the global test and find that some of the regression coefficients
MULTIPLE REGRESSION ANALYSIS 505
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 an- other 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 remaining 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 ex- plained (the R-square term). Two of the common methods are stepwise regression 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 inde- pendent variables.
Unfortunately, on occasion, the software may work “too hard” to find an equation that fits all the quirks of your particular data set. The suggested equation may not repre- sent the relationship in the population. Judgment is needed to choose among the equa- tions presented. Consider whether the results are logical. They should have a simple interpretation and be consistent with your knowledge of the application under study.
The regression output about eating places in Myrtle Beach is repeated below (see earlier self-reviews).
S E L F - R E V I E W 14–3
(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 con- sider eliminating? Use the .05 significance level.
(c) Outline a plan for possibly removing independent variables.
7. Given the following regression output,
E X E R C I S E S
506 CHAPTER 14
answer the following questions: a. Write the regression equation. b. If x1 is 4 and x2 is 11, what is the expected or predicted 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 coeffi-
cients 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 signifi- cance level. Which variable would you consider eliminating?
f. Outline a strategy for deleting independent variables in this case. 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.
x1 is the number of architects employed by the company. x2 is the number of engineers employed by the company. x3 is the number of years involved with health care projects. x4 is the number of states in which the firm operates. x5 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 coeffi-
cients 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 signifi- cance level. Which variable would you consider eliminating first?
e. Outline a strategy for deleting independent variables in this case.
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 evaluating each regression coefficient. We also discussed the decision-making process for including and excluding independent variables in the multiple regression equation.
It is important to know that the validity of the statistical global and individual tests rely on several assumptions. So if the assumptions are not true, the results might be bi- ased or misleading. However, strict adherence to the following assumptions is not al- ways possible. Fortunately, the statistical techniques discussed in this chapter are robust enough to work effectively even when one or more of the assumptions are vio- lated. Even if the values in the multiple regression equation are “off” slightly, our esti- mates using a multiple regression equation will be closer than any that could be made otherwise.
LO14-4 Evaluate the assumptions of multiple regression.
MULTIPLE REGRESSION ANALYSIS 507
In Chapter 13, we listed the necessary assumptions for regression when we con- sidered only a single independent variable. 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.
2. The variation in the residuals is the same for both large and small values of ŷ. To put it another way, (y − ŷ) is unrelated to whether ŷ 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 ŷ. So the term (y − ŷ) is computed for every observation in the data set. These residuals should approximately follow a normal probability distribution with a mean of 0.
4. The independent variables should not be correlated. That is, we would like to se- lect 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 addition, we provide methods to validate these assumptions and indicate the consequences if these assumptions cannot be met. For those interested in additional discussion, search on the term “Applied Linear Models.”
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 vari- ables and the dependent variable would form a three-dimensional space. The regres- sion equation would then form a plane as shown on page 490. 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 inde- pendent 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 heating 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.
0 20 40 60 Temperature
Scatterplot of Cost vs. Temp
80 0
200
300
400
100
Co st
0 5 10 Insulation
Scatterplot of Cost vs. Insul
15 0
200
300
400
100
Co st
Using Residual Plots Recall that a residual (y − ŷ) can be computed using the mul- tiple regression equation for each observation in a data set. In Chapter 13, we discussed
508 CHAPTER 14
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 observations above the regres- sion line (these residuals would have a positive sign) and a good number of the obser- vations below the line (these residuals would have a negative 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 portray the multiple regression. However, plots of the residuals can help us evaluate the linear- ity of the multiple regression equation. To investigate, the residuals are plotted on the vertical axis against the predicted variable, ŷ. In the following graphs, the left graph 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.
The plot on the right shows 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 non-linear pattern of the residuals. This indicates the relationship is probably not linear. In this case, we would evaluate different transformations of the variables in the equation as discussed in Chapter 13.
0 100 200 300 Predicted Values
Random Residual vs. Predicted Values
400 –100
100
0
Re si
du al
–50 0 50 100 Predicted Values
Nonrandom Residuals vs. Predicted Values
150 –30
60
0
30
Re si
du al
Variation in Residuals Same for Large and Small ŷ Values This requirement indicates that the variation in the residuals is constant, 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 variation in monthly income. We suspect that as age increases so does income, 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 50-year-olds than for 35-year-olds. 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 ŷ. This is the same graph that we used to evaluate the assumption of linearity. Based on the scatter dia- gram, it is reasonable to conclude that this assumption has not been violated.
MULTIPLE REGRESSION ANALYSIS 509
Distribution of Residuals To be sure that the inferences we make in the global and individual hypothesis tests are valid, we evaluate the distribution of residuals. Ideally, the residuals should follow a nor- mal probability distribution.
To evaluate this assumption, we can organize the residuals into a frequency distri- bution. The Histogram of Residuals graph is shown 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.
Another graph that helps to evaluate the assumption of normally distributed residu- als is called a Normal Probability Plot and is shown to the right of the histogram. This graphical analysis is often included in statistical software. If the plotted points are fairly close to a straight line drawn from the lower left to the upper right of the graph, the nor- mal probability plot supports the assumption of normally distributed residuals This plot supports the assumption of normally distributed residuals.
–76 –38 360 Residuals
Histogram of Residuals
76 0
8
6
4
2
Fr eq
ue nc
y
0 20 8040 60 Sample Percentile
Normal Probability Plot
100 0
450 400 350 300 250 200 150 100
50
Co st
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 indepen- dent variables make it difficult to make inferences about the individual regression coef- ficients and their individual effects on the dependent variable. In practice, it is 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 multicollinearity is important.
First, multicollinearity does not affect a multiple regression equation’s ability to pre- dict the dependent variable. However, when we are interested in evaluating the rela- tionship between each independent variable and the dependent variable, multicollinearity may show unexpected results.
For example, if we use two highly correlated independent variables, high school GPA and high school class rank, to predict the GPA of incoming college freshmen (de- pendent variable), we would expect that both independent variables would be posi- tively 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 to avoid correlated independent variables is they may lead to erroneous results in the hypothesis tests for the individual independent variables. This
510 CHAPTER 14
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 re- gression coefficient that is not significant.
2. A regression coefficient that should have a positive sign turns out to be negative, 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 problem using both of the in- dependent 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:
VARIANCE INFLATION FACTOR VIF = 1
1 − R 2j (14–7)
The term Rj 2 refers to the coefficient of determination, where the selected independent
variable is used as a dependent variable and the remaining independent variables are used as independent variables. A VIF greater than 10 is considered unsatisfactory, indi- cating that the independent variable should be removed from the analysis. The follow- ing example will explain the details of finding the VIF.
E X A M P L E
Refer to the data in Table 14–1, which relate 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.
S O L U T I O N
We begin by finding the correlation matrix for the dependent variable and the three independent variables. A correlation matrix shows the correlation between all pairs of the variables. A portion of that output follows:
Cost Temp Insul Age
Cost
Temp
Insul
Age
1.000
–0.812
–0.257
0.537 1.000
1.000
–0.103
–0.486
1.000
0.064
The highlighted area indicates the correlation among the independent variables. Because all of the correlations are between −.70 and .70, we do not suspect prob- lems with multicollinearity. The largest correlation among the independent vari- ables is −0.486 between age and temperature.
MULTIPLE REGRESSION ANALYSIS 511
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 the Regression Analysis in Excel to find the multiple coefficient of determination with temperature as the dependent variable and amount of insulation and age of the furnace as independent variables. The relevant regression output follows.
The coefficient of determination is .241, so inserting this value into the VIF formula:
VIF = 1
1 − R21 =
1 1 − .241
= 1.32
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, the R2 is .011 and, using formula (14–7), the VIF for insulation would be 1.011. To find the VIF for age, we would develop a regression equation with age as the dependent variable and temperature and insu- lation as the independent variables. For this equation, the R2 is .236 and, using formula (14–7), the VIF for age would be 1.310. All the VIF values are less than 10. Hence, we conclude there is not a problem with multicollinearity in this example.
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 residu- als 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 Agis 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 (remember 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 ŷ on the horizontal axis. Note the run of residuals above the mean of
512 CHAPTER 14
the residuals, followed by a run below the mean. A scatter plot such as this would indi- cate possible autocorrelation.
Re si
du al
s ( y
– y
)^
Fitted values y ̂
0
There is a test for autocorrelation, called the Durbin-Watson. It is discussed in Chapter 18.
QUALITATIVE INDEPENDENT VARIABLES In the previous example/solution regarding heating cost, the two independent variables outside 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 qual- ity or attribute, such as gender measured as male or female. To use a qualitative vari- able 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 are 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 at- tached garage, a 1 is used. We will refer to the “garage” variable as x4. The data from Table 14–2 are entered into the Excel system. Recall that the variable “age of the fur- nace” is not included in the analysis because we determined that it was not significantly related to heating cost.
The output from Excel is:
LO14-5 Use and interpret a qualitative, dummy variable in multiple regression.
STATISTICS IN ACTION
Multiple regression has been used in a variety of legal proceedings. It is par- ticularly useful in cases al- leging discrimination by gender or race. As an ex- ample, suppose that a woman alleges that Com- pany 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 response, 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 experienced than the male employees. In fact, the company might further argue that the cur- rent situation is actually due to its recent successful efforts to hire more women.
MULTIPLE REGRESSION ANALYSIS 513
What is the effect of the garage variable? Should it be included in the analysis? To show the effect of the variable, suppose we have two homes exactly alike next to each other in Buffalo, New York; one has an attached garage and the other does not. Both homes have 3 inches of insulation, and the mean January temperature in Buffalo is 20 degrees. For the house without an attached garage, a 0 is substituted for x4 in the regression equation. The estimated heating cost is $280.404, found by:
ŷ = 393.666 − 3.963x1 − 11.334x2 + 77.432x4 = 393.666 − 3.963(20) − 11.334(3) + 77.432(0) = 280.404
For the house with an attached garage, a 1 is substituted for x4 in the regression equa- tion. The estimated heating cost is $357.836, found by:
ŷ = 393.666 − 3.963x1 − 11.334x2 + 77.432x4 = 393.666 − 3.963(20) − 11.334(3) + 77.432(1) = 357.836
The difference between the estimated heating costs is $77.432 ($357.836 − $280.404). Hence, we can expect the cost to heat a house with an attached garage to be $77.432 more than the cost for an equivalent house without a garage.
We have shown the difference between the two types of homes to be $77.432, but is the difference significant? We conduct the following test of hypothesis.
H0: β4 = 0
H1: β4 ≠ 0
The information necessary to answer this question is in the output at the bottom of the previous page. The regression coefficient for the independent variable garage is $77.432, and the standard deviation of the sampling distribution is 22.783. We identify this as the fourth independent variable, so we use a subscript of 4. (Remember we
TABLE 14–2 Home Heating Costs, Temperature, Insulation, and Presence of a Garage for a Sample of 20 Homes
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
514 CHAPTER 14
dropped age of the furnace, the third independent variable.) Finally, we insert these values in formula (14–6).
t = b4 − 0
sb4 =
77.432 − 0 22.783
= 3.399
There are three independent variables in the analysis, so there are n − (k + 1) = 20 − (3 + 1) = 16 degrees of freedom. The critical value from Appendix B.5 is 2.120. The decision rule, using a two-tailed test and the .05 significance level, is to reject H0 if the computed t is to the left of −2.120 or to the right of 2.120. Because the computed value of 3.399 is to the right of 2.120, the null hypothesis is rejected. We conclude that the regression coefficient is not zero. The independent variable garage should be in- cluded in the analysis.
Using the p-value approach, the computed t value of 3.399 has a p-value of 0.004. This value is less than the .05 significance level. Therefore, we reject the null hypothe- sis. We conclude that the regression coefficient is not zero and the independent vari- able garage should be included in the analysis.
Is it possible to use a qualitative variable with more than two possible outcomes? 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, x1. To include the qualitative information regarding the quarter, we use three additional independent variables. For the variable x2, the five observations referring to the first quarter of each of the 5 years are coded 1 and the other quarters 0. Similarly, for x3 the five observations referring to the second quarter are coded 1 and the other quarters 0. For x4, 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 referring to this quarter is not necessary.
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 as- sociates 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 (female = 1 and male = 0).
Regression Analysis
Regression Statistics Multiple R 0.801 R Square 0.642 Adjusted R Square 0.600 Standard Error 3.219 Observations 20
ANOVA df SS MS F p-value Regression 2 315.9291 157.9645 15.2468 0.0002 Residual 17 176.1284 10.36049 Total 19 492.0575
Coefficients Standard Error t Stat p-value Intercept 15.7625 3.0782 5.121 .0001 Months 0.4415 0.0839 5.262 .0001 Gender 3.8598 1.4724 2.621 .0179
S E L F - R E V I E W 14–4
MULTIPLE REGRESSION ANALYSIS 515
(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 .05 significance level. What is your conclusion?
REGRESSION MODELS WITH INTERACTION In Chapter 12, we discussed interaction among independent variables. To explain, suppose 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, significant). We are interested in whether there is interaction among the independent variables. 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 ex- ercise 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 work- ers? Would workers produce more units in a quiet, cool room compared to a hot, noisy room?
In regression analysis, interaction is examined as a separate independent variable. An interaction prediction variable can be developed by multiplying the data values of one independent variable by the values of another independent variable, thereby cre- ating a new independent variable. A two-variable model that includes an interaction term is:
Y = α + β1X1 + β2X2 + β3X1X2
The term X1X2 is the interaction term. We create this variable by multiplying the values of X1 and X2 to create a third independent variable. We then develop a regression equa- tion using the three independent variables and test the significance of the third inde- pendent variable using the individual test for independent variables, described earlier in the chapter. An example will illustrate the details.
LO14-6 Include and interpret an interaction effect in a multiple regression analysis.
E X A M P L E
Refer to the heating cost example and the data in Table 14–1. Is there an inter- action between the outside temperature and the amount of insulation? If both variables are increased, is the effect on heating cost greater than the sum of savings from warmer temperature and the savings from increased insulation separately?
S O L U T I O N
The information from Table 14–1 for the independent variables temperature and insulation is repeated below. We create the interaction variable by multiplying the value of temperature by the value insulation for each observation in the data set. For the first sampled home, the value temperature is 35 degrees and insulation is
516 CHAPTER 14
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.
ŷ = 598.070 − 7.811x1 − 30.161x2 + 0.385x1x2 The question we wish to answer is whether the interaction variable is significant. Note we use the subscript, 1×2, to indicate the coefficient of the interaction of vari- ables 1 and 2. We will use the .05 significance level. In terms of a hypothesis:
H0: β1×2 = 0
H1: β1×2 ≠ 0
There are n − (k + 1) = 20 − (3 + 1) = 16 degrees of freedom. Using the .05 sig- nificance level and a two-tailed test, the critical values of t are −2.120 and 2.120. We reject the null hypothesis if t is less than −2.120 or t is greater than 2.120. From the output, b1×2 = 0.385 and sb1x2 = 0.291. To find the value of t, we use for- mula (14–6).
t = b1x2 − 0
sb1x2 =
0.385 − 0 0.291
= 1.324
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 .204 exceeds .05. We con- clude that there is not a significant interaction effect of temperature and insulation on home heating costs.
There are other situations that can occur when studying interaction among independent 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 nom- inal scale. In our heating cost example, we could have studied the interaction be- tween temperature and garage.
Studying all possible interactions can become very complex. However, careful consider- ation to possible interactions among independent variables can often provide useful insight into the regression models.
MULTIPLE REGRESSION ANALYSIS 517
STEPWISE REGRESSION In our heating cost example (see sample information in Table 14–1), we considered three independent variables: the mean outside temperature, the amount of insulation in the home, and the age of the furnace. To obtain the equation, we first ran a global or “all at once” test to determine if any of the regression coefficients were signifi- cant. When we found at least one to be significant, we tested the regression coeffi- cients individually to determine which were important. We kept the independent variables that had significant regression coefficients and left the others out. By re- taining the independent variables with significant coefficients, we found the regres- sion equation that used the fewest independent variables. This made the regression equation easier to interpret. Then we considered the qualitative variable, garage, and found that it was significantly related to heating cost. The variable, garage, was added to the equation.
Deciding the set of independent variables to include in a multiple regression equa- tion can be accomplished using a a technique called stepwise regression. This tech- nique efficiently builds an equation that only includes independent variables with significant regression coefficients.
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.
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 dependent variable. Stated differently, if we compute all the simple correlations be- tween 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 unexplained. 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 deter-
mination are shown.
Stepwise regression procedures are included in many statistical software packages. For example, Minitab’s stepwise regression analysis for the home heating cost prob- lem 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 variables that were included in our equation using the global test and the test for individual independent variables. The independent variable age, indicating the furnace’s age, is not included because it is not a significant predictor of cost.
LO14-7 Apply stepwise regression to develop a multiple regression model.
518 CHAPTER 14
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% of the variation in heating cost. The regression equation is:
ŷ = 388.8 − 4.93x1 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% to 80.46%. That is, by adding garage as an independent variable, we increase the coefficient of determination by 14.61 percentage points. The regression equation after step 2 is:
ŷ = 300.3 − 3.56x1 + 93.0x2 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 −4.93 to −3.56. This change is reflective of the added influence of the independent variable garage. Why did the stepwise method select the independent variable ga- rage instead of either insulation or age? The increase in R2, the coefficient of deter- mination, 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 − 3.96x1 + 77.0x2 − 11.3x3 At this point, 86.98% of the variation in heating cost is explained by the three inde-
pendent variables temperature, garage, and insulation. This is the same R2 value and regression equation we found on page 512 except for rounding differences.
4. Here, the stepwise procedure stops. This means the independent variable age does not add significantly to the coefficient of determination.
MULTIPLE REGRESSION ANALYSIS 519
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 advantage to the stepwise 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 variables and add one independent variable to the regression equation at each iteration. 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 criterion is to find the model with the largest R2 value, regardless of the number of independent variables. Also, each independent variable does not necessarily have a nonzero regres- sion coefficient. Since each independent variable could either be included or not in- cluded, there are 2k − 1 possible models, where k refers to the number of independent variables. In our heating cost example, we considered four independent variables so there are 15 possible regression models, found by 24 − 1 = 16 − 1 = 15. 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.
9. The 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 electrical repairmen employed by the company. An aptitude test is required by the human resources department to become an electrical repairman. The manager was able to get the score for each repairman in the sample. In addition, he deter- mined which of the repairmen were union members (code = 1) and which were not (code = 0). The sample information is reported below.
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 Householder 45 2 1 Lori 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.
E X E R C I S E S
520 CHAPTER 14
b. Comment on the regression equation. Be sure to include the coefficient of de- termination and the effect of union membership. Are these two variables effec- tive in explaining the variation in job performance?
c. Conduct a test of hypothesis to determine if union membership should be in- cluded as an independent variable.
d. Repeat the analysis considering possible interaction terms. 10. Cincinnati Paint Company sells quality brands of paints through hardware
stores throughout the United States. The company maintains a large sales force who call on existing customers and look for new business. The national sales man- ager is investigating 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
Develop a regression equation including an interaction term. Is there a significant interaction 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
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 bid- ders, the age of the painting, and the interaction between the number of bid- ders and the age of the painting. Which variables would you select?
MULTIPLE REGRESSION ANALYSIS 521
REVIEW OF MULTIPLE REGRESSION We described many topics involving multiple regression in this chapter. In this section of the chapter, we focus on a single example with a solution that reviews the procedure and guides your application of multiple regression analysis.
LO14-8 Apply multiple regression techniques to develop a linear model.
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 family (1 for yes, 0 for no), and the total years of education beyond high school for the husband 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.
E X A M P L E
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 house- hold, 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.
S O L U T I O N
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/Lind17e) and is identified as Bank of New England.
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
TABLE 14–3 Information on Sample of 25 Loans by the Bank of New England
522 CHAPTER 14
We begin by calculating the correlation matrix shown below. It shows the relation- ship between each of the independent variables and the dependent variable. It helps to identify the independent variables that are more closely related to the de- pendent variable (family income). The correlation matrix also reveals the indepen- dent variables that are highly correlated and possibly redundant.
1 10.720
0.188 0.243 0.116 0.486
Income Income Value Education Age Mortgage Gender
Value Education Age Mortgage Gender
–0.144 0.220 0.358 0.184
1 0.621
–0.210 0.062
1 –0.038
0.156 1
1–0.129
What can we learn from this correlation matrix?
1. The first column shows the correlations between each of the independent vari- ables and the dependent variable family income. Observe that each of the in- dependent variables is positively correlated with family income. The value of the home has the strongest correlation with family income. The level of educa- tion of the person applying for the loan and the current mortgage payment have a weak correlation with family income. These two variables are candi- dates to be dropped from the regression equation.
2. All possible correlations among the independent variables are identified with the green background. Our standard is to look for correlations that exceed an absolute value of .700. None of the independent variables are strongly cor- related with each other. This indicates that multicollinearity is not likely.
Next, we compute the multiple regression equation using all the independent vari- ables. The software output follows.
MULTIPLE REGRESSION ANALYSIS 523
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%, so the five independent variables account for three-quarters of the variation in family in- come. 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 vari- ables account for 68.4% 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.
ŷ = 70.606 + 0.072(Value) + 1.624(Education) − 0.122(Age) − 0.001(Mortgage) + 1.807(Gender)
Be careful in this interpretation. Both income and the value of the home are in thousands of dollars. Here is a summary:
1. An increase of $1,000 in the value of the home suggests an increase of $72 in family income. An increase of 1 year of education increases income by $1,624, another year older reduces income by $122, 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 in- versely related to family income. This is true because the sign of the regression coefficient 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.
H0: β1 = β2 = β3 = β4 = β5 = 0
H1: Not all the β’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. The p-values to test each regression coefficient are reported in cells E18 through E22 in the software output on the previous page. The null hypothesis and the alternate hypothesis are:
H0: βi = 0
H1: βi ≠ 0
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. The p-value for age and mortgage amount are greater than the significance level of .05, so we do not reject the null hypotheses for these variables. The regression coefficients are not different from zero and are not related to family income.
Based on the results of testing each of the regression coefficients, we con- clude that the variables age and mortgage amount are not effective predictors of family income. Thus, they should be removed from the multiple regression equa- tion. Remember 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 vari- able is mortgage amount. The result of the regression analysis without the mort- gage variable follows.
524 CHAPTER 14
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 significance 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 in- dependent 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.
MULTIPLE REGRESSION ANALYSIS 525
3. Reviewing the significance of the individual coefficients, the p-values associ- ated with each of the remaining independent variables are less than .05. We conclude that all the regression coefficients are different from zero. Each inde- pendent variable is a useful predictor of family income.
Our final step is to examine the regression assumptions (Evaluating the Assump- tions of Multiple Regression section on page 506) with our regression model. The first assumption is that there is a linear relationship between each independent variable and the dependent variable. 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.
92 50 100
Value ($000)
Scatterplot of Income vs. Value Scatterplot of Income vs. Education
Education
In co
m e
In co
m e
150 200
94 96 98
100 102 104 106
92 12 13 14 15 16 17
94 96 98
100 102 104 106
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 determine if the relationship is linear or not.
A plot of the residuals is also useful to evaluate the overall assumption of lin- earity. Recall that a residual is (y − ŷ), the difference between the actual value of the dependent variable (y) and the predicted value of the dependent variable (ŷ). As- suming a linear 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.
94 –3
0
3
96 98 100 Predicted Income
Re si
du al
s
Scatterplot of Residuals vs. Predicted Income
102 104
526 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 follow the normal probability distribution with a mean of zero. To evaluate this assumption, we will use a histogram and a normal probability plot.
92 0 20 40
Sample Percentile
Histogram of Residuals Normal Probability Plot
In co
m e
60 80 100
94 96 98
100 102 104 106
Fr eq
ue nc
y
Residuals
0 1 2 3 4 5 6 7 8 9
–1.25 –0.75 –.25 0.25 0.75 1.25
In general, the histogram on the left shows the major characteristics of a normal dis- tribution, 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 probability plot on the right is based on a cumulative normal probability distribution. The line shows the standardized cumulative normal distribution. The green dots show the cumulative distribution of the residuals. To confirm the normal distribution of the re- siduals, the green dots should be close to the line. This is true for most of the plot. However, we would note that there are departures and even perhaps a nonlinear 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 indepen- dent variables should not be highly correlated. We suggested a rule of thumb that multicollinearity would be a concern if the correlations among independent vari- ables were close to 0.7 or −0.7. There are no violations of this guideline.
There is a statistic that is used to more precisely evaluate multicollinearity, the variance inflation factor (VIF). To calculate the VIFs, we need to do a regression analysis for each independent variable as a function of the other independent vari- ables. From each of these regression analyses, we need the R2 to compute the VIF using formula (14–7). The following table shows the R2 for each regression analysis and the computed VIF. If the VIFs are less than 10, then multicollinearity is not a concern. In this case, the VIFs are all less than 10, so multicollinearity among the independent variables is not a concern.
Dependent Variable Independent Variables R-square VIF
Value Education and Gender 0.058 1.062 Education Gender and Value 0.029 1.030 Gender Value and Education 0.042 1.044
To summarize, the multiple regression equation is
ŷ = 74.527 + 0.063(Value) + 1.016(Education) + 1.770(Gender)
This equation explains 71.6% of the variation in family income. There are no major departures from the multiple regression assumptions of linearity, normally distrib- uted residuals, and multicollinearity.
MULTIPLE REGRESSION ANALYSIS 527
C H A P T E R S U M M A R Y
I. The general form of a multiple regression equation is:
ŷ = a + b1x1 + b2x2 + . . . + bkxk (14–1)
where a is the Y-intercept when all x’s are zero, bj refers to the sample regression coeffi- cients, and xj refers to the value of the various independent variables. 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 between the observed and predicted values of
the dependent variable. 3. It ranges from 0 to plus infinity. 4. It is calculated from the following equation.
sy.123...k = √ Σ ( y − ŷ )2
n − (k + 1) (14–2)
B. The coefficient of multiple determination reports the percent of the variation in the dependent variable explained by the variation in 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.
R2 = SSR
SS total (14–3)
4. When the number of independent variables is large, we adjust the coefficient of determination for the degrees of freedom as follows.
R2adj = 1 −
SSE n − (k + 1)
SS total n − 1
(14–4)
IV. A global test is used to investigate whether any of the independent variables have a re- gression coefficient that differs significantly from zero. 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 n − (k + 1) degrees of freedom in the de- nominator, where n is the sample size.
D. The formula to calculate the value of the test statistic for the global test is:
F = SSR∕k
SSE∕[n − (k + 1) ] (14–5)
V. The test for individual variables determines which independent variables have regression coefficients that differ significantly from zero. A. The variables that have zero regression coefficients are usually dropped from the
analysis. B. The test statistic is the t distribution with n − (k + 1) degrees of freedom.
528 CHAPTER 14
C. The formula to calculate the value of the test statistic for the individual test is:
t = bi − 0
sbi (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 ŷ. 1. Homoscedasticity means the variation is the same for all fitted values of the
dependent variable. 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 or a normal probability plot.
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 −0.70.
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.
VIF = 1
1 − R2j (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 to 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 x2) affects the rela- tionship with another independent variable (x1) 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.
P R O N U N C I A T I O N K E Y
SYMBOL MEANING PRONUNCIATION
b1 Regression coefficient for the b sub 1 first independent variable
bk Regression coefficient for any b sub k independent variable
sy.123...k Multiple standard error of s sub y dot 1, 2, 3 . . . k estimate
MULTIPLE REGRESSION ANALYSIS 529
C H A P T E R E X E R C I S E S
13. A multiple regression analysis yields the following partial results.
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 at least one of the regression coefficients is not equal to
zero. Let α = .05. 14. In a multiple regression analysis, two independent variables are considered, and the
sample size is 25. The regression coefficients and the standard errors are as follows.
b1 = 2.676 sb1 = 0.56 b2 = −0.880 sb2 = 0.71 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 regres- sion equation? Use the .05 significance level.
15. The following output was obtained from a multiple regression analysis.
a. What is the sample size? b. Compute the value of R2. c. Compute the multiple standard error of estimate. d. Conduct a global test of hypothesis to determine whether any of the regression
coefficients 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 analysis, k = 5 and n = 20, 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 coefficients 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 factors 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, consist- ing of several hundred stores, she selects a random sample of 30 stores. For each store, she gathered the following information. y = total sales last year (in $ thousands) x1 = number of competitors in the region x2 = population of the region (in millions) x3 = advertising expense (in $ thousands)
530 CHAPTER 14
The results of a multiple regression analysis, using Minitab, follow.
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 R2 value. c. Compute the multiple standard error of estimate. d. Conduct a global test of hypothesis to determine whether any of the regression
coefficients are not equal to zero. Use the .05 level of significance. e. Conduct tests of hypothesis 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 the total annual sales for each of the company’s regions. Five factors appear to be related to regional sales: the number of retail outlets in the region, the number of automobiles in the region registered as of April 1, the total personal income recorded in the first quarter of the year, the average age of the automobiles (years), and the number of sales supervisors in the region. The data for each region were gathered for last year. For example, see the following table. In 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 region’s 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 stron- gest correlation with the dependent variable? The correlations between the independent variables outlets and income and between outlets and number of automobiles are fairly strong. Could this be a problem? What is this condition called?
MULTIPLE REGRESSION ANALYSIS 531
sales outlets cars income age outlets 0.899 automobiles 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 shown below. What percent of the variation is ex- plained by the regression equation?
The regression equation is Sales = −19.7 − 0.00063 outlets + 1.74 autos + 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 automobiles 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 co- efficients 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 R2 changed from the previous analysis?
The regression equation is Sales = −18.9 + 1.61 autos + 0.400 income + 1.96 age
Predictor Coef SE Coef T P Constant −18.924 3.636 −5.20 0.002 automobiles 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
f. Following is a histogram of the residuals. Does the normality assumption appear rea- sonable? Why?
532 CHAPTER 14
g. Following is a plot of the fitted values of y (i.e., ŷ) and the residuals. What do you observe? Do you see any violations of the assumptions?
1.2
0
–1.2
8 16 24 32 40
Re si
du al
( y
– y
)
Fitted
^
ŷ
19. The administrator of a new paralegal program at Seagate Technical College wants to estimate 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
a. Consider the following correlation matrix. Which variable has the strongest correla- tion with the dependent variable? Some of the correlations among the independent variables are strong. Does this appear to be a problem?
Paralegal High School GPA GPA SAT Verbal High School GPA 0.911 SAT Verbal 0.616 0.609 SAT Math 0.487 0.636 0.599
b. Consider the following output. Compute the coefficient of multiple determination.
The regression equation is Paralegal GPA = −0.411 + 1.20 HSGPA + 0.00163 SAT_Verbal − 0.00194 SAT_Math
Predictor Coef SE Coef T P Constant −0.4111 0.7823 −0.53 0.622 HSGPA 1.2014 0.2955 4.07 0.010 SAT_Verbal 0.001629 0.002147 0.76 0.482 SAT_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 HSGPA 1 4.2061 SAT_Verbal 1 0.0303 SAT_Math 1 0.1231
MULTIPLE REGRESSION ANALYSIS 533
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 eliminating the variables “SAT_Verbal” and “SAT_Math”? Let α = .05.
e. The analysis has been rerun without “SAT_Verbal” and “SAT_Math.” See the follow- ing output. Compute the coefficient of determination. How much has R2 changed from the previous analysis?
The regression equation is Paralegal GPA = −0.454 + 1.16 HSGPA
Predictor Coef SE Coef T P Constant −0.4542 0.5542 −0.82 0.439 HSGPA 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
f. Following is a histogram of the residuals. Does the normality assumption for the re- siduals seem reasonable?
g. Following is a plot of the residuals and the ŷ values. Do you see any violation of the assumptions?
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
20. Mike Wilde is president of the teachers’ union for Otsego School District. In pre- paring for upcoming negotiations, he is investigating 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 teaching effectiveness rating given by the principal, and whether the teacher has a master’s degree. A random sample of 20 teachers resulted in the following data.
534 CHAPTER 14
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
*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 5 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 co- efficients differ from zero. Use the .05 significance level.
d. Conduct a test of hypothesis for the individual regression coefficients. Would you consider 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 ŷ values on the X-axis. Does the plot reveal any violations of the as- sumptions of regression?
21. A video media consultant collected the following data on popular LED televisions sold through on-line retailers.
Manufacturer Screen Price
Sharp 46 736.50 Samsung 52 1150.00 Samsung 46 895.00 Sony 40 625.00 Sharp 42 773.25 Samsung 46 961.25 Samsung 40 686.00 Sharp 37 574.75 Sharp 46 1000.00 Sony 40 722.25 Sony 52 1307.50 Samsung 32 373.75
Manufacturer Screen Price
Sharp 37 657.25 Sharp 32 426.75 Sharp 52 1389.00 Samsung 40 874.75 Sharp 32 517.50 Samsung 52 1475.00 Sony 40 954.25 Sony 52 1551.50 Sony 46 1303.00 Sony 46 1430.50 Sony 52 1717.00
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 normal
distribution. g. Plot the residuals versus the fitted values. Do they seem to have the same amount of
variation?
MULTIPLE REGRESSION ANALYSIS 535
22. A regional planner is studying the demographics of nine counties in the eastern region of an Atlantic seaboard state. She has gathered the following data:
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
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 anal-
ysis 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 normal
distribution. g. Plot the residuals versus the fitted values. Do they seem to have the same amount of
variation? 23. Great Plains Distributors, Inc. sells roofing and siding products to home improve-
ment retailers, such as Lowe’s and Home Depot, and commercial contractors. The owner is interested in studying the effects of several variables on the sales volume of fiber-cement siding products.
The company has 26 marketing districts across the United States. In each district, it collected information on the following variables: sales volume (in thousands of dollars), advertising dollars (in thousands), number of active accounts, number of competing brands, and a rating of market 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 9 200.1 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
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?
536 CHAPTER 14
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. A market researcher is studying on-line subscription services. She is particularly
interested in what variables relate to the number of subscriptions for a particular on-line service. She is able to obtain the following sample information on 25 on-line subscrip- tion services. The following notation is used:
Sub = Number of subscriptions (in thousands) Web page hits = Average monthly count (in thousands) Adv = The advertising budget of the service (in $ hundreds) Price = Average monthly subscription price ($)
Web Page Service Sub Hits Adv Price
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
a. Determine the regression equation. b. Conduct a global test of hypothesis to determine whether any of the regression
coefficients 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? 25. Fred G. Hire is the manager of human resources at Crescent Custom Steel Prod-
ucts. As part of his yearly report to the CEO, he is required to present an analysis of the salaried employees. For each of the 30 salaried employees, he records monthly salary; service at Crescent, in months; age; gender (1 = male, 0 = female); and whether the employee has a management or engineering position. Those employed in management are coded 0, and those in engineering are coded 1.
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
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 R2? Comment on this value. c. Conduct a global test of hypothesis to determine whether any of the independent
variables are different from 0.
MULTIPLE REGRESSION ANALYSIS 537
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 management or engineering position?
26. Many regions 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 indi- cate the amount families 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.
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
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 R2? 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? 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.
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
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?
538 CHAPTER 14
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
a. Develop a multiple regression equation with the amount of tips as the dependent variable and the amount of the bill and the number of diners as independent vari- ables. 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? 29. The president of Blitz Sales Enterprises sells kitchen products through cable tele-
vision 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)
20 3.2 15 2.6 25 3.4 10 1.8 18 2.2 18 2.4 15 2.4 12 1.5
Infomercials Sales ($000s)
22 2.5 15 2.4 25 3.0 16 2.7 12 2.0 20 2.6 25 2.8
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 for the 4th of July was predictive of attendance at the Fall Festival held in October. She gathered the following data to test her suspicion.
MULTIPLE REGRESSION ANALYSIS 539
4th of July ($000) Fall Festival (000)
10.6 8.8 8.5 6.4 12.5 10.8 9.0 10.2 5.5 6.0 12.0 11.1 8.0 7.5 7.5 8.4
4th of July ($000) Fall Festival (000)
9.0 9.5 10.0 9.8 7.5 6.6 10.0 10.1 6.0 6.1 12.0 11.3 10.5 8.8
Determine the regression equation. Is the amount spent on fireworks related to at- tendance at the Fall Festival? Evaluate the regression assumptions by examining the residuals.
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 1,021.23 58 69,000 90,600 836.46 3 35,600 104,500 1,056.37 22 63,000
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 de- scribes 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.
D A T A A N A L Y T I C S
(The data for these exercises are available at the text website: www.mhhe.com/Lind17e.)
33. The North Valley Real Estate data reports information on homes on the market. Use the selling price of the home as the dependent variable and determine the regression equa- tion using the size of the house, number of bedrooms, days on the market, and number of bathrooms as independent variables. a. Develop a correlation matrix. Which independent variables have strong or weak correla-
tions with the dependent variable? Do you see any problems with multicollinearity? b. Use a statistical software package to determine the multiple regression equation.
How did you select the variables to include in the equation? How did you use the in- formation from the correlation analysis? Show that your regression equation shows a significant relationship. Write out the regression equation and interpret its practical application. Report and interpret the R-square.
c. Using your results from part (b), evaluate the addition of the variables: pool or garage. Report your results and conclusions.
540 CHAPTER 14
d. Develop a histogram or a stem-and-leaf display of the residuals from the final regres- sion equation developed in part (c). Is it reasonable to conclude that the normality assumption has been met?
e. Plot the residuals against the fitted values from the final regression equation developed in part (c). Plot the residuals on the vertical axis and the fitted values on the horizontal axis.
34. Refer to the Baseball 2016 data, which report information on the 30 Major League Base- ball teams for the 2016 season. Let the number of games won be the dependent vari- able and the following variables be independent variables: team batting average, team Earned Run Average (ERA), number of home runs, and whether the team plays in the American or the National League. a. Develop a correlation matrix. Which independent variables have strong or weak
correlations with the dependent variable? Do you see any problems with multicol- linearity? Are you surprised that the correlation coefficient for ERA is negative?
b. Use a statistical software package to determine the multiple regression equation. How did you select the variables to include in the equation? How did you use the information from the correlation analysis? Show that your regression equation shows a significant relationship. Write out the regression equation and interpret its practical application. Report and interpret the R-square. Is the number of wins affected by whether the team plays in the National or the American League?
c. Conduct a global test on the set of independent variables. Interpret. d. Conduct a test of hypothesis on each of the independent variables. Would you
consider deleting any of the variables? If so, which ones? e. 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?
f. Plot the residuals against the fitted values from the final regression equation devel- oped in part (f ). Plot the residuals on the vertical axis and the fitted values on the horizontal axis.
35. Refer to the Lincolnville School District bus data. First, add a variable to change the type of engine (diesel or gasoline) to a qualitative variable. If the engine type is diesel, then set the qualitative variable to 0. If the engine type is gasoline, then set the qualitative variable to 1. Develop a regression equation using statistical software with maintenance cost as the dependent variable and age, odometer miles, miles since last maintenance, and engine type as the independent variables. a. Develop a correlation matrix. Which independent variables have strong or weak correla-
tions with the dependent variable? Do you see any problems with multicollinearity? b. Use a statistical software package to determine the multiple regression equation.
How did you select the variables to include in the equation? How did you use the information from the correlation analysis? Show that your regression equation shows a significant relationship. Write out the regression equation and interpret its practical application. Report and interpret the R-square.
c. 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?
d. Plot the residuals against the fitted values from the final regression equation devel- oped in part (f) against the fitted values of Y. Plot the residuals on the vertical axis and the fitted values on the horizontal axis.
A REVIEW OF CHAPTERS 13–14 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 is measured by the correlation coefficient. The correlation coefficient is designated by the letter r. It can assume any value between −1.00 and +1.00 inclusive. Coefficients of −1.00 and +1.00 indicate a perfect relationship, and 0 indicates no relationship. A value near 0, such as −.14 or .14, indicates a weak relationship. A value near −1 or +1, such as −.90 or +.90, indicates a strong rela- tionship. The coefficient of determination, also called R2, measures the proportion of the total variation in the dependent variable explained by the independent variable. It can be computed as the square of the correlation coefficient.
MULTIPLE REGRESSION ANALYSIS 541
Likewise, the strength of the relationship between several independent variables and a dependent variable is measured by the coefficient of multiple determination, R2. It measures the proportion of the variation in y explained by two or more independent variables.
The linear relationship in the simple case involving one independent variable and one dependent variable is described by the equation ŷ = a + bx. For k independent variables, x1, x2, . . . xk, the same multiple regression equation is
ŷ = a + b1x1 + b2x2 + . . . + bkxk Solving for b1, b2, . . . , bk 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 multiple standard error of estimate, the results of the global test, and the test of the individual variables, are reported in the output of most statistical software programs.
P R O B L E M S
1. The accounting department at Box and Go Apparel wishes to estimate the net 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. The data from two stores are:
Net Number Overhead Average Theft Profit of Cost Markup Loss ($ thousands) Employees ($ thousands) (percent) ($ thousands) Store ŷ x1 x2 x3 x4
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 ŷ = 67 + 8x1 − 10x2 +
0.004x3 − 3x4. What are the predicted sales for a store with 112 employees, an overhead cost of $65,000, a markup rate of 50%, and a loss from theft of $50,000?
d. Suppose R2 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
dollars 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:
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.
542 CHAPTER 14
3. The following ANOVA output is given.
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?
C A S E S
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 randomly 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 re- cent deliveries. Factors thought to be related to the cost of delivering a kit were:
Prep The time in minutes between when the custom- ized 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
Sample Number Cost Prep Delivery Mileage
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
MULTIPLE REGRESSION ANALYSIS 543
Sample Number Cost Prep Delivery Mileage
11 $28.17 5 35 16 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
Sample Number Cost Prep Delivery Mileage
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 de- livery and the other variables. Do these three vari- ables explain a reasonable amount of the variation in the dependent variable? Estimate the delivery cost for a kit that takes 10 minutes for preparation, takes 30 minutes to deliver, and must cover a dis- tance of 14 miles.
2. Test to determine if one or more regression coeffi- cient 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 re- gression equation until only significant variables are included.
3. Write a brief report interpreting the final regression equation.
P R A C T I C E T E S T
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?
(.65, −.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, Ŷ = 7 − .5X, and that the coefficient of determination is .81, answer questions 6, 7, and 8.
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 dependent 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.
544 CHAPTER 14
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 Ŷ is called a . 12. 13. For a dummy variable, such as gender, how many different values 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 regression analysis output:
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
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? 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 negative
association between the variables. 2. Given the following regression analysis output.
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
a. What 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.
15Nonparametric Methods: NOMINAL LEVEL HYPOTHESIS TESTS
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO15-1 Test a hypothesis about a population proportion.
LO15-2 Test a hypothesis about two population proportions.
LO15-3 Test a hypothesis comparing an observed set of frequencies to an expected frequency distribution.
LO15-4 Explain the limitations of using the chi-square statistic in goodness-of-fit tests.
LO15-5 Test a hypothesis that an observed frequency distribution is normally distributed.
LO15-6 Perform a chi-square test for independence on a contingency table.
FOR MANY YEARS, TV executives used the guideline that 30% of the audience were watching each of the traditional big three prime-time networks and 10% 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 24 and LO15-3.)
© Matilde Gattoni/Getty Images
546 CHAPTER 15
INTRODUCTION In Chapters 9 through 12 we describe tests of hypothesis for data of interval or ratio scale. Examples of interval and ratio scale data include the scores on the first statistics examination in your class, the incomes of corporate executive officers in technology companies, or years of employment of production workers at the BMW plant in Greer, South Carolina.
We conducted hypothesis tests about a single population mean (Chapter 10), about two population means (Chapter 11), and about three or more population means (Chapter 12). For these tests we use interval or ratio data and assume the populations follow the normal probability distribution. However, there are hypothesis tests that do not require any assumption regarding the shape of the population. Hence, the assumption of a normal population is not necessary. These tests are referred to as nonparametric hypothesis tests.
In this chapter, we begin with tests of hypothesis for nominal scale data. Recall that nominal scale data are simply classified into mutually exclusive categories. In the first two sections of this chapter we describe tests of proportions. In these tests individuals or objects are classified into one of two mutually exclusive groups. Examples include gender (male or female), quality (acceptable or unacceptable), diabetes (yes or no), and airline flight arrivals (on time or late).
We also expand the nominal scale tests to include situations where data are clas- sified into several mutually exclusive categories. The scale of measurement is still nominal, but there are several categories. Examples include the colors of M&M Plain Candies (red, green, blue, yellow, orange, and brown), brand of peanut butter purchased (Peter Pan, Jif, Skippy, and others), or days of the workweek (Monday, Tuesday, Wednesday, Thursday, and Friday). We introduce the chi-square distribution as a new test statistic. It is most often used when there are more than two nominal scale categories.
TEST A HYPOTHESIS OF A POPULATION PROPORTION Beginning on page 300 in Chapter 9, we discussed confidence intervals for propor- tions. We can also conduct a test of hypothesis for a proportion. Recall that a propor- tion 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 number of observations, so the propor- tion 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% 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 proportion increased?
• The American Association of Retired Persons (AARP) reports that 60% 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% 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% 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 population proportion. To test a hypothesis about a population proportion, a random sample is
LO15-1 Test a hypothesis about a population proportion.
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 547
chosen from the population. It is assumed that the binomial assumptions discussed in Chapter 6 are met: (1) the sample data collected are the result of counts; (2) the out- come 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 out- come of any other trial. This test is appropriate when both nπ and n(1 − π) are at least 5. n is the sample size, and π is the population proportion. It takes advantage of the fact that a binomial distribution can be approximated by the normal distribution.
E X A M P L E
A Republican governor of a western state is thinking about running for reelection. Historically, to be reelected, a Republican candidate needs to earn at least 80% of the vote in the northern section of the state. The governor hires a polling organiza- tion to survey the voters in the northern section of the state and determine what percent would vote for him. The polling organization will survey 2,000 voters. Use a statistical hypothesis-testing procedure to assess the governor’s chances of reelection.
S O L U T I O N
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 are 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 a normal approximation to the binomial distribution if both nπ and n(1 − π) exceed 5. In this case, n = 2,000 and π = 0.80. (π is the proportion of the vote in the northern part of the state, or 80%, needed to be elected.) Thus, nπ = 2,000(.80) = 1,600 and n(1 − π) = 2,000(1 − .80) = 400. Both 1,600 and 400 are clearly greater than 5.
Step 1: State the null hypothesis and the alternate hypothesis. The null hypothesis, H0, is that the population proportion π 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 be reelected. These hypotheses are written sym- bolically as:
H0: π ≥ .80
H1: π < .80
H1 states a direction. Thus, as noted previously, the test is one-tailed with the inequality sign pointing to the tail of the distribution contain- ing 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.
548 CHAPTER 15
Step 3: Select the test statistic. z is the appropriate statistic, found by:
where: π 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 dividing point or points between the regions where H0 is rejected and where it is not rejected. Because 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 15–1.) The significance level is .05. This probability is in the left tail and de- termines the region of rejection. The area between zero and the criti- cal value is .4500, found by .5000 – .0500. Referring to Appendix B.3, go to the column indicating a .05 significance level for a one- tailed test, find the row with infinite degrees of freedom, and read the z value of 1.645. The decision rule is, therefore: Reject the null hy- pothesis and accept the alternate hypothesis if the computed value of z falls to the left of –1.645; otherwise do not reject H0.
z = p − π
√ π(1 − π)
n
(15–1)TEST OF HYPOTHESIS, ONE PROPORTION
Step 5: Make a decision. Select a sample and make a decision about H0. A sample survey of 2,000 potential voters in the northern part of the state revealed that 1,550 planned to vote for the incumbent gover- nor. 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 er- ror? 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. π is .80, the hypothesized population proportion. z is a normally distributed test statistic. We can use it because the
normal approximation assumptions are true.
.4500 .5000
H0: π ≥ .80 H1: π < .80
0–1.645 Critical value
Scale of z
.05 Region of rejection
H0 is not rejected
CHART 15–1 Rejection Region for the .05 Level of Significance, One-Tailed Test
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 549
Using formula (15–1) and computing z gives
z = p − π
√ π(1 − π)
n
=
1,550 2,000
− .80
√ .80(1 − .80)
2,000
= .775 − .80 √.00008
= −2.80
The computed value of z (−2.80) is less than the critical value, so the null hypothesis is rejected at the .05 level. The difference of 2.5 percentage points between the sample percent (77.5%) and the hypothesized population percent in the northern part of the state necessary to carry the state (80%) is statistically significant. From Appendix B.3, the probability of a z value between zero and −2.80 is .4974. So the p-value is .0026, found by .5000 − .4974. Because the p-value is less than the significance level, the null hypothesis is rejected.
Step 6: Interpret the result. The governor can conclude that he does not have the necessary support in the northern section of the state to win reelection. To put it another way, the evidence at this point does not support the claim that the incumbent governor will return to the gov- ernor’s mansion for another 4 years.
A recent insurance industry report indicated that 40% of those persons involved in minor traffic accidents this year have been involved in at least one other traffic accident in the last 5 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 5 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.
S E L F - R E V I E W 15–1
1. The following hypotheses are given.
H0: π ≤ .70 H1: π > .70
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?
2. The following hypotheses are given.
H0: π = .40 H1: π ≠ .40
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 six-step hypothesis-testing procedure in solving the following problems.
E X E R C I S E S
550 CHAPTER 15
TWO-SAMPLE TESTS ABOUT PROPORTIONS In the previous section, we considered a test of a single population proportion. How- ever, we are often interested also in whether two sample proportions come from popu- lations that are equal. Here are several examples.
• The vice president of human resources wishes to know whether there is a differ- ence in the proportion of hourly employees who miss more than 5 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 millennials and another group of baby-boomers. 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 consultant wishes to know whether there is a difference in the pro- portion 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 classi- fied as “liking the new design” or “not liking the new design.” We then compare the proportion in the millennial group with the proportion in the baby-boomer 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 individuals or objects.
To conduct the test, we assume each sample is large enough that the normal distri- bution 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:
LO15-2 Test a hypothesis about two population proportions.
3. The U.S. Department of Transportation estimates that 10% of Americans carpool. Does that imply that 10% of cars will have two or more occupants? A sample of 300 cars traveling southbound on the New Jersey Turnpike yesterday revealed that 63 had two or more occupants. At the .01 significance level, can we conclude that 10% of cars traveling on the New Jersey Turnpike have two or more occupants?
4. A recent article 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 .01 significance level, can we conclude that a larger propor- tion of students at your school have jobs?
5. Chicken Delight claims that 90% 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% of the orders are delivered in less than 10 minutes?
6. Research at the University of Toledo indicates that 50% of students change their major area of study after their first year in a program. A random sample of 100 stu- dents 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 pro- gram? Test at the .05 level of significance.
z = p1 − p2
√ pc(1 − pc)
n1 +
pc(1 − pc) n2
(15–2)TWO-SAMPLE TEST OF PROPORTIONS
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 551
where: 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 pc = x1 + x2 n1 + n2
(15–3)
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.
E X A M P L E
Manelli Perfume Company recently devel- oped a new fragrance that it plans to mar- ket under the name Heavenly. A number of market studies indicate that Heavenly has very good market potential. The sales department at Manelli is particularly inter- ested in whether there is a difference in the proportions of working and stay-at- home women who would purchase Heav- enly if it were marketed. There are two independent populations, a population consisting of working women and a popu- lation consisting of stay-at-home women. Each sampled woman will be asked to smell Heavenly and indicate whether she likes the fragrance well enough to pur- chase a bottle.
S O L U T I O N
We will use the usual six-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 working and stay-at-home women who prefer Heavenly.” We designate π1 as the proportion of working women who would purchase Heavenly and π2 as the proportion of stay-at-home women who would purchase it. The alternate hypothe- sis is that the two proportions are not equal.
H0: π1 = π2 H1: π1 ≠ π2
Step 2: Select the level of significance. We choose the .05 significance level in this example.
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552 CHAPTER 15
Step 3: Determine the test statistic. The two samples are sufficiently large so we use the standard normal distribution as the test statistic. The value of the test statistic is computed using formula (15–2).
Step 4: Formulate the decision rule. Recall that the alternate hypothesis from Step 1 does not indicate a direction, so this is a two-tailed test. To find the critical value, go to Student’s t distribution (Appendix B.5). In the table headings, find the row labeled “Level of Significance for Two-Tailed Test” and select the column for an alpha of .05. Go to the bottom row with infinite degrees of freedom. The z critical value is 1.960, so the critical values are −1.960 and 1.960. As before, if the computed test statistic is less than −1.960 or greater than 1.960, the null hypothesis is rejected. This information is summa- rized in Chart 15–2.
Step 5: Select a sample and make a decision. A random sample of 100 working women revealed 19 liked the Heavenly fragrance well enough to purchase it. Similarly, a sample of 200 stay-at-home women revealed 62 liked the fragrance well enough to make a pur- chase. Let p1 refer to working women and p2 to stay-at-home women.
p1 = x1 n1
= 19
100 = .19 p2 =
x2 n2
= 62
200 = .31
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 proportion of working and stay-at-home women who like the Heavenly fragrance.
Next, we combine or pool the sample proportions. We use formula (15–3).
pc = x1 + x2 n1 + n2
= 19 + 62
100 + 200 =
81 300
= 0.27
Note that the pooled proportion is closer to .31 than to .19 because more stay-at-home women than working women were sampled.
We use formula (15–2) to find the value of the test statistic.
z = p1 − p2
√ pc(1 − pc)
n1 +
pc(1 − pc) n2
= .19 − .31
√ .27(1 − .27)
100 +
.27(1 − .27) 200
= −2.207
−1.960 1.960 Scale of z
H0 is rejected
H0 is not rejected
H0 is rejected
.95
.025.025
H0: p1 5 p2 H1: p1 Þ p2
CHART 15–2 Decision Rules for Heavenly Fragrance Test, .05 Significance Level
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 553
The computed value of −2.207 is in the area of rejection; that is, it is to the left of −1.960. Therefore, the null hypothesis is rejected at the .05 significance level. To put it another way, we reject the null hypoth- esis that the proportion of working women who would purchase Heavenly is equal to the proportion of stay-at-home women who would purchase Heavenly.
To find the p-value, we need to round the z test statistic from −2.207 to −2.21 so that we can use the table Areas under the Normal Curve in Appendix B.3. In the table, find the likelihood, or probability, of a z value less than −2.21 or greater than 2.21. The probability corresponding to 2.21 is .4864, so the likelihood of find- ing the value of the test statistic to be less than −2.21 or greater than 2.21 is:
p-value = 2(.5000 − .4864) = 2(.0136) = .0272
The p-value of .0272 is less than the significance level of .05, so our decision is to reject the null hypothesis.
Step 6: Interpret the result. The results of the hypothesis test indicate work- ing and stay-at-home women would purchase Heavenly at different rates or proportions.
The MegaStat add-in for Excel has a procedure to determine the value of the test statistic and compute the p-value. Notice that the MegaStat output includes the two sample proportions, the value of z, and the p-value. The difference in the p-value is rounding. The results follow.
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 proportion 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.
S E L F - R E V I E W 15–2
554 CHAPTER 15
7. The null and alternate hypotheses are:
H0: π1 ≤ π2 H1: π1 > π2
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 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?
8. The null and alternate hypotheses are:
H0: π1 = π2 H1: π1 ≠ π2
A sample of 200 observations from the first population indicated that x1 is 170. A sample 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 six-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. Like- wise, a sample of 400 vines sprayed with Action were checked. The results are:
E X E R C I S E S
Republicans Democrats
Number sampled 1,000 800 Number in favor 200 168
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 pro- portion of vines infested using Pernod 5 as opposed to Action?
10. GfK Research North America conducted identical surveys 5 years apart. One question asked of women was “Are most men basically kind, gentle, and thought- ful?” 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 com- pared 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:
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 555
Policy Type Percent
Whole life 40 Level term 25 Decreasing term 15 Other 20
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 re- vealed 120 had at least one accident in the previous three-year period. Similarly, a sample of 600 policies 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.
GOODNESS-OF-FIT TESTS: COMPARING OBSERVED AND EXPECTED FREQUENCY DISTRIBUTIONS Next, we discuss goodness-of-fit tests that compare an observed frequency distribu- tion to an expected frequency distribution for variables measured on a nominal or or- dinal scale. For example, a life insurance company classifies its policies into four categories using a nominal variable, policy type. Policy type has four categories: whole life, level term, decreasing term, and others. The table below shows the histor- ical relative frequency distribution of the policy types. These would be the expected frequencies.
LO15-3 Test a hypothesis comparing an observed set of frequencies to an expected frequency distribution.
The insurance company wishes to compare this historical distribution with an ob- served distribution of policy types for a sample of 2,000 current policies. The goodness- of-fit test would determine if the current distribution of policies “fits” the historical distribution or if it has changed. A goodness-of-fit test is one of the most commonly used statistical tests.
Hypothesis Test of Equal Expected Frequencies Our first illustration of a goodness-of-fit test involves the case where we choose the expected frequencies to be equal. As the full name implies, the purpose of the goodness- of-fit test is to compare an observed frequency distribution to an expected frequency distribution.
E X A M P L E
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 his or her favorite meal when dining out. The results are re- ported below.
556 CHAPTER 15
TABLE 15–2 Observed and Expected Frequencies for Survey of 120 Adults
Observed Expected Favorite Meal Frequency, fo Frequency, fe Chicken 32 30 Fish 24 30 Meat 35 30 Pasta 29 30
Total 120 120
TABLE 15–1 Favorite Entrée as Selected by a Sample of 120 Adults
Favorite Entrée Frequency
Chicken 32 Fish 24 Meat 35 Pasta 29
Total 120
Is it reasonable to conclude there is no preference among the four entrées?
S O L U T I O N
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 discrep- ancy 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 “classifica- tion” is based on the selected entrée. In addition, there is no natural order to the favorite entrée. No one entrée is assumed better than another. Therefore, the nom- inal 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 ex- pected frequency for each category, assuming there is no preference for any of the entrées. This information is summarized in Table 15–2. 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
© EQRoy/Shutterstock.com
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 557
times each entrée is selected due to chance, or should we conclude that the entrées are not equally preferred?
To investigate the issue, we use the six-step hypothesis-testing procedure.
Step 1: State the null hypothesis and the alternate hypothesis. The null hy- pothesis, H0, is that there is no difference between the set of ob- served frequencies and the set of expected frequencies. In other words, any difference between the two sets of frequencies is at- tributed to sampling error. The alternate hypothesis, H1, is that there is a difference between the observed and expected sets of frequen- cies. If the null hypothesis is rejected and the alternate hypothesis is accepted, we conclude the preferences are not equally distributed among the four categories.
H0: There is no difference in the proportion of adults selecting each entrée.
H1: There is a difference in the proportion of adults selecting each entrée.
Step 2: Select the level of significance. We selected the .05 signifi- cance 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 distri- bution, designated by χ2.
χ2 = Σ[ (fo − fe)2
fe ] (15–4)CHI-SQUARE TEST
STATISTIC
with k − 1 degrees of freedom, where: k is the number of categories. fo is an observed frequency in a particular category. fe is an expected frequency in a particular category.
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 hypoth- esis testing is the value that separates the region where we do not reject H0 from the region where H0 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 differ- ent 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 categories, there are k − 1 = 4 − 1 = 3 degrees of freedom. The critical value for 3 degrees of freedom and the .05 level of significance is found in Appendix B.7. A portion of that table is shown in Table 15–3. The critical value is 7.815, found by locating 3 degrees of freedom in the left margin and then moving horizontally (to the right) and read- ing the critical value in the .05 column.
558 CHAPTER 15
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 15–3 shows the decision rule.
The decision rule indicates that if there are large differences between the observed and expected frequencies, resulting in a computed χ2 of more than 7.815, the null hypothesis should be rejected. However, if the differences between fo and fe are small, the computed χ2 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 sample 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 15–1. The calculations for chi- square follow. (Note again that the expected frequencies are the same for each cell.)
Column D: Determine the differences between each fo and fe. That is, fo − fe. The sum of these differences is zero.
Column E: Square the difference between each observed and ex- pected frequency, that is, (fo − fe)2.
Column F: Divide the result for each observation by the expected fre- quency, that is, (fo − fe)2/fe. Finally, sum these values. The result is the value of χ2, which is 2.20.
TABLE 15–3 A Portion of the Chi-Square Table
Right-Tail Area Degrees 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
ty
Do not reject
H0
7.815 Scale of x2 Critical value
Region of rejection
.05
CHART 15–3 Chi-Square Probability Distribution for 3 Degrees of Freedom, Showing the Region of Rejection, .05 Level of Significance
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 559
The computed χ2 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.
Step 6: Interpret the results. We conclude that the differences between the observed and the expected frequencies could be due to chance. The data do not suggest that the preferences among the four entrées are different.
We can use MegaStat to compute the goodness-of-fit test as follows. The steps are shown in the Software Commands in Appendix C. 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.
For many years, researchers and statisticians believed that all variables were nor- mally distributed. In fact, it was generally assumed to be a universal law. How- ever, Karl Pearson observed that experimental data were not always normally distributed but there was no way to prove his obser- vations were correct. To solve this problem, Pearson discovered the chi-square statistic that basically com- pares an observed fre- quency distribution with an assumed or expected normal distribution. His dis- covery proved that all vari- ables were not normally distributed.
STATISTICS IN ACTION
The Chi-square distribution has many applications in statistics. Its characteristics are:
1. Chi-square values are never negative. This is because the difference between fo and fe is squared, that is, (fo − fe)2.
2. There is a family of chi-square distributions. There is a chi-square distribution 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. Therefore, 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 clas- sified into one of three categories—flight personnel, ground support, and adminis- trative 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 15–4 shows the distributions for selected degrees of freedom. No- tice that for 10 degrees of freedom the curve is approaching a normal distribution.
CHART 15–4 Chi-Square Distributions for Selected Degrees of Freedom
0 1 2 3 4 5 6 7 8 9 10
.40
.30
.20
.10
.00
Pr ob
ab ili
ty
Chi-square values (x2)
df = 1
df = 3
df = 5 df = 10
11 12 13 14 15 16 17 18 19
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
560 CHAPTER 15
The human resources director at Georgetown Paper Inc. is concerned about absenteeism among hourly workers. She decides to sample the company records to determine whether absenteeism is distributed evenly throughout the six-day workweek. The hypotheses are:
H0: Absenteeism is evenly distributed throughout the workweek. H1: Absenteeism is not evenly distributed throughout the workweek.
The sample results are:
S E L F - R E V I E W 15–3
Number Absent Number Absent
Monday 12 Thursday 10 Tuesday 9 Friday 9 Wednesday 11 Saturday 9
(a) What are the numbers 12, 9, 11, 10, 9, and 9 called? (b) How many categories 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% significance level? (f) Compute the χ2 test statistic. (g) What is the decision regarding the null hypothesis? (h) Specifically, what does this indicate to the human resources director?
13. In a particular chi-square goodness-of-fit test, there are four categories and 200 observations. Use the .05 significance level.
a. How many degrees of freedom are there? b. What is the critical value of chi-square?
14. In a particular chi-square goodness-of-fit test, there are six categories and 500 ob- servations. Use the .01 significance level.
a. How many degrees of freedom are there? b. What is the critical value of chi-square?
15. The null hypothesis and the alternate hypothesis are:
H0: The frequencies are equal. H1: 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 H0?
16. The null hypothesis and the alternate hypothesis are:
H0: The frequencies are equal. H1: The frequencies are not equal.
E X E R C I S E S
Category fo A 10 B 20 C 30 D 20
Category fo A 10 B 20 C 30
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 561
a. State the decision rule, using the .05 significance level. b. Compute the value of chi-square. c. What is your decision regarding H0?
17. 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?
18. Classic Golf Inc. manages five courses in the Jacksonville, Florida, area. The direc- tor of golf wishes to study the number of rounds of golf played per weekday at the five courses. He gathered the following sample information. At the .05 significance level, is there a difference in the number of rounds played by day of the week?
19. A group of department store buyers viewed a new line of dresses and gave their opinions of them. The results were:
Because the largest number (47) indicated the new line is outstanding, the head designer thinks that this is a mandate to go into mass production of the dresses. The head sweeper (who somehow became involved in this) believes that there is not a clear mandate and claims that the opinions are evenly distributed among the six categories. He further states that the slight differences among the various counts are probably due to chance. Test the null hypothesis that there is no signifi- cant difference among the opinions of the buyers at the .01 level of significance.
20. The safety director of a large steel mill took samples at random from company records of minor work-related accidents and classified them according to the time the accident took place.
Using the goodness-of-fit test and the .01 level of significance, determine whether the accidents are evenly distributed throughout the day. Write a brief explanation of your conclusion.
Opinion Number of Buyers Opinion Number of Buyers
Outstanding 47 Good 39 Excellent 45 Fair 35 Very good 40 Undesirable 34
Number of Number of Time Accidents Time Accidents
8 up to 9 a.m. 6 1 up to 2 p.m. 7 9 up to 10 a.m. 6 2 up to 3 p.m. 8 10 up to 11 a.m. 20 3 up to 4 p.m. 19 11 up to 12 p.m. 8 4 up to 5 p.m. 6
Outcome Frequency Outcome Frequency
1 3 4 3 2 6 5 9 3 2 6 7
Day Rounds
Monday 124 Tuesday 74 Wednesday 104 Thursday 98 Friday 120
562 CHAPTER 15
Hypothesis Test of Unequal Expected Frequencies The expected frequencies (fe) in the previous example/solution involving preferred entrées were all equal. According to the null hypothesis, it was expected that of the 120 adults in the study, an equal number would select each of the four entrées. So we expect 30 to select chicken, 30 to select fish, and so on. The chi-square test can also be used if the expected frequencies are not equal.
The following example illustrates the case of unequal frequencies and also gives a practical use of the chi-square goodness-of-fit test—namely, to find whether a local ex- perience differs from the national experience.
E X A M P L E
The American Hospital Administrators Association (AHAA) reports the following in- formation concerning the number of times senior citizens are admitted to a hospital during a one-year period. Forty percent are not admitted; 30% are admitted once; 20% are admitted twice, and the remaining 10% are admitted three or more times.
A survey of 150 residents of Bartow Estates, a community devoted to active seniors located in central Florida, revealed 55 residents were not admitted during the last year, 50 were admitted to a hospital once, 32 were admitted twice, and the rest of those in the survey were admitted three or more times. Can we conclude the survey at Bartow Estates is consistent with the information reported by the AHAA? Use the .05 significance level.
S O L U T I O N
We begin by organizing the above information into Table 15–4. Clearly, we cannot compare the percentages given in the AHAA study to the counts or frequencies reported for Bartow Estates residents. However, we can use the AHAA information to compute expected frequencies, fe, for the Bartow Estates residents. According to AHAA, 40% of the seniors in their survey did not require hospitalization. Thus, if there is no difference between the national experience and the Bartow Estates' study, then the expectation is that 40% of the 150 Bartow seniors surveyed, or fe = 60, would not have been hospitalized. Further, based on the AHAA information, 30% of the 150 Bartow seniors, or fe = 45, would be expected to be admitted once, and so on. The observed and expected frequencies for Bartow residents are given in Table 15–4.
TABLE 15–4 Summary of Study by AHAA and a Survey of Bartow Estates
Number of AHAA Relative Observed Frequency of Expected Frequency of Times Admitted Frequencies Bartow Residents (fo) Bartow Residents (fe)
0 40% 55 60 = (.40)(150) 1 30% 50 45 = (.30)(150) 2 20% 32 30 = (.20)(150) 3 or more 10% 13 15 = (.10)(150) Total 100 150
The null hypothesis and the alternate hypothesis are:
H0: There is no difference between local and national experience for hospital admissions.
H1: There is a difference between local and national experience for hospital admissions.
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 563
To find the decision rule, we use Appendix B.7 and the .05 significance level. There are four admitting categories, so the degrees of freedom are df = 4 − 1 = 3. The critical value is 7.815. Therefore, the decision rule is to reject the null hypothesis if χ2 > 7.815. The decision rule is portrayed in Chart 15–5.
Now to compute the chi-square test statistic:
The computed value of χ2 (1.3723) lies to the left of 7.815. Thus, we cannot reject the null hypothesis. We conclude that the survey results do not provide evidence of a difference between the local and national experience for hospital admissions.
Pr ob
ab ili
ty
Do not reject
H0 7.815 Scale of x2 Critical value
Region of rejection .05
df = 3
CHART 15–5 Decision Criteria for the Bartow Estates Research Study
Number of Times Admitted fo fe (fo − fe) (fo − fe)2 (fo − fe)2∕fe 0 55 60 −5 25 0.4167 1 50 45 5 25 0.5556 2 32 30 2 4 0.1333 3 or more 13 15 −2 4 0.2667 χ2 value Total 150 1.3723
Many state governments operate lotteries to help fund education. In many lotteries, numbered balls are mixed and selected by a machine. In a Select Three game, numbered balls are selected randomly from three groups of balls numbered zero through nine. Randomness would predict that the frequency of each number is equal. How would you test if the machine ensured a random selection process? A chi- square, goodness-of-fit test could be used to investi- gate this question.
STATISTICS IN ACTION
LIMITATIONS OF CHI-SQUARE If there is an unusually small expected frequency for a category, chi-square (if applied) might result in an erroneous conclusion. This can happen because fe appears in the denominator, and dividing by a very small number makes the quotient quite large! Two generally accepted policies regarding small category frequencies are:
1. If there are only two cells, the expected frequency in each category should be at least 5. The computation of chi-square would be permissible in the following prob- lem, involving a minimum fe of 6.
Individual fo fe Literate 641 642 Illiterate 7 6
2. For more than two categories, chi-square should not be used if more than 20% of the categories have expected frequencies less than 5. According to this policy, it would not be appropriate to use the goodness-of-fit test on the following data. Three of the seven categories, or 43%, have expected frequencies (fe) of less than 5.
LO15-4 Explain the limitations of using the chi-square statistic in goodness-of- fit tests.
564 CHAPTER 15
Level of Management fo fe Foreman 30 32 Supervisor 110 113 Manager 86 87 Middle management 23 24 Assistant vice president 5 2 Vice president 5 4 Senior vice president 4 1
Total 263 263
To show the reason for the 20% policy, we conducted the goodness-of-fit test on the above levels-of-management data. The MegaStat output follows.
For this test at the .05 significance level, H0 is rejected if the computed value of chi-square is greater than 12.592. The computed value is 14.008, so we reject the null hypothesis that the observed and expected frequency distributions are the same. However, examine the MegaStat output critically. More than 98% of the computed chi-square value is ac- counted for by the three vice president categories [(4.500 + .250 + 9.000)/14.008 = 0.9815]. Logically, too much weight is being given to these categories.
The issue can be resolved by combining categories if it is logical to do so. In the above example, we combine the three vice president categories, which satisfies the 20% policy. Note that the degrees of freedom for the goodness of fit test change from 6 to 4.
Level of Management fo fe Foreman 30 32 Supervisor 110 113 Manager 86 87 Middle management 23 24 Vice president 14 7
Total 263 263
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 565
The computed value of chi-square with the revised categories is 7.258. See the following MegaStat output. This value is less than the critical value of 9.488 (based on 4 degrees of freedom) for the .05 significance level. The null hypothesis is, therefore, not rejected at the .05 significance level. This indicates there is not a significant difference between the observed and expected distributions.
The American Accounting Association classifies accounts receivable as “current,” “late,” and “not collectible.” Industry figures show that 60% of accounts receivable are current, 30% are late, and 10% are not collectible. Massa and Barr, a law firm in Green- ville, Ohio, has 500 accounts receivable: 320 are current, 120 are late, and 60 are not collectible. Are these numbers in agreement with the industry distribution? Use the .05 significance level.
S E L F - R E V I E W 15–4
21. For a particular population, a hypothesis states:
H0: Forty percent of the observations are in category A, 40% are in B, and 20% are in C.
H1: The distribution of the observations is not as described in H0.
We took a sample of 60 observations from the population with the following results.
Category fo A 30 B 20 C 10
a. For the hypothesis test, state the decision rule using the .01 significance level.
b. Compute the value of chi-square. c. What is your decision regarding H0?
E X E R C I S E S
566 CHAPTER 15
TESTING THE HYPOTHESIS THAT A DISTRIBUTION IS NORMAL We use a goodness-of-fit test to compare an observed frequency distribution to an expected frequency distribution. In the example/solution regarding Bubba’s Fish and Pasta, the observed frequencies are the count of each entrée selected for a sample of 120 adults. We determine the expected frequencies by assuming there is no preference for any of the four entrées, so we expect that one-fourth of the sam- ple, or 30 adults, selects each entrée. In this section, we want to test a hypothesis that a distribution is normal by using the goodness-of-fit test to compare an ob- served frequency distribution to an expected frequency distribution that is normal. Why is this test important? In Chapter 11, when we tested for differences in two population means, we assumed the two populations followed the normal distribu- tion. We made the same assumption in Chapter 12 when we tested if several pop- ulation means were equal. In Chapter 13 we assume the distribution of the residuals in a least squares regression analysis follow the normal probability distribution.
The following example/solution provides the details of a goodness-of-fit test to in- vestigate the reasonableness of the normality assumption.
LO15-5 Test a hypothesis that an observed frequency distribution is normally distributed.
22. The chief of security for the Mall of the Dakotas directed a study of theft. He selected a sample of 100 boxes that had been tampered with and ascertained that, for 60 of the boxes, the missing pants, shoes, and so on were attributed to shoplifting. For 30 boxes, employees had stolen the goods, and for the remain- ing 10 boxes he blamed poor inventory control. In his report to the mall man- agement, can he say that shoplifting is twice as likely to be the cause of the loss as compared with either employee theft or poor inventory control and that em- ployee theft and poor inventory control are equally likely? Use the .02 signifi- cance level.
23. From experience, the bank credit card department of Carolina Bank knows that 5% of its card holders have had some high school, 15% have completed high school, 25% have had some college, and 55% have completed college. Of the 500 card holders whose cards have been called in for failure to pay their charges this month, 50 had some high school, 100 had completed high school, 190 had some college, and 160 had completed college. Can we conclude that the distribution of card holders who do not pay their charges is different from all others? Use the .01 signif- icance level.
24. For many years, TV executives used the guideline that 30% of the audience were watching each of the traditional big three prime-time networks and 10% 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, and 125 to the NBC affiliate, with the remainder viewing a cable sta- tion. At the .05 significance level, can we conclude that the guideline is still reasonable?
E X A M P L E
In Chapter 2 we use a frequency distribution to organize the profits from the Apple- wood Auto Group’s sale of 180 vehicles. The frequency distribution is repeated in Table 15–5.
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 567
Using statistical software, on page 64 in Chapter 3 we determined that the mean profit on a vehicle for the Applewood Auto Group was $1,843.17 and that the standard deviation was $643.63. Is it reasonable to conclude that the profit data is a sample obtained from a normal population? To put it another way, do the profit data follow a normal population? We use the .05 signif- icance level.
S O L U T I O N
To test for a normal distribution, we need to find the expected frequencies for each class in the distribution, assuming that the expected distribution follows a normal probability distribution. We start with the normal distribution by calculating probabil- ities for each class. Then we use these probabilities to compute the expected fre- quencies for each class.
To begin, we need to find the area, or probability, for each of the eight classes in Table 15–5, assuming a normal population with a mean of $1,843.17 and a stan- dard deviation of $643.63. To find this probability, we adapt formula (7–1) from Chapter 7 replacing μ with x and σ with s. So we use the following formula to deter- mine the various values of z.
z = x − x
s In this case, z is the value of the standard normal statistic; x , $1,843.17, is the sam- ple mean; and s, $643.63, is the sample standard deviation. To illustrate, we select class $200 up to $600 from Table 15–5. We want to determine the expected fre- quency in this class, assuming the distribution of profits follows a normal distribu- tion. First, we find the z value corresponding to $200.
z = x − x
s =
$200 − $1,843.17 $643.63
= −2.55
This indicates that the lower limit of this class is 2.55 standard deviations below the mean. From Appendix B.3, the probability of finding a z value less than −2.55 is .5000 − .4946 = .0054.
For the upper limit of the $200 up to $600 class:
z = x − x
s =
$600 − $1,843.17 $643.63
= −1.93
© Blend Images/Getty Images
TABLE 15–5 Frequency Distribution of Profits for Vehicles Sold Last Month by 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
568 CHAPTER 15
The area to the left of $600 is the probability of a z value less than −1.93. To find this value, we again use Appendix B.3 and reason that .5000 − .4732 = .0268.
Finally, to find the area between $200 and $600:
P($200 < x < $600) = P(−2.55 < z < −1.93) = .0268 − .0054 = .0214
That is, about 2.14% of the vehicles sold will result in a profit of between $200 and $600.
There is a chance that the profit earned is less than $200. To find this probability:
P(x < $200) = P(z < −2.55) = .5000 − .4946 = .0054
We enter these two probabilities in the second and third rows of column 3 in Table 15–6.
Logically, if we sold 180 vehicles, we would expect to earn a profit of between $200 and $600 on 3.85 vehicles, found by .0214(180). We would expect to sell 0.97 vehicle with a profit of less than $200, found by 180(.0054). We continue this process for the remaining classes. This information is summarized in Table 15–7. Don’t be concerned that we are reporting fractional vehicles.
Before continuing, we should emphasize one of the limitations of tests using chi-square as the test statistic. The second limitation on page 563 indicates that if more than 20% of the categories have expected frequencies of less than 5, some of the categories should be combined. In Table 15–6, there are three classes in which the expected fre- quencies are less than 5. Hence, we combine the “Under $200” class with the “$200
TABLE 15–7 Computations of the Chi-Square Statistic
Profit fo fe (fo − fe) (fo − fe) 2 (fo − fe)
2∕fe Under $600 8 4.82 3.18 10.1124 2.098 $ 600 up to $1,000 11 12.29 −1.29 1.6641 .135 1,000 up to 1,400 23 27.00 −4.00 16.0000 .593 1,400 up to 1,800 38 40.86 −2.86 8.1796 .200 1,800 up to 2,200 45 42.61 2.39 5.7121 .134 2,200 up to 2,600 32 31.00 1.00 1.0000 .032 2,600 up to 3,000 19 14.96 4.04 16.3216 1.091 3,000 and over 4 6.46 −2.46 6.0516 .937 Total 180 180.00 0 5.220
TABLE 15–6 Profits at Applewood Auto Group, z Values, Areas under the Normal Distribution, and Expected Frequencies
Profit z Values Area Found by Expected Frequency
Under $200 Under −2.55 .0054 0.5000 − 0.4946 0.97 $ 200 up to $ 600 −2.55 up to −1.93 .0214 0.4946 − 0.4732 3.85 600 up to 1,000 −1.93 up to −1.31 .0683 0.4732 − 0.4049 12.29 1,000 up to 1,400 −1.31 up to −0.69 .1500 0.4049 − 0.2549 27.00 1,400 up to 1,800 −0.69 up to −0.07 .2270 0.2549 − 0.0279 40.86 1,800 up to 2,200 −0.07 up to 0.55 .2367 0.0279 + 0.2088 42.61 2,200 up to 2,600 0.55 up to 1.18 .1722 0.3810 − 0.2088 31.00 2,600 up to 3,000 1.18 up to 1.80 .0831 0.4641 − 0.3810 14.96 3,000 up to 3,400 1.80 up to 2.42 .0281 0.4922 − 0.4641 5.06 3,400 or more 2.42 or more .0078 0.5000 − 0.4922 1.40 Total 1.0000 180.00
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 569
up to $600” class and the “$3,400 or more” class with the “$3,000 up to $3,400” class. So the expected frequency in the “Under $600” class is now 4.82, found by 0.97 plus 3.85. We do the same for the “$3,000 and over” class: 5.06 + 1.40 = 6.46. The results are shown in Table 15–7. The computed value of chi-square is 5.220.
Now let’s put this information into the formal hypothesis-testing format. The null and alternate hypotheses are:
H0: The population of profits follows the normal distribution. H1: The population of profits does not follow the normal distribution.
To determine the critical value of chi-square, we need to know the degrees of free- dom. In this case, there are 8 categories, or classes, so the degrees of freedom are k − 1 = 8 − 1 = 7. In addition, the values $1,843.17, the mean profit, and $643.63, the standard deviation of the Applewood Auto Group profits, were computed from a sam- ple. When we estimate population parameters from sample data, we lose a degree of freedom for each estimate. So we lose two more degrees of freedom for estimating the population mean and the population standard deviation. Thus, the number of degrees of freedom in this problem is 5, found by k − 2 − 1 = 8 − 2 − 1 = 5.
From Appendix B.7, using the .05 significance level, the critical value of chi- square is 11.070. Our decision rule is to reject the null hypothesis if the computed value of chi-square is more than 11.070.
Now, to compute the value of chi-square, we use formula (15–4):
χ2 = Σ (fo − fe)2
fe =
(8 − 4.82)2
4.82 + . . . +
(4 − 6.46)2
6.46 = 5.220
The values for each class are shown in the right-hand column of Table 15–7, as well as the column total, which is 5.220. Because the computed value of 5.220 is less than the critical value, we do not reject the null hypothesis. We conclude the evidence does not suggest the distribution of profits is other than normal.
To expand on the calculation of the number of degrees of freedom, if we know the mean and the standard deviation of a population and wish to find whether some sample data conform to a normal, the degrees of freedom are k − 1. On the other hand, sup- pose we have sample data grouped into a frequency distribution, but we do not know the value of the population mean and the population standard deviation. In this case, the degrees of freedom are k − 2 − 1. In general, when we use sample statistics to estimate population parameters, we lose a degree of freedom for each parameter we estimate. This is parallel to the situation on page 500 of Chapter 14, the chapter on multiple regression, where we lost a degree of freedom in the denominator of the F statistic for each independent variable considered.
25. The IRS is interested in the number of individual tax forms prepared by small ac- counting firms. The IRS randomly sampled 50 public accounting firms with 10 or fewer employees in the Dallas–Fort Worth area. The following frequency table reports the results of the study. Assume the sample mean is 44.8 clients and the sample standard deviation is 9.37 clients. Is it reasonable to conclude that the sample data are from a population that follows a normal probability distribution? Use the .05 significance level.
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
E X E R C I S E S
570 CHAPTER 15
CONTINGENCY TABLE ANALYSIS In Chapter 4, we discussed bivariate data, where we studied the relationship between two variables. We described a contingency table, which simultaneously summarizes two nominal-scale variables of interest. For example, a sample of students enrolled in the School of Business is classified by gender (male or female) and major (accounting, man- agement, finance, marketing, or business analytics). This classification is based on the nominal scale because there is no natural order to the classifications.
We discussed contingency tables in Chapter 5. On page 151, we illustrated the rela- tionship between the number of movies attended per month and the age of the attendee. We can use the chi-square distribution to test whether two nominal-scaled variables are related or not. To put it another way, is one variable independent of the other?
Here are some examples where we are interested in testing whether two nominal-scaled variables are related.
• Ford Motor Company operates an assembly plant in Dearborn, Michigan. The plant operates three shifts per day, 5 days a week. The quality control manager wishes to compare the quality level on the three shifts. Vehicles are classified by quality level (acceptable, unacceptable) and shift (day, afternoon, night). Is there a difference in the quality level on the three shifts? That is, is the quality of the product related to the shift when it was manufactured? Or is the quality of the product independent of the shift on which it was manufactured?
• A sample of 100 drivers who were stopped for speeding violations was classified by gender and whether or not they were wearing a seat belt. For this sample, is wearing a seatbelt related to gender?
• Does a male released from federal prison make a different adjustment to civilian life if he returns to his hometown or if he goes elsewhere to live? The two variables are adjustment to civilian life and place of residence. Note that both variables are mea- sured on the nominal scale.
The following example/solution provides the details of the analysis and possible conclusions.
LO15-6 Perform a chi-square test for independence on a contingency table.
26. 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. The mean expense is $52.0 million and the standard deviation is $11.32 million. Is it reasonable to con- clude the sample data are from a population that follows a normal probability dis- tribution? Use the .05 significance level.
Advertising Expense ($ Million) Number of 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
E X A M P L E
Rainbow Chemical, Inc. employs hourly and salaried employees. The vice presi- dent of human resources surveyed 380 employees about his/her satisfaction level with the current health care benefits program. The employees were then
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 571
At the .05 significance level, is it reasonable to conclude that pay type and level of satisfaction with the health care benefits are related?
S O L U T I O N
The first step is to state the null hypothesis and the alternate hypothesis.
H0: There is no relationship between level of satisfaction and pay type. H1: There is a relationship between level of satisfaction and pay type.
The significance level, as requested by the HR vice president, is .05. The level of measurement for pay type is the nominal scale. The satisfaction level with health benefits is actually the ordinal scale, but we use it as a nominal-scale variable. Each sampled employee is classified by two criteria: the level of satisfaction with benefits and pay type. The information is tabulated into Table 15–8, which is called a contin- gency table.
We use the chi-square distribution as the test statistic. To determine the critical value of chi-square, we calculate the degrees of freedom (df ) as:
df = (Number of rows − 1)(Number of columns − 1) = (r − 1)(c − 1)
In this example/solution there are 2 rows and 3 columns, so there are 2 degrees of freedom.
df = (r − 1)(c − 1) = (2 − 1)(3 − 1) = 2
To find the critical value for 2 degrees of freedom and the .05 level, refer to Appen- dix B.7. Move down the degrees of freedom column in the left margin to the row with 2 degrees of freedom. Move across this row to the column headed .05. At the intersection, the chi-square critical value is 5.991. The decision rule is to reject the null hypothesis if the computed value of χ2 is greater than 5.991. See Chart 15–6.
classified according to the pay type, i.e., salary or hourly. The results are shown in Table 15–8.
TABLE 15–8 Health Care Satisfaction Level for Rainbow Chemical Employees
Pay Type Satisfied Neutral Dissatisfied Total
Salary 30 17 8 55 Hourly 140 127 58 325
Total 170 144 66 380
Pr ob
ab ili
ty
Do not reject
H0 5.991 Scale of x2 Critical value
Region of rejection .05
df = 2
CHART 15–6 Chi-Square Distribution for 2 Degree of Freedom
572 CHAPTER 15
Next we compute the chi-square value χ2, using formula (15–4). The observed frequencies, fo, are shown in Table 15–9. How are the corresponding expected frequencies, fe, determined? To begin, notice from Table 15–8 that 55 of the 380 Rainbow Chemical employees sampled are salaried. So the fraction of salaried employees in the sample is 55/380 = .14474. If there is no relationship between pay type and level of satisfaction with the health care benefits program, we would expect about the same fraction of the employees who are satisfied with the health care to be salaried. There are 170 employees who are satisfied with the health care program, so the expected number of satisfied employees who are salaried is 24.61, found by (.14474)(170). Thus, the expected frequency for the upper-left cell is 24.61. Likewise, if there were no relationship between satisfaction level and pay type, we would expect .14474 of the 144 employees, or 20.84, who were neutral about the health care program to be salaried. We continue this process, filling in the remaining cells. It is not necessary to calculate each of these cell values. In fact we only need to calculate two cells. We can find the others by subtraction.
The expected frequency for any cell is determined by:
From this formula, the expected frequency for the upper-left cell in Table 15–8 is:
fe = (Row total) (Column total)
(Grand total) =
(55) (170) 380
= 24.61
The observed frequencies, fo, and the expected frequencies, fe, for all of the cells in the contingency table are listed in Table 15–9. Note there are slight differences due to rounding.
EXPECTED FREQUENCY fe = (Row total) (Column total)
(Grand total) (15–5)
TABLE 15–9 Observed and Expected Frequencies
Satisfaction Level with Health Care
Satisfied Neutral Dissatisfied
Pay Type fo fe fo fe fo fe Salary 30 24.61 17 20.84 8 9.55 Hourly 140 145.39 127 123.16 58 56.45
Total 170 170.00 144 144.00 66 66.00
We use formula (15–4) to determine the value of chi-square. Starting with the upper-left cell:
χ2 = Σ (fo − fe)2
fe =
(30 − 24.61)2
24.61 +
(17 − 20.84)2
20.84 + … +
(58 − 56.45)2
56.45
= 1.181 + .708 + … + .043 = 2.506
Because the computed value of chi-square (2.506) lies in the region to the left of 5.991, the null hypothesis is not rejected at the .05 significance level. What do we conclude? The sample data do not provide evidence that pay type and satisfaction level with health care benefits are related.
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 573
The following output is from the MegaStat Excel add-in.
Observe that the value of chi-square is the same as that computed earlier, 2.506. In addition, the p-value, .286, is reported. So the probability of finding a value of the test statistic as large or larger, assuming the null hypothesis is true, is .286. The p-value also results in the same decision: do not reject the null hypothesis.
Chi-square Contingency Table Test for Independence
Satisfaction Level with Health Care
Pay Type Satisfied Neutral Dissatisfied Total
Salary Observed 30 17 8 55 Expected 24.61 20.84 9.55 55.00 Hourly Observed 140 127 58 325 Expected 145.39 123.16 56.45 325.00 Total Observed 170 144 66 380 Expected 170.00 144.00 66.00 380.00
2.506 chi-square 2 df 0.286 p-value
A social scientist sampled 140 people and classified them according to income level and whether or not they played a state lottery in the last month. The sample information is re- ported below. Is it reasonable to conclude that playing the lottery is related to income level? Use the .05 significance level.
S E L F - R E V I E W 15–5
(a) What is this table called? (b) State the null hypothesis and the alternate hypothesis. (c) What is the decision rule? (d) Determine the value of chi-square. (e) Make a decision on the null hypothesis. Interpret the result.
Income
Low Middle High Total
Played 46 28 21 95 Did not play 14 12 19 45
Total 60 40 40 140
27. The director of advertising for the Carolina Sun Times, the largest newspaper in the Carolinas, is studying the relationship between the type of community in which a subscriber resides and the section of the newspaper he or she reads first. For a sample of readers, she collected the sample information in the following table.
E X E R C I S E S
National News Sports Food
City 170 124 90 Suburb 120 112 100 Rural 130 90 88
574 CHAPTER 15
At the .05 significance level, can we conclude there is a relationship between the type of community where the person resides and the section of the paper read first?
28. Four brands of lightbulbs are being considered for use in the final assembly area of the Ford F-150 truck plant in Dearborn, Michigan. The director of purchas- ing asked for samples of 100 from each manufacturer. The numbers of acceptable and unacceptable bulbs from each manufacturer are shown below. At the .05 sig- nificance level, is there a difference in the quality of the bulbs?
Manufacturer
A B C D
Unacceptable 12 8 5 11 Acceptable 88 92 95 89
Total 100 100 100 100
29. The quality control department at Food Town Inc., a grocery chain in upstate New York, conducts a monthly check on the comparison of scanned prices to posted prices. The chart below summarizes the results of a sample of 500 items last month. Company management would like to know whether there is any rela- tionship between error rates on regularly priced items and specially priced items. Use the .01 significance level.
Regular Price Special Price
Undercharge 20 10 Overcharge 15 30 Correct price 200 225
30. The use of cellular phones in automobiles has increased dramatically in the last few years. Of concern to traffic experts, as well as manufacturers of cellular phones, is the effect on accident rates. Is someone who is using a cellular phone more likely to be involved in a traffic accident? What is your conclusion from the following sample information? Use the .05 significance level.
Had Accident Did Not Have an Accident in the Last Year in the Last Year
Uses a cell phone 25 300 Does not use a cell phone 50 400
C H A P T E R S U M M A R Y
I. This chapter considered tests of hypothesis for nominal level data. II. When we sample from a single population and the variable of interest has only two pos-
sible outcomes, we call this a test of proportion. A. The binomial conditions must be met. B. Both nπ and n(1− π) must be at least 5. C. The test statistic is
z = p − π
√ π(1 − π)
n
(15–1)
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 575
III. 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:
pc = x1 + x2 n1 + n2
(15–3)
B. We compute the value of the test statistic from the following formula:
z = p1 − p2
√ pc(1 − pc)
n1 +
pc(1 − pc) n2
(15–2)
IV. The characteristics of the chi-square distribution are: A. The value of chi-square is never negative. B. The chi-square distribution is positively skewed. C. There is a family of chi-square distributions.
1. Each time the degrees of freedom change, a new distribution is formed. 2. As the degrees of freedom increase, the distribution approaches a normal
distribution. V. A goodness-of-fit test will show whether an observed set of frequencies could have
come from a hypothesized population distribution. A. The degrees of freedom are k − 1, where k is the number of categories. B. The formula for computing the value of chi-square is
χ2 = Σ (fo − fe)2
fe (15–4)
VI. A goodness-of-fit test can also be used to determine whether a sample of observations is from a normal population. A. First, calculate the mean and standard deviation of the sample data. B. Group the data into a frequency distribution. C. Convert the class limits to z values and find the standard normal probability distribu-
tion for each class. D. For each class, find the expected normally distributed frequency by multiplying the
standard normal probability distribution by the class frequency. E. Calculate the chi-square goodness-of-fit statistic based on the observed and ex-
pected class frequencies. F. Find the expected frequency in each cell by determining the product of the probabil-
ity of finding a value in each cell by the total number of observations. G. If we use the information on the sample mean and the sample standard deviation
from the sample data, the degrees of freedom are k − 3. VII. A contingency table is used to test whether two traits or characteristics are related.
A. Each observation is classified according to two traits. B. The expected frequency is determined as follows:
fe = (Row total) (Column total)
Grand total (15–5)
C. The degrees of freedom are found by:
df = (Rows − 1)(Columns − 1)
D. The usual hypothesis testing procedure is used.
P R O N U N C I A T I O N K E Y
SYMBOL MEANING PRONUNCIATION
pc Pooled proportion p sub c
χ2 Chi-square statistic ki square fo Observed frequency f sub oh
fe Expected frequency f sub e
576 CHAPTER 15
C H A P T E R E X E R C I S E S
31. 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 50 Super Bowl games, the coin toss resulted in 24 heads and 26 tails. However, the National Football Conference has correctly called the coin flip 34 times. Meanwhile, the American Football Confer- ence has correctly called the flip only 16 times. Use the six-step hypothesis-testing pro- cedure at the .01 significance level to test whether these data suggest that the National Football Conference has an advantage in calling the coin flip. a. Why can you use a z-statistic 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?
32. According to a study by the American Pet Food Dealers Association, 63% of U.S. house- holds own pets. A report is being prepared for an editorial in the San Francisco Chroni- cle. As a part of the editorial, a random sample of 300 households showed 210 own pets. Do these data disagree with the Pet Food Dealers Association’s data? Use a .05 level of significance.
33. 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% of the accounts are more than 3 months in arrears. A random sam- ple of 200 accounts showed that 140 were more than 3 months old. At the .01 signifi- cance level, can she conclude that more than 60% of the accounts are in arrears for more than three months?
34. The policy of the Suburban Transit Authority is to add a bus route if more than 55% 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.
35. Past experience at the Crowder Travel Agency indicated that 44% of those persons who wanted the agency to plan a vacation for them wanted to go to Europe. During the most recent season, a sampling of 1,000 persons was selected at random from the files. It was found that 480 persons wanted to go to Europe on vacation. Has there been a sig- nificant shift upward in the percentage of persons who want to go to Europe? Test at the .05 significance level.
36. Research in the gaming industry showed that 10% 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. At the .05 significance level, test whether these data contradict the research report. a. Why can you use a z-statistic 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?
37. An urban planner claims that, nationally, 20% 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 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.
38. After a losing season, there is a great uproar to fire the head football coach. In a random sample of 200 college alumni, 80 favor keeping the coach. Test at the .05 level of signif- icance whether the proportion of alumni who support the coach is less than 50%.
39. 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 are proof of a reduced fatality rate for lung cancer.
40. Each month the National Association of Purchasing Managers surveys purchasing manag- ers and publishes the NAPM index. One of the questions asked on the survey is: Do you
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 577
think the economy is contracting? Last month, of the 300 responding managers, 160 an- swered yes to the question. This month, 170 of the 290 managers indicated they felt the economy was contracting. At the .05 significance level, can we conclude that a larger proportion of the purchasing managers believe the economy is contracting this month?
41. As part of a recent survey among dual-wage-earner couples, an industrial psychologist 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 household 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?
42. There are two major cell phone providers in the Colorado Springs, Colorado area, one called HTC and the other, Mountain Communications. We want to investigate the “churn rate” for each provider. Churn is the number of customers or subscribers who cut ties with a company during a given time period. At the beginning of the month, HTC had 10,000 customers; at the end of the month, HTC had 9810 customers for a loss of 190. For the same month, Mountain Communications started with 12,500 customers and ended the month with 12,285 customers, for a loss of 215. At the .01 significance level, is there a difference in the churn rate for the two providers?
43. 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% of consumers said conditions were “good.” In the prior month, only 8.5% said they were “good.” Use the six-step hypothesis-testing method at the .05 level of significance to see whether you can determine if there is an increase in the share asserting conditions are “good.” Find the p-value and explain what it means.
44. 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 sam- ple of 270 American trade magazine advertisements, 56 were humorous. An indepen- dent random sample of 203 British trade magazines contained 52 humorous ads. Do these data provide evidence at the .05 significance level that there is a difference in the proportion of humorous ads in British versus American trade magazines?
45. 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 listeners than their spouses. At the .05 significance level, is there a difference between the responses of women and men?
46. The proportion of on-line shoppers who actually make a purchase appears to be rela- tively constant over time. In 2013, among a sample of 388 on-line shoppers, 160 pur- chased merchandise. In 2017, for a sample of 307 on-line shoppers, 144 purchased merchandise. At the .05 level of significance, did the proportion of on-line shoppers change from 2013 to 2017?
47. Vehicles heading west on Front Street may turn right, turn left, or go straight ahead at Elm Street. The city traffic engineer believes that half of the vehicles will continue straight through the intersection. Of the remaining half, equal proportions will turn right and left. Two hundred vehicles were observed, with the following results. Can we con- clude that the traffic engineer is correct? Use the .10 significance level.
Straight Right Turn Left Turn
Frequency 112 48 40
48. The publisher of a sports magazine plans to offer new subscribers one of three gifts: a sweatshirt with the logo of their favorite team, a coffee cup with the logo of their favorite team, or a pair of earrings also with the logo of their favorite team. In a sample of 500 new subscribers, the number selecting each gift is reported below. At the .05 significance level, is there a preference for the gifts or should we conclude that the gifts are equally well liked?
Gift Frequency
Sweatshirt 183 Coffee cup 175 Earrings 142
578 CHAPTER 15
49. In a particular metro area, there are three commercial television stations, each with its own news program from 6:00 to 6:30 p.m. According to a report in this morning’s local newspaper, a random sample of 150 viewers last night revealed 53 watched the news on WNAE (channel 5), 64 watched on WRRN (channel 11), and 33 on WSPD (channel 13). At the .05 significance level, is there a difference in the proportion of viewers watch- ing the three channels?
50. There are four entrances to the Government Center Building in downtown Phila- delphia. The building maintenance supervisor would like to know if the entrances are equally utilized. To investigate, 400 people were observed entering the building. The number using each entrance is reported below. At the .01 significance level, is there a difference in the use of the four entrances?
Entrance Frequency
Main Street 140 Broad Street 120 Cherry Street 90 Walnut Street 50 Total 400
51. The owner of a mail-order catalog would like to compare her sales with the geo- graphic distribution of the population. According to the U.S. Bureau of the Census, 21% of the population lives in the Northeast, 24% in the Midwest, 35% in the South, and 20% in the West. Listed below is a breakdown of a sample of 400 orders randomly selected from those shipped last month. At the .01 significance level, does the distribution of the orders reflect the population?
Region Frequency
Northeast 68 Midwest 104 South 155 West 73 Total 400
52. Banner Mattress and Furniture Company wishes to study the number of credit ap- plications received per day for the last 300 days. The sample information is reported below.
Number of Credit Frequency Applications (Number of Days)
0 50 1 77 2 81 3 48 4 31
5 or more 13
To interpret, there were 50 days on which no credit applications were received, 77 days on which only one application was received, and so on. Would it be reasonable to con- clude that the population distribution is Poisson with a mean of 2.0? Use the .05 signifi- cance level. (Hint: To find the expected frequencies use the Poisson distribution with a mean of 2.0. Find the probability of exactly one success given a Poisson distribution with a mean of 2.0. Multiply this probability by 300 to find the expected frequency for the number of days in which there was exactly one application. Determine the expected frequency for the other days in a similar manner.)
53. Each of the digits in a raffle is thought to have the same chance of occurrence. The table shows the frequency of each digit for consecutive drawings in a California lottery.
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 579
Perform the chi-square test to see if you reject the hypothesis at the .05 significance level that the digits are from a uniform population.
Digit Frequency Digit Frequency
0 44 5 24 1 32 6 31 2 23 7 27 3 27 8 28 4 23 9 21
54. John Isaac Inc., a designer and installer of industrial signs, employs 60 people. The company recorded the type of the most recent visit to a doctor by each employee. A recent national survey found that 53% of all physician visits were to primary care physi- cians, 19% to medical specialists, 17% to surgical specialists, and 11% to emergency departments. Test at the .01 significance level if Isaac employees differ significantly from the survey distribution. Here are their results:
Visit Type Number of Visits
Primary care 29 Medical specialist 11 Surgical specialist 16 Emergency 4
55. The Eckel Manufacturing Company believes that their hourly wages follow a nor- mal probability distribution. To confirm this, 270 employees were sampled and the re- sults organized into the following frequency distribution. Use the methods on pages 82–83 of Chapter 3 to find the mean and standard deviation of these data grouped into a frequency distribution. At the .10 significance level, is it reasonable to conclude that the distribution of hourly wages follows a normal distribution?
Hourly Wage Frequency
$5.50 up to $ 6.50 20 6.50 up to 7.50 24 7.50 up to 8.50 130 8.50 up to 9.50 68 9.50 up to 10.50 28 Total 270
56. The National Cable and Telecommunications Association recently reported that the mean number of HDTVs per household in the United States is 2.30 with a standard deviation of 1.474 sets. A sample of 100 homes in Boise, Idaho, revealed the following sample information.
Number of HDTVs Number of Households
0 7 1 27 2 28 3 18 4 10
5 or more 10
Total 100
At the .05 significance level, is it reasonable to conclude that the number of HDTVs per household follows a normal distribution? (Hint: Use limits such as 0.5 up to 1.5, 1.5 up to 2.5, and so on.)
580 CHAPTER 15
57. A survey investigated the public’s attitude toward the federal deficit. Each sampled citizen was classified as to whether he or she felt the government should reduce the deficit or increase the deficit, or if the individual had no opinion. The sample results of the study by gender are reported below.
Reduce the Increase the No Gender Deficit Deficit Opinion
Female 244 194 68 Male 305 114 25
At the .05 significance level, is it reasonable to conclude that gender is independent of a person’s position on the deficit?
58. A study regarding the relationship between age and the amount of pressure sales personnel feel in relation to their jobs revealed the following sample information. At the .01 significance level, is there a relationship between job pressure and age?
Degree of Job Pressure
Age (years) Low Medium High
Less than 25 20 18 22 25 up to 40 50 46 44 40 up to 60 58 63 59 60 and older 34 43 43
59. The claims department at Wise Insurance Company believes that younger drivers have more accidents and, therefore, should be charged higher insurance rates. Investi- gating a sample of 1,200 Wise policyholders revealed the following breakdown on whether a claim had been filed in the last 3 years and the age of the policyholder. Is it reasonable to conclude that there is a relationship between the age of the policyholder and whether or not the person filed a claim? Use the .05 significance level.
Age Group No Claim Claim
16 up to 25 170 74 25 up to 40 240 58 40 up to 55 400 44 55 or older 190 24
Total 1,000 200
60. A sample of employees at a large chemical plant was asked to indicate a prefer- ence for one of three pension plans. The results are given in the following table. Does it seem that there is a relationship between the pension plan selected and the job classi- fication of the employees? Use the .01 significance level.
Pension Plan
Job Class Plan A Plan B Plan C
Supervisor 10 13 29 Clerical 19 80 19 Labor 81 57 22
61. Did you ever purchase a bag of M&M’s candies and wonder about the distribution of colors? Did you know in the beginning they were all brown? Now, peanut M&M’s are 12% are brown, 15% yellow, 12% red, 23% blue, 23% orange, and 15% green. A 6-oz. bag purchased at the Book Store at Coastal Carolina University had 14 brown, 13 yel- low, 14 red, 12 blue, 7 orange, and 12 green. Is it reasonable to conclude that the actual distribution agrees with the expected distribution? Use the .05 significance level. Con- duct your own trial. Be sure to share with your instructor.
NONPARAMETRIC METHODS: NOMINAL LEVEL HYPOTHESIS TESTS 581
D A T A A N A L Y T I C S
(The data for these exercises are available at the text website: www.mhhe.com/Lind17e.)
62. The North Valley Real Estate data reports information on homes on the market. a. Determine the proportion of homes that have an attached garage. At the .05 signifi-
cance level, can we conclude that more than 60% of the homes have an attached garage? What is the p-value?
b. Determine the proportion of homes that have a pool. At the .05 significance level, can we conclude that more than 60% of the homes have a pool? What is the p-value?
c. Develop a contingency table that shows whether a home has a pool and the town- ship in which the house is located. Is there an association between the variables pool and township? Use the .05 significance level.
d. Develop a contingency table that shows whether a home has an attached garage and the township in which the home is located. Is there an association between the variables attached garage and township? Use the .05 significance level.
63. Refer to the Baseball 2016 data, which report information on the 30 Major League Base- ball teams for the 2016 season. Set up a variable that divides the teams into two groups, those that had a winning season and those that did not. There are 162 games in the sea- son, so define a winning season as having won 81 or more games. Next, find the median team salary and divide the teams into two salary groups. Let the 15 teams with the largest salaries be in one group and the 15 teams with the smallest salaries be in the other. At the .05 significance level, is there a relationship between salaries and winning?
64. Refer to the Lincolnwood School District bus data. a. Suppose we consider a bus “old” if it has been in service more than 8 years. At the
.01 significance level, can we conclude that less than 40% of the district’s buses are old? Report the p-value.
b. Find the median maintenance cost and the median age of the buses. Organize the data into a two-by-two contingency table, with buses above and below the median of each variable. Determine whether the age of the bus is related to the amount of the maintenance cost. Use the .05 significance level.
c. Is there a relationship between the maintenance cost and the manufacturer of the bus? Use the breakdown in part (b) for the buses above and below the median main- tenance cost and the bus manufacturers to create a contingency table. Use the .05 significance level.
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO16-1 Use the sign test to compare two dependent populations.
LO16-2 Test a hypothesis about a median using the sign test.
LO16-3 Test a hypothesis of dependent populations using the Wilcoxon signed-rank test.
LO16-4 Test a hypothesis of independent populations using the Wilcoxon rank-sum test.
LO16-5 Test a hypothesis of several independent populations using the Kruskal-Wallis test.
LO16-6 Test and interpret a nonparametric hypothesis test of correlation.
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 LO16-1.)
Nonparametric Methods:
ANALYSIS OF ORDINAL DATA 16
© Asia File/Alamy Stock Photo
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 583
INTRODUCTION In Chapter 15, we introduced tests of hypothesis for nominal-scale variables. Recall from Chapter 1 that the nominal level of measurement implies the data can only be classified into categories, and there is no particular order to the categories. The pur- pose of these tests is to determine whether an observed set of frequencies, fo, is signifi- cantly different from a corresponding set of expected frequencies, fe. Likewise, if you are interested in the relationship between two characteristics—such as gender 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 as- sumptions need to be made about the shape of the population. We do not have to as- sume, for example, that the population of interest follows the normal 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 require that the variables in the data can be sorted and ranked. The vari- ables must be measured with an ordinal, interval, or ratio scale. An example of an ordi- nal 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 coeffi- cient 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.
THE SIGN TEST The sign test is based on the sign of a difference between two related observations. 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 cholesterol level de- creases 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 choles- terol 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 the 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 auto repair shop wants to evaluate a new tune-up program for automo- biles. 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 coffee: decaffeinated and regular. Its market research de- partment wants to determine whether coffee drinkers prefer decaffeinated or regular coffee. Coffee drinkers are given two small, unmarked cups of coffee, and 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.
LO16-1 Use the sign test to compare two dependent populations.
© digitalreflections/Shutterstock.com
584 CHAPTER 16
We can best show the application of the sign test by an example. We will use a “before/after” experiment.
E X A M P L E
The director of information systems at Samuelson Chemicals recommended that an in-plant training program be instituted for certain managers. The objective is to im- prove the technology knowledge base in the Payroll, Accounting, and Production Planning Departments.
A sample of 15 managers is randomly selected from the three departments. The managers are rated on their technology knowledge based on an assessment of how they use technology to solve problems. Based on the results, they were rated as outstanding, excellent, good, fair, or poor. (See Table 16–1.) After the three-month training program, the same assessment rated each manager’s tech- nology knowledge again. The two ratings (before and after) are shown along with the sign of the difference. A “+” sign indicates improvement, and a “−” sign indi- cates that the manager’s competence using technology had declined 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
TABLE 16–1 Competence Before and After the Training Program
We are interested in whether the in-plant training program increased the man- agers’ technology knowledge. That is, are the managers more knowledgeable after the training program than before?
S O L U T I O N
We will use the six-step hypothesis-testing procedure.
Step 1: State the null hypothesis and the alternate hypothesis. H0: π ≤ .50 There has been no change in the technology knowl-
edge base of the managers as a result of the training program.
H1: π > .50 There has been an increase in the technology knowl- edge base of the managers as a result of the training program.
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 585
The symbol π refers to the proportion in the population with a particular characteristic. If we do not reject the null hypothesis, it will indicate the training program has produced no change in the knowl- edge 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 assump- tions, namely:
1. There are only two outcomes: a “success” and a “failure.” A man- ager 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 (man- agers 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 technology knowledge. (See Table 16–1.) He was, therefore, elimi- nated from the study because he could not be assigned to either group, so n = 14. From the binomial probability distribution table in Appendix B.1, for an n of 14 and a probability of .50, we copied the binomial probability distribution in Table 16–2. The number of suc- cesses 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 probabilities 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 direction. 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 probabilities 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 probabili- ties in Table 16–2, column 3. We read up from the bottom until we reach the cumulative probability nearest to but not exceeding the level of significance (.10). That cumulative probability is .090. The number of successes (plus signs) corresponding to .090 in column 1 is 10. Therefore, the decision 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 put it another way, we add the probabilities from the bottom-up 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 H0. The region of rejection is portrayed in Chart 16–1.
STATISTICS IN ACTION
A recent study of under- graduate 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? Students who were absent less than 10% 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.
586 CHAPTER 16
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 desired level of significance (α) as possible without exceeding it. In this example, α is .10. The probability of 3 or fewer successes is .029, found by .000 + .001 + .006 + .022. The probabil- ity 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 probabilities 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 technology knowl- edge. The number 11 is in the rejection region, which starts at 10, so H0 is rejected.
Step 6: Interpret the results. We conclude that the three-month training course was effective. It increased the managers’ level of technology knowledge.
Number of Probability of Cumulative Successes Success Probability
0 0.000 1.000 1 0.001 0.999 2 0.006 0.998 3 0.022 0.992 4 0.061 0.970 5 0.122 0.909 6 0.183 0.787 7 0.209 Add up 0.604 8 0.183 0.395 9 0.122 0.212 10 0.061 0.090 11 0.022 0.029 12 0.006 0.007 13 0.001 0.001 14 0.000 0.000
.000 + .001 + .006 + .022 + .061
TABLE 16–2 Binomial Probability Distribution for n = 14, π = .50
0.250
0.200
0.150
0.100
0.050
0.000 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Pr ob
ab ili
ty o
f a S
uc ce
ss
Number of + Signs
Rejection region 10 or more + signs
H0: p # .50 H1: p . .50
CHART 16–1 Binomial Distribution, n = 14, π = .50
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 587
If the alternate hypothesis does not give a direction—for example, H0: π = .50 and H1: π ≠ .50—the test of hypothesis is two-tailed. In such cases, there are two re- jection regions, one in the lower tail and one in the upper tail. If α = .10 and the test is two-tailed, the area in each tail is .05 (α/2 = .10/2 = .05). Self-Review 16–1 illus- trates this.
© Corbis All Rights Reserved
Recall the Taster’s Choice example described on page 583, involving a consumer test to determine the preference for decaffeinated versus regular coffee. Use the .10 significance level. The null and alternate hypotheses are:
H0: π = .50 n = 12 H1: π ≠ .50 (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
regular coffee a “−,” it was found that two customers preferred decaffeinated. What is your decision? Explain.
S E L F - R E V I E W 16–1
1. The following hypothesis-testing situation is given: H0: π ≤ .50 and H1: π > .50. The significance 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 hypothe-
sis? Explain. 2. The following hypothesis-testing situation is given: H0: π = .50 and H1: π ≠ .50. The
significance 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. Calorie Watchers claims that you can eat anything you want for the third meal and still lose at least 5 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 en- rollees are:
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 H0 and H1. b. Using the .05 level of significance, what is the decision rule? c. What is your conclusion about the Calorie Watchers program?
E X E R C I S E S
588 CHAPTER 16
Using the Normal Approximation to the Binomial If the number of observations in the sample is larger than 10, the normal distribution can be used to approximate the binomial. Using the normal distribution will simplify the sign test. In Chapter 6 on page 187, we computed the mean of the binomial distribution from μ = nπ and the standard deviation from σ = √nπ(1 − π) . In this case, π = .50, so the equations reduce to μ = .50n and σ = .50 √n, respectively.
The test statistic z is
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 mea- sured 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.
SIGN TEST, n > 10 z = (x ± .50) − μ
σ [16–1]
SIGN TEST, n > 10, + SIGNS MORE THAN n/2 z =
(x − .50) − μ σ
= (x − .50) − .50n
.50√n [16–2]
SIGN TEST, n > 10, + SIGNS LESS THAN n/2 z =
(x + .50) − μ σ
= (x + .50) − .50n
.50√n [16–3]
If the number of pluses or minuses is more than n/2, we use the following form as the test statistic:
If the number of pluses or minuses is less than n/2, the test statistic z is
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 Chapter 7 starting on page 230.
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 589
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.
E X A M P L E
The market research department of Cola, Inc. has the task of taste 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.
S O L U T I O N
Step 1: State the null and alternate hypotheses.
H0: π = .50 There is no preference. H1: π ≠ .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 (16–1).
z = (x ± .50) − μ
σ where μ = .50n and σ = .50 √n . Step 4: Formulate the decision rule. Using Appendix B.5, Student’s t Distribu-
tion, with infinite degrees of freedom, for a two-tailed test (because H1 states that π ≠ .50) and the .05 significance level, the critical values are + 1.960 and −1.960. Therefore, do not reject H0 if the computed z value is between +1.960 and −1.960. Otherwise, reject H0 and accept H1.
Step 5: Compute z, compare the computed value with the critical value, and make a decision regarding H0. Preference for cola A was given a “+” sign and preference for B a “−” sign. Out of the 64 in the sam- ple, 42 preferred the sweet taste, which is cola A. Therefore, there are 42 pluses. Since 42 is more than n/2 = 64/2 = 32, we use for- mula (16–2) for z:
z = (x − .50) − .50n
.50√n =
(42 − .50) − .50(64) .50√64
= 2.38
The computed z of 2.38 is beyond the critical value of 1.96. There- fore, the null hypothesis of no difference is rejected at the .05 signifi- cance level.
Step 6: Interpret the results. 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. To compute the p-value, we first use Appendix B.3, Areas under the Normal Curve, and find the probability that a z value greater than 2.38 is .5000 − .4913 = .0087. For a two-tailed test, this probability is multiplied by two; the p-value is .0174. So the probability of obtaining a sample statistic this extreme when the null hypothesis is true is less than 2%.
590 CHAPTER 16
The human resources department at Ford Motor Company began a pilot health education program at the beginning of the year. The study director randomly selected 100 employees to participate in the program. In the first week of January, the blood pressure of each em- ployee was recorded. To evaluate the effectiveness of the program, the blood pressure of the same 100 employees was recorded in July. Eighty employees showed a reduction in blood pressure. Can we conclude that the program was effective in reducing blood pressure? (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.
S E L F - R E V I E W 16–2
5. A sample of 45 overweight men participated in an exercise program. At the conclu- sion of the program, 32 had lost weight. At the .05 significance level, can we con- clude 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 contacted and asked whether the skills learned in the program were effective. A total of 42 responded yes. At the .05 significance level, can we con- clude 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 preferred the regular menu or the gourmet menu. Forty-three pre- ferred the gourmet menu. At the .02 level, can we conclude the customers pre- ferred the gourmet menu?
8. Assembly workers at Coastal Computers Inc. assemble just one or two subassem- blies 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 complete computer. A sample of 25 employees was selected to experi- ment 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 con- clude the employees preferred assembling the entire unit? Explain the steps you used to arrive at your decision.
E X E R C I S E S
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 Chapter 3 that the median is the value above which half of the observa- tions 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.
LO16-2 Test a hypothesis about a median using the sign test.
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 591
E X A M P L E
The U.S. Bureau of Labor Statistics reported in 2015 that the median amount spent eating out by American families is about $3,000 annually. The food editor of the Portland (Oregon) Tribune wishes to know if the citizens of Portland differ from this national value in the last year in terms of eating out. She selected a random sample of 102 couples and found 60 spent more than $3,000 last year eating out, 40 spent less than $3,000, and surprisingly 2 spent exactly $3,000. At the .10 signifi- cance level, is it reasonable to conclude that the median amount spent this year in Portland, Oregon, is not equal to $3,000?
S O L U T I O N
If the population median is $3,000, then we expect about half of the sampled cou- ples spent more than $3,000 last year and about half less than $3,000. After dis- carding the two couples that spent exactly $3,000, we would expect 50 to be above the median and 50 to be below the median. Is the difference between the 60 couples that spent more than $3,000 and the number expected to spend more than $3,000 attributable to chance? Is the median some value other than $3,000? The statistical test for the median will help answer this question.
The null and the alternate hypotheses are:
H0: Median = $3,000
H1: Median ≠ $3,000
This is a two-tailed test because the alternate hypothesis does not indicate a direction. That is, we are not interested in whether the median is less than or greater than $3,000, only that it is different from $3,000. 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, π = .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 $3,000 annually on eating out.
The usable sample size is 100 and π is .50, so nπ = 100(.50) = 50 and n(1 − π) = 100(1 − .50) = 50, which are both larger than 5, so we use the normal distribution to approximate the binomial. That is, we actually use the standard normal distribu- tion as the test statistic. The significance level is .10, so α/2 = .10/2 = .05, which is the area in each tail of a normal distribution. From Appendix B.5, in the row with infinite degrees of freedom, the critical values are −1.645 and 1.645. The decision rule is to reject the null hypothesis if z is less than −1.645 or greater than 1.645.
We use formula (16–2) for z because 60 is greater than n/2 or (100/2 = 50).
z = (x − .5) − .5n
.50√n =
(60 − .5) − 50(100) .50√100
= 1.90
The null hypothesis is rejected because the computed value of 1.90 is greater than the critical value of 1.645. The sample evidence indicates that the median amount spent annually is not $3,000. The food editor in Portland should conclude that there is a difference in the median amount spent annually last year in Portland as compared with that reported by the U.S. Bureau of Labor Statistics in 2015. The p-value is .0574, found by 2(.5000 − .4713). The p-value is smaller than the signif- icance level of .10 for this test. So based on the p-value and a significance level of .10, we also reject the null hypothesis and conclude that Portland couples spend an amount that is different from the national median value.
592 CHAPTER 16
After reading the results of the Portland, Oregon, study, the food editor of the Tampa Times decided to conduct a similar study. The Tampa food editor decides to alter the study slightly by investigating whether families in her region spend more than a median amount of $3,000 A sample of 64 Tampa couples revealed 42 spent more than $3,000 per year eat- ing out. Using the .05 significance level, what should the editor conclude?
S E L F - R E V I E W 16–3
9. The median salary for a chiropractor in the United States is $81,500 per year, ac- cording 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 gradu- ated, 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 from Chicago to Jackson Hole revealed 160 tickets were below $503. None of the tickets was exactly $503. Let α = .05.
a. State the null and alternate hypotheses. b. What is your decision regarding H0? Interpret.
E X E R C I S E S
WILCOXON SIGNED-RANK TEST FOR DEPENDENT POPULATIONS The paired t test (page 370), described in Chapter 11, has two requirements. First, the sam- ples must be dependent. Recall that dependent samples are characterized by a measure- ment, 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 6 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 measurement, an intervention, and then another measurement.
The second requirement for the paired t test is that the distribution of the differences follow the normal probability distribution. In the company wellness example, this would
require that the differences in the weights of the population of participants follow the normal probability distribution. In that case, this assumption is reason- able. However, there are instances when we want to study the differences be- tween 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
LO16-3 Test a hypothesis of dependent populations using the Wilcoxon signed-rank test.
© Blend Images/Getty Images
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 593
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 differences 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.
E X A M P L E
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 be sure that patrons will like the spicy flavor better.
To begin his taste test, Bernie selects a random sample of 15 customers. Each sampled customer 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 rating near 0 indicates they did not like the flavor. Next, the same 15 participants 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 be- low. Is it reasonable to conclude that the spicy flavor is preferred? Use the .05 sig- nificance level.
Spicy Current Participant Flavor Rating Flavor Rating
Arquette 14 12 Jones 8 16 Fish 6 2 Wagner 18 4 Badenhop 20 12 Hall 16 16 Fowler 14 5 Virost 6 16 Garcia 19 10 Sundar 18 10 Miller 16 13 Peterson 18 2 Boggart 4 13 Hein 7 14 Whitten 16 4
S O L U T I O N
Each participant is asked to rate both flavors of chicken. So the ratings are dependent or related and, for each participant, we compute the difference between the ratings for the spicy flavor and the current flavor. The resulting value shows the amount the participants favor one flavor over the other. If we choose to subtract the current flavor rating from the spicy flavor rating, a positive result is the “amount” the participant favors the spicy flavor. Negative differences indicate the participant favored the current flavor. Because of the subjective nature of the ratings, we are not confident that the distribution of the differences follows the normal distribution. We decide to use the nonparametric Wilcoxon signed-rank test.
594 CHAPTER 16
As usual, we will use the six-step hypothesis-testing procedure. The null hy- pothesis is that there is no difference in the rating of the chicken flavors by the participants. The alternate hypothesis is that the ratings are higher for the spicy flavor. More formally:
H0: There is no difference in the ratings of the two flavors. H1: 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 pop- ulation of customers like the new flavor better. The significance level is .05, as stated on the previous page.
The steps to conduct the Wilcoxon signed-rank test are as follows.
1. Compute the difference between the spicy flavor rating and the current flavor rating for each participant. For example, Arquette’s spicy flavor rating was 14 and current flavor rating 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 D of Table 16–3.
2. Only the positive and negative differences are considered further. That is, if the difference in flavor ratings is 0, that participant is dropped from further analysis and the number in the sample reduced. From Table 16–3, Hall, the sixth participant, rated 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 D. Recall that in an absolute difference we ignore the sign of the difference and focus on the magnitude of the differences in ratings. The absolute differences are shown in column E.
Smaller rank sum
A B C D E F G H Spicy Current Signed Flavor Flavor Difference Absolute Rank Participant Rating Rating in Ratings 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 0 * * 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
Sums 75 30
TABLE 16–3 Flavor Rating for Current and Spicy Flavors
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 595
4. Next, rank the absolute differences from smallest to largest. Arquette, the first participant, rated the spicy chicken a 14 and the current a 12. The difference of 2 in the two taste ratings 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 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 av- erage 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 F is then given the same sign as the original dif- ference, and the results are reported in column G or H. For example, the sec- ond participant has a difference of −8 and a rank of 6. So the value of 6 is recorded for Jones in the R− section of column H.
6. Finally, the R+ and R− 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.8. A portion of that table is shown below. The α row is used for one-tailed tests and the 2α 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 α 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.8 is the largest value in the rejection region. To put it another way, our decision rule is to reject H0 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 difference in the flavor ratings between the current and the spicy. The study has failed to show that customers prefer the new flavor. Mr. Frick should stay with the current flavor of chicken.
2α .15 .10 .05 .04 .03 .02 .01 n α .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
596 CHAPTER 16
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 improved 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.
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
(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 produc-
tion before and after redesign?
S E L F - R E V I E W 16–4
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 couples live in larger homes than their parents?
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:
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
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. To test whether the new procedure is superior to the old procedure, a random sam- ple of 15 assembly-line workers was selected. The number of units produced in an hour under the old procedure was determined, then the new procedure was
E X E R C I S E S
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 597
introduced. After an appropriate break-in period, their production was measured again using the new procedure. The results were:
Procedure Procedure
Employee Old New Employee Old New
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
At the .05 significance level, can we conclude the production is greater using the new procedure?
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 in- creased 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. A listing of the number of subassemblies produced per week before and after the new work environment for each employee follows.
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
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?
WILCOXON RANK-SUM TEST FOR INDEPENDENT POPULATIONS 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 starting on page 230 in Chapter 11. Recall that the t test requires that the two populations follow the normal distribution and have equal population variances. These conditions are not required for the Wilcoxon rank-sum test.
LO16-4 Test a hypothesis of independent populations using the Wilcoxon rank- sum test.
598 CHAPTER 16
The Wilcoxon rank-sum test is based on the sum of ranks. The data are ranked as if the observations were from a single population. 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 samples. 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 stan- dard normal distribution is used as the test statistic. We use the following formula to find the value of the test statistic.
WILCOXON RANK-SUM TEST z = W −
n1(n1 + n2 + 1) 2
√ n1n2(n1 + n2 + 1)
12
(16–4)
where: n1 is the number of observations from the first population. n2 is the number of observations from the second population. W is the sum of the ranks from the first population.
E X A M P L E
Dan Thompson, the president of OTG Airlines, recently noted an increase in the num- ber of bags that were checked-in at the gate (gate-checked bags) in Atlanta. He is par- ticularly interested in determining whether there are more gate-checked bags from Atlanta compared with flights leaving Chicago. A sample of nine flights from Atlanta and eight from Chicago are reported in Table 16–4. At the .05 significance level, can we conclude that there are more gate-checked bags for flights originating in Atlanta?
TABLE 16–4 Number of Bags Checked at the Gate
Atlanta Chicago
11 13 15 14 10 10 18 8 11 16 20 9 24 17 22 21 25
S O L U T I O N
If the populations of gate-checked bags follow the normal probability distribution and have equal variances, the two-sample t test is 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 gate-checked bags 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
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 599
put it another way, the average rank of the two groups will be about the same. If the number of gate-checked bags is not the same, we expect the average of the ranks to be quite different.
Mr. Thompson believes there are more gate-checked bags for Atlanta flights. Thus, a one-tailed test is appropriate. The null and alternate hypotheses are:
H0: The number of gate-checked bags for Atlanta is the same or less than the number of gate-checked bags for Chicago.
H1: The number gate-checked bags for Atlanta is more than the number of gate-checked bags for Chicago.
The test statistic follows the standard normal distribution. At the .05 significance level, we find from the last row in Appendix B.5 the critical value of z is 1.645. The null hypothesis is rejected if the computed value of z is greater than 1.645.
The alternate hypothesis is that there are more gate-checked bags in Atlanta, which means that the Atlanta distribution is located to the right of the Chicago dis- tribution. The details of rank assignment are shown in Table 16–5. We rank the observations from both samples as if they were a single group. The Chicago flight with only 8 gate-checked bags had the fewest, so it is assigned a rank of 1. The Chicago flight with 9 gate-checked bags is ranked 2, and so on. The Atlanta flight with 25 gate-checked bags 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 gate-checked bags. There are also two Atlanta flights with 11 gate-checked bags. How do we handle these ties? The solution is to average the ranks involved and assign the average rank to both flights. In the case involving 10 gate-checked bags, 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 gate-checked bags.
TABLE 16–5 Ranked Number of Gate-checked Bags
Atlanta Chicago
Gate-checked Bags Rank Gate-checked Bags 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
Atlanta Rank Sum
The sum of the ranks for the Atlanta flights is 96.5. This is the value of W in formula (16–4). From Table 16–5, there are nine flights originating in Atlanta and eight in Chicago, so n1 = 9 and n2 = 8. Computing z from formula (16–4) gives:
z = W −
n1(n1 + n2 + 1) 2
√ n1n2(n1 + n2 + 1)
12
= 96.5 −
9(9 + 8 + 1) 2
√ 9(8) (9 + 8 + 1)
12
= 1.49
600 CHAPTER 16
Because the computed z value (1.49) is less than 1.645, the null hypothesis is not rejected. The evidence does not show a difference in the distributions of the number of gate-checked bags. That is, it appears that the number of gate-checked bags is the same in Atlanta as in Chicago. The p-value of .0681, found by determin- ing the area to the right of 1.49 (.5000 − .4319), indicates the same result.
The Wilcoxon - Mann/Whitney test is available in MegaStat. It provides the following results. The p-value, .0679, is slightly different as the software cor- rects 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 identi- fied as population 1. If, in the gate-checked bags example, the population of Chicago was identified as number 1, the direction of the alternate hypothesis would be changed. The value of z would be the same but have the opposite sign.
H0: The population distribution of gate-checked bags is the same or larger for Chicago than for Atlanta.
H1: The population distribution of gate-checked bags is smaller for Chicago than for Atlanta.
The computed value of z is −1.49, found by:
z = W −
n1(n1 + n2 + 1) 2
√ n1n2(n1 + n2 + 1)
12
= 56.5 −
8(8 + 9 + 1) 2
√ 8(9) (8 + 9 + 1)
12
= −1.49
Our conclusion is the same as described earlier. There is no difference in the typical number of gate-checked bags for Chicago and Atlanta.
The research director for Top Flite wants to know whether there is a difference in the distri- bution 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:
XL-5000: 252, 263, 279, 273, 271, 265, 257, 280 D2: 262, 242, 256, 260, 258, 243, 239, 265
Do not assume the distributions of the distances traveled follow the normal probability dis- tribution. At the .05 significance level, is there a difference between the two distributions?
S E L F - R E V I E W 16–5
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 601
KRUSKAL-WALLIS TEST: ANALYSIS OF VARIANCE BY RANKS The analysis of variance (ANOVA) procedure discussed in Chapter 12 tests the hypothe- sis that several population means are equal. The data were interval or ratio level. Also, it was assumed the populations followed the normal probability distribution and their stan- dard deviations were equal. What if the data are ordinal scale and/or the populations do not follow a normal distribution? Then we would use the Kruskal-Wallis one-way analysis
LO16-5 Test a hypothesis of several independent populations using the Kruskal-Wallis test.
15. Eight observations were randomly selected from two populations (population A and population B) that were not 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
16. Nine observations were randomly selected from population A and eight ob- servations were randomly selected from population B. The populations are not nor- mally 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 12, 14, 15, 19, 23, 29, 33, 40, 51 Population B 13, 16, 19, 21, 22, 33, 35, 43
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 second 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 students revealed the following information.
Campus 28, 16, 42, 29, 31, 22, 50, 42, 23, 25 Online 26, 42, 65, 38, 29, 32, 59, 42, 27, 41, 46, 18
Do not assume the two distributions of study times (in hours) follow a normal distri- bution. At the .05 significance level, can we conclude the online students spend more time studying?
18. In recent times, with mortgage rates at low levels, financial institutions have had to provide more customer convenience. One of the innovations offered by Coastal National Bank and Trust is online mortgage applications. Listed below are the times, in minutes, for eight customers to complete the application process for a 15-year fixed-rate mortgage and the times for nine customers to complete an appli- cation for a 30-year fixed-rate mortgage.
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
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.
E X E R C I S E S
602 CHAPTER 16
of variance by ranks. It requires ordinal, interval, or ratio scaled variables that can be ranked. The analysis makes no assumptions about the shape of the population distributions.
For the Kruskal-Wallis test to be applied, the populations must be independent. For example, if samples from three populations—executives, staff, and supervisors—are 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.
E X A M P L E
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 Medical Center to the south, and Piedmont Hospital on the east side of town. The director of adminis- tration is concerned about the waiting time of patients with non-life-threatening injuries that arrive during weekday evenings at the three hospitals. Specifically, is there a difference in the waiting times at the three hospitals?
S O L U T I O N
To investigate, the director selected random samples of patients at the three locations and determined the time, in minutes, between entering the particular facility and when treatment was completed. The times in minutes are reported in Table 16–6.
TABLE 16–6 Waiting Times for Emergency Treatment at Hospital Systems of the Carolinas
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 89
From Table 16–6, we observe that the shortest waiting time is 35 minutes for the fifth sampled patient at Piedmont Hospital. The longest waiting time is 107 min- utes 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 on page 392 of Chapter 12. However, there are three requirements for this 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. However, assumptions two and three may not be true. Specifically, the variances of the three samples are:
Sample variances
St. Luke’s Memorial Swedish Medical Center Piedmont Hospital
163.57 577.36 486.67
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 603
Notice that the variance for the Swedish Medical Center and Piedmont Hospital are more than twice that of St. Luke’s. The equal variance assumption is tenuous. Also, with the small number of observations in each sample it would be difficult to support the assumption of normally distributed populations. Clearly, all the assump- tions required for ANOVA techniques are not supported. So, we should use the Kruskal-Wallis test. It does not require these assumptions.
The first step in conducting the Kruskal-Wallis test is to state the null and the alternate hypotheses.
H0: The population distributions of waiting times are the same for the three hospitals.
H1: 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:
ΣR1, ΣR2, . . . , ΣRk, are the sums of the ranks of samples 1, 2, . . . , k, respectively.
n1, n2, . . . , nk 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 follows the chi-square distribution with k − 1 degrees of freedom. We prefer that each sample include at least five obser- vations. 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 Piedmont 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.7. 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 hy- pothesis and accept the alternate hypothesis.
The next step is to determine the value of the test statistic. We assign the waiting times at the three hospitals with the corresponding ranks. Considering the waiting times as a single group, 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 wait- ing 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 16–7 on the following page.
Solving for H gives
H = 12
n(n + 1)[ (ΣR1)2
n1 +
(ΣR2)2
n2 +
(ΣR3)2
n3 ] − 3(n + 1)
= 12
21(21 + 1)[ 58.52
7 +
1202
8 +
52.52
6 ] − 3(21 + 1) = 5.38
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 distributions of waiting times at the three hospitals.
KRUSKAL-WALLIS TEST
H = 12
n(n + 1)[ (ΣR1)2
n1 +
(ΣR2)2
n2 + . . . +
(ΣRk)2
nk ] − 3(n + 1) (16–5)
604 CHAPTER 16
The Kruskal-Wallis procedure can be done using the MegaStat add-in for Excel. Output for the example regarding the hospital waiting time follows. The computed value of H is 5.39 and the p-value is .067. The values are slightly different because of rounding. Based on these results, the decision and conclusion are the same.
TABLE 16–7 Waiting Times for Emergency Treatment at Hospital Systems of the Carolinas
St. Luke’s Memorial Swedish Medical Center Piedmont Hospital
Time Rank Time Rank Time Rank
56 9 103 20 42 5.5 39 4 87 16 38 2.5 48 7 51 8 89 17.5 38 2.5 95 19 75 15 73 14 68 13 35 1 60 10 42 5.5 61 11 62 12 107 21 89 17.5
ΣR1 = 58.5 ΣR2 = 120 ΣR3 = 52.5
Waiting time rank sums
Recall from Chapter 12 that, for the analysis of variance technique to apply, we as- sume (1) the populations are normally distributed, (2) these populations have equal standard deviations, and (3) the samples are selected from independent populations. If these assumptions are met in the hospital waiting time example, 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 differences 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.
H0: μ1 = μ2 = μ3 H1: 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 critical 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.
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 605
Using the one-way ANOVA test, the computed value of F is 3.822 and the p-value is .041. Our decision is to reject the null hypothesis and 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.
This comparison of the Kruskal-Wallis and ANOVA analyses shows important differ- ences. The result of the Kruskal-Wallis test is to fail to reject the hypothesis that the samples are from identical populations. Remember that this test does not require any assumptions about the variances or distributions of the populations. The result of the ANOVA test is to reject the hypothesis and conclude that the population means are different. However, the assumption of equal population variances is most probably not true, and the assumption of normally distributed populations is difficult to validate. Therefore, we should suspect that the ANOVA results are not valid. Comparing the re- sults of the Kruskal-Wallis and ANOVA serves to show the importance of validating the assumptions required for ANOVA. If the ANOVA assumptions are not supported, the results are not reliable. In this case, the Kruskal-Wallis analysis should be used.
The regional bank manager of Statewide Financial Bank is interested in the number of transactions occurring in personal checking accounts at four of the bank’s branches. Each branch randomly samples a number of personal checking accounts and records the num- ber of transactions made in each account over the last six months. The results are in the table below. Using the .01 level and the Kruskal-Wallis test, determine whether there is a difference in the number personal checking account transactions 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
S E L F - R E V I E W 16–6
19. Under what conditions should the Kruskal-Wallis test be used instead of ANOVA? 20. Under what conditions should the Kruskal-Wallis test be used instead of the
Wilcoxon rank-sum test?
E X E R C I S E S
606 CHAPTER 16
21. The following sample data were obtained from three populations that did not follow a normal distribution.
Sample 1 Sample 2 Sample 3
50 48 39 54 49 41 59 49 44 59 52 47 65 56 51 57
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
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 to protect exhaust components from corrosion. Bill Davis, the owner, wishes to deter- mine whether the durability of the paint was equal for three different conditions: saltwater, freshwater 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. Five boats were tested for each condition.
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 number of hours the paint lasted 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 re- garding the three mixtures, no assumptions regarding the population distribution of weights exist. 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 hypothesis that there is no effect of feed mixture on weight.
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 607
RANK-ORDER CORRELATION In Chapter 13 we described the correlation coefficient. Recall that it measures the asso- ciation between two interval- or ratio-scaled variables. For example, the correlation co- efficient reports the association between the salary of executives and their years of experience, or the association between the number of miles a shipment had to travel and the number of days it took to arrive at its destination. The correlation coefficient is a very versatile measure of association. However, there are several conditions when it is not appropriate or can be misleading. Those conditions include:
1. When the scale of measurement of one of the two variables is ordinal (ranked). 2. When the relationship between the variables is not linear. 3. When one or more of the data points are quite different from the others.
Charles Spearman, a British statistician, introduced a measure of correlation for ranked 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?
This coefficient of rank correlation is also applied to conditions 2 and 3 above. Instead of using the actual values in the data set, we rank the sample data and compute a correlation between the ranked values. Like Pearson’s coefficient of correlation described in Chapter 13, it can range from −1.00 up to 1.00. Values of −1.00 or 1.00 indicate perfect association between the ranks. A value of 0 indicates no association between the ranks of the two variables. Values of −.84 and .84 both exhibit the same amount of association but −.84 indicates an inverse relationship and .84 a direct rela- tionship. We denote Spearman’s coefficient of rank correlation as rs.
We can use formula (13–1) on page 443 to find the coefficient of rank correlation. How- ever, we use the ranked values in place of the actual values. The values are always ranked from low to high. A simpler method is to use the following, which uses the actual ranks.
LO16-6 Test and interpret a nonparametric hypothesis test of correlation.
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
SPEARMAN’S COEFFICIENT OF RANK CORRELATION
rs = 1 − 6Σd2
n(n2 − 1) (16–6)
where:
d is the difference between the ranks for each pair. n is the number of paired observations.
The following example/solution provides the details of computing the coefficient of rank correlation.
608 CHAPTER 16
E X A M P L E
Recent studies focus on the relationship between the age of online shoppers and the number of minutes spent browsing on the Internet. Table 16–8 shows a sample of 15 online shoppers who actually made a purchase last week. Included is their age and the time, in minutes, spent browsing on the Internet last week.
TABLE 16–8 Age and Browsing Minutes for a Sample of Internet Shoppers
Shopper Age Browsing Time (minutes)
Spina, Sal 28 342 Gordon, Ray 50 125 Schnur, Roberta 44 121 Alvear, Jose 32 257 Myers, Tom 55 56 Lyons, George 60 225 Harbin, Joe 38 185 Bobko, Jack 22 141 Koppel, Marty 21 342 Rowatti, Marty 45 169 Monahan, Joyce 52 218 Lanoue, Bernie 33 241 Roll, Judy 19 583 Goodall, Jody 17 394 Broderick, Ron 21 249
1. Draw a scatter diagram; plot age on the horizontal axis and browsing minutes on the vertical axis.
2. What type of association do the sample data suggest? Strong or weak, direct or inverse?
3. Do you see any issues with the relationship between the variables? 4. Find the coefficient of rank correlation. 5. Conduct a test of hypothesis to determine if there is a negative association
between the ranks.
S O L U T I O N
Our first step is to create a scatter diagram, shown in Chart 16–2, to better under- stand the relationship between the two variables.
0 0 10 20 30 40
Age (years)
Br ow
si ng
T im
e (m
in ut
es )
50 60 70
100
200
300
400
500
600
700
Possible outlier points
CHART 16–2 Scatter Diagram of Age and Internet Browsing Minutes
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 609
There appears to be a fairly strong inverse relationship between the Internet shopper’s age and the amount of time spent browsing the Internet. That is, the younger shoppers seem to spend more time browsing. Not surprising. There are a couple of data points that seem quite different than the others. They are identified in Chart 16-2 on the previous page. Because of these points we decide to use the coefficient of rank correlation.
To calculate the coefficient of rank correlation, we first rank the age of the shoppers. The youngest shopper is Jody Goodall, who is 17, so she is ranked 1. The next youngest is Judy Roll, so she is ranked 2. There are two shoppers, Marty Koppel and Ron Broderick, who are 21 years old. This tie is resolved by giving each a rank of 3.5, which is the average of ranks 3 and 4. The oldest shopper in the sam- ple is George Lyons; he is given a rank of 15. The same ranking strategy is used for browsing minutes. Tom Myers spent the fewest minutes browsing the Internet, so he has the rank of 1. Roberta Schnur is next with 141 minutes, so she has a brows- ing rank of 2. All the ranking data are shown in Table 16–9.
TABLE 16–9 Age, Browsing Time, Ranks, Differences, and Differences Squared
Browsing Time Browsing Shopper Age Age Rank (minutes) Rank d d2
Spina, Sal 28 6.0 342 12.5 −6.50 42.25 Gordon, Ray 50 12.0 125 3.0 9.00 81.00 Schnur, Roberta 44 10.0 121 2.0 8.00 64.00 Alvear, Jose 32 7.0 257 11.0 −4.00 16.00 Myers, Tom 55 14.0 56 1.0 13.00 169.00 Lyons, George 60 15.0 225 8.0 7.00 49.00 Harbin, Joe 38 9.0 185 6.0 3.00 9.00 Bobko, Jack 22 5.0 141 4.0 1.00 1.00 Koppel, Marty 21 3.5 342 12.5 −9.00 81.00 Rowatti, Marty 45 11.0 169 5.0 6.00 36.00 Monahan, Joyce 52 13.0 218 7.0 6.00 36.00 Lanoue, Bernie 33 8.0 241 9.0 −1.00 1.00 Roll, Judy 19 2.0 583 15.0 −13.00 169.00 Goodall, Jody 17 1.0 394 14.0 −13.00 169.00 Broderick, Ron 21 3.5 249 10.0 −6.50 42.25 Sum d2 965.50
The coefficient of rank correlation is −.724, found by using formula (16–6).
rs = 1 − 6Σd2
n(n2 − 1) = 1 −
6(965.5)
15(152 − 1) = 1 − 1.724 = −0.724
The value of −0.724 indicates a fairly strong negative association between age of the Internet shopper and the minutes spent browsing. The younger shoppers tended to spend more time browsing the Internet.
Testing the Significance of rs Next we wish to investigate the question as to whether the degree of association be- tween the ranked variables occurred by chance. Recall, in Chapter 13 starting on page 447, we investigated this same issue. In that instance, we used interval or ratio data; now we use ranks. In the preceding example/solution we found a rank correlation of −0.724. Is it possible that the correlation of −0.724 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 this question.
610 CHAPTER 16
The null and the alternate hypotheses are:
H0: The rank correlation in the population is zero. H1: There is a negative association among the ranks.
We use the .05 significance level. The decision rule is to reject the null hypothesis if the computed value is less than −1.771. Go to Appendix B.5 and use a one-tailed test with 15 − 2 = 13 degrees of freedom.
t = rs √ n − 2 1 − r2s
= −0.724 √ 15 − 2
1 − (−0.724)2 = −3.784
The computed value of t is −3.784. The null hypothesis is rejected and the alternate hypothesis is accepted. The computed t of −3.784 is less than −1.771. There is evi- dence of a negative association between the age of the Internet shopper and the time spent browsing the Internet.
HYPOTHESIS TEST, RANK CORRELATION t = rs√ n − 2 1 − r 2s
(16–7)
A sample of individuals applying for manufacturing jobs at Davis Enterprises revealed the following scores on an eye perception test (X) and a mechanical aptitude test (Y):
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
(a) Compute the coefficient of rank correlation between eye perception and mechanical aptitude.
(b) At the .05 significance level, can we conclude that the correlation in the population is different from 0?
S E L F - R E V I E W 16–7
STATISTICS IN ACTION
Manatees are large mam- mals that like to float just below the water’s surface. Because they float just be- low the surface, they are in danger from powerboat propellers. A study of the correlation between the number of powerboat reg- istrations in coastal Florida counties and the number of accidental manatee deaths revealed a strong positive correlation. As a result, Florida created re- gions where powerboats are prohibited, so that manatees could thrive.
25. Do husbands and wives like the same TV shows? A recent study by Nielsen Media Research asked a young married couple to rank 14 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.
E X E R C I S E S
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 distribution with n − 2 degrees of freedom.
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 611
Program Male Ranking Female Ranking
60 Minutes 4 5 CSI — New York 6 4 Bones 7 8 SportsCenter 2 7 Late Show with Stephen Colbert 12 11 NBC Nightly News 8 6 Law and Order: Los Angeles 5 3 Miami Medical 3 9 Survivor 13 2 Parks and Recreation 14 10 American Idol 1 1 Grey’s Anatomy 9 13 Doc Martin 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 asso- ciation between the two rankings?
26. Far West University offers both day and evening classes in business adminis- tration. A survey of students inquires how they perceive the prestige associated with eight 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 Carrer Student Student Carrer 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
Find Spearman’s coefficient of rank correlation. 27. Ten new sales representatives for Clark Sprocket and Chain, Inc. were
required to attend a training program before being assigned to a regional sales of- fice. At the end of the program, the representatives took a series of tests and the scores were ranked. For example, Arden had the lowest test score and is ranked 1; Arbuckle had the highest test score and is ranked 10. At the end of the first sales year, the representatives’ ranks based on test scores were paired with their first year sales.
Annual Ranking in Annual Ranking in Sales Training Sales Training Representative ($ thousands) Program Representative ($ thousands) Program
Kitchen 319 8 Arden 300 1 Bond 150 2 Crane 280 6 Gross 175 5 Arthur 200 9 Arbuckle 460 10 Keene 190 4 Greene 348 7 Knopf 300 3
612 CHAPTER 16
a. Compute and interpret the coefficient of rank correlation between first-year sales and class 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 uni- versity. Each assistant coach attended three games and then ranked the 10 players with respect to potential. To explain, the first coach ranked Norma Tidwell as the best player among the 10 scouted and Jeannie Black the worst.
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
a. Determine Spearman’s rank correlation coefficient. b. At the .05 significance level, can we conclude there is a positive association
between the ranks?
C H A P T E R S U M M A R Y
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 distribu-
tion for the critical value. D. For a sample of 10 or more, use the standard normal distribution and the following
formula.
z = (x ± .50) − .50n
.50√n (16–2) (16–3)
II. The median test is used to test a hypothesis about a population median. A. Find μ and σ 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 ob-
servations above or below the median.
z = (x ± .50) − μ
σ (16–1)
III. The Wilcoxon signed-rank test is a nonparametric test for differences between two de- pendent populations. A. The assumption of normally distributed populations is not required. 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.8 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. The assumption of normally distributed populations is not required. B. The data must be at least ordinal scale.
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 613
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.
z = W −
n1(n1 + n2 + 1) 2
√ n1n2(n1 + n2 + 1)
12
(16–4)
G. The standard normal distribution, found in Appendix B.3, is the test statistic. V. The Kruskal-Wallis one-way ANOVA by ranks is used to test whether several population
distributions are the same. A. The assumption of normally distributed populations is not required. B. The populations 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
observations in each sample. E. The value of the test statistic is computed from the following:
H = 12
n(n + 1)[ (ΣR1)2
n1 +
(ΣR2)2
n2 + … +
(ΣRk)2
nk ] − 3(n + 1) (16–5)
VI. Spearman’s coefficient of rank correlation is a measure of the association between two ordinal-scale variables. 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 posi-
tive correlation. B. The value of rs is computed from the following formula.
rs = 1 − 6Σd 2
n(n2 − 1) (16–6)
C. Provided the sample size is at least 10, we can conduct a test of hypothesis using the following formula:
t = rs √ n − 2 1 − r 2s
(16–7)
1. The test statistic follows the t distribution. 2. There are n − 2 degrees of freedom.
P R O N U N C I A T I O N K E Y
SYMBOL MEANING PRONUNCIATION
(ΣR1)2 Square of the total of the first Sigma R sub 1 column ranks squared
rs Spearman’s coefficient of rank r sub s correlation
C H A P T E R E X E R C I S E S
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 possibilities 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.
614 CHAPTER 16
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 personnel, a preference test was 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 sampled employees 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 un- decided. 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 MP3 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 music track using the Sony player was played. Then the same track was played using the Panasonic player. A “+” in the following table indicates an individual’s preference for the Sony player, a “−” indicates preference for the Panasonic player, and a 0 signifies no preference.
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.
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 H0 and H1. 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 orange juice with or without pulp. A random sample of 212 consumers was selected. Each member of the sample tasted a small, unlabeled cup of juice with pulp and a cup of juice without pulp. Twelve consumers said they had no preference, 40 preferred juice with pulp, and the remainder liked the juice without pulp better. Test at the .05 level that the preferences for orange juice with pulp and orange juice without pulp are equal.
34. The objective of a community research project is to determine whether women are more community conscious before marriage or after 5 years of marriage. A test de- signed to measure community consciousness was administered to a sample of nine women before marriage, and the same test was given to them 5 years after marriage. The test scores are:
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. H0 is: There is no difference in community consciousness before and after marriage. H1 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 North- west? Test at the .05 level. Annual divorce rates per 1,000 population for five randomly
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 615
selected counties in the Southwest, six counties in the Southeast, and 5 counties in the Northwest are:
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 below is the number of idle minutes yesterday for the five day-shift workers and the six evening-shift workers. Use the .05 significance level.
Day Shift Evening Shift
92 96 103 114 116 80 81 82 89 88 91
37. Drs. Trythall and Kerns are studying the mobility of executives in selected indus- tries. 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 com- panies; the fewest, for changing jobs within a company and not moving. They randomly sampled five executives in the chemical industry, six in the retail industry, five in the Internet industry, and five in the aerospace industry. 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 signifi- cance level.
Chemical Retail Internet Aerospace
4 3 62 30 17 12 40 38
8 40 81 46 20 17 96 40 16 31 76 21 19
38. A series of questions on sports and world events was asked of 14 randomly se- lected young adult naturalized citizens. The results were translated into sports and world events “knowledge” scores. 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
616 CHAPTER 16
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 between the sports and world events “knowledge” scores 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 shortest time receive the highest grades and those who take the longest to com- plete 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 between the order of completion and the rank of the test score. At the .05 significance level, can Professor Forman conclude there is a positive association between the order of finish and the test scores?
D A T A A N A L Y T I C S
(The data for these exercises are available at the text website: www.mhhe.com/Lind17e.)
41. The North Valley Real Estate data report information on homes on the market. 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 six 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 distribu- tion of selling prices is not normally distributed.
c. Compare the distribution of FICO scores for fixed and variable mortgages. FICO scores are people's credit rating. Higher ratings indicate better credit. Can we con- clude there is a difference in the distributions of FICO scores based on the type of mortgage? Use the .05 significance level.
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 617
42. Refer to the Baseball 2016 data, which report information on the 2016 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 Lincolnville School District school bus data. a. Assume the distribution of the maintenance cost for the bus capacities 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 fuel types, diesel or gasoline, does not follow a normal distribution. Conduct a test of hypothesis at the .05 signifi- cance level to determine whether the distributions differ.
c. 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 signifi- cance level to determine whether the distributions differ.
A REVIEW OF CHAPTERS 15–16 In Chapters 15 and 16, we describe statistical methods to study data that are either the nominal or the ordinal scale of measurement. These methods are nonparametric or distribution-free statistics. They do not require assumptions regard- ing 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 15, we describe tests for the nominal level of data. We begin by studying one- and two-sample tests of proportions. A proportion is the fraction of individuals or objects that possess a certain characteristic. In a proportion the sampled individual or object either has the characteristic or does not. For example, in a one-sample test of propor- tions, we study 100 gasoline purchases at the local Kwick Fill station. The individual either purchased regular gasoline or did not. There are only two possible outcomes. In a two-sample test of proportions, we compare the proportion of purchasers who bought regular gasoline at the Corry, Texas Kwick Fill with the proportion of regular gasoline pur- chases at the Tyrone, Texas Kwick Fill. The test statistic is the standard normal distribution in both the one-sample and the two-sample tests.
We also use the chi-square distribution to compare an observed set of frequencies with the corresponding set of ex- pected frequencies. The level of measurement is either the nominal or ordinal scale. In the previous example, there were only two possible outcomes: the purchaser bought regular gasoline or did not buy it. We use the chi-square distribution to investigate an instance where there are several possible nominal-scale outcomes. A gasoline purchaser can buy regular, midgrade, or premium. 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 15, 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 relationship between the quality of the prod- uct (acceptable or unacceptable) and the shift when it was manufactured (day, afternoon, or night)? The chi-square distri- bution is used as the test statistic.
In Chapter 16, 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 number is reasonable or too large to have occurred by chance.
618 CHAPTER 16
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 values. It has its own sampling distribution, which is reported in Appendix B.8.
The Wilcoxon ranked-sum test assumes independent populations, but does not require the populations 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 populations have equal standard deviations.
The statistic Spearman’s coefficient of rank correlation is a special case of the Pearson coefficient 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 associa- tion between the ranks.
P R O B L E M S
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 the number of snow- cones sold on 20 randomly selected days. Is it reasonable to conclude that the median is actually greater than 60? 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
2. The manufacturer of children’s raincoats wants to know if there is a preference among children for blue, red, green, or yellow raincoats. The information below is the color preference for a sample of 50 children between the ages of 6 and 10. Use the .05 sig- nificance level to investigate.
Color Frequency
Blue 17 Red 8 Green 12 Yellow 13
3. Is there a difference in the length of suspension bridges in the northeast, southeast, and far west parts of the United States? The following table shows the lengths (in feet) of seven bridges in the northeast, nine bridges in the southeast, and eight bridges in the far west. Conduct an appropriate test of hypothesis on the following data. Do not assume the bridge lengths follow a normal probability distribution. Use the .05 significance level.
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
NONPARAMETRIC METHODS: ANALYSIS OF ORDINAL DATA 619
4. Research by the First Bank of Illinois revealed that 8% of its customers wait more than 5 minutes for a teller in a bank lobby. Management considers this reasonable and will not add more tellers unless the proportion waiting longer than 5 minutes becomes larger than 8%. 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 5 minutes. At the .01 significance level, is it reasonable to conclude that more than 8% of the customers wait more than 5 minutes?
C A S E S
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 Simulator. 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 turn- ing signals, not exercising caution on wet or snowy pave- ment, 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 Driving Simulator because subjects do not gain any advantage 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-min- ute period and return to the Rupple Driving Simulator for
another test drive. The number of driving errors after drink- ing 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, there- fore, 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.
P R A C T I C E T E S T
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 observation? 3. 4. In a contingency table, there are four rows and three columns. How many degrees
of freedom are there? 4.
620 CHAPTER 16
5. In a goodness-of-fit test, the critical value of chi-square is based on . (sample size, number of categories, number of variables, 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, 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, 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. 10. The Kruskal-Wallis test is used in 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 six-step hypothesis testing procedure.
1. A recent census report indicated that 65% of families have two parents present, 20% have only a mother present, 10% have only a father present, and 5% have no parent present. A random sample of 200 children from a large rural school district revealed the following frequencies of children with two parents, mother only, father only, no parent, and the total of 200.
Two Parents Mother Only Father Only No Parent Total
120 40 30 10 200
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 book selected for recreational reading by men and women. A ran- dom sample of 540 men and 500 women provided the following information regarding their preferences for mystery, romance, or self-help books. At the .05 significance level, should we conclude that gender is related or unrelated to type of book selected?
Mystery Romance Self-Help Total
Men 250 100 190 540 Women 130 170 200 500
3. An instructor has three sections of basic statistics: 8:00 a.m., 10:00 a.m., and 1:30 p.m. Listed below are the grades on the first exam for each section. 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?
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 87
4. According to a study in Health Magazine, one in three children in the United States is obese or overweight. A health practitioner in Louisiana sampled 500 children and found 210 who were obese or overweight. Does this evidence suggest that the actual proportion of obese or overweight children is more than one out of three? Use the .01 signif- icance level.
Index Numbers 17
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO17-1 Compute and interpret a simple, unweighted index.
LO17-2 Compute and interpret an unweighted aggregate index.
LO17-3 Compute and interpret a weighted aggregate index.
LO17-4 List and describe special-purpose indexes.
LO17-5 Apply the Consumer Price Index.
© Steve Cole/Photodisc/Getty Images
INFORMATION ON PRICES and quantities for margarine, shortening, milk, and potato chips for 2000 and 2016 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 LO17-1.)
622 CHAPTER 17
INTRODUCTION In this chapter, we will examine a useful descriptive tool called an index. An index ex- presses 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 federal government, by busi- ness publications such as Bloomberg Businessweek and Forbes, in most daily newspa- pers, 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 eco- nomics are very concerned 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 production.
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 up- dated 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 news- papers. Many are also reported in real time. Shown below are the Dow Jones Indus- trial Average, NASDAQ, and other indexes from the Yahoo.com website (http:// financeyahoo.com).
© Image Source/Getty Images
SIMPLE INDEX NUMBERS What is an index number? An index or index number measures the change in a particu- lar item (typically a product or service) between two time periods.
LO17-1 Compute and interpret a simple, unweighted index.
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 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.
INDEX NUMBERS 623
E X A M P L E
According to the Bureau of Labor Statistics, in 2000 the average hourly earnings of production workers was $14.02. In March 2016, it was $21.37. What is the index of hourly earnings of production workers for March 2016 based on 2000 data?
S O L U T I O N
It is 152.43, found by:
P = Average hourly earnings in 2016 Average hourly earnings in 2000
(100) = $21.37 $14.02
(100) = 152.43
Thus, the hourly earnings in 2016 compared to 2000 were 152.43%. This means there was a 52.43% increase in hourly earnings during the period, found by 152.43 − 100.0 = 52.43.
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.
624 CHAPTER 17
E X A M P L E
An index can also compare one item with another. The population of the Canadian province of British Columbia in 2014 was 4,657,947, and for Ontario it was 13,730,187. What is the population index of British Columbia compared to Ontario?
S O L U T I O N
The index of population for British Columbia is 33.9, found by:
P = Population of British Columbia
Population of Ontario (100) =
4,657,947 13,730,187
(100) = 33.9
This indicates that the population of British Columbia is 33.9% (about one-third) of the population of Ontario, or the population of British Columbia is 66.1% less than the population of Ontario (100 − 33.9 = 66.1).
E X A M P L E
The following chart shows the number of passengers (in millions) for the 10 busiest airports in the United States in 2014. Use the McCarran International Airport in Las Vegas as the base. What is the index for the other airports compared to Las Vegas?
S O L U T I O N
To find the 10 indexes, we divide the passengers for Las Vegas into the passengers for the other nine airports. So the index for Atlanta is 224.2, found by (96.2/42.9)*(100). The index for San Francisco International is 109.8, found by
1 2 3 4 5 6 7 8 9
10
96.2 70.7 70.0 63.6 53.5 53.3 47.1 44.4 44.3 42.9
0 20 40 60 80 100
224.2 164.8 163.2 148.3 124.7 124.2 109.8 103.5 103.3 100.0
Hartsfield-Jackson Atlanta International Airport Los Angeles International Airport Chicago O’Hare International Airport Dallas/Fort Worth International Airport Denver International Airport John F. Kennedy International Airport San Francisco International Airport Miami International Airport Charlotte Douglas International Airport McCarran International Airport
Hartsfield-Jackson Atlanta International Airport Los Angeles International Airport
Chicago O’Hare International Airport Dallas/Fort Worth International Airport
Denver International Airport John F. Kennedy International Airport
San Francisco International Airport Miami International Airport
Charlotte Douglas International Airport McCarran International Airport
Number of Passengers at the Ten Busiest US Airports
Passengers (mil)
Rank Airport Passengers (mil) Index
INDEX NUMBERS 625
(47.1/42.9)*(100). So Atlanta has 124.2% more passengers than Las Vegas and San Francisco has 9.8 more than Las Vegas. The chart summarizes the indexes. The chart reveals that in 2014, Miami and Charlotte served about the same number of passen- gers as Las Vegas (all three of the indexes are close to 100). Atlanta, on the other hand, served more than twice as many passengers as McCarran International.
Note from the previous discussion that:
1. The index of average hourly earnings of production workers (152.43) is a percent- age, but the percent symbol is usually omitted.
2. Indexes have either a base or a base period. In the example/solution regarding the average hourly earnings of production workers, we used 2000 as the base period. The Consumer Price Index uses 1982–84 as the base period. In contrast, in the example/solution about airport passengers, the McCarran International Airport in Las Vegas was used as the base for comparison.
3. Most business and economic indexes are reported to the nearest whole number, such as 214 or 96, or to the nearest tenth of a percent, such as 83.4 or 118.7.
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 200 categories of items summarized by 8 groups—food and beverages, housing, apparel, trans- portation, medical care, recreation, education and communication, and other goods and services. The prices of 80,000 goods and services in the 200 categories are collected. Prices are expressed in many different units such as dollars per pound or a dozen eggs. Only by summarizing these prices with an 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 com- posed of exceptionally large numbers. For example, the estimate of U.S. retail e-commerce sales in 2014 was $304,913,000 compared to $168,895,000 in 2010. This is an in- crease of $136,018,000 in the four-year period, but because of the very large numbers it is difficult to realize the increase. If we compute an index of 2014 sales based on 2010 sales, the index is 180.5. The index means that 2014 sales increased 80.5% com- pared to 2010 sales.
Index = 2014 e-commerce sales 2010 e-commerce sales
(1 00) = $304,913,000 $168,895,000
(100) = 180.5
Construction of Index Numbers We already discussed the construction of a simple price index. The price in a selected year (such as 2017) 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:
SIMPLE INDEX P = pt p0
× 100 (17–1)
626 CHAPTER 17
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 2017. What is the price index for 2017 using 2000 as the base period and 100 as the base value? It is 176.7, found by:
P = pt p0
(100) = $795 $450
(100) = 176.7
Interpreting this result, the price of the fall weekend package increased 76.7% from 2000 to 2017.
The base period need not be a single year. Note in Table 17–1 that if we use 2005 – 06 = 100, the base price for the stapler would be $21 [found by determining the mean price of 2005 and 2006, ($20 + $22)/2 = $21]. The prices $20, $22, and $23 are averaged if 2005—07 is selected as the base. The mean price would be $21.67. The indexes constructed using the three different base periods are presented in Table 17–1. (Note that when 2005 − 07 = 100, the index numbers for 2005, 2006, and 2007 average 100.0, as we would expect.) Logically, the index numbers for 2015 using the three different bases are not the same.
Price of Price Index Price Index Price Index Year Stapler (2005 = 100) (2005–06 = 100) (2005–07 = 100)
2000 $18 90.0 18 21
× 100 = 85.7 18
21.67 × 100 = 83.1
2005 20 100.0 20 21
× 100 = 95.2 20
21.67 × 100 = 92.3
2006 22 110.0 22 21
× 100 = 104.8 22
21.67 × 100 = 101.5
2007 23 115.0 23 21
× 100 = 109.5 23
21.67 × 100 = 106.1
2015 38 190.0 38 21
× 100 = 181.0 38
21.67 × 100 = 175.4
TABLE 17–1 Prices of a Benson Automatic Stapler, Model 3, Converted to Indexes Using Three Different Base Periods
1. Listed below are the top steel-producing countries for the year 2014 (http://www. worldsteel.org). Express the amount produced by China, Japan, India, and Russia as an index, using the United States as a base. What percent more steel does China produce than the United States?
S E L F - R E V I E W 17–1
Amount Country (millions of tons)
China 822.7 Japan 110.7 United States 88.2 India 86.5 Russia 71.5
2. The average hourly earnings of production workers for selected years are given below.
INDEX NUMBERS 627
Year Average Hourly Earnings
1995 $11.65 2000 14.02 2005 16.13 2013 19.97 2016 21.37
(a) Using 1995 as the base period and 100 as the base value, determine the indexes for the other years. Interpret the index.
(b) Use the average of 1995 and 2000 as the base and determine indexes for the other years. Interpret the index.
1. PNC Bank Inc., which has its headquarters in Pittsburgh, reported the follow- ing commercial loan totals for years 2008 through 2015. Using 2008 as the base, develop a simple index of commercial loans for the following years
Year Loans ($ Millions)
2008 69,220 2009 54,818 2010 55,177 2011 65,694 2012 83,040 2013 88,378 2014 97,420 2015 98,608
2. The table below reports the earnings per share of common stock for Home Depot Inc. for 2003 through 2015. Develop an index, with 2003 as the base, for earnings per share for years 2004 through 2015.
Year Earnings per Share
2003 1.56 2004 1.88 2005 2.26 2006 2.72 2007 2.79 2008 2.37 2009 1.34
Year Earnings per Share
2010 1.57 2011 2.01 2012 2.47 2013 3.76 2014 4.71 2015 5.46
3. Listed below are the net sales for a San Francisco–area mail-order retailer for the years 2006 to 2015. Use the mean sales for the earliest three years to determine a base and then find the index for 2014 and 2015. By how much have net sales increased from the base period?
Year Sales (millions) Year Sales (millions)
2006 $486.6 2011 $568.5 2007 506.8 2012 581.9 2008 522.2 2013 496.1 2009 574.6 2014 456.6 2010 580.7 2015 433.3
E X E R C I S E S
628 CHAPTER 17
UNWEIGHTED INDEXES In many situations, we wish to combine several items and develop an index to compare the cost of this aggregation of items in two different time periods. For example, we might be interested in an index for items that relate to the expense of operating 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 entertainment. There are several ways we can combine the items to determine the index.
Simple Average of the Price Indexes Table 17–2 reports the prices for several food items for 2003 and 2015. We would like to develop an index for this group of food items for 2015, using the 2003 prices as the base. This is written in the abbreviated code 2003 = 100.
LO17-2 Compute and interpret an unweighted aggregate index.
4. In January 2003, the price for a whole fresh chicken was $1.004 per pound. In February 2016, the price for the same chicken was $1.475 per pound. Use the January 2003 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?
Item 2003 Price 2015 Price Simple Index
Bread, white, cost per pound $ 1.042 $ 1.440 138.2 Eggs, dozen 1.175 2.133 181.5 Milk, gallon, white 2.686 3.463 128.9 Apples, Red Delicious, 1 pound 0.911 1.265 138.9 Orange juice, 12 oz. concentrate 1.848 2.678 144.9 Coffee, 100% ground roast, 1 pound 2.999 4.827 161.0
Total $10.661 $15.806
TABLE 17–2 Computation of the Index for Various Food Items, 2003 = 100
We could begin by computing a simple average of the price indexes for each item using 2003 as the base. The simple index for bread is 138.2, found by using formula (17–1).
P = pt p0
(100) = 1.440 1.042
(100) = 138.2
We compute the simple index for the other items in Table 17–2 similarly. The largest price increase was 81.5% for eggs, and coffee was second at 61.0%.
We can also determine the percentage change in the group of foods by averaging the simple indexes. The formula is:
where Pi refers to the simple index for each of the items and n the number of items. In our example, the index is 148.9, found by:
P = ΣPi n
= 138.2 + … + 161.0
6 =
893.4 6
= 148.9
P = ΣPi n
(17–2)SIMPLE AVERAGE OF THE PRICE RELATIVES
INDEX NUMBERS 629
This indicates that the mean price of the group of food items increased 48.9% from 2003 to 2015.
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 combined 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 negative 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 periods and then determine the index based on the totals. The formula is:
SIMPLE AGGREGATE INDEX P = Σpt Σp0
× 100 (17–3)
This is called a simple aggregate index. The index for the food items on the previous page is found by dividing the sum of the prices in 2015 by the sum of the prices in 2003. The sum of the prices for the base period is $10.661 and for the given period it is $15.806. The simple aggregate index is 148.3. This means that the aggregate group of prices had increased 48.3% in the 13-year period.
P = Σpt Σp0
100 = $15.806 $10.661
(100) = 148.3
Because the value of a simple aggregate index can be influenced by the units of mea- surement, 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. 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 the current price of coffee will drive the index much more than any other item. Therefore, we need a way to appropri- ately “weight” the items according to their relative importance.
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 quantities of the pur- chased items 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:
LO17-3 Compute and interpret a weighted aggregate index.
LASPEYRES PRICE INDEX P = Σpt q0 Σp0q0
× 100 (17–4)
630 CHAPTER 17
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.
Determine a weighted price index using the Laspeyres method. Interpret the result.
S O L U T I O N
First we determine the total amount spent for the six items in the base period, 2003. To find this value, we multiply the base period, 2003, price for bread $1.042 by the base period, 2003, quantity of 50. The result is $52.10. This indicates that a total of $52.10 was spent in the base period on bread. We continue that for all items and total the results. The base period total is $493.86. The current period total is computed in a similar fashion. For the first item, bread, we multiply the quan- tity in 2003 by the price of bread in 2015, that is, $1.440(50). The result is $72.00. We make the same calculation for each item and total the results. The total is $683.68. Because of the repetitive nature of these calculations, a spreadsheet is very useful for creating the following table.
2003 2015
Item Price Quantity Price Quantity
Bread, white, cost per pound $1.042 50 $1.440 55 Eggs, dozen 1.175 26 2.133 20 Milk, gallon, white 2.686 102 3.463 130 Apples, Red Delicious, 1 pound 0.911 30 1.265 40 Orange juice, 12 oz. concentrate 1.848 40 2.678 41 Coffee, 100% ground roast, 1 pound 2.999 12 4.827 12
TABLE 17–3 Price and Quantity of Food Items in 2003 and 2015
Laspeyres Index
2003 03 Price* 2015 15 Price*
Item Price Quantity 03 Quantity Price Quantity 03 Quantity
Bread, white, cost per pound $1.042 50 $ 52.10 $1.440 55 $ 72.00 Eggs, dozen $1.175 26 $ 30.55 $2.133 20 $ 55.46 Milk, gallon, white $2.686 102 $273.97 $3.463 130 $353.23 Apples, Red Delicious, 1 pound $0.911 30 $ 27.33 $1.265 40 $ 37.95 Orange Juice, 12 oz concentrate $1.848 40 $ 73.92 $2.678 41 $107.12 Coffee, 100% ground roast, 1 pound $2.999 12 $ 35.99 $4.827 12 $ 57.92
$493.86 $683.68 Index 138.44
E X A M P L E
The prices for the six food items from Table 17–2 are repeated on the following page in Table 17–3. Also included is the typical number of units consumed by a family in 2003 and 2015.
INDEX NUMBERS 631
The weighted price index for 2015 is 138.44, found by
P = Σpt q0 Σp0 q0
100 = $683.68 $493.86
(100) = 138.44
Based on this analysis, we conclude that the price of this group of items has in- creased 38.44% in the 12 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% of the weight in determining the in- dex. In the Laspeyres index, the item with the most weight is milk because the product of the price and the units sold is the largest.
Paasche Price Index The major disadvantage of the Laspeyres index is it assumes that the base-period quan- tities are still realistic in the given period. That is, the quantities used for the six items are about the same in 2003 as 2015. In this case, the quantity of eggs purchased declined by 23%, the quantity of milk increased by nearly 28%, and the quantity of apples in- creased by 33%.
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 2003 prices and the 2015 quantities. This has the advan- tage of using the more recent quantities. If there has been a change in the quantities consumed since the base period, such a change is reflected in the Paasche index.
PAASCHE PRICE INDEX P = Σpt qt Σp0qt
× 100 (17–5)
Paasche Index
2003 03 Price* 2015 15 Price*
Item Price Quantity 13 Quantity Price Quantity 13 Quantity
Bread, white, cost per pound $1.04 50 $ 57.31 $1.44 55 $ 79.20 Eggs, dozen $1.18 26 $ 23.50 $2.13 20 $ 42.66 Milk, gallon, white $2.69 102 $349.18 $3.46 130 $450.19 Apples, Red Delicious, 1 pound $0.91 30 $ 36.44 $1.27 40 $ 50.60 Orange Juice, 12 oz concentrate $1.85 40 $ 75.77 $2.68 41 $109.80 Coffee, 100% ground roast, 1 pound $3.00 12 $ 35.99 $4.83 12 $ 57.92
$578.19 $790.37 Index 136.7
E X A M P L E
Use the information from Table 17–3 to determine the Paasche index. Discuss which of the indexes should be used.
S O L U T I O N
The following table shows the calculations to determine the Paasche index. Be- cause of the repetitive calculations, Excel is very useful to compute the index.
632 CHAPTER 17
The Paasche index is 136.7, found by
P = Σpt qt Σp0qt
100 = $790.37 $578.19
(100) = 136.70
This result indicates that there has been an increase of 36.7% in the price of this “market basket” of goods between 2003 and 2015. That is, it costs 36.7% more to purchase these items in 2015 than it did in 2003.
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 cur-
rent 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 product of prices and quantities to be recomputed each year.
Fisher’s Ideal Index As noted earlier, Laspeyres’ index tends to overweight goods whose prices have in- creased. Paasche’s index, on the other hand, tends to overweight goods whose prices have decreased. 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 de- scribed the geometric mean in Chapter 3. It is determined by taking the kth root of the product of k positive numbers.
Fisher’s ideal index = √(Laspeyres index) (Paasche index) (17–6)
Fisher’s index seems to be theoretically ideal because it combines the best features of the Laspeyres and Paasche indexes. That is, it balances the effects of the two in- dexes. However, it is rarely used in practice because it has the same basic set of prob- lems as the Paasche index. It requires that a new set of quantities be determined for each period.
E X A M P L E
Determine Fisher’s ideal index for the data in Table 17–3.
S O L U T I O N
Fisher’s ideal index is 137.57.
Fisher’s ideal index = √(Laspeyres index) (Paasche index)
= √(138.44) (136.70) = 137.57
INDEX NUMBERS 633
Construct an index of clothing prices for 2016 based on 2000. The clothing items consid- ered 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.
S E L F - R E V I E W 17–2
(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.
2000 2016
Item Price Quantity Price Quantity
Dress (each) $75 500 $85 520 Shoes (pair) 40 1,200 45 1,300
For exercises 5–8: a. Determine the simple price indexes. b. Determine the simple aggregate price index for the two years. c. Determine Laspeyres price index. d. Determine the Paasche price index. e. Determine Fisher’s ideal index.
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 2017. Also included are the quantity purchased. Use August 2000 as the base.
August 2000 August 2017
Item Price Quantity Price Quantity
Toothpaste $2.49 6 $3.35 6 Shampoo 3.29 4 4.49 5 Cough drops 1.59 2 4.19 3 Antiperspirant 1.79 3 2.49 4
6. Fruit prices and the amounts consumed for 2000 and 2017 are below. Use 2000 as the base.
2000 2017
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 stamping plant are reported below. Use 2000 as the base.
E X E R C I S E S
634 CHAPTER 17
2000 2017
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 2017 for Kinzua Valley Geriatrics. Use 2000 as the base period.
2000 2017
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 Copier paper (box) 12.00 30 13.65 40
Value Index A value index measures changes in both the price and quantities involved. A value in- dex, 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:
VALUE INDEX V = Σpt qt Σp0q0
× 100 (17–7)
E X A M P L E
The prices and quantities sold at the Waleska Clothing Emporium for ties, suits, and shoes for May 2000 and May 2017 are:
2000 2017
Price Quantity Price Quantity Item po qo pt qt Ties (each) $ 1.00 1,000 $ 2 900 Suits (each) 30.00 100 40 120 Shoes (pair) 10.00 500 8 500
What is the index of value for May 2017 using May 2000 as the base period?
S O L U T I O N
Total sales in May 2017 were $10,600 and the comparable figure for 2000 is $9,000. (See Table 17–4.) Thus, the value index for May 2017 using 2000 = 100 is 117.8. The value of apparel sales in 2017 was 117.8% of the 2000 sales.
INDEX NUMBERS 635
To put it another way, the value of apparel sales increased 17.8% from May 2000 to May 2017.
V = Σpt qt Σp0q0
(100) = $10,600 $9,000
(100) = 117.8
TABLE 17–4 Construction of Value Index for 2017 (2000 = 100)
2000 2017
Price Quantity $ Price Quantity $ Item po qo poqo pt qt ptqt Ties (each) $ 1.00 1,000 $1,000 $ 2.00 900 $ 1,800.00 Suits (each) 30.00 100 3,000 40.00 120 4,800.00 Shoes (pair) 10.00 500 5,000 8.00 500 4,000.00
Total $9,000 Index = 117.8 $10,600.00
9. The prices and production of grains for 2002 and 2015 (http://www.ers. usda.gov) are listed below.
2002 2015 Production Production 2002 Price (millions of 2015 Price (millions of Grain Per Bushel bushels) Per Bushel bushels)
Oats $1.81 116 $2.09 90 Wheat 3.56 2 5.99 2 Corn 2.32 8,967 3.65 13,601 Barley 2.72 227 5.53 214
Using 2002 as the base period, find the value index of grains produced for 2015.
E X E R C I S E S
The number of items produced by Houghton Products for 2001 and 2017 and the whole- sale prices for the two periods are:
S E L F - R E V I E W 17–3
(a) Find the value index of production for 2017 using 2001 as the base period. (b) Interpret the index.
Price Number Produced
Item Produced 2001 2017 2001 2017
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
636 CHAPTER 17
SPECIAL-PURPOSE INDEXES Many important indexes are prepared and published by private organizations. J. D. Power & Associates surveys automobile purchasers to determine how satisfied custom- ers are with their vehicle after one year of ownership. This special index is called the Consumer Satisfaction Index. Financial institutions, utility companies, and university re- search 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 in- dexes prepared? An example will explain the details.
LO17-4 List and describe special- purpose indexes.
After review and consultation, the director assigned weights of 40% to depart- ment store sales, 30% to employment, 10% to freight car loadings, and 20% to exports. To develop the General Business Activity Index of the Northwest for
Department Index of Freight Car Year Store Sales Employment Loadings Exports
2005 20 100 50 500 2010 41 110 30 900 2016 44 125 18 700
TABLE 17–5 Data for Computation of the General Business Activity Index of the Northwest
10. Johnson Wholesale Company manufactures a variety of products. The prices and quantities produced for April 2000 and April 2017 are:
2000 2017 2000 2017 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 2000 as the base period, find the value index of goods produced for April 2017.
E X A M P L E
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.
S O L U T I O N
After considerable thought and research, the director concluded that four factors should be considered: the regional department store sales (which are reported in $ millions), the regional employment index (which has a 2005 base and is reported by the State of Washington), the freight car loadings (reported in millions), and ex- ports for the Seattle Harbor (reported in thousands of tons). Table 17–5 shows this information for years 2005, 2010, and 2016.
INDEX NUMBERS 637
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 http://www.bls.gov, then under Data Tools select Inflation and Prices, then select All Urban Consumers (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.
2016 using 2005 = 100, each 2016 value is expressed as a percentage, with the base-period value as the denominator. For illustration, department store sales for 2016 are converted to a percentage by ($44/$20)(100) = 220. This means that department store sales have increased 120% in the period. This per- centage 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 2010 and 2016 are shown below.
2010 2016
Department store sales [($41)∕($20)][100][.40] = 82.0 [($44)∕($20)][100][.40] = 88.0 Index of 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 Experts [(900)∕(500)][100][.20] = 36.0 [(700)∕(500)][100][.20] = 28.0 Total 157.0 157.1
The General Business Activity Index of the Northwest for 2010 is 157.0 and for 2016 it is 157.1. Interpreting, business activity has increased 57.0% from 2005 to 2010 and 57.1% from the base period of 2005 to 2016.
638 CHAPTER 17
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 because, over time, stocks split, companies merged, and stocks have been added or dropped. When changes in the 30 selected stocks occur, adjustments are made in the denominator used to compute 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. More information about the Dow Jones Industrial Average is available by going to the website: http://www.dowjones.com. You can find the current value of the DJIA by going to http://www.marketwatch.com. The default is to show the change in the Dow, NASDAQ, S&P 500, and others for the current day.
Producer Price Index Formerly called the Wholesale Price Index, it dates back to 1890 and is also published by the U.S. Bureau of Labor Statistics. It reflects the prices of over 3,400 commodities. 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 ac- cess this information, go to http://www.bls.gov, then Data Tools. Select Inflation and Prices, then select Commodity Data, Top Picks, and finally select Finished Goods. You may select to include different periods. Below is a recent output.
© Image Ideas Inc./Picture Quest
INDEX NUMBERS 639
By clicking on DJIA F you can find additional detail on changes. At the bottom center of the chart you can change the horizontal scale to show the changes for a day, 5 days, and so on up to 5 years. In this case we changed the scale to show the changes for a year. Information on the NASDAQ and the S&P 500 is available by clicking on them as well.
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 a weight of .10. The base period is 2006.
Year Cotton Price Automobiles Sold Money Turnover
2006 $0.20 1,000 80 2011 0.25 1,200 90 2016 0.50 900 75
(a) Construct the index for 2011 and 2016. (b) Interpret the index for 2011 and 2016.
S E L F - R E V I E W 17–4
640 CHAPTER 17
11. The index of leading economic indicators, compiled and published by the U.S. National Bureau of Economic Research, is composed of 12 time series, such as the average work hours of production in manufacturing, manufacturers’ new orders, and money supply. This index and similar indexes are designed to move up or down before the economy begins to move the same way. Thus, an economist has statistical evidence to forecast future trends.
You want to construct a leading indicator for Erie County in upstate New York. The index is to be based on 2000 data. Because of the time and work involved, you decide to use only four time series. As an experiment, you select these four series: unemployment in the county, a composite index of county stock prices, the County Price Index, and retail sales. Here are the figures for 2000 and 2016.
2000 2016
Unemployment rate (percent) 5.3 6.8 Composite county stocks 265.88 362.26 County Price Index (1982 = 100) 109.6 125.0 Retail sales ($ millions) 529,917.0 622,864.0
The weights you assign are unemployment rate 20%, stock prices 40%, County Price Index 25%, and retail sales 15%.
a. Using 2000 as the base period, construct a leading economic indicator for 2016.
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 fore- cast future economic trends in the state. You decide that several key factors should be included in the index: number of new businesses started during the year, num- ber of business failures, state income tax receipts, college enrollment, and the state sales tax receipts. Here are the data for 2000 and 2016.
2000 2016
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 2016. c. Interpret the indexes.
E X E R C I S E S
CONSUMER PRICE INDEX Frequent mention has been made of the Consumer Price Index (CPI) in the preceding 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 Statistics began publishing CPIs for two groups of the population. One index, called the Consumer Price Index—All Urban Consumers, covers about 87% of the total population. The other index is for urban wage earners and clerical workers and covers about 32% of the population.
LO17-5 Apply the Consumer Price Index.
INDEX NUMBERS 641
The CPI serves several major functions. It allows consumers to determine 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 payments 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 is based on the prices of 80,000 items collected monthly by about 250 agents. Prices are collected from thousands of U.S. retail stores, service establishments, rental units, and doctors’ offices (http://stats.bls.gov/cpi/cpifaq.htm). Bread, beer, gas- oline, 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 “market basket” of goods and services that a typical consumer purchases.
The CPI originated in 1913 and has been published regularly since 1921. The stan- dard reference period (the base period) has been updated periodically. The current 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 purchasing pat- terns that determine the “market basket” of goods and services change dramatically, and these changes must be reflected in the base period prices.
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 March 2016.
Item CPI-U
All items 238.132 Food and beverage 247.677 Apparel 127.427 Transportation 191.257 Medical care 458.620 Housing 241.485
A review of this list shows that a weighted index of all items has increased 138.132% since 1982–84; medical care has increased the most, 358.620%; and ap- parel went up the least, 27.427%.
Special Uses of the Consumer Price Index In addition to measuring changes in the prices of goods and services, both consumer price indexes have a number of other applications. The CPI is used to determine real disposable personal income, to deflate sales or other variables, to find the 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, as- sume 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 dou- bled 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 17–6 for computations.) In general:
STATISTICS IN ACTION
Does it seem that prices only increase? The Con- sumer Price Index, com- puted and reported by the U.S. Department of Labor, is a relative measure of price changes. It shows in- teresting price information for categories of products and services. For example, did you know that the CPI for personal computers and peripheral equipment was 45.269 in March 2016? Using a base of December 2007, this means that rela- tive prices for computers and peripherals have de- creased about 55% since December 2007.
REAL INCOME Real income = Money income
CPI × 100 (17–8)
642 CHAPTER 17
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 con- stant dollars. Thus, in Table 17–6, to determine whether Ms. Watts’ standard of living changed, her money income was converted to constant dollars. We found that her pur- chasing power, expressed in 1982–84 dollars (constant dollars), remained at $20,000.
Consumer Annual Price Index Computation Year Money Income (1982–84 = 100) of Real Income Real Income
1982–84 $20,000 100 $20,000
100 (100) $20,000
Present year 40,000 200 $40,000
200 (100) 20,000
TABLE 17–6 Computation of Real Income for 1982–84 and Present Year
The take-home pay of Jon Greene and the CPI for 2000 and 2016 are:
Take-Home CPI Year Pay (1982–84 = 100) 2000 $25,000 170.8 2016 41,200 238.132
(a) What was Jon’s real income in 2000? (b) What was his real income in 2016? (c) Interpret your findings.
Deflating Sales A price index can also be used to “deflate” sales or similar money series. Deflated sales are determined by
S E L F - R E V I E W 17–5
USING AN INDEX AS A DEFLATOR Deflated sales =
Actual sales An appropriate index
× 100 (17–9)
E X A M P L E
The sales of Hill Enterprises, a small injection molding company in upstate New York, increased from 1982 to 2015. The following table shows the increase.
Year $ Sales
1982 875,000 1990 1,482,000 1995 1,491,000 2000 1,502,000 2005 1.515,000 2010 1,596,000 2015 1,697,000
The owner, Harry Hill, realizes that the price of raw materials used in the process also has 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 1990, 1995, 2000, 2005, 2010, and 2015 expressed in constant 1982 dollars?
INDEX NUMBERS 643
E X A M P L E
Suppose the Consumer Price Index this month is 200.0 (1982–84 = 100). What is the purchasing power of the dollar?
S O L U T I O N
From formula (17–10), it is 50 cents, found by:
Purchasing power of dollar = $1
200.0 (100) = $0.50
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.
S O L U T I O N
The Producer Price Index (PPI) is an index released every month and published in the Monthly Labor Review; it is also available at the Bureau of Labor Statistics web- site. The prices included in the PPI reflect the prices the manufacturer pays for the metals, rubber, and other purchased raw materials. So the PPI seems an appropri- ate index to use to deflate the manufacturer’s sales. The manufacturer’s sales are listed in the second column of Table 17–7, 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.
TABLE 17–7 Calculation of Deflated Sales for Hill Enterprises
Year Sales PPI Constant Dollars Found by
1982 $ 875,000 100.0 $ 875,000.00 ($875,000/100.0)*(100) 1990 1,482,000 119.2 1.243.288.59 ($1,482,000/119.2)*(100) 1995 1,491,000 127.9 1,165,754.50 ($1,491,000/127.9)*(100) 2000 1,502,000 138.0 1,088,405.80 ($1,502,000/138.0)*(100) 2005 1,515,000 155.7 973,025.05 ($1,515,000/155.7)*(100) 2010 1,596,000 179.8 887,652.95 ($1,596,000/179.8)*(100) 2015 1,697,000 193.9 875,193.40 ($1,697,000/193.9)*(100)
Sales increased from 1982 through 2015; in fact, they increased by 93.9% [($1,697,000/$875,000)*100] − 100.0. However, if we compare the sales in 1982 and 2015, they are nearly the same, $875,000 versus $875,193.
Purchasing Power of the Dollar The Consumer Price Index is also used to deter- mine the purchasing power of the dollar.
USING AN INDEX TO FIND PURCHASING POWER Purchasing power of dollar =
$1 CPI
× 100 (17–10)
Cost-of-Living Adjustments The Consumer Price Index (CPI) is also the basis for cost-of-living adjustments, or COLAs, in many management–union contracts. The spe- cific clause in the contract is often referred to as the “escalator clause.” About 31 million
644 CHAPTER 17
Social Security beneficiaries, 2.5 million retired military and federal civil service employ- ees 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. A retiree receives a pension of $500 a month and the CPI increases 5 points from 165 to 170. Suppose for each percentage point the CPI in- creases, the pension benefits increase 1 percent, so the monthly increase in benefits will be $15.15, found by [(170 − 165)/165] × 500. Now the retiree will receive $515.15 per month.
The Consumer Price Index for the latest month is 238.132 (1982–84 = 100). What is the purchasing power of the dollar? Interpret.
S E L F - R E V I E W 17–6
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, apparel and upkeep, and medical care since the base pe- riod, 1982–84. Note in Table 17–8 that all of the consumer price indexes use the same base.
Year All Items Food and Beverages Housing Apparel Upkeep Medical 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 152.4 148.9 148.5 132 220.5 2000 172.2 168.4 169.6 129.6 260.8 2005 195.3 191.2 195.7 119.5 323.2 2010 218.056 219.984 216.256 119.503 388.436 2015 237.017 246.804 238.060 125.903 446.752
TABLE 17–8 Trend in Consumer Price to 2015 (1982–84 = 100)
Table 17–8 shows that the price of all consumer items combined increased 137.017% from the base period (1982–84) to the year 2015. (Beginning with Jan- uary 2007, the CPI is reported to three decimal places instead of one.) Likewise, food and beverage prices increased 146.804%, housing prices increased 138.060%, apparel and upkeep increased 25.903%, and medical care increased 346.752%.
A problem arises, however, when comparing two or more series that do not have the same base period. The following example compares the two most widely reported stock market indexes, the DJIA and NASDAQ.
E X A M P L E
We want to compare the opening prices of the Dow Jones Industrial Average (DJIA) and the NASDAQ Composite on the first trading day of the year for 2004 through 2016.
INDEX NUMBERS 645
Opening Price
Date DJIA NASDAQ
1/2/04 $10,452.74 $2,011.08 1/3/05 10,783.75 2,184.75 1/5/06 10,718.30 2,216.53 1/3/07 12,459.54 2,429.72 1/2/08 13,261.82 2,653.91 1/2/09 8,772.25 1,578.87 1/4/10 10,430.69 2,294.41 1/3/11 11,577.43 2,676.65 1/3/12 12,221.19 2,657.39 1/2/13 13,104.30 3,091.33 1/2/14 16,572.17 4160.03 1/2/15 17,823.07 4760.24 1/4/16 17,405.48 4897.65
S O L U T I O N
A direct comparison of the DJIA and NASDAQ opening prices is not appropriate. Because we want to compare changes in the opening prices for the two mar- kets, a logical approach is to compute indexes for each market using the 2004 opening price as the base. For the DJIA, the base is $10,452.74, and for the NASDAQ it is $2,011.08.
The calculation of the index for the DJIA in 2016 is:
Index = $17,405.48 $10,452.74
(100) = 166.52
The following table reports the complete set of indexes.
DJIA NASDAQ
Date Value Index Value Index
1/2/04 $10,452.74 100.0 $2,011.08 100.0 1/3/05 10,783.75 103.2 2,184.75 108.6 1/5/06 10,718.30 102.5 2,216.53 110.2 1/3/07 12,459.54 119.2 2,429.72 120.8 1/2/08 13,261.82 126.9 2,653.91 132.0 1/2/09 8,772.25 83.9 1,578.87 78.5 1/4/10 10,430.69 99.8 2,294.41 114.1 1/3/11 11,577.43 110.8 2,676.65 133.1 1/3/12 12,221.19 116.9 2,657.39 132.1 1/2/13 13,104.30 125.4 3,091.33 153.7 1/2/14 16,572.17 158.5 4160.03 206.9 1/2/15 17,823.07 170.5 4760.24 236.7 1/4/16 17,405.48 166.5 4897.65 243.5
We conclude that both indexes have increased over the period. The DJIA has increased 66.5% and the NASDAQ 143.5%.
The following chart shows the DJIA indexes in blue and NASDAQ indexes in brown. The graph shows the changes for both indexes starting from the base of January 2, 2004. From this graph we conclude that the NASDAQ reached its high at the start of the year 2016. The high for the Dow was in January 2016. In general, the two indexes seem to mirror each other rather closely. We should point out that
646 CHAPTER 17
The changes in industrial production and in the prices manufacturers paid for raw materials are to be compared. Unfortunately, the index of industrial production, which measures changes in production, and the Producer Price Index for raw materials have different base periods. The production index has a 2012 base period and the Producer Price Index uses 1982 as the base period. The following table reports the value of each index at the start of the year. Compare the two series using 2010 as the base. Interpret the results.
Industrial Production Producer Price Year Index (2012 = 100) Index (1982 = 100) 2007 104.98 162.9 2008 101.25 175.8 2009 89.60 167.1 2010 94.52 175.4 2011 97.28 189.1 2012 100.00 193.9 2013 101.91 193.9 2014 104.89 195.5 2015 105.22 185.6
S E L F - R E V I E W 17–7
if we select different periods as the base, the results may not be exactly the same. One should always be cautious of the base period selected for a chart or a graph.
0
20 04
20 05
20 06
20 07
20 08
20 09
20 10
20 11
20 13
20 14
20 15
20 12
20 16
50
100
150
200
250
300 DJIA
Line Chart Comparing Indexes for the DJIA and NASDAQ
In de
x
NASDAQ
13. In April 2013, the mean salary for a nurse manager with a bachelor’s degree was $89,673. The Consumer Price Index for March 2016 was 238.132 (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 March 2016? 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 2016.
Year Plumbers (1995 = 100) Electricians (1998 = 100) 2000 133.8 126.0 2016 159.4 158.7
E X E R C I S E S
INDEX NUMBERS 647
15. In 2000, the mean salary of classroom teachers in Tinora School District was $28,650. By 2010, the mean salary increased to $33,972, and further increased in 2016 to $37,382. The American Federation of Classroom Teachers maintains infor- mation on the trends throughout the United States in classroom teacher salaries. Its index, which has a base period of 2000, was 122.5 for 2010 and 136.9 for 2016. 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 2013 and 2016. Also included is an industry in- dex for Web page designers that reports the rate of wage inflation in the industry. This index has a base period of 2000.
Year Wage ($000) Index (2000 = 100) 2013 134.8 160.6 2014 145.2 173.6 2015 156.6 187.9 2016 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?
C H A P T E R S U M M A R Y
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.
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.
P = pt p0
× 100 (17–1)
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.
P = ΣPi n
(17–2)
3. In a simple aggregate price index, the price of the items in the group are totaled for both periods and compared.
P = Σpt Σp0
× 100 (17–3)
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.
P = Σpt q0 Σp0 q0
× 100 (17–4)
648 CHAPTER 17
2. In the Paasche method, current period quantities are used.
P = Σpt qt Σp0qt
× 100 (17–5)
3. Fisher’s ideal index is the geometric mean of the Laspeyres and Paasche indexes.
Fisher’s ideal index = √(Laspeyres index) (Paasche index) (17–6)
C. A value index uses both base period and current period prices and quantities.
V = Σptqt Σp0q0
× 100 (17–7)
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. CPI is used to compute “real” income and purchasing power and to adjust pensions
and tax brackets.
C H A P T E R E X E R C I S E S
For exercises 17 through 22, use the following information 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.
Domestic Sales International Sales Employees Year ($ million) ($ million) (thousands)
2000 17,316 11,856 100.9 2001 19,825 12,492 101.8 2002 22,455 13,843 108.3 2003 25,274 16,588 110.6 2004 27,770 19,578 109.9 2005 28,377 22,137 115.6 2006 29,775 23,549 122.2 2007 32,444 28,651 119.2 2008 32,309 31,438 118.7 2009 30,889 31,008 115.5 2010 29,437 32,124 114.0 2011 28,907 36,107 117.9 2012 29,830 37,394 127.6 2013 31,910 39,402 128.1 2014 34,782 39,548 126.5 2015 35,687 34,387 127.1
17. Using 2000 as the base period, compute a simple index of domestic sales for each year from 2004 until 2015. Interpret the trend in domestic sales.
18. Using the period 2000–02 as the base period, compute a simple index of domes- tic sales for each year from 2004 to 2015.
19. Using 2000 as the base period, compute a simple index of international sales for each year from 2004 until 2015. Interpret the trend in international sales.
20. Using the period 2000–02 as the base period, compute a simple index of interna- tional sales for each year from 2004 to 2015.
21. Using 2000 as the base period, compute a simple index of the number of employ- ees for each year from 2004 until 2015. Interpret the trend in the number of employees.
22. Using the period 2000–02 as the base period, compute a simple index of the number of employees for each year from 2004 to 2015.
INDEX NUMBERS 649
For exercises 23 through 26, use the following information from General Electric Corpo- ration’s annual reports.
Revenue Employees Year ($ million) (000)
2004 134 325 2005 152 307 2006 157 316 2007 168 319 2008 177 327 2009 183 323
Revenue Employees Year ($ million) (000)
2010 150 304 2011 147 287 2012 147 301 2013 146 307 2014 149 305 2015 151 333
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 em- ployees over the period?
For exercises 27 through 32, use the following information on food items for the years 2000 and 2016.
2000 2016
Item Price Quantity Price Quantity
Margarine (pound) $0.81 18 $2.00 27 Shortening (pound) 0.84 5 1.88 9 Milk (½ gallon) 1.44 70 2.89 65 Potato chips 2.91 27 3.99 33
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 2016 using 2000 as the base period. 30. Compute Paasche’s index for 2016 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 2016 using 2000 as the base period.
For exercises 33 through 38, use the following information. Betts Electronics purchases three replacement parts for robotic machines used in its manufacturing process. Infor- mation on the price of the replacement parts and the quantity purchased is given below.
Price Quantity
Part 2000 2016 2000 2016
RC-33 $0.50 $0.60 320 340 SM-14 1.20 0.90 110 130 WC50 0.85 1.00 230 250
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 2016. Use 2000 as the base period.
650 CHAPTER 17
35. Compute Laspeyres’ price index for 2016 using 2000 as the base period. 36. Compute Paasche’s index for 2016 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 2016 using 2000 as the base period.
For exercises 39 through 44, use the following price information for selected foods for 2000 and 2016 given in the following table.
Price Quantity
Item 2000 2016 2000 2016
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
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 2016 using 2000 as the base period. 42. Compute Paasche’s index for 2016 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 2016 using 2000 as the base period.
For exercises 45 through 50, use the following price information for selected items for 1990 and 2016. Production figures for those two periods are also given.
Price Quantity
Item 1990 2016 1990 2016
Aluminum (cents per pound) $ 0.287 $ 0.73 1,000 1,200 Natural gas (1,000 cu. ft.) 0.17 2.12 5,000 4,000 Petroleum (barrel) 3.18 44.08 60,000 60,000 Platinum (troy ounce) 133.00 1,904.30 500 600
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 2016 using 1990 as the base period. 48. Compute Paasche’s index for 2016 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 2016 using 1990 as the base period. 51. A special-purpose index is to be designed to monitor the overall economy of the
Southwest. Four key series were selected. After considerable deliberation, it was de- cided to weight retail sales 20%, total bank deposits 10%, industrial production in the area 40%, and nonagricultural employment 30%. The data for 1996 and 2016 are:
Bank Industrial Retail Sales Deposits Production Year ($ millions) ($ billions) (1990 = 100) Employment 1996 1,159.0 87 110.6 1,214,000 2016 1,971.0 91 114.7 1,501,000
Construct a special-purpose index for 2016 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
INDEX NUMBERS 651
following table that the CPI has a base period of 1967, employment is in millions of per- sons, 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
53. The management of Ingalls Super Discount stores, with several stores in the Okla- homa 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, inventory 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%, bank depos- its 35%, and employment 25%. Seasonally adjusted data for the first 3 months of the year are:
Retail Bank Deposits Employment Month Sales ($ millions) ($ billions) (thousands)
January 8.0 20 300 February 6.8 23 303 March 6.4 21 297
Construct an index of economic activity for each of the 3 months, using January as the base period.
54. The following table gives information on the Consumer Price Index (Base = 1982–84) and the monthly take-home pay of Bill Martin, an employee at Ford Motor Corporation.
Consumer Price Index Mr. Martin’s Monthly Year (1982–84 = 100) Take-Home Pay 1982–84 100.0 $ 600 2016 238.132 7,000
a. What is the purchasing power of the dollar in 2016, based on the period 1982–84? b. Determine Mr. Martin’s “real” monthly income for 2016.
55. Suppose that the Producer Price Index and the sales of Hoskin’s Wholesale Distributors for 2006 and 2016 are:
Year Producer Price Index Sales
2006 160.5 $2,400,000 2016 109.7 3,500,000
What are Hoskin’s real sales (also called deflated sales) for the 2 years?
652 CHAPTER 17
D A T A A N A L Y T I C S
(The data for this exercise is available at the text website: www.mhhe.com/lind17e.)
56. Refer to the Baseball 2016 data, which include information on the 2016 Major League Baseball season. The data also include the mean player salary since 1989. a. Use the year 2000 salary as the base period and 100 as the base value to develop a
simple index for the years since 2000. By what percent has the typical salary increased?
b. The Consumer Price Index values for 2001, 2003, 2011, 2012, and 2016 are 177.1, 184.0, 224.94, 229.594, and 238.132, respectively. What are the average player’s real (or deflated) salaries for those years? Describe the trend in the deflated salaries in a few sentences. Compare these results with your reply in part (a).
Time Series and Forecasting 18
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO18-1 Define and describe the components of a time series.
LO18-2 Smooth a time series by computing a moving average.
LO18-3 Smooth a time series by computing a weighted moving average.
LO18-4 Use regression analysis to fit a linear trend line to a time series.
LO18-5 Use regression analysis to fit a nonlinear time series.
LO18-6 Compute and apply seasonal indexes to make seasonally adjusted forecasts.
LO18-7 Deseasonalize a time series using seasonal indexes.
LO18-8 Conduct a hypothesis test of autocorrelation.
© Bob Levey/Getty Images
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 LO18-6.)
654 CHAPTER 18
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 daily reports of the Dow Jones Industrial Average over the last three months.
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 patterns 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 essential 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 dictated by the very nature of business organizations in the United States and around the world. Competition for the consum- er’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 of the future is necessary to plan for the raw materials, production facilities, and staff needed to meet the pro- jected demand.
This chapter shows how to use time-series data to forecast future events. First, we look at the components of a time series. Then, we examine some of the tech- niques used to analyze time-series data. Finally, we use these techniques to forecast future events.
COMPONENTS OF A TIME SERIES There are four components to a time series: secular trend, cyclical variation, seasonal variation, and irregular variation.
Secular Trend The trend of sales, unemployment, stock prices, and other business and economic series follow various patterns. Some move steadily upward, others decline, and still others stay the same over time. The change over time may be linear and follow a straight line, or it may increase at an exponential rate. The long-run change (or lack of change) is called the trend of the time series or, more precisely, the secular trend.
SECULAR TREND The smoothed long-term direction of a time series.
The following are several examples of a secular trend.
• Home Depot, Inc. was founded in 1978 and is the world’s largest home improve- ment retailer. The following chart shows the number of associates working for Home Depot, Inc. You can see the number of associates increased from just over 98,000 in 1996 to 364,400 in 2006. Since then, the number of associates decreased from 2007 through 2009. From 2009 through 2015, the number of associates has increased steadily. A time series graph is always used to summa- rize the data.
LO18-1 Define and describe the components of a time series.
© Flying Colors Ltd/Photodisc/Getty Images
TIME SERIES AND FORECASTING 655
1996 0
50 100 150 200 250 300 350 400 450
Number of Associates at Home Depot, Inc. (1996–2015)
1998 2000 2002 2004 2006 2008 2010 2012 2014
Year
As so
ci at
es (0
00 ’s
)
Year Associates
1996 98.1
1997 124.4
1998 156.7
1999 201.4
2000 227.3
2001 256.3
2002 280.9
2003 298.8
2004 323.1
2005 344.8
Year Associates
2006 364.4
2007 331.0
2008 322.0
2009 317.0
2010 321.0
2011 331.0
2012 340.0
2013 365.0
2014 371.0
2015 385.0
• The average price for a gallon of regular gasoline was below $2.00 until 2005. From 2005 until 2013, except for a period between 2009 and 2010, price increased steadily by almost $0.20 per year. Since 2013, price declined. This infor- mation is shown in the time series chart below.
Year Cost/Gallon
1996 1.20
1997 1.20
1998 1.03
1999 1.14
2000 1.48
2001 1.42
2002 1.35
2003 1.56
2004 1.85
2005 2.27
Year Cost/Gallon
2006 2.57
2007 2.80
2008 3.25
2009 2.35
2010 2.78
2011 3.52
2012 3.62
2013 3.49
2014 3.34
2015 2.40 1 99
8 20
00 20
02 20
04 20
06 20
08 20
10 20
12 20
14 19
96 $0.00 $0.50 $1.00 $1.50 $2.00 $2.50 $3.00 $3.50 $4.00
Year
Price/Gallon of Regular Gasoline (1996–2015)
Pr ic
e
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, produc- tion, 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.
CYCLICAL VARIATION The rise and fall of a time series over periods longer than 1 year.
Chart 18–1 shows the annual unit sales of batteries sold by National Battery Retail- ers, Inc. from 1996 through 2016. 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.
656 CHAPTER 18
STATISTICS IN ACTION
Statisticians, economists, and business executives are constantly looking for variables that will forecast the country’s economy. The production of crude oil, price of gold on world mar- kets, and the Dow Jones average, as well as many published government in- dexes, 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 suc- cessful is the price of scrap metal. Why? Scrap metal is the beginning of the manu- facturing chain. When its demand increases, this is an indication that manufac- turing is also increasing.
Seasonal Variation The third component of a time series is the seasonal variation. Many sales, production, and other series fluctuate with the seasons. The unit of time reported is usually quarterly or monthly but could be weekly.
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 women’s apparel, for example, have extremely high sales just prior to Christmas and relatively low sales just after Christmas and during the summer. Beach rental properties are another exam- ple. They have very low rental rates in the winter but during the summer when the tempera- ture is warm, rentals increase significantly. Toy sales are another example with an extreme seasonal pattern. More than half of the annual toy sales usually occur in the months of November and December. The lawn care business is seasonal 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, the operator of a lawn care business will at- tach a snowplow to the front of a truck in an effort to earn income in the off-season. At ski resorts throughout the country, you will often find golf courses nearby. The owners of the lodges attract 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 18–2 shows the quarterly sales, in millions of dollars, of Hercher Sporting 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 quarters 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.
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 initial
Battery Year Sales (000)
1996 24.0 1997 30.0 1998 31.0 1999 26.5 2000 27.0 2001 27.5 2002 34.0 2003 35.0 2004 31.0 2005 32.0 2006 35.5
Battery Year Sales (000)
2007 41.0 2008 42.0 2009 38.0 2010 39.0 2011 46.0 2012 52.0 2013 53.5 2014 47.0 2015 51.0 2016 48.0 1995
0.0
10.0
20.0
30.0
40.0
50.0
60.0
2000 2005
Prosperity Contraction
Long-term secular trend
Depression Recovery
2010 Year
Ba tte
ry S
al es
(0 00
)
2015 2020
CHART 18–1 Battery Sales for National Battery Retailers, Inc. 1996 to 2016
TIME SERIES AND FORECASTING 657
impact on the economy of a major labor 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 or noise, are unpredictable, and they cannot be identified. Of course, neither episodic nor residual variation can be projected into the future.
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 seasonal fluctuation, described later in the chapter. In contrast to the least squares method, which expresses the trend in terms of a mathematical equa- tion ( ŷ = 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 linear 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 ir- regular, abbreviated T, C, and I. There is no seasonal variation because the data are re- corded 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 7 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 7 years (1991–97 inclusive) are $22 million, found by 1 + 2 + 3 + 4 + 5 + 4 + 3. (See Table 18–1.) The total of $22 mil- lion 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, 1994, as shown in Table 18–1. Then the total sales for the next seven years (1992–98 inclusive) are determined. (A convenient way of doing this is to subtract the sales for 1991 [$1 million] from the first seven-year total [$22 million] and add the sales for 1998 [$2 million], to give the new total of $23 million.) The mean of this total, $3.286 million, is positioned opposite the middle year, 1995. The sales data and seven-year moving average are shown graphically in Chart 18–3.
Year Quarter Code Sales (mil)
2014 Q1 1 9.0 Q2 2 10.2 Q3 3 3.5 Q4 4 5.0 2015 Q1 5 12.0 Q2 6 14.1 Q3 7 4.4 Q4 8 7.3 2016 Q1 9 16.0 Q2 10 18.3 Q3 11 4.6 Q4 12 8.6
Q1
Q3 0 2 4 6 8 10 12 14
0.0 2.0 4.0 6.0 8.0
10.0 12.0 14.0 16.0 18.0 20.0
Sa le
s ($
m ill
io n) Trend Line
Quarters
CHART 18–2 Sales of Baseball and Softball Equipment, Hercher Sporting Goods 2014–2016 by Quarter
STATISTICS IN ACTION
Forecasts are not always correct. The reality is that a forecast may just be a best guess as to what will hap- pen. What are the reasons forecasts are not correct? One expert lists eight com- mon errors: (1) failure to carefully examine the assumptions, (2) limited expertise, (3) lack of imagi- nation, (4) neglect of con- straints, (5) excessive optimism, (6) reliance on mechanical extrapolation, (7) premature closure, and (8) overspecification.
LO18-2 Smooth a time series by computing a moving average.
658 CHAPTER 18
TABLE 18–1 Computation of Seven-Year Moving Average
Sales Seven-Year Seven-Year Year ($ Mil) Moving Total Moving Average
1991 1 1992 2 1993 3 1994 4 22 3.143 1995 5 23 3.286 1996 4 24 3.429 1997 3 25 3.571 1998 2 26 3.714 1999 3 27 3.857 2000 4 28 4.000 2001 5 29 4.143 2002 6 30 4.286 2003 5 31 4.429 2004 4 32 4.571 2005 3 33 4.714 2006 4 34 4.857 2007 5 35 5.000 2008 6 36 5.143 2009 7 37 5.286 2010 6 38 5.429 2011 5 39 5.571 2012 4 40 5.714 2013 5 41 5.857 2014 6 2015 7 2016 8
1985 0
1
2
3
4
5
6
7
8
9
1990 1995 2000 Year
Sa le
s
2005 2010 2015 2020
Sales
Seven Year Moving Average
CHART 18–3 Sales and Seven-Year Moving Average
The number of data values to include in a moving average depends on the data collected. If the data are quarterly, then four values is typical because there are four quarters in a year. If the data are daily, then seven values is appropriate because there are 7 days in a week. You might also use trial and error to determine a number that best levels out the chance fluctuations. For example, Table 18–2 and Chart 18–4 show three-year and five-year moving averages for a series of production data.
TIME SERIES AND FORECASTING 659
TABLE 18–2 Production, Three-Year Moving Average, and Five-Year Moving Average
Three-Year Three-Year Five-Year Five-Year Moving Moving Moving Moving Year Production Total Average Total Average
1998 5 1999 6 19 6.33 2000 8 24 8.00 34 6.80 2001 10 23 7.67 32 6.40 2002 5 18 6.00 33 6.60 2003 3 15 5.00 35 7.00 2004 7 20 6.67 37 7.40 2005 10 29 9.67 43 8.60 2006 12 33 11.00 49 9.80 2007 11 32 10.67 55 11.00 2008 9 33 11.00 60 12.00 2009 13 37 12.33 66 13.20 2010 15 46 15.33 70 14.00 2011 18 48 16.00 72 14.40 2012 15 44 14.67 73 14.60 2013 11 40 13.33 75 15.00 2014 14 42 14.00 79 15.80 2015 17 53 17.67 2016 22
1996 0
5
10
15
20
25
1998 2000 2002 2004 2006 Year
Pr od
uc tio
n
Production
3-Year
5-Year
2008 2010 2012 2014 2016 2018
CHART 18–4 Time Series Plot of Production, Three-Year Moving Average, and 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 identical amplitudes. Thus, in practice, the application of a moving average does not result precisely in a line. For example, the production series in Table 18–2 repeats about every 5 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 1998.
660 CHAPTER 18
Four-year, six-year, and other even-numbered-year moving averages present one minor problem regarding the centering of the moving totals and moving averages. Note in Table 18–3 below there is no center time period, so the moving totals are positioned between two time periods. The total for the first 4 years ($42) is positioned between 2009 and 2010. The total for the next 4 years is $43. The averages of the first four years and the second 4 years ($10.50 and $10.75, respectively) are averaged, and the resulting figure is centered on 2010. This procedure is repeated until all possible four- year averages are computed.
TABLE 18–3 Calculations of a Four-Year Moving Average
Four-Year Four-Year Centered Moving Moving Moving Year Sales Total Average Average
2008 8
2009 11 42 (8 + 11 + 9 + 14) 10.50 (42/4) 2010 9 10.625 43 (11 + 9 + 14 + 9) 10.75 (43/4) 2011 14 10.625 42 10.50 2012 9 10.625 43 10.75 2013 10 10.000 37 9.25 2014 10 9.625 40 10.00 2015 8
2016 12
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.
A natural extension of the weighted mean discussed in Chapter 3 is to compute a weighted moving average. This involves selecting a 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 forecast of the future. So, the most recent observation receives the most weight, and the weight decreases for older data values. Of course, the sum of the weights must be equal to 1.
Suppose, for example, we compute a two-year weighted moving average for the data in Table 18–3 giving twice as much weight to the most recent value. In other words, give a weight of two-thirds to the last year and one-third to the value immediately before that. Then “forecast” sales for 2010 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. Proceeding in the same fashion, the final, or 2016, weighted moving average would be (1/3)($10) + (2/3)($8) = $10.667. To summarize the technique of using moving averages, its purpose is to help identify the long-term trend in a time series (because it will smooth out short-term fluctuations). It is used to reveal any cyclical and seasonal fluctuations.
LO18-3 Smooth a time series by computing a weighted moving average.
TIME SERIES AND FORECASTING 661
E X A M P L E
Cedar Fair operates eleven amusement parks, three outdoor water parks, one indoor water park, and five hotels. Its combined attendance (in thousands) for the last 20 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.
© UKRID/Shutterstock.com
S O L U T I O N
The three-year moving average is:
Attendance 3-Year Year (000) Moving Average Found by
1996 7,445 1997 7,405 8766.67 (7,445 + 7,405 + 11,450)/3 1998 11,450 10026.33 (7,405 + 11,450 + 11,224)/3 1999 11,224 11459.00 2000 11,703 11605.67 2001 11,890 11991.00 2002 12,380 12150.33 2003 12,181 12372.67 2004 12,557 12479.33 2005 12,700 14852.33 2006 19,300 18033.33 2007 22,100 21373.33 2008 22,720 21985.33 2009 21,136 22213.67 2010 22,785 22432.67 2011 23,377 23154.00 2012 23,300 23392.33 2013 23,500 23366.67 2014 23,300 23733.33 (23,500 + 23,300 + 24,400)/3 2015 24,400
Attendance Year (000)
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 2010 22,785 2011 23,377 2012 23,300 2013 23,500 2014 23,300 2015 24,400
662 CHAPTER 18
The three-year weighted moving average is:
Attendance 3-Year Weighted Year (000) Moving Average Found by
1996 7,445 1997 7,405 9,435.50 .2(7,445) + .3(7,405) + .5(11,450) 1998 11,450 10,528.00 .2(7,405) + .3(11,450) + .5(11,224) 1999 11,224 11,508.70 2000 11,703 11,700.70 2001 11,890 12,097.60 2002 12,380 12,182.50 2003 12,181 12,408.80 2004 12,557 12,553.30 2005 12,700 15,971.40 2006 19,300 19,380.00 2007 22,100 21,850.00 2008 22,720 21,804.00 2009 21,136 22,277.30 2010 22,785 22,751.20 2011 23,377 23,220.10 2012 23,300 23,415.40 2013 23,500 23,360.00 2014 23,300 23,890.00 .2(23,500) + .3(23,300) + .5(24,400) 2015 24,400
Year
At te
nd an
ce (0
00 ’s
)
1996 1999 2002 2005 2008 2011 2014 0
5,000
10,000
15,000
20,000
25,000
30,000 Attendance (000’s)
3-year moving average
3-year weighted moving average
Attendance, three-year moving average, and three-year weighted moving average for Cedar Fair (1996–2015)
Study the graph carefully. You will see that the attendance trend is evenly upward with approximately 400,000 additional visitors each year. However, there is a “hop” of approximately 3 million per year between 1997 and 1998. At this time, Cedar Fair acquired Knott’s Berry Farm (located in Southern California) in late 1997, leading to a boost in attendance. A similar boost occurred in 2006 with the pur- chase of King’s Island near Cincinnati, Ohio. You can also observe the financial cri- sis in 2009 followed by a steady recovery in attendance.
TIME SERIES AND FORECASTING 663
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.
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)
2011 2 2014 5 2012 6 2015 3 2013 4 2016 10
S E L F - R E V I E W 18–1
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 thousands. Apply weights of .1, .2, .3, and .4, respectively, for the past three quarters and current quarter. In a few words, describe the trend in the number of subscribers.
1st Quarter 2015 28,766 2nd Quarter 2015 30,057 3rd Quarter 2015 31,336 4th Quarter 2015 33,240 1st Quarter 2016 34,610 2nd Quarter 2016 35,102 3rd Quarter 2016 35,308 4th Quarter 2016 35,203 1st Quarter 2017 34,386
2. Listed below is the number of movie tickets sold at the Library Cinema-Complex, in thousands, for the period from 2004 to 2016. Compute a five-year weighted moving average using weights of .1, .1, .2, .3, and .3, respectively. Describe the trend in yield.
2004 8.61 2005 8.14 2006 7.67 2007 6.59 2008 7.37 2009 6.88 2010 6.71
2011 6.61 2012 5.58 2013 5.87 2014 5.94 2015 5.49 2016 5.43
E X E R C I S E S
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 trend is:
LO18-4 Use regression analysis to fit a linear trend line to a time series.
LINEAR TREND EQUATION ŷ = a + bt (18–1)
664 CHAPTER 18
where: ŷ , 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 where the line crosses the y-axis,
or when t = 0. b is the slope of the line, or the average change in ŷ for each increase of one unit in t. t is any value of time that is selected.
To illustrate the meaning of ŷ, a, b, and t in a time-series problem, a line is drawn in Chart 18–5 to represent the typical trend of sales. Assume that this company started in business in 2008. This beginning year (2008) has been arbitrarily coded as time period 1. Note that sales increased an average of $2 million every year; that is, based on the straight line drawn through the sales data, sales increased from $3 million in 2008 to $5 million in 2009, to $7 million in 2010, to $9 million in 2011, and so on. The slope, or b, is therefore 2. The slope is the average change in sales for each unit increase in the time period. Note too that the line intercepts the y-axis (when t = 0, which is the year 2007) at $1 million. This point is a. Another way of determining b is to locate the starting place of the straight line in year 1. It is 3 for 2008 in this example. Then locate the value on the straight line for the last year. It is 19 for 2016.
CHART 18–5 A Straight Line Fitted to Sales Data
Time Period Year Sales Trend Line
1 2008 6 3 2 2009 2 5 3 2010 9 7 4 2011 8 9 5 2012 16 11 6 2013 10 13 7 2014 12 15 8 2015 21 17 9 2016 18 19
0 1
2008 2
2009 3
2010 4
2011 5
2012 6
2013 7
2014 8
2015 9
2016
5
10
15
20
25
Time Year
Sa le
s ($
m ill
io ns
) Sales
Trend
The equation for the line in Chart 18–5 is:
ŷ = 1 + 2t
where: ŷ 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 2007. t is the coded time period for each year.
In Chapter 13, we drew a line through points on a scatter diagram to approximate the regression line. We stressed, however, that this method for determining the regres- sion 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 18–5 might not be the best-fitting line. Because of the subjective judg- ment 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.
TIME SERIES AND FORECASTING 665
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 variables. 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 5 years starting in 2012 and continuing through 2016, we would code the year 2012 as 1, 2013 as 2, and 2016 as 5.
When the least squares method is used to find the trend line for a time series, the errors or residuals are usually correlated and not independent. Therefore, the results of the hypothesis tests for regression analysis presented in Chapter 13 may not be valid in time series analysis. However, simple linear regression can still be used to find a line of best fit for a time series.
STATISTICS IN ACTION
Investors frequently use re- gression analysis to study the relationship between a particular stock and the general condition of the market. The dependent variable is the monthly percentage 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 equa- tion is the particular stock’s beta coefficient, or just the beta. If b is greater than 1, the implication is that the stock is sensitive to market changes. If b is between 0 and 1, the implication is that the stock is not sensi- tive to market changes.
E X A M P L E
The sales of Jensen Foods, a small grocery chain located in southwest Texas, for 2012 through 2016 are:
Year Time (t) Sales ($ million)
2012 1 7.0 2013 2 10.0 2014 3 9.0 2015 4 11.0 2016 5 13.0
Determine the regression equation. How much are sales increasing each year? What is the sales forecast for 2018?
S O L U T I O N
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. Chart 18–6 shows the plot of sales and the fitted or trended sales for each year.
1 2012
2 2013
3 2014
Time Code Year
Sa le
s ($
m ill
io n)
6.0 7.0 8.0 9.0
10.0 11.0 12.0 13.0 14.0
4 2015
5 2016
Trend Equation
y = 6.1 + 1.3tˆ
CHART 18–6 Sales and Trend Line for Jensen Foods (2012 to 2016)
666 CHAPTER 18
From the regression analysis, the trend equation is ŷ = 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 2011, which is called the base year. For example, to determine the point on the line for 2015, insert the t value of 4 in the equation. Then ŷ = 6.1 + 1.3(4) = 11.3.
If sales, production, or other data approximate a linear trend, the equation de- veloped by the least squares technique can be used to estimate future values. For Jensen Foods, the sales data appears to follow a linear trend, so we can use the trend equation to forecast future sales.
See Table 18–4. The year 2012 is coded 1, the year 2014 is coded 3, and year 2016 is coded 5. Logically, we code 2018 as 7 and 2019 as 8. So we substitute 7 into the trend equation and solve for ŷ .
TABLE 18–4 Calculations for Determining the Points on Trend Line
Sales Code Year ($ million) (t) y Found by
2012 7.0 1 7.4 6.1 + 1.3(1) 2013 10.0 2 8.7 6.1 + 1.3(2) 2014 9.0 3 10.0 6.1 + 1.3(3) 2015 11.0 4 11.3 6.1 + 1.3(4) 2016 13.0 5 12.6 6.1 + 1.3(5)
ŷ = 6.1 + 1.3t = 6.1 + 1.3(7) = 15.2
Thus, on the basis of past sales, the estimate for 2018 is $15.2 million. In this time series example, there were 5 years of sales data. Based on those
five sales figures, we estimated sales for 2018. 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.
Annual production of king-size rockers by Wood Products Inc. for 2009 through 2016 follows.
Production Year Time (thousands)
2009 1 4 2010 2 8 2011 3 5 2012 4 8 2013 5 11 2014 6 9 2015 7 11 2016 8 14
(a) Plot the production data. (b) Determine the least squares equation using a software package. (c) Using the least squares equation, determine the points on the line for 2009 and 2016. (d) Based on the linear trend equation, what is the estimated production for 2019?
S E L F - R E V I E W 18–2
TIME SERIES AND FORECASTING 667
3. Listed below is the number of rooms rented at Plantation Resorts of Georgia for the years from 2006 to 2016. Remember to code the years starting at 1 for year 2006.
Year Rental Year Rental Year Rental
2006 6,714 2010 9,762 2014 6,162 2007 7,991 2011 10,180 2015 6,897 2008 9,075 2012 8,334 2016 8,285 2009 9,775 2013 8,272
Determine the least squares equation. According to this information, what is the estimated number of rentals for 2017?
4. Listed below are the net sales in $ million for Home Depot Inc. and its sub- sidiaries from 1993 to 2015. Remember to code the years starting at 1 for year 1993.
Year Net Sales Year Net Sales Year Net Sales
1993 $ 9,239 2001 53,553 2009 66,176 1994 12,477 2002 58,247 2010 67,997 1995 15,470 2003 64,816 2011 70,395 1996 19,535 2004 73,094 2012 74,754 1997 24,156 2005 81,511 2013 78,812 1998 30,219 2006 90,837 2014 83,176 1999 38,434 2007 77,349 2015 88,519 2000 45,738 2008 71,288
Determine the least squares equation. On the basis of this information, what are the estimated sales for 2016 and 2017?
5. The following table lists the annual amounts of glass cullet produced by Kimble Glass Works Inc. for 2013 through 2017.
Scrap Year Code (tons)
2013 1 2 2014 2 4 2015 3 3
Scrap Year Code (tons)
2016 4 5 2017 5 6
Determine the least squares trend equation. Estimate the amount of scrap for the year 2019.
6. The sales by Walker’s Milk and Dairy Products in millions of dollars for the period from 2011 to 2017 are reported in the following table.
Sales Year Code ($ millions)
2011 1 17.5 2012 2 19.0 2013 3 21.0 2014 4 22.7
Sales Year Code ($ millions)
2015 5 24.5 2016 6 26.7 2017 7 27.3
Determine the least squares regression trend equation. Estimate the sales for 2019.
E X E R C I S E S
668 CHAPTER 18
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 that increase (or decrease) by equal percents or proportions over a period of time appear curvilinear. (See Chart 18–7.)
CHART 18–7 Sales for Gulf Shores Importers (2003–2017)
Year Sales ($000)
2003 124.2 2004 175.6 2005 306.9 2006 524.2 2007 714.0 2008 1,052.0 2009 1,638.3 2010 2,463.2
Year Sales ($000)
2011 3,358.2 2012 4,181.3 2013 5,388.5 2014 8,027.4 2015 10,587.2 2016 13,537.4 2017 17,515.6
2002 0.0
2,000.0 4,000.0 6,000.0 8,000.0
10,000.0 12,000.0 14,000.0 16,000.0 18,000.0 20,000.0
2004 2006 Year
Sa le
s (0
00 )
2008 2010 2012 2014 2016 2018
The trend equation for a time series that does approximate a curvilinear trend, such as the one portrayed in Chart 18–7, is computed by using the logarithms of the data and the least squares method. The general equation for the logarithmic trend equation is:
LO18-5 Use regression analysis to fit a nonlinear time series.
LOG TREND EQUATION log ŷ = log a + log b(t) (18–2)
The logarithmic trend equation can be determined for the Gulf Shores Importers data in Chart 18–7 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. That is, use the log of each year’s data, then use the logs as the dependent variable and the coded year as the independent variable.
TIME SERIES AND FORECASTING 669
The regression equation is ŷ = 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 ŷ for each unit increase in t. This value is similar to the geometric mean described 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 geometric mean annual rate of increase from 2003 to 2017. We conclude that imports increased at a rate of 42.35% annually during the period.
We also can use the logarithmic trend equation to make estimates of future val- ues. Suppose we want to estimate the imports in the year 2019. The first step is to determine the code for the year 2019. It is 19. To explain, the year 2017 has a code of 15 and the year 2019 is 4 years later, so 15 + 4 = 19. The log of imports for the year 2019 is
ŷ = 2.053805 + 0.153357t = 2.053805 + 0.153357(19) = 4.967588
To find the estimated imports for the year 2019, we need the antilog of 4.967588. It is 92,809. This is our estimate of the number of imports for 2019. Recall that the data were in thousands of dollars, so the estimate is $92,809,000.
Sales at Tomlin Manufacturing from 2013 to 2017 are:
Sales Year ($ millions)
2013 2.13 2014 18.10 2015 39.80 2016 81.40 2017 112.00
(a) Determine the logarithmic trend equation for the sales data. (b) Sales increased by what percentage annually from 2013–2017? (c) What is the projected sales amount for 2018?
S E L F - R E V I E W 18–3
7. Sally’s Software Inc. is a rapidly growing supplier of computer software to the Sarasota area. Sales for the last 5 years, 2013 to 2017, are given below.
Sales Year ($ millions)
2013 1.1 2014 1.5 2015 2.0 2016 2.4 2017 3.1
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 2020.
E X E R C I S E S
670 CHAPTER 18
SEASONAL VARIATION We mentioned that seasonal variation is another of the components of a time series. Time series data, 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 sea- sonal fluctuations is to have a sufficient 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 pro- duce 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 essential 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 18–5. Each index represents the average sales for a period of several years. The actual sales for some months were above average (which is rep- resented 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% above an average month; the index of 86.0 for July indicates that department store sales for July are typically 14% below an average month.
TABLE 18–5 Typical Seasonal Indexes for U.S. Department Store Sales, Excluding Mail-Order Sales
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
Suppose an enterprising store manager, in an effort to stimulate sales during December, introduced a number of unique promotions, including bands of carolers
LO18-6 Compute and apply seasonal indexes to make seasonally adjusted forecasts.
8. It appears that the imports of carbon black have been increasing by about 10% annually.
Imports of Carbon Black Year (thousands of tons)
2010 92.0 2011 101.0 2012 112.0 2013 124.0
Imports of Carbon Black Year (thousands of tons)
2014 135.0 2015 149.0 2016 163.0 2017 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 2020.
© Goran Bogicevic/Shutterstock.com
TIME SERIES AND FORECASTING 671
strolling through the store singing holiday songs, large mechanical exhibits, and clerks dressed in Santa Claus costumes. When the index of sales was computed 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% below the average for the year. An index of 107.2 for October means that the variable is typically 7.2% above the annual average.
Several methods have been developed to measure the typical seasonal fluctuation in a time series. The method most commonly used to compute the typical seasonal pat- tern 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 variation. 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 the MegaStat add-in for Excel to calculate the seasonal indexes.
E X A M P L E
Table 18–6 shows the quarterly sales for Toys International for the years 2012 through 2017. The sales are reported in millions of dollars. Determine a quarterly seasonal index using the ratio-to-moving-average method.
TABLE 18–6 Toys International Quarterly Sales ($ million), 2012 to 2017
Year Winter Spring Summer Fall
2012 6.7 4.6 10.0 12.7 2013 6.5 4.6 9.8 13.6 2014 6.9 5.0 10.4 14.1 2015 7.0 5.5 10.8 15.0 2016 7.1 5.7 11.1 14.5 2017 8.0 6.2 11.4 14.9
S O L U T I O N
Chart 18–8 depicts the quarterly sales for Toys International over the six-year period. Notice the seasonal nature of the sales. For each year, the fall, 4th-quarter sales are the largest and the spring, 2nd-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 fall quarter sales values. Over the six-year period, the sales in the fall quarter increased. You would expect a similar seasonal pattern for 2018.
672 CHAPTER 18
There are six steps to determining the quarterly seasonal indexes.
Step 1: For the following discussion, refer to Table 18–7. The first step is to determine the four-quarter moving total for 2012. Starting with the winter quarter of 2012, 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 add- ing the spring, summer, and fall sales of 2012 to the winter sales of 2013. The total is $33.8 (million), found by 4.6 + 10.0 + 12.7 + 6.5. This procedure is continued for the quarterly sales for each of the six years. Column 2 of Table 18–7 shows all of the moving totals. Note that the moving total 34.0 is positioned between the spring and sum- mer sales of 2012. The next moving total, 33.8, is positioned be- tween sales for summer and fall 2012, and so on. Check the totals frequently to avoid arithmetic errors.
Step 2: Each quarterly moving total in column 2 is divided by 4 to give the four-quarter moving average. (See column 3.) All the moving aver- ages are still positioned between the quarters. For example, the first moving average (8.500) is positioned between spring and summer 2012.
Step 3: The moving averages are then centered. The first centered moving average is found by (8.500 + 8.450)/2 = 8.475 and centered op- posite summer 2012. The second moving average is found by (8.450 + 8.450)/2 = 8.450. The others are found similarly. Note in column 4 that a centered moving average is positioned on a par- ticular quarter.
Step 4: The specific seasonal index for each quarter is then computed by dividing the sales in column 1 by the centered moving aver- age in column 4. The specific seasonal index reports the ratio of the original time series value to the moving average. To explain further, if the time series is represented by TSCI and the moving average by TC, then, algebraically, if we compute TSCI/TC, the re- sult is the specified seasonal component SI. The specific seasonal index for the summer quarter of 2012 is 1.180, found by 10.0/8.475.
CHART 18–8 Toys International Quarterly Sales in ($ Million), 2012 to 2017
Sales Year Quarter Time ($ Million)
2012 Winter 1 6.7 Spring 2 4.6 Summer 3 10.0 Fall 4 12.7 2013 Winter 5 6.5 Spring 6 4.6 Summer 7 9.8 Fall 8 13.6 2014 Winter 9 6.9 Spring 10 5.0 Summer 11 10.4 Fall 12 14.1
Sales Year Quarter Time ($ Million)
2015 Winter 13 7.0 Spring 14 5.5 Summer 15 10.8 Fall 16 15.0 2016 Winter 17 7.1 Spring 18 5.7 Summer 19 11.1 Fall 20 14.5 2017 Winter 21 8.0 Spring 22 6.2 Summer 23 11.4 Fall 24 14.9
0 0 2 4 6 8
10 12 14 16
4 8 12 Time
Sa le
s ($
m ill
io ns
)
16 20 24
TIME SERIES AND FORECASTING 673
(1) (2) (3) (4) (5) Four-Quarter Centered Sales Four-Quarter Moving Moving Specific Year Quarter ($ millions) Total Average Average Seasonal
2012 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 2013 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 2014 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 2015 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 2016 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 2017 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 18–7 Computations Needed for the Specific Seasonal Indexes
674 CHAPTER 18
Step 5: The specific seasonal indexes are organized in Table 18–8. This table will help us locate the specific seasonals for the corresponding quar- ters. 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 rea- sonable 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.
TABLE 18–8 Calculations Needed for Typical Quarterly Indexes
Year Winter Spring Summer Fall
2012 1.180 1.503 2013 0.772 0.540 1.130 1.550 2014 0.775 0.553 1.141 1.535 2015 0.753 0.581 1.126 1.558 2016 0.733 0.590 1.143 1.466 2017 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
Step 6: The four quarterly means (0.767, 0.576, 1.144, and 1.522) should the- oretically 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 problem, the total of the means is 4.009. A correction factor is therefore applied to each of the four means to force them to total 4.00.
In this example,
Correction factor = 4.00
4.009 = 0.997755
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 18–8 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% 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 FOR ADJUSTING QUARTERLY MEANS
Correction factor = 4.00
Total of four means (18–3)
Statistical software can perform these calculations. For example, the output from the MegaStat add-in for Excel 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 an- swers, due to the number of digits carried in the calculations.
TIME SERIES AND FORECASTING 675
Now we briefly summarize the reasoning underlying the preceding calculations. The original data in column 1 of Table 18–7 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 18–7 are concerned with deriving the centered moving average given in column 4. Basically, we “average out” the seasonal and irregular fluc- tuations 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.
Centered Moving Average and Deseasonalization Centered Moving Ratio to Seasonal Deseasonalized t Year Quarter Sales Average CMA Indexes Sales
1 2012 1 6.70 0.765 8.759 2 2012 2 4.60 0.575 8.004 3 2012 3 10.00 8.475 1.180 1.141 8.761 4 2012 4 12.70 8.450 1.503 1.519 8.361
5 2013 1 6.50 8.425 0.772 0.765 8.498 6 2013 2 4.60 8.513 0.540 0.575 8.004 7 2013 3 9.80 8.675 1.130 1.141 8.586 8 2013 4 13.60 8.775 1.550 1.519 8.953
9 2014 1 6.90 8.900 0.775 0.765 9.021 10 2014 2 5.00 9.038 0.553 0.575 8.700 11 2014 3 10.40 9.113 1.141 1.141 9.112 12 2014 4 14.10 9.188 1.535 1.519 9.283
13 2015 1 7.00 9.300 0.753 0.765 9.151 14 2015 2 5.50 9.463 0.581 0.575 9.570 15 2015 3 10.80 9.588 1.126 1.141 9.462 16 2015 4 15.00 9.625 1.558 1.519 9.875
17 2016 1 7.10 9.688 0.733 0.765 9.282 18 2016 2 5.70 9.663 0.590 0.575 9.918 19 2016 3 11.10 9.713 1.143 1.141 9.725 20 2016 4 14.50 9.888 1.466 1.519 9.546
21 2017 1 8.00 9.988 0.801 0.765 10.459 22 2017 2 6.20 10.075 0.615 0.575 10.788 23 2017 3 11.40 1.141 9.988 24 2017 4 14.90 1.519 9.809
Calculation of Seasonal Indexes 1 2 3 4
2012 1.180 1.503 2013 0.772 0.540 1.130 1.550 2014 0.775 0.553 1.141 1.535 2015 0.753 0.581 1.126 1.558 2016 0.733 0.590 1.143 1.466 2017 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
676 CHAPTER 18
Finally, we take the mean of all the winter typical indexes, all the spring in- dexes, 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.
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 5 years, 2013 through 2017, follows.
Quarter
Year Winter Spring Summer Fall
2013 117.0 80.7 129.6 76.1 2014 118.6 82.5 121.4 77.0 2015 114.0 84.3 119.9 75.0 2016 120.7 79.6 130.7 69.6 2017 125.2 80.2 127.6 72.0
(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.
S E L F - R E V I E W 18–4
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 3 years, 2014 through 2016, he recorded the following number of em- ployee absences, in days, for each quarter.
Quarter
Year I II III IV
2014 4 10 7 3 2015 5 12 9 4 2016 6 16 12 4
Determine a typical seasonal index for each of the four quarters. 10. Appliance Center sells a variety of electronic equipment and home appli-
ances. For the last 4 years, 2013 through 2016, the following quarterly sales (in $ millions) were reported.
Quarter
Year I II III IV
2013 5.3 4.1 6.8 6.7 2014 4.8 3.8 5.6 6.8 2015 4.3 3.8 5.7 6.0 2016 5.6 4.6 6.4 5.9
Determine a typical seasonal index for each of the four quarters.
E X E R C I S E S
TIME SERIES AND FORECASTING 677
DESEASONALIZING DATA A set of typical indexes is very useful in adjusting a sales series, for example, for sea- sonal fluctuations. The resulting sales series is called deseasonalized sales or season- ally 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 proce- dure, the quarterly sales totals of Toys International from Table 18–6 are repeated in column 1 of Table 18–9.
TABLE 18–9 Actual and Deseasonalized Sales for Toys International
(1) (2) (3) Seasonal Deseasonalized Year Quarter Sales Index Sales
2012 Winter 6.7 0.765 8.759 Spring 4.6 0.575 8.004 Summer 10.0 1.141 8.761 Fall 12.7 1.519 8.361 2013 Winter 6.5 0.765 8.498 Spring 4.6 0.575 8.004 Summer 9.8 1.141 8.586 Fall 13.6 1.519 8.953 2014 Winter 6.9 0.765 9.021 Spring 5.0 0.575 8.700 Summer 10.4 1.141 9.112 Fall 14.1 1.519 9.283 2015 Winter 7.0 0.765 9.151 Spring 5.5 0.575 9.570 Summer 10.8 1.141 9.462 Fall 15.0 1.519 9.875 2016 Winter 7.1 0.765 9.282 Spring 5.7 0.575 9.918 Summer 11.1 1.141 9.725 Fall 14.5 1.519 9.546 2017 Winter 8.0 0.765 10.459 Spring 6.2 0.575 10.788 Summer 11.4 1.141 9.988 Fall 14.9 1.519 9.809
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 seasonal index for that quarter, that is, TSCI/S. For example, the actual sales for the first, winter quarter of 2012 were $6.7 million. The seasonal index for the winter quarter is 76.5%, using the results on page 674. The index of 76.5 indicates that sales for the winter quarter are typically 23.5% 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 winter quarter of 2012. It is $8,758,170, found by ($6,700,000/76.5)100. We continue this process for the other quarters in column 3 of Table 18–9, with the results reported in millions of dollars. Because the seasonal component has been removed (divided out) from the quarterly sales, the deseasonalized sales value contains only the trend (T ), cyclical (C ), and irreg- ular (I ) components. Scanning the deseasonalized sales in column 3 of Table 18–9, we see that the sales of toys showed a moderate increase over the six-year period. Chart 18–9 shows both the actual sales and the deseasonalized sales. It is clear that removing the seasonal factor allows us to focus on the overall long-term trend of sales.
LO18-7 Deseasonalize a time series using seasonal indexes.
678 CHAPTER 18
We will also be able to determine the regression equation of the trend data and use it to forecast future sales.
CHART 18–9 Actual and Deseasonalized Sales for Toys International (2012 to 2017)
Sales Deseasonalized Year Time ($ Million) Sales ($ Million)
2012 1 6.7 8.759 2 4.6 8.004 3 10.0 8.761 4 12.7 8.361 2013 5 6.5 8.498 6 4.6 8.004 7 9.8 8.586 8 13.6 8.953 2014 9 6.9 9.021 10 5.0 8.700 11 10.4 9.112 12 14.1 9.283
Sales Deseasonalized Year Time ($ Million) Sales ($ Million)
2015 13 7.0 9.151 14 5.5 9.570 15 10.8 9.462 16 15.0 9.875 2016 17 7.1 9.282 18 5.7 9.918 19 11.1 9.725 20 14.5 9.546 2017 21 8.0 10.459 22 6.2 10.788 23 11.4 9.988 24 14.9 9.809
0 0 2 4 6 8
10 12 14 16
4 8 12 Time
Sa le
s ($
m ill
io n)
Sales ($ million)
16 20 24
Deseasonalized Sales ($ million)
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.
E X A M P L E
Toys International would like to forecast its sales for each quarter of 2018. Use the information in Table 18–9 to determine the forecast.
S O L U T I O N
The deseasonalized data depicted in Chart 18–9 seem to follow a straight line. Hence, it is reasonable to develop a linear trend equation based on these data. The deseasonalized trend equation is:
ŷ = a + bt
where: ŷ 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 2012 is the first quarter, so it is coded 1, the spring quarter of 2012 is coded 2, and so on. The last quarter of 2017 is coded 24.
We use Excel 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.
TIME SERIES AND FORECASTING 679
The equation for the trend line is:
ŷ = 8.11043 + 0.08988t
The slope of the trend line is 0.08988. This shows that over the 24 quarters the deseasonalized sales increased at a rate of 0.08988 ($ million) per quarter, or $89,880 per quarter. The value of 8.11043 is the intercept of the trend line on the Y-axis (i.e., for t = 0).
7.5 0 4 8 12
Time
Sa le
s ($
m ill
io n)
16 20 24
8
8.5
9
9.5
10
10.5
11
Deseasonalized sales data
Trend equation
The coefficient of determination is .785. This value is computed using Excel and is shown in the text box on the graph. We can use this value as an indication of the fit of the data. Because this is not sample 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 2018 the value of t is 25. Therefore, the estimated sales of that period are 10.35743, found by
ŷ = 8.11043 + 0.08988t = 8.11043 + 0.08988(25) = 10.35743
The estimated deseasonalized sales for the winter quarter of 2018 are $10,357,430. This is the sales forecast, before we consider the effects of seasonality.
We use the same procedure and an Excel spreadsheet to determine a forecast for each of the four quarters of 2018. A partial Excel output follows.
Quarterly Forecast for Toys International 2018
Estimated Seasonal Quarterly Quarter Time Sales Index Forecast
Winter 25 10.35743 0.765 7.92343 Spring 26 10.44731 0.575 6.00720 Summer 27 10.53719 1.141 12.02293 Fall 28 10.62707 1.519 16.14252
Now that we have the forecasts for the four quarters of 2018, 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 2018 by 10.35743(.765) = 7.92343. The estimates for each of the four quarters of 2018 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.
680 CHAPTER 18
THE DURBIN-WATSON STATISTIC Time series data or observations collected successively over a period of time present 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 residuals are not highly correlated, and there are not long runs of positive or negative residuals. In Chart 18–10, the resid- uals are scaled on the vertical axis and the ŷ 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.
This condition is called autocorrelation or serial correlation.
LO18-8 Conduct a hypothesis test of autocorrelation.
Westberg Electric Company sells electric motors to customers in the Jamestown, New York, area. The monthly trend equation, based on 5 years of monthly data, is
ŷ = 4.4 + 0.5t
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.
S E L F - R E V I E W 18–5
11. The planning department of Padget and Kure Shoes, the manufacturer of an exclu- sive brand of women’s shoes, developed the following trend equation, in millions of pairs, based on 5 years of quarterly data.
ŷ = 3.30 + 1.75t
The following table gives the seasonal factors for each quarter.
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 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.
13. Refer to Exercise 9 regarding the absences at Anderson Belts, Inc. Use the sea- sonal indexes you computed to determine the deseasonalized absences. Deter- mine the linear trend equation based on the quarterly data for the 3 years. Forecast the seasonally adjusted absences for 2017.
14. Refer to Exercise 10, regarding sales at Appliance Center. Use the seasonal in- dexes you computed to determine the deseasonalized sales. Determine the linear trend equation based on the quarterly data for the 4 years. Forecast the seasonally adjusted sales for 2017.
E X E R C I S E S
AUTOCORRELATION Successive residuals are correlated.
TIME SERIES AND FORECASTING 681
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
Yt = α + β1Xt + εt 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 pre- diction 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 characteristics as the coefficient of correlation. That is, values close to −1.00 or 1.00 indicate a strong association, and values near 0 indicate 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 determin- ing the residuals for each observation. That is, et = (yt − ŷt). Next, we compute d using the following relationship.
Re si
du al
(0) e1
e2 e3
e4 e5 e6
e7 e8 e9
e10 e11
CHART 18–10 Correlated Residuals
DURBIN-WATSON STATISTIC d = ∑
n
t=2 (et − et−1)2
∑ n
t=1 (et)
2 (18–4)
To determine the numerator of formula (18–4), 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 is close
682 CHAPTER 18
to 0, this indicates positive autocorrelation. Values beyond 2 indicate negative autocor- relation. Negative autocorrelation seldom exists in practice. To occur, successive resid- uals 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 (ρ = 0)
H1: Positive residual correlation (ρ > 0)
Recall from the previous chapter that r refers to the sample correlation and that ρ is the correlation coefficient in the population. The critical values for d are reported in Appendix B.9. To determine the critical value, we need α (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. However, there is also a range of values where d is inconclusive. That is, in the incon- clusive 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 re- siduals are not correlated, there is no autocorrelation present, and the regression as- sumption has been met. There will not be any problem with the estimated value of the standard error of estimate. If the null hypothesis is rejected, then we conclude that au- tocorrelation 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.
E X A M P L E
Banner Rocker Company manufac- tures and markets rocking chairs. The company developed a special rocker for senior citizens, which it advertises extensively on TV. Ban- ner’s market for the special chair is the Carolinas, Florida, and Arizona where there are many senior citi- zens and retired people. The presi- dent of Banner Rocker is studying the association between his adver- tising expense (X ) and the number of rockers sold over the last 20
months (Y ). He collected the following 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 prob- lems with autocorrelation.
© Arthur Tilley/Getty Images
TIME SERIES AND FORECASTING 683
Advertising Month Sales (000) ($ millions)
1 153 $5.5 2 156 5.5 3 153 5.3 4 147 5.5 5 159 5.4 6 160 5.3 7 147 5.5 8 147 5.7 9 152 5.9 10 160 6.2
Advertising Month Sales (000) ($ millions)
11 169 6.3 12 176 5.9 13 176 6.1 14 179 6.2 15 184 6.2 16 181 6.5 17 192 6.7 18 205 6.9 19 215 6.5 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.
S O L U T I O N
The first step is to determine the regression equation.
The coefficient of determination is 68.5%. So we know there is a strong positive asso- ciation 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.
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 equation is:
ŷ = –43.80 + 35.95x
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.
684 CHAPTER 18
What about the potential issue of autocorrelation? Many software packages 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 (18–4), we use an Excel spreadsheet.
To investigate the possible autocorrelation, we need to determine the residuals for each observation. We find the fitted values—that is, the ŷ—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:
ŷ = −43.80 + 35.950x = −43.80 + 35.950(5.5) = 153.925
e1 = y1 − ŷ1 = 153 − 153.925 = −0.925
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 for months 7 through 11. In column F, we lag the residuals one period. In column G, we find the differ- ence between the current residual and the residual in the previous and square this difference. Using the values from the software:
(et − et−1)2 = (e2 − e2−1)2 = [2.075 − (−.925)]2 = (3.0000)2 = 9.0000
The other values in column G are found the same way. The values in column H are the squares of those in column E.
(e1) 2 = (−0.925)2 = 0.8556
To find the value of d, we need the sums of columns G and H. These sums are noted in blue in the spreadsheet.
TIME SERIES AND FORECASTING 685
d = ∑
n
t=2 (et − et−1)2
∑ n
t=1 (et)
2 =
2338.570 2744.269
= 0.8522
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.9, 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.
n k
uddl
21
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
d
dl
0
Positive autocorrelation Undetermined
InconclusiveReject Ho
0.85
HoAccept
1.20
du
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 regression model has not been correctly specified. It is likely we need to add one or more indepen- dent variables that have some time-ordered effects on the dependent variable. The simplest independent variable to add is one that represents the time periods.
686 CHAPTER 18
15. Recall Exercise 9 from Chapter 14 and the regression equation to predict job performance. See page 519.
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 equa- tion to predict commissions earned. See page 520.
a. Plot the residuals in the order in which the data are presented. b. Test for autocorrelation at the .01 significance level.
E X E R C I S E S
C H A P T E R S U M M A R Y
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 ŷ = a + bt, 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. The seasonal factor is 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 [18–4] is used to test for autocorrelation.
d = ∑
n
t=2 (et − et−1)2
∑ n
t=1 (et)
2 (18–4)
C H A P T E R E X E R C I S E S
17. The asset turnovers, excluding cash and short-term investments, for RNC Com- pany from 2007 to 2017 are:
2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017
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.
TIME SERIES AND FORECASTING 687
c. Calculate the points on the trend line for 2010 and 2015, and plot the line on the graph.
d. Estimate the asset turnover for 2022. e. How much did the asset turnover increase per year, on the average, from 2007 to
2017? 18. The sales, in billions of dollars, of Keller Overhead Door Inc. for 2012 to
2017 are:
Year Sales Year Sales
2012 7.45 2015 7.94 2013 7.83 2016 7.76 2014 8.07 2017 7.90
a. Plot the data. b. Determine the least squares trend equation. c. Use the trend equation to calculate the points for 2014 and 2017. Plot them on the
graph and draw the regression line. d. Estimate the net sales for 2020. e. By how much have sales increased (or decreased) per year on the average during
the period? 19. The number of employees, in thousands, of Keller Overhead Door Inc. for the
years 2012 to 2017 are:
Year Employees (000’s) Year Employees (000’s)
2012 45.6 2015 39.3 2013 42.2 2016 34.0 2014 41.1 2017 30.0
a. Plot the data. b. Determine the least squares trend equation. c. Use the trend equation to calculate the points for 2014 and 2017. Plot them on the
graph and draw the regression line. d. Estimate the number of employees in 2020. e. By how much has the number of employees increased (or decreased) per year, on
the average, during the period? 20. Listed below is the selling price for a share of PepsiCo Inc. at the close of each
year from 1998 to 2015.
Year Price Year Price Year Price Year Price
1998 40.6111 2003 46.62 2008 54.77 2013 82.94 1999 35.0230 2004 52.20 2009 60.80 2014 94.56 2000 49.5625 2005 59.85 2010 65.33 2015 99.92 2001 48.68 2006 62.00 2011 66.35 2002 42.22 2007 77.51 2012 68.43
a. Plot the data. b. Determine the least squares trend equation. c. Calculate the points for the years 2008 and 2013. d. Estimate the selling price in 2018. 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
during the period?
688 CHAPTER 18
21. 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, there- fore, a logarithmic equation should be used.
Sales Sales Year ($ millions) Year ($ millions)
2007 8.0 2013 39.4 2008 10.4 2014 50.5 2009 13.5 2015 65.0 2010 17.6 2016 84.1 2011 22.8 2017 109.0 2012 29.3
a. Determine the logarithmic equation. b. Determine the coordinates of the points on the logarithmic straight line for 2007
and 2014. c. By what percent did sales increase per year, on the average, during the period from
2007 to 2017? d. Based on the equation, what are the estimated sales for 2018?
22. Reported below are the amounts spent on advertising ($ millions) by a large firm from 2007 to 2017.
Year Amount Year Amount
2007 88.1 2013 132.6 2008 94.7 2014 141.9 2009 102.1 2015 150.9 2010 109.8 2016 157.9 2011 118.1 2017 162.6 2012 125.6
a. Determine the logarithmic trend equation. b. Estimate the advertising expenses for 2020. c. By what percent per year did advertising expense increase during the period?
23. Listed below is the selling price for a share of Oracle Inc. stock at the close of the year for 1998 through 2015.
Year Price Year Price Year Price Year Price
1998 7.1875 2003 13.23 2008 17.73 2013 37.78 1999 28.0156 2004 13.72 2009 24.53 2014 45.02 2000 29.0625 2005 12.21 2010 31.30 2015 36.01 2001 13.81 2006 19.11 2011 26.33 2002 10.80 2007 20.23 2012 34.08
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. Using the equation with the logarithm of price, calculate the points for the years
2001 and 2006. d. Using the equation with the logarithm of price, estimate the selling price in 2018.
Does this seem like a reasonable estimate based on the historical data? e. Using the equation with the logarithm of price, how much has the stock price in-
creased or decreased (per year) on average during the period?
TIME SERIES AND FORECASTING 689
24. The production of Reliable Manufacturing Company for 2012 and part of 2013 follows.
2012 2013 2012 2013 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 2012.
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 Sep- tember 2012, and determine the 12 typical seasonal indexes.
Year Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec.
2012 ? ? ? 92.1 106.5 92.9 2013 88.9 102.9 178.9 118.2 60.1 43.1 44.0 74.0 200.9 90.0 101.9 90.9 2014 87.6 103.7 170.2 125.9 59.4 48.6 44.2 77.2 196.5 89.6 113.2 80.6 2015 79.8 105.6 165.8 124.7 62.1 41.7 48.2 72.1 203.6 80.2 103.0 94.2 2016 89.0 112.1 182.9 115.1 57.6 56.9
c. Interpret the typical seasonal index. 25. The sales of Andre’s Boutique for 2012 and part of 2013 are:
2012 Sales 2013 Sales 2012 Sales 2013 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 2012.
b. Assume that the seasonal indexes in the following table are correct. Insert in the ta- ble the seasonal indexes you computed in part (a) for July, August, September, and October 2012, and determine the 12 typical seasonal indexes.
Year Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec.
2012 ? ? ? ? 123.6 150.9 2013 83.9 77.6 86.1 118.7 99.7 92.0 87.0 91.4 97.3 105.4 124.9 140.1 2014 86.7 72.9 86.2 121.3 96.6 92.0 85.5 93.6 98.2 103.2 126.1 141.7 2015 85.6 65.8 89.2 125.6 99.6 94.4 88.9 90.2 100.2 102.7 121.6 139.6 2016 77.3 81.2 85.8 115.7 100.3 89.7
c. Interpret the typical seasonal index.
690 CHAPTER 18
26. The quarterly production of pine lumber, in millions of board feet, by Northwest Lumber for 2012 through 2016 is:
Quarter
Year Winter Spring Summer Fall
2012 7.8 10.2 14.7 9.3 2013 6.9 11.6 17.5 9.3 2014 8.9 9.7 15.3 10.1 2015 10.7 12.4 16.8 10.7 2016 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 2017.
27. 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 for 2011 through 2016 are:
Quarter
I II III IV Year Jan.–Mar. Apr.–June July–Sept. Oct.–Dec.
2011 142 312 488 208 2012 146 318 512 212 2013 160 330 602 187 2014 158 338 572 176 2015 162 380 563 200 2016 162 362 587 205
a. Using the ratio-to-moving-average method, determine the four typical quarterly indexes.
b. Interpret the typical seasonal pattern. 28. Sales of roof material, by quarter, for 2010 through 2016, by Carolina Home Con-
struction Inc. are shown below (in $000).
Quarter
Year I II III IV
2010 210 180 60 246 2011 214 216 82 230 2012 246 228 91 280 2013 258 250 113 298 2014 279 267 116 304 2015 302 290 114 310 2016 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 sales for the four quarters of next year using the trend equation and season-
ally adjust these values to find the predicted sales for each quarter.
TIME SERIES AND FORECASTING 691
29. 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. The course earns a certain percentage of this total. The nonpackage play in- cludes play by local residents and visitors to the area who wish to play golf. The follow- ing data, beginning with July 2014 and ending with June 2017, report the package and nonpackage play by month, as well as the total amount, in thousands of dollars.
Year Month Package Local Total
2014 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 2015 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
2016 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 2017 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
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? 30. The following is the number of retirees receiving benefits from the State Teachers
Retirement System of Ohio from 1996 until 2015.
Year Retirees Year Retirees Year Retirees Year Retirees
1996 70,448 2001 83,918 2006 99,248 2011 117,138 1997 72,601 2002 86,666 2007 102,771 2012 112,136 1998 75,482 2003 89,257 2008 106,099 2013 127,797 1999 78,341 2004 92,574 2009 109,031 2014 130,521 2000 81,111 2005 95,843 2010 112,483 2015 136,018
a. Plot the data. b. Determine the least squares trend equation. Use a linear equation. c. Calculate the points for the years 2005 and 2015. d. Estimate the number of retirees that will be receiving benefits in 2018. 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 aver-
age during the period?
692 CHAPTER 18
31. Ray Anderson, owner of Anderson Ski Lodge in upstate New York, is interested in forecasting the number of visitors for the upcoming year. The following data are avail- able, by quarter, from the first quarter of 2010 to the fourth quarter of 2016. Develop a seasonal index for each quarter. How many visitors would you expect for each quarter of 2017, if Ray projects that there will be a 10% increase from the total number of visi- tors in 2016? Determine the trend equation, project the number of visitors for 2017, and seasonally adjust the forecast. Which forecast would you choose?
Year Quarter Visitors Year Quarter Visitors
2010 I 86 2014 I 188 II 62 II 172 III 28 III 128 IV 94 IV 198 2011 I 106 2015 I 208 II 82 II 202 III 48 III 154 IV 114 IV 220 2012 I 140 2016 I 246 II 120 II 240 III 82 III 190 IV 154 IV 252 2013 I 162 II 140 III 100 IV 174
32. The enrollment in the College of Business at Midwestern University by quarter from 2013 to spring 2017 is:
Quarter
Year Winter Spring Summer Fall
2013 2,033 1,871 714 2,318 2014 2,174 2,069 840 2,413 2015 2,370 2,254 927 2,704 2016 2,625 2,478 1,136 3,001 2017 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 enrollment for the next four quarters.
33. At the beginning of this chapter, we presented a graph showing the price per gal- lon of gasoline from 1996 until 2015. The actual data for each year are given below.
Year Cost/Gallon
1996 1.20 1997 1.20 1998 1.03 1999 1.14 2000 1.48 2001 1.42 2002 1.35 2003 1.56 2004 1.85 2005 2.27
Year Cost/Gallon
2006 2.57 2007 2.80 2008 3.25 2009 2.35 2010 2.78 2011 3.52 2012 3.62 2013 3.49 2014 3.34 2015 2.40
TIME SERIES AND FORECASTING 693
Determine the trend line using both linear and nonlinear equations. Which would you suggest? Why? Based on your recommendation as the better equation, what cost per gallon would you estimate for 2016 and 2017?
D A T A A N A L Y T I C S
(The data for this exercise is available at the text website: www.mhhe.com/lind17e.) Use the following information obtained from annual reports of Home Depot to solve
Exercises 34, 35, and 36. You will need the help of a software package such as Excel and perhaps a companion package such as MegaStat. For the years 1993 through 2015, the data include a coded time variable (1 through 23), the number of associates in thousands, net sales in millions of dollars, the mean dollar amount per transaction, and the Consumer Price Index (CPI) for each year.
Associates Net Sales Mean Amount Year Time (000) ($ million) per Transaction CPI
1993 1 50.6 9,239 39.13 144.500 1994 2 67.3 12,477 41.29 148.200 1995 3 80.8 15,470 41.78 152.400 1996 4 98.1 19,535 42.09 156.900 1997 5 124.4 24,156 43.63 160.500 1998 6 156.7 30,219 45.05 163.000 1999 7 201.4 38,454 47.87 166.000 2000 8 227.3 45,738 48.65 172.200 2001 9 256.3 53,553 48.64 177.100 2002 10 280.9 58,247 49.43 179.900 2003 11 298.8 64,816 51.15 184.000 2004 12 323.1 73,094 54.89 188.900 2005 13 344.8 81,511 57.98 195.300 2006 14 364.4 79,022 58.90 201.600 2007 15 331.0 77,349 57.48 207.342 2008 16 322.0 71,288 55.61 215.303 2009 17 317.0 66,176 51.76 214.537 2010 18 321.0 67,997 51.93 218.056 2011 19 331.0 70,395 53.28 224.939 2012 20 340.0 74,754 54.89 229.594 2013 21 365.0 78,812 56.76 232.957 2014 22 371.0 83,176 57.87 236.736 2015 23 385.0 88,519 58.77 237.017
34. Develop a trend equation for the net sales ($ million) for Home Depot. Consider both a linear trend and a nonlinear trend. Which one would you select and why? Given the trend equation you selected, forecast the net sales for 2016 and 2017.
35. Consider the variable mean amount per transaction in the Home Depot data presented above. This variable indicates, for example, that the average customer spent $39.13 on goods during a store visit in 1993. By 2012 this amount increased to $54.89. During that same period the Consumer Price Index (CPI) as reported by the Bureau of Labor Statistics increased from 144.5 to 229.594. Convert the CPI to a 1993 base, as de- scribed on page 644 in chapter 17, and convert the mean amount per transaction to 1993 dollars. Develop a linear trend equation for the constant 1993 dollars of the mean amount per transaction. Is it reasonable that the trend is linear? Can we conclude that the value of the amount the customer spent is less?
36. Use the variables mean amount per transaction and number of associates to forecast net sales. Are these two independent variables reasonable predictors of net sales?
694 CHAPTER 18
(Hints: What is the R2 value? Is it large? Look at the p-values for each of the independent variables. Is this value less than .05 for each?) Because all of these variables are associ- ated with time, it may be that there is autocorrelation. Conduct the appropriate test of hypothesis to determine if there is autocorrelation present. What is your conclusion? Again, use the .05 significance level.
37. Refer to the Baseball 2016 data, which include information on the 2016 Major League Baseball season. The data include the mean player salary since 2000. Plot the informa- tion and develop a linear trend equation. Write a brief report on your findings.
A REVIEW OF CHAPTERS 17–18 Chapter 17 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 number of sales representatives employed over the same period of time. A direct compar- ison is not meaningful because the units for one set of data are dollars and the other people. Index numbers also facili- tate the comparison of very large values, where the amount of change in the actual values is very large and therefore difficult to interpret.
There are two types of price indexes. In an unweighted price index, the quantities are not considered. To form an un- weighted 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 2011 and $18,600,000 in 2017, the simple unweighted price index for 2017 is:
p = pt p0
(100) = $18,600,000 $12,000,000
(100) = 155.0
We conclude there is a 55% increase in 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 2011. In 2017, the same store sold 60 computers at $1,200 and 230 DVDs at $175. The Laspeyres price index is:
p = Σpt q0 Σp0q0
(100) = $1,200 × 50 + $175 × 200 $1,000 × 50 + $150 × 200
(100) = $95,000 $80,000
(100) = 118.75
Notice the same base period quantities are used as weights in both the numerator and the denominator. The index indi- cates there has been an 18.75% 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 re- ported 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 18, we studied time series and forecasting. A time series is a collection of data over a period of time. The earn- ings per share of General Electric common stock over the last 10 years is an example of a time series. There are four components to a time series: the secular 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 several 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 during the winter and ski lodges in Wyoming are not used in the summer months. Hence, we say the rentals of beach front properties and ski lodges are seasonal.
TIME SERIES AND FORECASTING 695
The irregular component includes any unpredictable events. In other words, the irregular component includes events that cannot be forecast. There are two types of irregular components. Episodic variations are unpredictable, but can usually be identified. The Houston flooding in the spring of 2016 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 ŷ = a + bt, where ŷ is the estimated 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.
P R O B L E M S
1. Listed below are the net sales (in millions of euros) for the Adidas Group from 2008 to 2015.
Year Net Sales
2008 10,799 2009 10,381 2010 11,990 2011 13,332 2012 14,883 2013 14,203 2014 14,534 2015 16,915
a. Determine the index for 2015, using 2008 as the base period. b. Use the period 2008 to 2010 as the base period and find the index for 2015. c. With 2008 as the base year, use the least squares method to find the trend equation.
What is the estimated consolidated revenue for 2017? What is the rate of increase per year?
2. The table below shows the labor force and the unemployment rate for three counties in northwest Pennsylvania for 2004 and 2015.
2004 2015
Labor Percent Labor Percent County Force Unemployed Force Unemployed
Erie 134,538 6.1 134,873 5.3 Warren 20,158 5.0 19,980 4.8 McKean 19,302 5.7 18,987 5.9
a. In 2004, the national unemployment rate was 5.5%. For 2004, compute a simple av- erage unemployment index for the region using the national unemployment rate as the base. Interpret the simple average index.
b. In 2015, the national unemployment rate was 5.0%. For 2015, compute a simple av- erage 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 unem- ployment index using the Laspeyres method. Use 2004 as the base period. Interpret the index.
3. Based on five years of monthly data (the period from January 2011 to December 2015), the trend equation for a small company is ŷ = 3.5 + 0.7t. The seasonal index for January is 120 and for June it is 90. What are the seasonally adjusted sales forecasts for January 2016 and June 2016?
696 CHAPTER 18
P R A C T I C E T E S T
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 period, 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 the 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 5 years, 2012 through 2016.
Year Sales
2012 $130,000 2013 145,000 2014 120,000 2015 170,000 2016 190,000
a. Find the simple index for each year using 2012 as the base year. b. Find the simple index for each year using 2012–2013 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 for 2012 and 2016.
2012 2016
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
a. Determine the simple aggregate price index, with 2012 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:
ŷ = 5.50 + 1.25t
The equation is based on 4 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.
Statistical Process Control and Quality Management
19
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO19-1 Explain the purpose of quality control in production and service operations.
LO19-2 Define the two sources of process variation and explain how they are used to monitor quality.
LO19-3 Explain the use of charts to investigate the sources of process variation.
LO19-4 Compute control limits for mean and range control charts for a variable measure of quality.
LO19-5 Evaluate control charts to determine if a process is out of control.
LO19-6 Compute control limits of control charts for an attribute measure of quality.
LO19-7 Explain the process of acceptance sampling.
© EMMANUEL DUNAND/AFP/Getty Images
A BICYCLE MANUFACTURER randomly selects 10 frames each day and tests for defects. The numbers of defective frames found over the last 14 days are 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 LO19-6.)
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INTRODUCTION Throughout this text, we present many applications of hypothesis testing. In Chapter 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 populations. 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 qual- ity 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 developed against which the actual attributes of the product or service are compared.
A BRIEF HISTORY OF QUALITY CONTROL Prior to the 1900s, U.S. industry was largely characterized by small shops making rela- tively 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 selection of the materi- als, 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 individual worker no longer had complete control over the quality of the product. A semiprofes- sional staff, usually called the Inspection Department, became responsible for the qual- ity of the product. The quality responsibility was usually fulfilled by a 100% 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 controlling quality, Shewhart developed charting techniques for controlling in-process manufac- turing operations. In addition, he introduced the concept of statistical sample inspec- tion 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 production operation.
Statistical quality control came into its own during World War II. The need for mass-produced war-related items, such as bomb sights, accurate radar, and other elec- tronic equipment, at the lowest possible cost hastened the use of statistical sampling and quality control charts. Since World War II, these statistical techniques 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 U.S. Department of Agriculture, to help them develop an overall plan. In a series of seminars with Japanese planners, he stressed a philoso- phy 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 manufac- turer must be able, via market research, to anticipate the needs of customers. Senior management has the responsibility for long-term improvement. Another of his points, and one that the Japanese strongly endorsed, is that every member of the company
LO19-1 Explain the purpose of quality control in production and service operations.
STATISTICAL PROCESS CONTROL AND QUALITY MANAGEMENT 699
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 philos- ophies in the United States. Two areas where Deming’s ideas differed from U.S. man- agement philosophy were with production quotas and merit ratings. He believed 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 opportunity for improvement. On the other hand, bad news opens the door for new products 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 numerical
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 production through many stages of manufacturing. We use the tools of statistical quality 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 of control,” that is, when an excessive number of defective units are being produced.
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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% of the cost of production is currently spent find- ing 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%. In Japan, he indicated, this cost is about 3%!
In recent years, companies have been motivated to improve quality by the challenge of being recognized for their quality achieve- ments. The Malcolm Baldrige National Quality Award, established in 1988, is awarded annually to U.S. companies that demonstrate ex- cellence in quality achievement and management. The award cate- gories include manufacturing, 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 2016 winners were:
• Don Chalmers Ford, an independent business franchised by the Ford Motor Company and located in Rio Rancho, New Mexico won in the small business category. The company motto, “Real Value, Real People, Real Simple”, is imple- mented through a commitment to “customers, quality and community.”
• The Momentum Group, with headquarters in Irvine, California started in 1994 as a small contract textile distributor. Now they specialize in customized commercial in- teriors. In the last twenty years, Momentum Groups’ sales grew more than 400 percent.
• Kindred Nursing and Rehabilitation Center, located in Mountain View, California won the award in the health care category. One hundred percent of the residents and family members surveyed expressed either “very” or “extreme” overall satis- faction levels since 2013, well above the national average assessment of similar facilities.
• Memorial Hermann Sugar Land Hospital is the largest not-for-profit health system in Southeast Texas. The hospital’s focus on patient safety resulted in zero medical errors related to pressure ulcers, ventilator-associated pneumonia, transfusion re- actions, and deaths from normally low mortality conditions.
You can obtain more information on these and other winners by visiting the website http://www.nist.gov/baldrige.
Six Sigma Many service, manufacturing, and nonprofit organizations are committed to improving the quality of their services and products. “Six Sigma” is a name given to an organiza- tionwide program designed to improve quality and performance throughout an organi- zation. 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 organizational support pro- cesses. The ultimate outcomes of a Six Sigma program are to reduce the costs of de- fects and errors, increase customer satisfaction and sales of products and services, and increase profits.
Six Sigma gets its name from the normal distribution. The term sigma means stan- dard deviation, and “plus or minus” three standard deviations gives a total range of six
STATISTICS IN ACTION
Does excellence in quality management lead to higher financial performance? Re- cent research compared the financial performance of companies that received the Baldrige National Quality Award to similar companies that did not receive the award. The research showed that the companies receiving the award had an average of 39% higher operating in- come and 26% higher sales, and were 1.22% lower in their cost per sales dollar.
Courtesy of National Institute of Standards and Technology
STATISTICAL PROCESS CONTROL AND QUALITY MANAGEMENT 701
standard deviations. So Six Sigma means that a process should not generate 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 pro- cesses 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 improve- ments to reduce process variation, and implementing procedures for improving 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: histograms, 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 im- provement 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 example, the city reduced missed trash pickups by 50% 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 http://www.6sigma.us.
SOURCES 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 pound. Some will weigh more than 0.25 pound, 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” 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.
Internal machine friction, slight variations in material or process conditions (such as the temperature of the mold being used to make glass bottles), atmospheric condi- tions (such as temperature, humidity, and the dust content of the air), and vibrations transmitted to a machine from a passing forklift are a few examples of sources of chance variation.
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.
If the hole drilled in a piece of steel is too large due to a dull drill, the drill may be sharp- ened or a new drill inserted. An operator who continually sets up the machine incor- rectly 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.
ASSIGNABLE VARIATION Variation that is not random. It can be eliminated or reduced by investigating the problem and finding the cause.
LO19-2 Define the two sources of process variation and explain how they are used to monitor quality.
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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.
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 hap- pen within a product or service. The chart is named after a 19-century Italian scien- tist, 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% of the activity is caused by 20% of the factors. By concentrating on 20% of the factors, managers can attack 80% of the problem. For example, Emily’s Family Restaurant, located at the junction of Interstates 75 and 70, is investigating “customer com- plaints.” 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. The data show these two factors are more than 85% of the total complaints. Addressing these two complaints first will yield the largest reduction in complaints and the largest in- crease in customer satisfaction.
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.
LO19-3 Explain the use of charts to investigate the sources of process variation.
E X A M P L E
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 us- age in Grove City. To investigate, she selects a sample of 100 homes and deter- mines the typical daily water usage for various purposes. These sample results are as follows.
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
What is the area of greatest usage? Where should she concentrate her efforts to reduce the water usage?
STATISTICAL PROCESS CONTROL AND QUALITY MANAGEMENT 703
Below the chart, we list the activities, their frequency of occurrence, and the percent of the time each activity occurs. In the last row, we list the cumulative per- centage. This cumulative row will allow us to quickly determine which set of activi- ties account for most of the activity. These cumulative percents are plotted above the vertical bars. In the Grove City example, the activities of watering lawn, personal bathing, and pools account for 82.1% of the water usage. The city manager can attain the greatest gain by looking to reduce the water usage in these three areas.
S O L U T I O N
A Pareto chart is useful for identifying the major areas of water usage and focusing 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 gal- lons 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% of the amount of water used. The next largest category is personal bathing, which accounts for 31.4% of the water used. These two activities account for 73.8% of the water usage.
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 ac- tivity 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.
CHART 19–1 Pareto Chart for Water Usage in Grove City, Utah
704 CHAPTER 19
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 usually 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
Under each of the possible causes are subcauses that are identified and investi- gated. The subcauses are factors that may be producing the particular effect. Informa- tion 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.
To illustrate a fishbone diagram, we investigate the causes of cold food served at Emily’s Family Restaurant. Recall that a Pareto analysis showed that cold food was one of the top two complaints. In Chart 19–3, notice that each of the subcauses is listed as
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
CHART 19–3 Fishbone Diagram for a Restaurant Investigation of Cold Food Complaints
STATISTICAL PROCESS CONTROL AND QUALITY MANAGEMENT 705
an assumption. Each of these subcauses must be investigated to find the real problem regarding the cold food. In a fishbone diagram, there is no weighting of the subcauses.
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 replace- ment windows for older homes. The vinyl coating must have a thickness between cer- tain 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 conditions. 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 windows are produced.
LO19-4 Compute control limits for mean and range control charts for a variable measure of quality.
Develop a Pareto chart. What complaints would you suggest the administrator work on first to achieve the most significant improvement?
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, he or she is asked to complete a complaint form. Listed below is a summary of the complaint forms received during the last 12 months.
S E L F - R E V I E W 19–1
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
1. Tom Sharkey is the owner of Sharkey Chevy, Buick, GMC. 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. A summary of the complaints for the first six months follows. Develop a Pareto chart. How should Tom prioritize the complaints to improve the quality of service?
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.
E X E R C I S E S
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
706 CHAPTER 19
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 func- tion. These charts indicate to the workers, group leaders, quality con- trol 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 situ- ation; it is just a piece of paper with numbers and dots on it. Instead, the person responsible must adjust the machine manufacturing 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
bottle 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 prod- uct 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 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 pieces is selected each hour from the production process and the mean of each sample com- puted. The sample means are x1, x2, x3, and so on. The mean of these sample means is denoted as x . We use k to indicate the number of sample means. The overall or grand mean is found by:
© Bloomberg/Getty Images
GRAND MEAN x = Σ of the sample means
Number of sample means =
Σx k
(19–1)
STANDARD ERROR OF THE MEAN sx = s
√n (19–2)
The standard error of the distribution of the sample means is designated by sx . It is found by:
These relationships allow us to establish limits for 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.
E X A M P L E
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 im- possible to have every call answered immediately by a technical representative. The company knows that the length of waiting time is related to a customer’s perception of service quality. To understand its service-call process, Statistical Soft- ware 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
STATISTICAL PROCESS CONTROL AND QUALITY MANAGEMENT 707
by the caller. Yesterday, for the 16 hours of operation, five calls were sampled each hour and the total time to resolve a customer’s problem, in minutes, was recorded. This information is reported below.
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?
S O L U T I O N
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:
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
where s is an estimate of the standard deviation of the population, σ. Notice that in the calculation of the upper and lower control limits the number 3 appears. It rep- resents the 99.74% confidence limits. The limits are often called the 3-sigma limits. However, other levels of confidence (such as 90 or 95%) can be used.
This application was developed before computers were widely available, and computing standard deviations was difficult. Rather than calculate the standard de- viation 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 standard deviation, so we can use the following formulas to determine the 99.74% control limits for the mean. It can be demonstrated that the term 3(s∕√n) from for- mula (19–3) is equivalent to A2R in the following formula.
UCL = x + 3 s
√n and LCL = x − 3
s √n
(19–3)CONTROL LIMITS FOR THE MEAN
UCL = x + A2 R LCL = x − A2 R (19–4) CONTROL LIMITS FOR THE MEAN
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where: A2 is a constant used in computing the upper and the lower control limits. It
is based on the average range, R. The factors for various sample sizes are available in Appendix B.10. (Note: n in this table refers to the number of items in the sample.) A portion of Appendix B.10 is shown below. To locate the A2 factor for this problem, find the sample size for n in the left margin. It is 5. Then move horizontally to the A2 column and read the factor. It is 0.577.
x is the mean of the sample means, computed by Σ x /k where k is the number of samples selected. In this problem, a sample of five observations is taken each hour for 16 hours, so k = 16.
R is the mean of the ranges of the sample. It is ΣR/k. Remember the range is the difference between the largest and the smallest value in each sample. It de- scribes the variability occurring in that particular sample. (See Table 19–1.)
The centerline for the chart is x . It is 9.413 minutes, found by 150.6/16. The mean of the ranges (R ) is 6.375 minutes, found by 102/16. Thus, the upper control limit of the mean chart is:
UCL = x + A2R = 9.413 + 0.577(6.375) = 13.091
The lower control limit of the mean chart is:
LCL = x − A2R = 9.413 − 0.577(6.375) = 5.735
x , UCL, LCL, and the sample means are portrayed in Chart 19–4. The mean, x , is 9.413 minutes, the upper control limit is located at 13.091 minutes, and the lower
n A2 d2 D3 D4 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
TABLE 19–1 Sixteen Samples of Call Duration Data for Statistical Software, Inc.
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
STATISTICAL PROCESS CONTROL AND QUALITY MANAGEMENT 709
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% of the time the mean length of a sample of five calls will be between 5.735 minutes and 13.091 minutes.
Because the statistical theory is based on the normality of large samples, control charts should be based on a stable process, that is, a fairly large sample, taken over a long period of time. Before using control charts in practice, at least 25 samples need to be collected to establish control chart limits.
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 vari- ability of the operation.
The upper and lower control limits of the range chart are determined from the fol- lowing equations.
CHART 19–4 Control Chart for Mean Call Duration for Statistical Software Inc.
The values for D3 and D4, which reflect the usual three σ (sigma) limits for various sam- ple sizes, are found in Appendix B.10 or in the table on page 708.
CONTROL CHART FOR RANGES UCL = D4 R LCL = D3 R (19–5)
E X A M P L E
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?
STATISTICS IN ACTION
Control charts were used to help convict a person who bribed jai alai players to lose. Mean and range 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 assignable variation stopped, and prosecutors were able to tie those times to the arrest of the suspect.
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
x
LCL
13 12 11 10
9 8 7 6
710 CHAPTER 19
S O L U T I O N
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 min- utes. 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 R = 102/16. Referring to Appendix B.10 or the partial table on page 708, D3 and D4 are 0 and 2.115, respec- tively. The lower and upper control limits are 0 and 13.483.
UCL = D4 R = 2.115(6.375) = 13.483
LCL = D3 R = 0(6.375) = 0
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.
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.
Statistical software packages will compute all the statistics and draw control charts. The following is the Minitab analysis for the Statistical Software example showing the mean and range charts. The minor differences in the control limits are due to rounding.
STATISTICAL PROCESS CONTROL AND QUALITY MANAGEMENT 711
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 to- ward the out-of-control areas.
UCL
x
LCL
Mean Chart
UCL
R
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.
UCL
x
LCL
Mean Chart
UCL
R
LCL
Range Chart
LO19-5 Evaluate control charts to determine if a process is out of control.
712 CHAPTER 19
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.
UCL
x
LCL
Mean Chart
UCL
R
LCL
Range Chart
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 x and increasing, and, in fact, the last two observations are out of control. The fact that the sample means were increasing for six consecutive observations is very improbable and another indication that the process is out of control.
The manager of River City McDonald’s randomly selects four customers during three hours of the day: 9 a.m., 10 a.m., and 11 a.m. For these selected customers, she determines the time, in minutes, between order entry and order delivery. The results are shown below.
S E L F - R E V I E W 19–2
(a) Compute the mean service time and the mean range of service time, determine the control limits for the mean and the range, and create control charts.
(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
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 A2 factor used to determine the upper and lower con- trol limits for the mean?
b. What are the values of the D3 and D4 factors used to determine the lower and upper control limits for the range?
E X E R C I S E S
STATISTICAL PROCESS CONTROL AND QUALITY MANAGEMENT 713
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” condi- tion) 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 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 (proportion defective) and the c-bar chart (num- ber of defectives per unit).
p-Charts If the item recorded is the proportion of unacceptable parts made in a larger batch of parts, the appropriate control chart is the p-chart. This chart is based on the binomial distribution, discussed in Chapter 6, and proportions, discussed in Chapter 15. The centerline is at p, the mean proportion defective. The p replaces the x of the variable control chart. The mean proportion defective is found by:
LO19-6 Compute control limits of control charts for an attribute measure of quality.
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 expe- rience regarding the oven temperature, an inspector reads the temperature at four different places inside the oven each half hour starting at 8:00 a.m. The last reading was at 10:30 a.m., for a total of six samples. 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 computations easier.)
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 temperature. Determine the grand mean. Plot the results on a control 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 plot results on a control chart. b. Does there seem to be a time when there is too much variation in the
temperature?
p = Total number defective
Total number of items sampled (19–6)MEAN PROPORTION
DEFECTIVE
714 CHAPTER 19
The variation in the sample proportion is described by the standard error of a pro- portion. It is found by:
Hence, the upper control limit (UCL) and the lower control limit (LCL) are computed as the mean proportion defective plus or minus three times the standard error of the proportions. The formula for the control limits is:
sp = √ p(1 − p)
n (19–7)
STANDARD ERROR OF THE SAMPLE PROPORTION
LCL, UCL = p ± 3√ p(1 − p)
n (19–8)
CONTROL LIMITS FOR PROPORTIONS
An example will show the details of the calculations and the conclusions.
E X A M P L E
Jersey Glass Company Inc. produces small hand mirrors. Jersey Glass runs day and evening shifts each weekday. The quality assurance department (QA) monitors 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 unacceptable. Finally, QA counts the number of mirrors in the sample that do not conform to quality specifica- tions. Listed next are the results of these checks over the last 10 business days.
Number Date Sampled Defects
10-Oct 50 1 50 0 50 9 50 9 11-Oct 50 4 50 4 50 5 50 3 12-Oct 50 9 50 3 50 10 50 2 13-Oct 50 2 50 4 50 9 50 4 14-Oct 50 6 50 9 50 2 50 4
Number Date Sampled Defects
17-Oct 50 7 50 9 50 0 50 8 18-Oct 50 6 50 9 50 6 50 1 19-Oct 50 4 50 5 50 2 50 5 20-Oct 50 0 50 0 50 4 50 7 21-Oct 50 5 50 1 50 9 50 9
Construct a p-chart for this process. What are the upper and lower control limits? Interpret the results. Does it appear the process is out of control during the period?
STATISTICAL PROCESS CONTROL AND QUALITY MANAGEMENT 715
S O L U T I O N
The first step is to determine the overall proportion defective. We use formula (19–6).
p = Total number defective
Total number of items sampled =
196 2,000
= .098
So we estimate that .098 of the mirrors produced during the period do not meet specifications.
Number Proportion Date Sampled Defects Defective
10-Oct 50 1 0.02 50 0 0.00 50 9 0.18 50 9 0.18 11-Oct 50 4 0.08 50 4 0.08 50 5 0.10 50 3 0.06 12-Oct 50 9 0.18 50 3 0.06 50 10 0.20 50 2 0.04 18-Oct 50 6 0.12 50 9 0.18 50 6 0.12 50 1 0.02 19-Oct 50 4 0.08 50 5 0.10 50 2 0.04 50 5 0.10
Number Proportion Date Sampled Defects Defective
13-Oct 50 2 0.04 50 4 0.08 50 9 0.18 50 4 0.08 14-Oct 50 6 0.12 50 9 0.18 50 2 0.04 50 4 0.08 17-Oct 50 7 0.14 50 9 0.18 50 0 0.00 50 8 0.16 20-Oct 50 0 0.00 50 0 0.00 50 4 0.08 50 7 0.14 21-Oct 50 5 0.10 50 1 0.02 50 9 0.18 50 9 0.18 Total 2,000 196
The upper and lower control limits are computed by using formula (19–8).
LCL, UCL = p ± 3 √ p(1 − p)
n = .098 ± 3 √
.098(1 − .098) 50
= .098 ± .1261
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 limits indicates the quality level of the process has changed.
This information is summarized in Chart 19–6, which is output from the Minitab system.
CHART 19–6 p-Chart for Mirrors at Jersey Glass
716 CHAPTER 19
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.
Number Proportion Date Sampled Defects Defective
24-Oct 50 1 0.02 50 13 0.26 50 10 0.20 50 7 0.14 25-Oct 50 4 0.08 50 5 0.10 50 6 0.12 50 10 0.20 26-Oct 50 6 0.12 50 1 0.02 50 8 0.16 50 4 0.08
Number Proportion Date Sampled Defects Defective
27-Oct 50 2 0.04 50 1 0.02 50 7 0.14 50 12 0.24 28-Oct 50 5 0.10 50 5 0.10 50 10 0.20 50 9 0.18
The process was out of control on two occasions, on October 24 when the proportion of defects was 0.26 and again on October 27 when the proportion of defects was 0.24. QA should report this information to the production department for the appropriate action. The Minitab output follows.
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 Blue Sky 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 others, 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 c be the mean number of defects per unit. Thus, c is the mean number of bags mishandled by Blue Sky 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%, limits on a c-bar chart by:
LCL , UCL = c ± 3√c (19–9)CONTROL LIMITS FOR THE NUMBER OF DEFECTS PER UNIT
STATISTICAL PROCESS CONTROL AND QUALITY MANAGEMENT 717
E X A M P L E
The publisher of the Oak Harbor Daily Telegraph is concerned about the number of misspelled words in the daily newspaper. In an effort to control the problem and promote the need for correct spelling, a control chart will be used. The numbers of misspelled words found in the final edition of the paper for the last 10 days are 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?
S O L U T I O N
During the 10-day period, there were a total of 37 misspelled words. The mean number of misspelled words per edition is 3.7. The number of misspelled words per edition follows the Poisson probability distribution. The standard deviation is the square root of the mean.
c = Σx n
= 5 + 6 + . . . + 4
10 =
37 10
= 3.7 s = √c = √3.7 = 1.924
To find the upper control limit, we use formula (19–9). The lower control limit is zero.
UCL = c + 3√c = 3.7 + 3√3.7 = 3.7 + 5.77 = 9.47
The computed lower control limit would be 3.7 − 3(1.924) = −2.07. However, the number 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 re- sults 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
718 CHAPTER 19
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 numbers of defective batteries found over the last 12 shifts are 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.
S E L F - R E V I E W 19–3
9. Below is a p-chart for a manufacturing process.
a. What is the mean proportion 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 mon-
itor the proportion 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 p-chart and label the mean proportion 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 con-
trol for any of the months? 11. A bicycle manufacturer randomly selects 10 frames each day and tests for
defects. The numbers of defective frames found over the last 14 days are 3, 2, 1, 3, 2, 2, 8, 2, 0, 3, 5, 2, 0, and 4. Construct a control chart for this process and com- ment on whether the process is “in control.”
12. During the process of producing toilet paper, Scott Paper randomly selects a toilet paper roll 5 times throughout the day and subjects each roll to a stress test to see how often the paper tears. Over a 3-day period, the testing of 15 rolls found the following number of defectives in each roll: 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 monitors the checkout scanners by randomly examining the receipts for scanning errors. On October 27th, they recorded the following number of scanner errors on each receipt: 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 concerned 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 rea- sonable number of voids at his facilities: 3, 8, 3, 4, 6, 5, 0, 1, 2, 4? Construct a con- trol chart for this process and comment on whether the process is “in control.”
E X E R C I S E S
STATISTICAL PROCESS CONTROL AND QUALITY MANAGEMENT 719
ACCEPTANCE SAMPLING The previous section was concerned with maintaining the quality of the product as it is being produced. In many business situa- tions, 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 purchase order is for 100,000 DVDs, packaged 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% defective. He would like to develop a plan to inspect incom- ing 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% defective tubes. It would like to develop a sampling plan to determine which lots meet the criterion and which do not.
• 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% 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 product 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 pro- tection against accepting lots that are below the quality standard. The best protection against inferior quality is 100% inspection. Unfortunately, the cost of 100% inspection is often prohibitive. Another problem with checking each item is that the test may be de- structive. If all lightbulbs were tested until burning out before they were shipped, there would be none left to sell. Also, 100% inspection may not lead to the identification of all defects because boredom might cause a loss of perception on the part of the inspec- tors. Thus, complete inspection is rarely employed in practical situations.
To evaluate the quality of incoming parts, we use a statistical 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 sampling. The inspection will determine the number of defects in the sample. This number is compared with a predetermined num- ber called the critical number or the acceptance number. The acceptance number is usually designated c. If the number 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% inspection.
Acceptance sampling is a decision-making process. There are two possible deci- sions: accept or reject the lot. In addition, there are two situations under which the deci- sion 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 decision is made. However, there are two other possibilities. The lot may actually contain 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 during the sample inspection. This is called the producer’s risk. The following summary table for acceptance decisions shows these possibilities.
LO19-7 Explain the process of acceptance sampling.
© Ingram Publishing
720 CHAPTER 19
Notice how this discussion is very similar to the ideas of Type I and Type II errors dis- cussed in Chapter 10.
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. An OC curve reports the percent defective 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 distri- bution is used to develop the probabilities for an OC curve.
E X A M P L E
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% 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. De- velop an OC curve for this inspection plan. What is the probability of accepting a lot that is 10% defective?
S O L U T I O N
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 π represent the actual proportion defective in the population.
The lot is good if π ≤ .10. The lot is bad if π > .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% or fewer defective DVDs. If the lot is ac- ceptable when it has exactly 10% defectives, it would be even more acceptable if it contained fewer than 10% 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: the DVD is either acceptable or unacceptable.
2. There are a fixed number of trials. The number of trials is the sample 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.
STATISTICAL PROCESS CONTROL AND QUALITY MANAGEMENT 721
Appendix B.1 gives various binomial probabilities. However, the tables in Appendix B.1 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 π is equal to .05, .10, .15, .20, .25, and .30.
We need to convert the terms used in Chapter 6 to acceptance sampling vocabulary. We let π 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% defective. This means that π = .05, c = 2, and n = 20. From the Excel output, the likelihood of selecting a sample of 20 items from a shipment that contained 5% 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 notation as follows (recall that bar “|” means “given that”).
P(x ≤ 2 | π = .05 and n = 20) = .358 + .377 + .189 = .924
Continuing, the likelihood of accepting a lot that is actually 10% defective is .677. That is:
P(x ≤ 2 | π = .10 and n = 20) = .122 + .270 + .285 = .677
The complete OC curve in Chart 19–8 shows the smoothed curve for all values of π between 0 and about 30%. There is no need to show values larger than 30% because their probability is very close to 0. The likelihood of accepting lots of se- lected 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 evalu- ate the probabilities of various quality levels.
722 CHAPTER 19
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
t 0 5 10 15 20 25 30
Incoming lot percent defective
CHART 19–8 OC Curve for Sampling Plan (n = 20, c = 2)
Compute the probability of accepting a lot of DVDs that is actually 30% defective, using the sampling plan for Sims Software.
S E L F - R E V I E W 19–4
15. Determine the probability of accepting lots that are 10%, 20%, 30%, and 40% de- fective using a sample of size 12 and an acceptance number of 2.
16. Determine the probability of accepting lots that are 10%, 20%, 30%, and 40% de- fective 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 outgoing 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%, 20%, 30%, and 40% 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 sam- pling 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%, 20%, 30%, and 40%. You will need a statistical software package.
E X E R C I S E S
C H A P T E R S U M M A R Y
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% of the activity is caused by 20% 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, equip-
ment, 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.
STATISTICAL PROCESS CONTROL AND QUALITY MANAGEMENT 723
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 and minus 3 standard errors
from the mean. b. The formulas for the upper and lower control limits for the mean are:
UCL = x + A2 R LCL = x − A2 R (19–4)
c. The formulas for the upper and lower control limits for the range are:
UCL = D4 R LCL = D3 R (19–5)
2. A p-chart is an attribute chart that shows the proportion of the product or service that does not conform to the standard. a. The mean proportion defective is found by
p = Total number defective
Total number of items sampled (19–6)
b. The control limits for the proportion defective are determined from the equation
LCL, UCL = p ± 3 √ p(1 − p)
n (19–8)
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. c. The control limits are determined from the following equation.
LCL, UCL = c ± 3√c (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
distribution to determine the probability of accepting lots of various quality levels.
P R O N U N C I A T I O N K E Y
SYMBOL MEANING PRONUNCIATION
x Mean of the sample means x double bar
sx̄ Standard error of the mean s sub x bar
A2 Constant used to determine the upper A sub 2 and lower control limit for the mean
R Mean of the sample ranges R bar
D4 Constant used to determine the D sub 4 upper control limit for the range
c̄ Mean number of defects per unit c bar
724 CHAPTER 19
C H A P T E R E X E R C I S E S
19. The production supervisor at Westburg Electric Inc. noted an increase in the number of elec- tric 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 fail- ures found. Develop a Pareto chart to show the major problem areas.
Type of Nonconformity Frequency
Sole separation 34 Heel separation 98 Sole penetration 62
Type of Nonconformity Frequency
Lace breakage 14 Eyelet failure 10 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 ounce. Is the process in con- trol? Should other action be taken?
22. A new machine has just been installed to produce printed circuit boards. One of the critical measurements is the thickness of the resistance layer. The quality control inspector randomly selects five boards each half-hour, measures the thickness, and records the results. The measurements (in millimeters) for the period 8:00 a.m. to 10:30 a.m. follow.
Thickness (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 outside diameter and the range. c. Are there any points on the mean or the range chart that are out of control? Comment
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 se- lected from different shifts over the last month of operation. The tread wear is reported below in hundredths of an inch.
Sample Tread Wear
1 44 41 19 2 39 31 21 3 38 16 25 4 20 33 26 5 34 33 36 6 28 23 39 7 40 15 34 8 36 36 34 9 32 29 30 10 29 38 34
Sample Tread Wear
11 11 33 34 12 51 34 39 13 30 16 30 14 22 21 35 15 11 28 38 16 49 25 36 17 20 31 33 18 26 18 36 19 26 47 26 20 34 29 32
STATISTICAL PROCESS CONTROL AND QUALITY MANAGEMENT 725
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
throughout 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.
Loan Amount ($000)
Officer 1 2 3 4 5
Weinraub 59 74 53 48 65 Visser 42 51 70 47 67 Moore 52 42 53 87 85 Brunner 36 70 62 44 79 Wolf 34 59 39 78 61
Loan Amount ($000)
Officer 1 2 3 4 5
Bowyer 66 80 54 68 52 Kuhlman 74 43 45 65 49 Ludwig 75 53 68 50 31 Longnecker 42 65 70 41 52 Simonetti 43 38 10 19 47
25. During the 2016 Masters Golf tournament Jordan Spieth failed to repeat as champion. His scores relative to par in each round, on each hole are reported in the table below. Develop and review appropriate control charts for his performance. Report your findings.
Hole Round 1 Round 2 Round 3 Round 4
1 0 –1 0 0 2 0 0 –1 –1 3 –1 –1 0 0 4 0 0 0 0 5 0 2 0 1 6 –1 0 0 –1 7 0 0 1 –1 8 –1 –1 –1 –1 9 0 1 0 –1 10 –1 1 0 1 11 0 0 2 1 12 0 0 –1 4 13 –1 0 0 –1 14 0 0 –1 0 15 0 –1 –1 –1 16 0 1 0 0 17 0 1 1 1 18 –1 0 2 0
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 packages delivered each of the last 15 working days, the following numbers 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 qual- ity control inspector selects 50 bolts at random and determines how many are defec- tive. The numbers of defectives in the first 10 samples are 3, 5, 0, 4, 1, 2, 6, 5, 7, and 7.
726 CHAPTER 19
a. Design a p-chart. Insert the mean proportion defective, UCL, and LCL. b. Plot the proportion 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 package 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 for 25 samples are reported below.
Sample Weights
1 26.1 24.4 25.6 25.2 2 25.2 25.9 25.1 24.8 3 25.6 24.5 25.7 25.1 4 25.5 26.8 25.1 25.0 5 25.2 25.2 26.3 25.7 6 26.6 24.1 25.5 24.0 7 27.6 26.0 24.9 25.3 8 24.5 23.1 23.9 24.7 9 24.1 25.0 23.5 24.9 10 25.8 25.7 24.3 27.3 11 22.5 23.0 23.7 24.0 12 24.5 24.8 23.2 24.2 13 24.4 24.5 25.9 25.5
Sample Weights
14 23.1 23.3 24.4 24.7 15 24.6 25.1 24.0 25.3 16 24.4 24.4 22.8 23.4 17 25.1 24.1 23.9 26.2 18 24.5 24.5 26.0 26.2 19 25.3 27.5 24.3 25.5 20 24.6 25.3 25.5 24.3 21 24.9 24.4 25.4 24.8 22 25.7 24.6 26.8 26.9 23 24.8 24.3 25.0 27.2 24 25.4 25.9 26.6 24.8 25 26.2 23.5 23.7 25.0
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 partic- ular 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
Develop p-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.
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
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% 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% of her free throw attempts. In an effort to improve
STATISTICAL PROCESS CONTROL AND QUALITY MANAGEMENT 727
this statistic, she attended a summer camp devoted to shooting techniques. The next 20 days she shot 100 free throws each day. She carefully recorded the number of free throws that she made each day. The results are reported below.
55 61 52 59 67 57 61 59 69 58 57 66 63 63 63 65 63 68 64 67
To interpret, the first day she made 55 out of 100, or 55%. The second day she made 61 shots, the third day, 52 shots, the tenth day 58, the eleventh day 57 shots. The last day she made 67 out of 100, or 67%. 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 5 days of practice. Use the hypothesis testing procedure, formula (15-1), to determine if there is an improvement from 55%.
33. Eric’s Cookie House sells chocolate chip cookies in shopping malls. Of concern is the number of chocolate chips in each cookie. Eric, the owner and president, would like to es- tablish 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 control
in any of the cookies sampled? 34. The numbers of “near misses” recorded for the last 20 months at Lima Interna-
tional Airport are 3, 2, 3, 2, 2, 3, 5, 1, 2, 2, 4, 4, 2, 6, 3, 5, 2, 5, 1, and 3. Develop an ap- propriate 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 numbers of robberies were reported during the last 10 days to the robbery division of the Metro City Police: 10, 8, 8, 7, 8, 5, 8, 5, 4, and 7. Develop an appropriate control chart. Determine the mean number of robberies reported per day and determine the control limits. Are there any days when the number of robberies re- ported 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% defective will be
accepted? b. Design an OC curve for incoming lots that have zero, 10%, 20%, 30%, and 40%
defective stems. 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%, 20%, and 30% defective? b. Estimate the probability of accepting a lot that is 15% defective. c. John Brennen, owner of Team Sports, would like the probability of accepting a lot
that is 5% defective to be more than 90%. Does this appear to be the case with this sampling plan?
An Introduction to Decision Theory20
LEARNING OBJECTIVES When you have completed this chapter, you will be able to:
LO20-1 Identify and apply the three components of a decision.
LO20-2 Analyze a decision using expected monetary value.
LO20-3 Analyze a decision using opportunity loss.
LO20-4 Apply maximin, maximax, and minimax regret strategies to make a decision.
LO20-5 Compute and explain the expected value of perfect information.
LO20-6 Apply sensitivity analysis to evaluate a decision subject to uncertainty.
LO20-7 Use a decision tree to illustrate and analyze decision making under uncertainty.
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 LO20-2.)
© Mark Horn/Getty Images
AN INTRODUCTION TO DECISION THEORY 729
INTRODUCTION Decision theory is a branch of statistics that evaluates two or more decision alternatives in the face of an uncertain future. Probabilities are used to forecast the likelihood of future events. As the name implies, the focus is on the process of making decisions and explicitly includes the payoffs that may result from selecting a particular decision alter- native. In contrast, classical statistics focuses on estimating a parameter, such as the population mean, constructing a confidence interval, or conducting a hypothesis test. Classical statistics does not address the 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 follow- ing examples of decision-theory problems.
• Ford Motor Company must decide whether to purchase assembled door locks for the 2017 Ford F-150 truck or to manufacture and assemble the door locks at its Louisville, Kentucky, plant. If sales of the F-150 truck continue to in- crease, it will be more profitable to manufacture 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 re- gions 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 pro- portion of viewers who wear jackets. Should it purchase commercial television time?
• General Electric is considering three options regarding the prices of refrigerators for next year. GE could (1) raise the prices 5%, (2) raise the prices 2.5%, 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 manufacturers 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.
ELEMENTS OF A DECISION There are three components to any decision: (1) the choices available, or alternatives; (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 Louisville, or it can decide to purchase them. Banana Republic has two advertising alternatives: to purchase commer- cial television time or not. GE is considering three pricing alternatives regarding the marketing of refrigerators. To simplify our presentation, we assume the decision maker can select from a rather small number of outcomes. With the help of computers, how- ever, the decision alternatives can be expanded to a large number of possibilities.
LO20-1 Identify and apply the three components of a decision.
© Jeff Kowalsky/Bloomberg/Getty Images
730 CHAPTER 20
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 basketball final. GE does not know how competitors will price refrigerators.
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 Louisville 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 estimated to be $22,000. Banana Republic needs to estimate the payoffs for purchasing commercial television time given estimates of the potential audience for the rain jacket. GE needs to estimate the payoffs for each pricing alternative given estimates of competitors’ pricing.
The main elements of the decision under conditions of uncertainty are identified schematically:
Act
Event Uncertainty regarding future demand. State of nature (future demand) unknown. Decision maker has no control over state of nature.
h
h Pro�t.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.
h
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 subjective esti- mates. Ford may estimate the probability of continued high demand as .70. Banana Republic may estimate a probability of .75 that the commercial will reach the targeted audience. GE may estimate the probability to be .25 that Amana and other manufactur- ers will raise the prices of their refrigerators.
DECISION MAKING UNDER CONDITIONS OF UNCERTAINTY We begin this section with an example of decision making under uncertainty. The pur- pose of the example is to explain a logical process to evaluate a set of alternatives and select an alternative as the decision. 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. However, 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 regarding the value of his $1,100 investment for the three stocks for a bull market and for a bear mar- ket are shown in Table 20–1. This table is a payoff table.
LO20-2 Analyze a decision using expected monetary value.
AN INTRODUCTION TO DECISION THEORY 731
TABLE 20–1 Payoff Table for Three Common Stocks under Two Market Conditions
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
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 conser- vative 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 Chem- icals, 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. Suppose a study of these investment services 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 proba- bility estimates (.60 and .40) are combined to arrive at the expected payoff of buying each of the three stocks. Expected payoff is also called expected monetary value, shortened to EMV. It can also be described as the mean payoff. The calculations needed to determine the expected payoff for the act of purchasing Kayser Chemicals are shown in Table 20–2.
TABLE 20–2 Expected Payoff for the Act of Buying Kayser Chemicals, EMV (A1)
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
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% 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” pay- off in the long run.
These calculations are summarized as follows.
EXPECTED MONETARY VALUE EMV(Ai) = Σ[P(Sj) · V(Ai, Sj)] (20–1)
732 CHAPTER 20
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:
EMV(A1) = [P(S1) · V(A1, S1)] + [P(S2) · V(A1, S2)] = .60($2,400) + .40($1,000) = $1,840
Purchasing Kayser Chemicals stock is only one possible choice. The expected pay- offs for the acts of buying Kayser Chemicals, Rim Homes, and Texas Electronics are given in Table 20–3.
TABLE 20–3 Expected Payoffs for Three Stocks
Expected Purchase Payoff
Kayser Chemicals $1,840 Rim Homes 1,760 Texas Electronics 1,600
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 experience with respect to the rise and decline of stock prices. It should be emphasized that although purchas- ing Kayser stock represents the best action under the expected-value criterion, the in- vestor still might decide to buy Texas Electronics stock in order to minimize the risk of losing some of the $1,100 investment.
Verify the conclusion, shown in Table 20–3, that the expected payoff for the act of purchas- ing Rim Homes stock is $1,760.
S E L F - R E V I E W 20–1
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
E X E R C I S E S
AN INTRODUCTION TO DECISION THEORY 733
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 behavior) 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 com- mon 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 in- vestor 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 an- other way, the $200 represents the opportunity loss for not knowing the future 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 in- creased, 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.
TABLE 20–4 Opportunity Losses for Various Combinations of Stock Purchase and Market Movement
Opportunity Loss
Purchase Market Rise Market Decline
Kayser Chemicals $ 0 $150 Rim Homes 200 50 Texas Electronics 500 0
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.
LO20-3 Analyze a decision using opportunity loss.
2. Wilhelms Cola Company plans to market a new lime-flavored cola this summer. The decision is whether to package the cola in returnable or in nonreturnable bottles. Currently, the state legislature is considering eliminating nonreturnable bottles. Tybo Wilhelms, president of Wilhelms Cola Company, has discussed the problem with his state representative and established the probability to be .70 that nonre- turnable 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 nonre- turnable 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
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.
S E L F - R E V I E W 20–2
734 CHAPTER 20
Expected Opportunity Loss The opportunity losses in Table 20–4 again ignore the historical experience of market movements. Recall that the probability of a market rise is .60 and that of a market de- cline .40. These probabilities and the opportunity losses can be combined 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.
TABLE 20–5 Expected Opportunity Loss for the Act of Buying Rim Homes Stock
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
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 de- cided 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 mar- ket, 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.
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.
E X E R C I S E S
These calculations are summarized as follows:
EXPECTED OPPORTUNITY LOSS EOL(Ai) = Σ[P(Sj) · R(Ai, Sj)] (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 computed as follows:
EOL(A2) = [P(S1) · R(A2, S1)] + [P(S2) · R(A2, S2)]
= .60($200) + .40($50) = $140
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.
AN INTRODUCTION TO DECISION THEORY 735
TABLE 20–6 Expected Opportunity Losses for the Three Stocks
Expected Opportunity Purchase Loss
Kayser Chemicals $ 60 Rim Homes 140 Texas Electronics 300
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 ex- pected payoff ) will always lead to the same decision concerning which course of action to follow.
Referring to Table 20–6, verify that the expected opportunity loss for the act of purchasing Texas Electronics is $300.
S E L F - R E V I E W 20–3
5. Refer to Exercises 1 and 3. Compute the expected opportunity losses. 6. Refer to Exercises 2 and 4. Compute the expected opportunity losses.
E X E R C I S E S
MAXIMIN, MAXIMAX, AND MINIMAX REGRET STRATEGIES Suppose several financial advisors believe the decision to purchase Kayser Chemi- cals stock is 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 maximin strategy: it maximizes the minimum gain. On the basis of the payoff table (Table 20–1), they reason that the investor would be assured a profit of at least $1,150. 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 possibil- ity 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 ap- proach 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 opportunity losses! The maximum regrets were $200 for Rim Homes and $500 for Texas Electronics.
LO20-4 Apply maximin, maximax, and minimax regret strategies to make a decision.
736 CHAPTER 20
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 ad- vance information worth? The dollar value of this information is called the expected value of perfect information, written EVPI. In this example, if perfect information is available, Bob Hill would know what will happen in the future and whether the stock market will rise or decline. If Bob had this information, he would always make the best decision. The ques- tion is: What is the most that Bob should be willing to pay for perfect information?
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 mar- ket 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.
LO20-5 Compute and explain the expected value of perfect information.
VALUE OF PERFECT INFORMATION The difference between the maximum expected payoff under conditions of certainty and the maximum expected 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 optimum decision using the expected-value criterion.
To explain, 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.)
TABLE 20–7 Calculations for the Expected Payoff under Conditions of Certainty
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
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 infor- mation 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:
EXPECTED VALUE OF PERFECT INFORMATION
EVPI = Expected value under conditions of certainty − Expected value under conditions of uncertainty
(20–3)
AN INTRODUCTION TO DECISION THEORY 737
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 $1,900, and if the analyst asked $40 for the information, the investor would be as- sured 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 ser- vice would not be worth $100 because the investor could expect $1,840 on aver- age 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.
Payoff Table
Opportunity Loss Table
Purchase
Kayser Rim Texas
$2,400 2,200 1,900
$1,000 1,100 1,150
$1,840 1,760 1,600
Kayser Rim Texas
$ 0 200 500
$ 150 50
1,150
$ 60 140 760
Bull Market Bear Market Expected Value
Purchase Bull Market Bear Market Expected Value
The expected payoff and the expected opportunity loss are the same as re- ported in Table 20–3 and Table 20–6, respectively. The calculations in the preced- ing 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.
SENSITIVITY ANALYSIS In the foregoing stock selection situation, the set of probabilities applied to the payoff values was derived from historical experience with similar market conditions. Objec- tions 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 example, sup- pose the investor’s brother believes that instead of a 60% chance of a market rise and a 40% 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 probabil- ities suggested by the investor’s brother (center 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.
LO20-6 Apply sensitivity analysis to evaluate a decision subject to uncertainty.
738 CHAPTER 20
TABLE 20–8 Expected Payoffs for Three Sets of Probabilities
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
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.)
S E L F - R E V I E W 20–5
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.
S E L F - R E V I E W 20–4
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 likelihood of 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 probability estimates need to be in order to feel comfortable with your choice.
TABLE 20–9 Expected Values for Purchasing the Three Stocks
Expected Purchase Payoff
Kayser Chemicals $1,280 Rim Homes 1,320 Texas Electronics 1,300
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 and use Expected Monetary Value to evaluate the decision. Does this change the decision?
10. Refer to Exercise 2. Reverse the probabilities; that is, let P(S1) = .30 and P(S2) = .70 and use Expected Monetary Value to evaluate the decision. Does this alter your decision?
E X E R C I S E S
AN INTRODUCTION TO DECISION THEORY 739
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 alternatives under consideration. Referring to Chart 20–1, on the left is the box with three branches radiating from it, repre- senting the acts of purchasing Kayser Chemicals, Rim Homes, or Texas Electronics.
$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
3
$1,840
CHART 20–1 Decision Tree for the Investor’s Decision
The three nodes, or circles, numbered 1, 2, and 3, represent the expected payoff 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 paren- theses. The numbers at the extreme ends of the branches are the estimated future val- ues 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 is 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 an- ticipated 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 ex- pected value of $1,600 above circled node 3 as shown in Chart 20–1. Similarly, the inves- tor would determine the expected values for Rim Homes and Kayser Chemicals.
Assuming the investor wants to maximize the expected value of his stock purchase, $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 representing the two alter- natives 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 deci- sion tree analysis (see Chart 20–2). Recall that under conditions of certainty the 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
LO20-7 Use a decision tree to illustrate and analyze decision making under uncertainty.
740 CHAPTER 20
Electronics in a falling market, and the expected payoff would be $1,900. Again, back- ward induction would be used to arrive at the expected payoff of $1,900.
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
CHART 20–2 Decision Tree Given Perfect Information
The monetary difference based on perfect information in Chart 20–2 and the deci- sion 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 method to perform the calculations presented earlier in the chapter. Some managers find these graphic sketches help them in following the decision logic.
C H A P T E R S U M M A R Y
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 as-
signed to the states of nature. C. The consequence of a particular decision alternative and state of nature is 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. The expected monetary value (EMV) computes the expected value for each decision. If the payoffs are costs, the decision with the smallest EMV is selected. If the payoffs are profits, the decision with the largest EMV is selected.
B. A decision can be selected using an opportunity loss table. 1. To construct an opportunity loss table, first calculate the difference between the best
payoff and the payoffs of the other decision alternatives for each state of nature. 2. The difference between the optimal decision and any other decision is the oppor-
tunity 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 na- ture for each decision alternative to determine the expected opportunity loss.
C. A maximin strategy compares the lowest payoffs of the decision alternatives and se- lects the decision alternative with the maximum of these payoffs.
D. A maximax strategy selects the decision alternative with the highest payoff. E. A minimax regret strategy, first, converts payoffs to regrets or opportunity losses.
Then, decision alternatives are compared based on the maximum regrets and the al- ternative with the smallest of the maximum regrets is selected.
III. The expected value of perfect information (EVPI) is the difference between the best ex- pected 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.
AN INTRODUCTION TO DECISION THEORY 741
C H A P T E R E X E R C I S E S
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 minutes 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% 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 pres- ident 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 recession. 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%. If there is a recession, he predicts a 10% loss if he expands his production and a 5% loss if he invests in stocks. If there is not a recession, an expansion of production will result in a 15% gain, and stock investment will produce a 12% 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 alternatives: inspect all the parts or inspect none of the parts. The proportion of parts defective 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
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)
742 CHAPTER 20
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 vacations. Various surveys indicated, however, that there was a considerable interest in this type of family vacation, which includes horseback riding, cattle drives, swim- ming, fishing, and the like. Dude Ranches Incorporated bought a large farm near sev- eral 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 de- cided to enter into an agreement with Mobile Homes Manufacturing Company to sup- ply 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 pre- fer 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, however, the valuable experience that Dude Ranches Incorporated has acquired in the past 4 years (about 200 weeks) actually operating a dude ranch in the South- west. 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 facili- ties before renting. A frequency distribution showing the number of weeks in which 10, 11, . . . , 14 ranch cottages were rented during the 200-week period is found in the following table.
Number of Number of Cottages 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 con- sidering purchasing or leasing several snowmobiles for the use of guests. The owner found that other financial obligations made it impossible to purchase the machines. Snowmobiles Incorporated (SI) will lease a machine for $20 a week, including any needed maintenance. 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 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:
AN INTRODUCTION TO DECISION THEORY 743
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 equipment of this nature is available locally, and the management of the furniture store is considering 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 be- cause 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 season. Based on this suggestion, management must now decide on the most prof- itable number of complete sets to lease for the season.
The leasing firm in Boston also made available some additional information gath- ered from several rental firms similar to the newly formed subsidiary. Note in the follow- ing 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% of the days—these rental firms rented out 41 complete party sets. On 10% of the days during a typical summer, they rented 42 complete sets, and so on.
Number of Number of Sets Rented Days
40 0 41 360 42 600 43 840
Number of Number of Sets Rented Days
44 2,400 45 1,500 46 300 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
744 CHAPTER 20
opportunity loss table, and complete the table by computing the expected opportu- nity 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 ar- rangement with Landrum Leasing to purchase used cars at $6,000 each. His cars re- ceive 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 2 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 probabilities 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 in- crease 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% that the engine is out of tune. What should you do?
Appendixes Introduction
745
A P P E N D I X C : SOFTWARE COMMANDS
A P P E N D I X D : ANSWERS TO ODD-NUMBERED CHAPTER E X E R C I S E S & R E V I E W E X E R C I S E S & S O L U T I O N S T O P R A C T I C E T E S T S
A P P E N D I X E : ANSWERS TO SELF-REVIEW
A P P E N D I X A : D A T A S E T S
A.1 Data Set 1—North Valley Real Estate Data A.2 Data Set 2—Baseball Statistics, 2015 Season A.3 Data Set 3—Lincolnville School District Bus Data A.4 Data Set 4—Applewood Auto Group A.5 Banking Data Set—Century National Bank Case
A P P E N D I X B : T A B L E S
B.1 Binomial Probability Distribution B.2 Poisson Distribution B.3 Areas under the Normal Curve B.4 Table of Random Numbers B.5 Student’s t Distribution B.6A Critical Values of the F Distribution, (α = .05) B.6B Critical Values of the F Distribution, (α = .01) B.7 Critical Values of Chi-Square B.8 Wilcoxon T Values B.9A Critical Values for the Durbin-Watson d Statistic (α = .05) B.9B Critical Values for the Durbin-Watson d Statistic (α = .025) B.9C Critical Values for the Durbin-Watson d Statistic (α = .01) B.10 Factors for Control Charts
746
APPENDIX A
A.1 Data Set 1—North Valley Real Estate Data
Variables Record = Property identification number Agent = Name of the real estate agent assigned to the property Price = Market price in dollars Size = Livable square feet of the property Bedrooms = Number of bedrooms Baths = Number of bathrooms Pool = Does the home have a pool? (1 = yes, 0 = no) Garage = Does the home have an attached garage (1 = yes, 0 = no) Days = Number of days of the property on the market Township = Area where the property is located Mortgage type = Fixed or adjustable. The fixed mortgage is a 30 year, fixed interest rate loan. The adjustable rate loan begins with an introductory interest rate of 3% for the first five years, then the interest rate is based on the current interest rates plus 1% (i.e. the interest rate AND the payment is likely to change each year after the 5th year) Years = the number of years that the mortgage loan has been paid FICO = the credit score of the mortgage loan holder. The highest score is 850; an average score is 680, a low score is below 680. The score reflects a person’s ability to pay their debts. Default = Is the mortgage loan in default? (1 = yes, 0 = no)
Pool Garage Mortgage Default Record Agent Price Size Bedrooms Baths (Yes is 1) (Yes is 1) Days Township type Years FICO (Yes is 1)
1 Marty 206424 1820 2 1.5 1 1 33 2 Fixed 2 824 0 2 Rose 346150 3010 3 2 0 0 36 4 Fixed 9 820 0 3 Carter 372360 3210 4 3 0 1 21 2 Fixed 18 819 0 4 Peterson 310622 3330 3 2.5 1 0 26 3 Fixed 17 817 0 5 Carter 496100 4510 6 4.5 0 1 13 4 Fixed 17 816 0 6 Peterson 294086 3440 4 3 1 1 31 4 Fixed 19 813 0 7 Carter 228810 2630 4 2.5 0 1 39 4 Adjustable 10 813 0 8 Isaacs 384420 4470 5 3.5 0 1 26 2 Fixed 6 812 0 9 Peterson 416120 4040 5 3.5 0 1 26 4 Fixed 3 810 0 10 Isaacs 487494 4380 6 4 1 1 32 3 Fixed 6 808 0 11 Rose 448800 5280 6 4 0 1 35 4 Fixed 8 806 1 12 Peterson 388960 4420 4 3 0 1 50 2 Adjustable 9 805 1 13 Marty 335610 2970 3 2.5 0 1 25 3 Adjustable 9 801 1 14 Rose 276000 2300 2 1.5 0 0 34 1 Fixed 20 798 0 15 Rose 346421 2970 4 3 1 1 17 3 Adjustable 10 795 0 16 Isaacs 453913 3660 6 4 1 1 12 3 Fixed 18 792 0 17 Carter 376146 3290 5 3.5 1 1 28 2 Adjustable 9 792 1 18 Peterson 694430 5900 5 3.5 1 1 36 3 Adjustable 10 788 0 19 Rose 251269 2050 3 2 1 1 38 3 Fixed 16 786 0 20 Rose 547596 4920 6 4.5 1 1 37 5 Fixed 2 785 0 21 Marty 214910 1950 2 1.5 1 0 20 4 Fixed 6 784 0 22 Rose 188799 1950 2 1.5 1 0 52 1 Fixed 10 782 0 23 Carter 459950 4680 4 3 1 1 31 4 Fixed 8 781 0 24 Isaacs 264160 2540 3 2.5 0 1 40 1 Fixed 18 780 0 25 Carter 393557 3180 4 3 1 1 54 1 Fixed 20 776 0 26 Isaacs 478675 4660 5 3.5 1 1 26 5 Adjustable 9 773 0 27 Carter 384020 4220 5 3.5 0 1 23 4 Adjustable 9 772 1 28 Marty 313200 3600 4 3 0 1 31 3 Fixed 19 772 0 29 Isaacs 274482 2990 3 2 1 0 37 3 Fixed 5 769 0 30 Marty 167962 1920 2 1.5 1 1 31 5 Fixed 6 769 0
(continued )
747
A.1 Data Set 1—North Valley Real Estate Data (continued)
Pool Garage Mortgage Default Record Agent Price Size Bedrooms Baths (Yes is 1) (Yes is 1) Days Township type Years FICO (Yes is 1)
31 Isaacs 175823 1970 2 1.5 1 0 28 5 Adjustable 9 766 1 32 Isaacs 226498 2520 4 3 1 1 28 3 Fixed 8 763 1 33 Carter 316827 3150 4 3 1 1 22 4 Fixed 2 759 1 34 Carter 189984 1550 2 1.5 1 0 22 2 Fixed 17 758 0 35 Marty 366350 3090 3 2 1 1 23 3 Fixed 5 754 1 36 Isaacs 416160 4080 4 3 0 1 25 4 Fixed 12 753 0 37 Isaacs 308000 3500 4 3 0 1 37 2 Fixed 18 752 0 38 Rose 294357 2620 4 3 1 1 15 4 Fixed 10 751 0 39 Carter 337144 2790 4 3 1 1 19 3 Fixed 15 749 0 40 Peterson 299730 2910 3 2 0 0 31 2 Fixed 13 748 0 41 Rose 445740 4370 4 3 0 1 19 3 Fixed 5 746 0 42 Rose 410592 4200 4 3 1 1 27 1 Adjustable 9 741 1 43 Peterson 667732 5570 5 3.5 1 1 29 5 Fixed 4 740 0 44 Rose 523584 5050 6 4 1 1 19 5 Adjustable 10 739 0 45 Marty 336000 3360 3 2 0 0 32 3 Fixed 6 737 0 46 Marty 202598 2270 3 2 1 0 28 1 Fixed 10 737 0 47 Marty 326695 2830 3 2.5 1 0 30 4 Fixed 8 736 0 48 Rose 321320 2770 3 2 0 1 23 4 Fixed 6 736 0 49 Isaacs 246820 2870 4 3 0 1 27 5 Fixed 13 735 0 50 Isaacs 546084 5910 6 4 1 1 35 5 Adjustable 10 731 0 51 Isaacs 793084 6800 8 5.5 1 1 27 4 Fixed 6 729 0 52 Isaacs 174528 1600 2 1.5 1 0 39 2 Fixed 15 728 0 53 Peterson 392554 3970 4 3 1 1 30 4 Fixed 17 726 0 54 Peterson 263160 3060 3 2 0 1 26 3 Fixed 10 726 0 55 Rose 237120 1900 2 1.5 1 0 14 3 Fixed 18 723 0 56 Carter 225750 2150 2 1.5 1 1 27 2 Fixed 15 715 0 57 Isaacs 848420 7190 6 4 0 1 49 1 Fixed 5 710 0 58 Carter 371956 3110 5 3.5 1 1 29 5 Fixed 8 710 0 59 Carter 404538 3290 5 3.5 1 1 24 2 Fixed 14 707 0 60 Rose 250090 2810 4 3 0 1 18 5 Fixed 11 704 0 61 Peterson 369978 3830 4 2.5 1 1 27 4 Fixed 10 703 0 62 Peterson 209292 1630 2 1.5 1 0 18 3 Fixed 10 701 0 63 Isaacs 190032 1850 2 1.5 1 1 30 4 Adjustable 2 675 0 64 Isaacs 216720 2520 3 2.5 0 0 2 4 Adjustable 5 674 1 65 Marty 323417 3220 4 3 1 1 22 4 Adjustable 2 673 0 66 Isaacs 316210 3070 3 2 0 0 30 1 Adjustable 1 673 0 67 Peterson 226054 2090 2 1.5 1 1 28 1 Adjustable 6 670 0 68 Marty 183920 2090 3 2 0 0 30 2 Adjustable 8 669 1 69 Rose 248400 2300 3 2.5 1 1 50 2 Adjustable 4 667 0 70 Isaacs 466560 5760 5 3.5 0 1 42 4 Adjustable 3 665 0 71 Rose 667212 6110 6 4 1 1 21 3 Adjustable 8 662 1 72 Peterson 362710 4370 4 2.5 0 1 24 1 Adjustable 2 656 0 73 Rose 265440 3160 5 3.5 1 1 22 5 Adjustable 3 653 0 74 Rose 706596 6600 7 5 1 1 40 3 Adjustable 7 652 1 75 Marty 293700 3300 3 2 0 0 14 4 Adjustable 7 647 1 76 Marty 199448 2330 2 1.5 1 1 25 3 Adjustable 5 644 1 77 Carter 369533 4230 4 3 1 1 32 2 Adjustable 2 642 0 78 Marty 230121 2030 2 1.5 1 0 21 2 Adjustable 3 639 0 79 Marty 169000 1690 2 1.5 0 0 20 1 Adjustable 7 639 1 80 Peterson 190291 2040 2 1.5 1 1 31 4 Adjustable 6 631 1 81 Rose 393584 4660 4 3 1 1 34 3 Adjustable 7 630 1 82 Marty 363792 2860 3 2.5 1 1 48 5 Adjustable 3 626 0 83 Carter 360960 3840 6 4.5 0 1 32 2 Adjustable 5 626 1 84 Carter 310877 3180 3 2 1 1 40 1 Adjustable 6 624 1 85 Peterson 919480 7670 8 5.5 1 1 30 4 Adjustable 1 623 0 86 Carter 392904 3400 3 2 1 0 40 2 Adjustable 8 618 1 87 Carter 200928 1840 2 1.5 1 1 36 4 Adjustable 3 618 1
(continued )
748
Pool Garage Mortgage Default Record Agent Price Size Bedrooms Baths (Yes is 1) (Yes is 1) Days Township type Years FICO (Yes is 1)
88 Carter 537900 4890 6 4 0 1 23 1 Adjustable 7 614 0 89 Rose 258120 2390 3 2.5 0 1 23 1 Adjustable 6 614 1 90 Carter 558342 6160 6 4 1 1 24 3 Adjustable 7 613 0 91 Marty 302720 3440 4 2.5 0 1 38 3 Adjustable 3 609 1 92 Isaacs 240115 2220 2 1.5 1 0 39 5 Adjustable 1 609 0 93 Carter 793656 6530 7 5 1 1 53 4 Adjustable 3 605 1 94 Peterson 218862 1930 2 1.5 1 0 58 4 Adjustable 1 604 0
95 Peterson 383081 3510 3 2 1 1 27 2 Adjustable 6 601 1 96 Marty 351520 3380 3 2 0 1 35 2 Adjustable 8 599 1 97 Peterson 841491 7030 6 4 1 1 50 4 Adjustable 8 596 1 98 Marty 336300 2850 3 2.5 0 0 28 1 Adjustable 6 595 1 99 Isaacs 312863 3750 6 4 1 1 12 4 Adjustable 2 595 0 100 Carter 275033 3060 3 2 1 1 27 3 Adjustable 3 593 0 101 Peterson 229990 2110 2 1.5 0 0 37 3 Adjustable 6 591 1 102 Isaacs 195257 2130 2 1.5 1 0 11 5 Adjustable 8 591 1 103 Marty 194238 1650 2 1.5 1 1 30 2 Adjustable 7 590 1 104 Peterson 348528 2740 4 3 1 1 27 5 Adjustable 3 584 1 105 Peterson 241920 2240 2 1.5 0 1 34 5 Adjustable 8 583 1
A.1 Data Set 1—North Valley Real Estate Data (concluded)
749
A.2 Data Set 2—Baseball Statistics, 2015 Season
Variables Team = Team’s name League = American or National League Year Opened = First year the team’s stadium was used Team Salary = Total team salary expressed in millions of dollars Attendance = Total number of people attending regular season games Wins = Number of regular season games won ERA = Team earned run average BA = Team batting average HR = Team home runs Year = Year of operation Average salary = Average annual player salary in dollars
Team League Year Opened Team Salary Attendance Wins ERA BA HR
Arizona National 1998 65.80 2,080,145 79 4.04 0.264 154 Atlanta National 1996 89.60 2,001,392 67 4.41 0.251 100 Baltimore American 1992 118.90 2,281,202 81 4.05 0.250 217 Boston American 1912 168.70 2,880,694 78 4.31 0.265 161 Chicago Cubs National 1914 117.20 2,959,812 97 3.36 0.244 171 Chicago Sox American 1991 110.70 1,755,810 76 3.98 0.250 136 Cincinnati National 2003 117.70 2,419,506 64 4.33 0.248 167 Cleveland American 1994 87.70 1,388,905 81 3.67 0.256 141 Colorado National 1995 98.30 2,506,789 68 5.04 0.265 186 Detroit American 2000 172.80 2,726,048 74 4.64 0.270 151 Houston American 2000 69.10 2,153,585 86 3.57 0.250 230 Kansas City American 1973 112.90 2,708,549 95 3.73 0.269 139 LA Angels American 1966 146.40 3,012,765 85 3.94 0.246 176 LA Dodgers National 1962 230.40 3,764,815 92 3.44 0.250 187 Miami National 2012 84.60 1,752,235 71 4.02 0.260 120 Milwaukee National 2001 98.70 2,542,558 68 4.28 0.251 145 Minnesota American 2010 108.30 2,220,054 83 4.07 0.247 156 NY Mets National 2009 100.10 2,569,753 90 3.43 0.244 177 NY Yankees American 2009 213.50 3,193,795 87 4.05 0.251 212 Oakland American 1966 80.80 1,768,175 68 4.14 0.251 146 Philadelphia National 2004 133.00 1,831,080 63 4.69 0.249 130 Pittsburgh National 2001 85.90 2,498,596 98 3.21 0.260 140 San Diego National 2004 126.60 2,459,742 74 4.09 0.243 148 San Francisco National 2000 166.50 3,375,882 84 3.72 0.267 136 Seattle American 1999 123.20 2,193,581 76 4.16 0.249 198 St. Louis National 2006 120.30 3,520,889 100 2.94 0.253 137 Tampa Bay American 1990 74.80 1,287,054 80 3.74 0.252 167 Texas American 1994 144.80 2,491,875 88 4.24 0.257 172 Toronto American 1989 116.40 2,794,891 93 3.8 0.269 232 Washington National 2008 174.50 2,619,843 83 3.62 0.251 177
(continued )
750
Year Average salary
2000 1,988,034 2001 2,264,403 2002 2,383,235 2003 2,555,476 2004 2,486,609 2005 2,632,655 2006 2,866,544 2007 2,944,556 2008 3,154,845 2009 3,240,206 2010 3,297,828 2011 3,305,393 2012 3,440,000 2013 3,650,000 2014 3,950,000 2015 4,250,000
A.2 Data Set 2—Baseball Statistics, 2015 Season (concluded)
751
A.3 Data Set 3—Lincolnville School District Bus Data
Variables ID = Bus identification number Manufacturer = Source of the bus (Bluebird, Keiser, or Thompson) Engine type = If the engine is diesel then engine type = 0; if the engine is gasoline,
then engine type = 1) Capacity = number of seats on the bus Maintenance cost = dollars spent to maintain a bus last year Age = number of years since the bus left the manufacturer Odometer Miles = total number of miles traveled by a bus Miles = number of miles traveled since last maintenance
Engine Type Maintenance Odometer ID Manufacturer (0=diesel) Capacity cost Age Miles Miles 10 Keiser 1 14 4646 5 54375 11973 396 Thompson 0 14 1072 2 21858 11969 122 Bluebird 1 55 9394 10 116580 11967 751 Keiser 0 14 1078 2 22444 11948 279 Bluebird 0 55 1008 2 22672 11925 500 Bluebird 1 55 5329 5 50765 11922 520 Bluebird 0 55 4794 10 119130 11896 759 Keiser 0 55 3952 8 87872 11883 714 Bluebird 0 42 3742 7 73703 11837 875 Bluebird 0 55 4376 9 97947 11814 600 Bluebird 0 55 4832 10 119860 11800 953 Bluebird 0 55 5160 10 117700 11798 101 Bluebird 0 55 1955 4 41096 11789 358 Bluebird 0 55 2775 6 70086 11782 29 Bluebird 1 55 5352 6 69438 11781 365 Keiser 0 55 3065 6 63384 11778 162 Keiser 1 55 3143 3 31266 11758 686 Bluebird 0 55 1569 3 34674 11757 370 Keiser 1 55 7766 8 86528 11707 887 Bluebird 0 55 3743 8 93672 11704 464 Bluebird 1 55 2540 3 34530 11698 948 Keiser 0 42 4342 9 97956 11691 678 Keiser 0 55 3361 7 75229 11668 481 Keiser 1 6 3097 3 34362 11662 43 Bluebird 1 55 8263 9 102969 11615 704 Bluebird 0 55 4218 8 83424 11610 814 Bluebird 0 55 2028 4 40824 11576 39 Bluebird 1 55 5821 6 69444 11533 699 Bluebird 1 55 9069 9 98307 11518 75 Bluebird 0 55 3011 6 71970 11462 693 Keiser 1 55 9193 9 101889 11461 989 Keiser 0 55 4795 9 106605 11418 982 Bluebird 0 55 505 1 10276 11359 321 Bluebird 0 42 2732 6 70122 11358 724 Keiser 0 42 3754 8 91968 11344 732 Keiser 0 42 4640 9 101196 11342 880 Keiser 1 55 8410 9 97065 11336 193 Thompson 0 14 5922 11 128711 11248 884 Bluebird 0 55 4364 9 92457 11231 57 Bluebird 0 55 3190 7 79240 11222 731 Bluebird 0 42 3213 6 68526 11168 61 Keiser 0 55 4139 9 103536 11148
(continued )
752
Engine Type Maintenance Odometer ID Manufacturer (0=diesel) Capacity cost Age Miles Miles 135 Bluebird 0 55 3560 7 76426 11127 833 Thompson 0 14 3920 8 90968 11112 671 Thompson 1 14 6733 8 89792 11100 692 Bluebird 0 55 3770 8 93248 11048 200 Bluebird 0 55 5168 10 103700 11018 754 Keiser 0 14 7380 14 146860 11003 540 Bluebird 1 55 3656 4 45284 10945 660 Bluebird 1 55 6213 6 64434 10911 353 Keiser 1 55 4279 4 45744 10902 482 Bluebird 1 55 10575 10 116534 10802 398 Thompson 0 6 4752 9 95922 10802 984 Bluebird 0 55 3809 8 87664 10760 977 Bluebird 0 55 3769 7 79422 10759 705 Keiser 0 42 2152 4 47596 10755 767 Keiser 0 55 2985 6 71538 10726 326 Bluebird 0 55 4563 9 107343 10724 120 Keiser 0 42 4723 10 110320 10674 554 Bluebird 0 42 1826 4 44604 10662 695 Bluebird 0 55 1061 2 23152 10633 9 Keiser 1 55 3527 4 46848 10591 861 Bluebird 1 55 9669 10 106040 10551 603 Keiser 0 14 2116 4 44384 10518 156 Thompson 0 14 6212 12 140460 10473 427 Keiser 1 55 6927 7 73423 10355 883 Bluebird 1 55 1881 2 20742 10344 168 Thompson 1 14 7004 7 83006 10315 954 Bluebird 0 42 5284 10 101000 10235 768 Bluebird 0 42 3173 7 71778 10227 490 Bluebird 1 55 10133 10 106240 10210 725 Bluebird 0 55 2356 5 57065 10209 45 Keiser 0 55 3124 6 60102 10167 38 Keiser 1 14 5976 6 61662 10140 314 Thompson 0 6 5408 11 128117 10128 507 Bluebird 0 55 3690 7 72849 10095 40 Bluebird 1 55 9573 10 118470 10081 918 Bluebird 0 55 2470 5 53620 10075 387 Bluebird 1 55 6863 8 89960 10055 418 Bluebird 0 55 4513 9 104715 10000
A.3 Data Set 3—Lincolnville School District Bus Data (concluded)
753
A.4 Data Set 4—Applewood Auto Group
Age = the age of the buyer at the time of the purchase Profit = the amount earned by the dealership on the sale of each vehicle 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 customer
Age Profit Location Vehicle-Type Previous
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 40 1,485 Sheffield Compact 0
Age Profit Location Vehicle-Type Previous
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 46 1,938 Kane Sedan 0 46 1,940 Kane Truck 3
(continued )
754
Age Profit Location Vehicle-Type Previous
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 53 1,401 Tionesta SUV 2
Age Profit Location Vehicle-Type Previous
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
A.4 Data Set 4—Applewood Auto Group (concluded)
755
A.5 Banking Data Set—Century National Bank Case
Balance = Account balance in $ ATM = Number of ATM transactions in the month Services = Number of other bank services used Debit = Account has a debit card (1 = yes, 0 = no) Interest = Receives interest on the account (1 = yes, 0 = no) City = City where banking is done 60 Accounts
Balance ATM Services Debit Interest City
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
Balance ATM Services Debit Interest City
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
756
APPENDIX B: TABLES
B.1 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
(continued)
757
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
B.1 Binomial Probability Distribution (continued)
(continued)
758
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
B.1 Binomial Probability Distribution (continued)
(continued)
759
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
B.1 Binomial Probability Distribution (continued)
(continued)
760
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
B.1 Binomial Probability Distribution (concluded)
761
B.2 Poisson Distribution
μ
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
μ
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
762
B.3 Areas under the Normal Curve
Example: If z = 1.96, then P(0 to z ) = 0.4750.
z 0 1.96
0.4750
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 2.0 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 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
763
B.4 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
764
0 0
1 2a aa a
1 2
Confidence interval t–t 0
Left-tailed test –t t
Right-tailed test t
Two-tailed test t–t
B.5 Student’s t Distribution
Confidence Intervals, c
80% 90% 95% 98% 99% 99.9%
Level of Significance for One-Tailed Test, α df 0.10 0.05 0.025 0.01 0.005 0.0005
Level of Significance for Two-Tailed Test, α 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, α df 0.10 0.05 0.025 0.01 0.005 0.0005
Level of Significance for Two-Tailed Test, α 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)
765
B.5 Student’s t Distribution (concluded)
Confidence Intervals, c
80% 90% 95% 98% 99% 99.9%
Level of Significance for One-Tailed Test, α df 0.10 0.05 0.025 0.01 0.005 0.0005
Level of Significance for Two-Tailed Test, α 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, α df 0.10 0.05 0.025 0.01 0.005 0.0005
Level of Significance for Two-Tailed Test, α 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
766
B.6A Critical Values of the F Distribution (α = .05)
D eg
re es
o f F
re ed
om fo
r t he
D en
om in
at or
.05
F0
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
767
B.6B Critical Values of the F Distribution (α = .01)
.01
F0
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
D eg
re es
o f F
re ed
om fo
r t he
D en
om in
at or
768
B.7 Critical Values of Chi-Square
This table contains the values of χ2 that correspond to a specific right-tail area and specific number of degrees of freedom.
Degrees of Right-Tail Area 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
x20
Example: With 17 df and a .02 area in the upper tail, x2 5 30.995
769
B.8 Wilcoxon T Values
2α .15 .10 .05 .04 .03 .02 .01
α 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
770
B.9A Critical Values for the Durbin–Watson d Statistic (α = .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. 159–178. Reproduced by permission of the Biometrika Trustees.
771
B.9B Critical Values for the Durbin–Watson d Statistic (α = .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. 159–178. Reproduced by permission of the Biometrika Trustees.
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B.9C Critical Values for the Durbin–Watson d Statistic (α = .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. 159–178. Reproduced by permission of the Biometrika Trustees.
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B.10 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,
n
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CHAPTER 2 2–1. The Excel commands to use the PivotTable Wizard to create
the frequency table, bar chart, and pie chart on page 22 are: a. Open the Applewood Auto Group data file. b. Click on a cell somewhere in the data set, such as cell C5. c. Click on the Insert menu on the toolbar. Then click PivotTa-
ble on the far left of the Ribbon.
e. On the right-hand side of the spreadsheet, a PivotTable Field List will appear with a list of the data set variables. To summarize the “Vehicle-Type” variable, click on the “Vehicle-Type” variable and it will appear in the lower left box called Row Label. You will note that the frequency table is started in cell N1 with the rows labeled with the values of the variable “Vehicle-Type.” Next, return to the top box, and select and drag the “Vehicle-Type” variable to the “Σ Values” box. A column of frequencies will be added to the table. Note that you can format the table to center the values and also relabel the column headings as needed.
f. To create the bar chart, select any cell in the PivotTable. Next, select the Insert menu from the tool bar and within the Charts group, select a bar chart from the Column drop- down menu. A bar chart appears. Click on the chart head- ing and label the chart as needed.
g. To create the pie chart, the frequencies should be con- verted to relative frequencies. Click in the body of the Piv- otTable and the PivotTable Field List will appear to the right. In the “Σ Values” box, click on the pull-down menu for “Count of Vehicle Type” and select the Value Field Settings option. You will see a number of different selections that can be used to summarize the variables in a PivotTable. Click on the tab “Show Values As” and, in the pull-down menu, select “% of Grand Total.” The frequencies will be converted to relative frequencies.
To create the pie chart, select any cell in the PivotTable. Next, select the Insert menu from the tool bar, and within
d. The following screen will appear. Click on “Select a table or range” to select the data range as shown in the Table/ Range row. Next, click on “Existing Worksheet” and select a cell location, such a N1, and click OK.
the Charts group, select a pie chart from the Column drop- down menu. A pie chart appears. Click on the chart head- ing and label the chart as needed. To add the percentages, click on the pie chart and a menu will appear. Click on “Add Data Labels.”
2–2. The Excel commands to use the PivotTable Wizard to create the frequency and relative frequency distributions on page 30 and the histogram on page 33 follow.
a. Open the Applewood Auto Group data file. b. Click on a cell somewhere in the data set, such as cell C5. c. Click on the Insert menu on the toolbar. Then click on Pivot-
Table on the far left of the Ribbon. d. The following screen will appear. Click on “Select a table or
range” to select the data range as shown in the Table/ Range row. Next, click on “New Worksheet” and the Pivot- Table will be created in a new worksheet.
e. On the right-hand side of the spreadsheet, a PivotTable Field List will appear with a list of the data set variables. To summarize the “Profit” variable, click on the “Profit” vari- able and drag it to the “Row Labels” box. Then return to the top box, click on “Profits” again and drag it to the “Σ Values” box. Staying in this box, click on the pull-down menu for “Sum of Profit.” You will see a number of different selections that can be used to summarize the variables in a PivotTable. In the “Summarize Values As” tab, select “Count” to create frequencies for the variable “Profit.” A PivotTable will ap- pear in the new worksheet.
WHAT IS STATISTICS? 774APPENDIX C: SOFTWARE COMMANDS
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f. In the PivotTable, the left column shows each value of the variable “Profit.” To create classes for “Profit,” select any cell in the column and right-click. A menu appears. Select “Group” from the menu to create the classes. First, un- check both boxes. Then, in the dialogue box, enter the lower limit of the first class as the “Starting at” value. Enter the upper limit of the last class as the “Ending at” value. Then enter the class interval as the “By” value. Click OK. A frequency distribution appears.
g. To create a relative frequency distribution, point and click on one of the cells in the PivotTable and the “PivotTable Field List” appears to the right. Click and drag the variable “Profits” to the “Σ Values” box. A second “Counts of Profit” appears. In the “Σ Values,” click on the second “Counts of Profit” and select the “Value Fields Setting.” You will see a number of different selections that can be used to summa- rize the variables in a PivotTable. Click on the tab “Show Values As” and, in the pull-down menu, select “% of Grand Total.” The relative frequencies will be added to the table. You can format the table by relabeling the column head- ings such as “Frequency” and “Relative Frequency.”
h. To create a histogram, select a cell in the PivotTable, choose the Insert menu from the tool bar, and within the Charts group, select a column chart from the Column drop-down
menu. A histogram appears with both “Count of Profit” and “Count of Profit2.” On the “Count of Profit2” bubble at the top of the chart, right-click and select “Remove Field.” Then the chart and PivotTable only report the frequencies. To eliminate the space between the bars, select the entire chart area, and “PivotChart Tools” will appear at the top. Select “Design.” In the “Chart Layouts” choices, select the option that shows no spaces between the bars. The option is illustrated in the figure to the right. To add data labels, select the histo- gram, right-click, and select “Add Data Labels.” Relabel the chart and axes as needed.
CHAPTER 3 3–1. The Excel Commands for the descriptive statistics on page 64 are: a. From the website, www.mhhe.com/Lind17e, 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 statis- tics, 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.
CHAPTER 4 4–1. The Minitab commands for the dot plot on page 96 are: a. Enter the number of vehicles serviced at Tionesta Ford Lin-
coln 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 appropri- ate 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
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4–2. The Minitab commands for the stem-and-leaf display on page 99 are:
a. Import the data for Table 4–1 from www.mhhe.com/ lind17e.
b. Select Graph, and click on Stem-and-Leaf. c. Select the variable Spots, enter 10 for the Increment, and
then click OK. 4–3. The Minitab commands for the descriptive summary on
page 105 are: a. Input the data on the Smith Barney commissions from the
Example on page 103. b. From the toolbar, select Stat, Basic Statistics, and Display
Descriptive Statistics. In the dialog box, select Commis- sions as the Variable, and then click OK.
4–4. The Excel commands for the quartiles on page 105 follow. a. Input the data on the Smith Barney commissions from the
example on page 103 in column A. If you are using Excel 2010 and wish to compute quartiles us-
ing formula (4–1), the steps are: b. In cell C3 type Formula (4–1), in C4 write Quartile 1, and
in C6 type Quartile 3. c. In cell D4 type “=QUARTILE.EXC(A2:A16,1)” and hit Enter. In cell D6 type “=QUARTILE.EXC(A2:A16,3)” and hit Enter. If you are using either Excel 2007 or 2010 and wish to com-
pute quartiles using the Excel Method: b. In cell C8 type Excel Method, in C9 write Quartile 1, and in
C11 type Quartile 3. c. In cell D8 type “=QUARTILE(A2:A16,1)” and hit Enter. In
cell D11 type “=QUARTILE(A2:A16,3)” and hit Enter. 4–5. The Minitab commands for the box plot on page 108 are: a. Import the Applewood Auto Group data from www.mhhe.
com/lind17e.
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 appro- priate heading, and then click OK.
4–6. The Minitab commands for the descriptive summary on page 113 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.
4–7. The Excel commands for the scatter diagram on page 116 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 Pri- mary Vertical Axis Title, name the vertical axis Profit. Using the Primary Horizontal Axis Title, name the horizontal axis Age. Next, select Legend and select None.
CHAPTER 5 5–1. The Excel Commands to determine the number of permuta-
tions shown on page 165 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 cate-
gory, 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.
5–2. The Excel Commands to determine the number of combina- tions shown on page 165 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 cate- gory, 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 6 6–1. The Excel commands necessary to determine the binomial
probability distribution on page 189 are: a. On a blank Excel worksheet, write the word Success 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 cate- gory and BINOM.DIST in the function 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_s. 2. Enter 40 for the Trials. 3. Enter .09 for the probability of a success. 4. Enter the word false or the 0 for Cumulative and click
on OK. 5. Excel will compute the probability of 0 successes in
40 trials, with a .09 probability of success. The result, .02299618, is stored in cell B2.
as the variables, click on Statistics, select the desired sta- tistics to be output, and finally click OK twice.
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e. To complete the probability distribution for successes of 1 through 15, double-click on cell B2. The binomial function should appear. Replace the 0 to the right of the open pa- rentheses with the cell reference A2.
f. Move the mouse to the lower right corner of cell B2 until a solid black + symbol appears, then click and hold and high- light the B column to cell B17. The probability of a success for the various values of the random variable will appear.
6–2. The Excel commands necessary to determine the hypergeo- metric distribution on page 196 are:
a. On a blank Excel worksheet, write the word Union Members in cell A1 and the word Probability in B1. In cells A2 through A7, enter the numbers 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.
e. To complete the probability distribution for successes of 1 through 5, double-click on cell B2. The hypergeometric 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 high- light the B column to cell B7. The probability of a success for the various outcomes will appear.
c. In the first dialog box, select Statistical and HYPGEOM. DIST and then click OK.
d. In the second dialog box, enter the four items necessary to compute a hypergeometric probability.
1. Enter 0 for the Sample_s. 2. Enter 5 for the Number_sample. 3. Enter 40 for the Population_s. 4. Enter 50 for the Number_pop. 5. Enter 0 for the Cumulative and click OK. 6. Excel will compute the probability of 0 successes in 5
trials (.000118937) and store that result in cell B2.
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 cate- gory and POISSON.DIST in the function name category, then click OK.
6–3. The Excel commands necessary to determine the Poisson probability distribution on page 199 are:
a. On a blank Excel worksheet, write the word Success in cell A1 and the word Probability in B1. In cells A2 through A9, write the integers 0 to 7. Click on B2 as the active cell.
778
Also recall that Excel samples with replacement, so it is possible for a population value to appear more than once in the sample.
CHAPTER 9 9–1. The Minitab commands for the confidence interval for the
amount spent at the Inlet Square Mall on page 297 are: a. Enter the 20 amounts spent in column C1 and name the
variable Amount. 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.
CHAPTER 7 7–1. The Excel commands necessary to produce the output on
page 222 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 NORM.DIST and click OK.
b. In the dialog box, put 1100 in the box for X, 1000 for the Mean, 100 for the Standard_dev, and True in the Cumula- tive box, and click OK.
c. The result will appear in the dialog box. If you click OK, the answer appears in your spreadsheet.
7–2. The Excel commands necessary to produce the output on page 228 are:
a. Click the Formulas tab in the top menu, then, on the far left, select Insert Function fx. Then from the category box, se- lect Statistical and below that NORM.INV 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 an- swer is different from page ••• 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 8 8–1. The Excel commands to select a simple random sample from
the rental data on page 254 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 column 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 information. Note that your sample results will differ from those in the text.
d. In the second dialog box, enter the three items necessary to compute a Poisson probability.
1. Enter 0 for X. 2. Enter 0.3 for the Mean. 3. Enter the word false or the number 0 for Cumulative
and click OK. 4. Excel will compute the probability of 0 successes for a
Poisson probability distribution with a mean of 0.3. The result, .74081822, is stored in cell B2.
e. To complete the probability distribution for successes of 1 through 7, double-click on cell B2. The Poisson function should appear. Replace the 0 to the right of the open pa- rentheses with the cell reference A2.
f. Move the mouse to the lower right corner of cell B2 until a solid black + symbol appears, then click and hold and highlight the B column to cell B9. The probability of a suc- cess for the various values of the random variable will appear.
779
9–2. The Excel commands for the confidence interval for the amounts spent at the Inlet Square Mall on page 298 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 Confidence Level for Mean, and then click on OK.
CHAPTER 10 10–1. The Minitab commands for the histogram and the descriptive
statistics on page 335 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.
10–2. The Minitab commands for the one-sample t test on page 340 are:
a. Enter the sample data into column C1 and name the vari- able Minutes.
b. From the menu bar, select Stat, Basic Statistics, and 1-Sample t, and then hit Enter.
c. Select Minutes as the variable, select Perform hypothe- sized mean, insert the value 40. Click Options. Under Alternate, select greater than. Finally, click OK twice.
CHAPTER 11 11–1. The Excel commands for the two-sample t-test on page 364 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 en- ter 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.
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11–2. The Minitab commands for the two-sample t-test on page 368 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 col- umns, select C1 Store for the First column and C2 Name of the Second, click the box next to Assume equal variances, and click OK.
11–3. The Excel commands for the paired t-test on page 373 are: a. Enter the data into columns B and C (or any other two col-
umns) 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 Variable 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.
CHAPTER 12 12–1. The Excel commands for the test of variances on page 391 are: a. Enter the data for U.S. 25 in column A and for I-75 in col-
umn 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.
12–2. The Excel commands for the one-way ANOVA on page 400 are: a. Key in data into four columns labeled Northern, WTA, Po-
cono, 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 0.05, and finally select Output Range as F1 and click OK.
c. In the dialog box, indicate that the range of Variable 1 is from A1 to A6 and Variable 2 from B1 to B7, the Hypothe- sized Mean Difference is 0, click Labels, Alpha is 0.05, and the Output Range is D1. Click OK.
781
12–3. The Minitab commands for the pairwise comparisons on page 403 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, select “Response data are in a separate column for each factor level”, select and enter the variable names into the Responses box by click- ing on the variable names in the following order: Branson, Pocono, WTA, and Northern. Then select Comparisons and then select Fisher’s, individual error rate. Then OK.
12–4. The Excel commands for the two-way ANOVA on page 408 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 names 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, enter 0.05 for Alpha, select G3 for the Output Range, and then click OK.
12–5. The Excel commands for the two-way ANOVA with interaction on page 410 are:
a. Enter the data into Excel as shown on page 410. 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, for the Input Range highlight the entire range for the data including row and column labels, enter Rows per sample as 3, enter 0.05 for Alpha, Output Range and cell H1.
CHAPTER 13 13–1. The Excel commands for calculating the correlation coefficient
on page 445 are: a. Access the Applewood Auto Group dataset (www.mhhe.
com/lind17e). b. Select the Data tab on the top of the ribbon. Then, on the
far right, select Data Analysis. Select Correlation, and click OK.
c. For the Input Range, highlight the Age and Profit columns, including the labels in row 1. The data are grouped by Col- umns. Check the Labels in first row box. Select a cell in the worksheet as the beginning of the range to output the cor- relation. Click OK.
13–2. The computer commands for the Excel output on page 460 are:
a. Enter the variable names in row 1 of columns A, B, and C. Enter the data in rows 2 through 16 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.
782
13–3. The Minitab commands to the confidence intervals and predic- tion intervals on page 470 are:
a. Select Stat, Regression, and Fitted line plot. b. In the next dialog box, the Response (Y) is Copiers Sold
and Predictor (X) is Sales Calls. Select Linear for the type of regression model and then click on Options.
c. In the Options dialog box, click on Display confidence in- terval and prediction interval, use the 95.0 for confidence level, type an appropriate heading in the Title box, then click OK and then OK again.
CHAPTER 14 Note: We do not show steps for all the statistical software in Chapter 14. The following shows the basic steps. 14–1. The Excel commands to produce the multiple regression out-
put on page 492 are: a. Import the data from the text website: www.mhhe.com/
lind17e. 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 Output Range is F1, then click OK.
CHAPTER 15 15–1. The MegaStat commands for the two-sample test of propor-
tions on page 553 are: a. Select MegaStat from the Add-Ins tab. From the menu, se-
lect Hypothesis Tests, and then Compare Two Indepen- dent Proportions.
b. Enter the data. For Group 1, enter x as 19 and n as 100. For Group 2, enter x as 62 and n as 200. Select OK.
15–2. The MegaStat commands to create the chi-square goodness- of-fit test on page 559 are:
a. Enter the information from Table 15–2 into a worksheet as shown.
b. Select MegaStat, Chi-Square/Crosstabs, and Goodness of Fit Test and hit Enter.
c. In the dialog box, select B2:B5 as the Observed values, C2:C5 as the Expected values, and enter 0 as the Number of parameters estimated from the data. Click OK.
c. For our spreadsheet, we have Calls in column B and Sales in column C. The Input Y Range is C1:C16 and the Input X Range is B1:B16. Click on Labels, select E1 as the Output Range, and click OK.
783
15–3. The MegaStat commands to create the chi-square goodness- of-fit tests on pages 564 and 565 are the same except for the number of items in the observed and expected frequency col- umns. Only one dialog box is shown.
a. Enter the Levels of Management information shown on page 564.
b. Select MegaStat, Chi-Square/Crosstabs, and Goodness of Fit Test and hit Enter.
c. In the dialog box, select B1:B7 as the Observed values, C1:C7 as the Expected values, and enter 0 as the Number of parameters estimated from the data. Click OK.
15–4. The MegaStat commands for the contingency table analysis on page 573 are:
a. Enter Table 15–8 on page 571 into cells A1 through D3. Include the row and column labels. DO NOT include the Total column or row.
b. Select MegaStat from the Add-Ins tab. From the menu select Chi-square/Crosstab, then select Contingency Table.
c. For the Input Range, select cells A1 through D3. Check the chi-square and Expected values boxes. Select OK.
CHAPTER 16 16–1. The MegaStat for Excel commands necessary for the Wilcoxon
rank-sum test on page 600 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 (A1:A9) and for
Group 2 use the data on Chicago flights (B1:B8). Click on Correct for ties and one-tailed, and greater than as the Alternative, then click on OK.
16–2. The MegaStat commands for the Kruskal-Wallis Test on page 604 are:
a. Enter the data in Table 16–6 into an Excel worksheet in- cluding labels into columns A, B, and C starting in row 1.
b. Select MegaStat from the Add-Ins tab. From the menu select Nonparametric Tests, then select Kruskal-Wallis Test.
c. For the Input Range, select cells A1 through C9. Check the Correct for Ties box. Select OK.
16–3. The Excel commands for the one-way ANOVA on page 605 are: a. Enter the data in Table 16–6 into an Excel worksheet in-
cluding labels into columns A, B, and C starting in row 1. 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 18–1. The MegaStat commands for creating the seasonal indexes on
page 675 are: a. Enter the coded time period and the value of the time se-
ries in two columns. You may also want to include informa- tion on the years and quarters.
b. Select MegaStat, Time Series/Forecasting, and Desea- sonalization, and hit Enter.
c. Input only the range of the sales data, indicate the data are in the first quarter, and click OK.
CHAPTER 19 19–1. The Minitab commands for the Pareto chart on page 703 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.
784
19–2. The Minitab commands for the mean and range charts on page 610 are:
a. Enter the information in Table 19–1 or from the text website, www.mhhe.com/lind17e. The file name is Table 19–1.
b. Click on Stat, Control Charts, Variables Charts for Sub- groups, Xbar-R, and hit Enter.
c. Select All observations for a chart are in one column. Then in the box below, select the variable Minutes. For Subgroup sizes, enter the variable Time.
19–4. The Minitab commands for the c-bar chart on page 717 are: a. Enter the data on the number of misspelled words from
page 717. b. Click on Stat, Control Charts, Attribute Charts, C, and hit
Enter. c. Select the Variable indicating the number of misspelled
words, then click on Labels and type the title in the space provided, and click OK twice.
19–3. The Minitab commands for the p-chart on page 715 are: a. Enter the data on the number of defects from page 715. b. Click on Stat, Control Charts, Attribute Charts, P, and hit
Enter. c. Under Variables, select Defects, then enter 50 for Sub-
group sizes. Click on Labels, type in the title, and click OK twice.
c. Indicate Defects of attribute data as the variable Usage and Frequencies as the variable Gallons. Click on Options and type a chart title, and click OK.
785
CHAPTER 1 1. a. Interval d. Nominal b. Ratio e. Ordinal c. Nominal f. Ratio 3. Answers will vary. 5. Qualitative data are not numerical, whereas quantitative data are
numerical. Examples will vary by student. 7. A discrete variable may assume only certain values. A continu-
ous 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 dis- crete 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 Best Buy stores, then
the daily number sold last month would be a sample. However, if you considered the store as the only store of interest, then the daily number sold last month would be a population.
13.
Discrete Variable Continuous Variable
Qualitative b. Gender d. Soft drink preference g. Student rank in class h. Rating of a finance professor
Quantitative c. Sales volume of MP3 players a. Salary f. SAT scores e. Temperature i. Number of home computers
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 c. Sales volume of MP3 players a. Salary i. Number of home computers
15. According to the sample information, 120/300 or 40% would
accept a job transfer. 17. a.
Manufacturer Difference
General Motors Corp. 128,133 Chrysler 126,955 Ford Motor Company 112,975 Toyota Motor Sales USA Inc. 96,078 Nissan North America Inc. 72,146 Subaru of America Inc. 61,834 American Honda Motor Co Inc. 38,440 Kia Motors America Inc. 36,313
Manufacturer Difference
Hyundai Motor America 30,656 Mercedes-Benz 20,187 Audi of America Inc. 18,970 Mitsubishi Motors N A, Inc. 16,119 Land Rover 13,535 BMW of North America Inc. 12,202 Mazda Motor of America Inc. 7,407 Volvo 5,980 Mini 4,573 Porsche Cars NA Inc. 4,337 Tesla 1,850 Lamborghini 372 Ferrari 164 Rolls Royce 19 Bentley −351 Jaguar −633 Maserati −783 Smart −2,512 Fiat −3,650 Volkswagen of America Inc. −6,585
b. Percentage differences with top five and bottom five.
Manufacturer % change from 2014
Lamborghini 75% Land Rover 32% Mitsubishi Motors N A, Inc. 25% Subaru of America Inc. 15% Audi of America Inc. 13% Volvo 13% Tesla 12% Porsche Cars NA Inc. 11% Mini 10% Ferrari 9% Chrysler 8% Kia Motors America Inc. 7% Mercedes-Benz 7% Nissan North America Inc. 6% Ford Motor Company 5% General Motors Corp. 5% Hyundai Motor America 5% Toyota Motor Sales USA Inc. 5% BMW of North America Inc. 5% American Honda Motor Co Inc. 3% Mazda Motor of America Inc. 3% Rolls Royce 3% Volkswagen of America Inc. −2% Jaguar −5% Maserati −8% Fiat −9% Bentley −15% Smart −29%
APPENDIX D: ANSWERS TO ODD-NUMBERED CHAPTER EXERCISES & REVIEW EXERCISES & SOLUTIONS TO PRACTICE TESTS
Answers to Odd-Numbered Chapter Exercises
786
c.
3,000,000
2,000,000
2,500,000
1,500,000
1,000,000
500,000
Ge ne
ra l M
ot or
s
Fo rd
M ot
or
To yo
ta M
ot or
S al
es
Ch ry
sl er
Am er
ic an
H on
da
Ni ss
an N
or th
Hy un
da i M
ot or
Ki a
M ot
or s
Am er
ic a
Su ba
ru o
f A m
er ic
a
M er
ce de
s- Be
nz
Vo lk
sw ag
en o
f
BM W
o f N
or th
0
Unit Sales, October 2014 and 2015
Through October 2015 Through October 2014
16%
12% 14%
10%
0% 2% 4% 6% 8%
–2%
Ge ne
ra l M
ot or
s Co
rp .
% C
ha ng
e fro
m 2
01 4
to 2
01 5
Fo rd
M ot
or C
om pa
ny
To yo
ta M
ot or
S al
es U
SA
Ch ry
sl er
Am er
ic an
H on
da M
ot or
Ni ss
an N
or th
A m
er ic
a
Hy un
da i M
ot or
A m
er ic
a
Ki a
M ot
or s
Am er
ic a
In c.
Su ba
ru o
f A m
er ic
a In
c.
M er
ce de
s- Be
nz
Vo lk
sw ag
en o
f A m
er ic
a
BM W
o f N
or th
A m
er ic
a–4%
% Change 2014–2015 for the Top Twelve Manufacters
19. The graph shows a gradual increase from 2003 to 2006 with a large increase in 2008. However, during the recession of 2009, earnings decreased, then increased to 2008 levels over the years 2010 through 2012. Over 2013 and 2014, earnings were steady at a bit over $30 billion.
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% 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 1,300 1.00
b.
Fr eq
ue nc
y
Chart of Frequency vs. Color
500
400
300
200
100
0 Metallic Black
Color
Bright White
Magnetic Lime
Tangerine Orange
Fusion Red
c.
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
i ≥ $567 − $235
8 = 41 Class intervals of 45 or 50 would be
acceptable. 11. a. 24 = 16 Suggests 5 classes.
b. i ≥ 31 − 25
5 = 1.2 Use interval of 1.5.
c. 24 d.
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
e. The largest concentration is in the 27.0 up to 28.5 class (8). 13. a.
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
787
b. The largest group of shoppers (21) shop at the BiLo Super- market 3, 4, or 5 times during a month period. Some custom- ers visit the store only 1 time during the month, but others shop as many as 15 times.
c. 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
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.
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 d. X = 1.5, Y = 5 e.
–1.5
25 20 15 10 5 0Nu
m be
r o f e
m pl
oy ee
s
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.
Miles CF
Less than 3 5 Less than 6 17 Less than 9 40 Less than 12 48 Less than 15 50
c. 50
40
30
20
10
0
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. Quantita- tive variables are either discrete or continuous. There is a natural order to the results for a quantitative variable. Quanti- tative 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
c.
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. i ≥ 48 − 16
5 = 6.4 Use interval of 7.
c. 15 d.
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 25
e. It is fairly symmetric, with most of the values between 22 and 36.
788
31. a. 25 = 32, 26 = 64, 6 classes recommended.
b. i = 10 − 1
6 = 1.5 use an interval of 2.
c. 0 d.
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 10 up to 12 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. 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 1,000 2
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% of all plays.
35. a. 56 b. 10 (found by 60 − 50) c. 55 d. 17 37. a. Use $35 because the minimum is ($265 − $82)/6 = $30.5. b.
$ 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
c. 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. Bar charts are preferred when the goal is to compare the actual amount in each category.
Fuel 0
100 200 300 400 500 600 700 800 900
1000
Interest Repairs
Amount
Item
$
Insurance Depreciation
41.
Wages 73%
Other 3%Pensions
3%IRA 8%
Social Security
2%
Dividends 11%
SC Income Percent Cumulative
Wages 73 73 Dividends 11 84 IRA 8 92 Pensions 3 95 Social Security 2 97 Other 3 100
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 contributes roughly another 10%.
43. a. Since 26 = 64 < 70 < 128 = 27, 7 classes are recommended. The interval should be at least (1,002.2 − 3.3)/7 = 142.7. Use 150 as a convenient value.
b. Based on the histogram, the majority of people has less than $500,000 in their investment portfolio and may not have enough money for retirement. Merrill Lynch financial advisors need to pro- mote the importance of investing for retirement in this age group.
30
20
10
0 75 225 375 525 675 825 975
Value
Fr eq
ue nc
y
45. a. Pie chart b. 700, found by 0.7(1,000) c. Yes, 0.70 + 0.20 = 0.90 47. a.
Product
Ve hic
les
Ma ch
ine ry
Ele ctr
ica l
ma ch
ine ry
Mi ne
ral fu
el
an d o
il Pla sti
c 0
10 20 30 40 50 60 70
Top 5 U.S. Exports to Canada 2014
$ (B
ill io
ns )
789
b. 25.5%, found by (59.7 + 36.6)/376 c. 47.8% found by (59.7 + 36.6)/(63.3 + 59.7 + 36.6 + 24.8 + 17)) 49.
Brown 29%
Yellow 22%
Red 22%
Orange 8%
Blue 12%
Green 7%
M & M s
Brown, yellow, and red make up almost 75 percent of the candies. The other 25 percent is composed of blue, orange, and green.
51. There are many choices and possibilities here. For example you could choose to start the first class at 160,000 rather than 120,000. The choice is yours!
i > = (919,480 − 167,962)/7 = 107,360. Use intervals of 120,000
Selling Price (000) Frequency Cumulative Frequency
120 up to 240 26 26 240 up to 360 36 62 360 up to 480 27 89 480 up to 600 7 96 600 up to 720 4 100 720 up to 840 2 102 840 up to 960 1 105
a. Most homes (60%) sell between $240,000 and $480,000. b. The typical price in the first class is $180,000 and in the last
class it is $900,000 c.
Cu m
ul at
iv e
Nu m
be r o
f H om
es 120
100
80
60
40
Price ($000)
Selling Price
200100 300 400 500 600 1000700 800 900 0
20
Fifty percent (about 52) of the homes sold for about $320,000 or less. The top ten percent (about 90) of homes sold for at least $520,000 About 41 (about 41) percent of the homes sold for less than $300,000.
d.
8765432
25
20
15
10
5
0
Bedrooms
Co un
t
Chart of Bedrooms
2,3 and 4 bedroom houses are most common with about 25 houses each. 7 and 8 bedroom houses are rather rare. (LO2-3)
53. Since 26 = 64 < 80 < 128 = 27, use 7 classes. The interval should be at least (11973 − 10000)/7 = 281 miles. Use 300. The resulting frequency distribution is:
Class f
9900 up to 10200 8 10200 up to 10500 8 10500 up to 10800 11 10800 up to 11100 8 11110 up to 11400 13 11400 up to 11700 12 11700 up to 12000 20
a. The typical amount driven, or the middle of the distribution is about 11100 miles. Based on the frequency distribution, the range is from 9900 up to 12000 miles.
Miles Sinces Last Maintenance
Nu m
be r o
f B us
es
Miles Since Last Maintenance
0
5
10
15
20
25
99 00
–1 01
99
10 20
0– 10
49 9
10 50
0– 10
79 9
10 80
0– 11
09 9
11 10
0– 11
39 9
11 40
0– 11
69 9
11 70
0– 11
99 9
b. The distribution is somewhat “skewed” with a longer “tail” to the left and no outliers. (LO2-3)
c.
99 00
–1 01
99
10 20
0– 10
49 9
11 70
0– 11
99 9
11 40
0– 11
69 9
11 10
0– 11
39 9
10 80
0– 11
09 9
10 50
0– 10
79 9
120% 100%
80% 60% 40% 20%
0%
Miles since Last Maintenance
Relative Cumulative Miles since Last Maintenance
Forty percent of the buses were driven fewer than about 10800 miles. About 30% of the 80 busses (about 24) were driven less than 10500 miles. (LO2-3)
790
d. The difference between the highest number sold (10) and the smallest number sold (3) is 7. The typical squared devia- tion from 6 is 6.8.
37. a. 30, found by 54 − 24 b. 38, found by 380/10 c. 74.4, found by 744/10 d. The difference between 54 and 24 is 30. The average of the
squared deviations from 38 is 74.4. 39.
State Mean Median Range
California 33.10 34.0 32 Iowa 24.50 25.0 19
The mean and median ratings were higher, but there was also more variation in California.
41. a. 5 b. 4.4, found by
(8 − 5)2 + (3 − 5)2 + (7 − 5)2 + (3 − 5)2 + (4 − 5)2
5
43. a. $2.77 b. 1.26, found by
(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
5 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. Stan- dard deviation: 2.568, found by √6.5944.
b. Dennis has a higher mean return (11.76 > 6.94). However, Den- nis has greater spread in its returns on equity (16.89 > 6.59).
47. a. x = 4
s2 = (7 − 4)2 + . . . + (3 − 4)2
5 − 1 =
22 5 − 1
= 5.5
b. s = 2.3452 49. a. x = 38
s2 = (28 − 38)2 + … + (42 − 38)2
10 − 1
= 744
10 − 1 = 82.667
b. s = 9.0921
51. a. x = 951 10
= 95.1
s2 = (101 − 95.1)2 + … + (88 − 95.1)2
10 − 1
= 1,112.9
9 = 123.66
b. s = √123.66 = 11.12 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.
Class f M fM (M − x ) f (M − x )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
x = 3,310
70 = 47.29
s = √ 10,234.287
70 − 1 = 12.18
d. The first diagram shows that Bluebird makes about 59 per- cent of the busses, Keiser about 31% and Thompson only about 10 percent. The second chart shows that nearly 69% of the buses have 55 seats. (LO2-2)
CHAPTER 3 1. μ = 5.4, found by 27/5 3. a. x = 7.0, found by 28/4 b. (5 − 7) + (9 − 7) + (4 − 7) + (10 − 7) = 0 5. x = 14.58, found by 43.74/3 7. a. 15.4, found by 154/10 b. Population parameter, since it includes all the salespeople 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. x = Σx n
so
Σx = x · n = ($5,430) (30) = $162,900 13. a. No mode b. The given value would be the mode. c. 3 and 4 bimodal 15. a. Mean = 3.583 b. Median = 5 c. Mode = 5 17. a. Median = 2.9 b. Mode = 2.9
19. x = 647 11
= 58.82
Median = 58, Mode = 58 Any of the three measures would be satisfactory.
21. a. x = 90.4
12 = 7.53
b. Median = 7.45. There are several modes: 6.5, 7.3, 7.8, and 8.7.
c. x = 33.8
4 = 8.45,
Median = 8.7 About 1 percentage point higher in winter
23. $22.91, found by 300($20) + 400($25) + 400($23)
300 + 400 + 400 25. $17.75, found by ($400 + $750 + $2,400)/200 27. 12.8%, found by √5 (1.08) (1.12) (1.14) (1.26) (1.05) = 1.128 29. 12.28% increase, found by √5 (1.094) (1.138) (1.117) (1.119) (1.147) = 1.1228
31. 2.14%, found by √ 15 236.525
172.2 − 1
33. 58.95%, found by √ 15
752,000,000
720,000 − 1
35. a. 7, found by 10 − 3 b. 6, found by 30/5 c. 6.8, found by 34/5
Bus Seat Capacity
3.75%
6
14
42
55
68.75% 13.75%
13.75%
Pie Chart of Manufacturer
58.75%31.25%
10.00%
Bluebird
Keiser
Thompson
791
61. Number of Clients f M fM (M − x ) f (M − x )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
x = 2,240
50 = 44.8
s = √ 4,298 50 − 1
= 9.37
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. Σ(x − μ) = (6 − 5) + (4 − 5) + (3 − 5) + (7 − 5) + (5 − 5) = 0
65. x = 545 16
= 34.06
Median = 37.50 67. The mean is 35.675, found by 1,427/40. The median is 36,
found by sorting the data and averaging the 20th and 21st observations.
69. xw = $5.00(270) + $6.50(300) + $8.00(100)
270 + 300 + 100 = $6.12
71. xw = 15,300(4.5) + 10,400(3.0) + 150,600(10.2)
176,300 = 9.28
73. GM = √ 50 4,600,000
42,000 − 1 = 0.0985, So about 9.85%
75. a. 55, found by 72 − 17 b. 17.6245, found by the square root of 2795.6/9 77. a. This is a population because it includes all the public univer-
sities in Ohio. b. The mean is 25,165.4. c. The median is 20,595. d. The range is 60,560. e. The standard deviation is 16,344.9. 79. a. There were 13 flights, so all items are considered.
b. μ = 2,259
13 = 173.77
c. Range = 301 − 7 = 294
s = √ 133,846
13 = 101.47
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. x = 273 30
= 9.1, Median = 9
b. Range = 18 − 4 = 14
s = √ 368.7 30 − 1
= 3.57
c. 25 = 32, so suggest 5 classes
i = 18 − 4
5 = 2.8 use i = 3
Class M f fM M − x (M − x )2 f(M − x )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
d. x = 270 30
= 9.0
s = √ 366
30 − 1 = 3.552
The mean and standard deviation from grouped data are es- timates of the mean and standard deviations of the actual values.
85. x = 13 = 910 70
s = 5.228 = √1807.5∕69
87. a. 1. The mean team salary is $121.12 million and the median is $112.91 million. Since the distribution is skewed, the median value of $112.91 million is more typical.
2. The range is $154.29 million found by $223.35 million − $69.06. The population standard deviation is $39.66. At least 95% of the team salaries are between $74.97 million and $233.61 million; found by $154.29 million plus or minus 2($39.66 million).
b. 4.78% per year, found by √ 17 4.40
1.99 − 1 = 0.0478 = 4.78%
CHAPTER 4 1. In a histogram, observations are grouped so their individual iden-
tity 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 9.
Stem Leaves
0 5 1 28 2 3 0024789 4 12366 5 2
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)( 12 ) ∴ 6th value in from lowest Q1 = 49, found by (11 + 1)(
1 4 ) ∴ 3rd value in from lowest
Q3 = 55, found by (11 + 1)( 3 4 ) ∴ 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.
14.0 21.0 28.0 35.0 42.0 49.0 The distribution is somewhat positively skewed. Note that the
dashed line above 35 is longer than below 18. 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
s = √ 62.8
4 = 3.96
792
b. −0.15, found by 3(30.8 − 31.0)
3.96 c.
Salary ( (x − x )
s ) ( (x − x )
s ) 3
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
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
s = √ 6,283
14 = 21.18
b. 0.868, found by [3(21.93 − 15.8)]/21.18 c. 2.444, found by [15/(14 × 13)] × 29.658 23.
7
6
5
4
3
2
7 8 9 10 X
Y
Scatter Diagram of Y versus X
11 12 13
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% of the men ordered dessert, but only 15% of the women. 27. a. Dot plot b. 15 c. 5 29. Stem-and-leaf N = 23 3 3 222 5 3 77 11 4 000002 (6) 4 666666 6 5 222222 31. a. L50 = (20 + 1) 50100 = 10.50
Median = 83.7 + 85.6
2 = 84.65
L25 = (21) (.25) = 5.25 Q1 = 66.6 + .25(72.9 − 66.6) = 68.175 L75 = 21(.75) = 15.75 Q3 = 87.1 + .75(90.2 − 87.1) = 89.425 b. L26 = 21(.26) = 5.46 P26 = 66.6 + .46(72.9 − 66.6) = 69.498 L83 = 21(.83) = 17.43 P83 = 93.3 + .43(98.6 − 93.3) = 95.579 c.
64.0 72.0 80.0 88.0 96.0 C20
33. a. Q1 = 26.25, Q3 = 35.75, Median = 31.50
I +
+ + + + ++
I
24.5 28.0 31.5 35.0 38.5 42.0 b. Q1 = 33.25, Q3 = 38.75, Median = 37.50
I I
32.5 35.0 37.5 40.0 42.5 45.0 c. The median time for public transportation is about 6 minutes
less. There is more variation in public transportation. The dif- ference between Q1 and Q3 is 9.5 minutes for public trans- portation 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 lo- cated at $255. The median is about $50.
37. a.
Box Plot of Price
Price
* * *
0 10,000 20,000 30,000 40,000 50,000
Median is 3,733. First quartile is 1,478. Third quartile is 6,141. So prices over 13,135.5, found by 6,141 + 1.5 × (6,141 − 1,478), are outliers. There are three (13,925; 20,413; and 44,312).
b.
Box Plot of Size
Size 0 1 2 3 4 5
* * *
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).
c.
0
10,000
20,000
30,000
40,000
50,000
0 1 2 3 4 5
Size
Scatterplot of Price versus Size
Pr ic
e
There is a direct association between them. The first obser- vation is larger on both scales.
793
d. 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
The majority of the diamonds are round (23). Premium cut is most common (17). The Round Premium combination occurs most often (13).
39. sk = 0.065 or sk = 3(7.7143 − 8.0)
3.9036 = −0.22
41.
5
4
3
2
1
0
15.0 17.5 20.0 22.5 25.0 Age
Ac ci
de nt
s
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 (7,523/139,340)100 c. Men = 5.64% Women = 5.12% 45. a. Box plot of age assuming the current year is 2016.
Distribution of stadium is highly skewed to the right. Any sta- dium older than 47.625 years (Q3 + 1.5(Q3−Q1) = 26.25 + 1.5(26.25−12) is an outlier. Boston, Chicago Cubs, La Dodgers, Oakland, and LA Angels.
b.
The first quartile is $87.3 million and the third is $144.35 million. Outliers are greater than (Q3 + 1.5(Q3−Q1)) or 144.35 = 1.5*(144.35−87.3) = $229.925 million. The distribution is posi- tively skewed. However in 2016, there were no outliers. (LO4–4)
c.
Higher salaries do not necessarily lead to more wins. (LO4-6) d.
The distribution is fairly uniform between 59 and 103. (LO4-1)
CHAPTER 5 1.
Person
Outcome 1 2
1 A A 2 A F 3 F A 4 F F
3. a. .176, found by 6
34 b. Empirical
5. a. Empirical b. Classical
794
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. 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. Let A denote the event the fish is green and B be the event the fish is male.
a. P(A) = 80/140 = 0.5714 b. P(B) = 60/140 = 0.4286 c. P(A and B) = 36/140 = 0.2571 d. P(A or B) = P(A) + P(B) − P(A and B) = 80/140 + 60/140 −
36/140 = 104/140 = 0.7429 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:
Conditional Probabilities
Joint Probabilities
50/ 500
Bel ow
Av era
ge
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
31. a. Out of all 545 students, 171 prefer skiing. So the probability is 171/545, or 0.3138.
b. Out of all 545 students, 155 are in junior college. Thus, the probability is 155/545, or 0.2844.
c. Out of 210 four-year students, 70 prefer ice skating. So the probability is 70/210, or 0.3333.
d. Out of 211 students who prefer snowboarding, 68 are in junior college. So the probability is 68/211, or 0.3223.
e. Out of 180 graduate students, 74 prefer skiing and 47 prefer ice skating. So the probability is (74 + 47)/180 = 121/180, or 0.6722.
33. P(A1 ∣ B1) = P(A1) × P(B1 ∣ A1)
P(A1) × P(B1 ∣ A1) + P(A2) × P(B1 ∣ A2)
= .60 × .05
(.60 × .05) + (.40 × .10) = .4286
35. P(night ∣ win) = P(night)P(win ∣ night)
P(night)P(win ∣ night) + P(day)P(win ∣ day)
= (.70) (.50)
[ (.70) (.50)] + [ (.30) (.90)] = .5645
37. P(cash ∣ >$50)
= P(cash) P( >$50 ∣ cash) [P(cash) P( >$50 ∣ cash)
+ P(credit) P( >$50 ∣ credit) + P(debit) P( >$50 ∣ debit)]
= (.30) (.20)
(.30) (.20) + (.30) (.90) + (.40) (.60) = .1053
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. (4)(8)(3) = 96 combinations 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 (2/3) · (2/3) b. 3/4, because (3/4) · (2/3) = 0.5 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.
Female
Total 1.000
.80
.20
Male
College Attended .80 3 .90 = .720.90
.10
.78
.22
attended .80 3 .10 = .080
Attended .20 3 .78 = .156
Not
Not
JointSex
attended .20 3 .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
795
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. P(not perfect) = P(bad sector) + P(defective)
= 112
1,000 +
31 1,000
= .143
b. P(defective ∣ not perfect) = .031 .143
= .217
71. P(poor ∣ profit) = .10(.20)
.10(.20) + .60(.80) + .30(.60) = .0294 73. a. 0.1 + 0.02 = 0.12 b. 1 − 0.12 = 0.88 c. (0.88)3 = 0.6815 d. 1 − .6815 = 0.3185 75. Yes, 256 is found by 28. 77. .9744, found by 1 − (.40)4 79. a. 0.193, found by .15 + .05 − .0075 = .193 b. .0075, found by (.15)(.05) 81. a. P(F and >60) = .25, found by solving with the general rule of
multiplication: P(F) · P(>60|F) = (.5)(.5) b. 0 c. .3333, found by 1/3 83. 264 = 456,976 85. 1/3, 628,800 87. a. P(D) = .20(.03) + .30(.04) + .25(.07) + .25(.065) = .05175
b. P(Tyson ∣ defective) = .20(.03)
[.20(.03) + .30(.04) + .25(.07) + .25(.065)]
= .1159
Supplier Joint Revised
Tyson .00600 .1159 Fuji .01200 .2319 Kirkpatricks .01750 .3382 Parts .01625 .3140
.05175 1.0000
89. 0.512, found by (0.8)3
91. .525, found by 1 − (.78)3 93. a.
Winning Low Moderate High Season Attendance Attendance Attendance Total
No 7 6 1 14 Yes 1 9 6 16 Total 8 15 7 30
1. 0.5333 found by 16/30 2. 0.5667 found by 16/30 + 5/30 − 4/30 = 17/30 3. 0.8571 found by 6/7 4. 0.0333 found by 1/30 b.
Losing Winning Season Season Total
New 8 8 16 Old 6 8 14 Total 14 16 30
1. 0.4667 found by 14/30 2. 0.2667 found by 8/30 3. 0.8000 found by 16/30 + 16/30 − 8/30 = 24/30
CHAPTER 6 1. Mean = 1.3, variance = .81, found by: µ = 0(.20) + 1(.40) + 2(.30) + 3(.10) = 1.3 σ2 = (0 − 1.3)2(.2) + (1 − 1.3)2(.4) + (2 − 1.3)2(.3) + (3 − 1.3)2(.1) = .81 3. Mean = 14.5, variance = 27.25, found by:
μ = 5(.1) + 10(.3) + 15(.2) + 20(.4) = 14.5 σ2 = (5 − 14.5)2(.1) + (10 − 14.5)2(.3)
+ (15 − 14.5)2(.2) + (20 − 14.5)2(.4) = 27.25
5. a. Calls, x Frequency P (x) xP (x) (x − µ)2 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. µ = Σx · P(x) = 1.70 d. σ = √1.01 = 1.005 7.
Amount P (x) xP (x) (x − µ)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
a. µ = ΣxP(x) = 21 b. σ2 = Σ(x − µ)2P(x) = 259 σ = √259 = 16.093
9. a. P(2) = 4!
2!(4 − 2)! (.25)2(.75)4−2 = .2109
b. P(3) = 4!
3!(4 − 3)! (.25)3(.75)4−3 = .0469
11. a. x P (x)
0 .064 1 .288 2 .432 3 .216
b. μ = 1.8 σ2 = 0.72 σ = √0.72 = .8485
13. a. .2668, found by P(2) = 9!
(9 − 2)!2! (.3)2(.7)7
b. .1715, found by P(4) = 9!
(9 − 4)!4! (.3)4(.7)5
c. .0404, found by P(0) = 9!
(9 − 0)!0! (.3)0(.7)9
15. a. .2824, found by P(0) = 12!
(12 − 0)!0! (.1)0(.9)12
b. .3766, found by P(1) = 12!
(12 − 1)!1! (.1)1(.9)11
c. .2301, found by P(2) = 12!
(12 − 2)!2! (.1)2(.9)10
d. µ = 1.2, found by 12(.1) σ = 1.0392, found by √1.08
796
17. a. 0.1858, found by 15!
2!13! (0.23)2(0.77)13
b. 0.1416, found by 15!
5!10! (0.23)5(0.77)10
c. 3.45, found by (0.23)(15) 19. a. 0.296, found by using Appendix B.1 with n of 8, π 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.1 with n of 9, π 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. µ = 10.5, found by 15(0.7) and σ = √15(0.7) (0.3) = 1.7748
b. 0.2061, found by 15!
10!5! (0.7)10(0.3)5
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. P(2) = [6C2] [4C1]
10C3 =
15(4)
120 = .50
27. N is 10, the number of loans in the population; S is 3, the num- ber of underwater loans in the population; x is 0, the number of selected underwater loans in the sample; and n is 2, the size of the sample. Use formula (6–6) to find
P(0) = (7C2) (3C0)
10C2 =
21(1)
45 = 0.4667.
29. P(2) = [9C3][6C2]
[15C5] =
84(15)
3003 = .4196
31. a. .6703 b. .3297 33. a. .0613 b. .0803 35. µ = 6 P(x ≥ 5) = 1 − (.0025 + .0149 + .0446 + .0892 + .1339) = .7149 37. A random variable is an outcome that results from a chance
experiment. A probability distribution also includes the likelihood of each possible outcome.
39. µ = $1,000(.25) + $2,000(.60) + $5,000(.15) = $2,200 σ2 = (1,000 − 2,200)2 .25 + (2,000 − 2,200)2 .60 +
(5,000 − 2,200)2 .15 = 1,560,000 41. µ = 12(.25) + … + 15(.1) = 13.2 σ2 = (12 − 13.2)2.25 + … + (15 − 13.2)2.10 = 0.86 σ = √0.86 = .927 43. a. 5 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 15!
10!5! (0.4)10(0.6)5
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. µ = 20(0.075) = 1.5 σ = √20(0.075) (0.925) = 1.1779
b. 0.2103, found by 20!
0!20! (0.075)0(0.925)20
c. 0.7897, found by 1 − 0.2103
49. a. 0.1311, found by 16!
4!12! (0.15)4(0.85)12
b. 2.4, found by (0.15)(16) c. 0.2100, found by 1 − 0.0743 − 0.2097 − 0.2775 − 0.2285
51. 0.2784, found by 0.1472 + 0.0811 + 0.0348 + 0.0116 + 0.0030 + 0.0006 + 0.0001 + 0.0000
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. µ = 12(0.52) = 6.24 σ = √12(0.52) (0.48) = 1.7307 c. 0.1768 d. 0.3343, found by
0.0002 + 0.0019 + 0.0116 + 0.0418 + 0.1020 + 0.1768
55. a. P(1) = [7C2] [3C1]
[10C3] =
(21) (3)
120 = .5250
b. P(0) = [7C3] [3C0]
[10C3] =
(35) (1)
120 = .2917
P(x ≥ 1) = 1 − P(0) = 1 − .2917 = .7083
57. P(x = 0) = [8C4][4C0]
[12C4] =
70 495
= .141
59. a. .0498 b. .7746, found by (1 − .0498)5 61. a. .0183 b. .1954 c. .6289 d. .5665
63. a. 0.1733, found by (3.1)4 e−3.1
4!
b. 0.0450, found by (3.1)0 e−3.1
0! c. 0.9550, found by 1 − 0.0450
65. μ = nπ = 23 ( 2
113) = .407
P(2) = (.407)2e− .407
2! = 0.0551
P(0) = (.407)0 e− .407
0! = 0.6656
67. Let µ = nπ = 155(1∕3,709) = 0.042
P(4) = 0.0424 e−0.042
4! = 0.00000012
Very unlikely! 69. a. µ = nπ = 15(.67) = 10.05 σ = √nπ(1 − π) = √15(.67) (.33) = 1.8211 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 2.3, found. The aver-
age season home runs per team is 187. Then (187 × 2)/162 = 2.3.
a. P(0) = 2.30 e−2.3
0! = 0.1003
b. P(2) = 2.32 e−2.3
2! = 0.2652
c. P(x ≥ 4) = 0.1981, found by 1 − (0.1169 + 0.0538 + 0.0206 + 0.0068)
CHAPTER 7
1. a. b = 10, a = 6 b. μ = 6 + 10
2 = 8
c. σ = √ (10 − 6)2
12 = 1.1547
d. Area = 1
(10 − 6) ·
(10 − 6) 1
= 1
e. P(x > 7) = 1
(10 − 6) ·
10 − 7 1
= 3 4
= .75
f. P(7 ≤ x ≤ 9) = 1
(10 − 6) ·
(9 − 7) 1
= 2 4
= .50
797
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
b. μ = 0.5 + 3.00
2 = 1.75
σ = √ (3.00 − .50)2
12 = .72
c. P(x < 1) = 1
(3.0 − 0.5) ·
1 − .5 1
= .5
2.5 = 0.2
d. 0, found by 1
(3.0 − 0.5) (1.0 − 1.0)
1
e. P(x > 1.5) = 1
(3.0 − 0.5) ·
3.0 − 1.5 1
= 1.5 2.5
= 0.6
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. zRob = $50,000 − $60,000
$5,000 = −2
zRachel = $50,000 − $35,000
$8,000 = 1.875
Adjusting for their industries, Rob is well below average and Rachel well above.
13. a. 1.25, found by z = 25 − 20
4.0 = 1.25
b. 0.3944, found in Appendix B.3
c. 0.3085, found by z = 18 − 20
2.5 = −0.5
Find 0.1915 in Appendix B.3 for z = −0.5, then 0.5000 − 0.1915 = 0.3085
15. a. 0.3413, found by z = $24 − $20.50
$3.50 = 1.00, then find 0.3413
in Appendix B.3 for z = 1 b. 0.1587, found by 0.5000 − 0.3413 = 0.1587
c. 0.3336, found by z = $19.00 − $20.50
$3.50 = −0.43
Find 0.1664 in Appendix B.3, 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.3. 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. 0.2514: Begin by using formula (7–5) to find the z value for
$3,100, which is (3,100 − 2,800)∕450, or 0.67. Then see Appendix B.3 to find the area between 0 and 0.67, which is 0.2486. Finally, since the area of interest is beyond 0.67, subtract that probability from 0.5000. The result is 0.5000 − 0.2486, or 0.2514.
b. 0.1908: Use formula (7–5) to find the z value for $3,500, which is (3,500 − 2,800)/450, or 1.56. Then see Appendix B.3 for the area under the standard normal curve. That probabil- ity is 0.4406. Since the two points (1.56 and 0.66) are on the same side of the mean, subtract the smaller probability from the larger. The result is 0.4406 − 0.2486 = 0.1920.
c. 0.8294: Use formula (7–5) to find the z value for $2,250, which is −1.22, found by (2,250 − 2,800)∕450. The corre-
sponding area is 0.3888. Since 1.56 and −1.22 are on differ- ent sides of the mean, add the corresponding probabilities. Thus, we find 0.3888 + 0.4406 = 0.8294.
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% of the data to fall in- side the curve. Solving for x: 1.65 = (x − 50)∕4, so x = 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%, found by 12 + 1.65(18) 31. a. μ = nπ = 50(0.25) = 12.5 σ2 = nπ (1 − π) = 12.5(1 − 0.25) = 9.375 σ = √9.375 = 3.0619 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. μ = nπ = 80(0.07) = 5.6 σ = √5.208 = 2.2821 0.3483, found from z = (6.5 − 5.6)∕2.2821 = 0.39 with the corre-
sponding 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 cor-
responding 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: over-
weight 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 μ = 500(0.30) = 150 σ2 = 500(.30) (.70) = 105 σ = √105 = 10.24695
z = x − μ
σ =
174.5 − 150 10.24695
= 2.39
The area under the curve for 2.39 is 0.4916. Then 0.5000 − 0.4916 = 0.0084.
c. 0.8461, found by z = 139.5 − 150
10.24695 = −1.02
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. μ = 11.96 + 12.05
2 = 12.005
b. σ = √ (12.05 − 11.96)2
12 = .0260
c. P(x < 12) = 1
(12.05 − 11.96) 12.00 −11.96
1 =
.04
.09 = .44
798
59. 0.0968, found by: μ = 50(0.40) = 20 σ2 = 50(0.40) (0.60) = 12 σ = √12 = 3.46 z = (24.5 − 20)∕3.46 = 1.30. The area is 0.4032. Then, for 25 or more, 0.5000 − 0.4032 =
0.0968. 61. a. 1.65 = (45 − µ)∕5 µ = 36.75 b. 1.65 = (45 − µ)∕10 µ = 28.5 c. z = (30 − 28.5)∕10 = 0.15, then 0.5000 + 0.0596 = 0.5596 63. a. 21.19%, 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. σ = (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. 470 − μ
σ = 0.25
500 − μ σ
= 1.28 σ = 29,126 and
µ = 462,718
69. μ = 150(0.15) = 22.5 σ = √150(0.15) (0.85) = 4.37 z = (29.5 − 22.5)∕4.37 = 1.60 P(z > 1.60) = .05000 − 0.4452 = 0.0548 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. 0.0427, found by 0.5000 − 0.4573 with z = (3.500 − 2.439)/
0.618 = 1.72; leads to 1.3 teams, found by 30(0.0427). One team actually had attendance of more than 3.5 million. So the estimate is accurate.
b. 0.7019, found by 0.5000 + 0.2019 with z = (100 − 121)/ 40.0 = 0.53; leads to 21.1 teams with salaries of more than $100 mil- lion, found by 30(0.7019). 18 teams actually had salaries of more than $100 million. We may want to try another compari- son to assess if the normal probability distribution is a good description of salary.
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. York Automotive
Thayer Chevrolet Toyota Franklin Park Lincoln Mercury Mathews Ford Oregon Inc. Valiton Chrysler
d. P(x > 11.98) = 1
(12.05 − 11.96) (
12.05 − 11.98 1 )
= .07 .09
= .78
e. All cans have more than 11.00 ounces, so the probability is 100%.
43. a. μ = 4 + 10
2 = 7
b. σ = √ (10 − 4)2
12 = 1.732
c. P(x < 6) = 1
(10 − 4) · (
6 − 4 1 ) =
2 6
= .33
d. P(x > 5) = 1
(10 − 4) · (
10 − 5 1 ) =
5 6
= .83
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 deviation below the mean. Employees is 2.92 standard deviation above the mean.
c. 65.54% of the aluminum fabricators have greater net sales compared with Clarion, found by 0.1554 + 0.5000. Only 0.18% have more employees than Clarion, found by 0.5000 − 0.4982.
47. a. 0.5000, because z = 430 − 890
90 = −5.11
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 hours/week for women. 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. 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 for women.
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. 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: μ = 60(0.64) = 38.4 σ2 = 60(0.64) (0.36) = 13.824 σ = √13.824 = 3.72 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
799
.50
.40
.30
.20
.10
Pr ob
ab ili
ty
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
The population has more dispersion than the sample means. The sample means vary from 1.33 to 4.0. The population var- ies from 0 to 6.
11. a.
.1
.05
0 1 2 3 4 5 6 7 8 9
μ = 0 + 1 + . . . + 9
10 = 4.5
b. Sample Sum x
1 11 2.2 2 31 6.2 3 21 4.2 4 24 4.8 5 21 4.2
Sample Sum x
6 20 4.0 7 23 4.6 8 29 5.8 9 35 7.0
10 27 5.4
3
2
1
Fr eq
ue nc
y
Values 2.0 3 4 5 6 7 8
5. a. 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
b. μx = (12 + 13 + 14 + 13 + 14 + 15)∕6 = 13.5 μ = (12 + 12 + 14 + 16)∕4 = 13.5 c. More dispersion with population data compared to the sam-
ple means. The sample means vary from 12 to 15, whereas the population varies from 12 to 16.
7. a. 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
b. μx = (12.66 + . . . + 15.33 + 15.33)
10 = 14.6
μ = (12 + 12 + 14 + 15 + 20)∕5 = 14.6 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.
Sample Cases Sum Mean
Ruud, Wu, Sass 3, 6, 3 12 4.00 Ruud, Sass, Flores 3, 3, 3 9 3.00
⋮ ⋮ ⋮ ⋮ Sass, Flores, Schueller 3, 3, 1 7 2.33
c. μx = 2.67, found by 53.33
20 μ = 2.67, found by (3 + 6 + 3 + 3 + 0 + 1)∕6. They are equal. d.
.50
.40
.30
.20
.10
6
Pr ob
ab ili
ty
Number of cases
Population
0 2 m 4
800
d. μx = 79.67, found by 1,195∕15. μ = 79.67, found by 478∕6. They are equal. e. Answers will vary. Not likely as the student is not graded on
all available information. Based on these test sores however, this student has a 8/15 chance of receiving a higher grade with this method than the average and a 7/15 chance of re- ceiving a lower grade.
29. a. 10, found by 5C2 b.
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
c. μx = (3.5 + 4.5 + . . . + 2.5)∕10 = 3.4 μ = (4 + 3 + 5 + 3 + 2)∕5 = 3.4 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.
b. σx = 5.5 √25
= 1.1
c. z = 36 − 35 5.5∕√25
= 0.91
p = 0.1814, found by 0.5000 + 0.3186
d. z = 34.5 − 35 5.5∕√25
= −0.45
p = 0.6736, found by 0.5000 + 0.1736 e. 0.4922, found by 0.3186 + 0.1736
33. z = $335 − $350
$45∕√40 = −2.11
p = 0.9826, found by 0.5000 + 0.4826
35. z = 29.3 − 29 2.5∕√60
= 0.93
p = 0.8238, found by 0.5000 + 0.3238 37. Between 5,954 and 6,046, found by 6,000 ± 1.96 (150∕√40)
39. z = 900 − 947 205∕√60
= −1.78
p = 0.0375, found by 0.5000 − 0.4625 41. a. Alaska, Connecticut, Georgia, Kansas, Nebraska, South
Carolina, Virginia, Utah b. Arizona, Florida, Iowa, Massachusetts, Nebraska, North
Carolina, Rhode Island, Vermont
43. a. z = 600 − 510 14.28∕√10
= 19.9, P = 0.00, or virtually never
b. z = 500 − 510 14.28∕√10
= −2.21,
p = 0.4864 + 0.5000 = 0.9864
c. z = 500 − 510 14.28∕√10
= −2.21,
p = 0.5000 − 0.4864 = 0.0136
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 be- tween 4 and 5.
13. a.–c. Answers will vary depending on the coins in your possession.
15. a. z = 63 − 60 12∕√9
= 0.75
p = .2266, found by .5000 − .2734
b. z = 56 − 60 12∕√9
= −1.00
p = .1587, found by .5000 − .3413 c. p = .6147, found by 0.3413 + 0.2734
17. z = 1,950 − 2,200
250∕√50 = −7.07 p = 1, 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. 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
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 + … + 3.0)/9 = 3/9 = 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 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. (79+64+84+82+92+77)/6 =79.67% b. 15 found by 6C2 c.
Sample Value Sum Mean
1 79, 64 143 71.5 2 79, 84 163 81.5
⋮ ⋮ ⋮ ⋮ 15 92, 77 169 84.5
1,195.0
801
25. a. 577, found by n = 0.60(0.40)( 1.96 0.04)
2
= 576.24
b. 601, found by n = 0.50(0.50)( 1.96 0.04)
2
= 600.25
27. 33.41 and 36.59, found by
35 ± 2.030( 5
√36) √
300 − 36 300 − 1
29. 1.683 and 2.037, found by
1.86 ± 2.680( 0.5 √50)
√ 400 − 50 400 − 1
31. 6.13 years to 6.87 years, found by 6.5 ± 1.989(1.7∕√85) 33. a. Between $864.82 and 903.18, found by
884 ± 2.426( 50
√40) .
b. $950 is not reasonable because it is outside of the confi- dence interval.
35. a. The population mean is unknown. b. Between 7.50 and 9.14, found by 8.32 ± 1.685(3.07∕√40) c. 10 is not reasonable because it is outside the confidence
interval. 37. a. 65.49 up to 71.71 hours, found by
68.6 ± 2.680(8.2∕√50) b. The value suggested by the NCAA is included in the confi-
dence 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. 61.47, rounded to 62. Found by solving for n in the equation:
1.96(16∕√n) = 4 41. Between $55,461.23 up to $57,769.43 found by 55,051 ±
1.711 ( 7,568 √25 )
. 55,000 is reasonable because it is inside of the
confidence interval. 43. a. 82.58, found by 991/12 b. Between 80.54 and $84.62, found by
82.58 ± 1.796 ( 3.94 √12 )
.
c. 80 hours per week 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
89.4667 ± 2.145(8.08∕√15) 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 0.035 ± 2.576(√ 0.035(1 − 0.035)
400 ). It would not be
reasonable to conclude that fewer than 5% of the employees are now failing the test because 0.05 is inside the confidence interval.
49. Between 0.648 and 0.752, found by
0.70 ± 2.576 (√ 0.70(1 − 0.70)
500 )(√ 20,000 − 500
20,000 − 1 ).
Yes, because even the lower limit of the confidence interval is above 0.500.
51. $52.51 and $55.49, found by
$54.00 ± 2.032 $4.50
√35 √
(500 − 35) 500 − 1
53. 369, found by n = 0.60(1 − 0.60)(1.96∕0.05)2 55. 97, found by [(1.96 × 500)∕100]2 57. a. Between 7,849 and 8,151, found by
8,000 ± 2.756(300∕√30)
b. 554, found by n = ( (1.96) (300)
25 ) 2
45. a. σX = 2.1 √81
= 0.23
b. z = 7.0 − 6.5 2.1∕√81
= 2.14, z = 6.0 − 6.5 2.1∕√81
= −2.14,
p = .4838 + .4838 = .9676
c. z = 6.75 − 6.5
2.1∕√81 = 1.07, z =
6.25 − 6.5 2.1∕√81
= −1.07,
p = .3577 + .3577 = .7154 d. .0162, found by .5000 − .4838 47. Mean 2016 attendance is 2.439 million. Likelihood of a sample
mean this large or larger is 0.5359, found by 0.5000 + 0.0359,
where z = 2.439 − 2.45
0.71 √30
= −0.09.
CHAPTER 9 1. 51.314 and 58.686, found by 55 ± 2.58(10∕ √49) 3. a. 1.581, found by σx = 25∕√250 b. The population is normally distributed and the population
variance is known. In addition, the Central Limit Theorem says that the sampling distribution of sample means will be normally distributed.
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 $20 ± 1.96($5∕ √49). About
95% of the intervals similarly constructed will include the population mean.
7. a. 8.60 gallons b. 7.83 and 9.37, found by 8.60 ± 2.58(2.30∕√60) 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 20 ± 2.093(2∕√20) e. Neither value is reasonable because they are not inside the
interval. 13. Between 95.39 and 101.81, found by 98.6 ± 1.833(5.54∕√10) 15. a. 0.8, found by 80∕100 b. Between 0.72 and 0.88, found by
0.8 ± 1.96 (√ 0.8(1 − 0.8)
100 )
c. We are reasonably sure the population proportion is be- tween 72 and 88%.
17. a. 0.625, found by 250∕400 b. Between 0.563 and 0.687, found by
0.625 ± 2.58 (√ 0.625(1 − 0.625)
400 )
c. We are reasonably sure the population proportion is be- tween 56 and 69%. Because the estimated population pro- portion is more than 50%, the results indicate that Fox TV should schedule the new comedy show.
19. 97, found by n = ( 1.96 × 10
2 ) 2
= 96.04
21. 196, found by n = 0.15(0.85)( 1.96 0.05)
2
= 195.9216
23. 554, found by n = ( 1.96 × 3
0.25 ) 2
= 553.19
802
59. a. Between 75.44 and 80.56, found by 78 ± 2.010(9∕√50)
b. 220, found by n = ( (1.645) (9)
1.0 ) 2
61. a. 4, found by 24∕√36
b. Between $641.88 and $658.12, found by
650 ± 2.030( 24
√36) c. 23, found by n = {(1.96 × 24)∕10}2 = 22.13 63. 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 65. a. Between 0.156 and 0.184, found by
0.17 ± 1.96 √ (0.17) (1 − 0.17)
2700 b. Yes, because 18% are 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∕√352) 69. a. Answers will vary. b. Answers will vary. c. Answers will vary. d. Answers may vary. e. Select a different sample of 20 homes and compute a confi-
dence interval using the new sample. There is a 5% probabil- ity that a sample mean will be more than 1.96 standard errors from the mean. If this happens, the confidence interval will not include the population mean.
71. a. Between $4,033.1476 and $5,070.6274, found by 4,551.8875 ± 518.7399.
b. Between 71,040.0894 and 84,877.1106, found by 77,958.6000 ± 6,918.5106.
c. In general, the confidence intervals indicate that the average maintenance cost and the average odometer reading sug- gest an aging bus fleet.
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 z = (49 − 50)∕(5∕√36) = −1.2 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 z = (21 − 20)∕(5∕√36) d. Fail to reject H0 at the .05 significance level e. p = .1151, found by .5000 − .3849. An 11.51% chance of finding
a z value this large or larger. 5. a. H0: µ = 60,000 H1: µ ≠ 60,000 b. Reject H0 if z < −1.96 or z > 1.96. c. −0.69, found by:
z = 59,500 − 60,000
(5,000∕√48) 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: μ ≥ 6.8 H1: μ < 6.8 b. Reject H0 if z < −1.65
c. z = 6.2 − 6.8 1.8∕√36
= −2.0
d. H0 is rejected. e. p = 0.0228. The mean number of DVDs watched is less than
6.8 per month. If H0 is true, you will get a statistic this small less than one time out of 40 tests.
9. a. Reject H0 when t < 1.833
b. t = 12 − 10 (3∕√10)
= 2.108
c. Reject H0. The mean is greater than 10. 11. H0: µ ≤ 40 H1: µ > 40 Reject H0 if t > 1.703.
t = 42 − 40 (2.1∕√28)
= 5.040
Reject H0 and conclude that the mean number of calls is greater than 40 per week.
13. H0: µ ≤ 50,000 H1: µ > 50,000 Reject H0 if t > 1.833.
t = (60000 − 50000)
(10000∕√10) = 3.16
Reject H0 and conclude that the mean income in Wilmington is greater than $50,000.
15. a. Reject H0 if t < −3.747.
b. x = 17 and s = √ 50
5 − 1 = 3.536
t = 17 − 20
(3.536∕√5) = −1.90
c. Do not reject H0. We cannot conclude the population mean is less than 20.
d. Between .05 and .10, about .065 17. H0: µ ≤ 1.4 H1: µ >1.4 Reject H0 if t > 2.821.
t = 1.6 − 1.4
0.216∕√10 = 2.93
Reject H0 and conclude that water consumption has increased. The p-value is between 0.01 and 0.005. There is a slight proba- bility (between one chance in 100 and one chance in 200) this rise could have arisen by chance.
19. H0: µ ≤ 67 H1: µ > 67 Reject H0 if t > 1.796
t = (82.5 − 67) (59.5∕√12)
= 0.902
Fail to reject H0 and conclude that the mean number of text messages is not greater than 67. The p-value is greater than 0.05. There is a good probability (about 18%) this could happen by chance. (LO10-7)
21. 1.05, found by z = (9,992 − 9,880)∕(400∕√100). Then 0.5000 − 0.3531 = 0.1469, which is the probability of a Type II error.
23. H0: µ ≥ 60 H1: µ < 60 Reject H0 if z < −1.282; the critical value is 59.29.
z = 58 − 60
(2.7∕√24) = −3.629
Reject H0. The mean assembly time is less than 60 minutes. Using the sample mean, 58, as μ1, the z-score for 59.29 is 2.34.
So the probability for values between 58 and 59.29 is .4904. The Type II error is the area to the right of 59.29 or .5000 − .4904 = .0096.
25. H0: µ = $45,000 H1: µ ≠ $45,000 Reject H0 if z < −1.65 or z > 1.65.
z = $45,500 − $45,000
$3000∕√120 = 1.83
Reject H0. We can conclude that the mean salary is not $45,000. p-value 0.0672, found by 2(0.5000 − 0.4664).
27. H0: µ ≥ 10 H1: µ < 10 Reject H0 if z < −1.65.
z = 9.0 − 10.0 2.8∕√50
= −2.53
Reject H0. The mean weight loss is less than 10 pounds. p-value = 0.5000 − 0.4943 = 0.0057
803
29. H0: µ ≥ 7.0 H1: µ < 7.0 Assuming a 5% significance level, reject H0 if t < −1.677.
t = 6.8 − 7.0 0.9∕√50
= −1.57
Do not reject H0. West Virginia students are not sleeping less than 6 hours. p-value is between .05 and .10.
31. H0: µ ≥ 3.13 H1: µ < 3.13 Reject H0 if t < −1.711
t = 2.86 − 3.13 1.20∕√25
= −1.13
We fail to reject H0 and conclude that the mean number of resi- dents is not necessarily less than 3.13.
33. H0: µ ≤ $6,658 H1: µ > $6,658 Reject H0 if t > 1.796
x = 85,963
12 = 7,163.58 s = √
9,768,674.92 12 − 1
= 942.37
t = 7163.58 − 6,658
942.37∕√12 = 1.858
Reject H0. The mean interest paid is greater than $6,658. 35. H0: µ = 3.1 H1: µ ≠ 3.1 Assume a normal population. Reject H0 if t < −2.201 or t > 2.201.
x = 41.1 12
= 3.425
s = √ 4.0625 12 − 1
= .6077
t = 3.425 − 3.1 .6077∕√12
= 1.853
Do not reject H0. Cannot show a difference between senior citi- zens and the national average. p-value is about 0.09.
37. H0: µ ≥ 6.5 H1: µ < 6.5 Assume a normal population. Reject H0 if t < −2.718. x = 5.1667 s = 3.1575
t = 5.1667 − 6.5 3.1575∕√12
= −1.463
Do not reject H0. The p-value is greater than 0.05. 39. H0: µ = 0 H1: µ ≠ 0 Reject H0 if t < −2.110 or t > 2.110. x = −0.2322 s = 0.3120
t = −0.2322 − 0 0.3120∕√18
= −3.158
Reject H0. The mean gain or loss does not equal 0. The p-value is less than 0.01, but greater than 0.001.
41. H0: µ ≤ 100 H1: µ > 100 Assume a normal population. Reject H0 if t > 1.761.
x = 1,641
15 = 109.4
s = √ 1,389.6 15 − 1
= 9.9628
t = 109.4 − 100 9.9628∕√15
= 3.654
Reject H0. The mean number with the scanner is greater than 100. p-value is 0.001.
43. H0: µ = 1.5 H1: µ ≠ 1.5 Reject H0 if t > 3.250 or t < −3.250.
t = 1.3 − 1.5 0.9∕√10
= −0.703
Fail to reject H0. 45. H0: µ ≥ 30 H1: µ < 30 Reject H0 if t < −1.895.
x = 238.3
8 = 29.7875 s = √
5.889 8 − 1
= 0.9172
t = 29.7875 − 30
0.9172∕√8 = −0.655
Do not reject H0. The cost is not less than $30,000.
47. a. 9.00 ± 1.645(1∕√36) = 9.00 ± 0.274. So the limits are 8.726 and 9.274.
b. z = 8.726 − 8.6
1∕√36 = 0.756.
P(z < 0.756) = 0.5000 + 0.2764 = .7764
c. z = 9.274 − 9.6
1∕√36 = −1.956.
P(z > −1.96) = 0.4750 + 0.5000 = .9750
49. 50 + 2.33 10 √n
= 55 − .525 10 √n
n = (5.71)2 = 32.6
Let n = 33 51. H0: µ ≥ 8 H1: µ < 8 Reject H0 if t < −1.714.
t = 7.5 − 8 3.2∕√24
= −0.77
Do not reject the null hypothesis. The time is not less. 53. a. H0: µ = 100 H1: µ ≠ 100 Reject H0 if t is not between −2.045 and 2.045.
t = 121.12 − 100 39.66∕√30
= 2.92
Reject the null. The mean salary is probably not $100.0 million. b. H0: µ ≤ 2,000,000 H1: µ > 2,000,000 Reject H0 if t is > 1.699.
t = 2,438,636 − 2,000,000
617,670∕√30 = 3.89
Reject the null. The mean attendance was more than 2,000,000.
CHAPTER 11 1. a. Two-tailed test b. Reject H0 if z < −2.05 or z > 2.05
c. z = 102 − 99
√ 52
40 + 6
2
50
= 2.59
d. Reject H0 e. p-value = .0096, found by 2(.5000 −.4952) 3. Step 1 H0: µ1 ≥ µ2 H1: µ1 < µ2 Step 2 The .05 significance level was chosen. Step 3 Reject H0 if z < −1.65. Step 4 −0.94, found by:
z = 7.6 − 8.1
√ (2.3)2
40 +
(2.9)2
55
= −0.94
Step 5 Fail to reject H0. Step 6 Babies using the Gibbs brand did not gain less weight.
p-value = .1736, found by .5000 −.3264. 5. Step 1 H0: μmarried = μunmarried H1: μmarried ≠ μunmarried Step 2 The 0.05 significance level was chosen Step 3 Use a z-statistic as both population standard deviations
are known. Step 4 If z < −1.960 or z > 1.960, reject H0.
Step 5 z = 3.0 − 3.4
√ (1.2)2
45 +
(1.1)2
39
= −1.59
Fail to reject the null. Step 6 It is reasonable to conclude that the time that married and
unmarried women spend each week is not significantly different. The p-value is greater than 0.05. The difference of 0.4 hours per week could be explained by sampling error.
7. a. Reject H0 if t > 2.120 or t < −2.120. df = 10 + 8 − 2 = 16
b. s2p = (10 − 1) (4)2 + (8 − 1) (5)2
10 + 8 − 2 = 19.9375
c. t = 23 − 26
√19.9375( 1
10 +
1 8)
= −1.416
804
d. Reject H0. There are more defective parts produced on the day shift.
e. p-value is less than .005 but greater than .0005. 19. H0: µd ≤ 0 H1: µd > 0 d = 25.917 sd = 40.791 Reject H0 if t >1.796
t = 25.917
40.791∕√12 = 2.20
Reject H0. The incentive plan resulted in an increase in daily in- come. The p-value is about .025.
21. H0: µM = µW H1: µM ≠ µW Reject H0 if t < −2.645 or t > 2.645 (df = 35 + 40 − 2).
sp 2 =
(35 − 1) (4.48)2 + (40 − 1) (3.86)2
35 + 40 − 2 = 17.3079
t = 24.51 − 22.69
√17.3079( 1
35 +
1 40)
= 1.890
Do not reject H0. 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.
23. H0: µ1 = µ2 H1: µ1 ≠ µ2 Reject H0 if z < −1.96 or z > 1.96.
z = 4.77 − 5.02
√ (1.05)2
40 +
(1.23)2
50
= −1.04
H0 is not rejected. There is no difference in the mean number of calls. p-value = 2(.5000 −.3508) = .2984.
25. H0: µB ≤ µA H1: µB > µA Reject H0 if t > 1.668
t = $61,000 − $57,000
√ ($7,100)2
30 +
($9,200)2
40
= $4,000.00
$1,948.42 = 2.05
Reject H0. The mean income is larger for Plan B. The p-value = .5000 − .4798 = .0202.
27. a. df = (
0.3136 12
+ 0.0900
12 ) 2
( 0.3136
12 ) 2
12 − 1 +
( 0.0900
12 ) 2
12 − 1
= 0.0011
0.000062 + 0.0000051 = 16.37 → 16df
b. H0: µa = µw H1: µa ≠ µw Reject H0 if t > 2.120 or t < −2.120.
c. t = 1.65 − 2.20
√ 0.3136
12 +
0.0900 12
= −3.00
d. Reject the null hypothesis. There is a difference. 29. Assume equal population standard deviations. H0: µn = µs H1: µn ≠ µs Reject H0 if t < −2.086 or t > 2.086.
s2p = (10 − 1) (10.5)2 + (12 − 1) (14.25)2
10 + 12 − 2 = 161.2969
t = 83.55 − 78.8
√161.2969( 1
10 +
1 12)
= 0.874
p-value > .10. Do not reject H0. There is no difference in the mean number of hamburgers sold at the two locations.
d. Do not reject H0. e. p-value is greater than .10 and less than .20. 9. Step 1 H0: μPitchers = μPosition Players
H1: μPitchers ≠ μPosition Players Step 2 The 0.01 significance level was chosen. Step 3 Use a t-statistic assuming a pooled variance with the stan-
dard deviation unknown. Step 4 df = 12 + 13 − 2 = 23. Reject H0 if t is not between
−2.807 and 2.807.
s2p = (12 − 1) (8.597)2 + (13 − 1) (8.578)2
12 + 13 − 2 = 73.738
t = 6.091 − 10.684
√73.738( 1
12 +
1 13)
= −1.336
Step 5 Do not reject H0. Step 6 There is no difference in the mean salaries of pitchers
and position players. 11. Step 1 H0: µs ≤ µa H1: µs > µa Step 2 The .10 significance level was chosen. Step 3 df = 6 + 7 − 2 = 11 Reject H0 if t > 1.363.
Step 4 s2p = (6 − 1) (12.2)2 + (7 − 1) (15.8)2
6 + 7 − 2 = 203.82
t = 142.5 − 130.3
√203.82( 1 6
+ 1 7)
= 1.536
Step 5 Reject H0. Step 6 The mean daily expenses are greater for the sales staff.
The p-value is between .05 and .10.
13. a. df = (
25 15
+ 225 12 )
2
( 25 15 )
2
15 − 1 +
( 225 12 )
2
12 − 1
= 416.84
0.1984 + 31.9602
= 12.96 → 12df b. H0: µ1 = µ2 H1: µ1 ≠ µ2 Reject H0 if t > 2.179 or t < −2.179.
c. t = 50 − 46
√ 25 15
+ 225 12
= 0.8852
d. Fail to reject the null hypothesis.
15. a. df = (
697,225 16
+ 2,387,025
18 ) 2
( 697,225
16 ) 2
16 − 1 +
( 2,387,025
18 ) 2
18 − 1
= 26.7 → 26df
b. H0: µPrivate ≤ µPublic H1: µPrivate > µPublic Reject H0 if t >1.706.
c. t = 12,840 − 11,045
√ 2,387,025
18 +
697,225 16
= 4.276
d. Reject the null hypothesis. The mean adoption cost from a private agency is greater than the mean adoption cost from a public agency.
17. a. Reject H0 if t > 2.353.
b. d = 12 4
= 3.00 sd = √ 2 3
= 0.816
c. t = 3.00
0.816∕√4 = 7.35
805
Do not reject H0. There is no difference in the mean number of cars in the two lots.
45. H0: µd ≤ 0 H1: µd > 0 Reject H0 if t > 1.711. d = 2.8 sd = 6.59
t = 2.8
6.59∕√25 = 2.124
Reject H0. There are on average more cars in the US 17 lot. 47. a. Using statistical software, the result is that we fail to reject
the null hypothesis that the mean prices of homes with and without pools are equal. Assuming equal population vari- ances, the p-value is 0.4908.
b. Using statistical software, the result is that we reject the null hypothesis that the mean prices of homes with and without garages are equal. There is a large difference in mean prices between homes with and without garages. Assuming equal population variances, the p-value is less than 0.0001.
c. Using statistical software, the result is that we fail to reject the null hypothesis that the mean prices of homes are equal with mortgages in default and not in default. Assuming equal population variances, the p-value is 0.6980.
49. Using statistical software, the result is that we reject the null hypothesis that the mean maintenance cost of buses powered by diesel and gasoline engines is the same. Assuming equal population variances, the p-value is less than 0.0001.
CHAPTER 12 1. a. 9.01, from Appendix B.6 3. Reject H0 if F > 10.5, where degrees of freedom in the numerator
are 7 and 5 in the denominator. Computed F = 2.04, found by:
F = s21
s22 =
(10)2
(7)2 = 2.04
Do not reject H0. There is no difference in the variations of the two populations.
5. H0: σ 2 1 = σ22 H1: σ
2 1 ≠ σ22
Reject H0 where F > 3.10. (3.10 is about halfway between 3.14 and 3.07.) Computed F = 1.44, found by:
F = (12)2
(10)2 = 1.44
Do not reject H0. There is no difference in the variations of the two populations.
7. a. H0: µ1 = µ2 = µ3; H1: Treatment means are not all the same. b. Reject H0 if F > 4.26. c & d.
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 H0. The treatment means are not all the same. 9. H0: µ1 = µ2 = µ3; H1: Treatment means are not all the same. Reject
H0 if F > 4.26.
Source SS df MS F
Treatment 276.50 2 138.25 14.18 Error 87.75 9 9.75
Reject H0. The treatment means are not all the same. 11. a. H0: µ1 = µ2 = µ3; H1: Not all means are the same. b. Reject H0 if F > 4.26. c. SST = 107.20, SSE = 9.47, SS total = 116.67 d.
Source SS df MS F
Treatment 107.20 2 53.600 50.96 Error 9.47 9 1.052
Total 116.67 11
31. Assume equal population standard deviations. H0: µ1 = µ2 H1: µ1 ≠ µ2 Reject H0 if t > 2.819 or t < −2.819.
s2p = (10 − 1) (2.33)2 + (14 − 1) (2.55)2
10 + 14 − 2 = 6.06
t = 15.87 − 18.29
√6.06( 1
10 +
1 14)
= −2.374
Do not reject H0. There is no difference in the mean amount purchased.
33. Assume equal population standard deviations. H0: µ1 ≤ µ2 H1: µ1 > µ2 Reject H0 if t > 2.567.
s2p = (8 − 1) (2.2638)2 + (11 − 1) (2.4606)2
8 + 11 − 2 = 5.672
t = 10.375 − 5.636
√5.672( 1 8
+ 1 11)
= 4.28
Reject H0. The mean number of transactions by the young adults is more than for the senior citizens.
35. H0: µ1 ≤ µ2 H1: µ1 > µ2 Reject H0 if t > 2.650. x1 = 125.125 s1 = 15.094 x2 = 117.714 s2 = 19.914
s2p = (8 − 1) (15.094)2 + (7 − 1) (19.914)2
8 + 7 − 2 = 305.708
t = 125.125 − 117.714
√305.708( 1 8
+ 1 7)
= 0.819
H0 is not rejected. There is no difference in the mean number sold at the regular price and the mean number sold at the re- duced price.
37. H0: µd ≤ 0 H1: µd > 0 Reject H0 if t > 1.895. d = 1.75 sd = 2.9155
t = 1.75
2.9155∕√8 = 1.698
Do not reject H0. There is no difference in the mean number of absences. The p-value is greater than .05 but less than .10.
39. H0: µ1 = µ2 H1: µ1 ≠ µ2 Reject H0 if t < −2.024 or t > 2.204.
sp 2 =
(15 − 1) (40)2 + (25 − 1) (30)2
15 + 25 − 2 = 1,157.89
t = 150 − 180
√1,157.89( 1
15 +
1 25)
= −2.699
Reject the null hypothesis. The population means are different. 41. H0: µd ≤ 0 H1: µd > 0 Reject H0 if t > 1.895. d = 3.11 sd = 2.91
t = 3.11
2.91∕√8 = 3.02
Reject H0. The mean is lower. 43. H0: µO = µR, H1: µO ≠ µR df = 25 + 28 − 2 = 51 Reject H0 if t < −2.008 or t > 2.008. xO = 86.24, sO = 23.43 xR = 92.04, sR = 24.12
s2p = (25 − 1) (23.43)2 + (28 − 1) (24.12)2
25 + 28 − 2 = 566.335
t = 86.24 − 92.04
√566.335( 1
25 +
1 28)
= −0.886
806
21. a.
d-320 j-1000
M ea
n
Machine
44
46
42
38
40
34
36
32
30 uv-57
Interaction Plot (data means) for Sales
Position Inside Outside
Yes, there appears to be an interaction effect. Sales are different based on machine position, either in the inside or outside position.
b. 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
The position and the interaction of position and machine ef- fects are significant. The effect of machine on sales is not significant.
c. 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
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. Posi- tion the UV-57 machine inside.
23. H0: σ 2 1 ≤ σ22; H1: σ
2 1 > σ22. df1 = 21 − 1 = 20;
df2 = 18 − 1 = 17. H0 is rejected if F > 3.16.
F = (45,600)2
(21,330)2 = 4.57
Reject H0. There is more variation in the selling price of ocean- front homes.
e. Since 50.96 > 4.26, H0 is rejected. At least one of the means differs.
f. ( x1 − x2) ± t√MSE(1∕n1 + 1∕n2) = (9.667 − 2.20) ± 2.262√1.052(1∕3 + 1∕5) = 7.467 ± 1.69 = [5.777, 9.157] Yes, we can conclude that treatments 1 and 2 have different
means. 13. H0: µ1 = µ2 = µ3 = µ4; H1: Not all means are equal. H0 is rejected if F > 3.71.
Source SS df MS F
Treatment 32.33 3 10.77 2.36 Error 45.67 10 4.567
Total 78.00 13
Because 2.36 is less than 3.71, H0 is not rejected. There is no difference in the mean number of weeks.
15. a. H0: µ1 = µ2; H1: Not all treatment means are equal. b. Reject H0 if F > 18.5. c. H0: µ1 = µ2 = µ3; H1: Not all block means are equal. H0 is rejected if F > 19.0. d. SS total = (46.0 − 36.5)2 + . . . + (35− 36.5)2 = 289.5 SST = 3(42.33 − 36.5)2 + 3(30.67 − 36.5)2 = 204.167 SSB = 2(38.5 − 36.5)2 + 2(31.5 − 36.5)2 + 2(39.5 − 36.5)2 = 8 + 50 + 18 = 76 SSE = 289.50 − 204.1667 − 76 = 9.3333 e.
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. 43.75 > 18.5, so reject H0. There is a difference in the treat- ments. 8.14 < 19.0, so do not reject H0 for blocks. There is no difference among blocks.
17. 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
There is a difference in shifts, but not by employee. 19.
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
a. Since the p-value (0.18) is greater than 0.05, there is no differ- ence 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.
807
39. For color, the critical value of F is 4.76; for size, it is 5.14.
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
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 0.668. Do not reject H0. b. Critical value of F is 3.26. Computed F value is 100.204. Re-
ject H0 for block means. There is a difference in homes but not assessors. 43. For gasoline: H0: µ1 = µ2 = µ3; H1: Mean mileage is not the same. Reject H0 if F > 3.89. For automobile: H0: µ1 = µ2 = . . . = µ7; H1: Mean mileage is not the same. Reject H0 if F > 3.00.
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
There is a difference in both autos and gasoline. 45. H0: µ1 = µ2 = µ3 = µ4 = µ5 = µ6; H1: The treatment means are not
equal. Reject H0 if F > 2.37.
Source SS df MS F
Treatment 0.03478 5 0.00696 3.86 Error 0.10439 58 0.0018
Total 0.13917 63
H0 is rejected. There is a difference in the mean weight of the colors. 47. a.
b. Two-way ANOVA: Wage versus Gender, Sector
Private Sector
Public
1,250
1,200
1,150
1,100
1,050
1,000
M ea
n
Interaction Plot (data means) for Wage
Gender Men Women
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
25. Sharkey: n = 7 ss = 14.79 White: n = 8 sw = 22.95 H0: σ2w ≤ σ2s ; H1: σ2w > σ2s . dfs = 7 − 1 = 6; dfw = 8 − 1 = 7. Reject H0 if F > 8.26.
F = (22.95)2
(14.79)2 = 2.41
Cannot reject H0. There is no difference in the variation of the monthly sales.
27. a. H0: µ1 = µ2 = µ3 = µ4 H1: Treatment means are not all equal. b. α = .05 Reject H0 if F > 3.10. c.
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. 29. H0: µ1 = µ2 = µ3; 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
H0 is rejected. There is a difference in the treatment means. 31. H0: µ1 = µ2 = µ3 = µ4; H1: Not all means are equal. H0 is rejected if F > 3.10.
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: µ1 = µ2; H1: µ1 ≠ µ2. Critical value of F = 4.75.
Source SS df MS F
Treatment 219.43 1 219.43 23.10 Error 114.00 12 9.5
Total 333.43 13
b. t = 37 − 45
√9.5( 1 6
+ 1 8)
= −4.806
Since t2 = F. That is (−4.806)2 ≈ 23.10 (actually 23.098, differ- ence due to rounding). The p-value for this statistic is 0.0004 as well. Reject H0 in favor of the alternative.
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 mile per gallon are not the same.
37. H0: µ1 = µ2 = µ3 = µ4. H1: At least one mean is different. Reject H0 if F > 2.7395. Since 13.74 > 2.74, reject H0. You can also see this from the p-value of 0.0001 < 0.05. Priority mail express is faster than all three of the other classes, and priority mail is faster than either first-class or standard. However, first-class and stan- dard mail may be the same.
808
3. a. Sales. b.
30
20
10
0
Number of advertisements 1 2 3 4 5 6
Sa le
s ($
00 0)
y
x
c. Σ (x − x ) ( y − y ) = 36, n = 5, sx = 1.5811, sy = 6.1237
r = 36
(5 − 1) (1.5811) (6.1237) = 0.9295
d. There is a strong positive association between the variables. 5. a. Either variable could be independent. In the scatter plot, police
is the independent variable. b.
18.0
12.0
6.0
Police 12.0 16.0 20.0 24.0 28.0
Cr im
es
c. n = 8, Σ (x − x ) (y − y ) = −231.75, sx = 5.8737, sy = 6.4462
r = −231.75
(8 − 1) (5.8737) (6.4462) = −0.8744
d. Strong inverse relationship. As the number of police in- creases, the crime decreases or, as crime increases the num- ber of police decrease.
7. Reject H0 if t > 1.812.
t = .32√12 − 2 √1 − (.32)2
= 1.068
Do not reject H0. 9. H0: ρ ≤ 0; H1: ρ > 0. Reject H0 if t > 2.552. df = 18.
t = .78√20 − 2 √1 − (.78)2
= 5.288
Reject H0. There is a positive correlation between gallons sold and the pump price.
11. H0: ρ ≤ 0 H1: ρ > 0 Reject H0 if t > 2.650 with df = 13.
t = 0.667√15 − 2
√1 − 0.6672 = 3.228
Reject H0. There is a positive correlation between the number of passengers and plane weight.
13. a. ŷ = 3.7671 + 0.3630x
b = 0.7522( 1.3038 2.7019) = 0.3630
a = 5.8 − 0.3630(5.6) = 3.7671 b. 6.3081, found by ŷ = 3.7671 + 0.3630(7)
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.
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%
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%
d. The statistical results show that only sector, private or public, has a significant effect on the wages of accountants.
49. a. H0: σ 2 p = σ2np H1: σ2p ≠ σ2np.
Reject H0. The p-value is less than 0.05. There is a difference in the variance of average selling prices between houses with pools and houses without pools.
b. H0: σ 2 g = σ2ng H1: σ
2 g ≠ σ2ng
Reject H0. There is a difference in the variance of average selling prices between house with garages and houses with- out garages. The p-value is < 0.0001.
c. H0: µ1 = µ2 = µ3 = µ4 = µ5; H1: Not all treatment means are equal
Fail to reject H0. The p-value is much larger than 0.05. There is no statistical evidence of differences in the mean selling price between the five townships.
d. H0: μc = μi = μm = μp = μr H1: Not all treatment means are equal
Fail to reject H0. The p-value is much larger than 0.05. There is no statistical evidence of differences in the mean selling price between the five agents. Is fairness of assignment based on the overall mean price, or based on the compari- son of the means of the prices assigned to the agents?
While the p-value is not less than 0.05, it may indicate that the pairwise differences should be reviewed. These indicate that Marty’s comparisons to the other agents are significantly different.
e. The results show that the mortgage type is a significant effect on the mean years of occupancy (p=0.0227). The interaction effect is also significant (p=0.0026).
51. a. H0: μB = μK = μT H1: Not all treatment (manufacturer) mean maintenance costs, are equal.
Do not reject H0. (p = 0.7664). The mean maintenance costs by the bus manufacturer is not different.
b. H0: μB = μK = μT H1: Not all treatments have equal mean miles since the last maintenance.
Do not reject H0. The mean miles since the last maintenance by the bus manufacturer is not different. P-value = 0.4828.
CHAPTER 13 1. Σ (x − x ) ( y − y ) = 10.6, sx = 2.7, sy = 1.3
r = 10.6
(5 − 1) (2.709) (1.38) = 0.75
809
31. a. 6.308 ± (3.182) (.993) √ .2 + (7 − 5.6)2
29.2 = 6.308 ± 1.633 = [4.675, 7.941] b. 6.308 ± (3.182) (.993) √1 + 1∕5 + .0671 = [2.751, 9.865] 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. H0: ρ ≤ 0; H1: ρ > 0. Reject H0 if t > 1.714.
t = .94√25 − 2 √1 − (.94)2
= 13.213
Reject H0. There is a positive correlation between passengers and weight of luggage.
39. H0: ρ ≤ 0; H1: ρ > 0. Reject H0 if t > 2.764.
t = .47√12 − 2 √1 − (.47)2
= 1.684
Do not reject H0. There is not a positive correlation between engine size and performance. p-value is greater than .05 but less than .10.
41. a. The sales volume is inversely related to their market share.
20
25
30
35
40
45
50
5.0 7.5 10.0 12.5 15.0 17.5 Cars sold (millions)
Scatterplot of Percent GM vs Cars sold (millions)
Pe rc
en t G
M
b. The correlation coefficient is −0.876; there is an inverse linear relationship between the two variables.
c. H0: ρ ≥ 0 H1: ρ < 0 Reject H0 if t < −2.681 df = 12
t = − 0.876√14 − 2 √1 − (−0.876)2
= −6.29
Reject H0. There is a negative correlation between cars sold and market share.
d. 76.7%, found by (−0.876)2, of the variation in market share is accounted for by variation in cars sold.
43. a. r = −0.024 b. The coefficient of determination is 0.00058, found by squar-
ing (−0.024). c. H0: ρ ≥ 0 H1: ρ < 0 Reject H0 if t < −1.697
t = −0.024√32 − 2 √1 − (−0.024)2
= −0.13
Reject H0. There is a negative correlation between points scored and points allowed.
15. a. Σ (x − x ) ( y − y ) = 44.6, sx = 2.726, sy = 2.011
r = 44.6
(10 − 1) (2.726) (2.011) = .904
b = .904( 2.011 2.726) = 0.667
a = 7.4 − .677(9.1) = 1.333 b. Y^ = 1.333 + .667(6) = 5.335 17. a.
10
5
0
Ea rn
in gs
Sales 0 10 20 30 40 50 60 70 80 90
b. Σ (x − x ) (y − y ) = 629.64, sx = 26.17, sy = 3.248
r = 629.64
(12 − 1) (26.17) (3.248) = .6734
c. b = .6734( 3.248 26.170) = 0.0836
a = 64.1 12
− 0.0836( 501.10
12 ) = 1.8507
d. ŷ = 1.8507 + 0.0836(50.0) = 6.0307 ($ millions)
19. a. b = −.8744( 6.4462 5.8737) = −0.9596
a = 95 8
− (−0.9596)( 146 8 ) = 29.3877
b. 10.1957, found by 29.3877 − 0.9596(20) c. For each policeman added, crime goes down by almost one. 21. H0: β ≥ 0 H1: β < 0 df = n − 2 = 8 − 2 = 6 Reject H0 if t < −1.943. t = −0.96∕0.22 = −4.364 Reject H0 and conclude the slope is less than zero. 23. H0: β = 0 H1: β ≠ 0 df = n − 2 = 12 − 2 = 10 Reject H0 if t not between −2.228 and 2.228. t = 0.08∕0.03 = 2.667 Reject H0 and conclude the slope is different from zero.
25. The standard error of estimate is 3.378, found by √ 68.4814 8 − 2
.
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 √ 6.667 10 − 2
.
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. r2 = 1,000 1,500
= .6667
b. r = √.6667 = .8165
c. sy · x = √ 500 13
= 6.2017
810
51. a. 15
9
Da ys
Distance 600 700 800 900
4
There appears to be a relationship between the two variables. As the distance increases, so does the shipping time.
b. r = 0.692 H0: ρ ≤ 0; H1: ρ > 0. Reject H0 if t > 1.734.
t = 0.692√20 − 2 √1 − (0.692)2
= 4.067
H0 is rejected. There is a positive association between ship- ping distance and shipping time.
c. R2 = 0.479. Nearly half of the variation in shipping time is ex- plained by shipping distance.
d. sy · x = 2.004 53. a. b = 2.41 a = 26.8 The regression equation is: Price = 26.8 + 2.41 × Dividend. For
each additional dollar of dividend, the price increases by $2.41. b. To test the significance of the slope, we use n − 2, or
30 − 2 = 28 degrees of freedom. For a 0.05 level of signifi- cance, the critical values are −2.048 and 2.048. The t-test
statistic is t = b − 0
sb =
2.408 0.328
= 7.34. We reject the null
hypothesis that the slope is equal to zero.
c. R 2 = 5,057.6 7,682.7
= 0.658 Thus, 65.8% of the variation in
price is explained by the dividend. d. r = √.658 = 0.811 H0: ρ ≤ 0 H1: ρ > 0 At the 5% level, reject H0 when t > 1.701.
t = 0.811√30 − 2 √1 − (0.811)2
= 7.34
Thus, H0 is rejected. The population correlation is positive. 55. a. 35 b. sy · x = √29,778,406 = 5,456.96
c. r 2 = 13,548,662,082 14,531,349,474
= 0.932
d. r = √0.932 = 0.966 e. H0: ρ ≤ 0, H1: ρ > 0; reject H0 if t > 1.692.
t = .966√ 35 − 2 √1 − (.966)2
= 21.46
Reject H0. There is a direct relationship between size of the house and its market value.
57. a. The regression equation is Price = −386.5 + 704.0 Speed. b. The computers 2, 3, and 10 have errors in excess of $200.00.
d. For the National conference (NFC): H0: ρ ≥ 0 H1: ρ < 0 Reject H0 if t < −1.761.
t = − 0.139√16 − 2 √1 − (−0.139)2
= −0.53
Do not reject H0. We cannot say there is a negative correla- tion between points scored and points allowed in the NFC.
For the American conference (AFC): H0: ρ ≥ 0 H1: ρ < 0 Reject H0 if t < −1.761.
t = −0.292√16 − 2 √1 − (−0.292)2
= −1.142
Do not reject H0. We cannot say there is a negative correla- tion between points scored and points allowed in the AFC.
45. a. 12
10
8
6
4
2
Months 2 4 6 8 10 12
Ho ur
s
There is an inverse relationship between the variables. As the months owned increase, the number of hours exercised decreases.
b. r = −0.827 c. H0: ρ ≥ 0; H1: ρ < 0. Reject H0 if t < −2.896.
t = −0.827√10 − 2 √1 − (−0.827)2
= −4.16
Reject H0. There is a negative association between months owned and hours exercised.
47. a. Median age and population are directly related.
b. r = 11.93418
(10 − 1) (2.207) (1.330) = 0.452
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 year.
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, .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. H0: ρ = 0 H1: ρ ≠ 0 Reject H0 if t is not between −1.86 and 1.86.
df = 8 t = 0.452√10 − 2 √1 − (0.452)2
= 1.433 Do not reject H0.
There may be no relationship between age and population. 49. a. b = −0.4667, a = 11.2358 b. ŷ = 11.2358 − 0.4667(7.0) = 7.9689
c. 7.9689 ± (2.160) (1.114) √1 + 1
15 +
(7 − 7.1333)2
73.7333 = 7.9689 ± 2.4854 = [5.4835, 10.4543] d. R2 = 0.499. Nearly 50% of the variation in the amount of the
bid is explained by the number of bidders.
811
c. The correlation of Speed and Price is 0.835. H0: ρ ≤ 0 H1: ρ > 0 Reject H0 if t > 1.8125.
t = 0.835√12 − 2 √1 − (0.835)2
= 4.799
Reject H0. It is reasonable to say the population correlation is positive.
59. a. r = .987, H0: ρ ≤ 0, H1: ρ > 0. Reject H0 if t > 1.746.
t = .987√18 − 2 √1 − (.987)2
= 24.564
b. ŷ = −29.7 + 22.93x; an additional cup increases the dog’s weight by almost 23 pounds.
c. Dog number 4 is an overeater. 61. a.
100 0 500 1,000
Distance
Scatter Diagram of Fares and Distances
Fa re
1,500 2,000 2,500
150
200
250
300
350
The relationship is direct. Fares increase for longer flights. b. The correlation of Distance and Fare is 0.656.
H0: ρ ≤ 0 H1: ρ > 0 Reject H0 if t > 1.701. df = 28
t = 0.656√30 − 2 √1 − (0.656)2
= 4.599
Reject H0. There is a significant positive correlation between fares and distances.
c. 43%, found by (0.656)2, of the variation in fares is explained by the variation in distance.
d. The regression equation is Fare = 147.08 + 0.05265 Dis- tance. Each additional mile adds $0.05265 to the fare.
A 1,500-mile flight would cost $226.06, found by $147.08 + 0.05265(1,500).
e. A flight of 4,218 miles is outside the range of the sampled data, so the regression equation may not be useful.
63. a. There does seem to be a direct relationship between the variables.
1,000,000 50 75 100 125 150 175 200 225
1,500,000
2,000,000
2,500,000
3,000,000
3,500,000
4,000,000
Team salary
At te
nd an
ce
Scatterplot of Attendance vs Team Salary
b. Expected Attendance with a salary of $100 million is 2,224,268, found by 1,208,968 + 10,153(100)
c. Increasing the salary by 30 million will increase attendance by 304,590 on average, found by 10,153 (30). This is also the difference between the expected attendance with a salary of 130 and the expected attendance of 100 million.
d. The regression output from Excel is below.
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.652 R Square 0.425 Adjusted R Square 0.404 Standard Error 484846.605 Observations 30.000
ANOVA
Significance df SS MS F F
Regression 1 4863338397693.920 4863338397693.920 20.688 0.000 Residual 28 6582134442815.280 235076230100.546 Total 29 11445472840509.200
Coefficients Standard Error t Stat P-value
Intercept 1208968.430 284472.197 4.250 0.000 Team Salary 10152.694 2232.124 4.548 0.000
H0: β ≤ 0 H1: β > 0 df = n − 2 = 30 − 2 = 28 Reject H0 if t > 1.701 t = 10152.694/2232.124 = 4.548. Reject H0 and conclude the slope is positive.
e. 0.4249 or 42.49% of the variation in attendance is explained by variation in salary.
f. The correlation between attendance and batting average is 0.1472.
H0: ρ ≤ 0 H1: ρ > 0 At the 5% level, reject H0 if t > 1.701.
t = 0.1472√30 − 2 √1 − (0.1472)2
= 0.787
Fail to reject H0. p-value = 0.4377 The batting average and attendance are not positively
correlated. The correlation between attendance and ERA is –0.4745. The
correlation between attendance and ERA is stronger than the correlation between attendance and batting average.
H0: ρ ≥ 0 H1: ρ < 0 At the 5% level, reject H0 if t < − 1.701.
t = −0.4745√30 − 2 √1 − (−0.4745)2
= −2.8524
Fail to reject H0. p-value = 0.0081 The ERA and attendance are not negatively correlated.
CHAPTER 14 1. a. Multiple regression equation b. The y-intercept c. ŷ = 64,100 + 0.394(796,000) + 9.6(6,940)
− 11,600(6.0) = $374,748 3. a. 497.736, found by ŷ = 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.
812
5. a. sY · 12 = √ SSE
n − (k + 1) = √
583.693 65 − (2 + 1)
= √9.414 = 3.068 95% of the residuals will be between ±6.136, found by
2(3.068).
b. R2 = SSR
SS total =
77.907 661.6
= .118
The independent variables explain 11.8% of the variation.
c. R2adj = 1 −
SSE n − (k + 1)
SS total n − 1
= 1 −
583.693 65 − (2 + 1)
661.6 65 − 1
= 1 − 9.414
10.3375 = 1 − .911 = .089
7. a. ŷ = 84.998 + 2.391x1 − 0.4086x2 b. 90.0674, found by ŷ = 84.998 + 2.391(4) − 0.4086(11) c. n = 65 and k = 2 d. H0: β1 = β2 = 0 H1: Not all βs 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: β1 = 0 H0: β2 = 0 H1: β1 ≠ 0 H1: β2 ≠ 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 Apti-
tude + 22.1 Union 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
b. These variables are effective in predicting performance. They explain 53.3% of the variation in performance. In particular, union membership increases the typical performance by 22.1.
c. H0: β2 = 0 H1: β2 ≠ 0 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
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 be- tween aptitude and union membership when predicting job performance.
11. a. The regression equation is Price = 3,080 − 54.2 Bidders + 16.3 Age
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
The price decreases 54.2 as each additional bidder partici- pates. 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
c. R2 = 750
1,250 = .60
d. sy · 1234 = √500∕35 = 3.7796 e. H0: β1 = β2 = β3 = β4 = 0
H1: Not all the βs equal zero. H0 is rejected if F > 2.65.
F = 750∕4
500∕35 = 13.125
H0 is rejected. At least one βi does not equal zero. 15. a. n = 26 b. R2 = 100/140 = .7143 c. 1.4142, found by √2 d. H0: β1 = β2 = β3 = β4 = β5 = 0 H1: Not all the βs are 0. H0 is rejected if F > 2.71. Computed F = 10.0. Reject H0. At least one regression coeffi-
cient 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
b. R2 = SSR
SS total =
3,050 5,250
= .5809
c. 9.199, found by √84.62 d. H0 is rejected if F > 2.97 (approximately)
Computed F = 1,016.67 84.62
= 12.01
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. Com- puted t for x2 and x3 exceed the critical value. Thus, “popula- tion” and “advertising expenses” should be retained and “number of competitors,” x1, dropped.
19. a. The strongest correlation is between High School GPA and Paralegal GPA. No problem with multicollinearity.
813
b. R2 = 4.3595 5.0631
= .8610
c. H0 is rejected if F > 5.41.
F = 1.4532 0.1407
= 10.328
At least one coefficient is not zero. d. Any H0 is rejected if t < −2.571 or t > 2.571. It appears that
only High School GPA is significant. Verbal and math could be eliminated.
e. R2 = 4.2061 5.0631
= .8307
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 = −1242.1 + 50.671 (screen
size). For each inch increase in screen size, the price increases $50.671 on average.
d. Using a “dummy” variable for Sony, the regression equation is Price = 11145.6 + 46.955 (Screen) + 187.10 (Sony). If we set “Sony” = 0, then the manufacturer is Samsung and the price is predicted only by screen size. If we set “Sony” = 1, then the manufacturer is Sony. Therefore, Sony TV’s are, on average, $187.10 higher in price than Samsung TVs.
e. Here is some of the output.
Coefficients Term Coef SE Coef 95% CI T-Value P-Value
Constant −1145.6 220.7 (−1606.1, −685.2) −5.19 <0.0001 Screen 46.955 5.149 (36.215, 57.695) 9.12 <0.0001 Sony 1 187.10 71.84 (37.24, 336.96) 2.60 0.0170
Based on the p-values, screen size and manufacturer are both significant in predicting price.
f. A histogram of the residuals indicates they follow a normal distribution.
g. There is no apparent relationship in the residuals, but the re- sidual variation may be increasing with larger fitted values.
23. a.
300
2 4 6 8 30 40 50 60 70
200
100
300
200
100
0
0 4 6 8 10 12 5 10 15 20
CompetitorsS al
es
Potential
Advertising Accounts Scatter Diagram of Sales vs. Advertising, Accounts, Competitors, Potential
Sales seem to fall with the number of competitors and rise with the number of accounts and potential.
b. Pearson correlations
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
The number of accounts and the market potential are moder- ately correlated.
c. The regression equation is: Sales = 178 + 1.81 Advertising + 3.32 Accounts − 21.2
Competitors + 0.325 Potential
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%
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
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 ef- fective 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
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
Now advertising is not significant. That would also lead you to cut out the advertising variable and report that the polished regres- sion equation is: Sales = 187 + 3.41 Accounts − 21.2 Competitors
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
814
f.
Residual
Fr eq
ue nc
y
0
–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 inde- pendent 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. ŷ = 651.9 + 13.422x1 − 6.710x2 + 205.65x3 − 33.45x4 b. R2 = .433, which is somewhat low for this type of study. c. H0: β1 = β2 = β3 = β4 = 0; H1: Not all βs equal zero Reject H0 if F > 2.76.
F = 1,066,830∕4 1,398,651∕25
= 4.77
H0 is rejected. Not all the βis 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. Ser- vice and gender should remain in the analyses; age and job should be dropped.
e. Following is the computer output using the independent vari- ables service and gender.
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 differ- ence between management and engineering positions is not significant.
27. a. ŷ = 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 nor-
mal distribution.
f. There does not seem to be a problem with the plot of the residuals versus the fitted values.
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 Infomercials
Predictor Coef SE Coef T P Constant 1.0188 0.3105 3.28 0.006 Infomercials 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.
815
b.
20.4 20.2 0.0 0.2 0.4 RESI1
Histogram of RESI1 Fr
eq ue
nc y
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 = −118,929 + 1.63 Loan + 2.1 Monthly
Payment + 50 Payments Made
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
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 hy- pothesis that all the regression coefficients are zero can be rejected. At least one of the multiple regression coefficients is different from zero.
b. 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 null hypothesis is that the coefficient is zero in the indi- vidual 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 re- moved. However, the monthly payment and payments made variables would likely be removed.
c. The revised regression equation is: Auction Price = −119,893 + 1.67 Loan
33. a. The correlation matrix is as follows:
Size Days on Price Bedrooms (square feet) baths Market
Price 1.000 Bedrooms 0.844 1.000 Size (square feet) 0.952 0.877 1.000 Baths 0.825 0.985 0.851 1.000 Days on Market 0.185 0.002 0.159 –0.002 1
The correlations for strong, positive relationships between “Price” and the independent variables “Bedrooms”, “Size”, and “Baths”. There appears to be no relationship between “Price” and “Days on the Market”. The correlations among the independent variables are very strong. So, there would be a high degree of multicollinearity in a multiple regression
equation if all the variables were included. We will need to be careful in selecting the best independent variable to predict price.
b.
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.952 R Square 0.905 Adjusted R Square 0.905 Standard Error 49655.822 Observations 105.000
ANOVA
df SS MS F Significance F
Regression 1 2.432E+12 2.432E+12 9.862E+02 1.46136E-54 Residual 103 2.540E+11 2.466E+09 Total 104 2.686E+12
Coefficients Standard Error t Stat P-value
Intercept –15775.955 12821.967 –1.230 0.221 Size (square feet) 108.364 3.451 31.405 0.000
The regression analysis shows a significant relationship be- tween price and house size. The p-value of the F-statistic is 0.00, so the null hypothesis of “no relationship” is rejected. Also, the p-value associated with the regression coefficient of “size” is 0.000. Therefore, this coefficient is clearly differ- ent from zero.
The regression equation is: Price = –15775.995 + 108.364 Size.
In terms of pricing, the regression equation suggests that houses are priced at about $108 per square foot.
c. The regression analyses of price and size with the qualita- tive variables pool and garage follow. The results show that the variable “pool” is statistically significant in the equation. The regression coefficient indicates that if a house has a pool, it adds about $28,575 to the price. The analysis of including “garage” to the analysis indicates that it does not affect the pricing of the house.
Adding pool to the regression equation increases the R-square by about 1%.
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.955 R Square 0.913 Adjusted R Square 0.911 Standard Error 47914.856 Observations 105
ANOVA
Significance df SS MS F F
Regression 2.00 2451577033207.43 1225788516603.72 533.92 0.00 Residual 102.00 234175013207.24 2295833462.82 Total 104.00 2685752046414.68
Coefficients Standard Error t Stat P-value
Intercept –34640.573 13941.203 –2.485 0.015 Size (square feet) 108.547 3.330 32.595 0.000 Pool (yes is 1) 28575.145 9732.223 2.936 0.004
816
d. The following histogram was developed using the residuals from part c. The normality assumption is reasonable.
0
5
10
15
20
25
30
35
–112.50 –77.50 –42.50 –7.50 27.50 62.50 97.50 Residuals
Frequency of Residuals
e. The following scatter diagram is based on the residuals in part c with the predicted dependent variable on the horizon- tal axis and residuals on the vertical axis. There does appear that the variance of the residuals increases with higher val- ues of the predicted price. You can experiment with transfor- mations such as the Log of Price or the square root of price and observe the changes in the graphs of residuals. Note that the transformations will make the interpretation of the regression equation more difficult. (LO14-6)
0
0
10 00
00
20 00
00
30 00
00
40 00
00
50 00
00
60 00
00
70 00
00
80 00
00
90 00
00
50000
–50000
–100000
–150000
100000
150000 Residuals vs. Predicted
35. a.
Miles Maintenance Age Odometer since last cost($) (years) Miles Maintenance
Maintenance cost($) 1 Age(years) 0.710194278 1 Odometer Miles 0.700439797 0.990675674 1 Miles since last Maint. –0.160275988 –0.140196856 –0.118982823 1
The correlation analysis shows that age and odometer miles are positively correlated with cost and that “miles since last maintenance” shows that costs increase with fewer miles be- tween maintenance. The analysis also shows a strong cor- relation between age and odometer miles. This indicates the strong possibility of multicollinearity if age and odometer miles are included in a regression equation.
b. There are a number of analyses to do. First, using Age or Odometer Miles as an independent variable. When you re-
view these analyses, both result in significant relationships. However, Age has a slightly higher R2. So I would select age as the first independent variable. The interpretation of the coefficient using age is bit more useful for practical use. That is, we can expect about an average of $600 increase in maintenance costs for each additional year a bus ages. The results are:
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.708 R Square 0.501 Adjusted R Square 0.494 Standard Error 1658.097 Observations 80
ANOVA
df SS MS F Significance F
Regression 1 215003471.845 215003471.845 78.203 0.000 Residual 78 214444212.142 2749284.771 Total 79 429447683.988
Coefficients Standard Error t Stat P-value
Intercept 337.297 511.372 0.660 0.511 Age (years) 603.161 68.206 8.843 0.000
We can also explore including the variable “miles since last maintenance” with Age. Your analysis will show that “miles since last maintenance” is not significantly related to costs.
Last, it is possible that maintenance costs are different for diesel versus gasoline engines. So, adding this variable to the analysis shows:
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.960 R Square 0.922 Adjusted R Square 0.920 Standard Error 658.369 Observations 80
ANOVA
df SS MS F Significance F
Regression 2 396072093.763 198036046.881 456.884 0.000 Residual 77 33375590.225 433449.224 Total 79 429447683.988
Coefficients Standard Error t Stat P-value
Intercept –1028.539 213.761 –4.812 0.000 Age (years) 644.528 27.157 23.733 0.000 Engine Type (0=diesel) 3190.481 156.100 20.439 0.000
The results show that the engine type is statistically signifi- cant and increases the R2 to 92.2%. Now the practical inter- pretation of the analysis is that, on average, buses with gasoline engines cost about $3,190 more to maintain. Also, the maintenance costs increase with bus age at an average of $644 per year of bus age.
817
c. The normality conjecture appears realistic.
0
5
10
15
20
25
30
35
–1375 –825 –275 325 875 1425 1975
Frequency of Residuals
d. The plot of residuals versus predicted values shows the fol- lowing. There are clearly patterns in the graph that indicate that the residuals do not follow the assumptions required for the tests of hypotheses.
0
0–2000 2000 4000 6000 8000 10000
500
–500
–1500 –1000
–2000
1000 1500 2000 2500
Residuals
Lets remember the scatter plot of costs versus age. The graph clearly shows the effect of engine type on costs. So there are essentially two regression equations depending on the type of engine.
8
0 2000 4000 6000 8000 10000 12000
10
6
2 4
0
12 14 16
Age (Years)
So based on our knowledge of the data, lets create a resid- ual plot of costs for each engine type.
0
10000 2000 3000 4000 6000 70005000 8000
100
–100
–300 –200
–400 –500
200 300 400 500
Residuals for Diesel Engines
0
20 4 6 8 10 12
–500
–1000
–1500
500
1000 Residual Plot for Gasoline Engines
Re si
du al
s
The graphs show a much better distribution of residuals. (LO14-5)
CHAPTER 15 1. a. H0 is rejected if z > 1.65. b. 1.09, found by z = (0.75 − 0.70)∕√(0.70 × 0.30)∕100 c. H0 is not rejected. 3. Step 1: H0: π = 0.10 H1: π ≠ 0.10 Step 2: The 0.01 significance level was chosen Step 3: Use the z-statistic as the binomial distribution can be ap-
proximated by the normal distribution as nπ = 30 > 5 and n (1–π) = 270 > 5.
Step 4: Reject H0 if z > 2.326 Step 5:
z = { (63/300) − 0.10}
√{0.10(0.90)/300} = 6.35,
Reject H0. Step 6: We conclude that the proportion of carpooling cars on
the Turnpike is not 10%. 5. a. H0: π ≥ 0.90 H1: π < 0.90 b. H0 is rejected if z < −1.28. c. −2.67, found by z = (0.82 − 0.90)∕√(0.90 × 0.10)∕100 d. H0 is rejected. Fewer than 90% of the customers receive their
orders in less than 10 minutes. 7. a. H0 is rejected if z > 1.65.
b. 0.64, found by pc = 70 + 90
100 + 150
818
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
Since 0.6493 is not greater than 5.991, we fail to reject the null hypothesis. These data could be from a normal distribution.
27. H0: There is no relationship between community size and section read. H1: There is a relationship.
Reject H0 if χ2 > 9.488.
χ2 = (170 − 157.50)2
157.50 + . . . +
(88 − 83.62)2
83.62 = 7.340
Do not reject H0. There is no relationship between community size and section read.
29. H0: No relationship between error rates and item type. H1: There is a relationship between error rates and item type.
Reject H0 if π2 > 9.21.
χ2 = (20 − 14.1)2
14.1 + . . . +
(225 − 225.25)2
225.25 = 8.033
Do not reject H0. There is not a relationship between error rates and item type.
31. a. This is a binomial situation with both the mean number of successes and failures equal to 25, found by 0.5(50).
b. H0: π = 0.50 H1: π ≠ 0.50 c.
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. z =
34 50
− 0.5
√0.5(1 − 0.5) /50 = 2.55
We fail to reject the null hypothesis. This data does not prove that either league has an advantage.
e. The p-value is 0.0108, found by 2(0.5000 – 0.4946). A value this extreme will happen about once out of 100 times with a fair coin.
33. H0: π ≤ 0.60 H1: π > 0.60 H0 is rejected if z > 2.33.
z = .70 − .60
√ .60(.40)
200
= 2.89
H0 is rejected. Ms. Dennis is correct. More than 60% of the accounts are more than three months old.
c. 1.61, found by
z = 0.70 − 0.60
√[ (0.64 × 0.36)∕100] + [ (0.64 × 0.36)∕150] d. H0 is not rejected. 9. a. H0: π1 = π2 H1: π1 ≠ π2 b. H0 is rejected if z < −1.96 or z > 1.96.
c. pc = 24 + 40
400 + 400 = 0.08
d. −2.09, found by
z = 0.06 − 0.10
√[ (0.08 × 0.92)∕400] + [ (0.08 × 0.92)∕400] e. H0 is rejected. The proportion infested is not the same in the
two fields. 11. H0: πd ≤ πr H1: πd > πr H0 is rejected if z > 2.05.
pc = 168 + 200
800 + 1,000 = 0.2044
z = 0.21 − 0.20
√ (0.2044) (0.7956)
800 +
(0.2044) (0.7956)
1,000
= 0.52
H0 is not rejected. We cannot conclude that a larger proportion of Democrats favor lowering the standards. p-value = .3015.
13. a. 3 b. 7.815 15. a. Reject H0 if χ2 > 5.991
b. χ2 = (10 − 20)2
20 +
(20 − 20)2
20 +
(30 − 20)2
20 = 10.0
c. Reject H0. The proportions are not equal. 17. H0: The outcomes are the same; H1: The outcomes are not the
same. Reject H0 if χ2 > 9.236.
χ2 = (3 − 5)2
5 + . . . +
(7 − 5)2
5 = 7.60
Do not reject H0. Cannot reject H0 that outcomes are the same. 19. H0: There is no difference in the proportions. H1: There is a difference in the proportions. Reject H0 if χ2 > 15.086.
χ2 = (47 − 40)2
40 + . . . +
(34 − 40)2
40 = 3.400
Do not reject H0. There is no difference in the proportions. 21. a. Reject H0 if χ2 > 9.210.
b. χ2 = (30 − 24)2
24 +
(20 − 24)2
24 +
(10 − 12)2
12 = 2.50
c. Do not reject H0. 23. H0: Proportions are as stated; H1: Proportions are not as stated.
Reject H0 if χ2 > 11.345.
χ2 = (50 − 25)2
25 + . . . +
(160 − 275)2
275 = 115.22
Reject H0. The proportions are not as stated. 25.
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
The first and last class both have expected frequencies smaller than 5. They are combined with adjacent classes.
H0: The population of clients follows a normal distribution. H1: The population of clients does not follow a normal distribution. Reject the null if χ2 > 5.991.
819
49. H0: There is no preference with respect to TV stations. H1: There is a preference with respect to TV stations. df = 3 − 1 = 2. H0 is rejected if χ2 > 5.991.
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
H0 is rejected. There is a preference for TV stations. 51. H0: πn = 0.21, πm = 0.24, πs = 0.35, πw = 0.20 H1: The distribution is not as given. Reject H0 if χ2 > 11.345.
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
H0 is not rejected. The distribution of order destinations reflects the population.
53. H0: The proportions are the same. H1: The proportions are not the same. Reject H0 if χ2 > 16.919.
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
Do not reject H0. The digits are evenly distributed. 55.
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
The sample mean is 8.222, found by 2,220/270. The sample standard deviation is 1.003, found as the square
root of 270.7/269. H0: The population of wages follows a normal distribution. H1: The population of hourly wages does not follow a normal
distribution.
35. H0: π ≤ 0.44 H1: π > 0.44 H0 is rejected if z > 1.65.
z = 0.480 − 0.44
√(0.44 × 0.56)∕1.000 = 2.55
H0 is rejected. We conclude that there has been an increase in the proportion of people wanting to go to Europe.
37. H0: π ≤ 0.20 H1: π > 0.20 H0 is rejected if z > 2.33
z = (56∕200) − 0.20
√(0.20 × 0.80)∕200 = 2.83
H0 is rejected. More than 20% of the owners move during a par- ticular year. p-value = 0.5000 − 0.4977 = 0.0023.
39. H0: π ≥ 0.0008 H1: π < 0.0008 H0 is rejected if z < −1.645.
z = 0.0006 − 0.0008
√ 0.0008 (0.9992)
10,000
= −0.707 H0 is not rejected.
These data do not prove there is a reduced fatality rate. 41. H0: π1 ≤ π2 H1: π1 > π2 If z > 2.33, reject H0.
pc = 990 + 970
1,500 + 1,600 = 0.63
z = .6600 − .60625
√ .63(.37)
1,500 +
.63(.37)
1,600
= 3.10
Reject the null hypothesis. We can conclude the proportion of men who believe the division is fair is greater.
43. H0: π1 ≤ π2 H1: π1 > π2 H0 is rejected if z > 1.65.
pc = .091 + .085
2 = .088
z = 0.091 − 0.085
√ (0.088) (0.912)
5,000 +
(0.088) (0.912)
5,000
= 1.059
H0 is not rejected. There has not been an increase in the propor- tion calling conditions “good.” The p-value is .1446, found by .5000 − .3554. The increase in the percentages will happen by chance in one out of every seven cases.
45. H0: π1 = π2 H1: π1 ≠ π2 H0 is rejected if z is not between −1.96 and 1.96.
pc = 100 + 36
300 + 200 = .272
z =
100 300
− 36
200
√ (0.272) (0.728)
300 +
(0.272) (0.728)
200
= 3.775
H0 is rejected. There is a difference in the replies of the sexes. 47. H0: πs = 0.50, πr = πe = 0.25 H1: Distribution is not as given above. df = 2. Reject H0 if χ2 > 4.605.
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
H0 is not rejected. The proportions are as given in the null hypothesis.
820
b. Reject H0 because the cumulative probability associated with nine or more successes (.073) does not exceed the signifi- cance level (.10).
3. a. H0: π ≤ .50; H1: π > .50; n = 10 b. H0 is rejected if there are nine or more plus signs. A “+” rep-
resents a loss. c. Reject H0. It is an effective program because there were nine
people who lost weight. 5. a. H0: π ≤ .50 (There is no change in weight.) H1: π > .50 (There
is a loss of weight.) b. Reject H0 if z > 1.65.
c. z = (32 − .50) − .50(45)
.50√45 = 2.68
d. Reject H0. The weight loss program is effective. 7. H0: π ≥ .50, H1: π ≥ .50. H0 is rejected if z > 2.05.
z = 42.5 − 40.5
4.5 = .44
Because .44 < 2.05, do not reject H0. No preference. 9. a. H0: Median ≤ $81,500; H1: Median > $81,500 b. H0 is rejected if z > 1.65.
c. z = 170 − .50 − 100
7.07 = 9.83
H0 is rejected. The median income is greater than $81,500. 11.
Couple Difference Rank
1 550 7 2 190 5 3 250 6 4 −120 3 5 −70 1 6 130 4 7 90 2
Sums: −4, +24. So T = 4 (the smaller of the two sums). From Appendix B.8, .05 level, one-tailed test, n = 7, the critical value is 3. Since the T of 4 > 3, do not reject H0 (one-tailed test). There is no difference in square footage. Professional couples do not live in larger homes.
13. a. H0: The production is the same for the two systems. H1: Production using the new procedure is greater. b. H0 is rejected if T ≤ 21, n = 13. c. The calculations for the first three employees are:
Employee Old new d Rank R1 R2
A 60 64 4 6 6 B 40 52 12 12.5 12.5 C 59 58 −1 2 2
The sum of the negative ranks is 6.5. Since 6.5 is less than 21, H0 is rejected. Production using the new procedure is greater.
15. H0: The distributions are the same. H1: The distributions are not the same. Reject H0 if z, 21.96 or z > 1.96.
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
Reject the null if χ2 > 4.605.
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
Since 30.874 is greater than 4.605, we reject the null hypothesis not from a normal distribution.
57. H0: Gender and attitude toward the deficit are not related. H1: Gender and attitude toward the deficit are related. Reject H0 if χ2 > 5.991.
χ2 = (244 − 292.41)2
292.41 +
(194 − 164.05)2
164.05
+ (68 − 49.53)2
49.53 +
(305 − 256.59)2
256.59
+ (114 − 143.95)2
143.95 +
(25 − 43.47)2
43.47 = 43.578
Since 43.578 > 5.991, you reject H0. A person’s position on the deficit is influenced by his or her gender.
59. H0: Whether a claim is filed and age are not related. H1: Whether a claim is filed and age are related. Reject H0 if χ2 > 7.815.
χ2 = (170 − 203.33)2
203.33 + . . . +
(24 − 35.67)2
35.67 = 53.639
Reject H0. Age is related to whether a claim is filed. 61. H0: πBL = πO = .23, πY = πG = .15, πBR = πR = .12. H1: The proportions are not as given. Reject H0 if χ2 > 15.086.
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 H0. The color distribution agrees with the manufac- turer’s information.
63. H0: Salary and winning are not related H1: Salary and winning are related Reject H0 if χ2 > 3.841 with 1 degree of freedom.
Salary
Winning Lower half Top half Total
No 10 4 14 Yes 5 11 16 Total 15 15
χ2 = (10 − 7)2
7 +
(4 − 7)2
7 +
(5 − 8)2
8 +
(11 − 8)2
8 = 4.82
Reject H0. Conclude that salary and winning are related.
CHAPTER 16 1. a. If the number of pluses (successes) in the sample is 9 or
more, reject H0.
821
H = 12
15(16)[ (46)2
5 +
(39)2
5 +
(35)2
5 ] − 3(16) = 0.62
H0 is not rejected. There is no difference in the three distributions.
25. a.
0 0 2 4 6 8 10 12 14
2
4
6
8
10
12
14
Male
Fe m
al e
Scatter Diagram of Female versus Male
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
rs = 1 − 6(242)
14(142 − 1) = 0.47
c. H0: No correlation among the ranks. H1: A positive correlation among the ranks. Reject H0 if t > 1.782.
t = 0.47√ 14 − 2
1 − (0.47)2 = 1.84
H0 is rejected. We conclude the correlation in population among the ranks is positive. Husbands and wives generally like the same shows.
27. Representative Sales Rank Training Rank d d 2
1 319 8 8 0 0 2 150 1 2 1 1 3 175 2 5 3 9 4 460 10 10 0 0 5 348 9 7 −2 4 6 300 6.5 1 5.5 30.25 7 280 5 6 1 1 8 200 4 9 5 25 9 190 3 4 1 1
10 300 6.5 3 −3.5 12.25 83.50
a. rs = 1 − 6(83.5)
10(102 − 1) = 0.494
A moderate positive correlation
z = 86.5 −
8(8 + 8 + 1) 2
√ 8(8) (8 + 8 + 1)
12
= 1.943
H0 is not rejected. There is no difference in the two populations. 17. H0: The distributions are the same. H1: The distribution of Campus
is to the right. Reject H0 if z > 1.65.
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
z = 158.5 −
12(12 + 10 + 1) 2
√ 12(10) (12 + 10 + 1)
12
= 1.35
H0 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. H0: The three population distributions are equal. H1: Not all of the distributions are the same. b. Reject H0 if H > 5.991. c.
Sample 1 Sample 2 Sample 3 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
H = 12
16(16 + 1)[ (64)2
5 +
(53)2
6 +
(19)2
5 ] − 3(16 + 1)
= 59.98 − 51 = 8.98 d. Reject H0 because 8.98 > 5.991. The three distributions are
not equal. 23. H0: The distributions of the lengths of life are the same. H1: The distributions of the lengths of life are not the same. H0 is rejected if H > 9.210.
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
found by 69220/69220 found by 54818/69220 found by 55177/69220 found by 65694/69220 found by 83040/69220 found by 88378/69220 found by 97420/69220 found by 98608/69220
822
H0 is rejected if H > 7.815, from χ2 with 3 degrees of freedom.
H = 12
80(81) [
(132)2
3 +
(501)2
11 +
(349)2
11 +
(2258)2
55 ] − 3(81) = 2.186
Fail to reject H0. There is no difference in the maintenance cost for the four bus capacities.
b. H0: The distributions of maintenance costs by fuel type are the same.
H1: The distributions are different. Reject H0 if z < –1.96 or z > 1.96.
z = 1693 −
53(53 + 27 + 1) 2
√ (53) (27) (53 + 27 + 1)
12
= −4.614
We reject reject H0 and conclude that maintenance costs are different for diesel and gasoline fueled buses.
c. H0: The distributions of the maintenance costs are the same for the three bus manufacturers.
H1: The distributions of the costs are not the same Ho is rejected if H > 5.991, from χ2 with 2 degrees of freedom.
H = 12
80(81)[ (414)2
8 +
(1005)2
25 +
(1821)2
47 ] − 3(81) = 2.147
Ho is not rejected. There may be no difference in the mainte- nance cost for the three different manufacturers. The distri- butions could be the same.
CHAPTER 17 1.
Year Loans ($ Millions) Index (Base = 2008) 2008 69220 100.0 2009 54818 79.2 2010 55177 79.7 2011 65694 94.9 2012 83040 120.0 2013 88378 127.7 2014 97420 140.7 2015 98608 142.5
3. The mean sales for the earliest three years is $(486.6 + 506.8 + 522.2)/3 or $505.2.
2014: 90.4, found by (456.6/505.2) (100) 2015: 85.8, found by (433.3/505.2) (100)
Net sales decreased by 9.6 and 14.2 percent from the 2003– 2005 period to 2014 and 2015 respectively.
5. a. Pt = 3.35 2.49
(100) = 134.54 Ps = 4.49 3.29
(100) = 136.47
Pc = 4.19 1.59
(100) = 263.52 Pa = 2.49 1.79
(100) = 139.11
b. P = 14.52 9.16
(100) = 158.52
c. 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
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
e. l = √ (147.1) (150.2) = 148.64
7. a. PW = 0.10 0.07
(100) = 142.9 PC = 0.03 0.04
(100) = 75.0
PS = 0.15 0.15
(100) = 100 PH = 0.10 0.08
(100) = 125.0
b. P = 0.38 0.34
(100) = 111.8
b. H0: No correlation among the ranks. H1: A positive correlation among the ranks. Reject H0 if t > 1.860.
t = 0.494√ 10 − 2
1 − (0.494)2 = 1.607
H0 is not rejected. The correlation in population among the ranks could be 0.
29. H0: π 5 .50. H1: π fi .50. Use a software package to develop the binomial probability distribution for n = 19 and π 5 .50. H0 is rejected if there are either 5 or fewer “+” signs, or 14 or more. The total of 12 “+” signs falls in the acceptance region. H0 is not rejected. There is no preference between the two shows.
31. H0: π 5 .50 H1: π fi .50 H0 is rejected if there are 12 or more or 3 or fewer plus signs.
Because there are only 8 plus signs, H0 is not rejected. There is no preference with respect to the two brands of components.
33. H0: π = .50; H1: π fi .50. Reject H0 if z > 1.96 or z < −1.96.
z = 159.5 − 100
7.071 = 8.415
Reject H0. There is a difference in the preference for the two types of orange juice.
35. H0: Rates are the same; H1: The rates are not the same. H0 is rejected if H > 5.991. H = .082. Do not reject H0. 37. H0: The populations are the same. H1: The populations differ.
Reject H0 if H > 7.815. H = 14.30. Reject H0.
39. rs = 1 − 6(78)
12(122 − 1) = 0.727
H0: There is no correlation between the rankings of the coaches and of the sportswriters.
H1: There is a positive correlation between the rankings of the coaches and of the sportswriters. Reject H0 if t > 1.812.
t = 0.727√ 12 − 2
1 − (.727)2 = 3.348
H0 is rejected. There is a positive correlation between the sports- writers and the coaches.
41. a. H0: There is no difference in the distributions of the selling prices in the five townships.
H1: There is a difference in the distributions of the selling prices of the five townships.
H0 is rejected if H is greater than 9.488. The computed value of H is 2.70, so the null hypothesis is not rejected. The sam- ple data does not suggest a difference in the distributions of selling prices.
b. H0: There is no difference in the distributions of the selling prices depending on the number of bedrooms.
H1: There is a difference in the distributions of the selling prices depending on the number of bedrooms.
H0 is rejected if H is greater than 9.488. The computed value of H is 75.71 so the null hypothesis is rejected. The sample data indicates there is a difference in the distributions of sell- ing prices based on the number of bedr