statistics homework

Introduction to the Practice of Statistics

NINTH EDITION

David S. Moore George P. McCabe Bruce A. Craig Purdue University

Vice President, STEM: Ben Roberts Publisher: Terri Ward Senior Acquisitions Editor: Karen Carson Marketing Manager: Tom DeMarco Marketing Assistant: Cate McCaffery Development Editor: Jorge Amaral Senior Media Editor: Catriona Kaplan Assistant Media Editor: Emily Tenenbaum Director of Digital Production: Keri deManigold Senior Media Producer: Alison Lorber Associate Editor: Victoria Garvey Editorial Assistant: Katharine Munz Photo Editor: Cecilia Varas Photo Researcher: Candice Cheesman Director of Design, Content Management: Diana Blume Text and Cover Designer: Blake Logan Project Editor: Edward Dionne, MPS North America LLC Illustrations: MPS North America LLC Production Manager: Susan Wein Composition: MPS North America LLC Printing and Binding: LSC Communications Cover Illustration: Drawing Water: Spring 2011 detail (Midwest) by David Wicks “Look Back” Arrow: NewCorner/Shutterstock

Library of Congress Control Number: 2016946039

Student Edition Hardcover: ISBN-13: 978-1-319-01338-7 ISBN-10: 1-319-01338-4

Student Edition Loose-leaf: ISBN-13: 978-1-319-01362-2 ISBN-10: 1-319-01362-7

Instructor Complimentary Copy: ISBN-13: 978-1-319-01428-5 ISBN-10: 1-319-01428-3

© 2017, 2014, 2012, 2009 by W. H. Freeman and Company All rights reserved Printed in the United States of America First printing

W. H. Freeman and Company One New York Plaza Suite 4500 New York, NY 10004-1562 www.macmillanlearning.com

Brief Contents

To Teachers: About This Book To Students: What Is Statistics? About the Authors Data Table Index Beyond the Basics Index

PART I Looking at Data CHAPTER 1 Looking at Data—Distributions

CHAPTER 2 Looking at Data—Relationships

CHAPTER 3 Producing Data

PART II Probability and Inference CHAPTER 4 Probability: The Study of Randomness

CHAPTER 5 Sampling Distributions

CHAPTER 6 Introduction to Inference

CHAPTER 7 Inference for Means

CHAPTER 8 Inference for Proportions

PART III Topics in Inference CHAPTER 9 Inference for Categorical Data

CHAPTER 10 Inference for Regression

CHAPTER 11 Multiple Regression

CHAPTER 12 One-Way Analysis of Variance

CHAPTER 13 Two-Way Analysis of Variance Tables Answers to Odd-Numbered Exercises Notes and Data Sources Index

Contents

To Teachers: About This Book To Students: What Is Statistics? About the Authors Data Table Index Beyond the Basics Index

PART I Looking at Data CHAPTER 1 Looking at Data—Distributions Introduction

1.1 Data Key characteristics of a data set

Section 1.1 Summary Section 1.1 Exercises 1.2 Displaying Distributions with Graphs

Categorical variables: Bar graphs and pie charts Quantitative variables: Stemplots and histograms Histograms Data analysis in action: Don’t hang up on me Examining distributions Dealing with outliers Time plots

Section 1.2 Summary Section 1.2 Exercises 1.3 Describing Distributions with Numbers

Measuring center: The mean Measuring center: The median Mean versus median Measuring spread: The quartiles The five-number summary and boxplots The 1.5 × IQR rule for suspected outliers Measuring spread: The standard deviation Properties of the standard deviation Choosing measures of center and spread Changing the unit of measurement

Section 1.3 Summary Section 1.3 Exercises 1.4 Density Curves and Normal Distributions

Density curves

Measuring center and spread for density curves Normal distributions The 68–95–99.7 rule Standardizing observations Normal distribution calculations Using the standard Normal table Inverse Normal calculations Normal quantile plots

Beyond the Basics: Density estimation Section 1.4 Summary Section 1.4 Exercises Chapter 1 Exercises

CHAPTER 2 Looking at Data—Relationships Introduction

2.1 Relationships Examining relationships

Section 2.1 Summary Section 2.1 Exercises 2.2 Scatterplots

Interpreting scatterplots The log transformation Adding categorical variables to scatterplots Scatterplot smoothers Categorical explanatory variables

Section 2.2 Summary Section 2.2 Exercises 2.3 Correlation

The correlation r Properties of correlation

Section 2.3 Summary Section 2.3 Exercises 2.4 Least-Squares Regression

Fitting a line to data Prediction Least-squares regression Interpreting the regression line Facts about least-squares regression Correlation and regression Another view of r2

Section 2.4 Summary Section 2.4 Exercises 2.5 Cautions about Correlation and Regression

Residuals Outliers and influential observations

Beware of the lurking variable Beware of correlations based on averaged data Beware of restricted ranges

Beyond the Basics: Data mining Section 2.5 Summary Section 2.5 Exercises 2.6 Data Analysis for Two-Way Tables

The two-way table Joint distribution Marginal distributions Describing relations in two-way tables Conditional distributions Simpson’s paradox

Section 2.6 Summary Section 2.6 Exercises 2.7 The Question of Causation

Explaining association Establishing causation

Section 2.7 Summary Section 2.7 Exercises Chapter 2 Exercises

CHAPTER 3 Producing Data Introduction

3.1 Sources of Data Anecdotal data Available data Sample surveys and experiments

Section 3.1 Summary Section 3.1 Exercises 3.2 Design of Experiments

Comparative experiments Randomization Randomized comparative experiments How to randomize Randomization using software Randomization using random digits Cautions about experimentation Matched pairs designs Block designs

Section 3.2 Summary Section 3.2 Exercises 3.3 Sampling Design

Simple random samples How to select a simple random sample

Stratified random samples Multistage random samples Cautions about sample surveys

Beyond the Basics: Capture-recapture sampling Section 3.3 Summary Section 3.3 Exercises 3.4 Ethics

Institutional review boards Informed consent Confidentiality Clinical trials Behavioral and social science experiments

Section 3.4 Summary Section 3.4 Exercises Chapter 3 Exercises

PART II Probability and Inference CHAPTER 4 Probability: The Study of Randomness Introduction

4.1 Randomness The language of probability Thinking about randomness The uses of probability

Section 4.1 Summary Section 4.1 Exercises 4.2 Probability Models

Sample spaces Probability rules Assigning probabilities: Finite number of outcomes Assigning probabilities: Equally likely outcomes Independence and the multiplication rule Applying the probability rules

Section 4.2 Summary Section 4.2 Exercises 4.3 Random Variables

Discrete random variables Continuous random variables Normal distributions as probability distributions

Section 4.3 Summary Section 4.3 Exercises 4.4 Means and Variances of Random Variables

The mean of a random variable Statistical estimation and the law of large numbers

Thinking about the law of large numbers Beyond the Basics: More laws of large numbers

Rules for means The variance of a random variable Rules for variances and standard deviations

Section 4.4 Summary Section 4.4 Exercises 4.5 General Probability Rules

General addition rules Conditional probability General multiplication rules Tree diagrams Bayes’s rule Independence again

Section 4.5 Summary Section 4.5 Exercises Chapter 4 Exercises

CHAPTER 5 Sampling Distributions Introduction

5.1 Toward Statistical Inference Sampling variability Sampling distributions Bias and variability Sampling from large populations Why randomize?

Section 5.1 Summary Section 5.1 Exercises 5.2 The Sampling Distribution of a Sample Mean

The mean and standard deviation of x̅ The central limit theorem A few more facts

Beyond the Basics: Weibull distributions Section 5.2 Summary Section 5.2 Exercises 5.3 Sampling Distributions for Counts and Proportions

The binomial distributions for sample counts Binomial distributions in statistical sampling Finding binomial probabilities Binomial mean and standard deviation Sample proportions Normal approximation for counts and proportions The continuity correction Binomial formula The Poisson distributions

Section 5.3 Summary

Section 5.3 Exercises Chapter 5 Exercises

CHAPTER 6 Introduction to Inference Introduction Overview of inference 6.1 Estimating with Confidence

Statistical confidence Confidence intervals Confidence interval for a population mean How confidence intervals behave Choosing the sample size Some cautions

Section 6.1 Summary Section 6.1 Exercises 6.2 Tests of Significance

The reasoning of significance tests Stating hypotheses Test statistics P-values Statistical significance Tests for a population mean Two-sided significance tests and confidence intervals The P-value versus a statement of significance

Section 6.2 Summary Section 6.2 Exercises 6.3 Use and Abuse of Tests

Choosing a level of significance What statistical significance does not mean Don’t ignore lack of significance Statistical inference is not valid for all sets of data Beware of searching for significance

Section 6.3 Summary Section 6.3 Exercises 6.4 Power and Inference as a Decision

Power Increasing the power Inference as decision Two types of error Error probabilities The common practice of testing hypotheses

Section 6.4 Summary Section 6.4 Exercises Chapter 6 Exercises

CHAPTER 7 Inference for Means

Introduction

7.1 Inference for the Mean of a Population The t distributions The one-sample t confidence interval The one-sample t test Matched pairs t procedures Robustness of the t procedures

Beyond the Basics: The bootstrap Section 7.1 Summary Section 7.1 Exercises 7.2 Comparing Two Means

The two-sample z statistic The two-sample t procedures The two-sample t confidence interval The two-sample t significance test Robustness of the two-sample procedures Inference for small samples Software approximation for the degrees of freedom The pooled two-sample t procedures

Section 7.2 Summary Section 7.2 Exercises 7.3 Additional Topics on Inference

Choosing the sample size Inference for non-Normal populations

Section 7.3 Summary Section 7.3 Exercises Chapter 7 Exercises

CHAPTER 8 Inference for Proportions Introduction

8.1 Inference for a Single Proportion Large-sample confidence interval for a single proportion

Beyond the Basics: The plus four confidence interval for a single proportion Significance test for a single proportion Choosing a sample size for a confidence interval Choosing a sample size for a significance test

Section 8.1 Summary Section 8.1 Exercises 8.2 Comparing Two Proportions

Large-sample confidence interval for a difference in proportions Beyond the Basics: The plus four confidence interval for a difference in proportions

Significance test for a difference in proportions Choosing a sample size for two sample proportions

Beyond the Basics: Relative risk Section 8.2 Summary

Section 8.2 Exercises Chapter 8 Exercises

PART III Topics in Inference CHAPTER 9 Inference for Categorical Data Introduction

9.1 Inference for Two-Way Tables The hypothesis: No association Expected cell counts The chi-square test Computations Computing conditional distributions The chi-square test and the z test

Beyond the Basics: Meta-analysis Section 9.1 Summary Section 9.1 Exercises 9.2 Goodness of Fit Section 9.2 Summary Section 9.2 Exercises Chapter 9 Exercises

CHAPTER 10 Inference for Regression Introduction

10.1 Simple Linear Regression Statistical model for linear regression Preliminary data analysis and inference considerations Estimating the regression parameters Checking model assumptions Confidence intervals and significance tests Confidence intervals for mean response Prediction intervals Transforming variables

Beyond the Basics: Nonlinear regression Section 10.1 Summary Section 10.1 Exercises 10.2 More Detail about Simple Linear Regression

Analysis of variance for regression The ANOVA F test Calculations for regression inference Inference for correlation

Section 10.2 Summary Section 10.2 Exercises Chapter 10 Exercises

CHAPTER 11 Multiple Regression Introduction

11.1 Inference for Multiple Regression Population multiple regression equation Data for multiple regression Multiple linear regression model Estimation of the multiple regression parameters Confidence intervals and significance tests for regression coefficients ANOVA table for multiple regression Squared multiple correlation R2

Section 11.1 Summary Section 11.1 Exercises 11.2 A Case Study

Preliminary analysis Relationships between pairs of variables Regression on high school grades Interpretation of results Examining the residuals Refining the model Regression on SAT scores Regression using all variables Test for a collection of regression coefficients

Beyond the Basics: Multiple logistic regression Section 11.2 Summary Section 11.2 Exercises Chapter 11 Exercises

CHAPTER 12 One-Way Analysis of Variance Introduction

12.1 Inference for One-Way Analysis of Variance Data for one-way ANOVA Comparing means The two-sample t statistic An overview of ANOVA The ANOVA model Estimates of population parameters Testing hypotheses in one-way ANOVA The ANOVA table The F test Software

Beyond the Basics: Testing the equality of spread Section 12.1 Summary Section 12.1 Exercises 12.2 Comparing the Means

Contrasts

Multiple comparisons Power

Section 12.2 Summary Section 12.2 Exercises Chapter 12 Exercises

CHAPTER 13 Two-Way Analysis of Variance Introduction

13.1 The Two-Way ANOVA Model Advantages of two-way ANOVA The two-way ANOVA model Main effects and interactions

13.2 Inference for Two-Way ANOVA The ANOVA table for two-way ANOVA

Chapter 13 Summary Chapter 13 Exercises Tables Answers to Odd-Numbered Exercises Notes and Data Sources Index

To Teachers: About This Book

Statistics is the science of data. Introduction to the Practice of Statistics (IPS) is an introductory text based on this principle. We present methods of basic statistics in a way that emphasizes working with data and mastering statistical reasoning. IPS is elementary in mathematical level but conceptually rich in statistical ideas. After completing a course based on our text, we would like students to be able to think objectively about conclusions drawn from data and use statistical methods in their own work.

In IPS, we combine attention to basic statistical concepts with a comprehensive presentation of the elementary statistical methods that students will find useful in their work. IPS has been successful for several reasons:

1. IPS examines the nature of modern statistical practice at a level suitable for beginners. We focus on the production and analysis of data as well as the traditional topics of probability and inference.

2. IPS has a logical overall progression, so data production and data analysis are a major focus, while inference is treated as a tool that helps us draw conclusions from data in an appropriate way.

3. IPS presents data analysis as more than a collection of techniques for exploring data. We emphasize systematic ways of thinking about data. Simple principles guide the analysis: always plot your data; look for overall patterns and deviations from them; when looking at the overall pattern of a distribution for one variable, consider shape, center, and spread; for relations between two variables, consider form, direction, and strength; always ask whether a relationship between variables is influenced by other variables lurking in the background. We warn students about pitfalls in clear cautionary discussions.

4. IPS uses real examples to drive the exposition. Students learn the technique of least-squares regression and how to interpret the regression slope. But they also learn the conceptual ties between regression and correlation and the importance of looking for influential observations.

5. IPS is aware of current developments both in statistical science and in teaching statistics. Brief, optional Beyond the Basics sections give quick overviews of topics such as density estimation, scatterplot smoothers, data mining, nonlinear regression, and meta-analysis. Chapter 16 gives an elementary introduction to the bootstrap and other computer-intensive statistical methods.

The title of the book expresses our intent to introduce readers to statistics as it is used in practice. Statistics in practice is concerned with drawing conclusions from data. We focus on problem solving rather than on methods that may be useful in specific settings.

GAISE The College Report of the Guidelines for Assessment and Instruction in Statistics Education (GAISE) Project (www.amstat.org/education/gaise/) was funded by the American Statistical Association to make recommendations for how introductory statistics courses should be taught. This report and its update contain many interesting teaching suggestions, and we strongly recommend that you read it. The philosophy and approach of IPS closely reflect the GAISE recommendations. Let’s examine each of the latest recommendations in the context of IPS.

1. Teach statistical thinking. Through our experiences as applied statisticians, we are very familiar with the components that are needed for the appropriate use of statistical methods. We focus on formulating questions, collecting and finding data, evaluating the quality of data, exploring the relationships among variables, performing statistical analyses, and drawing conclusions. In examples and exercises throughout the text, we emphasize putting the analysis in the proper context and translating numerical and graphical summaries into conclusions.

2. Focus on conceptual understanding. With the software available today, it is very easy for almost anyone to apply a wide variety of statistical procedures, both simple and complex, to a set of data. Without a firm grasp of the concepts, such applications are frequently meaningless. By using the methods that we present on real sets of data, we believe that students will gain an excellent understanding of these concepts. Our emphasis is on the input (questions of interest, collecting or finding data, examining data) and the output (conclusions) for a statistical analysis. Formulas are given only where they will provide some insight into concepts.

3. Integrate real data with a context and a purpose. Many of the examples and exercises in IPS include data that we have obtained from collaborators or consulting clients. Other data sets have come from research related to these activities. We have also used the Internet as a data source, particularly for data related to social media and other topics of interest to undergraduates. Our emphasis on real data, rather than artificial data chosen to illustrate a

calculation, serves to motivate students and help them see the usefulness of statistics in everyday life. We also frequently encounter interesting statistical issues that we explore. These include outliers and nonlinear relationships. All data sets are available from the text website.

4. Foster active learning in the classroom. As we mentioned earlier, we believe that statistics is exciting as something to do rather than something to talk about. Throughout the text, we provide exercises in Use Your Knowledge sections that ask the students to perform some relatively simple tasks that reinforce the material just presented. Other exercises are particularly suited to being worked on and discussed within a classroom setting.

5. Use technology for developing concepts and analyzing data. Technology has altered statistical practice in a fundamental way. In the past, some of the calculations that we performed were particularly difficult and tedious. In other words, they were not fun. Today, freed from the burden of computation by software, we can concentrate our efforts on the big picture: what questions are we trying to address with a study and what can we conclude from our analysis?

6. Use assessments to improve and evaluate student learning. Our goal for students who complete a course based on IPS is that they are able to design and carry out a statistical study for a project in their capstone course or other setting. Our exercises are oriented toward this goal. Many ask about the design of a statistical study and the collection of data. Others ask for a paragraph summarizing the results of an analysis. This recommendation includes the use of projects, oral presentations, article critiques, and written reports. We believe that students using this text will be well prepared to undertake these kinds of activities. Furthermore, we view these activities not only as assessments but also as valuable tools for learning statistics.

Teaching Recommendations We have used IPS in courses taught to a variety of student audiences. For general undergraduates from mixed disciplines, we recommend covering Chapters 1 through 8 and Chapters 9, 10, or 12. For a quantitatively strong audience—sophomores planning to major in actuarial science or statistics—we recommend moving more quickly. Add Chapters 10 and 11 to the core material in Chapters 1 through 8. In general, we recommend deemphasizing the material on probability because these students will take a probability course later in their program. For beginning graduate students in such fields as education, family studies, and retailing, we recommend that the students read the entire text (Chapters 11 and 13 lightly), again with reduced emphasis on Chapter 4 and some parts of Chapter 5. In all cases, beginning with data analysis and data production (Part I) helps students overcome their fear of statistics and builds a sound base for studying inference. We believe that IPS can easily be adapted to a wide variety of audiences.

The Ninth Edition: What’s New? Chapter 1 now begins with a short section giving an overview of data. “Toward Statistical Inference” (previously Section 3.3), which introduces the concepts of statistical inference and sampling distributions, has been moved to Section 5.1 to better assist with the transition from a single data set to sampling distributions. Coverage of mosaic plots as a visual tool for relationships between two categorical variables has been added to Chapters 2 and 9. Chapter 3 now begins with a short section giving a basic overview of data sources. Coverage of equivalence testing has been added to Chapter 7. There is a greater emphasis on sample size determination using software in Chapters 7 and 8. Resampling and bootstrapping are now introduced in Chapter 7 rather than Chapter 6. “Inference for Categorical Data” is the new title for Chapter 9, which includes goodness of fit as well as inference for two-way tables. There are more JMP screenshots and updated screenshots of Minitab, Excel, and SPSS outputs. Design A new design incorporates colorful, revised figures throughout to aid the students’ understanding of text material. Photographs related to chapter examples and exercises make connections to real-life applications and provide a visual context for topics. More figures with software output have been included. Exercises and Examples More than 30% of the exercises are new or revised, and there are more than 1700 exercises total. Exercise sets have been added at the end of sections in Chapters 9 through 12. To maintain the attractiveness of the examples to students, we have replaced or updated a large number of them. More than 30% of the 430 examples are new or revised. A list of exercises and examples categorized by application area is provided on the inside of the front cover.

In addition to the new ninth edition enhancements, IPS has retained the successful pedagogical features from previous editions:

Look Back At key points in the text, Look Back margin notes direct the reader to the first explanation of a topic, providing page numbers for easy reference.

Caution Warnings in the text, signaled by a caution icon, help students avoid common errors and misconceptions.

Challenge Exercises More challenging exercises are signaled with an icon. Challenge exercises are varied: some are mathematical, some require open-ended investigation, and others require deeper thought about the basic concepts.

Applets Applet icons are used throughout the text to signal where related interactive statistical applets can be found on the IPS website and in LaunchPad. Use Your Knowledge Exercises We have found these exercises to be a very useful learning tool. They appear throughout each section and are listed, with page numbers, before the section-ending exercises. Technology output screenshots Most statistical analyses rely heavily on statistical software. In this book, we discuss the use of Excel 2013, JMP 12, Minitab 17, SPSS 23, CrunchIt, R, and a TI-83/-84 calculator for conducting statistical analysis. As specialized statistical packages, JMP, Minitab, and SPSS are the most popular software choices both in industry and in colleges and schools of business. R is an extremely powerful statistical environment that is free to anyone; it relies heavily on members of the academic and general statistical communities for support. As an all-purpose spreadsheet program, Excel provides a limited set of statistical analysis options in comparison. However, given its pervasiveness and wide acceptance in industry and the computer world at large, we believe it is important to give Excel proper attention. It should be noted that for users who want more statistical capabilities but want to work in an Excel environment, there are a number of commercially available add-on packages (if you have JMP, for instance, it can be invoked from within Excel). Finally, instructions are provided for the TI-83/-84 calculators.

Even though basic guidance is provided in the book, it should be emphasized that IPS is not bound to any of these programs. Computer output from statistical packages is very similar, so you can feel quite comfortable using any one these packages.

Acknowledgments We are pleased that the first eight editions of Introduction to the Practice of Statistics have helped to move the teaching of introductory statistics in a direction supported by most statisticians. We are grateful to the many colleagues and students who have provided helpful comments, and we hope that they will find this new edition another step forward. In particular, we would like to thank the following colleagues who offered specific comments on the new edition: Ali Arab, Georgetown University Tessema Astatkie, Dalhousie University Fouzia Baki, McMaster University Lynda Ballou, New Mexico Institute of Mining and Technology Sanjib Basu, Northern Illinois University David Bosworth, Hutchinson Community College

Max Buot, Xavier University Nadjib Bouzar, University of Indianapolis Matt Carlton, California Polytechnic State University–San Luis Obispo Gustavo Cepparo, Austin Community College Pinyuen Chen, Syracuse University Dennis L. Clason, University of Cincinnati–Blue Ash College Tadd Colver, Purdue University Chris Edwards, University of Wisconsin–Oshkosh Irina Gaynanova, Texas A&M University Brian T. Gill, Seattle Pacific University Mary Gray, American University Gary E. Haefner, University of Cincinnati Susan Herring, Sonoma State University Lifang Hsu, Le Moyne College Tiffany Kolba, Valparaiso University Lia Liu, University of Illinois at Chicago Xuewen Lu, University of Calgary Antoinette Marquard, Cleveland State University Frederick G. Schmitt, College of Marin James D. Stamey, Baylor University Engin Sungur, University of Minnesota–Morris Anatoliy Swishchuk, University of Calgary Richard Tardanico, Florida International University Melanee Thomas, University of Calgary Terri Torres, Oregon Institute of Technology Mahbobeh Vezvaei, Kent State University Yishi Wang, University of North Carolina–Wilmington John Ward, Jefferson Community and Technical College Debra Wiens, Rocky Mountain College Victor Williams, Paine College Christopher Wilson, Butler University Anne Yust, Birmingham-Southern College Biao Zhang, The University of Toledo Michael L. Zwilling, University of Mount Union

The professionals at Macmillan, in particular, Terri Ward, Karen Carson, Jorge Amaral, Emily Tenenbaum, Ed Dionne, Blake Logan, and Susan Wein, have contributed greatly to the success of IPS. In addition, we would like to thank Tadd Colver at Purdue University for his valuable contributions to the ninth edition, including authoring the back-of-book answers, solutions, and Instructor’s Guide. We’d also like to thank Monica Jackson at American University for accuracy reviewing the back-of-book answers and solutions and for authoring the test bank. Thanks also to Michael Zwilling at University of Mount Union for accuracy reviewing the test bank, Christopher Edwards at University of Wisconsin Oshkosh for authoring the lecture slides, and James Stamey at Baylor University for authoring the Clicker slides.

Most of all, we are grateful to the many friends and collaborators whose data and research questions have enabled us to gain a deeper understanding of the science of data. Finally, we would like to acknowledge the contributions of John W. Tukey, whose contributions to data analysis have had such a great influence on us as well as a whole generation of applied statisticians.

Media and Supplements

LaunchPad, our online course space, combines an interactive e-Book with high-quality multimedia content and ready- made assessment options, including LearningCurve adaptive quizzing. Content is easy to assign or adapt with your own material, such as readings, videos, quizzes, discussion groups, and more. LaunchPad also provides access to a Gradebook that offers a window into your students’ performance—either individually or as a whole. Use LaunchPad on its own or integrate it with your school’s learning management system so your class is always on the same page. To learn more about LaunchPad for Introduction to the Practice of Statistics, Ninth Edition, or to request access, go to

launchpadworks.com.

Assets integrated into LaunchPad include:

Interactive e-Book. Every LaunchPad e-Book comes with powerful study tools for students, video and multimedia content, and easy customization for instructors. Students can search, highlight, and bookmark, making it easier to study and access key content. And teachers can ensure that their classes get just the book they want to deliver: customize and rearrange chapters; add and share notes and discussions; and link to quizzes, activities, and other resources.

LearningCurve provides students and instructors with powerful adaptive quizzing, a game-like format, direct links to the e-Book, and instant feedback. The quizzing system features questions tailored specifically to the text and adapts to students’ responses, providing material at different difficulty levels and topics based on student performance.

JMP Student Edition (developed by SAS) is easy to learn and contains all the capabilities required for introductory statistics. JMP is the leading commercial data analysis software of choice for scientists, engineers, and analysts at companies throughout the world (for Windows and Mac). Register inside LaunchPad at no additional cost.

CrunchIt!® is a Web-based statistical program that allows users to perform all the statistical operations and graphing needed for an introductory statistics course and more. It saves users time by automatically loading data from IPS, 9e, and it provides the flexibility to edit and import additional data.

StatBoards Videos are brief whiteboard videos that illustrate difficult topics through additional examples, written and explained by a select group of statistics educators.

Stepped Tutorials are centered on algorithmically generated quizzing with step-by-step feedback to help students work their way toward the correct solution. These exercise tutorials (two to three per chapter) are easily assignable and assessable.

Statistical Video Series consists of StatClips, StatClips Examples, and Statistically Speaking “Snapshots.” View animated lecture videos, whiteboard lessons, and documentary-style footage that illustrate key statistical concepts and help students visualize statistics in real-world scenarios.

Video Technology Manuals, available for TI-83/84 calculators, Minitab, Excel, JMP, SPSS, R, Rcmdr, and CrunchIt! ®, provide brief instructions for using specific statistical software.

StatTutor Tutorials offer multimedia tutorials that explore important concepts and procedures in a presentation that combines video, audio, and interactive features. The newly revised format includes built-in, assignable assessments and a bright new interface.

Statistical Applets give students hands-on opportunities to familiarize themselves with important statistical concepts and procedures in an interactive setting that allows them to manipulate variables and see the results graphically. Icons in the textbook indicate when an applet is available for the material being covered. Applets are assessable and assignable in LaunchPad.

Stats@Work Simulations put students in the role of the statistical consultant, helping them better understand statistics interactively within the context of real-life scenarios.

EESEE Case Studies (Electronic Encyclopedia of Statistical Examples and Exercises), developed by The Ohio State University Statistics Department, teach students to apply their statistical skills by exploring actual case studies using real data.

SolutionMaster offers an easy-to-use web-based version of the instructor’s solutions, allowing instructors to generate a solution file for any set of homework exercises.

Data files are available in JMP, ASCII, Excel, TI, Minitab, SPSS (an IBM Company)*, R, and CSV formats.

Student Solutions Manual provides solutions to the odd-numbered exercises in the text and is available as a print supplement and electronically in LaunchPad.

Instructor’s Guide with Full Solutions includes teaching suggestions, chapter comments, and detailed solutions to all exercises and is available electronically in LaunchPad.

Test Bank offers hundreds of multiple-choice questions and is available in LaunchPad.

Lecture Slides offer a customizable, detailed lecture presentation of statistical concepts covered in each chapter of IPS, 9e. Image slides contain all textbook figures and tables. Lecture slides and images slides are available in LaunchPad.

WebAssign offers algorithmic questions from IPS, 9e, in a powerful online instructional system. WebAssign lets you easily create assignments, grade homework, and give your students instant feedback. Along with flexible features, class and question-level analytics are available for instructors and students. WebAssign Premium also includes the following resources described above: e-Book, data files, LearningCurve, StatTutor Tutorials, Statistical Videos, Video Technology Manuals, solutions manuals, lecture and image slides, i-Clicker slides, test bank, and practice quizzes.

Additional Resources Available with IPS, 9e Special Software Package A student version of JMP is available for packaging with the printed text. JMP is also available inside LaunchPad at no additional cost.

i-Clicker is a two-way radio-frequency classroom response solution developed by educators for educators. Each step of i-Clicker’s development has been informed by teaching and learning.

* SPSS was acquired by IBM in October 2009

To Students: What Is Statistics?

Statistics is the science of collecting, organizing, and interpreting numerical facts, which we call data. We are bombarded by data in our everyday lives. The news mentions movie box-office sales, the latest poll of the president’s popularity, and the average high temperature for today’s date. Advertisements claim that data show the superiority of the advertiser’s product. All sides in public debates about economics, education, and social policy argue from data. A knowledge of statistics helps separate sense from nonsense in this flood of data.

The study and collection of data are also important in the work of many professions, so training in the science of statistics is valuable preparation for a variety of careers. Each month, for example, government statistical offices release the latest numerical information on unemployment and inflation. Economists and financial advisers, as well as policymakers in government and business, study these data in order to make informed decisions. Doctors must understand the origin and trustworthiness of the data that appear in medical journals. Politicians rely on data from polls of public opinion. Business decisions are based on market research data that reveal consumer tastes and preferences. Engineers gather data on the quality and reliability of manufactured products. Most areas of academic study make use of numbers and, therefore, also make use of the methods of statistics. This means it is extremely likely that your undergraduate research projects will involve, at some level, the use of statistics.

Learning from Data The goal of statistics is to learn from data. To learn, we often perform calculations or make graphs based on a set of numbers. But to learn from data, we must do more than calculate and plot because data are not just numbers; they are numbers that have some context that helps us learn from them.

More than two-thirds of Americans are overweight or obese according to the Centers for Disease Control and Prevention (CDC) website (www.cdc.gov/nchs/nhanes.htm). What does it mean to be obese or to be overweight? To answer this question, we need to talk about body mass index (BMI). Your weight in kilograms divided by the square of your height in meters is your BMI. A man who is 6 feet tall (1.83 meters) and weighs 180 pounds (81.65 kilograms) will have a BMI of 81.65/(1.83)2 = 24.4 kg/m2. How do we interpret this number? According to the CDC, a person is classified as overweight if his or her BMI is between 25 and 29.9 kg/m2 and as obese if his or her BMI is 30 kg/m2 or more. Therefore, more than two-thirds of Americans have a BMI of 25 kg/m2 or more. The man who weighs 180 pounds and is 6 feet tall is not overweight or obese, but if he gains 5 pounds, his BMI would increase to 25.1, and he would be classified as overweight.

When you do statistical problems, even straightforward textbook problems, don’t just graph or calculate. Think about the context and state your conclusions in the specific setting of the problem. As you are learning how to do statistical calculations and graphs, remember that the goal of statistics is not calculation for its own sake but gaining understanding from numbers. The calculations and graphs can be automated by a calculator or software, but you must supply the understanding. This book presents only the most common specific procedures for statistical analysis. A thorough grasp of the principles of statistics will enable you to quickly learn more advanced methods as needed. On the other hand, a fancy computer analysis carried out without attention to basic principles will often produce elaborate nonsense. As you read, seek to understand the principles as well as the necessary details of methods and recipes.

The Rise of Statistics Historically, the ideas and methods of statistics developed gradually as society grew interested in collecting and using data for a variety of applications. The earliest origins of statistics lie in the desire of rulers to count the number of inhabitants or measure the value of taxable land in their domains. As the physical sciences developed in the seventeenth and eighteenth centuries, the importance of careful measurements of weights, distances, and other physical quantities grew. Astronomers and surveyors striving for exactness had to deal with variation in their measurements. Many measurements should be better than a single measurement, even though they vary among themselves. How can we best combine many varying observations? Statistical methods that are still important were invented in order to analyze scientific measurements.

By the nineteenth century, the agricultural, life, and behavioral sciences also began to rely on data to answer

fundamental questions. How are the heights of parents and children related? Does a new variety of wheat produce higher yields than the old, and under what conditions of rainfall and fertilizer? Can a person’s mental ability and behavior be measured just as we measure height and reaction time? Effective methods for dealing with such questions developed slowly and with much debate.

As methods for producing and understanding data grew in number and sophistication, the new discipline of statistics took shape in the twentieth century. Ideas and techniques that originated in the collection of government data, in the study of astronomical or biological measurements, and in the attempt to understand heredity or intelligence came together to form a unified “science of data.” That science of data—statistics—is the topic of this text.

The Organization of This Book Part I of this book, called simply “Looking at Data,” concerns data analysis and data production. The first two chapters deal with statistical methods for organizing and describing data. These chapters progress from simpler to more complex data. Chapter 1 examines data on a single variable; Chapter 2 is devoted to relationships among two or more variables. You will learn both how to examine data produced by others and how to organize and summarize your own data. These summaries will first be graphical, then numerical, and then, when appropriate, in the form of a mathematical model that gives a compact description of the overall pattern of the data. Chapter 3 outlines arrangements (called designs) for producing data that answer specific questions. The principles presented in this chapter will help you to design proper samples and experiments for your research projects and to evaluate other such investigations in your field of study.

Part II, consisting of Chapters 4 through 8, introduces statistical inference—formal methods for drawing conclusions from properly produced data. Statistical inference uses the language of probability to describe how reliable its conclusions are, so some basic facts about probability are needed to understand inference. Probability is the subject of Chapters 4 and 5. Chapter 6, perhaps the most important chapter in the text, introduces the reasoning of statistical inference. Effective inference is based on good procedures for producing data (Chapter 3), careful examination of the data (Chapters 1 and 2), and an understanding of the nature of statistical inference as discussed in Chapter 6. Chapters 7 and 8 describe some of the most common specific methods of inference, for drawing conclusions about means and proportions from one and two samples.

The five shorter chapters in Part III introduce somewhat more advanced methods of inference, dealing with relations in categorical data, regression and correlation, and analysis of variance. Four supplementary chapters, available from the text website, present additional statistical topics.

What Lies Ahead Introduction to the Practice of Statistics is full of data from many different areas of life and study. Many exercises ask you to express briefly some understanding gained from the data. In practice, you would know much more about the background of the data you work with and about the questions you hope the data will answer. No textbook can be fully realistic. But it is important to form the habit of asking, “What do the data tell me?” rather than just concentrating on making graphs and doing calculations.

You should have some help in automating many of the graphs and calculations. You should certainly have a calculator with basic statistical functions. Look for keywords such as “two-variable statistics” or “regression” when you shop for a calculator. More advanced (and more expensive) calculators will do much more, including some statistical graphs. You may be asked to use software as well. There are many kinds of statistical software, from spreadsheets to large programs for advanced users of statistics. The kind of computing available to learners varies a great deal from place to place—but the big ideas of statistics don’t depend on any particular level of access to computing.

Because graphing and calculating are automated in statistical practice, the most important assets you can gain from the study of statistics are an understanding of the big ideas and the beginnings of good judgment in working with data. Ideas and judgment can’t (at least yet) be automated. They guide you in telling the computer what to do and in interpreting its output. This book tries to explain the most important ideas of statistics, not just teach methods. Some examples of big ideas that you will meet are “always plot your data,” “randomized comparative experiments,” and “statistical significance.”

You learn statistics by doing statistical problems. “Practice, practice, practice.” Be prepared to work problems. The basic principle of learning is persistence. Being organized and persistent is more helpful in reading this book than knowing lots of math. The main ideas of statistics, like the main ideas of any important subject, took a long time to discover and take some time to master. The gain will be worth the pain.

About the Authors

David S. Moore is Shanti S. Gupta Distinguished Professor of Statistics, Emeritus, at Purdue University and was 1998 president of the American Statistical Association. He received his AB from Princeton and his PhD from Cornell, both in mathematics. He has written many research papers in statistical theory and served on the editorial boards of several major journals.

Professor Moore is an elected fellow of the American Statistical Association and of the Institute of Mathematical Statistics and is an elected member of the International Statistical Institute. He has served as program director for statistics and probability at the National Science Foundation.

In recent years, Professor Moore has devoted his attention to the teaching of statistics. He was the content developer for the Annenberg/Corporation for Public Broadcasting college-level telecourse, Against All Odds: Inside Statistics, and for the series of video modules, Statistics: Decisions through Data, intended to aid the teaching of statistics in schools. He is the author of influential articles on statistics education and of several leading texts. Professor Moore has served as president of the International Association for Statistical Education and has received the Mathematical Association of America’s national award for distinguished college or university teaching of mathematics.

George P. McCabe is Associate Dean for Academic Affairs in the College of Science and Professor of Statistics at Purdue University. In 1966, he received a BS degree in mathematics from Providence College and in 1970 a PhD in mathematical statistics from Columbia University. His entire professional career has been spent at Purdue, with sabbaticals at Princeton University, the Commonwealth Scientific and Industrial Research Organization (CSIRO) in Melbourne (Australia), the University of Berne (Switzerland), the National Institute of Standards and Technology (NIST) in Boulder, Colorado, and the National University of Ireland in Galway. Professor McCabe is an elected fellow of the American Association for the Advancement of Science and of the American Statistical Association; he was 1998 chair of its section on Statistical Consulting. In 2008–2010, he served on the Institute of Medicine Committee on Nutrition Standards for the National School Lunch and Breakfast Programs. He has served on the editorial boards of several statistics journals. He has consulted with many major corporations and has testified as an expert witness on the use of statistics in several cases.

Professor McCabe’s research interests have focused on applications of statistics. Much of his recent work has focused on problems in nutrition, including nutrient requirements, calcium metabolism, and bone health. He is the author or coauthor of more than 190 publications in many different journals.

Bruce A. Craig is Professor of Statistics and Director of the Statistical Consulting Service at Purdue University. He received his BS in mathematics and economics from Washington University in St. Louis and his PhD in statistics from the University of Wisconsin–Madison. He is an elected fellow of the American Association for the Advancement of Science and of the American Statistical Association and was chair of its section on Statistical Consulting in 2009. He has also been an active member of the Eastern North American Region of the International Biometrics Society and was elected by the voting membership to the Regional Committee between 2003 and 2006.

Professor Craig has served on the editorial board of several statistical journals and has been a member of several data and safety monitoring boards, including Purdue’s institutional review board.

Professor Craig’s research interests focus on the development of novel statistical methodology to address research questions in the life sciences. Areas of current interest are diagnostic testing, inter-rater agreement, and abundance estimation. He is an author or coauthor of more than 100 papers in more than 50 different journals. In 2005, he was named Purdue University Faculty Scholar.

Data Table Index

TABLE 1.1 IQ test scores for 60 randomly chosen fifth-grade students TABLE 1.2 Service times (seconds) for calls to a customer service center TABLE 1.3 Educational data for 78 seventh-grade students TABLE 2.1 Four data sets for exploring correlation and regression TABLE 2.2 Two measures of glucose level in diabetics TABLE 2.3 Dwelling permits, sales, and production for 21 countries TABLE 2.4 World record times for the 10,000-meter run TABLE 5.1 Length (in minutes) of 60 visits to a statistics help room TABLE 7.1 Monthly rates of return on a portfolio (%) TABLE 7.2 Parts measurements using optical software TABLE 7.3 DRP scores for third-graders TABLE 7.4 Seated systolic blood pressure (mm Hg) TABLE 7.5 Length (in seconds) of audio files sampled from an iPod TABLE 10.1 Annual number of tornadoes in the United States between 1953 and 2014 TABLE 10.2 In-state tuition and fees (in dollars) for 33 public universities TABLE 10.3 Sales price and assessed value (in thousands of $) of 35 homes in a midwestern city TABLE 10.4 Watershed area (km2), percent forest, and index of biotic integrity TABLE 13.1 Iron content (mg/100 g) of food cooked in different pots TABLE 13.2 Tool diameter data

Beyond the Basics Index

Chapter 1 Density estimation Chapter 2 Data mining Chapter 3 Capture-recapture sampling Chapter 4 More laws of large numbers Chapter 5 Weibull distributions Chapter 7 The bootstrap Chapter 8 The plus four confidence interval for a single proportion Chapter 8 The plus four confidence interval for a difference in proportions Chapter 8 Relative risk Chapter 9 Meta-analysis Chapter 10 Nonlinear regression Chapter 11 Multiple logistic regression Chapter 12 Testing the equality of spread

1.1 1.2 1.3 1.4

CHAPTER 1 Looking at Data—Distributions

Data Displaying Distributions with Graphs Describing Distributions with Numbers Density Curves and Normal Distributions

Introduction Statistics is the science of learning from data. Data are numerical or qualitative descriptions of the objects that we want to study. In this chapter, we will master the art of examining data.

We begin in Section 1.1 with some basic ideas about data. We will learn about the different types of data that are collected and how data sets are organized.

Section 1.2 starts our process of learning from data by looking at graphs. These visual displays give us a picture of the overall patterns in a set of data. We have excellent software tools that help us make these graphs. However, it takes

a little experience and a lot of judgment to study the graphs carefully and to explain what they tell us about our data. Section 1.3 continues our process of learning from data by computing numerical summaries. These sets of numbers

describe key characteristics of the patterns that we saw in our graphical summaries. The final section in this chapter helps us make the transition from data summaries to statistical models that are used

to draw conclusions and to make predictions. Specifically, we learn about using density curves to describe a set of data and are introduced to the Normal distributions. These distributions can be used to describe many sets of data that we will encounter. They also play a fundamental role in many of the methods of statistical analysis.

1.1 Data

When you complete this section, you will be able to:

Give examples of cases in a data set. Identify the variables in a data set. Demonstrate how a label can be used as a variable in a data set. Identify the values of a variable. Classify variables as categorical or quantitative. Describe the key characteristics of a set of data. Explain how a rate is the result of adjusting one variable to create another.

A statistical analysis starts with a set of data. We construct a set of data by first deciding what cases, or units, we want to study. For each case, we record information about characteristics that we call variables.

CASES, LABELS, VARIABLES, AND VALUES

Cases are the objects described by a set of data. Cases may be customers, companies, subjects in a study, units in an experiment, or other objects.

A label is a special variable used in some data sets to distinguish the different cases.

A variable is a characteristic of a case.

Different cases can have different values of the variables.

EXAMPLE 1.1

COUPONS

Restaurant discount coupons. A website offers coupons that can be used to get discounts for various items at local restaurants. Coupons for food are very popular. Figure 1.1 gives information for seven restaurant coupons that were available for a recent weekend. These are the cases. Data for each coupon are listed on a different line, and the first column has the coupons numbered from 1 to 7. The remaining columns gives the type of restaurant, the name of the restaurant, the item being discounted, the regular price, and the discount price.

FIGURE 1.1 Spreadsheet of food discount coupons, Example 1.1.

1.1

1.2

Some variables, like the type of restaurant, the name of the restaurant, and the item simply place coupons into categories. The regular price and discount price columns have numerical values for which we can do arithmetic. It makes sense to give an average of the regular prices, but it does not make sense to give an “average” type of restaurant. We can, however, do arithmetic to compare the regular prices classified by type of restaurant.

CATEGORICAL AND QUANTITATIVE VARIABLES

A categorical variable places a case into one of several groups or categories.

A quantitative variable takes numerical values for which arithmetic operations such as adding and averaging make sense.

EXAMPLE 1.2

COUPONS

Categorical and quantitative variables for coupons. The restaurant discount coupon file has six variables: coupon number, type of restaurant, name of restaurant, item, regular price, and discount price. The two price variables are quantitative variables. Coupon number, type of restaurant, name of restaurant, and item are categorical variables.

An appropriate label for your cases should be chosen carefully. In our food coupon example, a natural choice of a label would be the name of the restaurant. However, if there are two or more coupons available for a particular restaurant, or if a restaurant is a chain with different discounts offered at different locations, then the name of the restaurant would not uniquely label each of the coupons. In the restaurant discount coupon file, the first variable, ID, is a unique label for each coupon.

spreadsheet

The display in Figure 1.1 is from an Excel spreadsheet. Spreadsheets are very useful for doing the kind of simple computations that you will do in Exercise 1.2. You can type in a formula and have the same computation performed for each row.

Note that the names we have chosen for the variables in our spreadsheet do not have spaces. For example, instead of “Restaurant Name” for the name of the restaurant, we simply use Name. In some statistical software packages, however, spaces are not allowed in variable names. For this reason, when creating spreadsheets for eventual use with statistical software, it is best to avoid spaces in variable names. Another convention is to use an underscore (_) where you would normally use a space. For our data set, we could have used Regular_Price and Discount_Price for the two price variables.

USE YOUR KNOWLEDGE

Read the spreadsheet. Refer to Figure 1.1. Give the regular price and the discount price for the Smokey Grill ribs coupon.

How much is the discount worth? Refer to Example 1.1. Consider adding another column to the spreadsheet that gives the coupon savings. Explain how you would compute the entries in this column. Does the new column contain values for a categorical variable or for a quantitative variable? Explain your answer.

unit of measurement

Another important part of the description of any quantitative variable is its unit of measurement. For both RegPrice and DiscPrice, the unit of measurement is clearly dollars. In other settings, it may not be as obvious. For example, if we were measuring heights of children, we might choose to use either inches or centimeters. The units of measurement are an important part of the description of a quantitative variable.

Key characteristics of a data set In practice, any set of data is accompanied by background information that helps us understand the data. When you plan a statistical study or explore data from someone else’s work, ask yourself the following questions:

1. Who? What cases do the data describe? How many cases does the data set contain? 2. What? How many variables do the data contain? What are the exact definitions of these variables? What are

the units of measurement for each quantitative variable? 3. Why? What purpose do the data have? Do we hope to answer some specific questions? Do we want to draw

conclusions about cases other than the ones we actually have data for? Are the variables that are recorded suitable for the intended purpose?

EXAMPLE 1.3

Statistics class data. Suppose that you are a teaching assistant for a statistics class and one of your jobs is to keep track of the grades for students in two sections of the course. The cases are the students in the class. There are weekly homework assignments, two exams during the semester, and a final exam. Each of these components is given a numerical score, and the components are added to get a total score that can range from 0 to 1000. Cutoffs of 900, 800, 700, etc., are used to assign letter grades of A, B, C, etc.

The spreadsheet for this course will have seven variables:

An identifier for each student. The number of points earned for homework. The number of points earned for the first exam. The number of points earned for the second exam. The number of points earned for the final exam. The total number of points earned. The letter grade earned.

The student identifier is a label and the letter grade earned is a categorical variable. All the other variables are measured in “points.” Because we can do arithmetic with their values, these variables are quantitative variables.

In our example of statistics class data, the possible values for the grade variable are A, B, C, D, and F. When computing grade point averages, many colleges and universities translate these letter grades into numbers using A = 4, B = 3, C = 2, D = 1, and F = 0. The transformed variable with numeric values is considered to be quantitative because we can average the numerical values across different courses to obtain a grade point average.

Sometimes, experts argue about numerical scales such as this. They ask whether or not the difference between an A and a B is the same as the difference between a D and an F. Similarly, many questionnaires ask people to respond on a 1 to 5 scale, with 1 representing strongly agree, 2 representing agree, etc. Again we could ask whether or not the five possible values for this scale are equally spaced in some sense. From a practical point of view, the averages that can be computed when we convert categorical scales such as these to numerical values frequently provide a very useful way to summarize data.

1.3

1.4

EXAMPLE 1.4

Who, what, and why for the statistics class data. The data set in Example 1.3 was constructed to keep track of the grades for students in an introductory statistics course. The cases are the students in the class. There are seven variables in this data set. These include a label for each student and scores for the various course requirements. There are no units for the label and grade. The other variables all have “points” as the unit.

USE YOUR KNOWLEDGE

Who, what, and why? For the restaurant discount coupon data of Example 1.1 (page 2), what cases do the data describe? How many cases are there? How many variables are there? What are their definitions and units of measurement? What purpose do the data have?

EXAMPLE 1.5

Statistics class data for a different purpose. Suppose that the data for the students in the introductory statistics class were also to be used to study relationships between student characteristics and success in the course. Here, we have decided to focus on the TotalPoints and Grade as the outcomes of interest. Other variables of interest would have been included—for example, Sex, PrevStat (whether or not the student has taken a statistics course previously), and Year (student classification as first, second, third, or fourth year). ID is a categorical variable, TotalPoints is a quantitative variable, and the remaining variables are all categorical.

USE YOUR KNOWLEDGE

Apartment rentals. A data set lists apartments available for students to rent. Information provided includes the monthly rent, whether or not cable is included free of charge, whether or not pets are allowed, the number of bedrooms, and the distance to the campus. Describe the cases in the data set, give the number of variables, and specify whether each variable is categorical or quantitative.

instrument

Often, the variables in a statistical study are easy to understand: height in centimeters, study time in minutes, and so on. But each area of work also has its own special variables. A psychologist uses the Minnesota Multiphasic Personality Inventory (MMPI), and a physical fitness expert measures “VO2 max” (the volume of oxygen consumed per minute while exercising at your maximum capacity). Both of these variables are measured with special instruments. VO2 max is measured by exercising while breathing into a mouthpiece connected to an apparatus that measures oxygen consumed. Scores on the MMPI are based on a long questionnaire, which is also called an instrument.

Part of mastering your field of work is learning what variables are important and how they are best measured. Because details of particular measurements usually require knowledge of the particular field of study, we will say little about them.

rate

Be sure that each variable really does measure what you want it to. A poor choice of variables can lead to misleading conclusions. Often, for example, the rate at which something occurs is a more meaningful measure than a simple count of occurrences.

EXAMPLE 1.6

1.5

1.6

Comparing colleges based on graduates. Think about comparing colleges based on the numbers of graduates. This view tells you something about the relative sizes of different colleges. However, if you are interested in how well colleges succeed at graduating students they admit, it would be better to use a rate. For example, you can find data on the Internet on the six-year graduation rates of different colleges. These rates are computed by examining the progress of first-year students who enroll in a given year. Suppose that at College A there were 1000 first-year students in a particular year, and 800 graduated within six years. The graduation rate is

or 80%. College B has 2000 students who entered in the same year, and 1200 graduated within six years. The graduation rate is

or 60%. How do we compare these two colleges? College B has more graduates but College A has a better graduation rate.

adjusting one variable to create another

In Example 1.6, when we computed the graduation rate, we used the total number of students to adjust the number of graduates. We constructed a new variable by dividing the number of graduates by the total number of students. adjusting one variable to Computing a rate is just one of several ways of adjusting one variable to create another. We often divide one variable by another to compute a more meaningful variable to study. Example 1.20 (page 20) is another type of adjustment.

USE YOUR KNOWLEDGE

How should you express the change? Between the first exam and the second exam in your statistics course, you increased the amount of time that you spent working exercises. Which of the following three ways would you choose to express the results of your increased work: (a) give the grades on the two exams, (b) give the ratio of the grade on the second exam divided by the grade on the first exam, (c) take the difference between the grade on the second exam and the grade on the first exam, and express this as a percent of the grade on the first exam. Give reasons for your answer.

Which variable would you choose? Refer to Example 1.6 on colleges and their graduates. (a) Give a setting in which you would prefer to evaluate the colleges based on the numbers of graduates.

Give a reason for your choice.

(b) Give a setting in which you would prefer to evaluate the colleges based on the graduation rates. Give a reason for your choice.

Exercises 1.5 and 1.6 illustrate an important point about presenting the results of your statistical calculations. Always consider how to best communicate your results to a general audience. For example, the numbers produced by your calculator or by statistical software frequently contain more digits that are needed. Be sure that you do not include extra information generated by software that will distract from a clear explanation of what you have found.

SECTION 1.1 SUMMARY A data set contains information on a number of cases. Cases may be customers, companies, subjects in a study, units in an experiment, or other objects. For each case, the data give values for one or more variables. A variable describes some characteristic of a case, such as a person’s height, gender, or salary. Variables can have different values for different cases. A label is a special variable used to identify cases in a data set. Some variables are categorical and others are quantitative. A categorical variable places each individual into a category, such as male or female. A quantitative variable has numerical values that measure some characteristic of each case, such as height in centimeters or annual salary in dollars. The key characteristics of a data set answer the questions Who?, What?, and Why?

SECTION 1.1 EXERCISES For Exercises 1.1 and 1.2, see page 3; for Exercise 1.3, see page 5; for Exercise 1.4, see page 5; and for Exercises 1.5 and 1.6, see page 6.

1.7 How do you do online research? A study of 552 first-year college students asked about their favorite choice for doing online research. Possible choices were “Google or Google Scholar,” “Library database or website,” “Wikipedia or online encyclopedia,” and “Other.” Names of the students were not recorded, but the students were numbered from 1 to 552 in the data file. The researchers also recorded age, sex, and major area of study for each student.

(a) What are the cases?

(b) Identify the variables and their possible values.

(c) Classify each variable as categorical or quantitative. Be sure to include at least one of each.

(d) Was a label used? Explain your answer.

(e) Summarize the key characteristics of your data set.

1.8 Summer jobs. You are collecting information about summer jobs that are available for college students in your area. Describe a data set that you could use to organize the information that you collect.

(a) What are the cases?

(b) Identify the variables and their possible values.

(c) Classify each variable as categorical or quantitative. Be sure to include at least one of each.

(d) Use a label and explain how you chose it.

(e) Summarize the key characteristics of your data set.

1.9 Employee application data. The personnel department keeps records on all employees in a company. Here is the information that they keep in one of their data files: employee identification number, last name, first name, middle initial, department, number of years with the company, salary, education (coded as high school, some college, or college degree), and age.

(a) What are the cases for this data set?

(b) Describe each type of information as a label, a quantitative variable, or a categorical variable.

(c) Set up a spreadsheet that could be used to record the data. Give appropriate column headings and five sample cases.

1.10 How would you rank cities? Various organizations rank cities and produce lists of the 10 or the 100 best based on various measures. Create a list of criteria that you would use to rank cities. Include at least eight variables, and give reasons for your choices. Say whether each variable is quantitative or categorical.

1.11 Survey of students. A survey of students in an introductory statistics class asked the following questions: (1) age; (2) do you like to sing? (Yes, No); (3) can you play a musical instrument (not at all, a little, pretty well); (4) how much did you spend on food last week (in dollars); (5) height.

(a) Classify each of these variables as categorical or quantitative and give reasons for your answers.

(b) For each variable give the possible values.

1.12 What questions would you ask? Refer to the previous exercise. Make up your own survey with at least six questions. Include at least two categorical variables and at least two quantitative variables. Tell which variables are categorical and which are quantitative. Give reasons for your answers. For each variable, give the possible values.

1.13 How would you rate colleges? Popular magazines rank colleges and universities on their “academic quality” in serving undergraduate students. Describe five variables that you would like to see measured for each college if you were choosing where to study. Give reasons for each of your choices.

1.14 Attending college in your state or in another state. The U.S. Census Bureau collects a large amount of information concerning higher education.1 For example, the bureau provides a table that includes the following variables: state, number of students from the state who attend college, number of students who attend college in their home state.

(a) What are the cases for this set of data?

(b) Is there a label variable? If yes, what is it?

(c) Identify each variable as categorical or quantitative.

(d) Explain how you might use each of the quantitative variables to explain something about the states.

(e) Consider a variable computed as the number of students in each state who attend college in the state divided by the total number of students from the state who attend college. Explain how you would use this variable to explain something about the states.

1.15 Alcohol-impaired driving fatalities. A report on drunk-driving fatalities in the United States gives the number of alcohol-impaired driving fatalities for each state.2 Discuss at least three different ways that these numbers could be converted to rates. Give the advantages and disadvantages of each.

1.2 Displaying Distributions with Graphs

When you complete this section, you will be able to:

Analyze the distribution of a categorical variable using a bar graph. Analyze the distribution of a categorical variable using a pie chart. Analyze the distribution of a quantitative variable using a stemplot. Analyze the distribution of a quantitative variable using a histogram. Examine the distribution of a quantitative variable with respect to the overall pattern of the data and deviations from that pattern. Identify the shape, center, and spread of the distribution of a quantitative variable. Identify and describe any outliers in the distribution of a quantitative variable. Use a time plot to describe the distribution of a quantitative variable that is measured over time.

exploratory data analysis

Statistical tools and ideas help us examine data to describe their main features. This examination is called exploratory data analysis. Like an explorer crossing unknown lands, we want first to simply describe what we see. Here are two basic strategies that help us organize our exploration of a set of data:

Begin by examining each variable by itself. Then move on to study the relationships among the variables. Begin with a graph or graphs. Then add numerical summaries of specific aspects of the data.

We follow these principles in organizing our learning. This chapter presents methods for describing a single variable. We will study relationships among several variables in Chapter 2. Within each chapter, we will begin with graphical displays, then add numerical summaries for a more complete description.

Categorical variables: Bar graphs and pie charts distribution of a categorical variable

count percent proportion

The values of a categorical variable are labels for the categories, such as “yes” and “no.” The distribution of a categorical variable lists the categories and gives either the count or the percent of cases that fall in each category. An alternative to the percent is the proportion, the count divided by the sum of the counts. Note that the percent is simply the proportion times 100.

EXAMPLE 1.7

ONLINE

How do you do online research? A study of 552 first-year college students asked about their preferences for online resources. One question asked them to pick their favorite.3 Here are the results:

Resource Count (n) Google or Google Scholar 406 Library database or website 75 Wikipedia or online encyclopedia 52 Other 19 Total 552

Resource is the categorical variable in this example, and the values are the names of the online resources.

Note that the last value of the variable resource is “Other,” which includes all other online resources that were given as selection options. For data sets that have a large number of values for a categorical variable, we often create a category such as this that includes categories that have relatively small counts or percents. Careful judgment is needed when doing this. You don’t want to cover up some important piece of information contained in the data by combining data in this way.

EXAMPLE 1.8

ONLINE

Favorites as percents. When we look at the online resources data set, we see that Google is the clear winner. We see that 406 reported Google or Google Scholar as their favorite. To interpret this number, we need to know that the total number of students polled was 552. When we say that Google is the winner, we can describe this win by saying that 73.6% (406 divided by 552, expressed as a percent) of the students reported Google as their favorite. Here is a table of the preference percents:

Resource Percent(%) Google or Google Scholar 73.6 Library database or website 13.6 Wikipedia or online encyclopedia 9.4 Other 3.4 Total 100.0

The use of graphical methods allows us to see this information and other characteristics of the data easily. We now examine two types of graphs.

EXAMPLE 1.9

ONLINE

bar graph

Bar graph for the online resource preference data. Figure 1.2 displays the online resource preference data using a bar graph. The heights of the four bars show the percents of the students who reported each of the resources as their favorite.

FIGURE 1.2 Bar graph for the online resource preference data, Example 1.9.

The categories in a bar graph can be put in any order. In Figure 1.2, we ordered the resources based on their preference percents. For other data sets, an alphabetical ordering or some other arrangement might produce a more useful graphical display.

You should always consider the best way to order the values of the categorical variable in a bar graph. Choose an ordering that will be useful to you. If you have difficulty, ask a friend if your choice communicates what you expect. Note that a bar graph using counts will look the same as a bar graph using percents. A pie chart naturally uses percents.

EXAMPLE 1.10

ONLINE

pie chart

Pie chart for the online resource preference data. The pie chart in Figure 1.3 helps us see what part of the whole each group forms. Here it is very easy to see that Google is the favorite for about three-quarters of the students.

FIGURE 1.3 Pie chart for the online resource preference data, Example 1.10.

USE YOUR KNOWLEDGE

1.16

ONLINE

Compare the bar graph with the pie chart. Refer to the bar graph in Figure 1.2 and the pie chart in Figure 1.3 for the online resource preference data. Which graphical display does a better job of describing the data? Give reasons for your answer.

To make a pie chart, you must include all the categories that make up a whole. A category such as “Other” in this example can be used, but the sum of the percents for all the categories should be 100%. This constraint makes bar graphs more flexible.

Quantitative variables: Stemplots and histograms A stemplot (also called a stem-and-leaf plot) gives a quick picture of the shape of a distribution while including the actual numerical values in the graph. Stemplots work best for small numbers of observations that are all greater than 0.

STEMPLOT

To make a stemplot,

1. Separate each observation into a stem consisting of all but the final (rightmost) digit and a leaf, the final digit. Stems may have as many digits as needed, but each leaf contains only a single digit.

2. Write the stems in a vertical column with the smallest at the top, and draw a vertical line at the right of this column.

3. Write each leaf in the row to the right of its stem, in increasing order out from the stem.

EXAMPLE 1.11

STAT

Soluble corn fiber and calcium. Soluble corn fiber (SCF) has been promoted for various health benefits. One study examined the effect of SCF on the absorption of calcium of adolescent boys and girls. Calcium absorption is expressed as a percent of calcium in the diet. Here are the data for the condition where subjects consumed 12 grams per day (g/d) of SCF.4

50 43 43 44 50 44 35 49 54 76 31 48 61 70 62 47 42 45 43 59 53 53 73

To make a stemplot of these data, use the first digits as stems and the second digits as leaves. Figure 1.4 shows the steps in making the plot, We use the first digit of each value as the stem. Figure 1.4(a) shows the stems that have values 3, 4, 5, 6, and 7. The first entry in our data set is 50. This appears in Figure 1.4(b) on the 5 stem with a leaf of 0. Similarly, the second value, 43, appears in the 4 stem with a leaf of 3. The stemplot is completed in Figure 1.4(c), where the leaves are ordered from smallest to largest.

The center of the distribution is in the 40s, and the data are more stretched out toward high values than low values (the highest value is 76, while the lowest is 31). In the plot, we do not see any extreme values that lie far from the remaining data.

FIGURE 1.4 Making a stemplot of the data in Example 1.11. (a) Write the stems. (b) Go through the data and write each leaf on the proper stem. For example, the values on the 3-stem are 35 and 31 in the order given in the display for the example. (c) Arrange the leaves on each stem in order out from the stem. The 3-stem now has leaves 1 and 5.

1.17

USE YOUR KNOWLEDGE

STAT

Make a stemplot. Here are the scores on the first exam in an introductory statistics course for 30 students in one section of the course:

82 73 92 82 75 98 94 57 80 90 92 80 87 91 65 73 70 85 83 61 70 90 75 75 59 68 85 78 80 94

Use these data to make a stemplot. Then use the stemplot to describe the distribution of the first-exam scores for this course.

back-to-back stemplot

When you wish to compare two related distributions, a back-to-back stemplot with common stems is useful. The leaves on each side are ordered out from the common stem.

EXAMPLE 1.12

SCF

Soluble corn fiber and calcium. Refer to Example 1.11, which gives the data for subjects consuming 12 g/d of SCF. Here are the data for subjects under control conditions (0 g/d of SCF):

42 33 41 49 42 47 48 47 53 72 47 63 68 59 35 46 43 55 38 49 51 51 66

Figure 1.5 gives the back-to-back stemplot for the SCF and control conditions. The values on the left give absorption for the control condition, while the values on the right give absorption when SCF was consumed. The values for SCF appear to be somewhat higher than the controls.

FIGURE 1.5 A back-to-back stemplot to compare the distributions of calcium absorption under control and SCF conditions, Example 1.12.

splitting stems

trimming

There are two modifications of the basic stemplot that can be helpful in different situations. You can double the number of stems in a plot by splitting each stem into two: one with leaves 0 to 4 and the other with leaves 5 through 9.

1.18

1.19

When the observed values have many digits, it is often best to trim the numbers by removing the last digit or digits before making a stemplot. If you are using software, you can round the data, which is what was done for the data given in Example 1.11.

You must use your judgment in deciding whether to split stems and whether to trim or round, though statistical software will often make these choices for you. Remember that the purpose of a stemplot is to display the shape of a distribution. If there are many stems with no leaves or only one leaf, trimming will reduce the number of stems. Let’s take a look at the effect of splitting the stems for our SCF data.

EXAMPLE 1.13

SCF

Stemplot with split stems for SCF. Figure 1.6 presents the data from Example 1.12 in a stemplot with split stems.

FIGURE 1.6 A back-to-back stemplot with split stems to compare the distributions of calcium absorption under control and SCF conditions, Example 1.13.

USE YOUR KNOWLEDGE

Which stemplot do you prefer? Look carefully at the stemplots for the SCF data in Figures 1.5 and 1.6. Which do you prefer? Give reasons for your answer.

Why should you keep the space? Suppose that you had a data set similar to the one given in Example 1.12, but in which the control values of 66 and 68 were both changed to 64.

(a) Make a stemplot of these data using split stems.

(b) Should you use one stem or two stems for the 60s? Give a reason for your answer. (Hint: How would your choice reveal or conceal a potentially important characteristic of the data?)

TABLE 1.1

Histograms Stemplots display the actual values of the observations. This feature makes stemplots awkward for large data sets. Moreover, the picture presented by a stemplot divides the observations into groups (stems) determined by the number system rather than by judgment.

histogram

Histograms do not have these limitations. A histogram breaks the range of values of a variable into classes and displays only the count or percent of the observations that fall into each class. You can choose any convenient number of classes, but you should choose classes of equal width.

Making a histogram by hand requires more work than a stemplot. Histograms do not display the actual values observed. For these reasons, we prefer stemplots for small data sets.

The construction of a histogram is best shown by example. Most statistical software packages will make a histogram for you.

EXAMPLE 1.14

IQ

Distribution of IQ scores. You have probably heard that the distribution of scores on IQ tests is supposed to be roughly “bell-shaped.” Let’s look at some actual IQ scores. Table 1.1 displays the IQ scores of 60 fifth-grade students chosen at random from one school.

1. Divide the range of the data into classes of equal width. Let’s use

75 ≤ IQ score < 85

85 ≤ IQ score < 95

145 ≤ IQ score < 155

IQ Test Scores for 60 Randomly Chosen Fifth-Grade Students

145 139 126 122 125 130 96 110 118 118 101 142 134 124 112 109 134 113 81 113 123 94 100 136 109 131 117 110 127 124 106 124 115 133 116 102 127 117 109 137 117 90 103 114 139 101 122 105 97 89 102 108 110 128 114 112 114 102 82 101

Be sure to specify the classes precisely so that each individual falls into exactly one class. A student with IQ 84 would fall into the first class, but IQ 85 falls into the second.

frequency frequency table

2. Count the number of individuals in each class. These counts are called frequencies, and a table of frequencies for all classes is a frequency table.

Class Count 75 ≤ IQ score < 85 2 85 ≤ IQ score < 95 3

1.20

95 ≤ IQ score < 105 10 105 ≤ IQ score < 115 16 115 ≤ IQ score < 125 13 125 ≤ IQ score < 135 10 135 ≤ IQ score < 145 5 145 ≤ IQ score < 155 1

3. Draw the histogram. First, on the horizontal axis mark the scale for the variable whose distribution you are displaying. That’s the IQ score. The scale runs from 75 to 155 because that is the span of the classes we chose. The vertical axis contains the scale of counts. Each bar represents a class. The base of the bar covers the class, and the bar height is the class count. There is no horizontal space between the bars unless a class is empty, so its bar has height zero. Figure 1.7 is our histogram. It does look roughly “bell-shaped.”

FIGURE 1.7 Histogram of the IQ scores of 60 fifth-grade students, Example 1.14.

Large sets of data are often reported in the form of frequency tables when it is not practical to publish the individual observations. In addition to the frequency (count) for each class, we may be interested in the fraction or percent of the observations that fall in each class. A histogram of percents looks just like a frequency histogram such as Figure 1.7. Simply relabel the vertical scale to read in percents. Use histograms of percents for comparing several distributions that have different numbers of observations.

USE YOUR KNOWLEDGE

STAT

Make a histogram. Refer to the first-exam scores from Exercise 1.17 (page 12). Use these data to make a histogram with classes 50 to 59, 60 to 69, etc. Compare the histogram with the stemplot as a way of describing this distribution. Which do you prefer for these data?

Our eyes respond to the area of the bars in a histogram. Because the classes are all the same width, area is determined by height and all classes are fairly represented. There is no one right choice of the classes in a histogram. Too few classes will give a “skyscraper” graph, with all values in a few classes with tall bars. Too many will produce a “pancake” graph, with most classes having one or no observations. Neither choice will give a good picture of the shape of the distribution. You must use your judgment in choosing classes to display the shape. Statistical software will choose the classes for you. The software’s choice is often a good one, but you can change it if you want.

1.21

1.22

You should be aware that the appearance of a histogram can change when you change the classes. The histogram function in the One-Variable Statistical Calculator applet on the text website allows you to change the number of classes by dragging with the mouse, so that it is easy to see how the choice of classes affects the histogram.

USE YOUR KNOWLEDGE

Change the classes in the histogram. Refer to the first-exam scores from Exercise 1.17 (page 12) and the histogram that you produced in Exercise 1.20. Now make a histogram for these data using classes 40 to 59, 60 to 79, and 80 to 100. Compare this histogram with the one that you produced in Exercise 1.20. Which do you prefer? Give a reason for your answer.

STAT

Use smaller classes. Repeat the previous exercise using classes 55 to 59, 60 to 64, 65 to 69, etc. Of the three histograms, which do you prefer? Give reasons for your answer.

Although histograms resemble bar graphs, their details and uses are distinct. A histogram shows the distribution of counts or percents among the values of a single variable. A bar graph compares the counts or percents of different items. The horizontal axis of a bar graph need not have any measurement scale but simply identifies the items being compared.

Draw bar graphs with blank space between the bars to separate the items being compared. Draw histograms with no space, to indicate that all values of the variable are covered. Some spreadsheet programs, which are not primarily intended for statistics, will draw histograms as if they were bar graphs, with space between the bars. Often, you can tell the software to eliminate the space to produce a proper histogram.

TABLE 1.2

Data analysis in action: Don’t hang up on me Many businesses operate call centers to serve customers who want to place an order or make an inquiry. Customers want their requests handled thoroughly. Businesses want to treat customers well, but they also want to avoid wasted time on the phone. They therefore monitor the length of calls and encourage their representatives to keep calls short.

Service Times (Seconds) for Calls to a Customer Service Center

77 289 128 59 19 148 157 203 126 118 104 141 290 48 3 2 372 140 438 56 44 274 479 211 179 1 68 386 2631 90 30 57 89 116 225 700 40 73 75 51 148 9 115 19 76 138 178 76 67 102 35 80 143 951 106 55 4 54 137 367 277 201 52 9 700 182 73 199 325 75 103 64 121 11 9 88 1148 2 465 25

EXAMPLE 1.15

CALLS80

How long are customer service center calls? We have data on the lengths of all 31,492 calls made to the customer service center of a small bank in a month. Table 1.2 displays the lengths of the first 80 calls.5

Take a look at the data in Table 1.2. In this data set, the cases are calls made to the bank’s call center. The variable recorded is the length of each call. The units are seconds. We see that the call lengths vary a great deal. The longest call lasted 2631 seconds, almost 44 minutes. More striking is that 8 of these 80 calls lasted less than 10 seconds.

We started our study of the customer service center data by examining a few cases, the ones displayed in Table 1.2. It would be very difficult to examine all 31,492 cases in this way. How can we do this? Let’s try a histogram.

EXAMPLE 1.16

CALLS

Histogram for customer service center call lengths. Figure 1.8 is a histogram of the lengths of all 31,492 calls. We did not plot the few lengths greater than 1200 seconds (20 minutes). As expected, the graph shows that most calls last between about 1 and 5 minutes, with some lasting much longer when customers have complicated problems. More striking is the fact that 7.6% of all calls are no more than 10 seconds long.

FIGURE 1.8 The distribution of call lengths for 31,492 calls to a bank’s customer service center, Example 1.16. The data show a surprising number of very short calls. These are mostly due to representatives deliberately hanging up in order to bring down their average call length.

It turned out that the bank penalized representatives whose average call length was too long—so some representatives just hung up on customers to bring their average length down. Neither the customers nor the bank were happy about this. The bank changed its policy, and later data showed that calls under 10 seconds had almost disappeared.

tails

The extreme values of a distribution are in the tails of the distribution. The high values are in the upper, or right, tail and the low values are in the lower, or left, tail. The overall pattern in Figure 1.8 is made up of the many moderate call lengths and the long right tail of more lengthy calls. The striking deviation from the overall pattern is the surprising number of very short calls in the left tail.

Our examination of the call center data illustrates some important principles:

After you understand the background of your data (cases, variables, units of measurement), the first thing to do is plot your data. When you look at a plot, look for an overall pattern and also for any striking deviations from the pattern.

Examining distributions Making a statistical graph is not an end in itself. The purpose of the graph is to help us understand the data. After you make a graph, always ask, “What do I see?” Once you have displayed a distribution, you can see its important features as follows.

EXAMINING A DISTRIBUTION

In any graph of data, look for the overall pattern and for striking deviations from that pattern.

You can describe the overall pattern of a distribution by its shape, center, and spread.

An important kind of deviation is an outlier, an individual value that falls outside the overall pattern.

In Section 1.3, we will learn how to describe center and spread numerically. For now, we can describe the center of a distribution by its midpoint, the value with roughly half the observations taking smaller values and half taking larger values. We can describe the spread of a distribution by giving the smallest and largest values. Stemplots and histograms display the shape of a distribution in the same way. Just imagine a stemplot turned on its side so that the larger values lie to the right.

Some things to look for in describing shape are

modes unimodal

Does the distribution have one or several major peaks, called modes? A distribution with one major peak is called unimodal.

symmetric skewed

Is it approximately symmetric or is it skewed in one direction? A distribution is symmetric if the pattern of values smaller and larger than its midpoint are mirror images of each other. It is skewed to the right if the right tail (larger values) is much longer than the left tail (smaller values).

Some variables commonly have distributions with predictable shapes. Many biological measurements on specimens from the same species and sex—lengths of bird bills, heights of young women—have symmetric distributions. Money amounts, on the other hand, usually have right-skewed distributions. There are many moderately priced houses, for example, but the few very expensive mansions give the distribution of house prices a strong right-skew.

EXAMPLE 1.17

IQ

Examine the histogram of IQ scores. What does the histogram of IQ scores (Figure 1.7, page 15) tell us?

Shape: The distribution is roughly symmetric with a single peak in the center. We don’t expect real data to be perfectly symmetric, so in judging symmetry, we are satisfied if the two sides of the histogram are roughly similar in shape and extent.

Center: You can see from the histogram that the midpoint is not far from 110. Looking at the actual data shows that the midpoint is 114.

Spread: The histogram has a spread from 75 to 155. Looking at the actual data shows that the spread is from 81 to 145. There are no outliers or other strong deviations from the symmetric, unimodal pattern.

1.23

EXAMPLE 1.18

Examine the histogram of call lengths. The distribution of call lengths in Figure 1.8, on the other hand, is strongly skewed to the right. The midpoint, the length of a typical call, is about 115 seconds, or just under 2 minutes. The spread is very large, from 1 second to 28,739 seconds.

The longest few calls are outliers. They stand apart from the long right tail of the distribution, though we can’t see this from Figure 1.8, which omits the largest observations. The longest call lasted almost 8 hours—that may well be due to equipment failure rather than an actual customer call.

USE YOUR KNOWLEDGE

STAT

Describe the first-exam scores. Refer to the first-exam scores from Exercise 1.17 (page 12). Use your favorite graphical display to describe the shape, the center, and the spread of these data. Are there any outliers?

Dealing with outliers

In data sets smaller than the service call data, you can spot outliers by looking for observations that stand apart (either high or low) from the overall pattern of a histogram or stemplot. Identifying outliers is a matter for judgment. Look for points that are clearly apart from the body of the data, not just the most extreme observations in a distribution. You should search for an explanation for any outlier. Sometimes outliers point to errors made in recording the data. In other cases, the outlying observation may be caused by equipment failure or other unusual circumstances.

EXAMPLE 1.19

COLLEGE

College students. How does the number of undergraduate college students vary by state? Figure 1.9 is a histogram of the numbers of undergraduate students in each of the states.6 Notice that more than 50% of the states are included in the first bar of the histogram. These states have fewer than 300,000 undergraduates. The next bar includes another 30% of the states. These have between 300,000 and 600,000 students. The bar at the far right of the histogram corresponds to the state of California, which has 2,685,893 undergraduates. California certainly stands apart from the other states for this variable. It is an outlier.

FIGURE 1.9 The distribution of the numbers of undergraduate college students for the 50 states, Example 1.19.

The state of California is an outlier in the previous example because it has a very large number of undergraduate students. California has the largest population of all the states, so we might expect it to have a large number of college students. Let’s look at these data in a different way.

EXAMPLE 1.20

COLLEGE

College students per 1000. To account for the fact that there is large variation in the populations of the states, for each state we divide the number of undergraduate students by the population and then multiply by 1000. This gives the undergraduate college enrollment expressed as the number of students per 1000 people in each state. Figure 1.10

1.24

gives a stemplot of the distribution. California has 60 undergraduate students per 1000 people. This is one of the higher values in the distribution, but it is clearly not an outlier.

FIGURE 1.10 Stemplot of the numbers of undergraduate college students per 1000 people in each of the 50 states, Example 1.20.

USE YOUR KNOWLEDGE

COLLEGE

Four states with large populations. There are four states with populations greater than 15 million. (a) Examine the data file and report the names of these four states.

(b) Find these states in the distribution of number of undergraduate students per 1000 people. To what extent do these four states influence the distribution of number of undergraduate students per 1000 people?

In Example 1.19, we looked at the distribution of the number of undergraduate students, while in Example 1.20, we adjusted these data by expressing the counts as number per 1000 people in each state. Which way is correct? The answer depends upon why you are examining the data.

If you are interested in marketing a product to undergraduate students, the unadjusted numbers would be of interest because you want to reach the most people. On the other hand, if you are interested in comparing states with respect to how well they provide opportunities for higher education to their residents, the population-adjusted values would be more suitable. Always think about why you are doing a statistical analysis, and this will guide you in choosing an appropriate analytic strategy.

Here is an example with a different kind of outlier.

EXAMPLE 1.21

PTH

Healthy bones and PTH. Bones are constantly being built up (bone formation) and torn down (bone resorption). Young people who are growing have more formation than resorption. When we age, resorption increases to the point where it exceeds formation. (The same phenomenon occurs when astronauts travel in space.) The result is osteoporosis, a disease associated with fragile bones that are more likely to break. The underlying mechanisms that control these processes are complex and involve a variety of substances. One of these is parathyroid hormone (PTH). Here are the values of PTH measured on a sample of 29 boys and girls aged 12 to 15 years:7

39 59 30 48 71 31 25 31 71 50 38 63 49 45 31 33 28 40 127 49 59 50 64 28 46 35 28 19 29

The data are measured in picograms per milliliter (pg/ml) of blood. The original data were recorded with one digit after the decimal point. They have been rounded to simplify our presentation here. Figure 1.11 gives a stemplot of the data.

The observation 127 clearly stands out from the rest of the distribution. A PTH measurement on this individual taken on a different day was similar to the rest of the values in the data set. We conclude that this outlier was caused by a laboratory error or a recording error, and we are confident in discarding it for any additional analysis.

FIGURE 1.11 Stemplot of the values of PTH, Example 1.21.

Time plots

Whenever data are collected over time, it is a good idea to plot the observations in time order. Displays of the distribution of a variable that ignore time order, such as stemplots and histograms, can be misleading when there is systematic change over time.

TIME PLOT

A time plot of a variable plots each observation against the time at which it was measured. Always put time on the horizontal scale of your plot and the variable you are measuring on the vertical scale.

EXAMPLE 1.22

VITDS

Seasonal variation in vitamin D. Although we get some of our vitamin D from food, most of us get about 75% of what we need from the sun. Cells in the skin make vitamin D in response to sunlight. If people do not get enough exposure to the sun, they can become deficient in vitamin D, resulting in weakened bones and other health problems. The elderly, who need more vitamin D than younger people, and people who live in northern areas, where there is relatively little sunlight in the winter, are particularly vulnerable to these problems.

Figure 1.12 is a plot of the serum levels of vitamin D versus time of year for samples of subjects from Switzerland.8 The units measuring Vitamin D are nanomoles per liter (nmol/l) of blood. The observations are grouped into periods of two months for the plot. Means are marked by filled-in circles and are connected by a line in the plot. The effect of the lack of sunlight in the winter months on vitamin D levels is clearly evident in the plot.

FIGURE 1.12 Plot of vitamin D versus months of the year, Example 1.22.

The data described in the preceding example are based on a subset of the subjects in a study of 248 subjects. The researchers were particularly concerned about subjects whose levels were deficient, defined as a serum vitamin D level of less than 50 nmol/l. They found that there was a 3.8-fold higher deficiency rate in February-March than in August- September: 91.2% versus 24.3%. To ensure that individuals from this population have adequate levels of vitamin D, some form of supplementation is needed, particularly during certain times of the year.

SECTION 1.2 SUMMARY Exploratory data analysis uses graphs and numerical summaries to describe the variables in a data set and the relations among them. The distribution of a variable tells us what values it takes and how often it takes these values. Bar graphs and pie charts display the distributions of categorical variables. These graphs use the counts or percents of the categories. Stemplots and histograms display the distributions of quantitative variables. Stemplots separate each observation into a stem and a one-digit leaf. Histograms plot the frequencies (counts) or the percents of equal- width classes of values. When examining a distribution, look for shape, center, and spread and for clear deviations from the overall shape. Some distributions have simple shapes, such as symmetric or skewed. The number of modes (major peaks) is another aspect of overall shape. Not all distributions have a simple overall shape, especially when there are few observations. Outliers are observations that lie outside the overall pattern of a distribution. Always look for outliers and try to explain them. When observations on a variable are taken over time, make a time plot that graphs time horizontally and the values of the variable vertically. A time plot can reveal changes over time.

SECTION 1.2 EXERCISES For Exercise 1.16, see page 11; for Exercise 1.17, see page 12; for Exercises 1.18 and 1.19, see page 14; for Exercise 1.20, see page 16; for Exercises 1.21 and 1.22, see page 16; for Exercise 1.23, see page 19; and for Exercise 1.24, see page 21.

1.25 Your Facebook app can generate a million dollars a month. A report on Facebook suggests that Facebook apps can generate large amounts of money, as much as $1 million a month.9 The following table gives the numbers of Facebook users by country for the top 10 countries based on the number of users:10 FACEBK

Country Facebook users (in millions)

Brazil 29.30 India 37.38 Mexico 29.80 Germany 21.46 France 23.19 Philippines 26.87 Indonesia 40.52 United Kingdom 30.39 United States 155.74 Turkey 30.63

(a) Use a bar graph to describe the numbers of users in these countries.

(b) Do you think that the United States is an outlier in this data set? Explain your answer.

(c) Describe the major features of your graph in a short paragraph.

1.26 Facebook use increases by country. Refer to the previous exercise. The report also gave the increases in the number of Facebook users for a one-month period for the same countries: FACEBK

Country Increase in users (in millions)

Brazil 2.47 India 1.75 Mexico 0.84 Germany 0.51 France 0.38 Philippines 0.38 Indonesia 0.37 United Kingdom 0.22 United States 0.65 Turkey 0.09

(a) Use a bar graph to describe the increase in users in these countries.

(b) Describe the major features of your graph in a short paragraph.

(c) Do you think a stemplot would be a better graphical display for these data? Give reasons for your answer.

(d) Write a short paragraph about possible business opportunities suggested by the data you described in this exercise and the previous one.

1.27 The Titanic and class. On April 15, 1912, on her maiden voyage, the Titanic collided with an iceberg and sank. The ship was luxurious but did not have enough lifeboats for the 2224 passengers and crew. As a result of the collision, 1502 people died.11 The ship had three classes of passengers. The level of luxury and the price of the ticket varied with the class, with first class being the most luxurious. There were 323 passengers in first class, 277 in second class, and 709 in third class.12

TITANIC

(a) Make a bar graph of these data.

(b) Give a short summary of how the number of passengers varied with class.

(c) If you made a bar graph of the percent of passengers in each class, would the general features of the graph differ from the one you made in part (a)? Explain your answer.

1.28 Another look at the Titanic and class. Refer to the previous exercise. TITANIC

(a) Make a pie chart to display the data.

(b) Compare the pie chart with the bar graph. Which do you prefer? Give reasons for your answer.

1.29 Who survived? Refer to the two previous exercises. The number of first-class passengers who survived was 200. For second and third class, the numbers were 119 and 181, respectively. Create a graphical summary that shows how the survival of passengers depended on class. TITANIC

1.30 Potassium from potatoes. The 2015 Dietary Guidelines for Americans13 notes that the average potassium (K) intake for U.S. adults is about half of the recommended amount. A major source of potassium is potatoes. Nutrients in the diet can have different absorption depending on the source. One study looked at absorption of potassium from different sources. Participants ate a controlled diet for five days, and the amount of potassium absorbed was measured. Data for a diet that included 40 milliequivalents (mEq) of potassium were collected from 27 adult subjects.14 KPOT40

(a) Make a stemplot of the data.

(b) Describe the pattern of the distribution.

(c) Are there any outliers? If yes, describe them and explain why you have declared them to be outliers.

(d) Describe the shape, center, and spread of the distribution.

1.31 Potassium from a supplement. Refer to the previous exercise. Data were also recorded for 29 subjects who received a potassium salt supplement with 40 mEq of potassium. Answer the questions in the previous exercise for the supplemented subjects. KSUP40

1.32 Energy consumption. The U.S. Energy Information Administration reports data summaries of various energy statistics. Let’s look at the total amount of energy consumed, in quadrillions of British thermal units (Btu), for each month in a recent year. Here are the data:15 ENERGY

Month Energy (quadrillion Btu)

January 9.58 February 8.46 March 8.56 April 7.56 May 7.66 June 7.79 July 8.23 August 8.21 September 7.64 October 7.78 November 8.19 December 8.82

(a) Look at the table and describe how the energy consumption varies from month to month.

(b) Make a time plot of the data and describe the patterns.

(c) Suppose you wanted to communicate information about the month-to-month variation in energy consumption. Which would be more effective, the table of the data or the graph? Give reasons for your answer.

1.33 Energy consumption in a different year. Refer to the previous exercise. Here are the data for the previous year: ENERGY

Month Energy (quadrillion Btu)

January 8.99 February 8.02 March 8.38 April 7.52 May 7.62 June 7.72 July 8.27 August 8.17 September 7.64 October 7.72 November 8.14 December 9.08

(a) Analyze these data using the questions in the previous exercise as a guide.

(b) Compare the patterns across the two years. Describe any similarities and differences.

1.34 Favorite colors. What is your favorite color? One survey produced the following summary of responses to that question: blue, 42%; green, 14%; purple, 14%; red, 8%; black, 7%; orange, 5%; yellow, 3%; brown, 3%; gray, 2%; and white, 2%.16 Make a bar graph of the percents and write a short summary of the major features of your graph. FAVCOL

1.35 Least-favorite colors. Refer to the previous exercise. The same study also asked people about their least-favorite color. Here are the results: orange, 30%; brown, 23%; purple, 13%; yellow, 13%; gray, 12%; green, 4%; white, 4%; red, 1%; black, 0%; and blue, 0%. Make a bar graph of these percents and write a summary of the results. LFAVCOL

1.36 Garbage. The formal name for garbage is “municipal solid waste.” In the United States, approximately 250 million tons of garbage are generated in a year. Following is a breakdown of the materials that made up American municipal solid waste in 2012:17 GARBAGE

Material Weight (million tons) Percent of total

Food scraps 36.4 14.5 Glass 11.6 4.6 Metals 22.4 8.9 Paper, paperboard 68.6 27.4 Plastics 31.7 12.7 Rubber, leather 7.5 3.0 Textiles 14.3 5.7 Wood 15.8 6.3 Yard trimmings 34.0 13.5 Other 8.5 3.4 Total 250.9 100.0

(a) Add the weights. The sum is not exactly equal to the value of 250.9 million tons given in the table. Why?

(b) Make a bar graph of the percents. The graph gives a clearer picture of the main contributors to garbage if you order the bars from tallest to shortest.

(c) Also make a pie chart of the percents. Comparing the two graphs, notice that it is easier to see the small differences among “Food scraps,” “Plastics,” and “Yard trimmings” in the bar graph.

1.37 Vehicle colors. Vehicle colors differ among regions of the world. Here are data on the most popular colors for vehicles

TABLE 1.3

in North America:18 VCOLOR

Color (percent)

White 24 Black 19 Silver 16 Gray 15 Red 10 Blue 7 Brown 5 Other 4

(a) Describe these data with a bar graph.

(b) Describe these data with a pie chart.

(c) Which graphical summary do you prefer. Give reasons for your answer.

1.38 Sketch a skewed distribution. Sketch a histogram for a distribution that is skewed to the left. Suppose that you and your friends emptied your pockets of coins and recorded the year marked on each coin. The distribution of dates would be skewed to the left. Explain why.

1.39 Grades and self-concept. Table 1.3 presents data on 78 seventh-grade students in a rural midwestern school.19 The researcher was interested in the relationship between the students’ “self-concept” and their academic performance. The data we give here include each student’s grade point average (GPA), score on a standard IQ test, and gender, taken from school records. Gender is coded as F for female and M for male. The students are identified only by an observation number (OBS). The missing OBS numbers show that some students dropped out of the study. The final variable is each student’s score on the Piers- Harris Children’s Self-Concept Scale, a psychological test administered by the researcher. SEVENGR

(a) How many variables does this data set contain? Which are categorical variables and which are quantitative variables?

(b) Make a stemplot of the distribution of GPA, after rounding to the nearest tenth of a point.

(c) Describe the shape, center, and spread of the GPA distribution. Identify any suspected outliers from the overall pattern.

(d) Make a back-to-back stemplot of the rounded GPAs for female and male students. Write a brief comparison of the two distributions.

1.40 Describe the IQ scores. Make a graph of the distribution of IQ scores for the seventh-grade students in Table 1.3. Describe the shape, center, and spread of the distribution, as well as any outliers. IQ scores are usually said to be centered at 100. Is the midpoint for these students close to 100, clearly above, or clearly below? SEVENGR

Educational Data for 78 Seventh-Grade Students

OBS GPA IQ Gender Self-concept

001 7.940 111 M 67 002 8.292 107 M 43 003 4.643 100 M 52 004 7.470 107 M 66 005 8.882 114 F 58 006 7.585 115 M 51 007 7.650 111 M 71 008 2.412 97 M 51 009 6.000 100 F 49 010 8.833 112 M 51 011 7.470 104 F 35 012 5.528 89 F 54 013 7.167 104 M 54

014 7.571 102 F 64 015 4.700 91 F 56 016 8.167 114 F 69 017 7.822 114 F 55 018 7.598 103 F 65 019 4.000 106 M 40 020 6.231 105 F 66 021 7.643 113 M 55 022 1.760 109 M 20 024 6.419 108 F 56 026 9.648 113 M 68 027 10.700 130 F 69 028 10.580 128 M 70 029 9.429 128 M 80 030 8.000 118 M 53 031 9.585 113 M 65 032 9.571 120 F 67 033 8.998 132 F 62 034 8.333 111 F 39 035 8.175 124 M 71 036 8.000 127 M 59 037 9.333 128 F 60 038 9.500 136 M 64 039 9.167 106 M 71 040 10.140 118 F 72 041 9.999 119 F 54 043 10.760 123 M 64 044 9.763 124 M 58 045 9.410 126 M 70 046 9.167 116 M 72 047 9.348 127 M 70 048 8.167 119 M 47 050 3.647 97 M 52 051 3.408 86 F 46 052 3.936 102 M 66 053 7.167 110 M 67 054 7.647 120 M 63 055 0.530 103 M 53 056 6.173 115 M 67 057 7.295 93 M 61 058 7.295 72 F 54 059 8.938 111 F 60 060 7.882 103 F 60 061 8.353 123 M 63 062 5.062 79 M 30 063 8.175 119 M 54 064 8.235 110 M 66 065 7.588 110 M 44 068 7.647 107 M 49 069 5.237 74 F 44

071 7.825 105 M 67 072 7.333 112 F 64 074 9.167 105 M 73 076 7.996 110 M 59 077 8.714 107 F 37 078 7.833 103 F 63 079 4.885 77 M 36 080 7.998 98 F 64 083 3.820 90 M 42 084 5.936 96 F 28 085 9.000 112 F 60 086 9.500 112 F 70 087 6.057 114 M 51 088 6.057 93 F 21 089 6.938 106 M 56

1.41 Describe the self-concept scores. Based on a suitable graph, briefly describe the distribution of self-concept scores for the students in Table 1.3. Be sure to identify any suspected outliers. SEVENGR

1.42 The Boston Marathon. Women were allowed to enter the Boston Marathon in 1972. Here are the times (in minutes, rounded to the nearest minute) for the winning women from 1972 to 2015.

Make a graph that shows change over time. What overall pattern do you see? Have times stopped improving in recent years? If so, when did improvement end? MARATH

Year Time

1972 190 1973 186 1974 167 1975 162 1976 167 1977 168 1978 165 1979 155 1980 154 1981 147 1982 150 1983 143 1984 149 1985 154 1986 145 1987 146 1988 145 1989 144 1990 145 1991 144 1992 144 1993 145 1994 142 1995 145 1996 147

1997 146 1998 143 1999 143 2000 146 2001 144 2002 141 2003 145 2004 144 2005 145 2006 143 2007 149 2008 145 2009 152 2010 146 2011 142 2012 151 2013 146 2014 139 2015 145

1.3 Describing Distributions with Numbers

When you complete this section, you will be able to:

Describe the center of a distribution by using the mean. Describe the center of a distribution by using the median. Compare the mean and the median as measures of center for a particular set of data. Describe the spread of a distribution by using quartiles. Describe a distribution by using the five-number summary. Describe a distribution by using a boxplot. Compare one or more sets of data measured on the same variable by using side-by-side boxplots. Identify outliers by using the 1.5 x IQR rule. Describe the spread of a distribution by using the standard deviation. Choose measures of center and spread for a particular set of data. Compute the effects of a linear transformation on the mean, the median, the standard deviation, and the interquartile range.

We can begin our data exploration with graphs, but numerical summaries make our analysis more specific. For categorical variables, numerical summaries are the counts or percents that we use to construct pie charts or bar graphs. In this section, we focus on numerical summaries for quantitative variables. A brief description of the distribution of a quantitative variable should include its shape and numbers describing its center and spread. We describe the shape of a distribution based on inspection of a histogram or a stemplot. Now we will learn specific ways to use numbers to measure the center and spread of a distribution. We can calculate these numerical measures for any quantitative variable. But to interpret measures of center and spread, and to choose among the several measures we will learn, you must think about the shape of the distribution and the meaning of the data. The numbers, like graphs, are aids to understanding, not “the answer” in themselves.

EXAMPLE 1.23

TTS24

The distribution of business start times. An entrepreneur faces many bureaucratic and legal hurdles when starting a new business. The World Bank collects information about starting businesses throughout the world. They have determined the time, in days, to complete all the procedures required to start a business.20 Data for 189 countries are included in the data set, TTS. For this section, we examine data, rounded to integers, for a sample of 24 of these countries. Here are the data:

FIGURE 1.13 Stemplot for the sample of 24 business start times, Example 1.23.

16 4 5 6 5 7 12 19 10 2 25 19 38 5 24 8 6 5 53 32 13 49 11 17

The stemplot in Figure 1.13 shows us the shape, center, and spread of the business start times. The stems are tens of days and the leaves are days. The distribution is skewed to the right with a very long tail of high values. All but six of the times are less than 20 days. The center appears to be about 10 days, and the values range from 2 days to 53 days. There do not appear to be any outliers.

Measuring center: The mean Numerical description of a distribution begins with a measure of its center or average. The two common measures of center are the mean and the median. The mean is the “average value” and the median is the “middle value.” These are two different ideas for “center,” and the two measures behave differently. We need precise recipes for the mean and the median.

THE MEAN x̅

To find the mean x̅ of a set of observations, add their values and divide by the number of observations. If the n observations are x1, x2, . . . , xn, their mean is

or, in more compact notation,

The Σ (capital Greek sigma) in the formula for the mean is short for “add them all up.” The bar over the x indicates the mean of all the x-values. Pronounce the mean x̅ as “x-bar.” This notation is so common that writers who are discussing data use x̅, , etc., without additional explanation. The subscripts on the observations xi are a way of keeping the n observations separate.

EXAMPLE 1.24

TTS24

Mean time to start a business. The mean time to start a business is

The mean time to start a business for the 24 countries in our data set is 16.3 days. Note that we have rounded the answer. Our goal in using the mean to describe the center of a distribution is not to demonstrate that we can compute with great accuracy. The additional digits do not provide any additional useful information. In fact, they distract our attention from the important digits that are meaningful. Do you think it would be better to report the mean as 16 days?

The value of the mean will not necessarily be equal to the value of one of the observations in the data set. Our example of time to start a business illustrates this fact.

In practice, you can key the data into your calculator and hit the Mean key. You don’t have to actually add and divide. But you should know that this is what the calculator is doing.

USE YOUR KNOWLEDGE

1.43

1.44

TTS25

Include the outlier. For Example 1.23, a random sample of 24 countries was selected from a data set that included 189 countries. The South American country of Suriname, where the start time is 208 days, was not included in the random sample. Consider the effect of adding Suriname to the original set. Show that the mean for the new sample of 25 countries has increased to 24 days. (This is a rounded number. You should report the mean with two digits after the decimal to show that you have performed this calculation.)

STAT

Find the mean. Here are the scores on the first exam in an introductory statistics course for 10 students:

83 74 93 85 75 97 93 55 92 81

Find the mean first-exam score for these students.

resistant measure

Exercise 1.43 illustrates an important weakness of the mean as a measure of center: the mean is sensitive to the influence of a few extreme observations. These may be outliers, but a skewed distribution that has no outliers will also pull the mean toward its long tail. Because the mean cannot resist the influence of extreme observations, we say that it is not a resistant measure of center.

robust measure

A measure that is resistant does more than limit the influence of outliers. Its value does not respond strongly to changes in a few observations, no matter how large those changes may be. The mean fails this requirement because we can make the mean as large as we wish by making a large enough increase in just one observation. A resistant measure is sometimes called a robust measure.

Measuring center: The median We used the midpoint of a distribution as an informal measure of center in Section 1.2. The median is the formal version of the midpoint, with a specific rule for calculation.

THE MEDIAN M

The median M is the midpoint of a distribution. Half the observations are smaller than the median and the other half are larger than the median. Here is a rule for finding the median:

1. Arrange all observations in order of size, from smallest to largest. 2. If the number of observations n is odd, the median M is the center observation in the ordered list. Find the

location of the median by counting (n + 1)/2 observations up from the bottom of the list. 3. If the number of observations n is even, the median M is the mean of the two center observations in the

ordered list. The location of the median is again (n + 1)/2 from the bottom of the list.

Note that the formula (n + 1)/2 does not give the median, just the location of the median in the ordered list. Medians require little arithmetic, so they are easy to find by hand for small sets of data. Arranging even a moderate number of observations in order is tedious, however, so that finding the median by hand for larger sets of data is unpleasant. Even simple calculators have an x button, but you will need computer software or a graphing calculator to automate finding the median.

EXAMPLE 1.25

TTS24

Median time to start a business. To find the median time to start a business for our 24 countries, we first arrange the data in order from smallest to largest:

2 4 5 5 5 5 6 6 7 8 10 11 12 13 16 17 19 19 24 25 32 38 49 53

The count of observations n = 24 is even. The median, then, is the average of the two center observations in the ordered list. To find the location of the center observations, we first compute

location of

Therefore, the center observations are the 12 th and 13th observations in the ordered list. The median is

Note that you can use the stemplot in Figure 1.13 directly to compute the median. In the stemplot the cases are already ordered and you simply need to count from the top or the bottom to the desired location.

USE YOUR KNOWLEDGE

1.45

1.46

1.47

TTS25

Include the outlier. Include Suriname, where the start time is 208 days, in the data set, and show that the median is 12 days. Note that with this case included, the sample size is now 25 and the median is the 13 th observation in the ordered list. Write out the ordered list and circle the outlier. Describe the effect of the outlier on the median for this set of data.

CALLS80

Calls to a customer service center. The service times for 80 calls to a customer service center are given in Table 1.2 (page 17). Use these data to compute the median service time.

STAT

Find the median. Here are the scores on the first exam in an introductory statistics course for 10 students:

83 74 93 85 75 97 93 55 92 81

Find the median first-exam score for these students.

Mean versus median Exercises 1.43 and 1.45 illustrate an important difference between the mean and the median. Suriname is an outlier. It pulls the mean time to start a business up from 16 days to 24 days. The median increased slightly, from 11.5 days to 12 days.

The median is more resistant than the mean. If the largest start time in the data set was 1200 days, the median for all 25 countries would still be 12 days. The largest observation just counts as one observation above the center, no matter how far above the center it lies. The mean uses the actual value of each observation and so will chase a single large observation upward.

The best way to compare the response of the mean and median to extreme observations is to use an interactive applet that allows you to place points on a line and then drag them with your computer’s mouse. Exercises 1.83, 1.84, and 1.85 use the Mean and Median applet on the website for this text to compare the mean and the median.

The median and mean are the most common measures of the center of a distribution. The mean and median of a symmetric distribution are close together. If the distribution is exactly symmetric, the mean and median are exactly the same. In a skewed distribution, the mean is farther out in the long tail than is the median.

The endowment for a college or university is money set aside and invested. The income from the endowment is usually used to support various programs. The distribution of the sizes of the endowments of colleges and universities is strongly skewed to the right. Most institutions have modest endowments, but a few are very wealthy. The median endowment of colleges and universities in a recent year was $93 million—but the mean endowment was $498 million.21 The few wealthy institutions pull the mean up but do not affect the median. Don’t confuse the “average” value of a variable (the mean) with its “typical” value, which we might describe by the median.

We can now give a better answer to the question of how to deal with outliers in data. First, look at the data to identify outliers and investigate their causes. You can then correct outliers if they are wrongly recorded, delete them for good reason, or otherwise give them individual attention. The outlier in Example 1.21 (page 21) can be dropped from the data once we discover that it is an error. If you have no clear reason to drop outliers, you may want to use resistant measures in your analysis, so that outliers have little influence over your conclusions. The choice is often a matter for judgment.

Measuring spread: The quartiles A measure of center alone can be misleading. Two countries with the same median family income are very different if one has extremes of wealth and poverty and the other has little variation among families. A drug manufactured with the correct mean concentration of active ingredient is dangerous if some batches are much too high and others much too low.

We are interested in the spread or variability of incomes and drug potencies as well as their centers. The simplest useful numerical description of a distribution consists of both a measure of center and a measure of spread.

quartile

We can describe the spread or variability of a distribution by giving several percentiles. The median divides the data in two; half of the observations are above the median and half are below the median. We could call the median the 50th percentile. The upper quartile is the median of the upper half of the data. Similarly, the lower quartile is the median of the lower half of the data. With the median, the quartiles divide the data into four equal parts; 25% of the data are in each part.

percentile

We can do a similar calculation for any percent. The pth percentile of a distribution is the value that has p percent of the observations fall at or below it. To calculate a percentile, arrange the observations in increasing order and count up the required percent from the bottom of the list.

Our definition of percentiles is a bit inexact because there is not always a value with exactly p percent of the data at or below it. We will be content to take the nearest observation for most percentiles, but the quartiles are important enough to require an exact rule.

THE QUARTILES Q1 AND Q3

To calculate the quartiles:

1. Arrange the observations in increasing order and locate the median M in the ordered list of observations. 2. The first quartile Q1 is the median of the observations whose positions in the ordered list are to the left of

the location of the overall median. 3. The third quartile Q3 is the median of the observations whose positions in the ordered list are to the right

of the location of the overall median.

Here is an example.

EXAMPLE 1.26

TTS24

Finding the quartiles. Here is the ordered list of the times to start a business in our sample of 24 countries:

2 4 5 5 5 5 6 6 7 8 10 11 12 13 16 17 19 19 24 25 32 38 49 53

The count of observations n = 24 is even, so the median is at position (24 + 1)/2 = 12.5, that is, between the 12th and the 13th observation in the ordered list. There are 12 cases above this position and 12 below it. The first quartile is the median of the first 12 observations, and the third quartile is the median of the last 12 observations. Check that Q1 = 5.5 and Q3 = 21.5.

1.48

Notice that the quartiles are resistant. For example, Q3 would have the same value if the highest start time was 530 days rather than 53 days.

Be careful when several observations take the same numerical value. Write down all the observations and apply the rules just as if they all had distinct values.

USE YOUR KNOWLEDGE

STAT

Find the quartiles. Here are the scores on the first exam in an introductory statistics course for 10 students:

83 74 93 85 75 97 93 55 92 81

Find the quartiles for these first-exam scores.

There are several rules for calculating quartiles, which often give slightly different values. The differences are generally small. For describing data, just report the values that your software gives.

1.49

1.50

The five-number summary and boxplots In Section 1.2, we used the smallest and largest observations to indicate the spread of a distribution. These single observations tell us little about the distribution as a whole, but they give information about the tails of the distribution that is missing if we know only Q1, M, and Q3. To get a quick summary of both center and spread, use all five numbers.

THE FIVE-NUMBER SUMMARY

The five-number summary of a set of observations consists of the smallest observation, the first quartile, the median, the third quartile, and the largest observation, written in order from smallest to largest. In symbols, the five-number summary is

Minimum Q1 M Q3 Maximum

EXAMPLE 1.27

CALLS80

Service center call lengths. Table 1.2 (page 17) gives the service center call lengths for the sample of 80 calls that we discussed in Example 1.15. The five-number summary for these data is 1.0, 54.5, 103.5, 200, and 2631. The distribution is highly skewed. The mean is 197 seconds, a value that is very close to the third quartile.

USE YOUR KNOWLEDGE

CALLS80

Verify the calculations. Refer to the five-number summary and the mean for service center call lengths given in Example 1.27. Verify these results. Do not use software for this exercise and be sure to show all your work.

STAT

Find the five-number summary. Here are the scores on the first exam in an introductory statistics course for 10 students:

83 74 93 85 75 97 93 55 92 81

Find the five-number summary for these first-exam scores.

The five-number summary leads to another visual representation of a distribution, the boxplot.

1.51

BOXPLOT

A boxplot is a graph of the five-number summary.

A central box spans the quartiles Q1 and Q3.

A line in the box marks the median M. Lines extend from the box out to the smallest and largest observations.

whiskers box-and-whisker plots

The lines extending to the smallest and largest observations are sometimes called whiskers, and boxplots are sometimes called box-and-whisker plots. Software provides many varieties of boxplots, some of which use different choices for the placement of the whiskers.

When you look at a boxplot, first locate the median, which marks the center of the distribution. Then look at the spread. The quartiles show the spread of the middle half of the data, and the extremes (the smallest and largest observations) show the spread of the entire data set.

EXAMPLE 1.28

IQ

IQ scores. In Example 1.14 (page 14), we used a histogram to examine the distribution of a sample of 60 IQ scores. A boxplot for these data is given in Figure 1.14. Note that the mean is marked with a “+” and appears very close to the median. The two quartiles are each approximately the same distance from the median, and the two whiskers are approximately the same distance from the corresponding quartiles. All these characteristics are consistent with a symmetric distribution, as illustrated by the histogram in Figure 1.7.

FIGURE 1.14 Boxplot for sample of 60 IQ scores, Example 1.28.

USE YOUR KNOWLEDGE

STAT

Make a boxplot. Here are the scores on the first exam in an introductory statistics course for 10 students:

83 74 93 85 75 97 93 55 92 81

Make a boxplot for these first-exam scores.

The 1.5 × IQR rule for suspected outliers If we look at the data in Table 1.2 (page 17), we can spot a clear outlier, a call lasting 2631 seconds, more than twice the length of any other call. How can we describe the spread of this distribution? The smallest and largest observations are extremes that do not describe the spread of the majority of the data. The distance between the quartiles (the range of the center half of the data) is a more resistant measure of spread than the range. This distance is called the interquartile range.

THE INTERQUARTILE RANGE IQR

The interquartile range IQR is the distance between the first and third quartiles, IQR = Q3 − Q1

EXAMPLE 1.29

IQR for service center call length data. In Exercise 1.49 (page 34) you verified that the five-number summary for our data on service center call lengths was 1.0, 54.5, 103.5, 200, and 2631. Therefore, we calculate

The quartiles and the IQR are not affected by changes in either tail of the distribution. They are resistant, therefore, because changes in a few data points have no further effect once these points move outside the quartiles.

However, no single numerical measure of spread, such as IQR, is very useful for describing skewed distributions. The two sides of a skewed distribution have different spreads, so one number can’t summarize them. We can often detect skewness from the five-number summary by comparing how far the first quartile and the minimum are from the median (left tail) with how far the third quartile and the maximum are from the median (right tail). The interquartile range is mainly used as the basis for a rule of thumb for identifying suspected outliers.

THE 1.5 × IQR RULE FOR OUTLIERS

Call an observation a suspected outlier if it falls more than 1.5 × IQR above the third quartile or below the first quartile.

EXAMPLE 1.30

CALLS80

Suspected outliers for call length data. For the call length data in Table 1.2 (page 17), 1.5 × IQR = 1.5 × 145.5 = 218.25

1.52

Any values below 54.5 – 218.25 = −163.75 or above 200 + 218.25 = 418.25 are flagged as possible outliers. There are no low outliers, but the eight longest calls are flagged as possible high outliers. Their lengths are

438 465 479 700 700 951 1148 2631

It is difficult to imagine calls lasting this long.

USE YOUR KNOWLEDGE

STAT

Find the IQR. Here are the scores on the first exam in an introductory statistics course for 10 students:

83 74 93 85 75 97 93 55 92 81

Find the interquartile range and use the 1.5 × IQR rule to check for outliers. How low would the lowest score need to be for it to be an outlier according to this rule?

modified boxplot

Two variations on the basic boxplot can be very useful. The first, called a modified boxplot, uses the 1.5 × IQR rule. The lines that extend out from the quartiles are terminated in whiskers that are 1.5 × IQR in length. Points beyond the whiskers are plotted individually and are classified as outliers according to the 1.5 × IQR rule.

side-by-side boxplots

The other variation is to use two or more boxplots in the same graph to compare groups measured on the same variable. These are called side-by-side boxplots. The following example illustrates these two variations.

EXAMPLE 1.31

POETS

Do poets die young? According to William Butler Yeats, “She is the Gaelic muse, for she gives inspiration to those she persecutes. The Gaelic poets die young, for she is restless, and will not let them remain long on earth.” One study designed to investigate this issue examined the age at death for writers from different cultures and genders.22

Three categories of writers examined were novelists, poets, and nonfiction writers. We examine the ages at death for female writers in these categories from North America. Figure 1.15 shows modified side-by-side boxplots for the three categories of writers.

Displaying the boxplots for the three categories of writers lets us compare the three distributions. We see that nonfiction writers tend to live the longest, followed by novelists. The poets do appear to die young! There is one outlier among the nonfiction writers, which is plotted individually along with the value of its label. This writer died at the age of 40, young for a nonfiction writer, but not for a novelist or a poet!

FIGURE 1.15 Modified side-by-side boxplots for the data on writers’ age at death, for Example 1.31.

Measuring spread: The standard deviation The five-number summary is not the most common numerical description of a distribution. That distinction belongs to the combination of the mean to measure center and the standard deviation to measure spread, or variability. The standard deviation measures spread by looking at how far the observations are from their mean.

THE STANDARD DEVIATION s

The variance s2 of a set of observations is the average of the squares of the deviations of the observations from their mean. In symbols, the variance of n observations x1, x2, . . . , xn is

or, in more compact notation,

The standard deviation s is the square root of the variance s2:

The idea behind the variance and the standard deviation as measures of spread is as follows: The deviations xi − x̅ display the spread of the values xt about their mean x̅. Some of these deviations will be positive and some negative because some of the observations fall on each side of the mean. In fact, the sum of the deviations of the observations from their mean will always be zero. Squaring the deviations makes the negative deviations positive so that observations far from the mean in either direction have large positive squared deviations. The variance is the average squared deviation. Therefore, S2 and s will be large if the observations are widely spread about their mean and small if the observations are all close to the mean.

EXAMPLE 1.32

METABOL

Metabolic rate. A person’s metabolic rate is the rate at which the body consumes energy. Metabolic rate is important in studies of weight gain, dieting, and exercise. Here are the metabolic rates of seven men who took part in a study of dieting. (The units are calories per 24 hours. These are the same calories used to describe the energy content of foods.)

1792 1666 1362 1614 1460 1867 1439

Enter these data into your calculator or software and verify that

x̅ = 1600 calories s = 189.24 calories

Figure 1.16 plots these data as dots on the calorie scale, with their mean marked by an asterisk (*). The arrows mark two of the deviations from the mean. If you were calculating s by hand, you would find the first deviation as

x1 − x̅ = 1792 − 1600 = 192

1.53

FIGURE 1.16 Metabolic rates for seven men, with the mean (*) and the deviations of two observations from the mean, Example 1.32.

Exercise 1.80 asks you to calculate the seven deviations from Example 1.32, square them, and find s2 and s directly from the deviations. Working one or two short examples by hand helps you understand how the standard deviation is obtained. In practice, you will use either software or a calculator that will find s.

USE YOUR KNOWLEDGE

STAT

Find the variance and the standard deviation. Here are the scores on the first exam in an introductory statistics course for 10 students:

83 74 93 85 75 97 93 55 92 81

Find the variance and the standard deviation for these first-exam scores.

The idea of the variance is straightforward: it is the average of the squares of the deviations of the observations from their mean. The details we have just presented, however, raise some questions.

Why do we square the deviations?

First, the sum of the squared deviations of any set of observations from their mean is the smallest that the sum of squared deviations from any number can possibly be. This is not true of the unsquared distances. So squared deviations point to the mean as center in a way that distances do not. Second, the standard deviation turns out to be the natural measure of spread for a particularly important class of symmetric unimodal distributions, the Normal distributions. We will meet the Normal distributions in the next section.

Why do we emphasize the standard deviation rather than the variance?

One reason is that s, not s2, is the natural measure of spread for Normal distributions, which are introduced in the next section.

There is also a more general reason to prefer s to s2. Because the variance involves squaring the deviations, it does not have the same unit of measurement as the original observations. The variance of the metabolic rates, for example, is measured in squared calories. Taking the square root gives us a description of the spread of the distribution in the original measurement units.

Why do we average by dividing by n – 1 rather than n in calculating the variance?

Because the sum of the deviations is always zero, the last deviation can be found once we know the other n − 1. So we are not averaging n unrelated numbers. Only n − 1 of the squared deviations can vary freely, and we average by dividing the total by n − 1.

degrees of freedom

The number n – 1 is called the degrees of freedom of the variance or standard deviation. Many calculators offer a choice between dividing by n and dividing by n − 1, so be sure to use n − 1.

1.54

1.55

Properties of the standard deviation Here are the basic properties of the standard deviation s as a measure of spread.

PROPERTIES OF THE STANDARD DEVIATION

s measures spread about the mean and should be used only when the mean is chosen as the measure of center. s = 0 only when there is no spread. This happens only when all observations have the same value. Otherwise, s > 0. As the observations become more spread out about their mean, s gets larger. s, like the mean x̅, is not resistant. A few outliers can make s very large.

USE YOUR KNOWLEDGE

A standard deviation of zero. Construct a data set with 6 cases that has a variable with s = 0.

The use of squared deviations renders s even more sensitive than x̅ to a few extreme observations. For example, when we add Suriname to our sample of 24 countries for the analysis of the time to start a business (Exercise 1.43, page 29, and Exercise 1.45, page 31), we increase the standard deviation from 14.2 to 40.8! Distributions with outliers and strongly skewed distributions have standard deviations that do not give much helpful information about such distributions.

USE YOUR KNOWLEDGE

Effect of an outlier on the IQR. Find the IQR for the time to start a business with and without Suriname. What do you conclude about the sensitivity of this measure of spread to the inclusion of an outlier?

Choosing measures of center and spread

TTS24, TTS25

How do we choose between the five-number summary and x̅ and s to describe the center and spread of a distribution? Because the two sides of a strongly skewed distribution have different spreads, no single number such as s describes the spread well. The five-number summary, with its two quartiles and two extremes, does a better job.

CHOOSING A SUMMARY

The five-number summary is usually better than the mean and standard deviation for describing a skewed distribution or a distribution with strong outliers. Use x̅ and s for reasonably symmetric distributions that are free of outliers.

Remember that a graph gives the best overall picture of a distribution. Numerical measures of center and spread report specific facts about a distribution, but they do not describe its shape. Numerical summaries do not disclose the presence of multiple modes or gaps, for example. Always plot your data.

EXAMPLE 1.33

TTS24

Results from software. We prefer to examine the numerical summaries and graphical summaries together. Figure 1.17 gives (a) a boxplot, (b) a histogram, and (c) numerical summaries for the time to start a business from Example 1.23 (page 28) using Minitab. Similar displays are given for SPSS in Figure 1.18 (a), (b), and (c) and for JMP in Figure 1.19. Examine and compare the outputs carefully. Notice that they give different numbers of significant digits for some of these numerical summaries. There are also variations in how they make the boxplots and how they define classes for the histograms.

FIGURE 1.17 Graphical and numerical summaries from Minitab: (a) boxplot, (b) histogram, and (c) numerical summaries for the time to start a business, Example 1.33.

FIGURE 1.18 Graphical and numerical summaries from SPSS: (a) boxplot, (b) histogram, and (c) numerical summaries for the time to start a business, Example 1.33.

FIGURE 1.19 Graphical and numerical summaries from JMP for the time to start a business, Example 1.33.

Changing the unit of measurement The same variable can be recorded in different units of measurement. Americans commonly record distances in miles and temperatures in degrees Fahrenheit, while the rest of the world measures distances in kilometers and temperatures in degrees Celsius. Fortunately, it is easy to convert numerical descriptions of a distribution from one unit of measurement to another. This is true because a change in the measurement unit is a linear transformation of the measurements.

LINEAR TRANSFORMATIONS

A linear transformation changes the original variable x into the new variable xnew given by an equation of the form

xnew = a + bx

Adding the constant a shifts all values of x upward or downward by the same amount. In particular, such a shift changes the origin (zero point) of the variable. Multiplying by the positive constant b changes the size of the unit of measurement.

EXAMPLE 1.34

Change the units. (a) If a distance x is measured in kilometers, the same distance in miles is

xnew = 0.62x

For example, a 10-kilometer race covers 6.2 miles. This transformation changes the units without changing the origin— a distance of 0 kilometers is the same as a distance of 0 miles.

(b) A temperature x measured in degrees Fahrenheit must be reexpressed in degrees Celsius to be easily understood by the rest of the world. The transformation is

Thus, the high of 95°F on a hot American summer day translates into 35°C. In this case,

This linear transformation changes both the unit size and the origin of the measurements. The origin in the Celsius scale (0°C, the temperature at which water freezes) is 32° in the Fahrenheit scale.

Linear transformations do not change the shape of a distribution. If measurements on a variable x have a right- skewed distribution, any new variable xnew obtained by a linear transformation xnew = a + bx (for b > 0) will also have a right-skewed distribution. If the distribution of x is symmetric and unimodal, the distribution of xnew remains symmetric and unimodal.

Although a linear transformation preserves the basic shape of a distribution, the center and spread will change. Because linear changes of measurement scale are common, we must be aware of their effect on numerical descriptive measures of center and spread. Fortunately, the changes follow a simple pattern.

EXAMPLE 1.35

Use scores to find the points. In an introductory statistics course, homework counts for 300 points out of a total of 1000 possible points for all course requirements. During the semester, there were 12 homework assignments, and each was given a grade on a scale of 0 to 100. The maximum total score for the 12 homework assignments is therefore 1200. To convert the homework scores to final grade points, we need to convert the scale of 0 to 1200 to a scale of 0 to 300. We do this by multiplying the homework scores by 300/1200. In other words, we divide the

1.56

homework scores by 4. Here are the homework scores and the corresponding final grade points for five students:

Student 1 2 3 4 5 Score 1056 1080 900 1164 1020 Points 264 270 225 291 255

These two sets of numbers measure the same performance on homework for the course. Because we obtained the points by dividing the scores by 4, the mean of the points will be the mean of the scores divided by 4. Similarly, the standard deviation of points will be the standard deviation of the scores divided by 4.

USE YOUR KNOWLEDGE

Calculate the points for a student. Use the setting of Example 1.35 to find the points for a student whose score is 950.

Here is a summary of the rules for linear transformations:

EFFECT OF A LINEAR TRANSFORMATION

To see the effect of a linear transformation on measures of center and spread, apply these rules:

Multiplying each observation by a positive number b multiplies both measures of center (mean and median) and measures of spread (interquartile range and standard deviation) by b. Adding the same number a (either positive or negative) to each observation adds a to measures of center and to quartiles and other percentiles but does not change measures of spread.

In Example 1.35, when we converted from score to points, we described the transformation as dividing by 4. The multiplication part of the summary of the effect of a linear transformation applies to this case because division by 4 is the same as multiplication by 0.25. Similarly, the second part of the summary applies to subtraction as well as addition because subtraction is simply the addition of a negative number.

The measures of spread IQR and s do not change when we add the same number a to all the observations because adding a constant changes the location of the distribution but leaves the spread unaltered. You can find the effect of a linear transformation xnew = a + bx by combining these rules. For example, if x has mean x̅, the transformed variable xnew has mean a + b x̅.

SECTION 1.3 SUMMARY A numerical summary of a distribution should report its center and its spread or variability. The mean x̅ and the median M describe the center of a distribution in different ways. The mean is the arithmetic average of the observations, and the median is their midpoint. When you use the median to describe the center of a distribution, describe its spread by giving the quartiles. The first quartile Q1 has one-fourth of the observations below it, and the third quartile Q3 has three-fourths of the observations below it. The interquartile range is the difference between the quartiles. It is the spread of the center half of the data. The 1.5 × IQR rule flags observations more than 1.5 × IQR beyond the quartiles as possible outliers. The five-number summary consisting of the median, the quartiles, and the smallest and largest individual observations provides a quick overall description of a distribution. The median describes the center, and the quartiles and extremes show the spread. Boxplots based on the five-number summary are useful for comparing several distributions. The box spans the quartiles and shows the spread of the central half of the distribution. The median is marked within the box. Lines extend from the box to the extremes and show the full spread of the data. In a modified boxplot, points identified by the 1.5 × IQR rule are plotted individually. Side-by-side boxplots can be used to display boxplots for more than one group on the same graph. The variance s2 and especially its square root, the standard deviation s, are common measures of spread about the mean as center. The standard deviation s is zero when there is no spread and gets larger as the spread increases. A resistant measure of any aspect of a distribution is relatively unaffected by changes in the numerical value of a small proportion of the total number of observations, no matter how large these changes are. The median and quartiles are resistant, but the mean and the standard deviation are not. The mean and standard deviation are good descriptions for symmetric distributions without outliers. They are most useful for the Normal distributions introduced in the next section. The five-number summary is a better exploratory description for skewed distributions. Linear transformations have the form xnew = a + bx. A linear transformation changes the origin if a ≠ 0 and changes the size of the unit of measurement if b > 0. Linear transformations do not change the overall shape of a distribution. A linear transformation multiplies a measure of spread by b and changes a percentile or measure of center m into a + bm. Numerical measures of particular aspects of a distribution, such as center and spread, do not report the entire shape of most distributions. In some cases, particularly distributions with multiple peaks and gaps, these measures may not be very informative.

SECTION 1.3 EXERCISES For Exercises 1.43 and 1.44, see page 29; for Exercises 1.45 to 1.47, see page 31; for Exercise 1.48, see page 33; for Exercises 1.49 and 1.50, see page 34; for Exercise 1.51, see page 35; for Exercise 1.52, see page 37; for Exercise 1.53, see page 39; for Exercise 1.54, see page 40; for Exercise 1.55, see page 40; and for Exercise 1.56, see page 45.

1.57 Potassium from potatoes. Refer to Exercise 1.30 (page 24) where you examined the potassium absorption of a group of 27 adults who ate a controlled diet that included 40 mEq of potassium from potatoes for five days. KPOT40

(a) Compute the mean for these data.

(b) Compute the median for these data.

(c) Which measure do you prefer for describing the center of this distribution? Explain your answer. (You may include a graphical summary as part of your explanation.)

1.58 Potassium from a supplement. Refer to Exercise 1.31 (page 24) where you examined the potassium absorption of a group of 29 adults who ate a controlled diet that included 40 mEq of potassium from a supplement for five days. KSUP40

(a) Compute the mean for these data.

(b) Compute the median for these data.

(c) Which measure do you prefer for describing the center of this distribution? Explain your answer. (You may include a graphical summary as part of your explanation.)

1.59 Potassium from potatoes. Refer to Exercise 1.30 (page 24) where you examined the potassium absorption of a group of 27 adults who ate a controlled diet that included 40 mEq of potassium from potatoes for five days. KPOT40

(a) Compute the standard deviation for these data.

(b) Compute the quartiles for these data.

(c) Give the five-number summary and explain the meaning of each of the five numbers.

(d) Which numerical summaries do you prefer for describing the distribution, the mean, and the standard deviation of the five- number summary? Explain your answer. (You may include a graphical summary as part of your explanation.)

1.60 Potassium from a supplement. Refer to Exercise 1.31 (page 24) where you examined the potassium absorption of a group of 29 adults who ate a controlled diet that included 40 mEq of potassium from a supplement for five days. KSUP40

(a) Compute the standard deviation for these data.

(b) Compute the quartiles for these data.

(c) Give the five-number summary and explain the meaning of each of the five numbers.

(d) Which numerical summaries do you prefer for describing the distribution, the mean, and the standard deviation of the five- number summary? Explain your answer. (You may include a graphical summary as part of your explanation.)

1.61 Potassium from potatoes. Refer to Exercise 1.30 (page 24) where you examined the potassium absorption of a group of 27 adults who ate a controlled diet that included 40 mEq of potassium from potatoes for five days. In Exercise 1.30, you used a stemplot to examine the distribution of the potassium absorption. KPOT40

(a) Make a histogram and use it to describe the distribution of potassium absorption.

(b) Make a boxplot and use it to describe the distribution of potassium absorption.

(c) Compare the stemplot, the histogram, and the boxplot as graphical summaries of this distribution. Which do you prefer? Give reasons for your answer.

1.62 Potassium from a supplement. Refer to Exercise 1.31 (page 24) where you examined the potassium absorption of a group of 29 adults who ate a controlled diet that included 40 mEq of potassium from a supplement for five days. In Exercise 1.31, you used a stemplot to examine the distribution of the potassium absorption. KSUP40

(a) Make a histogram and use it to describe the distribution of potassium absorption.

(b) Make a boxplot and use it to describe the distribution of potassium absorption.

(c) Compare the stemplot, the histogram, and the boxplot as graphical summaries of this distribution. Which do you prefer? Give reasons for your answer.

1.63 Compare the potatoes with the supplement. Refer to Exercises 1.30 and 1.31 (page 24). Use a back-to-back stemplot to display the data for the two sources of potassium. Use the stemplot to compare the two distributions and write a short summary of your findings. KPS40

1.64 Potassium sources. Refer to Exercises 1.30 and 1.31 (page 24). Use side-by-side boxplots in to describe the distributions. KPS40

(a) Summarize what you see in the boxplots and compare it with what you saw in the stemplots.

(b) For comparing these two distributions, do you prefer back-to-back stemplots or side-by-side boxplots? Give reasons for your answer.

1.65 Gosset’s data on double stout sales. William Sealy Gosset worked at the Guinness Brewery in Dublin and made substantial contributions to the practice of statistics.23 In his work at the brewery, he collected and analyzed a great deal of data. Archives with Gosset’s handwritten tables, graphs, and notes have been preserved at the Guinness Storehouse in Dublin.24 In one study, Gosset examined the change in the double stout market before and after World War I (1914–1918). For various regions in England and Scotland, he calculated the ratio of sales in 1925, after the war, as a percent of sales in 1913, before the war. Here are the data: STOUT

Bristol 94 Cardiff 112 English Agents 78 English O 68 English P 46 English R 111 Glasgow 66 Liverpool 140 London 428 Manchester 190 Newcastle-on-Tyne 118 Scottish 24

(a) Compute the mean for these data.

(b) Compute the median for these data.

(c) Which measure do you prefer for describing the center of this distribution? Explain your answer. (You may include a graphical summary as part of your explanation.)

1.66 Measures of spread for the double stout data. Refer to the previous exercise. STOUT

(a) Compute the standard deviation for these data.

(b) Compute the quartiles for these data.

(c) Which measure do you prefer for describing the spread of this distribution? Explain your answer. (You may include a graphical summary as part of your explanation.)

1.67 Are there outliers in the double stout data? Refer to the previous two exercises. STOUT

(a) Find the IQR for these data.

(b) Use the 1.5 × IQR rule to identify and name any outliers.

(c) Make a boxplot for these data and describe the distribution using only the information in the boxplot.

(d) Make a modified boxplot for these data and describe the distribution using only the information in the boxplot.

(e) Make a stemplot for these data.

(f) Compare the boxplot, the modified boxplot, and the stemplot. Evaluate the advantages and disadvantages of each graphical summary for describing the distribution of the double stout data.

1.68 Smolts. Smolts are young salmon at a stage when their skin becomes covered with silvery scales and they start to migrate from freshwater to the sea. The reflectance of a light shined on a smolt’s skin is a measure of the smolt’s readiness for the migration. Here are the reflectances, in percents, for a sample of 50 smolts:25 SMOLTS

57.6 54.8 63.4 57.0 54.7 42.3 63.6 55.5 33.5 63.3 58.3 42.1 56.1 47.8 56.1 55.9 38.8 49.7 42.3 45.6 69.0 50.4 53.0 38.3 60.4 49.3 42.8 44.5 46.4 44.3 58.9 42.1 47.6 47.9 69.2 46.6 68.1 42.8 45.6 47.3 59.6 37.8 53.9 43.2 51.4 64.5 43.8 42.7 50.9 43.8

(a) Find the mean reflectance for these smolts.

(b) Find the median reflectance for these smolts.

(c) Do you prefer the mean or the median as a measure of center for these data? Give reasons for your preference.

1.69 Measures of spread for smolts. Refer to the previous exercise. SMOLTS

(a) Find the standard deviation of the reflectance for these smolts.

(b) Find the quartiles of the reflectance for these smolts.

(c) Compute the quartiles for these data.

(d) Do you prefer the standard deviation or the quartiles as a measure of spread for these data? Give reasons for your preference.

1.70 Are there outliers in the smolt data? Refer to the previous two exercises. SMOLTS

(a) Find the IQR for the smolt data.

(b) Use the 1.5 × IQR rule to identify any outliers.

(c) Make a boxplot for the smolt data and describe the distribution using only the information in the boxplot.

(d) Make a modified boxplot for these data and describe the distribution using only the information in the boxplot.

(e) Make a stemplot for these data.

(f) Compare the boxplot, the modified boxplot, and the stemplot. Evaluate the advantages and disadvantages of each graphical summary for describing the distribution of the smolt reflectance data.

1.71 Potatoes. A quality product is one that is consistent and has very little variability in its characteristics. Controlling variability can be more difficult with agricultural products than with those that are manufactured. The following table gives the weights, in ounces, of the 25 potatoes sold in a 10-pound bag: POTATO

7.6 7.9 8.0 6.9 6.7 7.9 7.9 7.9 7.6 7.8 7.0 4.7 7.6 6.3 4.7 4.7 4.7 6.3 6.0 5.3 4.3 7.9 5.2 6.0 3.7

(a) Summarize the data graphically and numerically. Give reasons for the methods you chose to use in your summaries.

(b) Do you think that your numerical summaries do an effective job of describing these data? Why or why not?

(c) There appear to be two distinct clusters of weights for these potatoes. Divide the sample into two subsamples based on the clustering. Give the mean and standard deviation for each subsample. Do you think that this way of summarizing these data is better than a numerical summary that uses all the data as a single sample? Give a reason for your answer.

1.72 The alcohol content of beer. Brewing beer involves a variety of steps that can affect the alcohol content. A website

gives the percent alcohol for 159 domestic brands of beer.26 BEER

(a) Use graphical and numerical summaries of your choice to describe the data. Give reasons for your choice.

(b) The data set contains an outlier. Explain why this particular beer is unusual.

(c) For the outlier, give a short description of how you think this particular beer should be marketed.

1.73 Outlier for alcohol content of beer. Refer to the previous exercise. BEER

(a) Calculate the mean with and without the outlier. Do the same for the median. Explain how these values change when the outliers is excluded.

(b) Calculate the standard deviation with and without the outlier. Do the same for the quartiles. Explain how these values change when the outlier is excluded.

(c) Write a short paragraph summarizing what you have learned in this exercise.

1.74 Calories in beer. Refer to the previous two exercises. The data set also lists calories per 12 ounces of beverage. BEER

(a) Analyze the data and summarize the distribution of calories for these 159 brands of beer.

(b) In the previous exercise, you identified one brand of beer as an outlier. To what extent is this brand an outlier in the distribution of calories? Explain your answer.

(c) Does the distribution of calories suggest marketing strategies for this brand of beer? Describe some marketing strategies.

1.75 Median versus mean for net worth. A report on the assets of American households says that the median net worth of U.S. families is $81,200. The mean net worth of these families is $534,600.27 What explains the difference between these two measures of center?

1.76 Create a data set. Create a data set with seven observations for which the median would change by a large amount if the smallest observation were deleted.

1.77 Mean versus median. A small accounting firm pays each of its seven clerks $55,000, three junior accountants $80,000 each, and the firm’s owner $650,000. What is the mean salary paid at this firm? How many of the employees earn less than the mean? What is the median salary?

1.78 Be careful about how you treat the zeros. In computing the median income of any group, some federal agencies omit all members of the group who had no income. Give an example to show that the reported median income of a group can go down even though the group becomes economically better off. Is this also true of the mean income?

1.79 How does the median change? The firm in Exercise 1.77 gives no raises to the clerks and junior accountants, while the owner’s take increases to $900,000. How does this change affect the mean? How does it affect the median?

1.80 Metabolic rates. Calculate the mean and standard deviation of the metabolic rates in Example 1.32 (page 38), showing each step in detail. First find the mean x̅ by summing the seven observations and dividing by 7. Then find each of the deviations xi – x̅ and their squares. Check that the deviations have sum 0. Calculate the variance as an average of the squared deviations (remember to divide by n – 1). Finally, obtain s as the square root of the variance. METABOL

1.81 Earthquakes. Each year there are about 900,000 earthquakes of magnitude 2.5 or less that are usually not felt. In contrast, there are about 10 of magnitude 7.0 that cause serious damage.28 Explain why the average magnitude of

earthquakes is not a good measure of their impact. IQ

1.82 IQ scores. Many standard statistical methods that you will study in Part II of this book are intended for use with distributions that are symmetric and have no outliers. These methods start with the mean and standard deviation, x̅ and s. For example, standard methods would typically be used for the IQ and GPA data in Table 1.3 (page 26). IQ

(a) Find x̅ and s for the IQ data. In large populations, IQ scores are standardized to have mean 100 and standard deviation 15. In what way does the distribution of IQ among these students differ from the overall population?

(b) Find the median IQ score. It is, as we expect, close to the mean.

(c) Find the mean and median for the GPA data. The two measures of center differ a bit. What feature of the data (see your stemplot in Exercise 1.39 or make a new stemplot) explains the difference?

1.83 Mean and median for two observations. The Mean and Median applet allows you to place observations on a line and see their mean and median visually. Place two observations on the line by clicking below it. Why does only one arrow appear?

1.84 Mean and median for four observations. In the Mean and Median applet, place four observations on the line by clicking below it, three close together near the center of the line and one somewhat to the right of these two.

(a) Pull the single rightmost observation out to the right. (Place the cursor on the point, hold down a mouse button, and drag the point.) How does the mean behave? How does the median behave? Explain briefly why each measure acts as it does.

(b) Now drag the rightmost point to the left as far as you can. What happens to the mean? What happens to the median as you drag this point past the other two (watch carefully)?

1.85 Mean and median for seven observations. Place seven observations on the line in the Mean and Median applet by clicking below it.

(a) Add one additional observation without changing the median. Where is your new point?

(b) Use the applet to convince yourself that when you add yet another observation (there are now nine in all), the median does not change no matter where you put the seventh point. Explain why this must be true.

1.86 Imputation. Various problems with data collection can cause some observations to be missing. Suppose a data set has 20 cases. Here are the values of the variable x for 10 of these cases: IMPUTE

17 6 12 14 20 23 9 12 16 21

The values for the other 10 cases are missing. One way to deal with missing data is called imputation. The basic idea is that missing values are replaced, or imputed, with values that are based on an analysis of the data that are not missing. For a data set with a single variable, the usual choice of a value for imputation is the mean of the values that are not missing. The mean for this data set is 15.

(a) Verify that the mean is 15 and find the standard deviation for the 10 cases for which x is not missing.

(b) Create a new data set with 20 cases by setting the values for the 10 missing cases to 15. Compute the mean and standard deviation for this data set.

(c) Summarize what you have learned about the possible effects of this type of imputation on the mean and the standard deviation.

1.87 A standard deviation contest. This is a standard deviation contest. You must choose four numbers from the whole numbers 10 to 20, with repeats allowed.

(a) Choose four numbers that have the smallest possible standard deviation.

(b) Choose four numbers that have the largest possible standard deviation.

(c) Is more than one choice possible in either part (a) or part (b)? Explain.

1.88 Longleaf pine trees. The Wade Tract in Thomas County, Georgia, is an old-growth forest of longleaf pine trees (Pinus palustris) that has survived in a relatively undisturbed state since before the settlement of the area by Europeans. A study collected data on 584 of these trees.29 One of the variables measured was the diameter at breast height (DBH). This is the diameter of the tree at 4.5 feet and the units are centimeters (cm). Only trees with DBH greater than 1.5 cm were sampled. Here are the diameters of a random sample of 40 of these trees: PINES

10.5 13.3 26.0 18.3 52.2 9.2 26.1 17.6 40.5 31.8 47.2 11.4 2.7 69.3 44.4 16.9 35.7 5.4 44.2 2.2 4.3 7.8 38.1 2.2 11.4 51.5 4.9 39.7 32.6 51.8 43.6 2.3 44.6 31.5 40.3 22.3 43.3 37.5 29.1 27.9

(a) Find the five-number summary for these data.

(b) Make a boxplot.

(c) Make a histogram.

(d) Write a short summary of the major features of this distribution. Do you prefer the boxplot or the histogram for these data?

1.89 Weight gain. A study of diet and weight gain deliberately overfed 15 volunteers for eight weeks. The mean increase in fat was x̅ = 2.41 kilograms, and the standard deviation was s = 1.25 kilograms. What are x̅ and s in pounds? (A kilogram is 2.2 pounds.)

1.90 Changing units from inches to centimeters. Changing the unit of length from inches to centimeters multiplies each length by 2.54 because there are 2.54 centimeters in an inch. This change of units multiplies our usual

measures of spread by 2.54. This is true of IQR and the standard deviation. What happens to the variance when we change units in this way?

1.91 A different type of mean. The trimmed mean is a measure of center that is more resistant than the mean but uses more of the available information than the median. To compute the 10% trimmed mean, discard the highest 10% and the lowest 10% of the observations and compute the mean of the remaining 80%. Trimming eliminates the effect of a small number of outliers. Compute the 10% trimmed mean of the service time data in Table 1.2 (page 17). Then compute the 20% trimmed mean. Compare the values of these measures with the median and the ordinary untrimmed mean.

1.92 Changing units from centimeters to inches. Refer to Exercise 1.88 (page 50). Change the measurements from centimeters to inches by multiplying each value by 0.39. Answer the questions from that exercise and explain the

effect of the transformation on these data.

1.4 Density Curves and Normal Distributions

When you complete this section, you will be able to:

Compare the mean and the median for symmetric and skewed distributions. Sketch a Normal distribution for any given mean and standard deviation. Apply the 68–95–99.7 rule to find proportions of observations within one, two, and three standard deviations of the mean for any Normal distribution. Transform values of a variable from a general Normal distribution to the standard Normal distribution. Compute areas under a Normal curve using software or Table A. Perform inverse Normal calculations to find values of a Normal variable corresponding to various areas. Assess the extent to which the distribution of a set of data can be approximated by a Normal distribution.

We now have a kit of graphical and numerical tools for describing distributions. What is more, we have a clear strategy for exploring data on a single quantitative variable:

1. Always plot your data: make a graph, usually a stemplot or a histogram. 2. Look for the overall pattern and for striking deviations such as outliers. 3. Calculate an appropriate numerical summary to briefly describe center and spread.

density curves

Technology has expanded the set of graphs that we can choose for Step 1. It is possible, though painful, to make histograms by hand. Using software, clever algorithms can describe a distribution in a way that is not feasible by hand, by fitting a smooth curve to the data in addition to or instead of a histogram. The curves used are called density curves. Before we examine density curves in detail, here is an example of what software can do.

EXAMPLE 1.36

TTS

Density curves for times to start a business and Titanic passenger ages. Figure 1.20 illustrates the use of density curves along with histograms to describe distributions. Figure 1.20(a) shows the distribution of the times to start a business for 189 countries (see Example 1.23. page 28). The outlier, Suriname, described in Exercise 1.43 (page 29) has been deleted from the data set. The distribution is highly skewed to the right. Most of the data are in the first two classes, with 40 or fewer days to start a business.

TITANIC

Exercise 1.27 (page 24) describes data on the class of the ticket of the Titanic passengers, and Figure 1.20(b) shows the distribution of the ages of these passengers. It has a single mode, a long right tail, and a relatively short left tail.

FIGURE 1.20 (a) The distribution of the time to start a business, Example 1.36. The distribution is pictured with both a histogram and a density curve. (b) The distribution of the ages of the Titanic passengers, Example 1.36. These distributions have a single mode with tails of two different lengths.

FIGURE 1.21 (a) The distribution of Iowa Test vocabulary scores for Gary, Indiana, seventh-graders, Example 1.37. The shaded bars in the histogram represent scores less than or equal to 6.0. (b) The shaded area under the Normal density curve also represents scores less than or equal to 6.0. This area is 0.293, close to the true 0.303 for the actual data.

A smooth density curve is an idealization that gives the overall pattern of the data but ignores minor irregularities. We first discuss density curves in general and then focus on a special class of density curves, the bell-shaped Normal curves.

Density curves One way to think of a density curve is as a smooth approximation to the irregular bars of a histogram. Figure 1.21 shows a histogram of the scores of all 947 seventh-grade students in Gary, Indiana, on the vocabulary part of the Iowa Test of Basic Skills. Scores of many students on this national test have a very regular distribution. The histogram is symmetric, and both tails fall off quite smoothly from a single center peak. There are no large gaps or obvious outliers. The curve drawn through the tops of the histogram bars in Figure 1.21 is a good description of the overall pattern of the data.

EXAMPLE 1.37

Vocabulary scores. In a histogram, the areas of the bars represent either counts or proportions of the observations. In Figure 1.21(a), we shaded the bars that represent students with vocabulary scores 6.0 or lower. There are 287 such students, who make up the proportion 287/947 = 0.303 of all Gary seventh-graders. The shaded bars in Figure 1.21(a) make up proportion 0.303 of the total area under all the bars. If we adjust the scale so that the total area of the bars is 1, the area of the shaded bars will also be 0.303.

In Figure 1.21(b), we shaded the area under the curve to the left of 6.0. If we adjust the scale so that the total area under the curve is exactly 1, areas under the curve will then represent proportions of the observations. That is, area = proportion. The curve is then a density curve. The shaded area under the density curve in Figure 1.21(b) represents the proportion of students with score 6.0 or lower. This area is 0.293, only 0.010 away from the histogram result. You can see that areas under the density curve give quite good approximations of areas given by the histogram.

DENSITY CURVE

A density curve is a curve that

Is always on or above the horizontal axis. Has area exactly 1 underneath it.

A density curve describes the overall pattern of a distribution. The area under the curve and above any range of values is the proportion of all observations that fall in that range.

The density curve in Figure 1.21 is a Normal curve. Density curves, like distributions, come in many shapes. Figure 1.22 shows two density curves, a symmetric Normal density curve and a right-skewed curve.

We will discuss Normal density curves in detail in this section because of the important role that they play in statistics. There are, however, many applications where the use of other families of density curves are essential.

A density curve of an appropriate shape is often an adequate description of the overall pattern of a distribution. Outliers, which are deviations from the overall pattern, are not described by the curve.

Measuring center and spread for density curves Our measures of center and spread apply to density curves as well as to actual sets of observations, but only some of these measures are easily seen from the curve. A mode of a distribution described by a density curve is a peak point of the curve, the location where the curve is highest. Because areas under a density curve represent proportions of the observations, the median is the point with half the total area on each side. You can roughly locate the quartiles by dividing the area under the curve into quarters as accurately as possible by eye. The IQR is the distance between the first and third quartiles. There are mathematical ways of calculating areas under curves. These allow us to locate the median and quartiles exactly on any density curve.

FIGURE 1.22 (a) A symmetric Normal density curve with its mean and median marked. (b) a right-skewed density curve with its mean and median marked.

FIGURE 1.23 The a density curve is the point at which it would balance.

What about the mean and standard deviation? The mean of a set of observations is their arithmetic average. If we think of the observations as weights strung out along a thin rod, the mean is the point at which the rod would balance. This fact is also true of density curves. The mean is the point at which the curve would balance if it were made out of solid material. Figure 1.23 illustrates this interpretation of the mean.

A symmetric curve, such as the Normal curve in Figure 1.22(a), balances at its center of symmetry. Half the area under a symmetric curve lies on either side of its center, so this is also the median.

For a right-skewed curve, such as those shown in Figures 1.22(b) and 1.23, the small area in the long right tail tips the curve more than the same area near the center. The mean (the balance point), therefore, lies to the right of the median. It is hard to locate the balance point by eye on a skewed curve. There are mathematical ways of calculating the mean for any density curve, so we are able to mark the mean as well as the median in Figure 1.22(b). The standard deviation can also be calculated mathematically, but it can’t be located by eye on most density curves.

MEDIAN AND MEAN OF A DENSITY CURVE

The median of a density curve is the equal-areas point, the point that divides the area under the curve in half.

mean µ standard deviation σ

The mean of a density curve is the balance point, at which the curve would balance if made of solid material.

The median and mean are the same for a symmetric density curve. They both lie at the center of the curve. The mean of a skewed curve is pulled away from the median in the direction of the long tail.

A density curve is an idealized description of a distribution of data. For example, the density curve in Figure 1.21 is exactly symmetric, but the histogram of vocabulary scores is only approximately symmetric. We therefore need to distinguish between the mean and standard deviation of the density curve and the numbers x̅ and s computed from the actual observations. The usual notation for the mean of an idealized distribution is µ (the Greek letter mu). We write the standard deviation of a density curve as σ (the Greek letter sigma). In Chapter 5, we refer to x̅ and s as statistics associated with a sample and to µ and σ as parameters associated with a population.

Normal distributions Normal curves

Normal distributions

One particularly important class of density curves has already appeared in Figures 1.21 and 1.22(a). These density curves are symmetric, unimodal, and bell-shaped. They are called Normal curves, and they describe Normal distributions. All Normal distributions have the same overall shape.

The exact density curve for a particular Normal distribution is specified by giving the distribution’s mean µ and its standard deviation σ. The mean is located at the center of the symmetric curve and is the same as the median. Changing µ without changing σ moves the Normal curve along the horizontal axis without changing its spread.

The standard deviation s controls the spread of a Normal curve. Figure 1.24 shows two Normal curves with different values of σ. The curve with the larger standard deviation is more spread out.

The standard deviation σ is the natural measure of spread for Normal distributions. Not only do µ and σ completely determine the shape of a Normal curve, but we can locate σ by eye on the curve. Here’s how. As we move out in either direction from the center µ, the curve changes from falling ever more steeply

The points at which this change of curvature takes place are located at distance σ on either side of the mean µ. You can feel the change as you run your finger along a Normal curve, and so find the standard deviation. Remember that µ and σ alone do not specify the shape of most distributions, and that the shape of density curves in general does not reveal σ. These are special properties of Normal distributions.

FIGURE 1.24 Two Normal curves, showing the mean µ and the standard deviation σ.

There are other symmetric bell-shaped density curves that are not Normal. The Normal density curves are specified by a particular equation. The height of the density curve at any point x is given by

We will not make direct use of this fact, although it is the basis of mathematical work with Normal distributions. Notice that the equation of the curve is completely determined by the mean µ and the standard deviation σ.

Why are the Normal distributions important in statistics? Here are three reasons.

1. Normal distributions are good descriptions for some distributions of real data. Distributions that are often close to Normal include scores on tests taken by many people (such as the Iowa Test of Figure 1.21, page 53), repeated careful measurements of the same quantity, and characteristics of biological populations (such as lengths of baby pythons and yields of corn).

2. Normal distributions are good approximations to the results of many kinds of chance outcomes, such as tossing a coin many times.

3. Many statistical inference procedures based on Normal distributions work well for other roughly symmetric distributions.

However, even though many sets of data follow a Normal distribution, many do not. Most income distributions, for example, are skewed to the right and so are not Normal. Non-Normal data, like nonnormal people, not only are common but are also sometimes more interesting than their Normal counterparts.

The 68–95–99.7 rule Although there are many Normal curves, they all have common properties. Here is one of the most important.

THE 68–95–99.7 RULE

In the Normal distribution with mean µ and standard deviation σ:

Approximately 68% of the observations fall within σ of the mean µ. Approximately 95% of the observations fall within 2σ of µ. Approximately 99.7% of the observations fall within 3σ of µ.

Figure 1.25 illustrates the 68–95–99.7 rule. By remembering these three numbers, you can think about Normal distributions without constantly making detailed calculations.

FIGURE 1.25 The 68–95–99.7 rule for Normal distributions.

EXAMPLE 1.38

Heights of young women. The distribution of heights of young women aged 18 to 24 is approximately Normal with mean µ = 64.5 inches and standard deviation σ = 2.5 inches. Figure 1.26 shows what the 68–95–99.7 rule says about this distribution.

Two standard deviations equals five inches for this distribution. The 95 part of the 68–95–99.7 rule says that the middle 95% of young women are between 64.5 – 5 and 64.5 + 5 inches tall, that is, between 59.5 and 69.5 inches. This fact is exactly true for an exactly Normal distribution. It is approximately true for the heights of young women because the distribution of heights is approximately Normal.

The other 5% of young women have heights outside the range from 59.5 to 69.5 inches. Because the Normal distributions are symmetric, half of these women are on the tall side. So the tallest 2.5% of young women are taller than 69.5 inches.

1.93

1.94

FIGURE 1.26 The 68–95–99.7 rule applied to the heights of young women, Example 1.38.

N(µ, σ)

Because we will mention Normal distributions often, a short notation is helpful. We abbreviate the Normal distribution with mean µ and standard deviation σ as N(µ, σ). For example, the distribution of young women’s heights is N(64.5, 2.5).

USE YOUR KNOWLEDGE

Test scores. Many states assess the skills of their students in various grades. One program that is available for this purpose is the National Assessment of Educational Progress (NAEP).30 One of the tests provided by the NAEP assesses the reading skills of 12th-grade students. In a recent year, the national mean score was 288 and the standard deviation was 38. Assuming that these scores are approximately Normally distributed, N(288, 38), use the 68–95–99.7 rule to give a range of scores that includes 95% of these students.

Use the 68–95–99.7 rule. Refer to the previous exercise. Use the 68–95–99.7 rule to give a range of scores that includes 99.7% of these students.

1.95

1.96

Standardizing observations As the 68–95–99.7 rule suggests, all Normal distributions share many properties. In fact, all Normal distributions are the same if we measure in units of size σ about the mean µ as center. Changing to these units is called standardizing. To standardize a value, subtract the mean of the distribution and then divide by the standard deviation.

STANDARDIZING AND z-SCORES

If x is an observation from a distribution that has mean µ and standard deviation σ, the standardized value of x is

A standardized value is often called a z-score.

A z-score tells us how many standard deviations the original observation falls away from the mean, and in which direction. Observations larger than the mean are positive when standardized, and observations smaller than the mean are negative.

To compare scores based on different measures, z-scores can be very useful. For example, see Exercise 1.124 (page 73), where you are asked to compare an SAT score with an ACT score.

EXAMPLE 1.39

Find some z-scores. The heights of young women are approximately Normal with µ = 64.5 inches and σ = 2.5 inches. The z-score for height is

A woman’s standardized height is the number of standard deviations by which her height differs from the mean height of all young women. A woman 68 inches tall, for example, has z-score

or 1.4 standard deviations above the mean. Similarly, a woman 5 feet (60 inches) tall has z-score

or 1.8 standard deviations less than the mean height.

USE YOUR KNOWLEDGE

Find the z-score. Consider the NAEP scores (see Exercise 1.93, page 59), which we assume are approximately Normal, N(288, 38). Give the z-score for a student who received a score of 350.

Find another z-score. Consider the NAEP scores, which we assume are approximately Normal, N(288, 38). Give the z-score for a student who received a score of 240. Explain why your answer is negative even though all the test scores are positive.

We need a way to write variables, such as “height” in Example 1.38, that follow a theoretical distribution such as a Normal distribution. We use capital letters near the end of the alphabet for such variables. If X is the height of a young woman, we can then shorten “the height of a young woman is less than 68 inches” to “X < 68.” We will use lowercase x to stand for any specific value of the variable X.

We often standardize observations from symmetric distributions to express them in a common scale. We might, for example, compare the heights of two children of different ages by calculating their z-scores. The standardized heights tell us where each child stands in the distribution for his or her age group.

Standardizing is a linear transformation that transforms the data into the standard scale of z-scores. We know that a

linear transformation does not change the shape of a distribution, and that the mean and standard deviation change in a simple manner. In particular, the standardized values for any distribution always have mean 0 and standard deviation 1.

If the variable we standardize has a Normal distribution, standardizing does more than give a common scale. It makes all Normal distributions into a single distribution, and this distribution is still Normal. Standardizing a variable that has any Normal distribution produces a new variable that has the standard Normal distribution.

THE STANDARD NORMAL DISTRIBUTION

The standard Normal distribution is the Normal distribution N(0, 1) with mean 0 and standard deviation 1.

If a variable X has any Normal distribution N(µ, σ) with mean µ and standard deviation σ, then the standardized variable

has the standard Normal distribution.

FIGURE 1.27 The cumulative proportion for a value x is the proportion of all observations from the distribution that are less than or equal to x. This is the area to the left of x under the Normal curve.

Normal distribution calculations cumulative proportion

Areas under a Normal curve represent proportions of observations from that Normal distribution. There is no formula for areas under a Normal curve. Calculations use either software that calculates areas or a table of areas. The table and most software calculate one kind of area: cumulative proportions. A cumulative proportion is the proportion of observations in a distribution that lie at or below a given value. When the distribution is given by a density curve, the cumulative proportion is the area under the curve to the left of a given value. Figure 1.27 shows the idea more clearly than words do.

The key to calculating Normal proportions is to match the area you want with areas that represent cumulative proportions. Then get areas for cumulative proportions either from software or (with an extra step) from a table. The following examples show the method in pictures.

EXAMPLE 1.40

NCAA eligibility for competition. To be eligible to compete in their first year of college, the National Collegiate Athletic Association (NCAA) requires Division I athletes to meet certain academic standards. These are based on their grade point average (GPA) in certain courses and combined scores on the SAT Critical Reading and Mathematics sections or the ACT composite score.31

For a student with a 3.0 GPA, the combined SAT score must be 800 or higher. Based on the distribution of SAT scores for college-bound students, we assume that the distribution of the combined Critical Reading and Mathematics scores is approximately Normal with mean 1010 and standard deviation 225.32 What proportion of college-bound students have SAT scores of 800 or more?

Here is the calculation in pictures: the proportion of scores above 800 is the area under the curve to the right of 800. That’s the total area under the curve (which is always 1) minus the cumulative proportion up to 800.

That is, the proportion of college-bound SAT takers with a 3.0 GPA who are eligible to compete is 0.8247, or about 82%.

There is no area under a smooth curve that is exactly over the point 800. Consequently, the area to the right of 800 (the proportion of scores > 800) is the same as the area at or to the right of this point (the proportion of scores ≥ 800). The actual data may contain a student who scored exactly 800 on the SAT. That the proportion of scores exactly equal to 800 is 0 for a Normal distribution is a consequence of the idealized smoothing of Normal distributions for data.

EXAMPLE 1.41

NCAA eligibility for aid and practice. The NCAA has a category of eligibility in which a first-year student may not compete but is still eligible to receive an athletic scholarship and to practice with the team. The requirements for this category are a 3.0 GPA and combined SAT Critical Reading and Mathematics scores of at least 620.

What proportion of college-bound students who take the SAT would be eligible to receive an athletic scholarship and to practice with the team but would not be eligible to compete? That is, what proportion have scores between 620 and 800? Here are the pictures:

About 13% of college-bound students with a 3.0 GPA have SAT scores between 620 and 800.

How do we find the numerical values of the areas in Examples 1.40 and 1.41? If you use software, just plug in mean 1010 and standard deviation 225. Then ask for the cumulative proportions for 800 and for 620. (Your software will probably refer to these as “cumulative probabilities.” We will learn in Chapter 4 why the language of probability fits.) Sketches of the areas that you want similar to the ones in Examples 1.40 and 1.41 are very helpful in making sure that you are doing the correct calculations.

You can use the Normal Curve applet on the text website to find Normal proportions. The applet is more flexible than most software—it will find any Normal proportion, not just cumulative proportions. The applet is an excellent way to understand Normal curves. But, because of the limitations of web browsers, the applet is not as accurate as statistical software.

If you are not using software, you can find cumulative proportions for Normal curves from a table. That requires an extra step, as we now explain.

Using the standard Normal table The extra step in finding cumulative proportions from a table is that we must first standardize to express the problem in the standard scale of z-scores. This allows us to get by with just one table, a table of standard Normal cumulative proportions. Table A in the back of the book gives standard Normal probabilities. The picture at the top of the table reminds us that the entries are cumulative proportions, areas under the curve to the left of a value z.

EXAMPLE 1.42

Find the proportion from z. What proportion of observations on a standard Normal variable Z take values less than 1.47? We need to find the area to the left of 1.47; locate 1.4 in the left-hand column of Table A and then locate the remaining digit 7 as .07 in the top row. The entry opposite 1.4 and under .07 is 0.9292. This is the cumulative proportion we seek. Figure 1.28 illustrates this area.

FIGURE 1.28 The area under a standard Normal curve to the left of the point z = 1.47 is 0.9292, Example 1.42.

Now that you see how Table A works, let’s redo the NCAA Examples 1.40 and 1.41 using the table.

EXAMPLE 1.43

Find the proportion from x. What proportion of college-bound students who take the SAT have scores of at least 800? The picture that leads to the answer is exactly the same as in Example 1.40. The extra step is that we first standardize to read cumulative proportions from Table A. If X is SAT score, we want the proportion of students for which X ≥ x where x = 800.

1. Standardize. Subtract the mean, then divide by the standard deviation, to transform the problem about X into a problem about a standard Normal Z:

2. Use the table. Look at the pictures in Example 1.40. From Table A, we see that the proportion of observations less than −0.93 is 0.1762. The area to the right of −0.93 is therefore 1 – 0.1762 = 0.8238. This is about 82%.

The area from the table in Example 1.43 (0.8238) is slightly less accurate than the area from software in Example 1.40 (0.8247) because we must round z to two places when we use Table A. The difference is rarely important in practice.

1.97

1.98

EXAMPLE 1.44

Eligibility for aid and practice. What proportion of all students who take the SAT would be eligible to receive athletic scholarships and to practice with the team but would not be eligible to compete in the eyes of the NCAA? That is, what proportion of students have SAT scores between 620 and 800? First, sketch the areas, exactly as in Example 1.41. We again use X as shorthand for an SAT score.

1. Standardize.

2. Use the table.

As in Example 1.41, about 13% of students would be eligible to receive athletic scholarships and to practice with the team.

Sometimes we encounter a value of z more extreme than those appearing in Table A. For example, the area to the left of z = −4 is not given in the table. The z-values in Table A leave only area 0.0002 in each tail unaccounted for. For practical purposes, we can act as if there is zero area outside the range of Table A.

USE YOUR KNOWLEDGE

Find the proportion. Consider the NAEP scores, which are approximately Normal, N(288, 38). Find the proportion of students who have scores less than 350. Find the proportion of students who have scores greater than or equal to 350. Sketch the relationship between these two calculations using pictures of Normal curves similar to the ones given in Example 1.40 (page 61).

Find another proportion. Consider the NAEP scores, which are approximately Normal, N(288, 38). Find the proportion of students who have scores between 300 and 350. Use pictures of Normal curves similar to the ones given in Example 1.41 (page 62) to illustrate your calculations.

1.99

Inverse Normal calculations Examples 1.40 to 1.44 illustrate the use of Normal distributions to find the proportion of observations in a given event, such as “SAT score between 620 and 800.” We may instead want to find the observed value corresponding to a given proportion.

Statistical software will do this directly. Without software, use Table A backward, finding the desired proportion in the body of the table and then reading the corresponding z from the left column and top row.

EXAMPLE 1.45

How high for the top 10%? Scores for college-bound students on the SAT Critical Reading test in recent years follow approximately the N(500, 120) distribution.33 How high must a student score to place in the top 10% of all students taking the SAT?

Again, the key to the problem is to draw a picture. Figure 1.29 shows that we want the score x with an area of 0.10 above it. That’s the same as area below x equal to 0.90.

FIGURE 1.29 Locating the point on a Normal curve with area 0.10 to its right, Example 1.45.

Statistical software has a function that will give you the x for any cumulative proportion you specify. The function often has a name such as “inverse cumulative probability.” Plug in mean 500, standard deviation 120, and cumulative proportion 0.9. The software tells you that x = 653.786. We see that a student must score at least 654 to place in the highest 10%.

Without software, first find the standard score z with cumulative proportion 0.9, then “unstandardize” to find x. Here is the two-step process:

1. Use the table. Look in the body of Table A for the entry closest to 0.9. It is 0.8997. This is the entry corresponding to z = 1.28. So z = 1.28 is the standardized value with area 0.9 to its left.

2. Unstandardize to transform the solution from z back to the original x scale. We know that the standardized value of the unknown x is z = 1.28. So x itself satisfies

Solving this equation for x gives

x = 500 + (1.28)(120) = 653.6

This equation should make sense: it finds the x that lies 1.28 standard deviations above the mean on this particular Normal curve. That is the “unstandardized” meaning of z = 1.28. The general rule for unstandardizing a z-score is

x = μ + zσ

USE YOUR KNOWLEDGE

What score is needed to be in the top 20%? Consider the NAEP scores, which are approximately Normal, N(288, 38). How high a score is needed to be in the top 20% of students who take this exam?

1.100 Find the score that 75% of students will exceed. Consider the NAEP scores, which are approximately Normal, N(288, 38). Seventy-five percent of the students will score above x on this exam. Find x.

Normal quantile plots The Normal distributions provide good descriptions of some distributions of real data, such as the Iowa Test vocabulary scores. The distributions of some other common variables are usually skewed and therefore distinctly non- Normal. Examples include economic variables such as personal income and gross sales of business firms, the survival times of cancer patients after treatment, and the service lifetime of mechanical or electronic components. While experience can suggest whether or not a Normal distribution is plausible in a particular case, it is risky to assume that a distribution is Normal without actually inspecting the data.

Normal quantile plot

A histogram or stemplot can reveal distinctly non-Normal features of a distribution, such as outliers, pronounced skewness, or gaps and clusters. If the stemplot or histogram appears roughly symmetric and unimodal, however, we need a more sensitive way to judge the adequacy of a Normal model. The most useful tool for assessing Normality is another graph, the Normal quantile plot.

Here is the basic idea of a Normal quantile plot. The graphs produced by software use more sophisticated versions of this idea. It is not practical to make Normal quantile plots by hand.

1. Arrange the observed data values from smallest to largest. Record what percentile of the data each value occupies. For example, the smallest observation in a set of 20 is at the 5% point, the second smallest is at the 10% point, and so on.

Normal scores

2. Do Normal distribution calculations to find the values of z corresponding to these same percentiles. For example, z = −1.645 is the 5% point of the standard Normal distribution, and z = −1.282 is the 10% point. We call these values of Z Normal scores.

3. Plot each data point x against the corresponding Normal score. If the data distribution is close to any Normal distribution, the plotted points will lie close to a straight line.

Any Normal distribution produces a straight line on the plot because standardizing turns any Normal distribution into a standard Normal distribution. Standardizing is a linear transformation that can change the slope and intercept of the line in our plot but cannot turn a line into a curved pattern.

USE OF NORMAL QUANTILE PLOTS

If the points on a Normal quantile plot lie close to a straight line, the plot indicates that the data are Normal. Systematic deviations from a straight line indicate a non-Normal distribution. Outliers appear as points that are far away from the overall pattern of the plot. An optional line can be drawn on the plot that corresponds to the Normal distribution with mean equal to the mean of the data and standard deviation equal to the standard deviation of the data.

Figures 1.30 and 1.31 are Normal quantile plots for data we have met earlier. The data x are plotted vertically against the corresponding standard Normal z-score plotted horizontally. The z-score scale generally extends from −3 to 3 because almost all of a standard Normal curve lies between these values. These figures show how Normal quantile plots behave.

EXAMPLE 1.46

IQ

IQ scores are approximately Normal. Figure 1.30 is a Normal quantile plot of the 60 fifth-grade IQ scores from Table 1.1 (page 14). The points lie very close to the straight line drawn on the plot. We conclude that the distribution of IQ data is approximately Normal.

FIGURE 1.30 Normal quantile plot of IQ scores, Example 1.46. This distribution is approximately Normal.

EXAMPLE 1.47

TTS

Times to start a business are skewed. Figure 1.31 is a Normal quantile plot of the data on times to start a business from Example 1.23. We have excluded Suriname, the outlier that you examined in Exercise 1.43 (page 29). The line drawn on the plot shows clearly that the plot of the data is curved. We conclude that these data are not Normally distributed. The shape of the curve is what we typically see with a distribution that is strongly skewed to the right.

FIGURE 1.31 Normal quantile plot of 188 times to start a business, with the outlier, Suriname, excluded, Example 1.47 This distribution is highly skewed.

Real data often show some departure from the theoretical Normal model. When you examine a Normal quantile plot, look for shapes that show clear departures from Normality. Don’t overreact to minor wiggles in the plot. When we discuss statistical methods that are based on the Normal model, we are interested in whether or not the data are sufficiently Normal for these procedures to work properly. We are not concerned about minor deviations from

Normality. Many common methods work well as long as the data are approximately Normal and outliers are not present.

BEYOND THE BASICS

Density Estimation

density estimator

A density curve gives a compact summary of the overall shape of a distribution. Many distributions do not have the Normal shape. There are other families of density curves that are used as mathematical models for various distribution shapes. Modern software offers more flexible options. A density estimator does not start with any specific shape, such as the Normal shape. It looks at the data and draws a density curve that describes the overall shape of the data. Density estimators join stemplots and histograms as useful graphical tools for exploratory data analysis.

Density estimates can capture other unusual features of a distribution. Here is an example.

EXAMPLE 1.48

STUBHUB

StubHub! StubHub! is a website where fans can buy and sell tickets to sporting events. Ticket holders wanting to sell their tickets provide the location of their seats and the selling price. People wanting to buy tickets can choose from among the tickets offered for a given event.34

Tickets for the 2015 NCAA women’s basketball tournament were available from StubHub! in a package deal that included the semifinal games and the championship game. On June 28, 2014, StubHub! listed 518 tickets for sale. A histogram of the distribution of ticket prices with a density estimate is given in Figure 1.32. The distribution has three peaks: one around $700, another around $2800, and the third around $4650. This is the identifying characteristic of a trimodal distribution. There appears to be three types of tickets. How would you name the three types?

FIGURE 1.32 Histogram of StubHub! price per seat for tickets to the 2015 NCAA Women’s Semifinal and Championship games, with a density estimate, Example 1.48.

trimodal distribution bimodal distribution

Many distributions that we have met have a single peak, or mode. The distribution described in Example 1.48 has three modes and is called a trimodal distribution. A distribution that has two modes is called a bimodal distribution.

The previous example reminds of a continuing theme for data analysis. We looked at a histogram and a density estimate and saw something interesting. This led us to speculation. Additional data on the type and location of the seats may explain more about the prices than we see in Figure 1.32.

SECTION 1.4 SUMMARY The overall pattern of a distribution can often be described compactly by a density curve. A density curve has total area 1 underneath it. Areas under a density curve give proportions of observations for the distribution. The mean μ (balance point), the median (equal-areas point), and the quartiles can be approximately located by eye on a density curve. The standard deviation σ cannot be located by eye on most density curves. The mean and median are equal for symmetric density curves, but the mean of a skewed curve is located farther toward the long tail than is the median. The Normal distributions are described by bell-shaped, symmetric, unimodal density curves. The mean μ and standard deviation σ completely specify the Normal distribution N(μ, σ). The mean is the center of symmetry, and s is the distance from m to the change-of-curvature points on either side. All Normal distributions satisfy the 68–95–99.7 rule. To standardize any observation x, subtract the mean of the distribution and then divide by the standard deviation. The resulting z-score z = (x – μ)/σ says how many standard deviations x lies from the distribution mean. All Normal distributions are the same when measurements are transformed to the standardized scale. If X has the N(μ, σ) distribution, then the standardized variable Z = (X – μ)/σ has the standard Normal distribution N(0, 1). Proportions for any Normal distribution can be calculated by software or from the standard Normal table (Table A), which gives the cumulative proportions of Z < z for many values of z. The adequacy of a Normal model for describing a distribution of data is best assessed by a Normal quantile plot, which is available in most statistical software packages. A pattern on such a plot that deviates substantially from a straight line indicates that the data are not Normal.

SECTION 1.4 EXERCISES For Exercises 1.93 and 1.94, see page 59; for Exercises 1.95 and 1.96, see page 60; for Exercises 1.97 and 1.98, see page 64; and for Exercises 1.99 and 1.100, see page 66.

1.101 Means and medians.

(a) Sketch a symmetric distribution that is not Normal. Mark the location of the mean and the median.

(b) Sketch a distribution that is skewed to the left. Mark the location of the mean and the median.

1.102 The effect of changing the standard deviation.

(a) Sketch a Normal curve that has mean 30 and standard deviation 8.

(b) On the same x axis, sketch a Normal curve that has mean 30 and standard deviation 12.

(c) How does the Normal curve change when the standard deviation is varied but the mean stays the same?

1.103 The effect of changing the mean.

(a) Sketch a Normal curve that has mean 30 and standard deviation 8.

(b) On the same x axis, sketch a Normal curve that has mean 40 and standard deviation 8.

(c) How does the Normal curve change when the mean is varied but the standard deviation stays the same?

1.104 NAEP music scores. In Exercise 1.93 (page 59) we examined the distribution of NAEP scores for the 12th-grade reading skills assessment. For eighth-grade students, the average music score is approximately Normal with mean 150 and standard deviation 35.

(a) Sketch this Normal distribution.

(b) Make a table that includes values of the scores corresponding to plus or minus one, two, and three standard deviations from the mean. Mark these points on your sketch along with the mean.

(c) Apply the 68–95–99.7 rule to this distribution. Give the ranges of reading score values that are within one, two, and three standard deviations of the mean.

1.105 NAEP U.S. history scores. Refer to the previous exercise. The scores for 12th-grade students on the U.S. history assessment are approximately N(288,32) Answer the questions in the previous exercise for this assessment.

1.106 Standardize some NAEP music scores. The NAEP music assessment scores for eighth-grade students are approximately N(150,35). Find z-scores by standardizing the following scores: 150, 140, 100, 180, 230.

1.107 Compute the percentile scores. Refer to the previous exercise. When scores such as the NAEP assessment scores are reported for individual students, the actual values of the scores are not particularly meaningful. Usually, they are transformed into percentile scores. The percentile score is the proportion of students who would score less than or equal to the score for the individual student. Compute the percentile scores for the five scores in the previous exercise. State whether you used software or Table A for these computations.

1.108 Are the NAEP U.S. history scores approximately Normal? In Exercise 1.105, we assumed that the NAEP U.S history scores for 12th-grade students are approximately Normal with the reported mean and standard

deviation, N(288,32). Let’s check that assumption. In addition to means and standard deviations, you can find selected percentiles for the NAEP assessments (see previous exercise). For the 12th-grade U.S. history scores, the following percentiles are reported:

Percentile Score

10% 246 25% 276 50% 290 75% 311 90% 328

Use these percentiles to assess whether or not the NAEP U.S History scores for 12th-grade students are approximately Normal.

Write a short report describing your methods and conclusions.

1.109 Are the NAEP mathematics scores approximately Normal? Refer to the previous exercise. For the NAEP mathematics scores for 12th-graders, the mean is 153 and the standard deviation is 34. Here are the reported

percentiles:

Percentile Score

10% 110 25% 130 50% 154 75% 177 90% 197

Is the N(153,34) distribution a good approximation for the NAEP mathematics scores? Write a short report describing your methods and conclusions.

1.110 Do women talk more? Conventional wisdom suggests that women are more talkative than men. One study designed to examine this stereotype collected data on the speech of 42 women and 37 men in the United States.35 TALK

(a) The mean number of words spoken per day by the women was 14,297 with a standard deviation of 6441. Use the 68–95– 99.7 rule to describe this distribution.

(b) Do you think that applying the rule in this situation is reasonable? Explain your answer.

(c) The men averaged 14,060 words per day with a standard deviation of 9056. Answer the questions in parts (a) and (b) for the men.

(d) Do you think that the data support the conventional wisdom? Explain your answer. Note that in Section 7.2 we will learn formal statistical methods to answer this type of question.

1.111 Data from Mexico. Refer to the previous exercise. A similar study in Mexico was conducted with 31 women and 20 men. The women averaged 14,704 words per day with a standard deviation of 6215. For men the mean was 15,022 and the standard deviation was 7864. TALKM

(a) Answer the questions from the previous exercise for the Mexican study.

(b) The means for both men and women are higher for the Mexican study than for the U.S. study. What conclusions can you draw from this observation?

1.112 A uniform distribution. If you ask a computer to generate “random numbers” between 0 and 1, you will get observations from a uniform distribution. Figure 1.33 graphs the density curve for a uniform distribution. Use areas under this density curve to answer the following questions.

(a) Why is the total area under this curve equal to 1?

(b) What proportion of the observations lie above 0.44?

(c) What proportion of the observations lie between 0.44 and 0.70?

1.113 Use a different range for the uniform distribution. Many random number generators allow users to specify the range of the random numbers to be produced. Suppose that you specify that the outcomes are to be distributed uniformly between 0 and 4. Then the density curve of the outcomes has constant height between 0 and 4, and height 0 elsewhere.

(a) What is the height of the density curve between 0 and 4? Draw a graph of the density curve.

(b) Use your graph from part (a) and the fact that areas under the curve are proportions of outcomes to find the proportion of outcomes that are more than 1.

(c) Find the proportion of outcomes that lie between 1.5 and 2.5.

FIGURE 1.33 The density curve of a uniform distribution, Exercise 1.122.

FIGURE 1.34 Three density curves, Exercise 1.115.

1.114 Find the mean, the median, and the quartiles. What are the mean and the median of the uniform distribution in Figure 1.33? What are the quartiles?

1.115 Three density curves. Figure 1.34 displays three density curves, each with three points marked on it. At which of these points on each curve do the mean and the median fall?

1.116 Use the Normal Curve applet. Use the Normal Curve applet for the standard Normal distribution to say how many standard deviations above and below the mean the quartiles of any Normal distribution lie.

1.117 Use the Normal Curve applet. The 68–95–99.7 rule for Normal distributions is a useful approximation. You can use the Normal Curve applet on the text website to see how accurate the rule is. Drag one flag across the other so that the applet shows the area under the curve between the two flags.

(a) Place the flags one standard deviation on either side of the mean. What is the area between these two values? What does the 68–95–99.7 rule say this area is?

(b) Repeat for locations two and three standard deviations on either side of the mean. Again compare the 68–95–99.7 rule with the area given by the applet.

1.118 Find some proportions. Using either Table A or your calculator or software, find the proportion of observations from a standard Normal distribution that satisfies each of the following statements. In each case, sketch a standard Normal curve and shade the area under the curve that is the answer to the question.

(a) Z > 1.75

(b) Z < 1.75

(c) Z > −0.80

(d) −0.80 < Z < 1.75

1.119 Find more proportions. Using either Table A or your calculator or software, find the proportion of observations from a standard Normal distribution for each of the following events. In each case, sketch a standard Normal curve and shade the area representing the proportion.

(a) Z ≤ −1.4

(b) Z ≥ −1.4

(c) Z > 2.0

(d) −1.4 < Z < 2.0

1.120 Find some values of z. Find the value z of a standard Normal variable Z that satisfies each of the following conditions. (If you use Table A, report the value of z that comes closest to satisfying the condition.) In each case, sketch a

standard Normal curve with your value of z marked on the axis.

(a) 38% of the observations fall below z

(b) 70% of the observations fall above z

1.121 Find more values of z. The variable Z has a standard Normal distribution.

(a) Find the number z that has cumulative proportion 0.88.

(b) Find the number z such that the event Z > z has proportion 0.12.

1.122 Find some values of z. The Wechsler Adult Intelligence Scale (WAIS) is the most common IQ test. The scale of scores is set separately for each age group, and the scores are approximately Normal with mean 100 and standard deviation 15. People with WAIS scores below 70 are considered developmentally disabled when, for example, applying for Social Security disability benefits. What percent of adults are developmentally disabled by this criterion?

1.123 High IQ scores. The Wechsler Adult Intelligence Scale (WAIS) is the most common IQ test. The scale of scores is set separately for each age group, and the scores are approximately Normal with mean 100 and standard deviation 15. The organization MENSA, which calls itself “the high-IQ society,” requires a WAIS score of 130 or higher for membership. What percent of adults would qualify for membership?

There are two major tests of readiness for college, the ACT and the SAT. ACT scores are reported on a scale from 1 to 36. The distribution of ACT scores is approximately Normal with mean μ = 21.5 and standard deviation σ = 5.4. SAT scores are reported on a scale from 600 to 2400. The distribution of SAT scores is approximately Normal with mean μ = 1498 and standard deviation σ = 316. Exercises 1.124 through 1.133 are based on this information.

1.124 Compare an SAT score with an ACT score. Jessica scores 1830 on the SAT. Ashley scores 27 on the ACT. Assuming that both tests measure the same thing, who has the higher score? Report the z-scores for both students.

1.125 Make another comparison. Joshua scores 16 on the ACT. Anthony scores 1050 on the SAT. Assuming that both tests measure the same thing, who has the higher score? Report the z-scores for both students.

1.126 Find the ACT equivalent. Jorge scores 2090 on the SAT. Assuming that both tests measure the same thing, what score on the ACT is equivalent to Jorge’s SAT score?

1.127 Find the SAT equivalent. Alyssa scores 30 on the ACT. Assuming that both tests measure the same thing, what score on the SAT is equivalent to Alyssa’s ACT score?

1.128 Find an SAT percentile. Reports on a student’s ACT or SAT results usually give the percentile as well as the actual score. The percentile is just the cumulative proportion stated as a percent: the percent of all scores that were lower than or equal to this one. Renee scores 2050 on the SAT. What is her percentile?

1.129 Find an ACT percentile. Reports on a student’s ACT or SAT results usually give the percentile as well as the actual score. The percentile is just the cumulative proportion stated as a percent: the percent of all scores that were lower than or equal to this one. Joshua scores 19 on the ACT. What is his percentile?

1.130 How high is the top 12%? What SAT scores make up the top 12% of all scores?

1.131 How low is the bottom 12%? What SAT scores make up the bottom 12% of all scores?

1.132 Find the ACT quintiles. The quintiles of any distribution are the values with cumulative proportions 0.20, 0.40, 0.60, and 0.80. What are the quintiles of the distribution of ACT scores?

1.133 Find the SAT quartiles. The quartiles of any distribution are the values with cumulative proportions 0.25 and 0.75. What are the quartiles of the distribution of SAT scores?

1.134 Do you have enough “good cholesterol?” High-density lipoprotein (HDL) is sometimes called the “good cholesterol” because low values are associated with a higher risk of heart disease. According to the American Heart Association, people over the age of 20 years should have at least 40 milligrams per deciliter (mg/dl) of HDL cholesterol.36 U.S. women aged 20 and over have a mean HDL of 55 mg/dl with a standard deviation of 15.5 mg/dl. Assume that the distribution is Normal.

(a) What percent of women have low values of HDL (40 mg/dl or less)?

(b) HDL levels of 60 mg/dl and higher are believed to protect people from heart disease. What percent of women have protective levels of HDL?

(c) Women with more than 40 mg/dl but less than 60 mg/dl of HDL are in the intermediate range, neither very good or very

bad. What proportion are in this category?

1.135 Men and HDL cholesterol. HDL cholesterol levels for men have a mean of 46 mg/dl with a standard deviation of 13.6 mg/dl. Answer the questions given in the previous exercise for the population of men.

1.136 Diagnosing osteoporosis. Osteoporosis is a condition in which the bones become brittle due to loss of minerals. To diagnose osteoporosis, an elaborate apparatus measures bone mineral density (BMD). BMD is usually reported in standardized form. The standardization is based on a population of healthy young adults. The World Health Organization (WHO) criterion for osteoporosis is a BMD 2.5 standard deviations below the mean for young adults. BMD measurements in a population of people similar in age and sex roughly follow a Normal distribution.

(a) What percent of healthy young adults have osteoporosis by the WHO criterion?

(b) Women aged 70 to 79 are of course not young adults. The mean BMD in this age is about −2 on the standard scale for young adults. Suppose that the standard deviation is the same as for young adults. What percent of this older population has osteoporosis?

1.137 Deciles of Normal distributions. The deciles of any distribution are the 10th, 20th, …, 90th percentiles. The first and last deciles are the 10th and 90th percentiles, respectively.

(a) What are the first and last deciles of the standard Normal distribution?

(b) The weights of 9-ounce potato chip bags are approximately Normal with mean 9.12 ounces and standard deviation 0.15 ounce. What are the first and last deciles of this distribution?

1.138 Quartiles for Normal distributions. The quartiles of any distribution are the values with cumulative proportions 0.25 and 0.75.

(a) What are the quartiles of the standard Normal distribution?

(b) Using your numerical values from part (a), write an equation that gives the quartiles of the N(μ, σ) distribution in terms of μ and σ.

1.139 IQR for Normal distributions. Continue your work from the previous exercise. The interquartile range IQR is the distance between the first and third quartiles of a distribution.

(a) What is the value of the IQR for the standard Normal distribution?

(b) There is a constant c such that IQR – cσ for any Normal distribution N(μ, σ). What is the value of c?

1.140 Outliers for Normal distributions. Continue your work from the previous two exercises. The percent of the observations that are suspected outliers according to the 1.5 × IQR rule is the same for any Normal distribution. What is

this percent?

1.141 Deciles of HDL cholesterol. The deciles of any distribution are the 10th, 20th, . . . , 90th percentiles. Refer to Exercise 1.134 where we assumed that the distribution of HDL cholesterol in U.S. women aged 20 and over is Normal with mean 55 mg/dl and standard deviation 15.5 mg/dl. Find the deciles for this distribution.

1.142 Longleaf pine trees. Exercise 1.88 (page 50) gives the diameter at breast height (DBH) for 40 longleaf pine trees from the Wade Tract in Thomas County, Georgia. Make a Normal quantile plot for these data and write a short paragraph interpreting what it describes. PINES

1.143 Potassium from potatoes. Refer to Exercise 1.30 (page 24) where you used a stemplot to examine the potassium absorption of a group of 27 adults who ate a controlled diet that included 40 mEq of potassium from potatoes for five days. In Exercise 1.61 (page 47), you compared the stemplot, the histogram, and the boxplot as graphical summaries of this distribution.

KPOT40

(a) Generate these three graphical summaries.

(b) Make a Normal quantile plot and interpret it.

1.144 Potassium from a supplement. Refer to Exercise 1.31 (page 24) where you used a stemplot to examine where you examined the potassium absorption of a group of 29 adults who ate a controlled diet that included 40 mEq of potassium from a supplement for five days. In Exercise 1.62 (page 47), you compared the stemplot, the histogram, and the boxplot as graphical summaries of this distribution. KSUP40

(a) Generate these three graphical summaries.

(b) Make a Normal quantile plot and interpret it.

CHAPTER 1 EXERCISES

1.145 Sources of energy consumed. Energy consumed in the United States can be classified as coming from one of three sources: fossil fuels, nuclear and electric power, and renewable energy. In 2014, the energy from these

three sources was 80.3, 8.3, and 9.6 quadrillion Btu, respectively. In 2004, the corresponding amounts were 85.8, 8.2, and 6.1.37 Write a description of the changes from 2004 to 2014 expressed in these data. Illustrate your summary with appropriate graphical summaries. Be sure to discuss both the amounts of energy from each source as well as the percents.

1.146 CO2 emissions in vehicles. Natural Resources Canada tests new vehicles each year and reports several variables related to fuel consumption for vehicles in different classes.38 For 2015, it provides data for 526 vehicles that use regular fuel. Two variables reported are carbon dioxide (CO2) emissions and highway fuel consumption. CO2 is measured in grams per kilometer (g/km), and highway fuel consumption measured in liters per 100 kilometers (L/km). Use graphical and numerical summaries to describe the distribution of CO2 emissions for these vehicles. Be sure to justify your choice of summaries. CANFREG

1.147 Highway fuel consumption. Refer to the previous exercise. Use graphical and numerical summaries to describe the distribution of highway fuel consumption for these vehicles. Be sure to justify your choice of summaries.

CANFREG

1.148 Jobs for business majors. What types of jobs are available for students who graduate with a business degree? The website careerbuilder.com lists job opportunities classified in a variety of ways. A recent posting had 25,120 jobs. The following table gives types of jobs and the numbers of postings listed under the classification “business administration” on a recent day:39 BUSJOBS

Type Number

Management 10916 Sales 5981 Information technology 4605 Customer service 4116 Marketing 3821 Finance 2339 Health care 2231 Accounting 2175 Human resources 1685

Describe these data using the methods you learned in this chapter, and write a short summary about jobs that are available for those who have a business degree. Include comments on the limitations that should be kept in mind when interpreting this particular set of data.

1.149 Flopping in the 2014 World Cup. Soccer players are often accused of spending an excessive amount of time dramatically falling to the ground followed by other activities, suggesting that a possible injury is very serious. It has been suggested that these tactics are often designed to influence the call of a referee or to take extra time off the clock. Recordings of the first 32 games of the 2014 World Cup were analyzed, and there were 302 times when the referee interrupted the match because of a possible injury. The number of injuries and the total time, in minutes, spent flopping for each of the 32 teams who participated in these matches was recorded.40 Here are the data: FLOPS

Country Injuries Time

Brazil 17 3.30 Chile 16 6.97 Honduras 15 7.67 Nigeria 15 6.42

Mexico 15 3.97 Costa Rica 13 3.80 USA 12 6.40 Ecuador 12 4.55 France 10 7.32 South Korea 10 4.52 Algeria 10 4.05 Iran 9 5.43 Russia 9 5.27 Ivory Coast 9 4.63 Croatia 9 4.32 Colombia 9 4.32 Uruguay 9 4.12 Greece 9 2.65 Cameroon 8 3.15 Germany 8 1.97 Spain 8 1.82 Belgium 7 3.38 Japan 7 2.08 Italy 7 1.60 Switzerland 7 1.35 England 7 3.13 Argentina 6 2.80 Ghana 6 1.85 Australia 6 1.83 Portugal 4 1.82 Netherlands 4 1.65 Bosnia and 2 0.40 Herzegovina

Describe these data using the methods you learned in this chapter, and write a short summary about flopping in the 2014 World Cup based on your analysis.

1.150 Twitter accounts. Twitter has more than 52,900,000 million users in the United States. A study of Twitter accounts classified users by age. Here are the numbers of users (in millions) for six age groups:41 TWIT

Age Number

18–24 11.7 25–34 13.3 35–44 8.7 45–54 6.7 55–64 4.1 65 and over 2.7

Describe these data using the methods you learned in this chapter, and write a short summary about the age distribution of Twitter users based on your analysis.

1.151 What graph would you use? What type of graph or graphs would you plan to make in a study of each of the following issues?

(a) What makes of cars do students drive? How old are their cars?

(b) How many hours per week do students study? How does the number of study hours change during a semester?

(c) Which radio stations are most popular with students?

(d) When many students measure the concentration of the same solution for a chemistry course laboratory assignment, do their measurements follow a Normal distribution?

1.152 Canadian international trade. The government organization Statistics Canada provides data on many topics related to Canada’s population, resources, economy, society, and culture. Go to the web page

statcan.gc.ca/start-debut-eng.html. Under the “Subject” tab, choose “International trade.” Pick some data from the resources listed and use the methods that you learned in this chapter to create graphical and numerical summaries. Write a report summarizing your findings that includes supporting evidence from your analyses.

1.153 Travel and tourism in Canada. Refer to the previous exercise. Under the “Subject” tab, choose “Travel and tourism.” Pick some data from the resources listed and use the methods that you learned in this chapter

to create graphical and numerical summaries. Write a report summarizing your findings that includes supporting evidence from your analyses.

1.154 Leisure time for college students. You want to measure the amount of “leisure time” that college students enjoy. Write a brief discussion of two issues:

(a) How will you define “leisure time”?

(b) Once you have defined leisure time, how will you measure it?

1.155 How much vitamin C do you need? The U.S. Food and Nutrition Board of the Institute of Medicine, working in cooperation with scientists from Canada, have used scientific data to answer this question for a variety

of vitamins and minerals.42 Their methodology assumes that needs, or requirements, follow a distribution. They have produced guidelines called dietary reference intakes for different gender-by-age combinations. For vitamin C, there are three dietary reference intakes: the estimated average requirement (EAR), which is the mean of the requirement distribution; the recommended dietary allowance (RDA), which is the intake that would be sufficient for 97% to 98% of the population; and the tolerable upper level (UL), the intake that is unlikely to pose health risks. For women aged 19 to 30 years, the EAR is 60 milligrams per day (mg/d), the RDA is 75 mg/d, and the UL is 2000 mg/d.43

(a) The researchers assumed that the distribution of requirements for vitamin C is Normal. The EAR gives the mean. From the definition of the RDA, let’s assume that its value is the 97.72 percentile. Use this information to determine the standard deviation of the requirement distribution.

(b) Sketch the distribution of vitamin C requirements for 19- to 30-year-old women. Mark the EAR, the RDA, and the UL on your plot.

1.156 How much vitamin C do men need? Refer to the previous exercise. For men aged 19 to 30 years, the EAR is 75 milligrams per day (mg/d), the RDA is 90 mg/d, and the UL is 2000 mg/d. Answer the questions in the

previous exercise for this population.

1.157 How much vitamin C do women consume? To evaluate whether or not the intake of a vitamin or mineral is adequate, comparisons are made between the intake distribution and the requirement distribution. Here is

some information about the distribution of vitamin C intake, in milligrams per day, for women aged 19 to 30 years:44

Mean

Percentile (mg/d)

1st 5th 19th 25th 50th 75th 90th 95th 99th

84.1 31 42 48 61 79 102 126 142 179

(a) Use the 5th, the 50th, and the 95th percentiles of this distribution to estimate the mean and standard deviation of this distribution assuming that the distribution is Normal. Explain your method for doing this.

(b) Sketch your Normal intake distribution on the same graph with a sketch of the requirement distribution that you produced in part (b) of Exercise 1.155.

(c) Do you think that many women aged 19 to 30 years are getting the amount of vitamin C that they need? Explain your answer.

1.158 How much vitamin C do men consume? To evaluate whether or not the intake of a vitamin or mineral is adequate, comparisons are made between the intake distribution and the requirement distribution. Here is

some information about the distribution of vitamin C intake, in milligrams per day, for men aged 19 to 30 years:

Percentile (mg/d)

Mean 1st 5th 19th 25th 50th 75th 90th 95th 99th

122.2 39 55 65 85 114 150 190 217 278

(a) Use the 5th, the 50th, and the 95th percentiles of this distribution to estimate the mean and standard deviation of this distribution assuming that the distribution is Normal. Explain your method for doing this.

(b) Sketch your Normal intake distribution on the same graph with a sketch of the requirement distribution that you produced in Exercise 1.156.

(c) Do you think that many men aged 19 to 30 years in the United States are getting the amount of vitamin C that they need? Explain your answer.

1.159 Time spent studying. Do women study more than men? We asked the students in a large first-year college class how many minutes they studied on a typical weeknight. Here are the responses of random samples of 30 women and 30 men from the class: STUDY

Women Men

170 120 180 360 240 80 120 30 90 200 120 180 120 240 170 90 45 30 120 75 150 120 180 180 150 150 120 60 240 300 200 150 180 150 180 240 60 120 60 30 120 60 120 180 180 30 230 120 95 150 90 240 180 115 120 0 200 120 120 180

(a) Examine the data. Why are you not surprised that most responses are multiples of 10 minutes? We eliminated one student who claimed to study 30,000 minutes per night. Are there any other responses that you consider suspicious?

(b) Make a back-to-back stemplot of these data. Report the approximate midpoints of both groups. Does it appear that women study more than men (or at least claim that they do)?

(c) Make side-by-side boxplots of these data. Compare the boxplots with the stemplot you made in part (b). Which to you prefer? Give reasons for your answer.

1.160 Product preference. Product preference depends in part on the age, income, and gender of the consumer. A market researcher selects a large sample of potential car buyers. For each consumer, she records gender, age, household income, and automobile preference. Which of these variables are categorical and which are quantitative?

1.161 Two distributions. If two distributions have exactly the same mean and standard deviation, must their histograms have the same shape? If they have the same five-number summary, must their histograms have the same shape? Explain.

1.162 Spam filters. A university department installed a spam filter on its computer system. During a 21-day period, 6693 messages were tagged as spam. How much spam you get depends on what your online habits are. Here are the counts for some students and faculty in this department (with log-in IDs changed, of course):

ID Count

AA 1818 EE 399 II 251 BB 1358 FF 389 JJ 178 CC 442 GG 304 KK 158 DD 416 HH 251 LL 103

All other department members received fewer than 100 spam messages. How many did the others receive in total? Make a graph and comment on what you learn from these data. SPAM

1.163 Phish. One of the most favored songs of the band Phish is “Divided Sky.” The band plays this song at many of their concerts. Frequently, after the main theme, Trey, the guitarist, pauses before playing the resolving

note.45 The data file PHISH gives the date of each concert where “Divided Sky” was played, the venue, and the length of the pause for 366 concerts. Analyze the data and write a report summarizing what you have found. Be sure to include graphical and numerical summaries. Include the rationale for decisions that you made in performing your analysis. For example, did you give any consideration to the relatively large number of zeros? PHISH

1.164 Visits to a help room for statistics. A help room staffed by graduate students provides assistance to students taking statistics courses. To justify the cost of providing this service, extensive records are kept. Each time

a student visits the help room, the student signs a sheet with several variables. These include the date of the visit, the course number that they are taking, the time they arrived at the room, and the time that they left the room. The length of time that the each student spent in the help room is computed from the two time variables. Data for 1268 visits are given in the file HELP.46 Analyze the data and write a report summarizing what you have found. Be sure to include graphical and numerical summaries. Include the rationale for the choices of methods that you chose for your analysis. There are some missing course numbers. How did you handle these? HELP

1.165 Blueberries and anthocyanins. Anthocyanins are compounds that have been associated with health benefits associated with the heart, bones, and the brain. Blueberries are a good source of many different

anthocyanins. Researchers at the Piedmont Research Station of North Carolina State University have assembled a database giving the concentrations of 18 different anthocyanins for 267 varieties of blueberries.47 Four of the anthocyanins measured are delphinidin-3-arabinoside, malvidin-3-arabinoside, cyanidin-3-galactoside, and delphinidin- 3-glucoside, all measured in units of mg/100g of berries. In the data file, we have simplified the names of these anthocyanins to Antho1, Antho2, Antho3, and Antho4. Figure 1.35 gives graphical and numeric summaries from JMP for Antho1. Use this output to write a summary of the distribution of Antho1 using the methods and ideas that you learned in this chapter. BERRIES

1.166 Blueberries and anthocyanins, Antho2. Refer to the previous exercise. Generate your own output for the analysis of Antho2 and use your output to write a summary of the distribution of Antho2 using the methods and

ideas that you learned in this chapter. BERRIES

1.167 Blueberries and anthocyanins, Antho3. Refer to Exercise 1.165. Figure 1.36 gives the JMP output for Antho3. Use this output to write a summary of the distribution of Antho3 using the methods and ideas that you

learned in this chapter. BERRIES

1.168 Blueberries and anthocyanins, Antho4. Refer to Exercise 1.165. Generate your own output for the analysis of Antho4 and use your output to write a summary of the distribution of Antho4 using the methods and

ideas that you learned in this chapter. BERRIES

FIGURE 1.35 JMP descriptive statistics for Antho1, Exercise 1.165.

FIGURE 1.36 JMP descriptive statistics for Antho3, Exercise 1.167.

1. See census.gov.

2. From State of Drunk Driving Fatalities in America 2010, available at responsibility.org.

3. James P. Purdy, “Why first-year college students select online research sources as their favorite,” First Monday, 17, No. 9 (September 3, 2012). See firstmonday.org.

4. Data collected in the lab of Connie Weaver, Department of Nutrition Sciences, Purdue University, and provided by Linda McCabe. For more information, see Corrie M. Whisner, et al., “Soluble maize fibre affects short-term calcium absorption in adolescent boys and girls: A randomized controlled trial using dual stable isotropic tracers,” British Journal of Nutritiion, 112 (2014), pp. 446–456.

5. Haipeng Shen, “Nonparametric regression for problems involving lognormal distributions,” PhD dissertation, University of Pennsylvania, 2003. Thanks to Haipeng Shen and Larry Brown for sharing the data.

6. From the Digest of Education Statistics at the website of the National Center for Education Statistics, nces.ed.gov/programs/digest.

7. See Note 4.

8. Based on Barbara Ernst et al., “Seasonal variation in the deficiency of 25–hydroxyvitamin D3 in mildly to extremely obese subjects,”

Obesity Surgery, 19 (2009), pp. 180–183.

9. See, for example, facebook.com/Million.Dollar.Application.

10. From socialbakers.com. The website says that the data are updated daily. These data were downloaded on June 15, 2014.

11. More information about the Titanic can be found at the website for the Titanic Project in Belfast, Ireland, at titanicbelfast.com.

12. Data describing the passengers on the Titanic can be found at cran.r-project.org/web/packages/titanic/titanic.pdf.

13. See health.gov/dietaryguidelines/2015/.

14. Data collected in the lab of Connie Weaver, Department of Nutrition Sciences, Purdue University and provided by Linda McCabe.

15. Data from Table 1.1 in the U.S. Energy Information Administration’s July 2015 Monthly Energy Review, available at eia.gov/totalenergy/data/monthly/pdf/mer.pdf.

16. From the Color Assignment website of Joe Hallock, joehallock.com/edu/COM498/index.html.

17. From the U.S. Environmental Protection Agency. See www.epa.gov/sites/production/files/2015-09/documents/2012_msw_fs.pdf.

18. See dupont.com/.

19. Data provided by Darlene Gordon, Purdue University.

20. Data for 1980 to 2013 are available from the World Bank at data.worldbank.org/indicator/IC.REG.DURS. Data for 2013 were used for this example.

21. See, for example, nacubo.org/Research.

22. The data were provided by James Kaufman. The study is described in James C. Kaufman, “The cost of the muse: Poets die young,” Death Studies, 27 (2003), pp. 813–821. The quote from Yeats appears in this article.

23. See, for example, the bibliographic entry for Gosset in the School of Mathematics and Statistics of the University of St. Andrews, Scotland, MacTutor History of Mathematics archive at www-history.mcs.st-and.ac.uk/Biographies/Gosset.html.

24. These and other data that were collected and used by Gosset can be found in the Guinness Archives in Dublin. See www.guinness- storehouse.com/en/archives.

25. These data were provided by Krista Nichols, Department of Biological Sciences, Purdue University.

26. From beer100.com/beercalories.htm on July 14, 2015.

27. Net worth from the Federal Reserve Bulletin, 100, No. 4 (2014), p. 12.

28. For more information about earthquakes, see the U.S. Geological Service website at usgs.gov.

29. See Noel Cressie, Statistics for Spatial Data, Wiley, 1993.

30. The National Assessment of Educational Progress (NAEP) is conducted by the National Center for Education Statistics (NCES). The NAEP is a large assessment of student knowledge in a variety of subjects. See nces.ed.gov/nationsreportcard/naepdata.

31. See the NCAA Eligibility Center Quick Reference Sheet, available at fs.ncaa.org/Docs/eligibility_center/Quick_Reference_Sheet.pdf.

32. Distributions for SAT scores can be found at the College Board website, research.collegeboard.org/content/sat-data-tables.

33. See Note 32.

34. See stubhub.com.

35. From Matthias R. Mehl et al., “Are women really more talkative than men?” Science, 317, No. 5834 (2007), p. 82. The raw data were provided by Matthias Mehl.

36. From the American Heart Association website, www.heart.org.

37. See eia.gov/totalenergy/.

38. From nrcan.gc.ca/energy/efficiency/11938.

39. Data from the careerbuilder.com website on July 3, 2014. See careerbuilder.com/jobs/keyword/business-administration.

40. See online.wsj.com/articles/the-world-rankings-of-flopping-1403660175.

41. Data for 2015 from statista.com/statistics/398152/us-twitter-user-age-groups/.

42. The Institute of Medicine website, www.iom.edu, provides links to reports related to dietary reference intakes as well as other health and nutrition topics.

43. Dietary Reference Intakes for Vitamin C, Vitamin E, Selenium and Carotenoids, National Academy of Sciences, 2000.

44. See Note 43.

45. See phish.net/song/divided-sky/history.

46. Data from Tadd Colver, Department of Statistics, Purdue University.

47. Data provided by Mary Ann Lila, Director, Plants for Human Health Institute, David H. Murdock Distinguished Professor, North Carolina Research Campus, North Carolina State University.

2.1 2.2 2.3 2.4 2.5 2.6 2.7

CHAPTER 2 Looking at Data—Relationships

Relationships Scatterplots Correlation Least-Squares Regression Cautions about Correlation and Regression Data Analysis for Two-Way Tables The Question of Causation

Introduction In Chapter 1, we learned to use graphical and numerical methods to describe the distribution of a single variable. Many of the interesting examples of the use of statistics involve relationships between pairs of variables. Learning ways to

describe relationships with graphical and numerical methods is the focus of this chapter. In Section 2.2, we focus on graphical descriptions. The scatterplot is our fundamental graphical tool for displaying

the relationship between two quantitative variables. Sections 2.3 and 2.4 move on to numerical summaries for these relationships. Cautions about the use of these methods are discussed in Section 2.5. Graphical and numerical methods for describing the relationship between two categorical variables are presented in Section 2.6. We conclude with Section 2.7, a brief overview of issues related to the distinction between associations and causation.

2.1

2.1 Relationships

When you complete this section, you will be able to:

Identify the key characteristics of a data set to be used to explore a relationship between two variables. Categorize variables as response variables or explanatory variables.

In Chapter 1 (page 2), we discussed the key characteristics of a data set. Cases are the objects described by a set of data, and a variable is a characteristic of a case. We also learned to categorize variables as categorical or quantitative. For Chapter 2, we focus on data sets that have pairs of variables that we want to study together. Here is an example.

EXAMPLE 2.1

College students cope with stress. Stress is a common problem for college students. Exploring factors that are associated with stress may lead to strategies that will help students to relieve some of the stress that they experience. A recent study found that students who experienced greater stress had less access to resources that would help them to cope with their stress.1 The two variables involved in the relationship here are perceived stress and resources to cope. The cases are the 97 students who are the subjects for a particular study.

When we study relationships between two variables, it is not sufficient to collect data on the two variables. A key idea for this chapter is that both variables must be measured on the same cases.

USE YOUR KNOWLEDGE

Facebook friends. Do people who have more Facebook friends spend more time on Facebook? In an introductory statistics class of 38 students, there were 32 users of Facebook. Each of these students was asked to report how many Facebook friends they had and the average amount of time that they spent on Facebook per week.

(a) Who are the cases for this study?

(b) What are the variables?

(c) Are the variables quantitative or categorical? Explain your answer.

We use the term associated to describe the relationship between two variables, such as stress and access to resources to cope in Example 2.1. Here is another example where two variables are associated.

EXAMPLE 2.2

Size and price of a coffee beverage. You visit a local Starbucks to buy a Mocha Frappuccino®. The barista explains that this blended coffee beverage comes in three sizes and asks if you want a Tall, a Grande, or a Venti. The prices are $3.95, $4.45, and $4.95, respectively. There is a clear association between the size of the Mocha Frappuccino® and its price.

ASSOCIATION BETWEEN VARIABLES

Two variables measured on the same cases are associated if knowing the values of one of the variables tells you something about the values of the other variable.

In the Mocha Frappuccino® example, knowing the size tells you the exact price, so the association here is very strong. Many statistical associations, however, are simply overall tendencies that allow exceptions. For example, it’s likely that some students in Example 2.1 are highly stressed and have a high level of resources to cope. Others experience little stress and have a low level of resources to cope. The association in that example is much weaker than the one in the Mocha Frappuccino® example.

2.2

2.3

Examining relationships To examine the relationship between two or more variables, we first need to know some basic characteristics of the data.

EXAMPLE 2.3

Stress and resources to cope. Refer to Example 2.1. The study asked 97 first-year college students about their stress (perceived stress) and the availability of resources to deal with stress (resources to cope).2 Perceived stress is based on responses to 10 questions that are summarized in a single variable. Therefore, we will treat the perceived stress as a quantitative variable. Resources to cope is constructed in a similar way summarizing the responses to 20 questions. We treat resources to cope as a quantitative variable also.

In many situations, we measure a collection of categorical variables and then combine them in a scale that can be viewed as a quantitative variable. The perceived stress and resources to cope are examples. We can also turn the tables in the other direction. Here is an example.

EXAMPLE 2.4

Hemoglobin and anemia. Hemoglobin is a measure of iron in the blood. The units are grams of hemoglobin per deciliter of blood (g/dl). Typical values depend on age and sex. Adult women typically have values between 12 and 16 g/dl.

Anemia is a major problem in developing countries, and many studies have been designed to address the problem. In these studies, computing the mean hemoglobin is not particularly useful. For studies like these, it is more appropriate to use a definition of severe anemia (a hemoglobin of less than 8 g/dl). Thus, for example, researchers can compare the proportions of subjects who are severely anemic for two treatments rather than the difference in the mean hemoglobin levels. In this situation, the categorical variable, severely anemic or not, is much more useful than the quantitative variable, hemoglobin.

When analyzing data to draw conclusions, it is important to carefully consider the best way to summarize the data. Just because a variable is measured as a quantitative variable, it does not necessarily follow that the best summary is based on the mean (or the median). As the previous example illustrates, converting a quantitative variable to a categorical variable is a very useful option to keep in mind.

USE YOUR KNOWLEDGE

Create a categorical variable from a quantitative variable. Consider the study described in Example 2.3. Suppose that we order the students based on the values of resources to cope from smallest to largest. Then, we define three resource groups: low resources, the first 32 students; medium resources, the next 33 students; and high resources, the remaining 32 students. If we compare the perceived stress of the three resource groups, are we using resource group as a quantitative variable or as a categorical variable? Explain your answer and describe some advantages to using the groups versus the original variable in explaining the results of a study such as this.

Replace names by ounces. In the Mocha Frappuccino® example, the variable size is categorical, with Tall, Grande, and Venti as the possible values. Suppose that you converted these values to the number of ounces: Tall is 12 ounces, Grande is 16 ounces, and Venti is 24 ounces. For studying the relationship between ounces and price, describe the cases and the variables and state whether each is quantitative or categorical.

2.4

When you examine the relationship between two variables, a new question becomes important: Is your purpose simply to explore the nature of the relationship, or do you hope to show that one of the variables

can explain variation in the other? In other words, is one of the variables a response variable and the other an explanatory variable?

RESPONSE VARIABLE, EXPLANATORY VARIABLE

A response variable measures an outcome of a study. An explanatory variable explains or causes changes in the response variable.

EXAMPLE 2.5

Stress and resources to cope. Refer to the study of stress and resources to cope in Example 2.3. Here, the explanatory variable is resources to cope and the response variable is perceived stress.

USE YOUR KNOWLEDGE

Stress and resources or resources and stress? Consider the scenario described in the previous example. Note that the variable, resources to cope, is constructed by summarizing the responses to 20 questions that include items measuring the skills that the student has developed to reduce stress. Make an argument for treating stress as the explanatory variable and resources to cope as the response variable.

In some studies, it is easy to identify explanatory and response variables. The following example illustrates one situation where this is true: when we actually set values of one variable to see how it affects another variable.

EXAMPLE 2.6

How much calcium do you need? Adolescence is a time when bones are growing very actively. If young people do not have enough calcium, their bones will not grow properly. How much calcium is enough? Research designed to answer this question has been performed for many years at events called “Camp Calcium.”3 At these camps, subjects eat controlled diets that are identical except for the amount of calcium. The amount of calcium retained by the body is the major response variable of interest. Because the amount of calcium consumed is controlled by the researchers, this variable is the explanatory variable.

When you don’t set the values of either variable but just observe both variables, there may or may not be explanatory and response variables. Whether there are depends on how you plan to use the data.

EXAMPLE 2.7

Student loans. A college student aid officer looks at the findings of the National Student Loan Survey. She notes data on the amount of debt of recent graduates, their current income, and how stressful they feel about college debt. She isn’t interested in predictions but is simply trying to understand the situation of recent college graduates.

A sociologist looks at the same data with an eye to using amount of debt and income, along with other variables, to explain the stress caused by college debt. Now, amount of debt and income are explanatory variables, and stress level is the response variable.

In many studies, the goal is to show that changes in one or more explanatory variables actually cause changes in a response variable. But many explanatory-response relationships do not involve direct causation. The SAT scores of

high school students help predict the students’ future college grades, but high SAT scores certainly don’t cause high college grades.

KEY CHARACTERISTICS OF DATA FOR RELATIONSHIPS

A description of the key characteristics of a data set that will be used to explore a relationship between two variables should include

Cases. Identify the cases and how many there are in the data set. Categorical or quantitative. Classify each variable as categorical or quantitative. Values. Identify the possible values for each variable. Explanatory or response. If appropriate, classify each variable as explanatory or response. Label. Identify what is used as a label variable if one is present.

independent variable

dependent variable

Some of the statistical techniques in this chapter require us to distinguish explanatory from response variables; others make no use of this distinction. You will often see explanatory variables called independent variables variable and response variables called dependent variables. These terms express mathematical ideas; they are not statistical terms. The concept that underlies this language is that the response depends on explanatory variables. Because the words “independent” and “dependent” have other meanings in statistics that are unrelated to the explanatory-response distinction, we prefer to avoid those words.

Most statistical studies examine data on more than one variable. Fortunately, statistical analysis of several-variable data builds on the tools used for examining individual variables. The principles that guide our work also remain the same:

Start with a graphical display of the data. Look for overall patterns and deviations from those patterns. Based on what you see, use numerical summaries to describe specific aspects of the data.

SECTION 2.1 SUMMARY To study relationships between variables, we must measure the variables on the same cases. It is also important to determine the number of cases and to classify each variable as categorical or quantitative. Two variables measured on the same cases are associated if knowing the values of one variable tells you something about the values of the other variable. If we think that a variable x may explain or even cause changes in another variable y, we call x an explanatory variable and y a response variable.

SECTION 2.1 EXERCISES For Exercise 2.1, see page 80; for Exercises 2.2 and 2.3, see page 82; and for Exercise 2.4, see page 82.

2.5 High click counts on Twitter. A study was done to identify variables that might produce high click counts on Twitter. You and nine of your friends collect data on all of your tweets for a week. You record the number of click counts, the time of day, the day of the week, the sex of the person posting the tweet, and the length of the tweet.

(a) What are the cases for this study?

(b) Classify each of the variables as categorical or quantitative.

(c) Classify each of the variables as explanatory, response, or neither. Explain your answers.

2.6 Explanatory or response? For each of the following scenarios, classify each of the pair of variables as explanatory or response or neither. Give reasons for your answers.

(a) Whether or not a person likes to sing and whether or not a person likes to dance.

(b) The number of pages in a textbook and the cost of a new copy of the textbook.

(c) The number of alcoholic drinks consumed and the blood alcohol content.

(d) In a study of adolescents, the dose of vitamin D given each day for a year (50, 100, or 200 international units) and the change in total bone mineral content from the beginning of the study to the end of the study.

2.7 Buy and sell prices of used textbooks. Think about a study designed to compare the prices of textbooks for third- and fourth-year college courses in five different majors. For the five majors, you want to examine the relationship between the difference in the price that you pay for a used textbook and the price that the seller gives back to you when you return the textbook. Describe a data set that could be used for this study, and give the key characteristics of the data.

2.8 Protein and fat. Think about a study designed to examine the relationship between protein intake and fat intake in the diets of first-year college students. Describe a data set that could be used for this study, and give the key characteristics of the data.

2.9 Soccer tickets and performance. For the teams in the Big Ten Conference last year, plan a study of the relationship between the average number of tickets sold for home soccer games and the percentage of games won. Give the key characteristics of the data that could be used for your study.

2.10

2.2 Scatterplots

When you complete this section, you will be able to:

Make a scatterplot to examine a relationship between two variables. Describe the overall pattern in a scatterplot and any striking deviations from that pattern. Use a scatterplot to describe the form, direction, and strength of a relationship. Use a scatterplot to identify outliers. Identify a linear pattern in a scatterplot. Explain the effect of a change of units on a scatterplot. Use a log transformation to change a curved relationship into a linear relationship. Use different plotting symbols to include information about a categorical variable in a scatterplot.

EXAMPLE 2.8

LAUNDRY

Laundry detergents. Consumers Union provides ratings on a large variety of consumer products. They use sophisticated testing methods as well as surveys of their members to create these ratings. The ratings are published in their magazine, Consumer Reports.4

One recent study rated 53 laundry detergents on a scale from 1 to 100. The scale summarizes washing performance under a variety of conditions. Price per load is given in cents.5 We will examine the relationship between rating and price per load for these laundry detergents. We expect that the higher-priced detergents will tend to have higher ratings.

USE YOUR KNOWLEDGE

LAUNDRY

Examine the spreadsheet. Examine the spreadsheet of the laundry detergent data. (a) How many cases are in the data set?

2.11

(b) Describe the labels, variables, and values.

(c) Which columns represent quantitative variables? Which columns give categorical variables.

(d) Is there an explanatory variable? A response variable? Explain your answer.

LAUNDRY

Use the data set. Using the data set from the previous exercise, create graphical and numerical summaries for the rating and for the price per load.

The most common way to display the relationship between two quantitative variables is a scatterplot.

SCATTERPLOT

A scatterplot shows the relationship between two quantitative variables measured on the same cases. The values of one variable appear on the horizontal axis, and the values of the other variable appear on the vertical axis. Each case in the data appears as the point in the plot determined by the values of both variables for that case.

EXAMPLE 2.9

LAUNDRY

Laundry detergents. A higher price for a product should be associated with a better product. Therefore, let’s treat price per load as the explanatory variable and rating as the response variable in our examination of the relationship between these two variables. We begin with a graphical display.

Figure 2.1 gives a scatterplot that displays the relationship between the response variable, rating, and the explanatory variable, price per load. The most striking feature that we see in the plot is a case that appears to be very different from the others. One of the laundry detergents has a rating that is about average (51), but the price per load (56 cents) is almost double that of the other products.

FIGURE 2.1 Scatterplot of price per load (in cents) versus rating for 53 laundry detergents, Example 2.9.

Cases that fall well outside the general pattern of the relationship are called outliers. We provide a more detailed description of these in Section 2.5. For now, we remove this case and focus on the relationship of the remaining data.

Figure 2.2 gives the scatterplot with the outlier removed. The relationship is weak. Paying a high price for your laundry detergent will not guarantee that you have selected a highly rated product.

2.12

2.13

FIGURE 2.2 Scatterplot of price per load (in cents) versus rating for 52 laundry detergents (with the outlier removed), Example 2.9.

Always plot the explanatory variable, if there is one, on the horizontal axis (the x axis) of a scatterplot. We usually call the explanatory variable x and the response variable y. If there is no explanatory-response distinction, either variable can go on the horizontal axis. Time plots such as the one in Figure 1.12 are special scatterplots where the explanatory variable x is a measure of time.

USE YOUR KNOWLEDGE

LAUND

Make a scatterplot. Let’s consider the laundry data with the outlier removed. (a) Make a scatterplot similar to Figure 2.2.

(b) Two of the laundry detergents cost 14 cents per load with a rating of 60. Mark the location of these items on your plot.

(c) Cases with identical values for both variables are generally indistinguishable in a scatterplot. To what extent do you think that this could give a distorted picture of the relationship between two variables for a data set that has a large number of duplicate values? Explain your answer. jitter

(d) An option called jitter is available with some statistical software that will add a little noise to each point so that points with identical values will appear to be different. If you have software that includes this option, apply it to your plot and summarize the effect of the jittering.

LAUNDRY

Change the units. Refer to the laundry data with the outlier. (a) Create a spreadsheet with the price per load expressed in dollars.

(b) Make a scatterplot for the data in your spreadsheet.

(c) Describe how this scatterplot differs from Figure 2.2.

Interpreting scatterplots To look more closely at a scatterplot such as Figure 2.2, apply the strategies of exploratory analysis learned in Chapter 1.

EXAMINING A SCATTERPLOT

In any graph of data, look for the overall pattern and for striking deviations from that pattern.

You can describe the overall pattern of a scatterplot by the form, direction, and strength of the relationship.

linear

relationship

The relationship in Figure 2.2 is difficult to see. Looking at it carefully suggests that its form is approximately linear. In other words, it may be appropriate to summarize the relationship with a straight line. To explore this possibility, we can use software to put a straight line through the data. We will see more details about how this is done in Section 2.4.

EXAMPLE 2.10

LAUND

Scatterplot with a straight line. Figure 2.3 plots the laundry detergent data with a straight line. The line helps us to see and to evaluate the linear form of the relationship. In Section 2.4 (page 107), we will learn how to determine this line.

FIGURE 2.3 Scatterplot of rating versus price per load (in cents), with a fitted straight line, Example 2.10.

There is a large amount of scatter about the line. We see that there are eight laundry detergents with a price of 14 cents per load. For these products, the variation in ratings is substantial, from 25 to 71. We do not see any additional outliers in this plot.

Although it is very weak, the relationship in Figure 2.3 has a direction, laundry detergents that cost more have somewhat higher ratings. This is a positive association between the two variables.

POSITIVE ASSOCIATION, NEGATIVE ASSOCIATION

Two variables are positively associated when above-average values of one tend to accompany above-average values of the other and below-average values also tend to occur together.

Two variables are negatively associated when above-average values of one tend to accompany below-average values of the other, and vice versa.

The strength of a relationship in a scatterplot is determined by how closely the points follow a clear form. The overall relationship in Figure 2.3 is weak. Here is an example of a stronger linear relationship.

EXAMPLE 2.11

EDSPEND

benchmarking

Education spending and population: Benchmarking. We expect that states with larger populations would spend more on education than states with smaller populations.6 What is the nature of this relationship? Can we use this relationship to evaluate whether some states are spending more than we expect or less than we expect? This type of exercise is called benchmarking. The basic idea is to compare processes or procedures of an organization with those of similar organizations.

Figure 2.4 is a spreadsheet giving the education spending and the populations of the 50 U.S. states for 2015. Figure 2.5 is a scatterplot of the education spending versus the population with a straight line. The scatterplot shows a strong positive relationship between these two variables.

FIGURE 2.4 State spending (in billions of dollars) and population (in millions) for the 50 U.S. states, Example 2.11.

2.14

FIGURE 2.5 Scatterplot of state spending (in billions of dollars) versus population for the 50 U.S. states, with a fitted straight line, Example 2.11.

USE YOUR KNOWLEDGE

Make a scatterplot. In our Mocha Frappuccino® example, the 12-ounce drink costs $3.95, the 16-ounce drink costs $4.45, and the 24-ounce drink costs $4.95. Explain which variable should be used as the explanatory variable, and make a scatterplot and include the fitted straight line if your software includes this option. Describe the scatterplot and the association between these two variables.

Of course, not all relationships are linear. Here is an example where a relationship is described by a curve.

EXAMPLE 2.12

CALCIUM

Calcium retention. Our bodies need calcium to build strong bones. How much calcium do we need? Does the amount that we need depend on our age? Questions like these are studied by nutrition researchers. One series of studies used the amount of calcium retained by the body as a response variable and the amount of calcium consumed as an explanatory variable.7

Figure 2.6 is a scatterplot of calcium retention in milligrams per day (mg/d) versus calcium intake (mg/d) for 56 children aged 11 to 15 years. A smooth curve generated by software helps us see the relationship between the two variables.

FIGURE 2.6 Scatterplot of calcium retention (mg/d) versus calcium intake (mg/d) for 56 children, with a fitted curve, Example 2.12. There is a positive relationship between these two variables, but it is not linear.

There is clearly a relationship here. As calcium intake increases, the body retains more calcium. However, the relationship is not linear. The curve is approximately linear for low values of intake, but then the line curves more and becomes almost level.

transformation

There are many kinds of curved relationships like that in Figure 2.6. For some of these, we can apply a transformation to the data that will make the relationship approximately linear. To do this, we replace the original values with the transformed values and then use the transformed values for our analysis.

Transforming data is common in statistical practice. There are systematic principles that describe how transformations behave and guide the search for transformations that will, for example, make a distribution more Normal or a curved relationship more linear.

The log transformation log transformation

The most important transformation that we will use is the log transformation. This transformation can be used for variables that have positive values only. Occasionally, we use it when there are zeros, but in this case we first replace the zero values by some small value, often one-half of the smallest positive value in the data set.

You have probably encountered logarithms in one of your high school mathematics courses as a way to do certain kinds of arithmetic. Logarithms are a powerful tool when used in statistical analyses. We will use natural logarithms. Statistical software and statistical calculators generally provide easy ways to perform this transformation.

Let’s try a log transformation on our calcium retention data. Here are the details.

EXAMPLE 2.13

CALCIUM

Calcium retention with logarithms. Figure 2.7 is a scatterplot of the log of calcium retention versus calcium intake. The plot includes a fitted straight line to help us see the relationship. We see that the transformation has worked. Our relationship is now approximately linear.

FIGURE 2.7 Scatterplot of log calcium retention versus calcium intake, with a fitted line, for 56 children, Example 2.13. The relationship is approximately linear.

Our analysis of the calcium retention data in Examples 2.12 and 2.13 reminds us of an important issue when describing relationships. In Example 2.12, we noted that the relationship appeared to become approximately flat. Biological processes are consistent with this observation. There is probably a point where additional intake does not result in any additional retention. With our transformed relationship in Figure 2.7, however, there is no leveling off as we saw in Figure 2.6, even though we appear to have a good fit to the data. The relationship and fit apply to the range of data that are analyzed. We cannot assume that the relationship extends beyond the range of the data.

For the calcium data, we used a log transformation to describe the curved relationship in Figure 2.6 as the linear relationship in Figure 2.7. Here is another application of a log transformation.

2.15

EXAMPLE 2.14

EDSPEND

Education spending and population with logarithms. Let’s examine the relationship between spending and population using logs for both variables. Figure 2.8 gives the plot with the fitted line.

FIGURE 2.8 Scatterplot of log spending versus log population for the 50 U.S. states, with a fitted line, Example 2.14. The relationship is approximately linear.

USE YOUR KNOWLEDGE

EDSPEND

Compare the plots. Compare the plot in Figure 2.8 with the one in Figure 2.5. Which one do you prefer? Give reasons for your answer.

Use of transformations and the interpretation of scatterplots are an art that requires judgment and knowledge about the variables that we are studying. Always ask yourself if the relationship that you see makes sense. If it does not, then additional analyses are needed to understand the data.

Adding categorical variables to scatterplots In Figure 2.3, we looked at the relationship between the rating and the price per load for 52 laundry detergents. A more detailed look at the data shows that there are two different types of laundry detergent included in this data set, liquid and powder. Let’s examine where these two types of laundry detergents are in our plot.

CATEGORICAL VARIABLES IN SCATTERPLOTS

To add a categorical variable to a scatterplot, use a different plot color or symbol for each category.

EXAMPLE 2.15

LAUND

Rating versus price and type of laundry detergent. In our scatterplot, we use the color blue for liquids and the color red for powders. The scatterplot is given in Figure 2.9. Separate lines are given for each type of laundry detergent. Most of the laundry detergents are liquids. There are three powders with somewhat low prices and four powders with relatively high prices. The prices of the powders are similar to the prices of the liquids.

FIGURE 2.9 Scatterplot of rating versus price per load (in cents), with fitted straight lines, for 52 laundry detergents, Example 2.15. The type of detergent is indicated by the color: blue for liquid and red for powder.

In this example, we used a categorical variable, type, to distinguish the two types of laundry detergents in our plot. Suppose that the additional variable that we want to investigate is quantitative. In this situation, we sometimes can combine the values into ranges of the quantitative variable—such as high, medium, and low—to create a categorical variable.

Careful judgment is needed in using this graphical method. Don’t be discouraged if your first attempt is not very successful. In performing a good data analysis, you will often produce several plots before you find the one that you believe to be the most effective in describing the data.8

Scatterplot smoothers smoothing

In Figure 2.6, we added a curve to our scatterplot to better understand the relationship between calcium retention and calcium intake. This curve helped us to see that the amount of calcium retained tends to level off as the intake increases. The method that we used to construct the curve is called smoothing.

Today, most statistical software includes options to perform the calculations needed for smoothing. The technical details vary, but the basic idea is that there is a smoothing parameter that controls the degree to which the relationship is smoothed. Here is another example.

EXAMPLE 2.16

LAUND

Laundry rating versus price with a smooth fit. Figure 2.2 gives the scatterplot for rating versus price for the remaining 52 laundry detergents that we studied in Example 2.9. In Figure 2.3, we added a straight line to the plot to help us see the relationship. Figure 2.10 shows the laundry detergent with two different smooth curves. The first (a) used a relatively small value of the smoothing parameter. The second (b) used a larger value, making the curve smoother. Overall, the relationship is very weak and there is no clear pattern in the plot.

FIGURE 2.10 Scatterplot of rating versus price per load (in cents), with smooth curves, Example 2.16: (a) with a small value of the smoothing parameter; (b) with a higher value of the smoothing parameter.

Scatterplot smoothers can help you to learn about relationships between two quantitative variables. They can confirm that there is a linear relationship, or they can suggest other features that are not evident in a casual look at the scatterplot. Here is an example of the latter scenario.

EXAMPLE 2.17

EDSPEND

A smooth fit for education spending and population with logs. Figure 2.11 gives the scatterplot of log education spending versus log population with a smooth curve. The curve suggests that the relationship is approximately linear except for states with relatively small populations. For these, the spending is relatively flat.

FIGURE 2.11 Scatterplot of log spending versus log population, with a smooth curve fitted to the data, for 50 U.S. states, Example 2.17. This smooth curve fits the data very well and suggests that the relationship is generally linear except for states with small populations.

Categorical explanatory variables Scatterplots display the association between two quantitative variables. To display a relationship between a categorical variable and a quantitative variable, make a side-by-side comparison of the distributions of the response for each category. Back-to-back stemplots (page 12) and side-by-side boxplots (page 37) are useful tools for this purpose.

We will study methods for describing the association between two categorical variables in Section 2.6 (page 136).

SECTION 2.2 SUMMARY A scatterplot displays the relationship between two quantitative variables. Mark values of one variable on the horizontal axis (x axis) and values of the other variable on the vertical axis (y axis). Plot each individual’s data as a point on the graph. Always plot the explanatory variable, if there is one, on the x axis of a scatterplot. Plot the response variable on the y axis. In examining a scatterplot, look for an overall pattern showing the form, direction, and strength of the relationship, and then for outliers or other deviations from this pattern. Form: Linear relationships, where the points show a straight-line pattern, are an important form of relationship between two variables. Curved relationships are other forms to watch for. Direction: If the relationship has a clear direction, we speak of either positive association (high values of the two variables tend to occur together) or negative association (high values of one variable tend to occur with low values of the other variable). Strength: The strength of a relationship is determined by how close the points in the scatterplot lie to a simple form such as a line. Plot points with different colors or symbols to see the effect of a categorical variable in a scatterplot. To display the relationship between a categorical explanatory variable and a quantitative response variable, make a graph that compares the distributions of the response for each category of the explanatory variable. A log transformation of one or both variables in a scatterplot can help us to understand the relationship between two quantitative variables. A scatterplot smoother is a tool to examine the relationship between two quantitative variables by fitting a smooth curve to the data. The amount of smoothing can be varied using a smoothing parameter.

SECTION 2.2 EXERCISES For Exercises 2.10 and 2.11, see page 85; for Exercises 2.12 and 2.13, see page 87; for Exercise 2.14, see page 90; and for Exercise 2.15, see page 92.

2.16 Make some sketches. For each of the following situations, make a scatterplot that illustrates the given relationship between two variables.

(a) No apparent relationship.

(b) A strong negative linear relationship.

(c) A weak positive relationship that is not linear.

(d) A more complicated relationship. Explain the relationship.

2.17 What’s wrong? Explain what is wrong with each of the following:

(a) If two variables are negatively associated, then low values of one variable are associated with low values of the other variable.

(b) A stemplot can be used to examine the relationship between two variables.

(c) In a scatterplot, we put the response variable on the x axis and the explanatory variable on the y axis.

2.18 Blueberries and anthocyanins. Anthocyanins are compounds that have been associated with health benefits associated with the heart, bones, and the brain. Blueberries are a good source of many different anthocyanins. Researchers at the Piedmont Research Station of North Carolina State University have assembled a database giving the concentrations of 18 different anthocyanins for 267 varieties of blueberries.9 Four of the anthocyanins measured are delphinidin-3-arabinoside, malvidin-3-arabinoside, cyanidin-3-galactoside, and delphinidin-3-glucoside, all measured in units of mg per 100g of berries (dry weight). In the data file, we have simplified the names of these anthocyanins to Antho1, Antho2, Antho3, and Antho4. In Exercises 1.167 and 1.168 (page 77), you examined the distributions of each Antho3 and Antho4. BERRIES

(a) Make a scatterplot of the data with Antho3 on the x axis and Antho4 on the y axis.

(b) Describe the form, direction, and strength of the relationship.

(c) Are there any outliers or unusual observations?

(d) Is it useful to add a straight line to your scatterplot? Explain your answer.

(e) If you have access to the appropriate software, explore the use of a scatterplot smoother to understand this relationship. Summarize what you have found using this method.

2.19 Blueberries and anthocyanins with logs. Refer to the previous exercise. Transform each of the variables with a log, make a scatterplot and answer the questions in the previous exercise for the transformed data. BERRIES

2.20 Blueberries and anthocyanins: Raw data or logs. Refer to Exercises 2.18 and 2.19.

(a) Compare your results from the two exercises.

(b) For exploring and explaining the relationship between Antho4 and Antho3, do you prefer the analysis you performed in Exercise 2.18 or the one you performed in Exercise 2.19? Give reasons for your answer. BERRIES

2.21 Fuel consumption. Natural Resources Canada tests new vehicles each year and reports several variables related to fuel consumption for vehicles in different classes.10 For 2015 they provide data for 527 vehicles that use regular fuel. Two variables reported are carbon dioxide (CO2) emissions and highway fuel consumption. CO2 is measured in grams per kilometer (g/km) and highway fuel consumption measured in liters per 100 kilometers (L/100km). CANFREG

(a) Make a scatterplot of the data with highway fuel consumption on the x axis and CO2 emissions on the y axis.

(b) Describe the form, direction, and strength of the relationship.

(c) Are there any outliers or unusual observations?

(d) Is it useful to add a straight line to your scatterplot? Explain your answer.

(e) If you have access to the appropriate software, explore the use of a scatterplot smoother to understand this relationship. Summarize what you find using this method.

2.22 Fuel consumption with a line. Refer to the previous exercise. CANFREG

(a) Add a line to the plot. To what extent to you think that the line does a good job of summarizing the relationship?

(b) If your have the appropriate software, use smooth curves to examine the relationship. Does your analysis support the idea of using a straight line to summarize the relationship? Explain your answer.

2.23 Fuel consumption for different types of vehicles. Refer to the previous two exercises. Those exercises examined data for vehicles that used regular fuel. Data are also available for vehicles that use several other types of fuel. There are 1067 vehicles in total. The variable Fuel has four different possible values: X, for regular fuel; Z, for premium fuel; D, for diesel; and E, for ethanol. CANFUEL

(a) Make a scatterplot of all of the data using different symbols or colors for the different fuel types.

(b) Does the relationship between CO2 and highway fuel consumption depend upon the type of fuel that the vehicle uses? Explain your answer.

2.24 Bone strength. Osteoporosis is a condition where bones become weak. It affects more than 200 million people worldwide. Exercise is one way to produce strong bones and to prevent osteoporosis. Because we use our dominant arm (the right arm for most people) more than our nondominant arm, we expect the bone in our dominant arm to be stronger than the bone in our nondominant arm. By comparing the strengths, we can get an idea of the effect that exercise can have on bone strength. Here are some data on the strength of bones, measured in Newton meters divided by 1000 (Nm/1000), for the arms of 15 young men:11 ARMSTR

ID Nondominant Dominant

1 15.7 16.3 2 25.2 26.9 3 17.9 18.7 4 19.1 22.0 5 12.0 14.8 6 20.0 19.8 7 12.3 13.1 8 14.4 17.5 9 15.9 20.1 10 13.7 18.7 11 17.7 18.7 12 15.5 15.2 13 14.4 16.2 14 14.1 15.0 15 12.3 12.9

Before attempting to compare the arm strengths of the nondominant and dominant arms, let’s take a careful look at the data for these two variables.

(a) Make a scatterplot of the data with the nondominant arm strength on the x axis and the dominant arm strength on the y axis.

(b) Describe the overall pattern in the scatterplot and any striking deviations from the pattern.

(c) Describe the form, direction, and strength of the relationship.

(d) Identify any outliers.

(e) Is the relationship approximately linear?

2.25 Bone strength for baseball players. Refer to the previous exercise. The study collected arm bone strength information for two groups of young men. The data in the previous exercise were for a control group. The second group in the study comprised men who played baseball. We know that these baseball players use their dominant arm in throwing (those who

throw with their nondominant arm were excluded), so they get more arm exercise than the controls. Here are the data for the baseball players:

ID Nondominant Dominant

16 17.0 19.3 17 16.9 19.0 18 17.7 25.2 19 21.2 37.7 20 21.0 40.3 21 14.6 20.8 22 31.5 36.9 23 14.9 21.2 24 15.1 19.4 25 13.5 20.4 26 13.6 17.1 27 20.3 26.5 28 17.3 30.3 29 14.6 17.4 30 22.6 35.0

Answer the questions in the previous exercise for the baseball players. ARMSTR

2.26 Compare the baseball players with the controls. Refer to the previous two exercises. ARMSTR

(a) Plot the data for the two groups on the same graph using different symbols for the baseball players and the controls.

(b) Use your plot to describe and compare the relationships for the two variables. Write a short paragraph summarizing what you have found.

2.27 Parents’ income and student loans. How well does the income of a college student’s parents predict how much the student will borrow to pay for college? We have data on parents’ income and college debt for a sample of 1200 recent college graduates. What are the explanatory and response variables? Are these variables categorical or quantitative? Do you expect a positive or negative association between these variables? Why?

2.28 What’s in the beer? The website beer100.com advertises itself as “Your Place for All Things Beer.” One of their “things” is a list of 159 domestic beer brands with the percent alcohol, calories per 12 ounces, and carbohydrates per 12 ounces (in grams).12 BEERD

(a) Figure 2.12 gives a scatterplot of calories versus percent alcohol. Give a short summary of what can be learned from the plot.

FIGURE 2.12 Scatterplot of calories versus percent alcohol for 159 domestic brands of beer, Exercise 2.28.

(b) One of the points is an outlier. Find the brand of the outlier. How is this brand of beer different from the other brands?

(c) Remove the outlier from the data set and generate a scatterplot of the remaining data.

(d) Describe the relationship between calories and percent alcohol based on what you see in your scatterplot.

2.29 More beer. Refer to the previous exercise. BEERD

(a) Make a scatterplot of calories versus percent alcohol using the data set without the outlier.

(b) Describe the relationship between these two variables. If your software is capable, use a line and smoothers to explore the relationship.

2.30 Imported beer. The beer100 website also gives data for imported beers. Describe the relationship between calories and percent alcohol for these imported beers. BEERI

2.31 Compare domestic with imported. Plot calories versus percent alcohol for domestic and imported beers on the same scatterplot. Use different colors or symbols for the two types of beers. Summarize what this plot tells you about the relationship and the difference between the two types of beer. In particular, note any characteristics that are better shown in this plot relative to what was learned in Exercises 2.28, 2.29, and 2.30. BEER

2.32 Decay of a radioactive element. Barium-137m is a radioactive form of the element barium that decays very rapidly. It is easy and safe to use for lab experiments in schools and colleges.13 In a typical experiment, the radioactivity of a sample of barium-137m is measured for one minute. It is then measured for three additional one-minute periods, separated by two minutes. So data are recorded at one, three, five, and seven minutes after the start of the first counting period. The measurement units are counts. Here are the data for one of these experiments:14 DECAY

Time 1 3 5 7 Count 578 317 203 118

(a) Make a scatterplot of the data. Give reasons for the choice of which variables to use on the x and y axes.

(b) Describe the overall pattern in the scatterplot and any striking deviations from the pattern.

(c) Describe the form, direction, and strength of the relationship.

(d) Identify any outliers.

(e) Is the relationship approximately linear?

2.33 Use a log for the radioactive decay. Refer to the previous exercise. Transform the counts using a log transformation. Then repeat parts (a) through (e) for the transformed data and compare your results with those from the previous exercise. DECAY

2.34 Internet use and babies. The World Bank collects data on many variables related to world development for countries throughout the world. Two of these are Internet use, in number of users per 100 people, and birthrate, in births per 1000 people.15 Figure 2.13 is a scatterplot of birthrate versus Internet use for the 106 countries that have data available for both variables. INBIRTH

(a) Describe the relationship between these two variables.

(b) A friend looks at this plot and concludes that using the Internet will decrease the number of babies born. Write a short paragraph explaining why the association seen in the scatterplot does not provide a reason to draw this conclusion.

2.35 Try a log. Refer to the previous exercise. INBIRTH

(a) Make a scatterplot of the log of births per 1000 people versus Internet users per 100 people.

(b) Describe the relationship that you see in this plot and compare it with Figure 2.13.

(c) Which plot do you prefer? Give reasons for your answer.

FIGURE 2.13 Scatterplot of births (per 1000 people) versus Internet users (per 100 people) for 106 countries, Exercise 2.34.

2.36 Make another plot. Refer to Exercise 2.34. INBIRTH

(a) Make a new data set that has Internet users expressed as users per 10,000 people and births as births per 10,000 people.

(b) Explain why these transformations to give new variables are linear transformations. (Hint: See linear transformations on page 44.)

(c) Make a scatterplot using the transformed variables.

(d) Compare your new plot with the one in Figure 2.13.

(e) Why do you think that the analysts at the World Bank chose to express births as births per 1000 people and Internet users as users per 100 people?

2.37 Body mass and metabolic rate. Metabolic rate, the rate at which the body consumes energy, is important in studies of weight gain, dieting, and exercise. The following table gives data on the lean body mass and resting metabolic

rate for 12 women and 7 men who are subjects in a study of dieting. Lean body mass, given in kilograms, is a person’s weight

leaving out all fat. Metabolic rate is measured in calories burned per 24 hours, the same calories used to describe the energy content of foods. The researchers believe that lean body mass is an important influence on metabolic rate. BMASS

Subject Sex Mass Rate

1 M 62.0 1792 2 M 62.9 1666 3 F 36.1 995 4 F 54.6 1425 5 F 48.5 1396 6 F 42.0 1418 7 M 47.4 1362 8 F 50.6 1502 9 F 42.0 1256 10 M 48.7 1614 11 F 40.3 1189 12 F 33.1 913 13 M 51.9 1460 14 F 42.4 1124 15 F 34.5 1052 16 F 51.1 1347 17 F 41.2 1204 18 M 51.9 1867 19 M 46.9 1439

(a) Make a scatterplot of the data, using different symbols or colors for men and women.

(b) Is the association between these variables positive or negative? What is the form of the relationship? How strong is the relationship?

(c) Does the pattern of the relationship differ for women and men? How do the male subjects as a group differ from the female subjects as a group?

2.3 Correlation

When you complete this section, you will be able to:

Use a correlation to describe the direction and strength of a linear relationship between two quantitative variables. Interpret the sign of a correlation. Identify situations in which the correlation is not a good measure of association between two quantitative variables. Identify a linear pattern in a scatterplot. For describing the relationship between two quantitative variables, identify the roles of the correlation, a numerical summary, and the scatterplot, a graphical summary.

A scatterplot displays the form, direction, and strength of the relationship between two quantitative variables. Linear (straight-line) relations are particularly important because a straight line is a simple pattern that is quite common. We say a linear relationship is strong if the points lie close to a straight line and weak if they are widely scattered about a line. Our eyes are not good judges of how strong a relationship is. The two scatterplots in Figure 2.14 depict exactly the same data, but the plot on the right is drawn smaller in a large field. The plot on the right seems to show a stronger relationship.

FIGURE 2.14 Two scatterplots of the same data. The linear pattern in the plot on the right appears stronger because of the surrounding space.

Our eyes can be fooled by changing the plotting scales or the amount of white space around the cloud of points in a scatterplot.16 We need to follow our strategy for data analysis by using a numerical measure to supplement the graph. Correlation is the measure we use.

2.38

2.39

The correlation r We have data on variables x and y for n individuals. Think, for example, of measuring height and weight for n people. Then x1 and y1 are your height and your weight, x2 and y2 are my height and my weight, and so on. For the ith individual, height xi goes with weight yi. Here is the definition of correlation.

CORRELATION

The correlation measures the direction and strength of the linear relationship between two quantitative variables. Correlation is usually written as r.

Suppose that we have data on variables x and y for n individuals. The means and standard deviations of the two variables are and sx for the x-values, and and sy for the y-values. The correlation r between x and y is

As always, the summation sign Σ means “add these terms for all the individuals.” The formula for the correlation r is a bit complex. It helps us see what correlation is but is not convenient for actually calculating r. In practice, you should use software or a calculator that computes r from the values of x and y pairs.

standardize, p. 59

The formula for r begins by standardizing the observations. Suppose, for example, that x is height in centimeters and y is weight in kilograms and that we have height and weight measurements for n people. Then and sx are the mean and standard deviation of the n heights, both in centimeters. The value

is the standardized height of the ith person. The standardized height says how many standard deviations above or below the mean a person’s height lies. Standardized values have no units—in this example, they are no longer measured in centimeters. You can standardize the weights also. The correlation r is an average of the products of the standardized height and the standardized weight for the n people.

USE YOUR KNOWLEDGE

LAUNDRY

Laundry detergents. Example 2.8 (page 85) describes data on the rating and price per load for 53 laundry detergents. Use these data to compute the correlation between rating and the price per load.

LAUNDRY

Change the units. Refer to the previous exercise. Express the price per load in dollars.

(a) Is the transformation from cents to dollars a linear transformation? Explain your answer.

(b) Compute the correlation between rating and price per load expressed in dollars.

(c) How does the correlation that you computed in part (b) compare with the one you computed in the previous exercise?

(d) What can you say in general about the effect of changing units using linear transformations on the size of the correlation?

Properties of correlation The formula for correlation helps us see that r is positive when there is a positive association between the variables. Height and weight, for example, have a positive association. People who are above average in height tend to also be above average in weight. Both the standardized height and the standardized weight for such a person are positive. People who are below average in height tend also to have below-average weight. Then both standardized height and standardized weight are negative. In both cases, the products in the formula for r are mostly positive, so r is positive. In the same way, we can see that r is negative when the association between x and y is negative. More detailed study of the formula gives more detailed properties of r.

Here is what you need to know to interpret correlation:

Correlation makes no use of the distinction between explanatory and response variables. It makes no difference which variable you call x and which you call y in calculating the correlation.

Correlation requires that both variables be quantitative. For example, we cannot calculate a correlation between the incomes of a group of people and what city they live in because city is a categorical variable. Because r uses the standardized values of the observations, r does not change when we change the units of measurement (a linear transformation) of x, y, or both. Measuring height in inches rather than centimeters and weight in pounds rather than kilograms does not change the correlation between height and weight. The correlation r itself has no unit of measurement; it is just a number. Positive r indicates positive association between the variables, and negative r indicates negative association. The correlation r is always a number between −1 and 1. Values of r near 0 indicate a very weak linear relationship. The strength of the relationship increases as r moves away from 0 toward either −1 or 1. Values of r close to −1 or 1 indicate that the points lie close to a straight line. The extreme values r = −1 and r = 1 occur only when the points in a scatterplot lie exactly along a straight line.

Correlation measures the strength of only the linear relationship between two variables. Correlation does not describe curved relationships between variables, no matter how strong they are.

resistant, p. 30

Like the mean and standard deviation, the correlation is not resistant: r is strongly affected by a few outlying observations. Use r with caution when outliers appear in the scatterplot.

The scatterplots in Figure 2.15 illustrate how values of r closer to 1 or −1 correspond to stronger linear relationships. To make the essential meaning of r clear, the standard deviations of both variables in these plots are equal, and the horizontal and vertical scales are the same. In general, it is not so easy to guess the value of r from the appearance of a scatterplot. Remember that changing the plotting scales in a scatterplot may mislead our eyes, but it does not change the standardized values of the variables and, therefore, cannot change the correlation. To explore how extreme observations can influence r, use the Correlation and Regression applet available on the text website. Also, see Exercises 2.56 and 2.57 (page 106).

FIGURE 2.15 How the correlation r measures the direction and strength of a linear association.

Finally, remember that correlation is not a complete description of two-variable data, even when the relationship between the variables is linear. You should give the means and standard deviations of both x and y along with the correlation. (Because the formula for correlation uses the means and standard deviations, these measures are the proper choices to accompany a correlation.) Conclusions based on correlations alone may require rethinking in the light of a more complete description of the data.

EXAMPLE 2.18

Scoring of figure skating in the Olympics. Until a scandal at the 2002 Olympics brought change, figure skating was scored by judges on a scale from 0.0 to 6.0. The scores were often controversial. We have the scores awarded by two judges, Pierre and Elena, to many skaters. How well do they agree? We calculate that the correlation

between their scores is r = 0.9. But the mean of Pierre’s scores is 0.8 point lower than Elena’s mean.

These facts in the example above do not contradict each other. They are simply different kinds of information. The mean scores show that Pierre awards lower scores than Elena. But because Pierre gives every skater a score about 0.8 point lower than Elena, the correlation remains high. Adding the same number to all values of either x or y does not change the correlation. If both judges score the same skaters, the competition is scored consistently because Pierre and Elena agree on which performances are better than others. The high r shows their agreement. But if Pierre scores some skaters and Elena others, we must add 0.8 point to Pierre’s scores to arrive at a fair comparison.

SECTION 2.3 SUMMARY The correlation r measures the direction and strength of the linear (straight line) association between two quantitative variables x and y. Although you can calculate a correlation for any scatterplot, r measures only linear relationships. Correlation indicates the direction of a linear relationship by its sign: r > 0 for a positive association and r < 0 for a negative association. Correlation always satisfies −1 ≤ r ≤ 1 and indicates the strength of a relationship by how close it is to −1 or 1. Perfect correlation, r = ±1, occurs only when the points lie exactly on a straight line. Correlation ignores the distinction between explanatory and response variables. The value of r is not affected by changes in the unit of measurement of either variable. Correlation is not resistant, so outliers can greatly change the value of r.

SECTION 2.3 EXERCISES For Exercises 2.38 and 2.39, see page 102.

2.40 Correlations and scatterplots. Explain why you should always look at a scatterplot when you want to use a correlation to describe the relationship between two quantitative variables.

2.41 Interpret some correlations. For each of the following correlations, describe the relationship between the two quantitative variables in terms of the direction and the strength of the linear relationship.

(a) r = 0.9.

(b) r = −0.9.

(c) r = −0.3.

(d) r = 0.0.

2.42 Blueberries and anthocyanins. In Exercise 2.18 (page 97), you examined the relationship between Antho4 and Antho3, two anthocyanins found in blueberries. BERRIES

(a) Find the correlation between these two anthocyanins.

(b) Look at the scatterplot for these data that you made in part (a) of Exercise 2.18 (or make one if you did not do that exercise). Is the correlation a good numerical summary of the graphical display in the scatterplot? Explain your answer.

(c) Does the size of the correlation suggest that the amounts of these two anthocyanins is approximately equal in these blueberries? Explain why or why not.

2.43 Blueberries and anthocyanins with logs. In Exercise 2.19 (page 97), you examined the relationship between Antho4 and Antho3, two anthocyanins found in blueberries, using logs for both variables. Answer the questions in the previous exercise for the variables transformed in this way. BERRIES

2.44 Fuel consumption. In Exercise 2.21 (page 97), you examined the relationship between CO2 emissions and highway fuel consumption for 527 vehicles that use regular fuel. Find the correlation between these two variables. Write a short paragraph describing the relationship using the scatterplot and the correlation. CANFREG

2.45 Fuel consumption for different types of vehicles. In Exercise 2.23 (page 97), you examined the relationship between CO2 emissions and highway fuel consumption for 1067 vehicles that use four different types of fuel. Find the correlations between CO2 and highway fuel consumption for each of these four categories of vehicle. Summarize your results explaining similarities and differences in the relationships among the four types of fuel. CANFUEL

2.46 Strong association but no correlation. Here is a data set that illustrates an important point about correlation: CORR

X 25 35 45 55 65 Y 10 30 50 30 10

(a) Make a scatterplot of Y versus X.

(b) Describe the relationship between Y and X. Is it weak or strong? Is it linear?

(c) Find the correlation between Y and X.

(d) What important point about correlation does this exercise illustrate?

2.47 Bone strength. Exercise 2.24 (page 97) gives the bone strengths of the dominant and the nondominant arms for 15 men who were controls in a study. ARMSTR

(a) Find the correlation between the bone strength of the dominant arm and the bone strength of the nondominant arm.

(b) Look at the scatterplot for these data that you made in part (a) of Exercise 2.24 (or make one if you did not do that exercise). Is the correlation a good numerical summary of the graphical display in the scatterplot? Explain your answer.

2.48 Bone strength for baseball players. Refer to the previous exercise. Similar data for baseball players are given in Exercise 2.25 (page 98). Answer parts (a) and (b) of the previous exercise for these data. ARMSTR

2.49 Student ratings of teachers. A college newspaper interviews a psychologist about student ratings of the teaching of faculty members. The psychologist says, “The evidence indicates that the correlation between the research productivity and teaching rating of faculty members is close to zero.” The paper reports this as “Professor McDaniel said that good researchers tend to be poor teachers, and vice versa.” Explain why the paper’s report is wrong. Write a statement in plain language (don’t use the word “correlation”) to explain the psychologist’s meaning.

2.50 Decay of a radioactive element. Data for an experiment on the decay of barium-137m is given in Exercise 2.32 (page 99). DECAY

(a) Find the correlation between the radioactive counts and the time after the start of the first counting period.

(b) Does the correlation give a good numerical summary of the relationship between these two variables? Explain your answer.

2.51 Decay in the log scale. Refer to the previous exercise and to Exercise 2.33 (page 99), where the counts were transformed by a log. DECAY

(a) Find the correlation between the log counts and the time after the start of the first counting period.

(b) Does the correlation give a good numerical summary of the relationship between these two variables? Explain your answer.

(c) Compare your results for this exercise with those from the previous exercise.

2.52 Brand names and generic products.

(a) If a store always prices its generic “store brand” products at 80% of the brand name products’ prices, what would be the correlation between the prices of the brand name products and the store brand products? (Hint: Draw a scatterplot for several prices.)

(b) If the store always prices its generic products $2 less than the corresponding brand name products, then what would be the correlation between the prices of the brand name products and the store brand products?

2.53 Alcohol and calories in beer. Figure 2.12 gives a scatterplot of the calories versus percent alcohol for 159 brands of domestic beer. BEERD

(a) Compute the correlation for these data.

(b) Does the correlation do a good job of describing the direction and strength of this relationship? Explain your answer.

2.54 Alcohol and calories in beer revisited. Refer to the previous exercise. The data that you used to compute the correlation includes an outlier. BEERD

(a) Remove the outlier and recompute the correlation.

(b) Write a short paragraph about the possible effects of outliers on a correlation using this example to illustrate your ideas.

2.55 Compare domestic with imported. In Exercise 2.31 (page 99), you compared domestic beers with imported beers with respect to the relationship between calories and percent alcohol. In that exercise, you used scatterplots to make the comparison. Compute the correlations for these two categories of beer and write a new summary of the comparison using correlations in addition to the scatterplots. BEERD, BEERI

2.56 Use the applet. Go to the Correlation and Regression applet. Click on the scatterplot to create a group of 12 points in the lower-right corner of the scatterplot with a strong straight-line negative pattern (correlation about −0.9).

(a) Add one point at the upper left that is in line with the first 12. How does the correlation change?

(b) Drag this last point down until it is opposite the group of 12 points. How small can you make the correlation? Can you make the correlation positive? A single outlier can greatly strengthen or weaken a correlation. Always plot your data to check for outlying points.

2.57 Use the applet. You are going to use the Correlation and Regression applet to make different scatterplots with 12

points that have correlation close to 0.8. Many patterns can have the same correlation. Always plot your data before you trust a correlation.

(a) Stop after adding the first two points. What is the value of the correlation? Why does it have this value no matter where the two points are located?

(b) Make a lower-left to upper-right pattern of 12 points with correlation about r = 0.8. (You can drag points up or down to adjust r after you have 12 points.) Make a rough sketch of your scatterplot.

(c) Make another scatterplot, this time with 11 points in a vertical stack at the left of the plot. Add one point far to the right and move it until the correlation is close to 0.8. Make a rough sketch of your scatterplot.

(d) Make yet another scatterplot, this time with 12 points in a curved pattern that starts at the lower left, rises to the right, then falls again at the far right. Adjust the points up or down until you have a quite smooth curve with correlation close to 0.8. Make a rough sketch of this scatterplot also.

2.58 An interesting set of data. Make a scatterplot of the following data: INTER

X 1 2 3 4 10 10 Y 1 3 3 5 1 10

Verify that the correlation is about 0.5. What feature of the data is responsible for reducing the correlation to this value despite a strong straight-line association between x and y in most of the observations?

2.59 Internet use and babies. Figure 2.13 is a scatterplot of the number of births per 1000 people rate versus Internet users per 100 people for 106 countries. In Exercise 2.34 (page 99), you described this relationship. INBIRTH

(a) Make a plot of the data similar to Figure 2.13 and report the correlation.

(b) Is the correlation a good numerical summary for this relationship? Explain your answer.

2.60 What’s wrong? Each of the following statements contains a blunder. Explain in each case what is wrong.

(a) There is a high correlation between the age of American workers and their occupation.

(b) We found a high correlation (r = 1.19) between students’ ratings of faculty teaching and ratings made by other faculty members.

(c) The correlation between the sex of a group of students and the color of their cell phone was r = 0.23.

2.4 Least-Squares Regression

When you complete this section, you will be able to:

Draw a straight line on a scatterplot of a set of data, given the equation of the line. Predict a value of the response variable y for a given value of the explanatory variable x using a regression equation. Explain the meaning of the term least squares. Calculate the equation of a least-squares regression line from the means and standard deviations of the explanatory and response variables and their correlation. Read the output of statistical software to find the equation of the least-squares regression line and the value of r2. Explain the meaning of r2 in the regression setting.

Correlation measures the direction and strength of the linear (straight-line) relationship between two quantitative variables. If a scatterplot shows a linear relationship, we would like to summarize this overall pattern by drawing a line on the scatterplot. A regression line summarizes the relationship between two variables, but only in a specific setting: when one of the variables helps explain or predict the other. That is, regression describes a relationship between an explanatory variable and a response variable.

REGRESSION LINE

A regression line is a straight line that describes how a response variable y changes as an explanatory variable x changes. We often use a regression line to predict the value of y for a given value of x. Regression, unlike correlation, requires that we have an explanatory variable and a response variable.

EXAMPLE 2.19

FIDGET

Fidgeting and fat gain. Does fidgeting keep you slim? Some people don’t gain weight even when they overeat. Perhaps fidgeting and other “nonexercise activity” (NEA) explains why—the body might spontaneously increase nonexercise activity when fed more. Researchers deliberately overfed 16 healthy young adults for eight weeks. They measured fat gain (in kilograms) and, as an explanatory variable, increase in energy use (in calories) from activity other than deliberate exercise—fidgeting, daily living, and the like. Here are the data:17

NEA increase (cal) −94 −57 −29 135 143 151 245 355 Fat gain (kg) 4.2 3.0 3.7 2.7 3.2 3.6 2.4 1.3 NEA increase (cal) 392 473 486 535 571 580 620 690 Fat gain (kg) 3.8 1.7 1.6 2.2 1.0 0.4 2.3 1.1

Figure 2.16 is a scatterplot of these data. The plot shows a moderately strong negative linear association with no outliers. The correlation is r = −0.7786. People with larger increases in nonexercise activity do indeed gain less fat. A line drawn through the points will describe the overall pattern well.

FIGURE 2.16 Fat gain after eight weeks of overeating plotted against the increase in nonexercise activity over the same period, Example 2.19.

Fitting a line to data fitting a line

When a scatterplot displays a linear pattern, we can describe the overall pattern by drawing a straight line through the points. Of course, no straight line passes exactly through all the points. Fitting a line to data means drawing a line that comes as close as possible to the points. The equation of a line fitted to the data gives a concise description of the relationship between the response variable y and the explanatory variable x. It is the numerical summary that supports the scatterplot, our graphical summary.

STRAIGHT LINES

Suppose that y is a response variable (plotted on the vertical axis) and x is an explanatory variable (plotted on the horizontal axis). A straight line relating y to x has an equation of the form

y = b0 + b1x

In this equation, b1 is the slope, the amount by which y changes when x increases by one unit. The number b0 is the intercept, the value of y when x = 0.

In practice, we will use software to obtain values of b0 and b1 for a given set of data.

EXAMPLE 2.20

FIDGET

Regression line for fat gain. Any straight line describing the nonexercise activity data has the form fat gain = b0 + (b1 × NEA increase)

In Figure 2.17, we have drawn the regression line with the equation

fat gain = 3.505 − (0.00344 × NEA increase)

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FIGURE 2.17 A regression line fitted to the nonexercise activity data and used to predict fat gain for an NEA increase of 400 calories, Examples 2.20 and 2.21.

The figure shows that this line fits the data well. The slope b1 = −0.00344 tells us that fat gained goes down by 0.00344 kilogram for each added calorie of NEA increase.

The slope b1 of a line y = b0 + b1x is the average change in the response y as the explanatory variable x changes. The slope of a regression line is an important numerical description of the relationship between the two variables. For Example 2.20, the intercept, b0 = 3.505 kilograms. This value is the estimated fat gain if NEA does not change. When we substitute the value zero for the NEA increase, the regression equation gives 3.505 (the intercept) as the predicted value of the fat gain.

USE YOUR KNOWLEDGE

FIDGET

Plot the line. Make a plot of the data in Example 2.19 and plot the line fat gain = 4.505 − (0.00344 × NEA increase)

on your sketch. Explain why this line does not give a good fit to the data.

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Prediction prediction

We can use a regression line to predict the response y for a specific value of the explanatory variable x. We can interpret the prediction as the average value of y corresponding to a collection of cases at the particular value of x or as our best guess at the value of y for an individual with the particular value of x.

EXAMPLE 2.21

FIDGET

Prediction for fat gain. Based on the linear pattern, we want to predict the fat gain for an individual whose NEA increases by 400 calories when she overeats. To use the fitted line to predict fat gain, go “up and over” on the graph in Figure 2.17. From 400 calories on the x axis, go up to the fitted line and over to the y axis. The graph shows that the predicted gain in fat is a bit more than 2 kilograms.

If we have the equation of the line, it is faster and more accurate to substitute x = 400 in the equation. The predicted fat gain is

fat gain = 3.505 − (0.00344 × 400) = 2.13 kilograms

The accuracy of predictions from a regression line depends on how much scatter about the line the data show. In Figure 2.17, fat gains for similar increases in NEA show a spread of 1 or 2 kilograms. The regression line summarizes the pattern but gives only roughly accurate predictions.

USE YOUR KNOWLEDGE

Predict the fat gain. Use the regression equation in Example 2.20 to predict the fat gain for a person whose NEA increases by 250 calories.

EXAMPLE 2.22

Is this prediction reasonable? Can we predict the fat gain for someone whose nonexercise activity increases by 1500 calories when she overeats? We can certainly substitute 1500 calories into the equation of the line. The prediction is

fat gain = 3.505 − (0.00344 × 1500) = −1.66 kilograms

That is, we predict that this individual loses fat when she overeats. This prediction is not trustworthy. Look again at Figure 2.17. An NEA increase of 1500 calories is far outside the range of our data. We can’t say whether increases this large ever occur, or whether the relationship remains linear at such extreme values. Predicting fat gain when NEA increases by 1500 calories extrapolates the relationship beyond what the data show.

EXTRAPOLATION

Extrapolation is the use of a regression line for prediction far outside the range of values of the explanatory variable x used to obtain the line. Such predictions are often not accurate and should be avoided.

2.63

USE YOUR KNOWLEDGE

Would you use the regression equation to predict? Consider the following values for NEA increase: −300, 300, 600, 800. For each, decide whether you would use the regression equation in Example 2.20 to predict fat gain or whether you would be concerned that the prediction would not be trustworthy because of extrapolation. Give reasons for your answers.

Least-squares regression Different people might draw different lines by eye on a scatterplot. This is especially true when the points are widely scattered. We need a way to draw a regression line that doesn’t depend on our guess as to where the line should go. No line will pass exactly through all the points, but we want one that is as close as possible. We will use the line to predict y from x, so we want a line that is as close as possible to the points in the vertical direction. That’s because the prediction errors we make are errors in y, which is the vertical direction in the scatterplot.

The line in Figure 2.17 predicts 2.13 kilograms of fat gain for an increase in nonexercise activity of 400 calories. If the actual fat gain turns out to be 2.3 kilograms, the error is

Errors are positive if the observed response lies above the line and negative if the response lies below the line. We want a regression line that makes these prediction errors as small as possible. Figure 2.18 illustrates the idea. For clarity, the plot shows only three of the points from Figure 2.17, along with the line, on an expanded scale. The line passes below two of the points and above one of them. The vertical distances of the data points from the line appear as vertical line segments. A “good” regression line makes these distances as small as possible. There are many ways to make “as small as possible” precise. The most common is the least-squares idea. The line in Figures 2.17 and 2.18 is, in fact, the least- squares regression line.

FIGURE 2.18 The least-squares idea: make the errors in predicting y as small as possible by minimizing the sum of their squares.

LEAST-SQUARES REGRESSION LINE

The least-squares regression line of y on x is the line that makes the sum of the squares of the vertical distances of the data points from the line as small as possible.

Here is the least-squares idea expressed as a mathematical problem. We represent n observations on two variables x and y as

(x1, y1), (x2, y2), …, (xn, yn)

If we draw a line y = b0 + b1x through the scatterplot of these observations, the line predicts the value of y corresponding to xi as . We write (read “y-hat”) in the equation of a regression line to emphasize that the line gives a predicted response for any x. The predicted response will usually not be exactly the same as the actually observed response y. The method of least squares chooses the line that makes the sum of the squares of these errors as small as possible. To find this line, we must find the values of the intercept b0 and the slope b1 that minimize

for the given observations x1 and y1. For the NEA data, for example, we must find the b0 and b1 that minimize

These values are the intercept and slope of the least-squares line. You will use software or a calculator with a regression function to find the equation of the least-squares regression

line from data on x and y. Therefore, we will give the equation of the least-squares line in a form that helps our understanding but is not efficient for calculation.

EQUATION OF THE LEAST-SQUARES REGRESSION LINE

We have data on an explanatory variable x and a response variable y for n individuals. The means and standard deviations of the sample data are and sx for x and and sy for y, and the correlation between x and y is r. The equation of the least-squares regression line of y on x is

with slope

and intercept

EXAMPLE 2.23

Check the calculations. Verify from the data in Example 2.19 that the mean and standard deviation of the 16 increases in NEA are

The mean and standard deviation of the 16 fat gains are

The correlation between fat gain and NEA increase is r = −0.7786. Therefore, the least-squares regression line of fat gain y on NEA increase x has slope

and intercept

The equation of the least-squares line is

When doing calculations like this by hand, you may need to carry extra decimal places in the preliminary calculations to get accurate values of the slope and intercept. Using software or a calculator with a regression function eliminates this worry.

Interpreting the regression line

The slope b1 = −0.00344 kilograms per calorie in Example 2.23 is the change in fat gain as NEA increases. The units “kilograms of fat gained per calorie of NEA” come from the units of y (kilograms) and x (calories). Although the correlation does not change when we change the units of measurement, the equation of the least-squares line does change. The slope in grams per calorie would be 1000 times as large as the slope in kilograms per calorie because there are 1000 grams in a kilogram. The small value of the slope, b1 = −0.00344, does not mean that the effect of increased NEA on fat gain is small—it just reflects the choice of kilograms as the unit for fat gain. The slope and intercept of the least-squares line depend on the units of measurement—you can’t conclude anything from their size.

EXAMPLE 2.24

FIDGET

Regression using software. Figure 2.19 displays the basic regression output for the nonexercise activity data from three statistical software packages. Other software produces very similar output. You can find the slope and intercept of the least-squares line, calculated to more decimal places than we need, in each output. The software also provides information that we do not yet need, including some that we trimmed from Figure 2.19.

FIGURE 2.19 Regression results for the nonexercise activity data from three statistical software packages: (a) SPSS; (b) Minitab; (c) JMP. Other software produces similar output.

Part of the art of using software is to ignore the extra information that is almost always present. Look for the results that you need. Once you understand a statistical method, you can read output from almost any software.

Facts about least-squares regression Regression is one of the most common statistical settings, and least squares is the most common method for fitting a regression line to data. Here are some facts about least-squares regression lines.

Fact 1. There is a close connection between correlation and the slope of the least-squares line. The slope is

This equation says that along the regression line, a change of one standard deviation in x corresponds to a change of r standard deviations in y. When the variables are perfectly correlated (r = 1 or r = −1), the change in the predicted response is the same (in standard deviation units) as the change in x. Otherwise, because −1 ≤ r ≤ 1, the change in is less than the change in x. As the correlation grows less strong, the prediction y moves less in response to changes in x. Note that if the correlation is zero, then the slope of the least-squares regression line will be zero.

Fact 2. The least-squares regression line always passes through the point ( , ) on the graph of y against x. So, the least-squares regression line of y on x is the line with slope rsy/sx that passes through the point ( , ). We can describe regression entirely in terms of the basic descriptive measures , sx, , sy, and r.

Fact 3. The distinction between explanatory and response variables is essential in regression. Least-squares regression looks at the distances of the data points from the line only in the y direction. If we reverse the roles of the two variables, we get a different least-squares regression line.

Correlation and regression Least-squares regression looks at the distances of the data points from the line only in the y direction. So the two variables x and y play different roles in regression.

EXAMPLE 2.25

FIDGET

Fidgeting and fat gain. Figure 2.20 is a scatterplot of the fidgeting and fat gain data described in Example 2.19 (page 107). There is a negative linear relationship. The two lines on the plot are the two least-squares regression lines. The regression line using nonexercise activity to predict fat gain is blue. The regression line using fat gain to predict nonexercise activity is red. Regression of fat gain on nonexercise activity and regression of nonexercise activity on fat gain give different lines. In the regression setting, you must decide which variable is explanatory.

FIGURE 2.20 Scatterplot of fat gain versus nonexercise activity for 16 subjects from Example 2.19. The two lines are the two least-squares regression lines, using nonexercise activity to predict fat gain (blue) and using fat gain to predict nonexercise activity (red), Example 2.25.

Even though the correlation r ignores the distinction between explanatory and response variables, there is a close connection between correlation and regression. We saw that the slope of the least-squares line involves r. Another connection between correlation and regression is even more important. In fact, the numerical value of r as a measure of the strength of a linear relationship is best interpreted by thinking about regression. Here is the fact we need.

r2 IN REGRESSION

The square of the correlation, r2, is the fraction of the variation in the values of y that is explained by the least- squares regression of y on x.

2.64

The correlation between NEA increase and fat gain for the 16 subjects in Example 2.19 (page 107) is r = −0.7786. Because r2 = 0.6062, the straight-line relationship between NEA and fat gain explains about 61% of the vertical scatter in fat gains in Figure 2.17.

When you report a regression, give r2 as a measure of how successfully the regression explains the response. All three software outputs in Figure 2.19 include r2, either in decimal form or as a percent.

When you see a correlation, square it to get a better feel for the strength of the association. Perfect correlation (r = −1 or r = 1) means that the points lie exactly on a line. Then r2 = 1 and all the variation in one variable is accounted for by the linear relationship with the other variable. If r = −0.7 or r = 0.7, r2 = 0.49 and about half the variation is accounted for by the linear relationship. In the r2 scale, correlation ±0.7 is about halfway between 0 and ±1.

USE YOUR KNOWLEDGE

What fraction of the variation is explained? Consider the following correlations: −0.9, −0.5, −0.2, 0, 0.2, 0.5, and 0.9. For each, give the fraction of the variation in y that is explained by the least-squares regression of y on x. Summarize what you have found from performing these calculations.

The use of r2 to describe the success of regression in explaining the response y is very common. It rests on the fact that there are two sources of variation in the responses y in a regression setting. Figure 2.17 gives a rough visual picture of the two sources. The first reason for the variation in fat gains is that there is a relationship between fat gain y and increase in NEA x. As x increases from −94 to 690 calories among the 16 subjects, it pulls fat gain y with it along the regression line in the figure. The linear relationship explains this part of the variation in fat gains.

The fat gains do not lie exactly on the line, however, but are scattered above and below it. This is the second source of variation in y, and the regression line tells us nothing about how large it is. The dashed lines in Figure 2.17 show a rough average for y when we fix a value of x. We use r2 to measure variation along the line as a fraction of the total variation in the fat gains. In Figure 2.17, about 61% of the variation in fat gains among the 16 subjects is due to the straight-line relationship between y and x. The remaining 39% is vertical scatter in the observed responses remaining after the line has fixed the predicted responses.

Another view of r2

Here is a more specific interpretation of r2. The fat gains y in Figure 2.17 range from 0.4 to 4.2 kilograms. The variance of these responses, a measure of how variable they are, is

variance of observed values y = 1.297

Much of this variability is due to the fact that as x increases from −94 to 690 calories, it pulls y along with it. If the only variability in the observed responses were due to the straight-line dependence of fat gain on NEA, the observed gains would lie exactly on the regression line. That is, they would be the same as the predicted gains . We can compute the predicted gains by substituting the NEA values for each subject into the equation of the least-squares line. Their variance describes the variability in the predicted responses. The result is

variance of predicted values = 0.786

This is what the variance would be if the responses fell exactly on the line; that is, if the linear relationship explained 100% of the observed variation in y. Because the responses don’t fall exactly on the line, the variance of the predicted values is smaller than the variance of the observed values. Here is the fact we need:

This fact is always true. The squared correlation gives the variance that the responses would have if there were no scatter about the least-squares line as a fraction of the variance of the actual responses. This is the exact meaning of “fraction of variation explained” as an interpretation of r2.

These connections with correlation are special properties of least-squares regression. They are not true for other methods of fitting a line to data. One reason that least squares is the most common method for fitting a regression line to data is that it has many convenient special properties.

SECTION 2.4 SUMMARY A regression line is a straight line that describes how a response variable y changes as an explanatory variable x changes. The most common method of fitting a line to a scatterplot is least squares. The least-squares regression line is the straight line = b0 + b1x that minimizes the sum of the squares of the vertical distances of the observed y- values from the line. You can use a regression line to predict the value of y for any value of x by substituting this x into the equation of the line. Extrapolation beyond the range of x-values spanned by the data is risky. The slope b1 of a regression line = b0 + b1x is the rate at which the predicted response changes along the line as the explanatory variable x changes. Specifically, b1 is the change in when x increases by 1. The numerical value of the slope depends on the units used to measure x and y. The intercept b0 of a regression line = b0 + b1x is the predicted response when the explanatory variable x = 0. This prediction is not particularly useful unless x can actually take values near 0. The least-squares regression line of y on x is the line with slope b1 = rsy/sx and intercept . This line always passes through the point ( , ). Correlation and regression are closely connected. The correlation r is the slope of the least-squares regression line when we measure both x and y in standardized units. The square of the correlation r2 is the fraction of the variance of one variable that is explained by least-squares regression on the other variable.

SECTION 2.4 EXERCISES For Exercise 2.61, see page 109; for Exercise 2.62, see page 110; for Exercise 2.63, see page 111; and for Exercise 2.64, see page 116.

2.65 Blueberries and anthocyanins. In Exercise 2.18 (page 97), you examined the relationship between Antho4 and Antho3, two anthocyanins found in blueberries. In Exercise 2.42 (page 105), you found the correlation between these two variables. BERRIES

(a) Find the equation of the least-squares regression line for predicting Antho4 from Antho3.

(b) Make a scatterplot of the data with the fitted line.

(c) How well does the line fit the data? Explain your answer.

(d) Use the line to predict the value of Antho4 when Antho3 is equal to 1.5.

2.66 Fuel consumption. In Exercise 2.21 (page 97), you examined the relationship between CO2 emissions and highway fuel consumption for 527 vehicles that use regular fuel. In Exercise 2.44 (page 105), you found the correlation between these two variables. CANFREG

(a) Find the equation of the least-squares regression line for predicting CO2 emissions from highway fuel consumption.

(b) Make a scatterplot of the data with the fitted line.

(c) How well does the line fit the data? Explain your answer.

(d) Use the line to predict the value of CO2 for vehicles that consume 8.0 liters per kilometer (L/km).

2.67 Fuel consumption for different types of vehicles. In Exercise 2.23 (page 97), you examined the relationship between CO2 emissions and highway fuel consumption for 1067 vehicles. You used different plotting symbols for the four different types of fuel used by these vehicles: regular, premium, diesel, and ethanol. CANFUEL

(a) Find the least-squares equation for predicting CO2 emissions from highway fuel consumption for all 1067 vehicles.

(b) Make a scatterplot of the data with the fitted line.

(c) Based on what you learned from Example 2.23, do you think that a single least-squares regression line provides a good fit for all four types of vehicles? Explain your answer.

2.68 Bone strength. Exercise 2.24 (page 97), gives the bone strengths of the dominant and the nondominant arms for 15 men who were controls in a study. ARMSTR

(a) Plot the data. Use the bone strength in the nondominant arm as the explanatory variable and bone strength in the dominant arm as the response variable.

(b) The least-squares regression line for these data is dominant = 2.74 + (0.936 × nondominant)

Add this line to your plot.

(c) Use the scatterplot (a graphical summary), with the least-squares line (a graphical display of a numerical summary) to write a short paragraph describing this relationship.

2.69 Bone strength for baseball players. Refer to the previous exercise. Similar data for baseball players are given in Exercise 2.25 (page 98). Here is the equation of the least-squares line for the baseball players:

dominant = 0.886 + (1.373 × nondominant)

Answer parts (a) and (c) of the previous exercise for these data. ARMSTR

2.70 Predict the bone strength. Refer to Exercise 2.68. A young male who is not a baseball player has a bone strength of 16.0 cm4/1000 in his nondominant arm. Predict the bone strength in the dominant arm for this person. ARMSTR

2.71 Predict the bone strength for a baseball player. Refer to Exercise 2.69. A young male who is a baseball player

has a bone strength of 16.0 cm4 ·1000 in his nondominant arm. Predict the bone strength in the dominant arm for this person. ARMSTR

2.72 Compare the predictions. Refer to the two previous exercises. You have predicted two dominant-arm bone strengths, one for a baseball player and one for a person who is not a baseball player. The nondominant bone strengths are both 16.0 cm4/1000. ARMSTR

(a) Compare the two predictions by computing the difference in means, baseball player minus control.

(b) Explain how the difference in the two predictions is an estimate of the effect of baseball throwing exercise on the strength of arm bones.

(c) For nondominant arm strengths of 12 cm4/1000 and 20 cm4/1000, repeat your predictions and take the differences. Make a table of the results of all three calculations (for 12, 16, and 20 cm4/1000).

(d) Write a short summary of the results of your calculations for the three different nondominant-arm strengths. Be sure to include an explanation of why the differences are not the same for the three nondominant-arm strengths.

2.73 Least-squares regression for radioactive decay. Refer to Exercise 2.32 (page 99) for the data on radioactive decay of barium-137m. Here are the data: DECAY

Time 1 3 5 7 Count 578 317 203 118

(a) Using the least-squares regression equation count = 602.8 − (74.7 × time)

find the predicted values for the counts.

(b) Compute the differences, observed count minus predicted count. How many of these are positive; how many are negative?

(c) Square and sum the differences that you found in part (b).

(d) Repeat the calculations that you performed in parts (a), (b), and (c) using the equation count = 500 − (100 × time)

(e) In a short paragraph, explain the least-squares idea using the calculations that you performed in this exercise.

2.74 Least-squares regression for the log counts. Refer to Exercise 2.33 (page 99), where you analyzed the radioactive decay of barium-137m data using log counts. Here are the data: DECAY

Time 1 3 5 7 Log count 6.35957 5.75890 5.31321 4.77068

(a) Using the least-squares regression equation log count = 6.593 − (0.2606 × time)

find the predicted values for the log counts.

(b) Compute the differences, observed count minus predicted count. How many of these are positive; how many are negative?

(c) Square and sum the differences that you found in part (b).

(d) Repeat the calculations that you performed in parts (a), (b), and (c) using the equation log count = 7 − (0.2 × time)

(e) In a short paragraph, explain the least-squares idea using the calculations that you performed in this exercise.

2.75 College students by state. How well does the population of a state predict the number of undergraduates? The National Center for Education Statistics collects data for each of the 50 U.S. states that we can use to address this question.18

COLLEGE

(a) Make a scatterplot with population on the x axis and number of undergraduates on the y axis.

(b) Describe the form, direction, and strength of the relationship. Are there any outliers?

(c) For the number of undergraduates, the mean is 302,136 and the standard deviation is 358,460, and for population, the mean is 5,955,551 and the standard deviation is 6,620,733. The correlation between the number of undergraduates and the population is 0.98367. Use this information to find the least-squares regression line. Show your work.

(d) Add the least-squares line to your scatterplot.

2.76 College students by state without the four largest states. Refer to the previous exercise. Let’s eliminate the four largest states, which have populations greater than 15 million. Here are the numerical summaries: for number of undergraduate college students, the mean is 220,134 and the standard deviation is 165,270; for population, the mean is 4,367,448 and the standard deviation is 3,310,957. The correlation between the number of undergraduate college students and the population is 0.97081. Use this information to find the least-squares regression line. Show your work. COL46

2.77 Make predictions and compare. Refer to the two previous exercises. Consider a state with a population of 4 million (this value is approximately the median population for the 50 states). COLLEGE

(a) Using the least-squares regression equation for all 50 states, find the predicted number of undergraduate college students.

(b) Do the same using the least-squares regression equation for the 46 states with populations less than 15 million.

(c) Compare the predictions that you made in parts (a) and (b). Write a short summary of your results and conclusions. Pay particular attention to the effect of including the four states with the largest populations in the prediction equation for a median-sized state.

2.78 College students by state. Refer to Exercise 2.75, where you examined the relationship between the number of undergraduate college students and the populations for the 50 states. Figure 2.21 gives the output from a software package for the regression. Use this output to answer the following questions: COLLEGE

(a) What is the equation of the least-squares regression line?

(b) What is the value of r2?

(c) Interpret the value of r2.

(d) Does the software output tell you that the relationship is linear and not, for example, curved? Explain your answer.

FIGURE 2.21 SPSS output for predicting number of undergraduate college students using population for the 50

states, Exercise 2.78.

FIGURE 2.22 SPSS output for predicting number of undergraduate college students using population, with the four largest states deleted, Exercise 2.79.

2.79 College students by state without the four largest states. Refer to Exercise 2.76, where you eliminated the four largest states that have populations greater than 15 million. Figure 2.22 gives software output for these data. Answer the questions in the previous exercise for the data set with the 46 states. COL46

2.80 Data generated by software. The following 20 observations on Y and X were generated by a computer program. GENDATA

X Y

22.06 34.38 19.88 30.38 18.83 26.13 22.09 31.85 17.19 26.77 20.72 29.00 18.10 28.92 18.01 26.30 18.69 29.49 18.05 31.36 17.75 27.07 19.96 31.17 17.87 27.74 20.20 30.01 20.65 29.61 20.32 31.78 21.37 32.93 17.31 30.29

TABLE 2.1

23.50 28.57 22.02 29.80

(a) Make a scatterplot and describe the relationship between Y and X.

(b) Find the equation of the least-squares regression line and add the line to your plot.

(c) What percent of the variability in Y is explained by X?

(d) Summarize your analysis of these data in a short paragraph.

2.81 Add an outlier. Refer to Exercise 2.80. Add an additional observation with y = 25 and x = 35 to the data set. Repeat the analysis that you performed in Exercise 2.80 and summarize your results, paying particular attention to the effect of this outlier.

GEN21A

2.82 Add a different outlier. Refer to Exercise 2.80 and the previous exercise. Add an additional observation with y = 36 and x = 30 to the original data set. GEN21B

(a) Repeat the analysis that you performed in Exercise 2.80 and summarize your results, paying particular attention to the effect of this outlier.

(b) In this exercise and in the previous one, you added an outlier to the original data set and reanalyzed the data. Write a short summary of the changes in correlations that can result from different kinds of outliers.

2.83 Alcohol and calories in beer. Figure 2.12 gives a scatterplot of calories versus percent alcohol in 159 brands of domestic beer. BEERD

(a) Find the equation of the least-squares regression line for these data.

(b) Find the value of r2 and interpret it in the regression context.

Four Data Sets for Exploring Correlation and Regression

Data Set A x 10 8 13 9 11 14 6 4 12 7 5 y 8.04 6.95 7.58 8.81 8.33 9.96 7.24 4.26 10.84 4.82 5.68 Data Set B x 10 8 13 9 11 14 6 4 12 7 5 y 9.14 8.14 8.74 8.77 9.26 8.10 6.13 3.10 9.13 7.26 4.74 Data Set C x 10 8 13 9 11 14 6 4 12 7 5 y 7.46 6.77 12.74 7.11 7.81 8.84 6.08 5.39 8.15 6.42 5.73 Data Set D x 8 8 8 8 8 8 8 8 8 8 19 y 6.58 5.76 7.71 8.84 8.47 7.04 5.25 5.56 7.91 6.89 12.50

(c) Write a short report on the relationship between calories and percent alcohol in beer. Include graphical and numerical summaries for each variable separately as well as graphical and numerical summaries for the relationship in your report.

2.84 Alcohol and calories in beer revisited. Refer to the previous exercise. The data that you used includes an outlier. BEERD

(a) Remove the outlier and answer parts (a), (b), and (c) for the new set of data.

(b) Write a short paragraph about the possible effects of outliers on a least-squares regression line and the value of r2, using this example to illustrate your ideas.

2.85 Always plot your data! Table 2.1 presents four sets of data prepared by the statistician Frank Anscombe to illustrate

the dangers of calculating without first plotting the data.19 ANSC

(a) Without making scatterplots, find the correlation and the least-squares regression line for all four data sets. What do you notice? Use the regression line to predict y for x = 10.

(b) Make a scatterplot for each of the data sets and add the regression line to each plot.

(c) In which of the four cases would you be willing to use the regression line to describe the dependence of y on x? Explain your answer in each case.

2.86 Progress in math scores. Every few years, the National Assessment of Educational Progress asks a national sample of eighth-graders to perform the same math tasks. The goal is to get an honest picture of progress in math. Here are the last few national mean scores, on a scale of 0 to 500:20 NAEP

Year 1990 1992 1996 2000 2003 2005 2008 2011 2013 Score 263 268 272 273 278 279 281 283 285

(a) Make a time plot of the mean scores, by hand. This is just a scatterplot of score against year. There is a slow linear increasing trend.

(b) Find the regression line of mean score on time step-by-step. First calculate the mean and standard deviation of each variable and their correlation (use a calculator with these functions). Then find the equation of the least-squares line from these. Draw the line on your scatterplot. What percent of the year-to-year variation in scores is explained by the linear trend?

(c) Now use software or the regression function on your calculator to verify your regression line.

2.87 The regression equation. The equation of a least-squares regression line is y = 15 − 2x.

(a) What is the value of y for x = 4?

(b) If x increases by one unit, what is the corresponding change in y?

(c) What is the intercept for this equation?

2.88 Metabolic rate and lean body mass. Compute the mean and the standard deviation of the metabolic rates and lean body masses in Exercise 2.37 (page 100) and the correlation between these two variables. Use these values to find the slope of the regression line of metabolic rate on lean body mass. Also find the slope of the regression line of lean body mass on metabolic rate. What are the units for each of the two slopes? BMASS

2.89 Use an applet for progress in math scores. Go to the Two-Variable Statistical Calculator. Enter the data for the progress in math scores from Exercise 2.86 using the “User-entered data” option in the “Data” tab. Explore the data

by clicking the other tabs in the applet. Using only the only the results provided by the applet, write a short report summarizing the analysis of these data.

2.90 A property of the least-squares regression line. Use the equation for the least-squares regression line to show that this line always passes through the point ( , ).

2.91 Class attendance and grades. A study of class attendance and grades among first-year students at a state university showed that, in general, students who missed a higher percent of their classes earned lower grades. Class attendance explained 16% of the variation in grade index among the students. What is the numerical value of the correlation between percent of classes attended and grade index?

2.5 Cautions about Correlation and Regression

When you complete this section, you will be able to:

Calculate the residuals for a set of data using the equation of the least-squares regression line and the observed values of the explanatory variable. Use a plot of the residuals versus the explanatory variable to assess the fit of a regression line. Identify outliers and influential observations by examining scatterplots and residual plots. Identify lurking variables that can influence the interpretation of relationships between two variables. Explain the difference between association and causality when interpreting the relationship between two variables.

Correlation and regression are among the most common statistical tools. They are used in more elaborate form to study relationships among many variables, a situation in which we cannot see the essentials by studying a single scatterplot. We need a firm grasp of the use and limitations of these tools, both now and as a foundation for more advanced statistics.

Residuals A regression line describes the overall pattern of a linear relationship between an explanatory variable and a response variable. Deviations from the overall pattern are also important. In the regression setting, we see deviations by looking at the scatter of the data points about the regression line. The vertical distances from the points to the least-squares regression line are as small as possible in the sense that they have the smallest possible sum of squares. Because they represent “leftover” variation in the response after fitting the regression line, these distances are called residuals.

RESIDUALS

A residual is the difference between an observed value of the response variable and the value predicted by the regression line. That is,

EXAMPLE 2.26

FIDGET

Residuals for fat gain. Example 2.19 (page 107) describes measurements on 16 young people who volunteered to overeat for eight weeks. Those whose nonexercise activity (NEA) spontaneously rose substantially gained less fat than others. Figure 2.23(a) is a scatterplot of these data. The pattern is linear. The least-squares line is

fat gain = 3.505 − (0.00344 × NEA increase)

One subject’s NEA rose by 135 calories. That subject gained 2.7 kilograms of fat. The predicted gain for 135 calories is

= 3.505 − (0.00344 × 135) = 3.04 kg

The residual for this subject is, therefore,

2.92

2.93

FIGURE 2.23 (a) Scatterplot of fat gain versus increase in nonexercise activity, with the least-squares regression line, Example 2.26. (b) Residual plot for the regression displayed in panel (a); the line at y = 0 marks the mean of the residuals.

Most regression software will calculate and store residuals for you.

USE YOUR KNOWLEDGE

Find the predicted value and the residual. Let’s say that we have an individual in the NEA data set who has NEA increase equal to 250 calories and fat gain equal to 2.4 kg. Find the predicted value of fat gain for this individual and then calculate the residual. Explain why this residual is negative.

Because the residuals show how far the data fall from our regression line, examining the residuals helps us assess how well the line describes the data. Although residuals can be calculated from any model fit to the data, the residuals from the least-squares line have a special property: the mean of the least-squares residuals is always zero.

USE YOUR KNOWLEDGE

FIDGET

Find the sum of the residuals. Here are the 16 residuals for the NEA data rounded to two decimal places:

0.37 −0.70 0.10 −0.34 0.19 0.61 −0.26 −0.98 1.64 −0.18 −0.23 0.54 −0.54 −1.11 0.93 −0.03

Find the sum of these residuals. Note that the sum is not exactly zero because of roundoff error.

You can see the residuals in the scatterplot of Figure 2.23(a) by looking at the vertical deviations of the points from the line. The residual plot in Figure 2.23(b) makes it easier to study the residuals by plotting them against the explanatory variable, increase in NEA.

RESIDUAL PLOTS

A residual plot is a scatterplot of the regression residuals against the explanatory variable. Residual plots help us assess the fit of a regression line.

Because the mean of the residuals is always zero, the horizontal line at zero in Figure 2.23(b) helps orient us. This line (residual = 0) corresponds to the fitted line in Figure 2.23(a). The residual plot magnifies the deviations from the line to make patterns easier to see. If the regression line catches the overall pattern of the data, there should be no pattern in the residuals. That is, the residual plot should show an unstructured horizontal band centered at zero. The residuals in Figure 2.23(b) do have this irregular scatter.

You can see the same thing in the scatterplot of Figure 2.23(a) and the residual plot of Figure 2.23(b). It’s just a bit easier in the residual plot. Deviations from an irregular horizontal pattern point out ways in which the regression line fails to catch the overall pattern. Here is an example.

EXAMPLE 2.27

INBIRTH

Patterns in birthrate and Internet user residuals. In Exercise 2.34 (page 99) we used a scatterplot to study the relationship between birthrate and Internet users for 106 countries. In this scatterplot, Figure 2.13, we see that there are many countries with low numbers of Internet users. In addition, the relationship between births and Internet users appears to be curved. For low values of Internet users, there is a clear relationship, while for higher values, the curve becomes relatively flat.

Figure 2.24(a) gives the data with the least-squares regression line, and Figure 2.24(b) plots the residuals. Look at the right part of Figure 2.24(b), where the values of Internet users are high. Here we see that the residuals tend to be positive.

TABLE 2.2

FIGURE 2.24 (a) Scatterplot of birthrate versus Internet users, with the least-squares regression line, Example 2.27. (b) Residual plot for the regression displayed in panel (a); the line at y = 0 marks the mean of the residuals.

The residual pattern in Figure 2.24(b) is characteristic of a simple curved relationship. There are many ways in which a relationship can deviate from a linear pattern. We now have an important tool for examining these deviations. Use it frequently and carefully when you study relationships.

Two Measures of Glucose Level in Diabetics

Subject HbA1c (%)

FPG (mg/ml)

1 6.1 141 2 6.3 158 3 6.4 112

4 6.8 153 5 7.0 134 6 7.1 95 7 7.5 96 8 7.7 78 9 7.9 148 10 8.7 172 11 9.4 200 12 10.4 271 13 10.6 103 14 10.7 172 15 10.7 359 16 11.2 145 17 13.7 147 18 19.3 255

Outliers and influential observations When you look at scatterplots and residual plots, look for striking individual points as well as for an overall pattern. Here is an example of data that contain some unusual cases.

EXAMPLE 2.28

HBA1C

Diabetes and blood sugar. People with diabetes must manage their blood sugar levels carefully. They measure their fasting plasma glucose (FPG) several times a day with a glucose meter. Another measurement, made at regular medical checkups, is called HbA1c. This is roughly the percent of red blood cells that have a glucose molecule attached. It measures average exposure to glucose over a period of several months.

This diagnostic test is becoming widely used and is sometimes called A1c by health care professionals. Table 2.2 gives data on both HbA1c and FPG for 18 diabetics five months after they completed a diabetes education class.21

Because both FPG and HbA1c measure blood glucose, we expect a positive association. The scatterplot in Figure 2.25(a) shows a surprisingly weak relationship, with correlation r = 0.4819. The line on the plot is the least-squares regression line for predicting FPG from HbA1c. Its equation is

FIGURE 2.25 (a) Scatterplot of fasting plasma glucose against HbA1c (which measures long-term blood glucose), with the least-squares regression line, Example 2.28. (b) Residual plot for the regression of fasting plasma glucose on HbA1c. Subject 15 is an outlier in fasting plasma glucose. Subject 18 is an outlier in HbA1c that may be influential but does not have a large residual.

It appears that one-time measurements of FPG can vary quite a bit among people with similar long-term levels, as measured by HbA1c. This is why A1c is an important diagnostic test.

Two unusual cases are marked in Figure 2.25(a). Subjects 15 and 18 are unusual in different ways. Subject 15 has dangerously high FPG and lies far from the regression line in the y direction. Subject 18 is close to the line but far out in the x direction. The residual plot in Figure 2.25(b) confirms that Subject 15 has a large residual and that Subject 18 does not.

Points that are outliers in the x direction, like Subject 18, can have a strong influence on the position of the regression line. Least-squares lines make the sum of squares of the vertical distances to the points as small as possible. A point that is extreme in the x direction with no other points near it pulls the line toward itself.

OUTLIERS AND INFLUENTIAL OBSERVATIONS IN REGRESSION

An outlier is an observation that lies outside the overall pattern of the other observations. Points that are outliers in the y direction of a scatterplot have large regression residuals, but other outliers need not have large residuals.

An observation is influential for a statistical calculation if removing it would markedly change the result of the calculation. Points that are outliers in the x direction of a scatterplot are often influential for the least-squares regression line.

Influence is a matter of degree—how much does a calculation change when we remove an observation? It is difficult to assess influence on a regression line without actually doing the regression both with and without the suspicious observation. A point that is an outlier in x is often influential. But if the point happens to lie close to the regression line calculated from the other observations, then its presence will move the line only a little and the point will not be influential.

The influence of a point that is an outlier in y depends on whether there are many other points with similar values of x that hold the line in place. Figures 2.25(a) and (b) identify two unusual observations. How influential are they?

EXAMPLE 2.29

HBA1C

Influential observations. Subjects 15 and 18 both influence the correlation between FPG and HbA1c, in opposite directions. Subject 15 weakens the linear pattern; if we drop this point, the correlation increases from r = 0.4819 to r = 0.5684. Subject 18 extends the linear pattern; if we omit this subject, the correlation drops from r = 0.4819 to r = 0.3837.

To assess influence on the least-squares line, we recalculate the line leaving out a suspicious point. Figure 2.26 shows three least-squares lines. The solid line is the regression line of FPG on HbA1c based on all 18 subjects. This is the same line that appears in Figure 2.25(a). The dotted line is calculated from all subjects except Subject 18. You see that point 18 does pull the line down toward itself. But the influence of Subject 18 is not very large—the dotted and solid lines are close together for HbA1c values between 6 and 14, the range of all except Subject 18.

FIGURE 2.26 Three regression lines for predicting fasting plasma glucose from HbA1c, Example 2.29. The solid line uses all 18 subjects. The dotted line leaves out Subject 18. The dashed line leaves out Subject 15. “Leaving one out” calculations are the surest way to assess influence.

The dashed line omits Subject 15, the outlier in y. Comparing the solid and dashed lines, we see that Subject 15 pulls the regression line up. The influence is again not large, but it exceeds the influence of Subject 18.

We did not need the distinction between outliers and influential observations in Chapter 1. A single large salary that pulls up the mean salary for a group of workers is an outlier because it lies far above the other salaries. It is also influential because the mean changes when it is removed. In the regression setting, however, not all outliers are influential. Because influential observations draw the regression line toward themselves, we may not be able to spot them by looking for large residuals.

Beware of the lurking variable Correlation and regression are powerful tools for measuring the association between two variables and for expressing the dependence of one variable on the other. These tools must be used with an awareness of their limitations. We have seen that

Correlation measures only linear association, and fitting a straight line makes sense only when the overall pattern of the relationship is linear. Always plot your data before calculating. Extrapolation (using a fitted model far outside the range of the data that we used to fit it) often produces unreliable predictions. Correlation and least-squares regression are not resistant. Always plot your data and look for potentially influential points.

Another caution is even more important: the relationship between two variables can often be understood only by taking other variables into account. Lurking variables can make a correlation or regression misleading.

LURKING VARIABLE

A lurking variable is a variable that is not among the explanatory or response variables in a study and yet may influence the interpretation of relationships among those variables.

EXAMPLE 2.30

Discrimination in medical treatment? Studies show that men who complain of chest pain are more likely to get detailed tests and aggressive treatment such as bypass surgery than are women with similar complaints. Is this association between sex and treatment due to discrimination?

Perhaps not. Men and women develop heart problems at different ages—women are, on the average, between 10 and 15 years older than men. Aggressive treatments are more risky for older patients, so doctors may hesitate to recommend them. Lurking variables—the patient’s age and condition—may explain the relationship between sex and doctors’ decisions.

Here is an example of a different type of lurking variable.

EXAMPLE 2.31

Gas and electricity bills. A single-family household receives bills for gas and electricity each month. The 12 observations for a recent year are plotted with the least-squares regression line in Figure 2.27. We have arbitrarily chosen to put the electricity bill on the x axis and the gas bill on the y axis. There is a clear negative association. Does this mean that a high electricity bill causes the gas bill to be low and vice versa?

To understand the association in this example, we need to know a little more about the two variables. In this household, heating is done by gas and cooling is done by electricity. Therefore, in the winter months the gas bill will be relatively high and the electricity bill will be relatively low. The pattern is reversed in the summer months. The association that we see in this example is due to a lurking variable: time of year.

FIGURE 2.27 Scatterplot with least-squares regression line for predicting a household’s monthly charges for gas using its monthly charges for electricity, Example 2.31.

Correlations that are due to lurking variables are sometimes called “nonsense correlations.” The correlation is real. What is nonsense is the suggestion that the variables are directly related so that changing one of the variables causes changes in the other. The question of causation is important enough to merit separate treatment in Section 2.7. For now, just remember that an association between two variables x and y can reflect many types of relationships among x, y, and one or more lurking variables.

ASSOCIATION DOES NOT IMPLY CAUSATION

An association between an explanatory variable x and a response variable y, even if it is very strong, is not by itself good evidence that changes in x actually cause changes in y.

Lurking variables sometimes create a correlation between x and y, as in Examples 2.30 and 2.31. When you observe an association between two variables, always ask yourself if the relationship that you see might be due to a lurking variable. As in Example 2.31, time is often a likely candidate.

Beware of correlations based on averaged data Regression or correlation studies sometimes work with averages or other measures that combine information from many individuals. For example, if we plot the average height of young children against their age in months, we will see a very strong positive association with correlation near 1. But individual children of the same age vary a great deal in height. A plot of height against age for individual children will show much more scatter and lower correlation than the plot of average height against age.

A correlation based on averages over many individuals is usually higher than the correlation between the same variables based on data for individuals. This fact reminds us again of the importance of noting exactly what variables a statistical study involves.

Beware of restricted ranges restricted-range problem

The range of values for the explanatory variable in a regression can have a large impact on the strength of the relationship. For example, if we use age as a predictor of reading ability for a sample of students in the third grade, we will probably see little or no relationship. However, if our sample includes students from grades 1 through 8, we would expect to see a relatively strong relationship. We call this phenomenon the restricted-range problem.

EXAMPLE 2.32

A test for job applicants. Your company gives a test of cognitive ability to job applicants before deciding whom to hire. Your boss has asked you to use company records to see if this test really helps predict the performance ratings of employees. The restricted-range problem may make it difficult to see a strong relationship between test scores and performance ratings. The current employees were selected by a mechanism that is likely to result in scores that tend to be higher than those of the entire pool of applicants.

BEYOND THE BASICS

Data Mining Chapters 1 and 2 of this text are devoted to the important aspect of statistics called exploratory data analysis (EDA). We use graphs and numerical summaries to examine data, searching for patterns and paying attention to striking deviations from the patterns we find. In discussing regression, we advanced to using the pattern we find (in this case, a linear pattern) for prediction.

data mining

Suppose now that we have a truly enormous database, such as all purchases recorded by the cash register scanners of a national retail chain during the past week. Surely this treasure chest of data contains patterns that might guide business decisions. If we could see clearly the types of activewear preferred in large California cities and compare the preferences of small Midwest cities—right now, not at the end of the season—we might improve profits in both parts of the country by matching stock with demand. This sounds much like EDA, and indeed it is. Exploring really large databases in the hope of finding useful patterns is called data mining. Here are some distinctive features of data mining:

When you have terrabytes of data, even straightforward calculations and graphics become very time- consuming. So efficient algorithms are very important. The structure of the database and the process of storing the data (the fashionable term is data warehousing), perhaps by unifying data scattered across many departments of a large corporation, require careful consideration. Data mining requires automated tools that work based on only vague queries by the user. The process is too complex to do step-by-step as we have done in EDA.

All these features point to the need for sophisticated computer science as a basis for data mining. Indeed, data mining is often viewed as a part of computer science. Yet many statistical ideas and tools—mostly tools for dealing with multidimensional data, not the sort of thing that appears in a first statistics course—are very helpful. Like many other modern developments, data mining crosses the boundaries of traditional fields of study.

Do remember that the perils we associate with blind use of correlation and regression are yet more perilous in data mining, where the fog of an immense database can prevent clear vision. Extrapolation, ignoring lurking variables, and confusing association with causation are traps for the unwary data miner.

SECTION 2.5 SUMMARY You can examine the fit of a regression line by plotting the residuals, which are the differences between the observed and predicted values of y. Be on the lookout for points with unusually large residuals and also for nonlinear patterns and uneven variation about the line. Also look for influential observations, individual points that substantially change the regression line. Influential observations are often outliers in the x direction, but they need not have large residuals. Correlation and regression must be interpreted with caution. Plot the data to be sure that the relationship is roughly linear and to detect outliers and influential observations. Lurking variables may explain the relationship between the explanatory and response variables. Correlation and regression can be misleading if you ignore important lurking variables. We cannot conclude that there is a cause-and-effect relationship between two variables just because they are strongly associated. High correlation does not imply causation. A correlation based on averages is usually higher than if we used data for individuals.

SECTION 2.5 EXERCISES For Exercise 2.92, see page 125; and for Exercise 2.93 see page 125.

2.94 Bone strength. Exercise 2.24 (page 97) gives the bone strengths of the dominant and the nondominant arms for 15 men who were controls in a study. The least-squares regression line for these data is

dominant = 2.74 + (0.936 × nondominant)

Here are the data for four cases:

ID Nondominant Dominant

5 12.0 14.8 6 20.0 19.8 7 12.3 13.1 8 14.4 17.5

Calculate the residuals for these four cases. ARMSTR

2.95 Bone strength for baseball players. Refer to the previous exercise. Similar data for baseball players is given in Exercise 2.25 (page 98). The equation of the least-squares line for the baseball players is

dominant = 0.886 + (1.373 × nondominant)

Here are the data for the first four cases:

ID Nondominant Dominant

20 21.0 40.3 21 14.6 20.8 22 31.5 36.9 23 14.9 21.2

Calculate the residuals for these four cases. ARMSTR

2.96 Least-squares regression for radioactive decay. Refer to Exercise 2.32 (page 99) for the data on radioactive decay of barium-137m. Here are the data: DECAY

Time 1 3 5 7 Count 578 317 203 118

(a) Using the least-squares regression equation count = 602.8 − (74.7 × time)

and the observed data, find the residuals for the counts.

(b) Plot the residuals versus time.

(c) Write a short paragraph assessing the fit of the least-squares regression line to these data based on your interpretation of the residual plot.

2.97 Least-squares regression for the log counts. Refer to Exercise 2.33 (page 99), where you analyzed the radioactive decay of barium-137m data using log counts. Here are the data: DECAY

Time 1 3 5 7 Log count 6.35957 5.75890 5.31321 4.77068

(a) Using the least-squares regression equation

log count = 6.593 − (0.2606 × time)

and the observed data, find the residuals for the counts.

(b) Plot the residuals versus time.

(c) Write a short paragraph assessing the fit of the least-squares regression line to these data based on your interpretation of the residual plot.

2.98 College students by state. Refer to Exercise 2.75 (page 119), where you examined the relationship between the number of undergraduate college students and the populations for the 50 states. COLLEGE

(a) Make a scatterplot of the data with the least-squares regression line.

(b) Plot the residuals versus population.

(c) Focus on California, the state with the largest population. Is this state an outlier when you consider only the distribution of population? Explain your answer and describe what graphical and numerical summaries you used as the basis for your conclusion.

(d) Is California an outlier in the distribution of undergraduate college students? Explain your answer and describe what graphical and numerical summaries you used as the basis for your conclusion.

(e) Is California an outlier when viewed in terms of the relationship between number of undergraduate college students and population? Explain your answer and describe what graphical and numerical summaries you used as the basis for your conclusion.

(f) Is California influential in terms of the relationship between number of undergraduate college students and population? Explain your answer and describe what graphical and numerical summaries you used as the basis for your conclusion.

2.99 College students by state using logs. Refer to the previous exercise. Answer parts (a) through (f) for that exercise using the logs of both variables. Write a short paragraph summarizing your findings and comparing them with those from the previous exercise. COLLEGE

2.100 Make some scatterplots. For each of the following scenarios, make a scatterplot with 10 observations that show a moderate positive association, plus one that illustrates the unusual case. Explain each of your answers.

(a) An outlier in x that is not influential for the regression.

(b) An outlier in x that is influential for the regression.

(c) An influential observation that is not an outlier in x.

(d) An observation that is influential for the intercept but not for the slope.

2.101 What’s wrong? Each of the following statements contains an error. Describe each error and explain why the statement is wrong.

(a) An influential observation will always have a large residual.

(b) High correlation is never present when there is causation.

(c) If we have data at values of x equal to 1, 2, 3, 4, and 5, and we try to predict the value of y for x = 2.5 using a least-squares regression equation, we are doing an extrapolation.

2.102 What’s wrong? Each of the following statements contains an error. Describe each error and explain why the statement is wrong.

(a) If the residuals are all negative, this implies that there is a negative relationship between the response variable and the explanatory variable.

(b) A strong negative relationship does not imply that there is an association between the explanatory variable and the response variable.

(c) A lurking variable is always something that can be measured.

2.103 Internet use and babies. Exercise 2.34 (page 99) explores the relationship between Internet use and birthrate for 106 countries. Figure 2.13 is a scatterplot of the data. It shows a negative association between these two variables. Do you think that this plot indicates that Internet use causes people to have fewer babies? Give another possible explanation for why these two variables are negatively associated. INBIRTH

2.104 A lurking variable. The effect of a lurking variable can be surprising when individuals are divided into groups. In recent years, the mean SAT score of all high school seniors has increased. But the mean SAT score has decreased for

students at each level of high school grades (A, B, C, and so on). Explain how grade inflation in high school (the lurking variable) can account for this pattern. A relationship that holds for each group within a population need not hold for the population as a whole. In fact, the relationship can even change direction.

2.105 How’s your self-esteem? People who do well tend to feel good about themselves. Perhaps helping people feel good about themselves will help them do better in their jobs and in life. For a time, raising self-esteem became a goal in many schools and companies. Can you think of explanations for the association between high self-esteem and good performance other than “Self-esteem causes better work”?

2.106 Are big hospitals bad for you? A study shows that there is a positive correlation between the size of a hospital (measured by its number of beds x) and the median number of days y that patients remain in the hospital. Does this mean that you can shorten a hospital stay by choosing a small hospital? Why?

2.107 Does herbal tea help nursing-home residents? A group of college students believes that herbal tea has remarkable powers. To test this belief, they make weekly visits to a local nursing home, where they visit with the residents and serve them herbal tea. The nursing-home staff reports that after several months many of the residents are healthier and more cheerful. We should commend the students for their good deeds but doubt that herbal tea helped the residents. Identify the explanatory and response variables in this informal study. Then explain what lurking variables account for the observed association.

2.108 Price and ounces. In Example 2.2 (page 80) and Exercise 2.3 (page 82), we examined the relationship between the price and the size of a Mocha Frappuccino®. The 12-ounce Tall drink costs $3.95, the 16-ounce Grande is $4.45, and the 24- ounce Venti is $4.95.

(a) Plot the data and describe the relationship. (Explain why you should plot size in ounces on the x axis.)

(b) Find the least-squares regression line for predicting the price using size. Add the line to your plot.

(c) Draw a vertical line from the least-squares line to each data point. This gives a graphical picture of the residuals.

(d) Find the residuals and verify that they sum to zero.

(e) Plot the residuals versus size. Interpret this plot.

2.109 Use the applet. It isn’t easy to guess the position of the least-squares line by eye. Use the Correlation and Regression applet to compare a line you draw with the least-squares line. Click on the scatterplot to create a group of 15

points from lower left to upper right with a clear, positive straight-line pattern (correlation around 0.6). Click the “Draw line” button and use the mouse to draw a line through the middle of the cloud of points from lower left to upper right. Note the “thermometer” that appears above the plot. The red portion is the sum of the squared vertical distances from the points in the plot to the least-squares line. The green portion is the “extra” sum of squares for your line—it shows by how much your line misses the smallest possible sum of squares.

(a) You drew a line by eye through the middle of the pattern. Yet the right-hand part of the bar is probably almost entirely green. What does that tell you?

(b) Now click the “Show least-squares line” box. Is the slope of the least-squares line smaller (the new line is less steep) or larger (line is steeper) than that of your line? If you repeat this exercise several times, you will consistently get the same result. The least-squares line minimizes the vertical distances of the points from the line. It is not the line through the “middle” of the cloud of points. This is one reason it is hard to draw a good regression line by eye.

2.110 Use the applet. Go to the Correlation and Regression applet. Click on the scatterplot to create a group of 12 points in the lower-right corner of the scatterplot with a strong straight-line pattern (correlation about −0.8). Now click

the “Show least-squares line” box to display the regression line.

(a) Add one point at the upper left that is far from the other 12 points but exactly on the regression line. Why does this outlier have no effect on the line even though it changes the correlation?

(b) Now drag this last point down until it is opposite the group of 12 points. You see that one end of the least-squares line chases this single point, while the other end remains near the middle of the original group of 12. What makes the last point so influential?

2.111 Education and income. There is a strong positive correlation between years of education and income for economists employed by business firms. (In particular, economists with doctorates earn more than economists with only a bachelor’s degree.) There is also a strong positive correlation between years of education and income for economists employed by colleges and universities. But when all economists are considered, there is a negative correlation between education and income. The explanation for this is that business pays high salaries and employs mostly economists with bachelor’s degrees,

while colleges pay lower salaries and employ mostly economists with doctorates. Sketch a scatterplot with two groups of cases (business and academic) that illustrates how a strong positive correlation within each group and a negative overall correlation can occur together.

2.112 Dangers of not looking at a plot. Table 2.1 (page 122) presents four sets of data prepared by the statistician Frank Anscombe to illustrate the dangers of calculating without first plotting the data.22 ANSC

(a) Use x to predict y for each of the four data sets. Find the predicted values and residuals for each of the four regression equations.

(b) Plot the residuals versus x for each of the four data sets.

(c) Write a summary of what the residuals tell you for each data set, and explain how the residuals help you to understand these data.

2.6 Data Analysis for Two-Way Tables

When you complete this section, you will be able to:

Identify the row variable, the column variable, and the cells in a two-way table. Find and interpret the joint distribution in a two-way table. Find and interpret the marginal distributions in a two-way table. Use the conditional distributions to describe the relationship displayed in a two-way table. Determine the joint distribution, the marginal distributions, and the conditional distributions in a two-way table from software output. Interpret examples of Simpson’s paradox.

quantitative and categorical variables, p. 3

When we study relationships between two variables, one of the first questions we ask is whether each variable is quantitative or categorical. For two quantitative variables, we use a scatterplot to examine the relationship, and we fit a line to the data if the relationship is approximately linear. If one of the variables is quantitative and the other is categorical, we can use the methods in Chapter 1 to describe the distribution of the quantitative variable for each value of the categorical variable. This leaves us with the situation where both variables are categorical. In this section, we discuss methods for studying these relationships.

Some variables—such as sex, race, and occupation—are inherently categorical. Other categorical variables are created by grouping values of a quantitative variable into classes. Published data are often reported in grouped form to save space. To describe categorical data, we use the counts (frequencies) or percents (relative frequencies) of individuals that fall into various categories.

2.113

The two-way table two-way table

A key idea in studying relationships between two variables is that both variables must be measured on the same individuals or cases. When both variables are categorical, the raw data are summarized in a two-way table that gives counts of observations for each combination of values of the two categorical variables. Here is an example.

EXAMPLE 2.33

IOM

Is the calcium intake adequate? Young children need calcium in their diet to support the growth of their bones. The Institute of Medicine provides guidelines for how much calcium should be consumed for people of different ages.23 One study examined whether or not a sample of children consumed an adequate amount of calcium, based on these guidelines. Because there are different requirements for children aged 5 to 10 years and for children aged 11 to 13 years of age, the children were classified into these two age groups. For each student, his or her calcium intake was classified as meeting or not meeting the requirement. There were 2029 children in the study. Here are the data:24

Two-way table for “met requirement” and age Age (years) Met requirement 5 to 10 11 to 13 No 194 557 Yes 861 417

We see that 194 children aged 5 to 10 did not meet the calcium requirement, and 861 children aged 5 to 10 years met the calcium requirement.

USE YOUR KNOWLEDGE

Read the table. Refer to the table in Example 2.33. How many children aged 11 to 13 met the requirement? How many did not?

row variable

column variable

cell

For the calcium requirement example, we could view age as an explanatory variable and “met requirement” as a response variable. This is why we put age in the columns (like the x axis in a scatterplot) and “met requirement” in the rows (like the y axis in a scatterplot). We call “met requirement” the row variable because each horizontal row in the table describes whether or not the requirement was met. Age is the column variable because each vertical column describes one age group. Each combination of values for these two variables is called a cell. For example, the cell corresponding to children who are 5 to 10 years old and who have not met the requirement contains the number 194. This table is called a 2 × 2 table because there are two rows and two columns.

To describe relationships between two categorical variables, we compute different types of percents. Our job is easier if we expand the basic two-way table by adding various totals. We illustrate the idea with our calcium requirement example.

2.114

EXAMPLE 2.34

IOM

Add the margins to the table. We expand the table in Example 2.33 by adding the totals for each row, for each column, and the total number of all the observations. Here is the result:

Two-way table for “met requirement” and age Age (years) Met requirement 5 to 10 11 to 13 Total No 194 557 751 Yes 861 417 1278 Total 1055 974 2029

In this study there were 1055 children aged 5 to 10. The total number of children who did not meet the calcium requirement is 751, and the total number of children in the study is 2029.

USE YOUR KNOWLEDGE

Read the margins of the table. How many children aged 11 to 13 were subjects in the calcium requirement study? What is the total number of children who met the calcium requirement?

In this example, be sure that you understand how the table is obtained from the raw data. Think about a data file with one line per subject. There would be 2029 lines or records in this data set. In the two-way table, each individual is counted once and only once. As a result, the sum of the counts in the table is the total number of individuals in the data set. Most errors in the use of categorical-data methods come from a misunderstanding of how these tables are constructed.

2.115

Joint distribution joint distribution

We are now ready to compute some proportions that help us understand the data in a two-way table. Suppose that we are interested in the children aged 5 to 10 years who do not meet the calcium requirement. The proportion of these is simply 194 divided by 2029, or 0.0956. We would estimate that 9.56% of children in the population from which this sample was drawn are 5- to 10-year-olds who do not meet the calcium requirement. For each cell, we can compute a proportion by dividing the cell entry by the total sample size. The collection of these proportions is the joint distribution of the two categorical variables.

EXAMPLE 2.35

IOM

The joint distribution. For the calcium requirement example, the joint distribution of age “met requirement” and age is

Joint distribution of “met requirement” and age Age (years) Met requirement 5 to 10 11 to 13 No 0.0956 0.2745 Yes 0.4243 0.2055

Because this is a distribution, the sum of the proportions should be 1. For this example the sum is 0.9999. The difference is due to roundoff error.

USE YOUR KNOWLEDGE

Explain the computation. Explain how the entry for the children aged 5 to 10 who met the calcium requirement in Example 2.35 is computed from the table in Example 2.34.

How might we use the information in the joint distribution for this example? Suppose that we were to develop an outreach unit to increase the consumption of calcium. The distribution suggests that the older students should be targeted if we have to make a choice because of limited funds. Children who are 11 to 13 years old and do not meet the calcium requirement are 27.45% of the total; however, children who are 5 to 10 years old and do not meet the requirement are only 9.56% of the total. For other uses of these data, we may need to calculate different numerical summaries. Let’s look at the distribution of age.

Marginal distributions marginal distribution

When we examine the distribution of a single variable in a two-way table, we are looking at a marginal distribution. There are two marginal distributions, one for each categorical variable in the two-way table. They are very easy to compute.

EXAMPLE 2.36

IOM

The marginal distribution of age. Look at the table in Example 2.34. The total numbers of children aged 5 to 10 and children aged 11 to 13 are given in the bottom row, labeled “Total.” Our sample has 1055 children aged 5 to 10 and 974 children aged 11 to 13. To find the marginal distribution of age, we simply divide these numbers by the total sample size, 2029. The marginal distribution of age is

Marginal distribution of age 5 to 10 11 to 13 Proportion 0.52 0.48

Note that the proportions sum to 1; there is no roundoff error.

Often, we prefer to use percents rather than proportions. Here is the marginal distribution of age described with percents:

Marginal distribution of age 5 to 10 11 to 13 Percent 52% 48%

Which form do you prefer? The percent of children in each age group is approximately the same. This is interesting because the first category

includes six ages (5, 6, 7, 8, 9, and 10); whereas the second includes only three ages (11, 12, and 13). Recall that the age categories were chosen in this way because the Institute of Medicine defined the calcium requirement differently for these age groups. In this study, the children were selected from grades 4, 5, and 6. The distribution of ages within these grades explains the marginal distribution of age for our sample.

The other marginal distribution for this example is the distribution of “met requirement.”

EXAMPLE 2.37

IOM

The marginal distribution of “met requirement.” Here is the marginal distribution of “met requirement,” in percents:

Marginal distribution of “met requirement”

2.116

No Yes Percent 37.01% 62.99%

USE YOUR KNOWLEDGE

Explain the marginal distribution. Explain how the marginal distribution of “met requirement” given in Example 2.37 is computed from the entries in the table given in Example 2.34.

bar graphs and pie charts, p. 9

Each marginal distribution from a two-way table is a distribution for a single categorical variable. We can use a bar graph or a pie chart to display such a distribution. For our two-way table, we will be content with numerical summaries: for example, 52% of the children are aged 5 to 10, and 37% of the children are not meeting their calcium requirement. When we have more rows or columns, the graphical displays are particularly useful.

2.117

Describing relations in two-way tables The table in Example 2.34 contains much more information than the two marginal distributions of age alone and “met requirement” alone. We need to do a little more work to examine the relationship. Relationships among categorical variables are described by calculating appropriate percents from the counts given. What percents do you think we should use to describe the relationship between age and meeting the calcium requirement?

EXAMPLE 2.38

IOM

Meeting the calcium requirement for children aged 5 to 10. What percent of the children aged 5 to 10 in our sample met the calcium requirement? This is the count of the children who are 5 to 10 years old and who met the calcium requirement as a percent of the number of children who are 5 to 10 years old:

USE YOUR KNOWLEDGE

Find the percent. Refer to the table in Example 2.34 (page 137). Show that the percent of children 11 to 13 years old who met the calcium requirement is 43%.

2.118

Conditional distributions conditional distribution

In Example 2.38, we looked at the children aged 5 to 10 alone and examined the distribution of the other categorical variable, “met requirement.” Another way to say this is that we conditioned on the value of age, 5 to 10 years old. Similarly, we can condition on the value of age being 11 to 13 years old. When we condition on the value of one variable and calculate the distribution of the other variable, we obtain a conditional distribution. Note that in Example 2.38, we calculated only the percent for children aged 5 to 10 years. The complete conditional distribution gives the proportions or percents for all possible values of the conditioning variable.

EXAMPLE 2.39

IOM

Conditional distribution of “met requirement” for children aged 5 to 10. For children aged 5 to 10 years, the conditional distribution of the “met requirement” variable in terms of percents is

Conditional distribution of “met requirement” for children aged 5 to 10 No Yes Percent 18.39% 81.61%

Note that we have included the percents for both of the possible values, Yes and No, of the “met requirement” variable. These percents sum to 100%.

USE YOUR KNOWLEDGE

IOM

A conditional distribution. Perform the calculations to show that the conditional distribution of “met requirement” for children aged 11 to 13 years is

Conditional distribution of “met requirement” for children aged 11 to 13 No Yes Percent 57.19% 42.81%

Comparing the conditional distributions (Example 2.39 and Exercise 2.118) reveals the nature of the association between age and meeting the calcium requirement. In this set of data, the older children are more likely to fail to meet the calcium requirement.

Bar graphs can help us to see relationships between two categorical variables. No single graph (such as a scatterplot) portrays the form of the relationship between categorical variables, and no single numerical measure (such as the correlation) summarizes the strength of an association. Bar graphs are flexible enough to be helpful, but you must think about what comparisons you want to display. For numerical measures, we must rely on well-chosen percents or on more advanced statistical methods.25

A two-way table contains a great deal of information in compact form. Making that information clear almost always requires finding percents. You must decide which percents you need. Of course, we prefer to use software to compute the joint, marginal, and conditional distributions.

EXAMPLE 2.40

IOM

Software output. Figure 2.28 gives computer output for the data in Example 2.33 using Minitab, SPSS, and JMP. There are minor variations among software packages, but these outputs are typical of what is usually produced. Each cell in the 2 × 2 table has four entries. These are the count (the number of observations in the cell), the conditional distributions for rows and columns, and the joint distribution. Note that all of these are expressed as percents rather than proportions. Marginal totals and distributions are given in the rightmost column and the bottom row.

Most software packages order the row and column labels numerically or alphabetically. In general, it is better to use words rather than numbers for the column labels. This sometimes involves some additional work, but it avoids the kind of confusion that can result when you forget the real values associated with each numerical value. You should verify that the entries in Figure 2.28 correspond to the calculations that we performed in Examples 2.34 through 2.39. In addition, verify the calculations for the conditional distributions of age for each value of “met requirement.”

FIGURE 2.28 Computer output for the calcium requirement study, Example 2.40: (a) Minitab; (b) SPSS; (c) JMP.

mosaic plot

The JMP output in Figure 2.28 includes a graphical display of the data called a mosaic plot. The sizes of the four boxes display the joint distribution. The narrow bar to the right shows the marginal distribution of Met and the widths of the vertical bars show the marginal distribution of Age. The conditional distribution of Met for each Age is represented in each of these vertical bars by the heights of the blue and red sections. Notice that they always add to one.

Simpson’s paradox As is the case with quantitative variables, the effects of lurking variables can strongly influence relationships between two categorical variables. Here is an example that demonstrates the surprises that can await the unsuspecting consumer of data.

EXAMPLE 2.41

CUSTSER

Which customer service representative is better? A customer service center has a goal of resolving customer questions in 10 minutes or less. Here are the records for two representatives:

Representative Goal met Alexis Peyton Yes 172 118 No 28 82 Total 200 200

Alexis has met the goal 172 times out of 200, a success rate of 86%. For Peyton, the success rate is 118 out of 200, or 59%. Alexis clearly has the better success rate.

Let’s look at the data in a little more detail. The data summarized come from two different weeks in the year.

EXAMPLE 2.42

Look at the data more carefully. Here are the counts broken down by week:

CUSTSER

Week 1 Week 2 Goal met Alexis Peyton Alexis Peyton Yes 162 19 10 99 No 18 1 10 81 Total 180 20 20 180

For Week 1, Alexis met the goal 90% of the time (162/180), while Peyton met the goal 95% of the time (19/20). Peyton had the better performance in Week 1. What about Week 2? Here, Alexis met the goal 50% of the time (10/20), while the success rate for Peyton was 55% (99/180). Peyton again had the better performance. How does this analysis compare with the analysis that combined the counts for the two weeks? That analysis clearly showed that Alexis had the better performance, 86% versus 59%.

These results can be explained by a lurking variable, Week. The first week was during a period when the product

had been in use for several months. Most of the calls to the customer service center concerned problems that had been encountered before. The representatives were trained to answer these questions and usually had no trouble in meeting the goal of resolving the problems quickly. On the other hand, the second week occurred shortly after the release of a new version of the product. Most of the calls during this week concerned new problems that the representatives had not yet encountered. Many more of these questions took longer than the 10-minute goal to resolve.

Look at the totals in the bottom row of the detailed table. During the first week, when calls were easy to resolve, Alexis handled 180 calls and Peyton handled 20. The situation was exactly the opposite during the second week, when the calls were difficult to resolve. There were 20 calls for Alexis and 180 for Peyton.

The original two-way table, which did not take account of week, was misleading. This example illustrates Simpson’s paradox.

SIMPSON’S PARADOX

An association or comparison that holds for all of several groups can reverse direction when the data are combined to form a single group. This reversal is called Simpson’s paradox.

The lurking variables in our Simpson’s paradox example, Week and problem difficulty, are categorical. That is, they break the observations into groups by workweek. Simpson’s paradox is an extreme form of the fact that observed associations can be misleading when there are lurking variables.

three-way table

aggregation

The data in Example 2.42 are given in a three-way table that reports counts for each combination of three categorical variables: week, representative, and whether or not the goal was met. In our example, we constructed the three-way table by constructing two two-way tables for representative by goal, one for each week. The original table in Example 2.41 can be obtained by adding the corresponding counts for these two tables. This process is called aggregating the data. When we aggregated data in Example 2.41, we ignored the variable week, which then became a lurking variable. Conclusions that seem obvious when we look only at aggregated data can become quite different when the data are examined in more detail.

SECTION 2.6 SUMMARY A two-way table of counts organizes data about two categorical variables. Values of the row variable label the rows that run across the table, and values of the column variable label the columns that run down the table. Two-way tables are often used to summarize large amounts of data by grouping outcomes into categories. The joint distribution of the row and column variables is found by dividing the count in each cell by the total number of observations. The row totals and column totals in a two-way table give the marginal distributions of the two variables separately. It is clearer to present these distributions as percents of the table total. Marginal distributions do not give any information about the relationship between the variables. To find the conditional distribution of the row variable for one specific value of the column variable, look only at that one column in the table. Find each entry in the column as a percent of the column total. There is a conditional distribution of the row variable for each column in the table. Comparing these conditional distributions is one way to describe the association between the row and the column variables. It is particularly useful when the column variable is the explanatory variable. When the row variable is explanatory, find the conditional distribution of the column variable for each row and compare these distributions. Bar graphs are a flexible means of presenting categorical data. There is no single best way to describe an association between two categorical variables. We present data on three categorical variables in a three-way table, printed as separate two-way tables for each level of the third variable. A comparison between two variables that holds for each level of a third variable can be changed or even reversed when the data are aggregated by summing over all levels of the third variable. Simpson’s paradox refers to the reversal of a comparison by aggregation. It is an example of the potential effect of lurking variables on an observed association.

SECTION 2.6 EXERCISES For Exercise 2.113, see page 137; for 2.114, see page 138; for 2.115, see page 138; for 2.116, see page 140; for 2.117, see page 140; and for 2.118, see page 141.

2.119 Does driver’s ed help? A study is planned to look at the effect of driver education programs on accidents. The driving records of all drivers under 18 in a given year will classify each driver as having taken a driver’s education course or not. The drivers will also classified with respect to the number of accidents that they had in the year after they received their license. The categories are zero, one, and two or more accidents.

(a) There are two variables in this study. Do you think one is an explanatory variable and the other is a response variable? Explain your answer.

(b) Sketch a two-way table that could be used to organize the data. Which variable is the row variable? Which variable is the column variable?

(c) How many cells are in the table? Describe in words what each of the cells will contain when the data are collected.

2.120 Music and video games. You are planning a study of undergraduates in which you will examine the relationship between listening to music and playing video games. The study subjects will be asked how much time they spend in each of these activities during a typical day. The choices for both activities will be a half hour or less, more than a half hour but less than an hour, and more than an hour.

(a) There are two variables in this study. Do you think one is an explanatory variable and the other is a response variable? Explain your answer.

(b) Sketch a two-way table that could be used to organize the data. Which variable is the row variable? Which variable is the column variable?

(c) How many cells are in the table? Describe in words what each of the cells will contain when the data are collected.

2.121 Eight is enough. A healthy body needs good food, and healthy teeth are needed to chew our food so that it can nourish our bodies. The U.S. Army has recognized this fact and requires recruits to pass a dental examination. If you wanted to be a soldier in the Spanish American War, which took place in 1898, you needed to have at least eight teeth. Here is the statement of the requirement: TEETH

Unless an applicant has at least four sound double teeth, one above and one below on each side of the mouth, and so opposed as to serve the purpose of mastication, he should be rejected.

A study reported the rejection data for enlistment candidates classified by age. Here are the data:26

Age

Rejected Under 20 20 to 25 25 to 30 30 to 35 35 to 40 Over 40

Yes 68 647 1114 1783 2887 3801 No 58,884 77,992 55,597 43,994 47,569 39,985

(a) Which variable is the explanatory variable? Which variable is the response variable? Give reasons for your answer.

(b) Find the joint distribution. Write a brief summary explaining the major features of this distribution.

(c) Find the two marginal distributions. Write a brief summary explaining the major features of these distributions.

(d) Which conditional distribution would you choose to explain the relationship between these two variables? Explain your answer.

(e) Find the conditional distribution that you chose in part (d), and write a summary that includes your interpretation of the relationship based on this conditional distribution.

2.122 Survival and class on the Titanic. In Exercise 1.27 (page 24), you created a graphical summary of the number of passengers who survived classified by the accommodations that they had on the ship: first, second, or third class. Let’s look at these data with a two-way table. TITANIC

(a) Create a two-way table that you could use to explore the relationship between survival and class.

(b) Which variable is the explanatory variable and which is the response variable? Give reasons for your answers.

(c) Find the two marginal distributions. Write a brief summary explaining the major features of these distributions.

(d) Which conditional distribution would you choose to explain the relationship between these two variables? Explain your answer.

FIGURE 2.29 Computer output for the lying to a teacher data, Exercise 2.123.

(e) Find the conditional distribution that you chose in part (d) and, write a summary that includes your interpretation of the relationship based on this conditional distribution.

2.123 Lying to a teacher. One of the questions in a survey of high school students asked about lying to teachers.27 The data set LYING gives the numbers of students who said that they lied to a teacher about something significant at least once during the past year, classified by sex. Figure 2.29 gives software output for these data. Use this output to analyze these data and write a report summarizing your work. Be sure to include a discussion of whether or not you consider this relationship to involve an explanatory variable and a response variable. LYING

2.124 Trust and honesty in the workplace. The students surveyed in the study described in the previous exercise were also asked whether they thought trust and honesty were essential in business and the workplace. Figure 2.30 gives software output for these data. Use this output to analyze these data and write a report summarizing your work. Be sure to include a discussion of whether or not you consider this relationship to involve an explanatory variable and a response variable. TRUST

2.125 Exercise and adequate sleep. A survey of 656 boys and girls, who were 13 to 18 years old, asked about adequate sleep and other health-related behaviors. The recommended amount of sleep is six to eight hours per night.28 In the survey, 59% of the respondents reported that they got less than this amount of sleep on school nights. An exercise scale was developed and used to classify the students as above or below the median in this domain. Here is the 2 × 2 table of counts with students classified as getting or not getting adequate sleep and by the exercise variable: SLEEP

FIGURE 2.30 Computer output for the trust and honesty in the workplace data, Exercise 2.124.

Exercise

Enough sleep High Low

Yes 151 115 No 148 242

(a) Find the distribution of adequate sleep for the high exercisers.

(b) Do the same for the low exercisers.

(c) Summarize the relationship between adequate sleep and exercise using the results of parts (a) and (b).

2.126 Adequate sleep and exercise. Refer to the previous exercise. SLEEP

(a) Find the distribution of exercise for those who get adequate sleep.

(b) Do the same for those who do not get adequate sleep.

(c) Write a short summary of the relationship between adequate sleep and exercise using the results of parts (a) and (b).

(d) Compare this summary with your summary from part (c) of the previous exercise. Which do you prefer? Give a reason for your answer.

2.127 Which hospital is safer? Insurance companies and consumers are interested in the performance of hospitals. The government releases data about patient outcomes in hospitals that can be useful in making informed health care decisions. Here is a two-way table of data on the survival of patients after surgery in two hospitals. All patients undergoing surgery in a recent time period are included. “Survived” means that the patient lived at least six weeks following surgery. HOSP

Hospital A Hospital B

Died 63 16 Survived 2037 784

Total 2100 800

What percent of Hospital A patients died? What percent of Hospital B patients died? These are the numbers one might see reported in the media.

2.128 Patients in “poor” or “good” condition. Refer to the previous exercise. Not all surgery cases are equally serious,

however. Patients are classified as being in either “poor” or “good” condition before surgery. Here are the data broken down by patient condition. The entries in the original two-way table are just the sums of the “poor” and “good” entries in this pair of tables. HOSP

Good condition

Hospital A Hospital B

Died 6 8 Survived 594 592

Total 600 600

Poor condition

Hospital A Hospital B

Died 57 8 Survived 1443 192

Total 1500 200

(a) Find the death rate for Hospital A patients who were classified as “poor” before surgery. Do the same for Hospital B. In which hospital do “poor” patients fare better?

(b) Repeat part (a) for patients classified as “good” before surgery.

(c) What is your recommendation to someone facing surgery and choosing between these two hospitals?

(d) How can Hospital A do better in both groups, yet do worse overall? Look at the data and carefully explain how this can happen.

2.129 Complete the table. Here are the row and column totals for a two-way table with two rows and two columns:

a b 300 c d 200

300 200 500

Find two different sets of counts a, b, c, and d for the body of the table that give these same totals. This shows that the relationship between two variables cannot be obtained from the two individual distributions of the variables.

2.130 Construct a table with no association. Construct a 3 × 4 table of counts where there is no apparent association between the row and column variables.

2.7 The Question of Causation

When you complete this section, you will be able to:

Identify the differences among causation, common response, and confounding in explaining an association. Apply the five criteria for establishing causation.

In many studies of the relationship between two variables, the goal is to establish that changes in the explanatory variable cause changes in the response variable. Even when a strong association is present, however, the conclusion that this association is due to a causal link between the variables is often hard to justify. What ties between two variables (and others lurking in the background) can explain an observed association? What constitutes good evidence for causation? We begin our consideration of these questions with a set of observed associations. In each case, there is a clear association between variable x and variable y. Moreover, the association is positive whenever the direction makes sense.

Explaining association

EXAMPLE 2.43

Observed associations. Here are some examples of observed association between x and y: 1. x = mother’s body mass index

y = daughter’s body mass index

2. x = amount of the artificial sweetener saccharin in a rat’s diet

y = count of tumors in the rat’s bladder

3. x = a student’s SAT score as a high school senior

y = a student’s first-year college grade point average

4. x = monthly flow of money into stock mutual funds

y = monthly rate of return for the stock market

5. x = whether a person regularly attends religious services

y = how long the person lives

6. x = the number of years of education a worker has

y = the worker’s income

Explaining association: Causation Figure 2.31 shows in outline form how a variety of underlying links between variables can explain association. The dashed double-arrow line represents an observed association between the variables x and y. Some associations are explained by a direct cause-and-effect link between these variables. The first diagram in Figure 2.31 shows “x causes y” by a solid arrow running from x to y.

Items 1 and 2 in Example 2.43 are examples of direct causation. Even when direct causation is present, very often it is not a complete explanation of an association between two variables. The best evidence for causation comes from experiments that actually change x while holding all other factors fixed. If y changes, we have good reason to think that x caused the change in y.

FIGURE 2.31 Possible explanations for an observed association. The dashed double-arrow lines show an association. The solid arrows show a cause-and-effect link. The variable x is explanatory, y is a response variable, and z is a lurking variable.

common response

Explaining association: Common response “Beware of the lurking variable” is good advice when thinking about an association between two variables. The second diagram in Figure 2.31 illustrates common response. The observed association between the variables x and y is explained by a lurking variable z. Both x and y change in response to changes in z. This common response creates an association even though there may be no direct causal link between x and y.

The third and fourth items in Example 2.43 illustrate how common response can create an association. What would be a good candidate for the variable z in these two examples?

Explaining association: Confounding For the first item in Example 2.43, we expect that inheritance explains part of the association between the body mass indexes (BMIs) of daughters and their mothers. Can we use r or r2 to say how much inheritance contributes to the daughters’ BMIs? No. It may well be that mothers who are overweight also set an example of little exercise, poor eating habits, and lots of television. Their daughters pick up these habits to some extent, so the influence of heredity is mixed up with influences from the girls’ environment. We call this mixing of influences confounding.

CONFOUNDING

Two variables are confounded when their effects on a response variable cannot be distinguished from each other. The confounded variables may be either explanatory variables or lurking variables or both.

When many uncontrolled variables are related to a response variable, you should always ask whether or not confounding of several variables prevents you from drawing conclusions about causation. The third diagram in Figure 2.31 illustrates confounding. Both the explanatory variable x and the lurking variable z may influence the response variable y. Because x is confounded with z, we cannot distinguish the influence of x from the influence of z. We cannot say how strong the direct effect of x on y is. In fact, it can be hard to say if x influences y at all.

The last two associations in Example 2.43 (Items 5 and 6) are explained in part by confounding. What would be a good candidate for the confounding variable z in these two examples?

Many observed associations are at least partly explained by lurking variables. Both common response and confounding involve the influence of a lurking variable (or variables) z on the response variable y. The distinction between these two types of relationship is less important than the common element, the influence of lurking variables. The most important lesson of these examples is one we have already emphasized: even a very strong association between two variables is not by itself good evidence that there is a cause-and-effect link between the variables.

Establishing causation How can a direct causal link between x and y be established? The best method—indeed, the only fully compelling method—of establishing causation is to conduct a carefully designed experiment in which the effects of possible lurking variables are controlled. Chapter 3 explains how to design convincing experiments.

Many of the sharpest disputes in which statistics plays a role involve questions of causation that cannot be settled by experiment. Does gun control reduce violent crime? Does living near power lines cause cancer? Has “outsourcing” work to overseas locations reduced overall employment in the United States? All these questions have become public issues. All concern associations among variables. And all have this in common: they try to pinpoint cause and effect in a setting involving complex relations among many interacting variables. Common response and confounding, along with the number of potential lurking variables, make observed associations misleading. Experiments are not possible for ethical or practical reasons. We can’t assign some people to live near power lines or compare the same nation with and without strong gun controls.

EXAMPLE 2.44

Power lines and leukemia. Electric currents generate magnetic fields. So living with electricity exposes people to magnetic fields. Living near power lines increases exposure to these fields. Really strong fields can disturb living cells in laboratory studies. Some people claim that the weaker fields we experience if we live near power lines cause leukemia in children.

It isn’t ethical to do experiments that expose children to magnetic fields. It’s hard to compare cancer rates among children who happen to live in more and less exposed locations because leukemia is rare and locations vary in many ways other than magnetic fields. We must rely on studies that compare children who have leukemia with children who don’t.

A careful study of the effect of magnetic fields on children took five years and cost $5 million. The researchers compared 638 children who had leukemia and 620 who did not. They went into the homes and actually measured the magnetic fields in the children’s bedrooms, in other rooms, and at the front door. They recorded facts about nearby power lines for the family home and also for the mother’s residence when she was pregnant. Result: no evidence of more than a chance connection between magnetic fields and childhood leukemia.29

“No evidence” that magnetic fields are connected with childhood leukemia doesn’t prove that there is no risk. It says only that a careful study could not find any risk that stands out from the play of chance that distributes leukemia cases across the landscape. Critics continue to argue that the study failed to measure some lurking variables or that the children studied don’t fairly represent all children. Nonetheless, a carefully designed study comparing children with and without leukemia is a great advance over haphazard and sometimes emotional counting of cancer cases.

EXAMPLE 2.45

Smoking and lung cancer. Despite the difficulties, it is sometimes possible to build a strong case for causation in the absence of experiments. The evidence that smoking causes lung cancer is about as strong as nonexperimental evidence can be.

Doctors had long observed that most lung cancer patients were smokers. Comparison of smokers and similar nonsmokers showed a very strong association between smoking and death from lung cancer. Could the association be

due to common response? Might there be, for example, a genetic factor that predisposes people both to nicotine addiction and to lung cancer? Smoking and lung cancer would then be positively associated even if smoking had no direct effect on the lungs. Or perhaps confounding is to blame. It might be that smokers live unhealthy lives in other ways (diet, alcohol, lack of exercise) and that some other habit confounded with smoking is a cause of lung cancer. How were these objections overcome?

Let’s answer this question in general terms: what are the criteria for establishing causation when we cannot do an experiment?

The association is strong. The association between smoking and lung cancer is very strong. The association is consistent. Many studies of different kinds of people in many countries link smoking to lung cancer. That reduces the chance that a lurking variable specific to one group or one study explains the association. Higher doses are associated with stronger responses. People who smoke more cigarettes per day or who smoke over a longer period get lung cancer more often. People who stop smoking reduce their risk. The alleged cause precedes the effect in time. Lung cancer develops after years of smoking. The alleged cause is plausible. Experiments show that tars from cigarette smoke cause cancer when applied to the backs of mice.

Medical authorities do not hesitate to say that smoking causes lung cancer. The U.S. Surgeon General states that cigarette smoking is “the largest avoidable cause of death and disability in the United States.”30 The evidence for causation is strong—but it is not as strong as the evidence provided by well-designed experiments.

SECTION 2.7 SUMMARY Some observed associations between two variables are due to a cause-and-effect relationship between these variables, but others are explained by lurking variables. The effect of lurking variables can operate through common response if changes in both the explanatory and the response variables are caused by changes in lurking variables. Confounding of two variables (either explanatory or lurking variables or both) means that we cannot distinguish their effects on the response variable. Establishing that an association is due to causation is best accomplished by conducting an experiment that changes the explanatory variable while controlling other influences on the response. In the absence of experimental evidence, be cautious in accepting claims of causation. Good evidence of causation requires (1) a strong association, (2) that appears consistently in many studies, (3) that has higher doses associated with stronger responses, (4) with the alleged cause preceding the effect in time, and (5) that is plausible.

SECTION 2.7 EXERCISES 2.131 Examples of association. Give three examples of association: one due to causation, one due to common response, and one due to confounding. Use your examples to write a short paragraph explaining the differences among these three explanations for an observed association.

2.132 The five criteria for establishing causation. Consider the five criteria for establishing causation. Explain how each of these, if not established seriously, weakens the case that an association is due to causation.

2.133 Iron and anemia. A lack of adequate iron in the diet is associated with anemia, a condition in which the body does not have enough red blood cells. However, anemia is also associated with malaria and infections with worms called helminths. Discuss these observed associations using the framework of Figure 2.31.

2.134 Stress and lack of sleep in college students. Studies of college students have shown that stress and lack of sleep are associated. Do you think that lack of sleep causes stress or that stress causes lack of sleep? Write a short paragraph summarizing your opinions.

2.135 Online courses. Many colleges offer online versions of some courses that are also taught in the classroom. It often happens that the students who enroll in the online version do better than the classroom students on the course exams. This does not show that online instruction is more effective than classroom teaching because the people who sign up for online courses are often quite different from the classroom students. Suggest some student characteristics that you think could be confounded with online versus classroom. Use a diagram like Figure 2.31(c) to illustrate your ideas.

2.136 Marriage and income. Data show that men who are married, and also divorced or widowed men, earn quite a bit more than men who have never been married. This does not mean that a man can raise his income by getting married. Suggest several lurking variables that you think are confounded with marital status and that help explain the association between marital status and income. Use a diagram like Figure 2.31(c) to illustrate your ideas.

2.137 Exercise and self-confidence. A college fitness center offers an exercise program for staff members who choose to participate. The program assesses each participant’s fitness, using a treadmill test, and also administers a personality questionnaire. There is a moderately strong positive correlation between fitness score and score for self-confidence. Is this good evidence that improving fitness increases self-confidence? Explain why or why not.

2.138 Computer chip manufacturing and miscarriages. A study showed that women who work in the production of computer chips have abnormally high numbers of miscarriages. The union claimed that exposure to chemicals used in production caused the miscarriages. Another possible explanation is that these workers spend most of their work time standing up. Illustrate these relationships in a diagram like one of those in Figure 2.31.

2.139 Hospital size and length of stay. A study shows that there is a positive correlation between the size of a hospital (measured by its number of beds x) and the median number of days y that patients remain in the hospital. Does this mean that you can shorten a hospital stay by choosing a small hospital? Use a diagram like one of those in Figure 2.31 to explain the association.

2.140 Watching TV and low grades. Children who watch many hours of television get lower grades in school, on average, than those who watch less TV. Explain clearly why this fact does not show that watching TV causes poor grades. In particular, suggest some other variables that may be confounded with heavy TV viewing and may contribute to poor grades.

2.141 Artificial sweeteners. People who use artificial sweeteners in place of sugar tend to be heavier than people who use sugar. Does this mean that artificial sweeteners cause weight gain? Give a more plausible explanation for this association.

2.142 Exercise and mortality. A sign in a fitness center says, “Mortality is halved for men over 65 who walk at least 2 miles a day.”

(a) Mortality is eventually 100% for everyone. What do you think “mortality is halved” means?

(b) Assuming that the claim is true, explain why this fact does not show that exercise causes lower mortality.

2.143 Effect of a math skills refresher initiative. Students enrolling in an elementary statistics course take a pretest that assesses their math skills. Those who receive low scores are given the opportunity to take three one-hour refresher sessions designed to review the basic math skills needed for the statistics course. Those who took the refresher sessions performed worse than those who did not on the final exam in the statistics course. Can you conclude that the refresher course has a negative impact on performance in the statistics course? Explain your answer.

TABLE 2.3

CHAPTER 2 EXERCISES

2.144 Dwelling permits and sales for 23 countries. The Organisation for Economic Co-operation and Development collects data on main economic indicators (MEIs) for many countries. Each variable is recorded as an index with the year 2000 serving as a base year. This means that the variable for each year is reported as a ratio of the value for the year divided by the value for 2000. Use of indices in this way makes it easier to compare values for different countries. Table 2.3 gives the values of three MEIs for 23 countries.31 MEIS

(a) Make a scatterplot with sales as the response variable and permits issued for new dwellings as the explanatory variable. Describe the relationship. Are there any outliers or influential observations?

(b) Find the least-squares regression line and add it to your plot.

(c) Interpret the slope of the line in the context of this exercise.

(d) Interpret the intercept of the line in the context of this exercise. Explain whether or not this interpretation is useful in explaining the relationship between these two variables.

(e) What is the predicted value of sales for a country that has an index of 224 for dwelling permits?

(f) Canada has an index of 224 for dwelling permits. Find the residual for this country.

(g) What percent of the variation in sales is explained by dwelling permits?

Dwelling Permits, Sales, and Production for 21 Countries

Country Dwelling permits Sales Production

Australia 116 137 109 Belgium 125 105 112 Canada 224 122 101 Czech Republic 178 134 162 Denmark 121 126 109 Finland 105 136 125 France 145 121 104 Germany 54 100 119 Greece 117 136 102 Hungary 109 140 155 Ireland 92 123 144 Japan 86 99 109 Korea 158 110 156 Luxembourg 145 161 118 Netherlands 160 107 109 New Zealand 127 139 112 Norway 125 136 94 Poland 163 139 159 Portugal 53 112 105 Spain 122 123 108 Sweden 180 142 116

2.145 Dwelling permits and production. Refer to the previous exercise. MEIS

(a) Make a scatterplot with production as the response variable and permits issued for new dwellings as the explanatory variable. Describe the relationship. Are there any outliers or influential observations?

(b) Find the least-squares regression line and add it to your plot.

(c) Interpret the slope of the line in the context of this exercise.

(d) Interpret the intercept of the line in the context of this exercise. Explain whether or not this interpretation is useful in explaining the relationship between these two variables.

(e) What is the predicted value of production for a country that has an index of 224 for dwelling permits?

(f) Canada has an index of 224 for dwelling permits. Find the residual for this country.

(g) What percent of the variation in production is explained by dwelling permits? How does this value compare with the value that you found in the previous exercise for the percent of variation in sales that is explained by building permits?

2.146 Sales and production. Refer to the previous two exercises. MEIS

(a) Make a scatterplot with sales as the response variable and production as the explanatory variable. Describe the relationship. Are there any outliers or influential observations?

(b) Find the least-squares regression line and add it to your plot.

(c) Interpret the slope of the line in the context of this exercise.

(d) Interpret the intercept of the line in the context of this exercise. Explain whether or not this interpretation is useful in explaining the relationship between these two variables.

(e) What is the predicted value of sales for a country that has an index of 109 for production?

(f) The Netherlands has an index of 109 for production. Find the residual for this country.

(g) What percent of the variation in sales is explained by production? How does this value compare with the percents of variation that you calculated in the two previous exercises?

FIGURE 2.32 Percent of the population over 65 years and percent of the population under 15 years in the 13 Canadian provinces and territories, Exercise 2.147.

2.147 Population in Canadian provinces and territories. Statistics Canada provides a great deal of demographic data organized in different ways.32 Figure 2.32 gives the percent of the population aged 65 years and older and the percent aged 15 years and younger for each of the 13 Canadian provinces and territories. Figure 2.33 is a scatterplot of the percent of the population over 65 versus the percent under 15. CANADAP

TABLE 2.4

(a) Write a short paragraph explaining what the plot tells you about these two demographic groups in the 13 Canadian provinces and territories.

(b) Find the correlation between the percent of the population over 65 and the percent under 15. Does the correlation give a good numerical summary of the strength of this relationship? Explain your answer.

FIGURE 2.33 Scatterplot of percent of the population over 65 years versus percent of the population under 15 years for the 13 Canadian provinces and territories, Exercise 2.147.

2.148 Nunavut. Refer to the previous exercise and Figures 2.32 and 2.33. CANADAP

(a) Do you think that Nunavut is an outlier?

(b) Make a residual plot for these data. Comment on the size of the residual for Nunavut. Use this information to expand on your answer to part (a).

(c) Find the value of the correlation without Nunavut. How does this compare with the value you computed in part (b) of the previous exercise?

(d) Write a short paragraph about Nunavut based on what you have found in this exercise and the previous one.

2.149 Compare the provinces with the territories. Refer to the previous exercise. The three Canadian territories are the Northwest Territories, Nunavut, and the Yukon Territories. All the other entries in Figure 2.32 are provinces.

CANADAP

(a) Generate a scatterplot of the Canadian demographic data similar to Figure 2.33 but with the points labeled “P” for provinces and “T” for territories.

(b) Use your new scatterplot to write a new summary of the demographics for the 13 Canadian provinces and territories.

2.150 Records for men and women in the 10K. Table 2.4 shows the progress of world record times (in seconds) for the 10,000-meter run for both men and women.33 TENK

World Record Times for the 10,000-Meter Run

Men Women

Record year Time (seconds) Record year Time (seconds) Record year Time (seconds)

1912 1880.8 1963 1695.6 1967 2286.4 1921 1840.2 1965 1659.3 1970 2130.5 1924 1835.4 1972 1658.4 1975 2100.4 1924 1823.2 1973 1650.8 1975 2041.4 1924 1806.2 1977 1650.5 1977 1995.1 1937 1805.6 1978 1642.4 1979 1972.5 1938 1802.0 1984 1633.8 1981 1950.8 1939 1792.6 1989 1628.2 1981 1937.2 1944 1775.4 1993 1627.9 1982 1895.3 1949 1768.2 1993 1618.4 1983 1895.0 1949 1767.2 1994 1612.2 1983 1887.6 1949 1761.2 1995 1603.5 1984 1873.8 1950 1742.6 1996 1598.1 1985 1859.4 1953 1741.6 1997 1591.3 1986 1813.7 1954 1734.2 1997 1587.8 1993 1771.8 1956 1722.8 1998 1582.7 1956 1710.4 2004 1580.3 1960 1698.8 2005 1577.5 1962 1698.2

(a) Make a scatterplot of world record time against year, using separate symbols for men and women. Describe the pattern for each sex. Then compare the progress of men and women.

(b) Women began running this long distance later than men, so we might expect their improvement to be more rapid. Moreover, it is often said that men have little advantage over women in distance running as opposed to sprints, where muscular strength plays a greater role. Do the data appear to support these claims?

2.151 Remote deposit capture. The Federal Reserve has called remote deposit capture (RDC) “the most important development the [U.S.] banking industry has seen in years.” This service allows users to scan checks and to transmit the scanned images to a bank for posting.34 In its annual survey of community banks, the American Bankers Association asked banks whether or not they offered this service.35 Here are the results classified by the asset size (in millions of dollars) of the bank: RDCSIZE

Offer RDC

Asset size Yes No

Under $100 63 309 $101 to $200 59 132 $201 or more 112 85

Summarize the results of this survey question numerically and graphically. Write a short paragraph explaining the relationship between the size of a bank, measured by assets, and whether or not RDC is offered.

2.152 How does RDC vary across the country? The survey described in the previous exercise also classified community banks by region. Here is the 6 × 2 table of counts:36 RDCREG

Offer RDC

Region Yes No

Northeast 28 38 Southeast 57 61 Central 53 84 Midwest 63 181 Southwest 27 51 West 61 76

Summarize the results of this survey question numerically and graphically. Write a short paragraph explaining the relationship between the location of a bank and whether or not RDC is offered.

2.153 Fields of study for college students. The following table gives the number of students (in thousands) graduating from college with degrees in several fields of study for seven countries:37 FOS

Field of study Canada France Germany Italy Japan U.K. U.S.

Social sciences, business, law 64 153 66 125 250 152 878 Science, mathematics, engineering 35 111 66 80 136 128 355 Arts and humanities 27 74 33 42 123 105 397 Education 20 45 18 16 39 14 167 Other 30 289 35 58 97 76 272

(a) Calculate the marginal totals and add them to the table.

(b) Find the marginal distribution of country and give a graphical display of the distribution.

(c) Do the same for the marginal distribution of field of study.

2.154 Fields of study by country for college students. In the previous exercise you examined data on fields of study for graduating college students from seven countries. FOS

(a) Find the seven conditional distributions giving the distribution of graduates in the different fields of study for each country.

(b) Display the conditional distributions graphically.

(c) Write a paragraph summarizing the relationship between field of study and country.

2.155 Graduation rates. One of the factors used to evaluate undergraduate programs is the proportion of incoming students who graduate. This quantity, called the graduation rate, can be predicted by other variables such as the SAT or ACT scores and the high school records of the incoming students. One of the components that U.S. News & World Report uses when evaluating colleges is the difference between the actual graduation rate and the rate predicted by a regression equation.38 In this chapter, we call this quantity the residual. Explain why the residual is a better measure to evaluate college graduation rates than the raw graduation rate.

2.156 Salaries and raises. For this exercise, we consider a hypothetical employee who starts working in Year 1 with a salary of $50,000. Each year her salary increases by approximately 5%. By Year 20, she is earning $126,000. The following table gives her salary for each year (in thousands of dollars): RAISES

Year Salary

1 50 2 53 3 56 4 58 5 61 6 63 7 67 8 70 9 74 10 78 11 81 12 85 13 90 14 93 15 99 16 104

17 109 18 114 19 120 20 126

FIGURE 2.34 Plot of salary versus year for an individual who receives approximately a 5% raise each year for 20 years, with the least-squares regression line, Exercise 2.156.

(a) Figure 2.34 is a scatterplot of salary versus year, with the least-squares regression line. Describe the relationship between salary and year for this person.

(b) The value of r2 for these data is 0.9832. What percent of the variation in salary is explained by year? Would you say that this is an indication of a strong linear relationship? Explain your answer.

2.157 Look at the residuals. Refer to the previous exercise. Figure 2.35 is a plot of the residuals versus year. RAISES

(a) Interpret the residual plot.

(b) Explain how this plot highlights the deviations from the least-squares regression line that you can see in Figure 2.34.

2.158 Try logs. Refer to the previous two exercises. Figure 2.36 is a scatterplot with the least-squares regression line for log salary versus year. For this model, r2 = 0.9995. RAISES

(a) Compare this plot with Figure 2.34. Write a short summary of the similarities and the differences.

(b) Figure 2.37 is a plot of the residuals for the model using year to predict log salary. Compare this plot with Figure 2.35 and summarize your findings.

FIGURE 2.35 Plot of residuals versus year for an individual who receives approximately a 5% raise each year for 20 years, Exercise 2.157.

FIGURE 2.36 Plot of log salary versus year for an individual who receives approximately a 5% raise each year for 20 years, with the least-squares regression line, Exercise 2.158.

2.159 Make some predictions. The individual whose salary we have been studying wants to do some financial planning. Specifically, she would like to predict her salary five years into the future, that is, for Year 25. She is willing to assume that her employment situation will be stable for the next five years and that it will be similar to the last 20 years.

RAISES

(a) Predict her salary for Year 25 using the least-squares regression equation constructed to predict salary from year.

(b) Predict her salary for Year 25 using the least-squares regression equation constructed to predict log salary from year. Note that you will need to take the predicted log salary and convert this value back to the predicted salary. Many calculators have a function that will perform this operation.

FIGURE 2.37 Plot of residuals, based on log salary, versus year for an individual who receives approximately a 5% raise each year for 20 years, Exercise 2.158.

(c) Which prediction do you prefer? Explain your answer.

(d) Someone looking at the numerical summaries and not the plots for these analyses says that because both models have very high values of r2, they should perform equally well in doing this prediction. Write a response to this comment.

(e) Discuss the value of graphical summaries and the problems of extrapolation using what you have learned in studying these salary data.

2.160 Faculty salaries. Here are the salaries for a sample of professors in a mathematics department at a large midwestern university for the academic years 2014–2015 and 2015–2016. FACULTY

2014–2015 salary ($)

2015–2016 salary ($)

145,700 147,700 112,700 114,660 109,200 111,400 98,800 101,900 112,000 113,000 111,790 113,800 103,500 105,700 149,000 150,900 136,650 138,650 132,160 134,150 74,290 76,590 74,500 77,000 83,000 85,400 141,850 143,830 122,500 124,510 115,100 117,100

(a) Construct a scatterplot with the 2015–2016 salaries on the vertical axis and the 2014–2015 salaries on the horizontal axis.

(b) Comment on the form, direction, and strength of the relationship in your scatterplot.

(c) What proportion of the variation in 2015–2016 salaries is explained by 2014–2015 salaries?

2.161 Find the line and examine the residuals. Refer to the previous exercise. FACULTY

(a) Find the least-squares regression line for predicting 2015–2016 salaries from 2014–2015 salaries.

(b) Analyze the residuals, paying attention to any outliers or influential observations. Write a summary of your findings.

2.162 Bigger raises for those earning less. Refer to the previous two exercises. The 2014–2015 salaries do an excellent job of predicting the 2015–2016 salaries. Is there anything more that we can learn from these data? In this department, there is a tradition of giving higher-than-average percent raises to those whose salaries are lower. Let’s see if we can find evidence to support this idea in the data. FACULTY

(a) Compute the percent raise for each faculty member. Take the difference between the 2015–2016 salary and the 2014–2015 salary, divide by the 2014–2015 salary, and then multiply by 100. Make a scatterplot with raise as the response variable and the 2014–2015 salary as the explanatory variable. Describe the relationship that you see in your plot.

(b) Find the least-squares regression line and add it to your plot.

(c) Analyze the residuals. Are there any outliers or influential cases? Make a graphical display and include this in a short summary of your conclusions.

(d) Is there evidence in the data to support the idea that greater percent raises are given to those with lower salaries? Include numerical and graphical summaries to support your conclusion.

2.163 Firefighters and fire damage. Someone says, “There is a strong positive correlation between the number of firefighters at a fire and the amount of damage the fire does. So sending lots of firefighters just causes more damage.” Explain why this reasoning is wrong.

2.164 Predicting text pages. The editor of a statistics text would like to plan for the next edition. A key variable is the number of pages that will be in the final version. Text files are prepared by the authors using a word processor called LaTeX, and separate files contain figures and tables. For the previous edition of the text, the number of pages in the LaTeX files can easily be determined, as well as the number of pages in the final version of the text. Here are the data:

TEXTP

Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13

LaTeX pages 77 73 59 80 45 66 81 45 47 43 31 46 26 Text pages 99 89 61 82 47 68 87 45 53 50 36 52 19

(a) Plot the data and describe the overall pattern.

(b) Find the equation of the least-squares regression line and add the line to your plot.

(c) Find the predicted number of pages for the next edition if the number of LaTeX pages is 62.

(d) Write a short report for the editor explaining to her how you constructed the regression equation and how she could use it to estimate the number of pages in the next edition of the text.

2.165 Plywood strength. How strong is a building material such as plywood? To be specific, support a 24-inch by 2-inch strip of plywood at both ends and apply force in the middle until the strip breaks. The modulus of

rupture (MOR) is the force needed to break the strip. We would like to be able to predict MOR without actually breaking the wood. The modulus of elasticity (MOE) is found by bending the wood without breaking it. Both MOE and MOR are measured in pounds per square inch. Here are data for 32 specimens of the same type of plywood:39 MOEMOR

MOE MOR

2,005,400 11,591 1,166,360 8,542 1,842,180 12,750 2,088,370 14,512

1,615,070 9,244 1,938,440 11,904 2,047,700 11,208 2,037,520 12,004 2,181,910 12,702 1,559,700 11,209 2,372,660 12,799 1,580,930 12,062 1,879,900 11,357 1,594,750 8,889 1,558,770 11,565 2,212,310 15,317 1,774,850 10,541 1,457,020 10,314 1,959,590 11,983 1,720,930 10,232 1,355,720 9,286 1,411,210 10,654 1,842,630 10,223 1,984,690 13,499 1,747,010 11,794 1,791,150 11,413 2,535,170 13,920 1,355,960 8,395 1,646,010 8,814 1,472,310 6,326 1,488,440 9,214 2,349,090 13,645

Can we use MOE to predict MOR accurately? Use the data to write a discussion of this question.

2.166 Distribution of the residuals. Some statistical methods require that the residuals from a regression line have a Normal distribution. The residuals for the nonexercise activity example are given in Exercise 2.93 (page 125). Is their distribution close to Normal? Make a Normal quantile plot to find out. FIDGET

2.167 An example of Simpson’s paradox. Mountain View University has professional schools in business and law. Here is a three-way table of applicants to these professional schools, categorized by sex, school, and admission decision:40 ADMITS

Business Law

Admit Admit

Sex Yes No Sex Yes No

Male 400 200 Male 90 110 Female 200 100 Female 200 200

(a) Make a two-way table of sex by admission decision for the combined professional schools by summing entries in the three-way table.

(b) From your two-way table, compute separately the percents of male and female applicants admitted. Male applicants are admitted to Mountain View’s professional schools at a higher rate than female applicants.

(c) Now compute separately the percents of male and female applicants admitted by the business school and by the law school.

(d) Explain carefully, as if speaking to a skeptical reporter, how it can happen that Mountain View appears to favor males when this is not true within each of the professional schools.

2.168 Simpson’s paradox and regression. Simpson’s paradox occurs when a relationship between variables within groups of observations reverses when all of the data are combined. The phenomenon is usually discussed in terms of categorical variables, but it also occurs in other settings. Here is an example: SIMREG

y x Group

10.1 1 1 8.9 2 1 8.0 3 1 6.9 4 1 6.1 5 1 18.3 6 2 17.1 7 2 16.2 8 2 15.1 9 2 14.3 10 2

(a) Make a scatterplot of the data for Group 1. Find the least-squares regression line and add it to your plot. Describe the relationship between y and x for Group 1.

(b) Do the same for Group 2.

(c) Make a scatterplot using all 10 observations. Find the least-squares line and add it to your plot.

(d) Make a plot with all of the data using different symbols for the two groups. Include the three regression lines on the plot. Write a paragraph about Simpson’s paradox for regression using this graphical display to illustrate your description.

2.169 Class size and class level. A university classifies its classes as either “small” (fewer than 40 students) or “large.” A dean sees that 62% of Department A’s classes are small, while Department B has only 40% small

classes. She wonders if she should cut Department A’s budget and insist on larger classes. Department A responds to the dean by pointing out that classes for third- and fourth-year students tend to be smaller than classes for first- and second- year students. The following three-way table gives the counts of classes by department, size, and student audience. Write a short report for the dean that summarizes these data. Start by computing the percents of small classes in the two departments and include other numerical and graphical comparisons as needed. Here are the numbers of classes to be analyzed: CSIZE

Department A Department B

Year Large Small Total Large Small Total

First 2 0 2 18 2 20 Second 9 1 10 40 10 50 Third 5 15 20 4 16 20 Fourth 4 16 20 2 14 16

2.170 More smokers live at least 20 more years! You can see the headlines: “More smokers than nonsmokers live at least 20 more years after being contacted for study!” A medical study contacted randomly chosen people in a district in England. Here are data on the 1314 women contacted who were either current smokers or who had never smoked. The tables classify these women by their smoking status and age at the time of the survey and whether they were still alive 20 years later.41 SMOKERS

Age 18 to 44 Age 45 to 64 Age 65+

Smoker Not Smoker Not Smoker Not Dead 19 13 78 52 42 165 Alive 269 327 167 147 7 28

(a) From these data, make a two-way table of smoking (yes or no) by dead or alive. What percent of the smokers stayed alive for 20 years? What percent of the nonsmokers survived? It seems surprising that a higher percent of smokers stayed alive.

(b) The age of the women at the time of the study is a lurking variable. Show that within each of the three age groups in the data, a higher percent of nonsmokers remained alive 20 years later. This is another example of Simpson’s paradox.

(c) The study authors give this explanation: “Few of the older women (over 65 at the original survey) were smokers, but many of them had died by the time of follow-up.” Compare the percent of smokers in the three age groups to verify the explanation.

2.171 Recycled product quality. Recycling is supposed to save resources. Some people think recycled products are lower in quality than other products, a fact that makes recycling less practical. People who actually use a recycled product may have different opinions from those who don’t use it. Here are data on attitudes toward coffee filters made of recycled paper among people who do and don’t buy these filters:42 RECYCLE

Think the quality of the recycled product is: Higher The same Lower

Buyers 20 7 9 Nonbuyers 29 25 43

(a) Find the marginal distribution of opinion about quality. Assuming that these people represent all users of coffee filters, what does this distribution tell us?

(b) How do the opinions of buyers and nonbuyers differ? Use conditional distributions as a basis for your answer. Include a mosaic plot if you have access to the needed software. Can you conclude that using recycled filters causes more favorable opinions? If so, giving away samples might increase sales.

2.172 Survival and sex on the Titanic. In Exercise 2.122, you examined the relationship between survival and class on the Titanic. The data file TITANIC contains data on the sex of the Titanic passengers. Examine the relationship between survival and sex and write a short summary of your findings. TITANIC

2.173 Survival, class, and sex on the Titanic. Refer to the previous exercise and Exercise 2.122 (page 146). When we looked at survival and class, we ignored sex. When we looked at survival and sex, we ignored class. Are we missing something interesting about these data when we choose this approach to the analysis? Here is one way to answer this question. TITANIC

(a) Create two separate two-way tables. One for survival and class for the women and another for survival and class for the men.

(b) Perform an analysis of the relationship between survival and class for the women. Summarize your findings.

(c) Perform an analysis of the relationship between survival and class for the men. Summarize your findings.

(d) Compare the analyses that you performed in parts (b) and (c). Write a short report on the relationship between survival and the two explanatory variables, class and sex.

2.174 Blueberries and anthocyanins. Refer to Exercises 1.165 and 1.166 (page 77). Figure 2.38 gives JMP output for examining the relationship between Antho2 and Antho1. Use this output to write a summary of this

relationship using the methods and ideas that you learned in this chapter. BERRIES

2.175 Averaged date for blueberries and anthocyanins. Refer to the previous exercise where you examined the relationship between Antho2 and Antho1. The variables Antho2M and Antho1M were computed by averaging Antho2 and Antho1 for values of Antho1 in the intervals (0, 0.5), [0.5, 1.0), [1.0, 1.5), [1.5, 2.0), [2.0, 2.5), [2.5, 3.0), and [3.0, 3.5). Analyze the relationship between Antho2M and Antho1M, and compare your results that you found in the previous exercise using Antho2 and Antho1. Summarize what the comparison tells you about relationships with averaged data. BERRIEM

2.176 Restricting the range for blueberries and anthocyanins. Refer to Exercise 2.174 where you examined the relationship between Antho2 and Antho1. The data file BERRIER was created from the data file

BERRIES by excluding cases with values of Antho1 that are less than 1.5 and cases with values of Antho1 that are greater than 3. Analyze the relationship between Antho2 and Antho1 for this restricted range data set, and compare your

results that you found in Exercise 2.174 for the complete data set. Summarize what the comparison tells you about relationships with a restricted range. BERRIER

FIGURE 2.38 Selected JMP output for examining the relationship between Antho2 and Antho1, Exercise 2.174.

1. Shana M. Wilson et al., “Prediction of emotional eating during adolescents’ transition to college: Does body mass index moderate the association between stress and emotional eating?” Journal of American College Health, 63, No. 3 (2015), pp. 163–170.

2. See Note 1.

3. See cfs.purdue.edu/fn/campcalcium/ for information about findings from these camps.

4. See consumerreports.org.

5. From consumerreports.org/cro/laundry-detergents.htm.

6. Data for 2014 from usgovernmentspending.com/compare_state_education_spend.

7. These studies were conducted by Connie Weaver, Department of Nutrition Science, Purdue University, over the past 20 years. The data for this example were provided by Linda McCabe. More details concerning this particular study and references to other related studies are given in Lu Wu et al. “Calcium requirements and metabolism in Chinese-American boys and girls,” Journal of Bone Mineral Research, 25, No. 8 (2010), pp. 1842–1849.

8. A sophisticated treatment of improvements and additions to scatterplots is W. S. Cleveland and R. McGill, “The many faces of a scatterplot,” Journal of the American Statistical Association, 79 (1984), pp. 807–822.

9. Data provided by Mary Ann Lila, Director, Plants for Human Health Institute, David H. Murdock Distinguished Professor, North Carolina Research Campus, North Carolina State University.

10. From nrcan.gc.ca/energy/efficiency/11938.

11. Stewart Warden et al., “Throwing induces substantial torsional adaption within the midshaft humerus of male baseball players,” Bone, 45 (2009), pp. 931–941. The data were provided by Stewart Warden, Department of Physical Therapy, School of Health and Rehabilitation Sciences, Indiana University.

12. See beer100.com/beercalories.htm.

13. See spectrumtechniques.com/isotope_generator.htm.

14. These data were collected under the supervision of Zach Grigsby, Science Express Coordinator, College of Science, Purdue University.

15. See worldbank.org.

16. A careful study of this phenomenon is W. S. Cleveland, P. Diaconis, and R. McGill, “Variables on scatterplots look more highly correlated when the scales are increased,” Science, 216 (1982), pp. 1138–1141.

17. Data from a plot in James A. Levine, Norman L. Eberhardt, and Michael D. Jensen, “Role of nonexercise activity thermogenesis in resistance to fat gain in humans,” Science, 283 (1999), pp. 212–214.

18. From the Digest of Education Statistics at the website of the National Center for Education Statistics, nces.ed.gov/programs/digest.

19. Frank J. Anscombe, “Graphs in statistical analysis,” American Statistician, 27 (1973), pp. 17–21.

20. From the website of the National Center for Education Statistics, nces.ed.gov.

21. Debora L. Arsenau, “Comparison of diet management instruction for patients with non-insulin dependent diabetes mellitus: Learning activity package vs. group instruction,” Master’s thesis, Purdue University, 1993.

22. See Note 19.

23. See iom.edu.

24. Based on a study described in Corby C. Martin et al., “Children in school cafeterias select foods containing more saturated fat and energy than the Institute of Medicine recommendations,” Journal of Nutrition, 140 (2010), pp. 1653–1660.

25. You can find a clear and comprehensive discussion of numerical measures of association for categorical data in Chapter 2 of Alan Agresti, Categorical Data Analysis, 2nd ed., Wiley, 2002.

26. Edward Bumgardner, “Loss of teeth as a disqualification for military service,” Transactions of the Kansas Academy of Science, 18 (1903), pp. 217–219.

27. Based on The Ethics of American Youth—2012, available from the Josephson Institute at charactercounts.org/wp- content/uploads/2014/02/ReportCard-2012-DataTables.pdf.

28. From M.-Y. Chen et al., “Adequate sleep among adolescents is positively associated with health status and health-related behaviors,” BMC Public Health, 6, No. 59 (2006); available from bmcpublichealth.biomedcentral.com/articles/10.1186/1471-2458-6-59.

29. M. S. Linet et al., “Residential exposure to magnetic fields and acute lymphoblastic leukemia in children,” New England Journal of Medicine, 337 (1997), pp. 1–7.

30. The Health Consequences of Smoking: 1983, U.S. Public Health Service, 1983.

31. OECD StatExtracts, Organisation for Economic Co-operation and Development, downloaded on June 29, 2008, from stats.oecd.org/wbos.

32. See www12.statcan.gc.ca/census-recensement/index-eng.cfm.

33. From en.wikipedia.org/wiki/10000_metres.

34. For an overview of remote deposit capture, see remotedepositcapture.com/overview/rdc.overview.aspx.

35. From the “Community Bank Competitiveness Survey,” 2008, ABA Banking Journal. The survey is available at nxtbook.com/nxtbooks/sb/ababj-compsurv08/index.php.

36. The counts reported were calculated using counts of the numbers of banks in the different regions and the percents given in the ABA report.

37. Education Indicators: An International Perspective, Institute of Education Studies, National Center for Education Statistics; see nces.ed.gov/surveys/international.

38. Information about this procedure was provided by Samuel Flanigan of U.S. News & World Report. See colleges.usnews.rankingsandreviews.com/best-colleges for a description of the variables used to construct the ranks and for the most recent ranks.

39. We thank Zhiyong Cai of Texas A&M University for providing the data. The data are from work performed in connection with his PhD dissertation in the Department of Forestry and Natural Resources, Purdue University.

40. Although these data are fictitious, similar though less simple situations occur. See P. J. Bickel and J. W. O’Connell, “Is there a sex bias in graduate admissions?” Science, 187 (1975), pp. 398–404.

41. Condensed from D. R. Appleton, J. M. French, and M. P. J. Vanderpump, “Ignoring a covariate: An example of Simpson’s paradox,” The American Statistician, 50 (1996), pp. 340–341.

42. Lien-Ti Bei, “Consumers’ purchase behavior toward recycled products: An acquisition-transaction utility theory perspective,” MS thesis, Purdue University, 1993.

3.1 3.2 3.3 3.4

CHAPTER 3 Producing Data

Sources of Data Design of Experiments Sampling Design Ethics

Introduction In Chapters 1 and 2, we learned some basic tools of data analysis. We used graphs and numbers to describe data. When we do exploratory data analysis, we rely heavily on plotting the data. We look for patterns that suggest interesting conclusions or questions for further study. However, exploratory analysis alone can rarely provide convincing evidence for its conclusions because striking patterns that we find in data can arise from many sources.

The validity of the conclusions that we draw from an analysis of data depends not only on the use of the best methods to perform the analysis, but also on the quality of the data. Therefore, Section 3.1 begins this chapter with a short overview on sources of data.

exploratory data analysis, p. 8

The two main sources for quality data are designed experiments and sample surveys. We study these two sources in Sections 3.2 and 3.3, respectively.

ethics

Should an experiment or sample survey that could possibly provide interesting and important information always be performed? How can we safeguard the privacy of subjects in a sample survey? What constitutes the mistreatment of people or animals who are studied in an experiment? These are questions of ethics. In Section 3.4, we address ethical issues related to the design of studies and the analysis of data.

3.1 Sources of Data

When you complete this section, you will be able to:

Identify anecdotal data and, using specific examples, explain why they have limited value. Identify available data and explain how they can be used in specific examples. Identify data collected from sample surveys and explain how they can be used in specific examples. Identify data collected from experiments and explain how they can be used in specific examples. Distinguish data that are from experiments, from observational studies that are sample surveys, and from observational studies that are not sample surveys. Identify the treatment in an experiment.

There are many sources of data. Some data are very easy to collect, but they may not be very useful. Other data require careful planning and need professional staff to gather. These can be much more useful. Whatever the source, a good statistical analysis will start with a careful study of the source of the data. Here is one type of source.

3.1

3.2

3.3

3.4

Anecdotal data It is tempting to simply draw conclusions from our own experience, making no use of more broadly representative data. A magazine article about Pilates says that men need this form of exercise even more than women do. The article describes the benefits that two men received from taking Pilates classes. A newspaper ad states that a particular brand of window is “considered to be the best” and says that “now is the best time to replace your windows and doors.” These types of stories, or anecdotes, sometimes provide data. However, this type of data does not give us a sound basis for drawing conclusions.

ANECDOTAL DATA

Anecdotal data represent individual cases, which often come to our attention because they are striking in some way. These cases are not necessarily representative of any larger group of cases.

USE YOUR KNOWLEDGE

The best instructor? A friend tells you that the instructor in her statistics class is the best teacher in the college. Can you conclude that this teacher is better than all of the other instructors in the college? Explain your answer.

Describe an anecdote. Find an example from some recent experience where anecdotal evidence was used to draw a conclusion that is not justified. Describe the example and explain why the anecdote should not be used in this way.

Opposition to a new requirement. Your student newspaper ran a story describing interviews with three students who were strongly opposed to a proposed new requirement that all students take a course on ethics. Can you conclude that most students are opposed to this requirement? Explain your answer.

Are all vehicles this good? A friend has driven a Toyota Camry for more than 200,000 miles and with only the usual service maintenance expenses. Explain why not all Camry owners can expect this kind of performance.

Not all anecdotal data are bad. The experiences of an individual or a small group of individuals might suggest an interesting study that could be performed using more carefully collected data.

Available data Occasionally, data are collected for a particular purpose but can also serve as the basis for drawing sound conclusions about other research questions. We use the term available data for this type of data.

AVAILABLE DATA

Available data are data that were produced for some other purpose but that may help answer a question of interest.

The library and the Internet can be good sources of available data. Because producing new data is expensive, we all use available data whenever possible. Here are two examples.

EXAMPLE 3.1

How Americans use their time. If you visit the U.S. Bureau of Labor Statistics website, bls.gov, you will find many interesting sets of data and statistical summaries. The American Time Use Survey1 recently reported that men spend an average of 5.71 hours per day on leisure and sports activities, while women spend an average of 4.93 hours on these activities.

EXAMPLE 3.2

Math skills. At the website of the National Center for Education Statistics, nces.ed.gov, you will find full details about the math skills of schoolchildren as determined by the latest National Assessment of Educational Progress (Figure 3.1). Mathematics scores have slowly but steadily increased since 1990. Across all racial/ethnic groups, both boys and girls in most states are getting better in math.

3.5

FIGURE 3.1 Websites of government statistical offices are prime sources of data. Here is a page from the website of the National Center for Education Statistics. (Source: U.S. Department of Education Institute of Education Sciences National Center for Education Statistics)

Many nations have a single national statistical office, such as Statistics Canada (statcan.gc.ca) and Mexico’s INEGI (www.inegi.org.mx). More than 70 different U.S. agencies collect data. You can reach most of them through the U.S. government’s FedStats site (fedstats.sites.usa.gov).

USE YOUR KNOWLEDGE

What more do you need? A website claims that Millennial generation consumers are very loyal to the brands that they prefer. What additional information do you need to evaluate this claim?

A survey of college athletes is designed to estimate the percent who gamble. Do restaurant patrons give higher tips when their server repeats their order carefully? The validity of our conclusions from the analysis of data collected to address these issues rests on a foundation of carefully collected data.

In this chapter, we will develop the skills needed to produce trustworthy data and to judge the quality of data produced by others. The techniques for producing data that we will study require no formulas, but they are among the most important ideas in statistics. Statistical designs for producing data rely on either sampling or experiments.

3.6

Sample surveys and experiments sample surveys

How have the attitudes of Americans, on issues ranging from abortion to work, changed over time? Sample surveys are the usual tool for answering questions like these.

EXAMPLE 3.3

The General Social Survey. One of the most important sample surveys is the General Social Survey (GSS) conducted by the National Opinion Research Center (NORC), an organization affiliated with the University of Chicago.2 The GSS interviews about 3000 adult residents of the United States every other year.

sample

population

The GSS selects a sample of adults to represent the larger population of all English-speaking adults living in the United States. The idea of sampling is to study a part in order to gain information about the whole. Data are often produced by sampling a population of people or things. Opinion polls, for example, may report the views of the entire country based on interviews with a sample of about 1000 people. Government reports on employment and unemployment are produced from a monthly sample of about 60,000 households. The quality of manufactured items is monitored by inspecting small samples each hour or each shift.

USE YOUR KNOWLEDGE

Check out the General Social Survey. Visit the General Social Survey website at gss.norc.org. Write a short summary of one of their reports, paying particular attention to the methods used to collect the data.

census

In all our examples, the expense of examining every item in the population makes sampling a practical necessity. Timeliness is another reason for preferring a sample to a census, which is an attempt to contact every individual in the population. We want information on current unemployment and public opinion next week, not next year. Moreover, a carefully conducted sample is often more accurate than a census. Accountants, for example, sample a firm’s inventory to verify the accuracy of the records. Attempting to count every last item in the warehouse would be not only expensive, but also inaccurate. Bored people do not count carefully.

If conclusions based on a sample are to be valid for the population, a sound design for selecting the sample is required. Sampling designs are the topic of Section 3.3.

A sample survey collects information about a population by selecting and measuring a sample from the population. The goal is a picture of the population, disturbed as little as possible by the act of gathering information. Sample surveys are one kind of observational study.

OBSERVATION VERSUS EXPERIMENT

In an observational study, we observe individuals and measure variables of interest but do not attempt to influence the responses.

In an experiment, we deliberately impose some condition on individuals and we observe their responses.

EXAMPLE 3.4

3.7

3.8

3.9

Baseball players have strong bones in their throwing arms. A study of young baseball players measured the strength of the bones in their throwing arms. A control group of subjects who were matched with the baseball players based on age were also measured. This is an example of an observational study that is not a sample survey. The study reported that bone strength was 30% higher in the baseball players.3

What can we conclude from this study? If you start to play baseball, will you have stronger bones in your throwing arm?

EXAMPLE 3.5

Is there a cause-and-effect relationship? Example 3.4 describes an observational study. People choose to participate in baseball or not. Is it possible that those who choose to play baseball have stronger arms than those who do not? The study does not address this question.

We can imagine an experiment that would remove these difficulties. From a large group of subjects, require some to play baseball and forbid the rest from playing. This is an experiment because the condition (playing baseball or not) is imposed on the subjects. Of course, this particular experiment is neither practical nor ethical.

EXAMPLE 3.6

Baseball and bones. Example 3.4 compared the arm bone strengths of baseball players with those of age- matched controls. Although the study tells us something about baseball players, the results are particularly interesting because they suggest that certain kinds of exercise can help us to build strong bones.

USE YOUR KNOWLEDGE

Available data. Can available data be from an observational study? Can available data be from an experiment? Explain your answers.

Picky eaters. A study of 2049 children in grades 4 to 6 in 33 schools recorded their behaviors in the lunchroom. One of the conclusions of the study was that girls discarded more food than boys.4 Is this an observational study or an experiment? Is it a sample survey? If it is an experiment, what is the condition? Explain your answers.

Automatic soap dispensers. A study compared several brands of automatic soap dispensers. For one test, the dispensers were run until their AA batteries failed. The times to failure were compared for the different brands.5 Is this an observational study or an experiment? Is it a sample survey? If it is an experiment, what is the condition? Explain your answers.

intervention

An observational study, even one based on a carefully chosen sample, is a poor way to determine what will happen if we change something. The best way to see the effects of a change is to do an intervention—where we actually impose the change. When our goal is to understand cause and effect, experiments are the only source of fully convincing data.

Confounding occurs when an explanatory variable is related to one or more other variables that have an influence on the response variable. When this happens, we sometimes attribute a relationship to an explanatory when the effect is fully or partly due to the confounding variables.

3.10

3.11

confounding, p. 150 explanatory variable, p. 82

In Example 3.4, the effect of baseball playing on arm bone strength is confounded with (mixed up with) other characteristics of the subjects in the study. Observational studies that examine the effect of a single variable on an outcome can be misleading when the effects of the explanatory variable are confounded with those of other variables.

Because experiments allow us to isolate the effects of specific variables, we generally prefer them. Here is an example.

EXAMPLE 3.7

Which web page design sells more? A company that sells products on the Internet wants to decide which of two possible web page designs to use. During a two-week period, they will use both designs and collect data on sales. They randomly select one of the designs to be used on the first day and then alternate the two designs on each of the following days. At the end of this period, they compare the sales for the two designs.

Experiments usually require some sort of randomization, as in this example. We begin the discussion of statistical designs for data collection in Section 3.2 with the principles underlying the design of experiments.

USE YOUR KNOWLEDGE

Software for teaching creative writing. An educational software company wants to compare the effectiveness of its computer animation for teaching creative writing with that of a textbook presentation. The company tests the creative-writing skills of a number of second-year college students and then randomly divides them into two groups. One group uses the animation, and the other studies the text. The company retests all the students and compares the increase in creative-writing skills in the two groups. Is this an experiment? Why or why not? What are the explanatory and response variables?

response variable, p. 82

Apples or apple juice? Food rheologists study different forms of foods and how the form of a food affects how full we feel when we eat it. One study prepared samples of apple juice and samples of apples with the same number of calories. Half of the subjects were fed apples on one day followed by apple juice on a later day; the other half received the apple juice followed by the apples. After eating, the subjects were asked about how full they felt. Is this an experiment? Why or why not? What are the explanatory and response variables?

SECTION 3.1 SUMMARY Anecdotal data come from stories or reports about cases that do not necessarily represent a larger group of cases. Available data are data that were produced for some other purpose but that may help answer a question of interest. A sample survey collects data from a sample of cases that represent some larger population of cases. A census collects data from all cases in the population of interest. In an experiment, a condition or intervention is imposed and the responses are recorded. Confounding occurs when the effects of two or more variables are related in such a way that we need to take care in assigning the effect on the response variable to one or to the other.

SECTION 3.1 EXERCISES For Exercises 3.1 to 3.4, see pages 164–165; for Exercise 3.5, see page 166; for Exercise 3.6, see page 167; for Exercises 3.7 to 3.9, see pages 168–169; and for Exercises 3.10 and 3.11, see pages 169–170.

In several of the following exercises, you are asked to identify the type of data that is described. Possible answers include anecdotal data, available data, observational data that are from sample surveys, observational data that are not from sample surveys, and experimental data. It is possible for some data to be classified in more than one category.

3.12 Not enough tuna. You like to eat tuna fish sandwiches. Recently you have noticed that there does not seem to be as much tuna as you expect when you open the can. Identify the type of data that this represents, and describe how it can or cannot be used to reach a conclusion about the amount of tuna in cans of tuna fish. Is this an experiment? If yes, what is the treatment?

3.13 More about tuna. According to a story in Consumer Reports, three major producers of canned tuna agreed to pay $3,300,000 to settle claims in California that the amount of tuna in their cans was less than the amount printed on the label of the cans.6 What kind of data do you think was used in this situation to convince the producers to pay this amount of money to settle the claims? Explain your answer fully.

3.14 What’s wrong? Explain what is wrong in each of the following statements.

(a) Available data is always anecdotal.

(b) A census collects information on a subset of the population of interest.

(c) A sample survey usually involves a treatment.

3.15 Satisfaction with allocation of concert tickets. Your college sponsored a concert that sold out.

(a) After the concert, an article in the student newspaper reported interviews with three students who were unable to get tickets and were very upset. What kind of data does this represent? Explain your answer.

(b) A week later the student organization that sponsored the concert set up a website where students could rank their satisfaction with the way that the tickets were allocated using a 5-point scale with values “very satisfied,” “satisfied,” “neither satisfied nor unsatisfied,” “dissatisfied,” and “very dissatisfied.” The website was open to any students who chose to provide their opinion. How would you classify these data? Give reasons for your answer.

(c) Suppose that the website in part (b) was changed so that only a sample of students from the college were invited by a text message to respond, and those who did not respond within three days were sent an additional text message reminding them to respond. How would your answer to part (b) change, if at all?

(d) Is the description in part (c) an experiment? If yes, what is the treatment?

(e) Write a short summary contrasting different types of data using your answers to parts (a), (b), (c), and (d) of this exercise.

3.16 Does echinacea reduce the severity of the common cold? In a study designed to evaluate the benefits of taking echinacea when you have a cold, 719 patients were randomly divided into four groups. The groups were (1) no pills, (2) pills that had no echinacea, (3) pills that had echinacea but the subjects did not know whether or not the pills contained echinacea, and (4) pills that had echinacea and the bottle containing the pills stated that the contents included echinacea. The outcome was a measure of the severity of the cold.7

(a) Identify the type of data collected in this study. Give reasons for your answer.

(b) Is this an experiment? If yes, what is the treatment?

3.2 Design of Experiments

When you complete this section, you will be able to:

Identify experimental units, subjects, treatments, and outcomes for an experiment. Identify a comparative experiment. Describe a placebo effect in an experiment. Identify bias in an experiment. Explain the need for a control group in an experiment. Explain the need for randomization in an experiment. When evaluating an experiment, apply the basic principles of experimental design: compare, randomize, and repeat. Use a table of random digits to randomly assign experimental units to treatments in an experiment. Use software to randomly assign experimental units to treatments in an experiment. Identify a matched pairs design. Identify a block design.

An experiment is a study in which we actually do something to people, animals, or objects in order to observe the response. Here is the basic vocabulary of experiments.

EXPERIMENTAL UNITS, SUBJECTS, TREATMENTS, AND OUTCOMES

The individuals on which the experiment is done are the experimental units. When the units are human beings, they are called subjects. Experimental conditions applied to the units are called treatments. The outcomes are the measured variables that are used to compare the treatments.

factors

level of a factor

Because the purpose of an experiment is to reveal the response of one variable to changes in one or more other variables, the distinction between explanatory and response variables is important. The explanatory variables in an experiment are often called factors. Many experiments study the joint effects of several factors. In such an experiment, each treatment is formed by combining a specific value (often called a level) of each of the factors.

EXAMPLE 3.8

Are smaller class sizes better? Do smaller classes in elementary school really benefit students in areas such as scores on standard tests, staying in school, and going on to college? We might do an observational study that compares students who happened to be in smaller classes with those who happened to be in larger classes in their early school years. Small classes are expensive, so they are more common in schools that serve richer communities. Students in small classes tend to also have other advantages: their schools have more resources, their parents are better educated, and so on. Confounding makes it impossible to isolate the effects of small classes.

The Tennessee STAR program was an experiment on the effects of class size. It has been called “one of the most important educational investigations ever carried out.” The subjects were 6385 students who were beginning kindergarten. Each student was assigned to one of three treatments: regular class (22 to 25 students) with one teacher, regular class (22 to 25 students) with a teacher and a full-time teacher’s aide, and small class (13 to 17 students). These treatments are levels of a single factor, the type of class. The students stayed in the same type of class for four years, then all returned to regular classes. In later years, students from the small classes had higher scores on the outcomes,

standard tests. The benefits of small classes were greatest for minority students.8

lurking variables, p. 130

Example 3.8 illustrates the big advantage of experiments over observational studies. In principle, experiments can give good evidence for causation. In an experiment, we study the specific factors we are interested in while controlling the effects of lurking variables. All the students in the Tennessee STAR program followed the usual curriculum at their schools. Because students were assigned to different class types within their schools, school resources and family backgrounds were not confounded with class type. The only systematic difference was the type of class. When students from the small classes did better than those in the other two types, we can be confident that class size made the difference.

EXAMPLE 3.9

Repeated exposure to advertising. What are the effects of repeated exposure to an advertising message? The answer may depend both on the length of the ad and on how often it is repeated. An experiment investigated this question using undergraduate students as subjects. All subjects viewed a 40-minute television program that included ads for a digital camera. Some subjects saw a 30-second commercial; others, a 90-second version. The same commercial was shown either one, three, or five times during the program.

This experiment has two factors: length of the commercial, with two levels, and repetitions, with three levels. The six combinations of one level of each factor form six treatments. Figure 3.2 shows the layout of the treatments. After viewing the TV program, all the subjects answered questions about their recall of the ad, their attitude toward the camera, and their intention to purchase it. These are the outcomes.

FIGURE 3.2 The treatments in the study of advertising, Example 3.9. Combining the levels of the two factors forms six treatments.

Example 3.9 shows how experiments allow us to study the combined effects of more than one factor. The interaction of several factors can produce effects that cannot be predicted from looking at the effects of each factor alone. Perhaps longer commercials increase interest in a product, and more commercials also increase interest, but if we both make a commercial longer and show it more often, viewers get annoyed and their interest in the product drops.

3.17

3.18

The two-factor experiment in Example 3.9 will help us find out.

USE YOUR KNOWLEDGE

Calcium and bones. Calcium is important for the growth of bone for children. In a study designed to understand how calcium is processed by the body, 40 young girls attended a summer camp where they were fed a controlled diet. The camp ran for two 3-week periods. For one period, the diet included a low amount of calcium. For the other period, there was a high amount of calcium in the diet. The researchers recorded the amount of calcium retained in the body for each girl. Explain why this study is an experiment and identify the experimental units, the treatments, and the response variable. Describe the factor and its levels.

Does echinacea reduce the severity of the common cold? In a study designed to evaluate the benefits of taking echinacea when you have a cold, 719 patients were randomly divided into four groups. The groups were (1) no pills, (2) pills that had no echinacea, (3) pills that had echinacea but the subjects did not know whether or not the pills contained echinacea, and (4) pills that had echinacea and the bottle containing the pills stated that the contents included echinacea. The outcome was a measure of the severity of the cold.9 Identify the experimental units, the treatments, and the outcome. Describe the factor and its levels. The study subjects were aged 12 to 80 years. To what extent do you think the results of this experiment can be generalized to young children?

Comparative experiments Laboratory experiments in science and engineering often have a simple design with only a single treatment, which is applied to all experimental units. The design of such an experiment can be outlined as

Treatment → Observe response For example, we may subject a beam to a load (treatment) and measure its deflection (observation). We rely on the controlled environment of the laboratory to protect us from lurking variables. When experiments are conducted outside the laboratory or with living subjects, such simple designs often yield invalid data. That is, we cannot tell whether the response was due to the treatment or to lurking variables.

EXAMPLE 3.10

Will writing about it reduce test anxiety? A study designed to reduce test anxiety had students write an essay about their feelings concerning an upcoming exam.10 The scores on this exam, the second of the semester, were compared with those on the first exam in the course. The mean scores on the second exam were higher than the mean scores on the first exam.

Write about feelings → Observe exam scores

The test anxiety experiment of Example 3.10 was poorly designed to evaluate the effect of the writing exercise. Perhaps exam scores would have increased on the second exam because the students became more familiar with the exam style of this particular instructor even without the writing exercise. Another possible explanation is that people typically respond to the personal attention that the students received by the person who explained how to write about their feelings regarding the exam.

placebo effect

In medical settings, this phenomenon is called the placebo effect. In medicine, a placebo is a dummy treatment, such as a sugar pill. People respond favorably to personal attention or to any treatment that they hope will help them. On the other hand, the writing exercise may have been very effective in improving exam scores. For this experiment, we don’t know whether the change was due to writing the essay, to the personal contacts with the study personnel, or to greater familiarity with the way the instructor designed exams.

comparative experiment

The test anxiety experiment gave inconclusive results because the effect of writing the essay was confounded with other factors that could have had an effect on exam scores. The best way to avoid confounding is to do a comparative experiment. Think about a study in which some students performed the writing exercise and others did not. A comparison of the exam scores of these two groups of students would provide an evaluation of the effect of the writing exercise.

control group

treatment group

In medical settings, it is standard practice to randomly assign patients either to a control group or a treatment group. All patients are treated the same in every way except that the treatment group receives the product that is being evaluated.

Uncontrolled experiments (that is, experiments that don’t include a control group) in medicine and the behavioral sciences can be dominated by such influences as the details of the experimental arrangement, the selection of subjects, and the placebo effect. The result is often bias.

BIAS

3.19

3.20

The design of a study is biased if it systematically favors certain outcomes.

An uncontrolled study of a new medical therapy, for example, is biased in favor of finding the treatment effective because of the placebo effect. Uncontrolled studies in medicine give new therapies a much higher success rate than proper comparative experiments do. Well-designed experiments usually compare several treatments.

USE YOUR KNOWLEDGE

Does aspirin cure headaches? A study enrolled 100 college students who had frequent headaches to participate in a study to examine the effects of aspirin on their headaches. The students were instructed to take aspirin when they had a headache and to report whether there was a substantial relief from the headache pain within an hour.

(a) Explain why this study is biased.

(b) How would you change the study to remove the bias? Explain your answer.

Are the teacher evaluations biased? The evaluations of two instructors by their students are compared when it is time to decide raises for the coming year. One teacher always hands out the evaluation forms in class when the grades on the first exam are given to the students. The other instructor always hands out the evaluation forms at the end of a class in which a very interesting film clip is shown. Discuss the possibility of bias in this context.

Randomization experimental design

The design of an experiment first describes the response variable or variables, the factors (explanatory variables), and the treatments, with comparison as the leading principle. Figure 3.2 illustrates this aspect of the design of a study of response to advertising. The second aspect of experimental design is how the experimental units are assigned to the treatments. Comparison of the effects of several treatments is valid only when all treatments are applied to similar groups of experimental units. If one corn variety is planted on more fertile ground or if one cancer drug is given to more seriously ill patients, comparisons among treatments are meaningless. If groups assigned to treatments are quite different in a comparative experiment, we should be concerned that our experiment will be biased. How can we assign experimental units to treatments in a way that is fair to all treatments?

matching

Experimenters often attempt to match groups by elaborate balancing acts. Medical researchers, for example, try to match the patients in a “new drug” experimental group and a “standard drug” control group by age, sex, physical condition, smoker or not, and so on. Matching is helpful but not adequate—there are too many lurking variables that might affect the outcome. The experimenter is unable to measure some of these variables and will not think of others until after the experiment.

Some important variables, such as how advanced a cancer patient’s disease is, are so subjective that they can’t be measured. In other cases, an experimenter might unconsciously bias a study by assigning those patients who seemed the sickest to a promising new treatment in the (unconscious) hope that it would help them.

The statistician’s remedy is to rely on chance to make an assignment that does not depend on any characteristic of the experimental units and that does not rely on the judgment of the experimenter in any way. The use of chance can be combined with matching, but the simplest experimental design creates groups by chance alone. Here is an example.

EXAMPLE 3.11

Which smartphone should be marketed? Two teams have each prepared a prototype for a new smartphone. Before deciding which one will be marketed, the smartphones will be evaluated by college students. Forty students will receive a new phone. They will use it for two weeks and then answer some questions about how well they like the phone. The 40 students will be randomized, with 20 receiving each phone.

This experiment has a single factor (prototype) with two levels. The researchers must divide the 40 student subjects into two groups of 20. To do this in a completely unbiased fashion, put the names of the 40 students in a hat, mix them up, and draw 20. These students will receive Phone 1, and the remaining 20 will receive Phone 2. Figure 3.3 outlines the design of this experiment.

FIGURE 3.3 Outline of a randomized comparative experiment, Example 3.11.

3.21

3.22

randomization

The use of chance to divide experimental units into groups is called randomization. The design in Figure 3.3 combines comparison and randomization to arrive at the simplest randomized comparative design. This “flowchart” outline presents all the essentials: randomization, the sizes of the groups and which treatment they receive, and the response variable. There are, as we will see later, statistical reasons for using treatment groups that are about equal in size.

USE YOUR KNOWLEDGE

Diagram the echinacea experiment. Refer to Exercise 3.16 (page 171). Draw a diagram similar to Figure 3.3 that describes the experiment.

Diagram the aspirin experiment. Draw a diagram similar to Figure 3.3 that describes the experiment you suggested in part (b) of Exercise 3.19 (page 175).

Randomized comparative experiments The logic behind the randomized comparative design in Figure 3.3 is as follows:

Randomization produces two groups of subjects that we expect to be similar in all respects before the treatments are applied. Comparative design helps ensure that influences other than the characteristics of the smartphone operate equally on both groups. Therefore, differences in the satisfaction with the smartphone must be due either to the characteristics of the phone or to the chance assignment of subjects to the two groups.

That “either-or” deserves more comment. We cannot say that all the difference in the satisfaction with the two smartphones is caused by the characteristics of the phones. There would be some difference even if both groups used the same phone. Some students would be more likely to be highly favorable of any new phone. Chance can assign more of these students to one of the phones so that there is a chance difference between the groups. We would not trust an experiment with just one subject in each group, for example. The results would depend too much on which phone got lucky and received the subject who was more likely to be highly satisfied. If we assign many students to each group, however, the effects of chance will average out. There will be little difference in the satisfaction between the two groups unless the phone characteristics causes a difference. “Use enough subjects to reduce chance variation” is the third big idea of statistical design of experiments.

PRINCIPLES OF EXPERIMENTAL DESIGN

The basic principles of statistical design of experiments are

1. Compare two or more treatments. This will control the effects of lurking variables on the response. 2. Randomize—use chance to assign experimental units to treatments. 3. Repeat each treatment on many units to reduce chance variation in the results.

How to randomize The idea of randomization is to assign subjects to treatments by drawing names from a hat. In practice, experimenters use software to carry out randomization. For example, most statistical software can choose five out of a list of 10 at random. The list might contain the names of 10 human subjects to be randomly assigned to two groups. The five chosen form one group, and the five that remain form the second group. The Simple Random Sample applet on the text website makes it particularly easy to choose treatment groups at random.

label

When we randomize, we first give a label to each in the collection of items to be randomized. The label could be the name of a subject in a clinical study or simply a numerical identification number. We then perform the randomization using software or a table of random numbers. To illustrate these methods, let’s randomize 10 subjects for a study that will compare a treatment with a placebo control. We will randomly select the five subjects for the treatment group, and the remaining subjects will receive the placebo. We start by labeling the subjects with the numbers 1 through 10.

Randomization using software Here is an example of one way to do the randomization using Excel. We start with a spreadsheet that has 10 rows corresponding to 10 subjects to be randomized to treatment or placebo.

uniform distribution, p. 71

The basic idea is that we generate a uniform random variable for each subject. In Excel, we use the RAND() function for this step. Then we sort the spreadsheet by the column with the uniform numbers and select the first five labels to be the treatment group and the remaining labels to be the placebo controls.

This process is essentially the same as writing the labels on a deck of 10 cards. We then shuffle the cards and deal five cards to form the treatment group.

EXAMPLE 3.12

Using software for the randomization. First create a data set with the numbers 1 to 10 in the first column. See Figure 3.4(a). Then we use RAND() to generate 10 random numbers in the second column. See Figure 3.4(b). Finally, we sort the data set based on the numbers in the second column. See Figure 3.4(c). The first five labels (7, 3, 4, 2, and 9) will receive the treatment. The remaining five labels (10, 6, 8, 1, and 5) will receive the placebo control.

FIGURE 3.4 Randomization of 10 experimental units using an Excel spreadsheet, Example 3.12: (a) labels; (b) random numbers; (c) sorted list of labels; (d) labels with group assignments.

If you want to save the uniform numbers that you generated in your file, you should copy them to another column using the “paste values” option before you perform the sort. Note that we have added a column called Group to the spreadsheet, which gives the group to which each subject is assigned. With this variable included, we can now sort the file on ID and delete the column with the random numbers. The result is shown in Figure 3.4(d). The spreadsheet in this form can now be used as a template for entering data.

Randomization using random digits You can randomize without software by using a table of random digits. Thinking about random digits helps you to understand randomization even if you will use software in practice. Table B at the back of the book is a table of random digits.

RANDOM DIGITS

A table of random digits is a list of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 that has the following properties:

1. The digit in any position in the list has the same chance of being any one of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. 2. The digits in different positions are independent in the sense that the value of one has no influence on the

value of any other.

You can think of Table B as the result of asking an assistant (or a computer) to mix the digits 0 to 9 in a hat, draw one, then replace the digit drawn, mix again, draw a second digit, and so on. The assistant’s mixing and drawing save us the work of mixing and drawing when we need to randomize. Table B begins with the digits 19223950340575628713. To make the table easier to read, the digits appear in groups of five and in numbered rows. The groups and rows have no meaning—the table is just a long list of digits having Properties 1 and 2 described earlier.

Our goal is to use random digits for experimental randomization. We need the following facts about random digits, which are consequences of Properties 1 and 2:

Any pair of random digits has the same chance of being any of the 100 possible pairs: 00, 01, 02, …, 98, 99. Any triple of random digits has the same chance of being any of the 1000 possible triples: 000, 001, 002, …, 998, 999. ... and so on for groups of four or more random digits.

EXAMPLE 3.13

Randomize the subjects. Let’s use random digits to perform the randomization that we performed using Excel in Example 3.12. Because the labels range from 1 to 10, we can use two digits for our labels

01, 02, 03, 04, 05, 06, 07, 08, 09, 10

when we select random digits from Table B. We could also have changed our labels to 0 through 9 and then we would only need to use single digits from Table B.

Start anywhere in Table B and read two-digit groups. Suppose we begin at line 175, which is

80011 09937 57195 33906 94831 10056 42211 65491

The first 10 two-digit groups in this line are

80 01 10 99 37 57 19 53 39 06

Each of these two-digit groups is a label. The labels 00 and 11 to 99 are not used in this example, so we ignore them. The first 10 labels between 01 and 10 that we encounter in the table choose subjects who will receive the treatment. Of the first 10 labels in line 175, we ignore seven because they are too high (over 10). The others are 01, 10, and 06. Continue across line 175 and 176 and verify that the next two subjects selected correspond to labels 03 and 04. Our randomization has selected subjects 1, 3, 4, 6, and 10 to receive the treatment. The remaining subjects, 2, 5, 7, 8, and 9 will receive the placebo control.

completely randomized design

When all experimental units are allocated at random among all treatments, as in Examples 3.12 and 3.13, the experimental design is completely randomized. Completely randomized designs can compare any number of treatments. The treatments can be formed by levels of a single factor or by more than one factor.

3.23

EXAMPLE 3.14

Randomization for the TV commercial experiment. Figure 3.2 displays six treatments formed by the two factors in an experiment on response to a TV commercial. Suppose that we have 150 students who are willing to serve as subjects. We must assign 25 students at random to each group. Figure 3.5 outlines the completely randomized design.

FIGURE 3.5 Outline of a completely randomized design comparing six treatments, Example 3.14.

To carry out the random assignment, label the 150 students 001 to 150. (Three digits are needed to label 150 subjects.) Using Excel, we would generate a uniform random variable for each label and sort the file as we did in Example 3.12. The first 25 students in this sorted file will receive Treatment 1, the next 25 will receive Treatment 2, etc.

Using random digits, we could enter Table B and read three-digit groups until you have selected 25 students to receive Treatment 1 (a 30-second ad shown once). If you start at line 140, the first few labels for Treatment 1 subjects are 129, 048, and 003.

Continue in Table B to select 25 more students to receive Treatment 2 (a 30-second ad shown three times). Then select another 25 for Treatment 3 and so on until you have assigned 125 of the 150 students to Treatments 1 through 5. The 25 students who remain get Treatment 6.

The randomization is straightforward but very tedious to do by using random digits. We strongly recommend that you use software, such as Excel or the Simple Random Sample applet. Exercise 3.37 (page 187) shows how to use the applet to do the randomization for this example.

USE YOUR KNOWLEDGE

Do the randomization. Use computer software to carry out the randomization in Example 3.14. Show your work by including the random uniform numbers in your final spreadsheet.

Cautions about experimentation double-blind

The logic of a randomized comparative experiment depends on our ability to treat all the experimental units identically in every way except for the actual treatments being compared. Good experiments, therefore, require careful attention to details. The ideal situation is where a study is double-blind—neither the subjects themselves nor the experimenters know which treatment any subject has received. The double-blind method avoids unconscious bias by, for example, a doctor who doesn’t think that “just a placebo” can benefit a patient.

Many—perhaps most—experiments have some weaknesses in detail. The environment of an experiment can influence the outcomes in unexpected ways. Although experiments are the gold standard for evidence of cause and effect, really convincing evidence usually requires that a number of studies in different places with different details produce similar results. Here are some brief examples of what can go wrong.

EXAMPLE 3.15

Placebo for a marijuana experiment. A study of the effects of marijuana recruited young men who used marijuana. Some were randomly assigned to smoke marijuana cigarettes, while others were given placebo cigarettes. This failed: the control group recognized that their cigarettes were phony and complained loudly. It may be quite common for blindness to fail because the subjects can tell which treatment they are receiving.11

lack of realism

The most serious potential weakness of experiments is lack of realism. The subjects or treatments or setting of an experiment may not realistically duplicate the conditions we really want to study. Here is an example.

EXAMPLE 3.16

Layoffs and feeling bad. How do layoffs at a workplace affect the workers who remain on the job? To try to answer this question, psychologists asked student subjects to proofread text for extra course credit, then “let go” some of the workers (who were actually accomplices of the experimenters). Some subjects were told that those let go had performed poorly (Treatment 1). Others were told that not all could be kept and that it was just luck that they were kept and others let go (Treatment 2). We can’t be sure that the reactions of the students are the same as those of workers who survive a layoff in which other workers lose their jobs. Many behavioral science experiments use student subjects in a campus setting. Do the conclusions apply to the real world?

Lack of realism can limit our ability to apply the conclusions of an experiment to the settings of greatest interest. Most experimenters want to generalize their conclusions to some setting wider than that of the actual experiment. Statistical analysis of an experiment cannot tell us how far the results will generalize to other settings. Nonetheless, the randomized comparative experiment, because of its ability to give convincing evidence for causation, is one of the most important ideas in statistics.

Matched pairs designs Completely randomized designs are the simplest statistical designs for experiments. They illustrate clearly the principles of control, randomization, and repetition. However, completely randomized designs are often inferior to more elaborate statistical designs. In particular, matching the subjects in various ways can produce more precise results than simple randomization.

matched pairs design

The simplest use of matching is a matched pairs design, which compares just two treatments. The subjects are matched in pairs. For example, an experiment to compare two advertisements for the same product might use pairs of subjects with the same age, sex, and income. The idea is that matched subjects are more similar than unmatched subjects so that comparing responses within a number of pairs is more efficient than comparing the responses of groups of randomly assigned subjects. Randomization remains important: which one of a matched pair sees the first ad is decided at random. One common variation of the matched pairs design imposes both treatments on the same subjects so that each subject serves as his or her own control. Here is an example.

EXAMPLE 3.17

Matched pairs for the smartphone prototype experiment. Example 3.11 describes an experiment to compare two prototypes of a new smartphone. The experiment compared two treatments: Phone 1 and Phone 2. The response variable is the satisfaction of the college student participant with the new smartphone. In Example 3.11, 40 student subjects were assigned at random, 20 students to each phone. This is a completely randomized design, outlined in Figure 3.3. Subjects differ in how satisfied they are with smartphones in general. The completely randomized design relies on chance to create two similar groups of subjects.

If we wanted to do a matched pairs version of this experiment, we would have each college student use each phone for two weeks. An effective design would randomize the order in which the phones are evaluated by each student. This will eliminate bias due to the possibility that the first phone evaluated will be systematically evaluated higher or lower than the second phone evaluated.

cross-over

The completely randomized design uses chance to decide which subjects will evaluate each smartphone prototype. The matched pairs design uses chance to decide which 20 subjects will evaluate Phone 1 first. The other 20 will evaluate Phone 2 first. This experiment is called a cross-over experiment. Situations where there are more than two treatments and all subjects receive all treatments can also be performed in this way.

Block designs The matched pairs design of Example 3.17 uses the principles of comparison of treatments, randomization, and repetition on several experimental units. However, the randomization is not complete (all subjects randomly assigned to treatment groups) but is restricted to assigning the order of the treatments for each subject. Block designs extend the use of “similar subjects” from pairs to larger groups.

BLOCK DESIGN

A block is a group of experimental units or subjects that are known before the experiment to be similar in some way that is expected to affect the response to the treatments. In a block design, the random assignment of units to treatments is carried out separately within each block.

Block designs can have blocks of any size. A block design combines the idea of creating equivalent treatment groups by matching with the principle of forming treatment groups at random. Blocks are another form of control. They control the effects of some outside variables by bringing those variables into the experiment to form the blocks. Here are some typical examples of block designs.

EXAMPLE 3.18

Blocking in a cancer experiment. The progress of a type of cancer differs in women and men. A clinical experiment to compare three therapies for this cancer then treats sex as a blocking variable. Two separate randomizations are done, one assigning the female subjects to the treatments and the other assigning the male subjects. Figure 3.6 outlines the design of this experiment. Note that there is no randomization involved in making up the blocks. They are groups of subjects who differ in some way (sex in this case) that is apparent before the experiment begins.

FIGURE 3.6 Outline of a block design, Example 3.18. The blocks consist of male and female subjects. The treatments are the three therapies for cancer.

EXAMPLE 3.19

Blocking in an agriculture experiment. The soil type and fertility of farmland differ by location. Because of this, a test of the effect of tillage type (two types) and pesticide application (three application schedules) on soybean yields uses small fields as blocks. Each block is divided into six plots, and the six treatments are randomly assigned to plots separately within each block.

EXAMPLE 3.20

Blocking in an education experiment. The Tennessee STAR class size experiment (Example 3.8, page 172) used a block design. It was important to compare different class types in the same school because the children in a school come from the same neighborhood, follow the same curriculum, and have the same school environment outside class. In all, 79 schools across Tennessee participated in the program. That is, there were 79 blocks. New kindergarten students were randomly placed in the three types of class separately within each school.

Blocks allow us to draw separate conclusions about each block, for example, about men and women in the cancer study in Example 3.18. Blocking also allows more precise overall conclusions because the systematic differences between men and women can be removed when we study the overall effects of the three therapies. The idea of blocking is an important additional principle of statistical design of experiments. A wise experimenter will form blocks based on the most important unavoidable sources of variability among the experimental units. Randomization will then average out the effects of the remaining variation and allow an unbiased comparison of the treatments.

SECTION 3.2 SUMMARY In an experiment, one or more treatments are imposed on the experimental units or subjects. Each treatment is a combination of levels of the explanatory variables, which we call factors. Outcomes are the measured variables that are used to compare the treatments. The design of an experiment refers to the choice of treatments and the manner in which the experimental units or subjects are assigned to the treatments. The basic principles of statistical design of experiments are compare, randomization, and repetition. The simplest form of control is comparison. Experiments should compare two or more treatments in order to prevent confounding the effect of a treatment with other influences, such as lurking variables. Randomization uses chance to assign subjects to the treatments. Randomization creates treatment groups that are similar (except for chance variation) before the treatments are applied. Randomization and comparison together prevent bias, or systematic favoritism, in experiments. You can carry out randomization by giving numerical labels to the experimental units and using a table of random digits to choose treatment groups. Repetition of the treatments on many units reduces the role of chance variation and makes the experiment more sensitive to differences among the treatments. Good experiments require attention to detail as well as good statistical design. Many behavioral and medical experiments are double-blind. Lack of realism in an experiment can prevent us from generalizing its results. In addition to comparison, a second form of control is to restrict randomization by forming blocks of experimental units that are similar in some way that is important to the response. Randomization is then carried out separately within each block. Matched pairs are a common form of blocking for comparing just two treatments. In some matched pairs designs, each subject receives both treatments in a random order. In others, the subjects are matched in pairs as closely as possible, and one subject in each pair receives each treatment.

SECTION 3.2 EXERCISES For Exercises 3.17 and 3.18, see page 173; for Exercises 3.19 and 3.20, see page 175; for Exercises 3.21 and 3.22, see page 176; and for Exercise 3.23, see page 181.

3.24 Blueberries and bones. A study of the effects of blueberries on the bones of mice compared diets containing no blueberries, blueberries as 5% of the diet, and blueberries as 10% of the diet. Ten mice were randomly assigned to each diet. The mice were fed the diets for 30 days, and the total body bone mineral density (TBBMD) was measured at the end of the feeding period. What are the experimental units, the treatments, and the outcomes for this experiment? Would you use the term subjects for the experimental units? Explain your answers.

3.25 Online homework. Thirty students participated in a study designed to evaluate a new online homework system. None of the students had used an online homework system in the past. After using the system for a month, they were asked to rate their satisfaction with the system using a five-point scale.

(a) What are the experimental units, the treatment, and the outcome for this experiment? Can we use the term subjects for the experimental units? Explain your answers.

(b) Is this a comparative experiment? If your answer is Yes, explain why. If your answer is No, describe how you would change the design so that it would be a comparative experiment.

(c) Suggest some different outcomes that you think would be appropriate for this experiment.

3.26 Do magnets reduce pain? Some claim that magnets can be used to reduce pain. Design a double-blind experiment to test this claim. Write a proposal requesting funding for your study giving all the important details, including the number of subjects, issues concerning randomization, and how you will make the study double-blind.

3.27 Online sales of running shoes. A company that sells running shoes online wants to compare two new marketing strategies. They will test the strategies on 10 weekdays. In the morning of each day, a web page describing the comfort of the running shoes will be displayed. In the afternoon of each day, a web page describing the discounted price for the shoes will be displayed. Sales of the featured running shoes in the morning will be compared with sales in the afternoon at the end of the experiment.

(a) What are the experimental units, the treatments, and the outcomes for this experiment? Explain your answers.

(b) Is this a comparative experiment? Why or why not?

(c) Could the experiment be improved by using randomization? Explain your answer.

(d) Could the experiment be improved by using a placebo treatment? Explain your answer.

3.28 Online sales of running shoes. Refer to the previous exercise. Suppose that for each day, you randomized the web pages, showing one in the morning and the other in the afternoon.

(a) Can you view this experiment as a block design? Explain your answer.

(b) Do you prefer this experiment or the one in the previous exercise? Give reasons for your answer.

3.29 Online sales of running shoes. Refer to Exercise 3.27. Here is another way in which the experiment could be designed. Suppose that you alternate the display each time a customer visits the website. Can you view this experiment as a matched pairs design? Explain your answer.

3.30 Randomize the web pages for the running shoes. Refer to Exercise 3.28. Use software or Table B to randomize the treatments. Give a step-by-step detailed description of how you performed the randomization.

3.31 What is needed? Explain what is deficient in each of the following proposed experiments and explain how you would improve the experiment.

(a) Two product promotion offers are to be compared. The first, which offers two items for $2, will be used in a store on Friday. The second, which offers three items for $3, will be used in the same store on Saturday.

(b) A study compares two marketing campaigns to encourage individuals to eat more fruits and vegetables. The first campaign is launched in Florida at the same time that the second campaign is launched in Minnesota.

(c) You want to evaluate the effectiveness of a new investment strategy. You try the strategy for one year and evaluate the performance of the strategy.

3.32 The Madden curse. Some people believe that individuals who appear on the cover of the football game Madden NFL will soon have a serious injury. Can you evaluate this belief with an experiment? Explain your answer.

3.33 Evaluate a new orientation program. Your company runs a two-day orientation program Monday and Tuesday each week for new employees. A new program is to be compared with the current one. Set up an experiment to compare the new program with the old. Be sure to provide details regarding randomization and what outcome variables you will measure.

3.34 What is wrong? Explain what is wrong with each of the following randomization procedures, and describe how you would do the randomization correctly.

(a) Twenty students are to be used to evaluate a new treatment. Ten men are assigned to receive the treatment, and 10 women are assigned to be the controls.

(b) Ten subjects are to be assigned to two treatments, five to each. For each subject, a coin is tossed. If the coin comes up heads, the subject is assigned to the first treatment; if the coin comes up tails, the subject is assigned to the second treatment.

(c) An experiment will assign 40 rats to four different treatment conditions. The rats arrive from the supplier in batches of 10, and the treatment lasts two weeks. The first batch of 10 rats is randomly assigned to one of the four treatments, and data for these rats are collected. After a one-week break, another batch of 10 rats arrives and is assigned to one of the three remaining treatments. The process continues until the last batch of rats is given the treatment that has not been assigned to the three previous batches.

3.35 Calcium and vitamin D. Vitamin D is needed for the body to use calcium. An experiment is designed to study the effects of calcium and vitamin D supplements on the bones of first-year college students. The outcome measure is the total body bone mineral content (TBBMC), a measure of bone health. Three doses of calcium will be used: 0, 250, and 500 milligrams per day (mg/day). The doses of vitamin D will be 0, 75, and 150 international units (IU) per day. The calcium and vitamin D will be given in a single tablet. All tablets, including those with no calcium and no vitamin D, will look identical. Subjects for the study will be 45 men and 45 women.

(a) What are the factors and the treatments for this experiment?

(b) Draw a picture explaining how you would randomize the 90 college students to the treatments.

(c) Use a spreadsheet to carry out the randomization.

(d) Is there a placebo in this experiment? Explain your answer.

3.36 Use the Simple Random Sample applet. You can use the Simple Random Sample applet to choose a group at random once you have labeled the subjects. Example 3.12 (page 178) uses Excel to choose five students from a group of

10 to receive a treatment in an experiment. The remaining five students will receive a placebo control.

(a) Use the applet to choose five students. Which students were selected?

(b) Compare using Excel, as we did in Example 3.12, with the applet that you used for this exercise. Which do you prefer? Give reasons for your answer.

3.37 Use the Simple Random Sample applet. The Simple Random Sample applet allows you to randomly assign experimental units to more than two groups without difficulty. Consider a randomized comparative experiment in which

100 students are randomly assigned to four groups of 25.

(a) Use the applet to randomly choose 25 out of 100 students to form the first group. Which students are in this group?

(b) The “population hopper” now contains the 75 students who were not chosen, in scrambled order. Click “Sample” again to choose 25 of these remaining students to make up the second group. Which students were chosen?

(c) Click “Sample” one more time to choose the third group. Don’t take the time to write this down. Check that there are only 25 students remaining in the “population hopper.” These subjects get Treatment 4. Which students are they?

3.38 Use the Simple Random Sample applet. The Simple Random Sample applet can demonstrate how randomization works to create similar groups for comparative experiments. Suppose that (unknown to the experimenters)

the 20 even-numbered students among the 40 subjects for the smartphone study in Example 3.11 (page 176) tend to send more text messages than the odd-numbered students. We would like the two groups to be similar with respect to text messaging. Use the applet to choose 10 samples of size 20 from the 40 students. (Be sure to click “Reset” after each sample.) Record the counts of even-numbered students in each of your 10 samples. You see that there is considerable chance variation but no systematic bias in favor of one or the other group in assigning the fast-reacting students. Larger samples from larger populations will, on the average, do a better job of making the two groups equivalent.

3.39 Health benefits of bee pollen. “Bee pollen is effective for combating fatigue, depression, cancer, and colon disorders.” So says a website that offers the pollen for sale. We wonder if bee pollen really does prevent colon disorders.

Here are two ways to study this question. Explain why the first design will produce more trustworthy data.

(a) Find 400 women who do not have colon disorders. Randomly assign 200 to take bee pollen capsules and the other 200 to take placebo capsules that are identical in appearance. Follow both groups for five years.

(b) Find 200 women who take bee pollen regularly. Match each with a woman of the same age, race, and occupation who does not take bee pollen. Follow both groups for five years.

3.40 Random digits. Table B is a table of random digits. Which of the following statements are true of a table of random digits, and which are false? Explain your answers.

(a) There are exactly four 0s in each row of 40 digits.

(b) Each pair of digits has chance 1/100 of being 00.

(c) The digits 0000 can never appear as a group because this pattern is not random.

3.41 Calcium and the bones of young girls. Calcium is important to the bone development of young girls. To study how the bodies of young girls process calcium, investigators used the setting of a summer camp. Calcium was given

in punch at either a high or a low level. The camp diet was otherwise the same for all girls. Suppose that there are 30 campers.

(a) Outline a completely randomized design for this experiment.

(b) Use software or Table B to do the randomization. Explain in step-by-step detail how you carried out how you performed the randomization.

(c) Make a table giving the treatment that each camper will receive.

3.42 Calcium and the bones of young girls. Refer to the previous exercise.

(a) Outline a matched pairs design in which each girl receives both levels of calcium (with a “washout period” in which no calcium supplementation was given between the two treatment periods).

(b) What is the advantage of the matched pairs design over the completely randomized design?

(c) The same randomization can be used in different ways for both designs. Explain why this is true.

(d) Use software or Table B to do the randomization. Explain what each subject will do for the matched pairs design.

3.43 Measuring water quality in streams and lakes. Water quality of streams and lakes is an issue of concern to the public. Although trained professionals typically are used to take reliable measurements, many volunteer groups are

gathering and distributing information based on data that they collect.12 You are part of a team to train volunteers to collect accurate water quality data. Design an experiment to evaluate the effectiveness of the training. Write a summary of your proposed design to present to your team. Be sure to include all the details that they will need to evaluate your proposal.

3.3 Sampling Design

When you complete this section, you will be able to:

Distinguish between a population and a sample. Use the response rate to evaluate a survey. Use Table B to generate a simple random sample (SRS). Use software to generate a simple random sample. Construct a stratified random sample using Table B or software to select the samples from the strata. Identify voluntary response samples, simple random samples, stratified random samples, and multistage random samples. Identify characteristics of samples that limit their usefulness, including undercoverage, nonresponse, response bias, and the wording of questions.

A political scientist wants to know what percent of college-age adults consider themselves conservatives. An automaker hires a market research firm to learn what percent of adults aged 18 to 35 recall seeing television advertisements for a new sports utility vehicle. Government economists inquire about average household income.

sample survey

In all these cases, we want to gather information about a large group of individuals. We will not, as in an experiment, impose a treatment in order to observe the response. Also, time, cost, and inconvenience forbid contacting every individual. In such cases, we gather information about only part of the group—a sample—in order to draw conclusions about the whole. Sample surveys are an important kind of observational study.

POPULATION AND SAMPLE

The entire group of individuals that we want information about is called the population.

A sample is a part of the population that we actually examine in order to gather information.

sample design

Notice that “population” is defined in terms of our desire for knowledge. If we wish to draw conclusions about all U.S. college students, that group is our population even if only local students are available for questioning. The sample is the part from which we draw conclusions about the whole. The design of a sample survey refers to the method used to choose the sample from the population.

EXAMPLE 3.21

The Reading Recovery program. The Reading Recovery (RR) program has specially trained teachers work one-on-one with at-risk first-grade students to help them learn to read. A study was designed to examine the relationship between the RR teachers’ beliefs about their ability to motivate students and the progress of the students whom they teach.13 The Reading Recovery International Data Evaluation Center website (www.idecweb.us) says that there are 13,823 RR teachers. The researchers send a questionnaire to a random sample of 200 of these. The population consists of all 13,823 RR teachers, and the sample is the 200 that were randomly selected.

3.44

3.45

Unfortunately, our idealized framework of population and sample does not exactly correspond to the situations that we face in many cases. In Example 3.21, the list of teachers was prepared at a particular time in the past. It is very likely that some of the teachers on the list are no longer working as RR teachers today. New teachers have been trained in RR methods and are not on the list. Despite these difficulties, we still view the list as the population. Also, we may have out-of-date addresses for some who are still working as RR teachers, and some teachers may choose not to respond to the survey questions.

response rate

In reporting the results of a sample survey, it is important to include all details regarding the procedures used. Follow-up mailings or phone calls to those who do not initially respond can help increase the response rate. The proportion of the original sample who actually provide usable data is called the response rate and should be reported for all surveys. If only 150 of the teachers who were sent questionnaires provided usable data, the response rate would be 150/200, or 75%.

USE YOUR KNOWLEDGE

Are they satisfied? An educational research team wanted to examine the relationship between faculty participation in decision making and job satisfaction in Mongolian public universities. They are planning to randomly select 250 faculty members from a list of 2000 faculty members in these universities. The Job Descriptive Index will be used to measure job satisfaction, and the Conway Adaptation of the Alutto-Belasco Decisional Participation Scale will be used to measure decision participation.

(a) Describe the population for this study.

(b) Describe the sample for this study.

(c) How would you determine the response rate for this study? Can you calculate it from the information given? If your answer is yes, calculate it.

What is the impact of the taxes? A study was designed to assess the impact of taxes on forest land usage in part of the Upper Wabash River Watershed in Indiana.14 A survey was sent to 772 forest owners from this region, and 348 were returned.

(a) What is the sample for this study?

(b) What is the population for this study?

(c) How would you determine the response rate for this study? Can you calculate it from the information given? If your answer is yes, calculate it.

Poor sample designs can produce misleading conclusions. Here is an example.

EXAMPLE 3.22

Sampling pieces of steel. A mill produces large coils of thin steel for use in manufacturing home appliances. The quality engineer wants to submit a sample of 5-centimeter squares to detailed laboratory examination. She asks a technician to cut a sample of 10 such squares. Wanting to provide “good” pieces of steel, the technician carefully avoids the visible defects in the coil material when cutting the sample. The laboratory results are wonderful, but the customers complain about the material they are receiving.

bias, p. 174

In Example 3.22, the sample was selected in a manner that guaranteed that it would not be representative of the entire population. This sampling scheme displays bias, or systematic error, in favoring some parts of the population over others.

Online polls use voluntary response samples, a particularly common form of biased sample. The sample who respond are not representative of the population at large. People who take the trouble to respond to an open invitation are not representative of the entire population.

VOLUNTARY RESPONSE SAMPLE

A voluntary response sample consists of people who choose themselves by responding to a general appeal. Voluntary response samples are biased because people with strong opinions, especially negative opinions, are most likely to respond.

The remedy for bias in choosing a sample is to allow chance to do the choosing so that there is neither favoritism by the sampler (Example 3.22) nor voluntary response (online opinion polls). Random selection of a sample eliminates bias by giving all individuals an equal chance to be chosen, just as randomization eliminates bias in assigning experimental units.

Simple random samples The simplest sampling design amounts to placing names in a hat (the population) and drawing out a handful (the sample). This is simple random sampling.

SIMPLE RANDOM SAMPLE

A simple random sample (SRS) of size n consists of n individuals from the population chosen in such a way that every set of n individuals has an equal chance to be the sample actually selected.

Each treatment group in a completely randomized experimental design is an SRS drawn from the available experimental units. We select an SRS by labeling all the individuals in the population and using software or a table of random digits to select a sample of the desired size, just as in experimental randomization. Notice that an SRS not only gives every possible sample an equal chance to be chosen, but also gives each individual an equal chance to be chosen. There are other random sampling designs that give each individual, but not each sample, an equal chance. One such design, systematic random sampling, is described in Exercise 3.64 (page 202).

How to select a simple random sample The basic ideas needed to select a simple random sample are very similar to those that we discussed when we randomized subjects to treatments (page 177). We first assign a label to each case in our population. Then we perform the randomization using software or random digits from Table B.

Selection of a simple random sample using software The World Bank collects information about starting businesses throughout the world. In Example 1.23 (page 26) and several other examples in Chapter 1, we examined the time to start a business in a subset of these countries. For those exercises, we used a subset of the data because it was easier to show some details about our calculations with a smaller amount of data.

Now, suppose we want to collect additional information about countries that would help us to understand the processes of starting a business. The complete data set contains entries for 189 countries, and the time required to collect the additional information on all these would be too much. Let’s use Excel to select a sample of 25 countries for a more detailed examination of these countries.

EXAMPLE 3.23

Select an SRS of countries using Excel. The data file TTS includes columns for the country name and a three-letter country code for each of the 189 countries. We could use either of these for our label. We will use the three letter codes in our screen shots to save space.

TTS

Figure 3.7(a) shows the codes for the first ten countries. In Figure 3.7(b), we show the uniform numbers generated with the RAND() function (and then pasted into column C). The first three countries are Afghanistan (AFG), Angola (AGO), and Albania (ALB).

FIGURE 3.7 Selection of a simple random sample of countries from the population of 189 countries, Example 3.23.

Figure 3.7(c) shows the file after we sort on the uniform random numbers in column C. Our sample is the first 25 countries in the sorted file. The first three selected are Chile (CHL), Belgium (BEL), and the Solomon Islands (SLB). Note that Excel does not display the last two digits for Belarus (BLR) because they are zero.

The Simple Random Sample applet on the text website is another convenient way to select an SRS.

3.46

3.47

3.48

3.49

USE YOUR KNOWLEDGE

Select an SRS. Use the Simple Random Sample applet or Excel to select an SRS of five countries from the TTS data file. Include a step-by-step detailed description of how you selected the countries.

TTS

Selection of a simple random sample using random digits We illustrate the procedure by selecting an SRS of countries from the population of 189 countries in the data file TTS. Recall that we used Excel to select such a sample in Example 3.23.

EXAMPLE 3.24

Select an SRS of countries using random digits. To use Table B, we need a numeric label. We could create such a label by adding a column to the data file TTS containing the numbers 1 to 189. An alternative requiring less work would be to use the numbers in the leftmost part of the spreadsheet. Notice in Figure 3.7(a), for example, that there is a 1 in the first row of the spreadsheet where we have entered the names of the variables in the columns. Therefore, the numbers corresponding to countries run from 2 through 190. We will use these numbers as our label.

We will examine the entries in Table B in sets of three. Three digit numbers between 2 and 190 will correspond to selected countries. We will ignore three digit numbers equal to 000, 001, or greater than 190. Let’s start our selection at line 106 in Table B. The entries on this line are

68417 35013 15529 72765 85089 57067 50211 47487

If we arrange these into sets of three, we have

684 173 501 315 529 727 658 508 057 067 502 114 748 7

The selected labels from this set of random digits are 173, 057, and 067. Checking the spreadsheet, we see that these numbers correspond to Turkey, France, and Greece.

Note that we do not use the last digit on line 106 to select the country with the label 7. We should combine this single digit with the first two digits from line 107 of Table B. This gives us the three-digit number 782, which is a number that we ignore. We complete our selection of the additional 22 countries that we need in our SRS using additional lines from Table B as needed.

USE YOUR KNOWLEDGE

Find the next three countries to be selected. Continue the process described in Example 3.24 to select the next three countries for the SRS. Show your work.

Listen to three rock songs. The walk to your statistics class takes about 10 minutes, about the amount of time needed to listen to three songs on your iPod or smartphone. You decide to take a simple random sample of songs from the top 10 listed on a Billboard Hot Rock Songs.15 Here is the list:

Shut Up and Dance Uma Thurman Renegades Ex’s & Oh’s Centuries Cecilia and the Satellite Tear in My Heart Brother Stressed Out Shots

Select your three hot rock songs using a simple random sample. Show your work.

Listen to three songs. Refer to the previous exercise. Suppose that you like to include more variety in your music, so you look at the Billboard Top 100 songs.16 Here are the top 10 on this list:

Cheerleader Can’t Feel My Face Watch Me Lean On The Hills Good for You Fight Song 679 Trap Queen Shut Up and Dance

Select the three songs for your iPod or smartphone using a simple random sample. Show your work.

EXAMPLE 3.25

Select an SRS of countries using JMP. Refer to Example 3.23, where we selected an SRS of countries from a population of 189 countries. Also see Example 3.24, where we used the random digits in Table B to select the SRS. We can also use JMP to select an SRS. The output for the first 17 countries selected is displayed in Figure 3.8. JMP provides a file with the selected countries with all columns in the original data file. Note that the selected files are listed in the order that they appear in the original file, which is alphabetically in this case.

FIGURE 3.8 JMP output for selecting an SRS of 25 countries from a population of 189 countries, Example 3.25. Only the first 17 are displayed.

Stratified random samples The general framework for designs that use chance to choose a sample is a probability sample.

PROBABILITY SAMPLE

A probability sample is a sample chosen by chance. We must know what samples are possible and what chance, or probability, each possible sample has.

Some probability sampling designs (such as an SRS) give each member of the population an equal chance to be selected. This may not be true in more elaborate sampling designs. In every case, however, the use of chance to select the sample is the essential principle of statistical sampling.

Designs for sampling from large populations spread out over a wide area are usually more complex than an SRS. For example, it is common to sample important groups within the population separately, then combine these samples. This is the idea of a stratified sample.

STRATIFIED RANDOM SAMPLE

To select a stratified random sample, first divide the population into groups of similar individuals, called strata. Then choose a separate SRS in each stratum and combine these SRSs to form the full sample.

Choose the strata based on facts known before the sample is taken. For example, a population of election districts might be divided into urban, suburban, and rural strata.

A stratified design can produce more exact information than an SRS of the same size by taking advantage of the fact that individuals in the same stratum are similar to one another. Think of the extreme case in which all individuals in each stratum are identical: just one individual from each stratum is then enough to completely describe the population.

Strata for sampling are similar to blocks in experiments. We have two names because the idea of grouping similar units before randomizing arose separately in sampling and in experiments.

EXAMPLE 3.26

A stratified sample of countries. In Examples 3.23 and 3.24, we selected SRSs of size 25 from the population of 189 countries in the World Bank file with data on starting businesses. Let’s think about using a stratified sample. You still want to select 25 companies to examine in detail.

Let’s classify each of the countries as located in Asia, Africa, Europe, North or South America, and Other. We have five strata, and we want a total of 25 countries to examine in detail. Therefore, we need to sample five countries from each stratum. We take an SRS of size 5 from each of these strata.

Multistage random samples multistage random sample

Another common means of restricting random selection is to choose the sample in stages. These designs are called multistage designs. They are widely used in national samples of households or people. For example, data on employment and unemployment are gathered by the government’s Current Population Survey, which conducts interviews in about 60,000 households each month. The cost of sending interviewers to the widely scattered households in an SRS would be too high. Moreover, the government wants data broken down by states and large cities.

clusters

Thus, the Current Population Survey uses a multistage random sampling design. The final sample consists of groups of nearby households, called clusters, that an interviewer can easily visit. Most opinion polls and other national samples are also multistage, though interviewing in most national samples today is done by telephone rather than in person, eliminating the economic need for clustering. The Current Population Survey sampling design is roughly as follows:17

Stage 1. Divide the United States into 2007 geographical areas called Primary Sampling Units, or PSUs. PSUs do not cross state lines. Select a sample of 754 PSUs. This sample includes the 428 PSUs with the largest population and a stratified sample of 326 of the others.

Stage 2. Divide each PSU selected into smaller areas called “blocks.” Stratify the blocks using ethnic and other information, and take a stratified sample of the blocks in each PSU.

Stage 3. Sort the housing units in each block into clusters of four nearby units. Interview the households in a probability sample of these clusters.

Analysis of data from sampling designs more complex than an SRS takes us beyond basic statistics. But the SRS is the building block of more elaborate designs, and analysis of other designs differs more in complexity of detail than in fundamental concepts.

Cautions about sample surveys Random selection eliminates bias in the choice of a sample from a list of the population. Sample surveys of large human populations, however, require much more than a good sampling design.18 To begin, we need an accurate and complete list of the population. Because such a list is rarely available, most samples suffer from some degree of undercoverage. A sample survey of households, for example, will miss not only homeless people, but also prison inmates and students in dormitories. An opinion poll conducted by telephone will miss the large number of American households without residential phones. The results of national sample surveys, therefore, have some bias if the people not covered—who most often are poor people—differ from the rest of the population.

A more serious source of bias in most sample surveys is nonresponse, which occurs when a selected individual cannot be contacted or refuses to cooperate. Nonresponse to sample surveys often reaches 50% or more, even with careful planning and several callbacks. Because nonresponse is higher in urban areas, most sample surveys substitute other people in the same area to avoid favoring rural areas in the final sample. If the people contacted differ from those who are rarely at home or who refuse to answer questions, some bias remains.

UNDERCOVERAGE AND NONRESPONSE

Undercoverage occurs when some groups in the population are left out of the process of choosing the sample.

Nonresponse occurs when an individual chosen for the sample can’t be contacted or does not cooperate.

EXAMPLE 3.27

Nonresponse in the Current Population Survey. How bad is nonresponse? The Current Population Survey (CPS) has the lowest nonresponse rate of any poll we know: only about 5% of the households in the CPS sample refuse to take part, and another 2% or 3% can’t be contacted.19 People are more likely to respond to a government survey such as the CPS, and the CPS contacts its sample in person before doing later interviews by phone.

The General Social Survey (Figure 3.9) is the nation’s most important social science research survey. The GSS also contacts its sample in person, and it is run by a university. Despite these advantages, its most recent survey had a 30% rate of nonresponse.20

FIGURE 3.9 Part of the home page for the General Social Survey (GSS). The GSS has assessed attitudes on a wide variety of topics since 1972. Its continuity over time makes the GSS a valuable source for studies of changing attitudes. (Source: GSS)

What about polls done by the media and by market research and opinion-polling firms? Often, we don’t know their rates of nonresponse because they won’t say. That itself is a bad sign.

EXAMPLE 3.28

Change in nonresponse in Pew surveys. The Pew Research Center conducts research using surveys on a variety of issues, attitudes, and trends.21 A study by the center examined the decline in the response rates to their surveys over time. The changes are dramatic, and there is a consistent pattern over time. Here are some data from the report:22

Year 1997 2000 2003 2006 2009 2012 Nonresponse rate 64% 72% 75% 79% 85% 91%

The center is devising alternative methods that show some promise of improving the response rates of their surveys.

Most sample surveys, and almost all opinion polls, are now carried out by telephone. This and other details of the interview method can affect the results. When presented with several options for a reply, such as “completely agree,” “mostly agree,” “mostly disagree,” and “completely disagree,” people tend to be a little more likely to respond to the first one or two options presented.

response bias

The behavior of the respondent or of the interviewer can cause response bias in sample results. Respondents may lie, especially if asked about illegal or unpopular behavior. The race or sex of the interviewer can influence responses to questions about race relations or attitudes toward feminism. Answers to questions that ask respondents to recall past events are often inaccurate because of faulty memory. For example, many people “telescope” events in the past, bringing them forward in memory to more recent time periods. “Have you visited a dentist in the last six months?” will often elicit a Yes from someone who last visited a dentist eight months ago.

wording of questions

The wording of questions is the most important influence on the answers given to a sample survey. Confusing or leading questions can introduce strong bias, and even minor changes in wording can change a survey’s outcome. Here are some examples.

EXAMPLE 3.29

The form of the question is important. In response to the question “Are you heterosexual, homosexual, or bisexual?” in a social science research survey, one woman answered, “It’s just me and my husband, so bisexual.” The issue is serious, even if the example seems silly: reporting about sexual behavior is difficult because people understand and misunderstand sexual terms in many ways.

How do Americans feel about government help for the poor? Only 13% think we are spending too much on “assistance to the poor,” but 44% think we are spending too much on “welfare.” How do the Scots feel about the movement to become independent from England? Well, 51% would vote for “independence for Scotland,” but only 34% support “an independent Scotland separate from the United Kingdom.” It seems that “assistance to the poor” and “independence” are nice, hopeful words. “Welfare” and “separate” are negative words.23

The statistical design of sample surveys is a science, but this science is only part of the art of sampling. Because of nonresponse, response bias, and the difficulty of posing clear and neutral questions, you should hesitate to fully trust reports about complicated issues based on surveys of large human populations. Insist on knowing the exact questions asked, the rate of nonresponse, and the date and method of the survey before you trust a poll result.

BEYOND THE BASICS

Capture-Recapture Sampling Sockeye salmon return to reproduce in the river where they were hatched four years earlier. How many salmon survived natural perils and heavy fishing to make it back this year? How many mountain sheep are there in Colorado? Are migratory songbird populations in North America decreasing or holding their own? These

questions concern the size of animal populations. Biologists address them with a special kind of repeated sampling, called capture-recapture sampling.

EXAMPLE 3.30

Estimate the number of least flycatchers. You are interested in the number of least flycatchers migrating along a major route in the north-central United States. You set up “mist nets” that capture the birds but do not harm them. The birds caught in the net are fitted with a small aluminum leg band and released. Last year, you banded and released 200 least flycatchers. This year, you repeat the process. Your net catches 120 least flycatchers, 12 of which have tags from last year’s catch.

The proportion of your second sample that have bands should estimate the proportion in the entire population that are banded. So if N is the unknown number of least flycatchers, we should have approximately

Solve for N to estimate that the total number of flycatchers migrating while your net was up this year is approximately

The capture-recapture idea extends the use of a sample proportion to estimate a population proportion. The idea works well if both samples are SRSs from the population and the population remains unchanged between samples. In practice, complications arise because, for example, some of the birds tagged last year died before this year’s migration.

Variations on capture-recapture samples are widely used in wildlife studies and are now finding other applications. One way to estimate the census undercount in a district is to consider the census as “capturing and marking” the households that respond. Census workers then visit the district, take an SRS of households, and see how many of those counted by the census show up in the sample. Capture-recapture estimates the total count of households in the district. As with estimating wildlife populations, there are many practical pitfalls. Our final word is as before: the real world is less orderly than statistics textbooks imply.

SECTION 3.3 SUMMARY A sample survey selects a sample from the population of all individuals about which we desire information. We base conclusions about the population on data about the sample. The design of a sample refers to the method used to select the sample from the population. Probability sampling designs use impersonal chance to select a sample. The basic probability sample is a simple random sample (SRS). An SRS gives every possible sample of a given size the same chance to be chosen. Choose an SRS using software. This can also be done using a table of random digits to select the sample. To choose a stratified random sample, divide the population into strata, groups of individuals that are similar in some way that is important to the response. Then choose a separate SRS from each stratum, and combine them to form the full sample. Multistage random samples select successively smaller groups within the population in stages, resulting in a sample consisting of clusters of individuals. Each stage may employ an SRS, a stratified sample, or another type of sample. Failure to use probability sampling often results in bias, or systematic errors in the way the sample represents the population. Voluntary response samples, in which the respondents choose themselves, are particularly prone to large bias. In human populations, even probability samples can suffer from bias due to undercoverage or nonresponse, from response bias due to the behavior of the interviewer or the respondent, or from misleading results due to poorly worded questions.

SECTION 3.3 EXERCISES For Exercises 3.44 and 3.45, see pages 189–190; for Exercise 3.46, see page 192; and for Exercises 3.47, 3.48, and 3.49, see page 193.

3.50 How many text messages? You would like to know something about how many text messages you will receive in the next 100 days. Counting the number for each of the 100 days would take more time than you would like to spend on this project, so you randomly select 10 days from the hundred to count.

(a) Describe the population for this setting.

(b) What is the sample?

3.51 Response rate? A survey designed to assess satisfaction with food items sold at a college’s football games was sent to 150 fans who had season tickets. The total number of fans who have season tickets is 5674. Responses to the survey were received from 98 fans.

(a) Describe the population for this survey.

(b) What is the sample?

(c) What is the response rate?

(d) What is the nonresponse rate?

(e) Suggest some ways that could be used in a future survey to increase the response rate.

3.52 Interview some students. You are a teaching assistant for an introductory statistics class. The instructor would like you to interview some of the students in the class to find out their opinion regarding some new interactive activities that she has introduced to the course. There are 123 students in the class, so you cannot interview all of them. You decide to select eight students to interview.

(a) What is the population for this setting?

(b) What is the sample?

(c) Make a spreadsheet with the numeric labels for the 123 students in the class.

(d) Use Excel to select the labels of the eight students to be interviewed from the spreadsheet.

(e) Explain the steps that you used in sufficient detail so that another person could repeat your work.

3.53 Interview some students. Refer to the previous exercise.

(a) Use Table B to select the students. Give details.

(b) Compare the use of Table B with software for selecting the students. Which do you prefer? Give reasons for your answer.

3.54 What kind of sample? In each of the following situations, identify the sample as an SRS, a stratified random sample, a multistage random sample, or a voluntary response sample. Explain your answers.

(a) There are seven sections of an introductory statistics course. A random sample of three sections is chosen, and then random samples of eight students from each of these sections are chosen.

(b) A student organization has 55 members. A table of random numbers is used to select a sample of five.

(c) An online poll asks people who visit this site to choose their favorite television show.

(d) Separate random samples of male and female first-year college students in an introductory psychology course are selected to receive a one-week alternative instructional method.

3.55 What’s wrong? Explain what is wrong in each of the following scenarios.

(a) The population consists of all individuals selected in a simple random sample.

(b) In a poll of an SRS of residents in a local community, respondents are asked to indicate the level of their concern about the dangers of dihydrogen monoxide, a substance that is a major component of acid rain and that, in its gaseous state, can cause severe burns. (Hint: Ask a friend who is majoring in chemistry about this substance or search the Internet for information about it.)

(c) Students in a class are asked to raise their hands if they have cheated on an exam one or more times within the past year.

3.56 What’s wrong? Explain what is wrong with each of the following random selection procedures, and explain how you would do the randomization correctly.

(a) To determine the reading level of an introductory statistics text, you evaluate all the written material in the third chapter.

(b) You want to sample student opinions about a proposed change in procedures for changing majors. You hand out questionnaires to 100 students as they arrive for class at 7:30 A.M.

(c) A population of subjects is put in alphabetical order, and a simple random sample of size 10 is taken by selecting the first 10 subjects in the list.

3.57 Importance of students as customers. A committee on community relations in a college town plans to survey local businesses about the importance of students as customers. From telephone book listings, the committee chooses 80 businesses at random. Of these, 46 return the questionnaire mailed by the committee.

(a) What is the population for this sample survey?

(b) What is the sample?

(c) What is the rate (percent) of nonresponse?

3.58 Identify the populations. For each of the following sampling situations, identify the population as exactly as possible. That is, say what kind of individuals the population consists of and say exactly which individuals fall in the population. If the information given is not complete, complete the description of the population in a reasonable way.

(a) A college has changed its core curriculum and wants to obtain detailed feedback information from the students during each of the first 12 weeks of the coming semester. Each week, a random sample of five students will be selected to be interviewed.

(b) The American Community Survey (ACS) replaced the census “long form” starting with the 2010 census. The ACS contacts 250,000 addresses by mail each month, with follow-up by phone and in person if there is no response. Each household answers questions about their housing, economic, and social status.

(c) An opinion poll contacts 1161 adults and asks them, “Which political party do you think has better ideas for leading the country in the twenty-first century?”

3.59 Interview residents of apartment complexes. You are planning a report on apartment living in a college town. You decide to select eight apartment complexes at random for in-depth interviews with residents. Select a simple random sample of eight of the following apartment complexes. If you use Table B, start at line 136. RESIDEN

Ashley Oaks Country View Mayfair Village Bay Pointe Country Villa Nobb Hill Beau Jardin Crestview Pemberly Courts Bluffs Del-Lynn Peppermill Brandon Place Fairington Pheasant Run Briarwood Fairway Knolls Richfield Brownstone Fowler Sagamore Ridge Burberry Franklin Park Salem Courthouse Cambridge Georgetown Village Manor Chauncey Village Greenacres Waterford Court Country Squire Lahr House Williamsburg

3.60 Using GIS to identify mint field conditions. A Geographic Information System (GIS) is to be used to distinguish different conditions in mint fields. Ground observations will be used to classify regions of each field as either healthy mint, diseased mint, or weed-infested mint. The GIS divides mint-growing areas into regions called pixels. An experimental area contains 100 pixels. For a random sample of 15 pixels, ground measurements will be made to determine the status of the mint, and these observations will be compared with information obtained by the GIS. Select the random sample. If you use Table B, start at line 132 and choose only the first 15 pixels in the sample.

3.61 Use the Simple Random Sample applet. After you have labeled the individuals in a population, the Simple Random Sample applet automates the task of choosing an SRS. Use the applet to choose the sample in the previous

exercise.

3.62 Select a simple random sample. There are 38 active telephone area codes in California. You want to choose an SRS of 10 of these area codes for a study of available telephone numbers. Label the codes 01 to 38 and use the Simple

Random Sample applet, Table B, or software to choose your sample. (If you use Table B, start at line 131.)

3.63 Stratified samples for attitudes about alcohol. At a party, there are 25 students over age 21 and 15 students under age 21. You choose at random five of those over 21 and separately choose at random three of those under 21 to

interview about attitudes toward alcohol. You have given every student at the party the same chance to be interviewed: what is that chance? Why is your sample not an SRS?

3.64 Systematic random samples. Systematic random samples are often used to choose a sample of apartments in a large building or dwelling units in a block at the last stage of a multistage sample. An example will illustrate the idea of a systematic sample. Suppose that we must choose five addresses out of 125. Because 125/5 = 25, we can think of the list as five lists of 25 addresses. Choose one of the first 25 at random, using software or Table B. The sample contains this address and the addresses 25, 50, 75, and 100 places down the list from it. If 13 is chosen, for example, then the systematic random sample consists of the addresses numbered 13, 38, 63, 88, and 113.

(a) A study of dating among college students wanted a sample of 200 of the 8000 single male students on campus. The sample consisted of every 40th name from a list of the 8000 students. Explain why the survey chooses every 40th name.

(b) Use software or Table B at line 112 to choose the starting point for this systematic sample.

3.65 Systematic random samples versus simple random samples. The previous exercise introduces systematic random samples. Explain carefully why a systematic random sample does give every individual the same chance to be

chosen but is not a simple random sample.

3.66 Random digit telephone dialing. An opinion poll in California uses random digit dialing to choose telephone numbers at random. Numbers are selected separately within each California area code. The size of the sample in each area code is proportional to the population living there. AREACOD

(a) What is the name for this kind of sampling design?

(b) California area codes, in rough order from north to south, are

209 213 310 323 341 369 408 415 424 442 510 530 559 562 619 626 627 628 650 657 661 669 707 714 747 752 760 764 805 818 831 858 909 916 925 935 949 951

Another California survey does not call numbers in all area codes but starts with an SRS of eight area codes. Choose such an SRS. If you use Table B, start at line 132.

3.67 Select club members to go to a convention. A club has 30 student members and 10 faculty members. The students are

Abel Fisher Huber Moran Reinmann Carson Golomb Jimenez Moskowitz Santos Chen Griswold Jones Neyman Shaw David Hein Kiefer O’Brien Thompson Deming Hernandez Klotz Pearl Utts Elashoff Holland Liu Potter Vlasic

and the faculty members are

Andrews Fernandez Kim Moore Rabinowitz Besicovitch Gupta Lightman Phillips Yang

The club can send seven students and three faculty members to a convention and decides to choose those who will go by random selection. Select a stratified random sample of seven students and three faculty members.

3.68 Stratified samples for accounting audits. Accountants use stratified samples during audits to verify a company’s records of such things as accounts receivable. The stratification is based on the dollar amount of the item and often includes

100% sampling of the largest items. One company reports 5000 accounts receivable. Of these, 100 are in amounts over $50,000; 500 are in amounts between $1000 and $50,000; and the remaining 4400 are in amounts under $1000. Using these groups as strata, you decide to verify all of the largest accounts and to sample 5% of the midsize accounts and 1% of the small accounts. How would you label the two strata from which you will sample? Use software or Table B, starting at line 125, to select the first six accounts from each of these strata.

3.69 The sampling frame. The list of individuals from which a sample is actually selected is called the sampling frame. Ideally, the frame should list every individual in the population, but in practice this is often difficult. A frame that leaves out part of the population is a common source of undercoverage.

(a) Suppose that a sample of households in a community is selected at random from the telephone directory. What households are omitted from this frame? What types of people do you think are likely to live in these households? These people will probably be underrepresented in the sample.

(b) It is usual in telephone surveys to use random digit dialing equipment that selects the last four digits of a telephone number at random after being given the area code and the exchange. The exchange is the first three digits of the telephone number. Which of the households that you mentioned in your answer to part (a) will be included in the sampling frame by random digit dialing?

3.70 Survey questions. Comment on each of the following as a potential sample survey question. Is the question clear? Is it slanted toward a desired response?

(a) “Some cell phone users have developed brain cancer. Should all cell phones come with a warning label explaining the danger of using cell phones?”

(b) “Do you agree that a national system of health insurance should be favored because it would provide health insurance for everyone and would reduce administrative costs?”

(c) “In view of escalating environmental degradation and incipient resource depletion, would you favor economic incentives for recycling of resource-intensive consumer goods?”

3.4 Ethics

When you complete this section, you will be able to:

Describe the purpose of an institutional review board and describe what kinds of expertise its members require. Describe informed consent and evaluate whether or not it has been given in specific examples. Determine when data have been kept confidential in a study. Evaluate a clinical trial from the viewpoint of ethics.

The production and use of data, like all human endeavors, raise ethical questions. We won’t discuss the telemarketer who begins a telephone sales pitch with “I’m conducting a survey.” Such deception is clearly unethical. It enrages legitimate survey organizations, which find the public less willing to talk with them. Neither will we discuss those few researchers who, in the pursuit of professional advancement, publish fake data. There is no ethical question here— faking data to advance your career is just wrong. It will end your career when uncovered.

But just how honest must researchers be about real, unfaked data? Here is an example that suggests the answer is “More honest than they often are.”

EXAMPLE 3.31

Provide all the critical information. Papers reporting scientific research are supposed to be short, with no extra baggage. But brevity can allow the researchers to avoid complete honesty about their data. Did they choose their subjects in a biased way? Did they report data on only some of their subjects? Did they try several statistical analyses and report only the ones that looked best?

The statistician John Bailar screened more than 4000 medical papers in more than a decade as consultant to the New England Journal of Medicine. He says, “When it came to the statistical review, it was often clear that critical information was lacking, and the gaps nearly always had the practical effect of making the authors’ conclusions look stronger than they should have.”24 The situation is no doubt worse in fields that screen published work less carefully.

The most complex issues of data ethics arise when we collect data from people. The ethical difficulties are more severe for experiments that impose some treatment on people than for sample surveys that simply gather information. Trials of new medical treatments, for example, can do harm as well as good to their subjects. Here are some basic standards of data ethics that must be obeyed by any study that gathers data from human subjects, whether sample survey or experiment.

BASIC DATA ETHICS

The organization that carries out the study must have an institutional review board that reviews all planned studies in advance in order to protect the subjects from possible harm.

All individuals who are subjects in a study must give their informed consent before data are collected.

All individual data must be kept confidential. Only statistical summaries for groups of subjects may be made public.

The law requires that studies funded by the federal government obey these principles. But neither the law nor the consensus of experts is completely clear about the details of their application.

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Institutional review boards The purpose of an institutional review board is not to decide whether a proposed study will produce valuable information or whether it is statistically sound. The board’s purpose is, in the words of one university’s board, “to protect the rights and welfare of human subjects (including patients) recruited to participate in research activities.” When protocols are greater than minimal risk, a statistician is often included on the board to help determine benefits.

The board reviews the plan of the study and can require changes. It reviews the consent form to be sure that subjects are informed about the nature of the study and about any potential risks. Once research begins, the board monitors its progress at least once a year.

The most pressing issue concerning institutional review boards is whether their workload has become so large that their effectiveness in protecting subjects drops. There are shorter review procedures for projects that involve only minimal risks to subjects, such as most sample surveys. When a board is overloaded, there is a temptation to put more proposals in the minimal-risk category to speed the work.

USE YOUR KNOWLEDGE

The exercises in this section on ethics are designed to help you think about the issues that we are discussing and to formulate some opinions. In general, there are no wrong or right answers, but you need to give reasons for your answers.

Do these proposals involve minimal risk? You are a member of your college’s institutional review board. You must decide whether several research proposals qualify for lighter review because they involve only minimal risk to subjects. Federal regulations say that “minimal risk” means that the risks are no greater than “those ordinarily encountered in daily life or during the performance of routine physical or psychological examinations or tests.” That’s vague. Which of these do you think qualifies as “minimal risk”? Explain your choices.

(a) Draw a drop of blood by pricking a finger in order to measure blood sugar.

(b) Draw blood from the arm for a full set of blood tests.

(c) Insert a tube that remains in the arm so that blood can be drawn regularly.

Who should be on an institutional review board? Government regulations require that institutional review boards consist of at least five people, including at least one scientist, one nonscientist, and one person from outside the institution. Most boards are larger, but many contain just one outsider.

(a) Why should review boards contain people who are not scientists?

(b) Do you think that one outside member is enough? How would you choose that member? (For example, would you prefer a medical doctor? A member of the clergy? An activist for patients’ rights?)

Informed consent Both words in the phrase “informed consent” are important, and both can be controversial. Subjects must be informed in advance about the nature of a study and any risk of harm it may bring. In the case of a sample survey, physical harm is not possible. The subjects should be told what kinds of questions the survey will ask and about how much of their time it will take. Experimenters must tell subjects the nature and purpose of the study and outline possible risks. Subjects must then consent in writing.

EXAMPLE 3.32

Who can give informed consent? Are there some subjects who can’t give informed consent? It was once common, for example, to test new vaccines on prison inmates who gave their consent in return for good-behavior credit. Now we worry that prisoners are not really free to refuse, and the law forbids medical experiments in prisons.

Young children can’t give fully informed consent, so the usual procedure is to ask their parents. A study of new ways to teach reading is about to start at a local elementary school, so the study team sends consent forms home to parents. Many parents don’t return the forms. Can their children take part in the study because the parents did not say No, or should we allow only children whose parents returned the form and said Yes?

What about research into new medical treatments for people with mental disorders? What about studies of new ways to help emergency room patients who may be unconscious or have suffered a stroke? In most cases, there is not time even to get the consent of the family. Does the principle of informed consent bar realistic trials of new treatments for unconscious patients?

These are questions without clear answers. Reasonable people differ strongly on all of them. There is nothing simple about informed consent.25

The difficulties of informed consent do not vanish even for capable subjects. Some researchers, especially in medical trials, regard consent as a barrier to getting patients to participate in research. They may not explain all possible risks; they may not point out that there are other therapies that might be better than those being studied; they may be too optimistic in talking with patients even when the consent form has all the right details.

On the other hand, mentioning every possible risk leads to very long consent forms that really are barriers. “They are like rental car contracts,” one lawyer said. Some subjects don’t read forms that run five or six printed pages. Others are frightened by the large number of possible (but unlikely) disasters that might happen and so refuse to participate. Of course, unlikely disasters sometimes happen. When they do, lawsuits follow and the consent forms become yet longer and more detailed.

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Confidentiality confidentiality

Ethical problems do not disappear once a study has been cleared by the review board, has obtained consent from its subjects, and has actually collected data about the subjects. Confidentiality means that only the researchers can identify responses of individual subjects. The report of an opinion poll may say what percent of the 1500 respondents felt that legal immigration should be reduced. It may not report what you said about this or any other issue.

anonymity

Confidentiality is not the same as anonymity. Anonymity means that subjects are anonymous—their names are not known even to the director of the study. Anonymity is rare in statistical studies. Even where anonymity is possible (mainly in surveys conducted by mail), it prevents any follow-up to improve nonresponse or inform subjects of results.

Any breach of confidentiality is a serious violation of data ethics. The best practice is to separate the identity of the subjects from the rest of the data at once. Sample surveys, for example, use the identification only to check on who did or did not respond. In an era of advanced technology, however, it is no longer enough to be sure that each individual set of data protects people’s privacy.

The government, for example, maintains a vast amount of information about citizens in many separate databases— census responses, tax returns, Social Security information, data from surveys such as the Current Population Survey, and so on. Many of these databases can be searched by computers for statistical studies.

A clever computer search of several databases might be able, by combining information, to identify you and learn a great deal about you even if your name and other identification have been removed from the data available for search. A colleague from Germany once remarked that “female full professor of statistics with a PhD from the United States” was enough to identify her among all the citizens of Germany. Privacy and confidentiality of data are hot issues among statisticians in the computer age.

EXAMPLE 3.33

Data collected by the government. Citizens are required to give information to the government. Think of tax returns and Social Security contributions. The government needs these data for administrative purposes—to see if we paid the right amount of tax and how large a Social Security benefit we are owed when we retire. Some people feel that individuals should be able to forbid any other use of their data, even with all identification removed. This would prevent using government records to study, say, the ages, incomes, and household sizes of Social Security recipients. Such a study could well be vital to debates on reforming Social Security.

USE YOUR KNOWLEDGE

How can we obtain informed consent? A researcher suspects that traditional religious beliefs tend to be associated with an authoritarian personality. She prepares a questionnaire that measures authoritarian tendencies and also asks many religious questions. Write a description of the purpose of this research to be read by subjects in order to obtain their informed consent. You must balance the conflicting goals of not deceiving the subjects as to what the questionnaire will tell about them and of not biasing the sample by scaring off religious people.

Should we allow this personal information to be collected? In which of the following circumstances would you allow collecting personal information without the subjects’ consent?

(a) A government agency takes a random sample of income tax returns to obtain information on the average income of people in different occupations. Only the incomes and occupations are recorded from the returns, not the names.

(b) A social psychologist attends public meetings of a religious group to study the behavior patterns of members.

(c) A social psychologist pretends to be converted to membership in a religious group and attends private meetings to study the behavior patterns of members.

Clinical trials Clinical trials are experiments that study the effectiveness of medical treatments on actual patients. Medical treatments can harm as well as heal, so clinical trials spotlight the ethical problems of experiments with human subjects. Here are the starting points for a discussion:

Randomized comparative experiments are the only way to see the true effects of new treatments. Without them, risky treatments that are no better than placebos will become common. Clinical trials produce great benefits, but most of these benefits go to future patients. The trials also pose risks, and these risks are borne by the subjects of the trial. So we must balance future benefits against present risks. Both medical ethics and international human rights standards say that “the interests of the subject must always prevail over the interests of science and society.”

The quoted words are from the 1964 Helsinki Declaration of the World Medical Association, the most respected international standard. The most outrageous examples of unethical experiments are those that ignore the interests of the subjects.

EXAMPLE 3.34

The Tuskegee study. In the 1930s, syphilis was common among black men in the rural South, a group that had almost no access to medical care. The Public Health Service Tuskegee study recruited 399 poor black sharecroppers with syphilis and 201 others without the disease in order to observe how syphilis progressed when no treatment was given. Beginning in 1943, penicillin became available to treat syphilis. The study subjects were not treated. In fact, the Public Health Service prevented any treatment until word leaked out and forced an end to the study in the 1970s.

The Tuskegee study is an extreme example of investigators following their own interests and ignoring the well- being of their subjects. A 1996 review said, “It has come to symbolize racism in medicine, ethical misconduct in human research, paternalism by physicians, and government abuse of vulnerable people.” In 1997, President Clinton formally apologized to the surviving participants in a White House ceremony.26

Because “the interests of the subject must always prevail,” medical treatments can be tested in clinical trials only when there is reason to hope that they will help the patients who are subjects in the trials. Future benefits aren’t enough to justify experiments with human subjects. Of course, if there is already strong evidence that a treatment works and is safe, it is unethical not to give it.

Here is the view of Dr. Charles Hennekens of the Harvard Medical School, who directed the large clinical trial that showed that aspirin reduces the risk of heart attacks:27 A clinical trial is justified if there is some evidence that the treatment will be effective. This evidence, however, is not sufficiently strong, to conclude that we would be harming the subjects who would receive the placebo.

Why is it ethical to give a control group of patients a placebo? Well, we know that placebos often work. What is more, placebos have no harmful side effects. In fact, the placebo group may be getting a better treatment than the drug group. If we knew which treatment was better, we would give it to everyone. When we don’t know, it is ethical to try

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both and compare them. The idea of using a control or placebo is a fundamental principle to be considered in designing experiments. In

many situations, deciding what to use as an appropriate control requires some careful thought.

The choice of the control can have a substantial impact on how the results of an experiment are interpreted. Here is an example.

EXAMPLE 3.35

Attentiveness improves by nearly 20%. The manufacturer of a breakfast cereal designed for children claims that eating this cereal has been clinically shown to improve attentiveness by nearly 20%. The study used two groups of children who were tested before and after breakfast. One group received the cereal for breakfast, while breakfast for the control group was water. The results of the tests taken three hours after breakfast were used in the claim.

The Federal Trade Commission investigated the marketing of this product. It charged that the claim was false and violated federal law. The charges were settled and the company agreed to not use misleading claims in its advertising.28

It is not sufficient to obtain appropriate controls. The data must be collected from all groups in the same way. Here is an example of this type of flawed design.

EXAMPLE 3.36

Accurate identification of ovarian cancer. Two scientists published a paper claiming to have developed an exciting new method to detect ovarian cancer using blood samples. When other scientists were unable to reproduce the results in different labs, the original work was examined more carefully. In the original study, there were samples for women with ovarian cancer and for healthy controls. The blood samples were all analyzed using a mass spectrometer. The control samples were analyzed on one day, and the cancer samples were analyzed on the next day. This design was flawed in that it could not control for changes over time in the measuring instrument.29

USE YOUR KNOWLEDGE

Is this study ethical? Researchers on aging proposed to investigate the effect of supplemental health services on the quality of life of older people. Eligible patients on the rolls of a large medical clinic were to be randomly assigned to treatment and control groups. The treatment group would be offered hearing aids, dentures, transportation, and other services not available without charge to the control group. The review board felt that providing these services to some but not other persons in the same institution raised ethical questions. Do you agree?

Should the treatments be given to everyone? Effective drugs for treating AIDS are very expensive, so most African nations cannot afford to give them to large numbers of people. Yet AIDS is more common in parts of Africa than anywhere else. Several clinical trials are looking at ways to prevent pregnant mothers infected with HIV from passing the infection to their unborn children, a major source of HIV infections in Africa. Some people say these trials are unethical because they do not give effective AIDS drugs to their subjects, as would be required in rich nations. Others reply that the trials are looking for treatments that can work in the real world in Africa and that they promise benefits at least to the children of their subjects. What do you think?

Behavioral and social science experiments When we move from medicine to the behavioral and social sciences, the direct risks to experimental subjects are less acute, but so are the possible benefits to the subjects. Consider, for example, the experiments conducted by psychologists in their study of human behavior.

EXAMPLE 3.37

Personal space. Psychologists observe that people have a “personal space” and get annoyed if others come too close to them. We don’t like strangers to sit at our table in a coffee shop if other tables are available, and we see people move apart in elevators if there is room to do so. Americans tend to require more personal space than people in most other cultures. Can violations of personal space have physical, as well as emotional, effects?

Investigators set up shop in a men’s public restroom. They blocked off urinals to force men walking in to use either a urinal next to an experimenter (treatment group) or a urinal separated from the experimenter (control group). Another experimenter, using a periscope from a toilet stall, measured how long the subject took to start urinating and how long he kept at it.30

This personal space experiment illustrates the difficulties facing those who plan and review behavioral studies:

There is no risk of harm to the subjects, although they would certainly object to being watched through a periscope. What should we protect subjects from when physical harm is unlikely? Possible emotional harm? Undignified situations? Invasion of privacy? What about informed consent? The subjects in Example 3.37 did not even know they were participating in an experiment. Many behavioral experiments rely on hiding the true purpose of the study. The subjects would change their behavior if told in advance what the investigators were looking for. Subjects are asked to consent on the basis of vague information. They receive full information only after the experiment.

The “Ethical Principles” of the American Psychological Association require consent unless a study merely observes behavior in a public place. They allow deception only when it is necessary to the study, does not hide information that might influence a subject’s willingness to participate, and is explained to subjects as soon as possible. The personal space study (from the 1970s) does not meet current ethical standards.

We see that the basic requirement for informed consent is understood differently in medicine and psychology. Here is an example of another setting with yet another interpretation of what is ethical. The subjects get no information and give no consent. They don’t even know that an experiment may be sending them to jail for the night.

EXAMPLE 3.38

Domestic violence. How should police respond to domestic violence calls? In the past, the usual practice was to remove the offender and order him to stay out of the household overnight. Police were reluctant to make arrests because the victims rarely pressed charges. Women’s groups argued that arresting offenders would help prevent future violence even if no charges were filed. Is there evidence that arrest will reduce future offenses? That’s a question that experiments have tried to answer.

A typical domestic violence experiment compares two treatments: arrest the suspect and hold him overnight, or warn the suspect and release him. When police officers reach the scene of a domestic violence call, they calm the participants and investigate. Weapons or death threats require an arrest. If the facts permit an arrest but do not require it, an officer radios headquarters for instructions. The person on duty opens the next envelope in a file prepared in advance by a statistician. The envelopes contain the treatments in random order. The police either arrest the suspect or warn and release him, depending on the contents of the envelope. The researchers then watch police records and visit the victim to see if the domestic violence reoccurs.

The first such experiment appeared to show that arresting domestic violence suspects does reduce their future violent behavior. As a result of this evidence, arrest has become the common police response to domestic violence.

The domestic violence experiments shed light on an important issue of public policy. Because there is no informed consent, the ethical rules that govern clinical trials and most social science studies would forbid these experiments.

They were cleared by review boards because, in the words of one domestic violence researcher, “These people became subjects by committing acts that allow the police to arrest them. You don’t need consent to arrest someone.”

SECTION 3.4 SUMMARY Approval of an institutional review board is required for studies that involve humans or animals as subjects. Human subjects must give informed consent if they are to participate in experiments. Data on human subjects must be kept confidential.

SECTION 3.4 EXERCISES For Exercises 3.71 and 3.72, see pages 204–205; for Exercises 3.73 and 3.74, see page 207; and for Exercises 3.75 and 3.76, see page 209.

3.77 Did you give informed consent? You were asked to participate in a study by a friend who is recruiting subjects. You trust your friend and you tell her that you are willing to do whatever is needed for the study. Have you given informed consent? Explain your answer.

3.78 Are the data confidential? You have participated in a study, and the results were published in an article in a very prestigious journal. Only summary information was published. The policy of the journal requires that all data used in the articles it publishes be available to the public, and it archives the data on a website. When you examine the data, you realize that you have a unique set of characteristics that would allow someone who knows you very well to identify which data are from you. Someone who does not know you would not be able to do this. Are the data confidential? Explain your answer.

3.79 Is the IRB responsible? An institutional review board (IRB) approved an experimental cancer vaccine for use in a clinical trial. The subjects were patients who had advanced disease and had received standard treatments with no success. Of the 94 subjects who received the vaccine, 26 died during the study. Their deaths were not due to the vaccine. Some family members of the subjects sued the hospital, the study director, the company that made the vaccine, a university official, individual members of the IRB, and the university bioethicist who consulted with the IRB.31 Discuss this case from the point of view of ethics. Include any additional information that you would need to form your opinion.

3.80 What is wrong? Explain what is wrong in each of the following scenarios.

(a) Clinical trials are always ethical as long as they randomly assign patients to the treatments.

(b) The job of an institutional review board is complete when it decides to allow a study to be conducted.

(c) A treatment that has no risk of physical harm to subjects is always ethical.

3.81 How should the samples have been analyzed? Refer to the ovarian cancer diagnostic test study in Example 3.36 (page 209). Describe how you would process the samples through the mass spectrometer.

3.82 The Vytorin controversy. Vytorin is a combination pill designed to lower cholesterol. It consists of a relatively inexpensive and widely used drug, Zocor, and a newer drug called Zetia. Early study results suggested that Vytorin was no more effective than Zetia. Critics claimed that the makers of the drugs tried to change the response variable for the study, and two congressional panels investigated why there was a two-year delay in the release of the results. Use the Internet to search for more information about this controversy and write a report about what you find. Include an evaluation in the framework of ethical use of experiments and data. A good place to start your search would be to look for the phrase “Vytorin’s shortcomings.”

3.83 The General Social Survey. One of the most important nongovernment surveys in the United States is the National Opinion Research Center’s General Social Survey. The GSS regularly monitors public opinion on a wide variety of political and social issues. Interviews are conducted in person in the subject’s home. Are a subject’s responses to GSS questions anonymous, confidential, or both? Explain your answer.

3.84 Anonymity and confidentiality in health screening. Texas A&M, like many universities, offers free screening for HIV, the virus that causes AIDS. The announcement says, “Persons who sign up for the HIV Screening will be assigned a number so that they do not have to give their name.” They can learn the results of the test by telephone, still without giving their name. Does this practice offer anonymity or just confidentiality?

3.85 Anonymity and confidentiality in mail surveys. Some common practices may appear to offer anonymity while actually delivering only confidentiality. Market researchers often use mail surveys that do not ask the respondent’s identity but contain hidden codes on the questionnaire that identify the respondent. A false claim of anonymity is clearly unethical. If only confidentiality is promised, is it also unethical to say nothing about the identifying code, perhaps causing respondents to believe their replies are anonymous?

3.86 Use of stored blood. Long ago, doctors drew a blood specimen from you as part of treating minor anemia. Unknown to you, the sample was stored. Now researchers plan to use stored samples from you and many other people to look for genetic factors that may influence anemia. It is no longer possible to ask your consent. Modern technology can read your entire genetic makeup from the blood sample.

(a) Do you think it violates the principle of informed consent to use your blood sample if your name is on it but you were not told that it might be saved and studied later?

(b) Suppose that your identity is not attached. The blood sample is known only to come from (say) “a 20-year-old white female being treated for anemia.” Is it now OK to use the sample for research?

(c) Perhaps we should use biological materials such as blood samples only from patients who have agreed to allow the material to be stored for later use in research. It isn’t possible to say in advance what kind of research, so this falls short of the usual standard for informed consent. Is it nonetheless acceptable, given complete confidentiality and the fact that using the sample can’t physically harm the patient?

3.87 Political polls. The presidential election campaign is in full swing, and the candidates have hired polling organizations to take regular polls to find out what the voters think about the issues.

(a) What information should the pollsters be required to give out?

(b) What does the standard of informed consent require the pollsters to tell potential respondents?

(c) The standards accepted by polling organizations also require giving respondents the name and address of the organization that carries out the poll. Why do you think this is required?

(d) The polling organization usually has a professional name such as “Samples Incorporated,” so respondents don’t know that the poll is being paid for by a political party or candidate. Would revealing the sponsor to respondents bias the poll? Should the sponsor always be announced whenever poll results are made public?

3.88 Should poll results be made public? Some people think that the law should require that all political results be made public. Otherwise, the possessors of poll results can use the information to their own advantage. They can act on the information, release only selected parts of it, or time the release for best effect. A candidate’s organization replies that it is paying for the poll to gain information for its own use, not to amuse the public. Do you favor requiring complete disclosure of political poll results? What about other private surveys, such as market research surveys of consumer tastes?

3.89 Informed consent to take blood samples. Researchers from Yale, working with medical teams in Tanzania, wanted to know how common infection with the AIDS virus is among pregnant women in that country. To do this, they planned to test blood samples drawn from pregnant women. Yale’s institutional review board insisted that the researchers get the informed consent of each woman and tell her the results

of the test. This is the usual procedure in developed nations. The Tanzanian government did not want to tell the women why blood was drawn or tell them the test results. The government feared panic if many people turned out to have an incurable disease for which the country’s medical system could not provide care. The study was canceled. Do you think that Yale was right to apply its usual standards for protecting subjects?

CHAPTER 3 EXERCISES

3.90 Visit statistics and the news. STATS is an organization concerned about the appropriate reporting of statistical knowledge in the news media. Visit the website stats.org/blog. Some recent postings include discussions of deflategate, health claims for coffee, and the graph that launched a thousand news stories. Read one of the articles posted on this site and then write a short report summarizing the major ideas in the article.

3.91 Online behavioral advertising. The Federal Trade Commission Staff Report, “Self-Regulatory Principles for Online Behavioral Advertising,” defines behavioral advertising as “the tracking of a consumer’s online activities over time—including the searches the consumer has conducted, the Web pages visited and the content viewed—in order to deliver advertising targeted to the individual consumer’s interests.” The report suggests four governing concepts for their proposals. These are (1) transparency and control: when companies collect information from consumers for advertising, they should tell consumers how the data will be collected, and consumers should be given a choice about whether to allow the data to be collected; (2) security and data retention: data should be kept secure and should be retained only as long as they are needed; (3) privacy: before data are used in a way that differs from promises made when they were collected, consent should be obtained from the consumer; and (4) sensitive data: affirmative express consent should be obtained before using any sensitive data.32 Write a report discussing your opinions concerning online behavioral advertising and the four governing concepts. Pay particular attention to issues related to the ethical collection and use of statistical data.

3.92 Confidentiality at NORC. The National Opinion Research Center conducts a large number of surveys and has established procedures for protecting the confidentiality of its survey participants. For its Survey of Consumer Finances, NORC provides a pledge to participants regarding confidentiality. This pledge is available at scf.norc.org/confidentiality.html. Review the pledge and summarize its key parts. Do you think that the pledge adequately addresses issues related to the ethical collection and use of data? Explain your answer.

3.93 Make it an experiment! In the following observational studies, describe changes that could be made to the data collection process that would result in an experiment rather than an observational study. Also, offer suggestions about unseen biases or lurking variables that may be present in the studies as they are described here.

(a) A friend of yours likes to play Texas hold ’em. Every time that he tells you about his playing, he says that he won.

(b) In an introductory statistics class, you notice that the students who sit in the first two rows of seats had higher scores on the first exam than the other students in the class.

3.94 Name the designs. What is the name for each of these study designs?

(a) A study to compare two methods of preserving wood started with boards of southern white pine. Each board was ripped from end to end to form two edge-matched specimens. One was assigned to Method A; the other, to Method B.

(b) A survey on youth and smoking contacted by telephone 300 smokers and 300 nonsmokers, all 14 to 22 years of age.

(c) Does air pollution induce DNA mutations in mice? Starting with 40 male and 40 female mice, 20 of each sex were housed in a polluted industrial area downwind from a steel mill. The other 20 of each sex were housed at an unpolluted rural location 30 kilometers away.

3.95 Price promotions and consumer expectations. A researcher studying the effect of price promotions on consumer expectations makes up two different histories of the store price of a hypothetical brand of laundry detergent for the past year. Students in a marketing course view one or the other price history on a computer. Some students see a steady price, while others see regular promotions that temporarily cut the price. The students are then asked what price they would expect to pay for the detergent. Is this study an experiment? Why? What are the explanatory and response variables?

3.96 Calcium and healthy bones. Adults need to eat foods or supplements that contain enough calcium to maintain healthy bones. Calcium intake is generally measured in milligrams per day (mg/d), and one measure of healthy bones is total body bone mineral density measured in grams per centimeter squared (TBBMD, g/cm2). Suppose that you want to study the relationship between calcium intake and TBBMD.

(a) Design an observational study to study the relationship.

(b) Design an experiment to study the relationship.

(c) Compare the relative merits of your two designs. Which do you prefer? Give reasons for your answer.

3.97 Choose the type of study. Give an example of a question about pets and their owners, their behavior, or their opinions that would best be answered by

(a) a sample survey.

(b) an observational study that is not a sample survey.

(c) an experiment.

3.98 Compare the fries. Do consumers prefer the fries from Burger King or from McDonald’s? Design a blind test in which the source of the fries is not identified. Describe briefly the design of a matched pairs experiment to investigate this question. How will you use randomization?

3.99 Bicycle gears. How does the time it takes a bicycle rider to travel 100 meters depend on which gear is used and how steep the course is? It may be, for example, that higher gears are faster on level ground, but lower gears are faster on steep inclines. Discuss the design of a two-factor experiment to investigate this issue, using one bicycle with three gears and one rider. How will you use randomization?

3.100 Design an experiment. The previous two exercises illustrate the use of statistically designed experiments to answer questions that arise in everyday life. Select a question of interest to you that an experiment

might answer, and carefully discuss the design of an appropriate experiment.

3.101 Design a survey. You want to investigate the attitudes of students at your school about the faculty’s commitment to teaching. The student government will pay the costs of contacting about 500 students.

(a) Specify the exact population for your study; for example, will you include part-time students?

(b) Describe your sample design. Will you use a stratified sample?

(c) Briefly discuss the practical difficulties that you anticipate; for example, how will you contact the students in your sample?

3.102 Compare two doses of a drug. A drug manufacturer is studying how a new drug behaves in patients. Investigators compare two doses: 5 milligrams (mg) and 10 mg. The drug can be administered by injection, by a skin patch, or by intravenous drip. Concentration in the blood after 30 minutes (the response variable) may depend both on the dose and on the method of administration.

(a) Make a sketch that describes the treatments formed by combining dose and method. Then use a diagram to outline a completely randomized design for this two-factor experiment.

(b) “How many subjects?” is a tough issue. We will explain the basic ideas in Chapter 6. What can you say now about the advantage of using larger groups of subjects?

3.103 Would the results be different for men and women? The drug that is the subject of the experiment in Exercise 3.102 may behave differently in men and women. How would you modify your experimental design to take this into account?

3.104 Informed consent. The requirement that human subjects give their informed consent to participate in an experiment can greatly reduce the number of available subjects. For example, a study of new teaching methods

asks the consent of parents for their children to be randomly assigned to be taught by either a new method or the standard method. Many parents do not return the forms, so their children must continue to be taught by the standard method. Why is it not correct to consider these children as part of the control group along with children who are randomly assigned to the standard method?

3.105 Two ways to ask sensitive questions. Sample survey questions are usually read from a computer screen. In a Computer-aided personal interview (CAPI), the interviewer reads the questions and enters the

responses. In a Computer-aided self interview (CASI), the interviewer stands aside and the respondent reads the questions and enters responses. One method almost always shows a higher percent of subjects admitting use of illegal drugs. Which method? Explain why.

3.106 Your institutional review board. Your college or university has an institutional review board that screens all studies that use human subjects. Get a copy of the document that describes this board (you can probably find it online).

(a) According to this document, what are the duties of the board?

(b) How are members of the board chosen? How many members are not scientists? How many members are not employees of the college? Do these members have some special expertise, or are they simply members of the “general public”?

3.107 Use of data produced by the government. Data produced by the government are often available free or at low cost to private users. For example, satellite weather data produced by the U.S. National Weather Service are available free to TV stations for their weather reports and to anyone on the Internet. Opinion 1: Government data should be available to everyone at minimal cost. Opinion 2: The satellites are expensive, and the TV stations are making a profit from their weather services, so they should share the cost. European governments, for example, charge TV stations for weather data. Which opinion do you support, and why?

3.108 Should we ask for the consent of the parents? The Centers for Disease Control and Prevention, in a survey of teenagers, asked the subjects if they were sexually active. Those who said Yes were then asked, “How old were you when you had sexual intercourse for the first time?” Should consent of parents be required to ask minors about sex, drugs, and other such issues, or is consent of the minors themselves enough? Give reasons for your opinion.

3.109 A theft experiment. Students sign up to be subjects in a psychology experiment. When they arrive, they are told that interviews are running late and are taken to a waiting room. The experimenters then stage a theft of a valuable object left in the waiting room. Some subjects are alone with the thief, and others are in pairs—these are the treatments being compared. Will the subject report the theft? The students had agreed to take part in an unspecified study, and the true nature of the experiment is explained to them afterward. Do you think this study is ethically OK?

3.110 A cheating experiment. A psychologist conducts the following experiment. She measures the attitude of subjects toward cheating and then has them play a game rigged so that winning without cheating is impossible. The computer that organizes the game also records—unknown to the subjects—whether or not they cheat. Then attitude toward cheating is retested. Subjects who cheat tend to change their attitudes to find cheating more acceptable. Those who resist the temptation to cheat tend to condemn cheating more strongly on the second test of attitude. These results confirm the psychologist’s theory. This experiment tempts subjects to cheat. The subjects are led to believe that they can cheat secretly when, in fact, they are observed. Is this experiment ethically objectionable? Explain your position.

1. See the news release of June 24, 2015, concerning the 2014 results for the American Time Use Survey, Table 11, at bls.gov/news.release/pdf/atus.pdf.

2. See norc.uchicago.edu.

3. Stewart Warden et al., “Throwing induces substantial torsional adaption within the midshaft humerus of male baseball players,” Bone, 45 (2009), pp. 931–941. The data were provided by Stewart Warden, Department of Physical Therapy, School of Health and Rehabilitation Sciences, Indiana University.

4. Corby C. Martin et al., “Children in school cafeterias select foods containing more saturated fat and energy than the Institute of Medicine recommendations,” Journal of Nutrition, 140 (2010), pp. 1653–1660.

5. Based on “Look, no hands: Automatic soap dispensers,” Consumer Reports, February 2013, p. 11.

6. From “Did you know,” Consumer Reports, February 2013, p. 10.

7. Bruce Barrett et al., “Echinacea for treating the common cold,” Annals of Internal Medicine, 153 (2010), pp. 769–777.

8. For a full description of the STAR program and its follow-up studies, go to heros-inc.org/star.htm.

9. See Note 6.

10. Based on Gerardo Ramirez and Sian L. Beilock, “Writing about testing worries boosts exam performance in the classroom,” Science, 331 (2011), p. 2011. Although we describe the experiment as not including a control group, the researchers who conducted this study did, in fact, use one.

11. A general discussion of failures of blinding is Dean Ferguson et al., “Turning a blind eye: The success of blinding reported in a random sample of randomised, placebo controlled trials,” British Medical Journal, 328 (2004), p. 432.

12. Based on a study conducted by Sandra Simonis under the direction of Professor Jon Harbor from the Purdue University Department of Earth, Atmospheric, and Planetary Sciences.

13. Based on a study conducted by Tammy Younts directed by Professor Deb Bennett of the Purdue University Department of Educational Studies. For more information about Reading Recovery, see readingrecovery.org/.

14. Based on a study conducted by Rajendra Chaini under the direction of Professor Bill Hoover of the Purdue University Department of Forestry and Natural Resources.

15. From the Hot Rock Songs list at billboard.com for the week of September 5, 2015.

16. From the Hot 100 list at billboard.com for the week of September 5, 2015.

17. From the online version of the Bureau of Labor Statistics, Handbook of Methods, modified April 17, 2003, at bls.gov. The details of the design are more complicated than we describe.

18. For more detail on the material of this section and complete references, see P. E. Converse and M. W. Traugott, “Assessing the accuracy of polls and surveys,” Science, 234 (1986), pp. 1094–1098.

19. From www.census.gov/programs-surveys/cps/technical-documentation/methodology/non-response-rates.html on January 29, 2013.

20. From www3.norc.org/GSS+Website/FAQs on January 29, 2013.

21. See pewresearch.org/about.

22. See “Assessing the representativeness of public opinion surveys,” May 15, 2012, from people-press.org/2012/05/15.

23. Sex: Tom W. Smith, “The JAMA controversy and the meaning of sex,” Public Opinion Quarterly, 63 (1999), pp. 385–400. Welfare: From a New York Times/CBS News Poll reported in the New York Times, July 5, 1992. Scotland: “All set for independence?” Economist, September 12, 1998. Many other examples appear in T. W. Smith, “That which we call welfare by any other name would smell sweeter,” Public Opinion Quarterly, 51 (1987), pp. 75–83.

24. John C. Bailar III, “The real threats to the integrity of science,” Chronicle of Higher Education, April 21, 1995, pp. B1–B2.

25. The difficulties of interpreting guidelines for informed consent and for the work of institutional review boards in medical research are a main theme of Beverly Woodward, “Challenges to human subject protections in U.S. medical research,” Journal of the American Medical Association, 282 (1999), pp. 1947–1952. The references in this paper point to other discussions.

26. Quotation from the Report of the Tuskegee Syphilis Study Legacy Committee, May 20, 1996. A detailed history is James H. Jones, Bad Blood: The Tuskegee Syphilis Experiment, Free Press, 1993.

27. Dr. Hennekens’s words are from an interview in the Annenberg/Corporation for Public Broadcasting video series Against All Odds: Inside Statistics.

28. See ftc.gov/opa/2009/04/kellogg.shtm.

29. On February 12, 2012, the CBS show 60 Minutes reported the latest news on this study, which was published in the Journal of Clinical Oncology in 2007. See cbsnews.com/video/watch/?id=7398476n.

30. R. D. Middlemist, E. S. Knowles, and C. F. Matter, “Personal space invasions in the lavatory: Suggestive evidence for arousal,” Journal of Personality and Social Psychology, 33 (1976), pp. 541–546.

31. From Randi Zlotnik Shaul et al., “Legal liabilities in research: Early lessons from North America,” BMJ Medical Ethics, 6, No. 4 (2005), pp. 1–4.

32. The report was issued in February 2009 and is available from www.ftc.gov/sites/default/files/documents/reports/federal-trade- commission-staff-report-self-regulatory-principles-online-behavioral-advertising/p085400behavadreport.pdf.

4.1 4.2 4.3 4.4 4.5

CHAPTER 4 Probability: The Study of

Randomness

Randomness Probability Models Random Variables Means and Variances of Random Variables General Probability Rules

Introduction The reasoning of statistical inference rests on asking, “How often would this method give a correct answer if I used it very many times?” When we produce data by random sampling or randomized comparative experiments, the laws of probability answer the question, “What would happen if we did this many times?” Games of chance like Texas hold

’em are exciting because the outcomes are determined by the rules of probability.

4.1 Randomness

When you complete this section, you will be able to:

Identify random phenomena. Interpret the term probability for particular examples. Identify trials as independent or not.

Toss a coin, or choose a simple random sample (SRS). The result can’t be predicted in advance because the result will vary when you toss the coin or choose the sample repeatedly. But there is, nonetheless, a regular pattern in the results, a pattern that emerges clearly only after many repetitions. This remarkable fact is the basis for the idea of probability.

EXAMPLE 4.1

Toss a coin 5000 times. When you toss a coin, there are only two possible outcomes, heads or tails. Figure 4.1 shows the results of tossing a coin 5000 times twice (Trial A and Trial B). For each number of tosses from 1 to 5000, we have plotted the proportion of those tosses that gave a head.

FIGURE 4.1 The proportion of tosses of a coin that give a head varies as we make more tosses. Eventually, however, the proportion approaches 0.5, the probability of a head. This figure shows the results of two trials of 5000 tosses each, Example 4.1.

Trial A (red line) begins tail, head, tail, tail. You can see that the proportion of heads for Trial A starts at 0 on the first toss, rises to 0.5 when the second toss gives a head, then falls to 0.33 and 0.25 as we get two more tails. Trial B (blue dotted line), on the other hand, starts with five straight heads, so the proportion of heads is 1 until the sixth toss.

The proportion of tosses that produce heads is quite variable at first. Trial A starts low and Trial B starts high. As we make more and more tosses, however, the proportion of heads for each trial gets close to 0.5 and stays there.

probability

If we made yet a third trial at tossing the coin a great many times, the proportion of heads would again settle down to 0.5 in the long run. We say that 0.5 is the probability of a head. The probability 0.5 appears as a horizontal line on the graph.

The Probability applet on the text website animates Figure 4.1. It allows you to choose the probability of a head and simulate any number of tosses of a coin with that probability. Try it. You will see that the proportion of heads gradually settles down close to the chosen probability. Equally important, you will also see that the proportion in a small or moderate number of tosses can be far from this probability. Probability describes only what happens in the long run. Most people expect chance outcomes to show more short-term regularity than is actually true.

EXAMPLE 4.2

Significance testing and Type I errors. In Chapter 6, we will learn about significance testing and Type I errors. When we perform a significance test, we have the possibility of making a Type I error under certain circumstances. The significance-testing procedure is set up so that the probability of making this kind of error is small, usually 5%. If we perform a large number of significance tests under this set of circumstances, the proportion of times that we will make a Type I error is 0.05.

fair coin

In the coin toss setting, the probability of a head is a characteristic of the coin being tossed. A coin is called fair if the probability of a head is 0.5; that is, it is equally likely to come up heads or tails. If we toss a coin five times and it comes up heads for all five tosses, we might suspect that the coin is not fair. Is this outcome likely if, in fact, the coin is fair? This is what happened in Trials A and B of Example 4.1. We will learn a lot more about significance testing in later chapters. For now, we are content with some very general ideas.

When the Type I error of a statistical significance procedure is set at 0.05, this probability is a characteristic of the procedure. If we roll a pair of dice once, we do not know whether the sum of the faces will be seven or not. Similarly, if we perform a significance test once, we do not know if we will make a Type I error or not. However, if the procedure is designed to have a Type I error probability of 0.05, then we are much less likely than not to make a Type I error.

4.1

The language of probability “Random” in statistics is not a synonym for “unpredictable” but a description of a kind of order that emerges in the long run. We often encounter the unpredictable side of randomness in our everyday experience, but we rarely see enough repetitions of the same random phenomenon to observe the long-term regularity that probability describes. You can see that regularity emerging in Figure 4.1. In the very long run, the proportion of tosses that give a head is 0.5. This is the intuitive idea of probability. Probability 0.5 means “occurs half the time in a very large number of trials.”

RANDOMNESS AND PROBABILITY

We call a phenomenon random if individual outcomes are uncertain but there is, nonetheless, a regular distribution of outcomes in a large number of repetitions.

The probability of any outcome of a random phenomenon is the proportion of times the outcome would occur in a very long series of repetitions.

Not all coins are fair. In fact, most real coins have bumps and imperfections that make the probability of heads a little different from 0.5. The probability might be 0.499999 or 0.500002. For our study of probability in this chapter, we will assume that we know the actual values of probabilities. Thus, we assume things like fair coins, even though we know that real coins are not exactly fair. We do this to learn what kinds of outcomes we are likely to see when we make such assumptions. When we study statistical inference in later chapters, we look at the situation from the opposite point of view: given that we have observed certain outcomes, what can we say about the probabilities that generated these outcomes?

USE YOUR KNOWLEDGE

Use Table B. We can use the random digits in Table B in the back of the book to simulate tossing a fair coin. Start at line 131 and read the numbers from left to right. If the number is 0, 2, 4, 6, or 8, you will say that the coin toss resulted in a head; if the number is a 1, 3, 5, 7, or 9, the outcome is tails. Use the first 10 random digits on line 131 to simulate 10 tosses of a fair coin. What is the actual proportion of heads in your simulated sample? Explain why you did not get exactly five heads.

Probability describes what happens in very many trials, and we must actually observe many trials to pin down a probability. In the case of tossing a coin, some diligent people have in fact made thousands of tosses.

EXAMPLE 4.3

Many tosses of a coin. The French naturalist Count Buffon (1707–1788) tossed a coin 4040 times. Result: 2048 heads, or proportion 2048/4040 = 0.5069 for heads.

Around 1900, the English statistician Karl Pearson heroically tossed a coin 24,000 times. Result: 12,012 heads, a proportion of 0.5005.

While imprisoned by the Germans during World War II, the South African statistician John Kerrich tossed a coin 10,000 times. Result: 5067 heads, proportion of heads 0.5067.

Thinking about randomness That some things are random is an observed fact about the world. The outcome of a coin toss, the time between emissions of particles by a radioactive source, and the sexes of the next litter of lab rats are all random. So is the outcome of a random sample or a randomized experiment. Probability theory is the branch of mathematics that describes random behavior. Of course, we can never observe a probability exactly. We could always continue tossing the coin, for example. Mathematical probability is an idealization based on imagining what would happen in an indefinitely long series of trials.

The best way to understand randomness is to observe random behavior—not only the long-run regularity but the unpredictable results of short runs. You can do this with physical devices such as coins and dice, but software simulations of random behavior allow faster exploration. As you explore randomness, remember:

independence

You must have a long series of independent trials. That is, the outcome of one trial must not influence the outcome of any other. Imagine a crooked gambling house where the operator of a roulette wheel can stop it where she chooses—she can prevent the proportion of “red” from settling down to a fixed number. These trials are not independent. The idea of probability is empirical. Simulations start with given probabilities and imitate random behavior, but we can estimate a real-world probability only by actually observing many trials. Nonetheless, simulations are very useful because we need long runs of trials. In situations such as coin tossing, the proportion of an outcome often requires several hundred trials to settle down to the probability of that outcome. The kinds of physical random devices suggested in the exercises are too slow to make performing so many trials practical. Short runs give only rough estimates of a probability.

The uses of probability Probability theory originated in the study of games of chance. Tossing dice, dealing shuffled cards, and spinning a roulette wheel are examples of deliberate randomization. In that respect, they are similar to random sampling. Although games of chance are ancient, they were not studied by mathematicians until the sixteenth and seventeenth centuries.

It is only a mild simplification to say that probability as a branch of mathematics arose when seventeenth-century French gamblers asked the mathematicians Blaise Pascal and Pierre de Fermat for help. Gambling is still with us, in casinos and state lotteries. We will make use of games of chance as simple examples that illustrate the principles of probability.

Careful measurements in astronomy and surveying led to further advances in probability in the eighteenth and nineteenth centuries because the results of repeated measurements are random and can be described by distributions much like those arising from random sampling. Similar distributions appear in data on human life span (mortality tables) and in data on lengths or weights in a population of skulls, leaves, or cockroaches.1

Now, we employ the mathematics of probability to describe the flow of traffic through a highway system, the Internet, or a computer processor; the genetic makeup of individuals or populations; the energy states of subatomic particles; the spread of epidemics or tweets; and the rate of return on risky investments. Although we are interested in probability because of its usefulness in statistics, the mathematics of chance is important in many fields of study.

SECTION 4.1 SUMMARY A random phenomenon has outcomes that we cannot predict but that nonetheless have a regular distribution in very many repetitions. The probability of an event is the proportion of times the event occurs in many repeated trials of a random phenomenon. Trials are independent if the outcome of one trial does not influence the outcome of any other trial.

SECTION 4.1 EXERCISES For Exercise 4.1, see page 218.

4.2 Are these phenomena random? Identify each of the following phenomena as random or not. Give reasons for your answers.

(a) The outside temperature in your town at noon on Groundhog Day, February 2.

(b) The first digit in your student identification number.

(c) You draw an ace from a well-shuffled deck of 52 cards.

4.3 Interpret the probabilities. Refer to the previous exercise. In each case, interpret the term probability for the phenomena that are random. For those that are not random, explain why the term probability does not apply.

4.4 Are the trials independent? For each of the following situations, identify the trials as independent or not. Explain your answers.

(a) You record the outside temperature in your town at noon on Groundhog Day, February 2, each year for the next five years.

(b) The number of tweets that you receive on the next 10 Mondays.

(c) Your grades in the five courses that you are taking this semester.

4.5 Winning at craps. The game of craps starts with a “come-out” roll, in which the shooter rolls a pair of dice. If the total of the “spots” on the up-faces is 7 or 11, the shooter wins immediately (there are ways that the shooter can win on later rolls if other numbers are rolled on the come-out roll). Roll a pair of dice 25 times and estimate the probability that the shooter wins immediately on the come-out roll. For a pair of perfectly made dice, the probability is 0.2222.

4.6 Use the Probability applet. The idea of probability is that the proportion of heads in many tosses of a balanced coin eventually gets close to 0.5. But does the actual count of heads get close to one-half the number of tosses? Let’s find

out. Set the “Probability of Heads” in the Probability applet to 0.5 and the number of tosses to 50. You can extend the number of tosses by clicking “Toss” again to get 50 more. Don’t click “Reset” during this exercise.

(a) After 50 tosses, what is the proportion of heads? What is the count of heads? What is the difference between the count of heads and 25 (one-half the number of tosses)?

(b) Keep going to 200 tosses. Again record the proportion and count of heads and the difference between the count and 100 (half the number of tosses).

(c) Keep going. Stop at 300 tosses and again at 400 tosses to record the same facts. Although it may take a long time, the laws of probability say that the proportion of heads will always get close to 0.5 and also that the difference between the count of heads and half the number of tosses will always grow without limit.

4.7 A question about dice. Here is a question that a French gambler asked the mathematicians Fermat and Pascal at the very beginning of probability theory: what is the probability of getting at least one 6 in rolling four dice? The Law of

Large Numbers applet allows you to roll several dice and watch the outcomes. (Ignore the title of the applet for now.) Because simulation—just like real random phenomena—often takes very many trials to estimate a probability accurately, let’s simplify the question: is this probability clearly greater than 0.5, clearly less than 0.5, or quite close to 0.5? Use the applet to roll four dice until you can confidently answer this question. You will have to set “Rolls” to 1 so that you have time to look at the four up-faces. Keep clicking “Roll dice” to roll again and again. How many times did you roll four dice? What percent of your rolls produced at least one 6?

4.2 Probability Models

When you complete this section, you will be able to:

Describe a sample space from a description of a random phenomenon. Apply the four probability rules. Identify random phenomena that have equally likely outcomes and distinguish them from those that do not.

probability model

The idea of probability as a proportion of outcomes in very many repeated trials guides our intuition but is hard to express in mathematical form. A description of a random phenomenon in the language of mathematics is called a probability model. To see how to proceed, think first about a very simple random phenomenon, tossing a coin once. When we toss a coin, we cannot know the outcome in advance. What do we know? We are willing to say that the outcome will be either heads or tails. Because the coin appears to be balanced, we believe that each of these outcomes has probability 1/2. This description of coin tossing has two parts:

A list of possible outcomes. A probability for each outcome.

This two-part description is the starting point for a probability model. We will begin by describing the outcomes of a random phenomenon and then learn how to assign probabilities to the outcomes.

Sample spaces A probability model first tells us what outcomes are possible.

SAMPLE SPACE

The sample space S of a random phenomenon is the set of all possible outcomes.

The name “sample space” is natural in random sampling, where each possible outcome is a sample and the sample space contains all possible samples. To specify S, we must state what constitutes an individual outcome and then state which outcomes can occur. We often have some freedom in defining the sample space, so the choice of S is a matter of convenience as well as correctness. The idea of a sample space, and the freedom we may have in specifying it, are best illustrated by examples.

EXAMPLE 4.4

Sample space for tossing a coin. Toss a coin. There are only two possible outcomes, and the sample space is

or, more briefly, S = {H, T}.

EXAMPLE 4.5

Sample space for random digits. Let your pencil point fall blindly into Table B of random digits. Record the value of the digit it lands on. The possible outcomes are

EXAMPLE 4.6

Sample space for tossing a coin four times. Toss a coin four times and record the results. That’s a bit vague. To be exact, record the results of each of the four tosses in order. A typical outcome is then HTTH. Counting shows that there are 16 possible outcomes. The sample space S is the set of all 16 strings of four H’s and T’s.

Suppose that our only interest is the number of heads in four tosses. Now we can be exact in a simpler fashion. The random phenomenon is to toss a coin four times and count the number of heads. The sample space contains only five outcomes:

This example illustrates the importance of carefully specifying what constitutes an individual outcome.

Although these examples seem remote from the practice of statistics, the connection is surprisingly close. Suppose that in conducting an opinion poll you select four people at random from a large population and ask each if he or she favors reducing federal spending on low-interest student loans. The answers are Yes or No. The possible outcomes— the sample space—are exactly as in Example 4.6 if we replace heads by Yes and tails by No. Similarly, the possible outcomes of an SRS of 1500 people are the same, in principle, as the possible outcomes of tossing a coin 1500 times. One of the great advantages of mathematics is that the essential features of quite different phenomena can be described by the same probability model.

4.8

4.9

USE YOUR KNOWLEDGE

When were you born? A student is asked “In what month were you born? Set up an appropriate sample space for this setting.

The sample spaces described in Examples 4.4, 4.5, and 4.6 correspond to categorical variables where we can list all the possible values. Other sample spaces correspond to quantitative variables. Here is an example.

EXAMPLE 4.7

Using software. Most statistical software has a function that will generate a random number between 0 and 1. The sample space is

This S is a mathematical idealization. Any specific random number generator produces numbers with some limited number of decimal places so that, strictly speaking, not all numbers between 0 and 1 are possible outcomes. For example, Minitab generates random numbers like 0.736891, with six decimal places. The entire interval from 0 to 1 is easier to think about. It also has the advantage of being a suitable sample space for different software systems that produce random numbers with different numbers of digits.

USE YOUR KNOWLEDGE

How many hours do you text? You record the number of hours per week that a randomly selected student spends texting. What is the sample space?

A sample space S lists the possible outcomes of a random phenomenon. To complete a mathematical description of the random phenomenon, we must also give the probabilities with which these outcomes occur.

The true long-term proportion of any outcome—say, “exactly two heads in four tosses of a coin”—can be found only empirically, and then only approximately. How then can we describe probability mathematically? Rather than immediately attempting to give “correct” probabilities, let’s confront the easier task of laying down rules that any assignment of probabilities must satisfy. We need to assign probabilities not only to single outcomes but also to sets of outcomes.

EVENT

An event is an outcome or a set of outcomes of a random phenomenon. That is, an event is a subset of the sample space.

EXAMPLE 4.8

Exactly one head in four tosses. Take the sample space S for four tosses of a coin to be the 16 possible outcomes in the form HTHH. Then “exactly one head” is an event. Call this event A. The event A expressed as a set of outcomes is

In a probability model, events have probabilities. What properties must any assignment of probabilities to events have? Here are some basic facts about any probability model. These facts follow from the idea of probability as “the long-run proportion of repetitions on which an event occurs.”

1. Any probability is a number between 0 and 1. Any proportion is a number between 0 and 1, so any probability is also a number between 0 and 1. An event with probability 0 never occurs, and an event with probability 1 occurs on every trial. An event with probability 0.5 occurs in half the trials in the long run.

2. All possible outcomes together must have probability 1. Because every trial will produce an outcome, the sum of the probabilities for all possible outcomes must be exactly 1.

3. If two events have no outcomes in common, the probability that one or the other occurs is the sum of their individual probabilities. If one event occurs in 40% of all trials, a different event occurs in 25% of all trials, and the two can never occur together, then one or the other occurs in 65% of all trials because 40% + 25% = 65%.

4. The probability that an event does not occur is 1 minus the probability that the event does occur. If an event occurs in (say) 70% of all trials, it fails to occur in the other 30%. The probability that an event occurs and the probability that it does not occur always add to 100%, or 1.

Probability rules Formal probability uses mathematical notation to state Facts 1 through 4 more concisely. We use capital letters near the beginning of the alphabet to denote events. If A is any event, we write its probability as P(A). Here are our probability facts in formal language. As you apply these rules, remember that they are just another form of intuitively true facts about long-run proportions.

FIGURE 4.2 Venn diagram showing disjoint events A and B. Disjoint events have no common outcomes.

FIGURE 4.3 Venn diagram showing the complement Ac of an event A. The complement consists of all outcomes that are not in A.

PROBABILITY RULES

Rule 1. The probability P(A) of any event A satisfies 0 ≤ P(A) ≤ 1.

Rule 2. If S is the sample space in a probability model, then P(S) = 1.

Rule 3. Two events A and B are disjoint if they have no outcomes in common and so can never occur together. If A and B are disjoint,

This is the addition rule for disjoint events.

Rule 4. The complement of any event A is the event that A does not occur, written as Ac. The complement rule states that

Venn diagram

You may find it helpful to draw a picture to remind yourself of the meaning of complements and disjoint events. A picture like Figure 4.2 that shows the sample space S as a rectangular area and events as areas within S is called a Venn diagram. The events A and B in Figure 4.2 are disjoint because they do not overlap. As Figure 4.3 shows, the complement Ac contains exactly the outcomes that are not in A.

EXAMPLE 4.9

Favorite vehicle colors. What is your favorite color for a vehicle? Our preferences can be related to our personality, our moods, or particular objects. Here is a probability model for color preferences.2

Color White Black Silver Gray Probability 0.24 0.19 0.16 0.15

Color Red Blue Brown Other Probability 0.10 0.07 0.05 0.04

Each probability is between 0 and 1. The probabilities add to 1 because these outcomes together make up the sample space S. Our probability model corresponds to selecting a person at random and asking what is their favorite color.

Let’s use the probability Rules 3 and 4 to find some probabilities for favorite vehicle colors.

EXAMPLE 4.10

Black or silver? What is the probability that a person’s favorite vehicle color is black or silver? If the favorite is black, it cannot be silver, so these two events are disjoint. Using Rule 3, we find

There is a 35% chance that a randomly selected person will choose black or silver as their favorite color. Suppose that we want to find the probability that the favorite color is not blue.

EXAMPLE 4.11

Use the complement rule. To solve this problem, we could use Rule 3 and add the probabilities for white, black, silver, gray, red, brown and other. However, it is easier to use the probability that we have for blue and Rule 4. The event that the favorite is not blue is the complement of the event that the favorite is blue. Using our notation for

4.10 4.11

events, we have

We see that 93% of people have a favorite vehicle color that is not blue.

USE YOUR KNOWLEDGE

Red or brown. Find the probability that the favorite color is red or brown. White, black, silver, gray, or red. Find the probability that the favorite color is white, black, silver, gray, or red using Rule 4. Explain why this calculation is easier than finding the answer using Rule 3.

Assigning probabilities: Finite number of outcomes The individual outcomes of a random phenomenon are always disjoint. So the addition rule provides a way to assign probabilities to events with more than one outcome: start with probabilities for individual outcomes and add to get probabilities for events. This idea works well when there are only a finite (fixed and limited) number of outcomes.

PROBABILITIES IN A FINITE SAMPLE SPACE

Assign a probability to each individual outcome. These probabilities must be numbers between 0 and 1 and must have sum 1.

The probability of any event is the sum of the probabilities of the outcomes making up the event.

EXAMPLE 4.12

Benford’s law

Benford’s law. Faked numbers in tax returns, payment records, invoices, expense account claims, and many other settings often display patterns that aren’t present in legitimate records. Some patterns, such as too many round numbers, are obvious and easily avoided by a clever crook. Others are more subtle. It is a striking fact that the first digits of numbers in legitimate records often follow a distribution known as Benford’s law. Here it is (note that a first digit can’t be 0):3

First digit 1 2 3 4 5 6 7 8 9 Probability 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046

Benford’s law usually applies to the first digits of the sizes of similar quantities, such as invoices, expense account claims, and county populations. Investigators can detect fraud by comparing the first digits in records such as invoices paid by a business with these probabilities.

EXAMPLE 4.13

Find some probabilities for Benford’s law. Consider the events

From the table of probabilities in Example 4.12,

Note that P(B) is not the same as the probability that a first digit is strictly more than 7. The probability P(7) that a first digit is 7 is included in “7 or more” but not in “more than 7.”

USE YOUR KNOWLEDGE

4.12

4.13

Benford’s law. Using the probabilities for Benford’s law, find the probability that a first digit is anything other than 1.

Use the addition rule. Use the addition rule with the probabilities for the events A and B from Example 4.13 to find the probability that a first digit is either 4 or 7 or more.

Be careful to apply the addition rule only to disjoint events.

EXAMPLE 4.14

Find more probabilities for Benford’s law. Check that the probability of the event C that a first digit is odd is

The probability

is not the sum of P(B) and P(C) because events B and C are not disjoint. Outcomes 7 and 9 are common to both events.

4.14

Assigning probabilities: Equally likely outcomes Assigning correct probabilities to individual outcomes often requires long observation of the random phenomenon. In some circumstances, however, we are willing to assume that individual outcomes are equally likely because of some balance in the phenomenon. Ordinary coins have a physical balance that should make heads and tails equally likely, for example, and the table of random digits comes from a deliberate randomization.

EXAMPLE 4.15

First digits that are equally likely. You might think that first digits are distributed “at random” among the digits 1 to 9 in business records. The nine possible outcomes would then be equally likely. The sample space for a single digit is

Because the total probability must be 1, the probability of each of the nine outcomes must be 1/9. That is, the assignment of probabilities to outcomes is

First digit 1 2 3 4 5 6 7 8 9 Probability 1/9 1/9 1/9 1/9 1/9 1/9 1/9 1/9 1/9

The probability of the event B that a randomly chosen first digit is 7 or more is

Compare this with the Benford’s law probability in Example 4.13. A person who fakes data by using “random” digits will end up with too many first digits that are 7 or more.

In Example 4.15, all outcomes have the same probability. Because there are nine equally likely outcomes, each must have probability 1/9. Because exactly three of the nine equally likely outcomes are 7 or more, the probability of this event is 3/9. In the special situation where all outcomes are equally likely, we have a simple rule for assigning probabilities to events.

EQUALLY LIKELY OUTCOMES

If a random phenomenon has k possible outcomes, all equally likely, then each individual outcome has probability 1/k. The probability of any event A is

Most random phenomena do not have equally likely outcomes, so the general rule for finite sample spaces (page 225) is more important than the special rule for equally likely outcomes.

USE YOUR KNOWLEDGE

Possible outcomes for rolling a die. A die has six sides with one to six spots on the sides. Give the probability distribution for the six possible outcomes that can result when a perfect die is rolled.

Independence and the multiplication rule Rule 3, the addition rule for disjoint events, describes the probability that one or the other of two events A and B will occur in the special situation when A and B cannot occur together because they are disjoint. Our final rule describes the probability that both events A and B occur, again only in a special situation. More general rules appear in Section 4.5, but in our study of statistics, we will need only the rules that apply to special situations.

Suppose that you toss a fair coin twice. You are counting heads, so two events of interest are

The events A and B are not disjoint. They occur together whenever both tosses give heads. We want to compute the probability of the event {A and B} that both tosses are heads. The Venn diagram in Figure 4.4 illustrates the event {A and B} as the overlapping area that is common to both A and B.

The coin tossing of Buffon, Pearson, and Kerrich described in Example 4.3 makes us willing to assign probability 1/2 to a head when we toss a coin. So

What is P(A and B)? Our common sense says that it is 1/4. The first toss will give a head half the time and the second toss will give a head half the time, so both tosses will give heads on 1/2 × 1/2 = 1/4 of all trials in the long run. This reasoning assumes that the second toss still has probability 1/2 of a head after the first has given a head. This is true— we can verify it by tossing a coin twice many times and observing the proportion of heads on the second toss after the first toss has produced a head. We say that the events “head on the first toss” and “head on the second toss” are independent. Here is our final probability rule.

FIGURE 4.4 Venn diagram showing the event {A and B}. This event consists of outcomes common to A and B.

MULTIPLICATION RULE FOR INDEPENDENT EVENTS

Rule 5. Two events A and B are independent if knowing that one occurs does not change the probability that the other occurs. If A and B are independent,

This is the multiplication rule for independent events.

Our definition of independence is rather informal. We will make this informal idea precise in Section 4.5. In practice, though, we rarely need a precise definition of independence because independence is usually assumed as part of a probability model when we want to describe random phenomena that seem to be physically unrelated to each other. Here is an example of independence.

EXAMPLE 4.16

4.15

4.16

Coins do not have memory. Because a coin has no memory, we assume that coin tosses are independent. For a fair coin, this means that the outcome of the first toss does not influence the outcome of any other toss.

USE YOUR KNOWLEDGE

A head and then a tail in two tosses. What is the probability of obtaining a head and then a tail on two tosses of a fair coin?

Here is an example of a situation where there are dependent events.

EXAMPLE 4.17

Dependent events in cards. The colors of successive cards dealt from the same deck are not independent. A standard 52-card deck contains 26 red and 26 black cards. For the first card dealt from a shuffled deck, the probability of a red card is 26/52 = 0.50 because the 52 possible cards are equally likely. Once we see that the first card is red, we know that there are only 25 reds among the remaining 51 cards. The probability that the second card is red is therefore only 25/51 = 0.49. Knowing the outcome of the first deal changes the probabilities for the second.

USE YOUR KNOWLEDGE

The probability of a second ace. A deck of 52 cards contains four aces, so the probability that a card drawn from this deck is an ace is 4/52. If we know that the first card drawn is an ace, what is the probability that the second card drawn is also an ace? Using the idea of independence, explain why this probability is not 4/52.

Here is another example of a situation where events are dependent.

EXAMPLE 4.18

Taking a test twice. If you take an IQ test or other mental test twice in succession, the two test scores are not independent. The learning that occurs on the first attempt influences your second attempt. If you learn a lot, then your second test score might be a lot higher than your first test score.

When independence is part of a probability model, the multiplication rule applies. Here is an example.

EXAMPLE 4.19

Mendel’s peas. Gregor Mendel used garden peas in some of the experiments that revealed that inheritance operates randomly. The seed color of Mendel’s peas can be either green or yellow. Two parent plants are “crossed” (one pollinates the other) to produce seeds.

Each parent plant carries two genes for seed color, and each of these genes has probability 0.5 of being passed to a seed. The two genes that the seed receives, one from each parent, determine its color. The parents contribute their genes independently of each other.

Suppose that both parents carry the G and the Y genes. The seed will be green if both parents contribute a G gene; otherwise, it will be yellow. If M is the event that the male contributes a G gene and F is the event that the female contributes a G gene, then the probability of a green seed is

In the long run, 1/4 of all seeds produced by crossing these plants will be green.

The multiplication rule applies only to independent events; you cannot use it if events are not independent. Here is a distressing example of misuse of the multiplication rule.

EXAMPLE 4.20

Sudden infant death syndrome. Sudden infant death syndrome (SIDS) causes babies to die suddenly (often in their cribs) with no explanation. Deaths from SIDS have been greatly reduced by placing babies on their backs, but as yet no cause is known.

When more than one SIDS death occurs in a family, the parents are sometimes accused. One “expert witness” popular with prosecutors in England told juries that there is only a 1 in 73 million chance that two children in the same family could have died from SIDS. Here’s his calculation: the rate of SIDS in a nonsmoking middle-class family is 1 in 8500. So the probability of two deaths is

Several women were convicted of murder on this basis, without any direct evidence that they harmed their children. As the Royal Statistical Society said, this reasoning is nonsense. It assumes that SIDS deaths in the same family are

independent events. The cause of SIDS is unknown: “There may well be unknown genetic or environmental factors that predispose families to SIDS, so that a second case within the family becomes much more likely.”4 The British government decided to review the cases of 258 parents convicted of murdering their babies.

mosaic plot, p. 143

The multiplication rule P(A and B) = P(A)P(B) holds if A and B are independent but not otherwise. The addition rule P(A or B) = P(A) + P(B) holds if A and B are disjoint but not otherwise. Resist the temptation to use these simple formulas when the circumstances that justify them are not present. You must also be certain not to confuse disjointness and independence. Disjoint events cannot be independent. If A and B are disjoint, then the fact that A occurs tells us that B cannot occur—look again at Figure 4.2. Unlike disjointness or complements, independence cannot be pictured by a Venn diagram because it involves the probabilities of the events rather than just the outcomes that make up the events. However, it could be displayed in a mosaic plot.

Applying the probability rules If two events A and B are independent, then their complements Ac and Bc are also independent and Ac is independent of Bc. Suppose, for example, that 75% of all registered voters in a suburban district are Republicans. If an opinion poll interviews two voters chosen independently, the probability that the first is a Republican and the second is not a Republican is (0.75)(0.25) = 0.1875.

The multiplication rule also extends to collections of more than two events, provided that all are independent. Independence of events A, B, and C means that no information about any one or any two can change the probability of the remaining events. The formal definition is a bit messy. Fortunately, independence is usually assumed in setting up a probability model. We can then use the multiplication rule freely.

By combining the rules we have learned, we can compute probabilities for rather complex events. Here is an example.

EXAMPLE 4.21

HIV testing. Many people who come to clinics to be tested for HIV, the virus that causes AIDS, don’t come back to learn the test results. Clinics now use “rapid HIV tests” that give a result in a few minutes. The false-positive rate for a diagnostic test is the probability that a person with no disease will have a positive test result. For the rapid HIV tests, the Food and Drug Administration (FDA) has established 2% as the maximum false-positive rate allowed.5 If a clinic uses a test that matches the FDA standard and tests 50 people who are free of HIV antibodies, what is the probability that at least one false-positive will occur?

It is reasonable to assume as part of the probability model that the test results for different individuals are independent. The probability that the test is positive for a single person is 0.02, so the probability of a negative result is 1 − 0.02 = 0.98 by the complement rule. The probability of at least one false-positive among the 50 people tested is, therefore,

There is approximately a 64% chance that at least 1 of the 50 people will test positive for HIV even though none of them has the virus.

Concern about excessive numbers of false-positives led the New York City Department of Health and Mental Hygiene to suspend the use of one particular rapid HIV test.6

SECTION 4.2 SUMMARY A probability model for a random phenomenon consists of a sample space S and an assignment of probabilities P. The sample space S is the set of all possible outcomes of the random phenomenon. Sets of outcomes are called events. P assigns a number P(A) to an event A as its probability. The complement Ac of an event A consists of exactly the outcomes that are not in A. Events A and B are disjoint if they have no outcomes in common. Events A and B are independent if knowing that one event occurs does not change the probability we would assign to the other event. Any assignment of probability must obey the rules that state the basic properties of probability: Rule 1. 0 ≤ P(A) ≤ 1 for any event A. Rule 2. P(S) = 1. Rule 3. Addition rule: If events A and B are disjoint, then P(A or B) = P(A) + P(B). Rule 4. Complement rule: For any event A, P(Ac) = 1 − P(A). Rule 5. Multiplication rule: If events A and B are independent, then P(A and B) = P(A)P(B).

SECTION 4.2 EXERCISES For Exercise 4.8, see page 222; for Exercise 4.9, see page 222; for Exercises 4.10 and 4.11, see page 225; for Exercises 4.12 and 4.13, see page 226; for Exercise 4.14, see page 228; for Exercise 4.15, see page 229; and for Exercise 4.16, see page 229.

4.17 What is the sample space? For each of the following questions, define a sample space for the associated random phenomenon. Explain your answers. Be sure to specify units if that is appropriate.

(a) Will it rain tomorrow?

(b) How many times do you tweet in a typical day?

(c) What is the average age of your Facebook friends?

(d) What are the majors for students at your college?

4.18 Probability rules. For each of the following situations, state the probability rule or rules that you would use and apply it or them. Write a sentence explaining how the situation illustrates the use of the probability rules.

(a) The probability of event A is 0.417. What is the probability that event A does not occur?

(b) A coin is tossed four times. The probability of zero heads is 1/16 and the probability of zero tails is 1/16. What is the probability that all four tosses result in the same outcome?

(c) Refer to part (b). What is the probability that there is at least one head and at least one tail?

(d) The probability of event A is 0.4 and the probability of event B is 0.8. Events A and B are disjoint. Can this happen?

(e) Event A is very rare. Its probability is −0.04. Can this happen?

4.19 Equally likely events. For each of the following situations, explain why you think that the events are equally likely or not. Explain your answers.

(a) The outcome of the next tennis match for Sloane Stevens is either a win or a loss. (You might want to check the Internet for information about this tennis player.)

(b) You roll a fair die and get a 3 or a 4.

(c) You are observing turns at an intersection. You classify each turn as a right turn or a left turn.

(d) For college basketball games, you record the times that the home team wins and the number of times that the home team loses.

4.20 The multiplication rule for independent events. The probability that a randomly selected person prefers the vehicle color white is 0.24. Can you apply the multiplication rule for independent events in the situations described in parts (a) and (b)? If your answer is Yes, apply the rule.

(a) Two people are chosen at random from the population. What is the probability that both prefer white?

(b) Two people who are sisters are chosen. What is the probability that both prefer white?

(c) Write a short summary about the multiplication rule for independent events using your answers to parts (a) and (b) to illustrate the basic idea.

4.21 What’s wrong? In each of the following scenarios, there is something wrong. Describe what is wrong and give a reason for your answer.

(a) If two events are disjoint, we can multiply their probabilities to determine the probability that they will both occur.

(b) If the probability of A is 0.7 and the probability of B is 0.5, the probability of both A and B happening is 1.2.

(c) If the probability of A is 0.45, then the probability of the complement of A is −0.45.

4.22 What’s wrong? In each of the following scenarios, there is something wrong. Describe what is wrong and give a reason for your answer.

(a) If the sample space consists of two outcomes, then each outcome has probability 0.5.

(b) If we select a digit at random, then the probability of selecting a 3 is 0.3.

(c) If the probability of A is 0.3, the probability of B is 0.4, and the probability of A and B is 0.5, then A and B are independent.

4.23 Evaluating web page designs. You are a web page designer and you set up a page with four different links. A user of the page can click on one of the links or he or she can leave that page. Describe the sample space for the outcome of someone visiting your web page.

4.24 Record the length of time spent on the page. Refer to the previous exercise. You also decide to measure the length of time a visitor spends on your page. Give the sample space for this measure.

4.25 Distribution of blood types. All human blood can be “ABO-typed” as one of O, A, B, or AB, but the distribution of the types varies a bit among groups of people. Here is the distribution of blood types for a randomly chosen person in the United States:7

Blood type A B AB O

U.S. probability 0.42 0.11 ? 0.44

(a) What is the probability of type AB blood in the United States?

(b) Maria has type B blood. She can safely receive blood transfusions from people with blood types O and B. What is the probability that a randomly chosen person from the United States can donate blood to Maria?

4.26 Blood types in Ireland. The distribution of blood types in Ireland differs from the U.S. distribution given in the previous exercise:

Blood type A B AB O

Ireland probability 0.35 0.10 0.03 0.52

Choose a person from the United States and a person from Ireland at random, independently of each other. What is the probability that both have type O blood? What is the probability that both have the same blood type?

4.27 Are the probabilities legitimate? In each of the following situations, state whether or not the given assignment of probabilities to individual outcomes is legitimate—that is, it satisfies the rules of probability. If not, give specific reasons for your answer.

(a) Choose a college student at random and record gender and enrollment status: P(female full-time) = 0.44, P(female part- time) = 0.56, P(male full-time) = 0.46, P(male part-time) = 0.54.

(b) Deal a card from a shuffled deck: P(clubs) = 16/52, P(diamonds) = 12/52, P(hearts) = 12/52, P(spades) = 12/52.

(c) Roll a die and record the count of spots on the up-face: P(1) = 1/3, P(2) = 0, P(3) = 1/6, P(4) = 1/3, P(5) = 1/6, P(6) = 0.

4.28 French and English in Canada. Canada has two official languages, English and French. Choose a Canadian at random and ask, “What is your mother tongue?” Here is the distribution of responses, combining many separate languages from the broad Asian/Pacific region:8

Language English French Asian/Pacific Other

Probability 0.59 ? 0.07 0.11

(a) What probability should replace “?” in the distribution?

(b) What is the probability that a Canadian’s mother tongue is not English? Explain how you computed your answer.

4.29 Education levels of young adults. Choose a young adult (age 25 to 34 years) at random. The probability is 0.12 that the person chosen did not complete high school, 0.31 that the person has a high school diploma but no further education, and 0.29 that the person has at least a bachelor’s degree.

(a) What must be the probability that a randomly chosen young adult has some education beyond high school but does not have a bachelor’s degree?

(b) What is the probability that a randomly chosen young adult has at least a high school education?

4.30 Loaded dice. There are many ways to produce crooked dice. To load a die so that 6 comes up too often and 1 (which is opposite 6) comes up too seldom, add a bit of lead to the filling of the spot on the 1 face. Because the spot is

solid plastic, this works even with transparent dice. If a die is loaded so that 6 comes up with probability 0.24 and the probabilities of the 2, 3, 4, and 5 faces are not affected, what is the assignment of probabilities to the six faces?

4.31 Rh blood types. Human blood is typed as O, A, B, or AB and also as Rh-positive or Rh-negative. ABO type and Rh- factor type are independent because they are governed by different genes. In the American population, 84% of people are Rh- positive. Use the information about ABO type in Exercise 4.25 to give the probability distribution of blood type (ABO and Rh) for a randomly chosen American.

4.32 Roulette. A roulette wheel has 38 slots, numbered 0, 00, and 1 to 36. The slots 0 and 00 are colored green, 18 of the others are red, and 18 are black. The dealer spins the wheel and, at the same time, rolls a small ball along the wheel in the opposite direction. The wheel is carefully balanced so that the ball is equally likely to land in any slot when the wheel slows. Gamblers can bet on various combinations of numbers and colors.

(a) What is the probability that the ball will land in any one slot?

(b) If you bet on “red,” you win if the ball lands in a red slot. What is the probability of winning?

(c) The slot numbers are laid out on a board on which gamblers place their bets. One column of numbers on the board contains all multiples of 3, that is, 3, 6, 9, . . . , 36. You place a “column bet” that wins if any of these numbers comes up. What is your probability of winning?

4.33 Winning the lottery. A state lottery’s Pick 3 game asks players to choose a three-digit number, 000 to 999. The state chooses the winning three-digit number at random so that each number has probability 1/1000. You win if the winning number contains the digits in your number, in any order.

(a) Your number is 059. What is your probability of winning?

(b) Your number is 223. What is your probability of winning?

4.34 PINs. The personal identification numbers (PINs) for automatic teller machines usually consist of four digits. You notice that most of your PINs have at least one 0, and you wonder if the issuers use lots of 0s to make the numbers easy to remember. Suppose that PINs are assigned at random, so that all four-digit numbers are equally likely.

(a) How many possible PINs are there?

(b) What is the probability that a PIN assigned at random has at least one 0?

4.35 Universal blood donors. People with type O-negative blood are universal donors. That is, any patient can receive a transfusion of O-negative blood. Only 7% of the American population have O-negative blood. If eight people appear at random to give blood, what is the probability that at least one of them is a universal donor?

4.36 Axioms of probability. Show that any assignment of probabilities to events that obeys Rules 2 and 3 on page 224 automatically obeys the complement rule (Rule 4). This implies that a mathematical treatment of probability can start

from just Rules 1, 2, and 3. These rules are sometimes called axioms of probability.

4.37 Independence of complements. Show that if events A and B obey the multiplication rule, P(A and B) = P(A)P(B), then A and the complement Bc of B also obey the multiplication rule, P(A and Bc) = P(A)P(Bc). That is, if

events A and B are independent, then A and Bc are also independent. (Hint: Start by drawing a Venn diagram and noticing that the events “A and B” and “A and Bc” are disjoint.)

Mendelian inheritance. Some traits of plants and animals depend on inheritance of a single gene. This is called Mendelian inheritance, after Gregor Mendel (1822–1884). Exercises 4.38 through 4.41 are based on the following information about Mendelian inheritance of blood type. Each of us has an ABO blood type, which describes whether two characteristics, called A and B, are present. Every one of us

has two blood type alleles (gene forms), one inherited from our mother and one from our father. Each of these alleles can be A, B, or O. Which two we inherit determines our blood type. Here is a table that shows what our blood type is for each combination of two alleles:

Alleles inherited Blood type

A and A A A and B AB A and O A B and B B B and O B O and O O

We inherit each of a parent’s two alleles with probability 0.5. We inherit independently from our mother and father.

4.38 Blood types of children. Emily and Michael both have alleles O and O.

(a) What blood types can their children have?

(b) What is the probability that their next child has each of these blood types?

4.39 Parents with alleles B and O. Andreona and Caleb both have alleles B and O.

(a) What blood types can their children have?

(b) What is the probability that their next child has each of these blood types?

4.40 Two children. Samantha has alleles B and O. Dylan has alleles A and B. They have two children. What is the probability that both children have blood type A? What is the probability that both children have the same blood type?

4.41 Three children. Anna has alleles B and O. Nathan has alleles A and O.

(a) What is the probability that a child of these parents has blood type O?

(b) If Anna and Nathan have three children, what is the probability that all three have blood type O? What is the probability that the first child has blood type O and the next two do not?

4.3 Random Variables

When you complete this section, you will be able to:

Describe the probability distribution of a discrete random variable. Use a probability histogram to provide a graphical description of the probability distribution of a discrete random variable. Use the distribution of a discrete random variable to calculate probabilities of events. Find probabilities of events for the uniform and normal distributions.

Sample spaces need not consist of numbers. When we toss a coin four times, we can record the outcome as a string of heads and tails, such as HTTH. In statistics, however, we are most often interested in numerical outcomes such as the count of heads in the four tosses. It is convenient to use a shorthand notation: Let X be the number of heads. If our outcome is HTTH, then X = 2. If the next outcome is TTTH, the value of X changes to X = 1. The possible values of X are 0, 1, 2, 3, and 4. Tossing a coin four times will give X one of these possible values. Tossing four more times will give X another and probably different value. We call X a random variable because its values vary when the coin tossing is repeated.

RANDOM VARIABLE

A random variable is a variable whose value is a numerical outcome of a random process.

In our earlier coin-tossing example, the process is the tossing of a coin four times. The random variable is the number of heads in the four tosses.

We usually denote random variables by capital letters near the end of the alphabet, such as X or Y. Of course, the random variables of greatest interest to us are outcomes such as the mean x̄ of a random sample, for which we will keep the familiar notation.9 As we progress from general rules of probability toward statistical inference, we will concentrate on random variables.

When a random variable X describes a random process, the sample space S just lists the possible values of the random variable. We usually do not mention S separately. There remains the second part of any probability model, the assignment of probabilities to events. There are two main ways of assigning probabilities to the values of a random variable. The two types of probability models that result will dominate our application of probability to statistical inference.

4.42

Discrete random variables We have learned several rules of probability, but only one method of assigning probabilities: state the probabilities of the individual outcomes and assign probabilities to events by summing over the outcomes. The outcome probabilities must be between 0 and 1 and have sum 1. When the outcomes are numerical, they are values of a random variable. We will now attach a name to random variables having probability assigned in this way.10

DISCRETE RANDOM VARIABLE

A discrete random variable X has possible values that can be given in an ordered list. The probability distribution of X lists the values and their probabilities:

Value of X x1 x2 x3 . . .

Probability p1 p2 p3 . . .

The probabilities pi must satisfy two requirements:

1. Every probability pi is a number between 0 and 1.

2. p1 + p2 + · · · = 1.

Find the probability of any event by adding the probabilities pi of the particular values xi that make up the event.

In most discrete random variable situations that we will study, the number of possible values is a finite number, k. For example, in our example on the number of heads in four tosses of a coin, there are k = 5 possible values: 0, 1, 2, 3, and 4.

There are, however, settings in which the number of possible values can be infinite. Think about tossing a fair coin until you get a head. The number of possible tosses is any positive integer.

EXAMPLE 4.22

Grade distributions. A liberal arts college posts the grade distributions for its courses. In a recent semester, students in one section of English 130 received 32% A’s, 42% B’s, 19% C’s, 3% D’s, and 4% F’s. Choose an English 130 student at random. To “choose at random” means to give every student the same chance to be chosen. The student’s grade on a five-point scale (with A = 4) is a random variable X.

The value of X changes when we repeatedly choose students at random, but it is always one of 0, 1, 2, 3, or 4. Here is the distribution of X:

Value of X 0 1 2 3 4 Probability 0.04 0.03 0.19 0.42 0.32

The probability that the student got a B or better is the sum of the probabilities of an A and a B. In the language of random variables,

USE YOUR KNOWLEDGE

Will the course satisfy the requirement? Refer to Example 4.22. Suppose that a grade of D or F in

English 130 does not satisfy a requirement for a major in linguistics. What is the probability that a randomly selected student will not satisfy this requirement?

probability histogram

We can use histograms to show probability distributions as well as distributions of data. Figure 4.5 displays probability histograms that compare the probability model for equally likely random digits (Example 4.15, page 227) with the model given by Benford’s law (Example 4.12, page 226). The height of each bar shows the probability of the outcome at its base. Because the heights are probabilities, they add to 1. As usual, all the bars in a histogram have the same width. So the areas also display the assignment of probability to outcomes. Think of these histograms as idealized pictures of the results of very many trials. The histograms make it easy to quickly compare the two distributions.

FIGURE 4.5 Probability histograms for (a) equally likely random digits 1 to 9 and (b) Benford’s law. The height of each bar shows the probability assigned to a single outcome.

EXAMPLE 4.23

Number of heads in four tosses of a coin. What is the probability distribution of the discrete random variable X that counts the number of heads in four tosses of a coin? We can derive this distribution if we make two reasonable assumptions:

The coin is balanced, so it is fair and each toss is equally likely to give H or T. The coin has no memory, so tosses are independent.

The outcome of four tosses is a sequence of heads and tails such as HTTH. There are 16 possible outcomes in all. Figure 4.6 lists these outcomes along with the value of X for each outcome. The multiplication rule for independent events tells us that, for example,

FIGURE 4.6 Possible outcomes in four tosses of a coin, Example 4.23. The outcomes are arranged by the values of the random variable X, the number of heads.

Each of the 16 possible outcomes similarly has probability 1/16. That is, these outcomes are equally likely. The number of heads X has possible values 0, 1, 2, 3, and 4. These values are not equally likely. As Figure 4.6

shows, there is only one way that X = 0 can occur: namely, when the outcome is TTTT. So

The event {X = 2} can occur in six different ways, so that

We can find the probability of each value of X from Figure 4.6 in the same way. Here is the result:

Value of X 0 1 2 3 4 Probability 0.0625 0.25 0.375 0.25 0.0625

FIGURE 4.7 Probability histogram for the number of heads in four tosses of a coin (Example 4.23).

Figure 4.7 is a probability histogram for the distribution in Example 4.23. The probability distribution is exactly symmetric. The probabilities (bar heights) are idealizations of the proportions after very many tosses of four coins. The actual distribution of proportions observed would be nearly symmetric but is unlikely to be exactly symmetric.

EXAMPLE 4.24

Probability of at least three heads. Any event involving the number of heads observed can be expressed in terms of X, and its probability can be found from the distribution of X. For example, the probability of tossing at least

4.43

three heads is

The probability of at least one head is most simply found by use of the complement rule:

Recall that tossing a coin n times is similar to choosing an SRS of size n from a large population and asking a Yes or No question. We will extend the results of Example 4.23 when we return to sampling distributions in the next chapter.

USE YOUR KNOWLEDGE

Two tosses of a fair coin. Find the probability distribution for the number of heads that appear in two tosses of a fair coin.

Continuous random variables When we use the table of random digits to select a digit between 0 and 9, the result is a discrete random variable. The probability model assigns probability 1/10 to each of the 10 possible outcomes. Suppose that we want to choose a number at random between 0 and 1, allowing any number between 0 and 1 as the outcome. Software random number generators will do this.

FIGURE 4.8 A spinner that generates a random number between 0 and 1.

You can visualize such a random number by thinking of a spinner (Figure 4.8) that turns freely on its axis and slowly comes to a stop. The pointer can come to rest anywhere on a circle that is marked from 0 to 1. The sample space is now an entire interval of numbers:

density curve, p. 51

How can we assign probabilities to events such as {0.3 ≤ x ≤ 0.7}? As in the case of selecting a random digit, we would like all possible outcomes to be equally likely. But we cannot assign probabilities to each individual value of x and then sum because there are too many possible values. Instead, we use a new way of assigning probabilities directly to events—as areas under a density curve. Any density curve has area exactly 1 underneath it, corresponding to total probability 1.

EXAMPLE 4.25

uniform distribution

Uniform random numbers. The random number generator will spread its output uniformly across the entire interval from 0 to 1 as we allow it to generate a long sequence of numbers. The results of many trials are represented by the density curve of a uniform distribution.

This density curve appears in red in Figure 4.9. It has height 1 over the interval from 0 to 1, and height 0 everywhere else. The area under the density curve is 1: the area of a square with base 1 and height 1. The probability of any event is the area under the density curve and above the event in question.

4.44

FIGURE 4.9 Assigning probabilities for generating a random number between 0 and 1, Example 4.25. The probability of any interval of numbers is the area above the interval and under the density curve.

As Figure 4.9(a) illustrates, the probability that the random number generator produces a number X between 0.3 and 0.7 is

because the area under the density curve and above the interval from 0.3 to 0.7 is 0.4. The height of the density curve is 1, and the area of a rectangle is the product of height and length, so the probability of any interval of outcomes is just the length of the interval.

Similarly,

Notice that the last event consists of two nonoverlapping intervals, so the total area above the event is found by adding two areas, as illustrated by Figure 4.9(b). This assignment of probabilities obeys all of our rules for probability.

USE YOUR KNOWLEDGE

Find the probability. For the uniform distribution described in Example 4.25, find the probability that X is between 0.3 and 0.9.

Probability as area under a density curve is a second important way of assigning probabilities to events. Figure 4.10 illustrates this idea in general form. We call X in Example 4.25 a continuous random variable because its values are not isolated numbers but an entire interval of numbers.

CONTINUOUS RANDOM VARIABLE

A continuous random variable X takes all values in an interval of numbers. The probability distribution of X is described by a density curve. The probability of any event is the area under the density curve and above the values of X that make up the event.

The probability model for a continuous random variable assigns probabilities to intervals of outcomes rather than to individual outcomes. In fact, all continuous probability distributions assign probability 0 to every individual outcome. Only intervals of values have positive probability. To see that this is true, consider a specific outcome such

as P(X = 0.8) in the context of Example 4.25. The probability of any interval is the same as its length. The point 0.8 has no length, so its probability is 0.

FIGURE 4.10 The probability distribution of a continuous random variable assigns probabilities as areas under a density curve. The total area under any density curve is 1.

Although this fact may seem odd, it makes intuitive, as well as mathematical, sense. The random number generator produces a number between 0.79 and 0.81 with probability 0.02. An outcome between 0.799 and 0.801 has probability 0.002. A result between 0.799999 and 0.800001 has probability 0.000002. You see that as we approach 0.8, the probability gets closer to 0.

To be consistent, the probability of an outcome exactly equal to 0.8 must be 0. Because there is no probability exactly at X = 0.8, the two events {X > 0.8} and {X ≥ 0.8} have the same probability. We can ignore the distinction between > and ≥ when finding probabilities for continuous (but not discrete) random variables.

Normal distributions as probability distributions

Normal distributions, p. 56

The density curves that are most familiar to us are the Normal curves. Because any density curve describes an assignment of probabilities, Normal distributions are probability distributions. Recall that N(μ, σ) is our shorthand for the Normal distribution having mean μ and standard deviation σ. In the language of random variables, if X has the N(μ, σ) distribution, then the standardized variable

is a standard Normal random variable having the distribution N(0, 1).

EXAMPLE 4.26

Texting while driving. Texting while driving can be dangerous, but young people want to remain connected. Suppose that 26% of teen drivers text while driving. If we take a sample of 500 teen drivers, what percent would we expect to say that they text while driving?11

The proportion p = 0.26 is a number that describes the population of teen drivers. The proportion p̂ of the sample who say that they text while driving is used to estimate p. The proportion p̂ is a random variable because repeating the SRS would give a different sample of 500 teen drivers and a different value of p̂.

We will see in the next chapter that in this setting, with teen drivers answering honestly, p̂ has approximately the N(0.26, 0.0196) distribution. The mean 0.26 of this distribution is the same as the population proportion because p̂ is an unbiased estimate of p. The standard deviation is controlled mainly by the size of the sample.

What is the probability that the survey result differs from the truth about the population by no more than 3 percentage points? We can use what we learned about Normal distribution calculations to answer this question. Because p = 0.26, the survey misses by no more than 3 percentage points if the sample proportion is between 0.23 and 0.29.

FIGURE 4.11 Probability as area under a Normal density curve, Example 4.26.

Figure 4.11 shows this probability as an area under a Normal density curve. You can find it by software or by standardizing and using Table A. From Table A,

About 87% of the time, the sample p̂ will be within 3 percentage points of the proportion p.

We began this chapter with a general discussion of the idea of probability and the properties of probability models. Two very useful specific types of probability models are distributions of discrete and continuous random variables. In our study of statistics, we will employ only these two types of probability models.

SECTION 4.3 SUMMARY A random variable is a variable taking numerical values determined by the outcome of a random phenomenon. The probability distribution of a random variable X tells us what the possible values of X are and how probabilities are assigned to those values. A random variable X and its distribution can be discrete or continuous. A discrete random variable has possible values that can be given in an ordered list. The probability distribution assigns each of these values a probability between 0 and 1 such that the sum of all the probabilities is exactly 1. The probability of any event is the sum of the probabilities of all the values that make up the event. A continuous random variable takes all values in some interval of numbers. A density curve describes the probability distribution of a continuous random variable. The probability of any event is the area under the curve and above the values that make up the event. Uniform distributions are continuous probability distributions that are very similar to equally likely discrete distributions. Normal distributions are one type of continuous probability distribution. You can picture a probability distribution by drawing a probability histogram in the discrete case or by graphing the density curve in the continuous case.

SECTION 4.3 EXERCISES For Exercise 4.42, see page 237; for Exercise 4.43, see page 239; and for Exercise 4.44, see page 241.

4.45 How many courses? At a small liberal arts college, students can register for one to six courses. Let X be the number of courses taken in the fall by a randomly selected student from this college. In a typical fall semester, 5% take one course, 5% take two courses, 13% take three courses, 26% take four courses, 36% take five courses, and 15% take six courses. Let X be the number of courses taken in the fall by a randomly selected student from this college. Describe the probability distribution of this random variable.

4.46 Make a graphical display. Refer to the previous exercise. Use a probability histogram to provide a graphical description of the distribution of X.

4.47 Find some probabilities. Refer to Exercise 4.45.

(a) Find the probability that a randomly selected student takes three or fewer courses.

(b) Find the probability that a randomly selected student takes four or five courses.

(c) Find the probability that a randomly selected student takes eight courses.

4.48 Use the uniform distribution. Suppose that a random variable X follows the uniform distribution described in Example 4.25 (page 240). For each of the following events, find the probability and illustrate your calculations with a sketch of the density curve similar to the ones in Figure 4.9.

(a) The probability that X is less than 0.2.

(b) The probability that X is greater than or equal to 0.7.

(c) The probability that X is less than 0.8 and greater than 0.4.

(d) The probability that X is 0.7.

4.49 What’s wrong? In each of the following scenarios, there is something wrong. Describe what is wrong and give a reason for your answer.

(a) The possible values for a discrete random variable can’t be negative.

(b) A continuous random variable can take any value between 0 and 1.

(c) Normal distributions are discrete random variables.

4.50 Use of Twitter. Suppose that the population proportion of Internet users who say that they use Twitter or another service to post updates about themselves or to see updates about others is 19%.12 Think about selecting random samples from a population in which 19% are Twitter users.

(a) Describe the sample space for selecting a single person.

(b) If you select three people, describe the sample space.

(c) Using the results of part (b), define the sample space for the random variable that expresses the number of Twitter users in the sample of size 3.

(d) What information is contained in the sample space for part (b) that is not contained in the sample space for part (c)? Do you think this information is important? Explain your answer.

4.51 Use of Twitter. Refer to the previous exercise. Find the probabilities for the number of Twitter users in a sample of size 2.

4.52 Households and families in government data. In government data, a household consists of all occupants of a dwelling unit, while a family consists of two or more persons who live together and are related by blood or marriage. So all families form households, but some households are not families. Here are the distributions of household size and of family size in the United States:

Number of persons 1 2 3 4 5 6 7

Household probability 0.27 0.33 0.16 0.14 0.06 0.03 0.01 Family probability 0 0.44 0.22 0.20 0.09 0.03 0.02

Make probability histograms for these two discrete distributions, using the same scales. What are the most important differences between the sizes of households and families?

4.53 Discrete or continuous. In each of the following situations, decide whether the random variable is discrete or continuous and give a reason for your answer.

(a) Your web page has five different links, and a user can click on one of the links or can leave the page. You record the length of time that a user spends on the web page before clicking one of the links or leaving the page.

(b) You record the number of hits per day on your web page.

(c) You record the yearly income of a visitor to your web page.

4.54 Texas hold ’em. The game of Texas hold ’em starts with each player receiving two cards. Here is the probability distribution for the number of aces in two-card hands:

Number of aces 0 1 2

Probability 0.8507 0.1448 0.0045

(a) Verify that this assignment of probabilities satisfies the requirement that the sum of the probabilities for a discrete distribution must be 1.

(b) Make a probability histogram for this distribution.

(c) What is the probability that a hand contains at least one ace? Show two different ways to calculate this probability.

4.55 Tossing two dice. Some games of chance rely on tossing two dice. Each die has six faces, marked with one, two, . . . , six spots called pips. The dice used in casinos are carefully balanced so that each face is equally likely to come up. When two dice are tossed, each of the 36 possible pairs of faces is equally likely to come up. The outcome of interest to a gambler is the sum of the pips on the two up-faces. Call this random variable X.

(a) Write down all 36 possible pairs of up-faces.

(b) If all pairs have the same probability, what must be the probability of each pair?

(c) Write the value of X next to each pair of up-faces and use this information with the result of part (b) to give the probability distribution of X. Draw a probability histogram to display the distribution.

(d) One bet available in the game called craps wins if a 7 or an 11 comes up on the next roll of two dice. What is the probability of rolling a 7 or an 11 on the next roll?

(e) Several bets in craps lose if a 7 is rolled. If any outcome other than 7 occurs, these bets either win or continue to the next roll. What is the probability that anything other than a 7 is rolled?

4.56 Nonstandard dice. Nonstandard dice can produce interesting distributions of outcomes. You have two balanced, six-sided dice. One is a standard die, with faces having one, two, three, four, five, and six spots. The other die has three

faces with one spot and three faces with six spots. Find the probability distribution for the total number of spots Y on the up- faces when you roll these two dice.

4.57 Spell-checking software. Spell-checking software catches “nonword errors,” which are strings of letters that are not words, as when “the” is typed as “eth.” When undergraduates are asked to write a 250-word essay (without spell-checking), the number X of nonword errors has the following distribution:

Value of X 0 1 2 3 4

Probability 0.1 0.3 0.3 0.2 0.1

(a) Sketch the probability distribution for this random variable.

(b) Write the event “at least one nonword error” in terms of X. What is the probability of this event?

(c) Describe the event X ≤ 3 in words. What is its probability? What is the probability that X < 3?

4.58 Find the probabilities. Let the random variable X be a random number with the uniform density curve in Figure 4.9. Find the following probabilities:

(a) P(X ≥ 0.35).

(b) P(X = 0.35).

(c) P(0.35 < X < 1.35).

(d) P(0.18 ≤ X ≤ 0.25 or 0.4 ≤ X ≤ 0.5).

(e) X is not in the interval 0.4 to 0.8.

4.59 Uniform numbers between 0 and 2. Many random number generators allow users to specify the range of the random numbers to be produced. Suppose that you specify that the range is to be all numbers between 0 and 2. Call the random number generated Y. Then the density curve of the random variable Y has constant height between 0 and 2, and height 0 elsewhere.

(a) What is the height of the density curve between 0 and 2? Draw a graph of the density curve.

(b) Use your graph from part (a) and the fact that probability is area under the curve to find P(Y ≤ 1.6).

(c) Find P(0.5 < Y < 1.7).

(d) Find P(Y ≥ 0.95).

4.60 The sum of two uniform random numbers. Generate two random numbers between 0 and 1 and take Y to be their sum. Then Y is a continuous random variable that can take any value between 0 and 2. The density curve of Y is the

triangle shown in Figure 4.12.

(a) Verify by geometry that the area under this curve is 1.

(b) What is the probability that Y is less than 1? [Sketch the density curve, shade the area that represents the probability, then find that area. Do this for part (c) also.]

(c) What is the probability that Y is greater than 1.5?

(d) What is the probability that Y is greater than 0.5?

FIGURE 4.12 The density curve for the sum Y of two random numbers, Exercise 4.60.

4.61 How many close friends? How many close friends do you have? Suppose that the number of close friends adults claim to have varies from person to person with mean μ = 9 and standard deviation σ = 2.4. An opinion poll asks this question of an SRS of 1100 adults. We will see in the next chapter that, in this situation, the sample mean response x̄ has approximately the Normal distribution with mean 9 and standard deviation 0.0724. What is , the probability that x̄ estimates μ to within ±1?

4.62 Normal approximation for a sample proportion. A sample survey contacted an SRS of 700 registered voters in Oregon shortly after an election and asked respondents whether they had voted. Voter records show that 56% of registered voters had actually voted. We will see in the next chapter that, in this situation, the proportion p̂ of the sample who voted has approximately the Normal distribution with mean μ = 0.56 and standard deviation σ = 0.019.

(a) If the respondents answer truthfully, what is ? This is the probability that p̂ estimates 0.56 within plus or minus 0.04.

(b) In fact, 72% of the respondents said they had voted . If respondents answer truthfully, what is ? This probability is so small that it is good evidence that some people who did not vote claimed that they did vote.

4.4 Means and Variances of Random Variables

When you complete this section, you will be able to:

Use a probability distribution to find the mean of a discrete random variable. Apply the law of large numbers to describe the behavior of the sample mean as the sample size increases. Find means using the rules for means of linear transformations, sums, and differences. Use a probability distribution to find the variance and the standard deviation of a discrete random variable. Find variances and standard deviations using the rules for variances and standard deviations for linear transformations. Find variances and standard deviations using the rules for variances and standard deviations for sums of and differences between two random variables and for uncorrelated and for correlated random variables.

The probability histograms and density curves that picture the probability distributions of random variables resemble our earlier pictures of distributions of data. In describing data, we moved from graphs to numerical measures such as means and standard deviations. Now we will make the same move to expand our descriptions of the distributions of random variables. We can speak of the mean winnings in a game of chance or the standard deviation of the randomly varying number of calls a travel agency receives in an hour. In this section, we will learn more about how to compute these descriptive measures and about the laws they obey.

4.63

The mean of a random variable In Chapter 1 (page 28), we learned that the mean x̄ is the average of the observations in a sample. Recall that a random variable X is a numerical outcome of a random process. Think about repeating the random process many times and recording the resulting values of the random variable. You can think of the value of a random variable as the average of a very large sample where the relative frequencies of the values are the same as their probabilities.

If we think of the random process as corresponding to the population, then the mean of the random variable is a characteristic of this population. Here is an example.

EXAMPLE 4.27

The Tri-State Pick 3 lottery. Most states and Canadian provinces have government-sponsored lotteries. Here is a simple lottery wager from the Tri-State Pick 3 game that New Hampshire shares with Maine and Vermont. You choose a three-digit number, 000 to 999. The state chooses a three-digit winning number at random and pays you $500 if your number is chosen.

Because there are 1000 three-digit numbers, you have probability 1/1000 of winning. Taking X to be the amount your ticket pays you, the probability distribution of X is

Payoff X $0 $500 Probability 0.999 0.001

The random process consists of drawing a three-digit number. The population consists of the numbers 000 to 999. Each of these possible outcomes is equally likely in this example. In the setting of sampling in Chapter 3 (page 191), we can view the random process as selecting an SRS of size 1 from the population. The random variable X is 1 if the selected number is equal to the one that you chose and 0 if it is not.

What is your average payoff from many tickets? The ordinary average of the two possible outcomes $0 and $500 is $250, but that makes no sense as the average because $500 is much less likely than $0. In the long run, you receive $500 once in every 1000 tickets and $0 on the remaining 999 of 1000 tickets. The long-run average payoff is

or 50 cents. That number is the mean of the random variable X. (Tickets cost $1, so in the long run, the state keeps half the money you wager.)

If you play Tri-State Pick 3 several times, we would—as usual—call the mean of the actual amounts you win x̄. The mean in Example 4.27 is a different quantity—it is the long-run average winnings you expect if you play a very large number of times.

USE YOUR KNOWLEDGE

Find the mean of the probability distribution. You toss a fair coin. If the outcome is heads, you win $10.00; if the outcome is tails, you win nothing. Let X be the amount that you win in a single toss of a coin. Find the probability distribution of this random variable and its mean.

mean μ

Just as probabilities are an idealized description of long-run proportions, the mean of a probability distribution describes the long-run average outcome. We can’t call this mean x̄, so we need a different symbol. The common symbol for the mean of a probability distribution is μ, the Greek letter mu. We used μ in Chapter 1 for the mean of a Normal distribution, so this is not a new notation. We will often be interested in several random variables, each having a different probability distribution with a different mean.

expected value

To remind ourselves that we are talking about the mean of X, we often write μX rather than simply μ. In Example

4.27, μX = $0.50. Notice that, as often happens, the mean is not a possible value of X. You will often find the mean of a random variable X called the expected value of X. This term can be misleading because we don’t necessarily expect one observation on X to be close to its expected value.

The mean of any discrete random variable is found just as in Example 4.27. It is an average of the possible outcomes, but a weighted average in which each outcome is weighted by its probability. Because the probabilities add to 1, we have total weight 1 to distribute among the outcomes. An outcome that occurs half the time has probability one-half and gets one-half the weight in calculating the mean. Here is the general definition.

MEAN OF A DISCRETE RANDOM VARIABLE

Suppose that X is a discrete random variable whose distribution is

Value of X x1 x2 x3 . . .

Probability p1 p2 p3 . . .

To find the mean of X, multiply each possible value by its probability, then add all the products:

EXAMPLE 4.28

The mean of equally likely first digits. If first digits in a set of data all have the same probability, the probability distribution of the first digit X is then

First digit X 1 2 3 4 5 6 7 8 9 Probability 1/9 1/9 1/9 1/9 1/9 1/9 1/9 1/9 1/9

The mean of this distribution is

Suppose that the random digits in Example 4.28 had a different probability distribution. In Example 4.12 (page 226), we described Benford’s law as a probability distribution that describes first digits of numbers in many real situations. Let’s calculate the mean for Benford’s law.

EXAMPLE 4.29

The mean of first digits that follow Benford’s law. Here is the distribution of the first digit for data that follow Benford’s law. We use the letter V for this random variable to distinguish it from the one that we studied in Example 4.28. The distribution of V is

First digit V 1 2 3 4 5 6 7 8 9 Probability 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046

The mean of V is

The mean reflects the greater probability of smaller first digits under Benford’s law than when first digits 1 to 9 are equally likely.

Figure 4.13 locates the means of X and V on the two probability histograms. Because the discrete uniform distribution of Figure 4.13(a) is symmetric, the mean lies at the center of symmetry. We can’t locate the mean of the right-skewed distribution of Figure 4.13(b) by eye—calculation is needed.

What about continuous random variables? The probability distribution of a continuous random variable X is described by a density curve. Chapter 1 (page 54) showed how to find the mean of the distribution: it is the point at which the area under the density curve would balance if it were made out of solid material. The mean lies at the center of symmetric density curves such as the Normal curves. Exact calculation of the mean of a distribution with a skewed density curve requires advanced mathematics.13 The idea that the mean is the balance point of the distribution applies to discrete random variables as well, but in the discrete case, we have a formula that gives us this point.

FIGURE 4.13 Locating the mean of a discrete random variable on the probability histogram for (a) digits between 1 and 9 chosen at random; (b) digits between 1 and 9 chosen from records that obey Benford’s law.

Statistical estimation and the law of large numbers We would like to estimate the mean height μ of the population of all American women between the ages of 18 and 24 years. This μ is the mean μX of the random variable X obtained by choosing a young woman at random and measuring her height. To estimate μ, we choose an SRS of young women and use the sample mean x̄ to estimate the unknown population mean μ. In the language of Section 5.1 (page 282), μ is a parameter and x̄ is a statistic.

Statistics obtained from probability samples are random variables because their values vary in repeated sampling. The sampling distributions of statistics are just the probability distributions of these random variables.

It seems reasonable to use x̄ to estimate μ. An SRS should fairly represent the population, so the mean x̄ of the sample should be somewhere near the mean μ of the population. Of course, we don’t expect x̄ to be exactly equal to μ, and we realize that if we choose another SRS, the luck of the draw will probably produce a different x̄.

If x̄ is rarely exactly right and varies from sample to sample, why is it nonetheless a reasonable estimate of the population mean μ? If we keep on adding observations to our random sample, the statistic x̄ is guaranteed to get as close as we wish to the parameter μ and then stay that close. We have the comfort of knowing that if we can afford to keep on measuring more women, eventually we will estimate the mean height of all young women very accurately. This remarkable fact is called the law of large numbers. It is remarkable because it holds for any population, not just for some special class such as Normal distributions.

LAW OF LARGE NUMBERS

Draw independent observations at random from any population with finite mean μ. Decide how accurately you would like to estimate μ. As the number of observations drawn increases, the mean x̄ of the observed values eventually approaches the mean μ of the population as closely as you specified and then stays that close.

The behavior of x̄ is similar to the idea of probability. In the long run, the proportion of outcomes taking any value gets close to the probability of that value, and the average outcome gets close to the distribution mean. Figure 4.1 shows how proportions approach probability in one example. Here is an example of how sample means approach the distribution mean.

EXAMPLE 4.30

Heights of young women. The distribution of the heights of all young women is close to the Normal distribution with mean 64.5 inches and standard deviation 2.5 inches. Suppose that μ = 64.5 were exactly true.

4.64

FIGURE 4.14 The law of large numbers in action, Example 4.30. As we take more observations, the sample mean always approaches the mean of the population.

Figure 4.14 shows the behavior of the mean height x̄ of n women chosen at random from a population whose heights follow the N(64.5, 2.5) distribution. The graph plots the values of x̄ as we add women to our sample. The first woman drawn had height 64.21 inches, so the line starts there. The second had height 64.35 inches, so for n = 2 the mean is

This is the second point on the line in the graph. At first, the graph shows that the mean of the sample changes as we take more observations. Eventually, however,

the mean of the observations gets close to the population mean μ = 64.5 and settles down at that value. The law of large numbers says that this always happens.

USE YOUR KNOWLEDGE

Use the Law of Large Numbers applet. The Law of Large Numbers applet animates a graph like Figure 4.14 for rolling dice. Use it to better understand the law of large numbers by making a similar graph.

The mean μ of a random variable is the average value of the variable in two senses. By its definition, μ is the average of the possible values, weighted by their probability of occurring. The law of large numbers says that μ is also the long-run average of many independent observations on the variable. The law of large numbers can be proved mathematically starting from the basic laws of probability.

Thinking about the law of large numbers The law of large numbers says broadly that the average results of many independent observations are stable and predictable. The gamblers in a casino may win or lose, but the casino will win in the long run because the law of large numbers says what the average outcome of many thousands of bets will be. An insurance company deciding how much to charge for life insurance and a fast-food restaurant deciding how many beef patties to prepare also rely on the fact that averaging over many individuals produces a stable result. It is worth the effort to think a bit more closely about so important a fact.

The “law of small numbers” Both the rules of probability and the law of large numbers describe the regular behavior of chance phenomena in the long run. Psychologists have discovered that our intuitive understanding of randomness is quite different from the true laws of chance.14 For example, most people believe in an incorrect “law of small numbers.” That is, we expect even short sequences of random events to show the kind of average behavior that in fact appears only in the long run.

Some teachers of statistics begin a course by asking students to toss a coin 50 times and bring the sequence of heads and tails to the next class. The teacher then announces which students just wrote down a random-looking sequence rather than actually tossing a coin. The faked tosses don’t have enough “runs” of consecutive heads or consecutive tails. Runs of the same outcome don’t look random to us but are, in fact, common. For example, the probability of a run of three or more consecutive heads or tails in just 10 tosses is greater than 0.8.15 The runs of consecutive heads or consecutive tails that appear in real coin tossing (and that are predicted by the mathematics of probability) seem surprising to us. Because we don’t expect to see long runs, we may conclude that the coin tosses are not independent or that some influence is disturbing the random behavior of the coin.

EXAMPLE 4.31

The “hot hand” in basketball. Belief in the law of small numbers influences behavior. If a basketball player makes several consecutive shots, both the fans and her teammates believe that she has a “hot hand” and is more likely to make the next shot. This is doubtful.

Careful study suggests that runs of baskets made or missed are no more frequent in basketball than would be expected if each shot were independent of the player’s previous shots. Baskets made or missed are just like heads and tails in tossing a coin. (Of course, some players make 30% of their shots in the long run and others make 50%, so a coin-toss model for basketball must allow coins with different probabilities of a head.) Our perception of hot or cold streaks simply shows that we don’t perceive random behavior very well.16

Our intuition doesn’t do a good job of distinguishing random behavior from systematic influences. This is also true when we look at data. We need statistical inference to supplement exploratory analysis of data because probability calculations can help verify that what we see in the data is more than a random pattern.

How large is a large number? The law of large numbers says that the actual mean outcome of many trials gets close to the distribution mean μ as more trials are made. It doesn’t say how many trials are needed to guarantee a mean outcome close to μ. That depends on the variability of the random outcomes. The more variable the outcomes, the more trials are needed to ensure that the mean outcome x̄ is close to the distribution mean μ. Casinos understand this: the outcomes of games of chance are variable enough to hold the interest of gamblers. Only the casino plays often enough to rely on the law of large numbers. Gamblers get entertainment; the casino has a business.

BEYOND THE BASICS

More Laws of Large Numbers The law of large numbers is one of the central facts about probability. It helps us understand the mean μ of a random variable. It explains why gambling casinos and insurance companies make money. It assures us that statistical estimation will be accurate if we can afford enough observations. The basic law of large numbers applies to independent observations that all have the same distribution. Mathematicians have extended the law to many

more general settings. Here are two of these.

Is there a winning system for gambling? Serious gamblers often follow a system of betting in which the amount bet on each play depends on the outcome of previous plays. You might, for example, double your bet on each spin of the roulette wheel until you win—or, of course, until your fortune is exhausted. Such a system tries to take advantage of the fact that you have a memory even though the roulette wheel does not. Can you beat the odds with a system based on the outcomes of past plays? No. Mathematicians have established a stronger version of the law of large numbers that says that, if you do not have an infinite fortune to gamble with, your long-run average winnings μ remain the same as long as successive trials of the game (such as spins of the roulette wheel) are independent.

What if observations are not independent? You are in charge of a process that manufactures video screens for computer monitors. Your equipment measures the tension on the metal mesh that lies behind each screen and is critical to its image quality. You want to estimate the mean tension μ for the process by the average x̄ of the measurements. Alas, the tension measurements are not independent. If the tension on one screen is a bit too high, the tension on the next is more likely to also be high. Many real-world processes are like this—the process stays stable in the long run, but two observations made close together are likely to both be above or both be below the long-run mean. Again the mathematicians come to the rescue: as long as the dependence dies out fast enough as we take measurements farther and farther apart in time, the law of large numbers still holds.

Rules for means You are studying flaws in the painted finish of refrigerators made by your firm. Dimples and paint sags are two kinds of surface flaw. Not all refrigerators have the same number of dimples: many have none, some have one, some two, and so on. You ask for the average number of imperfections on a refrigerator. The inspectors report finding an average of 0.7 dimple and 1.4 sags per refrigerator. How many total imperfections of both kinds (on the average) are there on a refrigerator? That’s easy: if the average number of dimples is 0.7 and the average number of sags is 1.4, then counting both gives an average of 0.7 + 1.4 = 2.1 flaws.

In more formal language, the number of dimples on a refrigerator is a random variable X that varies as we inspect one refrigerator after another. We know only that the mean number of dimples is μX = 0.7. The number of paint sags is a second random variable Y having mean μX = 1.4. (As usual, the subscripts keep straight which variable we are talking about.) The total number of both dimples and sags is another random variable, the sum X + Y. Its mean μX+Y is the average number of dimples and sags together. It is just the sum of the individual means μX and μY. That’s an important rule for how means of random variables behave.

Here’s another rule. The crickets living in a field have mean length of 1.2 inches. What is the mean in centimeters? There are 2.54 centimeters in an inch, so the length of a cricket in centimeters is 2.54 times its length in inches. If we multiply every observation by 2.54, we also multiply their average by 2.54. The mean in centimeters must be 2.54 × 1.2, or about 3.05 centimeters. More formally, the length in inches of a cricket chosen at random from the field is a random variable X with mean μX. The length in centimeters is 2.54X, and this new random variable has mean 2.54μX.

The point of these examples is that means behave like averages. Here are the rules we need.

RULES FOR MEANS OF LINEAR TRANSFORMATIONS, SUMS, AND DIFFERENCES

Rule 1. If X is a random variable and a and b are fixed numbers, then

Rule 2. If X and Y are random variables, then

Rule 3. If X and Y are random variables, then

linear transformation, p. 44

Note that a + bX is a linear transformation of the random variable X.

EXAMPLE 4.32

4.65

4.66

How many courses? In Exercise 4.45 (page 244) you described the probability distribution of the number of courses taken in the fall by students at a small liberal arts college. Here is the distribution:

Courses in the fall 1 2 3 4 5 6 Probability 0.05 0.05 0.13 0.26 0.36 0.15

For the spring semester, the distribution is a little different.

Courses in the spring 1 2 3 4 5 6 Probability 0.06 0.08 0.15 0.25 0.34 0.12

For a randomly selected student, let X be the number of courses taken in the fall semester, and let Y be the number of courses taken in the spring semester. The means of these random variables are

The mean course load for the fall is 4.28 courses, and the mean course load for the spring is 4.09 courses. We assume that these distributions apply to students who earned credit for courses taken in the fall and the spring semesters. The mean of the total number of courses taken for the academic year is X + Y. Using Rule 2, we calculate the mean of the total number of courses:

Note that it is not possible for a student to take 8.37 courses in an academic year. This number is the mean of the probability distribution.

EXAMPLE 4.33

What about credit hours? In the previous exercise, we examined the number of courses taken in the fall and in the spring at a small liberal arts college. Suppose that we were interested in the total number of credit hours earned for the academic year. We assume that for each course taken at this college, three credit hours are earned. Let T be the mean of the distribution of the total number of credit hours earned for the academic year. What is the mean of the distribution of T? To find the answer, we can use Rule 1 with a = 0 and b = 3. Here is the calculation:

The mean of the distribution of the total number of credit hours earned is 25.11.

USE YOUR KNOWLEDGE

Find μY. The random variable X has mean μX = 12. If Y = 12 + 6X, what is μY?

Find μW. The random variable U has mean μU = 25, and the random variable V has mean μV = 25. If W = 0.5U + 0.5V, find μW.

The variance of a random variable The mean is a measure of the center of a distribution. A basic numerical description requires, in addition, a measure of the spread or variability of the distribution. The variance and the standard deviation are the measures of spread that accompany the choice of the mean to measure center. Just as for the mean, we need a distinct symbol to distinguish the variance of a random variable from the variance s2 of a data set. We write the variance of a random variable X as . Once again, the subscript reminds us which variable we have in mind. The definition of the variance of a random variable is similar to the definition of the variance s2 given in Chapter 1 (page 38). That is, the variance is an average value of the squared deviation (X − μX)

2 of the variable X from its mean μX. As for the mean, the average we use is a weighted average in which each outcome is weighted by its probability in order to take account of outcomes that are not equally likely. Calculating this weighted average is straightforward for discrete random variables but requires advanced mathematics in the continuous case. Here is the definition.

VARIANCE OF A DISCRETE RANDOM VARIABLE

Suppose that X is a discrete random variable whose distribution is

Value of X x1 x2 x3 . . .

Probability p1 p2 p3 . . .

and that μX is the mean of X. The variance of X is

The standard deviation σX of X is the square root of the variance.

EXAMPLE 4.34

Find the mean and the variance. In Example 4.32 (pages 254–255), we saw that the distribution of the number X of fall courses taken by students at a small liberal arts college is

Courses in the fall 1 2 3 4 5 6 Probability 0.05 0.05 0.13 0.26 0.36 0.15

We can find the mean and variance of X by arranging the calculation in the form of a table. Both μX and are sums of columns in this table.

xi pi xi pi (xi − μX) 2pi

1 0.05 0.05 (1 − 4.28)2(0.05) = 0.53792

2 0.05 0.10 (2 − 4.28)2(0.05) = 0.25992

3 0.13 0.39 (3 − 4.28)2 (0.13) = 0.21299

4 0.26 1.04 (4 − 4.28)2(0.26) = 0.02038

5 0.36 1.80 (5 − 4.28)2(0.36) = 0.18662

6 0.15 0.90 (6 − 4.28)2(0.15) = 0.44376

μX = 4.28

4.67

We see that . The standard deviation of X is . The standard deviation is a measure of the variability of the number of fall courses taken by the students at the small liberal arts college. As in the case of distributions for data, the standard deviation of a probability distribution is easiest to understand for Normal distributions.

USE YOUR KNOWLEDGE

Find the variance and the standard deviation. The random variable X has the following probability distribution:

Value of X 0 3 Probability 0.3 0.7

Find the variance and the standard deviation σX for this random variable.

Rules for variances and standard deviations

What are the facts for variances that parallel Rules 1, 2, and 3 for means? The mean of a sum of random variables is always the sum of their means, but this addition rule is true for variances only in special situations. To understand why, take X to be the percent of a family’s after-tax income that is spent, and take Y to be the percent that is saved. When X increases, Y decreases by the same amount. Though X and Y may vary widely from year to year, their sum X + Y is always 100% and does not vary at all. It is the association between the variables X and Y that prevents their variances from adding.

independence

If random variables are independent, this kind of association between their values is ruled out and their variances do add. Two random variables X and Y are independent if knowing that any event involving X alone did or did not occur tells us nothing about the occurrence of any event involving Y alone.

Probability models often assume independence when the random variables describe outcomes that appear unrelated to each other. You should ask in each instance whether the assumption of independence seems reasonable.

correlation

When random variables are not independent, the variance of their sum depends on the correlation between them as well as on their individual variances. In Chapter 2, we met the correlation r between two observed variables measured on the same individuals. We defined (page 101) the correlation r as an average of the products of the standardized x and y observations. The correlation between two random variables is defined in the same way, once again using a weighted average with probabilities as weights. We won’t give the details—it is enough to know that the correlation between two random variables has the same basic properties as the correlation r calculated from data. We use ρ, the Greek letter rho, for the correlation between two random variables. The correlation ρ is a number between −1 and 1 that measures the direction and strength of the linear relationship between two variables. The correlation between two independent random variables is zero.

Returning to family finances, if X is the percent of a family’s after-tax income that is spent and Y is the percent that is saved, then Y = 100 − X. This is a perfect linear relationship with a negative slope, so the correlation between X and Y is ρ = −1. With the correlation at hand, we can state the rules for manipulating variances.

RULES FOR VARIANCES AND STANDARD DEVIATIONS OF LINEAR TRANSFORMATIONS, SUMS, AND DIFFERENCES

Rule 1. If X is a random variable and a and b are fixed numbers, then

Rule 2. If X and Y are independent random variables, then

This is the addition rule for variances of independent random variables.

Rule 3. If X and Y have correlation ρ, then

This is the general addition rule for variances of random variables.

To find the standard deviation, take the square root of the variance.

Because a variance is the average of squared deviations from the mean, multiplying X by a constant b multiplies by the square of the constant. Adding a constant a to a random variable changes its mean but does not change its

variability. The variance of X + a is, therefore, the same as the variance of X. Because the square of −1 is 1, the addition rule says that the variance of a difference between independent random variables is the sum of the variances. For independent random variables, the difference X − Y is more variable than either X or Y alone because variations in both X and Y contribute to variation in their difference.

As with data, we prefer the standard deviation to the variance as a measure of the variability of a random variable. Rule 2 for variances implies that standard deviations of independent random variables do not add. To combine standard deviations, use the rules for variances. For example, the standard deviations of 2X and −2X are both equal to 2σX because this is the square root of the variance .

EXAMPLE 4.35

Payoff in the Tri-State Pick 3 lottery. The payoff X of a $1 ticket in the Tri-State Pick 3 game is $500 with probability 1/1000 and 0 the rest of the time. Here is the combined calculation of mean and variance:

xi pi xi pi (xi − μX) 2pi

0 0.999 0 (0 − 0.5)2(0.999) = 0.24975

500 0.001 0.5 (500 − 0.5)2(0.001) = 249.50025

μX = 0.5

The mean payoff is 50 cents. The standard deviation is . It is usual for games of chance to have large standard deviations because large variability makes gambling exciting.

If you buy a Pick 3 ticket, your winnings are W = X − 1 because the dollar you paid for the ticket must be subtracted from the payoff. Let’s find the mean and variance for this random variable.

EXAMPLE 4.36

Winnings in the Tri-State Pick 3 lottery. By the rules for means, the mean amount you win is

That is, you lose an average of 50 cents on a ticket. The rules for variances remind us that the variance and standard deviation of the winnings W = X − 1 are the same as those of X. Subtracting a fixed number changes the mean but not the variance.

Suppose now that you buy a $1 ticket on each of two different days. The payoffs X and Y on the two tickets are independent because separate drawings are held each day. Your total payoff is X + Y. Let’s find the mean and standard deviation for this payoff.

EXAMPLE 4.37

Two tickets. The mean for the payoff for the two tickets is

Because X and Y are independent, the variance of X + Y is

The standard deviation of the total payoff is

This is not the same as the sum of the individual standard deviations, which is $15.80 + $15.80 = $31.60. Variances of independent random variables add; standard deviations do not.

When we add random variables that are correlated, we need to use the correlation for the calculation of the variance but not for the calculation of the mean. Here is an example.

EXAMPLE 4.38

Utility bills. Consider a household where the monthly bill for natural-gas averages $125 with a standard deviation of $75, while the monthly bill for electricity averages $174 with a standard deviation of $41. The correlation between the two bills is −0.55.

Let’s compute the mean and standard deviation of the sum of the natural-gas bill and the electricity bill. We let X stand for the natural-gas bill and Y stand for the electricity bill. Then the total is X + Y. Using the rules for means, we have

To find the standard deviation, we first find the variance and then take the square root to determine the standard deviation. From the general addition rule for variances of random variables,

Therefore, the standard deviation is

The total of the natural-gas bill and the electricity bill has mean $299 and standard deviation $63.

The negative correlation in Example 4.38 is due to the fact that, in this household, natural gas is used for heating and electricity is used for air-conditioning. So, when it is warm, the electricity charges are high and the natural-gas charges are low. When it is cool, the reverse is true. This causes the standard deviation of the sum to be less than it would be if the two bills were uncorrelated (see Exercise 4.79, page 263).

There are situations where we need to combine several of our rules to find means and standard deviations. Here is an example.

EXAMPLE 4.39

Calcium intake. To get enough calcium for optimal bone health, tablets containing calcium are often recommended to supplement the calcium in the diet. One study designed to evaluate the effectiveness of a supplement followed a group of young people for seven years. Each subject was assigned to take either a tablet containing 1000 milligrams of calcium per day (mg/d) or a placebo tablet that was identical except that it had no calcium.17 A major problem with studies like this one is compliance: subjects do not always take the treatments assigned to them.

In this study, the compliance rate declined to about 47% toward the end of the seven-year period. The standard deviation of compliance was 22%. Calcium from the diet averaged 850 mg/d with a standard deviation of 330 mg/d. The correlation between compliance and dietary intake was 0.68. Let’s find the mean and standard deviation for the total calcium intake. We let S stand for the intake from the supplement and D stand for the intake from the diet.

We start with the intake from the supplement. Because the compliance is 47% and the amount in each tablet is 1000

mg, the mean for S is

Because the standard deviation of the compliance is 22%, the variance of S is

The standard deviation is

Be sure to verify which rules for means and variances are used in these calculations. We can now find the mean and standard deviation for the total intake. The mean is

the variance is

and the standard deviation is

The mean of the total calcium intake is 1320 mg/d and the standard deviation is 506 mg/d.

The correlation in this example illustrates an unfortunate fact about compliance and having an adequate diet. Some of the subjects in this study have diets that provide an adequate amount of calcium while others do not. The positive correlation between compliance and dietary intake tells us that those who have relatively high dietary intakes are more likely to take the assigned supplements. On the other hand, those subjects with relatively low dietary intakes, the ones who need the supplement the most, are less likely to take the assigned supplements.

SECTION 4.4 SUMMARY The probability distribution of a random variable X, like a distribution of data, has a mean μX and a standard deviation σX.

The law of large numbers says that the average of the values of X observed in many trials must approach μ. The mean μ is the balance point of the probability histogram or density curve. If X is discrete with possible values xi having probabilities pi, the mean is the average of the values of X, each weighted by its probability:

The variance is the average squared deviation of the values of the variable from their mean. For a discrete random variable,

The standard deviation σX is the square root of the variance. The standard deviation measures the variability of the distribution about the mean. It is easiest to interpret for Normal distributions. The mean and variance of a continuous random variable can be computed from the density curve, but to do so requires more advanced mathematics. The means and variances of random variables obey the following rules. If a and b are fixed numbers, then

If X and Y are any two random variables having correlation ρ, then

If X and Y are independent, then ρ = 0. In this case,

To find the standard deviation, take the square root of the variance.

SECTION 4.4 EXERCISES For Exercise 4.63, see page 247; for Exercise 4.64, see page 251; for Exercises 4.65 and 4.66, see page 255; and for Exercise 4.67, see page 257.

4.68 Find the mean of the random variable. A random variable X has the following distribution.

X −1 0 1 2

Probability 0.2 0.3 0.2 0.3

Find the mean for this random variable. Show your work.

4.69 Explain what happens when the sample size gets large. Consider the following scenarios: (1) You take a sample of two observations on a random variable and compute the sample mean, (2) you take a sample of 100 observations on the same random variable and compute the sample mean, (3) you take a sample of 1000 observations on the same random variable and compute the sample mean. Explain in simple language how close you expect the sample mean to be to the mean of the random variable as you move from Scenario 1 to Scenario 2 to Scenario 3.

4.70 Find some means. Suppose that X is a random variable with mean 30 and standard deviation 4. Also suppose that Y is a random variable with mean 50 and standard deviation 8. Find the mean of the random variable Z for each of the following cases. Be sure to show your work.

(a) Z = 35 − 10X.

(b) Z = 12X − 5.

(c) Z = X + Y.

(d) Z = X − Y.

(e) Z = −2X + 2Y.

4.71 Find the variance and the standard deviation. A random variable X has the following distribution.

X −1 0 1 2

Probability 0.3 0.2 0.3 0.2

Find the variance and the standard deviation for this random variable. Show your work.

4.72 Find some variances and standard deviations. Suppose that X is a random variable with mean 30 and standard deviation 4. Also suppose that Y is a random variable with mean 50 and standard deviation 8. Assume that the correlation between X and Y is zero. Find the variance and the standard deviation of the random variable Z for each of the following cases. Be sure to show your work.

(a) Z = 35 − 10X.

(b) Z = 12X − 5.

(c) Z = X + Y.

(d) Z = X − Y.

(e) Z = −2X + 2Y.

4.73 What happens if the correlation is not zero? Suppose that X is a random variable with mean 30 and standard deviation 4. Also suppose that Y is a random variable with mean 50 and standard deviation 8. Assume that the correlation between X and Y is 0.5. Find the variance and the standard deviation of the random variable Z for each of the following cases. Be sure to show your work.

(a) Z = 35 − 10X.

(b) Z = 12X − 5.

(c) Z = X + Y.

(d) Z = X − Y.

(e) Z = −2X + 2Y.

4.74 What’s wrong? In each of the following scenarios, there is something wrong. Describe what is wrong and give a reason for your answer.

(a) If you toss a fair coin three times and get heads all three times, then the probability of getting a tail on the next toss is much greater than one-half.

(b) If you multiply a random variable by 10, then the mean is multiplied by 10 and the variance is multiplied by 10.

(c) When finding the mean of the sum of two random variables, you need to know the correlation between them.

4.75 Servings of fruits and vegetables. The following table gives the distribution of the number of servings of fruits and vegetables consumed per day in a population.

Number of servings X 0 1 2 3 4 5

Probability 0.3 0.1 0.1 0.2 0.2 0.1

Find the mean for this random variable.

4.76 Mean of the distribution for the number of aces. In Exercise 4.54 (page 244) you examined the probability distribution for the number of aces when you are dealt two cards in the game of Texas hold ’em. Let X represent the number of aces in a randomly selected deal of two cards in this game. Here is the probability distribution for the random variable X:

Value of X 0 1 2

Probability 0.8507 0.1448 0.0045

Find μX, the mean of the probability distribution of X.

4.77 Standard deviation of the number of aces. Refer to Exercise 4.76. Find the standard deviation of the number of aces.

4.78 Standard deviation for fruits and vegetables. Refer to Exercise 4.75. Find the variance and the standard deviation for the distribution of the number of servings of fruits and vegetables.

4.79 Suppose that the correlation is zero. Refer to Example 4.38 (page 259).

(a) Recompute the standard deviation for the total of the natural-gas bill and the electricity bill, assuming that the correlation is zero.

(b) Is this standard deviation larger or smaller than the standard deviation computed in Example 4.38? Explain why.

4.80 Find the mean of the sum. Figure 4.12 displays the density curve of the sum Y = X1 + X2 of two independent random numbers, each uniformly distributed between 0 and 1.

(a) The mean of a continuous random variable is the balance point of its density curve. Use this fact to find the mean of Y from Figure 4.12.

(b) Use the same fact to find the means of X1 and X2. (They have the density curve pictured in Figure 4.9, page 240.) Verify that the mean of Y is the sum of the mean of X1 and the mean of X2.

4.81 Calcium supplements and calcium in the diet. Refer to Example 4.39 (page 260). Suppose that people who have high intakes of calcium in their diets are more compliant than those who have low intakes. What effect would this have on the calculation of the standard deviation for the total calcium intake? Explain your answer.

4.82 Toss a four-sided die twice. Role-playing games like Dungeons & Dragons use many different types of dice. Suppose that a four-sided die has faces marked 1, 2, 3, and 4. The intelligence of a character is determined by rolling this

die twice and adding 1 to the sum of the spots. The faces are equally likely, and the two rolls are independent. What is the average (mean) intelligence for such characters? How spread out are their intelligences, as measured by the standard deviation of the distribution?

4.83 Means and variances of sums. The rules for means and variances allow you to find the mean and variance of a sum of random variables without first finding the distribution of the sum, which is usually much harder to do.

(a) A single toss of a balanced coin has either 0 or 1 head, each with probability 1/2. What are the mean and standard deviation of the number of heads?

(b) Toss a coin four times. Use the rules for means and variances to find the mean and standard deviation of the total number of heads.

(c) Example 4.23 (page 238) finds the distribution of the number of heads in four tosses. Find the mean and standard deviation from this distribution. Your results in parts (b) and (c) should agree.

4.84 What happens when the correlation is 1? We know that variances add if the random variables involved are uncorrelated (ρ = 0), but not otherwise. The opposite extreme is perfect positive correlation (ρ = 1). Show by using the

general addition rule for variances that, in this case, the standard deviations add. That is, if ρXY = 1.

4.85 Will you assume independence? In which of the following games of chance would you be willing to assume independence of X and Y in making a probability model? Explain your answer in each case.

(a) In blackjack, you are dealt two cards and examine the total points X on the cards (face cards count 10 points). You can choose to be dealt another card and compete based on the total points Y on all three cards.

(b) In craps, the betting is based on successive rolls of two dice. X is the sum of the faces on the first roll, and Y the sum of the faces on the next roll.

4.86 Transform the distribution of heights from centimeters to inches. A report of the National Center for Health Statistics says that the heights of 20-year-old men have mean 176.8 centimeters (cm) and standard deviation 7.2 cm. There are 2.54 centimeters in an inch. What are the mean and standard deviation in inches?

4.87 Fire insurance. An insurance company looks at the records for millions of homeowners and sees that the mean loss from fire in a year is μ = $300 per person. (Most of us have no loss, but a few lose their homes. The $300 is the average loss.) The company plans to sell fire insurance for $300 plus enough to cover its costs and profit. Explain clearly why it would be stupid to sell only 10 policies. Then explain why selling thousands of such policies is a safe business.

4.88 Mean and standard deviation for 10 and for 12 policies. In fact, the insurance company sees that in the entire population of homeowners, the mean loss from fire is μ = $300 and the standard deviation of the loss is σ = $400. What are the mean and standard deviation of the average loss for 10 policies? (Losses on separate policies are independent.) What are the mean and standard deviation of the average loss for 12 policies?

4.5 General Probability Rules

When you complete this section, you will be able to:

Apply the five rules of probability. Apply the general addition rule for unions of two or more events. Find conditional probabilities. Apply the multiplication rule. Use a tree diagram to find probabilities. Use Bayes’s rule to find probabilities. Determine whether or not two events that both have positive probability are independent.

Our study of probability has concentrated on random variables and their distributions. Now we return to the laws that govern any assignment of probabilities. The purpose of learning more laws of probability is to be able to give probability models for more complex random phenomena. We have already met and used five rules.

PROBABILITY RULES

Rule 1. 0 ≤ P(A) ≤ 1 for any event A

Rule 2. P(S) = 1

Rule 3. Addition rule: If A and B are disjoint events, then

Rule 4. Complement rule: For any event A,

Rule 5. Multiplication rule: If A and B are independent events, then

General addition rules Probability has the property that if A and B are disjoint events, then P(A or B) = P(A) + P(B). What if there are more than two events or if the events are not disjoint? These circumstances are covered by more general addition rules for probability.

UNION

The union of any collection of events is the event that at least one of the collection occurs.

For two events A and B, the union is the event {A or B} that A or B or both occur. From the addition rule for two disjoint events, we can obtain rules for more general unions. Suppose first that we have several events—say, A, B, and C—that are disjoint in pairs. That is, no two can occur simultaneously. The Venn diagram in Figure 4.15 illustrates three disjoint events. The addition rule for two disjoint events extends to the following law.

FIGURE 4.15 The addition rule for disjoint events: P(A or B or C) = P(A) + P(B) + P(C) when events A, B, and C are disjoint.

ADDITION RULE FOR DISJOINT EVENTS

If events A, B, and C are disjoint in the sense that no two have any outcomes in common, then

This rule extends to any number of disjoint events.

EXAMPLE 4.40

Probabilities as areas. Generate a random number X between 0 and 1. What is the probability that the first digit after the decimal point will be a 3, a 6, or a 9? The random number X is a continuous random variable whose density curve has constant height 1 between 0 and 1 and is 0 elsewhere. The event that the first digit of X is odd is the union of five disjoint events. These events are

Figure 4.16 illustrates the probabilities of these events as areas under the density curve. Each area is 0.1. Therefore, the union of the three has probability equal to the sum, or 0.3.

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FIGURE 4.16 The probability that the first digit after the decimal point of a random number is a 3, a 6, or a 9 is the sum of the probabilities of the three disjoint events shown, Example 4.40.

FIGURE 4.17 The union of two events that are not disjoint. The general addition rule says that P(A or B) = P(A) + P(B) − P(A and B).

USE YOUR KNOWLEDGE

Probability that you roll a 3 or a 4 or a 5. If you roll a die, the probability of each of the six possible outcomes (1, 2, 3, 4, 5, 6) is 1/6. What is the probability that you roll a 3 or a 4 or a 5?

If events A and B are not disjoint, they can occur simultaneously. The probability of their union is then less than the sum of their probabilities. As Figure 4.17 suggests, the outcomes common to both are counted twice when we add probabilities, so we must subtract this probability once. Here is the addition rule for the union of any two events, disjoint or not.

GENERAL ADDITION RULE FOR UNIONS OF TWO EVENTS

For any two events A and B,

If A and B are disjoint, the event {A and B} that both occur has no outcomes in it. This empty event is the complement of the sample space S and must have probability 0. So the general addition rule includes Rule 3, the addition rule for disjoint events.

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EXAMPLE 4.41

Adequate sleep and exercise. Suppose that 40% of adults get enough sleep and 46% exercise regularly. What is the probability that an adult gets enough sleep or exercises regularly? To find this probability, we also need to know the percent who get enough sleep and exercise. Let’s assume that 24% do both.

We will use the notation of the general addition rule for unions of two events. Let A be the event that an adult gets enough sleep, and let B be the event that a person exercises regularly. We are given that P(A) = 0.40, P(B) = 0.46, and P(A and B) = 0.24. Therefore,

The probability that an adult gets enough sleep or exercises regularly is 0.62, or 62%.

USE YOUR KNOWLEDGE

Probability that your roll is even or greater than 5. If you roll a die, the probability of each of the six possible outcomes (1, 2, 3, 4, 5, 6) is 1/6. What is the probability that your roll is even or greater than 5?

FIGURE 4.18 Venn diagram and probabilities, Example 4.41.

Venn diagrams are a great help in finding probabilities for unions because you can just think of adding and subtracting areas. Figure 4.18 shows some events and their probabilities for Example 4.41. What is the probability that an adult gets adequate sleep and does not exercise?

The Venn diagram shows that the probability that an adult gets adequate sleep minus the probability that an adult gets adequate sleep and exercises regularly is 0.40 − 0.24 = 0.16. Similarly, the probability that an adult does not get adequate sleep and exercises regularly is 0.46 − 0.24 = 0.22. The four probabilities that appear in the figure add to 1

because they refer to four disjoint events whose union is the entire sample space.

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Conditional probability The probability we assign to an event can change if we know that some other event has occurred. This idea is the key to many applications of probability.

EXAMPLE 4.42

Probability of being dealt a heart. Doyle is a professional poker player. He stares at the dealer, who prepares to deal. What is the probability that the card dealt to Doyle is a heart? There are 52 cards in the deck. Because the deck was carefully shuffled, the next card dealt is equally likely to be any of the cards that Doyle has not seen. Thirteen of the 52 cards are hearts. So

This calculation assumes that Doyle knows nothing about any cards already dealt. Suppose now that he is looking at four cards already in his hand and that they are all hearts. He knows nothing about the other 48 cards except that exactly nine (13 − 4) hearts are among them. Doyle’s probability of being dealt a heart given what he knows is now

Knowing that there are four hearts among the four cards Doyle can see changes the probability that the next card dealt is a heart.

conditional probability

The new notation P(A | B) is a conditional probability. That is, it gives the probability of one event (the next card dealt is a heart) under the condition that we know another event (exactly one of the four visible cards is a heart). You can read the bar | as “given the information that.”

MULTIPLICATION RULE

The probability that both of two events A and B happen together can be found by

Here P(B | A) is the conditional probability that B occurs, given the information that A occurs.

USE YOUR KNOWLEDGE

The probability of a heart. Refer to Example 4.42. Suppose that none of the four cards in Doyle’s hand are hearts. What is the probability that the next card dealt to him is a heart?

EXAMPLE 4.43

Downloading music from the Internet. The multiplication rule is just common sense made formal. For example, suppose that 30% of Internet users download music files, and 70% of downloaders say they don’t care if the music is copyrighted. So the percent of Internet users who download music (event A) and don’t care about copyright (event B) is 70% of the 30% who download, or

The multiplication rule expresses this as

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Here is another example that uses conditional probability.

EXAMPLE 4.44

Probability of a favorable draw. Doyle is still at the poker table. At the moment, he has two cards and they are both hearts. He has seen 24 cards and none of other players have any hearts. What is the chance that the next three cards he draws will be hearts? The full deck of 52 cards contains 13 hearts. Therefore, 11 of the unseen cards are hearts. There are 28 (52 − 24) unseen cards. To find Doyle’s probability of drawing three hearts, we first calculate

Doyle finds both probabilities by counting cards. The probability that the first card drawn is a heart is 11/28 because 11 of the 28 unseen cards are hearts. If the first card is a heart, that leaves 10 hearts among the 27 remaining cards. So the conditional probability of another diamond is 10/27. The multiplication rule now says that

We again apply the multiplication rule for the third card. The probability that the next three draws are hearts is equal to the probability that the first two draws are hearts times the probability that the third card is a heart given that the first two draws are hearts. This probability is

It is very unlikely that Doyle’s next three cards will be hearts, even though his hearts are the only ones that he has seen.

USE YOUR KNOWLEDGE

The probability that the next two cards are hearts. In the setting of Example 4.44, suppose that Doyle’s third card is a heart, so he now has three hearts, and that none of the five additional cards that he sees are hearts What is the probability that the next two cards dealt to Doyle will be hearts?

If P(A) and P(A and B) are given, we can rearrange the multiplication rule to produce a definition of the conditional probability P(B | A) in terms of unconditional probabilities.

DEFINITION OF CONDITIONAL PROBABILITY

When P(A) > 0, the conditional probability of B given A is

Be sure to keep in mind the distinct roles in P(B | A) of the event B whose probability we are computing and the event A that represents the information we are given. The conditional probability P(B | A) makes no sense if the event A

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can never occur, so we require that P(A) > 0 whenever we talk about P(B | A).

EXAMPLE 4.45

College students. Here is the distribution of U.S. college students classified by age and full-time or part-time status:

Age (years) Full-time Part-time 15 to 19 0.21 0.02 20 to 24 0.32 0.07 25 to 34 0.10 0.10 30 and over 0.05 0.13

Let’s compute the probability that a student is aged 20 to 24, given that the student is full-time. We know that the probability that a student is part-time and aged 20 to 24 is 0.32 from the table of probabilities. But what we want here is a conditional probability, given that a student is full-time. Rather than asking about age among all students, we restrict our attention to the subpopulation of students who are full-time. Let

Our formula is

We read P(A and B) = 0.32 from the table as we mentioned previously. What about P(B)? This is the probability that a student is full-time. Notice that there are four groups of students in our table that fit this description. To find the probability needed, we add the entries:

We are now ready to complete the calculation of the conditional probability:

The probability that a student is 20 to 24 years of age, given that the student is full-time, is 0.47.

Here is another way to give the information in the last sentence of this example: 47% of full-time college students are 20 to 24 years old. Which way do you prefer?

USE YOUR KNOWLEDGE

What rule did we use? In Example 4.45, we calculated P(B). What rule did we use for this calculation? Explain why this rule applies in this setting.

Find the conditional probability. Refer to Example 4.45. What is the probability that a student is part- time, given that the student is 20 to 24 years old? Explain in your own words the difference between this calculation and the one that we did in Example 4.45.

General multiplication rules The definition of conditional probability reminds us that, in principle, all probabilities—including conditional probabilities—can be found from the assignment of probabilities to events that describe random phenomena. More often, however, conditional probabilities are part of the information given to us in a probability model, and the multiplication rule is used to compute P(A and B). This rule extends to more than two events.

The union of a collection of events is the event that any of them occur. Here is the corresponding term for the event that all of them occur.

INTERSECTION

The intersection of any collection of events is the event that all the events occur.

To extend the multiplication rule to the probability that all of several events occur, the key is to condition each event on the occurrence of all the preceding events. For example, the intersection of three events A, B, and C has probability

EXAMPLE 4.46

High school athletes and professional careers. Only 5% of male high school basketball, baseball, and football players go on to play at the college level. Of these, only 1.7% enter major league professional sports. About 40% of the athletes who compete in college and then reach the pros have a career of more than three years. Define these events:

What is the probability that a high school athlete competes in college and then goes on to have a pro career of more than three years? We know that

Therefore, the probability we want is

Only about 3 of every 10,000 high school athletes can expect to compete in college and have a professional career of more than three years. High school students would be wise to concentrate on studies rather than on unrealistic hopes of fortune from pro sports.

Tree diagrams Probability problems often require us to combine several of the basic rules into a more elaborate calculation. Here is an example that illustrates how to solve problems that have several stages.

EXAMPLE 4.47

Online chat rooms. Online chat rooms are dominated by the young. Teens are the biggest users. If we look only at adult Internet users (aged 18 and over), 47% of the 18 to 29 age group chat, as do 21% of the 30 to 49 age group and just 7% of those 50 and over. To learn what percent of all Internet users participate in chat, we also need the age breakdown of users. Here it is: 29% of adult Internet users are 18 to 29 years old (event A1), another 47% are 30 to 49 (event A2), and the remaining 24% are 50 and over (event A3).

tree diagram

What is the probability that a randomly chosen adult user of the Internet participates in chat rooms (event C)? To find out, use the tree diagram in Figure 4.19 to organize your thinking. Each segment in the tree is one stage of the problem. Each complete branch shows a path through the two stages. The probability written on each segment is the conditional probability of an Internet user following that segment, given that he or she has reached the node from which it branches.

FIGURE 4.19 Tree diagram, Example 4.47. The probability P(C) is the sum of the probabilities of the three branches marked with asterisks (*).

Starting at the left, an Internet user falls into one of the three age groups. The probabilities of these groups

mark the leftmost branches in the tree. Conditional on being 18 to 29 years old, the probability of participating in chat is P(C | A1) = 0.47. So the conditional probability of not participating is

These conditional probabilities mark the paths branching out from the A1 node in Figure 4.19. The other two age group nodes similarly lead to two branches marked with the conditional probabilities of chatting or not. The

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probabilities on the branches from any node add to 1 because they cover all possibilities, given that this node was reached.

There are three disjoint paths to C, one for each age group. By the addition rule, P(C) is the sum of their probabilities. The probability of reaching C through the 18 to 29 age group is

Follow the paths to C through the other two age groups. The probabilities of these paths are

The final result is

About 25% of all adult Internet users take part in chat rooms.

It takes longer to explain a tree diagram than it does to use it. Once you have understood a problem well enough to draw the tree, the rest is easy. Tree diagrams combine the addition and multiplication rules. The multiplication rule says that the probability of reaching the end of any complete branch is the product of the probabilities written on its segments. The probability of any outcome, such as the event C that an adult Internet user takes part in chat rooms, is then found by adding the probabilities of all branches that are part of that event.

USE YOUR KNOWLEDGE

Draw a tree diagram. Refer to Doyle’s chances of five hearts in Example 4.44 (page 268). Draw a tree diagram to describe the outcomes for the three cards that he will be dealt. At the first stage, his draw can be a heart or a nonheart. At the second and third stages, he has the same possible outcomes but the probabilities are different.

Bayes’s rule There is another kind of probability question that we might ask in the context of thinking about online chat. What percent of adult chat room participants are aged 18 to 29?

EXAMPLE 4.48

Conditional versus unconditional probabilities. In the notation of Example 4.47, this is the conditional probability P(A1 | C). Start from the definition of conditional probability and then apply the results of Example 4.47:

More than half of adult chat room participants are between 18 and 29 years old. Compare this conditional probability with the original information (unconditional) that 29% of adult Internet users are between 18 and 29 years old. Knowing that a person chats increases the probability that he or she is young.

We know the probabilities P(A1), P(A2), and P(A3) that give the age distribution of adult Internet users. We also know the conditional probabilities P(C | A1), P(C | A2), and P(C | A3) that a person from each age group chats. Example 4.47 shows how to use this information to calculate P(C). The method can be summarized in a single expression that adds the probabilities of the three paths to C in the tree diagram:

In Example 4.48, we calculated the “reverse” conditional probability P(A1 | C). The denominator 0.2518 in that example came from the previous expression. Put in this general notation, we have another probability law.

BAYES’S RULE

Suppose that A1, A2, . . . , Ak are disjoint events whose probabilities are not 0 and add to exactly 1. That is, any outcome is in exactly one of these events. Then if C is any other event whose probability is not 0 or 1,

The numerator in Bayes’s rule is always one of the terms in the sum that makes up the denominator. The rule is named after Thomas Bayes, who wrestled with arguing from outcomes like C back to the Ai in a book published in 1763. It is far better to think your way through problems like Examples 4.47 and 4.48 than to memorize these formal expressions.

Independence again The conditional probability P(B | A) is generally not equal to the unconditional probability P(B). That is because the occurrence of event A generally gives us some additional information about whether or not event B occurs. If knowing that A occurs gives no additional information about B, then A and B are independent events. The formal definition of independence is expressed in terms of conditional probability.

INDEPENDENT EVENTS

Two events A and B that both have positive probability are independent if

This definition makes precise the informal description of independence given in Section 4.2 (page 229). We now see that the multiplication rule for independent events, P(A and B) = P(A)P(B), is a special case of the general multiplication rule, P(A and B) = P(A)P(B | A), just as the addition rule for disjoint events is a special case of the general addition rule.

SECTION 4.5 SUMMARY The complement Ac of an event A contains all outcomes that are not in A. The union {A or B} of events A and B contains all outcomes in A, in B, and in both A and B. The intersection {A and B} contains all outcomes that are in both A and B, but not outcomes in A alone or B alone. The conditional probability P(B | A) of an event B, given an event A, is defined by

when P(A) > 0. In practice, conditional probabilities are most often found from directly available information. The essential general rules of elementary probability are

Legitimate values: 0 ≤ P(A) ≤ 1 for any event A Total probability 1: P(S) = 1

Complement rule: P(Ac) = 1 − P(A) Addition rule: P(A or B) = P(A) + P(B) − P(A and B) Multiplication rule: P(A and B) = P(A)P(B | A)

If A and B are disjoint, then P(A and B) = 0. The general addition rule for unions then becomes the special addition rule, P(A or B) = P(A) + P(B). A and B are independent when P(B | A) = P(B). The multiplication rule for intersections then becomes P(A and B) = P(A)P(B). In problems with several stages, draw a tree diagram to organize use of the multiplication and addition rules.

SECTION 4.5 EXERCISES For Exercise 4.89, see page 266; for Exercise 4.90, see page 266; for Exercise 4.91, see page 268; for Exercise 4.92, see page 269; for Exercises 4.93 and 4.94, see page 270; and for Exercise 4.95, see page 273.

4.96 Find and explain some probabilities.

(a) Can we have an event A that has negative probability? Explain your answer.

(b) Suppose P(A) = 0.3 and P(B) = 0.5. Explain what it means for A and B to be disjoint. Assuming that they are disjoint, find the probability that A or B occurs.

(c) Explain in your own words the meaning of the rule P(S) = 1.

(d) Consider an event A. What is the name for the event that A does not occur? If P(A) = 0.4, what is the probability that A does not occur?

(e) Suppose that A and B are independent and that P(A) = 0.8 and P(B) = 0.3. Explain the meaning of the event {A and B}, and find its probability.

4.97 Unions.

(a) Assume that P(A) = 0.2, P(B) = 0.4, and P(C) = 0.1. If the events A, B, and C are disjoint, find the probability that the union of these events occurs.

(b) Draw a Venn diagram to illustrate your answer to part (a).

(c) Find the probability of the complement of the union of A, B, and C.

4.98 Conditional probabilities. Suppose that P(A) = 0.4, P(B) = 0.3, and P(B | A) = 0.4.

(a) Find the probability that both A and B occur.

(b) Use a Venn diagram to explain your calculation.

(c) What is the probability of the event that B occurs and A does not?

4.99 Find the probabilities. Suppose that the probability that A occurs is 0.5 and the probability that A and B occur is 0.2.

(a) Find the probability that B occurs given that A occurs.

(b) Illustrate your calculations in part (a) using a Venn diagram.

4.100 Why not? Suppose that P(B) = 0.6. Explain why P(A and B) cannot be 0.7.

4.101 Is the calcium intake adequate? In the population of young children eligible to participate in a study of whether or not their calcium intake is adequate, 52% are 5 to 10 years of age and 48% are 11 to 13 years of age. For those who are 5 to 10 years of age, 18% have inadequate calcium intake. For those who are 11 to 13 years of age, 57% have inadequate calcium intake.18

(a) Use letters to define the events of interest in this exercise.

(b) Convert the percents given to probabilities of the events you have defined.

(c) Use a tree diagram similar to Figure 4.19 to calculate the probability that a randomly selected child from this population has an inadequate intake of calcium.

4.102 Use Bayes’s rule. Refer to the previous exercise. Use Bayes’s rule to find the probability that a child from this population who has inadequate intake is 11 to 13 years old.

4.103 Are the events independent? Refer to the previous two exercises. Are the age of the child and whether or not the child has adequate calcium intake independent? Calculate the probabilities that you need to answer this question and write a short summary of your conclusion.

4.104 What’s wrong? In each of the following scenarios, there is something wrong. Describe what is wrong and give a reason for your answer.

(a) P(A or B) is always equal to the sum of P(A) and P(B).

(b) The probability of an event minus the probability of its complement is always equal to 1.

(c) Two events are disjoint if P(B | A) = P(B).

4.105 Exercise and sleep. Suppose that 42% of adults get enough sleep, 39% get enough exercise, and 28% do both. Find the probabilities of the following events:

(a) Enough sleep and not enough exercise.

(b) Not enough sleep and enough exercise.

(c) Not enough sleep and not enough exercise.

(d) For each of parts (a), (b), and (c), state the rule that you used to find your answer.

4.106 Exercise and sleep. Refer to the previous exercise. Draw a Venn diagram showing the probabilities for exercise and sleep.

4.107 Lying to a teacher. Suppose that 53% of high school students would admit to lying at least once to a teacher during the past year and that 24% of students are male and would admit to lying at least once to a teacher during the past year.19 Assume that 44% of the students are male. What is the probability that a randomly selected student is either male or would admit to lying to a teacher during the past year? Be sure to show your work and indicate all the rules that you use to find your answer.

4.108 Lying to a teacher. Refer to the previous exercise. Suppose that you select a student from the subpopulation of those who would admit to lying to a teacher during the past year. What is the probability that the student is female? Be sure to show your work and indicate all the rules that you use to find your answer.

4.109 Attendance at two-year and four-year colleges. In a large national population of college students, 61% attend four-year institutions and the rest attend two-year institutions. Males make up 44% of the students in the four-year institutions and 41% of the students in the two-year institutions.

(a) Find the four probabilities for each combination of gender and type of institution in the following table. Be sure that your probabilities sum to 1.

Men Women

Four-year institution Two-year institution

(b) Consider randomly selecting a female student from this population. What is the probability that she attends a four-year institution?

4.110 Draw a tree diagram. Refer to the previous exercise. Draw a tree diagram to illustrate the probabilities in a situation where you first identify the type of institution attended and then identify the gender of the student.

4.111 Draw a different tree diagram for the same setting. Refer to the previous two exercises. Draw a tree diagram to illustrate the probabilities in a situation where you first identify the gender of the student and then identify the type of institution attended. Explain why the probabilities in this tree diagram are different from those that you used in the previous exercise.

4.112 Education and income. Call a household prosperous if its income exceeds $100,000. Call the household educated if the householder completed college. Select an American household at random, and let A be the event that the selected household is prosperous and B the event that it is educated. According to the Current Population Survey, P(A) = 0.138, P(B) = 0.261, and the probability that a household is both prosperous and educated is P(A and B) = 0.082. What is the probability P(A or B) that the household selected is either prosperous or educated?

4.113 Find a conditional probability. In the setting of the previous exercise, what is the conditional probability that a household is prosperous, given that it is educated? Explain why your result shows that events A and B are not independent.

4.114 Draw a Venn diagram. Draw a Venn diagram that shows the relation between the events A and B in Exercise 4.112. Indicate each of the following events on your diagram and use the information in Exercise 4.112 to calculate the probability of each event. Finally, describe in words what each event is.

(a) {A and B}.

(b) {Ac and B}.

(c) {A and Bc}.

(d) {Ac and Bc}.

4.115 Sales of cars and light trucks. Motor vehicles sold to individuals are classified as either cars or light trucks (including SUVs) and as either domestic or imported. In a recent year, 69% of vehicles sold were light trucks, 78% were domestic, and 55% were domestic light trucks. Let A be the event that a vehicle is a car and B the event that it is imported. Write each of the following events in set notation and give its probability.

(a) The vehicle is a light truck.

(b) The vehicle is an imported car.

4.116 Job offers. Emily is graduating from college. She has studied biology, chemistry, and computing and hopes to work as a forensic scientist applying her science background to crime investigation. Late one night she thinks about some jobs she has applied for. Let A, B, and C be the events where Emily is offered a job by

A = the Connecticut Office of the Chief Medical Examiner

B = the New Jersey Division of Criminal Justice

C = the federal Disaster Mortuary Operations Response Team

Julie writes down her personal probabilities for being offered these jobs:

P(A) = 0.6 P(B) = 0.5 P(C) = 0.3 P(A and B) = 0.3 P(A and C) = 0.1 P(B and C) = 0.1 P(A and B and C) = 0

Make a Venn diagram of the events A, B, and C. As in Figure 4.18, mark the probabilities of every intersection involving these events and their complements. Use this diagram for Exercises 4.117, 4.118, and 4.119.

4.117 Find the probability of at least one offer. What is the probability that Julie is offered at least one of the three jobs?

4.118 Find the probability of another event. What is the probability that Julie is offered both the Connecticut and New Jersey jobs, but not the federal job?

4.119 Find a conditional probability. If Julie is offered the federal job, what is the conditional probability that she is also offered the New Jersey job? If Julie is offered the New Jersey job, what is the conditional probability that she is also offered the federal job?

4.120 Conditional probabilities and independence. Using the information in Exercise 4.115, answer these questions.

(a) Given that a vehicle is imported, what is the conditional probability that it is a light truck?

(b) Are the events “vehicle is a light truck” and “vehicle is imported” independent? Justify your answer.

Genetic counseling. Conditional probabilities and Bayes’s rule are a basis for counseling people who may have genetic defects that can be passed to their children. Exercises 4.121, 4.112, and 4.123 concern genetic counseling settings.

4.121 Albinism. People with albinism have little pigment in their skin, hair, and eyes. The gene that governs albinism has two forms (called alleles), which we denote by a and A. Each person has a pair of these genes, one inherited from each parent. A child inherits one of each parent’s two alleles independently with probability 0.5. Albinism is a recessive trait, so a person is albino only if the inherited pair is aa.

(a) Beth’s parents are not albino but she has an albino brother. This implies that both of Beth’s parents have type Aa. Why?

(b) Which of the types aa, Aa, AA could a child of Beth’s parents have? What is the probability of each type?

(c) Beth is not albino. What are the conditional probabilities for Beth’s possible genetic types, given this fact? (Use the definition of conditional probability.)

4.122 Find some conditional probabilities. Beth knows the probabilities for her genetic types from part (c) of the previous exercise. She marries Bob, who is albino. Bob’s genetic type must be aa.

(a) What is the conditional probability that a child of Beth and Bob is non-albino if Beth has type Aa? What is the conditional probability of a non-albino child if Beth has type AA?

(b) Beth and Bob’s first child is non-albino. What is the conditional probability that Beth is a carrier, type Aa?

4.123 Muscular dystrophy. Muscular dystrophy is an incurable muscle-wasting disease. The most common and serious type, called DMD, is caused by a sex-linked recessive mutation. Specifically, women can be carriers but do not get the disease; a son of a carrier has probability 0.5 of having DMD; a daughter has probability 0.5 of being a carrier. As many as one-third of DMD cases, however, are due to spontaneous mutations in sons of mothers who are not carriers. Toni has one son, who has DMD. In the absence of other information, the probability is 1/3 that the son is the victim of a spontaneous mutation and 2/3 that

Toni is a carrier. There is a screening test called the CK test that is positive with probability 0.7 if a woman is a carrier and with probability 0.1 if she is not. Toni’s CK test is positive. What is the probability that she is a carrier?

CHAPTER 4 EXERCISES

4.124 Repeat the experiment many times. Here is a probability distribution for a random variable X:

Value of X −3 4

Probability 0.3 0.7

A single experiment generates a random value from this distribution. If the experiment is repeated many times, what will be the approximate proportion of times that the value is −3? Give a reason for your answer.

4.125 Repeat the experiment many times and take the mean. Here is a probability distribution for a random variable X:

Value of X −8 5

Probability 0.6 0.4

A single experiment generates a random value from this distribution. If the experiment is repeated many times, what will be the approximate value of the mean of these random variables? Give a reason for your answer.

4.126 Work with a transformation. Here is a probability distribution for a random variable X:

Value of X 2 3

Probability 0.2 0.8

(a) Find the mean and the standard deviation of this distribution.

(b) Let Y = 5X −1. Use the rules for means and variances to find the mean and the standard deviation of the distribution of Y.

(c) For part (b), give the rules that you used to find your answer.

4.127 A different transformation. Refer to the previous exercise. Now let Y = 5X2 −1.

(a) Find the distribution of Y.

(b) Find the mean and standard deviation for the distribution of Y.

(c) Explain why the rules that you used for part (b) of the previous exercise do not work for this transformation.

4.128 Roll a pair of dice two times. Consider rolling a pair of fair dice two times. Let A be the total on the up-faces for the first roll and let B be the total on the up-faces for the second roll. For each of the following pairs of events, tell whether they are disjoint, independent, or neither.

(a) A = {2 on the first roll}, B = {8 or more on the first roll}.

(b) A = {2 on the first roll}, B = {8 or more on the second roll}.

(c) A = {5 or less on the second roll}, B = {4 or less on the first roll}.

(d) A = {5 or less on the second roll}, B = {4 or less on the second roll}.

4.129 Find the probabilities. Refer to the previous exercise. Find the probabilities for each event.

4.130 Some probability distributions. Here is a probability distribution for a random variable X:

Value of X 2 3 4

Probability 0.4 0.3 0.3

(a) Find the mean and standard deviation for this distribution.

(b) Construct a different probability distribution with the same possible values, the same mean, and a larger standard deviation. Show your work and report the standard deviation of your new distribution.

(c) Construct a different probability distribution with the same possible values, the same mean, and a smaller standard deviation. Show your work and report the standard deviation of your new distribution.

4.131 A fair bet at craps. Almost all bets made at gambling casinos favor the house. In other words, the difference between the amount bet and the mean of the distribution of the payoff is a positive number. An exception is “taking the odds” at the game of craps, a bet that a player can make under certain circumstances. The bet becomes available when a shooter throws a 4, 5, 6, 8, 9, or 10 on the initial roll. This number is called the “point”; when a point is rolled, we say that a point has been established. If a 4 is the point, an odds bet can be made that wins if a 4 is rolled before a 7 is rolled. The probability of winning this bet is 1/3, and the same payoff for a $10 bet is $20 (you keep the $10 you bet and you receive an additional $20). The same probability of winning and payoff apply for an odds bet on a 10. For an initial roll of 5 or 9, the odds bet has a winning probability of 2/5, and the payoff for a $10 bet is $15. Similarly, when the initial roll is 6 or 8, the odds bet has a winning probability of 5/11, and the payoff for a $10 bet is $12.

(a) Find the mean of the payoff distribution for each of these bets.

(b) Confirm that the bets are fair by showing that the difference between the amount bet and the mean of the distribution of the payoff is zero.

4.132 An interesting case of independence. Independence of events is not always obvious. Toss two balanced coins independently. The four possible combinations of heads and tails in order each have probability 0.25. The events

A = head on the first toss

B = both tosses have the same outcome

may seem intuitively related. Show that P(B | A) = P (B), so that A and B are, in fact, independent.

4.133 Wine tasters. Two wine tasters rate each wine they taste on a scale of 1 to 5. From data on their ratings of a large number of wines, we obtain the following probabilities for both tasters’ ratings of a randomly chosen wine:

Taster 2

Taster 1 1 2 3 4 5

1 0.03 0.02 0.01 0.00 0.00 2 0.02 0.07 0.06 0.02 0.01 3 0.01 0.05 0.25 0.05 0.01 4 0.00 0.02 0.05 0.20 0.02 5 0.00 0.01 0.01 0.02 0.06

(a) Why is this a legitimate assignment of probabilities to outcomes?

(b) What is the probability that the tasters agree when rating a wine?

(c) What is the probability that Taster 1 rates a wine higher than 3? What is the probability that Taster 2 rates a wine higher than 3?

4.134 Wine tasting. In the setting of the previous exercise, Taster 1’s rating for a wine is 3. What is the conditional probability that Taster 2’s rating is higher than 3?

4.135 Lottery tickets. Michael buys a ticket in the Tri-State Pick 3 lottery every day, always betting on 812. He will win something if the winning number contains 8, 1, and 2 in any order. Each day, Michael has probability

0.006 of winning, and he wins (or not) independently of other days because a new drawing is held each day. What is the probability that Michael’s first winning ticket comes on the 10th day?

4.136 Higher education at two-year and four-year institutions. The following table gives the counts of U.S. institutions of higher education classified as public or private and as two-year or four-year:20

Public Private

Two-year 1000 721 Four-year 2774 672

Convert the counts to probabilities and summarize the relationship between these two variables using conditional probabilities.

4.137 Odds bets at craps. Refer to the odds bets at craps in Exercise 4.131. Suppose that whenever the shooter has an initial roll of 4, 5, 6, 8, 9, or 10, you take the odds. Here are the probabilities for these initial rolls:

Point 4 5 6 8 9 10

Probability 3/36 4/36 5/36 5/36 4/36 3/36

Draw a tree diagram with the first stage showing the point rolled and the second stage showing whether the point is again rolled before a 7 is rolled. Include a first-stage branch showing the outcome that a point is not established. In this case, the amount bet is zero and the distribution of the winnings is the special random variable that has P (X = 0) = 1. For the combined betting system where the player always makes a $10 odds bet when it is available, show that the game is fair.

4.138 Sample surveys for sensitive issues. It is difficult to conduct sample surveys on sensitive issues because many people will not answer questions if the answers might embarrass them. Randomized response is an

effective way to guarantee anonymity while collecting information on topics such as student cheating or sexual behavior. Here is the idea. To ask a sample of students whether they have plagiarized a term paper while in college, have each student toss a coin in private. If the coin lands heads and they have not plagiarized, they are to answer No. Otherwise, they are to give Yes as their answer. Only the student knows whether the answer reflects the truth or just the coin toss, but the researchers can use a proper random sample with follow-up for nonresponse and other good sampling practices. Suppose that, in fact, the probability is 0.3 that a randomly chosen student has plagiarized a paper. Draw a tree

diagram in which the first stage is tossing the coin and the second is the truth about plagiarism. The outcome at the end of each branch is the answer given to the randomized-response question. What is the probability of a No answer in the randomized-response poll? If the probability of plagiarism were 0.2, what would be the probability of a No response on the poll? Now suppose that you get 39% No answers in a randomized-response poll of a large sample of students at your college. What do you estimate to be the percent of the population who have plagiarized a paper?

4.139 Find some conditional probabilities. Choose a point at random in the square with sides 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. This means that the probability that the point falls in any region within the square is the area of that region. Let X be the x coordinate and Y the y coordinate of the point chosen. Find the conditional probability P (Y < 1/3 | Y > X). (Hint: Sketch the square and the events Y < 1/3 and Y > X.)

1. An informative and entertaining account of the origins of probability theory is Florence N. David, Games, Gods and Gambling, Charles Griffin, London, 1962.

2. See dupont.com/.

3. You can find a mathematical explanation of Benford’s law in Ted Hill, “The first-digit phenomenon,” American Scientist, 86 (1996), pp. 358–363; and Ted Hill, “The difficulty of faking data,” Chance, 12, No. 3 (1999), pp. 27–31. Applications in fraud detection are discussed in the second paper by Hill and in Mark A. Nigrini, “I’ve got your number,” Journal of Accountancy, May 1999, available online at www.journalofaccountancy.com/issues/1999/may/nigrini.html.

4. Royal Statistical Society news release, “Royal Statistical Society concerned by issues raised in Sally Clark case,” October 23, 2001, at www.rss.org.uk. For background, see an editorial and article in The Economist, January 22, 2004. The editorial is entitled “The probability of injustice.”

5. See cdc.gov/mmwr/preview/mmwrhtml/mm57e618a1.htm.

6. See the Note 5.

7. See bloodbook.com/world-abo.html for the distribution of blood types for various groups of people.

8. From Statistics Canada, www.statcan.ca.

9. We use x both for the random variable, which takes different values in repeated sampling, and for the numerical value of the random variable in a particular sample. Similarly, s and stand both for random variables and for specific values. This notation is mathematically imprecise but statistically convenient.

10. We will consider only the case in which X takes a finite number of possible values. The same ideas, implemented with more advanced mathematics, apply to random variables with an infinite but still countable collection of values.

11. Based on a Pew Internet report, “Teens and distracted driving,” available from pewinternet.org/Reports/2009/Teens-and- Distracted-Driving.aspx.

12. See pewinternet.org/Reports/2009/17-Twitter-and-Status-Updating-Fall-2009.aspx.

13. The mean of a continuous random variable X with density function f(x) can be found by integration:

This integral is a kind of weighted average, analogous to the discrete-case mean

The variance of a continuous random variable X is the average squared deviation of the values of X from their mean, found by the integral

14. See A. Tversky and D. Kahneman, “Belief in the law of small numbers,” Psychological Bulletin, 76 (1971), pp. 105–110, and other writings of these authors for a full account of our misperception of randomness.

15. Probabilities involving runs can be quite difficult to compute. That the probability of a run of three or more heads in 10 independent tosses of a fair coin is (1/2) 1 (1/128) 5 0.508 can be found by clever counting. A general treatment using advanced methods appears in Section XIII.7 of William Feller, An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd ed., Wiley, 1968.

16. R. Vallone and A. Tversky, “The hot hand in basketball: On the misperception of random sequences,” Cognitive Psychology, 17 (1985), pp. 295–314. A later series of articles that debate the independence question is A. Tversky and T. Gilovich, “The cold facts about the ‘hot hand’ in basketball,” Chance, 2, No. 1 (1989), pp. 16–21; P. D. Larkey, R. A. Smith, and J. B. Kadane, “It’s OK to believe in the ‘hot hand,’” Chance, 2, No. 4 (1989), pp. 22–30; and A. Tversky and T. Gilovich, “The ‘hot hand’: Statistical reality or cognitive illusion?” Chance, 2, No. 4 (1989), pp. 31–34.

17. Based on a study discussed in S. Atkinson, G. McCabe, C. Weaver, S. Abrams, and K. O’Brien, “Are current calcium recommendations for adolescents higher than needed to achieve optimal peak bone mass? The controversy,” Journal of Nutrition, 138, No. 6 (2008), pp. 1182–1186.

18. Based on a study described in Corby C. Martin et al., “Children in school cafeterias select foods containing more saturated fat and energy than the Institute of Medicine recommendations,” Journal of Nutrition, 140 (2010), pp. 1653–1660.

19. Based on The Ethics of American Youth—2012, available from the Josephson Institute, charactercounts.org/wp- content/uploads/2014/02/ReportCard-2012-DataTables.pdf.

20. See nces.ed.gov/programs/digest. Data are from the 2012 Digest of Education Statistics.

5.1 5.2 5.3

CHAPTER 5 Sampling Distributions

Toward Statistical Inference The Sampling Distribution of a Sample Mean Sampling Distributions for Counts and Proportions

Introduction Statistical inference draws conclusions about a population or process from data. It emphasizes substantiating these conclusions via probability calculations because probability allows us to take chance variation into account. We have already examined data and arrived at conclusions many times. How do we move from summarizing a single data set to formal inference involving probability calculations?

The foundation for statistical inference is described in Section 5.1. There, we not only discuss the use of statistics as estimates of population parameters, but also describe the chance variation of a statistic when the data are produced by random sampling or randomized experimentation.

The sampling distribution of a statistic shows how the statistic would vary in identical repeated data collections.

That is, the sampling distribution is a probability distribution that answers the question, “What would happen if we did this experiment or sampling many times?” It is these distributions that provide the necessary link between probability and the data in your sample or from your experiment. They are the key to understanding statistical inference.

The last two sections of this chapter study the sampling distributions of two common statistics: the sample mean (for quantitative data) and the sample proportion or count (for categorical data). The general framework for constructing a sampling distribution is the same for all statistics, so we focus on those statistics commonly used in inference. As part of this study, we revisit the Normal distributions and are introduced to two common discrete probability distributions, the binomial and Poisson distributions.

5.1 Toward Statistical Inference

When you complete this section, you will be able to:

Identify parameters, populations, statistics, and samples and the relationships among these items. Use simulation to study a sampling distribution. Interpret and use a sampling distribution to describe a property of a statistic. Identify bias in a statistic by examining its sampling distribution and characterize an unbiased estimator of a parameter. Describe the relationship between the sample size and the variability of a statistic. Identify ways to reduce bias and variability of a statistic. Use the margin of error to describe the variability of a statistic.

A market research firm interviews a random sample of 1200 undergraduates enrolled in four-year colleges and universities throughout the United States. One result: the average number of hours spent online weekly is 19.0 hours. That’s the truth about the 1200 students in the sample. What is the truth about the millions of undergraduates who make up this population?

Because the sample was chosen at random, it’s reasonable to think that these 1200 students represent the entire population fairly well. So the market researchers turn the fact that the sample mean is x̄ = 19.0 hours into an estimate that the average time spent online weekly in the population of undergraduates enrolled in four-year colleges and universities is 19.0 hours.

statistical inference

That’s a basic idea in statistics: use a fact about a sample to estimate the truth about the whole population. We call this statistical inference because we infer conclusions about the larger population from data on selected individuals.

To think about inference, we must keep straight whether a number describes a sample or a population. Here is the vocabulary we use.

PARAMETERS AND STATISTICS

A parameter is a number that describes the population. A parameter is a fixed number, but in practice, we do not know its value.

A statistic is a number that describes a sample. The value of a statistic is known when we have taken a sample, but it can change from sample to sample. We often use a statistic to estimate an unknown parameter.

EXAMPLE 5.1

Understanding the college student market. Since 1987, Student Monitor has published an annual market research study that provides clients with information about the college student market. The firm uses a random sample of 1200 students located throughout the United States.1 One phase of the research focuses on computing and technology. The firm reports that undergraduates spend an average of 19.0 hours per week on the Internet and that 88% own a cell phone.

sample proportion

The sample mean x̅ = 19.0 hours is a statistic. The corresponding parameter is the average (call it μ) of all undergraduates enrolled in four-year colleges and universities. Similarly, the proportion of the sample who own a cell phone

5.1

population proportion

is a statistic. The corresponding parameter is the proportion (call it p) of all undergraduates at four-year colleges and universities who own a cell phone. We don’t know the values of the parameters μ and p, so we use the statistics x̄ and P̂, respectively, to estimate them.

USE YOUR KNOWLEDGE

Street harassment. A large-scale survey of 16,607 women from 42 cities around the world reports that 84% of women experience their first street harassment before the age of 17.2 Describe the statistic, population, and population parameter for this setting.

Sampling variability sampling variability

If Student Monitor took a second random sample of 1200 students, the new sample would have different undergraduates in it. It is almost certain that the sample mean x̄ would not again be 19.0. Likewise, we would not expect there to be exactly 1056 students who own a cell phone. In other words, the value of a statistic will vary from sample to sample. This basic fact is called sampling variability: the value of a statistic varies in repeated random sampling.

Random samples eliminate any preferences or favoritism from the act of choosing a sample, but they can still be misleading because of this variability that results when we choose at random. For example, what if a second random sample of 1200 undergraduates resulted in only 57% of the students owning a cell phone? Do these two results, 88% and 57%, leave you more or less confident in the value of the true population proportion? When sampling variability is too great, we can’t trust the results of any one sample.

bias, p. 174

We can assess this variability by using the second advantage of random samples (the first advantage being the elimination of bias). Specifically, the fact that if we take lots of random samples of the same size from the same population, the variation from sample to sample will follow a predictable pattern. All of statistical inference is based on one idea: to see how trustworthy a procedure is, ask what would happen if we repeated it many times.

To understand why sampling variability is not fatal, we ask, “What would happen if we took many samples?” Here’s how to answer that question for any statistic:

Take a large number of random samples of size n from the same population. Calculate the statistic for each sample. Make a histogram of the values of the statistic. Examine the distribution displayed in the histogram for shape, center, and spread, as well as outliers or other deviations.

simulation

In practice, it is too expensive to take many samples from a large population such as all undergraduates enrolled in four-year colleges and universities. But we can imitate taking many samples by using random digits from a table or computer software to emulate chance behavior. This is called simulation.

EXAMPLE 5.2

random digits, p. 179

Simulate a random sample.Let’s simulate drawing simple random samples (SRSs) of size 100 from the population of undergraduates. Suppose that, in fact, 90% of the population owns a cell phone. Then the true value of the parameter we want to estimate is p = 0.9. (Of course, we would not sample in practice if we already knew that p = 0.9. We are sampling here to understand how the statistic P̂ behaves.)

For cell phone ownership, we can imitate the population by a table of random digits, with each entry standing for a person. Nine of the 10 digits (say, 0 to 8) stand for students who own a cell phone. The remaining digit, 9, stands for those who do not. Because all digits in a random number table are equally likely, this assignment produces a population proportion of cell phone owners equal to p = 0.9. We then simulate an SRS of 100 students from the population by taking 100 consecutive digits from Table B. The statistic P̂ is the proportion of 0s to 8s in the sample.

Here are the first 100 entries in Table B with digits 0 to 8 highlighted:

5.2

19223 95034 05756 28713 96409 12531 42544 82853 73676 47150 99400 01927 27754 42648 82425 36290 45467 71709 77558 00095

There are 90 digits between 0 and 8, so P̂ = 90/100 = 0.90. We are fortunate here that our estimate is the true population value p = 0.9. A second SRS based on the second 100 entries in Table B gives a different result, P̂ = 0.86. The third SRS gives the result P̂ = 0.92. All three sample results are different. That’s sampling variability.

USE YOUR KNOWLEDGE

Using a random numbers table. In Example 5.2, we considered p = 0.9 and used each entry in Table B as a person for our simulations. Suppose instead that p = 0.85. How might we use Table B for simulations in this setting?

Sampling distributions Simulation is a powerful tool for studying chance variation. Now that we see how simulation works, it is faster to abandon Table B and to use a computer to generate random numbers. This also allows us to study other statistics, such as the sample mean, when the population cannot be easily imitated by a table of random numbers. We address the sampling distribution of x̄ in the next section.

EXAMPLE 5.3

Take many random samples. Figure 5.1 illustrates the process of choosing many samples and finding the statistic P̂ for each one. Follow the flow of the figure from the population at the left, to choosing an SRS and finding the P̂ for this sample, to collecting together the P̂‘s from many samples. The histogram at the right of the figure shows the distribution of the values of P̂ from 1000 separate SRSs of size 100 drawn from a population with p = 0.9.

histogram, p. 14

FIGURE 5.1 The results of many SRSs have a regular pattern, Example 5.3. Here we draw 1000 SRSs of size 100 from the same population. The population parameter is p = 0.9. The histogram shows the distribution of 1000 sample proportions.

Of course, Student Monitor samples 1200 students, not just 100. Figure 5.2 is parallel to Figure 5.1. It shows the process of choosing 1000 SRSs, each of size 1200, from a population in which the true proportion is p = 0.9. The 1000 values of P̂ from these samples form the histogram at the right of the figure. Figures 5.1 and 5.2 are drawn on the same scale. Comparing them shows what happens when we increase the size of our samples from 100 to 1200. These histograms display the sampling distribution of the statistic P̂ for two sample sizes.

5.3

FIGURE 5.2 The distribution of the sample proportion for 1000 SRSs of size 1200 drawn from the same population as in Figure 5.1. The two histograms have the same scale. The statistic from the larger sample is less variable.

SAMPLING DISTRIBUTION

The sampling distribution of a statistic is the distribution of values taken by the statistic in all possible samples of the same size from the same population.

Strictly speaking, the sampling distribution is the ideal pattern that would emerge if we looked at all possible samples of size n (here, 100 or 1200) from our population. A distribution obtained from a fixed number of trials, like the 1000 trials in Figures 5.1 and 5.2, is only an approximation to the sampling distribution. We will see that probability theory, the mathematics of chance behavior, can sometimes describe sampling distributions exactly. The interpretation of a sampling distribution is the same, however, whether we obtain it by simulation or by the mathematics of probability.

USE YOUR KNOWLEDGE

Poker winnings. Doug plays poker with the same group of friends once a week for three hours. At the end of each night, he records how much he won or lost in an Excel spreadsheet. Does this collection of amounts represent an approximation to a sampling distribution of his weekly winnings? Explain your answer.

We can use the tools of data analysis to describe any distribution. Let’s apply those tools to Figures 5.1 and 5.2.

Shape: The histograms look Normal. Figure 5.3 is a Normal quantile plot of the values of P̂ for our samples of size 100. It confirms that the distribution in Figure 5.1 is close to Normal. The 1000 values for samples of size 1200 in Figure 5.2 are even closer to Normal. The Normal curves drawn through the histograms describe the overall shape quite well. Center: In both cases, the values of the sample proportion P̂ vary from sample to sample, but the values are centered at 0.9. Recall that p = 0.9 is the true population parameter. Some samples have a P̂ less than 0.9 and some greater, but there is no tendency to be always low or always high. That is, P̂ has no bias as an estimator of p. This is true for both large and small samples. (Want the details? The mean of the 1000 values of P̂ is 0.8985 for samples of size 100 and 0.8994 for samples of size 1200. The median value of is exactly 0.9 for samples of both sizes.) Spread: The values of P̂ from samples of size 1200 are much less spread out than the values from samples of size 100. In fact, the standard deviations are 0.0304 for Figure 5.1 and 0.0083 for Figure 5.2.

FIGURE 5.3 Normal quantile plot of the sample proportions in Figure 5.1. The distribution is close to Normal except for some clustering due to the fact that the sample proportions from a sample of size 100 can take only values that are a multiple of 0.01.

Although these results describe just two sets of simulations, they reflect facts that are true whenever we use random sampling.

Bias and variability Our simulations show that a sample of size 1200 will almost always give an estimate P̂ that is close to the truth about the population. Figure 5.2 illustrates this fact for just one value of the population proportion (p = 0.9), but it is true for any proportion. That is a primary reason Student Monitor uses a sample of size of 1200. There is more sampling variability the smaller the sample size. Samples of size 100, for example, might give an estimate of 83% or 97% when the truth is 90%.

Thinking about Figures 5.1 and 5.2 helps us restate the idea of bias when we use a statistic like P̂ to estimate a parameter like p. It also reminds us that variability matters as much as bias.

BIAS AND VARIABILITY

Bias concerns the center of the sampling distribution. A statistic used to estimate a parameter is an unbiased estimator if the mean of its sampling distribution is equal to the true value of the parameter being estimated.

The variability of a statistic is described by the spread of its sampling distribution. This spread is determined by the sampling design and the sample size n. Statistics from larger probability samples have smaller spreads.

The margin of error is a numerical measure of the spread of a sampling distribution. It can be used to set bounds on the size of the likely error in using the statistic as an estimator of a population parameter.

We can think of the true value of the population parameter as the bull’s-eye on a target and of the sample statistic as an arrow fired at the bull’s-eye. Bias and variability describe what happens when an archer fires many arrows at the target. Bias means that the aim is off, and the arrows do not center about the bull’s-eye. Large variability means that arrows are widely scattered about the target. In other words, there is a lack of precision, or consistency, among the arrows. Figure 5.4 shows this target illustration of the two types of error.

Notice that small variability (repeated shots are close together) can accompany large bias (the arrows are consistently away from the bull’s-eye in one direction). And small bias (the arrows center on the bull’s-eye) can accompany large variability (repeated shots are widely scattered). A good sampling scheme, like a good archer, must have both small bias and small variability. Here’s how we do this.

FIGURE 5.4 Bias and variability in shooting arrows at a target. Bias means the shots do not center around the bull’s-eye. Variability means that the shots are scattered.

MANAGING BIAS AND VARIABILITY

To reduce bias, use random sampling. When we start with a list of the entire population, simple random sampling produces unbiased estimates—the values of a statistic computed from an SRS neither consistently overestimate nor consistently underestimate the value of the population parameter.

To reduce the variability of a statistic from an SRS, use a larger sample. You can make the variability as small as you want by taking a large enough sample.

In practice, Student Monitor takes only one random sample. We don’t know how close to the truth the estimate from this one sample is because we don’t know what the true population parameter value is. But large random samples almost always give an estimate that is close to the truth. Looking at the pattern of many samples when n = 1200 shows that we can trust the result of one sample.

Similarly, the Current Population Survey’s sample of about 60,000 households estimates the national unemployment rate very accurately. Of course, only probability samples carry this guarantee. Using a probability sampling design and taking care to deal with practical difficulties reduce bias in a sample.

The size of the sample then determines how close to the population truth the sample result is likely to fall. Results from a sample survey usually come with a margin of error that sets bounds on the size of the likely error. The margin of error directly reflects the variability of the sample statistic, so it is smaller for larger samples. We will describe the details of its calculation in later chapters.

5.4

5.5

USE YOUR KNOWLEDGE

Bigger is better? Radio talk shows often report opinion polls based on tens of thousands of listeners. These sample sizes are typically much larger than those used in opinion polls that incorporate probability sampling. Does a larger sample size mean more trustworthy results? Explain your answer.

Effect of sample size on the sampling distribution. You are planning an opinion study and are considering taking an SRS of either 200 or 600 people. Explain how the sampling distributions of the population proportion p would differ in terms of center and spread for these two scenarios.

Sampling from large populations Student Monitor’s sample of 1200 students is only about 1 out of every 90,000 undergraduate students in the United States. Does it matter whether we sample 1-in-1000 individuals in the population or 1-in-90,000?

LARGE POPULATIONS DO NOT REQUIRE LARGE SAMPLES

The variability of a statistic from a random sample depends little on the size of the population, as long as the population is at least 20 times larger than the sample.

Why does the size of the population have little influence on the behavior of statistics from random samples? To see why this is plausible, imagine sampling harvested corn by thrusting a scoop into a lot of corn kernels. The scoop doesn’t know whether it is surrounded by a bag of corn or by an entire truckload. As long as the corn is well mixed (so that the scoop selects a random sample), the variability of the result depends only on the size of the scoop.

The fact that the variability of sample results is controlled by the size of the sample has important consequences for sampling design. An SRS of size 1200 from the 10.5 million undergraduates gives results as precise as an SRS of size 1200 from the roughly 156,000 inhabitants of San Francisco between the ages of 20 and 29. This is good news for designers of national samples but bad news for those who want accurate information about these citizens of San Francisco. If both use an SRS, both must use the same size sample to obtain equally trustworthy results.

Why randomize? Why randomize? The act of randomizing guarantees that the results of analyzing our data are subject to the laws of probability. The behavior of statistics is described by a sampling distribution. The form of the distribution is known and, in many cases, is approximately Normal. Often, the center of the distribution lies at the true parameter value so that the notion that randomization eliminates bias is made more explicit. The spread of the distribution describes the variability of the statistic and can be made as small as we wish by choosing a large enough sample. In a randomized experiment, we can reduce variability by choosing larger groups of subjects for each treatment.

These facts are at the heart of formal statistical inference. The remainder of this chapter has much to say in more technical language about sampling distributions. Later chapters describe the way statistical conclusions are based on them. What any user of statistics must understand is that all the technical talk has its basis in a simple question: What would happen if the sample or the experiment were repeated many times? The reasoning applies not only to an SRS, but also to the complex sampling designs actually used by opinion polls and other national sample surveys. The same conclusions hold as well for randomized experimental designs. The details vary with the design, but the basic facts are true whenever randomization is used to produce data.

Remember that proper statistical design is not the only aspect of a good sample or experiment. The sampling distribution shows only how a statistic varies due to the operation of chance in randomization. It reveals nothing about possible bias due to undercoverage or nonresponse in a sample (page 196) or to lack of realism in an experiment. The actual error in estimating a parameter by a statistic can be much larger than the sampling distribution suggests. What is worse, there is no way to say how large the added error is. The real world is less orderly than statistics textbooks imply.

In the next two sections, we will study the sampling distributions of two common statistics, the sample mean and the sample proportion. The focus will be on the important features of these distributions so that we can quickly describe and use them in the later chapters on statistical inference. We will see that, in each case, the sampling distribution depends on both the population and the way we collect the data from the population.

SECTION 5.1 SUMMARY A number that describes a population is a parameter. A number that describes a sample (is computed from the sample data) is a statistic. The purpose of sampling or experimentation is usually inference: use sample statistics to make statements about unknown population parameters. A statistic from a probability sample or a randomized experiment has a sampling distribution that describes how the statistic varies in repeated data productions. The sampling distribution answers the question “What would happen if we repeated the sample or experiment many times?” Formal statistical inference is based on the sampling distributions of statistics. A statistic as an estimator of a parameter may suffer from bias or from high variability. Bias means that the center of the sampling distribution is not equal to the true value of the parameter. The variability of the statistic is described by the spread of its sampling distribution. Variability is usually reported by giving a margin of error for conclusions based on sample results. Properly chosen statistics from randomized data production designs have no bias resulting from the way the sample is selected or the way the experimental units are assigned to treatments. We can reduce the variability of the statistic by increasing the size of the sample or the size of the experimental groups.

SECTION 5.1 EXERCISES For Exercise 5.1, see page 283; for Exercise 5.2, see page 284; for Exercise 5.3, see page 286; and for Exercises 5.4 and 5.5, see page 289.

5.6 Web polls. If you connect to the website peopleschoice.com/pca/polls/polls.jsp, you are given the opportunity to vote on various entertainment questions. Can you apply the ideas about populations and samples to these polls? Explain why or why not.

5.7 What population and sample? Thirty students from your college who are majoring in English are randomly selected to be on a committee to evaluate immediate changes in the statistics requirement for the major. There are 153 English majors at your college. The current rules say that a statistics course is one of three options for a quantitative competency requirement. The proposed change would be to require a statistics course. Each of the committee members is asked to vote Yes or No on the new requirement.

(a) Describe the population for this setting.

(b) What is the sample?

(c) Describe the statistic and how it would be calculated.

(d) What is the population parameter?

(e) Write a short summary based on your answers to parts (a) through (d) using this setting to explain population, sample, parameter, statistic, and the relationships among these items.

5.8 What’s wrong? State what is wrong in each of the following scenarios.

(a) A parameter describes a sample.

(b) Bias and variability are two names for the same thing.

(c) Large samples are always better than small samples.

(d) A sampling distribution is something generated by a computer.

5.9 Describe the population and the sample. For each of the following situations, describe the population and the sample.

(a) A survey of 17,096 students in U.S. four-year colleges reported that 19.4% were binge drinkers.

(b) In a study of work stress, 100 restaurant workers were asked about the impact of work stress on their personal lives.

(c) A tract of forest has 584 longleaf pine trees. The diameters of 40 of these trees were measured.

5.10 Is it unbiased? A statistic has a sampling distribution that is somewhat skewed. The mean is 17, the median is 15, the quartiles are 13 and 19.

(a) If the population parameter is 15, is the estimator unbiased?

(b) If the population parameter is 17, is the estimator unbiased?

(c) If the population parameter is 16, is the estimator unbiased?

(d) Write a short summary of your results in parts (a), (b), and (c) and include a discussion of bias and unbiased estimators.

5.11 Constructing a sampling distribution. Refer to Example 5.1 (page 283). Suppose Student Monitor also reported that the median number of hours per week spent on the Internet was 12.5 hours.

(a) Explain why we’d expect the population median to be less than the population mean in this setting by drawing the distribution of times spent on the Internet for all undergraduates. This is called the population distribution.

(b) Using Figure 5.2 as a guide and your distribution from part (a), describe how to approximate the sampling distribution of the sample median in this setting.

5.12 Bias and variability. Figure 5.5 shows histograms of four sampling distributions of statistics intended to estimate the same parameter. Label each distribution relative to the others as high or low bias and as high or low variability.

5.13 Constructing sampling distributions. The Probability applet simulates tossing a coin, with the advantage that you can choose the true long-term proportion, or probability, of a head. Suppose that we have a population in which

proportion p = 0.4 (the parameter) plan to vote in the next election. Tossing a coin with probability p = 0.4 of a head simulates this situation: each head is a person who plans to vote, and each tail is a person who does not. Set the “Probability of heads” in the applet to 0.4 and the number of tosses to 25. This simulates an SRS of size 25 from this population. By alternating between “Toss” and “Reset,” you can take many samples quickly.

FIGURE 5.5 Determine which of these sampling distributions displays high or low bias and high or low variability, Exercise 5.12.

(a) Take 50 samples, recording the number of heads in each sample. Make a histogram of the 50 sample proportions (count of heads divided by 25). You are constructing the sampling distribution of this statistic.

(b) Another population contains only 20% who plan to vote in the next election. Take 50 samples of size 25 from this population, record the number in each sample who approve, and make a histogram of the 50 sample proportions.

5.14 Comparing sampling distributions. Refer to the previous exercise.

(a) How do the centers of your two histograms reflect the differing truths about the two populations?

(b) Describe any differences in the shapes of the two histograms. Is one more skewed than the other?

(c) Compare the spreads of the two histograms. For which population is there less sampling variability?

(d) Suppose instead that the population proportions were 0.6 and 0.8, respectively. Describe how the sampling distributions of P̂ would differ from those constructed in Exercise 5.13.

5.15 Use the Simple Random Sample applet. The Simple Random Sample applet can illustrate the idea of a sampling distribution. Form a population labeled 1 to 100. We will choose an SRS of 15 of these numbers. That is, in this

exercise, the numbers themselves are the population, not just labels for 100 individuals. The mean of the whole numbers 1 to 100 is 50.5. This is the parameter, the mean of the population.

(a) Use the applet to choose an SRS of size 15. Which 15 numbers were chosen? What is their mean? This is a statistic, the sample mean x̄.

(b) Although the population and its mean 50.5 remain fixed, the sample mean changes as we take more samples. Take another SRS of size 15. (Use the “Reset” button to return to the original population before taking the second sample.) What are the 15 numbers in your sample? What is their mean? This is another value of x̄.

(c) Take 18 more SRSs from this same population and record their means. You now have 20 values of the sample mean x̄ from 20 SRSs of the same size from the same population. Make a histogram of the 20 values and mark the population mean 50.5 on the horizontal axis. Are your 20 sample values roughly centered at the population value? (If you kept going forever, your x̄-values would form the sampling distribution of the sample mean; the population mean would indeed be the center of this distribution.)

5.16 Use the Simple Random Sample applet, continued. Refer to the previous exercise.

(a) Suppose instead that a sample size of n = 10 was used. Based on what you know about the effect of the sample size on the sampling distribution, which sampling distribution should have the smaller variability?

(b) Repeat the previous exercise using n = 10. Did your simulations confirm your answer in part (a)? Explain your answer.

(c) Write a short paragraph about the effect of the sample size on the variability of a sampling distribution using these simulations to illustrate the basic idea.

5.2 The Sampling Distribution of a Sample Mean

When you complete this section, you will be able to:

Explain the difference between the sampling distribution of x̄ and the population distribution. Determine the mean and standard deviation of x̄ for an SRS of size n from a population with mean μ and standard deviation σ. Describe how much larger n has to be for an SRS to reduce the standard deviation of x̄ by a certain factor. Utilize the central limit theorem to approximate the sampling distribution of x̄ and perform probability calculations based on this approximation.

A variety of statistics are used to describe quantitative data. The sample mean, median, and standard deviation are all examples of statistics based on quantitative data. Statistical theory describes the sampling distributions of these statistics. However, the general framework for constructing a sampling distribution is the same for all statistics. In this section, we will concentrate on the sample mean. Because sample means are just averages of observations, they are among the most frequently used statistics.

Suppose that you plan to survey 1000 undergraduates enrolled in four-year U.S. universities about their sleeping habits. The sampling distribution of the average hours of sleep per night describes what this average would be if many simple random samples of 1000 students were drawn from the population of students in the United States. In other words, it gives you an idea of what you are likely to see from your survey. It tells you whether you should expect this average to be near the population mean and whether the variation of the statistic is roughly ±2 hours or ±2 minutes.

density curves, p. 51

Before constructing this distribution, however, we need to consider another set of probability distributions that also plays a role in statistical inference. Any quantity that can be measured on each member of a population is described by the distribution of its values for all members of the population. This is the context in which we first met distributions, as density curves that provide models for the overall pattern of data.

Imagine choosing one individual at random from a population and measuring a quantity. The quantities obtained from repeated draws of one individual from a population have a probability distribution that is the distribution of the population.

EXAMPLE 5.4

Total sleep time of college students. A recent survey describes the distribution of total sleep time among college students as approximately Normal with a mean of 6.78 hours and standard deviation of 1.24 hours.3 Suppose that we select a college student at random and obtain his or her sleep time. This result is a random variable X because, prior to the random sampling, we don’t know the sleep time. We do know, however, that in repeated sampling, X will have the same N(6.78, 1.24) distribution that describes the pattern of sleep time in the entire population. We call N(6.78, 1.24) the population distribution.

POPULATION DISTRIBUTION

The population distribution of a variable is the distribution of its values for all members of the population. The population distribution is also the probability distribution of the variable when we choose one individual at random from the population.

5.17

In this example, the population of all college students actually exists so that we can, in principle, draw an SRS of students from it. Sometimes, our population of interest does not actually exist. For example, suppose that we are interested in studying final-exam scores in a statistics course, and we have the scores of the 34 students who took the course last semester. For the purposes of statistical inference, we might want to consider these 34 students as part of a hypothetical population of similar students who would take this course. In this sense, these 34 students represent not only themselves, but also a larger population of similar students. The key idea is to think of the observations that you have as coming from a population with a probability distribution.

USE YOUR KNOWLEDGE

Time spent using apps on a mobile device. Nielsen has installed, with permission, Mobile Netview 3 on approximately 5000 cell phones to gather information on mobile app usage among adults in the United States. Nielsen reported that 18–24 year olds spend an average of 37 hours and 6 minutes a month using mobile apps.4 State the population that this survey describes, the statistic, and some likely values from the population distribution.

Now that we have made the distinction between the population distributions and sampling distributions, we can proceed with an in-depth study of the sampling distribution of a sample mean x̄.

EXAMPLE 5.5

HELP60

Sample means are approximately Normal. Figure 5.6 illustrates two striking facts about the sampling distribution of a sample mean. Figure 5.6(a) displays the distribution of student visit lengths (in minutes) to a statistics help room at a large midwestern university. Students visiting the help room were asked to sign in upon arrival and then sign out when leaving. During the school year, there were 1838 visits to the help room but only 1264 recorded visit lengths. This is because many visiting students forgot to sign out. We also omitted a few large outliers (visits lasting more than 10 hours).5 The distribution is strongly skewed to the right. The population mean is μ = 61.28 minutes.

FIGURE 5.6 (a) The distribution of visit lengths to a statistics help room during the school year, Example 5.5. (b) The distribution of the sample means x̅ for 500 random samples of size 60 from this population. The scales and

TABLE 5.1

histogram classes are exactly the same in both panels.

Length (in Minutes) of 60 Visits to a Statistics Help Room

10 14 15 16 18 20 20 20 23 25 28 30 30 30 30 30 31 33 35 35 46 48 50 50 50 50 51 54 55 55 60 60 60 60 60 60 60 65 65 65 75 77 80 80 84 85 88 98 100 100 105 105 105 115 120 135 135 136 157 210

Table 5.1 contains the lengths of a random sample of 60 visits from this population. The mean of these 60 visits is x̄ = 63.45 minutes. If we were to take another sample of size 60, we would likely get a different value of x̄. This is because this new sample would contain a different set of visits. To find the sampling distribution of x̄, we take many SRSs of size 60 and calculate x̄ for each sample. Figure 5.6(b) is the distribution of the values of x̄ for 500 random samples. The scales and choice of classes are exactly the same as in Figure 5.6(a) so that we can make a direct comparison.

The sample means are much less spread out than the individual visit lengths. What is more, the Normal quantile plot in Figure 5.7 confirms that the distribution in Figure 5.6(b) is close to Normal.

FIGURE 5.7 Normal quantile plot of the 500 sample means in Figure 5.6(b). The distribution is close to Normal.

This example illustrates two important facts about sample means that we will discuss in this section.

FACTS ABOUT SAMPLE MEANS

1. Sample means are less variable than individual observations. 2. Sample means are more Normal than individual observations.

These two facts contribute to the popularity of sample means in statistical inference.

The mean and standard deviation of x̅ The sample mean x̄ from a sample or an experiment is an estimate of the mean μ of the underlying population. The sampling distribution of x̄ is determined by

the design used to produce the data, the sample size n, and the population distribution.

Select an SRS of size n from a population, and measure a variable X on each individual in the sample. The n measurements are values of n random variables X1, X2,. . ., Xn. A single Xi is a measurement on one individual selected at random from the population and, therefore, has the distribution of the population. If the population is large relative to the sample, we can consider X1, X2, . . . , Xn to be independent random variables, each having the same distribution. This is our probability model for measurements on each individual in an SRS.

The sample mean of an SRS of size n is

rules for means, p. 254

If the population has mean μ, then μ is the mean of the distribution of each observation Xi. To get the mean of x̄, we use the rules for means of random variables. Specifically,

rules for variances, p. 258

That is, the mean of x̄ is the same as the mean of the population. The sample mean x̄ is, therefore, an unbiased estimator of the unknown population mean μ.

The observations are independent, so the addition rule for variances also applies:

With n in the denominator, the variability of x̄ about its mean decreases as the sample size grows. Thus, a sample mean from a large sample will usually be very close to the true population mean μ. Here is a summary of these facts.

MEAN AND STANDARD DEVIATION OF A SAMPLE MEAN

Let x̄ be the mean of an SRS of size n from a population having mean μ and standard deviation σ. The mean and standard deviation of x̄ are

5.18

5.19

How precisely does a sample mean x̄ estimate a population mean μ? Because the values of x̄ vary from sample to sample, we must give an answer in terms of the sampling distribution. We know that x̄ is an unbiased estimator of μ, so its values in repeated samples are not systematically too high or too low. Most samples will give an x̄-value close to μ if the sampling distribution is concentrated close to its mean μ. So the precision of estimation depends on the spread of the sampling distribution.

Because the standard deviation of x̄ is σ/ , the standard deviation of the statistic decreases in proportion to the square root of the sample size. This means, for example, that a sample size must be multiplied by 4 in order to divide the statistic’s standard deviation in half. By comparison, a sample size must be multiplied by 100 in order to reduce the standard deviation by a factor of 10.

EXAMPLE 5.6

Standard deviations for sample means of visit lengths. The standard deviation of the population of visit lengths in Figure 5.6(a) is σ = 41.84 minutes. The length of a single visit will often be far from the population mean. If we choose an SRS of 15 visits, the standard deviation of their mean length is

Averaging over more visits reduces the variability and makes it more likely that x̄ is close to μ. Our sample size of 60 visits is 4 times 15, so the standard deviation will be half as large:

USE YOUR KNOWLEDGE

Find the mean and the standard deviation of the sampling distribution. Compute the mean and standard deviation of the sampling distribution of the sample mean when you plan to take an SRS of size 64 from a population with mean 44 and standard deviation 16.

The effect of increasing the sample size. In the setting of the previous exercise, repeat the calculations for a sample size of 576. Explain the effect of the sample size increase on the mean and standard deviation of the sampling distribution.

5.20

The central limit theorem We have described the center and spread of the probability distribution of a sample mean x̄, but not its shape. The shape of the distribution of x̄ depends on the shape of the population distribution. Here is one important case: if the population distribution is Normal, then so is the distribution of the sample mean.

SAMPLING DISTRIBUTION OF A SAMPLE MEAN

If a population has the N(μ, σ) distribution, then the sample mean x̄ of n independent observations has the N(μ, σ/ ) distribution.

This is a somewhat special result. Many population distributions are not Normal. The help room visit lengths in Figure 5.6(a), for example, are strongly skewed. Yet Figures 5.6(b) and 5.7 show that means of samples of size 60 are close to Normal.

central limit theorem

One of the most famous facts of probability theory says that, for large sample sizes, the distribution of x̄ is close to a Normal distribution. This is true no matter what shape the population distribution has, as long as the population has a finite standard deviation σ. This is the central limit theorem. It is much more useful than the fact that the distribution of x̄ is exactly Normal if the population is exactly Normal.

CENTRAL LIMIT THEOREM

Draw an SRS of size n from any population with mean μ and finite standard deviation σ. When n is large, the sampling distribution of the sample mean x̄ is approximately Normal:

EXAMPLE 5.7

How close will the sample mean be to the population mean? With the Normal distribution to work with, we can better describe how precisely a random sample of 60 visits estimates the mean length of all visits to the help room. The population standard deviation for the 1264 visits in the population of Figure 5.6(a) is σ = 41.84 minutes. From Example 5.6 we know minutes. By the 95 part of the 68–95–99.7 rule, about 95% of all samples will have mean x̄ within two standard deviations of μ, that is, within ±10.8 minutes of μ.

68–95–99.7 rule, p. 57

USE YOUR KNOWLEDGE

Use the 68–95–99.7 rule. You take an SRS of size 64 from a population with mean 82 and standard deviation 24. According to the central limit theorem, what is the approximate sampling distribution of the sample mean? Use the 95 part of the 68–95–99.7 rule to describe the variability of x̄.

5.21

For the sample size of n = 60 in Example 5.7, the sample mean is not very precise. The population of help room visit lengths is very spread out, so the sampling distribution of x̄ has a large standard deviation.

EXAMPLE 5.8

How can we reduce the standard deviation? In the setting of Example 5.7, if we want to reduce the standard deviation of x̄ by a factor of 2, we must take a sample four times as large, n = 4 × 60, or 240. Then

For samples of size 240, about 95% of the sample means will be within twice 2.70, or 5.40 minutes, of the population mean μ.

finite population correction factor

The standard deviation computed in Example 5.8 is actually too large. This is due to the fact that the population size, N = 1264, is not at least 20 times larger than the sample size, n = 240. In these settings, it is better to adjust the standard deviation of x̄ to reflect only the variance remaining in the population that is not in the sample. This is done by

multiplying the unadjusted standard deviation by the finite population correction factor. This quantity is and moves the standard deviation of x̄ toward 0 as n moves toward N. Applying this correction to Example 5.8, the standard deviation of x̄ is reduced 10% to

Thus, for samples of size 240, about 95% of the sample means will be within twice 2.43, or 4.86 minutes, of the population mean m, rather than the 5.40 minutes reported in Example 5.8.

USE YOUR KNOWLEDGE

The effect of increasing the sample size. In the setting of Exercise 5.20, suppose that we increase the sample size to 2304. Use the 95 part of the 68–95–99.7 rule to describe the variability of this sample mean. Compare your results with those you found in Exercise 5.20.

Example 5.8 reminds us that if the population is very spread out, the in the formula for the deviation of x̄ implies that very large samples are needed to estimate the population mean precisely. The main point of the example, however, is that the central limit theorem allows us to use Normal probability calculations to answer questions about sample means even when the population distribution is not Normal.

How large a sample size n is needed for x̄ to be close to Normal depends on the population distribution. More observations are required if the shape of the population distribution is far from Normal. For the very skewed visit length population, samples of size 60 are large enough. Further study would be needed to see if the distribution of x̄ is close to Normal for smaller samples like n = 20 or n = 40. Here is a more detailed study of another skewed distribution.

EXAMPLE 5.9

The central limit theorem in action. Figure 5.8 shows the central limit theorem in action for another very non-Normal population. Figure 5.8(a) displays the density curve of a single observation from the population. The distribution is strongly right-skewed, and the most probable outcomes are near 0. The mean μ of this distribution is 1, and its standard deviation s is also 1. This particular continuous distribution is called an exponential distribution. Exponential distributions are used as models for how long an iPhone will function properly and for the time between snaps you receive on Snapchat.

Exponential distributions

FIGURE 5.8 The central limit theorem in action: the sampling distribution of sample means from a strongly non- Normal population becomes more Normal as the sample size increases, Example 5.9. (a) The distribution of 1 observation. (b) The distribution of for 2 observations. (c) The distribution of for 10 observations. (d) The distribution of for 25 observations.

Figures 5.8(b), (c), and (d) are the density curves of the sample means of 2, 10, and 25 observations from this population. As n increases, the shape becomes more Normal. The mean remains at μ = 1, but the standard deviation decreases, taking the value 1/ . The density curve for 10 observations is still somewhat skewed to the right but already resembles a Normal curve having μ = 1 and σ = 1/ = 0.32. The density curve for n = 25 is yet more Normal. The contrast between the shape of the population distribution and of the distribution of the mean of 10 or 25 observations is striking.

You can also use the Central Limit Theorem applet to study the sampling distribution of x̄. From one of three population distributions, 10,000 SRSs of a user-specified sample size n are generated, and a histogram of the sample means is constructed. You can then compare this estimated sampling distribution with the Normal curve that is based on the central limit theorem.

EXAMPLE 5.10

Using the Central Limit Theorem applet. In Example 5.9, we considered sample sizes of n = 2, 10, and 25 from an exponential distribution. Figure 5.9 shows a screenshot of the Central Limit Theorem applet for the exponential distribution when n = 10. The mean and standard deviation of this sampling distribution are 1 and ,

5.22

5.23

respectively. From the 10,000 SRSs, the mean is estimated to be 1.001 and the estimated standard deviation is 0.319. These are both quite close to the true values. In Figure 5.8(c), we saw that the density curve for 10 observations is still somewhat skewed to the right. We can see this same behavior in Figure 5.9 when we compare the histogram with the Normal curve based on the central limit theorem.

FIGURE 5.9 Screenshot of the Central Limit Theorem applet for the exponential distribution when n = 10, Example 5.10.

Try using the applet for the other sample sizes in Example 5.9. You should get histograms shaped like the density curves shown in Figure 5.8. You can also consider other sample sizes by sliding n from 1 to 100. As you increase n, the shape of the histogram moves closer to the Normal curve that is based on the central limit theorem.

USE YOUR KNOWLEDGE

Use the Central Limit Theorem applet. Let’s consider the uniform distribution between 0 and 10. For this distribution, all intervals of the same length between 0 and 10 are equally likely. This distribution has a mean of 5 and standard deviation of 2.89.

(a) Approximate the population distribution by setting n = 1 and clicking the “Generate samples” button.

(b) What are your estimates of the population mean and population standard deviation based on the 10,000 SRSs? Are these population estimates close to the true values?

(c) Describe the shape of the histogram and compare it with the Normal curve.

Use the Central Limit Theorem applet again. Refer to the previous exercise. In the setting of Example 5.9, let’s approximate the sampling distribution for samples of size n = 2, 10, and 25 observations.

(a) For each sample size, compute the mean and standard deviation of .

(b) For each sample size, use the applet to approximate the sampling distribution. Report the estimated mean and standard deviation. Are they close to the true values calculated in part (a)?

(c) For each sample size, compare the shape of the sampling distribution with the Normal curve based on the central limit theorem.

(d) For this population distribution, what sample size do you think is needed to make you feel comfortable using the central limit theorem to approximate the sampling distribution of ? Explain your answer.

Now that we know that the sampling distribution of the sample mean is approximately Normal for a sufficiently large n, let’s consider some probability calculations.

EXAMPLE 5.11

Time between snaps. Snapchat has more than 100 million daily users sending well over 400 million snaps a day.6 Suppose that the time X between snaps received is governed by the exponential distribution with mean μ = 15 minutes and standard deviation σ = 15 minutes. You record the next 50 times between snaps. What is the probability that their average exceeds 13 minutes?

The central limit theorem says that the sample mean time (in minutes) between snaps has approximately the Normal distribution with mean equal to the population mean μ = 15 minutes and standard deviation

The sampling distribution of is, therefore, approximately N(15,2.12). Figure 5.10 shows this Normal curve (solid) and also the actual density curve of (dashed).

The probability we want is . This is the area to the right of 13 under the solid Normal curve in Figure 5.10. A Normal distribution calculation gives

FIGURE 5.10 The exact distribution (dashed) and the Normal approximation from the central limit theorem (solid) for the average time between snaps received, Example 5.11.

The exactly correct probability is the area under the dashed density curve in the figure. It is 0.8265. The central limit theorem Normal approximation is off by only about 0.0001.

We can also use this sampling distribution to talk about the total time between the 1st and 51st snap received.

EXAMPLE 5.12

Convert the results to the total time. There are 50 time intervals between the 1st and 51st snap. According to the central limit theorem calculations in Example 5.11,

5.24

We know that the sample mean is the total time divided by 50, so the event is the same as the event . We can say that the probability is 0.8264 that the total time is 50(13.0) = 650 minutes (10.8

hours) or greater.

USE YOUR KNOWLEDGE

Find a probability. Refer to Example 5.11. Find the probability that the mean time between snaps is less than 15 minutes. The exact probability is 0.5188. Compare your answer with the exact one.

Figure 5.11 summarizes the facts about the sampling distribution of in a way that emphasizes the big idea of a sampling distribution. The general framework for constructing the sampling distribution of is shown on the left.

Take many random samples of size n from a population with mean μ and standard deviation σ. Find the sample mean for each sample. Collect all the ’s and display their distribution.

The sampling distribution of is shown on the right. Keep this figure in mind as you go forward.

FIGURE 5.11 The sampling distribution of a sample mean has mean μ and standard deviation . The sampling distribution is Normal if the population distribution is Normal; it is approximately Normal for large samples in any case.

A few more facts The central limit theorem is the big fact of probability theory in this section. Here are three additional facts related to our investigations that will be useful in describing methods of inference in later chapters.

The fact that the sample mean of an SRS from a Normal population has a Normal distribution is a special case of a more general fact: any linear combination of independent Normal random variables is also Normally distributed. That is, if X and Y are independent Normal random variables and a and b are any fixed numbers, aX + bY is also Normally distributed, and this is true for any number of Normal random variables. In particular, the sum or difference of independent Normal random variables has a Normal distribution. The mean and standard deviation of aX + bY are found as usual from the rules for means and variances. These facts are often used in statistical calculations. Here is an example.

rules for means, p. 254 rules for variances, p. 258

EXAMPLE 5.13

Getting to and from campus. You live off campus and take the shuttle, provided by your apartment complex, to and from campus. Your time on the shuttle in minutes varies from day to day. The time going to campus X has the N(20,4) distribution, and the time returning from campus Y varies according to the N(18, 8) distribution. If they vary independently, what is the probability that you will be on the shuttle for less time going to campus?

The difference in times X − Y is Normally distributed, with mean and variance

Because , X − Y has the N(2, 8.94) distribution. Figure 5.12 illustrates the probability computation:

Although, on average, it takes longer to go to campus than return, the trip to campus will take less time on roughly two of every five days.

FIGURE 5.12 The Normal probability calculation, Example 5.13. The difference in times going to campus and returning from campus (X − Y) is Normal with mean 2 minutes and standard deviation 8.94 minutes.

The second useful fact is that more general versions of the central limit theorem say that the distribution of a sum or average of many small random quantities is close to Normal. This is true even if the quantities are not independent (as long as they are not too highly correlated) and even if they have different distributions (as long as no single random quantity is so large that it dominates the others). These more general versions of the central limit theorem suggest why the Normal distributions are common models for observed data. Any variable that is a sum of many small random influences will have approximately a Normal distribution.

Finally, the central limit theorem also applies to discrete random variables. An average of discrete random variables will never result in a continuous sampling distribution, but the Normal distribution often serves as a good approximation. In Section 5.3, we will discuss the sampling distribution and Normal approximation for counts and proportions. This Normal approximation is just an example of the central limit theorem applied to these discrete random variables.

BEYOND THE BASICS

Weibull Distributions Our discussion of sampling distributions so far has concentrated on the Normal model to approximate the sampling distribution of the sample mean . This model is important in statistical practice because of the central limit theorem and the fact that sample means are among the most frequently used statistics. Simplicity also contributes to its popularity. The parameter μ is easy to understand, and to estimate it, we use a statistic that is also easy to understand and compute.

There are, however, many other probability distributions that are used to model data in various circumstances. The time that a product, such as a computer hard drive, lasts before failing rarely has a Normal distribution. Earlier, we mentioned the use of the exponential distribution to model time to failure. Another class of continuous distributions, the Weibull distributions, is more commonly used in these situations.

Weibull distributions

EXAMPLE 5.14

Weibull density curves. Figure 5.13 shows the density curves of three members of the Weibull family. Each describes a different type of distribution for the time to failure of a product.

1. The top curve in Figure 5.13 is a model for infant mortality. This describes products that often fail immediately, prior to delivery to the customer. However, if the product does not fail right away, it will likely last a long time. For products like this, a manufacturer might test them and ship only the ones that do not fail immediately.

2. The middle curve in Figure 5.13 is a model for early failure. These products do not fail immediately, but many fail early in their lives after they are in the hands of customers. This is disastrous—the product or the process that makes it must be changed at once.

3. The bottom curve in Figure 5.13 is a model for old-age wear-out. Most of these products fail only when they begin to wear out, and then many fail at about the same age.

FIGURE 5.13 Density curves for three members of the Weibull family of distributions, Example 5.14.

A manufacturer certainly wants to know to which of these classes a new product belongs. To find out, engineers operate a random sample of products until they fail. From the failure time data, we can estimate the parameter (called the “shape parameter”) that distinguishes among the three Weibull distributions in Figure 5.13. The shape parameter has no simple definition like that of a population proportion or mean, and it cannot be estimated by a simple statistic such as or .

Two things save the situation. First, statistical theory provides general approaches for finding good estimates of any parameter. These general methods not only tell us how to use in the Normal settings, but also how to estimate the Weibull shape parameter. Second, software can calculate the estimate from data even though there is no algebraic formula that we can write for the estimate. Statistical practice often relies on both mathematical theory and methods of computation more elaborate than the ones we will meet in this book. Fortunately, big ideas such as sampling distributions carry over to more complicated situations.7

SECTION 5.2 SUMMARY The population distribution of a variable is the distribution of its values for all members of the population. The sample mean of an SRS of size n drawn from a large population with mean μ and standard deviation σ has a sampling distribution with mean and standard deviation

The sample mean is an unbiased estimator of the population mean m and is less variable than a single observation. The standard deviation decreases in proportion to the square root of the sample size n. This means that to reduce the standard deviation by a factor of C, we need to increase the sample size by a factor of C2.

The central limit theorem states that, for large n, the sampling distribution of is approximately for any population with mean m and finite standard deviation s. This allows us to approximate

probability calculations of using the Normal distribution. Linear combinations of independent Normal random variables have Normal distributions. In particular, if the population has a Normal distribution, so does .

SECTION 5.2 EXERCISES For Exercise 5.17, see page 294; for Exercises 5.18 and 5.19, see page 297; for Exercise 5.20, see page 298; for Exercise 5.21, see page 299; for Exercises 5.22 and 5.23, see pages 301–302; and for Exercise 5.24, see page 303.

5.25 What is wrong? Explain what is wrong in each of the following statements.

(a) If the population standard deviation is 10, then the standard deviation of for an SRS of 10 observations is 10/10 = 1.

(b) When taking SRSs from a population, larger sample sizes will result in larger standard deviations of .

(c) For an SRS from a population, both the mean and the standard deviation of depend on the sample size n.

(d) The larger the population size N, the larger the sample size n needs to be for a desired standard deviation of .

5.26 What is wrong? Explain what is wrong in each of the following statements.

(a) The central limit theorem states that for large n, the population mean μ is approximately Normal.

(b) For large n, the distribution of observed values will be approximately Normal.

(c) For sufficiently large n, the 68–95–99.7 rule says that should be within μ ± 2σ about 95% of the time.

(d) As long as the sample size n is less than half the population size N, the standard deviation of is .

5.27 Generating a sampling distribution. Let’s illustrate the idea of a sampling distribution in the case of a very small sample from a very small population. The population is the 10 scholarship players currently on your women’s basketball team. For convenience, the 10 players have been labeled with the integers 0 to 9. For each player, the total amount of time spent (in minutes) on Twitter during the last week is recorded in the following table.

Player 0 1 2 3 4 5 6 7 8 9

Total time (min) 98 63 137 210 52 88 151 133 105 168

The parameter of interest is the average amount of time on Twitter. The sample is an SRS of size n = 3 drawn from this population of players. Because the players are labeled 0 to 9, a single random digit from Table B chooses one player for the sample.

(a) Find the mean for the 10 players in the population. This is the population mean μ.

(b) Use Table B to draw an SRS of size 3 from this population. (Note: You may sample the same player’s time more than once.) Write down the three times in your sample and calculate the sample mean . This statistic is an estimate of μ.

(c) Repeat this process nine more times using different parts of Table B. Make a histogram of the 10 values of . You are approximating the sampling distribution of .

(d) Is the center of your histogram close to μ? Explain why you’d expect it to get closer to μ the more times you repeated this sampling process.

5.28 Total sleep time of college students. In Example 5.4, the total sleep time per night among college students was approximately Normally distributed with mean μ = 6.78 hours and standard deviation σ = 1.24 hours. You plan to take an SRS of size n = 120 and compute the average total sleep time.

(a) What is the standard deviation for the average time?

(b) Use the 95 part of the 68–95–99.7 rule to describe the variability of this sample mean.

(c) What is the probability that your average will be below 6.9 hours?

5.29 Determining sample size. Refer to the previous exercise. You want to use a sample size such that about 95% of the averages fall within ±5 minutes (0.08 hour) of the true mean μ = 6.78.

(a) Based on your answer to part (b) in Exercise 5.28, should the sample size be larger or smaller than 120? Explain.

(b) What standard deviation of do you need such that approximately 95% of all samples will have a mean within 5 minutes of μ?

(c) Using the standard deviation you calculated in part (b), determine the number of students you need to sample.

5.30 Music file size on a tablet PC. A tablet PC contains 3217 music files. The distribution of file size is highly skewed with many small file sizes. Assume that the standard deviation for this population is 3.25 megabytes (MB).

(a) What is the standard deviation of the average file size when you take an SRS of 25 files from this population?

(b) How many files would you need to sample if you wanted the standard deviation of to be no larger than 0.50 MB?

5.31 Bottling an energy drink. A bottling company uses a filling machine to fill cans with an energy drink. The cans are supposed to contain 250 milliliters (ml). The machine, however, has some variability, so the standard deviation of the volume is σ = 0.4 ml. A sample of five cans is inspected each hour for process control purposes, and records are kept of the sample mean volume. If the process mean is exactly equal to the target value, what is the mean and standard deviation of the numbers recorded?

5.32 Average file size on a tablet. Refer to Exercise 5.30. Suppose that the true mean file size of the music and video files on the tablet is 2.35 MB and you plan to take an SRS of n = 50 files.

(a) Explain why it may be reasonable to assume that the average is approximately Normal even though the population distribution is highly skewed.

(b) Sketch the approximate Normal curve for the sample mean, making sure to specify the mean and standard deviation.

(c) What is the probability that your sample mean will differ from the population mean by more than 0.15 MB?

5.33 Can volumes. Averages are less variable than individual observations. It is reasonable to assume that the can volumes in Exercise 5.31 vary according to a Normal distribution. In that case, the mean of an SRS of cans also has a Normal distribution.

(a) Make a sketch of the Normal curve for a single can. Add the Normal curve for the mean of an SRS of five cans on the same sketch.

(b) What is the probability that the volume of a single randomly chosen can differs from the target value by 0.1 ml or more?

(c) What is the probability that the mean volume of an SRS of five cans differs from the target value by 0.1 ml or more?

5.34 Number of friends on Facebook. To commemorate Facebook’s 10-year milestone, Pew Research reported several facts about Facebook obtained from its Internet Project survey. One was that the average adult user of Facebook has 338 friends. This population distribution takes only integer values, so it is certainly not Normal. It is also highly skewed to the right, with a reported median of 200 friends.8 Suppose that σ = 380 and you take an SRS of 80 adult Facebook users.

(a) For your sample, what are the mean and standard deviation of , the mean number of friends per adult user?

(b) Use the central limit theorem to find the probability that the average number of friends for 80 Facebook users is greater than 350.

(c) What are the mean and standard deviation of the total number of friends in your sample?

(d) What is the probability that the total number of friends among your sample of 80 Facebook users is greater than 28,000?

5.35 Cholesterol levels of teenagers. A study of the health of teenagers plans to measure the blood cholesterol level of an SRS of 13- to 16-year olds. The researchers will report the mean from their sample as an estimate of the mean

cholesterol level μ in this population.

(a) Explain to someone who knows no statistics what it means to say that is an “unbiased” estimator of μ.

(b) The sample result is an unbiased estimator of the population truth μ no matter what size SRS the study chooses. Explain to someone who knows no statistics why a large sample gives more trustworthy results than a small sample.

5.36 Grades in a math course. Indiana University posts the grade distributions for its courses online.9 Students in one section of Math 118 in the fall semester received 18% A’s, 31% B’s, 26% C’s, 13% D’s and 12% F’s.

(a) Using the common scale A = 4, B = 3, C = 2, D = 1, F = 0, take X to be the grade of a randomly chosen Math 118 student. Use the definitions of the mean (page 248) and standard deviation (page 256) for discrete random variables to find the mean μ and the standard deviation σ of grades in this course.

(b) Math 118 is a large enough course that we can take the grades of an SRS of 25 students and not worry about the finite population correction factor. If is the average of these 25 grades, what are the mean and standard deviation of ?

(c) What is the probability that a randomly chosen Math 118 student gets a B or better, P(X ≥ 3)?

(d) What is the approximate probability that the grade point average for 25 randomly chosen Math 118 students is B or better,

?

5.37 Monitoring the emerald ash borer. The emerald ash borer is a beetle that poses a serious threat to ash trees. Purple traps are often used to detect or monitor populations of this pest. In the counties of your state where the beetle is present, thousands of traps are used to monitor the population. These traps are checked periodically. The distribution of beetle counts per trap is discrete and strongly skewed. A majority of traps have no beetles, and only a few will have more than two beetles. For this exercise, assume that the mean number of beetles trapped is 0.4 with a standard deviation of 0.9.

(a) Suppose that your state does not have the resources to check all the traps, so it plans to check only an SRS of n = 100 traps. What are the mean and standard deviation of the average number of beetles in 100 traps?

(b) Use the central limit theorem to find the probability that the average number of beetles in 100 traps is greater than 0.5.

(c) Do you think it is appropriate in this situation to use the central limit theorem? Explain your answer.

5.38 Risks and insurance. The idea of insurance is that we all face risks that are unlikely but carry high cost. Think of a fire destroying your home. So we form a group to share the risk: we all pay a small amount, and the insurance policy pays a large amount to those few of us whose homes burn down. An insurance company looks at the records for millions of homeowners and sees that the mean loss from fire in a year is μ = $500 per house and that the standard deviation of the loss is σ = $10,000. (The distribution of losses is extremely right-skewed: most people have $0 loss, but a few have large losses.) The company plans to sell fire insurance for $500 plus enough to cover its costs and profit.

(a) Explain clearly why it would be unwise to sell only 100 policies. Then explain why selling many thousands of such policies is a safe business.

(b) Suppose the company sells the policies for $600. If the company sells 50,000 policies, what is the approximate probability that the average loss in a year will be greater than $600?

5.39 Weights of airline passengers. In 2005, the Federal Aviation Administration (FAA) updated its passenger weight standards to an average of 190 pounds in the summer (195 in the winter). This includes clothing and carry-on baggage. The FAA, however, did not specify a standard deviation. A reasonable standard deviation is 35 pounds. Weights are not Normally distributed, especially when the population includes both men and women, but they are not very non-Normal. A commuter plane carries 25 passengers. What is the approximate probability that, in the summer, the total weight of the passengers exceeds 5200 pounds? (Hint: To apply the central limit theorem, restate the problem in terms of the mean weight.)

5.40 Iron depletion without anemia and physical performance. Several studies have shown a link between iron depletion without anemia (IDNA) and physical performance. In one recent study, the physical performance of 24 female collegiate rowers with IDNA was compared with 24 female collegiate rowers with normal iron status.10 Several different measures of physical performance were studied, but we’ll focus here on training-session duration. Assume that training-session duration of female rowers with IDNA is Normally distributed, with mean 58 minutes and standard deviation 11 minutes. Training-session duration of female rowers with normal iron status is Normally distributed, with mean 69 minutes and standard deviation 18 minutes.

(a) What is the probability that the mean duration of the 24 rowers with IDNA exceeds 63 minutes?

(b) What is the probability that the mean duration of the 24 rowers with normal iron status is less than 63 minutes?

(c) What is the probability that the mean duration of the 24 rowers with IDNA is greater than the mean duration of the 24 rowers with normal iron status?

5.41 Treatment and control groups. The previous exercise illustrates a common setting for statistical inference. This exercise gives the general form of the sampling distribution needed in this setting. We have a sample of n

observations from a treatment group and an independent sample of m observations from a control group. Suppose that the response to the treatment has the N(μX, σX) distribution and that the response of control subjects has the N(μY, σY) distribution. Inference about the difference μY − μX between the population means is based on the difference between the sample means in the two groups.

(a) Under the assumptions given, what is the distribution of ? Of ?

(b) What is the distribution of ?

5.42 Investments in two funds. Jennifer invests her money in a portfolio that consists of 70% Fidelity Spartan 500 Index Fund and 30% Fidelity Diversified International Fund. Suppose that, in the long run, the annual real return X on the

500 Index Fund has mean 10% and standard deviation 15%, the annual real return Y on the Diversified International Fund has mean 9% and standard deviation 19%, and the correlation between X and Y is 0.6.

(a) The return on Jennifer’s portfolio is R = 0.7X + 0.3Y. What are the mean and standard deviation of R?

(b) The distribution of returns is typically roughly symmetric but with more extreme high and low observations than a Normal distribution. The average return over a number of years, however, is close to Normal. If Jennifer holds her portfolio for 20 years, what is the approximate probability that her average return is less than 5%?

(c) The calculation you just made is not overly helpful because Jennifer isn’t really concerned about the mean return . To see why, suppose that her portfolio returns 12% this year and 6% next year. The mean return for the two years is 9%. If Jennifer starts with $1000, how much does she have at the end of the first year? At the end of the second year? How does this amount compare with what she would have if both years had the mean return, 9%? Over 20 years, there may be a large difference between the ordinary mean and the geometric mean, which reflects the fact that returns in successive years multiply rather than add.

5.3 Sampling Distributions for Counts and Proportions

When you complete this section, you will be able to:

Determine when a count X can be modeled using the binomial distribution. Determine when the sampling distribution of a count can be modeled using the binomial distribution. Calculate the mean and standard deviation of X when it has the B(n, p) distribution. Explain the difference between the sampling distribution of a count X and the sampling distribution of the sample proportion . Determine when one can approximate the sampling distribution of a count using the Normal distribution. Determine when one can approximate the sampling distribution of the sample proportion using the Normal distribution. Use the Normal approximation for counts and proportions to perform probability calculations about the statistics.

In the previous section, we discussed the probability distribution of the sample mean, which meant a focus on population values that were quantitative. We will now shift our focus to population values that are categorical. Counts and proportions are common discrete statistics that describe categorical data.

categorical variable, p. 3

In Section 5.1 (pages 283–287), we discussed the use of simulation to study the sampling distribution of the sample proportion. In this section, we will use probability theory to more precisely describe the sampling distributions of the sample count and proportion. Let’s start with an example.

EXAMPLE 5.15

Work hours make it difficult to spend time with children. A sample survey asks 1006 British parents whether they think long working hours are making it difficult to spend enough time with their children.11 We would like to view the responses of these parents as representative of a larger population of British parents who hold similar beliefs. That is, we will view the responses of the sampled parents as an SRS from a population.

When there are only two possible outcomes for a random variable, we can summarize the results by giving the count for one of the possible outcomes. We let n represent the sample size, and we use X to represent the random variable that gives the count for the outcome of interest.

5.43

5.44

5.45

EXAMPLE 5.16

The random variable of interest. In this sample survey of British parents, n = 1006. The parents in the sample were asked if they agree with the statement “These days, long working hours make it difficult for parents to spend enough time with their children.” The variable X is the number of parents who agreed with the statement. In this case, X = 755.

In our example, we chose the random variable X to be the number of parents who think that long working hours make it difficult to spend enough time with their children. We could have chosen X to be the number of parents who do not think that long working hours make it difficult to spend enough time with their children. The choice is yours. Often, we make the choice based on how we would like to describe the results in a summary. Which choice do you prefer in this case?

sample proportion, p. 283

When a random variable has only two possible outcomes, it is more common to use the sample proportion as the summary rather than the count X.

EXAMPLE 5.17

The sample proportion. The sample proportion of parents surveyed who think that long working hours make it difficult to spend enough time with their children is

Notice that this summary takes into account the sample size n. We need to know n in order to properly interpret the meaning of the random variable X. For example, the conclusion we would draw about parent opinions in this survey would be quite different if we had observed X = 755 from a sample twice as large, n = 2012.

USE YOUR KNOWLEDGE

Sexual harassment in middle school. A survey of 1391 students in grades 5 to 8 reports that 26% of the students say they have encountered some type of sexual harassment while at school.12 Give the sample size n, the count X, and the sample proportion for this survey.

High school graduates who took a statistics course. In a random sample of n = 4012 high school graduates, 10.8% reported that they had taken a statistics course.13 Give the sample size n, the count X, and the sample proportion for this setting.

Use of the Internet to find a place to live. A poll of 1500 college students asked whether or not they have used the Internet to find a place to live sometime within the past year. There were 1234 students who answered Yes; the other 266 answered No.

(a) What is the sample size n?

(b) Choose one of the two possible outcomes to define the random variable, X. Give a reason for your choice.

(c) What is the value of the count X?

(d) Find the sample proportion, .

Just like the sample mean, sample counts and sample proportions are commonly used statistics, and understanding their sampling distributions is important for statistical inference. These statistics, however, are discrete random variables, so their sampling distributions introduce us to a new family of probability distributions.

The binomial distributions for sample counts The distribution of a count X depends on how the data are produced. Here is a simple but common situation.

THE BINOMIAL SETTING

1. There is a fixed number of observations n. 2. The n observations are all independent. 3. Each observation falls into one of just two categories, which for convenience we call “success” and

“failure.” 4. The probability of a success, call it p, is the same for each observation.

Think of tossing a coin n times as an example of the binomial setting. Each toss gives either heads or tails, and the outcomes of successive tosses are independent. If we call heads a success, then p is the probability of a head and remains the same as long as we toss the same coin. The number of heads we count is a random variable X. The distribution of X (and, more generally, the distribution of the count of successes in any binomial setting) is completely determined by the number of observations n and the success probability p.

BINOMIAL DISTRIBUTIONS

The distribution of the count X of successes in the binomial setting is called the binomial distribution with parameters n and p. The parameter n is the number of observations, and p is the probability of a success on any one observation. The possible values of X are the whole numbers from 0 to n. As an abbreviation, we say that the distribution of X is B(n, p).

The binomial distributions are an important class of discrete probability distributions. Later in this section, we will learn how to assign probabilities to outcomes and how to find the mean and standard deviation of binomial distributions. The most important skill for using binomial distributions is the ability to recognize situations to which they do and do not apply. This can be done by checking all the facets of the binomial setting.

EXAMPLE 5.18

Binomial examples? (a) Genetics says that children receive genes from their parents independently. Each child of a particular pair of parents has probability 0.25 of having type O blood. If these parents have three children, the number who have type O blood is the count X of successes in three independent trials with probability 0.25 of a success on each trial. So X has the B(3, 0.25) distribution.

(b) Engineers define reliability as the probability that an item will perform its function under specific conditions for a specific period of time. Replacement heart valves made of animal tissue, for example, have probability 0.77 of performing well for 15 years.14 The probability of failure within 15 years is, therefore, 0.23. It is reasonable to assume that valves in different patients fail (or not) independently of each other. The number of patients in a group of 500 who will need another valve replacement within 15 years has the B(500, 0.23) distribution.

(c) A multicenter trial is designed to assess a new surgical procedure. A total of 540 patients will undergo the procedure, and the count of patients X who suffer a major adverse cardiac event (MACE) within 30 days of surgery will be recorded. Because these patients will receive this procedure from different surgeons at different hospitals, it may not be true that the probability of a MACE is the same for each patient. Thus, X may not have the binomial distribution.

5.46

5.47

USE YOUR KNOWLEDGE

Genetics and blood types. Genetics says that children receive genes from each of their parents independently. Suppose that each child of a particular pair of parents has probability 0.375 of having type AB blood. If these parents have three children, what is the distribution of the number who have type AB blood? Explain your answer.

Tossing a coin. Suppose you plan to toss a coin 20 times and record X, the number of heads that you observe. If the coin is fair (p = 0.5), what is the distribution of X? Also, explain why this distribution is also the sampling distribution of X.

Binomial distributions in statistical sampling The binomial distributions are important in statistics when we wish to make inferences about the proportion p of “successes” in a population. Here is a typical example.

EXAMPLE 5.19

Audits of financial records. The financial records of businesses are often audited by state tax authorities to test compliance with tax laws. Suppose that for one retail business, 800 of the 10,000 sales are incorrectly classified as subject to state sales tax. It would be too time-consuming for authorities to examine all these sales. Instead, an auditor examines an SRS of sales records. Is the count X of misclassified records in an SRS of 150 records a binomial random variable?

Choosing an SRS from a population is not quite a binomial setting. Removing one record in Example 5.19 changes the proportion of bad records in the remaining population, so the state of the second record chosen is not independent of the first. Because the population is large, however, removing a few items has a very small effect on the composition of the remaining population. Successive inspection results are very nearly independent. The population proportion of misclassified records is

If the first record chosen is bad, the proportion of bad records remaining is 799/9999 = 0.079908. If the first record is good, the proportion of bad records left is 800/9999 = 0.080008. These proportions are so close to 0.08 that, for practical purposes, we can act as if removing one record has no effect on the proportion of misclassified records remaining. We act as if the count X of misclassified sales records in the audit sample has the binomial distribution B(150, 0.08).

Populations like the one described in Example 5.19 often contain a relatively small number of items with very large values. For this example, these values would be very large sale amounts and likely represent an important group of items to the auditor. An SRS taken from such a population will likely include very few items of this type. Therefore, it is common to use a stratified sample in settings like this. Strata are defined based on dollar value of the sale, and within each stratum, an SRS is taken. The results are then combined to obtain an estimate for the entire population.

stratified random sample, p. 194

SAMPLING DISTRIBUTION OF A COUNT

A population contains proportion p of successes. If the population is much larger than the sample, the count X of successes in an SRS of size n has approximately the binomial distribution B(n, p).

The accuracy of this approximation improves as the size of the population increases relative to the size of the sample. As a rule of thumb, we will use the binomial sampling distribution for counts when the population is at least 20 times as large as the sample.

Finding binomial probabilities We will later give a formula for the probability that a binomial random variable takes any of its values. In practice, you will rarely have to use this formula for calculations because some calculators and most statistical software packages will calculate binomial probabilities for you.

EXAMPLE 5.20

Probabilities for misclassified sales records. In the audit setting of Example 5.19, what is the probability that the audit finds exactly 10 misclassified sales records? What is the probability that the audit finds no more than 10 misclassified records? Figure 5.14 shows the output from one statistical software system. You see that if the count X has the B(150, 0.08) distribution,

FIGURE 5.14 Binomial probabilities, Example 5.20: output from the Minitab statistical software.

It was easy to request these calculations in the software’s menus. For the TI-83/84 calculator, the functions binompdf and binomcdf would be used. In R, the functions dbinom and pbinom would be used. Typically, the output supplies more decimal places than we need and uses labels that may not be helpful (for example, “Probability Density Function” when the distribution is discrete, not continuous). But, as usual with software, we can ignore distractions and find the results we need.

If you do not have suitable computing facilities, you can still shorten the work of calculating binomial probabilities for some values of n and p by looking up probabilities in Table C in the back of this book. The entries in the table are the probabilities P(X = k) of individual outcomes for a binomial random variable X.

EXAMPLE 5.21

The probability histogram. Suppose that the audit in Example 5.19 chose just 15 sales records. What is the

probability that no more than one of the 15 is misclassified? The count X of misclassified records in the sample has approximately the B(15, 0.08) distribution. Figure 5.15 is a probability histogram for this distribution. The distribution is strongly skewed. Although X can take any whole-number value from 0 to 15, the probabilities of values larger than 5 are so small that they do not appear in the histogram.

FIGURE 5.15 Probability histogram for the binomial distribution with n = 15 and p = 0.08, Example 5.21.

We want to calculate

P(X ≤ 1) = P(X = 0) + P(X = 1)

when X has the B(15, 0.08) distribution. To use Table C for this calculation, look opposite n = 15 and under p = 0.08. The entries in the rows for each k are P(X = k). Blank cells in the table are 0 to four decimal places. You see that

About two-thirds of all samples will contain no more than one bad record. In fact, almost 29% of the samples will contain no bad records. The sample of size 15 cannot be trusted to provide adequate evidence about misclassified sales records. A larger number of observations is needed.

p n k .08 15 0 .2863 1 .3734 2 .2273 3 .0857 4 .0223 5 .0043 6 .0006 7 .0001 8 9

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The values of p that appear in Table C are all 0.5 or smaller. When the probability of a success is greater than 0.5, restate the problem in terms of the number of failures. The probability of a failure is less than 0.5 when the probability of a success exceeds 0.5. When using the table, always stop to ask whether you must count successes or failures.

EXAMPLE 5.22

Falling asleep in class. In the survey of 4513 college students described in Example 5.4, 46% of the respondents reported falling asleep in class due to poor sleep. You randomly sample 10 students in your dormitory, and eight state that they fell asleep in class during the last week due to poor sleep. Relative to the survey results, is this an unusually high number of students?

To answer this question, assume that the students’ actions (falling asleep or not) are independent, with the probability of falling asleep equal to 0.46. This independence assumption may not be reasonable if the students study and socialize together or if there is a loud student in the dormitory who keeps everyone up. We’ll assume this is not an issue here, so the number X of students who fell asleep in class out of 10 students has the B(10, 0.46) distribution.

We want the probability of classifying at least eight students as having fallen asleep in class. Using software, we find

We would expect to find eight or more students falling asleep in class about 3% of the time or in fewer than one of every 30 surveys of 10 students. This is a pretty rare outcome and falls outside the range of the usual chance variation due to random sampling.

USE YOUR KNOWLEDGE

Free-throw shooting. April is a college basketball player who makes 80% of her free throws. In a recent game, she had 10 free throws and missed three of them. How unusual is this outcome? Using software, calculator, or Table C, compute 1 2 P(X ≤ 2), where X is the number of free throws missed in 10 shots. Explain your answer.

Find the probabilities. (a) Suppose that X has the B(8, 0.3) distribution. Use software, calculator, or Table C to find P(X = 0) and

P(X ≥ 6).

(b) Suppose that X has the B(8, 0.7) distribution. Use software, calculator, or Table C to find P(X = 8) and P(X ≤ 2).

(c) Explain the relationship between your answers to parts (a) and (b) of this exercise.

Binomial mean and standard deviation If a count X has the B(n, p) distribution, what are the mean μX and the standard deviation σX? We can guess the mean. If we expect 46% of the students to have fallen asleep in class due to poor sleep, the mean number in 10 students should be 46% of 10, or 4.6. That’s μX when X has the B(10, 0.46) distribution.

Intuition suggests more generally that the mean of the B(n, p) distribution should be np. Can we show that this is correct and also obtain a short formula for the standard deviation? Because binomial distributions are discrete probability distributions, we could find the mean and variance by using the definitions in Section 4.4. Here is an easier way.

mean and variance of a discrete random variable, pp. 248, 256

A binomial random variable X is the count of successes in n independent observations that each have the same probability p of success. Let the random variable Si indicate whether the ith observation is a success or failure by taking the values Si = 1 if a success occurs and Si = 0 if the outcome is a failure. The Si are independent because the observations are, and each Si has the same simple distribution:

Outcome 1 0 Probability p 1 − p

From the definition of the mean of a discrete random variable, we know that the mean of each Si is

μS = (1)(p) + (0)(1 − p) = p

Similarly, the definition of the variance shows that Because each Si is 1 for a success and 0 for a failure, to find the total number of successes X we add the Si’s:

X = S1 + S2 + · · · + Sn

Apply the addition rules for means and variances to this sum. To find the mean of X we add the means of the Si’s:

addition rules for means and variances, pp. 254, 258

Similarly, the variance is n times the variance of a single S, so that The standard deviation σX is the square root of the variance. Here is the result.

BINOMIAL MEAN AND STANDARD DEVIATION

If a count X has the binomial distribution B(n, p), then

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EXAMPLE 5.23

The Helsinki Heart Study. The Helsinki Heart Study asked whether the anticholesterol drug gemfibrozil reduces heart attacks. In planning such an experiment, the researchers must be confident that the sample sizes are large enough to enable them to observe enough heart attacks. The Helsinki study planned to give gemfibrozil to about 2000 men aged 40 to 55 and a placebo to another 2000. The probability of a heart attack during the five-year period of the study for men this age is about 0.04. What are the mean and standard deviation of the number of heart attacks that will be observed in one group if the treatment does not change this probability?

There are 2000 independent observations, each having probability p = 0.04 of a heart attack. The count X of heart attacks has the B(2000, 0.04) distribution, so that

The expected number of heart attacks is large enough to permit conclusions about the effectiveness of the drug. In fact, there were 84 heart attacks among the 2035 men actually assigned to the placebo, quite close to the mean. The gemfibrozil group of 2046 men suffered only 56 heart attacks. This is evidence that the drug reduces the chance of a heart attack. In a later chapter, we will learn how to determine if this is strong enough evidence to conclude the drug is effective.

USE YOUR KNOWLEDGE

Free-throw shooting. Refer to Exercise 5.48 (page 317). If April takes 85 free throws in the upcoming season, what are the mean and standard deviation of the number of free throws made?

Find the mean and standard deviation

(a) Suppose that X has the B(8, 0.3) distribution. Compute the mean and standard deviation of X.

(b) Suppose that X has the B(8, 0.7) distribution. Compute the mean and standard deviation of X.

(c) Explain the relationship between your answers to parts (a) and (b) of this exercise.

Sample proportions What proportion of a company’s sales records have an incorrect sales tax classification? What percent of adults favor stronger laws restricting firearms? In statistical sampling, we often want to estimate the proportion p of “successes” in a population. Our estimator is the sample proportion of successes:

population proportion, p. 283

Be sure to distinguish between the proportion and the count X. The count takes whole-number values between 0 and n, but a proportion is always a number between 0 and 1. In the binomial setting, the count X has a binomial distribution. The proportion does not have a binomial distribution. We can, however, do probability calculations about by restating them in terms of the count X and using binomial methods. In Example 5.12 (page 303), we took a similar approach for the sum, restating the problem in terms of the sample mean and then using the Normal distribution to calculate the probability.

EXAMPLE 5.24

Shopping online. A survey by the Consumer Reports National Research Center revealed that 84% of all respondents were very satisfied with their online shopping experience.15 It was also reported, however, that people over the age of 40 were generally more satisfied than younger respondents. You decide to take a nationwide random sample of 2500 college students and ask if they agree or disagree that “I am very satisfied with my online shopping experience.” Suppose that 60% of all college students would agree if asked this question. What is the probability that the sample proportion who agree is at least 58%?

The count X who agree has the binomial distribution B(2500, 0.6). The sample proportion = X/2500 does not have a binomial distribution because it is not a count. But we can translate any question about a sample proportion into a question about the count X. Because 58% of 2500 is 1450,

This is a rather elaborate calculation. We must add more than 1000 binomial probabilities. Software tells us that P( ≥ 0.58) = 0.9802. But what do we do if we don’t have access to software?

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As a first step, find the mean and standard deviation of a sample proportion. We know the mean and standard deviation of a sample count, so apply the rules from Section 4.4 for the mean and variance of a constant times a random variable. Here is the result.

rules for means and variances, pp. 254, 258

MEAN AND STANDARD DEVIATION OF A SAMPLE PROPORTION

Let be the sample proportion of successes in an SRS of size n drawn from a large population having population proportion p of successes. The mean and standard deviation of are

The formula for is exactly correct in the binomial setting. It is approximately correct for an SRS from a large population. We will use it when the population is at least 20 times as large as the sample

Let’s now use these formulas to calculate the mean and standard deviation for Example 5.24.

EXAMPLE 5.25

The mean and the standard deviation. The mean and standard deviation of the proportion of the survey respondents in Example 5.24 who are satisfied with their online shopping experience are

USE YOUR KNOWLEDGE

Find the mean and the standard deviation. If we toss a fair coin 150 times, the number of heads is a random variable that is binomial.

(a) Find the mean and the standard deviation of the sample proportion of heads.

(b) Is your answer to part (a) the same as the mean and the standard deviation of the sample count of heads in 150 throws? Explain your answer.

The fact that the mean of is p states in statistical language that the sample proportion in an SRS is an unbiased estimator of the population proportion p. When a sample is drawn from a new population having a different value of the population proportion p, the sampling distribution of the unbiased estimator changes so that its mean moves to the new value of p. We observed this fact empirically in Section 5.1 and have now verified it from the laws of probability.

The variability of about its mean, as described by the variance or standard deviation, gets smaller as the sample size increases. So a sample proportion from a large sample will usually lie quite close to the population proportion p. We observed this in the simulation experiment on page 285 in Section 5.1. Now we have discovered exactly how the variability decreases: the standard deviation is Similar to what we observed in the previous section, the in the denominator means that the sample size must be multiplied by 4 if we wish to divide the standard

deviation in half.

Normal approximation for counts and proportions Using simulation, we discovered in Section 5.1 that the sampling distribution of a sample proportion is close to Normal. Now we know that the distribution of is that of a binomial count divided by the sample size n. This seems at first to be a contradiction. To clear up the matter, look at Figure 5.16. This is a probability histogram of the exact distribution of the proportion of very satisfied shoppers , based on the binomial distribution B(2500, 0.6). There are hundreds of narrow bars, one for each of the 2501 possible values of . Most have probabilities too small to show in a graph. The probability histogram looks very Normal! In fact, both the count X and the sample proportion are approximately Normal in large samples.

We also know this to be true as a result of the central limit theorem discussed in the previous section (page 298). Recall that we can consider the count X as a sum

X = S1 + S2 + · · · 1 Sn

of independent random variables Si that take the value 1 if a success occurs on the ith trial and the value 0 otherwise. The proportion of successes = X/n can then be thought of as the sample mean of the Si and, like all sample means, is approximately Normal when n is large. Given that is approximately Normal, the count will also be approximately Normal because it is just a constant n times , an approximately Normal random variable.

FIGURE 5.16 Probability histogram of the sample proportion based on a binomial count with n = 2500 and p = 0.6. The distribution is very close to Normal.

NORMAL APPROXIMATION FOR COUNTS AND PROPORTIONS

Draw an SRS of size n from a large population having population proportion p of successes. Let X be the count of successes in the sample and = X/n be the sample proportion of successes. When n is large, the sampling distributions of these statistics are approximately Normal:

As a rule of thumb, we will use this approximation for values of n and p that satisfy np ≥ 10 and n(1-p) ≥ 10.

These Normal approximations are easy to remember because they say that and X are Normal, with their usual means and standard deviations. Whether or not you use the Normal approximations should depend on how accurate your calculations need to be. For most statistical purposes, great accuracy is not required. Our “rule of thumb” for use of the Normal approximations reflects this judgment.

The accuracy of the Normal approximations improves as the sample size n increases. They are most accurate for any fixed n when p is close to 0.5, and least accurate when p is near 0 or 1. You can compare binomial distributions with their Normal approximations by using the Normal Approximation to Binomial applet. This applet allows you to change n or p while watching the effect on the binomial probability histogram and the Normal curve that approximates it.

Figure 5.17 summarizes the distribution of a sample proportion in a form that emphasizes the big idea of a sampling distribution. Just as with Figure 5.11, the general framework for constructing a sampling distribution is shown on the left.

Take many random samples of size n from a population that contains proportion p of successes. Find the sample proportion for each sample. Collect all the ’s and display their distribution.

FIGURE 5.17 The sampling distribution of a sample proportion is approximately Normal with mean p and standard deviation

The sampling distribution of is shown on the right. Keep this figure in mind as you move toward statistical inference.

EXAMPLE 5.26

Compare the Normal approximation with the exact calculation. Let’s compare the Normal approximation for the calculation of Example 5.24 with the exact calculation from software. We want to calculate P( ≥ 0.58) when the sample size is n = 2500 and the population proportion is p = 0.6. Example 5.25 shows that

Act as if were Normal with mean 0.6 and standard deviation 0.0098. The approximate probability, as illustrated in Figure 5.18, is

FIGURE 5.18 The Normal probability calculation, Example 5.26.

That is, about 98% of all samples have a sample proportion that is at least 0.58. Because the sample was large, this Normal approximation is quite accurate. It misses the software value 0.9802 by only 0.0009.

EXAMPLE 5.27

Using the Normal approximation. The audit described in Example 5.19 examined an SRS of 150 sales records for compliance with sales tax laws. In fact, 8% of all the company’s sales records have an incorrect sales tax classification. The count X of bad records in the sample has approximately the B(150, 0.08) distribution.

According to the Normal approximation to the binomial distributions, the count X is approximately Normal with mean and standard deviation

The Normal approximation for the probability of no more than 10 misclassified records is the area to the left of X = 10 under the Normal curve. Using Table A,

Software tells us that the actual binomial probability that no more than 10 of the records in the sample are misclassified is P(X ≤ 10) = 0.3384. The Normal approximation is only roughly accurate. Because np = 12, this combination of n and p is close to the border of the values for which we are willing to use the approximation.

The distribution of the count of bad records in a sample of 15 is distinctly non-Normal, as Figure 5.15 showed. When we increase the sample size to 150, however, the shape of the binomial distribution becomes roughly Normal. Figure 5.19 displays the probability histogram of the binomial distribution with the density curve of the approximating Normal distribution superimposed. Both distributions have the same mean and standard deviation, and both the area under the histogram and the area under the curve are 1. The Normal curve fits the histogram reasonably well. Look closely: the histogram is slightly skewed to the right, a property that the symmetric Normal curve can’t match.

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FIGURE 5.19 Probability histogram and Normal approximation for the binomial distribution with n = 150 and p = 0.08, Example 5.27.

USE YOUR KNOWLEDGE

Use the Normal approximation. Suppose that we toss a fair coin 150 times. Use the Normal approximation to find the probability that the sample proportion of heads is

(a) between 0.4 and 0.6.

(b) between 0.45 and 0.55.

The continuity correction Figure 5.20 illustrates an idea that greatly improves the accuracy of the Normal approximation to binomial probabilities. The binomial probability P(X ≤ 10) is the area of the histogram bars for values 0 to 10. The bar for X = 10) actually extends from 9.5 to 10.5. Because the discrete binomial distribution puts probability only on whole numbers, the probabilities P(X ≤ 10) and P(X ≤ 10.5) are the same. The Normal distribution spreads probability continuously, so these two Normal probabilities are different. The Normal approximation is more accurate if we consider X = 10) to extend from 9.5 to 10.5, matching the bar in the probability histogram.

The event {X ≤ 10} includes the outcome X = 10). Figure 5.20 shades the area under the Normal curve that matches all the histogram bars for outcomes 0 to 10, bounded on the right not by 10, but by 10.5. So P(X ≤ 10) is calculated as P(X ≤ 10.5). On the other hand, P(X < 10) excludes the outcome X = 10), so we exclude the entire interval from 9.5 to 10.5 and calculate P(X ≤ 9.5) from the Normal table. Here is the result of the Normal calculation in Example 5.27 improved in this way:

continuity correction

The improved approximation misses the binomial probability by only 0.012. Acting as though a whole number occupies the interval from 0.5 below to 0.5 above the number is called the continuity correction continuity correction to the Normal approximation. If you need accurate values for binomial probabilities, try to use software to do exact calculations. If no software is available, use the continuity correction unless n is very large. Because most statistical purposes do not require extremely accurate probability calculations, we do not emphasize use of the continuity correction.

FIGURE 5.20 Area under the Normal approximation curve for the probability in Example 5.27.

Binomial formula We can find a formula for the probability that a binomial random variable takes any value by adding probabilities for the different ways of getting exactly that many successes in n observations. Here is the example we will use to show the idea.

EXAMPLE 5.28

Blood types of children. Each child born to a particular set of parents has probability 0.25 of having blood type O. If these parents have five children, what is the probability that exactly two of them have type O blood?

The count of children with type O blood is a binomial random variable X with n = 5 tries and probability p = 0.25 of a success on each try. We want P(X = 2).

Because the method doesn’t depend on the specific example, we will use “S” for success and “F” for failure. In Example 5.28, “S” would stand for type O blood. Do the work in two steps.

Step 1: Find the probability that a specific two of the five tries give successes—say, the first and the third. This is the outcome SFSFF. The multiplication rule for independent events tells us that

Step 2: Observe that the probability of any one arrangement of two S’s and three F’s has this same probability. That’s true because we multiply together 0.25 twice and 0.75 three times whenever we have two S’s and three F’s. The probability that X = 2 is the probability of getting two S’s and three F’s in any arrangement whatsoever. Here are all the possible arrangements:

SSFFF SFSFF SFFSF SFFFS FSSFF

FSFSF FSFFS FFSSF FFSFS FFFSS

There are 10 of them, all with the same probability. The overall probability of two successes is, therefore,

P(X = 2) = 10(0.25)2(0.75)3 = 0.2637

The pattern of this calculation works for any binomial probability. To use it, we need to be able to count the number of arrangements of k successes in n observations without actually listing them. We use the following fact to do the counting.

BINOMIAL COEFFICIENT

The number of ways of arranging k successes among n observations is given by the binomial coefficient

for k = 0, 1, 2, . . . , n.

factorial

The formula for binomial coefficients uses the factorial notation. The factorial n! for any positive whole number n is

n! = n × (n − 1) × (n − 2) × · · · × 3 × 2 × 1

Also, 0! = 1. Notice that the larger of the two factorials in the denominator of a binomial coefficient will cancel much of the n! in the numerator. For example, the binomial coefficient we need for Example 5.28 is

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This agrees with our previous calculation.

The notation is not related to the fraction . A helpful way to remember its meaning is to read it as

“binomial coefficient n choose k.” Binomial coefficients have many uses in mathematics, but we are interested in them

only as an aid to finding binomial probabilities. The binomial coefficient counts the number of ways in which k

successes can be distributed among n observations. The binomial probability P(X = k) is this count multiplied by the probability of any specific arrangement of the k successes. Here is the formula we seek.

BINOMIAL PROBABILITY

If X has the binomial distribution B(n, p) with n observations and probability p of success on each observation, the possible values of X are 0, 1, 2, . . . , n. If k is any one of these values, the binomial probability is

Here is an example of the use of the binomial probability formula.

EXAMPLE 5.29

Using the binomial probability formula. The number X of misclassified sales records in the auditor’s sample in Example 5.21 (page 316) has the B(15,0.08) distribution. The probability of finding no more than one misclassified record is

The calculation used the facts that 0! 5 1 and that a0 = 1 for any number a ≠ 0. The result agrees with that obtained from Table C in Example 5.21.

USE YOUR KNOWLEDGE

An unfair coin. A coin is slightly bent, and as a result, the probability of a head is 0.53. Suppose that you toss the coin five times.

(a) Use the binomial formula to find the probability of three or more heads.

(b) Compare your answer with the one that you would obtain if the coin were fair.

The Poisson distributions A count X has a binomial distribution when it is produced under the binomial setting. If one or more facets of this setting do not hold, the count X will have a different distribution. In this subsection, we discuss one of these distributions.

Frequently, we meet counts that are open-ended; that is, they are not based on a fixed number of n observations: the number of customers at a popular café between 12:00 P.M. and 1:00 P.M.; the number of dings on your car door; the number of reported pedestrian/bicyclist collisions on campus during the academic year. These are all counts that could be 0, 1, 2, 3, and so on indefinitely.

The Poisson distribution is another model for a count and can often be used in these open-ended situations. The count represents the number of events (call them “successes”) that occur in some fixed unit of measure such as a period of time or region of space. The Poisson distribution is appropriate under the following conditions.

THE POISSON SETTING 1. number of successes that occur in two nonoverlapping units of measure are independent. 2. The probability that a success will occur in a unit of measure is the same for all units of equal size and is

proportional to the size of the unit. 3. The probability that more than one event occurs in a unit of measure is negligible for very small-sized units.

In other words, the events occur one at a time.

For binomial distributions, the important quantities were n, the fixed number of observations, and p, the probability of success on any given observation. For Poisson distributions, the only important quantity is the mean number of successes μ occurring per unit of measure.

POISSON DISTRIBUTION The distribution of the count X of successes in the Poisson setting is the Poisson distribution with mean μ. The parameter μ is the mean number of successes per unit of measure. The possible values of X are the whole numbers 0, 1, 2, 3, …. If k is any whole number, then*

The standard deviation of the distribution is .

EXAMPLE 5.30

Number of dropped calls. Suppose that the number of dropped calls on your cell phone varies, with an average of 2.1 calls per day. If we assume that the Poisson setting is reasonable for this situation, we can model the daily count of dropped calls X using the Poisson distribution with μ = 2.1 What is the probability of having no more than two dropped calls tomorrow?

We can calculate P(X ≤ 2) either using software or the Poisson probability formula. Using the probability formula:

Using the R software, the probability is

dpois(0,2.1) + dpois(1,2.1) + dpois(2,2.1) [1] 0.6496314

These two answers differ slightly due to roundoff error in the hand calculation. There is roughly a 65% chance that you will have no more than two dropped calls tomorrow.

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5.56

Similar to the binomial, Poisson probability calculations are rarely done by hand if the event includes numerous possible values for X. Most software provides functions to calculate P(X = k) and the cumulative probabilities of the form P(X ≤ k). These cumulative probability calculations make solving many problems less tedious. Here’s an example.

EXAMPLE 5.31

Counting software remote users. Your university supplies online remote access to various software programs used in courses. Suppose that the number of students remotely accessing these programs in any given hour can be modeled by a Poisson distribution with μ = 17.2. What is the probability that more than 25 students will remotely access these programs in the next hour?

Calculating this probability requires two steps.

1. Write P(X > 25) as an expression involving a cumulative probability:

P(X > 25) = 1 − P(X ≤ 25)

2. Obtain P(X ≤ 25) and subtract the value from 1. Again using R,

1-ppois(25,17.2) [1] 0.02847261

The probability that more than 25 students will use this remote access in the next hour is only 0.028. Relying on software to get the cumulative probability is much quicker and less prone to error than the method of Example 5.30. For this case, that method would involve determining 26 probabilities and then summing their values.

Under the Poisson setting, this probability of 0.028 applies not only to the next hour, but also to any other hour in the future. The probability does not change because the units of measure are the same size and nonoverlapping.

USE YOUR KNOWLEDGE

Number of aphids. The milkweed aphid is a common pest to many ornamental plants. Suppose that the number of aphids on a shoot of a Mexican butterfly weed follows a Poisson distribution with μ = 4.4 aphids.

(a) What is the probability of observing exactly five aphids on a shoot?

(b) What is the probability of observing five or fewer aphids on a shoot?

Number of aphids, continued. Refer to the previous exercise. (a) What proportion of shoots would you expect to have no aphids present?

(b) If you do not observe any aphids on a shoot, is the probability that a nearby shoot has no aphids smaller than, equal to, or larger than your answer in part (a)? Explain your reasoning.

If we add counts from successive nonoverlapping areas of equal size, we are just counting the successes in a larger area. That count still meets the conditions of the Poisson setting. However, because our unit of measure has doubled, the mean of this new count is twice as large. Put more formally, if X is a Poisson random variable with mean μX and Y is a Poisson random variable with mean μY and Y is independent of X, then X + Y is a Poisson random variable with mean μX + μY This fact means that we can combine areas or look at a portion of an area and still use Poisson distributions to model the count.

EXAMPLE 5.32

Number of potholes. The Automobile Association (AA) in Britain had member volunteers make a 60-minute, two-mile walk around their neighborhoods and survey the condition of their roads and sidewalks. One outcome was the number of potholes, defined as being at least 2 inches deep and at least 6 inches in diameter, in their roads.16 It was reported that Scotland averages 8.9 potholes per mile of road and London averages 4.9 potholes per mile of road. Suppose that the number of potholes per mile in each of these two regions follow the Poisson distribution. Then

The number of potholes per 20 miles of road in Scotland is a Poisson random variable with mean 20 × 8.9 = 178. The number of potholes per half mile of road in London is a Poisson random variable with mean 0.5 × 4.9 = 2.45. The number of potholes per 500 miles of road in Scotland is a Poisson random variable with mean 500 × 8.9 = 4450. If we examined 2 miles of road in Scotland and 5 miles of road in London, the total number of potholes would be a Poisson random variable with mean 2 × 8.9 + 5 × 4.9 = 42.3.

When the mean of the Poisson distribution is large, it may be difficult to calculate Poisson probabilities using a calculator or software. Fortunately, when μ is large, Poisson probabilities can be approximated using the Normal distribution with mean μ and standard deviation . Here is an example.

EXAMPLE 5.33

Number of snaps received. In Example 5.11, it was reported that Snapchat has more than 100 million daily users who send over 400 million snaps a day. Suppose that the number of snaps you receive per day follows a Poisson distribution with mean 12. What is the probability that, over a week, you would receive more than 100 snaps?

To answer this using software, we first compute the mean number of snaps sent per week. Because there are seven days in a week, the mean is 7 × 12 = 84. Plugging this into R tells us that there is slightly less than an 4% chance of receiving this many snaps:

1-ppois(100,84) [1] 0.03891883

For the Normal approximation we compute

The approximation is quite accurate, differing from the actual probability by only 0.0012.

While the Normal approximation is adequate for many practical purposes, we recommend using statistical software when possible so you can get exact Poisson probabilities.

There is one other approximation associated with the Poisson distribution that is worth mentioning. It is related to the binomial distribution. Previously, we recommended using the Normal distribution to approximate the binomial distribution when n and p satisfy np ≥ 10 and n (1 − p) ≥ 10. In cases where n is large but p is so small that np < 10, the Poisson distribution with μ = np yields more accurate results. For example, suppose that you wanted to calculate P(X ≤ 2) when X has the B(1000,.001) distribution. Using R, the actual binomial probability and the Poisson approximation are

pbinom(2,1000,.001) ppois(2,1)

[1] 0.9197907 [1] 0.9196986

The Poisson approximation gives a very accurate probability calculation for the binomial distribution in this case.

SECTION 5.3 SUMMARY A count X of successes has the binomial distribution B(n, p) in the binomial setting: there are n trials, all independent, each resulting in a success or a failure, and each having the same probability p of a success. The binomial distribution B(n, p) is a good approximation to the sampling distribution of the count of successes in an SRS of size n from a large population containing proportion p of successes. We will use this approximation when the population is at least 20 times larger than the sample. The sample proportion of successes = X/n is an estimator of the population proportion p. It does not have a binomial distribution, but we can do probability calculations about by restating them in terms of X. Binomial probabilities are most easily found by software. There is an exact formula that is practical for calculations when n is small. Table C contains binomial probabilities for some values of n and p. For large n, you can use the Normal approximation. The mean and standard deviation of a binomial count X and a sample proportion = X/n are

The sample proportion is, therefore, an unbiased estimator of the population proportion p. The Normal approximation to the binomial distribution says that if X is a count having the B(n,p) distribution, then when n is large,

We will use this approximation when np ≥ 10 and n(1 − p) ≥ 10. It allows us to approximate probability calculations about X and using the Normal distribution. The continuity correction improves the accuracy of the Normal approximations. The exact binomial probability formula is

where the possible values of X are k = 0,1, . . . , n. The binomial probability formula uses the binomial coefficient

Here the factorial n! is n! = n × (n − 1) × (n − 2) × . . . × 3 × 2 ×1

for positive whole numbers n and 0! 5 1. The binomial coefficient counts the number of ways of distributing k successes among n trials. A count X of successes has a Poisson distribution in the Poisson setting: the number of successes that occur in two nonoverlapping units of measure are independent; the probability that a success will occur in a unit of measure is the same for all units of equal size and is proportional to the size of the unit; the probability that more than one event occurs in a unit of measure is negligible for very small-sized units. In other words, the events occur one at a time. If X has the Poisson distribution with mean μ, then the standard deviation of X is , and the possible values of X are the whole numbers 0, 1, 2, 3, and so on. The Poisson probability that X takes any of these values is

Sums of independent Poisson random variables also have the Poisson distribution. For example, in a Poisson model with mean μ per unit of measure, the count of successes in a units is a Poisson random variable with mean aμ.

SECTION 5.3 EXERCISES For Exercises 5.43, 5.44, and 5.45, see page 312; for Exercises 5.46 and 5.47, see page 313; for Exercises 5.48 and 5.49, see page 317; for Exercises 5.50 and 5.51, see page 319; for Exercise 5.52, see page 320; for Exercise 5.53, see page 325; for Exercise 5.54, see page 328; and for Exercises 5.55 and 5.56, see page 330.

Most binomial probability calculations required in these exercises can be done by using Table C or the Normal approximation. Your instructor may request that you use the binomial probability formula or software. In exercises requiring the Normal approximation, you should use the continuity correction if you studied that topic.

5.57 What is wrong? Explain what is wrong in each of the following scenarios.

(a) If you toss a fair coin four times and a head appears each time, then the next toss is more likely to be a tail than a head.

(b) If you toss a fair coin four times and observe the pattern HTHT, then the next toss is more likely to be a head than a tail.

(c) The quantity is one of the parameters for a binomial distribution.

(d) The binomial distribution can be used to model the daily number of pedestrian/cyclist near-crash events on campus.

5.58 What is wrong? Explain what is wrong in each of the following scenarios.

(a) In the binomial setting, X is a proportion.

(b) The variance for a binomial count is

(c) The Normal approximation to the binomial distribution is always accurate when n is greater than 1000.

(d) We can use the binomial distribution to approximate the sampling distribution of when we draw an SRS of size n = 50 students from a population of 500 students.

5.59 Should you use the binomial distribution? In each of the following situations, is it reasonable to use a binomial distribution for the random variable X? Give reasons for your answer in each case. If a binomial distribution applies, give the values of n and p.

(a) A poll of 200 college students asks whether or not you usually feel irritable in the morning. X is the number who reply that they do usually feel irritable in the morning.

(b) You toss a fair coin until a head appears. X is the count of the number of tosses that you make.

(c) Most calls made at random by sample surveys don’t succeed in talking with a person. Of calls to New York City, only one-twelfth succeed. A survey calls 500 randomly selected numbers in New York City. X is the number of times that a person is reached.

(d) You deal 10 cards from a shuffled deck of standard playing cards and count the number X of black cards.

5.60 Should you use the binomial distribution? In each of the following situations, is it reasonable to use a binomial distribution for the random variable X? Give reasons for your answer in each case.

(a) In a random sample of students in a fitness study, X is the mean daily exercise time of the sample.

(b) A manufacturer of running shoes picks a random sample of 20 shoes from the production of shoes each day for a detailed inspection. X is the number of pairs of shoes with a defect.

(c) A nutrition study chooses an SRS of college students. They are asked whether or not they usually eat at least five servings of fruits or vegetables per day. X is the number who say that they do.

(d) X is the number of days during the school year when you skip a class.

5.61 Stealing from a store. A survey of more than 20,000 U.S. high school students revealed that 20% of the students say that they stole something from a store in the past year.17 This is down 7% from the last survey, which was performed two years earlier. You decide to take a random sample of 10 high school students from your city and ask them this question.

(a) If the high school students in your city match this 20% rate, what is the distribution of the number of students who say that they stole something from a store in the past year? What is the distribution of the number of students who do not say that they stole something from a store in the past year?

(b) What is the probability that four or more of the 10 students in your sample say that they stole something from a store in

the past year?

5.62 Illegal downloading. New regulations in Canada require all Internet service providers (ISPs) to send a notice to subscribers who are downloading files illegally asking them to stop. This “notice and notice” system was already in place with Rogers Cable. That company says that prior to these new regulations, 67% of its subscribers who received a notice did not reoffend.18 Consider a random sample of 50 of these Rogers subscribers who received a first notice.

(a) What is the distribution of the number X of subscribers who reoffend? Explain your answer.

(b) What is the probability that at least 18 of the 50 subscribers in your sample reoffend?

5.63 Stealing from a store, continued. Refer to Exercise 5.61.

(a) What is the expected number of students in your sample who say that they stole something from a store in the past year? What is the expected number of students who do not say that they stole? You should see that these two means add to 10, the total number of students.

(b) What is the standard deviation s of the number of students in your sample who say that they stole something?

(c) Suppose that you live in a city where only 10% of the high school students say that they stole something from a store in the past year. What is s in this case? What is s if p = 0.01? What happens to the standard deviation of a binomial distribution as the probability of a success gets close to 0?

5.64 Illegal downloading, continued. Refer to Exercise 5.62. Given the new regulations, suppose that 75% of the Canadian ISP subscribers will not reoffend after receiving a notice.

(a) If you choose at random 15 subscribers who received a notice, what is the mean of the count X who will not reoffend? What is the mean of the proportion in your sample who will not reoffend?

(b) Repeat the calculations in part (a) for samples of size 150 and 1500. What happens to the mean count of successes as the sample size increases? What happens to the mean proportion of successes?

5.65 More on illegal downloading. Consider the settings of Exercises 5.62 and 5.64.

(a) Using the 67% rate of Rogers subscribers prior to the new regulations, what is the smallest number m out of n = 15 Canadian ISP subscribers who receive a notice such that P(X ≥ m) is no larger than 0.05? You might consider m or more subscribers as evidence that the rate in your sample is larger than 67%.

(b) Now using the 75% rate of Canadian ISP subscribers after the new regulations and your answer to part (a), what is P(X ≥ m)? This represents the chance of obtaining enough evidence given that the rate is 75%.

(c) If you were to increase the sample size from n = 15 to n = 100 and repeat parts (a) and (b), would you expect the probability in part (b) to increase or decrease? Explain your answer.

5.66 Attitudes toward drinking and studies of behavior. Some of the methods in this section are approximations rather than exact probability results. We have given rules of thumb for safe use of these approximations.

(a) You are interested in attitudes toward drinking among the 75 members of a fraternity. You choose 30 members at random to interview. One question is “Have you had five or more drinks at one time during the last week?” Suppose that, in fact, 30% of the 75 members would say Yes. Explain why you cannot safely use the B(30, 0.3) distribution for the count X in your sample who say Yes.

(b) The National AIDS Behavioral Surveys found that 0.2% (that’s 0.002 as a decimal fraction) of adult heterosexuals had both received a blood transfusion and had a sexual partner from a group at high risk of AIDS. Suppose that this national proportion holds for your region. Explain why you cannot safely use the Normal approximation for the sample proportion who fall in this group when you interview an SRS of 1000 adults.

5.67 Random digits. Each entry in a table of random digits like Table B has probability 0.1 of being any given digit, and digits are independent of each other.

(a) What is the probability that a group of six digits from the table will contain at least one digit greater than 5?

(b) What is the mean number of digits greater than 5 in lines 40 digits long?

5.68 Use the Probability applet. The Probability applet simulates tosses of a coin. You can choose the number of tosses n and the probability p of a head. You can therefore use the applet to simulate binomial random variables.

The count of misclassified sales records in Example 5.21 has the binomial distribution with n = 15 and p = 0.08. Set these values for the number of tosses and probability of heads in the applet. Table C shows that the probability of getting a sample with exactly 0 misclassified records is 0.2863. This is the long-run proportion of samples with no bad records. Click “Toss” and “Reset” repeatedly to simulate 25 samples of 15 tosses. Record the number of bad records (the count of heads) in each of

the 25 samples.

(a) What proportion of the 25 samples had exactly 0 bad records? Do you think this sample proportion is close to the probability?

(b) Remember that this probability of 0.2863 tells us only what happens in the long run. Here we’re considering only 25 samples. If X is the number of samples out of 25 with exactly 0 misclassified records, what is the distribution of X?

(c) Explain how to use the distribution in part (b) to describe the sampling distribution of in part (a).

5.69 Cyberbullying. An online survey, in partnership with Habbo, was conducted to study cyberbullying among 13- to 25- year-olds in the United Kingdom. It was reported that 62% of the young people had received nasty private messages on a smartphone social network app.19 You randomly sample four young people from the United Kingdom and ask them if they’ve received nasty messages. Let X be the number who say Yes.

(a) What are n and p in the binomial distribution of X?

(b) Find the probability of each possible value of X, and draw a probability histogram for this distribution.

(c) Find the mean number of positive responders and mark the location of this value on your histogram.

5.70 The ideal number of children. “What do you think is the ideal number of children for a family to have?” A Gallup Poll asked this question of 1020 randomly chosen adults. Slightly less than half (48%) thought that a total of two children was ideal.20 Suppose that p = 0.48 is exactly true for the population of all adults. Gallup announced a margin of error of ±4 percentage points for this poll. What is the probability that the sample proportion for an SRS of size n = 1020 falls between 0.44 and 0.52? You see that it is likely, but not certain, that polls like this give results that are correct within their margin of error. We say more about margins of error in Chapter 6.

5.71 Cyberbullying, continued. Refer to Exercise 5.69. Assume instead that that you sample n = 500 young people from the United Kingdom.

(a) What is the probability that the sample proportion of those who received nasty messages is between 0.59 and 0.65 if the population proportion is p = 0.62?

(b) What is the probability that the sample proportion is between 0.87 and 0.93 if the population proportion is p = 0.90?

(c) Using the results from parts (a) and (b), how does the probability that falls within 60.03 of the true p change as p gets closer to 1?

5.72 How do the results depend on the sample size? Return to the Gallup Poll setting of Exercise 5.70. We are supposing that the proportion of all adults who think that having two children is ideal is p = 0.48. What is the probability that a sample proportion falls between 0.44 and 0.52 (that is, within 64 percentage points of the true p) if the sample is an SRS of size n = 300? Of size n = 5000? Combine these results with your work in Exercise 5.70 to make a general statement about the effect of larger samples in a sample survey.

5.73 Shooting free throws. Since the mid-1960s, the overall free-throw percent at all college levels, for both men and women, has remained pretty consistent. For men, players have been successful on roughly 69% of these free throws, with the season percent never falling below 67% or above 70%.21 Assume that 300,000 free throws will be attempted in the upcoming season.

(a) What are the mean and standard deviation of if the population proportion is p = 0.69?

(b) Using the 68–95–99.7 rule, we expect to fall between what two percents about 95% of the time?

(c) Given the width of the interval in part (b) and the range of season percents, do you think that it is reasonable to assume that the population proportion has been the same over the last 50 seasons? Explain your answer.

5.74 Online learning. The U.S. Department of Education released a report on online learning stating that blended instruction, a combination of conventional face-to-face and online instruction, appears more effective in terms of student performance than conventional teaching.22 You decide to poll incoming students at your institution to see if they prefer courses that blend face-to-face instruction with online components. In an SRS of 400 incoming students, you find that 373 prefer this type of course.

(a) What is the sample proportion of incoming students at your school who prefer this type of blended instruction?

(b) Assume the population proportion for all students nationwide is 85%. Assuming this is true for your insitution too, what is the standard deviation of ?

(c) Using the 68–95–99.7 rule, you would expect to fall between what two percents about 95% of the time?

(d) Based on your result in part (a), do you think that the incoming students at your institution prefer this type of instruction more, less, or about the same as students nationally? Explain your answer.

5.75 Binge drinking. The Centers for Disease Control and Prevention finds that 28% of people aged 18 to 24 years binge drank. Those who binge drank averaged 9.3 drinks per episode and 4.2 episodes per month. The study took a

sample of over 18,000 people aged 18 to 24 years, so the population proportion of people who binge drank is very close to p = 0.28.23 The administration of your college surveys an SRS of 200 students and finds that 56 binge drink.

(a) What is the sample proportion of students at your college who binge drink?

(b) If, in fact, the proportion of all students on your campus who binge drink is the same as the national 28%, what is the probability that the proportion in an SRS of 200 students is as large or larger than the result of the administration’s sample?

(c) A writer for the student paper says that the percent of students who binge brink is higher on your campus than nationally. Write a short letter to the editor explaining why the survey does not support this conclusion.

5.76 How large a sample is needed? The changing probabilities you found in Exercises 5.70 and 5.72 are due to the fact that the standard deviation of the sample proportion gets smaller as the sample size n increases. If the population

proportion is p = 0.48, how large a sample is needed to reduce the standard deviation of to (The 68–95–99.7 rule then says that about 95% of all samples will have within 0.01 of the true p.)

5.77 A test for ESP. In a test for ESP (extrasensory perception), the experimenter looks at cards that are hidden from the subject. Each card contains either a star, a circle, a wave, or a square. As the experimenter looks at each of 20 cards in turn, the subject names the shape on the card.

(a) If a subject simply guesses the shape on each card, what is the probability of a successful guess on a single card? Because the cards are independent, the count of successes in 20 cards has a binomial distribution.

(b) What is the probability that a subject correctly guesses at least 10 of the 20 shapes?

(c) In many repetitions of this experiment with a subject who is guessing, how many cards will the subject guess correctly on the average? What is the standard deviation of the number of correct guesses?

(d) A standard ESP deck actually contains 25 cards. There are five different shapes, each of which appears on five cards. The subject knows that the deck has this makeup. Is a binomial model still appropriate for the count of correct guesses in one pass through this deck? If so, what are n and p? If not, why not?

5.78 Admitting students to college. A selective college would like to have an entering class of 1000 students. Because not all students who are offered admission accept, the college admits more than 1000 students. Past experience shows that about 83% of the students admitted will accept. The college decides to admit 1200 students. Assuming that students make their decisions independently, the number who accept has the B(1200, 0.83) distribution. If this number is less than 1000, the college will admit students from its waiting list.

(a) What are the mean and the standard deviation of the number X of students who accept?

(b) Use the Normal approximation to find the probability that at least 800 students accept.

(c) The college does not want more than 1000 students. What is the probability that more than 1000 will accept?

(d) If the college decides to decrease the number of admission offers to 1150, what is the probability that more than 1000 will accept?

5.79 Is the ESP result better than guessing? When the ESP study of Exercise 5.77 discovers a subject whose performance appears to be better than guessing, the study continues at greater length. The experimenter looks at many

cards bearing one of five shapes (star, square, circle, wave, and cross) in an order determined by random numbers. The subject cannot see the experimenter as the experimenter looks at each card in turn, in order to avoid any possible nonverbal clues. The answers of a subject who does not have ESP should be independent observations, each with probability 1/5 of success. We record 900 attempts.

(a) What are the mean and the standard deviation of the count of successes?

(b) What are the mean and the standard deviation of the proportion of successes among the 900 attempts?

(c) What is the probability that a subject without ESP will be successful in at least 24% of 900 attempts?

(d) The researcher considers evidence of ESP to be a proportion of successes so large that there is only probability 0.01 that a subject could do this well or better by guessing. What proportion of successes must a subject have to meet this standard? (Example 1.45, on pages 65–66, shows how to do an inverse calculation for the Normal distribution that is similar to the

type required here.)

5.80 Show that these facts are true. Use the definition of binomial coefficients to show that each of the following facts is true. Then restate each fact in words in terms of the number of ways that k successes can be distributed among n

observations.

.

5.81 English Premier League Goals. The total number of goals scored per soccer match in the English Premier League (EPL) often follows the Poisson distribution. In one recent season, the average number of goals scored per match (over 380 games played) was 2.768. Compute the following probabilities.

(a) What is the probability that three or more goals are scored in a game?

(b) What is the probability that a game will end in a 020 tie?

(c) Explain why you cannot compute the probability that a game will end in a 121 tie but can provide an upper bound on this probability.

5.82 Number of colony-forming units. In microbiology, colony-forming units (CFUs) are used to measure the number of microorganisms present in a sample. To determine the number of CFUs, the sample is prepared, spread uniformly on an agar plate, and then incubated at some suitable temperature. Suppose that the number of CFUs that appear after incubation follows a Poisson distribution with μ = 15.

(a) If the area of the agar plate is 75 square centimeters (cm2), what is the probability of observing fewer than 4 CFUs in a 25 cm2 area of the plate?

(b) If you were to count the total number of CFUs in five plates, what is the probability you would observe more than 90 CFUs? Use the Poisson distribution to obtain this probability.

(c) Repeat the probability calculation in part (b), but now use the Normal approximation. How close is your answer to your answer in part (b)?

5.83 Metal fatigue. Metal fatigue refers to the gradual weakening and eventual failure of metal that undergoes cyclic loads. The wings of an aircraft, for example, are subject to cyclic loads when in the air, and cracks can form. It is thought that these cracks start at large particles found in the metal. Suppose that the number of particles large enough to initiate a crack follows a Poisson distribution with mean μ = 0.5 per square centimeter (cm2).

(a) What is the mean of the Poisson distribution if we consider a 100 cm2 area?

(b) Using the Normal approximation, what is the probability that this section has more than 60 of these large particles?

CHAPTER 5 EXERCISES

5.84 The cost of Internet access. In Canada, households spent an average of $80.63 CDN monthly for high-speed broadband access.24 Assume that the standard deviation is $27.32. If you ask an SRS of 500 Canadian households with high-speed broadband access how much they pay, what is the probability that the average amount will exceed $85?

5.85 Dust in coal mines. A laboratory weighs filters from a coal mine to measure the amount of dust in the mine atmosphere. Repeated measurements of the weight of dust on the same filter vary Normally with standard deviation σ = 0.09 milligram (mg) because the weighing is not perfectly precise. The dust on a particular filter actually weighs 137 mg.

(a) The laboratory reports the mean of three weighings of this filter. What is the distribution of this mean?

(b) What is the probability that the laboratory reports a weight of 140 mg or higher for this filter?

5.86 The effect of sample size on the standard deviation. Assume that the standard deviation in a very large population is 100.

(a) Calculate the standard deviation for the sample mean for samples of size 1, 4, 25, 100, 250, 500, 1000, and 5000.

(b) Graph your results with the sample size on the x axis and the standard deviation on the y axis.

(c) Summarize the relationship between the sample size and the standard deviation that your graph shows.

5.87 Marks per round in cricket. Cricket is a dart game that uses the numbers 15 to 20 and the bull’s-eye. Each time you hit one of these regions, you score either 0, 1, 2 or 3 marks. Thus, in a round of three throws, a person can score 0 to 9 marks. Lex plans to play 20 games. Her distribution of marks per round is discrete and strongly skewed. A majority of her rounds result in 0, 1, or 2 marks and only a few are more than 4 marks. Assume that her mean is 2.07 marks per round with a standard deviation of 2.11.

(a) Her 20 games involve 140 rounds of three throws each. What are the mean and standard deviation of the average number of marks x̅ in 140 rounds?

(b) Using the central limit theorem, what is the probability that she averages fewer than 2 marks per round?

(c) Do you think that the central limit theorem can be used in this setting? Explain your answer.

5.88 Common last names. The U.S. Census Bureau says that the 10 most common names in the United States are (in order) Smith, Johnson, Williams, Brown, Jones, Miller, Davis, Garcia, Rodriguez, and Wilson.25 These names account for 4.9% of all U.S. residents. Out of curiosity, you look at the authors of the textbooks for your current courses. There are 12 authors in all. Would you be surprised if none of the names of these authors were among the 10 most common? Give a probability to support your answer and explain the reasoning behind your calculation.

5.89 Benford’s law. It is a striking fact that the first digits of numbers in legitimate records often follow Benford’s law (see Example 4.12, page 226). Here it is:

First digit 1 2 3 4 5 6 7 8 9

Proportion 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046

Fake records usually have fewer first digits 1, 2, and 3. What is the approximate probability, if Benford’s law holds, that among 1000 randomly chosen invoices there are 575 or fewer in amounts with first digit 1, 2, or 3?

5.90 Genetics of peas. According to genetic theory, the blossom color in the second generation of a certain cross of sweet peas should be red or white in a 3:1 ratio. That is, each plant has probability 3/4 of having red blossoms, and the blossom colors of separate plants are independent.

(a) What is the probability that exactly 8 out of 10 of these plants have red blossoms?

(b) What is the mean number of red-blossomed plants when 130 plants of this type are grown from seeds?

(c) What is the probability of obtaining at least 90 red-blossomed plants when 130 plants are grown from seeds?

5.91 Leaking gas tanks. Leakage from underground gasoline tanks at service stations can damage the environment.

It is estimated that 25% of these tanks leak. You examine 15 tanks chosen at random, independently of each other.

(a) What is the mean number of leaking tanks in such samples of 15?

(b) What is the probability that 10 or more of the 15 tanks leak?

(c) Now you do a larger study, examining a random sample of 2000 tanks nationally. What is the probability that at least 540 of these tanks are leaking?

5.92 A roulette payoff. A $1 bet on a single number on a casino’s roulette wheel pays $35 if the ball ends up in the number slot you choose. Here is the distribution of the payoff X:

Payoff X $0 $35

Probability 0.974 0.026

Each spin of the roulette wheel is independent of other spins.

(a) What are the mean and standard deviation of X?

(b) Sam comes to the casino weekly and bets on 10 spins of the roulette wheel. What does the law of large numbers say about the average payoff Sam receives from his bets each visit?

(c) What does the central limit theorem say about the distribution of Sam’s average payoff after betting on 520 spins in a year?

(d) Sam comes out ahead for the year if his average payoff is greater than $1 (the amount he bet on each spin). What is the probability that Sam ends the year ahead? The true probability is 0.396. Does using the central limit theorem provide a reasonable approximation?

5.93 A roulette payoff revisited. Refer to the previous exercise. In part (d), the central limit theorem was used to approximate the probability that Sam ends the year ahead. The estimate was about 0.10 too large. Let’s see if we can get closer using the Normal approximation to the binomial with the continuity correction.

(a) If Sam plans to bet on 520 roulette spins, he needs to win at least $520 to break even. If each win gives him $35, what is the minimum number of wins m he must have?

(b) Given p = 1/38 = 0.026, what are the mean and standard deviation of X, the number of wins in 520 roulette spins?

(c) Use the information in the previous two parts to compute P(X ≥ m) with the continuity correction. Does your answer get closer to the exact probability 0.396?

5.94 Learning a foreign language. Does delaying oral practice hinder learning a foreign language? Researchers randomly assigned 25 beginning students of Russian to begin speaking practice immediately and

another 25 to delay speaking for four weeks. At the end of the semester both groups took a standard test of comprehension of spoken Russian. Suppose that in the population of all beginning students, the test scores for early speaking vary according to the N(32, 6) distribution and scores for delayed speaking have the N(29, 5) distribution.

(a) What is the sampling distribution of the mean score x̅ in the early-speaking group in many repetitions of the experiment? What is the sampling distribution of the mean score in the delayed-speaking group?

(b) If the experiment were repeated many times, what would be the sampling distribution of the difference y̅ − x̅ between the mean scores in the two groups?

(c) What is the probability that the experiment will find (misleadingly) that the mean score for delayed speaking is at least as large as that for early speaking?

5.95 Summer employment of college students. Suppose (as is roughly true) that 88% of college men and 82% of college women were employed last summer. A sample survey interviews SRSs of 400 college men and 400 college women. The two samples are of course independent.

(a) What is the approximate distribution of the proportion of women who worked last summer? What is the approximate distribution of the proportion of men who worked?

(b) The survey wants to compare men and women. What is the approximate distribution of the difference in the proportions who worked, ? Explain the reasoning behind your answer.

(c) What is the probability that in the sample a higher proportion of women than men worked last summer?

5.96 Income of working couples. A study of working couples measures the income X of the husband and the

income Y of the wife in a large number of couples in which both partners are employed. Suppose that you knew the means μX and μY and the variances and of both variables in the population.

(a) Is it reasonable to take the mean of the total income X + Y to be μX + μY? Explain your answer.

(b) Is it reasonable to take the variance of the total income to be and ? Explain your answer.

5.97 A random walk. A particle moves along the line in a random walk. That is, the particle starts at the origin (position 0) and moves either right or left in independent steps of length 1. If the particle moves to the right with

probability 0.6, its movement at the ith step is a random variable Xi with distribution

The position of the particle after k steps is the sum of these random movements,

Use the central limit theorem to find the approximate probability that the position of the particle after 500 steps is at least 200 to the right.

5.98 A lottery payoff. A $1 bet in a state lottery’s Pick 3 game pays $500 if the three-digit number you choose exactly matches the winning number, which is drawn at random. Here is the distribution of the payoff X:

Payoff X $0 $500

Probability 0.999 0.001

Each day’s drawing is independent of other drawings.

(a) Joe buys a Pick 3 ticket twice a week. The number of times he wins follows a B(104, 0.001) distribution. Using the Poisson approximation to the binomial, what is the probability that he wins at least once?

(b) The exact binomial probability is 0.0988. How accurate is the Poisson approximation here?

(c) If Joe pays $5 a ticket, he needs to win at least twice a year to come out ahead. Using the Poisson approximation, what is the probability that Joe comes out ahead?

5.99 Poisson distribution? Suppose you find in your spam folder an average of two spam emails every 10 minutes. Furthermore, you find that the rate of spam mail from midnight to 6 A.M. is twice the rate during other parts of the day. Explain whether or not the Poisson distribution is an appropriate model for the spam process.

5.100 Tossing a die. You are tossing a balanced die that has probability 1/6 of coming up 1 on each toss. Tosses are independent. We are interested in how long we must wait to get the first 1.

(a) The probability of a 1 on the first toss is 1/6. What is the probability that the first toss is not a 1 and the second toss is a 1?

(b) What is the probability that the first two tosses are not 1s and the third toss is a 1? This is the probability that the first 1 occurs on the third toss.

(c) Now you see the pattern. What is the probability that the first 1 occurs on the fourth toss? On the fifth toss?

5.101 The geometric distribution. Generalize your work in Exercise 5.100. You have independent trials, each resulting in a success or a failure. The probability of a success is p on each trial. The binomial distribution

describes the count of successes in a fixed number of trials. Now the number of trials is not fixed; instead, continue until you get a success. The random variable Y is the number of the trial on which the first success occurs. What are the possible values of Y? What is the probability P(Y = k) for any of these values? (Comment: The distribution of the number of trials to the first success is called a geometric distribution.)

5.102 Wi-fi interruptions. Suppose that the number of wi-fi interruptions on your home network follows the Poisson distribution with an average of 0.9 wi-fi interruptions per day.

(a) Show that the probability of no interruptions on a given day is 0.4066.

(b) Treating each day as a trial in a binomial setting, use the binomial formula to compute the probability of no interruptions in a week.

(c) Now, instead of using the binomial model, let’s use the Poisson distribution exclusively. What is the mean number

of wi-fi interruptions during a week?

(d) Based on the Poisson mean of part (c), use the Poisson distribution to compute the probability of no interruptions in a week. Confirm that this probability is the same as found part (b). Explain in words why the two ways of computing no interruptions in a week give the same result.

(e) Explain why using the binomial distribution to compute the probability that only one day in the week will not be interruption free would not give the same probability had we used the Poisson distribution to compute that only one interruption occurs during the week.

* The e in the Poisson probability formula is a mathematical constant equal to 2.71828 to six decimal places. Many calculators have an ex function.

1. See the 2015 press release from the Student Monitor, at www.studentmonitor.com.

2. 2015 study conducted by Dr. Beth Livingston and graduate assistants Maria Grillo and Rebecca Paluch, Cornell University ILR School in partnership with Hollaback!

3. K. M. Orzech et al., “The state of sleep among college students at a large public university,” Journal of American College Health, 59 (2011), pp. 612–619.

4. Findings can be found at www.nielsen.com/us/en/insights/news/2014/smartphones-so-many-apps--so-much-time.html.

5. Haipeng Shen, “Nonparametric regression for problems involving lognormal distributions,” PhD dissertation, University of Pennsylvania, 2003. Thanks to Haipeng Shen and Larry Brown for sharing the data.

6. Findings from a 2015 DMR article titled “By the numbers: 60 amazing Snapchat statistics.”

7. Statistical methods for dealing with time-to-failure data, including the Weibull model, are presented in Wayne Nelson, Applied Life Data Analysis, Wiley, 1982.

8. Statistics are from Pew Research Center’s article titled “6 new facts about Facebook,” posted February 3, 2014, on www.pewresearch.org.

9. From the grade distribution database of the Indiana University Office of the Registrar, gradedistribution. registrar.indiana.edu.

10. Diane M. Dellavalle and Jere D. Haas, “Iron status is associated with endurance performance and training in female rowers,” Medicine and Science in Sports and Exercise, 44, No. 8 (2012), pp. 1552–1559.

11. Results of this and other questions from this survey can be found at www.mumsnet.com/surveys/pressure-on-children-and- parents.

12. Results are from S. Rinehart et al., “Sexual harassment and sexual violence experiences among middle school youth,” presented at the 2014 American Educational Research Association annual meeting.

13. U.S. Department of Education, National Center for Education Statistics, “The Condition of Education 2015” (NCES 2015-144), High School Coursetaking, 2015.

14. S. A. Rahimtoola, “Outcomes 15 years after valve replacement with a mechanical vs. a prosthetic valve: Final report of the Veterans Administration randomized trial,” American College of Cardiology, content.onlinejacc.org/article.aspx?articleid=1126703.

15. Based on the article “E-retailers beat stores in customer satisfaction study,” posted on February 19, 2015, by Internet Retailer.

16. The results of this 2012 survey can be found at www.theaa.com/newsroom/news-2012/streetwatch-october-2012-fewer- potholes.html.

17. The results of this 2012 survey can be found at josephsoninstitute.org.

18. Results from the Global News article “New regulations about illegal downloading go into effect,” posted January 2, 2015, and found at globalnews.ca/news/1752246/new-regulations-about-illegal-downloading-go-into-effect/.

19. “The Wireless Report 2014” can be found at www.ditchthelabel.org/the-wireless-report-2014.

20. A summary over time can be found at www.gallup.com/poll/1588/children-violence.aspx.

21. A summary of Larry Wright’s study can be found at www.nytimes.com/2009/03/04/sports/basketball/04freethrow.html.

22. Barbara Means et al., “Evaluation of evidence-based practices in online learning: A meta-analysis and review of online learning studies,” U.S. Department of Education, Office of Planning, Evaluation, and Policy Development, 2010.

23. Dafna Kanny et al., “Vital signs: Binge drinking among women and high school girls—United States, 2011,” Morbidity and Mortality Weekly Report, January 8, 2013.

24. Information was obtained from “Price comparisons of wireline, wireless and internet services in Canada and with foreign jurisdictions,” Canadian Radio-Television and Telecommunications Commission, April 6, 2012.

25. This information can be found at www.census.gov/topics/population/genealogy/data/2000_surnames.html.

6.1 6.2 6.3 6.4

CHAPTER 6 Introduction to Inference

Estimating with Confidence Tests of Significance Use and Abuse of Tests Power and Inference as a Decision

Introduction Statistical inference draws conclusions about a population or process from sample data. It also provides a statement of how much confidence we can place in our conclusions. Although there are numerous methods for inference, there are only a few general types of statistical inference. This chapter introduces the two most common types: confidence intervals and tests of significance.

Because the underlying reasoning for these two types of inference remains the same across different settings, this chapter considers just one simple setting that is closely related to our study of the sampling distributions of in Section 5.2 (page 293): inference about the mean of a large population whose standard deviation is known. This setting,

although unrealistic, allows us to focus on the underlying rationale of statistical inference rather than the calculations. Later chapters present inference methods to use in most of the settings we met in learning to explore data. In fact,

there are libraries—both of books and of computer software—full of more elaborate statistical techniques. Informed use of any of these methods, however, requires a firm understanding of the underlying reasoning. That is the goal of this chapter. A computer or calculator will do the arithmetic, but you must exercise sound judgment based on understanding.

Overview of Inference The purpose of statistical inference is to draw conclusions from data. Formal inference emphasizes substantiating our conclusions via probability calculations. Probability allows us to take chance variation into account. Here is an example.

EXAMPLE 6.1

WADE

Clustering of trees in a forest. The Wade Tract in Thomas County, Georgia, is an old-growth forest of longleaf pine trees (Pinus palustris) that has survived in a relatively undisturbed state since before the settlement of the area by Europeans. Foresters who study these trees are interested in how the trees are distributed in the forest. Is there some sort of clustering, resulting in regions of the forest with more trees than others? Or are the tree locations random, resulting in no particular patterns? Figure 6.1 gives a plot of the locations of all 584 longleaf pine trees in a 200-meter by 200-meter region in the Wade Tract.1

FIGURE 6.1 The distribution of longleaf pine trees, Example 6.1.

Do the locations appear to be random, or do there appear to be clusters of trees? One approach to the analysis of these data indicates that a pattern as clustered as, or more clustered than, the one in Figure 6.1 would occur only 4% of the time if, in fact, the locations of longleaf pine trees in the Wade Tract are random. Because this chance is fairly

small, we conclude that there is some clustering of these trees.

This probability calculation helps us to distinguish between patterns that are consistent or inconsistent with the random location scenario. Here is an example assessing a new oral antibiotic for acne—with a different conclusion.

EXAMPLE 6.2

Effectiveness of a new oral antibiotic. Researchers want to know if a new oral antibiotic is more effective in relieving acne than a popular topical (on the skin) antibiotic. Twenty patients are randomly assigned to receive the oral medication, and another 20 receive the topical medication. Fifteen (75%) of those taking the oral medication find satisfactory symptom relief versus only 11 (55%) of the topical medication patients.

Our unaided judgment suggests that the oral medication is better, 75% to 55%. However, probability calculations tell us that a difference this large or larger between the results in the two groups of 20 patients would occur about one time in five simply because of chance variation. In this case, it is better to conclude that the data fail to establish a real difference between the two treatments. This probability (nearly 0.19) is too large to ignore.

In this chapter, we introduce the two most frequently used types of statistical inference. Section 6.1 concerns confidence intervals for estimating the value of a population parameter. Section 6.2 presents tests of significance, which assess the evidence for a claim, such as those in Examples 6.1 and 6.2.

sampling distribution, p. 286

Both types of inference are based on the sampling distributions of statistics. That is, both report probabilities that state what would happen if we used the inference method many times. This kind of probability statement is characteristic of standard statistical inference. Users of statistics must understand the nature of this reasoning and the meaning of the probability statements that appear, for example, online and in journal articles and statistical software output.

Because the methods of formal inference are based on sampling distributions, they require a probability model for the data. Trustworthy probability models can arise in many ways, but the model is most secure and inference is most reliable when the data are produced by a properly randomized design.

When you use statistical inference, you are acting as if the data come from a random sample or a randomized experiment. If this is not true, your conclusions may be open to challenge. Do not be overly impressed by the complex details of formal inference. This elaborate machinery cannot remedy basic flaws in producing the data such as voluntary response samples and confounded experiments. Use the common sense developed in your study of the first three chapters of this book, and proceed to detailed formal inference only when you are satisfied that the data deserve such analysis.

6.1 Estimating with Confidence

When you complete this section, you will be able to:

Describe a level C confidence interval for a population parameter in terms of an estimate and its margin of error. Construct a level C confidence interval for μ from a simple random sample (SRS) of size n from a large population having known standard deviation σ. Explain how the margin of error changes with a change in the confidence level C. Determine the sample size needed to obtain a specified margin of error for a level C confidence interval for μ. Identify situations where inference about m based on the confidence interval

may be suspect.

linear transformations, p. 44

The SAT is a widely used measure of readiness for college study. It consists of two sections, one for mathematical reasoning ability (SATM), one for reading and writing ability (SATV). Possible scores on each section range from 200 to 800, for a total range of 400 to 1600. Since 1995, section scores have been recentered so that the mean is approximately 500 with a standard deviation of 100 in a large “standardized group.” This scale has been maintained so that scores have a constant interpretation.

EXAMPLE 6.3

Estimating the mean SATM score for seniors in California. Suppose that you want to estimate the mean SATM score for the 485,264 high school seniors in California.2 You know better than to trust data from the students who choose to take the SAT. Only about 38% of California students typically take the SAT. These self- selected students are planning to attend college and are not representative of all California seniors. At considerable effort and expense, you give the test to an SRS of 500 California high school seniors. The mean score for your sample is = 495. What can you say about the mean score μ in the population of all 485,264 seniors?

unbiased estimator, p. 287

law of large numbers, p. 250

The sample mean is the natural estimator of the unknown population mean μ. We know that is an unbiased estimator of μ. More important, the law of large numbers says that the sample mean must approach the population mean as the size of the sample grows. The value = 495, therefore, appears to be a reasonable estimate of the mean score μ that all 485,264 students would achieve if they took the test.

But how reliable is this estimate? A second sample of 500 students would surely not give a sample mean of 495 again. Unbiasedness says only that there is no systematic tendency to underestimate or overestimate the truth. Could we plausibly get a sample mean of 485 and a sample mean of 520 in repeated samples? An estimate without an indication of its variability is of little value.

Statistical confidence

central limit theorem, p. 298

The unbiasedness of an estimator concerns the center of its sampling distribution, but questions about variation are answered by looking at its spread. The central limit theorem says that if the entire population of SATM scores has mean μ and standard deviation σ, then in repeated SRSs of size 500, the sample mean is approximately

. Let us suppose that we know that the standard deviation σ of SATM scores in our California population is σ = 100. (We will see in the next chapter how to proceed when σ is not known. For now, we are more interested in statistical reasoning than in details of realistic methods.) This means that in repeated sampling the sample mean has an approximately Normal distribution centered at the unknown population mean μ and a standard deviation of

FIGURE 6.2 Distribution of the sample mean, Example 6.3. lies within ±9 points of μ in 95% of all samples. This also means that μ is within ±9 points of in those samples.

Now we are ready to proceed. Consider this line of thought, which is illustrated in Figure 6.2:

The 68–95–99.7 rule says that the probability is about 0.95 that will be within 9 points (that is, two standard deviations of ) of the population mean score μ. To say that lies within 9 points of μ is the same as saying that μ is within 9 points of . So about 95% of all samples will contain the true μ in the interval from − 9 to + 9.

We have simply restated a fact about the sampling distribution of . The language of statistical inference uses this fact about what would happen in the long run to express our confidence in the results of any one sample. Our sample gave = 495. We say that we are 95% confident that the unknown mean score for all California seniors lies between

− 9 = 495 − 9 = 486

and

+ 9 = 495 + 9 = 504

Be sure you understand the grounds for our confidence. There are only two possibilities for our SRS:

1. The interval between 486 and 504 contains the true μ.

6.1

6.2

6.3

2. The interval between 486 and 504 does not contain the true μ.

We cannot know whether our sample is one of the 95% for which the interval ± 9 contains μ or one of the unlucky 5% for which it does not contain μ. The statement that we are 95% confident is shorthand for saying, “We arrived at these numbers by a method that gives correct results 95% of the time.”

USE YOUR KNOWLEDGE

How much do you spend on lunch? The average amount you spend on a lunch during the week is not known. Based on past experience, you are willing to assume that the standard deviation is $2.10. If you take a random sample of 28 lunches, what is the value of the standard deviation of ?

Applying the 68–95–99.7 rule. In the setting of the previous exercise, the 68–95–99.7 rule says that the probability is about 0.95 that is within $________ of the population mean μ. Fill in the blank.

Constructing a 95% confidence interval. In the setting of the previous two exercises, about 95% of all samples will capture the true mean in the interval plus or minus $________. Fill in the blank.

Confidence intervals In the setting of Example 6.3, the interval of numbers between the values ± 9 is called a 95% confidence interval for μ. Like most confidence intervals we will discuss, this one has the form

estimate ± margin of error

margin of error, p. 287

The estimate ( = 495 in this case) is our guess for the value of the unknown parameter. The margin of error (9 here) reflects how accurate we believe our guess is, based on the variability of the estimate, and how confident we are that the procedure will produce an interval that will contain the true population mean μ.

Figure 6.3 illustrates the behavior of 95% confidence intervals in repeated sampling from a Normal distribution with mean μ. The center of each interval (marked by a dot) is at and varies from sample to sample. The sampling distribution of (also Normal) appears at the top of the figure to show the long-term pattern of this variation.

The 95% confidence intervals, ± margin of error, from 25 SRSs appear below the sampling distribution. The arrows on either side of the dot ( ) span the confidence interval. All except one of the 25 intervals contain the true value of μ. In those intervals that contain μ, sometimes μ is near the middle of the interval and sometimes it is closer to one of the ends. This again reflects the variation of . In practice, we don’t know the value of μ, but we have a method such that, in a very large number of samples, 95% of the confidence intervals will contain μ.

FIGURE 6.3 Twenty-five samples from the same population gave these 95% confidence intervals. In the long run, 95% of all samples give an interval that covers μ. The sampling distribution of is shown at the top.

6.4

6.5

We can construct confidence intervals for many different parameters based on a variety of designs for data collection. We will learn the details of a number of these in later chapters. Two important things about a confidence interval are common to all settings:

1. It is an interval of the form (a, b), where a and b are numbers computed from the sample data.

confidence level

2. It has a property called a confidence level that gives the probability of producing an interval that contains the unknown parameter.

Users can choose the confidence level, but 95% is the standard for most situations. Occasionally, 90% or 99% is used. We use C to stand for the confidence level in decimal form. For example, a 95% confidence level corresponds to C = 0.95.

CONFIDENCE INTERVAL

A level C confidence interval for a parameter is an interval computed from sample data by a method that has probability C of producing an interval containing the true value of the parameter.

With the Confidence Interval applet, you can construct diagrams similar to the one displayed in Figure 6.3. The only difference is that the applet displays the Normal population distribution at the top rather than the Normal sampling distribution of . You choose the confidence level C, the sample size n, and whether you want to generate 1 or 25 samples at a time. A running total (and percent) of the number of intervals that contain μ is displayed so you can consider a larger number of samples.

When generating single samples, the data for the latest SRS are shown below the confidence interval. The spread in these data reflects the spread of the population distribution. This spread is assumed known, and it does not change with sample size. What does change, as you vary n, is the margin of error, since it reflects the uncertainty in the estimate of μ. As you increase n, you’ll find that the span of the interval gets smaller.

USE YOUR KNOWLEDGE

Generating a single confidence interval. Using the default settings in the Confidence Interval applet (95% confidence level and n = 20), click “Sample” to choose an SRS and display its confidence interval.

(a) Is the spread in the data, shown as yellow dots below the confidence interval, larger than the span of the confidence interval? Explain why this would typically be the case.

(b) For the same data set, you can compare the span of the confidence interval for different values of C by sliding the confidence level to a new value. For the SRS you generated in part (a), what happens to the span of the interval when you move C to 99%? What about 90%? Describe the relationship you find between the confidence level C and the span of the confidence interval.

80% confidence intervals. The idea of an 80% confidence interval is that the interval captures the true parameter value in 80% of all samples. That’s not high enough confidence for practical use, but 80% hits and 20% misses make it easy to see how a confidence interval behaves in repeated samples from the same population.

(a) Set the confidence level in the Confidence Interval applet to 80%. Click “Sample 25” to choose 25 SRSs

and display their confidence intervals. How many of the 25 intervals contain the true mean μ? What proportion contain the true mean?

(b) We can’t determine whether a new SRS will result in an interval that contains μ or not. The confidence level only tells us what percent will contain μ in the long run. Click “Sample 25” again to get the confidence intervals from 50 SRSs. What proportion hit? Keep clicking “Sample 25” and record the proportion of hits among 100, 200, 300, 400, and 500 SRSs. As the number of samples increases, we expect the percent of captures to get closer to the confidence level, 80%. Do you find this pattern in your results?

Confidence interval for a population mean

central limit theorem, p. 298

We now construct a level C confidence interval for the mean μ of a population when the data are an SRS of size n. The construction is based on the sampling distribution of the sample mean . This distribution is exactly when the population has the N(μ, σ) distribution. The central limit theorem says that this same sampling distribution is approximately correct for large samples whenever the population mean and standard deviation are μ and σ. For now, we will assume we are in one of these two situations. We discuss what we mean by “large sample” after we briefly study these intervals.

Our construction of a 95% confidence interval for the mean SATM score began by noting that any Normal distribution has probability about 0.95 within ±2 standard deviations of its mean. To construct a level C confidence interval we first catch the central C area under a Normal curve. That is, we must find the number z* such that any Normal distribution has probability C within ±z* standard deviations of its mean.

Because all Normal distributions have the same standardized form, we can obtain everything we need from the standard Normal curve. Figure 6.4 shows how C and z* are related. Values of z* for many choices of C appear in the row labeled z* at the bottom of Table D. Here are the most important entries from that row:

z* 1.645 1.960 2.576 C 90% 95% 99%

Notice that for 95% confidence the value 2 obtained from the 68–95–99.7 rule is replaced with the more precise 1.96.

FIGURE 6.4 To construct a level C confidence interval, we must find the number z*. The area between −z* and z* under the standard Normal curve is C.

As Figure 6.4 reminds us, any Normal curve has probability C between the point z* standard deviations below the mean and the point z* standard deviations above the mean. The sample mean has the Normal distribution with mean μ and standard deviation so there is probability C that lies between

This is exactly the same as saying that the unknown population mean μ lies between

That is, there is probability C that the interval contains μ. This is our confidence interval. The estimate

of the unknown μ is , and the margin of error is .

CONFIDENCE INTERVAL FOR A POPULATION MEAN

Choose an SRS of size n from a population having unknown mean μ and known standard deviation σ. The margin of error for a level C confidence interval for μ is

Here, z* is the value on the standard Normal curve with area C between the critical points −z* and z*. The level C confidence interval for μ is

± m

The confidence level of this interval is exactly C when the population distribution is Normal and is approximately C when n is large in other cases.

Starting in 2008, Sallie Mae, a major provider of education loans and savings programs, has conducted an annual study titled “How America Pays for College.” In the 2015 survey, 1600 randomly selected individuals (800 parents of undergraduate students and 800 undergraduate students) were surveyed by telephone.3

Many of the survey questions focus on the composition of funding sources used to pay for college, so the undergraduates in the survey are often responding for their parents. For example, each participant is asked to report how much of the parent’s current income is used to pay for college. Do you think it is wise to combine responses across the parents and undergraduates? Are you fully aware of how much money your parents are spending and borrowing for college? The authors report overall averages and percents in their report. We will also consider this a sample from one population but this is certainly debatable.

EXAMPLE 6.4

Average college savings fund contribution. One survey question asked how much money from a college savings fund, such as a 529 plan, is used to pay for college. Of the 1600 who were surveyed, n = 1593 provided an answer. Nonresponse should always be considered as a source of bias. In this case, the nonresponse is very low, so we’ll proceed by treating the n = 1593 sample as if it were an unbiased sample.

The average amount is $1768. It’s very likely that this distribution is highly skewed to the right with many small amounts and a few very large amounts. Nevertheless, because the sample size is quite large, we can rely on the central limit theorem to assure us that the confidence interval based on the Normal distribution will be a good approximation.

Let’s compute an approximate 95% confidence interval for the true mean amount contributed from a college savings fund among all undergraduates. We’ll assume that the standard deviation for the population of college savings fund contributions is $1483. For 95% confidence, we see from Table D that z* = 1.960. The margin of error for the 95% confidence interval for μ is, therefore,

We have computed the margin of error with more digits than we really need. Our mean is rounded to the nearest $1, so we will do the same for the margin of error. Keeping additional digits would provide no additional useful information. Therefore, we will use m = 37. The approximate 95% confidence interval is

We are 95% confident that the mean amount contributed from a college savings fund among all undergraduates is between $1731 and $1805.

6.6

6.7

6.8

Suppose that the researchers who designed this study had used a different sample size. How would this affect the confidence interval? We can answer this question by changing the sample size in our calculations and assuming that the sample mean is the same.

EXAMPLE 6.5

How sample size affects the confidence interval. As in Example 6.4, the sample mean of the college savings fund contribution is $1768 and the population standard deviation is $1483. Suppose that the sample size is only 177 but still large enough for us to rely on the central limit theorem. In this case, the margin of error for 95% confidence is

and the approximate 95% confidence interval is

Notice that the margin of error for this example is three times as large as the margin of error that we computed in Example 6.4. The only change that we made was to assume that the sample size is 177 rather than 1593. This sample size is one-ninth of the original 1593. Thus, we triple the margin of error when we reduce the sample size to one-ninth of the original value. Figure 6.5 illustrates the effect in terms of the intervals.

FIGURE 6.5 Confidence intervals for n = 1593 and n = 177, Examples 6.4 and 6.5. A sample size nine times as large results in a confidence interval that is one-third as wide.

USE YOUR KNOWLEDGE

Average amount paid for college. Refer to Example 6.4. The average annual amount the n = 1593 families paid for college was $24,164.4 If the population standard deviation is $8500, give the 95% confidence interval for μ, the average annual amount a family pays for a college undergraduate.

Changing the sample size. In the setting of the previous exercise, would the margin of error for 95% confidence be roughly doubled or halved if the sample size were raised to n = 6375? Verify your answer by performing the calculations.

Changing the confidence level. In the setting of Exercise 6.7, would the margin of error for 99% confidence be larger or smaller? Verify your answer by performing the calculations.

The argument leading to the form of confidence intervals for the population mean μ rested on the fact that the statistic used to estimate μ has a Normal distribution. Because many sample estimates have Normal distributions (at least approximately), it is useful to notice that the confidence interval has the form

estimate ± z*σestimate

The estimate based on the sample is the center of the confidence interval. The margin of error is z*σestimate. The desired confidence level determines z* from Table D. The standard deviation of the estimate is found from knowledge of the sampling distribution in a particular case. When the estimate is from an SRS, the standard deviation of the estimate is . We return to this general form numerous times in the following chapters.

How confidence intervals behave The margin of error for the mean of a Normal population illustrates several important properties that are shared by all confidence intervals in common use. The user chooses the confidence level, and the margin of error follows from this choice.

Both high confidence and a small margin of error are desirable characteristics of a confidence interval. High confidence says that our method almost always gives correct answers. A small margin of error says that we have pinned down the parameter quite precisely.

Suppose that in planning a study you calculate the margin of error and decide that it is too large. Here are your choices to reduce it:

Use a lower level of confidence (smaller C). Choose a larger sample size (larger n). Reduce σ.

For most problems, you would choose a confidence level of 90%, 95%, or 99%, so z* will be 1.645, 1.960, or 2.576, respectively. Figure 6.4 shows that z* will be smaller for lower confidence (smaller C). The bottom row of Table D also shows this. If n and σ are unchanged, a smaller z* leads to a smaller margin of error.

EXAMPLE 6.6

How the confidence level affects the confidence interval. Suppose that for the college saving fund contribution data in Example 6.4 (page 350), we wanted 99% confidence. Table D tells us that for 99% confidence, z* = 2.576. The margin of error for 99% confidence based on 1593 observations is

and the 99% confidence interval is

Requiring 99%, rather than 95%, confidence has increased the margin of error from 37 to 96. Figure 6.6 compares the two intervals.

FIGURE 6.6 Confidence intervals, Examples 6.4 and 6.6. The larger the value of C, the wider the interval.

Similarly, choosing a larger sample size n reduces the margin of error for any fixed confidence level. The square root in the formula implies that we must multiply the number of observations by 4 in order to cut the margin of error in half. Likewise, if we want to reduce the standard deviation of by a factor of 4, we must take a sample 16 times as large.

The standard deviation σ measures the variation in the population. You can think of the variation among individuals in the population as noise that obscures the average value μ. It is harder to pin down the mean μ of a highly variable population; that is why the margin of error of a confidence interval increases with σ.

In practice, we can sometimes reduce σ by carefully controlling the measurement process. We also might change the mean of interest by restricting our attention to only part of a large population. Focusing on a subpopulation will

often result in a smaller σ. This is why many medical studies only use healthy male subjects. The tradeoff, however, is less generalizable results.

Choosing the sample size A wise user of statistics never plans data collection without, at the same time, planning the inference. You can arrange to have both high confidence and a small margin of error. The margin of error of the confidence interval for a population mean is

Notice once again that it is the size of the sample that determines the margin of error. The size of the population (as long as the population is much larger than the sample) does not influence the sample size we need.

To obtain a desired margin of error m, plug in the value of σ and the value of z* for your desired confidence level, and solve for the sample size n. Here is the result.

SAMPLE SIZE FOR DESIRED MARGIN OF ERROR

The confidence interval for a population mean will have a specified margin of error m when the sample size is

This formula does not account for collection costs. In practice, taking observations costs time and money. The required sample size may be impossibly expensive. In those situations, you might consider a larger margin of error and/or a lower confidence level to find a workable sample size.

EXAMPLE 6.7

How many undergraduates should we survey? Suppose that we are planning a survey similar to the one described in Example 6.4 (page 350). If we want the margin of error for the average amount contributed from a college savings plan to be $30 with 95% confidence, what sample size n do we need? For 95% confidence, Table D gives z* = 1.960. For σ we will use the value from the previous study, $1483. If the margin of error is $30, we have

Because 9387 measurements will give a slightly wider interval than desired and 9388 measurements a slightly narrower interval, we should choose n = 9388. We need information from 9388 undergraduates to determine an estimate of mean college savings fund contribution with the desired margin of error.

It is always safe to round up to the next higher whole number when finding n because this will give us a smaller margin of error. The purpose of this calculation is to determine a sample size that is sufficient to provide useful results, but the determination of what is useful is a matter of judgment.

Would we need a much larger sample size to obtain a margin of error of $25? Here is the calculation:

A sample of n = 13,519 is much larger, and the costs of such a large sample may be prohibitive.

Unfortunately, the actual number of usable observations is often less than what we plan at the beginning of a study. This is particularly true of data collected in surveys but is an important consideration in most studies. Careful study designers often assume a nonresponse rate or dropout rate that specifies what proportion of the originally planned sample will fail to provide data. We use this information to calculate the sample size to be used at the start of the study.

For example, if in Example 6.7 we expect only 50% of those contacted to respond, we would need to start with a

6.9

6.10

sample size of 2 × 9388 = 18,776 to obtain usable information from 9388 undergraduates and parents of undergraduates.

USE YOUR KNOWLEDGE

Starting salaries. You are planning a survey of starting salaries for recent computer science majors. In the latest survey by the National Association of Colleges and Employers, the average starting salary was reported to be $61,287.5 If you assume that the standard deviation is $3850, what sample size do you need to have a margin of error equal to $500 with 95% confidence?

Changes in sample size. Suppose that in the setting of the previous exercise you have the resources to contact 300 recent graduates. If all respond, will your margin of error be larger or smaller than $500? What if only 50% respond? Verify your answers by performing the calculations.

Some cautions

We have already seen that small margins of error and high confidence can require large numbers of observations. You should also be keenly aware that any formula for inference is correct only in specific circumstances. If the government required statistical procedures to carry warning labels like those on drugs, most inference methods would have long labels. Our formula for estimating a population mean comes with the following list of warnings for the user:

The data should be an SRS from the population. We are completely safe if we actually did a randomization and drew an SRS. We are not in great danger if the data can plausibly be thought of as independent observations from a population. That is the case in Examples 6.4 through 6.7, provided the undergraduates and parents can be considered one population. The formula is not correct for probability sampling designs more complex than an SRS. Correct methods for other designs are available. We will not discuss confidence intervals based on multistage or stratified samples (page 195). If you plan such samples, be sure that you (or your statistical consultant) know how to carry out the inference you desire. There is no correct method for inference from data haphazardly collected with bias of unknown size. Fancy formulas cannot rescue badly produced data.

resistant measure, p. 30

Because is not a resistant measure, outliers can have a large effect on the confidence interval. You should search for outliers and try to correct them or justify their removal before computing the interval. If the outliers cannot be removed, ask your statistical consultant about procedures that are not sensitive to outliers. If the sample size is small and the population is not Normal, the true confidence level will be different from the value C used in computing the interval. Prior to any calculations, examine your data carefully for skewness and other signs of non-Normality. Remember though that the interval relies only on the distribution of , which even for quite small sample sizes is much closer to Normal than is the distribution of the individual observations. When n ≥ 15, the confidence level is not greatly disturbed by non-Normal populations unless extreme outliers or quite strong skewness are present. Our college fund contribution data in Example 6.4 are very likely skewed, but because of the large sample size, we are confident that the distribution of the sample mean will be approximately Normal.

standard deviation s, p. 38

The interval assumes that the standard deviation σ of the population is known. This unrealistic requirement renders the interval of little use in statistical practice. We will learn in the next chapter what to do when σ is unknown. If, however, the sample is large, the sample standard deviation σ will be close to the unknown σ. The interval is then an approximate confidence interval for μ.

The most important caution concerning confidence intervals is a consequence of the first of these warnings. The margin of error in a confidence interval covers only random sampling errors. The margin of error is obtained from the sampling distribution and indicates how much error can be expected because of chance variation in randomized data production.

6.11

Practical difficulties such as undercoverage and nonresponse in a sample survey cause additional errors. These errors can be larger than the random sampling error. This often happens when the sample size is large (so that is small). Remember this unpleasant fact when reading the results of an opinion poll or other sample survey. The practical conduct of the survey influences the trustworthiness of its results in ways that are not included in the announced margin of error.

Every inference procedure that we will meet has its own list of warnings. Because many of the warnings are similar to those we have mentioned, we will not print the full warning label each time. It is easy to state (from the mathematics of probability) conditions under which a method of inference is exactly correct. These conditions are never fully met in practice.

For example, no population is exactly Normal. Deciding when a statistical procedure should be used in practice often requires judgment assisted by exploratory analysis of the data. Mathematical facts are, therefore, only a part of statistics. The difference between statistics and mathematics can be stated thusly: mathematical theorems are true; statistical methods are often effective when used with skill.

Finally, you should understand what statistical confidence does not say. Based on our SRS in Example 6.3, we are 95% confident that the mean SATM score for the California students lies between 486 and 504. This says that this interval was calculated by a method that gives correct results in 95% of all possible samples. It does not say that the probability is 0.95 that the true mean falls between 486 and 504. No randomness remains after we draw a particular sample and compute the interval. The true mean either is or is not between 486 and 504. The probability calculations of standard statistical inference describe how often the method, not a particular sample, gives correct answers.

USE YOUR KNOWLEDGE

Nonresponse in a survey. In earlier versions of the Sallie Mae survey of Example 6.4 (page 350), participants were asked to report the undergraduate’s outstanding credit card balance. Only about a third reported this amount. Provide a couple of reasons why a survey respondent might not provide an amount. Based on these reasons, do you think the sample mean using just the reported amounts is biased? Is the margin of error based just on the reported amounts a good measure of precision? Explain your answers.

SECTION 6.1 SUMMARY The purpose of a confidence interval is to estimate an unknown parameter with an indication of how accurate the estimate is and of how confident we are that the result is correct. Any confidence interval has two parts: an interval computed from the data and a confidence level. The interval often has the form

estimate ± margin of error

The confidence level states the probability that the method will give a correct answer. That is, if you use 95% confidence intervals, in the long run 95% of your intervals will contain the true parameter value. When you apply the method once (that is, to a single sample), you do not know if your interval gave a correct answer (this happens 95% of the time) or not (this happens 5% of the time). The margin of error for a level C confidence interval for the mean μ of a Normal population with known standard deviation σ, based on an SRS of size n, is given by

Here z* is obtained from the row labeled z* at the bottom of Table D. The probability is C that a standard Normal random variable takes a value between −z* and z*. The confidence interval is

± m

If the population is not Normal and n is large, the confidence level of this interval is approximately correct.

Other things being equal, the margin of error of a confidence interval decreases as the confidence level C decreases, the sample size n increases, and the population standard deviation σ decreases.

The sample size n required to obtain a confidence interval of specified margin of error m for a population mean is

where z* is the critical point for the desired level of confidence.

A specific confidence interval formula is correct only under specific conditions. The most important conditions concern the method used to produce the data. Other factors such as the form of the population distribution may also be important. These conditions should be investigated prior to any calculations.

SECTION 6.1 EXERCISES For Exercises 6.1 through 6.3, see pages 345–346; for Exercises 6.4 and 6.5, see pages 347–348; for Exercises 6.6 through 6.8, see page 351; for Exercises 6.9 and 6.10, see page 354; and for Exercise 6.11, see page 356.

6.12 Margin of error and the confidence interval. A study of stress on the campus of your university reported a mean stress level of 78 (on a 0 to 100 scale with a higher score indicating more stress) with a margin of error of 5 for 95% confidence. The study was based on a random sample of 64 undergraduates.

(a) Give the 95% confidence interval.

(b) If you wanted 99% confidence for the same study, would your margin of error be greater than, equal to, or less than 5? Explain your answer.

6.13 Changing the sample size. Consider the setting of the previous exercise. Suppose that the sample mean is again 78 and the population standard deviation is 20. Make a diagram similar to Figure 6.5 that illustrates the effect of sample size on the width of a 95% interval. Use the following sample sizes: 9, 25, 81, and 100. Summarize what the diagram shows.

6.14 Changing the confidence level. Consider the setting of the previous two exercises. Suppose that the sample mean is still 78, the sample size is 64, and the population standard deviation is 20. Make a diagram similar to Figure 6.6 that illustrates the effect of the confidence level on the width of the interval. Use 80%, 90%, 95%, and 99%. Summarize what the diagram shows.

6.15 Confidence interval mistakes and misunderstandings. Suppose that 500 randomly selected alumni of the University of Okoboji were asked to rate the university’s academic advising services on a 1 to 10 scale. The sample mean was found to be 8.6. Assume that the population standard deviation is known to be σ = 2.2.

(a) Ima Bitlost computes the 95% confidence interval for the average satisfaction score as 8.6 ± 1.96(2.2). What is her mistake?

(b) After correcting her mistake in part (a), she states, “I am 95% confident that the sample mean falls between 8.4 and 8.8.” What is wrong with this statement?

(c) She quickly realizes her mistake in part (b) and instead states, “The probability that the true mean is between 8.4 and 8.8 is 0.95.” What misinterpretation is she making now?

(d) Finally, in her defense for using the Normal distribution to determine the confidence interval she says, “Because the sample size is quite large, the population of alumni ratings will be approximately Normal.” Explain to Ima her misunderstanding and correct this statement.

6.16 More confidence interval mistakes and misunderstandings. Suppose that 100 randomly selected members of the Karaoke Channel were asked how much time they typically spend on the site during the week.6 The sample mean was found to be 3.8 hours. Assume that the population standard deviation is known to be σ = 2.9.

(a) Cary Oakey computes the 95% confidence interval for the average time on the site as 3.8 ± 1.96(2.9/100). What is his mistake?

(b) He corrects this mistake and then states that “95% of the members spend between 3.23 and 4.37 hours a week on the site.” What is wrong with his interpretation of this interval?

(c) The margin of error is slightly larger than half an hour. To reduce this to roughly 15 minutes, Cary says that the sample size needs to be doubled to 200. What is wrong with this statement?

6.17 The state of stress in the United States. Since 2007, the American Psychological Association has supported an annual nationwide survey to examine stress across the United States.7 This year, a total of 720 millennials (18- to 33-year-olds) were asked to indicate their average stress level (on a 10-point scale) during the past month. The mean score was 5.5. Assume that the population standard deviation is 2.8.

(a) Give the margin of error and find the 95% confidence interval for this sample.

(b) Repeat these calculations for a 99% confidence interval. How do the results compare with those in part (a)?

6.18 Inference based on integer values. Refer to Exercise 6.17. The data for this study are integer values between 1 and 10. Explain why the confidence interval based on the Normal distribution should be a good approximation.

6.19 Mean TRAP in young women. For many important processes that occur in the body, direct measurement of characteristics of the process is not possible. In many cases, however, we can measure a biomarker, a biochemical substance

that is relatively easy to measure and is associated with the process of interest. Bone turnover is the net effect of two processes: the breaking down of old bone, called resorption, and the building of new bone, called formation. One biochemical measure of bone resorption is tartrate-resistant acid phosphatase (TRAP), which can be measured in blood. In a study of bone turnover in young women, serum TRAP was measured in 31 subjects.8 The mean was 13.2 units per liter (U/l). Assume that the standard deviation is known to be 6.5 U/l. Give the margin of error and find a 95% confidence interval for the mean TRAP amount in young women represented by this sample.

6.20 Mean OC in young women. Refer to the previous exercise. A biomarker for bone formation measured in the same study was osteocalcin (OC), measured in the blood. For the 31 subjects in the study, the mean was 33.4 nanograms per milliliter (ng/ml). Assume that the standard deviation is known to be 19.6 ng/ml. Report the 95% confidence interval.

6.21 Populations sampled and margins of error. Consider the following two scenarios. (A) Take a simple random sample of 200 freshman students at your college or university. (B) Take a simple random sample of 200 students at your college or university. For each of these samples, you will record the amount spent on textbooks used for classes during the fall semester. Which sample should have the smaller margin of error? Explain your answer.

6.22 Average starting salary. The National Association of Colleges and Employers (NACE) Spring Salary Survey shows that the current class of college graduates received an average starting-salary offer of $48,127.9 Your institution

collected an SRS (n = 300) of its recent graduates and obtained a 95% confidence interval of ($46,382, $48,008). What can we conclude about the difference between the average starting salary of recent graduates at your institution and the overall NACE average? Write a short summary.

6.23 Consumption of sweet snacks. A recent study reported that the U.S. per capita consumption of sweet snacks among healthy weight children aged 12 to 19 years is 251.2 kilocalories per day (kcal/d).10 This was based on 24-hour dietary recall records of n = 2265 adolescents.

(a) Suppose that the population distribution is heavily skewed, with a standard deviation equal to 540 kcal/d. What is the margin of error for a 95% confidence interval of the per capita consumption of sweet snacks?

(b) A future study is being planned and the goal is to have the margin of error no more than 15 kcal/d. Based on your answer to part (a), will this study require an examination of more or fewer recall records? Explain your answer without calculations.

(c) Compute the sample size necessary for the study described in part (b).

6.24 Total sleep time of college students. In Example 5.4 (page 293), the total sleep time per night among college students was approximately Normally distributed with mean μ = 6.78 hours and standard deviation σ = 1.24 hours. You initially plan to take an SRS of size n = 175 and compute the average total sleep time.

(a) What is the standard deviation for the average time in hours? in minutes?

(b) Use the 95 part of the 68–95–99.7 rule to describe the variability of this sample mean.

(c) What is the probability that your average will be below 6.9 hours?

6.25 Determining sample size. Refer to the previous exercise. You really want to use a sample size such that about 95% of the averages fall within ±5 minutes of the true mean μ = 6.78.

(a) Based on your answer to part (b) in Exercise 6.24, should the sample size be larger or smaller than 175? Explain.

(b) What standard deviation of do you need such that 95% of all samples will have a mean within 5 minutes of μ?

(c) Using the standard deviation you calculated in part (b), determine the number of students you need to sample.

6.26 Inference based on skewed data. The mean OC for the 31 subjects in Exercise 6.20 was 33.4 ng/ml. In our calculations, we assumed that the standard deviation was known to be 19.6 ng/ml. Use the 68–95–99.7 rule from Chapter

1 (page 57) to find the approximate bounds on the values of OC that would include these percents of the population. If the assumed standard deviation is correct, this distribution may be highly skewed. Why? (Hint: The measured values for a variable such as this are all positive.) Do you think that this skewness will invalidate the use of the Normal confidence interval in this case? Explain your answer.

6.27 Average hours per week listening to the radio. The Student Monitor surveys 1200 undergraduates from four- year colleges and universities throughout the United States semiannually to understand trends among college students.11 Recently, the Student Monitor reported that the average amount of time listening to the radio per week was 11.5 hours. Of the 1200 students surveyed, 83% said that they listened to the radio, so this collection of listening times has around 204 (17% × 1200) zeros. Assume that the standard deviation is 8.3 hours.

(a) Give a 95% confidence interval for the mean time spent per week listening to the radio.

(b) Is it true that 95% of the 1200 students reported weekly times that lie in the interval you found in part (a)? Explain your answer.

(c) It appears that the population distribution has many zeros and is skewed to the right. Explain why the confidence interval based on the Normal distribution should nevertheless be a good approximation.

6.28 Average minutes per week listening to the radio. Refer to the previous exercise.

(a) Give the mean and standard deviation in minutes.

(b) Calculate the 95% confidence interval in minutes from your answer to part (a).

(c) Explain how you could have directly calculated this interval from the 95% interval that you calculated in the previous exercise.

6.29 Outlook on life. Since 2008, the Gallup-Healthways Well-Being Index tracks how people feel about their daily lives. In 2014, 54.1% of the respondents were classified as “thriving.” This classification is based on how a respondent rates his or her current and future lives. This is the highest percent of respondents in this category since the index started. Material provided with the results noted:

Results are based on telephone interviews . . . with a random sample of 176,903 adults, living in all 50 U.S. states and the District of Columbia. For results based on the total sample of national adults, the margin of sampling error is ±1 percentage points at the 95% confidence level.12

The poll uses a complex multistage sample design, but the sample percent has approximately a Normal sampling distribution.

(a) The announced poll result was 54.1% ± 1%. Can we be certain that the true population percent falls in this interval? Explain your answer.

(b) Explain to someone who knows no statistics what the announced result 54.1% ± 1% means.

(c) This confidence interval has the same form we have met earlier: estimate ± z* σestimate

What is the standard deviation σestimate of the estimated percent?

(d) Does the announced margin of error include errors due to practical problems such as nonresponse? Explain your answer.

6.30 Fuel efficiency. Computers in some vehicles calculate various quantities related to performance. One of these is the fuel efficiency, or gas mileage, usually expressed as miles per gallon (mpg). For one vehicle equipped in this way, the miles per gallon were recorded each time the gas tank was filled, and the computer was then reset.13 Here are the mpg values for a random sample of 20 of these records: MPG

41.5 50.7 36.6 37.3 34.2 45.0 48.0 43.2 47.7 42.2 43.2 44.6 48.4 46.4 46.8 39.2 37.3 43.5 44.3 43.3

Suppose that the standard deviation is known to be σ = 3.5 mpg.

(a) What is σ , the standard deviation of ?

(b) Examine the data for skewness and other signs of non-Normality. Show your plots and numerical summaries. Do you think it is reasonable to construct a confidence interval based on the Normal distribution? Explain your answer.

(c) Give a 95% confidence interval for μ, the mean miles per gallon for this vehicle.

6.31 Fuel efficiency in metric units. In the previous exercise, you found an estimate with a margin of error for the average miles per gallon. Convert your estimate and margin of error to the metric units kilometers per liter (kpl). To change mpg to kpl, use the fact that 1 mile = 1.609 kilometers and 1 gallon = 3.785 liters.

6.32 How many “hits”? The Confidence Interval applet lets you simulate large numbers of confidence intervals quickly. Select 95% confidence and then sample 50 intervals. Record the number of intervals that cover the true value

(this appears in the “Hit” box in the applet). Press the “Reset” button and repeat 30 times. Make a stemplot of the results and find the mean. Describe the results. If you repeated this experiment very many times, what would you expect the average number of hits to be?

6.33 Required sample size for specified margin of error. A new bone study is being planned that will measure the biomarker TRAP described in Exercise 6.19. Using the value of σ given there, 6.5 U/l, find the sample size required to provide

an estimate of the mean TRAP with a margin of error of 1.5 U/l for 95% confidence.

6.34 Adjusting required sample size for dropouts. Refer to the previous exercise. In similar previous studies, about 20% of the subjects drop out before the study is completed. Adjust your sample size requirement so that you will

have enough subjects at the end of the study to meet the margin of error criterion.

6.35 Radio poll. A national public radio (NPR) station invites listeners to enter a dispute about a proposed “pay as you throw” waste collection program. The station asks listeners to call in and state how much each 10 gallon bag of trash should cost. A total of 179 listeners call in. The station calculates the 95% confidence interval for the average fee to be $0.53 to $1.39. Is this result trustworthy? Explain your answer.

6.36 Accuracy of a laboratory scale. To assess the accuracy of a laboratory scale, a standard weight known to weigh 10 grams is weighed repeatedly. The scale readings are Normally distributed with unknown mean (this mean is 10 grams if the scale has no bias). The standard deviation of the scale readings is known to be 0.0002 gram.

(a) The weight is measured six times. The mean result is 10.0023 grams. Give a 99% confidence interval for the mean of repeated measurements of the weight.

(b) Based on the interval in part (a), do you think the scale is accurate? Explain your answer.

(c) How many measurements must be averaged to get a margin of error of ±0.0001 with 99% confidence?

6.37 More than one confidence interval. As we prepare to take a sample and compute a 95% confidence interval, we know that the probability that the interval we compute will cover the parameter is 0.95. That’s the meaning of 95%

confidence. If we plan to use several such intervals, however, our confidence that all of them will give correct results is less than 95%. Suppose that we plan to take independent samples each month for five months and report a 95% confidence interval for each set of data.

(a) What is the probability that all five intervals will cover the true means? This probability (expressed as a percent) is our overall confidence level for the five simultaneous statements.

(b) Suppose we instead considered individual 99% confidence intervals. Now, what is the overall confidence level for the five simultaneous statements?

(c) Based on the results of parts (a) and (b), how could you keep the overall confidence level near 95% if you were considering 10 simultaneous intervals?

6.2 Tests of Significance

When you complete this section, you will be able to:

Outline the four steps common to all tests of significance. Formulate the null and alternative hypotheses of a significance test. Describe a common form for the test statistic in terms of the parameter estimate, its standard deviation, and the hypothesized value. Define what a P-value is and explain whether a small P-value provides evidence for or against the null hypothesis. Draw a conclusion from a test of significance based on the test’s P-value and significance level α. Describe the relationship between a level α two-sided significance test for μ and the 1 − α confidence interval.

The confidence interval is appropriate when our goal is to estimate population parameters. The second common type of inference is directed at a quite different goal: to assess the evidence provided by the data in favor of some claim about the population parameters.

The reasoning of significance tests A significance test is a formal procedure for comparing observed data with a hypothesis whose truth we want to assess. The hypothesis is a statement about the population parameters. The results of a test are expressed in terms of a probability that measures how well the data and the hypothesis agree. We use the following examples to illustrate these concepts.

EXAMPLE 6.8

Scholarship amount by borrower status. One purpose of Sallie Mae’s annual study described in Example 6.4 (page 350) is to allow comparisons of different subgroups. For example, in the latest report, 980 of the 1593 participants (61.5%) did not borrow any money to pay for college. The average scholarship amount among these participants was $3925. The average scholarship among those who did borrow was $4350. The difference of $425 is fairly large, but we know that these numbers are estimates of the population means. If we took different samples, we would get different estimates.

Can we conclude from these data that the average scholarship amounts in these two groups are different? One way to answer this question is to compute the probability of obtaining a difference as large or larger than the observed $425 assuming that, in fact, there is no difference in the population means. This probability is 0.23. Because this probability is not particularly small, we conclude that observing a difference of $425 is not very surprising when the population means are equal. The data do not provide enough evidence for us to conclude that the average scholarship amount for borrowers and non-borrowers differ.

Here is an example with a different conclusion.

EXAMPLE 6.9

Parent income contribution by school type. Sallie Mae’s study also reports that the parents’ current income contribution among undergraduates going to a four-year public or four-year private college. The parents’ contribution averages $4444 among undergraduates at public colleges, while it is $6083 among undergraduates at private schools. Do parents pay more of their current income for undergraduates going to private schools? The observed difference is $1639, but as we learned in the previous example, an observed difference in means is not necessarily sufficient for us to conclude that the population means are different.

Again, we answer this question with a probability calculated under the assumption that there is no difference in the population means. The probability is 0.0001 of observing a difference in mean contributions that is $1639 or more when there really is no difference. Because this probability is so small, we have sufficient evidence in the data to conclude that the average current income contribution of parents is higher for undergraduates going to a private school than undergraduates going to a public school.

What are the key steps in these examples?

We started each with a question about the difference between two means. In Example 6.8, we compare borrowers with nonborrowers. In Example 6.9, we compare undergraduates attending private and public four-year colleges. In both cases, we ask whether or not the data are compatible with “no difference,” that is, a difference of $0. Next we compared the difference given by the data, $425 in the first case and $1639 in the second, with the value assumed in the question, $0. The results of the comparisons are probabilities, 0.23 in the first case and 0.0001 in the second.

The 0.23 probability is not particularly small, so we have limited evidence to question the possibility that the true difference is zero. In the second case, however, the probability is very small. Something that happens with probability 0.0001 occurs only about 1 time out of 10,000. In this case we have two possible explanations:

1. We have observed something that is very unusual. 2. The assumption that underlies the calculation, no difference in mean balance, is not true.

Because this probability is so small, we prefer the second conclusion: the average current income contribution from parents for undergraduates attending public colleges and for undergraduates attending private colleges is different, with the private school group contribution higher than that of the public school group.

The probabilities in Examples 6.8 and 6.9 are measures of the compatibility of the data (a difference in means of $425 and $1639) with the null hypothesis that there is no difference in the true means. Figures 6.7 and 6.8 compare the two results graphically. For each, a Normal curve centered at 0 is the sampling distribution. You can see from Figure 6.7 that we should not be particularly surprised to observe the difference $425, but the difference $1639 in Figure 6.8 is clearly an unusual observation. We will now consider some of the formal aspects of significance testing.

FIGURE 6.7 Comparison of the sample mean in Example 6.8 with the null hypothesized value 0.

FIGURE 6.8 Comparison of the sample mean in Example 6.9 with the null hypothesized value 0.

Stating hypotheses In Examples 6.8 and 6.9, we asked whether the difference in the observed means is reasonable if, in fact, there is no difference in the population means. To answer this, we begin by supposing that the statement following the “if” in the previous sentence is true. In other words, we suppose that the true difference is $0. We then ask whether the data provide evidence against the supposition we have made. If so, we have evidence in favor of an effect (the means are different) we are seeking. Often, the first step in a test of significance is to state a claim that we will try to find evidence against.

NULL HYPOTHESIS

The statement being tested in a test of significance is called the null hypothesis. The test of significance is designed to assess the strength of the evidence against the null hypothesis. Usually, the null hypothesis is a statement of “no effect” or “no difference.”

We abbreviate “null hypothesis” as H0. A null hypothesis is a statement about the population parameters. For example, our null hypothesis for Example 6.8 is

H0: there is no difference in the population means

or equivalently,

H0: the difference in population means is zero

Note that the null hypothesis refers to the population means for all undergraduates, including those for whom we do not have data.

alternative hypothesis

It is convenient also to give a name to the statement we hope or suspect is true instead of H0. This is called the alternative hypothesis and is abbreviated as Ha. In Example 6.8, the alternative hypothesis states that the means are different. We write this as

Ha: the population means are not the same

or equivalently,

Ha: the difference in population means is not zero

Hypotheses always refer to some populations or a model, not to a particular outcome. For this reason, we must state H0 and Ha in terms of population parameters.

one-sided or two-sided alternatives

Because Ha expresses the effect that we hope to find evidence for, we will sometimes begin with Ha and then set up H0 as the statement that the hoped-for effect is not present. Stating Ha, however, is often the more difficult task. It is not always clear, in particular, whether Ha should be one-sided or two-sided, which refers to whether a parameter differs from its null hypothesis value in a specific direction or in either direction.

The alternative hypothesis should express the hopes or suspicions we bring to the data. It is cheating to first look at the data and then frame Ha to fit what the data show. If you do not have a specific direction firmly in mind in advance, you must use a two-sided alternative. Moreover, some users of statistics argue that we should always use a two-sided alternative.

6.38

6.39

USE YOUR KNOWLEDGE

Dining court survey. The dining court closest to your university residence has been redesigned. A survey is planned to assess whether or not students think that the new design is an improvement. It will contain eight questions; a seven-point scale will be used for the answers, with scores less than 4 favoring the previous design and scores greater than 4 favoring the new design (to varying degrees). The average of these eight questions will be used as the student’s response. State the null and alternative hypotheses you would use for examining whether or not the new design is viewed more favorably.

DXA scanners. A dual-energy X-ray absorptiometry (DXA) scanner is used to measure bone mineral density for people who may be at risk for osteoporosis. One researcher believes that her scanner is not giving accurate readings. To assess this, the researcher uses an object called a “phantom” that has known mineral density μ = 1.4 grams per square centimeter. The researcher scans the phantom 10 times and compares the sample mean reading with the theoretical mean μ using a significance test. State the null and alternative hypotheses for this test.

Test statistics We will learn the form of significance tests in a number of common situations. Here are some principles that apply to most tests and that help in understanding these tests:

The test is based on a statistic that estimates the parameter that appears in the hypotheses. Usually, this is the same estimate we would use in a confidence interval for the parameter. When H0 is true, we expect the estimate to take a value near the parameter value specified by H0. We call this specified value the hypothesized value.

Values of the estimate far from the hypothesized value give evidence against H0. The alternative hypothesis determines which directions count against H0.

To assess how far the estimate is from the hypothesized value, standardize the estimate. In many common situations the test statistic has the form

test statistic

A test statistic measures compatibility between the null hypothesis and the data. We use it for the probability calculation that we need for our test of significance. It is a random variable with a distribution that we know.

Let’s return to our comparison of the scholarship amount among borrowers and nonborrowers and specify the hypotheses as well as calculate the test statistic.

EXAMPLE 6.10

Average scholarship amount of borrowers and nonborrowers: The hypotheses. In Example 6.8, the hypotheses are stated in terms of the difference in the average scholarship amount between borrowers and nonborrowers:

H0: there is no difference in the population means

Ha: there is a difference in the population means

Because Ha is two-sided, large values of both positive and negative differences count as evidence against the null hypothesis.

We can also state the null hypothesis as H0: the true mean difference is 0. This statement makes it more clear that the hypothesized value for this comparison of average scholarship amounts is 0.

EXAMPLE 6.11

Average scholarship amount of borrowers and nonborrowers: The test statistic. In Example 6.8, the estimate of the difference is $425. Using methods that we will discuss in detail later, we can determine that the standard deviation of the estimate is $353. For this problem the test statistic is

For our data,

We have observed a sample estimate that is one and one-fifth standard deviations away from the hypothesized value of the parameter.

standard Normal distribution, p. 60

Because the sample sizes are sufficiently large for us to conclude that the distribution of the sample estimate is approximately Normal, the standardized test statistic z will have approximately the N(0, 1) distribution. We will use facts about the Normal distribution in what follows.

P-values If all test statistics were Normal, we could base our conclusions on the value of the z test statistic. In fact, the Supreme Court of the United States has said that “two or three standard deviations” (z = 2 or 3) is its criterion for rejecting H0 (see Exercise 6.44 on page 370), and this is the criterion used in most applications involving the law. But because not all test statistics are Normal, we use the language of probability to express the meaning of a test statistic.

A test of significance finds the probability of getting an outcome as extreme or more extreme than the actually observed outcome. “Extreme” means “far from what we would expect if H0 were true.” The direction or directions that count as “far from what we would expect” are determined by Ha and H0.

P-VALUE

The probability, assuming H0 is true, that the test statistic would take a value as extreme or more extreme than that actually observed is called the P-value of the test. The smaller the P-value, the stronger the evidence against H0 provided by the data.

The key to calculating the P-value is the sampling distribution of the test statistic. For the problems we consider in this chapter, we need only the standard Normal distribution for the test statistic z.

In Example 6.8, we want to know if the average scholarship amount for borrowers differs from the average scholarship amount for non-borrowers. The difference we calculated based on our sample is $425, which corresponds to 1.20 standard deviations away from zero—that is, z = 1.20. Because we are using a two-sided alternative for this problem, the evidence against H0 is measured by the probability that we observe a value of Z as extreme or more extreme than 1.20.

EXAMPLE 6.12

Average scholarship amount of borrowers and nonborrowers: The P-value. In Example 6.11, we found that the test statistic for testing

H0: the true mean difference is 0

versus

Ha: there is a difference in the population means

is

If H0 is true, then z is a single observation from the standard Normal, N(0, 1), distribution. Figure 6.9 illustrates this calculation. The P-value is the probability of observing a value of Z at least as extreme as the one that we observed, z = 1.20. From Table A, our table of standard Normal probabilities, we find

P(Z ≥ 1.20) = 1 − 0.8849 = 0.1151

The probability for being extreme in the negative direction is the same:

P(Z ≤ −1.20) = 0.1151

So the P-value is

P = 2P(Z ≥ 1.20) = 2(0.1151) = 0.2302

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FIGURE 6.9 The P-value, Example 6.12. The P-value is the probability (when H0 is true) that takes a value as extreme or more extreme than the actual observed value, z = 1.20. Because the alternative hypothesis is two-sided, we use both tails of the distribution.

This is the value that we reported on page 361. There is a 23% chance of observing a difference as extreme as the $425 in our sample if the true population difference is zero. This P-value tells us that our outcome is not particularly extreme. In other words, the data do not provide substantial evidence for us to doubt the validity of the null hypothesis.

USE YOUR KNOWLEDGE

Normal curve and the P-value. A test statistic for a two-sided significance test for a population mean is z = 2.47. Sketch a standard Normal curve and mark this value of z on it. Find the P-value and shade the appropriate areas under the curve to illustrate your calculations.

More on the Normal curve and the P-value. A test statistic for a two-sided significance test for a population mean is z = −1.57. Sketch a standard Normal curve and mark this value of z on it. Find the P-value and shade the appropriate areas under the curve to illustrate your calculations.

Statistical significance We started our discussion of the reasoning of significance tests with the statement of null and alternative hypotheses. We then learned that a test statistic is the tool used to examine the compatibility of the observed data with the null hypothesis. Finally, we translated the test statistic into a P-value to quantify the evidence against H0. One important final step is needed: to state our conclusion.

significance level

We can compare the P-value we calculated with a fixed value that we regard as decisive. This amounts to announcing in advance how much evidence against H0 we will require to reject H0. The decisive value is called the significance level. It is commonly denoted by α (the Greek letter alpha). If we choose α = 0.05, we are requiring that the data give evidence against H0 so strong that it would happen no more than 5% of the time (1 time in 20) when H0 is true. If we choose α = 0.01, we are insisting on stronger evidence against H0, evidence so strong that it would appear only 1% of the time (1 time in 100) if H0 is in fact true.

STATISTICAL SIGNIFICANCE

If the P-value is as small or smaller than α, we say that the data are statistically significant at level α.

“Significant” in the statistical sense does not mean “important.” The original meaning of the word is “signifying something.” In statistics, the term is used to indicate only that the evidence against the null hypothesis has reached the standard set by α. For example, significance at level 0.01 is often expressed by the statement “The results were significant (P < 0.01).” Here, P stands for the P-value. The P-value is more informative than a statement of significance because we can then assess significance at any level we choose. For example, a result with P = 0.03 is significant at the α = 0.05 level but is not significant at the α = 0.01 level. We discuss this in more detail at the end of this section.

EXAMPLE 6.13

Average scholarship amount of borrowers and nonborrowers: The conclusion. In Example 6.12, we found that the P-value is

P = 2P(Z ≥ 1.20) = 2(0.1151) = 0.2302

There is an 23% chance of observing a difference as extreme as the $425 in our sample if the true population difference is zero. Because this P-value is larger than the α = 0.05 significance level, we conclude that our test result is not significant. We could report the result as “the data fail to provide evidence that would cause us to conclude that there is a difference in average scholarship amount between borrowers and nonborrowers (z = 1.20, P = 0.23).”

This statement does not mean that we conclude that the null hypothesis is true, only that the level of evidence we require to reject the null hypothesis is not met. Our criminal court system follows a similar procedure in which a defendant is presumed innocent (H0) until proven guilty. If the level of evidence presented is not strong enough for the jury to find the defendant guilty beyond a reasonable doubt, the defendant is acquitted. Acquittal does not imply innocence, only that the degree of evidence was not strong enough to prove guilt.

If the P-value is small, we reject the null hypothesis. Here is the conclusion for our second example.

EXAMPLE 6.14

Parent income contribution by school type: The conclusion. In Example 6.9, we found that the

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difference in the average parent current income contribution between undergraduates going to a private college versus public college was $1639. Because the cost of tuition at a private college is typically higher than the cost at a public college,14 we had a prior expectation that the parental current income contribution would be higher for undergraduates going to a private college. It is appropriate to use a one-sided alternative in this situation. So, our hypotheses are

H0: the true mean difference is 0

versus

Ha: the difference between the average parent income contribution of undergraduates at a private college and public college is positive

The standard deviation is $428 (again, we defer details regarding this calculation), and the test statistic is

Because only positive differences in parental contributions count against the null hypothesis, the one-sided alternative leads to the calculation of the P-value using the upper tail of the Normal distribution. In Table A, the largest z is 3.49. This means that for z = 3.81, P < 0.0002. Using software, we can be more precise. The P-value is

The calculation is illustrated in Figure 6.10. There is about a 1-in-10,000 chance of observing a difference as large or larger than the $1639 in our sample if the true population difference is zero. This P-value tells us that our outcome is extremely rare. We conclude that the null hypothesis must be false. Because the observed difference is positive, here is one way to report the result: “The data clearly show that the average parent income contribution for undergraduates at a private college is larger than the average parent income contribution for undergraduates at a public college (z = 3.81, P = 0.0001).”

FIGURE 6.10 The P-value, Example 6.14. The P-value is the probability (when H0 is true) that takes a value as extreme or more extreme than the actual observed value, z = 3.81. We look at only the right tail because we are considering the one-sided (>) alternative.

USE YOUR KNOWLEDGE

Finding significant z-scores. Consider a two-sided significance test for a population mean. (a) Sketch a Normal curve similar to that shown in Figure 6.9, but find the value z such that P = 0.05.

(b) Based on your curve from part (a), what values of the z statistic are statistically significant at the α = 0.05 level?

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More on finding significant z-scores. Consider a one-sided significance test for a population mean, where the alternative is “greater than.”

(a) Sketch a Normal curve similar to that shown in Figure 6.10, but find the value z such that P = 0.05.

(b) Based on your curve from part (a), what values of the z statistic are statistically significant at the α = 0.05 level?

The Supreme Court speaks. The Supreme Court has said that z-scores beyond 2 or 3 are generally convincing statistical evidence. For a two-sided test, what significance level corresponds to z = 2? To z = 3?

A test of significance is a process for assessing the significance of the evidence provided by data against a null hypothesis. The four steps common to all tests of significance are as follows:

1. State the null hypothesis H0 and the alternative hypothesis Ha. The test is designed to assess the strength of the evidence against H0; Ha is the statement that we will accept if the evidence enables us to reject H0.

2. Calculate the value of the test statistic on which the test will be based. This statistic usually measures how far the data are from H0.

3. Find the P-value for the observed data. This is the probability, calculated assuming that H0 is true, that the test statistic will weigh against H0 at least as strongly as it does for these data.

4. State a conclusion. One way to do this is to choose a significance level α, how much evidence against H0 you regard as decisive. If the P-value is less than or equal to α, you conclude that the alternative hypothesis is true; if it is greater than α, you conclude that the data do not provide sufficient evidence to reject the null hypothesis. Your conclusion is a sentence or two that summarizes what you have found by using a test of significance.

We will learn the details of many tests of significance in the following chapters. The proper test statistic is determined by the hypotheses and the data collection design. We use computer software or a calculator to find its numerical value and the P-value. The computer will not formulate your hypotheses for you, however. Nor will it decide if significance testing is appropriate or help you to interpret the P-value that it presents to you. These steps require judgment based on a sound understanding of this type of inference.

Tests for a population mean Our discussion has focused on the reasoning of statistical tests, and we have outlined the key ideas for one type of procedure. Our examples focused on the comparison of two population means. Here is a summary for a test about one population mean.

We want to test the hypothesis that a parameter has a specified value. This is the null hypothesis. For a test of a population mean μ, the null hypothesis is

H0: the true population mean is equal to μ0

which often is expressed as

H0: μ = μ0

where μ0 is the hypothesized value of μ that we would like to examine. The test is based on data summarized as an estimate of the parameter. For a population mean this is the sample

mean . Our test statistic measures the difference between the sample estimate and the hypothesized parameter in terms of standard deviations of the test statistic:

distribution of sample mean, p. 298

Recall from Chapter 5 that the standard deviation of is . Therefore, the test statistic is

central limit theorem, p. 298

Again recall from Chapter 5 that, if the population is Normal, then will be Normal and z will have the standard Normal distribution when H0 is true. By the central limit theorem, both distributions will be approximately Normal when the sample size is large even if the population is not Normal. We’ll assume that we’re in one of these two settings for now.

Suppose that we have calculated a test statistic z = 1.7. If the alternative is one-sided on the high side, then the P- value is the probability that a standard Normal random variable Z takes a value as large or larger than the observed 1.7. That is,

Similar reasoning applies when the alternative hypothesis states that the true μ lies below the hypothesized μ0 (one- sided). When Ha states that μ is simply unequal to μ0 (two-sided), values of z away from zero in either direction count against the null hypothesis. The P-value is the probability that a standard Normal Z is at least as far from zero as the observed z. Again, if the test statistic is z = 1.7, the two-sided P-value is the probability that Z ≤ −1.7 or Z ≥ 1.7. Because the standard Normal distribution is symmetric, we calculate this probability by finding P(Z ≥ 1.7) and doubling it:

We would make exactly the same calculation if we observed z = −1.7. It is the absolute value |z| that matters, not whether z is positive or negative. Here is a statement of the test in general terms.

z TEST FOR A POPULATION MEAN

To test the hypothesis H0: μ = μ0 based on an SRS of size n from a population with unknown mean μ and known standard deviation σ, compute the test statistic

In terms of a standard Normal random variable Z, the P-value for a test of H0 against

These P-values are exact if the population distribution is Normal and are approximately correct for large n in other cases.

EXAMPLE 6.15

Energy intake from sugar-sweetened beverages. Consumption of sugar-sweetened beverages (SSBs) has been positively associated with weight gain and obesity and negatively associated with the intake of important

micronutrients. One study used data from the National Health and Nutrition Examination Survey (NHANES) to estimate SSB consumption among adolescents (aged 12 to 19 years). More than 2400 individuals provided data for this study.15 The mean consumption was 298 calories per day.

You survey 100 students at your large university and find the average consumption of SSBs per day to be 262 calories. Is there evidence that the average calories per day from SSBs at your university differs from this large U.S. survey average?

The null hypothesis is “no difference” from the published mean μ0 = 298. The alternative is two-sided because you did not have a particular direction in mind before examining the data. So the hypotheses about the unknown mean μ of the students at your university are

H0: μ = 298 Hα: μ ≠ 298

As usual in this chapter, we make the unrealistic assumption that the population standard deviation is known. In this case, we’ll use the standard deviation from the large national study, σ = 435 calories.

We compute the test statistic:

Figure 6.11 illustrates the P-value, which is the probability that a standard Normal variable Z takes a value at least 0.83 away from zero. From Table A, we find that this probability is

P = 2P(Z ≥ 0.83) = 2(1 − 0.7967) = 0.4066

FIGURE 6.11 Sketch of the P-value calculation for the two-sided test, Example 6.15. The test statistic is z = −0.83.

That is, if the population mean were 298, more than 40% of the time an SRS of size 100 from the students at your university would have a mean consumption from SSBs at least as far from 298 as that of this sample. The observed = 262 is, therefore, not strong evidence that the student population mean at your university differs from that of the large population of adolescents.

This z test requires that the 100 students in the sample are an SRS from the population of students at your university. We will assume that the students in the sample were selected in a proper random manner. We’ll also assume that n = 100 is sufficiently large that we can rely on the central limit theorem to assure us that the P-value based on the Normal distribution will be a good approximation.

The data in Example 6.15 do not establish that the mean consumption μ for the students at your university is 298 calories. We sought evidence that μ differed from 298 and failed to find convincing evidence. That is all we can say. No doubt the mean amount at your university is not exactly equal to 298 calories. A large enough sample would give evidence of the difference, even if it is very small.

Tests of significance assess the evidence against H0. If the evidence is strong, we can confidently reject H0 in favor of the alternative. Failing to find evidence against H0 means only that the data are consistent with H0, not that we have clear evidence that H0 is true.

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EXAMPLE 6.16

Significance test of the mean SATM score. In a discussion of SAT Mathematics (SATM) scores, someone comments: “Because only a select minority of California high school students take the test, the scores overestimate the ability of typical high school seniors. I think that if all seniors took the test, the mean score would be no more than 485.” You do not agree with this claim and decide to use the SRS of 500 seniors from Example 6.3 (page 344) to assess the degree of evidence against it. Those 500 seniors had a mean SATM score of = 495. Is this strong enough evidence to conclude that this person’s claim is wrong?

Because the claim states that the mean is “no more than 485,” the alternative hypothesis is one-sided. The hypotheses are

H0: μ = 485

Ha: μ > 485

As we did in the discussion following Example 6.3, we assume that σ = 100. The z statistic is

Because Ha is one-sided on the high side, large values of z count against H0. From Table A, we find that the P-value is

P = P(Z ≥ 2.24) = 1 − 0.9875 = 0.0125

Figure 6.12 illustrates this P-value. A mean score as large as that observed would occur roughly 12 times in 1000 samples if the population mean were 485. This is convincing evidence that the mean SATM score for all California high school seniors is higher than 485. You can confidently tell this person that his or her claim is incorrect.

FIGURE 6.12 Sketch of the P-value calculation for the one-sided test, Example 6.16. The test statistic is z = 2.24.

USE YOUR KNOWLEDGE

Computing the test statistic and P-value. You will perform a significance test of H0: μ = 30 based on an SRS of n = 49. Assume that σ = 14.

(a) If = 33.5, what is the test statistic z?

(b) What is the P-value if Ha: μ > 30?

(c) What is the P-value if Ha: μ ≠ 30?

Testing a random number generator. Statistical software often has a “random number generator” that is supposed to produce numbers uniformly distributed between 0 and 1. If this is true, the numbers generated

come from a population with μ = 0.5. A command to generate 100 random numbers gives outcomes with mean = 0.469 and s = 0.286. Because the sample is reasonably large, take the population standard deviation also to be σ = 0.286. Do we have evidence that the mean of all numbers produced by this software is not 0.5?

Two-sided significance tests and confidence intervals Recall the basic idea of a confidence interval, discussed in Section 6.1. We constructed an interval that would include the true value of μ with a specified probability C. Suppose that we use a 95% confidence interval (C = 0.95). Then the values of μ0 that are not in our interval would seem to be incompatible with the data. This sounds like a significance test with α = 0.05 (or 5%) as our standard for drawing a conclusion. The following examples demonstrate that this is correct.

EXAMPLE 6.17

PBTEST

Water quality testing. The Deely Laboratory is a drinking-water testing and analysis service. One of the common contaminants it tests for is lead. Lead enters drinking water through corrosion of plumbing materials, such as lead pipes, fixtures, and solder. The service knows that their analysis procedure is unbiased but not perfectly precise, so the laboratory analyzes each water sample three times and reports the mean result. The repeated measurements follow a Normal distribution quite closely. The standard deviation of this distribution is a property of the analytic procedure and is known to be σ = 0.25 parts per billion (ppb).

The Deely Laboratory has been asked by a university to evaluate a claim that the drinking water in the Student Union has a lead concentration above the Environmental Protection Agency’s (EPA) action level of 15 ppb. Because the true concentration of the sample is the mean μ of the population of repeated analyses, the hypotheses are

H0: μ = 15

Ha: μ ≠ 15

We use the two-sided alternative here because there is no prior evidence to substantiate a one-sided alternative. The lab chooses the 1% level of significance, α = 0.01.

Three analyses of one specimen give concentrations

15.84 15.33 15.58

The sample mean of these readings is

The test statistic is

Because the alternative is two-sided, the P-value is

P = 2P(Z ≥ 4.02)

We cannot find this probability in Table A. The largest value of z in that table is 3.49. All that we can say from Table A is that P is less than 2P(Z ≥ 3.49) = 2(1 − 0.9998) = 0.0004. Software or a calculator could be used to give an accurate value of the P-value. However, because the P-value is clearly less than the lab’s standard of 1%, we reject H0. Because is larger than 15.00, we can conclude that the true concentration level of lead in this one specimen is higher than the EPA’s action level.

We can compute a 99% confidence interval for the same data to get a likely range for the actual mean concentration m of this specimen.

EXAMPLE 6.18

PBTEST

99% confidence interval for the mean concentration. The 99% confidence interval for μ in Example 6.17 is

The hypothesized value μ0 = 15.00 in Example 6.17 falls outside the confidence interval we computed in Example 6.18. In other words, it is in the region we are 99% confident that μ is not in. Thus, we can reject

H0: μ = 15.00

at the 1% significance level. On the other hand, we cannot reject

H0: μ = 15.30

at the 1% level in favor of the two-sided alternative Ha: μ ≠ 15.30, because 15.30 lies inside the 99% confidence interval for μ. Figure 6.13 illustrates both cases.

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FIGURE 6.13 The link between two-sided significance tests and confidence intervals. For the study described in Examples 6.17 and 6.18, values of μ falling outside a 99% confidence interval can be rejected at the 1% significance level; values falling inside the interval cannot be rejected. This holds for any significance level α and 1 − α confidence interval.

The calculation in Example 6.17 for a 1% significance test is very similar to the calculation for a 99% confidence interval. In fact, a two-sided test at significance level α can be carried out directly from a confidence interval with confidence level C = 1 − α.

TWO-SIDED SIGNIFICANCE TESTS AND CONFIDENCE INTERVALS

A level α two-sided significance test rejects a hypothesis H0: μ = μ0 exactly when the value μ0 falls outside a level 1 − α confidence interval for μ.

USE YOUR KNOWLEDGE

Two-sided significance tests and confidence intervals. The P-value for a two-sided test of the null hypothesis H0: μ = 30 is 0.037.

(a) Does the 95% confidence interval include the value 30? Explain.

(b) Does the 99% confidence interval include the value 30? Explain.

More on two-sided tests and confidence intervals. A 95% confidence interval for a population mean is (29, 58).

(a) Can you reject the null hypothesis that μ = 50 against the two-sided alternative at the 5% significance level? Explain.

(b) Can you reject the null hypothesis that μ = 60 against the two-sided alternative at the 5% significance level? Explain.

The P-value versus a statement of significance The observed result in Example 6.17 was z = 4.02. The conclusion that this result is significant at the 1% level does not tell the whole story. The observed z is far beyond the z corresponding to 1%, and the evidence against H0 is far stronger than 1% significance suggests. The actual P-value

2P(Z ≥ 4.02) = 0.000058

gives a better sense of how strong the evidence is. The P-value is the smallest level α at which the data are significant. Knowing the P-value allows us to assess significance at any level.

EXAMPLE 6.19

Test of the mean SATM score: Significance. In Example 6.16, we tested the hypotheses H0: μ = 485

Ha: μ ≥ 485

concerning the mean SAT Mathematics score μ of California high school seniors. The test had the P-value P = 0.0125. This result is significant at the α = 0.05 level because 0.0125 ≤ 0.05. It is not significant at the α = 0.01 level, because the P-value is larger than 0.01. See Figure 6.14.

FIGURE 6.14 Link between the P-value and the significance level α. An outcome with P-value P is significant at all levels a at or above P and is not significant at smaller levels α.

critical value

A P-value is more informative than a reject-or-not finding at a fixed significance level. But assessing significance at a fixed level α is easier because no probability calculation is required. You need only look up a number in a table. A value z* with a specified area to its right under the standard Normal curve is called a critical value of the standard Normal distribution. Because the practice of statistics almost always employs computer software or a calculator that calculates P-values automatically, the use of tables of critical values is becoming outdated. We include the usual tables of critical values (such as Table D) at the end of the book for learning purposes and to rescue students without good computing facilities. The tables can be used directly to carry out fixed α tests. They also allow us to approximate P- values quickly without a probability calculation. The following example illustrates the use of Table D to find an approximate P-value.

EXAMPLE 6.20

Average scholarship amount of borrowers and nonborrowers: Assessing significance. In Example 6.11 (page 365), we found the test statistic z = 1.20 for testing the null hypothesis that there was no difference in the mean scholarship amount between borrowers and nonborrowers. The alternative was two-sided. Under the null hypothesis, z has a standard Normal distribution, and from the last row in Table D, we can see that there is a 95% chance that z is between ±1.96. Therefore, we reject H0 in favor of Ha whenever z is outside this range. Because our calculated value is 1.20, we are within the range and we do not reject the null hypothesis at the 5% level of significance.

USE YOUR KNOWLEDGE

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6.51

P-value and the significance level. The P-value for a significance test is 0.033. (a) Do you reject the null hypothesis at level α = 0.05?

(b) Do you reject the null hypothesis at level α = 0.01?

(c) Explain how you determined your answers to parts (a) and (b).

More on P-value and the significance level. The P-value for a significance test is 0.069. (a) Do you reject the null hypothesis at level α = 0.05?

(b) Do you reject the null hypothesis at level α = 0.01?

(c) Explain how you determined your answers to parts (a) and (b).

One-sided and two-sided P-values. The P-value for a two-sided significance test is 0.076. (a) State the P-values for the two one-sided tests.

(b) What additional information do you need to properly assign these P-values to the > and < (one-sided) alternatives?

SECTION 6.2 SUMMARY A test of significance is intended to assess the evidence provided by data against a null hypothesis H0 in favor of an alternative hypothesis Ha.

The hypotheses are stated in terms of population parameters. Usually, H0 is a statement that no effect or no difference is present, and Ha says that there is an effect or difference in a specific direction (one-sided alternative) or in either direction (two-sided alternative). The test is based on a test statistic. The P-value is the probability, computed assuming that H0 is true, that the test statistic will take a value at least as extreme as that actually observed. Small P-values indicate strong evidence against H0. Calculating P-values requires knowledge of the sampling distribution of the test statistic when H0 is true.

If the P-value is as small or smaller than a specified value α, the data are statistically significant at significance level α. Significance tests for the hypothesis H0: μ = μ0 concerning the unknown mean μ of a population are based on the z statistic:

The z test assumes an SRS of size n, known population standard deviation σ, and either a Normal population or a large sample. P-values are computed from the Normal distribution (Table A). Fixed α tests use the table of standard Normal critical values (Table D).

SECTION 6.2 EXERCISES For Exercises 6.38 and 6.39, see page 364; for Exercises 6.40 and 6.41, see page 367; for Exercises 6.42 through 6.44, see page 370; for Exercises 6.45 and 6.46, see pages 374–375; for Exercises 6.47 and 6.48, see page 377; and for Exercises 6.49 through 6.51, see pages 378–379.

6.52 What’s wrong? Here are several situations where there is an incorrect application of the ideas presented in this section. Write a short paragraph explaining what is wrong in each situation and why it is wrong.

(a) A researcher tests the following null hypothesis: H0: = 23.

(b) A random sample of size 30 is taken from a population that is assumed to have a standard deviation of 5. The standard deviation of the sample mean is 5/30.

(c) A study with = 45 reports statistical significance for Ha: μ > 50.

(d) A researcher tests the hypothesis H0: μ = 350 and concludes that the population mean is equal to 350.

6.53 What’s wrong? Here are several situations where there is an incorrect application of the ideas presented in this section. Write a short paragraph explaining what is wrong in each situation and why it is wrong.

(a) A significance test rejected the null hypothesis that the sample mean is equal to 500.

(b) A test preparation company wants to test that the average score of its students on the ACT is better than the national average score of 21.2. The company states its null hypothesis to be H0: μ > 21.2.

(c) A study summary says that the results are statistically significant and the P-value is 0.98.

(d) The z test statistic is equal to 0.018. Because this is less than α = 0.05, the null hypothesis was rejected.

6.54 Determining hypotheses. State the appropriate null hypothesis H0 and alternative hypothesis Ha in each of the following cases.

(a) A 2015 study reported that 96% of students owned a cell phone. You plan to take an SRS of students to see if the percent has increased.

(b) The examinations in a large freshman chemistry class are scaled after grading so that the mean score is 75. The professor thinks that students who attend early-morning recitation sections will have a higher mean score than the class as a whole. Her students in these sections this semester can be considered a sample from the population of all students who might attend an early-morning section, so she compares their mean score with 75.

(c) The student newspaper at your college recently changed the format of its opinion page. You want to test whether students find the change an improvement. You take a random sample of students and select those who regularly read the newspaper. They are asked to indicate their opinions on the changes using a five-point scale: −2 if the new format is much worse than the old, −1 if the new format is somewhat worse than the old, 0 if the new format is the same as the old, +1 if the new format is somewhat better than the old, and +2 if the new format is much better than the old.

6.55 More on determining hypotheses. State the null hypothesis H0 and the alternative hypothesis Ha in each case. Be sure to identify the parameters that you use to state the hypotheses.

(a) A university gives credit in first-year calculus to students who pass a placement test. The mathematics department wants to know if students who get credit in this way differ in their success with second-year calculus. Scores in second-year calculus are scaled so the average each year is equivalent to a 77. This year, 21 students who took second-year calculus passed the placement test.

(b) Experiments on learning in animals sometimes measure how long it takes a mouse to find its way through a maze. The mean time is 20 seconds for one particular maze. A researcher thinks that playing rap music will cause the mice to complete the maze more slowly. She measures how long each of 12 mice takes with the rap music as a stimulus.

(c) The average square footage of one-bedroom apartments in a new student-housing development is advertised to be 880 square feet. A student group thinks that the apartments are smaller than advertised. They hire an engineer to measure a sample of apartments to test their suspicion.

6.56 Even more on determining hypotheses. In each of the following situations, state an appropriate null hypothesis H0 and alternative hypothesis Ha. Be sure to identify the parameters that you use to state the hypotheses. (We have not yet learned how to test these hypotheses.)

(a) A sociologist asks a large sample of high school students which television channel they like best. She suspects that a higher percent of males than of females will name MTV as their favorite channel.

(b) An education researcher randomly divides sixth-grade students into two groups for physical education class. He teaches both groups basketball skills, using the same methods of instruction in both classes. He encourages Group A with compliments and other positive behavior but acts cool and neutral toward Group B. He hopes to show that positive teacher attitudes result in a higher mean score on a test of basketball skills than do neutral attitudes.

(c) An education researcher believes that, among college students, there is a negative correlation between time spent at social network sites and self-esteem, measured on a 0 to 100 scale. To test this, she gathers social-networking information and self-esteem data from a sample of students at your college.

6.57 Translating research questions into hypotheses. Translate each of the following research questions into appropriate H0 and Ha.

(a) U.S. Census Bureau data show that the mean household income in the area served by a shopping mall is $42,800 per year. A market research firm questions shoppers at the mall to find out whether the mean household income of mall shoppers is higher than that of the general population.

(b) Last year, your online registration technicians took an average of 0.4 hour to respond to trouble calls from students trying to register. Do this year’s data show a different average response time?

6.58 Computing the P-value. A test of the null hypothesis H0: μ = μ0 gives test statistic z = 1.89.

(a) What is the P-value if the alternative is Ha: μ > μ0?

(b) What is the P-value if the alternative is Ha: μ < μ0?

(c) What is the P-value if the alternative is Ha: μ ≠ μ0?

6.59 More on computing the P-value. A test of the null hypothesis H0: μ = μ0 gives test statistic z = −1.33.

(a) What is the P-value if the alternative is Ha: μ > μ0?

(b) What is the P-value if the alternative is Ha: μ < μ0?

(c) What is the P-value if the alternative is Ha: μ ≠ μ0?

6.60 Timing of food intake and weight loss. A study found that a large group of late lunch eaters lost less weight over a 20-week observation period than a large group of early lunch eaters (P = 0.002).16 Explain what this P = 0.002 means in a way that could be understood by someone who has not studied statistics.

6.61 Average starting salary. Refer to Exercise 6.22 (page 358). Use the information presented in the exercise to test that the average income of graduates from your institution is different from the national average (α = 0.01). Write a short paragraph summarizing your conclusions.

6.62 Change in consumption of sweet snacks? Refer to Exercise 6.23 (page 358). A similar study performed four years earlier reported the average consumption of sweet snacks among healthy weight children aged 12 to 19 years to be 369.4 kilocalaries per day (kcal/d). Does this current study suggest a change in the average consumption? Perform a significance test using the 5% significance level. Write a short paragraph summarizing the results.

6.63 Peer pressure and choice of major. A study followed a cohort of students entering a business/economics program.17 All students followed a common track during the first three semesters and then chose to specialize in either business or economics. Through a series of surveys, the researchers were able to classify roughly 50% of the students as either peer driven (ignored abilities and chose major to follow peers) or ability driven (ignored peers and chose major based on ability). When looking at entry wages after graduation, the researchers conclude that a peer-driven student can expect an average wage that is 13% less than that of an ability-driven student. The report states that the significance level is P = 0.09. Can you be confident of the researchers’ conclusion statement regarding the wage decrease? Explain your answer.

6.64 Symbol of wealth in ancient China? Every society has its own symbols of wealth and prestige. In ancient China, it appears that owning pigs was such a symbol. Evidence comes from examining burial sites. If the skulls of sacrificed pigs tend to appear along with expensive ornaments, that suggests that the pigs, like the ornaments, signal the wealth and prestige of the person buried. A study of burials from around 3500 B.C. concluded that “there are striking differences in grave goods between burials with pig skulls and burials without them... A test indicates that the two samples of total artifacts are significantly different at the 0.01 level.”18 Explain clearly why “significantly different at the 0.01 level” gives good reason to think that there really is a systematic difference between burials that contain pig skulls and those that lack them.

6.65 Alcohol awareness among college students. A study of alcohol awareness among college students reported a higher awareness for students enrolled in a health and safety class than for those enrolled in a statistics class.19 The difference is described as being statistically significant. Explain what this means in simple terms and offer an explanation for why the health and safety students had a higher mean score.

6.66 Change in eighth-grade average mathematics score. A report based on the 2015 National Assessment of Educational Progress (NAEP)20 states that the average score on their mathematics test for eighth-grade students attending public schools is significantly higher than in 2011. The report also states that the average score for eighth-grade students attending private schools is not significantly different from the average score in 2011. A footnote states that comparisons are determined by two-sided statistical tests with 0.05 as the level of significance. Explain what this footnote means in language understandable to someone who knows no statistics. Do not use the word “significance” in your answer.

6.67 More on change in eighth-grade average mathematics score. Refer to the previous exercise. On the basis of the NAEP study, a friend who works for the school newspaper wants to report that between 2011 and 2013 the average mathematics score improved for students attending public schools but stayed the same for students attending private schools. Do you agree with this statement? Explain your answer.

6.68 Background television in homes of U.S. children. In one study, U.S. parents were surveyed to determine the amount of background television their children were exposed to. A total of n = 1454 families with one child between the ages of 8 months and 8 years participated.21 For those families in which the caregiver had a high school degree or less, the child was exposed to an average of 313.0 minutes of background television per day. For those families in which the caregiver had some college or a college degree, the child was exposed to an average of 218.8 minutes per day. These average times were reported to be significantly different with P < 0.05. The actual P-value is 0.003. Explain why the actual P-value is more informative than the statement of significance at the 0.05 level.

6.69 Sleep quality and elevated blood pressure. A study looked at n = 238 adolescents, all free of severe illness.22 Subjects wore a wrist actigraph, which allowed the researchers to estimate sleep patterns. Those subjects classified as having low sleep efficiency had an average systolic blood pressure that was 5.8 millimeters of mercury (mm Hg) higher than that of other adolescents. The standard deviation of this difference is 1.4 mm Hg. Based on these results, test whether this difference is significant at the 0.01 level.

6.70 Are the pine trees randomly distributed from north to south? In Example 6.1 (page 342), we looked at the distribution of longleaf pine trees in the Wade Tract. One way to formulate hypotheses about whether or not the trees

are randomly distributed in the tract is to examine the average location in the north–south direction. The values range from 0 to 200, so if the trees are uniformly distributed in this direction, any difference from the middle value (100) should be due to chance variation. The sample mean for the 584 trees in the tract is 99.74. A theoretical calculation based on the assumption that the trees are uniformly distributed gives a standard deviation of 58. Carefully state the null and alternative hypotheses in terms of this variable. Note that this requires that you translate the research question about the random distribution of the trees into specific statements about the mean of a probability distribution. Test your hypotheses, report your results, and write a short summary of what you have found.

6.71 Are the pine trees randomly distributed from east to west? Answer the questions in the previous exercise for the east–west direction, for which the sample mean is 113.8.

6.72 Who is the author? Statistics can help decide the authorship of literary works. Sonnets by a certain Elizabethan poet are known to contain an average of μ = 8.9 new words (words not used in the poet’s other works). The standard deviation of the number of new words is σ = 2.5. Now a manuscript with six new sonnets has come to light, and scholars are debating whether it is the poet’s work. The new sonnets contain an average of = 10.2 words not used in the poet’s known works. We expect poems by another author to contain more new words, so to see if we have evidence that the new sonnets are not by our poet we test

H0: μ = 8.9

Ha: μ > 8.9

Give the z test statistic and its P-value. What do you conclude about the authorship of the new poems?

6.73 Attitudes toward school. The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures the motivation, attitude toward school, and study habits of students. Scores range from 0 to 200. The mean score for U.S. college students is about 95, and the standard deviation is about 20. A teacher who suspects that older students have better attitudes toward school gives the SSHA to 25 students who are at least 30 years of age. Their mean score is = 103.3.

(a) Assuming that σ = 30 for the population of older students, carry out a test of H0: μ = 95

Ha: μ > 95

Report the P-value of your test, and state your conclusion clearly.

(b) Your test in part (a) required two important assumptions in addition to the assumption that the value of σ is known. What are they? Which of these assumptions is most important to the validity of your conclusion in part (a)?

6.74 Nutritional intake among Canadian high-performance athletes. Since previous studies have reported that elite athletes are often deficient in their nutritional intake (for example, total calories, carbohydrates, protein), a group of researchers decided to evaluate Canadian high-performance athletes.23 A total of n = 324 athletes from eight Canadian sports centers participated in the study. One reported finding was that the average caloric intake among the n = 201 women was 2403.7 kilocalories per day (kcal/d). The recommended amount is 2811.5 kcal/d. Is there evidence that female Canadian athletes are deficient in caloric intake?

(a) State the appropriate H0 and Ha to test this.

(b) Assuming a standard deviation of 880 kcal/d, carry out the test. Give the P-value, and then interpret the result in plain language.

6.75 Are the measurements similar? Refer to Exercise 6.30 (page 360). In addition to the computer’s calculations of miles per gallon, the driver also recorded the miles per gallon by dividing the miles driven by the number of gallons at each fill- up. The following data are the differences between the computer’s and the driver’s calculations for that random sample of 20 records. The driver wants to determine if these calculations are different. Assume that the standard deviation of a difference is σ = 3.0. MPGDIFF

5.0 6.5 −0.6 1.7 3.7 4.5 8.0 2.2 4.9 3.0 4.4 0.1 3.0 1.1 1.1 5.0 2.1 3.7 −0.6 −4.2

(a) State the appropriate H0 and Ha to test this suspicion.

(b) Carry out the test. Give the P-value, and then interpret the result in plain language.

6.76 Impact of on significance. The Statistical Significance applet illustrates statistical tests with a fixed level of significance for Normally distributed data with known standard deviation. Open the applet and keep the default settings

for the null (μ = 0) and the alternative (μ > 0) hypotheses, the sample size (n = 10), the standard deviation (σ = 1), and the significance level (α = 0.05). In the “I have data, and the observed is =” box, enter the value 1. Is the difference between and μ0 significant at the 5% level? Repeat for equal to 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9. Make a table giving and the results of the significance tests. What do you conclude?

6.77 Effect of changing α on significance. Repeat the previous exercise with significance level α = 0.01. How does the choice of α affect which values of are far enough away from μ0 to be statistically significant?

6.78 Changing to a two-sided alternative. Repeat the previous exercise but with the two-sided alternative hypothesis. How does this change affect which values of are far enough away from μ0 to be statistically significant at

the 0.01 level?

6.79 Changing the sample size. Refer to Exercise 6.76. Suppose that you increase the sample size n from 10 to 50. Again, make a table giving and the results of the significance tests at the 0.05 significance level. What do you

conclude?

6.80 Impact of on the P-value. We can also study the P-value using the Statistical Significance applet. Reset the applet to the default settings for the null (μ = 0) and the alternative (μ > 0) hypotheses, the sample size (n = 10), the

standard deviation (σ = 1), and the significance level (α = 0.05). In the “I have data, and the observed is =” box, enter the value 1. What is the P-value? It is shown at the top of the blue vertical line. Repeat for equal to 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9. Make a table giving and P-values. How does the P-value change as moves farther away from μ0?

6.81 Changing to a two-sided alternative, continued. Repeat the previous exercise but with the two-sided alternative hypothesis. How does this change affect the P-values associated with each ? Explain why the P-values

change in this way.

6.82 Other changes and the P-value. Refer to the previous exercise.

(a) What happens to the P-values when you change the significance level α to 0.01? Explain the result.

(b) What happens to the P-values when you change the sample size n from 10 to 50? Explain the result.

6.83 Understanding levels of significance. Explain in plain language why a significance test that is significant at the 1% level must always be significant at the 5% level.

6.84 More on understanding levels of significance. You are told that a significance test is significant at the 5% level. From this information, can you determine whether or not it is significant at the 1% level? Explain your answer.

6.85 Test statistic and levels of significance. Consider a significance test for a null hypothesis versus a two-sided alternative. Give a value of z that will give a result significant at the 1% level but not at the 0.5% level.

6.86 Using Table D to find a P-value. You have performed a two-sided test of significance and obtained a value of z = 2.08. Use Table D to find the approximate P-value for this test.

6.87 More on using Table D to find a P-value. You have performed a one-sided test of significance and obtained a value of z = 1.03. Use Table D to find the approximate P-value for this test when the alternative is greater than.

6.88 Using Table A and Table D to find a P-value. Consider a significance test for a null hypothesis versus a two- sided alternative. Between what values from Table D does the P-value for an outcome z = 1.88 lie? Calculate the P-value using Table A and verify that it lies between the values you found from Table D.

6.89 More on using Table A and Table D to find a P-value. Refer to the previous exercise. Find the P-value for z = −1.88.

6.3 Use and Abuse of Tests

When you complete this section, you will be able to:

Explain why it is important to report the P-value and not just report whether the result is statistically significant or not. Discriminate between practical (or scientific) significance and statistical significance. Identify poorly designed studies where formal statistical inference is suspect. Understand the consequences of searching solely for statistical significance, whether through the investigation of multiple tests or by identifying and testing using the same data set.

Carrying out a test of significance is often quite simple, especially if the P-value is given effortlessly by a computer. Using tests wisely is not so simple. Each test is valid only in certain circumstances, with properly produced data being particularly important.

The z test, for example, should bear the same warning label that was attached in Section 6.1 to the corresponding confidence interval (page 355). Similar warnings accompany the other tests that we will learn. There are additional caveats that concern tests more than confidence intervals, enough to warrant this separate section. Some hesitation about the unthinking use of significance tests is a sign of statistical maturity.

The reasoning of significance tests has appealed to researchers in many fields, so that tests are widely used to report research results. In this setting Ha is a “research hypothesis” asserting that some effect or difference is present. The null hypothesis H0 says that there is no effect or no difference. A low P-value represents good evidence that the research hypothesis is true. Here are some comments on the use of significance tests, with emphasis on their use in reporting scientific research.

Choosing a level of significance

The intention of a test of significance is to give a clear statement of the degree of evidence provided by the sample against the null hypothesis. The P-value does this. It is common practice to report P-values and to describe results as statistically significant whenever P ≤ 0.05. H owever, there is no sharp border between “significant” and “not significant,” only increasingly strong evidence as the P-value decreases. Having the P-value with a description of the effect that we have found allows us to draw better conclusions from our data.

EXAMPLE 6.21

Information provided by the P-value. Suppose that the test statistic for a two-sided significance test for a population mean is z = 1.95. From Table A we can calculate the P-value. It is

P = 2[1 − P(Z ≤ 1.95)] = 2(1 − 0.9744) = 0.0512

We have failed to meet the standard of evidence for α = 0.05. However, with the information provided by the P- value, we can see that the result just barely missed the standard. If the effect in question is interesting and potentially important, we might want to design another study with a larger sample to investigate it further.

Here is another example where the P-value provides useful information beyond that provided by the statement that we reject or fail to reject the null hypothesis.

EXAMPLE 6.22

More on information provided by the P-value. We have a test statistic of z = −4.66 for a two-sided significance test on a population mean. Software tells us that the P-value is 0.000003. This means that there are 3 chances in 1,000,000 of observing a sample mean this far or farther away from the null hypothesized value of μ. This kind of event is virtually impossible if the null hypothesis is true. There is no ambiguity in the result; we can clearly reject the null hypothesis.

We frequently report small P-values such as that in the previous example as P < 0.001. This corresponds to a chance of 1 in 1000 and is sufficiently small to lead us to a clear rejection of the null hypothesis.

One reason for the common use of α = 0.05 is the great influence of Sir R. A. Fisher, the inventor of formal statistical methods for analyzing experimental data. Here is his opinion on choosing a level of significance: “A scientific fact should be regarded as experimentally established only if a properly designed experiment rarely fails to give this level of significance.”24

What statistical significance does not mean

When a null hypothesis (“no effect” or “no difference”) can be rejected at the usual level α = 0.05, there is good evidence that an effect is present. That effect, however, can be extremely small. When large samples are available, even tiny deviations from the null hypothesis will be statistically significant.

EXAMPLE 6.23

It’s significant but is it important? Suppose that we are testing the null hypothesis of no correlation between two variables. With 400 observations, an observed correlation of only r = 0.1 is significant evidence at the α = 0.05 level that the correlation in the population is not zero. Figure 6.15 is an example of 400 (x, y) pairs that have an observed correlation of 0.10. The low significance level does not mean that there is a strong association, only that there is strong evidence of some association. The proportion of the variability in one of the variables explained by the other is r2 = 0.01, or 1%.

FIGURE 6.15 Scatterplot of n = 400 observations with an observed correlation of 0.10, Example 6.23. There is not a strong association between the two variables even though there is significant evidence (P < 0.05) that the population correlation is not zero.

For practical purposes, we might well decide to ignore this association. Statistical significance is not the same as practical significance. Statistical significance rarely tells us about the importance of the experimental results. This depends on the context of the experiment.

6.90

The remedy for attaching too much importance to statistical significance is to pay attention to the actual experimental results as well as to the P-value. Plot your data and examine them carefully. Beware of outliers. The user of statistics who feeds the data to a computer without exploratory analysis will often be embarrassed. It is usually wise to give a confidence interval for the parameter in which you are interested. Confidence intervals are not used as often as they should be, while tests of significance are overused.

USE YOUR KNOWLEDGE

Is it significant? More than 200,000 people worldwide take the GMAT examination each year when they apply for MBA programs. Their scores vary Normally with mean μ = 540 and standard deviation σ = 100. One hundred students go through a rigorous training program designed to raise their GMAT scores. Test the following hypotheses about the training program

H0: μ = 540

Ha: μ > 540

in each of the following situations.

(a) The students’ average score is = 556.4. Is this result significant at the 5% level?

(b) Now suppose that the average score is = 556.5. Is this result significant at the 5% level?

(c) Explain how you would reconcile this difference in significance, especially if any increase greater than 15 points is considered a success.

Don’t ignore lack of significance There is a tendency to conclude that there is no effect whenever a P-value fails to attain the usual 5% standard. A provocative editorial in the British Medical Journal entitled “Absence of Evidence Is Not Evidence of Absence” deals with this issue.25 Here is one of the examples they cite.

EXAMPLE 6.24

Interventions to reduce HIV-1 transmission. A randomized trial of interventions for reducing transmission of HIV-1 reported an incident rate ratio of 1.00, meaning that the intervention group and the control group both had the same rate of HIV-1 infection. The 95% confidence interval was reported as 0.63 to 1.58.26 The editorial notes that a summary of these results that says the intervention has no effect on HIV-1 infection is misleading. The confidence interval indicates that the intervention may be capable of achieving a 37% decrease in infection; it might also be harmful and produce a 58% increase in infection. Clearly, more data are needed to distinguish between these possibilities.

The situation can be worse. Research in some fields has rarely been published unless significance at the 0.05 level is attained.

EXAMPLE 6.25

Journal survey of reported significance results. A survey of four journals published by the American Psychological Association showed that of 294 articles using statistical tests, only eight reported results that did not attain the 5% significance level.27 It is very unlikely that these were the only eight studies of scientific merit that did not attain significance at the 0.05 level. Manuscripts describing other studies were likely rejected because of a lack of statistical significance or never submitted in the first place due to the expectation of rejection.

In some areas of research, small effects that are detectable only with large sample sizes can be of great practical significance. Data accumulated from a large number of patients taking a new drug may be needed before we can conclude that there are life-threatening consequences for a small number of people.

On the other hand, sometimes a meaningful result is not found significant.

EXAMPLE 6.26

A meaningful but statistically insignificant result. A sample of size 10 gave a correlation of r = 0.5 between two variables. The P-value is 0.102 for a two-sided significance test. In many situations, a correlation this large would be interesting and worthy of additional study. When it takes a lot of effort (say, in terms of time or money) to obtain samples, researchers often use small studies like these as pilot projects to gain interest from various funding sources. With financial support, a larger, more powerful study can then be run.

Another important aspect of planning a study is to verify that the test you plan to use does have high probability of detecting an effect of the size you hope to find. This probability is the power of the test. Power calculations are discussed in Section 6.4.

6.91

Statistical inference is not valid for all sets of data

design of experiments, p. 171

In Chapter 3, we learned that badly designed surveys or experiments often produce invalid results. Formal statistical inference cannot correct basic flaws in the design.

EXAMPLE 6.27

confounding, p. 150

English vocabulary and studying a foreign language. There is no doubt that there is a significant difference in English vocabulary scores between high school seniors who have studied a foreign language and those who have not. But because the effect of actually studying a language is confounded with the differences between students who choose language study and those who do not, this statistical significance is hard to interpret. The most plausible explanation is that students who were already good at English chose to study another language. A randomized comparative experiment would isolate the actual effect of language study and so make significance meaningful. Do you think it would be ethical to do such a study?

Tests of significance and confidence intervals are based on the laws of probability. Randomization in sampling or experimentation ensures that these laws apply. But we must often analyze data that do not arise from randomized samples or experiments. To apply statistical inference to such data, we must have confidence in a probability model for the data. The diameters of successive holes bored in auto engine blocks during production, for example, may behave like independent observations from a Normal distribution. We can check this probability model by examining the data. If the Normal distribution model appears approximately correct, we can apply the methods of this chapter to do inference about the process mean diameter μ.

USE YOUR KNOWLEDGE

Home security systems. A recent TV advertisement for home security systems said that homes without an alarm system are three times more likely to be broken into. Suppose that this conclusion was obtained by examining an SRS of police records of break-ins and determining whether the percent of homes with alarm systems was significantly smaller than 50%. Explain why the significance of this study is suspect and propose an alternative study that would help clarify the importance of an alarm system.

Beware of searching for significance

Statistical significance is an outcome much desired by researchers. It means (or ought to mean) that you have found an effect that you were looking for. The reasoning behind statistical significance works well if you decide what effect you are seeking, design an experiment or sample to search for it, and use a test of significance to weigh the evidence you get. But because a successful search for a new scientific phenomenon often ends with statistical significance, it is all too tempting to make significance itself the object of the search. There are several ways to do this, none of them acceptable in polite scientific society.

EXAMPLE 6.28

Genomics studies. In genomics experiments, it is common to assess the differences in expression for tens of thousands of genes. If each of these genes was examined separately and statistical significance declared for all that had P-values that pass the 0.05 standard, we would have quite a mess. In the absence of any real biological effects, we would expect that, by chance alone, approximately 5% of these tests will show statistical significance. Much research in genomics is directed toward appropriate ways to deal with this situation.28

We do not mean that searching data for suggestive patterns is not proper scientific work. It certainly is. Many important discoveries have been made by accident rather than by design. Exploratory analysis of data is an essential part of statistics. We do mean that the usual reasoning of statistical inference does not apply when the search for a pattern is successful. You cannot legitimately test a hypothesis on the same data that first suggested that hypothesis. The remedy is clear. Once you have a hypothesis, design a study to search specifically for the effect you now think is there. If the result of this study is statistically significant, you have real evidence.

SECTION 6.3 SUMMARY P-values are more informative than the reject-or-not result of a level α test. Beware of placing too much weight on traditional values of α, such as α = 0.05. Very small effects can be highly significant (small P), especially when a test is based on a large sample. A statistically significant effect need not be practically important. Plot the data to display the effect you are seeking, and use confidence intervals to estimate the actual values of parameters. On the other hand, lack of significance does not imply that H0 is true, especially when the test has a low probability of detecting an effect. Significance tests are not always valid. Faulty data collection, outliers in the data, and testing a hypothesis on the same data that suggested the hypothesis can invalidate a test. Many tests run at once will probably produce some significant results by chance alone, even if all the null hypotheses are true.

SECTION 6.3 EXERCISES For Exercise 6.90, see page 386; and for Exercise 6.91, see page 388.

6.92 A role as a statistical consultant. You are the statistical expert for a graduate student planning her PhD research. After you carefully present the mechanics of significance testing, she suggests using α = 0.20 for the study because she would be more likely to obtain statistically significant results and she really needs significant results to graduate. Explain in simple terms why this would not be a good use of statistical methods.

6.93 What do you know? A research report described two results that both achieved statistical significance at the 5% level. The P-value for the first is 0.048; for the second it is 0.0002. Do the P-values add any useful information beyond that conveyed by the statement that both results are statistically significant? Write a short paragraph explaining your views on this question.

6.94 Selective publication based on results. In addition to statistical significance, selective publication can also be due to the observed outcome. A recent review of 74 studies of antidepressant agents found 38 studies with positive results and 36 studies with negative or questionable results. All but one of the 38 positive studies were published. Of the remaining 36, 22 were not published and 11 were published in such a way as to convey a positive outcome.29 Describe how this selective reporting can have adverse consequences on health care.

6.95 What a test of significance can answer. Explain whether a test of significance can answer each of the following questions.

(a) Is the sample or experiment properly designed?

(b) Is the observed effect compatible with the null hypothesis?

(c) Is the observed effect important?

6.96 Vitamin C and colds. In a study to investigate whether vitamin C will prevent colds, 400 subjects are assigned at random to one of two groups. The experimental group takes a vitamin C tablet daily, while the control group takes a placebo. At the end of the experiment, the researchers calculate the difference between the percents of subjects in the two groups who were free of colds. This difference is statistically significant (P = 0.03) in favor of the vitamin C group. Can we conclude that vitamin C has a strong effect in preventing colds? Explain your answer.

6.97 How far do rich parents take us? How much education children get is strongly associated with the wealth and social status of their parents, termed “socioeconomic status,” or SES. The SES of parents, however, has little influence on whether children who have graduated from college continue their education. One study looked at whether college graduates took the graduate admissions tests for business, law, and other graduate programs. The effects of the parents’ SES on taking the LSAT test for law school were “both statistically insignificant and small.”

(a) What does “statistically insignificant” mean?

(b) Why is it important that the effects were small in size as well as statistically insignificant?

6.98 Do you agree? State whether or not you agree with each of the following statements and provide a short summary of the reasons for your answers.

(a) If the P-value is larger than 0.05, the null hypothesis is true.

(b) Practical significance is not the same as statistical significance.

(c) We can perform a statistical analysis using any set of data.

(d) If you find an interesting pattern in a set of data, it is appropriate to then use a significance test to determine its significance.

(e) It’s always better to use a significance level of α = 0.05 than to use α = 0.01 because it is easier to find statistical significance.

6.99 Practical significance and sample size. Every user of statistics should understand the distinction between statistical significance and practical importance. A sufficiently large sample will declare very small effects statistically significant. Consider the study of elite female Canadian athletes in Exercise 6.74 (page 382). Female athletes were consuming an average of 2403.7 kcal/d with a standard deviation of 880 kcal/d. Suppose that a nutritionist is brought in to implement a new health program for these athletes. This program should increase mean caloric intake but not change the standard deviation. Given the standard deviation and how calorie deficient these athletes are, a change in the mean of 50 kcal/d to 2453.7 is of little importance. However, with a large enough sample, this change can be significant. To see this, calculate the P-value for the test

of

H0: μ = 2403.7

Ha: μ > 2403.7

in each of the following situations:

(a) A sample of 100 athletes; their average caloric intake is = 2453.7.

(b) A sample of 500 athletes; their average caloric intake is = 2453.7.

(c) A sample of 2500 athletes; their average caloric intake is = 2453.7.

6.100 Statistical versus practical significance. A study with 7500 subjects reported a result that was statistically significant at the 5% level. Explain why this result might not be particularly important.

6.101 More on statistical versus practical significance. A study with 14 subjects reported a result that failed to achieve statistical significance at the 5% level. The P-value was 0.051. Write a short summary of how you would interpret these findings.

6.102 Find journal articles. Find two journal articles that report results with statistical analyses. For each article, summarize how the results are reported and write a critique of the presentation. Be sure to include details regarding use

of significance testing at a particular level of significance, P-values, and confidence intervals.

6.103 Create an example of your own. For each of the following cases, provide an example and an explanation as to why it is appropriate.

(a) A set of data or an experiment for which statistical inference is not valid.

(b) A set of data or an experiment for which statistical inference is valid.

6.104 Predicting success of trainees. What distinguishes managerial trainees who eventually become executives from those who, after expensive training, don’t succeed and leave the company? We have abundant data on past trainees

—data on their personalities and goals, their college preparation and performance, even their family backgrounds and their hobbies. Statistical software makes it easy to perform dozens of significance tests on these dozens of variables to see which ones best predict later success. We find that future executives are significantly more likely than washouts to have an urban or suburban upbringing and an undergraduate degree in a technical field. Explain clearly why using these “significant” variables to select future trainees is not wise. Then suggest a follow-up study

using this year’s trainees as subjects that should clarify the importance of the variables identified by the first study.

6.105 Searching for significance. Give an example of a situation where searching for significance would lead to misleading conclusions.

6.106 More on searching for significance. You perform 1000 significance tests using α = 0.05. Assuming that all null hypotheses are true, about how many of the test results would you expect to be statistically significant? Explain how you obtained your answer.

6.107 Interpreting a very small P-value. Assume that you are performing a large number of significance tests. Let n be the number of these tests. How large would n need to be for you to expect about one P-value to be 0.00001 or smaller? Use this information to write an explanation of how to interpret a result that has P = 0.00001 in this setting.

6.108 An adjustment for multiple tests. One way to deal with the problem of misleading P-values when performing more than one significance test is to adjust the criterion you use for statistical significance. The Bonferroni

procedure does this in a simple way. If you perform two tests and want to use the α = 5% significance level, you would require a P-value of 0.05/2 = 0.025 to declare either one of the tests significant. In general, if you perform k tests and want protection at level α, use α/k as your cutoff for statistical significance. You perform six tests and obtain individual P-values of 0.075, 0.021, 0.285, 0.002, 0.015, and <0.001. Which of these are statistically significant using the Bonferroni procedure with α = 0.05?

6.109 Significance using the Bonferroni procedure. Refer to the previous exercise. A researcher has performed 12 tests of significance and wants to apply the Bonferroni procedure with α = 0.05. The calculated P-values are 0.141,

0.519, 0.186, 0.753, 0.001, 0.008, 0.646, 0.038, 0.898, 0.013, <0.002, and 0.538. Which of these tests reject their null hypotheses with this procedure?

6.4 Power and Inference as a Decision

When you complete this section, you will be able to:

Define what is meant by the power of a test. Determine the power of a test to detect an alternative for a given sample size n. Describe the two types of possible errors when performing a test that focuses on deciding between two hypotheses. Relate the two errors to the significance level and power of the test.

Although we prefer to use P-values rather than the reject-or-not view of the level α significance test, the latter view is very important for planning studies and for understanding statistical decision theory. We will discuss these two topics in this section.

Power Level α significance tests are closely related to confidence intervals—in fact, we saw that a two-sided test can be carried out directly from a confidence interval (pages 353–354). The significance level, like the confidence level, says how reliable the method is in repeated use. If we use 5% significance tests repeatedly when H0 is, in fact, true, we will be wrong (the test will reject H0) 5% of the time and right (the test will fail to reject H0) 95% of the time.

The ability of a test to detect that H0 is false is measured by the probability that the test will reject H0 when an alternative is true. The higher this probability is, the more sensitive the test is.

POWER

The probability that a level α significance test will reject H0 when a particular alternative value of the parameter is true is called the power of the test to detect that alternative.

EXAMPLE 6.29

The power of a TBBMC significance test. Can a six-month exercise program increase the total body bone mineral content (TBBMC) of young women? A team of researchers is planning a study to examine this question. Based on the results of a previous study, they are willing to assume that σ = 2 for the percent change in TBBMC over the six- month period. They also believe that a change in TBBMC of 1% is important, so they would like to have a reasonable chance of detecting a change this large or larger. Is 25 subjects a large enough sample for this project?

We will answer this question by calculating the power of the significance test that will be used to evaluate the data to be collected. The calculation consists of three steps:

1. State H0, Ha (the particular alternative we want to detect), and the significance level α.

2. Find the values of that will lead us to reject H0.

3. Calculate the probability of observing these values of when the alternative is true.

Step 1. The null hypothesis is that the exercise program has no effect on TBBMC. In other words, the mean percent change is zero. The alternative is that exercise is beneficial; that is, the mean change is positive. Formally, we have

H0: μ = 0

Ha: μ > 0

The alternative of interest is μ = 1% increase in TBBMC. A 5% test of significance will be used.

Step 2. The z test rejects H0 at the α = 0.05 level whenever

Be sure you understand why we use 1.645. Rewrite this in terms of :

Because the significance level is α = 0.05, this event has probability 0.05 of occurring when the population mean m is 0.

Step 3. The power to detect the alternative μ = 1% is the probability that H0 will be rejected when in fact μ = 1%. We calculate this probability by standardizing , using the value μ = 1, the population standard deviation σ = 2, and the sample size n = 25. The power is

FIGURE 6.16 The sampling distributions of when μ = 0 and when μ = 1, Example 6.29. The power is the probability that the test rejects H0 when the alternative is true.

Figure 6.16 illustrates the power with the sampling distribution of when μ = 1. This significance test rejects the null hypothesis that exercise has no effect on TBBMC 80% of the time if the true effect of exercise is a 1% increase in TBBMC. If the true effect of exercise is a greater percent increase, the test will have greater power; it will reject with a higher probability.

Here is another example of a power calculation, this time for a two-sided z test.

EXAMPLE 6.30

Power of the lead concentration test. Example 6.17 (page 375) presented a test of H0: μ = 15.00

Ha: μ ≠ 15.00

at the 1% level of significance. What is the power of this test against the specific alternative μ = 15.50? The test rejects H0 when |z| ≥ 2.576. The test statistic is

Some arithmetic shows that the test rejects when either of the following is true:

z ≥ 2.576 (in other words, ≥ 15.37)

z ≤ −2.576 (in other words, ≤ 14.63)

These are disjoint events, so the power is the sum of their probabilities, computed assuming that the alternative μ = 15.50 is true. We find that

Figure 6.17 illustrates this calculation. A power of about 0.82, we are quite confident that the test will reject H0 when this alternative is true.

FIGURE 6.17 The power, Example 6.30. Unlike Figure 6.16, only the sampling distribution under the alternative is shown.

High power is desirable. Along with 95% confidence intervals and 5% significance tests, 80% power is becoming a standard. Many U.S. government agencies that provide research funds require that the sample size for the funded studies be sufficient to detect important results 80% of the time using a 5% test of significance.

EXAMPLE 6.31

Constructing a power curve. Example 6.30 considered one specific alternative, μ = 15.50. Often, it is helpful to consider the power for a range of alternatives. Fortunately, most statistical software saves us from having to do these calculations manually. Figure 6.18 shows Minitab output for the power over the range 15.00 ppm to 15.80 ppm. The power calculation of Example 6.30 is represented by a dot on the curve at a difference of 15.50 − 15.00 = 0.50. This curve is very informative. We see that with a sample size of three, the power is greater than 80% only for differences larger than about 0.48. If it is important to detect differences less than this, the Deely Laboratory needs to consider ways to increase the power.

FIGURE 6.18 Minitab output (a power curve) for the one-sample power calculation, Example 6.31.

Increasing the power Suppose that you have performed a power calculation and found that the power is too small. What can you do to increase it? Here are four ways. Note the similarity between these and the choices to reduce the margin of error (page 352).

Increase α. A 5% test of significance will have a greater chance of rejecting the alternative than a 1% test because the strength of evidence required for rejection is less. Consider a particular alternative that is farther away from μ0. Values of μ that are in Ha but lie close to the hypothesized value μ0 are harder to detect (lower power) than values of μ that are far from μ0.

Increase the sample size. More data will provide more information about so we have a better chance of distinguishing values of μ. Decrease σ. This has the same effect as increasing the sample size: more information about μ. Improving the measurement process and restricting attention to a subpopulation are possible ways to decrease σ.

Power calculations are important in planning studies. Using a significance test with low power makes it unlikely that you will find a significant effect even if the truth is far from the null hypothesis. A null hypothesis that is, in fact, false can become widely believed if repeated attempts to find evidence against it fail because of low power. The following example illustrates this point.

EXAMPLE 6.32

Are stock markets efficient? The “efficient market hypothesis” for the time series of stock prices says that future stock prices (when adjusted for inflation) show only random variation. No information available now will help us predict stock prices in the future because the efficient working of the market has already incorporated all available information in the present price. Many studies have tested the claim that one or another kind of information is helpful. In these studies, the efficient market hypothesis is H0, and the claim that prediction is possible is Ha. Almost all the studies have failed to find good evidence against H0. As a result, the efficient market hypothesis is quite popular. But an examination of the significance tests employed finds that the power is generally low. Failure to reject H0 when using tests of low power is not evidence that H0 is true. As one expert says, “The widespread impression that there is strong evidence for market efficiency may be due just to a lack of appreciation of the low power of many statistical tests.”30

Inference as decision We have presented tests of significance as methods for assessing the strength of evidence against the null hypothesis. This assessment is made by the P-value, which is a probability computed under the assumption that H0 is true. The alternative hypothesis (the statement we seek evidence for) enters the test only to help us see what outcomes count against the null hypothesis.

acceptance sampling

There is another way to think about these issues. Sometimes, we are really concerned about making a decision or choosing an action based on our evaluation of the data. Acceptance sampling is one such circumstance. A producer of bearings and a skateboard manufacturer agree that each carload lot of bearings shall meet certain quality standards. When a carload arrives, the manufacturer chooses a sample of bearings to be inspected. On the basis of the sample outcome, the manufacturer will either accept or reject the carload. Let’s examine how the idea of inference as a decision changes the reasoning used in tests of significance.

Two types of error Tests of significance concentrate on H0, the null hypothesis. If a decision is called for, however, there is no reason to single out H0. There are simply two hypotheses, and we must accept one and reject the other. It is convenient to call the two hypotheses H0 and Ha, but H0 no longer has the special status (the statement we try to find evidence against) that it had in tests of significance. In the acceptance sampling problem, we must decide between

H0: the lot of bearings meets standards

Ha: the lot does not meet standards

on the basis of a sample of bearings. We hope that our decision will be correct, but sometimes it will be wrong. There are two types of incorrect

decisions. We can accept a bad lot of bearings, or we can reject a good lot. Accepting a bad lot injures the consumer, while rejecting a good lot hurts the producer. To help distinguish these two types of error, we give them specific names.

FIGURE 6.19 The two types of error in testing hypotheses.

FIGURE 6.20 The two types of error in the acceptance sampling setting.

TYPE I AND TYPE II ERRORS

If we reject H0 (accept Ha) when in fact H0 is true, this is a Type I error. If we accept H0 (reject Ha) when in fact Ha is true, this is a Type II error.

The possibilities are summed up in Figure 6.19. If H0 is true, our decision either is correct (if we accept H0) or is a

Type I error. If Ha is true, our decision either is correct or is a Type II error. Only one error is possible at one time. Figure 6.20 applies these ideas to the acceptance sampling example.

Error probabilities Any rule for making decisions is assessed in terms of the probabilities of the two types of error. This is in keeping with the idea that statistical inference is based on probability. We cannot (short of inspecting the whole lot) guarantee that good lots of bearings will never be rejected and bad lots never be accepted. But by random sampling and the laws of probability, we can say what the probabilities of both kinds of error are.

Significance tests with fixed level α give a rule for making decisions because the test either rejects H0 or fails to reject it. If we adopt the decision-making way of thought, failing to reject H0 means deciding that H0 is true. We can then describe the performance of a test by the probabilities of Type I and Type II errors.

EXAMPLE 6.33

Outer diameter of a skateboard bearing. The mean outer diameter of a skateboard bearing is supposed to be 22.000 millimeters (mm). The outer diameters vary Normally with standard deviation σ = 0.010 mm. When a lot of the bearings arrives, the skateboard manufacturer takes an SRS of five bearings from the lot and measures their outer diameters. The manufacturer rejects the bearings if the sample mean diameter is significantly different from 22 mm at the 5% significance level.

This is a test of the hypotheses

H0: μ = 22

Ha: μ ≠ 22

To carry out the test, the manufacturer computes the z statistic:

and rejects H0 if

z < −1.96 or z > 1.96

A Type I error is to reject H0 when in fact μ = 22. What about Type II errors? Because there are many values of μ in Ha, we will concentrate on one value. The

producer and the manufacturer agree that a lot of bearings with mean 0.015 mm away from the desired mean 22.000 should be rejected. So a particular Type II error is to accept H0 when in fact μ = 22.015.

Figure 6.21 shows how the two probabilities of error are obtained from the two sampling distributions of , for μ = 22 and for μ = 22.015. When μ = 22, H0 is true and to reject H0 is a Type I error. When μ = 22.015, accepting H0 is a Type II error. We will now calculate these error probabilities.

FIGURE 6.21 The two error probabilities, Example 6.33. The probability of a Type I error (yellow area) is the probability of rejecting H0: μ = 22 when, in fact, μ = 22. The probability of a Type II error (blue area) is the probability of accepting H0 when, in fact, μ = 22.015.

The probability of a Type I error is the probability of rejecting H0 when it is really true. In Example 6.33, this is the probability that |z| ≥ 1.96 when μ = 22. But this is exactly the significance level of the test. The critical value 1.96 was chosen to make this probability 0.05, so we do not have to compute it again. The definition of “significant at level 0.05” is that sample outcomes this extreme will occur with probability 0.05 when H0 is true.

SIGNIFICANCE AND TYPE I ERROR

The significance level α of any fixed level test is the probability of a Type I error. That is, α is the probability that the test will reject the null hypothesis H0 when H0 is in fact true.

The probability of a Type II error for the particular alternative μ = 22.015 in Example 6.33 is the probability that the test will fail to reject H0 when μ has this alternative value. The power of the test to detect the alternative μ = 22.015 is just the probability that the test does reject H0. By following the method of Example 6.30, we can calculate that the power is about 0.92. The probability of a Type II error is therefore 1 − 0.92, or 0.08.

POWER AND TYPE II ERROR

The power of a fixed level test to detect a particular alternative is 1 minus the probability of a Type II error for that alternative.

The two types of error and their probabilities give another interpretation of the significance level and power of a test. The distinction between tests of significance and tests as rules for deciding between two hypotheses does not lie in the calculations but in the reasoning that motivates the calculations. In a test of significance, we focus on a single hypothesis (H0) and a single probability (the P-value). The goal is to measure the strength of the sample evidence against H0. Calculations of power are done to check the sensitivity of the test. If we cannot reject H0, we conclude only

that there is not sufficient evidence against H0, not that H0 is actually true. If the same inference problem is thought of as a decision problem, we focus on two hypotheses and give a rule for

deciding between them based on the sample evidence. Therefore, we must focus equally on two probabilities, the probabilities of the two types of error. We must choose one hypothesis and cannot abstain on grounds of insufficient evidence.

The common practice of testing hypotheses Such a clear distinction between the two ways of thinking is helpful for understanding. In practice, the two approaches often merge. We continued to call one of the hypotheses in a decision problem H0. The common practice of testing hypotheses mixes the reasoning of significance tests and decision rules as follows:

1. State H0 and Ha just as in a test of significance.

2. Think of the problem as a decision problem, so that the probabilities of Type I and Type II errors are relevant. 3. Because of Step 1, Type I errors are more serious. So choose an α (significance level) and consider only tests

with probability of a Type I error no greater than α. 4. Among these tests, select one that makes the probability of a Type II error as small as possible (that is, power as

large as possible). If this probability is too large, you will have to take a larger sample to reduce the chance of an error.

Testing hypotheses may seem to be a hybrid approach. It was, historically, the effective beginning of decision- oriented ideas in statistics. An impressive mathematical theory of hypothesis testing was developed between 1928 and 1938 by Jerzy Neyman and Egon Pearson. The decision-making approach came later (1940s). Because decision theory in its pure form leaves you with two error probabilities and no simple rule on how to balance them, it has been used less often than either tests of significance or tests of hypotheses. Decision ideas have been applied in testing problems mainly by way of the Neyman-Pearson hypothesis-testing theory. That theory asks you first to choose α, and the influence of Fisher has often led users of hypothesis testing comfortably back to α = 0.05 or α = 0.01. Fisher, who was exceedingly argumentative, violently attacked the Neyman-Pearson decision-oriented ideas, and the argument still continues.

SECTION 6.4 SUMMARY The power of a significance test measures its ability to detect an alternative hypothesis. The power to detect a specific alternative is calculated as the probability that the test will reject H0 when that alternative is true. This calculation requires knowledge of the sampling distribution of the test statistic under the alternative hypothesis. Increasing the size of the sample increases the power when the significance level remains fixed. An alternative to significance testing regards H0 and Ha as two statements of equal status that we must decide between. This decision theory point of view regards statistical inference in general as giving rules for making decisions in the presence of uncertainty. In the case of testing H0 versus Ha, decision analysis chooses a decision rule on the basis of the probabilities of two types of error. A Type I error occurs if H0 is rejected when it is in fact true. A Type II error occurs if H0 is accepted when in fact Ha is true.

In a fixed level α significance test, the significance level α is the probability of a Type I error, and the power to detect a specific alternative is 1 minus the probability of a Type II error for that alternative.

SECTION 6.4 EXERCISES 6.110 Make a recommendation. Your manager has asked you to review a research proposal that includes a section on sample size justification. A careful reading of this section indicates that the power is 18% for detecting an effect that would be considered important. Write a short report for your manager explaining what this means and make a recommendation on whether or not this study should be run.

6.111 Explain power and sample size. Two studies are identical in all respects except for the sample sizes. Consider the power versus a particular sample size. Will the study with the larger sample size have more power or less power than the one with the smaller sample size? Explain your answer in terms that could be understood by someone with very little knowledge of statistics.

6.112 Power for a different alternative. The power for a two-sided test of the null hypothesis μ = 0 versus the alternative μ = 6 is 0.83. What is the power versus the alternative μ = −6? Explain your answer.

6.113 More on the power for a different alternative. A one-sided test of the null hypothesis μ = 20 versus the alternative μ = 30 has power equal to 0.73. Will the power for the alternative μ = 35 be higher or lower than 0.73? Draw a picture and use this to explain your answer.

6.114 Effect of changing the alternative μ on power. The Statistical Power applet illustrates the power calculation similar to that in Figure 6.16. Open the applet and keep the default settings for the null (μ = 0) and the

alternative (μ > 0) hypotheses, the sample size (n = 10), the standard deviation (σ = 1), and the significance level (α = 0.05). In the “alt μ =” box, enter the value 1. What is the power? Repeat for alternative μ equal to 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9. Make a table giving μ and the power. What do you conclude?

6.115 Other changes and the effect on power. Refer to the previous exercise. For each of the following changes, explain what happens to the power for each alternative μ in the table.

(a) Change to the two-sided alternative.

(b) Decrease σ to 0.5.

(c) Increase n from 10 to 30.

6.116 Power of the random north–south distribution of trees test. In Exercise 6.70 (page 382), you performed a two-sided significance test of the null hypothesis that the average north–south location of the longleaf pine

trees sampled in the Wade Tract was μ = 100. There were 584 trees in the sample and the standard deviation was assumed to be 58. The sample mean in that analysis was = 99.74. Use the Statistical Power applet to compute the power for the alternative μ = 99 using a two-sided test at the 5% level of significance.

6.117 Power of the random east–west distribution of trees test. Refer to the previous exercise. Note that in the east–west direction, the average location was 113.8. Use the Statistical Power applet to find the power for the

alternative μ = 110.

6.118 Planning another test to compare consumption. Example 6.15 (page 372) gives a test of a hypothesis about the mean consumption of sugar-sweetened beverages at your university based on a sample of size n = 100. The

hypotheses are

H0: μ = 286

Ha: μ ≠ 286

While the result was not statistically significant, it did provide some evidence that the mean was smaller than 286. Thus, you plan to recruit another sample of students from your university, but this time use a one-sided alternative. You were thinking of surveying n = 100 students but now wonder if this sample size gives adequate power to detect a decrease of 15 calories per day to μ = 271.

(a) Given α = 0.05, for what values of z will you reject the null hypothesis?

(b) Using σ = 155 and μ = 286, for what values of will you reject H0?

(c) Using σ = 155 and μ = 271, what is the probability that will fall in the region defined in part (b)?

(d) Will a sample size of n = 100 give you adequate power? Or do you need to find ways to increase the power? Explain your answer.

(e) Use the Statistical Power applet or other statistical software to determine the sample size n that gives you power near 0.80.

6.119 Planning the dining court survey. Exercise 6.38 (page 364) describes a survey to assess whether a newly designed dining court is viewed more favorably than the old design. The organizers are considering randomly surveying n = 100 student patrons but would like some statistical advice. The hypotheses are

H0: μ = 4

Ha: μ > 4

and they’ve decided they want adequate power to detect a mean of at least 4.25.

(a) The organizers have no idea of σ. You suggest a small pilot study, which gives s = 1.73. Based on this result, you decide to use σ = 2. Provide an explanation for this choice to the organizers.

(b) Given α = 0.05, for what values of will you reject H0?

(c) Using μ = 4.25, what is the probability that will fall in the region defined in part (b)?

(d) Will a sample size of n = 100 give you adequate power? Explain your answer.

(e) Use the Statistical Power applet or statistical software to determine the sample size n that gives you power near 0.80.

6.120 Choose the appropriate distribution. You must decide which of two discrete distributions a random variable X has. We will call the distributions p0 and p1. Here are the probabilities they assign to the values x of X:

x 0 1 2 3 4 5 6

p0 0.1 0.1 0.2 0.3 0.1 0.1 0.1

p1 0.1 0.3 0.2 0.1 0.1 0.1 0.1

You have a single observation on X and wish to test

H0: p0 is correct

Ha: p1 is correct

One possible decision procedure is to reject H0 only if X ≤ 1.

(a) Find the probability of a Type I error, that is, the probability that you reject H0 when p0 is the correct distribution.

(b) Find the probability of a Type II error.

6.121 Power of the mean SATM score test. Example 6.16 (page 374) gives a test of a hypothesis about the SATM scores of California high school students based on an SRS of 500 students. The hypotheses are

H0: μ = 485

Ha: μ > 485

Assume that the population standard deviation is σ = 100. The test rejects H0 at the 1% level of significance when z ≥ 2.326, where

Is this test sufficiently sensitive to usually detect an increase of 14 points in the population mean SATM score? Answer this question by calculating the power of the test to detect the alternative μ = 499.

6.122 More on choosing the appropriate distribution. Refer to Exercise 6.120. Suppose that instead of a single observation X, you obtained two observations and use the decision rule to reject when ≤ 1.

(a) Under this scenario, would you expect the probabilities of a Type I and Type II errors to increase, decrease, or stay at the same values of Exercise 6.120? Explain your answer.

(b) Verify your answer to part (a) by computing the probabilities of a Type I and Type II error.

6.123 Computer-assisted career guidance systems. A wide variety of computer-assisted career guidance systems have been developed over the last decade. These programs use factors such as student interests, aptitude, skills, personality, and family history to recommend a career path. For simplicity, suppose that a program recommends a high school graduate either to

go to college or to join the workforce.

(a) What are the two hypotheses and the two types of error that the program can make?

(b) The program can be adjusted to decrease one error probability at the cost of an increase in the other error probability. Which error probability would you choose to make smaller, and why? (This is a matter of judgment. There is no single correct answer.)

CHAPTER 6 EXERCISES

6.124 Telemarketing wages. An advertisement in the student newspaper asks you to consider working for a telemarketing company. The ad states, “Earn between $500 and $1000 per week.” Do you think that the ad is describing a confidence interval? Explain your answer.

6.125 Exercise and statistics exams. A study examined whether light exercise performed an hour before the final exam in statistics affects how students perform on the exam. The P-value was given as 0.13.

(a) State null and alternative hypotheses that could be used for this study. (Note: There is more than one correct answer.)

(b) Do you reject the null hypothesis? State your conclusion in plain language.

(c) What other facts about the study would you like to know for a proper interpretation of the results?

6.126 Roulette. A roulette wheel has 18 red slots among its 38 slots. You observe many spins and record the number of times that red occurs. Now you want to use these data to test whether the probability of a red has the value that is correct for a fair roulette wheel. State the hypotheses H0 and Ha that you will test.

6.127 Food selection by children in school cafeterias. A group of researchers examined whether children’s food selection in a school cafeteria met the standards set by the School Meals Initiative. They measured food selection and food intake of 2049 fourth- through sixth-grade students in 33 schools over a three-day period using digital photography. The following table summarizes some of the food intake measurements.31

Boys n = 852

Girls n = 1197

Food intake Mean St. Dev. Mean St. Dev.

Energy (kilojoules) 2448 717 2170 693 Protein (g) 24.5 7.5 22.1 7.7 Calcium (mg) 324.1 130.6 265.0 128.9

Given the large sample sizes, we can assume that the sample standard deviations are the population standard deviations.

(a) Compute 95% confidence intervals for all three intake measures for the boys.

(b) Compute 95% confidence intervals for all three intake measures for the girls.

(c) In the next chapter, we will describe the confidence interval for the difference between two means. For now, let’s compare the boy and girl confidence intervals for each food intake measure. Do you think these pairs of intervals provide strong evidence against the null hypothesis that the boys and girls consume, on average, the same amount? Explain your answer.

6.128 Coverage percent of 95% confidence interval. For this exercise, you will use the Confidence Interval applet. Set the confidence level at 95% and click the “Sample” button 10 times to simulate 10

confidence intervals. Record the percent hit. Simulate another 10 intervals by clicking another 10 times (do not click the “Reset” button). Record the percent hit for your 20 intervals. Repeat the process of simulating 10 additional intervals and recording the results until you have a total of 200 intervals. Create a time plot of your results and write a summary of what you have found.

6.129 Coverage percent of 90% confidence interval. Refer to the previous exercise. Do the simulations and report the results for 90% confidence.

6.130 Effect of sample size on significance. You are testing the null hypothesis that μ = 0 versus the alternative μ > 0 using α = 0.05. Assume that σ = 16. Suppose that = 8 and n = 10. Calculate the test statistic and

its P-value. Repeat assuming the same value of but with n = 20. Do the same for sample sizes of 30, 40, and 50. Plot the values of the test statistic versus the sample size. Do the same for the P-values. Summarize what this demonstration shows about the effect of the sample size on significance testing.

6.131 Survey response and margin of error. Suppose that a business conducts a marketing survey. As is often

done, the survey is conducted by telephone. As it turns out, the business was only able to elicit responses from less than 10% of the randomly chosen customers. The low response rate is attributable to many factors, including caller ID screening. Undaunted, the marketing manager was pleased with the sample results because the margin of error was quite small, and thus the manager felt that the business had a good sense of the customers perceptions on various issues. Do you think the small margin of error is a good measure of the accuracy of the surveys results? Explain.

6.132 Reporting margins of error. A U.S. News & World Report article from July 17, 2014, reported Commerce Department estimates of changes in the construction industry:

Construction fell 9.3 percent last month to a seasonally adjusted annual rate of 893,000 homes.

If we turn to the original Commerce Department report (released the same day), it states:

Privately owned housing starts in June were at a seasonally adjusted annual rate of 893,000. This is 9.3 percent (10.3%) below the revised May estimate of 985,000.

(a) The 10.3% figure is the margin of error based on a 90% level of confidence. Given that fact, what is the 90% confidence interval for the percent change in housing starts from May to June?

(b) Explain why a credible media report should state:

The Commerce Department has no evidence that privately owned housing starts rose or fell in June from the previous month.

6.133 Blood phosphorus level in dialysis patients. Patients with chronic kidney failure may be treated by dialysis, in which a machine removes toxic wastes from the blood, a function normally performed by the kidneys.

Kidney failure and dialysis can cause other changes, such as retention of phosphorus, that must be corrected by changes in diet. A study of the nutrition of dialysis patients measured the level of phosphorus in the blood of several patients on six occasions. Here are the data for one patient (in milligrams of phosphorus per deciliter of blood):32

5.4 5.2 4.5 4.9 5.7 6.3

The measurements are separated in time and can be considered an SRS of the patient’s blood phosphorus level. Assume that this level varies Normally with σ = 0.9 mg/dl. PMGDL

(a) Give a 95% confidence interval for the mean blood phosphorus level.

(b) The normal range of phosphorus in the blood is considered to be 2.6 to 4.8 mg/dl. Is there strong evidence that this patient has a mean phosphorus level that exceeds 4.8?

6.134 Cellulose content in alfalfa hay. An agronomist examines the cellulose content of a variety of alfalfa hay. Suppose that the cellulose content in the population has standard deviation σ = 8 milligrams per gram (mg/g). A sample of 15 cuttings has mean cellulose content = 145 mg/g.

(a) Give a 90% confidence interval for the mean cellulose content in the population.

(b) A previous study claimed that the mean cellulose content was μ = 140 mg/g, but the agronomist believes that the mean is higher than that figure. State H0 and Ha and carry out a significance test to see if the new data support this belief.

(c) The statistical procedures used in parts (a) and (b) are valid when several assumptions are met. What are these assumptions?

6.135 Odor threshold of future wine experts. Many food products contain small quantities of substances that would give an undesirable taste or smell if they are present in large amounts. An example is the “off-odors” caused by sulfur compounds in wine. Oenologists (wine experts) have determined the odor threshold, the lowest concentration of a compound that the human nose can detect. For example, the odor threshold for dimethyl sulfide (DMS) is given in the oenology literature as 25 micrograms per liter of wine (μg/l). Untrained noses may be less sensitive, however. Here are the DMS odor thresholds for 10 beginning students of oenology:

31 31 43 36 23 34 32 30 20 24

Assume (this is not realistic) that the standard deviation of the odor threshold for untrained noses is known to be σ = 7 μg/l. ODOR

(a) Make a stemplot to verify that the distribution is roughly symmetric with no outliers. (A Normal quantile plot confirms that there are no systematic departures from Normality.)

(b) Give a 95% confidence interval for the mean DMS odor threshold among all beginning oenology students.

(c) Are you convinced that the mean odor threshold for beginning students is higher than the published threshold, 25 μg/l? Carry out a significance test to justify your answer.

6.136 Where do you buy? Consumers can purchase nonprescription medications at food stores, mass merchandise stores such as Target and Walmart, or pharmacies. About 45% of consumers make such purchases at

pharmacies. What accounts for the popularity of pharmacies, which often charge higher prices? A study examined consumers’ perceptions of overall performance of the three types of stores, using a long

questionnaire that asked about such things as “neat and attractive store,” “knowledgeable staff,” and “assistance in choosing among various types of nonprescription medication.” A performance score was based on 27 such questions. The subjects were 201 people chosen at random from the Indianapolis telephone directory. Here are the means and standard deviations of the performance scores for the sample:33

Store type s

Food stores 18.67 24.95 Mass merchandisers 32.38 33.37 Pharmacies 48.60 35.62

We do not know the population standard deviations, but a sample standard deviation s from so large a sample is usually close to σ. Use s in place of the unknown σ in this exercise.

(a) What population do you think the authors of the study want to draw conclusions about? What population are you certain they can draw conclusions about?

(b) Give 95% confidence intervals for the mean performance for each type of store.

(c) Based on these confidence intervals, are you convinced that consumers think that pharmacies offer higher performance than the other types of stores? (In Chapter 12, we will study a statistical method for comparing the means of several groups.)

6.137 CEO pay. A study of the pay of corporate chief executive officers (CEOs) examined the increase in cash compensation of the CEOs of 104 companies, adjusted for inflation, in a recent year. The mean increase in real compensation was = 6.9%, and the standard deviation of the increases was s = 55 %. Is this good evidence that the mean real compensation μ of all CEOs increased that year? The hypotheses are

H0: μ = 0 (no increase)

Ha: μ > 0 (an increase)

Because the sample size is large, the sample s is close to the population σ, so take σ = 55 %.

(a) Sketch the Normal curve for the sampling distribution of when H0 is true. Shade the area that represents the P- value for the observed outcome = 6.9%.

(b) Calculate the P-value.

(c) Is the result significant at the α = 0.05 level? Do you think the study gives strong evidence that the mean compensation of all CEOs went up?

6.138 Meaning of “statistically significant.” When asked to explain the meaning of “statistically significant at the α = 0.01 level,” a student says, “This means there is only probability 0.01 that the null hypothesis is true.” Is this an essentially correct explanation of statistical significance? Explain your answer.

6.139 More on the meaning of “statistically significant.” Another student, when asked why statistical significance appears so often in research reports, says, “Because saying that results are significant tells us that they cannot easily be explained by chance variation alone.” Do you think that this statement is essentially correct? Explain your answer.

6.140 Increasing the power. Refer to Example 6.17 (page 375). Suppose the Deely Laboratory wants to make sure the power is at least 80% for μ = 15.30 ppm. It cannot reduce σ, so the options are changing the significance level α and/or sample size n.

(a) With α = 0.01, what sample size n is needed to have at least 80% power?

(b) With α = 0.05, what sample size n is needed to have at least 80% power?

(c) Which of these options do you think the Deely Laboratory should choose? Explain your reasoning.

6.141 Simulation study of the confidence interval. Use a computer to generate n = 15 observations from a Normal distribution with mean 20 and standard deviation 5: N(20, 5). Find the 95% confidence interval for μ.

Repeat this process 100 times and then count the number of times that the confidence interval includes the value μ = 20. Explain your results.

6.142 Simulation study of a test of significance. Use a computer to generate n = 15 observations from a Normal distribution with mean 20 and standard deviation 5: N(20, 5). Test the null hypothesis that μ = 20 using a

two-sided significance test. Repeat this process 100 times and then count the number of times that you reject H0. Explain your results.

6.143 Another simulation study of a test of significance. Use the same procedure for generating data as in the previous exercise. Now test the null hypothesis that μ = 24. Explain your results.

6.144 Simulation study of power. Refer to the previous two exercises. What is the power of detecting a difference of four units (H0: μ = 24 versus μ = 20) in this setting? Compare this power with the proportion of times

you rejected H0 in the previous exercise. Explain your results.

6.145 Find published studies with confidence intervals. Search the Internet or some journals that report research in your field and find two reports that provide an estimate with a margin of error or a confidence interval.

For each report,

(a) describe the method used to collect the data.

(b) describe the variable being studied.

(c) give the estimate and the confidence interval.

(d) describe any practical difficulties that may have led to errors in addition to the sampling errors quantified by the margin of error.