Discussion 05.1: Muddiest Point

Conceptual Understanding Students need to be equipped with both the methods and conceptual understanding of statistics. MyStatLab offers a full question library of over 1,000 conceptual-based questions to help tighten the comprehension of statistical concepts.

Real-World Statistics MyStatLab video resources help foster conceptual understanding. StatTalk Videos, hosted by fun-loving statistician, Andrew Vickers, demonstrate important statistical concepts through interesting stories and real-life events. This series of 24 videos includes assignable questions built in MyStatLab and an instructor’s guide.

Visit www.mystatlab.com and click Get Trained to make sure you’re getting the most out of MyStatLab.

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BIOSTATISTICS FOR THE BIOLOGICAL AND HEALTH SCIENCES

MARC M. TRIOLA, MD, FACP New York University School of Medicine

MARIO F. TRIOLA Dutchess Community College

JASON ROY, PHD University of Pennsylvania

Perelman School of Medicine

SECOND EDITION

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To Ginny Dushana and Marisa Trevor and Mitchell

Director, Portfolio Management Deirdre Lynch Senior Portfolio Manager Suzy Bainbridge Portfolio Management Assistant Justin Billing Content Producer Peggy McMahon Managing Producer Karen Wernholm Courseware QA Manager Mary Durnwald Senior Producer Vicki Dreyfus Product Marketing Manager Yvonne Vannatta Field Marketing Manager Evan St. Cyr

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Copyright © 2018, 2006 by Pearson Education, Inc. All Rights Reserved. Printed in the United States of America. This publica- tion is protected by copyright, and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise. For information regarding permissions, request forms and the appropriate contacts within the Pearson Education Global Rights & Permissions department, please visit www.pearsoned.com/permissions/.

Attributions of third party content appear on page 683–684, which constitutes an extension of this copyright page.

PEARSON, ALWAYS LEARNING, and MYSTATLAB are exclusive trademarks owned by Pearson Education, Inc. or its affiliates in the U.S. and/or other countries.

Unless otherwise indicated herein, any third-party trademarks that may appear in this work are the property of their respective own- ers and any references to third-party trademarks, logos or other trade dress are for demonstrative or descriptive purposes only. Such references are not intended to imply any sponsorship, endorsement, authorization, or promotion of Pearson’s products by the owners of such marks, or any relationship between the owner and Pearson Education, Inc. or its affiliates, authors, licensees or distributors.

MICROSOFT AND>OR ITS RESPECTIVE SUPPLIERS MAKE NO REPRESENTATIONS ABOUT THE SUITABILITY OF THE INFORMATION CONTAINED IN THE DOCUMENTS AND RELATED GRAPHICS PUBLISHED AS PART OF THE SERVICES FOR ANY PURPOSE. ALL SUCH DOCUMENTS AND RELATED GRAPHICS ARE PROVIDED “AS IS” WITHOUT WARRANTY OF ANY KIND. MICROSOFT AND>OR ITS RESPECTIVE SUPPLIERS HEREBY DISCLAIM ALL WARRANTIES AND CONDITIONS WITH REGARD TO THIS INFORMATION, INCLUDING ALL WARRANTIES AND CONDITIONS OF MERCHANTABILITY, WHETHER EXPRESS, IMPLIED OR STATUTORY, FITNESS FOR A PARTICULAR PURPOSE, TITLE AND NON-INFRINGEMENT. IN NO EVENT SHALL MICROSOFTAND>OR ITS RESPEC- TIVE SUPPLIERS BE LIABLE FOR ANY SPECIAL, INDIRECT OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN AN ACTION OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING OUT OF OR IN CONNECTION WITH THE USE OR PERFORMANCE OF INFORMATION AVAILABLE FROM THE SERVICES.

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Library of Congress Cataloging-in-Publication Data

Names: Triola, Marc M. | Triola, Mario F. | Roy, Jason (Jason Allen) Title: Biostatistics for the biological and health sciences. Description: Second edition / Marc M. Triola, New York University,

Mario F. Triola, Dutchess Community College, Jason Roy, University of Pennsylvania. | Boston : Pearson, [2018] | Includes bibliographical references and index.

Identifiers: LCCN 2016016759| ISBN 9780134039015 (hardcover) | ISBN 0134039017 (hardcover)

Subjects: LCSH: Biometry. | Medical statistics. Classification: LCC QH323.5 .T75 2018 | DDC 570.1/5195–dc23 LC record available at https://lccn.loc.gov/2016016759 1 16

ISBN 13: 978-0-13-403901-5 ISBN 10: 0-13-403901-7

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iii

Marc Triola, MD, FACP is the Associate Dean for Educational Informatics at NYU School of Medicine, the founding director of the NYU Langone Medical Center Institute for Innovations in Medical Education (IIME), and an Associate Professor of Medicine. Dr. Triola’s research experience and expertise focus on the disruptive effects of the present revolution in educa- tion, driven by technological advances, big data, and learn- ing analytics. Dr. Triola has worked to create a “learning

ecosystem” that includes interconnected computer-based e-learning tools and new ways to effectively integrate growing amounts of electronic data in educational re- search. Dr. Triola and IIME have been funded by the National Institutes of Health, the Integrated Advanced Information Management Systems program, the National Science Foundation Advanced Learning Technologies program, the Josiah Macy, Jr. Foundation, the U.S. Department of Education, and the American Medical As- sociation Accelerating Change in Medical Education program. He chairs numer- ous committees at the state and national levels focused on the future of health professions educational technology development and research.

Mario F. Triola is a Professor Emeritus of Mathematics at Dutchess Community College, where he has taught statistics for over 30 years. Marty is the author of Elementary Statistics, 13th edition, Essentials of Sta- tistics, 5th edition, Elementary Statistics Using Excel, 6th edi- tion, and Elementary Statis- tics Using the TI-83>84 Plus Calculator, 4th edition, and he is a co-author of Statistical Reasoning for Everyday Life, 5th edition. Elementary Statis- tics is currently available as an

International Edition, and it has been translated into several foreign languages. Marty designed the original Statdisk statistical software, and he has written several manuals and workbooks for technology supporting statistics education.

ABOUT THE AUTHORS

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iv About the Authors

He has been a speaker at many conferences and colleges. Marty’s consulting work includes the design of casino slot machines and the design of fishing rods. He has worked with attorneys in determining probabilities in paternity lawsuits, analyz- ing data in medical malpractice lawsuits, identifying salary inequities based on gender, and analyzing disputed election results. He has also used statistical meth- ods in analyzing medical school surveys and in analyzing survey results for the New York City Transit Authority. Marty has testified as an expert witness in the New York State Supreme Court.

Jason Roy, PhD, is Associate Professor of Biostatistics in the Department of Biostatistics and Epidemiology, Perelman School of Medicine, Univer- sity of Pennsylvania. He re- ceived his PhD in Biostatistics in 2000 from the University of Michigan. He was recipi- ent of the 2002 David P. Byar Young Investigator Award from the American Statistical Asso- ciation Biometrics Section. His statistical research interests are in the areas of causal inference, missing data, and prediction

modeling. He is especially interested in the statistical challenges with analyzing data from large health care databases. He collaborates in many different disease areas, including chronic kidney disease, cardiovascular disease, and liver diseases. Dr Roy is Associate Editor of Biometrics, Journal of the American Statistical Association, and Pharmacoepidemiology & Drug Safety, and has over 90 peer- reviewed publications.

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v

CONTENTS

1 INTRODUCTION TO STATISTICS 1 1-1 Statistical and Critical Thinking 4 1-2 Types of Data 13 1-3 Collecting Sample Data 24

2 EXPLORING DATA WITH TABLES AND GRAPHS 40 2-1 Frequency Distributions for Organizing and Summarizing Data 42 2-2 Histograms 51 2-3 Graphs That Enlighten and Graphs That Deceive 56 2-4 Scatterplots, Correlation, and Regression 65

3 DESCRIBING, EXPLORING, AND COMPARING DATA 75 3-1 Measures of Center 77 3-2 Measures of Variation 89 3-3 Measures of Relative Standing and Boxplots 102

4 PROBABILITY 118 4-1 Basic Concepts of Probability 120 4-2 Addition Rule and Multiplication Rule 131 4-3 Complements, Conditional Probability, and Bayes’ Theorem 144 4-4 Risks and Odds 153 4-5 Rates of Mortality, Fertility, and Morbidity 162 4-6 Counting 167

5 DISCRETE PROBABILITY DISTRIBUTIONS 180 5-1 Probability Distributions 182 5-2 Binomial Probability Distributions 193 5-3 Poisson Probability Distributions 206

6 NORMAL PROBABILITY DISTRIBUTIONS 216 6-1 The Standard Normal Distribution 218 6-2 Real Applications of Normal Distributions 231 6-3 Sampling Distributions and Estimators 241 6-4 The Central Limit Theorem 252 6-5 Assessing Normality 261 6-6 Normal as Approximation to Binomial 269

7 ESTIMATING PARAMETERS AND DETERMINING SAMPLE SIZES 282 7-1 Estimating a Population Proportion 284 7-2 Estimating a Population Mean 299 7-3 Estimating a Population Standard Deviation or Variance 315 7-4 Bootstrapping: Using Technology for Estimates 324

8 HYPOTHESIS TESTING 336 8-1 Basics of Hypothesis Testing 338 8-2 Testing a Claim About a Proportion 354 8-3 Testing a Claim About a Mean 366 8-4 Testing a Claim About a Standard Deviation or Variance 377

9 INFERENCES FROM TWO SAMPLES 392 9-1 Two Proportions 394 9-2 Two Means: Independent Samples 406 9-3 Two Dependent Samples (Matched Pairs) 418 9-4 Two Variances or Standard Deviations 428

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vi Contents

10 CORRELATION AND REGRESSION 442 10-1 Correlation 444 10-2 Regression 462 10-3 Prediction Intervals and Variation 474 10-4 Multiple Regression 481 10-5 Dummy Variables and Logistic Regression 489

11 GOODNESS-OF-FIT AND CONTINGENCY TABLES 502 11-1 Goodness-of-Fit 503 11-2 Contingency Tables 514

12 ANALYSIS OF VARIANCE 531 12-1 One-Way ANOVA 533 12-2 Two-Way ANOVA 547

13 NONPARAMETRIC TESTS 560 13-1 Basics of Nonparametric Tests 562 13-2 Sign Test 564 13-3 Wilcoxon Signed-Ranks Test for Matched Pairs 575 13-4 Wilcoxon Rank-Sum Test for Two Independent Samples 581 13-5 Kruskal-Wallis Test for Three or More Samples 586 13-6 Rank Correlation 592

14 SURVIVAL ANALYSIS 603 14-1 Life Tables 604 14-2 Kaplan-Meier Survival Analysis 614

APPENDIX A TABLES 625 APPENDIX B DATA SETS 638 APPENDIX C WEBSITES AND BIBLIOGRAPHY OF BOOKS 645 APPENDIX D ANSWERS TO ODD-NUMBERED SECTION EXERCISES 646

(and all Quick Quizzes, all Review Exercises, and all Cumulative Review Exercises)

Credits 683 Index 685

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PREFACE

Statistics permeates nearly every aspect of our lives, and its role has become partic- ularly important in the biological, life, medical, and health sciences. From opinion polls to clinical trials in medicine and analysis of big data from health applications, statistics inf luences and shapes the world around us. Biostatistics for the Health and Biological Sciences forges the relationship between statistics and our world through extensive use of a wide variety of real applications that bring life to theory and methods.

Goals of This Second Edition ■ Incorporate the latest and best methods used by professional statisticians.

■ Include features that address all of the recommendations included in the Guide- lines for Assessment and Instruction in Statistics Education (GAISE) as recom- mended by the American Statistical Association.

■ Provide an abundance of new and interesting data sets, examples, and exercises.

■ Foster personal growth of students through critical thinking, use of technology, collaborative work, and development of communication skills.

■ Enhance teaching and learning with the most extensive and best set of supple- ments and digital resources.

Audience ,Prerequisites Biostatistics for the Health and Biological Sciences is written for students major- ing in the biological and health sciences, and it is designed for a wide variety of students taking their first statistics course. Algebra is used minimally, and calculus is not required. It is recommended that students have completed at least an elemen- tary algebra course or that students should learn the relevant algebra components through an integrated or co-requisite course. In many cases, underlying theory is included, but this book does not require the mathematical rigor more appropriate for mathematics majors.

Hallmark Features Great care has been taken to ensure that each chapter of Biostatistics for the Health and Biological Sciences will help students understand the concepts presented. The following features are designed to help meet that objective.

Real Data

Hundreds of hours have been devoted to finding data that are real, meaningful, and interesting to students. Fully 87% of the examples are based on real data, and 89% of the exercises are based on real data. Some exercises refer to the 18 data sets listed in Appendix B, and 12 of those data sets are new to this edition. Exercises requiring use of the Appendix B data sets are located toward the end of each exercise set and are marked with a special data set icon .

Real data sets are included throughout the book to provide relevant and interesting real-world statistical applications, including biometric security, body measurements, brain sizes and IQ scores, and data from births. Appendix B includes descriptions of

vii

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viii Preface

the 18 data sets that can be downloaded from the companion website www.pearson- highered.com/triola, the author maintained www.TriolaStats.com and MyStatLab.

TriolaStats.com includes downloadable data sets in formats for technologies including Excel, Minitab, JMP, SPSS, and TI@83>84 Plus calculators. The data sets are also included in the free Statdisk software, which is also available on the website.

Readability

Great care, enthusiasm, and passion have been devoted to creating a book that is readable, understandable, interesting, and relevant. Students pursuing any major in the biological, life, medical, or health fields are sure to find applications related to their future work.

Website

This textbook is supported by www.TriolaStats.com, and www.pearsonhighered.com/ triola which are continually updated to provide the latest digital resources, including:

■ Statdisk: A free, robust statistical software package designed for this book.

■ Downloadable Appendix B data sets in a variety of technology formats.

■ Downloadable textbook supplements including Glossary of Statistical Terms and Formulas and Tables.

■ Online instructional videos created specifically for this book that provide step- by-step technology instructions.

■ Triola Blog, which highlights current applications of statistics, statistics in the news, and online resources.

Chapter Features

Chapter Opening Features

■ Chapters begin with a Chapter Problem that uses real data and motivates the chapter material.

■ Chapter Objectives provide a summary of key learning goals for each section in the chapter.

Exercises

Many exercises require the interpretation of results. Great care has been taken to ensure their usefulness, relevance, and accuracy. Exercises are arranged in order of increasing difficulty, and they begin with Basic Skills and Concepts. Most sections include additional Beyond the Basics exercises that address more difficult concepts or require a stronger mathematical background. In a few cases, these exercises introduce a new concept.

End-of-Chapter Features

■ Chapter Quick Quiz provides review questions that require brief answers.

■ Review Exercises offer practice on the chapter concepts and procedures.

■ Cumulative Review Exercises reinforce earlier material.

■ Technology Project provides an activity that can be used with a variety of technologies.

■ From Data to Decision is a capstone problem that requires critical thinking and writing.

■ Cooperative Group Activities encourage active learning in groups.

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Preface ix

Other Features

Margin Essays There are 57 margin essays designed to highlight real-world topics and foster student interest.

Flowcharts The text includes flowcharts that simplify and clarify more complex con- cepts and procedures. Animated versions of the text’s flowcharts are available within MyStatLab and MathXL.

Quick-Reference Endpapers Tables A-2 and A-3 (the normal and t distributions) are reproduced on the rear inside cover pages.

Detachable Formula and Table Card This insert, organized by chapter, gives students a quick reference for studying, or for use when taking tests (if allowed by the instruc- tor). It also includes the most commonly used tables. This is also available for download at www.TriolaStats.com, www.pearsonhighered.com/triola and in MyStatLab.

Technology Integration

As in the preceding edition, there are many displays of screens from technology through- out the book, and some exercises are based on displayed results from technology. Where appropriate, sections include a reference to an online Tech Center subsection that in- cludes detailed instructions for Statdisk, Minitab®, Excel®, StatCrunch, or a TI@83>84 Plus® calculator. (Throughout this text, “TI-83>84 Plus” is used to identify a TI-83 Plus or TI-84 Plus calculator). The end-of-chapter features include a Technology Project.

The Statdisk statistical software package is designed specifically for this textbook and contains all Appendix B data sets. Statdisk is free to users of this book, and it can be downloaded at www.statdisk.org.

Changes in This Edition New Features

Chapter Objectives provide a summary of key learning goals for each section in the chapter.

Larger Data Sets: Some of the data sets in Appendix B are much larger than in the previous edition. It is no longer practical to print all of the Appendix B data sets in this book, so the data sets are described in Appendix B, and they can be downloaded at www.TriolaStats.com, www.pearsonhighered.com/triola, and MyStatLab.

New Content: New examples, new exercises, and Chapter Problems provide relevant and interesting real-world statistical applications, including biometric security, drug testing, gender selection, and analyzing ultrasound images.

Number New to This Edition Use Real Data

Exercises 1600 85% 89%

Examples 200 84% 87%

Major Organization Changes

All Chapters

■ New Chapter Objectives: All chapters now begin with a list of key learning goals for that chapter. Chapter Objectives replaces the former Overview numbered sec- tions. The first numbered section of each chapter now covers a major topic.

Chapter 1

■ New Section 1-1: Statistical and Critical Thinking

■ New Subsection 1-3, Part 2: Big Data and Missing Data: Too Much and Not Enough

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x Preface

Chapters 2 and 3

■ Chapter Partitioned: Chapter 2 (Describing, Exploring, and Comparing Data) from the first edition has been partitioned into Chapter 2 (Summarizing and Graph- ing) and Chapter 3 (Statistics for Describing, Exploring, and Comparing Data).

■ New Section 2-4: Scatterplots, Correlation, and Regression This new section includes scatterplots in Part 1, the linear correlation coefficient r in Part 2, and linear regression in Part 3. These additions are intended to greatly facilitate cover- age for those professors who prefer some early coverage of correlation and regres- sion concepts. Chapter 10 includes these topics discussed with much greater detail.

Chapter 4

■ Combined Sections: Section 3-3 (Addition Rule) and Section 3-4 (Multiplication Rule) from the first edition are now combined into one section: 4-2 (Addition Rule and Multiplication Rule).

■ New Subsection 4-3, Part 3: Bayes’ Theorem

Chapter 5

■ Combined Sections: Section 4-3 (Binomial Probability Distributions) and Section 4-4 (Mean, Variance, and Standard Deviation for the Binomial Distribu- tion) from the first edition are now combined into one section: 5-2 (Binomial Probability Distributions).

Chapter 6

■ Switched Sections: Section 6-5 (Assessing Normality) now precedes Section 6-6 (Normal as Approximation to Binomial).

Chapter 7

■ Combined Sections: Sections 6-4 (Estimating a Population Mean: s Known) and 6-5 (Estimating a Population Mean: s Not Known) from the first edition have been combined into one section: 7-2 (Estimating a Population Mean). The coverage of the s known case has been substantially reduced and it is now lim- ited to Part 2 of Section 7-2.

■ New Section 7-4: Bootstrapping: Using Technology for Estimates

Chapter 8

■ Combined Sections: Sections 7-4 (Testing a Claim About a Population Mean: s Known) and 7-5 (Testing a Claim About a Population Mean: s Not Known) from the first edition have been combined into one section: 8-3 (Testing a Claim About a Mean). Coverage of the s known case has been substantially reduced and it is now limited to Part 2 of Section 8-3.

Chapter 10

■ New Section: 10-5 Dummy Variables and Logistic Regression

Chapter 11

■ New Subsection: Section 11-2, Part 2 Test of Homogeneity, Fisher’s Exact Test, and McNemar’s Test for Matched Pairs

Chapter 14

■ Combined Sections: Section 13-2 (Elements of a Life Table) and Section 13-3 (Applications of Life Tables) from the first edition have been combined into Section 14-1 (Life Tables).

■ New Section: 14-2 Kaplan-Meier Survival Analysis

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Preface xi

Flexible Syllabus This book’s organization reflects the preferences of most statistics instructors, but there are two common variations:

■ Early Coverage of Correlation and Regression: Some instructors prefer to cover the basics of correlation and regression early in the course. Section 2-4 now includes basic concepts of scatterplots, correlation, and regression without the use of formulas and greater depth found in Sections 10-1 (Correlation) and 10-2 (Regression).

■ Minimum Probability: Some instructors prefer extensive coverage of probability, while others prefer to include only basic concepts. Instructors preferring mini- mum coverage can include Section 4-1 while skipping the remaining sections of Chapter 4, as they are not essential for the chapters that follow. Many instructors prefer to cover the fundamentals of probability along with the basics of the addi- tion rule and multiplication rule (Section 4-2).

GAISE This book reflects recommendations from the American Statistical Association and its Guidelines for Assessment and Instruction in Statistics Education (GAISE). Those guidelines suggest the following objectives and strategies.

1. Emphasize statistical literacy and develop statistical thinking: Each section exercise set begins with Statistical Literacy and Critical Thinking exercises. Many of the book’s exercises are designed to encourage statistical thinking rather than the blind use of mechanical procedures.

2. Use real data: 87% of the examples and 89% of the exercises use real data.

3. Stress conceptual understanding rather than mere knowledge of procedures: Instead of seeking simple numerical answers, most exercises and examples involve conceptual understanding through questions that encourage practical interpretations of results. Also, each chapter includes a From Data to Decision project.

4. Foster active learning in the classroom: Each chapter ends with several Cooperative Group Activities.

5. Use technology for developing conceptual understanding and analyzing data: Computer software displays are included throughout the book. Special Tech Center subsections are available online, and they include instruction for using the software. Each chapter includes a Technology Project. When there are dis- crepancies between answers based on tables and answers based on technology, Appendix D provides both answers. The websites www.TriolaStats.com and www.pearsonhighered.com/triola as well as MyStatLab include free text-specific software (Statdisk), data sets formatted for several different technologies, and instructional videos for technologies.

6. Use assessments to improve and evaluate student learning: Assessment tools include an abundance of section exercises, Chapter Quick Quizzes, Review Exercises, Cumulative Review Exercises, Technology Projects, From Data to Decision projects, and Cooperative Group Activities.

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xii Preface

Acknowledgments We would like to thank the many statistics professors and students who have contrib- uted to the success of this book. We thank the reviewers for their suggestions for this second edition:

James Baldone, Virginia College Naomi Brownstein, Florida State University Christina Caruso, University of Guelph Erica A. Corbett, Southeastern Oklahoma State University Xiangming Fang, East Carolina University Phil Gona, UMASS Boston Sharon Homan, University of North Texas Jackie Milton, Boston University Joe Pick, Palm Beach State College Steve Rigdon, St. Louis University Brian Smith, Black Hills State University Mahbobeh Vezvaei, Kent State University David Zeitler, Grand Valley State University

We also thank Paul Lorczak, Joseph Pick and Erica Corbett for their help in checking the accuracy of the text and answers.

Marc Triola Mario Triola

Jason Roy September 2016

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MyStatLab® Online Course for Biostatistics: For the Biological and Health Sciences, 2e by Marc M. Triola, Mario F. Triola and Jason Roy (access code required) MyStatLab is available to accompany Pearson’s market leading text offerings. To give students a consistent tone, voice, and teaching method each text’s flavor and ap- proach is tightly integrated throughout the accompanying MyStatLab course, making learning the material as seamless as possible.

Real-World Data Examples - Help understand how statistics applies to everyday life through the extensive current, real-world data examples and exercises provided throughout the text.

MathXL coverage - MathXL is a market-leading text-specific autograded homework system built to improve student learning outcomes.

Enhanced video program to meet Introductory Statistics needs: • New! Tech-Specific Video Tutorials - These

short, topical videos address how to use varying technologies to complete exercises.

• Updated! Section Lecture Videos - Watch author, Marty Triola, work through examples and elaborate on key objectives of the chapter.

Resources for Success

www.mystatlab.com xiii

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xiv Preface

Supplements For the Student

Student’s Solutions Manual, by James Lapp (Colorado Mesa University) provides detailed, worked-out solutions to all odd-numbered text exercises. (ISBN-13: 978-0-13-403909-1; ISBN-10: 0-13-403909-2)

Student Workbook for the Triola Statistics Series, by Laura lossi (Broward College) offers additional exam- ples, concept exercises, and vocabulary exercises for each chapter. (ISBN-13: 978-0-13-446423-7; ISBN 10: 0-13-446423-0)

The following technology manuals, available in MyStatLab, include instructions, examples from the main text, and interpretations to complement those given in the text.

Excel Student Laboratory Manual and Workbook (Download Only), by Laurel Chiappetta (University of Pittsburgh). (ISBN-13: 978-0-13-446427-5; ISBN-10: 0-13-446427-3)

MINITAB Student Laboratory Manual and Work- book (Download Only), by Mario F. Triola. (ISBN-13: 978-0-13-446418-3; ISBN-10: 0-13-446418-4)

Graphing Calculator Manual for the TI-83 Plus, TI-84 Plus, TI-84 Plus C and TI-84 Plus CE (Down- load Only), by Kathleen McLaughlin (University of Connecticut) & Dorothy Wakefield (University of Con- necticut Health Center). (ISBN-13: 978-0-13-446414-5; ISBN 10: 0-13-446414-1)

Statdisk Student Laboratory Manual and Workbook (Download Only), by Mario F. Triola. These files are available to instructors and students through the Triola Sta- tistics Series website, www.pearsonhighered.com/triola, and MyStatLab.

SPSS Student Laboratory Manual and Workbook (Download Only), by James J. Ball (Indiana State Uni- versity). These files are available to instructors and stu- dents through the Triola Statistics Series website, www. pearsonhighered.com/triola, and MyStatLab.

For the Instructor

Instructor’s Solutions Manual (Download Only), by James Lapp (Colorado Mesa University) contains so- lutions to all the exercises. These files are available to qualified instructors through Pearson Education’s on- line catalog at www.pearsonhighered.com/irc or within MyStatLab.

Insider’s Guide to Teaching with the Triola Statistics Series, by Mario F. Triola, contains sample syllabi and tips for incorporating projects, as well as lesson overviews, extra examples, minimum outcome objectives, and recom- mended assignments for each chapter. (ISBN-13: 978-0-13-446425-1; ISBN-10: 0-13-446425-7)

TestGen® Computerized Test Bank (www.pearsoned. com/testgen) enables instructors to build, edit, print, and administer tests using a computerized bank of questions developed to cover all the objectives of the text. TestGen is algorithmically based, allowing instructors to create mul- tiple but equivalent versions of the same question or test with the click of a button. Instructors can also modify test bank questions or add new questions. The software and tes- tbank are available for download from Pearson Education’s online catalog at www.pearsonhighered.com. A test bank (Download Only) is also available from the online catalog.

Learning Catalytics: Learning Catalytics is a web-based engagement and assessment tool. As a “bring-your-own- device” direct response system, Learning Catalytics offers a diverse library of dynamic question types that allow stu- dents to interact with and think critically about statistical concepts. As a real-time resource, instructors can take ad- vantage of critical teaching moments both in the classroom or through assignable and gradeable homework.

Technology Resources The following resources can be found on the Triola Statistics Series website (http://www.pearsonhighered.com/triola), the author maintained www.triolastats.com, and MyStatLab

■ Appendix B data sets formatted for Minitab, SPSS, SAS, Excel, JMP, and as text files. Additionally, these data sets are available as an APP for the TI-83>84 Plus calculators, and supplemental programs for the TI-83>84 Plus calculator are also available.

■ Statdisk statistical software instructions for down- load. New features include the ability to directly use lists of data instead of requiring the use of their sum- mary statistics.

■ Extra data sets, an index of applications, and a sym- bols table.

Video resources have been expanded, updated and now supplement most sections of the book, with many topics presented by the author. The videos aim to support both instructors and students through lecture, reinforcing sta- tistical basics through technology, and applying concepts:

■ Section Lecture Videos

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Preface xv

■ New! Technology Video Tutorials - These short, topical videos address how to use Excel, Statdisk, and the TI graphing calculator to complete exercises.

■ StatTalk Videos: 24 Conceptual Videos to Help You Actually Understand Statistics. Fun-loving statistician Andrew Vickers takes to the streets of Brooklyn, NY, to demonstrate important statistical concepts through interesting stories and real-life events. These fun and engaging videos will help students actually understand statistical concepts. Available with an instructors user guide and assess- ment questions.

MyStatLab™ Online Course (access code required) MyStatLab is a course management system that delivers proven results in helping individual students succeed.

■ MyStatLab can be successfully implemented in any environment—lab-based, hybrid, fully online, traditional—and demonstrates the quantifiable differ- ence that integrated usage has on student retention, subsequent success, and overall achievement.

■ MyStatLab’s comprehensive online gradebook au- tomatically tracks students’ results on tests, quizzes, homework, and in the study plan. Instructors can use the gradebook to provide positive feedback or inter- vene if students have trouble. Gradebook data can be easily exported to a variety of spreadsheet programs, such as Microsoft Excel. You can determine which points of data you want to export, and then analyze the results to determine success.

MyStatLab provides engaging experiences that personal- ize, stimulate, and measure learning for each student. In addition to the resources below, each course includes a full interactive online version of the accompanying textbook.

■ Tutorial Exercises with Multimedia Learning Aids: The homework and practice exercises in MyStatLab align with the exercises in the textbook, and they regenerate algorithmically to give students unlim- ited opportunity for practice and mastery. Exercises offer immediate helpful feedback, guided solutions, sample problems, animations, videos, and eText clips for extra help at point-of-use.

■ Getting Ready for Statistics: A library of questions now appears within each MyStatLab course to offer the developmental math topics students need for the course. These can be assigned as a prerequisite to other assignments, if desired.

■ Conceptual Question Library: In addition to algo- rithmically regenerated questions that are aligned with

your textbook, there is a library of 1000 Conceptual Questions available in the assessment manager that re- quire students to apply their statistical understanding.

■ StatCrunch™: MyStatLab integrates the web-based statistical software, StatCrunch, within the online as- sessment platform so that students can easily analyze data sets from exercises and the text. In addition, MyStatLab includes access to www.StatCrunch.com, a website where users can access more than 15,000 shared data sets, conduct online surveys, perform complex analyses using the powerful statistical software, and generate compelling reports.

■ Statistical Software Support: Knowing that students often use external statistical software, we make it easy to copy our data sets, both from the ebook and the MyStatLab questions, into software such as StatCrunch, Minitab, Excel, and more. Students have access to a variety of support tools—Technology Tutorial Videos, Technology Study Cards, and Tech- nology Manuals for select titles—to learn how to effectively use statistical software.

MathXL® for Statistics Online Course (access code required) MathXL® is the homework and assessment engine that runs MyStatLab. (MyStatLab is MathXL plus a learning management system.)

With MathXL for Statistics, instructors can:

■ Create, edit, and assign online homework and tests using algorithmically generated exercises correlated at the objective level to the textbook.

■ Create and assign their own online exercises and import TestGen tests for added flexibility.

■ Maintain records of all student work, tracked in MathXL’s online gradebook.

With MathXL for Statistics, students can:

■ Take chapter tests in MathXL and receive personal- ized study plans and>or personalized homework assignments based on their test results.

■ Use the study plan and>or the homework to link directly to tutorial exercises for the objectives they need to study.

■ Students can also access supplemental animations and video clips directly from selected exercises.

■ Knowing that students often use external statistical software, we make it easy to copy our data sets, both from the ebook and the MyStatLab questions, into software like StatCrunch™, Minitab, Excel, and more.

A01_TRIO9015_02_SE_FM_i-xvi.indd 15 03/11/16 4:02 PM

xvi Preface

MathXL for Statistics is available to qualified adopters. For more information, visit our website at www.mathxl .com, or contact your Pearson representative.

StatCrunch™ StatCrunch is powerful, web-based statistical software that allows users to perform complex analyses, share data sets, and generate compelling reports. A vibrant online community offers more than 15,000 data sets for students to analyze.

■■ Collect. Users can upload their own data to StatCrunch or search a large library of publicly shared data sets, spanning almost any topic of interest. Also, an online survey tool allows users to quickly collect data via web-based surveys.

■■ Crunch. A full range of numerical and graphical methods allow users to analyze and gain insights from any data set. Interactive graphics help users understand statistical concepts and are available for export to enrich reports with visual representations of data.

■■ Communicate. Reporting options help users create a wide variety of visually appealing representations of their data.

Full access to StatCrunch is available with MyStatLab and StatCrunch is available by itself to qualified adopt- ers. StatCrunch Mobile is now available to access from your mobile device. For more information, visit our web- site at www.StatCrunch.com, or contact your Pearson representative.

Minitab® 17 and Minitab Express™ make learning sta- tistics easy and provide students with a skill-set that’s in demand in today’s data driven workforce. Bundling Minitab® software with educational materials ensures stu- dents have access to the software they need in the class- room, around campus, and at home. And having 12 month versions of Minitab 17 and Minitab Express available ensures students can use the software for the duration of their course. ISBN 13: 978-0-13-445640-9 ISBN 10: 0-13-445640-8 (Access Card only; not sold as stand alone.)

JMP Student Edition, Version 12 is an easy-to-use, stream- lined version of JMP desktop statistical discovery software from SAS Institute, Inc., and is available for bundling with the text. (ISBN-13: 978-0-13-467979-2 ISBN-10: 0-13-467979-2)

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Statistical and Critical Thinking

Types of Data

Collecting Sample Data

1-1

1-2

1-3

Survey Question: Do You Need Caffeine to Start Up Your Brain for the Day?CHAPTER PROBLEM

Introduction to Statistics1

Surveys provide data that enable us to improve products or

services. Surveys guide political candidates, shape business

practices, identify effective medical treatments, and affect

many aspects of our lives. Surveys give us insight into the

opinions and behaviors of others. As an example, the National

Health and Nutrition Examination Survey (NHANES) is part

1

of a research program that studies the health and nutrition of

thousands of adults and children in the United States.

Let’s consider one USA Today survey in which respondents

were asked if they need caffeine to start up their brain for the

day. Among 2,006 respondents, 74% said that they did need the

caffeine. Figure 1-1 includes graphs that depict these results.

M01_TRIO9015_02_SE_C01_001-039.indd 1 02/08/16 4:46 PM

The survey results suggest that people overwhelmingly need caffeine to start up their brains

for the day. The graphs in Figure 1-1 visually depict the survey results. One of the most impor-

tant objectives of this book is to encourage the use of critical thinking so that such results are

not blindly accepted. We might question whether the survey results are valid. Who conducted

the survey? How were respondents selected? Do the graphs in Figure 1-1 depict the results

well, or are those graphs somehow misleading?

The survey results presented here have major flaws that are among the most common, so

they are especially important to recognize. Here are brief descriptions of each of the major flaws:

Flaw 1: Misleading Graphs The bar chart in Figure 1-1(a) is very deceptive. By using a vertical scale that does not start at zero, the difference between the two percentages is grossly

exaggerated. Figure 1-1(a) makes it appear that approximately eight times as many people

need the caffeine. However, with 74% needing caffeine and 26% not needing caffeine, the

ratio is actually about 3:1, rather than the 8:1 ratio that is suggested by the graph.

The illustration in Figure 1-1(b) is also deceptive. Again, the difference between the actual

response rates of 74% (needing caffeine) and 26% (not needing caffeine) is a difference that

is grossly distorted. The picture graph (or “pictograph”) in Figure 1-1(b) makes it appear that

2 CHAPTER 1 Introduction to Statistics

FIGURE 1-1(a) Survey Results

FIGURE 1-1(b) Survey Results

People Needing Ca�eine to Start Up Brain for the Day

People Not Needing Ca�eine to Start Up Brain for the Day

M01_TRIO9015_02_SE_C01_001-039.indd 2 02/08/16 4:46 PM

the ratio of people needing caffeine to people not needing caffeine is roughly 9:1 instead of

the correct ratio of about 3:1. (Objects with area or volume can distort perceptions because

they can be drawn to be disproportionately larger or smaller than the data indicate.) Decep-

tive graphs are discussed in more detail in Section 2-3, but we see here that the illustrations in

Figure 1-1 grossly exaggerate the number of people needing caffeine.

Flaw 2: Bad Sampling Method The aforementioned survey responses are from a USA Today survey of Internet users. The survey question was posted on a website and Internet

users decided whether to respond. This is an example of a voluntary response sample—a

sample in which respondents themselves decide whether to participate. With a voluntary

response sample, it often happens that those with a strong interest in the topic are more likely

to participate, so the results are very questionable. For example, people who strongly feel that

they cannot function without their morning cup(s) of coffee might be more likely to respond to

the caffeine survey than people who are more ambivalent about caffeine or coffee. When using

sample data to learn something about a population, it is extremely important to obtain sample

data that are representative of the population from which the data are drawn. As we proceed

through this chapter and discuss types of data and sampling methods, we should focus on

these key concepts:

• Sample data must be collected in an appropriate way, such as through a process of

random selection.

• If sample data are not collected in an appropriate way, the data may be so completely

useless that no amount of statistical torturing can salvage them.

It would be easy to accept the preceding survey results and blindly proceed with calcula-

tions and statistical analyses, but we would miss the critical two flaws described above. We

could then develop conclusions that are fundamentally wrong and misleading. Instead, we

should develop skills in statistical thinking and critical thinking so that we are better prepared

to analyze such data.

Chapter Objectives 3

The single most important concept presented in this chapter is this: When using meth- ods of statistics with sample data to form conclusions about a population, it is absolutely essential to collect sample data in a way that is appropriate. Here are the main chapter objectives:

Statistical and Critical Thinking

• Analyze sample data relative to context, source, and sampling method. • Understand the difference between statistical significance and practical significance. • Define and identify a voluntary response sample and know that statistical conclu-

sions based on data from such a sample are generally not valid.

1-1

CHAPTER OBJECTIVES

> > >

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4 CHAPTER 1 Introduction to Statistics

Because populations are often very large, a common objective of the use of statis- tics is to obtain data from a sample and then use those data to form a conclusion about the population.

Types of Data

• Distinguish between a parameter and a statistic. • Distinguish between quantitative data and categorical (or qualitative or attribute) data. • Distinguish between discrete data and continuous data. • Determine whether basic statistical calculations are appropriate for a particular data set.

Collecting Sample Data

• Define and identify a simple random sample. • Understand the importance of sound sampling methods and the importance of

good design of experiments.

1-2

1-3

Types of Data

• Distinguish between a parameter and a parameter and a parameter statistic. • Distinguish between quantitative data and categorical (or categorical (or categorical qualitative or attribute) data. • Distinguish between discrete data and continuous data. • Determine whether basic statistical calculations are appropriate for a particular data set.

Collecting Sample Data

• Define and identify a simple random sample. • Understand the importance of sound sampling methods and the importance of

good design of experiments.

Key Concept In this section we begin with a few very basic definitions, and then we consider an overview of the process involved in conducting a statistical study. This process consists of “prepare, analyze, and conclude.” “Preparation” involves consid- eration of the context, the source of data, and sampling method. In future chapters we construct suitable graphs, explore the data, and execute computations required for the statistical method being used. In future chapters we also form conclusions by deter- mining whether results have statistical significance and practical significance.

Statistical thinking involves critical thinking and the ability to make sense of results. Statistical thinking demands so much more than the ability to execute complicated cal- culations. Through numerous examples, exercises, and discussions, this text will help you develop the statistical thinking skills that are so important in today’s world.

We begin with some very basic definitions.

1-1 Statistical and Critical Thinking

DEFINITIONS

Data are collections of observations, such as measurements, or survey responses. (A single data value is called a datum, a term rarely used. The term “data” is plural, so it is correct to say “data are…” not “data is…”)

Statistics is the science of planning studies and experiments; obtaining data; and organizing, summarizing, presenting, analyzing, and interpreting those data and then drawing conclusions based on them.

A population is the complete collection of all measurements or data that are be- ing considered. Typically, the population is the complete collection of data that we would like to make inferences about.

A census is the collection of data from every member of the population.

A sample is a subcollection of members selected from a population.

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1-1 Statistical and Critical Thinking 5

We now proceed to consider the process involved in a statistical study. See Figure 1-2 for a summary of this process and note that the focus is on critical thinking, not mathe- matical calculations. Thanks to wonderful developments in technology, we have power- ful tools that effectively do the number crunching so that we can focus on understanding and interpreting results.

EXAMPLE 1 Residential Carbon Monoxide Detectors

In the journal article “Residential Carbon Monoxide Detector Failure Rates in the United States” (by Ryan and Arnold, American Journal of Public Health, Vol. 101, No. 10), it was stated that there are 38 million carbon monoxide detectors installed in the United States. When 30 of them were randomly selected and tested, it was found that 12 of them failed to provide an alarm in hazardous carbon monoxide conditions. In this case, the population and sample are as follows:

Population: All 38 million carbon monoxide detectors in the United States

Sample: The 30 carbon monoxide detectors that were selected and tested

The objective is to use the sample data as a basis for drawing a conclusion about the population of all carbon monoxide detectors, and methods of statistics are helpful in drawing such conclusions.

Conclude 1. Significance

• Do the results have statistical significance? • Do the results have practical significance?

Analyze 1. Graph the Data 2. Explore the Data

• Are there any outliers (numbers very far away from almost all of the other data)? • What important statistics summarize the data (such as the mean and standard deviation described in Chapter 3)? • How are the data distributed? • Are there missing data? • Did many selected subjects refuse to respond?

3. Apply Statistical Methods • Use technology to obtain results.

Prepare 1. Context

• What do the data represent? • What is the goal of study?

2. Source of the Data • Are the data from a source with a special interest so that there is pressure to obtain results that are favorable to the source?

3. Sampling Method • Were the data collected in a way that is unbiased, or were the data collected in a

way that is biased (such as a procedure in which respondents volunteer to participate)?

FIGURE 1-2 Statistical Thinking

Survivorship Bias

In World War

II, statisti-

cian Abraham

Wald saved

many lives

with his work

on the Applied

Mathematics Panel. Military

leaders asked the panel how they

could improve the chances of

aircraft bombers returning after

missions. They wanted to add

some armor for protection, and

they recorded locations on the

bombers where damaging holes

were found. They reasoned that

armor should be placed in loca-

tions with the most holes, but

Wald said that strategy would be

a big mistake. He said that armor

should be placed where returning

bombers were not damaged. His

reasoning was this: The bombers

that made it back with damage

were survivors, so the damage

they suffered could be survived.

Locations on the aircraft that

were not damaged were the most

vulnerable, and aircraft suffer-

ing damage in those vulnerable

areas were the ones that did

not make it back. The military

leaders would have made a big

mistake with survivorship bias by

studying the planes that survived

instead of thinking about the

planes that did not survive.

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6 CHAPTER 1 Introduction to Statistics

Prepare Context Figure 1-2 suggests that we begin our preparation by considering the context of the data, so let’s start with context by considering the data in Table 1-1. (The data are from Data Set 9 “IQ and Brain Size” in Appendix B.) The data in Table 1-1 consist of measured IQ scores and measured brain volumes from 10 different subjects. The data are matched in the sense that each individual “IQ>brain volume” pair of values is from the same person. The first subject had a measured IQ score of 96 and a brain volume of 1005 cm3. The format of Table 1-1 suggests the following goal: Determine whether there is a relationship between IQ score and brain volume. This goal suggests a possible hypothesis: People with larger brains tend to have higher IQ scores.

Source of the Data The data in Table 1-1 were provided by M. J. Tramo, W. C. Loftus, T. A. Stukel, J. B. Weaver, and M. S. Gazziniga, who discuss the data in the article “Brain Size, Head Size, and IQ in Monozygotic Twins,” Neurology, Vol. 50. The researchers are from reputable medical schools and hospitals, and they would not gain by presenting the results in way that is misleading. In contrast, Kiwi Brands, a maker of shoe polish, commissioned a study that resulted in this statement, which was printed in some newspapers: “According to a nationwide survey of 250 hiring profes- sionals, scuffed shoes was the most common reason for a male job seeker’s failure to make a good first impression.”

When physicians who conduct clinical experiments on the efficacy of drugs re- ceive funding from drug companies, they have an incentive to obtain favorable results. Some professional journals, such as the Journal of the American Medical Association, now require that physicians report sources of funding in journal articles. We should be skeptical of studies from sources that may be biased.

Sampling Method Figure 1-2 suggests that we conclude our preparation by consid- ering the sampling method. The data in Table 1-1 were obtained from subjects whose medical histories were reviewed in an effort to ensure that no subjects had neurologic or psychiatric disease. In this case, the sampling method appears to be sound, but we cannot be sure of that without knowing how the subjects were recruited and whether any payments may have affected participation in the study.

Sampling methods and the use of randomization will be discussed in Section 1-3, but for now, we stress that a sound sampling method is absolutely essential for good results in a statistical study. It is generally a bad practice to use voluntary response (or self-selected) samples, even though their use is common.

TABLE 1-1 IQ Scores and Brain Volumes (cm3)

IQ 96 87 101 103 127 96 88 85 97 124

Brain Volume (cm3) 1005 1035 1281 1051 1034 1079 1104 1439 1029 1160

DEFINITION

A voluntary response sample (or self-selected sample) is one in which the respondents themselves decide whether to be included.

The following types of polls are common examples of voluntary response samples. By their very nature, all are seriously flawed because we should not make conclusions about a population on the basis of samples with a strong possibility of bias:

■ Internet polls, in which people online can decide whether to respond

■ Mail-in polls, in which people decide whether to reply

Origin of “Statistics”

The word

statistics is

derived from

the Latin word

status (mean-

ing “state”).

Early uses of

statistics involved compilations

of data and graphs describing

various aspects of a state or

country. In 1662, John Graunt

published statistical information

about births and deaths. Graunt’s

work was followed by studies

of mortality and disease rates,

population sizes, incomes, and

unemployment rates. House-

holds, governments, and busi-

nesses rely heavily on statistical

data for guidance. For example,

unemployment rates, inflation

rates, consumer indexes, and

birth and death rates are carefully

compiled on a regular basis,

and the resulting data are used

by business leaders to make

decisions affecting future hiring,

production levels, and expansion

into new markets.

M01_TRIO9015_02_SE_C01_001-039.indd 6 02/08/16 4:46 PM

1-1 Statistical and Critical Thinking 7

■ Telephone call-in polls, in which newspaper, radio, or television announcements ask that you voluntarily call a special number to register your opinion

The Chapter Problem involves a USA Today survey with a voluntary response sample. See also the following Example 2.

EXAMPLE 2 Voluntary Response Sample

USA Today posted this question on the electronic edition of their newspaper: “Have you ever been bitten by an animal?” Internet users who saw that question then de- cided themselves whether to respond. Among the 2361 responses, 65% said “yes” and 35% said “no.” Because the 2361 subjects themselves chose to respond, they are a voluntary response sample and the results of the survey are highly question- able. It would be much better to get results through a poll in which the pollster ran- domly selects the subjects, instead of allowing the subjects to volunteer themselves.

Analyze Figure 1-2 indicates that after completing our preparation by considering the context, source, and sampling method, we begin to analyze the data.

Graph and Explore An analysis should begin with appropriate graphs and explora- tions of the data. Graphs are discussed in Chapter 2, and important statistics are dis- cussed in Chapter 3.

Apply Statistical Methods Later chapters describe important statistical methods, but application of these methods is often made easy with technology (calculators and>or statistical software packages). A good statistical analysis does not require strong computational skills. A good statistical analysis does require using common sense and paying careful attention to sound statistical methods.

Conclude Figure 1-2 shows that the final step in our statistical process involves conclusions, and we should develop an ability to distinguish between statistical significance and practi- cal significance.

Statistical Significance Statistical significance is achieved in a study when we get a result that is very unlikely to occur by chance. A common criterion is that we have statistical significance if the likelihood of an event occurring by chance is 5% or less.

■ Getting 98 girls in 100 random births is statistically significant because such an extreme outcome is not likely to result from random chance.

■ Getting 52 girls in 100 births is not statistically significant because that event could easily occur with random chance.

Practical Significance It is possible that some treatment or finding is effective, but common sense might suggest that the treatment or finding does not make enough of a difference to justify its use or to be practical, as illustrated in Example 3 which follows.

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8 CHAPTER 1 Introduction to Statistics

Analyzing Data: Potential Pitfalls Here are a few more items that could cause problems when analyzing data.

Misleading Conclusions When forming a conclusion based on a statistical analy- sis, we should make statements that are clear even to those who have no understand- ing of statistics and its terminology. We should carefully avoid making statements not justified by the statistical analysis. For example, later in this book we introduce the concept of a correlation, or association between two variables, such as smoking and pulse rate. A statistical analysis might justify the statement that there is a cor- relation between the number of cigarettes smoked and pulse rate, but it would not justify a statement that the number of cigarettes smoked causes a person’s pulse rate to change. Such a statement about causality can be justified by physical evidence, not by statistical analysis.

Correlation does not imply causation.

Sample Data Reported Instead of Measured When collecting data from people, it is better to take measurements yourself instead of asking subjects to report results. Ask people what they weigh and you are likely to get their desired weights, not their actual weights. People tend to round, usually down, sometimes way down. When asked, someone with a weight of 187 lb might respond that he or she weighs 160 lb. Accurate weights are collected by using a scale to measure weights, not by asking people what they weigh.

Loaded Questions If survey questions are not worded carefully, the results of a study can be misleading. Survey questions can be “loaded” or intentionally worded to elicit a desired response. Here are the actual rates of “yes” responses for the two dif- ferent wordings of a question:

97% yes: “Should the President have the line item veto to eliminate waste?”

57% yes: “Should the President have the line item veto, or not?”

Order of Questions Sometimes survey questions are unintentionally loaded by such factors as the order of the items being considered. See the following two

EXAMPLE 3 Statistical Significance Versus Practical Significance

ProCare Industries once supplied a product named Gender Choice that supposedly increased the chance of a couple having a baby with the gender that they desired. In the absence of any evidence of its effectiveness, the product was banned by the Food and Drug Administration (FDA) as a “gross deception of the consumer.” But suppose that the product was tested with 10,000 couples who wanted to have baby girls, and the results consist of 5200 baby girls born in the 10,000 births. This re- sult is statistically significant because the likelihood of it happening due to chance is only 0.003%, so chance doesn’t seem like a feasible explanation. That 52% rate of girls is statistically significant, but it lacks practical significance because 52% is only slightly above 50%. Couples would not want to spend the time and money to increase the likelihood of a girl from 50% to 52%. (Note: In reality, the likelihood of a baby being a girl is about 48.8%, not 50%.)

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1-1 Statistical and Critical Thinking 9

questions from a poll conducted in Germany, along with the very different response rates:

“Would you say that tra¦c contributes more or less to air pollution than indus- try?” (45% blamed tra¦c; 27% blamed industry.)

“Would you say that industry contributes more or less to air pollution than traf- fic?” (24% blamed tra¦c; 57% blamed industry.)

In addition to the order of items within a question, as illustrated above, the order of separate questions could also affect responses.

Nonresponse A nonresponse occurs when someone either refuses to respond to a survey question or is unavailable. When people are asked survey questions, some firmly refuse to answer. The refusal rate has been growing in recent years, partly be- cause many persistent telemarketers try to sell goods or services by beginning with a sales pitch that initially sounds as though it is part of an opinion poll. (This “selling under the guise” of a poll is called sugging.) In Lies, Damn Lies, and Statistics, author Michael Wheeler makes this very important observation:

People who refuse to talk to pollsters are likely to be different from those who do not. Some may be fearful of strangers and others jealous of their privacy, but their refusal to talk demonstrates that their view of the world around them is markedly different from that of those people who will let poll-takers into their homes.

Percentages Some studies cite misleading or unclear percentages. Note that 100% of some quantity is all of it, but if there are references made to percentages that exceed 100%, such references are often not justified. If a medical researcher claims that she has developed a treatment for migraine headaches and the treatment results in a 150% reduction in those headaches, that researcher cannot be correct, because totally elimi- nating all migraine headaches would be a 100% reduction. It is impossible to reduce the number of migraine headaches by more than 100%.

When working with percentages, we should know that % or “percent” really means “divided by 100.” Here is a principle used often in this book.

Percentage of: To find a percentage of an amount, replace the % symbol with division by 100, and then interpret “of” to be multiplication. The following calculation shows that 6% of 1200 is 72:

6% of 1200 responses = 6

100 * 1200 = 72

Statistical Literacy and Critical Thinking

1. Online Medical Info USA Today posted this question on its website: “How often do you seek medical information online?” Of 1072 Internet users who chose to respond, 38% of them responded with “frequently.” What term is used to describe this type of survey in which the people surveyed consist of those who decided to respond? What is wrong with this type of sampling method?

2. Reported Versus Measured In a survey of 1046 adults conducted by Bradley Corpora- tion, subjects were asked how often they wash their hands when using a public restroom, and 70% of the respondents said “always.”

a. Identify the sample and the population.

b. Why would better results be obtained by observing the hand washing instead of asking about it?

1-1 Basic Skills and Concepts

questions from a poll conducted in Germany, along with the very different response Publication Bias

There is a “pub-

lication bias”

in professional

journals. It is

the tendency to

publish positive

results (such

as showing that some treatment

is effective) much more often

than negative results (such as

showing that some treatment has

no effect). In the article “Regis-

tering Clinical Trials” (Journal of

the American Medical Asso-

ciation, Vol. 290, No. 4), authors

Kay Dickersin and Drummond

Rennie state that “the result of

not knowing who has performed

what (clinical trial) is loss and

distortion of the evidence, waste

and duplication of trials, inability

of funding agencies to plan, and

a chaotic system from which

only certain sponsors might

benefit, and is invariably against

the interest of those who offered

to participate in trials and of

patients in general.” They sup-

port a process in which all clinical

trials are registered in one central

system, so that future research-

ers have access to all previous

studies, not just the studies that

were published.

as showing that some treatment

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10 CHAPTER 1 Introduction to Statistics

3. Statistical Significance Versus Practical Significance When testing a new treatment, what is the difference between statistical significance and practical significance? Can a treat- ment have statistical significance, but not practical significance?

4. Correlation One study showed that for a recent period of 11 years, there was a strong cor- relation (or association) between the numbers of people who drowned in swimming pools and the amounts of power generated by nuclear power plants (based on data from the Centers for Disease Control and Prevention and the Department of Energy). Does this imply that increas- ing power from nuclear power plants is the cause of more deaths in swimming pools? Why or why not?

Consider the Source. In Exercises 5–8, determine whether the given source has the potential to create a bias in a statistical study.

5. Physicians Committee for Responsible Medicine The Physicians Committee for Re- sponsible Medicine tends to oppose the use of meat and dairy products in our diets, and that organization has received hundreds of thousands of dollars in funding from the Foundation to Support Animal Protection.

6. Arsenic in Rice Amounts of arsenic in samples of rice grown in Texas were measured by the Food and Drug Administration (FDA).

7. Brain Size A data set in Appendix B includes brain volumes from 10 pairs of monozygotic (identical) twins. The data were collected by researchers at Harvard University, Massachusetts General Hospital, Dartmouth College, and the University of California at Davis.

8. Chocolate An article in Journal of Nutrition (Vol. 130, No. 8) noted that chocolate is rich in flavonoids. The article notes “regular consumption of foods rich in flavonoids may reduce the risk of coronary heart disease.” The study received funding from Mars, Inc., the candy com- pany, and the Chocolate Manufacturers Association.

Sampling Method. In Exercises 9–12, determine whether the sampling method appears to be sound or is flawed.

9. Nuclear Power Plants In a survey of 1368 subjects, the following question was posted on the USA Today website: “In your view, are nuclear plants safe?” The survey subjects were Internet users who chose to respond to the question posted on the electronic edition of USA Today.

10. Clinical Trials Researchers at Yale University conduct a wide variety of clinical trials by using subjects who volunteer after reading advertisements soliciting paid volunteers.

11. NHANES Examinations In a recent year, the National Health and Nutrition Examina- tion Survey (NHANES), sponsored by the National Center for Health Statistics, selected more than 9000 subjects who were given physical exams. Subjects were selected through a somewhat complicated procedure designed to obtain results that are representative of the population.

12. Health In a survey of 3014 randomly selected U.S. adults, 45% reported that they have at least one chronic health condition, such as diabetes or high blood pressure. The survey was conducted by Princeton Survey Research Associates International.

Statistical Significance and Practical Significance. In Exercises 13–16, determine whether the results appear to have statistical significance, and also determine whether the results appear to have practical significance.

13. Diet and Exercise Program In a study of the Kingman diet and exercise program, 40  subjects lost an average of 22 pounds. There is about a 1% chance of getting such results with a program that has no effect.

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1-1 Statistical and Critical Thinking 11

14. MCAT The Medical College Admissions Test (MCAT) is commonly used as part of the decision-making process for determining which students to accept into medical schools. To test the effectiveness of the Siena MCAT preparation course, 16 students take the MCAT test, then they complete the preparatory course, and then they retake the MCAT test, with the result that the average (mean) score for this group rises from 25 to 30. There is a 0.3% chance of getting those results by chance. Does the course appear to be effective?

15. Gender Selection In a study of the Gender Aide method of gender selection used to increase the likelihood of a baby being born a girl, 2000 users of the method gave birth to 980 boys and 1020 girls. There is about a 19% chance of getting that many girls if the method had no effect.

16. Systolic Blood Pressure High systolic blood pressure is 140 mm Hg or higher. (Normal values are less than 120 mm Hg, and prehypertension levels are between 120 mm Hg and 139  mm Hg.) Subjects with high blood pressure are encouraged to take action to lower it. A pharmaceutical company develops a new medication designed to lower blood pressure, and tests on 25 subjects result in an average (mean) decrease of 2 mm Hg. Analysis of the results shows that there is a 15% chance of getting such results if the medication has no effect.

In Exercises 17–20, refer to the sample of body temperatures (degrees Fahrenheit) in the table below. (The body temperatures are recorded on the same day from a sample of five randomly selected males listed in a data set in Appendix B.)

Subject

1 2 3 4 5

8 AM 97.0 98.5 97.6 97.7 98.7

12 AM 97.6 97.8 98.0 98.4 98.4

17. Context of the Data Refer to the table of body temperatures. Is there some meaning- ful way in which each body temperature recorded at 8 AM is matched with the 12 AM temperature?

18. Source The listed body temperatures were obtained from Dr. Steven Wasserman, Dr. Philip Mackowiak, and Dr. Myron Levine, who were researchers at the University of Maryland. Is the source of the data likely to be biased?

19. Conclusion Given the body temperatures in the table, what issue can be addressed by con- ducting a statistical analysis of the data?

20. Conclusion If we analyze the listed body temperatures with suitable methods of statistics, we conclude that when the differences are found between the 8 AM body temperatures and the 12 AM body temperatures, there is a 64% chance that the differences can be explained by random results obtained from populations that have the same 8 AM and 12 AM body tempera- tures. What should we conclude about the statistical significance of those differences?

In Exercises 21–24, refer to the data in the table below. The entries are white blood cell counts (1000 cells ,ML) and red blood cell counts (million cells ,ML) from male subjects examined as part of a large health study conducted by the National Center for Health Statis- tics. The data are matched, so that the first subject has a white blood cell count of 8.7 and a red blood cell count of 4.91, and so on.

Subject

1 2 3 4 5

White 8.7 5.9 7.3 6.2 5.9

Red 4.91 5.59 4.44 4.80 5.17

continued

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12 CHAPTER 1 Introduction to Statistics

21. Context Given that the data (on the bottom of the preceding page) are matched and consid- ering the units of the data, does it make sense to use the difference between each white blood cell count and the corresponding red blood cell count? Why or why not?

22. Analysis Given the context of the data in the table (on the bottom of the preceding page), what issue can be addressed by conducting a statistical analysis of the measurements?

23. Source of the Data Considering the source of the data (on the bottom of the preceding page), does that source appear to be biased in some way?

24. Conclusion If we analyze the sample data (on the bottom of the preceding page) and conclude that there is a correlation between white and red blood cell counts, does it follow that higher white are the cause of higher red blood cell counts?

What’s Wrong? In Exercises 25–28, identify what is wrong.

25. Potatoes In a poll sponsored by the Idaho Potato Commission, 1000 adults were asked to select their favorite vegetables, and the favorite choice was potatoes, which were selected by 26% of the respondents.

26. Healthy Water In a USA Today online poll, 951 Internet users chose to respond, and 57% of them said that they prefer drinking bottled water instead of tap water.

27. Cheese and Bedsheet Deaths In recent years, there has been a strong correlation be- tween per capita consumption of cheese in the United States and the numbers of people who died from being tangled in their bedsheets. Really. Therefore, consumption of cheese causes bedsheet entanglement fatalities.

28. Smokers The electronic cigarette maker V2 Cigs sponsored a poll showing that 55% of smokers surveyed say that they feel ostracized “sometimes,” “often,” or “always.”

Percentages. In Exercises 29 and 30, answer the given questions, which are related to percentages.

29. Health It was noted in Exercise 12 “Health” that in a survey of 3014 randomly selected U.S. adults, 45% reported that they have at least one chronic health condition, such as diabetes or high blood pressure.

a. What is 45% of 3014 adults?

b. Could the result from part (a) be the actual number of survey subjects who have at least one chronic condition?

c. What is the actual number of survey subjects who have at least one chronic condition?

d. Among those surveyed, 1808 were called by landline and 1206 were called by cell phone. What percentage of the survey subjects were called by cell phone?

30. Chillax USA Today reported results from a Research Now for Keurig survey in which 1458 men and 1543 women were asked this: “In a typical week, how often can you kick back and relax?”

a. Among the women, 19% responded with “rarely, if ever.” What is the exact value that is 19% of the number of women surveyed?

b. Could the result from part (a) be the actual number of women who responded with “rarely, if ever”? Why or why not?

c. What is the actual number of women who responded with “rarely, if ever”?

d. Among the men who responded, 219 responded with “rarely, if ever.” What is the percentage of men who responded with “rarely, if ever”?

e. Consider the question that the subjects were asked. Is that question clear and unambiguous so that all respondents will interpret the question the same way? How might the survey be improved?

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1-2 Types of Data 13

If we have more than one statistic, we have “statistics.” Another meaning of “statis- tics” was given in Section 1-1, where we defined statistics to be the science of plan- ning studies and experiments; obtaining data; organizing, summarizing, presenting, analyzing, and interpreting those data; and then drawing conclusions based on them. We now have two different definitions of statistics, but we can determine which of these two definitions applies by considering the context in which the term statistics is used, as in the following example.

31. What’s Wrong with This Picture? The Newport Chronicle ran a survey by asking read- ers to call in their response to this question: “Do you support a ban on electronic cigarettes, which foster smoking among our children?” It was reported that 20 readers responded and that 87% said “no,” while 13% said “yes.” Identify four major flaws in this survey.

32. Falsifying Data A researcher at the Sloan-Kettering Cancer Research Center was once criticized for falsifying data. Among his data were figures obtained from 6 groups of mice, with 20 individual mice in each group. The following values were given for the percentage of successes in each group: 53%, 58%, 63%, 46%, 48%, 67%. What’s wrong with those values?

1-1 Beyond the Basics

DEFINITIONS

A parameter is a numerical measurement describing some characteristic of a population.

A statistic is a numerical measurement describing some characteristic of a sample.

HINT The alliteration in “population parameter” and “sample statistic” helps us remember the meanings of these terms.

EXAMPLE 1 Parameter ,Statistic There are 17,246,372 high school students in the United States. In a study of 8505 U.S. high school students 16 years of age or older, 44.5% of them said that they texted while driving at least once during the previous 30 days (based on data in

Key Concept A major use of statistics is to collect and use sample data to make con- clusions about populations. We should know and understand the meanings of the terms statistic and parameter, as defined below. In this section we describe a few different types of data. The type of data is one of the key factors that determine the statistical methods we use in our analysis.

In Part 1 of this section we describe the basics of different types of data, and then in Part 2 we consider “big data” and missing data.

PA RT 1 Basic Types of Data Parameter ,Statistic

1-2 Types of Data

continued

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14 CHAPTER 1 Introduction to Statistics

Quantitative ,Categorical Some data are numbers representing counts or measurements (such as a systolic blood pressure of 118 mm Hg), whereas others are attributes (such as eye color of green or brown) that are not counts or measurements. The terms quantitative data and cat- egorical data distinguish between these types.

“Texting While Driving and Other Risky Motor Vehicle Behaviors Among U.S. High School Students,” by Olsen, Shults, Eaton, Pediatrics, Vol. 131, No. 6).

1. Parameter: The population size of all 17,246,372 high school students is a parameter, because it is the size of the entire population of all high school students in the United States. If we somehow knew the percentage of all 17,246,372 high school students who reported they had texted while driving, that percentage would also be a parameter.

2. Statistic: The value of 44.5% is a statistic, because it is based on the sample, not on the entire population.

DEFINITIONS

Quantitative (or numerical) data consist of numbers representing counts or mea- surements.

Categorical (or qualitative or attribute) data consist of names or labels (not num- bers that represent counts or measurements).

CAUTION Categorical data are sometimes coded with numbers, with those num- bers replacing names. Although such numbers might appear to be quantitative, they are actually categorical data. See the third part of Example 2.

Include Units of Measurement With quantitative data, it is important to use the appropriate units of measurement, such as dollars, hours, feet, or meters. We should carefully observe information given about the units of measurement, such as “all amounts are in thousands of dollars,” or “all units are in kilograms.” Ignoring such units of measurement can be very costly. The National Aeronautics and Space Admin- istration (NASA) lost its $125 million Mars Climate Orbiter when the orbiter crashed because the controlling software had acceleration data in English units, but they were incorrectly assumed to be in metric units.

EXAMPLE 2 Quantitative ,Categorical 1. Quantitative Data: The ages (in years) of subjects enrolled in a clinical trial

2. Categorical Data as Labels: The genders (male>female) of subjects enrolled in a clinical trial

3. Categorical Data as Numbers: The identification numbers 1, 2, 3, . . . , 25 are assigned randomly to the 25 subjects in a clinical trial. Those numbers are substitutes for names. They don’t measure or count anything, so they are categorical data, not quantitative data.

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1-2 Types of Data 15

Discrete ,Continuous Quantitative data can be further described by distinguishing between discrete and con- tinuous types.

DEFINITIONS

Discrete data result when the data values are quantitative and the number of values is finite or “countable.” (If there are infinitely many values, the collection of values is countable if it is possible to count them individually, such as the number of tosses of a coin before getting tails or the number of births in Houston before getting a male.)

Continuous (numerical) data result from infinitely many possible quantitative values, where the collection of values is not countable. (That is, it is impossible to count the individual items because at least some of them are on a continuous scale, such as the lengths of distances from 0 cm to 12 cm.)

CAUTION The concept of countable data plays a key role in the preceding defini- tions, but it is not a particularly easy concept to understand. Continuous data can be measured, but not counted. If you select a particular data value from continuous data, there is no “next” data value. See Example 3.

Continuous Data Discrete Data

EXAMPLE 3 Discrete ,Continuous 1. Discrete Data of the Finite Type: Each of several physicians plans to count

the number of physical examinations given during the next full week. The data are discrete data because they are finite numbers, such as 27 and 46 that result from a counting process.

2. Discrete Data of the Infinite Type: Researchers plan to test the accuracy of a blood typing test by repeating the process of submitting a sample of the same blood (Type O+) until the test yields an error. It is possible that each research- er could repeat this test forever without ever getting an error, but they can still count the number of tests as they proceed. The collection of the numbers of tests is countable, because you can count them, even though the counting could go on forever.

3. Continuous Data: When the typical patient has blood drawn as part of a routine examination, the volume of blood drawn is between 0 mL and 50 mL. There are infinitely many values between 0 mL and 50 mL. Because it is im- possible to count the number of di®erent possible values on such a continuous scale, these amounts are continuous data.

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16 CHAPTER 1 Introduction to Statistics

Levels of Measurement Another common way of classifying data is to use four levels of measurement: nomi- nal, ordinal, interval, and ratio, all defined below. (Also see Table 1-2 for brief de- scriptions of the four levels of measurements.) When we are applying statistics to real problems, the level of measurement of the data helps us decide which procedure to use. There will be references to these levels of measurement in this book, but the important point here is based on common sense: Don’t do computations and don’t use statistical methods that are not appropriate for the data. For example, it would not make sense to compute an average (mean) of Social Security numbers, because those numbers are data used for identification, and they don’t represent measurements or counts of anything.

GRAMMAR: FEWER VERSUS LESS When describing smaller amounts, it is correct grammar to use “fewer” for discrete amounts and “less” for continuous amounts. It is correct to say that we drank fewer cans of cola and that, in the pro- cess, we drank less cola. The numbers of cans of cola are discrete data, whereas the volume amounts of cola are continuous data.

DEFINITION

The nominal level of measurement is characterized by data that consist of names, labels, or categories only. It is not possible to arrange the data in some order (such as low to high).

EXAMPLE 4 Nominal Level

Here are examples of sample data at the nominal level of measurement.

1. Yes ,No ,Undecided: Survey responses of yes, no, and undecided 2. Coded Survey Responses: For an item on a survey, respondents are given a

choice of possible answers, and they are coded as follows: “I agree” is coded as 1; “I disagree” is coded as 2; “I don’t care” is coded as 3; “I refuse to answer” is coded as 4; “Go away and stop bothering me” is coded as 5. The numbers 1, 2, 3, 4, 5 don’t measure or count anything.

Because nominal data lack any ordering or numerical significance, they should not be used for calculations. Numbers such as 1, 2, 3, and 4 are sometimes assigned to the different categories (especially when data are coded for computers), but these numbers have no real computational significance and any average (mean) calculated from them is meaningless and possibly misleading.

DEFINITION

Data are at the ordinal level of measurement if they can be arranged in some order, but differences (obtained by subtraction) between data values either cannot be determined or are meaningless.

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1-2 Types of Data 17

Ordinal data provide information about relative comparisons, but not the magni- tudes of the differences. Usually, ordinal data should not be used for calculations such as an average (mean), but this guideline is sometimes disregarded (such as when we use letter grades to calculate a grade-point average).

EXAMPLE 5 Ordinal Level

Here is an example of sample data at the ordinal level of measurement.

Course Grades: A biostatistics professor assigns grades of A, B, C, D, or F. These grades can be arranged in order, but we can’t determine differences between the grades. For example, we know that A is higher than B (so there is an ordering), but we cannot subtract B from A (so the difference cannot be found).

DEFINITION

Data are at the interval level of measurement if they can be arranged in order, and differences between data values can be found and are meaningful; but data at this level do not have a natural zero starting point at which none of the quantity is present.

EXAMPLE 6 Interval Level

These examples illustrate the interval level of measurement.

1. Temperatures: Body temperatures of 98.2°F and 98.8°F are examples of data at this interval level of measurement. Those values are ordered, and we can determine their di®erence of 0.6°F. However, there is no natural starting point. The value of 0°F might seem like a starting point, but it is arbitrary and does not represent the total absence of heat.

2. Years: The years 1492 and 1776 can be arranged in order, and the di®erence of 284 years can be found and is meaningful. However, time did not begin in the year 0, so the year 0 is arbitrary instead of being a natural zero starting point representing “no time.”

DEFINITION

Data are at the ratio level of measurement if they can be arranged in order, differ- ences can be found and are meaningful, and there is a natural zero starting point (where zero indicates that none of the quantity is present). For data at this level, dif- ferences and ratios are both meaningful.

EXAMPLE 7 Ratio Level

The following are examples of data at the ratio level of measurement. Note the pres- ence of the natural zero value, and also note the use of meaningful ratios of “twice” and “three times.”

1. Heights of Students: Heights of 180 cm and 90 cm for a high school student and a preschool student (0 cm represents no height, and 180 cm is twice as tall as 90 cm.)

2. Class Times: The times of 50 min and 100 min for a statistics class (0 min represents no class time, and 100 min is twice as long as 50 min.)

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18 CHAPTER 1 Introduction to Statistics

See Table 1-2 for brief descriptions of the four levels of measurements.

TABLE 1-2 Levels of Measurement

Level of Measurement

Brief Description

Example

Ratio There is a natural zero starting point and ratios make sense.

Heights, lengths, distances, volumes

Interval Differences are meaningful, but there is no natural zero starting point and ratios are meaningless.

Body temperatures in degrees Fahrenheit or Celsius

Ordinal Data can be arranged in order, but dif- ferences either can’t be found or are meaningless.

Ranks of colleges in U.S. News & World Report

Nominal Categories only. Data cannot be arranged in order.

Eye colors

HINT The distinction between the interval and ratio levels of measurement can be a bit tricky. Here are two tools for help with that distinction:

1. Ratio Test Focus on the term “ratio” and know that the term “twice” describes the ratio of one value to be double the other value. To distinguish between the interval and ratio levels of measurement, use a “ratio test” by asking this question: Does use of the term “twice” make sense? “Twice” makes sense for data at the ratio level of measurement, but it does not make sense for data at the interval level of mea- surement.

2. True Zero For ratios to make sense, there must be a value of “true zero,” where the value of zero indicates that none of the quantity is present, and zero is not simply an arbitrary value on a scale. The temperature of 0°F is arbitrary and does not indicate that there is no heat, so temperatures on the Fahrenheit scale are at the interval level of measurement, not the ratio level.

EXAMPLE 8 Distinguishing Between the Ratio Level and Interval Level

For each of the following, determine whether the data are at the ratio level of mea- surement or the interval level of measurement:

a. Times (minutes) it takes to complete a statistics test.

b. Body temperatures (Celsius) of statistics students.

S O L U T I O N

a. Apply the “ratio test” described in the preceding hint. If one student completes the test in 40 minutes and another student completes the test in 20 min, does it make sense to say that the first student used twice as much time? Yes! So the times are at the ratio level of measurement. Also, a time of 0 minutes does repre- sent “no time,” so the value of 0 is a true zero indicating that no time was used.

b. Apply the “ratio test” described in the preceding hint. If one student has a body temperature of 40°C and another student has a body temperature of 20°C, does it make sense to say that the first student is twice as hot as the

TA

Survey Pitfalls

Surveys con-

stitute a huge

and growing

business in the

United States,

but survey

results can be

compromised by many factors.

A growing number of people

refuse to respond; the average

response rate is now about 22%,

compared to 36% around the

year 2000. A growing number of

people are more difficult to reach

because they use cell phones

(no directories); about 15% of

adults now have cell phones and

no landlines, and they tend to

be younger than average. There

are obvious problems associated

with surveys that ask respon-

dents about drug use, theft, or

sexual behavior, and a social

desirability bias occurs when sur-

vey respondents are not honest

because they don’t want to be

viewed negatively by the person

conducting the interview.

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1-2 Types of Data 19

PA RT 2 Big Data and Missing Data: Too Much and Not Enough

When working with data, we might encounter some data sets that are excessively large, and we might also encounter some data sets with individual elements missing. Here in Part 2 we briefly discuss both cases.

Big Data Edward Snowden used his employment at the NSA (National Security Agency) to re- veal substantial top secret documents that led to the realization that the NSA was con- ducting telephone and Internet surveillance of U.S. citizens as well as world leaders. The NSA was collecting massive amounts of data that were analyzed in an attempt to prevent terrorism. Monitoring telephone calls and Internet communications is made possible with modern technology. The NSA can compile big data, and such ginormous data sets have led to the birth of data science. There is not universal agreement on the following definitions, and various other definitions can be easily found elsewhere.

second student? (Ignore subjective amounts of attractiveness and consider only science.) No! So the body temperatures are not at the ratio level of measurement. Because the di®erence between 40°C and 20°C is the same as the di®erence between 90°C and 70°C, the di®erences are meaningful, but be- cause ratios do not make sense, the body temperatures are at the interval level of measurement. Also, the temperature of 0°C does not represent “no heat” so the value of 0 is not a true zero indicating that no heat is present.

DEFINITIONS

Big data refers to data sets so large and so complex that their analysis is beyond the capabilities of traditional software tools. Analysis of big data may require soft- ware simultaneously running in parallel on many different computers.

Data science involves applications of statistics, computer science, and software en- gineering, along with some other relevant fields (such as biology and epidemiology).

Examples of Data Set Magnitudes We can see from the above definition of big data that there isn’t a fixed number that serves as an exact boundary for determining whether a data set qualifies as being big data, but big data typically involves amounts of data such as the following.

■ Terabytes (1012 or 1,000,000,000,000 bytes) of data

■ Petabytes (1015 bytes) of data

■ Exabytes (1018 bytes) of data

■ Zettabytes (1021 bytes) of data

■ Yottabytes (1024 bytes) of data

Examples of Applications of Big Data The following are a few examples involv- ing big data:

■ Attempt to forecast flu epidemics by analyzing Internet searches of flu symptoms.

■ The Spatio Temporal Epidemiological Modeler developed by IBM is providing a means for using a variety of data that are correlated with disease data.

the di®erence between 90°C and 70°C, the di®erences are meaningful, but be- cause ratios do not make sense, the body temperatures are at the interval level of measurement. Also, the temperature of 0°C does not represent “no heat” so

Big Data Instead of a Clinical Trial

Nicholas

Tatonetti of

Columbia

University

searched Food

and Drug

Administration

databases for

adverse reactions in patients that

resulted from different pairings

of drugs. He discovered that

the paroxetine drug for depres-

sion and the pravastatin drug

for high cholesterol interacted

to create increases in glucose

(blood sugar) levels. When taken

separately by patients, neither

drug raised glucose levels, but

the increase in glucose levels

occurred when the two drugs

were taken together. This finding

resulted from a general database

search of interactions from many

pairings of drugs, not from a

clinical trial involving patients

using Paxil and pravastatin.

continued

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20 CHAPTER 1 Introduction to Statistics

■ A National Electronic Disease Surveillance System is used to monitor disease trends and identify outbreaks of infectious disease.

■ Google provides live traffic maps by recording and analyzing GPS (global posi- tioning system) data collected from the smartphones of people traveling in their vehicles.

■ Amazon monitors and tracks 1.4 billion items in its store that are distributed across hundreds of fulfillment centers around the world.

Examples of Jobs According to Analytic Talent, there are 6000 companies hiring data scientists, and here are some job posting examples:

■ Facebook: Data Scientist

■ IBM: Data Scientist

■ PayPal: Data Scientist

■ The College Board: SAS Programmer>Data Scientist ■ Netflix: Senior Data Engineer>Scientist

Statistics in Data Science The modern data scientist has a solid background in statistics and computer systems as well as expertise in fields that extend beyond sta- tistics. The modern data scientist might be skilled with Hadoop software, which uses parallel processing on many computers for the analysis of big data. The modern data scientist might also have a strong background in some other field, such as psychology, biology, medicine, chemistry, or economics. Because of the wide range of disciplines required, a data science project might typically involve a team of collaborating indi- viduals with expertise in different fields. An introductory statistics course is a great first step in becoming a data scientist.

Missing Data When collecting sample data, it is quite common to find that some values are miss- ing. Ignoring missing data can sometimes create misleading results. If you make the mistake of skipping over a few different sample values when you are manually typ- ing them into a statistics software program, the missing values are not likely to have a serious effect on the results. However, if a survey includes many missing salary en- tries because those with very low incomes are reluctant to reveal their salaries, those missing low values will have the serious effect of making salaries appear higher than they really are.

For an example of missing data, see the following table. The body temperature for Subject 2 at 12 AM on day 2 is missing. (The table below includes the first three rows of data from Data Set 2 “Body Temperatures” in Appendix B.)

Body Temperatures (in degrees Fahrenheit) of Healthy Adults

Temperature Day 1

Temperature Day 2

Subject Age Sex Smoke 8 AM 12 AM 8 AM 12 AM

1 22 M Y 98.0 98.0 98.0 98.6

2 23 M Y 97.0 97.6 97.4 ----

3 22 M Y 98.6 98.8 97.8 98.6

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1-2 Types of Data 21

There are different categories of missing data. See the following definitions.

DEFINITION

A data value is missing completely at random if the likelihood of its being miss- ing is independent of its value or any of the other values in the data set. That is, any data value is just as likely to be missing as any other data value.

(Note: More complete discussions of missing data will distinguish between missing completely at random and missing at random, which means that the likelihood of a value being missing is independent of its value after controlling for another variable. There is no need to know this distinction in this book.)

Example of Missing Data—Random When using a keyboard to manually enter ages of survey respondents, the operator is distracted by a colleague singing “Day- dream Believer” and makes the mistake of failing to enter the age of 37 years. This data value is missing completely at random.

DEFINITION

A data value is missing not at random if the missing value is related to the reason that it is missing.

Example of Missing Data—Not at Random A survey question asks each respon- dent to enter his or her annual income, but respondents with very low incomes skip this question because they find it embarrassing.

Biased Results? Based on the above two definitions and examples, it makes sense to conclude that if we ignore data missing completely at random, the remaining values are not likely to be biased and good results should be obtained. However, if we ignore data that are missing not at random, it is very possible that the remaining values are biased and results will be misleading.

Correcting for Missing Data There are different methods for dealing with missing data.

1. Delete Cases: One very common method for dealing with missing data is to delete all subjects having any missing values.

■ If the data are missing completely at random, the remaining values are not likely to be biased and good results can be obtained, but with a smaller sam- ple size.

■ If the data are missing not at random, deleting subjects having any missing values can easily result in a bias among the remaining values, so results can be misleading.

2. Impute Missing Values: We impute missing data values when we substitute values for them. There are different methods of determining the replacement values, such as using the mean of the other values, or using a randomly selected value from other similar cases, or using a method based on regression analysis (which will make more sense after studying Chapter 10).

Measuring Disobedience

How are data

collected about

something that

doesn’t seem

to be measur-

able, such as

people’s level of

disobedience?

Psychologist Stanley Milgram

devised the following experi-

ment: A researcher instructed a

volunteer subject to operate a

control board that gave increas-

ingly painful “electrical shocks”

to a third person. Actually, no real

shocks were given, and the third

person was an actor. The volun-

teer began with 15 volts and was

instructed to increase the shocks

by increments of 15 volts. The

disobedience level was the point

at which the subject refused to

increase the voltage. Surpris-

ingly, two-thirds of the subjects

obeyed orders even when the

actor screamed and faked a

heart attack.

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22 CHAPTER 1 Introduction to Statistics

In this book we do not work much with missing data, but it is important to under- stand this:

When analyzing sample data with missing values, try to determine why they are missing, and then decide whether it makes sense to treat the remaining values as being representative of the population. If it appears that there are missing values that are missing not at random (that is, their values are related to the reasons why they are missing), know that the remaining data may well be biased and any conclusions based on those remaining values may well be misleading.

Statistical Literacy and Critical Thinking

1. Health Survey In a survey of 1020 adults in the United States, 44% said that they wash their hands after riding public transportation (based on data from KRC Research).

a. Identify the sample and the population.

b. Is the value of 44% a statistic or a parameter?

2. Health Survey For the same survey from Exercise 1, answer the following.

a. What is the level of measurement of the value of 44%? (nominal, ordinal, interval, ratio)

b. Are the numbers of subjects in such surveys discrete or continuous?

c. The responses are “yes,” “no,” “not sure,” or “refused to answer.” Are these responses quan- titative data or categorical data?

3. Quantitative ,Categorical Data Identify each of the following as quantitative data or cat- egorical data.

a. The platelet counts of exam subjects in Data Set 1 “Body Data” in Appendix B

b. The names of the pharmaceutical companies that manufacture aspirin tablets

c. The colors of pills

d. The weights of aspirin tablets

4. Discrete ,Continuous Data Which of the following describe discrete data? a. The numbers of people surveyed in each of the next several National Health and Nutrition Examination Surveys

b. The exact foot lengths (cm) of a random sample of statistics students

c. The exact times that randomly selected drivers spend texting while driving during the past 7 days

In Exercises 5–12, identify whether the given value is a statistic or a parameter.

5. Brain Volume The average (mean) volume of the brains included in Data Set 9 “IQ and Brain Size” in Appendix B is 1126.0 cm3.

6. CHIS A recent California Health Interview Survey (CHIS) included 2799 adolescent resi- dents of California.

7. Cigarettes A data set in Appendix B includes measurements from 25 king-size cigarettes, and the average (mean) amount of nicotine in those 25 cigarettes is 1.26 mg.

8. Triangle Fire Fatalities A deadly disaster in the United States was the Triangle Shirtwaist Factory Fire in New York City. A population of 146 garment workers died in that fire.

1-2 Basic Skills and Concepts

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1-2 Types of Data 23

9. Birth Weight In a study of 400 babies born at four different hospitals in New York State, it was found that the average (mean) weight at birth was 3152.0 grams.

10. Birth Genders In the same study cited in the preceding exercise, 51% of the babies were girls.

11. Titanic A study was conducted of all 2223 passengers aboard the Titanic when it sank.

12. Periodic Table The average (mean) atomic weight of all elements in the periodic table is 134.355 unified atomic mass units.

In Exercises 13–20, determine whether the data are from a discrete or continuous data set.

13. Freshman 15 In a study of weight gains by college students in their freshman year, re- searchers record the amounts of weight gained by randomly selected students (as in Data Set 10 “Freshman 15” in Appendix B).

14. Births Data Set 3 “Births” in Appendix B includes the length of stay (in days) for each baby in a sample of babies born in New York State. The first few values are 2, 2, 36, 5, and 2.

15. CHIS Among the subjects surveyed as part of the California Health Interview Survey (CHIS), several subjects are randomly selected and their heights are recorded.

16. Arm Circumference From Data Set 1 “Body Data” in Appendix B we see that a female had an arm circumference of 32.49 cm.

17. Families A sample of married couples is randomly selected and the number of children in each family is recorded.

18. Criminal Forensics When studying the relationship between lengths of feet and heights so that footprint evidence at a crime scene can be used to estimate the height of the suspect, a researcher records the exact lengths of feet from a large sample of random subjects.

19. Stitch In Time The Emergency Room of the Albany Medical Center records the numbers of stitches used for patients in a week.

20. Texting Fatalities The Insurance Institute for Highway Safety collects data consisting of the numbers of motor vehicle fatalities caused by driving while texting.

In Exercises 21–28, determine which of the four levels of measurement (nominal, ordinal, interval, ratio) is most appropriate.

21. Brain Volumes Volumes (cm3) of brains listed in Data Set 9 “IQ and Brain Size” in Appendix B

22. Blood Lead Level Blood lead levels of low, medium, and high used to describe the sub- jects in Data Set 8 “IQ and Lead” in Appendix B

23. Body Temperatures Body temperatures (in degrees Fahrenheit) listed in Data Set 2 “Body Temperatures” in Appendix B.

24. Privacy Codes Instead of using actual names, subjects included in the National Health and Nutrition Examination Survey are coded with consecutive numbers.

25. Hospitals A research project on the effectiveness of heart transplants begins with a compi- lation of the U.S. hospitals that provide heart transplants.

26. Hospital Charges A research project on the effectiveness of heart transplants begins with a compilation of the charges (dollars) for heart transplant procedures that were conducted within the past year.

27. Physician Ranks A research project on the effectiveness of heart transplants includes rankings (scale of 1, 2, 3, 4, 5) of physicians who perform those procedures.

28. Pharmaceuticals Pfizer records the years in which new products were launched, beginning with 1849.

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24 CHAPTER 1 Introduction to Statistics

In Exercises 29–32, identify the level of measurement of the data as nominal, ordinal, inter- val, or ratio. Also, explain what is wrong with the given calculation.

29. Hospital ID The four hospitals included in Data Set 3 “Births” in Appendix B are coded as follows: Albany Medical Center (1); Bellevue Hospital Center (1438); Olean General Hospital (66); Strong Memorial Hospital (413). The average (mean) of those numbers is 479.5.

30. Social Security Numbers As part of a clinical study, the Social Security number of each subject is recorded and the average (mean) of the individual digits is computed to be 4.7.

31. Temperatures A person has a body temperature of 98.0°F during the time when the out- side air temperature is 49.0°F, so the person is twice as warm as the outside air.

32. Medical School Ranks As of this writing, U.S. News & World Report ranked medical schools, including these results: Harvard (1), Stanford (2), Johns Hopkins (3), University of California at San Francisco (4), and University of Pennsylvania (5). The difference between Harvard and Stanford is the same as the difference between Johns Hopkins and University of California at San Francisco.

33. Countable For each of the following, categorize the nature of the data using one of these three descriptions: (1) discrete because the number of possible values is finite; (2) discrete because the number of possible values is infinite but countable; (3) continuous because the number of possible values is infinite and not countable.

a. Exact lengths of the feet of members of the band the Monkees

b. Shoe sizes of members of the band the Monkees (such as 9, 9½, and so on)

c. The number of albums sold by the Monkees band

d. The numbers of monkeys sitting at keyboards before one of them randomly types the lyrics for the song “Daydream Believer”

1-2 Beyond the Basics

Key Concept When using statistics in a study, planning is very important, and it is essential to use an appropriate method for collecting the sample data. This section includes comments about various methods and sampling procedures. Of particular im- portance is the method of using a simple random sample. We will make frequent use of this sampling method throughout the remainder of this book.

As you read this section, remember this:

If sample data are not collected in an appropriate way, the data may be so utterly useless that no amount of statistical torturing can salvage them.

PA RT 1 Basics of Design of Experiments and Collecting Sample Data

The Gold Standard Randomization with placebo>treatment groups is sometimes called the “gold standard” because it is so effective. (A placebo such as a sugar pill has no medicinal effect.) The following example describes how the gold standard was used in the largest health experiment ever conducted.

1-3 Collecting Sample Data

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1-3 Collecting Sample Data 25

Example 1 describes an experiment because subjects were given a treatment, but ethi- cal, cost, time, and other considerations sometimes prohibit the use of an experiment. We would never want to conduct a driving/texting experiment in which we ask sub- jects to text while driving—some of them could die. It would be far better to observe past crash results to understand the effects of driving while texting. See the following definitions.

EXAMPLE 1 The Salk Vaccine Experiment

In 1954, an experiment was designed to test the effectiveness of the Salk vaccine in preventing polio, which had killed or paralyzed thousands of children. By random selection, 401,974 children were randomly assigned to two groups: (1) 200,745 children were given a treatment consisting of Salk vaccine injections; (2) 201,229 children were injected with a placebo that contained no drug. Children were as- signed to the treatment or placebo group through a process of random selection, equivalent to flipping a coin. Among the children given the Salk vaccine, 33 later developed paralytic polio, and among the children given a placebo, 115 later devel- oped paralytic polio.

DEFINITIONS

In an experiment, we apply some treatment and then proceed to observe its effects on the individuals. (The individuals in experiments are called experimental units, and they are often called subjects when they are people.)

In an observational study, we observe and measure specific characteristics, but we don’t attempt to modify the individuals being studied.

Experiments are often better than observational studies because well-planned experi- ments typically reduce the chance of having the results affected by some variable that is not part of a study. A lurking variable is one that affects the variables included in the study, but it is not included in the study.

EXAMPLE 2 Ice Cream and Drownings

Observational Study: Observe past data to conclude that ice cream causes drown- ings (based on data showing that increases in ice cream sales are associated with increases in drownings). The mistake is to miss the lurking variable of temperature and the failure to see that as the temperature increases, ice cream sales increase and drownings increase because more people swim.

Experiment: Conduct an experiment with one group treated with ice cream while another group gets no ice cream. We would see that the rate of drowning victims is about the same in both groups, so ice cream consumption has no effect on drownings.

Here, the experiment is clearly better than the observational study.

Design of Experiments Good design of experiments includes replication, blinding, and randomization.

■ Replication is the repetition of an experiment on more than one individual. Good use of replication requires sample sizes that are large enough so that we can see

In 1954, an experiment was designed to test the effectiveness of the Salk vaccine in Clinical Trials vs. Observational Studies

In a New York

Times article

about hormone

therapy for

women, reporter

Denise Grady

wrote about

randomized

clinical trials that involve subjects

who were randomly assigned to

a treatment group and another

group not given the treatment.

Such randomized clinical trials

are often referred to as the “gold

standard” for medical research.

In contrast, observational studies

can involve patients who decide

themselves to undergo some

treatment. Subjects who decide

themselves to undergo treat-

ments are often healthier than

other subjects, so the treatment

group might appear to be more

successful simply because it

involves healthier subjects, not

necessarily because the treat-

ment is effective. Researchers

criticized observational studies of

hormone therapy for women by

saying that results might appear

to make the treatment more ef-

fective than it really is.

M01_TRIO9015_02_SE_C01_001-039.indd 25 19/08/16 2:16 PM

26 CHAPTER 1 Introduction to Statistics

effects of treatments. In the Salk experiment in Example 1, the experiment used sufficiently large sample sizes, so the researchers could see that the Salk vaccine was effective.

■ Blinding is used when the subject doesn’t know whether he or she is receiving a treatment or a placebo. Blinding is a way to get around the placebo effect, which occurs when an untreated subject reports an improvement in symptoms. (The reported improvement in the placebo group may be real or imagined.) The Salk experiment in Example 1 was double-blind, which means that blinding occurred at two levels: (1) The children being injected didn’t know whether they were getting the Salk vaccine or a placebo, and (2) the doctors who gave the injec- tions and evaluated the results did not know either. Codes were used so that the researchers could objectively evaluate the effectiveness of the Salk vaccine.

■ Randomization is used when individuals are assigned to different groups through a process of random selection, as in the Salk vaccine experiment in Example 1. The logic behind randomization is to use chance as a way to create two groups that are similar. The following definition refers to one common and effective way to collect sample data in a way that uses randomization.

DEFINITION

A simple random sample of n subjects is selected in such a way that every pos- sible sample of the same size n has the same chance of being chosen. (A simple random sample is often called a random sample, but strictly speaking, a random sample has the weaker requirement that all members of the population have the same chance of being selected. That distinction is not so important in this text. (See Exercise 38 “Simple Random Sample vs. Random Sample.”)

Throughout, we will use various statistical procedures, and we often have a requirement that we have collected a simple random sample, as defined above.

Unlike careless or haphazard sampling, random sampling usually requires very careful planning and execution.

Other Sampling Methods In addition to simple random sampling, here are some other sampling methods commonly used for surveys. Figure 1-3 illustrates these dif- ferent sampling methods.

DEFINITIONS

In systematic sampling, we select some starting point and then select every kth (such as every 50th) element in the population.

With convenience sampling, we simply use data that are very easy to get.

In stratified sampling, we subdivide the population into at least two different subgroups (or strata) so that subjects within the same subgroup share the same characteristics (such as gender). Then we draw a sample from each subgroup (or stratum).

In cluster sampling, we first divide the population area into sections (or clusters). Then we randomly select some of those clusters and choose all the members from those selected clusters.

Hawthorne and Experimenter Effects

The well-

known

placebo effect

occurs when

an untreated

subject incor-

rectly believes

that he or she is receiving a

real treatment and reports an

improvement in symptoms. The

Hawthorne effect occurs when

treated subjects somehow re-

spond differently, simply because

they are part of an experiment.

(This phenomenon was called

the “Hawthorne effect” because

it was first observed in a study

of factory workers at Western

Electric’s Hawthorne plant.) An

experimenter effect (sometimes

called a Rosenthal effect) occurs

when the researcher or experi-

menter unintentionally influences

subjects through such factors as

facial expression, tone of voice,

or attitude.

M01_TRIO9015_02_SE_C01_001-039.indd 26 02/08/16 4:46 PM

1-3 Collecting Sample Data 27

Multistage Sampling Professional pollsters and government researchers often collect data by using some combination of the preceding sampling methods. In a multistage sample design, pollsters select a sample in different stages, and each stage might use different methods of sampling, as in the following example.

M A

IN

CENTER HeritageSchool

P ar

k S

t. N

or th

S t.

1s t S

t.

2n d

S t.

3r d

S t.

82nd St.

52nd St.

36th St. 43rd St.

A St. B St.

C St. D St.

E St. F St.

Way St.

4t h

S t.

5t h

S t.

MLK PKWY

555-867-5309 555-606-0842 555-777-9311

Simple Random Sample A sample of n subjects is selected so that every sample of the same size n has the same chance of being selected.

Stratified Sample Subdivide population into strata (groups) with the same characteristics, then randomly sample within those strata.

Cluster Sample Partition the population in clusters (groups), then randomly select some clusters, then select all members of the selected clusters.

Systematic Sample Select every kth subject.

Convenience Sample Use data that are very easy to get.

Men Women

3rd 6th

FIGURE 1-3 Common Sampling Methods

EXAMPLE 3 Multistage Sample Design

The U.S. government’s unemployment statistics are based on surveys of house- holds. It is impractical to personally survey each household in a simple random sample, because they would be scattered all over the country. Instead, the U.S. Census Bureau and the Bureau of Labor Statistics collaborate to conduct a survey called the Current Population Survey. A recent survey incorporates a multistage sample design, roughly following these steps:

1. The entire United States is partitioned into 2,007 di®erent regions called primary sampling units (PSUs). The primary sampling units are metropolitan areas, large counties, or combinations of smaller counties. The 2,007 primary sampling units are then grouped into 824 di®erent strata.

Value of a Statistical Life

The value of a

statistical life

(VSL) is a mea-

sure routinely

calculated and

used for making

decisions in

fields such as

medicine, insurance, environ-

mental health, and transportation

safety. As of this writing, the

value of a statistical life is

$6.9 million.

Many people oppose the con-

cept of putting a value on a hu-

man life, but the word statistical

in the “value of a statistical life”

is used to ensure that we don’t

equate it with the true worth

of a human life. Some people

legitimately argue that every life

is priceless, but others argue that

there are conditions in which it

is impossible or impractical to

save every life, so a value must

be somehow assigned to a hu-

man life in order that sound and

rational decisions can be made.

Not far from the author’s home, a

parkway was modified at a cost

of about $3 million to improve

safety at a location where car

occupants had previously died

in traffic crashes. In the cost-

benefit analysis that led to this

improvement in safety, the value

of a statistical life was surely

considered.

continued

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28 CHAPTER 1 Introduction to Statistics

PA RT 2 Beyond the Basics of Design of Experiments and Collecting Sample Data

Observational Studies In Part 2 of this section, we discuss different types of ob- servational studies and different ways of designing experiments. The following defi- nitions identify the standard terminology used in professional journals for different types of observational studies. These definitions are illustrated in Figure 1-4.

2. In each of the 824 di®erent strata, one of the primary sampling units is selected so that the probability of selection is proportional to the size of the population in each primary sampling unit.

3. In each of the 824 selected primary sampling units, census data are used to identify a census enumeration district, with each containing about 300 house- holds. Enumeration districts are then randomly selected.

4. In each of the selected enumeration districts, clusters of about four addresses (contiguous whenever possible) are randomly selected.

5. A responsible person in each of the 60,000 selected households is interviewed about the employment status of each household member of age 16 or older.

This multistage sample design includes a combination of random, stratified, and cluster sampling at different stages. The end result is a very complicated sampling design, but it is much more practical, less expensive, and faster than using a simpler design, such as a simple random sample.

When are the

observations made?

Observational Study: Observe and measure,

but do not modify.

One point in time

Retrospective (or case-control) study:

Go back in time to collect data over some

past period.

Cross-sectional study:

Data are measured at one

point in time.

Prospective (or longitudinal or cohort) study: Go forward in time and observe groups sharing common factors,

such as smokers and nonsmokers.

Forward in timePast period of time

FIGURE 1-4 Types of Observational Studies

DEFINITIONS

In a cross-sectional study, data are observed, measured, and collected at one point in time, not over a period of time.

In a retrospective (or case-control) study, data are collected from a past time pe- riod by going back in time (through examination of records, interviews, and so on).

In a prospective (or longitudinal or cohort) study, data are collected in the future from groups that share common factors (such groups are called cohorts).

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1-3 Collecting Sample Data 29

Experiments In a study, confounding occurs when we can see some effect, but we can’t identify the specific factor that caused it, as in the ice cream and drowning observational study in Example 2. See also the bad experimental design illustrated in Figure 1-5(a), where confounding can occur when the treatment group of women shows strong positive results. Because the treatment group consists of women and the placebo group consists of men, confounding has occurred because we cannot deter- mine whether the treatment or the gender of the subjects caused the positive results. The Salk vaccine experiment in Example 1 illustrates one method for controlling the effect of the treatment variable: Use a completely randomized experimental design, whereby randomness is used to assign subjects to the treatment group and the placebo group. A completely randomized experimental design is one of the following methods that are used to control effects of variables.

Completely Randomized Experimental Design: Assign subjects to different treat- ment groups through a process of random selection, as illustrated in Figure 1-5(b).

Treatment Group: Women

Bad experimental design: Treat all women subjects and give the men a placebo. (Problem: We don’t know if e�ects are due to sex or to treatment.)

Completely randomized experimental design: Use randomness to determine who gets the treatment and who gets the placebo.

Treat all women subjects.

Placebo Group: Men

Give all men a placebo

Treat these randomly selected subjects and give the others a placebo.

(a) (b)

Before After

Alex

Bob

Chris

Block of Women

Randomized block design: 1. Form a block of women and a block of men. 2. Within each block, randomly select subjects to be treated.

Matched pairs design: Get measurements from the same subjects before and after some treatment.

Treat randomly selected women.

Block of Men

Treat randomly selected men.

(c) (d)

FIGURE 1-5 Designs of Experiments

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30 CHAPTER 1 Introduction to Statistics

Experimental design requires much more thought and care than we can describe in this relatively brief section. Taking a complete course in the design of experiments is a good start in learning so much more about this important topic.

Randomized Block Design: See Figure 1-5c. A block is a group of subjects that are similar, but blocks differ in ways that might affect the outcome of the experiment. Use the following procedure, as illustrated in Figure 1-5(c):

1. Form blocks (or groups) of subjects with similar characteristics.

2. Randomly assign treatments to the subjects within each block.

For example, in designing an experiment to test the effectiveness of aspirin treatments on heart disease, we might form a block of men and a block of women, because it is known that the hearts of men and women can behave differently. By controlling for gender, this randomized block design eliminates gender as a possible source of con- founding.

A randomized block design uses the same basic idea as stratified sampling, but randomized block designs are used when designing experiments, whereas stratified sampling is used for surveys.

Matched Pairs Design: Compare two treatment groups (such as treatment and pla- cebo) by using subjects matched in pairs that are somehow related or have similar characteristics, as in the following cases.

■ Before/After: Matched pairs might consist of measurements from subjects before and after some treatment, as illustrated in Figure 1-5(d) on the preceding page. Each subject yields a “before” measurement and an “after” measurement, and each before/after pair of measurements is a matched pair.

■ Twins: A test of Crest toothpaste used matched pairs of twins, where one twin used Crest and the other used another toothpaste.

Rigorously Controlled Design: Carefully assign subjects to different treatment groups, so that those given each treatment are similar in the ways that are important to the experiment. This can be extremely difficult to implement, and often we can never be sure that we have accounted for all of the relevant factors.

Sampling Errors In statistics, you could use a good sampling method and do everything correctly, and yet it is possible to get wrong results. No matter how well you plan and execute the sample collection process, there is likely to be some error in the results. The different types of sampling errors are described here.

DEFINITIONS

A sampling error (or random sampling error) occurs when the sample has been selected with a random method, but there is a discrepancy between a sample result and the true population result; such an error results from chance sample fluctuations.

A nonsampling error is the result of human error, including such factors as wrong data entries, computing errors, questions with biased wording, false data provided by respondents, forming biased conclusions, or applying statistical methods that are not appropriate for the circumstances.

A nonrandom sampling error is the result of using a sampling method that is not random, such as using a convenience sample or a voluntary response sample.

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1-3 Collecting Sample Data 31

Statistical Literacy and Critical Thinking

1. Back Pain Treatment In a study designed to test the effectiveness of paracetamol (also known as acetaminophen) as a treatment for lower back pain, 1643 patients were randomly assigned to one of three groups: (1) the 547 subjects in the placebo group were given pills containing no medication; (2) 550 subjects were in a group given pills with paracetamol taken at regular intervals; (3) 546 subjects were in a group given pills with paracetamol to be taken when needed for pain relief. (See “Efficacy of Paracetamol for Acute Low-Back Pain,” by Williams et al., Lancet.) Is this study an experiment or an observational study? Explain.

2. Blinding What does it mean when we say that the study cited in Exercise 1 was “double-blind”?

3. Replication In what specific way was replication applied in the study cited in Exercise 1?

4. Sampling Method The patients included in the study cited in Exercise 1 were those “who sought care for low-back pain directly or in response to a community advertisement.” What type of sampling best describes the way in which the 1634 subjects were chosen: simple ran- dom sample, systematic sample, convenience sample, stratified sample, cluster sample? Does the method of sampling appear to adversely affect the quality of the results?

Exercises 5–8 refer to the study of an association between which ear is used for cell phone calls and whether the subject is left-handed or right-handed. The study is reported in “Hemi- spheric Dominance and Cell Phone Use,” by Seidman et al., JAMA Otolaryngology—Head & Neck Surgery, Vol. 139, No. 5. The study began with a survey e-mailed to 5000 people belonging to an otology online group, and 717 surveys were returned. (Otology relates to the ear and hearing.)

5. Sampling Method What type of sampling best describes the way in which the 717 subjects were chosen: simple random sample, systematic sample, convenience sample, stratified sample, cluster sample? Does the method of sampling appear to adversely affect the quality of the results?

6. Experiment or Observational Study Is the study an experiment or an observational study? Explain.

7. Response Rate What percent of the 5000 surveys were returned? Does that response rate appear to be low? In general, what is a problem with a very low response rate?

8. Sampling Method Assume that the population consists of all students currently in your statistics class. Describe how to obtain a sample of six students so that the result is a sample of the given type.

a. Simple random sample

b. Systematic sample

c. Stratified sample

d. Cluster sample

In Exercises 9–20, identify which of these types of sampling is used: random, systematic, convenience, stratified, or cluster.

9. Cormorant Density Cormorant bird population densities were studied by using the “line transect method” with aircraft observers flying along the shoreline of Lake Huron and collecting sample data at intervals of every 20 km (based on data from Journal of Great Lakes Research).

1-3 Basic Skills and Concepts

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32 CHAPTER 1 Introduction to Statistics

10. Sexuality of Women The sexuality of women was discussed in Shere Hite’s book Women and Love: A Cultural Revolution. Her conclusions were based on sample data that consisted of 4500 mailed responses from 100,000 questionnaires that were sent to women.

11. Acupuncture Study In a study of treatments for back pain, 641 subjects were randomly assigned to the four different treatment groups of individualized acupuncture, standardized acupuncture, simulated acupuncture, and usual care (based on data from “A Randomized Trial Comparing Acupuncture, Simulated Acupuncture, and Usual Care for Chronic Low Back Pain,” by Cherkin et al., Archives of Internal Medicine, Vol. 169, No. 9).

12. Class Survey A professor surveys her statistics class by identifying groups of males and females, then randomly selecting five students from each of those two groups.

13. Class Survey A professor conducts a survey by randomly selecting three different classes and surveying all of the students as they left those classes.

14. Exercise Program In a study designed to test the effectiveness of exercise in lowering blood pressure, 532 subjects were randomly assigned to these two different groups: (1) group given regular exercise programs; (2) group given no exercise programs.

15. Hospital Survey A researcher collects sample data by randomly selecting 20 hospital employees from each of the categories of physician, nurse, and administrator.

16. Deforestation Rates Satellites are used to collect sample data for estimating deforesta- tion rates. The Forest Resources Assessment of the United Nations (UN) Food and Agriculture Organization uses a method of selecting a sample of a 10-km-wide square at every 1° intersec- tion of latitude and longitude.

17. Testing Lipitor In a clinical trial of the cholesterol drug Lipitor (atorvastatin), subjects were partitioned into groups given a placebo or Lipitor doses of 10 mg, 20 mg, 40 mg, or 80 mg. The subjects were randomly assigned to the different treatment groups (based on data from Pfizer, Inc.).

18. Blood Drives A researcher for the American Red Cross randomly selected five different blood donor sites and then interviewed all blood donors as they left the sites.

19. Smoking Prevalence A medical student collects sample data on the prevalence of smok- ing among adults by surveying all of the patients she encounters in the clinic where she is doing her residency.

20. Health Survey The Texas Health and Human Services Commission obtains an alphabeti- cal listing of all 20,126,759 adults and constructs a sample by selecting every 10,000th name on that list.

Critical Thinking: What’s Wrong? In Exercises 21–28, determine whether the study is an experiment or an observational study, and then identify a major problem with the study.

21. Online Medical Information In a survey conducted by USA Today, 1072 Internet users chose to respond to this question posted on the USA Today electronic edition: “How often do you seek medical information online?” 38% of the respondents said “frequently.”

22. Physicians’ Health Study The Physicians’ Health Study involved 22,071 male physi- cians. Based on random selections, 11,037 of them were treated with aspirin and the other 11,034 were given placebos. The study was stopped early because it became clear that aspirin reduced the risk of myocardial infarctions by a substantial amount.

23. Drinking and Driving A researcher for a consortium of insurance companies plans to test for the effects of drinking on driving ability by randomly selecting 1000 drivers and then randomly assigning them to two groups: One group of 500 will drive in New York City after no alcohol consumption, and the second group will drive in New York City after consuming three shots of Jim Beam bourbon whiskey.

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1-3 Collecting Sample Data 33

24. Blood Pressure A medical researcher tested for a difference in systolic blood pressure levels between male and female students who are 20 years of age. She randomly selected four males and four females for her study.

25. Salt Deprivation In a program designed to investigate the effects of salt deprivation in diets, the original plan was to use a sample of 500 adults randomly selected throughout the country. The program managers know that they would get a biased sample if they limit their study to adults in New York City, so they planned to compensate for that bias by using a larger sample of 2000 adults in New York City.

26. Atkins Weight Loss Program An independent researcher tested the effectiveness of the Atkins weight loss program by randomly selecting 1000 subjects using that program. Each of the subjects was called to report his or her weight before the diet and after the diet.

27. Crime Research A researcher has created a brief survey to be given to 2000 adults ran- domly selected from the U.S. population. Here are her first two questions: (1) Have you ever been the victim of a felony crime? (2) Have you ever been convicted of a felony?

28. Medications The Pharmaceutical Research and Manufacturers of America wants infor- mation about the consumption of various medications. An independent researcher conducts a survey by mailing 10,000 questionnaires to randomly selected adults in the United States, and she receives 152 responses.

In Exercises 29–32, indicate whether the observational study used is cross-sectional, retrospective, or prospective.

29. Nurses’ Health Study II Phase II of the Nurses’ Health Study was started in 1989 with 116,000 female registered nurses. The study is ongoing.

30. Heart Health Study Samples of subjects with and without heart disease were selected, then researchers looked back in time to determine whether they took aspirin on a regular basis.

31. Marijuana Study Researchers from the National Institutes of Health want to determine the current rates of marijuana consumption among adults living in states that have legalized the use of marijuana. They conduct a survey of 500 adults in those states.

32. Framingham Heart Study The Framingham Heart Study was started in 1948 and is ongo- ing. Its focus is on heart disease.

In Exercises 33–36, identify which of these designs is most appropriate for the given experiment: completely randomized design, randomized block design, or matched pairs design.

33. Lunesta Lunesta (eszopiclone) is a drug designed to treat insomnia. In a clinical trial of Lunesta, amounts of sleep each night are measured before and after subjects have been treated with the drug.

34. Lipitor A clinical trial of Lipitor treatments is being planned to determine whether its effects on diastolic blood pressure are different for men and women.

35. West Nile Vaccine Currently, there is no approved vaccine for the prevention of West Nile virus infection. A clinical trial of a possible vaccine is being planned to include subjects treated with the vaccine while other subjects are given a placebo.

36. HIV Vaccine The HIV Trials Network is conducting a study to test the effectiveness of two different experimental HIV vaccines. Subjects will consist of 80 pairs of twins. For each pair of twins, one of the subjects will be treated with the DNA vaccine and the other twin will be treated with the adenoviral vector vaccine.

1-3 Beyond the Basics

M01_TRIO9015_02_SE_C01_001-039.indd 33 02/08/16 4:46 PM

34 CHAPTER 1 Introduction to Statistics

37. Sample Design Literacy In “Cardiovascular Effects of Intravenous Triiodothyronine in Patients Undergoing Coronary Artery Bypass Graft Surgery” (Journal of the American Medi- cal Association, Vol. 275, No. 9), the authors explain that patients were assigned to one of three groups: (1) a group treated with triiodothyronine, (2) a group treated with normal saline bolus and dopamine, and (3) a placebo group given normal saline. The authors summarize the sample design as a “prospective, randomized, double-blind, placebo-controlled trial.” Describe the meaning of each of those terms in the context of this study.

38. Simple Random Sample vs. Random Sample Refer to the definition of simple random sample in this section and the accompanying definition of random sample enclosed within pa- rentheses. Determine whether each of the following is a simple random sample and a random sample.

a. A statistics class with 36 students is arranged so that there are 6 rows with 6 students in each row, and the rows are numbered from 1 through 6. A die is rolled and a sample consists of all students in the row corresponding to the outcome of the die.

b. For the same class described in part (a), the 36 student names are written on 36 individual index cards. The cards are shuffled and six names are drawn from the top.

c. For the same class described in part (a), the six youngest students are selected.

1. Clinical Study When conducting a clinical study, it is common to maintain the privacy of subjects by assigning them number codes that will be used instead of their actual names. Sev- eral subjects are assigned these codes: 1, 2, 3, 5, 6, 9, 11, 13, 16, 20, 22, 26, 32, and 40. Does it make sense to calculate the average (mean) of these numbers?

2. Clinical Study Which of the following best describes the level of measurement of the data listed in Exercise 1: nominal, ordinal, interval, ratio?

3. Waist Data Set 1 “Body Data” includes measurements of waist circumferences. Are waist circumferences values that are discrete or continuous?

4. Waist Are the waist circumferences described in Exercise 3 quantitative data or categorical data?

5. Waist Which of the following best describes the level of measurement of the waist circumferences described in Exercise 3: nominal, ordinal, interval, ratio?

6. Waist If you construct a sample by selecting every sixth waist circumference from those listed in Data Set 1 “Body Data,” is the result a simple random sample of the listed waist circumferences?

7. Gallup Poll In a recent Gallup poll, pollsters randomly selected adults and asked them whether they smoke. Because the subjects agreed to respond, is the sample a voluntary re- sponse sample?

8. Parameter and Statistic In a recent Gallup poll, pollsters randomly selected adults and asked them whether they smoke. Among the adults who responded to the survey question, 21% said that they did smoke. Is that value of 21% an example of a statistic or a parameter?

9. Observational Study or Experiment Are the data described in Exercise 8 the result of an observational study or an experiment?

10. Statistical Significance and Practical Significance True or false: If data lead to a con- clusion with statistical significance, then the results also have practical significance.

Chapter Quick Quiz

M01_TRIO9015_02_SE_C01_001-039.indd 34 02/08/16 4:46 PM

1. Hospitals Currently, there are 5723 registered hospitals in the United States.

a. Are the numbers of hospitals in different states discrete or continuous?

b. What is the level of measurement for the numbers of hospitals in different years? (nominal, ordinal, interval, ratio)

c. If a survey is conducted by randomly selecting 10 patients in every hospital, what type of sampling is used? (random, systematic, convenience, stratified, cluster)

d. If a survey is conducted by randomly selecting 20 hospitals and interviewing all of the mem- bers of each board of directors, what type of sampling is used? (random, systematic, conve- nience, stratified, cluster)

e. What is wrong with surveying patient satisfaction by mailing questionnaires to 10,000 ran- domly selected patients?

2. What’s Wrong? A survey sponsored by the American Laser Centers included responses from 575 adults, and 24% of the respondents said that the face is their favorite body part (based on data from USA Today). What is wrong with this survey?

3. What’s Wrong? A survey included 2028 responses from Internet users who decided to respond to a question posted by AOL. Here is the question: “How often do you drink soda?” Among the respondents, 33% said that they drink soda almost every day. What is wrong with this survey?

4. Sampling Seventy-two percent of Americans squeeze their toothpaste tube from the top. This and other not-so-serious findings are included in The First Really Important Survey of American Habits. Those results are based on 7000 responses from the 25,000 questionnaires that were mailed.

a. What is wrong with this survey?

b. As stated, the value of 72% refers to all Americans, so is that 72% a statistic or a parameter? Explain.

c. Does the survey constitute an observational study or an experiment?

5. Percentages

a. The labels on U-Turn protein energy bars include the statement that these bars contain “125% less fat than the leading chocolate candy brands” (based on data from Consumer Reports magazine). What is wrong with that claim?

b. In a Pew Research Center poll on driving, 58% of the 1182 respondents said that they like to drive. What is the actual number of respondents who said that they like to drive?

c. In a Pew Research Center poll on driving, 331 of the 1182 respondents said that driving is a chore. What percentage of respondents said that driving is a chore?

6. Simple Random Sample Which of the following is>are simple random samples? a. As Lipitor pills are being manufactured, a quality control plan is to select every 500th pill and test it to confirm that it contains 80 mg of atorvastatin.

b. To test for a gender difference in the way that men and women make online purchases, Gallup surveys 500 randomly selected men and 500 randomly selected women.

c. A list of all 10,877 adults in Trinity County, California, is obtained; the list is numbered from 1 to 10,877; and then a computer is used to randomly generate 250 different numbers between 1 and 10,877. The sample consists of the adults corresponding to the selected numbers.

Review Exercises

CHAPTER 1 Review Exercises 35

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36 CHAPTER 1 Introduction to Statistics

7. Statistical Significance and Practical Significance The Gengene Research Group has developed a procedure designed to increase the likelihood that a baby will be born a girl. In a clinical trial of their procedure, 112 girls were born to 200 different couples. If the method has no effect, there is about a 4% chance that such extreme results would occur. Does the procedure appear to have statistical significance? Does the procedure appear to have practical signifi- cance?

8. Marijuana Survey In a recent Pew poll of 1500 adults, 52% of the respondents said that the use of marijuana should not be made legal. In the same poll, 23% of the respondents said that the use of marijuana for medical purposes should not be legal.

a. The sample of 1500 adults was selected from the population of all adults in the United States. The method used to select the sample was equivalent to placing the names of all adults in a giant bowl, mixing the names, and then drawing 1500 names. What type of sampling is this? (random, systematic, convenience, stratified, cluster)

b. If the sampling method consisted of a random selection of 30 adults from each of the 50 states, what type of sampling would this be? (random, systematic, convenience, stratified, cluster)

c. What is the level of measurement of the responses of yes, no, don’t know, and refused to respond?

d. Is the given value of 52% a statistic or a parameter? Why?

e. What would be wrong with conducting the survey by mailing a questionnaire that respon- dents could complete and mail back?

9. Marijuana Survey Identify the type of sampling (random, systematic, convenience, strati- fied, cluster) used when a sample of the 1500 survey responses is obtained as described. Then determine whether the sampling scheme is likely to result in a sample that is representative of the population of all adults.

a. A complete list of all 241,472,385 adults in the United States is compiled, and every 150,000th name is selected until the sample size of 1500 is reached.

b. A complete list of all 241,472,385 adults in the United States is compiled, and 1500 adults are randomly selected from that list.

c. The United States is partitioned into regions with 100 adults in each region. Then 15 of those regions are randomly selected, and all 100 people in each of those regions are surveyed.

d. The United States is partitioned into 150 regions with approximately the same number of adults in each region; then 10 people are randomly selected from each of the 150 regions.

e. A survey is mailed to 10,000 randomly selected adults, and the 1500 responses are used.

10. Marijuana Survey Exercise 8 referred to a Pew poll of 1500 adults, and 52% of the respondents said that the use of marijuana should not be made legal.

a. Among the 1500 adults who responded, what is the number of respondents who said that the use of marijuana should not be made legal?

b. In the same poll of 1500 adults, 345 of the respondents said that the use of marijuana for medical purposes should not be legal. What is the percentage of respondents who said that the use of marijuana for medical purposes should not be legal?

c. In this survey of 1500 adults, 727 are men and 773 are women. Find the percentage of respondents who are men, and then find the percentage of respondents who are women.

d. Does the difference between the two percentages from part (c) appear to have statistical significance?

e. Does the difference between the two percentages from part (c) appear to have practical significance?

M01_TRIO9015_02_SE_C01_001-039.indd 36 02/08/16 4:46 PM

For Chapter 2 through Chapter 14, the Cumulative Review Exercises include topics from preceding chapters. For this chapter, we present a few calculator warm-up exercises, with expressions similar to those found throughout this book. Use your calculator to find the indicated values.

1. Birth Weights Listed below are the weights (grams) of newborn babies from Albany Medi- cal Center Hospital. What value is obtained when those weights are added and the total is di- vided by the number of weights? (This result, called the mean, is discussed in Chapter 3.) What is notable about these values, and what does it tell us about how the weights were measured?

3600 1700 4000 3900 3100 3800 2200 3000

2. Six Children Jule Cole is a founder of Mabel’s Labels, and she is the mother of six chil- dren. The probability that six randomly selected children are all girls is found by evaluating 0.56. Find that value.

3. Tallest Person Robert Wadlow (1918–1940) is the tallest known person to have lived. The expression below converts his height of 272 cm to a standardized score. Find this value and round the result to two decimal places. Such standardized scores are considered to be signifi- cantly high if they are greater than 2 or 3. Is the result significantly high?

272 - 176 6

4. Body Temperature The given expression is used for determining the likelihood that the av- erage (mean) human body temperature is different from the value of 98.6°F that is commonly used. Find the given value and round the result to two decimal places.

98.2 - 98.6 0.62

2106 5. Determining Sample Size The given expression is used to determine the size of the sam- ple necessary to estimate the proportion of college students who have the profound wisdom to take a statistics course. Find the value and round the result to the nearest whole number.

1.962 # 0.25 0.032

6. Standard Deviation One way to get a very rough approximation of the value of a standard deviation of sample data is to find the range, then divide it by 4. The range is the difference be- tween the highest sample value and the lowest sample value. In using this approach, what value is obtained from the sample data listed in Exercise 1 “Birth Weights”?

7. Standard Deviation The standard deviation is an extremely important concept introduced in Chapter 3. Using the sample data from Exercise 1 “Birth Weights,” part of the calculation of the standard deviation is shown in the expression below. Evaluate this expression. (Fortunately, calculators and software are designed to automatically execute such expressions, so our future work with standard deviations will not be burdened with cumbersome calculations.)

13600 - 3162.522 7

8. Standard Deviation The given expression is used to compute the standard deviation of three randomly selected body temperatures. Perform the calculation and round the result to two decimal places.

B 198.4 - 98.622 + 198.6 - 98.622 + 198.8 - 98.622

3 - 1

Cumulative Review Exercises

CHAPTER 1 Cumulative Review Exercises 37

M01_TRIO9015_02_SE_C01_001-039.indd 37 02/08/16 4:46 PM

38 CHAPTER 1 Introduction to Statistics

Scientific Notation. In Exercises 9–12, the given expressions are designed to yield re- sults expressed in a form of scientific notation. For example, the calculator-displayed re- sult of 1.23E5 can be expressed as 123,000, and the result of 1.23E-4 can be expressed as 0.000123. Perform the indicated operation and express the result as an ordinary number that is not in scientific notation.

9. 0.48 10. 911 11. 614 12. 0.312

Technology Project

Missing Data The focus of this project is to download a data set and manipulate it to work around missing data.

a. First, download Data Set 2 “Body Temperatures” in Appendix B from www.TriolaStats.com. Choose the download format that matches your technology. (If you have no preferred technol- ogy, you can download a free copy of Statdisk (from www.statdisk.org), which is designed for this book and contains all Appendix B data sets.)

b. Some statistical procedures, such as those involved with correlation and regression (dis- cussed in later chapters) require data that consist of matched pairs of values, and those proce- dures ignore pairs in which at least one of the data values in a matched pair is missing. Assume that we want to conduct analyses for correlation and regression on the last two columns of data in Data Set 2: body temperatures measured at 8 AM on day 2 and again at 12 AM on day 2. For those last two columns, identify the rows with at least one missing value. Note that in some technologies, such as TI-83>84 Plus calculators, missing data must be represented by a constant such as - 9 or 999.

c. Here are two different strategies for reconfiguring the data set to work around the missing data in the last two columns (assuming that we need matched pairs of data with no missing values):

i. Manual Deletion Highlight rows with at least one missing value in the last two columns, then delete those rows. This can be tedious if there are many rows with missing data and those rows are interspersed throughout instead of being adjacent rows.

ii. Sort Most technologies have a Sort feature that allows you to rearrange all rows using one particular column as the basis for sorting (TI-83>84 Plus calculators do not have this type of sort feature). The result is that all rows remain the same but they are in a different order. First use the technology’s Sort feature to rearrange all rows using the “8 AM day 2” column as the basis for sorting (so that all missing values in the “8 AM day 2” column are at the beginning); then highlight and delete all of those rows with missing values in the “8 AM day 2” column. Next, use the technology’s Sort feature to rearrange all rows using the “12 AM day 2” column as the basis for sorting (so that all missing values in the “12 AM day 2” column are at the beginning); then highlight and delete all of those rows with missing values in the “12 AM day 2” column. The remaining rows will include matched pairs of body temperatures, and those rows will be suitable for analyses such as correlation and regression. Print the resulting reconfigured data set.

M01_TRIO9015_02_SE_C01_001-039.indd 38 19/08/16 2:16 PM

Cooperative Group Activities

1. In-class activity Working in groups of three or four, design an experiment to determine whether pulse rates of college students are the same while the students are standing and sitting. Conduct the experiment and collect the data. Save the data so that they can be analyzed with methods presented in the following chapters.

2. In-class activity Working in groups of three or four, construct a brief survey that includes only a few questions that can be quickly asked. Include some objective questions along with some that are biased, such as the first question below.

•  Should your college force all students to pay a $100 activity fee?

•  Should your college fund activities by collecting a $100 fee?

Conduct the survey and try to detect the effect that the biased wording has on the responses.

3. In-class activity Identify problems with a mailing from Consumer Reports magazine that included an annual questionnaire about cars and other consumer products. Also included were a request for a voluntary contribution of money and a voting ballot for the board of directors. Responses were to be mailed back in envelopes that required postage stamps.

4. Out-of-class activity Find a report of a survey that used a voluntary response sample. De- scribe how it is quite possible that the results do not accurately reflect the population.

5. Out-of-class activity Find a professional journal with an article that includes a statistical analysis of an experiment. Describe and comment on the design of the experiment. Identify one particular issue addressed by the study, and determine whether the results were found to be statistically significant. Determine whether those same results have practical significance.

FROM DATA TO DECISION

Critical Thinking: Do Male Symphony Conductors Really Live Longer? Several media reports made the interesting observation that male symphony conductors live longer than other males. John Amaral wrote in Awaken that orchestra conductors “live longer than almost any other group of people by three to seven years.” Robert Levine wrote in Polyphonic.org that

they live longer “because they stand up while working.” Some provided other explanations for this phenomenon, often referring to cardiovascular activity. But do male sym- phony conductors really live longer than other groups of males? The Internet can be researched for possible answers. Let’s also consider the following.

Analysis

1. Consider the statement that “male symphony conductors live longer.” Identify the specific group that they supposedly live longer than. Does that other group consist of males ran- domly selected from the general population?

2. It is reasonable to assume that males do not become sym- phony conductors until they have reached at least the age of 40 years. When comparing life spans of male conduc- tors, should we compare them to other males in the general population, or should we compare them to other males who lived until at least 40 years of age? Explain.

3. Without any disabilities, males qualify for Medicare if they are 65 or older and meet a few other requirements. If we compare life spans of males on Medicare to life spans of males randomly selected from the general population, why would we find that males on Medicare have longer life spans?

4. Explain in detail how to design a study for collecting data to determine whether it is misleading to state that male sym- phony conductors live longer. Should the study be an experi- ment or an observational study?

CHAPTER 1 Cooperative Group Activities 39

M01_TRIO9015_02_SE_C01_001-039.indd 39 02/08/16 4:46 PM

40

Frequency Distributions for Organizing and Summarizing Data

Histograms

Graphs That Enlighten and Graphs That Deceive

Scatterplots, Correlation, and Regression

2-1

2-2

2-3

2-4

Does Exposure to Lead Affect IQ Scores?CHAPTER PROBLEM

Exploring Data with Tables and Graphs

Data Set 8 “IQ and Lead” in Appendix B includes full IQ scores

from three groups of children who lived near a lead smelter.

The children in Group 1 had low levels of measured lead in

their blood (with blood levels less than 40 micrograms>100 mL in each of two years). Group 2 had medium levels of measured

lead in their blood (with blood levels of at least

40 micrograms/100 mL in exactly one of two years). Group 3

had high levels of measured lead in their blood (with blood lev-

els of at least 40 micrograms>100 mL in each of two years). Let’s consider the measured full IQ scores from Group 1

(low lead level) and Group 3 (high lead level), as listed in

Table 2-1. It is an exceptionally rare person who can look

at both lists of IQ scores and form meaningful conclusions.

Almost all of us must work at describing, exploring, and

2

M02_TRIO9015_02_SE_C02_040-074.indd 40 02/08/16 4:44 PM

comparing the two sets of data. In this chapter we pres-

ent methods that focus on summarizing the data and using

graphs that enable us to understand important characteris-

tics of the data, especially the distribution of the data. These

methods will help us compare the two sets of data so that we

can determine whether the IQ scores of the low lead group

are somehow different from the IQ scores of the high lead

group. Such comparisons will be helpful as we try to address

this important and key issue: Does exposure to lead have an

effect on IQ score?

This chapter and the following chapter focus on important characteristics of data, including the following:

Characteristics of Data

1. Center: A representative value that shows us where the middle of the data set is located.

2. Variation: A measure of the amount that the data values vary.

3. Distribution: The nature or shape of the spread of the data over the range of values (such as bell-shaped).

4. Outliers: Sample values that lie very far away from the vast majority of the other sample values. (Later, a more objective definition of “outlier” will be given.)

5. Time: Any change in the characteristics of the data over time.

This chapter provides tools that enable us to gain insight into data by organizing, sum- marizing, and representing them in ways that enable us to see important characteristics of the data. Here are the chapter objectives:

Frequency Distributions for Organizing and Summarizing Data

• Develop an ability to summarize data in the format of a frequency distribution and a relative frequency distribution.

• For a frequency distribution, identify values of class width, class midpoint, class lim- its, and class boundaries.

2-1

Chapter Objectives 41

CHAPTER OBJECTIVES

> > >

TABLE 2-1 Full IQ Scores of the Low Lead Group and the High Lead Group

Low Lead Level (Group 1)

70 85 86 76 84 96 94 56 115 97 77 128 99 80 118 86

141 88 96 96 107 86 80 107 101 91 125 96 99 99 115 106

105 96 50 99 85 88 120 93 87 98 78 100 105 87 94 89

80 111 104 85 94 75 73 76 107 88 89 96 72 97 76 107

104 85 76 95 86 89 76 96 101 108 102 77 74 92

High Lead Level (Group 3)

82 93 85 75 85 80 101 89 80 94 88 104 88 88 83 104

96 76 80 79 75

M02_TRIO9015_02_SE_C02_040-074.indd 41 19/08/16 2:34 PM

42 CHAPTER 2 Exploring Data with Tables and Graphs

Histograms

• Develop the ability to picture the distribution of data in the format of a histogram or relative frequency histogram.

• Examine a histogram and identify common distributions, including a uniform distribu- tion and a normal distribution.

Graphs That Enlighten and Graphs That Deceive

• Develop an ability to graph data using a dotplot, stemplot, time-series graph, Pareto chart, pie chart, and frequency polygon.

• Determine when a graph is deceptive through the use of a nonzero axis or a pictograph that uses an object of area or volume for one-dimensional data.

Scatterplots, Correlation, and Regression

• Develop an ability to construct a scatterplot of paired data. • Analyze a scatterplot to determine whether there appears to be a correlation

between two variables.

2-2

2-3

2-4

Histograms

• Develop the ability to picture the distribution of data in the format of a histogram or relative frequency histogram.

• Examine a histogram and identify common distributions, including a uniform distribu- tion and a normal distribution.

Graphs That Enlighten and Graphs That Deceive

• Develop an ability to graph data using a dotplot, stemplot, time-series graph, Pareto chart, pie chart, and frequency polygon.

• Determine when a graph is deceptive through the use of a nonzero axis or a pictograph that uses an object of area or volume for one-dimensional data.

Scatterplots, Correlation, and Regression

• Develop an ability to construct a scatterplot of paired data. • Analyze a scatterplot to determine whether there appears to be a correlation

between two variables.

Key Concept When working with large data sets, a frequency distribution (or frequency table) is often helpful in organizing and summarizing data. A frequency distribution helps us to understand the nature of the distribution of a data set.

2-1

Frequency Distributions for Organizing and Summarizing Data

DEFINITION

A frequency distribution (or frequency table) shows how data are partitioned among several categories (or classes) by listing the categories along with the num- ber (frequency) of data values in each of them.

Consider the IQ scores of the low lead group listed in Table 2-1. Table 2-2 is a fre- quency distribution summarizing those IQ scores. The frequency for a particular class is the number of original values that fall into that class. For example, the first class in Table 2-2 has a frequency of 2, so 2 of the IQ scores are between 50 and 69 inclusive.

The following standard terms are often used in constructing frequency distributions and graphs.

TABLE 2-2 IQ Scores of the Low Lead Group

IQ Score Frequency

50–69 2

70–89 33

90–109 35

110–129 7

130–149 1 DEFINITIONS

Lower class limits are the smallest numbers that can belong to each of the differ- ent classes. (Table 2-2 has lower class limits of 50, 70, 90, 110, and 130.)

Upper class limits are the largest numbers that can belong to each of the different classes. (Table 2-2 has upper class limits of 69, 89, 109, 129, and 149.)

Class boundaries are the numbers used to separate the classes, but without the gaps created by class limits. In Figure 2-1 we see that the values of 69.5, 89.5, 109.5, and 129.5 are in the centers of those gaps, and following the pattern of those class boundaries, we see that the lowest class boundary is 49.5 and the

M02_TRIO9015_02_SE_C02_040-074.indd 42 11/9/16 4:45 PM

2-1 Frequency Distributions for Organizing and Summarizing Data 43

Procedure for Constructing a Frequency Distribution We construct frequency distributions to (1) summarize large data sets, (2) see the dis- tribution and identify outliers, and (3) have a basis for constructing graphs (such as histograms, introduced in Section 2-2). Technology can generate frequency distribu- tions, but here are the steps for manually constructing them:

1. Select the number of classes, usually between 5 and 20. The number of classes might be affected by the convenience of using round numbers.

2. Calculate the class width.

Class width ≈ 1maximum data value2 - 1minimum data value2

number of classes

Round this result to get a convenient number. (It’s usually best to round up.) Using a specific number of classes is not too important, and it’s usually wise to change the number of classes so that they use convenient values for the class limits.

3. Choose the value for the first lower class limit by using either the minimum value or a convenient value below the minimum.

highest class boundary is 149.5. Thus the complete list of class boundaries is 49.5, 69.5, 89.5, 109.5, 129.5, and 149.5.

Class midpoints are the values in the middle of the classes. Table 2-2 has class midpoints of 59.5, 79.5, 99.5, 119.5, and 139.5. Each class midpoint is computed by adding the lower class limit to the upper class limit and dividing the sum by 2.

Class width is the difference between two consecutive lower class limits (or two consecutive lower class boundaries) in a frequency distribution. Table 2-2 uses a class width of 20. (The first two lower class boundaries are 50 and 70, and their dif- ference is 20.)

CAUTION Finding the correct class width can be tricky. For class width, don’t make the most common mistake of using the difference between a lower class limit and an upper class limit. See Table 2-2 and note that the class width is 20, not 19.

69.5

49.5

50 69

149.5

STEP 1: List the class limits from Table 2-2.

STEP 2: Split the di�erence as shown.

STEP 3: Find the first and last values of 49.5 and 149.5 by projecting the same pattern.

70 89

89.5

90 109

109.5

110 129

129.5

130 149

FIGURE 2-1 Finding Class Boundaries from Class Limits in Table 2-2

Growth Charts Updated

Pediatricians

typically use

standardized

growth charts to

compare their

patient’s weight

and height

to a sample of other children.

Children are considered to be in

the normal range if their weight

and height fall between the 5th

and 95th percentiles. If they fall

outside that range, they are often

given tests to ensure that there

are no serious medical problems.

Pediatricians became increas-

ingly aware of a major problem

with the charts: Because they

were based on children living be-

tween 1929 and 1975, the growth

charts had become inaccurate.

To rectify this problem, the charts

were updated in 2000 to reflect

the current measurements of

millions of children. The weights

and heights of children are good

examples of populations that

change over time. This is the

reason for including changing

characteristics of data over time

as an important consideration for

a population.

to a sample of other children.

CAUTION For class boundaries, remember that they split the difference between the end of one class and the beginning of the next class, as shown in Figure 2-1.

continued

M02_TRIO9015_02_SE_C02_040-074.indd 43 26/10/16 3:31 PM

44 CHAPTER 2 Exploring Data with Tables and Graphs

4. Using the first lower class limit and the class width, list the other lower class limits. (Do this by adding the class width to the first lower class limit to get the second lower class limit. Add the class width to the second lower class limit to get the third lower class limit, and so on.)

5. List the lower class limits in a vertical column and then determine and enter the upper class limits.

6. Take each individual data value and put a tally mark in the appropriate class. Add the tally marks to find the total frequency for each class.

When constructing a frequency distribution, be sure the classes do not overlap. Each of the original values must belong to exactly one class. Include all classes, even those with a frequency of zero. Try to use the same width for all classes, although it is sometimes impossible to avoid open-ended intervals, such as “65 years or older.”

EXAMPLE 1 IQ Scores of Low Lead Group

Using the IQ scores of the low lead group in Table 2-1, follow the above procedure to construct the frequency distribution shown in Table 2-2. Use five classes.

S O L U T I O N

Step 1: Select 5 as the number of desired classes.

Step 2: Calculate the class width as shown below. Note that we round 18.2 up to 20, which is a much more convenient number.

Class width ≈ 1maximum data value2 - 1minimum data value2

number of classes

= 141 - 50

5 = 18.2 ≈ 20 1rounded up to a convenient number2

Step 3: The minimum data value is 50 and it is a convenient starting point, so use 50 as the first lower class limit. (If the minimum value had been 52 or 53, we would have rounded down to the more convenient starting point of 50.)

Step 4: Add the class width of 20 to 50 to get the second lower class limit of 70. Continue to add the class width of 20 until we have five lower class limits. The lower class limits are therefore 50, 70, 90, 110, and 130.

Step 5: List the lower class limits vertically, as shown in the margin. From this list, we identify the corresponding upper class limits as 69, 89, 109, 129, and 149.

Step 6: Enter a tally mark for each data value in the appropriate class. Then add the tally marks to find the frequencies shown in Table 2-2.

50–

70–

90–

110–

130–

Categorical Data So far we have discussed frequency distributions using only quan- titative data sets, but frequency distributions can also be used to summarize categori- cal (or qualitative or attribute) data, as illustrated in Example 2.

EXAMPLE 2 Emergency Room Visits for Injuries from Sports and Recreation

Table 2-3 lists data for the highest seven sources of injuries resulting in a visit to a hospital emergency room (ER) in a recent year (based on data from the Centers for Disease Control and Prevention). The activity names are categorical data at

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2-1 Frequency Distributions for Organizing and Summarizing Data 45

Relative Frequency Distribution A variation of the basic frequency distribution is a relative frequency distribution or percentage frequency distribution, in which each class frequency is replaced by a relative frequency (or proportion) or a percentage. In this text we use the term “rela- tive frequency distribution” whether we use relative frequencies or percentages. Rela- tive frequencies and percentages are calculated as follows.

Relative frequency for a class = frequency for a class

sum of all frequencies

Percentage for a class = frequency for a class

sum of all frequencies * 100%

Table 2-4 is an example of a relative frequency distribution. It is a variation of Table 2-2 in which each class frequency is replaced by the corresponding percent- age value. Because there are 78 data values, divide each class frequency by 78, and then multiply by 100%. The first class of Table 2-2 has a frequency of 2, so divide 2 by 78 to get 0.0256, and then multiply by 100% to get 2.56%, which we rounded to 2.6%. The sum of the percentages should be 100%, with a small discrepancy al- lowed for rounding errors, so a sum such as 99% or 101% is acceptable. The sum of the percentages in Table 2-4 is 100.1%.

The sum of the percentages in a relative frequency distribution must be very close to 100%.

Cumulative Frequency Distribution Another variation of a frequency distribution is a cumulative frequency distribu- tion in which the frequency for each class is the sum of the frequencies for that class and all previous classes. Table 2-5 is a cumulative frequency distribution based on Table 2-2. Using the original frequencies of 2, 33, 35, 7, and 1, we add 2 + 33 to get the second cumulative frequency of 35; then we add 2 + 33 + 35 to get the third; and so on. See Table 2-5, and note that in addition to the use of cumulative frequen- cies, the class limits are replaced by “less than” expressions that describe the new ranges of values.

TABLE 2-3 Annual ER Visits for Injuries from Sports and Recreation

Activity Frequency

Bicycling 26,212

Football 25,376

Playground 16,706

Basketball 13,987

Soccer 10,436

Baseball 9,634

All-terrain vehicle 6,337

the nominal level of measurement, but we can create the frequency distribution as shown. It might be surprising to see that bicycling is at the top of this list, but this doesn’t mean that bicycling is the most dangerous of these activities; many more people bicycle than play football or ride an all-terrain vehicle or do any of the other listed activities.

TABLE 2-4 Relative Frequency Distribution of IQ Scores of Low Lead Group

IQ Score Frequency

50–69 2.6%

70–89 42.3%

90–109 44.9%

110–129 9.0%

130–149 1.3%

TABLE 2-5 Cumulative Frequency Distribution of IQ Scores of Low Lead Group

IQ Score

Cumulative Frequency

Less than 70 2

Less than 90 35

Less than 110 70

Less than 130 77

Less than 150 78

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46 CHAPTER 2 Exploring Data with Tables and Graphs

Critical Thinking: Using Frequency Distributions to Understand Data At the beginning of this section we noted that a frequency distribution can help us un- derstand the distribution of a data set, which is the nature or shape of the spread of the data over the range of values (such as bell-shaped). In statistics we are often interested in determining whether the data have a normal distribution. (Normal distributions are discussed extensively in Chapter 6.) Data that have an approximately normal distribu- tion are characterized by a frequency distribution with the following features:

Normal Distribution

1. The frequencies start low, then increase to one or two high frequencies, and then decrease to a low frequency.

2. The distribution is approximately symmetric: Frequencies preceding the maximum frequency should be roughly a mirror image of those that follow the maximum frequency.

Table 2-6 satisfies these two conditions. The frequencies start low, increase to the max- imum of 56, and then decrease to a low frequency. Also, the frequencies of 1 and 10 that precede the maximum are a mirror image of the frequencies 10 and 1 that follow the maximum. Real data sets are usually not so perfect as Table 2-6, and judgment must be used to determine whether the distribution comes “close enough” to satisfying the above two conditions. (There are more objective procedures included later.)

TABLE 2-6 Frequency Distribution Showing a Normal Distribution

Score Frequency Normal Distribution

50–69 1 d Frequencies start low, . . .

70–89 10

90–109 56 d Increase to a maximum, . . .

110–129 10

130–149 1 d Decrease to become low again.

Analysis of Last Digits Example 3 illustrates this principle:

Frequencies of last digits sometimes reveal how the data were collected or measured.

EXAMPLE 3 Exploring Data: How Were the Weights Obtained in California?

When collecting weights of people, it’s better to actually weigh people than to ask them what they weigh. People often tend to round way down, so that a weight of 196 lb might be reported as 170 lb. Table 2-7 summarizes the last digits of the weights of 100 people used in the California Health Interview Survey. If people are actually weighed on a scale, the last digits of weights tend to have frequencies that are approximately the same, but Table 2-6 shows that the vast majority of weights have last digits of 0 or 5, and this is strong evidence that people reported their weights and were not physically weighed. (Also, the word “interview” in the title of the California Health Interview Survey reveals that people were interviewed and were not physically measured.)

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2-1 Frequency Distributions for Organizing and Summarizing Data 47

Gaps Example 4 illustrates this principle:

The presence of gaps can suggest that the data are from two or more di�erent populations.

The converse of this principle is not true, because data from different populations do not necessarily result in gaps.

TABLE 2-7 Last Digits of Weights from the California Health Interview Survey

Last Digit of Weight Frequency

0 46

1 1

2 2

3 3

4 3

5 30

6 4

7 0

8 8

9 3

EXAMPLE 4 Exploring Data: What Does a Gap Tell Us?

Table 2-8 is a frequency distribution of the heights (in.) of males. Examination of the frequencies reveals a large gap between the shortest males and the tallest males. This can be explained by the fact that half of the males are 7 years old and the other half are adults, so we really have samples from two different populations.

TABLE 2-8 Heights of Males

Height (in.) Frequency

40–44 3

45–49 17

50–54 29

55–59 1

60–64 0

65–69 24

70–74 23

75–79 3

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48 CHAPTER 2 Exploring Data with Tables and Graphs

TABLE 2-9 IQ Scores from the Low Lead Group and the High Lead Group

IQ Score Low Lead Group High Lead Group

50–69 2.6%

70–89 42.3% 71.4%

90–109 44.9% 28.6%

110–129 9.0%

130–149 1.3%

EXAMPLE 5 Comparing IQ Scores of the Low Lead Group and the High Lead Group

Table 2-1, which is given with the Chapter Problem at the beginning of this chapter, lists IQ scores from the low lead group and the high lead group. Because the sample sizes of 78 and 21 are so different, a comparison of frequency distributions is not easy, but Table 2-9 shows the relative frequency distributions for those two groups. By comparing those relative frequencies, we see that the majority of children in the low lead group had IQ scores of 90 or higher, but the majority of children in the high lead group had IQ scores below 90. This suggests that perhaps high lead expo- sure has a detrimental effect on IQ scores.

Frequency Distributions Access tech instructions, videos, and data sets at www.TriolaStats.com

TECH CENTER

Statistical Literacy and Critical Thinking

1. Cotinine in Smokers Refer to the accompanying table summarizing measured amounts of serum cotinine (ng/mL) from a sample of smokers (from Data Set 14 “Passive and Active Smoke” in Appendix B). When nicotine is absorbed by the body, cotinine is produced. How many subjects are included in the summary? Is it possible to identify the exact values of all of the original cotinine measurements?

2-1 Basic Skills and Concepts

Cotinine (ng ,mL) Frequency 0–99 11

100–199 12

200–299 14

300–399 1

400-499 2 2. Cotinine in Smokers Refer to the accompanying frequency distribution. What problem is created by using classes of 0–100, 100–200, . . . ?

3. Relative Frequency Distribution Use percentages to construct the relative frequency dis- tribution corresponding to the accompanying frequency distribution for cotinine amounts.

4. What’s Wrong? Heights of adult males are known to have a normal distribution, as de- scribed in this section. A researcher claims to have randomly selected adult males and mea- sured their heights with the resulting relative frequency distribution as shown here. Identify two major flaws with theses results.

Height (cm)

Relative Frequency

130–144 23%

145–159 25%

160–174 22%

175–189 27%

190–204 28%

Comparisons Example 5 illustrates this principle:

Combining two or more relative frequency distributions in one table makes comparisons of data much easier.

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2-1 Frequency Distributions for Organizing and Summarizing Data 49

In Exercises 5–8, identify the class width, class midpoints, and class boundaries for the given frequency distribution. The frequency distributions are based on real data from Appendix B.

5. 

Cotinine (NonSmokers Exposed to Smoke

in ng ,mL)

Frequency

0–99 34

100–199 2

200–299 1

300–399 1

400–499 0

500–599 2

6. 

Brain Volume (cm3) Frequency

960–1049 6

1050–1139 7

1140–1229 3

1230–1319 2

1320–1409 1

1410–1499 1

7. 

Blood Platelet Count of Males

Frequency

0–99 1

100–199 51

200–299 90

300–399 10

400–499 0

500–599 0

600–699 1

8. 

Blood Platelet Count of Females

Frequency

100–199 25

200–299 92

300–399 28

400–499 0

500–599 2

Normal Distributions. In Exercises 9–12, answer the given questions, which are related to normal distributions.

9. Cotinine Determine whether the frequency distribution given in Exercise 5 is approximately a normal distribution. Explain.

10. Brain Volume Refer to the frequency distribution given in Exercise 6 and ignore the given fre- quencies. Assume that the first three frequencies are 1, 3, and 6, respectively. Assuming that the dis- tribution of the 20 sample values is a normal distribution, identify the remaining three frequencies.

11. Normal Distribution Refer to the frequency distribution given in Exercise 7 and ignore the given frequencies. Assume that the first three frequencies are 2, 12, and 18, respectively. Assuming that the distribution of the 153 sample values is a normal distribution, identify the remaining four frequencies.

12.  Normal Distribution Refer to the frequency distribution given in Exercise 8 and deter- mine whether it appears to be a normal distribution. Explain.

Constructing Frequency Distributions. In Exercises 13–22, use the indicated data and construct the frequency distribution. (The data for Exercises 13–22 can be downloaded at TriolaStats.com.)

13. Pulse Rates of Males Refer to Data Set 1 “Body Data” in Appendix B and use the pulse rates (beats per minute) of males. Begin with a lower class limit of 40 and use a class width of 10. Do the pulse rates of males appear to have a normal distribution?

14.  Pulse Rates of Females Refer to Data Set 1 “Body Data” in Appendix B and use the pulse rates (beats per minute) of females. Begin with a lower class limit of 30 and use a class width of 10. Do the pulse rates of females appear to have a normal distribution?

15.  Lead and IQ Refer to Data Set 8 “IQ and Lead” in Appendix B and use the verbal IQ scores of the low lead group. Begin with a lower class limit of 50 and use a class width of 10. Do these IQ scores appear to be normally distributed?

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50 CHAPTER 2 Exploring Data with Tables and Graphs

16.  Lead and IQ Refer to Data Set 8 “IQ and Lead” in Appendix B and use the verbal IQ scores of the high lead group. Begin with a lower class limit of 60 and use a class width of 10. Do these IQ scores appear to be normally distributed?

17. Male Red Blood Cell Counts Refer to Data Set 1 “Body Data” in Appendix B and use the red blood cell counts (million cells>mL) for males. Begin with a lower class limit of 3.00 and use a class width of 0.50. Using a very loose interpretation of the requirements for a nor- mal distribution, do the red blood cell counts appear to be normally distributed?

18. Female Red Blood Cell Counts Repeat the preceding exercise using the red blood cell counts for females.

19. Freshman 15 Refer to Data Set 10 “Freshman 15” in Appendix B and use the weights (kg) of males in September of their freshman year. Begin with a lower class limit of 50 kg and use a class width of 10 kg.

20.  Freshman 15 Repeat the preceding exercise using the weights (kg) of males in April. Compare the result to the frequency distribution from the preceding exercise. Does it appear that males gain 15 lb (or 6.8 kg) during their freshman year?

21. Analysis of Last Digits Heights of statistics students were obtained by one of the authors as part of an experiment conducted for class. The last digits of those heights are listed below. Construct a frequency distribution with 10 classes. Based on the distribution, do the heights ap- pear to be reported or actually measured? What do you know about the accuracy of the results?

0 0 0 0 0 0 0 0 0 1 1 2 3 3 3 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 8 8 8 9

22. Analysis of Last Digits Listed below are the last digits of weights of subjects. After con- structing the frequency distribution, does it appear that the weights were reported or physically measured? Explain.

2 7 7 3 2 8 5 9 7 2 8 6 9 2 7 5 6 4 0 7 6 8 4 0 4

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

Relative Frequencies for Comparisons. In Exercises 23 and 24, find the relative fre- quencies and answer the given questions.

23. Cotinine Construct one table (similar to Table 2-9 on page 48) that includes relative fre- quencies based on the frequency distributions from Exercise 1 (smokers) and Exercise 5 (non- smokers exposed to smoke), and then compare them. Are there notable differences?

24. Blood Platelet Counts Construct one table (similar to Table 2-9 on page 48) that includes relative frequencies based on the frequency distributions from Exercises 7 and 8, and then com- pare them. Are there notable differences?

Cumulative Frequency Distributions. In Exercises 25 and 26, construct the cumulative frequency distribution that corresponds to the frequency distribution in the exercise indicated.

25. Exercise 5

26. Exercise 6

27. Interpreting Effects of Outliers Exercise 5 in this section involved cotinine levels of nonsmokers who were exposed to tobacco smoke. (See the middle column in Data Set 14 “Passive and Active Smoke” in Appendix B.)

a. Identify any outliers.

b. After adding another value of 999 to the cotinine levels of nonsmokers exposed to smoke, construct the frequency distribution as in Exercise 5. How is the frequency distribution affected by the addition of the outlier 999? State a generalization about the effect of an outlier on a frequency distribution.

2-1 Beyond the Basics

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2-2 Histograms 51

Important Uses of a Histogram

■ Visually displays the shape of the distribution of the data

■ Shows the location of the center of the data

■ Shows the spread of the data

■ Identifies outliers

A histogram is basically a graph of a frequency distribution. For example, Figure 2-2 shows the histogram corresponding to the frequency distribution given in Table 2-2 on page 42.

Class frequencies should be used for the vertical scale and that scale should be la- beled as in Figure 2-2. There is no universal agreement on the procedure for selecting which values are used for the bar locations along the horizontal scale, but it is com- mon to use class boundaries (as shown in Figure 2-2) or class midpoints or class limits or something else. It is often easier for us mere mortals to use class midpoints for the horizontal scale. Histograms can usually be generated using technology.

Relative Frequency Histogram A relative frequency histogram has the same shape and horizontal scale as a histo- gram, but the vertical scale uses relative frequencies (as percentages or proportions) instead of actual frequencies. Figure 2-3 is the relative frequency histogram corre- sponding to Figure 2-2.

PA RT 1 Basic Concepts of Histograms Key Concept While a frequency distribution is a useful tool for summarizing data and investigating the distribution of data, an even better tool is a histogram, which is a graph that is easier to interpret than a table of numbers.

2-2 Histograms

DEFINITION

A histogram is a graph consisting of bars of equal width drawn adjacent to each other (unless there are gaps in the data). The horizontal scale represents classes of quantitative data values, and the vertical scale represents frequencies. The heights of the bars correspond to frequency values.

FIGURE 2-2 Histogram FIGURE 2-3 Relative Frequency Histogram

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52 CHAPTER 2 Exploring Data with Tables and Graphs

Critical Thinking: Interpreting Histograms The ultimate objective of a histogram is to understand characteristics of the data. Ex- plore the data by analyzing the histogram to see what can be learned about “CVDOT”: the center of the data, the variation (which will be discussed at length in Section 3-2), the shape of the distribution, whether there are any outliers (values far away from the other values), and time (whether there is any change in the characteristics of the data over time). Examining Figure 2-2, we see that the histogram is centered close to 90, the values vary from around 50 to 150, and the distribution is roughly bell-shaped. There aren’t any outliers and any changes in time are irrelevant for these data.

Common Distribution Shapes The histograms shown in Figure 2-4 depict four common distribution shapes.

Normal Distribution

When graphed as a histogram, data with a normal distribution have a “bell” shape similar to the one superimposed in Figure 2-5. Many collections of data have a dis- tribution that is approximately normal. Many statistical methods require that sample data come from a population having a distribution that is approximately a normal dis- tribution, and we can often use a histogram to judge whether this requirement is satis- fied. There are more advanced and less subjective methods for determining whether the distribution is a normal distribution. Normal quantile plots are very helpful for assessing normality: see Part 2 of this section.

FIGURE 2-4 Common Distributions

(a) (b)

(c) (d)

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2-2 Histograms 53

Uniform Distribution

The different possible values occur with approximately the same frequency, so the heights of the bars in the histogram are approximately uniform, as in Figure 2-4(b). Figure 2-4(b) depicts outcomes of last digits of weights from a large sample of ran- domly selected subjects, and such a graph is helpful in determining whether the sub- jects were actually weighed or whether they reported their weights.

Population sizes of an organism are often uniformly distributed when they are found in equally sized areas of a region where they must compete for a limited re- source. For example, redwood trees must compete for light, and numbers of redwood trees in equally sized areas of a region tend to be uniformly distributed.

Skewness A distribution of data is skewed if it is not symmetric and extends more to one side than to the other. Data skewed to the right (also called positively skewed) have a longer right tail, as in Figure 2-4(c). Annual incomes of adult Americans are skewed to the right; death rates of nations are skewed to the right. Data skewed to the left (also called negatively skewed) have a longer left tail, as in Figure 2-4(d). Life span data in humans are skewed to the left. (Here’s a mnemonic for remembering skew- ness: A distribution skewed to the right resembles the toes on your right foot, and one skewed to the left resembles the toes on your left foot.) Distributions skewed to the right are more common than those skewed to the left because it’s often easier to get exceptionally large values than values that are exceptionally small. With annual incomes, for example, it’s impossible to get values below zero, but there are a few people who earn millions or billions of dollars in a year. Annual incomes therefore tend to be skewed to the right.

PA RT 2 Assessing Normality with Normal Quantile Plots

Some methods presented in later chapters have a requirement that sample data must be from a population having a normal distribution. Histograms can be helpful in de- termining whether the normality requirement is satisfied, but they are not very help- ful with small data sets. Section 6-5 discusses methods for assessing normality—that is, determining whether the sample data are from a normally distributed population. Section 6-5 includes a procedure for constructing normal quantile plots, which are

FIGURE 2-5 Bell-Shaped Distribution Because this histogram is roughly bell-shaped, we say that the data have a normal distribution. (A more rigorous definition will be given in Chapter 6.)

Remembering Skewness: Skewed Left: Resembles

toes on left foot

Skewed Right: Resembles toes on right foot

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54 CHAPTER 2 Exploring Data with Tables and Graphs

easy to generate using technology such as Statdisk, SPSS, JMP, Minitab, XLSTAT, StatCrunch, or a TI-83>84 Plus calculator. Interpretation of a normal quantile plot is based on the following criteria:

Criteria for Assessing Normality with a Normal Quantile Plot

Normal Distribution: The population distribution is normal if the pattern of the points in the normal quantile plot is reasonably close to a straight line, and the points do not show some systematic pattern that is not a straight-line pattern.

Not a Normal Distribution: The population distribution is not normal if the normal quantile plot has either or both of these two conditions:

•  The points do not lie reasonably close to a straight-line pattern. •  The points show some systematic pattern that is not a straight-line pattern.

The following are examples of normal quantile plots. Procedures for creating such plots are described in Section 6-5.

Normal Distribution: The points are reasonably close to a straight-line pattern, and there is no other systematic pattern that is not a straight-line pattern.

Not a Normal Distribution: The points do not lie reasonably close to a straight line.

Not a Normal Distribution: The points show a systematic pattern that is not a straight-line pattern.

Histograms Access tech instructions, videos, and data sets at www.TriolaStats.com

TECH CENTER

Statistical Literacy and Critical Thinking

1. Histogram Table 2-2 is a frequency distribution summarizing the IQ scores of the low lead group listed in Table 2-1 on page 41, and Figure 2-2 on page 51 is a histogram depicting that same data set. When trying to better understand the IQ data, what is the advantage of examin- ing the histogram instead of the frequency distribution?

2. Voluntary Response Sample The histogram in Figure 2-2 on page 51 is constructed from a simple random sample of children. If you construct a histogram with data collected from a voluntary response sample, will the distribution depicted in the histogram reflect the true dis- tribution of the population? Why or why not?

3. Blood Platelet Counts Listed below are blood platelet counts (1000 cells>mL) randomly selected from adults in the United States. Why does it not make sense to construct a histogram for this data set?

191 286 263 193 193 215 162 646 250 386

2-2 Basic Skills and Concepts

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2-2 Histograms 55

4. Normal Distribution When it refers to a normal distribution, does the term “normal” have the same meaning as in ordinary language? What criterion can be used to determine whether the data depicted in a histogram have a distribution that is approximately a normal distribution? Is this criterion totally objective, or does it involve subjective judgment?

Interpreting a Histogram. In Exercises 5–8, answer the questions by referring to the fol- lowing histogram, which represents the sepal widths (mm) of a sample of irises. (See Data Set 16 “Iris Measurements” in Appendix B.)

5. Sample Size Based on the histogram, what is the approximate number of irises in the sample?

6. Class Width and Class Limits What is the class width? What are the approximate lower and upper class limits of the first class?

7. Outlier? What is the largest possible value? Would that value be an outlier?

8. Normal Distribution Does it appear that the sample is from a population having a normal distribution?

Constructing Histograms. In Exercises 9–18, construct the histograms and answer the given questions. Use class midpoint values for the horizontal scale.

9.  Pulse Rates of Males Use the frequency distribution from Exercise 13 in Section 2-1 on page 49 to construct a histogram. Do the pulse rates of males appear to have a normal distribution?

10. Pulse Rates of Females Use the frequency distribution from Exercise 14 in Section 2-1 on page 49 to construct a histogram. Do the pulse rates of females appear to have a normal distribution?

11. Lead and IQ Use the frequency distribution from Exercise 15 in Section 2-1 on page 49 to construct a histogram. Do the IQ scores appear to have a normal distribution?

12. Lead and IQ Use the frequency distribution from Exercise 16 in Section 2-1 on page 50 to construct a histogram. Do the IQ scores appear to have a normal distribution?

13.  Male Red Blood Cell Counts Use the frequency distribution from Exercise 17 in Section 2-1 on page 50 to construct a histogram. Do the red blood cell counts appear to have a normal distribution?

14.  Female Red Blood Cell Counts Use the frequency distribution from Exercise 18 in Section 2-1 on page 50 to construct a histogram. Do the red blood cell counts appear to have a normal distribution?

15. Freshman 15 Use the frequency distribution from Exercise 19 in Section 2-1 on page 50 to construct a histogram.

16. Freshman 15 Use the frequency distribution from Exercise 20 in Section 2-1 on page 50 to construct a histogram.

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56 CHAPTER 2 Exploring Data with Tables and Graphs

17. Last Digit Analysis Use the frequency distribution from Exercise 21 in Section 2-1 on page 50 to construct a histogram. What does the histogram suggest about the method used to collect the heights?

18. Last Digit Analysis Use the frequency distribution from Exercise 22 in Section 2-1 on page 50 to construct a histogram. What does the histogram suggest about the method used to collect the weights?

2-2 Beyond the Basics

Key Concept Section 2-2 introduced the histogram, and this section introduces other common graphs that foster understanding of data. We also discuss some graphs that are deceptive because they create impressions about data that are somehow mislead- ing or wrong.

The era of charming and primitive hand-drawn graphs has passed, and technol- ogy now provides us with powerful tools for generating a wide variety of graphs. Here we go.

Graphs That Enlighten Dotplots

A dotplot consists of a graph of quantitative data in which each data value is plotted as a point (or dot) above a horizontal scale of values. Dots representing equal values are stacked.

Features of a Dotplot

■ Displays the shape of the distribution of data.

■ It is usually possible to recreate the original list of data values.

2-3 Graphs That Enlighten and Graphs That Deceive

19. Interpreting Normal Quantile Plots Which of the following normal quantile plots appear to represent data from a population having a normal distribution? Explain.

(a) (b)

(c) (d)

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2-3 Graphs That Enlighten and Graphs That Deceive 57

Stemplots

A stemplot (or stem-and-leaf plot) represents quantitative data by separating each value into two parts: the stem (such as the leftmost digit) and the leaf (such as the rightmost digit). Better stemplots are often obtained by first rounding the original data values. Also, stemplots can be expanded to include more rows and can be condensed to include fewer rows.

Features of a Stemplot

■ Shows the shape of the distribution of the data.

■ Retains the original data values.

■ The sample data are sorted (arranged in order).

FIGURE 2-6 Dotplot of Pulse Rates of Males

EXAMPLE 1 Dotplot of Pulse Rates of Males

Figure 2-6 shows a dotplot of the pulse rates (beats per minute) of males from Data Set 1 “Body Data” in Appendix B. The two stacked dots above the position at 50 in- dicate that two of the pulse rates are 50. (In this dotplot, the horizontal scale allows even numbers only, but the original pulse rates are all even numbers.)

EXAMPLE 2 Stemplot of Male Pulse Rates

The following stemplot displays the pulse rates of the males in Data Set 1 “Body Data” in Appendix B. The lowest pulse rate of 40 is separated into the stem of 4 and the leaf of 0. The stems and leaves are arranged in increasing order, not the order in which they occur in the original list. If you turn the stemplot on its side, you can see the distribution of the IQ scores in the same way you would see it in a histogram or dotplot.

Pulse rates are 40 and 42

Pulse rates are 90, 92, 94, 96, 96

Time-Series Graph

A time-series graph is a graph of time-series data, which are quantitative data that have been collected at different points in time, such as monthly or yearly. An advan- tage of a time-series graph is that it reveals information about trends over time.

Features of a Time-series Graph

■ Reveals information about trends over time

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58 CHAPTER 2 Exploring Data with Tables and Graphs

Bar Graphs

A bar graph uses bars of equal width to show frequencies of categories of categori- cal (or qualitative) data. The bars may or may not be separated by small gaps.

Feature of a Bar Graph

■ Shows the relative distribution of categorical data so that it is easier to compare the different categories

Pareto Charts

A Pareto chart is a bar graph for categorical data, with the added stipulation that the bars are arranged in descending order according to frequencies, so the bars decrease in height from left to right.

Features of a Pareto Chart

■ Shows the relative distribution of categorical data so that it is easier to compare the different categories

■ Draws attention to the more important categories

FIGURE 2-7 Time-Series Graph of Law Enforcement Fatalities

EXAMPLE 3 Time-Series Graph of Fatalities of Law Enforcement Officers

The time-series graph shown in Figure 2-7 depicts the yearly number of fatalities of law enforcement officers in the United States. See that a spike occurred in 2001, the year of the September 11, 2001 terrorist attacks. Except for the data from 2001, there appears to be a slight downward trend.

EXAMPLE 4 Pareto Chart of Causes of Accidental Deaths

For the accidental deaths in a recent year, Figure 2-8 shows the most common causes. We can see that deaths from poison represent the most serious problem. (Deaths from poison include deaths from drug overdoses.)

The Power of a Graph

With annual

sales around

$13 billion and

with roughly

50 million

people using

it, Pfizer’s

prescription drug Lipitor (ator-

vastatin) has become the most

profitable and most widely used

prescription drug ever marketed.

In the early stages of its develop-

ment, Lipitor was compared to

other drugs (Zocor [simvastatin],

Mevacor [lovastatin], Lescol

[fluvastatin], and Pravachol

pravastatin) in a process that

involved controlled trials. The

summary report included a graph

showing a Lipitor curve that had

a steeper rise than the curves for

the other drugs, visually showing

that Lipitor was more effective

in reducing cholesterol than the

other drugs. Pat Kelly, who was

then a senior marketing execu-

tive for Pfizer, said, “I will never

forget seeing that chart…. It was

like ‘Aha!’ Now I know what this

is about. We can communicate

this!” The Food and Drug Admin-

istration approved Lipitor and al-

lowed Pfizer to include the graph

with each prescription. Pfizer

sales personnel also distributed

the graph to physicians.

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2-3 Graphs That Enlighten and Graphs That Deceive 59

Pie Charts

A pie chart is a very common graph that depicts categorical data as slices of a circle, in which the size of each slice is proportional to the frequency count for the category. Although pie charts are very common, they are not as effective as Pareto charts.

Feature of a Pie Chart

■ Shows the distribution of categorical data in a commonly used format

FIGURE 2-8 Pareto Chart of Causes of Accidental Deaths

EXAMPLE 5 Pie Chart of Causes of Accidental Deaths

Figure 2-9 is a pie chart of the same cause of death data from Example 4. Construc- tion of a pie chart involves slicing up the circle into the proper proportions that rep- resent relative frequencies. For example, the category poison accounts for 34% of the total, so the slice representing poison should be 34% of the total (with a central angle of 0.34 * 360° = 122°).

FIGURE 2-9 Pie Chart of Causes of Accidental Deaths

The Pareto chart in Figure 2-8 and the pie chart in Figure 2-9 depict the same data in different ways, but the Pareto chart does a better job of showing the relative sizes of the different components. Graphics expert Edwin Tufte makes the following suggestion:

Never use pie charts because they waste ink on components that are not data, and they lack an appropriate scale.

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60 CHAPTER 2 Exploring Data with Tables and Graphs

Frequency Polygon

A frequency polygon uses line segments connected to points located directly above class midpoint values. A frequency polygon is very similar to a histogram, but a fre- quency polygon uses line segments instead of bars.

A variation of the basic frequency polygon is the relative frequency polygon, which uses relative frequencies (proportions or percentages) for the vertical scale. An advantage of relative frequency polygons is that two or more of them can be combined on a single graph for easy comparison, as in Figure 2-11.

FIGURE 2-10 Frequency Polygon of Full IQ Scores of Low Lead Group

EXAMPLE 6 Frequency Polygon of Full IQ Scores of Low Lead Group

See Figure 2-10 for the frequency polygon corresponding to the full IQ scores of the low lead group summarized in the frequency distribution of Table 2-2 on page 42 (from Data Set 8 in Appendix B). The heights of the points correspond to the class frequencies, and the line segments are extended to the right and left so that the graph begins and ends on the horizontal axis. The points are plotted directly above class midpoint values.

EXAMPLE 7 Relative Frequency Polygon: IQ Scores of Lead Groups

Figure 2-11 shows the relative frequency polygons for the full IQ scores of two groups: (1) group with low blood lead levels; (2) group with high blood lead levels. Here, relative frequency polygons are much better than frequency polygons because the different sample sizes of 21 and 78 would have made a comparison difficult, but that difficulty is removed by using relative percentages.

Figure 2-11 shows that the group with high blood lead levels has full IQ scores that are somewhat lower than those in the low blood level group. This suggests that exposure to lead has an effect on IQ scores. Later chapters will provide us with more tools that allow us to examine this issue beyond the subjective interpretation of a graph.

Frequency

A class midpoint values. A frequency polygon is very similar to a histogram, but a fre quency polygon uses line segments instead of bars.

Florence Nightingale

Florence

Nightingale

(1820–1910)

is known to

many as the

founder of

the nursing

profession, but she also saved

thousands of lives by using

statistics. When she encountered

an unsanitary and undersup-

plied hospital, she improved

those conditions and then used

statistics to convince others of

the need for more widespread

medical reform. She developed

original graphs to illustrate that

during the Crimean War, more

soldiers died as a result of

unsanitary conditions than were

killed in combat. Florence Night-

ingale pioneered the use of social

statistics as well as graphics

techniques.

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2-3 Graphs That Enlighten and Graphs That Deceive 61

Graphs That Deceive Deceptive graphs are commonly used to mislead people, and we really don’t want statistics students to be among those susceptible to such deceptions. Graphs should be constructed in a way that is fair and objective. The readers should be allowed to make their own judgments, instead of being manipulated by misleading graphs. We present two of the ways in which graphs are commonly used to misrepresent data.

Nonzero Vertical Axis

A common deceptive graph involves using a vertical scale that starts at some value greater than zero to exaggerate differences between groups.

FIGURE 2-11 Relative Frequency Polygons for Full IQ Scores of High and Low Lead Groups

NONZERO AXIS: Always examine a graph carefully to see whether a vertical axis begins at some point other than zero so that differences are exaggerated.

EXAMPLE 8 Nonzero Axis

Figure 2-12(a) and Figure 2-12(b) are based on the same data from a clinical trial of OxyContin (oxycodone), a drug used to treat moderate to severe pain. The results of that clinical trial included the percentage of subjects who experienced nausea in an OxyContin treatment group and the percentage in a group given a placebo.

FIGURE 2-12 Nausea in a Clinical Trial

(a) (b)

continued

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62 CHAPTER 2 Exploring Data with Tables and Graphs

Pictographs

Drawings of objects, called pictographs, are often misleading. Data that are one- dimensional in nature (such as budget amounts) are often depicted with two-dimensional objects (such as dollar bills) or three-dimensional objects (such as stacks of coins, homes, or barrels). With pictographs, artists can create false impressions that grossly distort differences by using these simple principles of basic geometry: (1) When you double each side of a square, its area doesn’t merely double; it increases by a factor of four. (2) When you double each side of a cube, its volume doesn’t merely double; it increases by a factor of eight.

By using a vertical scale that starts at 10% instead of 0%, Figure 2-12(a) grossly exaggerates the difference between the two groups. Figure 2-12(a) makes it appear that those using OxyContin experience nausea at a rate that is about 12 times higher than the rate for those using a placebo, but Figure 2-12(b) shows that the true ratio is about 2:1, not 12:1. Perhaps someone wants to discourage recreational use of OxyContin by misleading people into thinking that the problem with nausea is much greater than it really is. The objective might be sincere, but the use of a mis- leading graph is not the way to achieve that objective.

PICTOGRAPHS: When examining data depicted with a pictograph, determine whether the graph is misleading because objects of area or volume are used to depict amounts that are actually one-dimensional. (Histograms and bar charts represent one-dimensional data with two-dimensional bars, but they use bars with the same width so that the graph is not misleading.)

EXAMPLE 9 Pictograph of Cigarette Smokers

Refer to Figure 2-13 and see that the larger cigarette is about twice as long, twice as tall, and twice as deep as the smaller cigarette, so the volume of the larger cigarette is about eight times the volume of the smaller cigarette. (The data are from the Cen- ters for Disease Control and Prevention.) The larger cigarette appears to be eight times as large as the smaller cigarette, but the actual percentages show that the 37% smoking rate in 1970 is about twice that of the 18% rate in 2013.

FIGURE 2-13 Smoking by U.S. Adults 1970: 37% of U.S. adults smoked. 2013: 18% of U.S. adults smoked.

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2-3 Graphs That Enlighten and Graphs That Deceive 63

Concluding Thoughts In addition to the graphs we have discussed in this section, there are many other useful graphs—some of which have not yet been created. The world desperately needs more people who can create original graphs that enlighten us about the nature of data. In The Visual Display of Quantitative Information, Edward Tufte offers these principles:

■ For small data sets of 20 values or fewer, use a table instead of a graph.

■ A graph of data should make us focus on the true nature of the data, not on other elements, such as eye-catching but distracting design features.

■ Do not distort data; construct a graph to reveal the true nature of the data.

■ Almost all of the ink in a graph should be used for the data, not for other design elements.

Graphing Capabilities Access tech instructions, videos, and data sets at www.TriolaStats.com

TECH CENTER

Statistical Literacy and Critical Thinking

1. Body Temperatures Listed below are body temperatures (°F) of healthy adults. Why is it that a graph of these data would not be very effective in helping us understand the data?

98.6 98.6 98.0 98.0 99.0 98.4 98.4 98.4 98.4 98.6

2. Voluntary Response Data If we have a large voluntary response sample consisting of weights of subjects who chose to respond to a survey posted on the Internet, can a graph help to overcome the deficiency of having a voluntary response sample?

3. Ethics There are data showing that smoking is detrimental to good health. Given that people could be helped and lives could be saved by reducing smoking, is it ethical to graph the data in a way that is misleading by exaggerating the health risks of smoking?

4. CVDOT Section 2-1 introduced important characteristics of data summarized by the acro- nym CVDOT. What characteristics do those letters represent, and which graph does the best job of giving us insight into the last of those characteristics?

Dotplots. In Exercises 5 and 6, construct the dotplot.

5. Pulse Rates Listed below are pulse rates (beats per minute) of females selected from Data Set 1 “Body Data” in Appendix B. All of those pulse rates are even numbers. Is there a pulse rate that appears to be an outlier? What is its value?

80 94 58 66 56 82 78 86 88 56 36 66 84 76 78 64 66 78 60 64

6. Diastolic Blood Pressure Listed below are diastolic blood pressure measurements (mm Hg) of females selected from Data Set 1 “Body Data” in Appendix B. All of the values are even numbers. Are there any outliers? If so, identify their values.

62 70 72 88 70 66 68 70 82 74 90 62 70 76 90 86 60 78 82 78 84 76 60 64

Stemplots. In Exercises 7 and 8, construct the stemplot.

7. Pulse Rates Refer to the data listed in Exercise 5. How are the data sorted in the stemplot?

8. Diastolic Blood Pressure Refer to the data listed in Exercise 6. Identify the two values that are closest to the middle when the data are sorted in order from lowest to highest. (These values are often used to find the median, which is defined in Section 3-1.)

2-3 Basic Skills and Concepts

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64 CHAPTER 2 Exploring Data with Tables and Graphs

Time-Series Graphs. In Exercises 9 and 10, construct the time-series graph.

9. Triplets Listed below are the numbers of triplets born in the United States each year beginning with 1995. Is there a trend?

4551 5298 6148 6919 6742 6742 6885 6898 7110

6750 6208 6118 5967 5877 5905 5153 5137 4598

10. Drunk Driving Fatalities Listed below are annual fatality rates (per 100,000 population) from drunk driving. The first entry represents the year 1991. Is there a trend? Any explanation?

6.3 5.5 5.3 5.1 5.1 5.1 4.8 4.6 4.6 4.7 4.7

4.7 4.5 4.5 4.6 4.5 4.3 3.9 3.5 3.3 3.2 3.3

Pareto Charts. In Exercises 11 and 12 construct the Pareto chart.

11. Journal Retractions In a study of retractions in biomedical journals, 436 were due to error, 201 were due to plagiarism, 888 were due to fraud, 291 were duplications of publica- tions, and 287 had other causes (based on data from “Misconduct Accounts for the Majority of Retracted Scientific Publications,” by Fang, Steen, Casadevall, Proceedings of the National Academy of Sciences of the United States of America, Vol. 110, No. 3). Among such retrac- tions, does misconduct (fraud, duplication, plagiarism) appear to be a major factor?

12. Getting a Job In a survey, subjects seeking a job were asked to whom they send a thank- you note after having a job interview. Results were as follows: 40 said only the person they spent the most time with, 40 said only the most senior-level person, 396 said everyone that they met, 15 said the person that they had the best conversation with, and 10 said that they don’t send thank-you notes (based on data from TheLadders.com). Comment on the results.

Pie Charts. In Exercises 13 and 14, construct the pie chart.

13. Journal Retractions Use the data from Exercise 11 “Journal Retractions.”

14. Getting a Job Use the data from Exercise 12 “Getting a Job.”

Frequency Polygon. In Exercises 15 and 16, construct the frequency polygons.

15. Pulse Rates of Males Use the frequency distribution for the pulse rates of males from Exercise 13 in Section 2-1 on page 49 to construct a frequency polygon. Comment on the shape of the distribution.

16. Pulse Rates of Females Use the frequency distribution for the pulse rates of females from Exercise 14 in Section 2-1 on page 49 to construct a frequency polygon. Comment on the shape of the distribution.

Deceptive Graphs. In Exercises 17–18, identify how the graph is deceptive.

17. Self-Driving Vehicles In a survey of adults, subjects were asked if they felt comfortable being in a self-driving vehicle. The accompanying graph depicts the results (based on data from TE Connectivity).

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2-4 Scatterplots, Correlation, and Regression 65

18. Cost of Giving Birth According to the Agency for Healthcare Research and Quality Healthcare Cost and Utilization Project, the typical cost of a C-section baby delivery is $4500, and the typical cost of a vaginal delivery is $2600. See the accompanying illustration.

Cost of C-Section Delivery: $4500 Cost of Vaginal Delivery: $2600

Key Concept This section introduces the analysis of paired (or “bivariate”) sample data, which are data from two different variables that are paired in some way, such as the variables of heights and weights from subjects. In Part 1 of this section we dis- cuss correlation and the role of a graph called a scatterplot. In Part 2 we provide an introduction to the use of the linear correlation coefficient. In Part 3 we provide a very brief discussion of linear regression, which involves the equation and graph of the straight line that best fits the sample paired data.

All of the principles discussed in this section are discussed more fully in Chapter 10, but this section serves as a quick introduction to some important concepts of correlation and regression. This section does not include details for executing manual calculations, and those calculations are rarely done. Instructions for using technology to obtain results can be found at www.TriolaStats.com; refer to the instructions for Chapter 10.

PA RT 1 Scatterplot and Correlation Our objective in this section is to explore whether there is a correlation, or associa- tion, between two variables. We begin with basic definitions.

2-4 Scatterplots, Correlation, and Regression

DEFINITIONS

A correlation exists between two variables when the values of one variable are somehow associated with the values of the other variable.

A linear correlation exists between two variables when there is a correlation and the plotted points of paired data result in a pattern that can be approximated by a straight line.

A scatterplot or scatter diagram is a plot of paired (x, y) quantitative data with a horizontal x-axis and a vertical y-axis. The horizontal axis is used for the first vari- able (x), and the vertical axis is used for the second variable (y).

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66 CHAPTER 2 Exploring Data with Tables and Graphs

A scatterplot can be used as a visual aid in determining whether there is a correlation (or relationship) between the two variables. (This issue is discussed at length when the topic of correlation is considered in Section 10-1.)

CAUTION: The presence of a correlation between two variables is not evidence that one of the variables causes the other. We might find a correlation between beer consumption and weight, but we cannot conclude from the statistical evidence that drinking beer has a direct effect on weight.

Correlation does not imply causality!

EXAMPLE 1 Correlation: Waist and Arm Circumference

Data Set 1 “Body Data” in Appendix B includes waist circumferences (cm) and arm circumferences (cm) of randomly selected adult subjects. Figure 2-14 is a scatter- plot of the paired waist>arm measurements. The points show a pattern of increasing values from left to right. This pattern suggests that there is a correlation or relation- ship between waist circumferences and arm circumferences.

EXAMPLE 2 No Correlation: Weight and Pulse Rate

Data Set 1 “Body Data” in Appendix B includes weights (kg) and pulse rates (beats per minute) of randomly selected adult subjects. Figure 2-15 is a scatterplot of the paired weight>pulse rate measurements. The points in Figure 2-15 do not show any obvious pattern, and this lack of a pattern suggests that there is no correlation or re- lationship between weights and pulse rates.

FIGURE 2-14 Waist and Arm Circumferences Correlation: The distinct straight-line pattern of the plotted points suggests that there is a correlation between waist circumferences and arm circumference.

FIGURE 2-15 Weights and Pulse Rates No Correlation: The plotted points do not show a distinct pattern, so it appears that there is no correlation between weights and pulse rates.

The preceding two examples involve making decisions about a correlation based on subjective judgments of scatterplots, but Part 2 introduces the linear corre- lation coefficient as a numerical measure that can help us make such decisions more objectively. Using paired data, we can calculate the value of the linear correlation coefficient r.

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2-4 Scatterplots, Correlation, and Regression 67

PA RT 2 Linear Correlation Coefficient r Using paired data, we can calculate the value of the linear correlation coefficient r.

DEFINITION

The linear correlation coefficient is denoted by r, and it measures the strength of the linear association between two variables.

The value of a linear correlation coefficient r can be manually computed by applying Formula 10-1 or Formula 10-2 found in Section 10-1 on page 447, but in practice, r is almost always found by using technology.

Using r for Determining Correlation

The computed value of the linear correlation coefficient is always between - 1 and 1. A value of exactly - 1 or 1 implies that all of the data fall exactly on a line, which reflects a perfect correlation. If r is close to - 1 or close to 1, there appears to be a strong correlation, but if r is close to 0, there appears to be a weak or no linear cor- relation. For the data depicted in the scatterplot of Figure 2-14, r = 0.802 (somewhat close to 1), and the data in the scatterplot of Figure 2-15 result in r = 0.082 (pretty close to 0). These descriptions of “close to” - 1 or 1 or 0 are vague, but there are other objective criteria discussed in Chapter 10. See the following example illustrating the interpretation of the linear correlation coefficient r.

TABLE 2-10 Shoe Print Lengths and Heights of Males

Shoe Print Length (cm) 29.7 29.7 31.4 31.8 27.6

Height (cm) 175.3 177.8 185.4 175.3 172.7

EXAMPLE 3 Correlation Between Shoe Print Lengths and Heights?

Consider the data in Table 2-10 (using data from Data Set 7 “Foot and Height” in Appendix B). From the accompanying scatterplot of the paired data in Table 2-10, it isn’t very clear whether there is a linear correlation. The Statdisk display of the results shows that the linear correlation coefficient has the value of r = 0.591 (rounded).

Police Deaths in Car Chases

USA Today

investigated

the annual

reporting of

the numbers of

police who were

killed during

car chases. It was found that the

Federal Bureau of Investigation

(FBI) counted 24 deaths in the

past 35 years, but other records

show that there were 371 deaths

during that time period. USA

Today reporter Thomas Frank

wrote that “the undercount is one

of the most extreme examples of

the federal government’s inability

to accurately track violent deaths

and has led the FBI to minimize

the danger of police chasing

motorists.” Apparently, the FBI

was categorizing these deaths

as automobile accidents instead

of designating them as police

deaths that occurred during a car

chase.

Statdisk

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68 CHAPTER 2 Exploring Data with Tables and Graphs

In Example 3, we know from the Statdisk display that using the five pairs of data from Table 2-10, the linear correlation coefficient is computed to be r = 0.591. The value of r = 0.591 is not very close to 0 or 1, so based on that value and the displayed scat- terplot, it does not appear that there is a strong correlation between shoeprint lengths and heights of males.

EXAMPLE 4 Correlation Between Shoe Print Lengths and Heights?

Example 3 used only five pairs of data from Data Set 7 “Foot and Height” in Appendix B. If we use the shoe print lengths and heights from all of the 40 subjects listed in Data Set 7 in Appendix B, we get the scatterplot shown in Figure 2-16 and we get the Minitab results shown in the accompanying display. The scatterplot does show a distinct pattern instead of having points scattered about willy-nilly. Also, we see that the value of the linear correla- tion coefficient is r = 0.813. Because r = 0.813 is reasonably close to 1 and because of the pattern of points in the scatterplot, it appears that there is a linear correlation between shoe print lengths and heights.

In Example 3 with only five pairs of data, we did not have enough evidence to conclude that there is a linear correlation, but in this example with 40 pairs of data, it does appear that there is a linear correlation between shoe print lengths and heights.

PA RT 3 Regression When we do conclude that there appears to be a linear correlation between two vari- ables (as in Example 4), we can find the equation of the straight line that best fits the sample data, and that equation can be used to predict the value of one variable when given a specific value of the other variable. Based on the results from Example 4, we can predict someone’s height given the length of their shoe print (which may have been found at a crime scene).

Instead of using the straight-line equation format of y = mx + b that we have all learned in prior math courses, we use the format that follows.

FIGURE 2-16 Scatterplot of 40 Pairs of Data

Minitab

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2-4 Scatterplots, Correlation, and Regression 69

The regression equation

yn = b0 + b1x

algebraically describes the regression line.

Section 10-2 gives a good reason for using the format of yn = b0 + b1x instead of the format of y = mx + b. Section 10-2 also provides formulas that could be used to identify the values of the y-intercept b0 and the slope b1, but those values are usually found by using technology.

DEFINITION

Given a collection of paired sample data, the regression line (or line of best fit or least-squares line) is the straight line that “best” fits the scatterplot of the data. (The specific criterion for the “best”-fitting straight line is the “least squares” property described in Section 10-2.)

FIGURE 2-17 Regression Line

EXAMPLE 5 Regression Line

Example 4 included a scatterplot of the 40 pairs of shoe print lengths and heights from Data Set 7 “Foot and Height” in Appendix B. Figure 2-17 shown here is that same scatterplot with the graph of the regression line included. Also shown is the Statdisk display from the 40 pairs of data.

From the Statdisk display, we see that the general form of the regression equa- tion has a y-intercept of b0 = 80.9 (rounded) and slope b1 = 3.22 (rounded), so the equation of the regression line shown in Figure 2-17 is yn = 80.9 + 3.22x. It might be helpful to express that equation more clearly by using the names of the variables:

Height = 80.9 + 3.22 1Shoe Print Length2 Note that the equation shows the y-intercept of 80.9 that does not appear on the ver- tical scale in the graph. The leftmost vertical scale in Figure 2-19 is not the actual y-axis that passes through 0 on the x-axis. If the graph were extended to the left, the regression line would intercept the actual y-axis at the height of y = 80.9 cm.

Statdisk

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70 CHAPTER 2 Exploring Data with Tables and Graphs

Statistical Literacy and Critical Thinking

1. Linear Correlation In this section we use r to denote the value of the linear correlation co- efficient. Why do we refer to this correlation coefficient as being linear?

2. Causation A study has shown that there is a correlation between body weight and blood pressure. Higher body weights are associated with higher blood pressure levels. Can we con- clude that gaining weight is a cause of increased blood pressure?

3. Scatterplot What is a scatterplot and how does it help us?

4. Estimating r For each of the following, estimate the value of the linear correlation coeffi- cient r for the given paired data obtained from 50 randomly selected adults.

a. Their heights are measured in inches (x) and those same heights are recorded in centimeters (y).

b. Their IQ scores (x) are measured and their heights (y) are measured in centimeters.

c. Their pulse rates (x) are measured and their IQ scores are measured (y).

d. Their heights (x) are measured in centimeters and those same heights are listed again, but with negative signs (y) preceding each of these second listings.

Scatterplot. In Exercises 5–8, use the sample data to construct a scatterplot. Use the first vari- able for the x-axis. Based on the scatterplot, what do you conclude about a linear correlation?

5. Brain Volume and IQ The table lists brain volumes (cm3) and IQ scores of five males (from Data Set 9 “IQ and Brain Size” in Appendix B).

Brain volume (cm3) 1173 1067 1347 1029 1204

IQ 101 93 94 97 113

6. Bear Measurements The table lists chest sizes (distance around chest in inches) and weights (pounds) of anesthetized bears that were measured (from Data Set 11 in Appendix B).

Chest (in.) 26 45 54 49 35 41 41

Weight (lb) 80 344 416 348 166 220 262

7. Body Temperatures The table lists body temperatures (°F) of seven healthy adults at 8 AM on one day and at 8 AM on the following day (from Data Set 2 “Body Temperatures” in Appendix B).

Day 1 98.6 97.4 98.2 98.2 98.2 96.6 97.4

Day 2 97.8 97.0 97.0 96.6 97.0 96.8 96.6

8. Heights of Fathers and Sons The table lists heights (in.) of fathers and the heights (in.) of their first sons (from Francis Galton).

Height of father (in.) 73.0 75.5 75.0 75.0 75.0 74.0 74.0 73.0 73.0 78.5

Height of first son (in.) 74.0 73.5 71.0 70.5 72.0 76.5 74.0 71.0 72.0 73.2

Linear Correlation Coefficient. In Exercises 9–12, the linear correlation coefficient r is provided. What do you conclude about a linear correlation?

9. Using the data from Exercise 5 “Brain Volume and IQ,” the linear correlation coefficient is r = 0.127.

10. Using the data from Exercise 6 “Bear Measurements,” the linear correlation coefficient is r = 0.980.

2-4 Basic Skills and Concepts

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11. Using the data from Exercise 7 “Body Temperatures,” the linear correlation coefficient is r = 0.520.

12. Using the data from Exercise 8 “Heights of Fathers and Sons,” the linear correlation coef- ficient is r = - 0.017.

Chapter Quick Quiz

1. BAC When constructing a table representing the frequency distribution of blood alcohol content (g>dL) of drunk drivers involved in fatal car crashes, the first two classes of a fre- quency distribution are 0.08 – 0.11 and 0.12 – 0.15. What is the class width?

2. BAC Using the same first two classes from Exercise 1, identify the class boundaries of the first class.

3. BAC The first class described in Exercise 1 has a frequency of 36. If you know only the class limits given in Exercise 1 and the frequency of 36, can you identify the original 36 data values?

4. BAC A stemplot is created from the ages of drunk drivers involved in fatal car crashes and the first row is 1 | 67889. Identify the values represented by that row.

5. Reaction Times A large sample is randomly selected from a normally distributed popula- tion of reaction times, and a histogram is constructed from a frequency distribution. What is the shape of the histogram?

6. Tylenol In testing samples of regular Tylenol pills to verify that they have close to the de- sired amount of 325 mg of acetaminophen, which important characteristic of data is missing from this list: center, distribution, outliers, changes over time?

7. Tylenol A quality control manager wants to monitor the production of regular Tylenol pills to be sure that the mean amount of acetaminophen does not change over time. Which of the fol- lowing graphs is most helpful for that purpose: histogram, Pareto chart, pie chart, scatterplot, time-series graph, dotplot?

8. Blood Pressure In an investigation of the relationship between systolic blood pressure and diastolic blood pressure, which of the following graphs is most helpful: histogram; pie chart; scatterplot; stemplot; dotplot?

9. Blood Pressure Thing The W. A. Baum Company manufactures sphygmomanometers used to measure blood pressure. Quality control managers at such companies monitor defects and identify various causes, including worn machinery, human error, bad supplies, and packag- ing mistreatment. Which of the following graphs would be best for describing the causes of defects: histogram, scatterplot, Pareto chart, dotplot, stemplot?

10. Frequency Distribution and Histogram What is the basic difference between a fre- quency distribution and a histogram?

1. Frequency Distribution of Body Temperatures Construct a frequency distribution of the 20 body temperatures 1°F2 listed below. (These data are from Data Set 2 “Body Temperatures” in Appendix B.) Use a class width of 0.5°F and a starting value of 97.0°F.

97.1 97.2 97.5 97.6 97.6 97.8 98.0 98.0 98.2 98.2

98.2 98.3 98.4 98.6 98.6 98.7 98.7 98.9 99.1 99.4

Review Exercises

CHAPTER 2 Review Exercises 71

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72 CHAPTER 2 Exploring Data with Tables and Graphs

2. Histogram of Body Temperatures Construct the histogram that corresponds to the fre- quency distribution from Exercise 1. Use class midpoint values for the horizontal scale. Does the histogram suggest that the data are from a population having a normal distribution? Why or why not?

3. Dotplot of Body Temperatures Construct a dotplot of the body temperatures listed in Exercise 1. Which does a better job of illustrating the distribution of the data: the histogram from Exercise 2 or the dotplot?

4. Stemplot of Body Temperatures Construct a stemplot of the body temperatures listed in Exercise 1. Are there any outliers?

5. Bears Listed below are the neck sizes (in.) and weights (lb) of bears (from Data Set 11 “Bear Measurements” in Appendix B). Construct a scatterplot. Based on the graph, does there appear to be a relationship between neck sizes and weights of bears?

Neck Size (in.) 16 28 31 31.5 22 21 26.5 27 20 18

Weight (lb) 80 344 416 348.0 166 220 262.0 360 204 144

6. Monitoring Weight

a. After collecting the average (mean) weight of adult males in the United States for each of the most recent 100 years, we want to construct the graph that is most appropriate for these data. Which graph is best?

b. After collecting the average (mean) weight and height of males for the most recent 100 years, we want to construct a graph to investigate the association between those two variables. Which graph is best?

c. An investigation of health problems associated with overweight males includes heart dis- ease, stroke, high blood pressure, diabetes, and breathing problems. If we want to construct a graph that illustrates the relative importance of these adverse effects, which graph is best?

7. Medical School Enrollees The accompanying graph illustrates male and female enrollees in U.S. medical schools in a recent year. What is wrong with the graph?

Cumulative Review Exercises

1. Hygiene Listed below are times (minutes) spent on hygiene and grooming in the morning by randomly selected subjects (based on data from a Svenska Cellulosa Aktiebolaget survey). Construct a table representing the frequency distribution. Use the classes 0–9, 10–19, and so on.

0 5 12 15 15 20 22 24 25 25 25 27 27 28 30 30 35 35 40 45

2. Hygiene Histogram Use the frequency distribution from Exercise 1 to construct a histo- gram. Use class midpoint values for the horizontal scale. Based on the result, do the data ap- pear to be from a population with a normal distribution? Explain.

3. Hygiene Stemplot Use the data from Exercise 1 to construct a stemplot.

4. Analysis of Last Digits Use the data from Exercise 1 and construct a frequency distribution of the last digits of the grooming times. What does the result suggest about the grooming times?

5. Hygiene Refer to the grooming times given in Exercise 1.

a. What is the level of measurement of those times? (nominal, ordinal, interval, ratio)

b. Are the exact unrounded grooming times discrete data or continuous data?

c. Are the grooming times categorical data?

d. The average (mean) of the grooming times is 24.3 minutes. Is that value a statistic or a parameter?

M02_TRIO9015_02_SE_C02_040-074.indd 72 02/08/16 4:45 PM

6. Mother ,Daughter Heights Refer to the following list of heights of mothers and the heights of their first daughters (from Data Set 6 “Family Heights” in Appendix B). What issue would be investigated with these data? Construct the best graph for investigating that issue. What does that graph suggest?

Mother’s Height (in.) 67.0 66.5 64 64 58.5 68.0 62 66.5 65.0 64.5

Daughter’s Height (in.) 69.2 65.5 68 67 66.5 70.5 68 66.7 68.7 66.5

Technology Project

It was stated in this chapter that the days of charming and primitive hand-drawn graphs are well behind us, and technology now provides us with powerful tools for generating a wide variety of different graphs. This project therefore serves as a good preparation for professional presenta- tions that will be inevitably made in the future.

The complete data sets in Appendix B can be downloaded from www.TriolaStats.com. They can be opened by statistical software packages, such as Minitab, Excel, SPSS, and JMP. Statdisk already includes the data sets. Use a statistical software package to open Data Set 1 “Body Data.” Use this statistical software with the methods of this chapter to describe, explore, and compare the ages of males and females. Does there appear to be a difference? Reports of randomized clinical trials typically include “baseline characteristics” of the subjects in the different groups so that we can see whether the groups are similar in ways that are important. Based on ages, do the males and females in Data Set 1 appear to be similar? (Later chapters will present more formal methods for making such comparisons.)

FROM DATA TO DECISION

Car crash fatalities are tragic losses of lives and they are devastating to the families involved. Listed below are the ages of 100 randomly selected drivers who were killed in car

crashes. Also given is a frequency distribution of licensed drivers by age (based on recent data from the Insurance Insti- tute for Highway Safety).

Ages (in years) of Drivers Killed in Car Crashes Age

Licensed Drivers (millions)

41 43 38 31 57 29 65 18 42 47 16–19 9.7

69 50 22 60 30 30 34 18 18 42 20–29 33.6

18 16 74 25 41 43 50 34 54 45 30–39 40.2

32 20 50 36 27 59 19 23 57 74 40–49 40.3

27 38 29 24 56 72 21 22 74 20 50–59 29.6

43 34 38 62 39 45 56 70 68 75 60–69 18.3

37 49 25 24 21 25 31 21 76 69 70–79 13.4

28 62 69 26 22 62 64 24 56 70 80–89 5.4

21 52 32 30 38 73 35 52 38 29

23 17 44 25 24 70 16 49 45 34

Analysis Convert the given frequency distribution to a relative fre- quency distribution, then create a relative frequency distri- bution using the 100 ages of drivers killed in car crashes. Compare the two relative frequency distributions. Which age categories appear to have substantially greater proportions

of fatalities than the proportions of licensed drivers? If you were responsible for establishing the rates for auto insurance, which age categories would you select for higher rates? Con- struct a graph that is effective in identifying age categories that are more prone to fatal car crashes.

CHAPTER 2 Technology Project 73

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74 CHAPTER 2 Exploring Data with Tables and Graphs

Cooperative Group Activities

1. In-class activity In class, each student should record two pulse rates by counting the number of heartbeats in 1 minute. The first pulse rate should be measured while the student is seated, and the second pulse rate should be measured while the student is standing. Us- ing the pulse rates measured while seated, construct a frequency distribution and histogram for the pulse rates of males, and then construct another frequency distribution and histo- gram for the pulse rates of females. Using the pulse rates measured while standing, construct a frequency distribution and histogram for the pulse rates of males, and then construct another frequency distribution and histogram for the pulse rates of females. Compare the results. Do males and females appear to have different pulse rates? Do pulse rates measured while seated appear to be different from pulse rates measured while standing? Use an appropriate graph to determine whether there is a relationship between sitting pulse rate and standing pulse rate.

2. In-class activity Given below are the ages of motorcyclists at the time they were fatally injured in traffic accidents (based on data from the U.S. Department of Transportation). If your objective is to dramatize the dangers of motorcycles for young people, which graph would be most effective: histogram, Pareto chart, pie chart, dotplot, stemplot, frequency polygon, time- series graph? Construct the graph that best meets that objective. Is it okay to deliberately distort data if the objective is one such as saving lives of motorcyclists?

17 38 27 14 18 34 16 42 28 24 40 20 23 31

37 21 30 25 17 28 33 25 23 19 51 18 29

3. Out-of-class activity In each group of three or four students, select one of the following items and construct a graph that is effective in addressing the question:

a. Is there a difference between the body mass index (BMI) values for men and for women? (See Data Set 1 “Body Data” in Appendix B.)

b. Is there a relationship between the heights of sons (or daughters) and the heights of their fathers (or mothers)? (See Data Set 6 “Family Heights” in Appendix B.)

4. Out-of-class activity Search the Internet to find an example of a graph that is misleading. Describe how the graph is misleading. Redraw the graph so that it depicts the information cor- rectly. If possible, please submit your graph to www.TriolaStats.com.

5. Out-of-class activity Find Charles Joseph Minard’s graph describing Napoleon’s march to Moscow and back, and explain why Edward Tufte says that “it may well be the best graphic ever drawn.” (See The Visual Display of Quantitative Information by Edward Tufte, Graphics Press). Minard’s graph can be seen at www.TriolaStats.com under “Textbook Supplements.”

6. Out-of-class activity In The Visual Display of Quantitative Information by Edward Tufte (Graphics Press), find the graph that appeared in American Education, and explain why Tufte says that “this may well be the worst graphic ever to find its way into print.” The graph can be seen at www.TriolaStats.com under “Textbook Supplements.” Construct a graph that is effec- tive in depicting the same data.

M02_TRIO9015_02_SE_C02_040-074.indd 74 02/08/16 4:45 PM

75

CHAPTER PROBLEM

Data Set 1 “Body Data” in Appendix B includes pulse rates

of men and women. The full data set contains measurements

from 300 adults, and the first 5 cases are printed in Appendix B.

Figure 3-1 shows dotplots of the pulse rates categorized ac-

cording to gender. Close examination of Figure 3-1 reveals

that the pulse rates consist of even numbers only. This sug-

gests that the pulse rates were measured for 30 seconds,

and the result was doubled to provide a pulse rate in beats

per minute. Examine Figure 3-1 closely to see that the pulse

rates of males tend to be generally a little lower (farther to

the left) than the pulse rates of females. This observation

suggests a hypothesis: Males have lower pulse rates than

females. A conclusion about such a hypothesis should not

be made on the basis of a graph alone. We should consider

Do men and women have the same pulse rates?

Measures of Center

Measures of Variation

Measures of Relative Standing and Boxplots

3-1

3-2

3-3

Describing, Exploring, and Comparing Data3

M03_TRIO9015_02_SE_C03_075-117.indd 75 02/08/16 4:52 PM

> > >

whether the sample data were collected with an appropriate

method. We should also consider whether the apparent dif-

ference between male pulse rates and female pulse rates is

actually a significant difference and not just a random chance

anomaly.

Instead of relying solely on subjective interpretations of a

graph like Figure 3-1, this chapter introduces measures that

are essential to any study of statistics. This chapter introduces

the mean, median, standard deviation, and variance, which

are among the most important statistics presented in this book,

and they are among the most important statistics in the study

of statistics. We will use these statistics for describing, explor-

ing, and comparing the measured pulse rates for males and

females in Data Set 1 “Body Data.”

CHAPTER OBJECTIVES

Critical Thinking and Interpretation: Going Beyond Formulas and Arithmetic

In this modern biostatistics course, it isn’t so important to memorize formulas or manu- ally do messy arithmetic. We can get results with a calculator or software so that we can focus on making practical sense of results through critical thinking. Although this chapter includes detailed steps for important procedures, it isn’t always necessary to master those steps. It is, however, generally helpful to perform a few manual calcula- tions before using technology, so that understanding is enhanced.

The methods and tools presented in this chapter are often called methods of descriptive statistics, because they summarize or describe relevant characteristics of data. In later chapters we use inferential statistics to make inferences, or generaliza- tions, about populations. Here are the chapter objectives:

Measures of Center

• Develop the ability to measure the center of data by finding the mean, median, mode, and midrange.

• Determine whether an outlier has a substantial effect on the mean and median.

Measures of Variation

• Develop the ability to measure variation in a set of sample data by finding values of the range, variance, and standard deviation.

3-1

3-2

FIGURE 3-1 Dotplot of Pulse Rates of Males and Females

76 CHAPTER 3 Describing, Exploring, and Comparing Data

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3-1 Measures of Center 77

There are different approaches for measuring the center, so we have different defi- nitions for those different approaches. We begin with the mean.

Mean

The mean (or arithmetic mean) is generally the most important of all numerical mea- surements used to describe data, and it is what many people call an average.

Key Concept The focus of this section is to obtain a value that measures the center of a data set. We present measures of center, including mean and median. Our objective here is not only to find the value of each measure of center, but also to interpret and make sense of those values. Part 1 of this section includes core concepts that should be understood before considering Part 2.

PA RT 1 Basic Concepts of Measures of Center In Part 1 of this section, we introduce the mean, median, mode, and midrange as dif- ferent measures of center. Measures of center are widely used to provide representa- tive values that “summarize” data sets.

3-1 Measures of Center

DEFINITION

A measure of center is a value at the center or middle of a data set.

DEFINITION

The mean (or arithmetic mean) of a set of data is the measure of center found by adding all of the data values and dividing the total by the number of data values.

Important Properties of the Mean

■ Sample means drawn from the same population tend to vary less than other mea- sures of center.

■ The mean of a data set uses every data value.

• Develop the ability to interpret values of the standard deviation by applying the range rule of thumb to determine whether a particular value is significantly low or significantly high.

Measures of Relative Standing and Boxplots

• Develop the ability to compute a z score and use the result to determine whether a given value x is significantly low or significantly high.

• Identify percentile values and quartile values from a set of data. • Develop the ability to construct a boxplot from a set of data.

3-3

• Develop the ability to interpret values of the standard deviation by applying the range rule of thumb to determine whether a particular value is significantly low or significantly low or significantly low significantly high.

Measures of Relative Standing and Boxplots

• Develop the ability to compute a z score and use the result to determine whether a z score and use the result to determine whether a z given value x is x is x significantly low or significantly low or significantly low significantly high.

• Identify percentilepercentileper values and quartile values from a set of data. • Develop the ability to construct a boxplot from a set of data.

continued

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78 CHAPTER 3 Describing, Exploring, and Comparing Data

Calculation and Notation of the Mean

The definition of the mean can be expressed as Formula 3-1, in which the Greek letter Σ (uppercase sigma) indicates that the data values should be added, so Σ x represents the sum of all data values. The symbol n denotes the sample size, which is the number of data values.

FORMULA 3-1

Mean = Σ x n

d sum of all data values d number of data values

If the data are a sample from a population, the mean is denoted by x (pronounced “x-bar”); if the data are the entire population, the mean is denoted by m (lowercase Greek mu).

DEFINITION

A statistic is resistant if the presence of extreme values (outliers) does not cause it to change very much.

■ A disadvantage of the mean is that just one extreme value (outlier) can change the value of the mean substantially. (Using the following definition, we say that the mean is not resistant.)

NOTATION Hint: Sample statistics are usually represented by English letters, such as x, and population parameters are usually represented by Greek letters, such as m.

Σ denotes the sum of a set of data values. x is the variable usually used to represent the individual data values. n represents the number of data values in a sample. N represents the number of data values in a population.

x = Σ x n

is the mean of a set of sample values.

m = Σ x N

is the mean of all values in a population.

EXAMPLE 1 Mean

Data Set 1 “Body Data” in Appendix B includes measures of pulse rates. Find the mean of the first five pulse rates for males: 84, 74, 50, 60, 52 (all in beats per minute, or BPM).

S O L U T I O N

The mean is computed by using Formula 3-1. First add the data values, then divide by the number of data values:

x = Σx n

= 84 + 74 + 50 + 60 + 52

5 =

320 5

= 64.0 BPM

The mean of the first five male pulse rates is 64.0 BPM.

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3-1 Measures of Center 79

Median

The median can be thought of loosely as a “middle value” in the sense that about half of the values in a data set are less than the median and half are greater than the me- dian. The following definition is more precise.

CAUTION Never use the term average when referring to a measure of center. The word average is often used for the mean, but it is sometimes used for other measures of center. The term average is not used by statisticians, it is not used in professional journals, and it will not be used throughout the remainder of this book when referring to a specific measure of center.

DEFINITION

The median of a data set is the measure of center that is the middle value when the original data values are arranged in order of increasing (or decreasing) magnitude.

Important Properties of the Median

■ The median does not change by large amounts when we include just a few ex- treme values, so the median is a resistant measure of center.

■ The median does not directly use every data value. (For example, if the largest value is changed to a much larger value, the median does not change.)

Calculation and Notation of the Median

The median of a sample is sometimes denoted by x∼ (pronounced “x-tilde”) or M or Med; there isn’t a commonly accepted notation and there isn’t a special symbol for the median of a population. To find the median, first sort the values (arrange them in order), and then follow one of these two procedures:

1. If the number of data values is odd, the median is the number located in the ex- act middle of the sorted list.

2. If the number of data values is even, the median is found by computing the mean of the two middle numbers in the sorted list.

EXAMPLE 2 Median with an Odd Number of Data Values

Find the median of the first five pulse rates for males: 84, 74, 50, 60, 52 (all in BPM).

S O L U T I O N

First sort the data values by arranging them in ascending order, as shown below:

50 52 60 74 84

Because there are 5 data values, the number of data values is an odd number (5), so the median is the number located in the exact middle of the sorted list, which is 60.0 BPM. The median is therefore 60.0 BPM. Note that the median of 60.0 BPM is different from the mean of 64.0 BPM found in Example 1.

What the Median Is Not

Harvard

biologist Ste-

phen Jay Gould

wrote, “The

Median Isn’t the

Message.” In

it, he describes

how he learned that he had ab-

dominal mesothelioma, a form of

cancer. He went to the library to

learn more, and he was shocked

to find that mesothelioma was

incurable, with a median survival

time of only eight months after

it was discovered. Gould wrote

this: “I suspect that most people,

without training in statistics,

would read such a statement as

‘I will probably be dead in eight

months’ the very conclusion that

must be avoided, since it isn’t

so, and since attitude (in fighting

the cancer) matters so much.”

Gould went on to carefully

interpret the value of the median.

He knew that his chance of liv-

ing longer than the median was

good because he was young, his

cancer was diagnosed early, and

he would get the best medical

treatment. He also reasoned that

some could live much longer

than eight months, and he saw

no reason why he could not be

in that group. Armed with this

thoughtful interpretation of the

median and a strong positive

attitude, Gould lived for 20 years

after his diagnosis. He died of

another cancer not related to the

mesothelioma.

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80 CHAPTER 3 Describing, Exploring, and Comparing Data

Mode

The mode isn’t used much with quantitative data, but it’s the only measure of center that can be used with qualitative data (consisting of names, labels, or categories only).

EXAMPLE 3 Median with an Even Number of Data Values

Repeat Example 2 after including the sixth pulse rate of 62 BPM. That is, find the median of these pulse rates: 84, 74, 50, 60, 52, 62 (all in BPM).

S O L U T I O N

First arrange the values in ascending order: 50, 52, 60, 62, 74, 84.

Because the number of data values is an even number (6), the median is found by computing the mean of the two middle numbers, which are 60 and 62.

Median = 60 + 62

2 =

122 2

= 61.0 BPM

The median is 61.0 BPM.

DEFINITION

The mode of a data set is the value(s) that occurs with the greatest frequency.

Important Properties of the Mode

■ The mode can be found with qualitative data.

■ A data set can have no mode or one mode or multiple modes.

Finding the Mode: A data set can have one mode, more than one mode, or no mode.

■ When two data values occur with the same greatest frequency, each one is a mode and the data set is said to be bimodal.

■ When more than two data values occur with the same greatest frequency, each is a mode and the data set is said to be multimodal.

■ When no data value is repeated, we say that there is no mode.

EXAMPLE 4 Mode

Find the mode of these pulse rates (in BPM):

58 58 58 58 60 60 62 64

S O L U T I O N

The mode is 58 BPM, because it is the pulse rate occurring most often (four times).

In Example 4, the mode is a single value. Here are other possible circumstances:

Two modes: The pulse rates (BPM) of 58, 58, 58, 60, 60, 60, 62, 64 have two modes: 58 BPM and 60 BPM.

No mode: The pulse rates of 58, 60, 64, 68, 72 have no mode because no value is repeated.

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3-1 Measures of Center 81

Midrange

Another measure of center is the midrange.

DEFINITION

The midrange of a data set is the measure of center that is the value midway be- tween the maximum and minimum values in the original data set. It is found by add- ing the maximum data value to the minimum data value and then dividing the sum by 2, as in the following formula:

Midrange = maximum data value + minimum data value

2

Important Properties of the Midrange

■ Because the midrange uses only the maximum and minimum values, it is very sensitive to those extremes so the midrange is not resistant.

■ In practice, the midrange is rarely used, but it has three redeeming features:

1. The midrange is very easy to compute.

2. The midrange helps reinforce the very important point that there are several different ways to define the center of a data set.

3. The value of the midrange is sometimes used incorrectly for the median, so con- fusion can be reduced by clearly defining the midrange along with the median.

EXAMPLE 5 Midrange

Find the midrange of these five pulse rates for males used in Example 1: 84, 74, 50, 60, 52 (BPM).

S O L U T I O N

The midrange is found as follows:

Midrange = maximum data value + minimum data value

2

= 84 + 50

2 = 67.0 BPM

The midrange is 67.0 BPM.

Rounding Measures of Center

When calculating measures of center, we often need to round the result. We use the following rule.

ROUND-OFF RULES FOR MEASURES OF CENTER:

■ For the mean, median, and midrange, carry one more decimal place than is present in the original set of values.

■ For the mode, leave the value as is without rounding (because values of the mode are the same as some of the original data values).

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82 CHAPTER 3 Describing, Exploring, and Comparing Data

When applying any rounding rules, round only the final answer, not intermedi- ate values that occur during calculations. For example, the mean of 2, 3, and 5 is 3.333333. . . , which is rounded to 3.3, which has one more decimal place than the origi- nal values of 2, 3, and 5. As another example, the mean of 80.4 and 80.6 is 80.50 (one more decimal place than was used for the original values). Because the mode is one or more of the original data values, we do not round values of the mode; we simply use the same original values that are modes.

Critical Thinking

We can always calculate measures of center from a sample of numbers, but we should always think about whether it makes sense to do that. In Section 1-2 we noted that it makes no sense to do numerical calculations with data at the nominal level of mea- surement, because those data consist of names, labels, or categories only, so statistics such as the mean and median are meaningless for such data. We should also think about the sampling method used to collect the data. If the sampling method is not sound, the statistics we obtain may be very misleading.

EXAMPLE 6 Critical Thinking and Measures of Center

Each of the following illustrates data for which the mean and median are not mean- ingful statistics.

a. Zip codes of the hospitals in the United States. (The zip codes don’t measure or count anything. The numbers are just labels for geographic locations.)

b. Ranks of selected medical schools: 2, 3, 7, 10, 14. (The ranks reflect an order- ing, but they don’t measure or count anything.)

c. Numbers on the jerseys of the starting defense for the Seattle Seahawks when they won Super Bowl XLVIII: 31, 28, 41, 56, 25, 54, 69, 50, 91, 72, 29. (The numbers on the football jerseys don’t measure or count anything; they are just substitutes for names.)

d. Top 5 incomes of hospital chief executive o¤cers. (Such “top 5” or “top 10” lists include data that are not at all representative of the larger population.)

e. The 50 mean ages computed from the means in each of the 50 states. (If you calculate the mean of those 50 values, the result is not the mean age of people in the entire United States. The population sizes of the 50 di¥erent states must be taken into account, as described in the weighted mean introduced in Part 2 of this section.)

In the spirit of describing, exploring, and comparing data, we provide Table 3-1, which summarizes the different measures of center for the 300 pulse rates referenced in the Chapter Problem. Figure 3-1 on page 76 suggests that males have lower pulse rates, and comparison of the means and medians in Table 3-1 also suggests that males have lower pulse rates. The following chapters will describe other tools that can be used for an effective comparison.

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3-1 Measures of Center 83

PA RT 2 Beyond the Basics of Measures of Center Calculating the Mean from a Frequency Distribution

Formula 3-2 is the same calculation for the mean that was presented in Part 1, but it incorporates this approach: When working with data summarized in a frequency dis- tribution, we make calculations possible by pretending that all sample values in each class are equal to the class midpoint. Formula 3-2 is not really a new concept; it is simply a variation of Formula 3-1 for the mean.

FORMULA 3-2 MEAN FROM A FREQUENCY DISTRIBUTION

First multiply each frequency and class midpoint; then add the products.

T

x = Σ 1f ·x2 Σ f

1Result is an approximation2

c Sum of frequencies

(equal to n)

Example 7 illustrates the procedure for finding the mean from a frequency distribution.

TABLE 3-1 Male and Female Pulse Rates

Male Female

Mean 69.6 74.0

Median 68.0 74.0

Mode 66 72, 74, 82

Midrange 72.0 70.0

EXAMPLE 7 Computing the Mean from a Frequency Distribution

The first two columns of Table 3-2 on the next page constitute a frequency distribution summarizing the pulse rates from males in Data Set 1 “Body Data” from Appendix B. Use the frequency distribution in the first two columns to find the mean.

S O L U T I O N

Remember, when working with data summarized in a frequency distribution, we make calculations possible by pretending that all sample values in each class are equal to the class midpoint. For example, see Table 3-2 and consider the first class interval of 40–54 with a frequency of 15. We pretend that each of the 15 pulse rates is 47 (the class midpoint). With the pulse rate of 47 repeated 15 times, we have a total of 47 # 15 = 705, as shown in the last column of Table 3-2. We can then add those results to find the sum of all sample values.

The bottom row of Table 3-2 shows the two components we need for the cal- culation of the mean (as in Formula 3-2): Σf = 153 and Σ1f # x2 = 10,611. We calculate the mean using Formula 3-2 as follows:

x = Σ1 f # x2

Σf =

10,611 153

= 69.4 BPM continued

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84 CHAPTER 3 Describing, Exploring, and Comparing Data

Calculating a Weighted Mean

When different x data values are assigned different weights w, we can compute a weighted mean. Formula 3-3 can be used to compute the weighted mean.

FORMULA 3-3

Weighted mean: x = Σ(w # x) Σw

Formula 3-3 tells us to first multiply each weight w by the corresponding value x, then to add the products, and then finally to divide that total by the sum of the weights, Σw.

The result of x = 69.4 BPM is an approximation because it is based on the use of class midpoint values instead of the original list of pulse rates. The mean of 69.6 BPM found by using all of the original pulse rates for males is a more accu- rate result.

TABLE 3-2 Pulse Rates (BPM) of Males

Pulse Rate Frequency f Class Midpoint x f ~ x

40–54 15 47 705

55–69 63 62 3906

70–84 62 77 4774

85–99 11 92 1012

100–114 2 107 214

Totals: �f = 153 � 1f ~ x2 = 10,611

EXAMPLE 8 Computing Grade-Point Average

In her first semester of college, a student of one of the authors took five courses. Her final grades along with the number of credits for each course were A (3 cred- its), A (4 credits), B (3 credits), C (3 credits), and F (1 credit). The grading system assigns quality points to letter grades as follows: A = 4; B = 3; C = 2; D = 1; F = 0. Compute her grade-point average.

S O L U T I O N

Use the numbers of credits as weights: w = 3, 4, 3, 3, 1. Replace the letter grades of A, A, B, C, and F with the corresponding quality points: x = 4, 4, 3, 2, and 0. We now use Formula 3-3 as shown below. The result is a first-semester grade-point average of 3.07. (In using the preceding round-off rule, the result should be rounded to 3.1, but it is common to round grade-point averages to two decimal places.)

x = Σ1w # x2

Σw

= 13 * 42 + 14 * 42 + 13 * 32 + 13 * 22 + 11 * 02

3 + 4 + 3 + 3 + 1

= 43 14

= 3.07

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3-1 Measures of Center 85

Descriptive Statistics Access tech instructions, videos, and data sets at www.TriolaStats.com

TECH CENTER

Statistical Literacy and Critical Thinking

1. Average A report includes a statement that the “average” Medical College Admission Test (MCAT) score of applicants to medical schools is 28.4. What is the role of the term average in statistics? Should another term be used in place of average?

2. What’s Wrong? The Centers for Disease Control and Prevention (CDC) publishes a list of smoking rates in each state. If we add the 50 percentages and then divide by 50, we get 19.67%. Is the value of 19.67% the mean smoking rate for all of the United States? Why or why not?

3. Measures of Center In what sense are the mean, median, mode, and midrange measures of “center”?

4. Resistant Measures Here are five pulse rates (BPM) of females: 80, 94, 58, 66, 56. Find the mean and median of these five values. Then find the mean and median after including a sixth value of 740, which is an outlier. (One of the female pulse rates is 74, but 740 is used here as an error resulting from an incorrect data entry.) Compare the two sets of results. How much was the mean affected by the inclusion of the outlier? How much is the median affected by the inclusion of the outlier?

Critical Thinking. Each of Exercises 5–16 involves some feature that is somewhat tricky. Find the (a) mean, (b) median, (c) mode, and (d) midrange, and then answer the given question.

5. Charges for Births Data Set 3 “Births” in Appendix B includes total charges for births at four hospitals in New York State, and the top 10 highest amounts (in dollars) are listed below. What do the results tell us about the population of all such charges?

471,062 359,091 290,837 271,863 255,788

247,477 232,782 197,912 183,271 155,857

6. MCAT Score Listed below are mean MCAT scores listed in order by year, starting with the year 2002. What important feature of the data is not revealed by any of the measures of center?

27.0 26.8 27.1 27.3 27.4 27.7 28.1 27.9 28.3 28.2 28.3 28.4

7. Football Player Numbers Listed below are the jersey numbers of 11 players randomly selected from the roster of the Seattle Seahawks when they won Super Bowl XLVIII. What do the results tell us?

89 91 55 7 20 99 25 81 19 82 60

8. Football Player Weights Listed below are the weights in pounds of 11 players randomly selected from the roster of the Seattle Seahawks when they won Super Bowl XLVIII (the same players from the preceding exercise). Are the results likely to be representative of all National Football League (NFL) players?

189 254 235 225 190 305 195 202 190 252 305

9.  Peas in a Pod Biologists conducted experiments to determine whether a deficiency of carbon dioxide in the soil affects the phenotypes of peas. Listed below are the pheno- type codes, where 1 = smooth@yellow, 2 = smooth@green, 3 = wrinkled@yellow, and

3-1 Basic Skills and Concepts

continued

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86 CHAPTER 3 Describing, Exploring, and Comparing Data

4 = wrinkled@green. Can the measures of center be obtained for these values? Do the results make sense?

2 1 1 1 1 1 1 4 1 2 2 1 2 3 3 2 3 1 3 1 3 1 3 2 2

10. TV Prices A physician plans to buy a television for her large waiting room. Listed below are selling prices in dollars of TVs that are 60 inches or larger and rated as a “best buy” by Consumer Reports magazine. Are the resulting statistics representative of the population of all TVs that are 60 inches and larger? If you decide to buy one of these TVs, what statistic is most relevant, other than the measures of central tendency?

1800 1500 1200 1500 1400 1600 1500 950 1600 1150 1500 1750

11. Cell Phone Radiation Listed below are the measured radiation absorption rates (in W>kg) corresponding to these cell phones: iPhone 5S, BlackBerry Z30, Sanyo Vero, Optimus V, Droid Razr, Nokia N97, Samsung Vibrant, Sony Z750a, Kyocera Kona, LG G2, and Virgin Mobile Supreme. The data are from the Federal Communications Commission (FCC). The media often report about the dangers of cell phone radiation as a cause of cancer. The FCC has a standard that a cell phone absorption rate must be 1.6 W>kg or less. If you are planning to purchase a cell phone, are any of the measures of center the most important statistic? Is there another statistic that is most relevant? If so, which one?

1.18 1.41 1.49 1.04 1.45 0.74 0.89 1.42 1.45 0.51 1.38

12. Caffeine in Soft Drinks Listed below are measured amounts of caffeine (mg per 12 oz of drink) obtained in one can from each of 20 brands (7-UP, A&W Root Beer, Cherry Coke, . . . , Tab). Are the statistics representative of the population of all cans of the same 20 brands consumed by Americans?

0 0 34 34 34 45 41 51 55 36 47 41 0 0 53 54 38 0 41 47

13. Firefighter Fatalities Listed below are the numbers of heroic firefighters who lost their lives in the United States each year while fighting forest fires. The numbers are listed in order by year, starting with the year 2000. What important feature of the data is not revealed by any of the measures of center?

20 18 23 30 20 12 24 9 25 15 8 11 15 34

14. Foot Lengths Listed below are foot lengths in inches of randomly selected Army women measured in the 1988 Anthropometric Survey (ANSUR). Are the statistics representative of the current population of all Army women?

10.4 9.3 9.1 9.3 10.0 9.4 8.6 9.8 9.9 9.1 9.1

15. Medical School Tuition Listed below in dollars are the annual costs of tuition at the 10 most expensive private medical schools in the United States for a recent year (based on data from U.S. News & World Report). What does this “top 10” list tell us about those costs for the population of all U.S. private medical school tuitions?

57,261 56,784 55,196 54,976 54,653 54,528 54,268 54,050 53,581 53,323

16. California Smokers In the California Health Interview Survey, randomly selected adults are interviewed. One of the questions asks how many cigarettes are smoked per day, and results are listed below for 50 randomly selected respondents. How well do the results reflect the smoking behavior of California adults?

9 10 10 20 40 50 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0

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3-1 Measures of Center 87

In Exercises 17–20, find the mean and median for each of the two samples, then compare the two sets of results.

17.  Blood Pressure A sample of blood pressure measurements is taken from Data Set 1 “Body Data” in Appendix B, and those values (mm Hg) are listed below. The values are matched so that 10 subjects each have systolic and diastolic measurements. (Systolic is a mea- sure of the force of blood being pushed through arteries, but diastolic is a measure of blood pressure when the heart is at rest between beats.) Are the measures of center the best statistics to use with these data? What else might be better?

Systolic: 118 128 158 96 156 122 116 136 126 120

Diastolic: 80 76 74 52 90 88 58 64 72 82

18.  White>Red Blood Counts Listed below are white blood cell counts (1000 cells>mL) and red blood cell counts (million cells>mL) from different subjects (from Data Set 1 “Body Data” in Appendix B). The values are matched so that each of the 12 subjects has a white blood cell count and a red blood cell count. Are the measures of center the best statistics to use with these data? What else might be better?

White: 8.7 4.9 6.9 7.5 6.1 5.7 4.1 8.1 8.0 5.6 8.3 6.9

Red: 4.8 4.7 4.5 4.3 5.0 4.0 4.7 4.6 4.1 5.5 4.4 4.2

19.  White Blood Counts Listed below are white blood cell counts (1000 cells>mL) from males and females (from Data Set 1 “Body Data” in Appendix B). Do they appear to be different?

Female: 8.7 6.9 8.1 8.0 6.9 8.1 6.4 6.3 10.9 4.8 5.9 7.2

Male: 4.9 7.5 6.1 5.7 4.1 5.6 8.3 5.1 9.5 6.1 5.7 5.4

20. Queues A Providence Hospital experiment involves two different waiting line configura- tions for patients arriving for admission. The waiting times (in seconds) are recorded with a single line configuration that feeds four stations and another configuration with individual lines at the four stations. Determine whether there is a difference between the two data sets that is not apparent from a comparison of the measures of center. If so, what is it?

Single Line 390 396 402 408 426 438 444 462 462 462

Individual Lines 252 324 348 372 402 462 462 510 558 600

Large Data Sets from Appendix B. In Exercises 21–24, refer to the indicated data set in Appendix B. Use software or a calculator to find the means and medians.

21. HDL Use the high-density lipoprotein (HDL) cholesterol measurements (mg>dL) from the 300 subjects included in Data Set 1 “Body Data” in Appendix B. Identify the highest value. Does it appear to be an outlier? Do the mean and median change much when that highest value is deleted?

22. LDL Repeat the preceding exercise using the low-density lipoprotein (LDL) measurements (mg>dL). 23. Body Temperatures Refer to Data Set 2 “Body Temperatures” in Appendix B and use the body temperatures for 12:00 AM on day 2. Do the results support or contradict the common belief that the mean body temperature is 98.6oF?

24.  Births Use the birth weights (grams) of the 400 babies listed in Data Set 3 “Births” in Appendix B. Examine the list of birth weights to make an observation about those numbers. How does that observation affect the way that the results should be rounded?

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88 CHAPTER 3 Describing, Exploring, and Comparing Data

In Exercises 25 and 26, find the mean of the data summarized in the frequency distribution. Also, compare the computed means to the actual means obtained by using the original list of data values, which are as follows: (Exercise 25) 224.3; (Exercise 26) 255.1.

25.  Blood Platelet

Count of Males

Frequency

0–99 1

100–199 51

200–299 90

300–399 10

400–499 0

500–599 0

600–699 1

26.  Blood Platelet Count

of Females

Frequency

100–199 25

200–299 92

300–399 28

400–499 0

500–599 2

27.  Weighted Mean A student of one of the authors earned grades of A, C, B, A, and D. Those courses had these corresponding numbers of credit hours: 3, 3, 3, 4, and 1. The grad- ing system assigns quality points to letter grades as follows: A = 4; B = 3; C = 2; D = 1; F = 0. Compute the grade-point average (GPA) and round the result with two decimal places. If the dean’s list requires a GPA of 3.00 or greater, did this student make the dean’s list?

28. Weighted Mean A student of one of the authors earned grades of 63, 91, 88, 84, and 79 on her five regular statistics tests. She earned a grade of 86 on the final exam and 90 on her class projects. Her combined homework grade was 70. The five regular tests count for 60% of the final grade, the final exam counts for 10%, the project counts for 15%, and homework counts for 15%. What is her weighted mean grade? What letter grade did she earn (A, B, C, D, or F)? Assume that a mean of 90 or above is an A, a mean of 80 to 89 is a B, and so on.

29. Degrees of Freedom Five pulse rates randomly selected from Data Set 1 “Body Data” in Appendix B have a mean of 78.0 beats per minute. Four of the pulse rates are 82, 78, 56, and 84.

a. Find the missing value.

b. We need to create a list of n values that have a specific known mean. We are free to select any values we desire for some of the n values. How many of the n values can be freely assigned before the remaining values are determined? (The result is referred to as the number of degrees of freedom.)

30. Censored Data Recently, five U.S. presidents were still alive and after their first inaugura- tion, they have lived 37 years, 25 years, 21 years, 13 years, and 5 years so far. We might use the values of 37 + , 25 + , 21 + , 13 + , and 5 + , where the positive signs indicate that the actual value is equal to or greater than the current value. (These values are said to be censored at the current time that this list was compiled.) If we ignore the presidents who took office because of an as- sassination or resignation and if we ignore the five presidents who are still alive, the mean of the 33 remaining presidents is 15.0 years. What do we know about the mean if we include the censored values of 37 + , 25 + , 21 + , 13 + , and 5 + ? Do the two results differ by much?

31. Trimmed Mean Because the mean is very sensitive to extreme values, we say that it is not a resistant measure of center. By deleting some low values and high values, the trimmed mean (or truncated mean) is more resistant. To find the 10% trimmed mean for a data set, first arrange the data in order, next delete the bottom 10% of the values and delete the top 10% of the values, and then calculate the mean of the remaining values. Use the LDL measurements of the 300 subjects from Data Set 1 “Body Data” in Appendix B. Compare the mean, the 10% trimmed mean, and the 20% trimmed mean.

3-1 Beyond the Basics

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3-2 Measures of Variation 89

PA RT 1 Basic Concepts of Variation To visualize the property of variation, see Figure 3-2, which illustrates pulse rates (beats per minute or BPM) for subjects given a treatment and subjects given a pla- cebo. (A high priority is placed on using real data, but these pulse rates are fabricated for the purposes of making an important point here.) Verify this important observa- tion: The pulse rates in the treatment group (top dotplot) have more variation than those in the placebo group (bottom dotplot). Both sets of pulse rates have the same mean of 70.2 BPM, they have the same median of 70.0 BPM, and they have the same mode of 70 BPM. Those measures of center do not “see” the difference in variation.

To keep our round-off rules as consistent and as simple as possible, we will round the measures of variation using this rule:

Key Concept Variation is the single most important topic in statistics, so this is the single most important section in this book. This section presents three important measures of variation: range, standard deviation, and variance. These statistics are numbers, but our focus is not just computing those numbers but developing the abil- ity to interpret and understand them. This section is not a study of arithmetic; it is about understanding and interpreting measures of variation, especially the standard deviation.

3-2 Measures of Variation

STUDY HINT: Part 1 of this section presents basic concepts of variation, and Part 2 presents additional concepts related to the standard deviation. Part 1 and Part 2 both include formulas for computation, but do not spend too much time memorizing formulas or doing arithmetic calculations. Instead, focus on understanding and interpreting values of standard deviation.

ROUND-OFF RULE FOR MEASURES OF VARIATION When rounding the value of a measure of variation, carry one more decimal place than is present in the original set of data.

FIGURE 3-2 Dotplots of Pulse Rates

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90 CHAPTER 3 Describing, Exploring, and Comparing Data

Range

Let’s begin with the range because it is quick and easy to compute, but it is not as im- portant as other measures of variation.

DEFINITION

The range of a set of data values is the difference between the maximum data value and the minimum data value.

Range = 1maximum data value2 − 1minimum data value2

Important Properties of the Range

■ The range uses only the maximum and the minimum data values, so it is very sensitive to extreme values. The range is not resistant.

■ Because the range uses only the maximum and minimum values, it does not take every value into account and therefore does not truly reflect the variation among all of the data values.

EXAMPLE 1 Range

Find the range of the first five pulse rates for males from Data Set 1 “Body Data” in Appendix B: 84, 74, 50, 60, 52 (all in BPM).

S O L U T I O N

The range is found by subtracting the lowest value from the largest value, so we get

Range = 1maximum value2 - 1minimum value2 = 84 - 50 = 34.0 BPM The range of 34.0 BPM is shown with one more decimal place than is present in the original data values.

Standard Deviation of a Sample

The standard deviation is the measure of variation most commonly used in statistics.

DEFINITION

The standard deviation of a set of sample values, denoted by s, is a measure of how much data values deviate away from the mean. It is calculated by using Formula 3-4 or 3-5. Formula 3-5 is just a different version of Formula 3-4; both for- mulas are algebraically the same.

The standard deviation found from sample data is a statistic denoted by s, but the stan- dard deviation found from population data is a parameter denoted by s. The formula for s is slightly different with division by the population size N used instead of divi- sion by n - 1. The population standard deviation s will be discussed later.

Notation

s = sample standard deviation s = population standard deviation

Range

Let’s begin with the range because it is quick and easy to compute, but it is not as im portant as other measures of variation.

Got a Second?

The time unit

of 1 second

is defined

to be “the

duration of

9,192,631,770

periods of the

radiation corresponding to the

transition between the two hy-

perfine levels of the ground state

of the cesium-133 atom.” That

definition redefines time to be

based on the behavior of atoms

instead of the earth’s motion. It

results in accuracy of ±1 second

in 10,000,000 years, which is the

most accurate measurement we

use. Because it is so accurate, the

second is being used to define

other quantities, such as the me-

ter. The meter was once defined

as 1>10,000,000 of the distance along the surface of the earth

between the North Pole and the

equator (passing through Paris).

The meter is now defined as the

length of the distance traveled by

light in a vacuum during a time

interval of 1>299,792,458 sec. When dealing with time mea-

surement devices, the traditional

standard deviation has been

found to be poor because of a

trend in which the mean changes

over time. Instead, other special

measures of variation are used,

such as Allan variance, total vari-

ance, and TheoH.

Unrelated to statistics but

nonetheless interesting is the

fact that ads for watches usually

show a watch with a time close

to 10:10. That time allows the

brand name to be visible, and it

creates a subliminal image of a

happy face. The time of 10:10

has been the industry standard

since the 1940s.

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3-2 Measures of Variation 91

FORMULA 3-4

s = B Σ1x - x22

n - 1 sample standard deviation

FORMULA 3-5

s = B n1Σ x22 - 1Σ x22

n1n - 12 shortcut formula for sample standard

deviation (used by calculators and software)

Later we give the reasoning behind these formulas, but for now we recommend that you use Formula 3-4 for an example or two, and then learn how to find standard deviation values using a calculator or software.

Important Properties of Standard Deviation

■ The standard deviation is a measure of how much data values deviate away from the mean.

■ The value of the standard deviation s is never negative. It is zero only when all of the data values are exactly the same.

■ Larger values of s indicate greater amounts of variation.

■ The standard deviation s can increase dramatically with one or more outliers.

■ The units of the standard deviation s (such as minutes, feet, pounds) are the same as the units of the original data values.

■ The sample standard deviation s is a biased estimator of the population standard deviation s, which means that values of the sample standard deviation s do not center around the value of s. (This is explained in Part 2.)

Example 2 illustrates a calculation using Formula 3-4 because that formula better illustrates that the standard deviation is based on deviations of sample values away from the mean.

EXAMPLE 2 Calculating Standard Deviation with Formula 3-4

Use Formula 3-4 to find the standard deviation of the first five pulse rates for males from Data Set 1 “Body Data” in Appendix B: 84, 74, 50, 60, 52 (all in BPM).

S O L U T I O N

The left column of Table 3-3 on the next page summarizes the general procedure for finding the standard deviation using Formula 3-4, and the right column illus- trates that procedure using the sample values 84, 74, 50, 60, 52. The result shown in Table 3-3 is 14.6 BPM, which is rounded to one more decimal place than is present in the original list of sample values. Also, the units for the standard devia- tion are the same as the units of the original data. Because the original data values are all in units of BPM, the standard deviation is 14.6 BPM.

Variation in Faces

Researchers

commented

that “if everyone

looked more or

less the same,

there would be

total chaos.”

They studied

human body measurements and

found that facial traits varied

more than other body traits,

and the greatest variation oc-

curred within the triangle formed

by the eyes and mouth. They

learned that facial traits vary

independently of each other. For

example, there is no relationship

between the distance between

your eyes and how big your

mouth is. The researchers stated

that our facial variation played an

important role in human evolu-

tion. (See “Morphological and

Population Genomic Evidence

That Human Faces Have Evolved

to Signal Individual Identity,” by

Sheehan and Nachman, Nature

Communications, Vol. 5,

No. 4800.)

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92 CHAPTER 3 Describing, Exploring, and Comparing Data

TABLE 3-3

General Procedure for Finding Standard Deviation with Formula 3-4

Specific Example Using These Sample Values: 84, 74, 50, 60, 52

Step 1: Compute the mean x. The sum of 84, 74, 50, 60, 52 is 320; therefore:

x = Σ x n

= 84 + 74 + 50 + 60 + 52

5

= 320

5 = 64.0

Step 2: Subtract the mean from each individual sample value. [The result is a list of deviations of the form 1x - x2.]

Subtract the mean of 64.0 from each sample value to get these deviations away from the mean: 20, 10, −14, −4, −12.

Step 3: Square each of the deviations obtained from Step 2. [This produces numbers of the form 1x - x22.]

The squares of the deviations from Step 2 are: 400, 100, 196, 16, 144.

Step 4: Add all of the squares obtained from Step 3. The result is Σ1x - x22.

The sum of the squares from Step 3 is 856.

Step 5: Divide the total from Step 4 by the num- ber n - 1, which is 1 less than the total number of sample values present.

With n = 5 data values, n - 1 = 4, so we divide 856 by 4 to get this result:

856 4

= 214

Step 6: Find the square root of the result of Step 5. The result is the standard deviation, denoted by s.

The standard deviation is 2214 = 14.6287. Rounding the result, we get s = 14.6 BPM.

EXAMPLE 3 Calculating Standard Deviation with Formula 3-5

Use Formula 3-5 to find the standard deviation of the first five pulse rates of males from Data Set 1 “Body Data”: 84, 74, 50, 60, 52.

S O L U T I O N

Here are the components needed in Formula 3-5.

n = 5 (because there are 5 values in the sample) Σ x = 320 (found by adding the original sample values) Σ x2 = 21,336 (found by adding the squares of the sample values, as in

842 + 742 + 502 + 602 + 522 = 21,336)

Using Formula 3-5, we get

s = B n1Σ x22 - 1Σ x22

n1n - 12 = B 5121,3362 - 132022

515 - 12 = B 4280 20

= 14.6 BPM

The result of s = 14.6 BPM is the same as the result in Example 2.

Range Rule of Thumb for Understanding Standard Deviation

The range rule of thumb is a crude but simple tool for understanding and interpreting standard deviation. It is based on the principle that for many data sets, the vast major- ity (such as 95%) of sample values lie within 2 standard deviations of the mean. We could improve the accuracy of this rule by taking into account such factors as the size of the sample and the distribution, but here we sacrifice accuracy for the sake of sim- plicity. The concept of significance as given below will be enhanced in later chapters, especially those that include the topic of hypothesis tests, which are also called tests

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3-2 Measures of Variation 93

of significance. The following range rule of thumb is based on the population mean m and the population standard deviation s, but for large and representative samples, we could use x and s instead.

Range Rule of Thumb for Identifying Significant Values

Significantly low values are m - 2s or lower. Significantly high values are m + 2s or higher. Values not significant: Between 1m - 2s2 and 1m + 2s2

See Figure 3-3, which illustrates the above criteria.

Values not significant Significantly low values

Significantly high values

mm − 2s m + 2s

FIGURE 3-3 Range Rule of Thumb for Identifying Significant Values

Range Rule of Thumb for Estimating a Value of the Standard Deviation s To roughly estimate the standard deviation from a collection of known sample data, use

s ≈ range

4

EXAMPLE 4 Range Rule of Thumb for Interpreting s

Using the 153 pulse rates of males listed in Data Set 1 “Body Data” in Appendix B, the mean is x = 69.6 BPM and the standard deviation is s = 11.3 BPM. Use x and s as estimates of m and s, and use the range rule of thumb to find the limits separat- ing values that are significantly low or significantly high. Then determine whether a male pulse rate of 100 BPM is significantly high.

S O L U T I O N

With a mean of 69.6 and a standard deviation of 11.3, we use the range rule of thumb to find values that are significantly low or significantly high as follows:

Significantly low values are 169.6 - 2 * 11.32 or lower, so significantly low values are 47.0 BPM or lower.

Significantly high values are 169.6 + 2 * 11.32 or higher, so significantly high values are 92.2 BPM or higher.

Values not significant: Between 47.0 and 92.2 BPM

I N T E R P R E TAT I O N

Based on these results, we expect that typical pulse rates of males are between 47.0 BPM and 92.2 BPM. Because the given value of 100 BPM falls above 92.2 BPM, we consider it to be a significantly high pulse rate for a male.

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94 CHAPTER 3 Describing, Exploring, and Comparing Data

Standard Deviation of a Population

The definition of standard deviation and Formulas 3-4 and 3-5 apply to the standard deviation of sample data. A slightly different formula is used to calculate the standard deviation s (lowercase sigma) of a population: Instead of dividing by n - 1, we di- vide by the population size N, as shown here:

Population standard deviation s = B Σ1x - m22

N

Because we generally deal with sample data, we will usually use Formula 3-4, in which we divide by n - 1. Many calculators give both the sample standard deviation and the population standard deviation, but they use a variety of different notations.

EXAMPLE 5 Range Rule of Thumb for Estimating s

Use the range rule of thumb to estimate the standard deviation of the sample of 153 pulse rates of males listed in Data Set 1 “Body Data” in Appendix B. Those 153 values have a minimum of 40 BPM and a maximum of 104 BPM.

S O L U T I O N

The range rule of thumb indicates that we can estimate the standard deviation by finding the range and dividing it by 4. With a minimum of 40 BPM and a maximum of 104 BPM, the range rule of thumb can be used to estimate the standard deviation s as follows:

s ≈ range

4 =

104 - 40 4

= 16.0 BPM

I N T E R P R E TAT I O N

The actual value of the standard deviation is s = 11.3 BPM, so the estimate of 16.0 BPM is very roughly in the general neighborhood of the exact result. Because this estimate is based on only the minimum and maximum values, it might be off by a considerable amount.

CAUTION When using a calculator to find standard deviation, identify the notation used by your particular calculator so that you get the sample standard deviation, not the population standard deviation.

Variance of a Sample and a Population

So far, we have used the term variation as a general description of the amount that val- ues vary among themselves. (The terms dispersion and spread are sometimes used in- stead of variation.) Unlike the term variation, the term variance has a specific meaning.

DEFINITION

The variance of a set of values is a measure of variation equal to the square of the standard deviation.

•  Sample variance: s2 = square of the standard deviation s.

•  Population variance: s2 = square of the population standard deviation s.

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3-2 Measures of Variation 95

Notation Here is a summary of notation for the standard deviation and variance:

s = sample standard deviation s2 = sample variance s = population standard deviation s2 = population variance

Note: Articles in professional journals and reports often use SD for standard deviation and VAR for variance.

Important Properties of Variance

■ The units of the variance are the squares of the units of the original data values. (If the original data values are in feet, the variance will have units of ft2; if the original data values are in seconds, the variance will have units of sec2.)

■ The value of the variance can increase dramatically with the inclusion of outliers. (The variance is not resistant.)

■ The value of the variance is never negative. It is zero only when all of the data values are the same number.

■ The sample variance s2 is an unbiased estimator of the population variance s2, as described in Part 2 of this section. (The sample standard deviation s is a biased estimator of the population standard deviation s.)

The variance is a statistic used in some statistical methods, but for our present pur- poses, the variance has the serious disadvantage of using units that are different than the units of the original data set. This makes it difficult to understand variance as it relates to the original data set. Because of this property, it is better to first focus on the standard deviation when trying to develop an understanding of variation.

PA RT 2 Beyond the Basics of Variation In Part 2, we focus on making sense of the standard deviation so that it is not some mysterious number devoid of any practical significance. We begin by addressing com- mon questions that relate to the standard deviation.

Why Is Standard Deviation Defined as in Formula 3-4?

In measuring variation in a set of sample data, it makes sense to begin with the indi- vidual amounts by which values deviate from the mean. For a particular data value x, the amount of deviation is x - x. It makes sense to somehow combine those devia- tions into one number that can serve as a measure of the variation. Adding the devia- tions isn’t good, because the sum will always be zero. To get a statistic that measures variation, it’s necessary to avoid the canceling out of negative and positive numbers. One approach is to add absolute values, as in Σ� x - x �. If we find the mean of that sum, we get the mean absolute deviation (or MAD), which is the mean distance of the data from the mean:

Mean absolute deviation = Σ� x - x �

n

Why Not Use the Mean Absolute Deviation Instead of the Standard Deviation? Computation of the mean absolute deviation uses absolute values, so it uses an operation that is not “algebraic.” (The algebraic operations include

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96 CHAPTER 3 Describing, Exploring, and Comparing Data

addition, multiplication, extracting roots, and raising to powers that are integers or fractions.) The use of absolute values would be simple, but it would create algebraic difficulties in inferential methods of statistics discussed in later chap- ters. The standard deviation has the advantage of using only algebraic opera- tions. Because it is based on the square root of a sum of squares, the standard deviation closely parallels distance formulas found in algebra. There are many instances where a statistical procedure is based on a similar sum of squares. Consequently, instead of using absolute values, we square all deviations 1x - x2 so that they are nonnegative, and those squares are used to calculate the standard deviation.

Why Divide by n − 1? After finding all of the individual values of 1x - x22 we combine them by finding their sum. We then divide by n - 1 because there are only n - 1 values that can assigned without constraint. With a given mean, we can use any numbers for the first n - 1 values, but the last value will then be automatically deter- mined. With division by n - 1, sample variances s2 tend to center around the value of the population variance s2; with division by n, sample variances s2 tend to underesti- mate the value of the population variance s2.

Comparing Variation in Different Samples or Populations

It’s a good practice to compare two sample standard deviations only when the sample means are approximately the same. When comparing variation in samples or popula- tions with very different means, it is better to use the coefficient of variation. Also use the coefficient of variation to compare variation from two samples or populations with different scales or units of values, such as the comparison of variation of pulse rates of men and heights of men. (See Example 6.)

DEFINITION

The coefficient of variation (or CV) for a set of nonnegative sample or popula- tion data, expressed as a percent, describes the standard deviation relative to the mean, and is given by the following:

Sample Population

CV = s x # 100% CV = s

m # 100%

ROUND-OFF RULE FOR THE COEFFICIENT OF VARIATION Round the coefficient of variation to one decimal place (such as 25.3%).

EXAMPLE 6 Pulse Rates and Heights

Compare the variation of the 153 male pulse rates listed in Data Set 1 “Body Data” in Appendix B and the heights of the same males. For the male pulse rates, x = 69.6 BPM and s = 11.3 BPM; for their heights, x = 174.12 cm and s = 7.10 cm. Note that we want to compare variation among pulse rates to varia- tion among heights.

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3-2 Measures of Variation 97

Biased and Unbiased Estimators

The sample standard deviation s is a biased estimator of the population standard de- viation s, which means that values of the sample standard deviation s do not tend to center around the value of the population standard deviation s. While individual values of s could equal or exceed s, values of s generally tend to underestimate the value of s. For example, consider an IQ test designed so that the population standard deviation is 15. If you repeat the process of randomly selecting 100 subjects, giving them IQ tests, and calculating the sample standard deviation s in each case, the sample standard deviations that you get will tend to be less than 15, which is the population standard deviation. There is no correction that allows us to fix the bias for all distribu- tions of data. There is a correction that allows us to fix the bias for normally distrib- uted populations, but it is rarely used because it is too complex and makes relatively minor corrections.

The sample variance s2 is an unbiased estimator of the population variance s2, which means that values of s2 tend to center around the value of s2 instead of system- atically tending to overestimate or underestimate s2. Consider an IQ test designed so that the population variance is 225. If you repeat the process of randomly selecting 100 subjects, giving them IQ tests, and calculating the sample variance s2 in each case, the sample variances that you obtain will tend to center around 225, which is the population variance.

Biased estimators and unbiased estimators will be discussed more in Section 6-3.

S O L U T I O N

We can compare the standard deviations if the same scales and units are used and the two means are approximately equal, but here we have different scales and differ- ent units of measurement, so we use the coefficients of variation:

Male Pulse Rates: CV = s x # 100% = 11.3 BPM

69.6 BPM # 100% = 16.2%

Male Heights: CV = s x # 100% = 7.10 cm

174.12 cm # 100% = 4.1%

We can now see that the male pulse rates (with CV = 16.2%) vary more than male heights (with CV = 4.1%).

Measures of Variation Access tech instructions, videos, and data sets at www.TriolaStats.com

TECH CENTER

Statistical Literacy and Critical Thinking

1.  Range Rule of Thumb for Estimating s The 20 brain volumes (cm3) from Data Set 9 “IQ and Brain Size” in Appendix B vary from a low of 963 cm3 to a high of 1439 cm3. Use the range rule of thumb to estimate the standard deviation s and compare the result to the exact standard deviation of 124.9 cm3.

3-2 Basic Skills and Concepts

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98 CHAPTER 3 Describing, Exploring, and Comparing Data

2. Range Rule of Thumb for Interpreting s The 20 brain volumes (cm3) from Data Set 9 “IQ and Brain Size” in Appendix B have a mean of 1126.0 cm3 and a standard deviation of 124.9 cm3. Use the range rule of thumb to identify the limits separating values that are signifi- cantly low or significantly high. For such data, would a brain volume of 1440 cm3 be signifi- cantly high?

3.  Variance The 20 subjects used in Data Set 9 “IQ and Brain Size” in Appendix B have weights with a standard deviation of 20.0414 kg. What is the variance of their weights? Include the appropriate units with the result.

4. Symbols Identify the symbols used for each of the following: (a) sample standard devia- tion; (b) population standard deviation; (c) sample variance; (d) population variance.

In Exercises 5–20, find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as “minutes”) in your results. (The same data were used in Section 3-1 where we found measures of center. Here we find measures of varia- tion.) Then answer the given questions.

5. Charges for Births Data Set 3 “Births” in Appendix B includes total charges for births at four hospitals in New York State, and the top 10 highest amounts (in dollars) are listed below. What do the results tell us about the population of all such charges?

471,062 359,091 290,837 271,863 255,788

247,477 232,782 197,912 183,271 155,857

6. MCAT Score Listed below are mean MCAT scores listed in order by year, starting with the year 2002. What important feature of the data is not revealed by any of the measures of center?

27.0 26.8 27.1 27.3 27.4 27.7 28.1 27.9 28.3 28.2 28.3 28.4

7. Football Player Numbers Listed below are the jersey numbers of 11 players randomly selected from the roster of the Seattle Seahawks when they won Super Bowl XLVIII. What do the results tell us?

89 91 55 7 20 99 25 81 19 82 60

8. Football Player Weights Listed below are the weights (lb) of 11 players randomly selected from the roster of the Seattle Seahawks when they won Super Bowl XLVIII (the same players from the preceding exercise). Are the results likely to be representative of all NFL players?

189 254 235 225 190 305 195 202 190 252 305

9.  Peas in a Pod Biologists conducted experiments to determine whether a deficiency of carbon dioxide in the soil affects the phenotypes of peas. Listed below are the pheno- type codes, where 1 = smooth @ yellow, 2 = smooth @ green, 3 = wrinkled @yellow, and 4 = wrinkled @ green. Do the results make sense?

2 1 1 1 1 1 1 4 1 2 2 1 2 3 3 2 3 1 3 1 3 1 3 2 2

10. TV Prices A physician plans to buy a television for her large waiting room. Listed below are selling prices (in dollars) of TVs that are 60 inches or larger and rated as a “best buy” by Consumer Reports magazine. Are the resulting statistics representative of the population of all TVs that are 60 inches and larger?

1800 1500 1200 1500 1400 1600 1500 950 1600 1150 1500 1750

11. Cell Phone Radiation Listed below are the measured radiation absorption rates (in W>kg) corresponding to these cell phones: iPhone 5S, BlackBerry Z30, Sanyo Vero, Optimus V, Droid Razr, Nokia N97, Samsung Vibrant, Sony Z750a, Kyocera Kona, LG G2, and Virgin Mobile Supreme. The data are from the Federal Communications Commission. If one of each model of cell phone is measured for radiation and the results are used to find the measures of variation, are the results typical of the population of cell phones that are in use?

1.18 1.41 1.49 1.04 1.45 0.74 0.89 1.42 1.45 0.51 1.38

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3-2 Measures of Variation 99

12. Caffeine in Soft Drinks Listed below are measured amounts of caffeine (mg per 12 oz of drink) obtained in one can from each of 20 brands (7-UP, A&W Root Beer, Cherry Coke, . . . , Tab). Are the statistics representative of the population of all cans of the same 20 brands consumed by Americans?

0 0 34 34 34 45 41 51 55 36 47 41 0 0 53 54 38 0 41 47

13. Firefighter Fatalities Listed below are the numbers of heroic firefighters who lost their lives in the United States each year while fighting forest fires. The numbers are listed in order by year, starting with the year 2000. What important feature of the data is not revealed by any of the measures of variation?

20 18 23 30 20 12 24 9 25 15 8 11 15 34

14. Foot Lengths Listed below are foot lengths in inches of randomly selected Army women measured in the 1988 Anthropometric Survey (ANSUR). Are the statistics representative of the current population of all Army women?

10.4 9.3 9.1 9.3 10.0 9.4 8.6 9.8 9.9 9.1 9.1

15. Medical School Tuition Listed below in dollars are the annual costs of tuition at the 10 most expensive private medical schools in the United States for a recent year (based on data from U.S. News & World Report). What does this “top 10” list tell us about those costs for the population of all U.S. private medical school tuitions?

57,261 56,784 55,196 54,976 54,653 54,528 54,268 54,050 53,581 53,323

16. California Smokers In the California Health Interview Survey, randomly selected adults are interviewed. One of the questions asks how many cigarettes are smoked per day, and results are listed below for 50 randomly selected respondents. How well do the results reflect the smoking behavior of California adults?

9 10 10 20 40 50 1Plus 44 other values that are all 02 In Exercises 17–20, find the coefficient of variation for each of the two samples; then com- pare the variation. (The same data were used in Section 3-1.)

17.  Blood Pressure A sample of blood pressure measurements is taken from Data Set 1 “Body Data” in Appendix B, and those values (mm Hg) are listed below. The values are matched so that 10 subjects each have systolic and diastolic measurements.

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

Systolic: 118 128 158 96 156 122 116 136 126 120

Diastolic: 80 76 74 52 90 88 58 64 72 82

18. White ,Red Blood Counts Listed below are white blood cell counts (1000 cells>mL) and red blood cell counts (million cells>mL) from different subjects (from Data Set 1 “Body Data” in Appendix B). The values are matched so that each of the 12 subjects has a white blood cell count and a red blood cell count.

White: 8.7 4.9 6.9 7.5 6.1 5.7 4.1 8.1 8.0 5.6 8.3 6.9

Red: 4.8 4.7 4.5 4.3 5.0 4.0 4.7 4.6 4.1 5.5 4.4 4.2

19.  White Blood Counts Listed below are white blood cell counts (1000 cells>mL) from males and females (from Data Set 1 “Body Data” in Appendix B). Do they appear to be different?

Female: 8.7 6.9 8.1 8.0 6.9 8.1 6.4 6.3 10.9 4.8 5.9 7.2

Male: 4.9 7.5 6.1 5.7 4.1 5.6 8.3 5.1 9.5 6.1 5.7 5.4

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100 CHAPTER 3 Describing, Exploring, and Comparing Data

20. Queues A Providence Hospital experiment involves two different waiting line configura- tions for patients arriving for admission. The waiting times (in seconds) are recorded with a single line configuration that feeds four stations and another configuration with individual lines at the four stations.

Single Line 390 396 402 408 426 438 444 462 462 462

Individual Lines 252 324 348 372 402 462 462 510 558 600

Large Data Sets from Appendix B. In Exercises 21–24, refer to the indicated data set in Appendix B. Use software or a calculator to find the range, variance, and standard devia- tion. Express answers using appropriate units, such as “minutes.”

21. HDL Use the HDL cholesterol measurements (mg>dL) from the 300 subjects included in Data Set 1 “Body Data” in Appendix B. Identify the highest value. Does it appear to be an out- lier? Do the measures of variation change much when that highest value is deleted?

22. LDL Repeat the preceding exercise using the LDL measurements (mg>dL). 23. Body Temperatures Refer to Data Set 2 “Body Temperatures” in Appendix B and use the body temperatures for 12:00 AM on day 2.

24.  Births Use the birth weights (grams) of the 400 babies listed in Data Set 3 “Births” in Appendix B. Examine the list of birth weights to make an observation about those numbers. How does that observation affect the way that the results should be rounded?

Estimating Standard Deviation with the Range Rule of Thumb. In Exercises 25–28, refer to the data in the indicated exercise. After finding the range of the data, use the range rule of thumb to estimate the value of the standard deviation. Compare the result to the standard deviation computed with all of the data.

25. HDL Exercise 21

26. LDL Exercise 22

27. Body Temperatures Exercise 23

28. Births Exercise 24

Identifying Significant Values with the Range Rule of Thumb. In Exercises 29–32, use the range rule of thumb to identify the limits separating values that are significantly low or significantly high.

29.  Pulse Rates of Females Based on Data Set 1 “Body Data” in Appendix B, females have pulse rates with a mean of 74.0 beats per minute and a standard deviation of 12.5 beats per minute. Is a female pulse rate of 44 beats per minute significantly low or significantly high? (All of these pulse rates are measured at rest.)

30.  Pulse Rates of Males Based on Data Set 1 “Body Data” in Appendix B, males have pulse rates with a mean of 69.6 beats per minute and a standard deviation of 11.3 beats per minute. Is a male pulse rate of 50 beats per minute significantly low or significantly high? (All of these pulse rates are measured at rest.) Explain.

31. Foot Lengths Based on Data Set 7 “Foot and Height” in Appendix B, adult males have foot lengths with a mean of 27.32 cm and a standard deviation of 1.29 cm. Is the adult male foot length of 30 cm significantly low or significantly high? Explain.

32.  Body Temperatures Based on Data Set 2 “Body Temperatures” in Appendix B, body temperatures of adults have a mean of 98.20oF and a standard deviation of 0.62oF. Is an adult body temperature of 100oF significantly low or significantly high?

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3-2 Measures of Variation 101

Finding Standard Deviation from a Frequency Distribution. In Exercises 33 and 34, refer to the frequency distribution in the given exercise and find the standard deviation by using the formula below, where x represents the class midpoint, f represents the class fre- quency, and n represents the total number of sample values. Also, compare the computed standard deviations to these standard deviations obtained by using Formula 3-4 with the original list of data values: (Exercise 33) 59.5; (Exercise 34) 65.4.

s = B n3Σ1f # x224 - 3Σ1f # x242

n1n - 12

33.  Blood Platelet

Count of Males

Frequency

0–99 1

100–199 51

200–299 90

300–399 10

400–499 0

500–599 0

600–699 1

34.  Blood Platelet

Count of Females

Frequency

100–199 25

200–299 92

300–399 28

400–499 0

500–599 2

35. Why Divide by n − 1? Let a population consist of these values: 9 cigarettes, 10 cigarettes, and 20 cigarettes smoked in a day (based on data from the California Health Interview Survey). Assume that samples of two values are randomly selected with replace- ment from this population. (That is, a selected value is replaced before the second selection is made.)

a. Find the variance s2 of the population {9 cigarettes, 10 cigarettes, 20 cigarettes}.

b. After listing the nine different possible samples of two values selected with replacement, find the sample variance s2 (which includes division by n - 1) for each of them; then find the mean of the nine sample variances s2.

c. For each of the nine different possible samples of two values selected with replacement, find the variance by treating each sample as if it is a population (using the formula for population variance, which includes division by n); then find the mean of those nine population variances.

d. Which approach results in values that are better estimates of s2: part (b) or part (c)? Why? When computing variances of samples, should you use division by n or n - 1?

e. The preceding parts show that s2 is an unbiased estimator of s2. Is s an unbiased estimator of s? Explain.

36. Mean Absolute Deviation Use the same population of {9 cigarettes, 10 cigarettes, 20 cigarettes} from Exercise 35. Show that when samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population?

3-2 Beyond the Basics

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102 CHAPTER 3 Describing, Exploring, and Comparing Data

Key Concept This section introduces measures of relative standing, which are num- bers showing the location of data values relative to the other values within the same data set. The most important concept in this section is the z score, which will be used often in following chapters. We also discuss percentiles and quartiles, which are com- mon statistics, as well as another statistical graph called a boxplot.

PA RT 1 Basics of z Scores, Percentiles, Quartiles, and Boxplots

z Scores

A z score is found by converting a value to a standardized scale, as given in the fol- lowing definition. This definition shows that a z score is the number of standard devia- tions that a data value is away from the mean. The z score is used often in Chapter 6 and later chapters.

3-3 Measures of Relative Standing and Boxplots

DEFINITION

A z score (or standard score or standardized value) is the number of standard deviations that a given value x is above or below the mean. The z score is calcu- lated by using one of the following:

Sample Population

z = x - x

s or z =

x - m s

ROUND-OFF RULE FOR z SCORES Round z scores to two decimal places (such as 2.31).

This round-off rule is motivated by the format of standard tables in which z scores are expressed with two decimal places, as in Table A-2 in Appendix A. Example 1 il- lustrates how z scores can be used to compare values, even if they come from different populations.

Important Properties of z Scores

1. A z score is the number of standard deviations that a given value x is above or below the mean.

2. z Scores are expressed as numbers with no units of measurement.

3. A data value is significantly low if its z score is less than or equal to - 2 or the value is significantly high if its z score is greater than or equal to + 2.

4. If an individual data value is less than the mean, its corresponding z score is a negative number.

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3-3 Measures of Relative Standing and Boxplots 103

Using z Scores to Identify Significant Values In Section 3-2 we used the range rule of thumb to conclude that a value is significantly low or significantly high if it is at least 2 standard deviations away from the mean. It follows that significantly low values have z scores less than or equal to - 2 and significantly high values have z scores greater than or equal to + 2, as illustrated in Figure 3-4. Using this criterion with the two individual values used in Example 1 above, we see that neither value is significant because both z scores are between - 2 and + 2.

EXAMPLE 1 Comparing a Baby’s Weight and Adult Body Temperature

Which of the following two data values is more extreme relative to the data set from which it came?

■ The 4000 g weight of a newborn baby (among 400 weights with sample mean x = 3152.0 g and sample standard deviation s = 693.4 g)

■ The 99°F temperature of an adult (among 106 adults with sample mean x = 98.20°F and sample standard deviation s = 0.62°F)

S O L U T I O N

The 4000 g weight and the 99°F body temperature can be standardized by convert- ing each of them to z scores as shown below.

4000 g birth weight:

z = x - x

s =

4000 g - 3152.0 g 693.4 g

= 1.22

99°F body temperature:

z = x - x

s =

99°F - 98.20°F 0.62°F

= 1.29

I N T E R P R E TAT I O N

The z scores show that the 4000 g birth weight is 1.22 standard deviations above the mean, and the 99°F body temperature is 1.29 standard deviations above the mean. Because the body temperature is farther above the mean, it is the more extreme value, but not by much. A 99°F body temperature is slightly more extreme than a birth weight of 4000 g.

Values not significant Significantly low values

Significantly high values

z

−2 −1 2−3 30 1

FIGURE 3-4 Interpreting z Scores Significant values are those with z scores … - 2.00 or Ú 2.00.

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104 CHAPTER 3 Describing, Exploring, and Comparing Data

A z score is a measure of position, in the sense that it describes the location of a value (in terms of standard deviations) relative to the mean. Percentiles and quartiles are other measures of position useful for comparing values within the same data set or between different sets of data.

Percentiles

Percentiles are one type of quantiles—or fractiles—which partition data into groups with roughly the same number of values in each group.

EXAMPLE 2 Is a Platelet Count of 75 Significantly Low?

The lowest platelet count in Data Set 1 “Body Data” in Appendix B is 75. (The platelet counts are measured in 1000 cells>mL). Is that value significantly low? Based on the platelet counts from Data Set 1 in Appendix B, assume that platelet counts have a mean of x = 239.4 and a standard deviation of s = 64.2.

S O L U T I O N

The platelet count of 75 is converted to a z score as shown below:

z = x - x

s =

75 - 239.4 64.2

= - 2.56

I N T E R P R E TAT I O N

The platelet count of 75 converts to the z score of - 2.56. Refer to Figure 3-4 to see that z = - 2.56 is less than - 2, so the platelet count of 75 is significantly low. (Low platelet counts are called thrombocytopenia. What a wonderful name.)

DEFINITION

Percentiles are measures of location, denoted P1, P2, . . . , P99, which divide a set of data into 100 groups with about 1% of the values in each group.

The 50th percentile, denoted P50, has about 50% of the data values below it and about 50% of the data values above it, so the 50th percentile is the same as the median. There is not universal agreement on a single procedure for calculating percentiles, but we will describe relatively simple procedures for (1) finding the percentile of a data value and (2) converting a percentile to its corresponding data value. We begin with the first procedure.

Finding the Percentile of a Data Value

The process of finding the percentile that corresponds to a particular data value x is given by the following (round the result to the nearest whole number):

Percentile of value x = number of values less than x

total number of values # 100

Detecting Phony Data

A class is

given the

homework

assignment of

recording the

results when a

coin is tossed

500 times. One dishonest stu-

dent decides to save time by just

making up the results instead of

actually flipping a coin. Because

people generally cannot make

up results that are really random,

we can often identify such phony

data. With 500 tosses of an ac-

tual coin, it is extremely likely that

you will get a run of six heads or

six tails, but people almost never

include such a run when they

make up results.

Another way to detect fab-

ricated data is to establish that

the results violate Benford’s law:

For many collections of data, the

leading digits are not uniformly

distributed. Instead, the leading

digits of 1, 2,…, 9 occur with

rates of 30%, 18%, 12%, 10%,

8%, 7%, 6%, 5%, and 5%, re-

spectively. (See “The Difficulty of

Faking Data,” by Theodore Hill,

Chance, Vol. 12, No. 3.)

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3-3 Measures of Relative Standing and Boxplots 105

Example 3 shows how to convert from a given sample value to the corresponding percentile. There are several different methods for the reverse procedure of converting a given percentile to the corresponding value in the data set. The procedure we will use is summarized in Figure 3-5 on the next page, which uses the following notation.

Notation

n total number of values in the data set

k percentile being used (Example: For the 25th percentile, k = 25.) L locator that gives the position of a value (Example: For the 12th value in

the sorted list, L = 12.) Pk kth percentile (Example: P25 is the 25th percentile.)

TABLE 3-4 Sorted Cotinine Measures of Smokers

0 1 1 3 17 32 35 44 48 86

87 103 112 121 123 130 131 149 164 167

173 173 198 208 210 222 227 234 245 250

253 265 266 277 284 289 290 313 477 491

EXAMPLE 3 Finding a Percentile

Table 3-4 lists the 40 cotinine measures (ng>mL) of smokers from Data Set 14 “Passive and Active Smoke” in Appendix B, and they are listed in order. Find the percentile for the cotinine level of 198 ng>mL.

S O L U T I O N

From the sorted list of cotinine levels in Table 3-4, we see that there are 22 values less than 198 ng>mL, so

Percentile of 198 ng>mL = 22 40

# 100 = 55

I N T E R P R E TAT I O N

A cotinine level of 198 ng>mL is in the 55th percentile. This can be interpreted loosely as this: A cotinine level of 198 ng>mL separates the lowest 55% of values from the highest 45% of values. We have P55 = 198 ng>mL.

EXAMPLE 4 Converting a Percentile to a Data Value

Refer to the sorted cotinine levels of smokers in Table 3-4 and use the procedure in Figure 3-5 to find the value of the 33rd percentile, P33.

continued

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106 CHAPTER 3 Describing, Exploring, and Comparing Data

The value of Pk is the Lth value, counting from

the lowest.

Is L a whole number?

Yes

No

Change L by rounding it up to the next larger

whole number.

Compute

n 5 number of values k 5 percentile in question

Sort the data. (Arrange the data in order

of lowest to highest.)

Start

L 5 n where k

100

The value of the kth percentile is midway between the Lth value and the next value in the sorted set of data. Find Pk by adding

the Lth value and the next value and dividing the total by 2.

FIGURE 3-5 Converting from the kth percentile to the corresponding data value

S O L U T I O N

From Figure 3-5, we see that the sample data are already sorted, so we can proceed to find the value of the locator L. In this computation we use k = 33 because we are trying to find the value of the 33rd percentile. We use n = 40 because there are 40 data values.

L = k

100 # n = 33

100 # 40 = 13.2

Since L = 13.2 is not a whole number, we proceed to the next lower box in Figure 3-5 where we change L by rounding it up from 13.2 to the next larger whole number: 14. (In this book we typically round off the usual way, but this is one of two cases where we round up instead of rounding off.) From the bottom box we see that the value of P33 is the 14th value, counting from the lowest. In Table 3-4, the 14th value is 121. That is, P33 = 121 ng>mL. Roughly speaking, about 33% of the cotinine levels in Table 3-4 are less than 121 ng>mL and 67% of them are more than 121 ng>mL.

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3-3 Measures of Relative Standing and Boxplots 107

Quartiles

Just as there are 99 percentiles that divide the data into 100 groups, there are three quartiles that divide the data into four groups.

EXAMPLE 5 Converting a Percentile to a Data Value

Refer to the sorted pulse rates in Table 3-4. Use Figure 3-5 to find the 25th percen- tile, denoted by P25.

S O L U T I O N

Referring to Figure 3-5, we see that the sample data are already sorted, so we can proceed to compute the value of the locator L. In this case, we use k = 25 because we are attempting to find the value of the 25th percentile, and we use n = 40 be- cause there are 40 data values.

L = k

100 # n = 25

100 # 40 = 10

Since L = 10 is a whole number, we proceed to the box in Figure 3-5 located at the right. We now see that the value of the 25th percentile is midway between the Lth (10th) value and the next higher value in the original set of data. That is, the value of the 25th percentile is midway between the 10th value and the 11th value. The 10th value in Table 3-4 is 86 and the 11th value is 87, so the value midway between them is 86.5 ng>mL. We conclude that the 25th percentile is P25 = 86.5 ng>mL.

DEFINITION

Quartiles are measures of location, denoted Q1, Q2, and Q3,which divide a set of data into four groups with about 25% of the values in each group.

Here are descriptions of quartiles that are more accurate than those given in the preceding definition:

Q1 (First quartile): Same value as P25. It separates the bottom 25% of the sorted values from the top 75%. (To be more precise, at least 25% of the sorted values are less than or equal to Q1, and at least 75% of the values are greater than or equal to Q1.)

Q2 (Second quartile): Same as P50 and same as the median. It separates the bottom 50% of the sorted values from the top 50%.

Q3 (Third quartile): Same as P75. It separates the bottom 75% of the sorted val- ues from the top 25%. (To be more precise, at least 75% of the sorted values are less than or equal to Q3, and at least 25% of the values are greater than or equal to Q3.)

Finding values of quartiles can be accomplished with the same procedure used for finding percentiles. Simply use the relationships shown in the margin. In Example 4 we found that P25 = 86.5 ng>mL, so it follows that Q1 = 86.5 ng>mL.

Q1 = P25

Q2 = P50

Q3 = P75

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108 CHAPTER 3 Describing, Exploring, and Comparing Data

In earlier sections of this chapter we described several statistics, including the mean, median, mode, range, and standard deviation. Some other statistics are defined using quartiles and percentiles, as in the following:

Interquartile range (or IQR) = Q3 - Q1

Semi-interquartile range = Q3 - Q1

2

Midquartile = Q3 + Q1

2

10–90 percentile range = P90 - P10

5-Number Summary and Boxplot

The values of the minimum, maximum and three quartiles 1Q1, Q2, Q32 are used for the 5-number summary and the construction of boxplot graphs.

CAUTION Just as there is not universal agreement on a procedure for finding percentiles, there is not universal agreement on a single procedure for calculating quartiles, and different technologies often yield different results. If you use a calculator or software for exercises involving quartiles, you may get results that differ somewhat from the answers obtained by using the procedures described here.

DEFINITION

For a set of data, the 5-number summary consists of these five values:

1. Minimum

2. First quartile, Q1 3. Second quartile, Q2 (same as the median)

4. Third quartile, Q3 5. Maximum

EXAMPLE 6 Finding a 5-Number Summary

Use the cotinine measurements in Table 3-4 to find the 5-number summary.

S O L U T I O N

Because the cotinine measurements in Table 3-4 are sorted, it is easy to see that the minimum is 0 ng>mL and the maximum is 491 ng>mL. The value of the first quartile is Q1 = 86.5 ng>mL (from Example 5). The median is equal to Q2, and it is 170.0 ng>mL. Also, we can find that Q3 = 251.5 ng>mL by using the procedure for finding P75 (as summarized in Figure 3-5). The 5-number summary is therefore 0, 86.5, 170.0, 251.5, and 491 (all in units of ng>mL).

DEFINITION

A boxplot (or box-and-whisker diagram) is a graph of a data set that consists of a line extending from the minimum value to the maximum value, and a box with lines drawn at the first quartile Q1, the median, and the third quartile Q3. (See Figure 3-6.)

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3-3 Measures of Relative Standing and Boxplots 109

Procedure for Constructing a Boxplot

1. Find the 5-number summary (minimum value, Q1, Q2, Q3, maximum value).

2. Construct a line segment extending from the minimum data value to the maxi- mum data value.

3. Construct a box (rectangle) extending from Q1 to Q3, and draw a line in the box at the value of Q2 (median).

CAUTION Because there is not universal agreement on procedures for finding quartiles, and because boxplots are based on quartiles, different technologies may yield different boxplots.

EXAMPLE 7 Constructing a Boxplot

Use the cotinine measurements listed in Table 3-4 to construct a boxplot.

S O L U T I O N

The boxplot uses the 5-number summary found in Example 6: 0, 86.5, 170.0, 251.5, and 491 (all in units of ng>mL). Figure 3-6 is the boxplot representing the cotinine measurements listed in Table 3-4.

FIGURE 3-6 Boxplot of Cotinine Measurements (ng ,mL)

Skewness A boxplot can often be used to identify skewness (discussed in Section 2-2). The boxplot in Figure 3-6 isn’t exactly symmetric; it shows that the data are slightly skewed to the right.

Because the shape of a boxplot is determined by the five values from the 5-number summary, a boxplot is not a graph of the distribution of the data, and it doesn’t show as much detailed information as a histogram or stemplot. However, boxplots are often great for comparing two or more data sets. When using two or more boxplots for com- paring different data sets, graph the boxplots on the same scale so that comparisons can be easily made. Methods discussed later in this book allow us to analyze com- parisons of data sets more formally than subjective conclusions based on a graph. It is always wise to construct suitable graphs, such as histograms, dotplots, and boxplots, but we should not rely solely on subjective judgments based on graphs.

EXAMPLE 8 Comparing the Pulse Rates of Men and Women

The Chapter Problem involves pulse rates of men and women, and the data are found in Data Set 1 “Body Data” in Appendix B. Construct boxplots of those two different sets of pulse rates.

continued

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110 CHAPTER 3 Describing, Exploring, and Comparing Data

Outliers

When analyzing data, it is important to identify and consider outliers because they can strongly affect values of some important statistics (such as the mean and standard deviation), and they can also strongly affect important methods discussed later in this book. In Chapter 2 we described outliers as sample values that lie very far away from the vast majority of the other values in a set of data, but that description is vague and it does not provide specific objective criteria. Part 2 of this section includes a descrip- tion of modified boxplots along with a more precise definition of outliers used in the context of creating modified boxplots.

S O L U T I O N

The Statdisk-generated boxplots are shown in Figure 3-7. The three quartiles for males are all lower than the corresponding three quartiles for females, which sug- gests that males generally have lower pulse rates than females. The minimums and maximums are not very different in the two boxplots, and we shouldn’t place too much importance on those differences because they are not very reliable measures.

FIGURE 3-7 Boxplots of Pulse Rates of Men and Women

CAUTION When analyzing data, always identify outliers and consider their effects, which can be substantial.

PA RT 2 Outliers and Modified Boxplots We noted that the description of outliers is somewhat vague, but for the purposes of constructing modified boxplots, we can consider outliers to be data values meeting specific criteria based on quartiles and the interquartile range. (The interquartile range is often denoted by IQR, and IQR = Q3 - Q1.)

Identifying Outliers for Modified Boxplots

1. Find the quartiles Q1, Q2, and Q3.

2. Find the interquartile range (IQR), where IQR = Q3 - Q1. 3. Evaluate 1.5 * IQR. 4. In a modified boxplot, a data value is an outlier if it is

above Q3, by an amount greater than 1.5 : IQR or below Q1, by an amount greater than 1.5 : IQR

Modified Boxplots

The boxplots described earlier in Part 1 are called skeletal (or regular) boxplots, but some statistical software packages provide modified boxplots, which represent outliers as

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3-3 Measures of Relative Standing and Boxplots 111

special points. A modified boxplot is a regular boxplot constructed with these modifi- cations: (1) A special symbol (such as an asterisk or point) is used to identify outliers as defined above, and (2) the solid horizontal line extends only as far as the minimum data value that is not an outlier and the maximum data value that is not an outlier. (Note: Exercises involving modified boxplots are found in the “Beyond the Basics” exercises only.)

EXAMPLE 9 Constructing Modified Boxplots

Use the pulse rates of males in Data Set 1 “Body Data” from Appendix B to con- struct a modified boxplot. The five-number summary is 40, 62.0, 68.0, 76.0, 104 (all in BPM).

S O L U T I O N

Let’s begin with the four steps for identifying outliers in a modified boxplot.

1. Using the pulse rates of males, the three quartiles are Q1 = 62.0, the median is Q2 = 68.0, and Q3 = 76.0.

2. The interquartile range is IQR = Q3 - Q1 = 76.0 - 62.0 = 14.0. 3. 1.5 * IQR = 1.5 * 14.0 = 21.0.

4. Any outliers are

■ Greater than Q3 = 76.0 by more than 21.0 or ■ Less than Q1 = 62.0 by more than 21.0.

This means that any outliers are greater than 97.0 or less than 41.0. We can now examine the original pulse rates of males to identify any that are greater than 97.0 or less than 41.0. We find that the pulse rates of 102 and 104 are greater than 97.0, and the pulse rate of 40 is less than 41.0. The outliers are 102, 104, and 40.

We can now construct the modified boxplot shown in Figure 3-8. In Figure 3-8, the three outliers (40, 102, 104) are identified as special points, the three quartiles (62.0, 68.0, 76.0) are shown as in a regular boxplot, and the horizontal line extends from the lowest data value that is not an outlier (42) to the highest data value that is not an outlier (96).

FIGURE 3-8 Modified Boxplot of Male Pulse Rates (BPM)

CAUTION Because there is not universal agreement on procedures for finding quartiles, and because modified boxplots are based on quartiles, different technologies may yield different modified boxplots.

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112 CHAPTER 3 Describing, Exploring, and Comparing Data

Boxplots, 5-Number Summary, Outliers Access tech instructions, videos, and data sets at www.TriolaStats.com

TECH CENTER

Statistical Literacy and Critical Thinking

1.  z Scores LeBron James, one of the most successful basketball players of all time, has a height of 6 feet 8 inches, or 203 cm. Based on statistics from Data Set 1 “Body Data” in Appendix B, his height converts to the z score of 4.07. How many standard deviations is his height above the mean?

2. Heights The boxplot shown below results from the heights (cm) of males listed in Data Set 1 “Body Data” in Appendix B. What do the numbers in that boxplot tell us?

3. Boxplot Comparison Refer to the boxplots shown below that are drawn on the same scale. One boxplot represents weights of men and the other boxplot represents weights of women. Which boxplot represents weights of women? Explain.

4. z Scores If your score on your next statistics test is converted to a z score, which of these z scores would you prefer: - 2.00, - 1.00, 0, 1.00, 2.00? Why?

z Scores. In Exercises 5–8, express all z scores with two decimal places.

5. Female Pulse Rates For the pulse rates of females listed in Data Set 1 “Body Data” in Appendix B, the mean is 74.0 BPM, the standard deviation is 12.5 BPM, and the maximum is 104 BPM.

a. What is the difference between the maximum and the mean?

b. How many standard deviations is that [the difference found in part (a)]?

c. Convert the maximum pulse rate to a z score.

d. If we consider pulse rates that convert to z scores between - 2 and 2 to be neither signifi- cantly low nor significantly high, is the maximum pulse rate significant?

6.  Female Pulse Rates For the pulse rates of females listed in Data Set 1 “Body Data” in Appendix B, the mean is 74.0 BPM, the standard deviation is 12.5 BPM, and the minimum is 36 BPM.

a. What is the difference between the minimum and the mean?

b. How many standard deviations is that [the difference found in part (a)]?

3-3 Basic Skills and Concepts

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3-3 Measures of Relative Standing and Boxplots 113

c. Convert the minimum pulse rate to a z score.

d. If we consider pulse rates that convert to z scores between - 2 and 2 to be neither significantly low nor significantly high, is the minimum pulse rate significantly low or significantly high?

7. Body Temperatures For the body temperatures at 12 AM on day 2 (listed in Data Set 2 “Body Temperatures” in Appendix B), the mean is 98.20oF, the standard deviation is 0.62oF, and the minimum is 96.5oF.

a. What is the difference between the minimum and the mean?

b. How many standard deviations is that [the difference found in part (a)]?

c. Convert the minimum temperature to a z score.

d. If we consider body temperatures that convert to z scores between - 2 and 2 to be neither significantly low nor significantly high, is the minimum body temperature significant?

8. Body Temperatures For the body temperatures at 12 AM on day 2 (listed in Data Set 2 “Body Temperatures” in Appendix B), the mean is 98.20°F, the standard deviation is 0.62°F, and Q3 = 98.60°F.

a. What is the difference between Q3 and the mean?

b. How many standard deviations is that [the difference found in part (a)]?

c. Convert Q3 to a z score.

d. If we consider temperatures that convert to z scores between - 2 and 2 to be neither signifi- cantly low nor significantly high, is Q3 significant?

Significant Values. In Exercises 9–12, consider a value to be significantly low if its z score is less than or equal to − 2 or consider the value to be significantly high if its z score is greater than or equal to 2.

9. ACT The ACT test is used to assess readiness for college. In a recent year, the mean ACT score was 21.1 and the standard deviation was 5.1. Identify the ACT scores that are signifi- cantly low or significantly high.

10. MCAT In a recent year, scores on the MCAT had a mean of 25.2 and a standard deviation of 6.4. Identify the MCAT scores that are significantly low or significantly high.

11. Birth Weights Data Set 3 “Births” lists birth weights (g) of 400 babies. Those weights have a mean of 3152.0 g and a standard deviation of 693.4 g. Identify birth weights that are significantly low or significantly high.

12. Ergonomics in Aircraft Seats In the process of designing aircraft seats, it was found that men have hip breadths with a mean of 36.6 cm and a standard deviation of 2.5 cm (based on anthropometric survey data from Gordon, Clauser, et al.). Identify the hip breadths of men that are significantly low or significantly high.

Comparing Values. In Exercises 13–16, use z scores to compare the given values.

13.  Tallest and Shortest Men The tallest living man at the time of this writing is Sultan Kosen, who has a height of 251 cm. The shortest living man is Chandra Bahadur Dangi, who has a height of 54.6 cm. Heights of men have a mean of 174.12 cm and a standard deviation of 7.10 cm. Which of these two men has the height that is more extreme?

14. Red Blood Cell Counts Based on Data Set 1 “Body Data” in Appendix B, males have red blood cell counts with a mean of 4.719 and a standard deviation of 0.490, while females have red blood cell counts with a mean of 4.349 and a standard deviation of 0.402. Who has the higher count relative to the sample from which it came: a male with a count of 5.58 or a female with a count of 5.23? Explain.

15. Birth Weights Based on Data Set 3 “Births” in Appendix B, newborn males have weights with a mean of 3272.8 g and a standard deviation of 660.2 g. Newborn females have weights with a mean of 3037.1 g and a standard deviation of 706.3 g. Who has the weight that is more

continued

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114 CHAPTER 3 Describing, Exploring, and Comparing Data

extreme relative to the group from which they came: a male who weighs 1500 g or a female who weighs 1500 g? Who has the larger weight relative to the group from which they came?

16. Oscars In the 87th Academy Awards, Eddie Redmayne won for best actor at the age of 33 and Julianne Moore won for best actress at the age of 54. For all best actors, the mean age is 44.1 years and the standard deviation is 8.9 years. For all best actresses, the mean age is 36.2 years and the standard deviation is 11.5 years. (All ages are determined at the time of the awards ceremony.) Relative to their genders, who had the more extreme age when winning the Oscar: Eddie Redmayne or Julianne Moore? Explain.

Percentiles. In Exercises 17–20, use the following lengths (inches) of bears (from Data Set 11 “Bear Measurements” in Appendix B). Find the percentile corresponding to the given length.

36.0 37.0 40.0 40.0 41.0 43.0 43.5 46.0 46.0 47.0 48.0 48.0

49.0 50.0 52.0 52.5 53.0 53.0 54.0 57.3 57.5 58.0 59.0 59.0

59.0 60.0 60.5 61.0 61.0 61.5 62.0 63.0 63.0 63.0 63.5 64.0

64.0 64.0 65.0 65.0 66.5 67.0 67.5 68.5 70.0 70.5 72.0 72.0

72.0 72.0 73.0 73.5 75.0 76.5

17. 61.0 in. 18. 47.0 in. 19. 70.0 in. 20. 58.0 in.

In Exercises 21–28, use the same list of bear lengths (in.) given for Exercises 17–20. Find the indicated percentile or quartile.

21. P60 22. Q1

23. Q3 24. P40

25. P50 26. P75

27. P25 28. P85

Boxplots. In Exercises 29–32, use the given data to construct a boxplot and identify the 5-number summary.

29. Foot Lengths The following are the foot lengths (cm) of 19 males (from Data Set 7 “Foot and Height” in Appendix B).

25.1 25.4 25.7 25.9 26.4 26.7 26.7 26.7 26.8 27.5

27.8 27.9 27.9 28.1 28.6 28.7 28.8 29.2 29.2

30.  Cell Phone Radiation Listed below are the measured radiation absorption rates (in W>kg) corresponding to these cell phones: iPhone 5S, BlackBerry Z30, Sanyo Vero, Optimus V, Droid Razr, Nokia N97, Samsung Vibrant, Sony Z750a, Kyocera Kona, LG G2, and Virgin Mobile Supreme. The data are from the Federal Communications Commission.

1.18 1.41 1.49 1.04 1.45 0.74 0.89 1.42 1.45 0.51 1.38

31. Radiation in Baby Teeth Listed below are amounts of strontium-90 (in millibecquerels, or mBq) in a simple random sample of baby teeth obtained from Pennsylvania residents born after 1979 (based on data from “An Unexpected Rise in Strontium-90 in U.S. Deciduous Teeth in the 1990s,” by Mangano et. al., Science of the Total Environment).

128 130 133 137 138 142 142 144 147 149 151 151 151 155

156 161 163 163 166 172

32. Blood Pressure Measurements Fourteen different second-year medical students at Bellevue Hospital measured the blood pressure of the same person. The systolic readings (mm Hg) are listed below.

138 130 135 140 120 125 120 130 130 144 143 140 130 150

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Boxplots from Large Data Sets in Appendix B. In Exercises 33 and 34, use the given data sets from Appendix B. Use the boxplots to compare the two data sets.

33. BMI Use the body mass indexes (BMI) for males and use the BMI measures for females listed in Data Set 1 “Body Data.”

34. Lead and IQ Use the same scale to construct boxplots for the full IQ scores (IQF) for the low lead level group and the high lead level group in Data Set 8 “IQ and Lead” in Appendix B.

35. Outliers and Modified Boxplots Repeat Exercise 33 “BMI” using modified boxplots. Identify any outliers as defined in Part 2 of this section.

3-3 Beyond the Basics

Chapter Quick Quiz

1.  Sleep Mean As part of the National Health and Nutrition Examination Survey, subjects were asked how long they slept the preceding night and the following times (hours) were re- ported: 8, 7, 5, 7, 4, 7, 6, 7, 8, 8, 8, 6. Find the mean.

2. Sleep Median What is the median of the sample values listed in Exercise 1?

3. Sleep Mode What is the mode of the sample values listed in Exercise 1?

4.  Sleep Variance The standard deviation of the sample values in Exercise 1 is 1.3 hours. What is the variance (including units)?

5. Sleep Outlier If the sleep time of 0 hours is included with the sample data given in Exercise 1, is it an outlier? Why or why not?

6. Sleep z Score A larger sample of 50 sleep times (hours) has a mean of 6.3 hours and a stan- dard deviation of 1.4 hours. What is the z score for a sleep time of 5 hours?

7. Sleep Q3 For a sample of 80 sleep times, approximately how many of those times are less than Q3?

8.  Sleep 5-Number Summary For a sample of 100 sleep times, give the names of the val- ues that constitute the 5-number summary. (The actual values can’t be identified; just give the names of those values.)

9. Estimating s A large sample of sleep times includes values ranging from a low of 4 hours to a high of 10 hours. Use the range rule of thumb to estimate the standard deviation.

10. Sleep Notation Consider a sample of sleep times taken from the population of all adults living in Alaska. Identify the symbols used for the sample mean, population mean, sample stan- dard deviation, population standard deviation, sample variance, and the population variance.

Review Exercises

1. Ergonomics When designing an eye-recognition security device, engineers must consider the eye heights of standing women. (It’s easy for men to bend lower, but it’s more difficult for women to rise higher.) Listed below are the eye heights (in millimeters) obtained from a simple random sample of standing adult women (based on anthropometric survey data from Gordon, Churchill, et al.). Use the given eye heights to find the (a) mean; (b) median; (c) mode; (d) mid- range; (e) range; (f) standard deviation; (g) variance.

1550 1642 1538 1497 1571

CHAPTER 3 Review Exercises 115

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116 CHAPTER 3 Describing, Exploring, and Comparing Data

2. z Score Using the sample data from Exercise 1, find the z score corresponding to the eye height of 1642 mm. Is that eye height significantly low or significantly high? Why or why not?

3. ER Codes In an analysis of activities that resulted in brain injuries presenting at hospital emergency rooms (ERs), the following activities were identified by the code shown in paren- theses: bicycling (12); football (14); playground (22); basketball (27); swimming (40). Find the mean of 12, 14, 22, 27, and 40. What is wrong with this result?

4. Comparing Birth Weights The birth weights of a sample of males have a mean of 3273 g and a standard deviation of 660 g. The birth weights of a sample of females have a mean of 3037 g and a standard deviation of 706 g (based on Data Set 3 “Births” in Appendix B). When consid- ered among members of the same gender, which baby has the relatively larger birth weight: a boy with a birth weight of 3400 g or a girl with a birth weight of 3200 g? Why?

5. Effects of an Outlier Listed below are platelet counts (1000 cells>mL) from subjects in- cluded in Data Set 1 “Body Data.” Identify the outlier and then comment on the effect it has on the mean and standard deviation by finding the values of those statistics with the outlier included and then with the outlier excluded.

263 206 185 246 188 191 308 262 198 253 646

6. Interpreting a Boxplot Shown below is a boxplot of a sample of 30 maximal skull breadths (mm) measured from Egyptian skulls from around 4000 B.C. What do the numbers in the box- plot represent?

7. Interpreting Standard Deviation A physician routinely makes physical examinations of children. She is concerned that a three-year-old girl has a height of only 87.8 cm. Heights of three-year-old girls have a mean of 97.5 cm and a standard deviation of 6.9 cm (based on data from the National Health and Nutrition Examination Survey). Use the range rule of thumb to find the limits separating heights of three-year-old girls that are significantly low or signifi- cantly high. Based on the result, is the height of 87.8 cm significant? Should the physician be concerned?

8. Mean or Median? A biostatistics class consists of 30 students with no income, 10 students with small incomes from part-time jobs, plus a professor with a very large income that is well deserved. Which is better for describing the income of a typical person in this class: mean or median? Explain.

Cumulative Review Exercises

1. Arsenic in Rice Listed below are measured amounts (mg per serving) of arsenic in a sample of servings of brown rice [data from the Food and Drug Administration (FDA)]. Construct a frequency distribution. Use a class width of 2 mg, and use 0 mg as the lower class limit of the first class.

6.1 5.4 6.9 4.9 6.6 6.3 6.7 8.2 7.8 1.5 5.4 7.3

2.  Histogram Use the frequency distribution from Exercise 1 to construct a histogram. Use class midpoint values for the horizontal scale.

3. Stemplot Use the amounts of arsenic from Exercise 1 to construct a stemplot.

4. Descriptive Statistics Use amounts of arsenic in Exercise 1 and find the following: (a) mean; (b) median; (c) standard deviation; (d) variance; (e) range. Include the appropriate units of measurement.

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CHAPTER 3 Cooperative Group Activities 117

5. a. A medical researcher has a collection of data at the nominal level of measurement and she wants to obtain a representative data value. Which of the following is most appropriate: mean, median, mode, or midrange? Why?

b. A botanist wants to obtain data about the plants being grown in homes. If a sample is ob- tained by telephoning the first 250 people listed in the local telephone directory, what type of sampling is being used? (random, stratified, systematic, cluster, convenience)

c. A botanist is experimenting with fertilizer sticks used for growing plants. She finds that the amounts of fertilizer placed in the sticks are not very consistent, so that some fertilization lasts longer than claimed, but others don’t last long enough. She wants to improve quality by making the amounts of fertilizer in the sticks more consistent. When analyzing the amounts of fertil- izer for that purpose, which of the following statistics is most relevant: mean, median, mode, midrange, standard deviation, first quartile, third quartile? Should the value of that statistic be raised, lowered, or left unchanged?

Technology Project

Freshman 15 Refer to Data Set 10 “Freshman 15” in Appendix B, which includes results from a study of the legend that college freshmen tend to gain around 15 pounds (or 6.8 kilograms) during their freshman year. That data set includes 5 columns of data from 67 subjects. Use the methods of this chapter to make relevant comparisons, then form a subjective conclusion about the 15-pound (or 6.8-kilogram) weight gain. Write a brief report including your conclusions with supporting graphs and statistics.

FROM DATA TO DECISION

Second-Hand Smoke Data Set 14 “Passive and Active Smoke” in Appendix B lists measures of cotinine from three groups of subjects: (1) smokers; (2) nonsmokers exposed to environmental to- bacco smoke; and (3) nonsmokers not exposed to environ- mental tobacco smoke. Cotinine is an indicator of nicotine absorption.

Critical Thinking Use the methods from this chapter to explore and compare the cotinine measures in the three groups. Are there any notable differences? Are there any outliers? What do you conclude about the effects that smokers have on nonsmok- ers? Write a brief report of your conclusions, and provide supporting statistical evidence.

Cooperative Group Activities

1. In-class activity In class, each student should record two pulse rates by counting the num- ber of heartbeats in 1 minute. The first pulse rate should be measured while the student is seated, and the second pulse rate should be measured while the student is standing. Use the methods of this chapter to compare results. Do males and females appear to have different pulse rates? Do pulse rates measured while seated appear to be different from pulse rates mea- sured while standing?

2.  Out-of-class activity Appendix B includes many real and interesting data sets. In each group of three or four students, select a data set from Appendix B and analyze it using the methods discussed so far in this book. Write a brief report summarizing key conclusions.

3. Out-of-class activity In each group of three or four students, collect an original data set of values at the interval or ratio level of measurement. Provide the following: (1) a list of sample values; (2) printed software results of descriptive statistics and graphs; and (3) a written de- scription of the nature of the data, the method of collection, and important characteristics.

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118

Basic Concepts of Probability

Addition Rule and Multiplication Rule

Complements, Conditional Probability, and Bayes’ Theorem

Risks and Odds

Rates of Mortality, Fertility, and Morbidity

Counting

4-1

4-2

4-3

4-4

4-5

4-6

Drug Testing of Job ApplicantsCHAPTER PROBLEM

Probability

Approximately 85% of U. S. companies test employees

and>or job applicants for drug use. A common and inexpen- sive (around $50) urine test is the EMIT (enzyme multiplied

immunoassay technique) test, which tests for the presence of

any of five drugs: marijuana, cocaine, amphetamines, opiates,

or phencyclidine. Most companies require that positive test

results be confirmed by a more reliable GC-MS (gas chroma-

tography mass spectrometry) test.

Like nearly all medical tests, drug tests are sometimes

wrong. Wrong results are of two different types: (1) false posi-

tive results and (2) false negative results. In today’s society,

these terms should be clearly understood. A job applicant or

4

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employee who gets a false positive result is someone who

incorrectly appears to be using drugs when he or she is not

actually using drugs. This type of mistake can unfairly result in

job denial or termination of employment.

Analyzing the Results Table 4-1 includes results from 555 adults in the United States.

If one of the subjects from Table 4-1 is randomly selected

from those who do not use drugs, what is the probability of a

false positive result? If one of the subjects from Table 4-1 is

randomly selected from those who do not use drugs, what is

the probability of a true negative result? We will address such

questions in this chapter.

• Prevalence: Proportion of the population having the condi-

tion (such as drug use or disease) being considered.

• False positive: Wrong test result that incorrectly indicates

that the subject has a condition when the subject does not

have that condition.

• False negative: Wrong test result that incorrectly indicates

that the subject does not have a condition when the subject

does have that condition.

• True positive: Correct test result that indicates that a

subject has a condition when the subject does have the

condition.

• True negative: Correct test result that indicates that a sub-

ject does not have a condition when the subject does not

have the condition.

• Test sensitivity: The probability of a true positive test

result, given that the subject actually has the condition

being tested.

• Test specificity: The probability of a true negative test

result, given that the subject does not have the condition

being tested.

• Positive predictive value: Probability that a subject actu-

ally has the condition, given that the test yields a positive

result (indicating that the condition is present).

• Negative predictive value: Probability that the subject

does not actually have the condition, given that the test

yields a negative result (indicating that the condition is not

present).

Chapter Objectives 119

TABLE 4-1 Results from Drug Tests of Job Applicants

Positive Test Result (Test shows drug use.)

Negative Test Result (Test shows no drug use.)

Subject Uses Drugs 45 (True Positive)

5 (False Negative)

Subject Does Not Use Drugs 25 (False Positive)

480 (True Negative)

The main objective of this chapter is to develop a sound understanding of probability values, because those values constitute the underlying foundation on which methods of inferential statistics are built. The important methods of hypothesis testing com- monly use P-values, which are probability values expressed as numbers between 0 and 1, inclusive. Smaller probability values, such as 0.01, correspond to events that are very unlikely. Larger probability values, such as 0.99, correspond to events that are very likely. Here are the chapter objectives:

Basic Concepts of Probability

• Identify probabilities as values between 0 and 1, and interpret those values as expressions of likelihood of events.

• Develop the ability to calculate probabilities of events. • Define the complement of an event and calculate the probability of that

complement.

4-1

CHAPTER OBJECTIVES

> > >

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120 CHAPTER 4 Probability

Key Concept The single most important objective of this section is to learn how to interpret probability values, which are expressed as values between 0 and 1. A small probability, such as 0.001, corresponds to an event that rarely occurs.

Role of Probability in Statistics

Probability plays a central role in the important statistical method of hypothesis testing introduced later in Chapter 8. Statisticians make decisions using data by rejecting explanations (such as chance) based on very low probabilities. See the following example illustrating the role of probability and a fundamental way that statisticians think.

4-1 Basic Concepts of Probability

Addition Rule and Multiplication Rule

• Develop the ability to calculate the probability that in a single trial, some event A oc- curs or some event B occurs or they both occur. Apply the addition rule by correctly adjusting for events that are not disjoint (or are overlapping).

• Develop the ability to calculate the probability of an event A occurring in a first trial and an event B occurring in a second trial. Apply the multiplication rule by adjusting for events that are not independent.

• Distinguish between independent events and dependent events.

Complements, Conditional Probability, and Bayes’ Theorem

• Compute the probability of “at least one” occurrence of an event A. • Apply the multiplication rule by computing the probability of some event, given that

some other event has already occurred.

Risks and Odds

• Compare two probabilities using measures of absolute risk reduction and relative risk.

• Obtain a measure of risk by calculating the odds ratio. • Measure the practical effectiveness of a treatment by determining the “number

needed to treat,” which is the number of subjects that must be treated in order to prevent one occurrence of some event.

Rates of Mortality, Fertility, and Morbidity

• Use rates to describe the likelihood of an event. • Determine mortality rates, fertility rates, and morbidity rates.

Counting

• Develop the ability to apply the fundamental counting rule, factorial rule, permuta- tions rule, and combinations rule.

• Distinguish between circumstances requiring the permutations rule and those requiring the combinations rule.

4-2

4-3

4-4

4-5

4-6

Addition Rule and Multiplication Rule

• Develop the ability to calculate the probability that in a single trial, some event A oc- curs or some event B occurs or they both occur. Apply the addition rule by correctly adjusting for events that are not disjoint (or are overlapping).

• Develop the ability to calculate the probability of an event A occurring in a first trial and an event B occurring in a second trial. Apply the multiplication rule by adjusting for events that are not independent.

• Distinguish between independent events and dependent events.

Complements, Conditional Probability, and Bayes’ Theorem

• Compute the probability of “at least one” occurrence of an event A. • Apply the multiplication rule by computing the probability of some event, given that

some other event has already occurred.

Risks and Odds

• Compare two probabilities using measures of absolute risk reduction and relative risk.

• Obtain a measure of risk by calculating the odds ratio. • Measure the practical effectiveness of a treatment by determining the “number

needed to treat,” which is the number of subjects that must be treated in order to prevent one occurrence of some event.

Rates of Mortality, Fertility, and Morbidity

• Use rates to describe the likelihood of an event. • Determine mortality rates, fertility rates, and morbidity rates.

Counting

• Develop the ability to apply the fundamental counting rule, factorial rule, permuta- tions rule, and combinations rule.

• Distinguish between circumstances requiring the permutations rule and those requiring the combinations rule.

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4-1 Basic Concepts of Probability 121

I N T E R P R E TAT I O N

Among the 100 babies, 75 girls and 55 girls are both greater than the 50 girls that we typically expect, but only the event of 75 girls leads us to believe that the gender selection method is effective. Even though there is a chance of getting 75 girls (or more) in 100 births with no special treatment, the probability of that happening is so small (0.0000003) that we should reject chance as a reasonable explanation. Instead, it would be generally recognized that the results provide strong support for the claim that the gender selection method is effective. This is exactly how statisticians think: They reject explanations (such as chance) based on very low probabilities.

EXAMPLE 1 Analyzing a Claim

Researchers have made this claim (really, they have):

Claim: “We have developed a gender selection method that greatly increases the likelihood of a baby being a girl.”

Hypothesis Used When Testing the Preceding Claim: The method of gender selection has no e�ect, so that for couples using this method, about 50% of the births result in girls.

(The probability of a girl in the United States is actually 0.488, but here we assume that boys and girls are equally likely.)

Figure 4-1 shows the sample data from two tests of 100 couples using the gender selection method and the conclusion reached for each test.

75 Girls

Statisticians reject explanations based on very low probabilities

Test A Result

Probability of 75 (or more) Girls by chance 5 3 in 10,000,000

5 0.0000003

Chance rejected as reasonable explanation

Gender selection method appears to be e�ective

25 Boys

55 Girls

Test B Result

Probability of 55 (or more) Girls by chance 5 184 in 1,000

5 0.184

Chance not rejected as reasonable explanation

Cannot conclude gender selection method is e�ective

45 Boys

Di�erent Gender Selection Methods Tested with 100 Births

FIGURE 4-1 Gender Selection Method Test Data and Conclusions

Basics of Probability

In probability, we deal with procedures (such as generating male>female births or manufacturing defective>nondefective pregnancy test kits) that produce outcomes.

Probabilities That Challenge Intuition

In certain cases,

our subjective

estimates of

probability val-

ues are dramati-

cally different

from the actual

probabilities.

Here is a classical example: If

you take a deep breath, there is

better than a 99% chance that

you will inhale a molecule that

was exhaled in dying Caesar’s

last breath. In that same morbid

and unintuitive spirit, if Socrates’

fatal cup of hemlock was mostly

water, then the next glass of

water you drink will likely contain

one of those same molecules.

Here’s another, less morbid

example that can be verified: In

classes of 25 students, there is

better than a 50% chance that

at least 2 students will share the

same birthday (day and month).

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122 CHAPTER 4 Probability

Example 2 illustrates the concepts defined above.

DEFINITIONS

An event is any collection of results or outcomes of a procedure.

A simple event is an outcome or an event that cannot be further broken down into simpler components.

The sample space for a procedure consists of all possible simple events. That is, the sample space consists of all outcomes that cannot be broken down any further.

EXAMPLE 2 Simple Event and Sample Spaces

In the following display, we use “b” to denote a baby boy and “g” to denote a baby girl.

Procedure

Example of Event

Sample Space: Complete List of Simple Events

Single birth 1 girl (simple event) {b, g}

3 births 2 boys and 1 girl (bbg, bgb, and gbb are all simple events resulting in 2 boys and 1 girl)

{bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg}

Simple Events:

■ With one birth, the result of 1 girl is a simple event and the result of 1 boy is another simple event. They are individual simple events because they cannot be broken down any further.

■ With three births, the result of 2 girls followed by a boy (ggb) is a simple event.

■ When rolling a single die, the outcome of 5 is a simple event, but the outcome of an even number is not a simple event.

Not a Simple Event: With three births, the event of “2 girls and 1 boy” is not a simple event because it can occur with these di�erent simple events: ggb, gbg, bgg.

Sample Space: With three births, the sample space consists of the eight di�erent simple events listed in the above table.

Three Common Approaches to Finding the Probability of an Event

We first list some basic notation, and then we present three common approaches to finding the probability of an event.

Notation for Probabilities

P denotes a probability.

A, B, and C denote specific events.

P(A) denotes the “probability of event A occurring.”

The following three approaches for finding probabilities result in values between 0 and 1: 0 … P1A2 … 1. Figure 4-2 shows the possible values of probabilities and the more familiar and common expressions of likelihood.

Certain

Likely

50–50 Chance

Unlikely

Impossible0

0.5

1

FIGURE 4-2 Possible Values for Probabilities

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4-1 Basic Concepts of Probability 123

1. Relative Frequency Approximation of Probability Conduct (or observe) a procedure and count the number of times that event A occurs. P(A) is then ap- proximated as follows:

P1A2 = number of times A occurred number of times the procedure was repeated

When referring to relative frequency approximations of probabilities, this text will not distinguish between results that are exact probabilities and those that are approximations, so an instruction to “find the probability” could actually mean “estimate the probability.”

2. Classical Approach to Probability (Requires Equally Likely Outcomes) If a procedure has n different simple events that are equally likely, and if event A can occur in s different ways, then

P1A2 = number of ways A occurs number of different simple events

= s n

CAUTION When using the classical approach, always confirm that the outcomes are equally likely.

3. Subjective Probabilities P(A), the probability of event A, is estimated by using knowledge of the relevant circumstances.

Figure 4-3 illustrates the approaches of the preceding three definitions.

1. Relative Frequency Approach: When trying to de- termine the probability that an individual car crashes in a year, we must examine past results to determine the num- ber of cars in use in a year and the number of them that crashed; then we find the ratio of the number of cars that crashed to the total number of cars. For a recent year, the result is a probability of 0.0480. (See Example 3.)

2. Classical Approach: When trying to determine the probability of randomly selecting three children who are of the same gender, there are two ways to get the same genders (boy/boy/boy and girl/girl/girl) among the eight equally likely outcomes, so the probability is 2/8 or 1/4. (See Example 4.)

3. Subjective Probability: When trying to estimate the probability of someone with an appendix getting acute appendicitis in the next year, we know from personal ex- perience that the probability is quite small. Let’s estimate it to be, say, 0.001 (equivalent to 1 chance in 1000). (See Example 5.)

FIGURE 4-3 Three Approaches to Finding a Probability

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124 CHAPTER 4 Probability

Simulations Sometimes none of the preceding three approaches can be used. A simu- lation of a procedure is a process that behaves in the same ways as the procedure itself so that similar results are produced. Probabilities can sometimes be found by using a simulation. See the Technology Project near the end of this chapter.

Rounding Probabilities Although it is difficult to develop a universal rule for round- ing off probabilities, the following guide will apply to most problems in this text.

ROUNDING PROBABILITIES

When expressing the value of a probability, either give the exact fraction or deci- mal or round off final decimal results to three significant digits. (Suggestion: When a probability is not a simple fraction such as 2>3 or 5>9, express it as a decimal so that the number can be better understood.) All digits in a number are significant ex- cept for the zeros that are included for proper placement of the decimal point. See the following examples.

■ The probability of 0.4450323339 (from Example 6) has ten significant digits (4450323339), and it can be rounded to three significant digits as 0.445.

■ The probability of 1>3 can be left as a fraction or rounded to 0.333. (Do not round to 0.3.)

■ The probability of 2>8 can be expressed as 1>4 or 0.25. (Because 0.25 is exact, there’s no need to express it with three significant digits as 0.250.)

Probabilities Expressed as Percentages? Mathematically, a probability of 0.25 is equivalent to 25%, but there are good reasons for sticking with fractions and decimals and not using percentages. Professional journals almost universally express probabili- ties as decimals, not as percentages. Later in this book, we will use probability values generated from statistical software, and they will always be in the form of decimals.

When finding probabilities with the relative frequency approach, we obtain an ap- proximation instead of an exact value. As the total number of observations increases, the corresponding approximations tend to get closer to the actual probability. This property is commonly referred to as the law of large numbers.

LAW OF LARGE NUMBERS

As a procedure is repeated again and again, the relative frequency probability of an event tends to approach the actual probability.

The law of large numbers tells us that relative frequency approximations tend to get better with more observations. This law reflects a simple notion supported by common sense: A probability estimate based on only a few trials can be off by a substantial amount, but with a very large number of trials, the estimate tends to be much more accurate.

CAUTIONS

1. The law of large numbers applies to behavior over a large number of trials, and it does not apply to any one individual outcome. Gamblers sometimes foolishly lose large sums of money by incorrectly thinking that a string of losses increases the chances of a win on the next bet, or that a string of wins is likely to continue.

2. If we know nothing about the likelihood of different possible outcomes, we should not assume that they are equally likely. For example, we should not think that the probability of passing the next statistics test is 1>2, or 0.5 (because we either pass the test or do not). The actual probability depends on factors such as the amount of preparation and the difficulty of the test.

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4-1 Basic Concepts of Probability 125

EXAMPLE 3 Relative Frequency Probability: Skydiving

Find the probability of dying when making a skydiving jump.

S O L U T I O N

In a recent year, there were about 3,000,000 skydiving jumps and 21 of them resulted in deaths. We use the relative frequency approach as follows:

P1skydiving death2 = number of skydiving deaths total number of skydiving jumps

= 21

3,000,0000 = 0.000007

Here the classical approach cannot be used because the two outcomes (dying, surviving) are not equally likely. A subjective probability can be estimated in the absence of historical data.

EXAMPLE 4 Classical Probability: Three Children of the Same Gender

When three children are born, the sample space of genders is as shown in Example 1: {bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg}. If boys and girls are equally likely, then those eight simple events are equally likely. Assuming that boys and girls are equally likely, find the probability of getting three children all of the same gender when three children are born. (In reality, the probability of a boy is 0.512 instead of 0.5.)

S O L U T I O N

The sample space {bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg} includes eight equally likely outcomes, and there are exactly two outcomes in which the three children are of the same gender: bbb and ggg. We can use the classical approach to get

P1three children of the same gender2 = 2 8 =

1 4

or 0.25

EXAMPLE 5 Subjective Probability: Acute Appendicitis

What is the probability that you will get acute appendicitis next year?

S O L U T I O N

We could probably find past results and use the relative frequency approach, but for now, in the absence of historical data on acute appendicitis, we make a subjective estimate. Experience suggests that the probability is quite small. Let’s estimate it to be, say, 0.001 (equivalent to 1 chance in 1000). Depending on our knowledge of the relevant circumstances, that subjective estimate might be reasonably accurate or it might be grossly wrong.

CAUTION Don’t make the common mistake of finding a probability value by mindlessly dividing a smaller number by a larger number. Instead, think carefully about the numbers involved and what they represent. Carefully identify the total number of items being considered, as illustrated in Example 6.

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126 CHAPTER 4 Probability

Instead of trying to determine an answer directly from the given statement, first sum- marize the information in a format that allows clear understanding, such as this format:

3785 texted while driving

4720 did not text while driving

8505 total number of drivers in the sample

We can now use the relative frequency approach as follows:

P1texting while driving2 = number of drivers who texted while driving total number of drivers in the sample

= 3785 8505

= 0.445

I N T E R P R E TAT I O N

There is a 0.445 probability that if a high school driver is randomly selected, he or she texted while driving during the previous 30 days.

EXAMPLE 6 Texting and Driving

In a study of U.S. high school drivers, it was found that 3785 texted while driving during the previous 30 days, and 4720 did not text while driving during that same time period (based on data from “Texting While Driving . . . . ,” by Olsen, Shults, Eaton, Pediatrics, Vol. 131, No. 6). Based on these results, if a high school driver is randomly selected, find the probability that he or she texted while driving during the previous 30 days.

S O L U T I O N

CAUTION A common mistake is to blindly plug in numbers to get the wrong probability of 3785>4720 = 0.802. We should think about what we are doing, as follows.

EXAMPLE 7 Thanksgiving Day

If a year is selected at random, find the probability that Thanksgiving Day in the United States will be (a) on a Wednesday or (b) on a Thursday.

S O L U T I O N

a. In the United States, Thanksgiving Day always falls on the fourth Thursday in November. It is therefore impossible for Thanksgiving to be on a Wednesday. When an event is impossible, its probability is 0. P(Thanksgiving on Wednesday) = 0.

b. It is certain that a Thanksgiving Day in the United States will be on a Thursday. When an event is certain to occur, its probability is 1. P(Thanksgiving on Thursday) = 1.

Because any event imaginable is impossible, certain, or somewhere in between, it follows that the mathematical probability of any event A is 0, 1, or a number between 0 and 1 (as shown in Figure 4-2). That is, 0 … P1A2 … 1.

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4-1 Basic Concepts of Probability 127

Complementary Events Sometimes we need to find the probability that an event A does not occur.

DEFINITION

The complement of event A, denoted by A, consists of all outcomes in which event A does not occur.

EXAMPLE 8 Complement of Death from Skydiving

Example 3 shows that in a recent year, there were 3,000,000 skydiving jumps and 21 of them resulted in death. Find the probability of not dying when making a skydiving jump.

S O L U T I O N

Among 3,000,000 jumps there were 21 deaths, so it follows that the other 2,999,979 jumps were survived. We get

P1not dying when making a skydiving jump2 = 2,999,979 3,000,000

= 0.999993

I N T E R P R E TAT I O N

The probability of not dying when making a skydiving jump is 0.999993.

Relationship Between P1A2 and P1A2 If we denote the event of dying in a skydiving jump by D, Example 3 showed that P1D2 = 0.000007 and Example 8 showed that P1D2 = 0.999993. The probability of P1D2 could be found by just sub- tracting P(D) from 1.

Identifying Significant Results with Probabilities: The Rare Event Rule for Inferential Statistics

If, under a given assumption, the probability of a particular observed event is very small and the observed event occurs significantly less than or significantly greater than what we typically expect with that assumption, we conclude that the assumption is probably not correct.

We can use probabilities to identify values that are significantly low or significantly high as follows.

Using Probabilities to Determine When Results Are Significantly High or Significantly Low

■ Significantly high number of successes: x successes among n trials is a signifi- cantly high number of successes if the probability of x or more successes is un- likely with a probability of 0.05 or less. That is, x is a significantly high number of successes if P(x or more) … 0.05.*

■ Significantly low number of successes: x successes among n trials is a sig- nificantly low number of successes if the probability of x or fewer successes is unlikely with a probability of 0.05 or less. That is, x is a significantly low number of successes if P(x or fewer) … 0.05.*

*The value 0.05 is not absolutely rigid. Other values, such as 0.01, could be used to distinguish between results

that can easily occur by chance and events that are significant.

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128 CHAPTER 4 Probability

See Example 1 on page 121, which illustrates the following:

■ Among 100 births, 75 girls is significantly high because the probability of 75 or more girls is 0.0000003, which is less than or equal to 0.05 (so the gender selection method appears to be effective).

■ Among 100 births, 55 girls is not significantly high because the probability of 55 or more girls is 0.184, which is greater than 0.05 (so the gender selection does not appear to be effective).

Probability Review

Important Principles and Notation for Probability

■ The probability of an event is a fraction or decimal number between 0 and 1 inclusive.

■ The probability of an impossible event is 0.

■ The probability of an event that is certain to occur is 1.

■ Notation: P1A2 = the probability of event A. ■ Notation: P1A2 = the probability that event A does not occur.

Statistical Literacy and Critical Thinking

1.  Probability Rewrite the following statement with the probability expressed as a number with a decimal format: “The probability of selecting someone with blue eyes is 35%.”

2. Probability Given that the following statement is incorrect, rewrite it correctly: “The prob- ability of a baby being born a boy is 50-50.”

3. Births In Example 4 “Three Children of the Same Gender” it was noted that in reality, the probability of a boy is 0.512 instead of 0.5. Let A denote the event of getting a boy when a baby is born. What is the value of P1A2? 4. Subjective Probability Estimate the probability that the next time a physician walks into a patient’s room and turns on a light switch, she discovers that the light bulb does work.

5. Identifying Probability Values Which of the following are probabilities?

0 3>5 5>3 - 0.25 250% 7:3 1 50@50 5:1 0.135 2.017 6. Penicillin “Who discovered penicillin: Sean Penn, William Penn, Penn Jillette, Alexander Fleming, or Louis Pasteur?” If you make a random guess for the answer to that question, what is the probability that your answer is the correct answer of Alexander Fleming?

7. Avogadro Constant If you are asked on a quiz to give the first (leftmost) nonzero digit of the Avogadro constant and, not knowing the answer, you make a random guess, what is the probability that your answer is the correct answer of 6?

8. Births Example 2 in this section includes the sample space for genders from three births. Identify the sample space for the genders from two births.

In Exercises 9–12, assume that 50 births are randomly selected. Use subjective judgment to describe the given number of girls as (a) significantly low, (b) significantly high, or (c) nei- ther significantly low nor significantly high.

9. 47 girls. 10. 26 girls. 11. 23 girls. 12. 5 girls.

4-1 Basic Skills and Concepts

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4-1 Basic Concepts of Probability 129

In Exercises 13–20, express the indicated degree of likelihood as a probability value between 0 and 1.

13. Testing If you make a random guess for the answer to a true>false test question, there is a 50-50 chance of being correct.

14. MCAT Test When making a random guess for an answer to a multiple-choice question on an MCAT test, the possible answers are a, b, c, d, e, so there is 1 chance in 5 of being correct.

15. Genes One of the four DNA bases of A, G, C, and T is randomly selected, and the result is G. Assume that the four DNA bases are equally likely.

16.  Sleepwalking Based on a report in Neurology magazine, 29.2% of survey respondents have sleepwalked.

17.  Randomness When using a computer to randomly generate the last digit of a phone number to be called for a survey, there is 1 chance in 10 that the last digit is zero.

18.  Job Applicant Mistakes Based on an Adecco survey of hiring managers who were asked to identify the biggest mistakes that job candidates make during an interview, there is a 50-50 chance that they will identify “inappropriate attire.”

19. Square Peg Sydney Smith wrote in “On the Conduct of the Understanding” that it is im- possible to fit a square peg in a round hole.

20. Death and Taxes Benjamin Franklin said that death is a certainty of life.

In Exercises 21–24, refer to the sample data in Table 4-1, which is included with the Chapter Problem. Assume that 1 of the 555 subjects included in Table 4-1 is randomly selected.

TABLE 4-1 Results from Drug Tests of Job Applicants

Positive Test Result (Test shows drug use.)

Negative Test Result (Test shows no drug use.)

Subject Uses Drugs 45 (True Positive)

5 (False Negative)

Subject Does Not Use Drugs 25 (False Positive)

480 (True Negative)

21. Drug Testing Job Applicants Find the probability of selecting someone who got a re- sult that is a false negative. Who would suffer from a false negative result? Why?

22. Drug Testing Job Applicants Find the probability of selecting someone who got a re- sult that is a false positive. Who would suffer from a false positive result? Why?

23.  Drug Testing Job Applicants Find the probability of selecting someone who uses drugs. Does the result appear to be reasonable as an estimate of the “prevalence rate” described in the Chapter Problem?

24. Drug Testing Job Applicants Find the probability of selecting someone who does not use drugs. Does the result appear to be reasonable as an estimate of the proportion of the adult population that does not use drugs?

In Exercises 25–32, find the probability and answer the questions.

25. XSORT Gender Selection MicroSort’s XSORT gender selection technique was designed to increase the likelihood that a baby will be a girl. At one point before clinical trials of the XSORT gender selection technique were discontinued, 945 births consisted of 879 baby girls and 66 baby boys (based on data from the Genetics & IVF Institute). Based on these results, what is the probability of a girl born to a couple using MicroSort’s XSORT method? Does it ap- pear that the technique is effective in increasing the likelihood that a baby will be a girl?

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130 CHAPTER 4 Probability

26.  YSORT Gender Selection MicroSort’s YSORT gender-selection technique is de- signed to increase the likelihood that a baby will be a boy. At one point before clinical trials of the YSORT gender-selection technique were discontinued, 291 births consisted of 239 baby boys and 52 baby girls (based on data from the Genetics & IVF Institute). Based on these results, what is the probability of a boy born to a couple using MicroSort’s YSORT method? Does it appear that the technique is effective in increasing the likelihood that a baby will be a boy?

27.  Mendelian Genetics When Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 428 green peas and 152 yellow peas. Based on those results, estimate the probability of getting an offspring pea that is green. Is the result reasonably close to the expected value of 3>4, as Mendel claimed? 28. Guessing Birthdays On their first date, Kelly asks Mike to guess the date of her birth, not including the year.

a. What is the probability that Mike will guess correctly? (Ignore leap years.)

b. Would it be unlikely for him to guess correctly on his first try?

c. If you were Kelly, and Mike did guess correctly on his first try, would you believe his claim that he made a lucky guess, or would you be convinced that he already knew when you were born?

d. If Kelly asks Mike to guess her age, and Mike’s guess is too high by 15 years, what is the probability that Mike and Kelly will have a second date?

29.  Online Medicine In a survey, 933 respondents say that they seek medical information online and 139 other respondents say that they never seek medical information online. What is the probability that a randomly selected person never seeks medical information online? Is it unlikely for someone to never seek medical information online? How are these results affected by the fact that the responses are from subjects who decided to respond to the survey posted on the Internet by AOL?

30.  Car Rollovers In a recent year in the United States, 83,600 passenger cars rolled over when they crashed, and 5,127,400 passenger cars did not roll over when they crashed. Find the probability that a randomly selected passenger car crash results in a rollover. Is it unlikely for a car to roll over in a crash?

31. Genetics: Eye Color Each of two parents has the genotype brown>blue, which con- sists of the pair of alleles that determine eye color, and each parent contributes one of those alleles to a child. Assume that if the child has at least one brown allele, that color will domi- nate and the eyes will be brown. (The actual determination of eye color is more complicated than that.)

a. List the different possible outcomes. Assume that these outcomes are equally likely.

b. What is the probability that a child of these parents will have the blue>blue genotype? c. What is the probability that the child will have brown eyes?

32. X-Linked Genetic Disease Men have XY (or YX) chromosomes and women have XX chromosomes. X-linked recessive genetic diseases (such as juvenile retinoschisis) occur when there is a defective X chromosome that occurs without a paired X chromosome that is not defec- tive. In the following, represent a defective X chromosome with lowercase x, so a child with the xY or Yx pair of chromosomes will have the disease and a child with XX or XY or YX or xX or Xx will not have the disease. Each parent contributes one of the chromosomes to the child.

a. If a father has the defective x chromosome and the mother has good XX chromosomes, what is the probability that a son will inherit the disease?

b. If a father has the defective x chromosome and the mother has good XX chromosomes, what is the probability that a daughter will inherit the disease?

continued

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4-2 Addition Rule and Multiplication Rule 131

c. If a mother has one defective x chromosome and one good X chromosome and the father has good XY chromosomes, what is the probability that a son will inherit the disease?

d. If a mother has one defective x chromosome and one good X chromosome and the father has good XY chromosomes, what is the probability that a daughter will inherit the disease?

Probability from a Sample Space. In Exercises 33–36, use the given sample space or construct the required sample space to find the indicated probability.

33. Three Children Use this sample space listing the eight simple events that are possible when a couple has three children (as in Example 2 on page 122): {bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg}. Assume that boys and girls are equally likely, so that the eight simple events are equally likely. Find the probability that when a couple has three children, there is exactly one girl.

34. Three Children Using the same sample space and assumption from Exercise 33, find the probability that when a couple has three children, there are exactly two girls.

35. Four Children Exercise 33 lists the sample space for a couple having three children. After identifying the sample space for a couple having four children, find the probability of getting three girls and one boy (in any order).

36. Four Children Using the same sample space and assumption from Exercise 35, find the probability that when a couple has four children, all four are of the same gender.

Using Probability to Form Conclusions. In Exercises 37–40, use the given probability value to determine whether the sample results could easily occur by chance, then form a conclusion.

37.  Predicting Gender A study addressed the issue of whether pregnant women can cor- rectly predict the gender of their baby. Among 104 pregnant women, 57 correctly predicted the gender of their baby (based on data from “Are Women Carrying ‘Basketballs’…,” by Perry, DiPietro, Constigan, Birth, Vol. 26, No. 3). If pregnant women have no such ability, there is a 0.327 probability of getting such sample results by chance. What do you conclude?

38. Clinical Trial of Tamiflu Clinical trials involved the use of Tamiflu (oseltamivir phos- phate) for treating flu patients. Among 724 patients treated with Tamiflu, 72 (or about 10%) experienced nausea. (An untreated group experienced a 6% rate of nausea.) If Tamiflu really has no effect on nausea, there is a 0.00000246 probability of getting these sample results by chance. What do you conclude about the effect of Tamiflu on nausea?

39.  Sleepiness In a clinical trial of OxyContin (oxycodone) used for pain relief, 227 sub- jects were treated with OxyContin and 52 of them experienced sleepiness (based on data from Purdue Pharma L.P.). If OxyContin has no effect on sleepiness, the probability of getting these sample results by chance is less than 0.001 (when comparing this sample group with another group not treated with OxyContin). What do you conclude?

40. Cell Phones and Cancer A study of 420,095 Danish cell phone users resulted in 135 who developed cancer of the brain or nervous system (based on data from the Journal of the National Cancer Institute). When comparing this sample group to another group of people who did not use cell phones, it was found that there is a probability of 0.512 of getting such sample results by chance. What do you conclude?

Key Concepts In this section we present the addition rule as a tool for finding P(A or B), which is the probability that either event A occurs or event B occurs (or they both occur) as the single outcome of a procedure. To find P(A or B), we begin by add- ing the number of ways that A can occur and the number of ways that B can occur, but add without double counting. The word “or” in the addition rule is associated with the addition of probabilities.

4-2 Addition Rule and Multiplication Rule

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132 CHAPTER 4 Probability

This section also presents the basic multiplication rule used for finding P(A and B), which is the probability that event A occurs and event B occurs. If the outcome of event A somehow affects the probability of event B, it is important to adjust the prob- ability of B to reflect the occurrence of event A. The rule for finding P(A and B) is called the multiplication rule because it involves the multiplication of the probability of event A and the probability of event B (where, if necessary, the probability of event B is adjusted because of the outcome of event A). The word “and” in the multiplica- tion rule is associated with the multiplication of probabilities.

In Section 4-1 we considered only simple events, but in this section we consider compound events.

DEFINITION

A compound event is any event combining two or more simple events.

Addition Rule Notation for Addition Rule

P1A or B2 = P1in a single trial, event A occurs or event B occurs or they both occur2

The word “or” used in the preceding notation is the inclusive or, which means ei- ther one or the other or both. The formal addition rule is often presented as a formula, but blind use of formulas is not recommended. Instead, understand the spirit of the rule and use that understanding, as in the intuitive addition rule that follows.

INTUITIVE ADDITION RULE

To find P(A or B), add the number of ways event A can occur and the number of ways event B can occur, but add in such a way that every outcome is counted only once. P(A or B) is equal to that sum, divided by the total number of outcomes in the sample space.

FORMAL ADDITION RULE

P1A or B2 = P1A2 + P1B2 - P1A and B2 where P(A and B) denotes the probability that A and B both occur at the same time as an outcome in a trial of a procedure.

One way to apply the addition rule is to add the probability of event A and the probability of event B and, if there is any overlap that causes double-counting, com- pensate for it by subtracting the probability of outcomes that are included twice. This approach is reflected in the above formal addition rule.

EXAMPLE 1 Drug Testing of Job Applicants

Refer to Table 4-1, reproduced here for your convenience and viewing pleasure. If 1 subject is randomly selected from the 555 subjects given a drug test, find the probability of selecting a subject who had a positive test result or uses drugs.

which is the probability that event event ability of

Proportions of Males ,Females

It is well

known that

when a baby is

born, boys and

girls are not

equally likely. It

is currently be-

lieved that 105 boys are born for

every 100 girls, so the probability

of a boy is 0.512. Kristen Navara

of the University of Georgia

conducted a study showing that

around the world, more boys are

born than girls, but the difference

becomes smaller as people are

located closer to the equator.

She used latitudes, tempera-

tures, unemployment rates, and

gross national products from

200 countries and conducted a

statistical analysis showing that

the proportions of boys appear

to be affected only by latitude

and its related weather. So far, no

one has identified a reasonable

explanation for this phenomenon.

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4-2 Addition Rule and Multiplication Rule 133

Disjoint Events and the Addition Rule

The addition rule is simplified when the events are disjoint.

TABLE 4-1 Results from Drug Tests of Job Applicants

Positive Test Result (Test shows drug use.)

Negative Test Result (Test shows no drug use.)

Subject Uses Drugs 45 (True Positive)

5 (False Negative)

Subject Does Not Use Drugs 25 (False Positive)

480 (True Negative)

*Numbers in red correspond to positive test results or subjects who use drugs, and the total of those numbers is 75.

S O L U T I O N

Refer to Table 4-1 and carefully count the number of subjects who tested positive (first column) or use drugs (first row), but be careful to count subjects exactly once, not twice. When adding the frequencies from the first column and the first row, include the frequency of 45 only once. In Table 4-1, there are 45 + 25 + 5 = 75 subjects who had positive test results or use drugs. We get this result:

P1positive test result or subject uses drugs2 = 75>555 = 0.135

DEFINITION

Events A and B are disjoint (or mutually exclusive) if they cannot occur at the same time. (That is, disjoint events do not overlap.)

EXAMPLE 2 Disjoint Events

Disjoint events: Event A—Randomly selecting someone for a clinical trial who is a male

Event B—Randomly selecting someone for a clinical trial who is a female

(The selected person cannot be both.)

Events that are not disjoint: Event A—Randomly selecting someone taking a statistics course

Event B—Randomly selecting someone who is a female

(The selected person can be both.)

Whenever A and B are disjoint, P(A and B) becomes zero in the formal addition rule, so for disjoint events A and B we have P1A or B2 = P1A2 + P1B2. But again, instead of blind use of a formula, it is better to understand and use the intuitive addi- tion rule.

Here is a summary of the key points of the addition rule:

1. To find P(A or B), first associate the word or with addition.

2. To find the value of P(A or B), add the number of ways A can occur and the number of ways B can occur, but be careful to add without double counting.

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134 CHAPTER 4 Probability

Complementary Events and the Addition Rule

In Section 4-1 we used A to indicate that event A does not occur. Common sense dic- tates this principle: We are certain (with probability 1) that either an event A occurs or it does not occur, so it follows that P1A or A2 = 1. Because events A and A must be disjoint, we can use the addition rule to express this principle as follows:

P1A or A2 = P1A2 + P1A2 = 1 This result of the addition rule leads to the following three expressions that are “equiv- alent” in the sense that they are just different forms of the same principle.

RULE OF COMPLEMENTARY EVENTS

P1A2 + P1A2 = 1 P1A2 = 1 - P1A2 P1A2 = 1 - P1A2

EXAMPLE 3 Sleepwalking

Based on a journal article, the probability of randomly selecting someone who has sleepwalked is 0.292, so P(sleepwalked) = 0.292 (based on data from “Prevalence and Comorbidity of Nocturnal Wandering in the U.S. General Population,” by Ohayon et al., Neurology, Vol. 78, No. 20). If a person is randomly selected, find the probability of getting someone who has not sleepwalked.

S O L U T I O N

Using the rule of complementary events, we get

P1has not sleepwalked2 = 1 - P1sleepwalked2 = 1 - 0.292 = 0.708 The probability of randomly selecting someone who has not sleepwalked is 0.708.

Multiplication Rule Notation for Multiplication Rule

We begin with basic notation followed by the multiplication rule. We strongly suggest using the intuitive multiplication rule, because it is based on understanding instead of blind use of a formula.

Notation

P1A and B2 = P1event A occurs in one trial and event B occurs in a different trial2 P1B � A2 represents the probability of event B occurring after it is assumed that event A has already occurred. (Interpret B � A as “event B occurs after event A has already occurred.”)

CAUTION The notation P(A and B) has two meanings, depending on its context. For the multiplication rule, P(A and B) denotes that event A occurs in one trial and event B occurs in another trial; for the addition rule we use P(A and B) to denote that events A and B both occur in the same trial.

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4-2 Addition Rule and Multiplication Rule 135

INTUITIVE MULTIPLICATION RULE

To find the probability that event A occurs in one trial and event B occurs in an- other trial, multiply the probability of event A by the probability of event B, but be sure that the probability of event B is found by assuming that event A has already occurred.

FORMAL MULTIPLICATION RULE

P1A and B2 = P1A2 # P1B � A2

Independence and the Multiplication Rule

When applying the multiplication rule and considering whether the probability of event B must be adjusted to account for the previous occurrence of event A, we are focusing on whether events A and B are independent.

DEFINITIONS

Two events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other. (Several events are independent if the occurrence of any does not affect the probabilities of the occurrence of the oth- ers.) If A and B are not independent, they are said to be dependent.

CAUTION Don’t think that dependence of two events means that one is the direct cause of the other. Having a working light in your kitchen and having a working light in your bedroom are dependent events because they share the same power source. One of the lights may stop working for many reasons, but if one light is out, there is a higher probability that the other light will be out (because of the common power source).

Example 4 illustrates the basic multiplication rule, with independent events in part (a) and dependent events in part (b).

EXAMPLE 4 Drug Screening and the Basic Multiplication Rule

Let’s use only the 50 test results from the subjects who use drugs (from Table 4-1), as shown below:

Positive Test Results: 45

Negative Test Results: 5

Total: 50

a. If 2 of these 50 subjects are randomly selected with replacement, find the probability that the first selected person had a positive test result and the sec- ond selected person had a negative test result.

b. Repeat part (a) by assuming that the two subjects are selected without replacement.

continued

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136 CHAPTER 4 Probability

The key point of part (b) in Example 4 is this: We must adjust the probability of the second event to reflect the outcome of the first event. Because selection of the second subject is made without replacement of the first subject, the second probability must take into account the fact that the first selection removed a subject who tested positive, so only 49 subjects are available for the second selection, and 5 of them had a negative test result. Part (a) of Example 4 involved sampling with replacement, so the events are independent; part (b) of Example 4 involved sampling without replacement, so the events are dependent. See the following.

Sampling In the wonderful world of statistics, sampling methods are critically impor- tant, and the following relationships hold:

■ Sampling with replacement: Selections are independent events.

■ Sampling without replacement: Selections are dependent events.

Exception: Treating Dependent Events as Independent

Some cumbersome calculations can be greatly simplified by using the common prac- tice of treating events as independent when small samples are drawn without replace- ment from large populations. (In such cases, it is rare to select the same item twice.) Here is a common guideline routinely used with applications such as analyses of sur- vey results:

TREATING DEPENDENT EVENTS AS INDEPENDENT:

5% GUIDELINE FOR CUMBERSOME CALCULATIONS

When sampling without replacement and the sample size is no more than 5% of the size of the population, treat the selections as being independent (even though they are actually dependent).

S O L U T I O N

a. With Replacement: First selection (with 45 positive results among 50 total results):

P1positive test result2 = 45 50

Second selection (with 5 negative test results among the same 50 total results):

P1negative test result2 = 5 50

We now apply the multiplication rule as follows:

P11st selection is positive and 2nd is negative2 = 45 50

# 5 50

= 0.0900

b. Without Replacement: Without replacement of the first subject, the calcula- tions are the same as in part (a), except that the second probability must be adjusted to reflect the fact that the first selection was positive and is not avail- able for the second selection. After the first positive result is selected, we have 49 test results remaining, and 5 of them are negative. The second probability is therefore 5>49, as shown below:

P11st selection is positive and 2nd is negative2 = 45 50

# 5 49

= 0.0918

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4-2 Addition Rule and Multiplication Rule 137

Example 5 illustrates use of the 5% guideline for cumbersome calculations and it also illustrates that the basic multiplication rule extends easily to three or more events.

EXAMPLE 5 Drug Screening and the 5% Guideline for Cumbersome Calculations

Assume that three adults are randomly selected without replacement from the 247,436,830 adults in the United States. Also assume that 10% of adults in the United States use drugs. Find the probability that the three selected adults all use drugs.

S O L U T I O N

Because the three adults are randomly selected without replacement, the three events are dependent, but here we can treat them as being independent by applying the 5% guideline for cumbersome calculations. The sample size of 3 is clearly no more than 5% of the population size of 247,436,830. We get

P1all 3 adults use drugs2 = P1first uses drugs and second uses drugs and third uses drugs2

= P1first uses drugs2 # P1second uses drugs2 # P1third uses drugs2

= 10.10210.10210.102 = 0.00100 There is a 0.00100 probability that all three selected adults use drugs.

CAUTION In any probability calculation, it is extremely important to carefully identify the event being considered. See Example 6, where parts (a) and (b) might seem quite similar but their solutions are very different.

EXAMPLE 6 Birthdays

When two different people are randomly selected from those in your class, find the indicated probability by assuming that birthdays occur on the days of the week with equal frequencies.

a. Find the probability that the two people are born on the same day of the week.

b. Find the probability that the two people are both born on Monday.

In Example 5, if we treat the events as dependent without using the 5% guideline, we get the following cumbersome calculation that begins with 247,436,830 adults, with 10% of them (or 24,743,683) using drugs:

a 24,743,683 247,436,830

ba 24,743,682 247,436,829

ba 24,743,681 247,436,828

b = 0.0009999998909

= 0.00100 1rounded2 Just imagine randomly selecting 1000 adults instead of just 3, as is commonly done in typical polls. Extending the above calculation to include 1000 factors instead of 3 factors would be what statisticians refer to as “painful.”

continued

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138 CHAPTER 4 Probability

Redundancy: Important Application of Multiplication Rule The principle of redundancy is used to increase the reliability of many systems. Our eyes have passive redundancy in the sense that if one of them fails, we continue to see. An important finding of modern biology is that genes in an organism can often work in place of each other. Engineers often design redundant components so that the whole system will not fail because of the failure of a single component, as in the following example.

S O L U T I O N

a. Because no particular day of the week is specified, the first person can be born on any one of the seven weekdays. The probability that the second person is born on the same day as the first person is 1>7. The probability that two people are born on the same day of the week is therefore 1>7.

b. The probability that the first person is born on Monday is 1>7 and the prob- ability that the second person is also born on Monday is 1>7. Because the two events are independent, the probability that both people are born on Monday is

1 7 # 1 7 =

1 49

WATCH YOUR LANGUAGE! Example 6 illustrates that finding correct or relevant probability values often requires greater language skills than computational skills. In Example 6, what exactly do we mean by “same day of the week”? See how parts (a) and (b) in Example 6 are very different.

EXAMPLE 7 Airbus 310: Redundancy for Better Safety

Modern aircraft are now highly reliable, and one design feature contributing to that reliability is the use of redundancy, whereby critical components are duplicated so that if one fails, the other will work. For example, the Airbus 310 twin-engine air- liner has three independent hydraulic systems, so if any one system fails, full flight control is maintained with another functioning system. For this example, we will as- sume that for a typical flight, the probability of a hydraulic system failure is 0.002.

a. If the Airbus 310 were to have one hydraulic system, what is the probability that the aircraft’s flight control would work for a flight?

b. Given that the Airbus 310 actually has three independent hydraulic systems, what is the probability that on a typical flight, control can be maintained with a working hydraulic system?

S O L U T I O N

a. The probability of a hydraulic system failure is 0.002, so the probability that it does not fail is 0.998. That is, the probability that flight control can be main- tained is as follows:

P11 hydraulic system does not fail2 = 1 - P1failure2 = 1 - 0.002 = 0.998

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4-2 Addition Rule and Multiplication Rule 139

Rationale for the Multiplication Rule To see the reasoning that underlies the multiplication rule, consider a pop quiz consist- ing of these two questions:

1. True or false: A pound of feathers is heavier than a pound of gold.

2. Who said, “By a small sample, we may judge of the whole piece”? (a) Judge Judy; (b) Judge Dredd; (c) Miguel de Cervantes; (d) George Gallup; (e) Gandhi

The answers are T (true) and c. (The first answer is true, because weights of feath- ers are in avoirdupois units where a pound is 453.59 g, but weights of gold and other precious metals are in troy units where a pound is 373.24 g. The second answer is from Don Quixote by Cervantes.)

Here is the sample space for the different possible answers:

Ta Tb Tc Td Te Fa Fb Fc Fd Fe

If both answers are random guesses, then the above 10 possible outcomes are equally likely, so

P1both correct2 = P1T and c2 = 1 10

= 0.1

With P1T and c2 = 1>10, P1T2 = 1>2, and P1c2 = 1>5, we see that 1 10

= 1 2 # 1 5

A tree diagram is a graph of the possible outcomes of a procedure, as in Figure 4-4. Figure 4-4 shows that if both answers are random guesses, all 10 branches are equally likely and the probability of getting the correct pair (T, c) is 1>10. For each response to the first question, there are 5 responses to the second. The total number of outcomes is 5 taken 2 times, or 10. The tree diagram in Figure 4-4 therefore provides a visual illustra- tion for using multiplication.

b. With three independent hydraulic systems, flight control will be maintained if the three systems do not all fail. The probability of all three hydraulic systems failing is 0.002 # 0.002 # 0.002 = 0.000000008. It follows that the probabil- ity of maintaining flight control is as follows:

P1it does not happen that all three hydraulic systems fail2 = 1 - 0.000000008 = 0.999999992

I N T E R P R E TAT I O N

With only one hydraulic system we have a 0.002 probability of failure, but with three independent hydraulic systems, there is only a 0.000000008 probability that flight control cannot be maintained because all three systems failed. By using three hydraulic systems instead of only one, risk of failure is decreased not by a factor of 1>3, but by a factor of 1>250,000. By using three independent hydraulic systems, risk is dramatically decreased and safety is dramatically increased.

Ta Tb Tc Td Te Fa Fb Fc Fd Fe

a b c d e a b c d e

T

F

105 52 3

FIGURE 4-4 Tree Diagram of Test Answers

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140 CHAPTER 4 Probability

Summary of Addition Rule and Multiplication Rule

Addition Rule for P(A or B): The word or suggests addition, and when adding P(A) and P(B), we must add in such a way that every outcome is counted only once.

Multiplication Rule for P(A and B): The word and for two trials suggests multiplication, and when multiplying P(A) and P(B), we must be sure that the probability of event B takes into account the previous occurrence of event A.

Statistical Literacy and Critical Thinking

1. Notation When randomly selecting an adult, A denotes the event of selecting someone with blue eyes. What do P(A) and P1A2 represent? 2.  Notation When randomly selecting adults, let M denote the event of randomly selecting a male and let B denote the event of randomly selecting someone with blue eyes. What does P1M 0B2 represent? Is P1M 0B2 the same as P1B 0M2? 3. Sample for a Poll There are 15,524,971 adults in Florida. If The Gallup organization ran- domly selects 1068 adults without replacement, are the selections independent or dependent? If the selections are dependent, can they be treated as being independent for the purposes of calculations?

4.  Rule of Complements When randomly selecting an adult, let B represent the event of randomly selecting someone with Group B blood. Write a sentence describing what the rule of complements is telling us: P1B or B2 = 1. Finding Complements. In Exercises 5–8, find the indicated complements.

5. LOL A U.S. Cellular survey of smartphone users showed that 26% of respondents answered “yes” when asked if abbreviations (such as LOL) are annoying when texting. What is the prob- ability of randomly selecting a smartphone user and getting a response other than “yes”?

6. Color Blindness Women have a 0.25% rate of red>green color blindness. If a woman is randomly selected, what is the probability that she does not have red>green color blindness? 7. Clinical Test When the drug Viagra (sildenafil citrate) was clinically tested, 117 patients reported headaches and 617 did not. If one of these patients is randomly selected, find the prob- ability of getting one who did not report a headache.

8.  Sobriety Checkpoint When one of the authors observed a sobriety checkpoint con- ducted by the Dutchess County Sheriff Department, he saw that 676 drivers were screened and 6 were arrested for driving while intoxicated. Based on those results, we can estimate that P1I2 = 0.000888, where I denotes the event of screening a driver and getting someone who is intoxicated. What does P1I2 denote, and what is its value? In Exercises 9–20, use the data in the following table, which summarizes blood groups and Rh types for randomly selected subjects. Assume that subjects are randomly selected from those included in the table.

O A B AB

Type Rh+ 59 53 12 6

Rh− 9 8 3 2

4-2 Basic Skills and Concepts

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4-2 Addition Rule and Multiplication Rule 141

9. Blood Groups and Types If one person is selected, find the probability of getting some- one who is not Group A.

10.  Blood Groups and Types If one person is selected, find the probability of getting someone who is not type Rh+.

11.  Blood Groups and Types If one person is selected, find the probability of getting someone who is Group A or type Rh+. Are the events of selecting someone who is Group A and the event of someone who is type Rh+ disjoint events?

12.  Blood Groups and Types If one person is selected, find the probability of getting someone who is type Rh− or Group AB. Are the events of selecting someone who is type Rh−

and the event of someone who is Group AB disjoint events?

13. Blood Groups and Types If two people are selected, find the probability that they are both Group B.

a. Assume that the selections are made with replacement. Are the events independent?

b. Assume that the selections are made without replacement. Are the events independent?

14. Blood Groups and Types If two people are selected, find the probability that they are both type Rh−.

a. Assume that the selections are made with replacement. Are the events independent?

b. Assume that the selections are made without replacement. Are the events independent?

15. Blood Groups and Types If two people are selected, find the probability that they are both type Rh+.

a. Assume that the selections are made with replacement. Are the events independent?

b. Assume that the selections are made without replacement. Are the events independent?

16. Blood Groups and Types If two people are selected, find the probability that they are both Group AB.

a. Assume that the selections are made with replacement. Are the events independent?

b. Assume that the selections are made without replacement. Are the events independent?

17.  Blood Groups and Types If one person is selected, find the probability of getting someone who is Group A or Group B or type Rh-.

18.  Blood Groups and Types If one person is selected, find the probability of getting someone who is Group O or Group AB or type Rh+.

19. Blood Groups and Types If three different people are selected, find the probability that they are all Group A.

20. Blood Groups and Types If three different people are selected, find the probability that they are all type Rh-.

In Exercises 21–24, use these results from the “1-Panel-THC” test for marijuana use, which is provided by the company Drug Test Success: Among 143 subjects with positive test results, there are 24 false positive results; among 157 negative results, there are 3 false negative results. (Hint: Construct a table similar to Table 4-1, which is included with the Chapter Problem.)

21. Testing for Marijuana Use

a. How many subjects are included in the study?

b. How many of the subjects had a true negative result?

c. What is the probability that a randomly selected test subject had a true negative result?

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142 CHAPTER 4 Probability

22.  Testing for Marijuana Use If one of the test subjects is randomly selected, find the probability that the subject tested negative or used marijuana.

23.  Testing for Marijuana Use If one of the test subjects is randomly selected, find the probability that the subject tested positive or did not use marijuana.

24.  Testing for Marijuana Use If one of the test subjects is randomly selected, find the probability that the subject used marijuana. Do you think that the result reflects the general population rate of subjects who use marijuana?

Redundancy. Exercises 25 and 26 involve redundancy.

25.  Redundancy in Computer Hard Drives It is generally recognized that it is wise to back up computer data. Assume that there is a 3% rate of disk drive failure in a year (based on data from various sources including lifehacker.com).

a. If you store all of your computer data on a single hard disk drive, what is the probability that the drive will fail during a year?

b. If all of your computer data are stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that both drives will fail during a year?

c. If copies of all of your computer data are stored on three independent hard disk drives, what is the probability that all three will fail during a year?

d. Describe the improved reliability that is gained with backup drives.

26. Redundancy in Hospital Generators Hospitals typically require backup generators to provide electricity in the event of a power outage. Assume that emergency backup generators fail 22% of the times when they are needed (based on data from Arshad Mansoor, senior vice president with the Electric Power Research Institute). A hospital has two backup generators so that power is available if one of them fails during a power outage.

a. Find the probability that both generators fail during a power outage.

b. Find the probability of having a working generator in the event of a power outage. Is that probability high enough for the hospital?

Acceptance Sampling. With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the entire batch is accepted if every item in the sample is found to be okay (or conforming). Exercises 27 and 28 involve acceptance sampling.

27. Defective Pacemakers Among 8834 cases of heart pacemaker malfunctions, 504 were found to be caused by firmware, which is software programmed into the device (based on data from “Pacemaker and ICD Generator Malfunctions,” by Maisel et al., Journal of the American Medical Association, Vol. 295, No. 16). If the firmware is tested in three different pacemak- ers randomly selected from this batch of 8834 and the entire batch is accepted if there are no failures, what is the probability that the firmware in the entire batch will be accepted? Is this procedure likely to result in the entire batch being accepted?

28.  Defective Ultrasound Transducers Among 676 ultrasound transducers tested, 269 were defective with transducer errors (based on data from “High Incidence of Defective Ultra- sound Transducers in Use in Routine Clinical Practice,” by Martensson et al., European Jour- nal of Cardiology, Vol. 10). If four different units are randomly selected and tested, what is the probability that the entire batch will be accepted? Does that probability seem adequate?

In Exercises 29 and 30, find the probabilities and indicate when the “5% guideline for cum- bersome calculations” is used.

29.  Medical Helicopters In a study of helicopter usage and patient survival, results were obtained from 47,637 patients transported by helicopter and 111,874 patients transported by ground (based on data from “Association Between Helicopter vs Ground Emergency Medical

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4-2 Addition Rule and Multiplication Rule 143

Services and Survival for Adults with Major Trauma,” by Galvagno et al., Journal of the Ameri- can Medical Association, Vol. 307, No. 15).

a. If 1 of the 159,511 patients in the study is randomly selected, what is the probability that the subject was transported by helicopter?

b. If 5 of the subjects in the study are randomly selected without replacement, what is the prob- ability that all of them were transported by helicopter?

30. Medical Helicopters In the same study cited in the preceding exercise, among the 47,637 patients transported by helicopter, 188 of them left the treatment center against medical advice, and the other 47,449 did not leave against medical advice. If 40 of the subjects transported by helicopter are randomly selected without replacement, what is the probability that none of them left the treatment center against medical advice?

31. MRI Reliability Refer to the accompanying figure showing surge protectors p and q used to protect an expensive magnetic resonance imaging (MRI) scanner used in a hospital. If there is a surge in the voltage, the surge protector reduces it to a safe level. Assume that each surge protector has a 0.985 probability of working correctly when a voltage surge occurs.

a. If the two surge protectors are arranged in series, what is the probability that a voltage surge will not damage the MRI? (Do not round the answer.)

b. If the two surge protectors are arranged in parallel, what is the probability that a voltage surge will not damage the MRI? (Do not round the answer.)

c. Which arrangement should be used for better protection?

Series configuration

p q MRI

Parallel configuration

p

q MRI

32. Same Birthdays If 25 people are randomly selected, find the probability that no 2 of them have the same birthday. Ignore leap years.

33. Exclusive Or The exclusive or means either one or the other events occurs, but not both.

a. For the formal addition rule, rewrite the formula for P(A or B) assuming that the addition rule uses the exclusive or instead of the inclusive or.

b. Repeat Exercise 11 “Blood Groups and Types” using the exclusive or instead of the inclu- sive or.

34. Complements and the Addition Rule Refer to the table of blood groups and types used for Exercises 9–20. Assume that one subject is randomly selected. Let A represent the event of getting someone with Group A blood and let B represent the event of getting someone with Group B blood. Find P1A or B2, find P1A or B2, and then compare the results. In general, does P1A or B2 = P1A or B2?

4-2 Beyond the Basics

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144 CHAPTER 4 Probability

Key Concept In Part 1 of this section we extend the use of the multiplication rule to include the probability that among several trials, we get at least one of some specified event. In Part 2 we consider conditional probability: the probability of an event occur- ring when we have additional information that some other event has already occurred. In Part 3 we provide a brief introduction to the use of Bayes’ theorem.

PA RT 1 Complements: The Probability of “At Least One”

When finding the probability of some event occurring “at least once,” we should un- derstand the following:

■ “At least one” has the same meaning as “one or more.”

■ The complement of getting “at least one” particular event is that you get no occurrences of that event.

For example, not getting at least 1 girl in 10 births is the same as getting no girls, which is also the same as getting 10 boys.

Not getting at least 1 girl in 10 births = Getting no girls = Getting 10 boys

The following steps describe the details of this backward method of finding the probability of getting at least one of some event:

Finding the probability of getting at least one of some event:

1. Let A = getting at least one of some event.

2. Then A = getting none of the event being considered.

3. Find P1A2 = probability that event A does not occur. (This is relatively easy using the multiplication rule.)

4. Subtract the result from 1. That is, evaluate this expression:

P1at least one occurrence of event A2 = 1 − P1no occurrences of event A2

4-3 Complements, Conditional Probability,

and Bayes’ Theorem

EXAMPLE 1 At Least One Subject with Group AB Blood

The probability of randomly selecting someone with Group AB blood is 0.0526 (based on the table given with Exercises 9–20 in the preceding section). A re- searcher needs at least one subject having Group AB blood. If 20 subjects are ran- domly selected, find the probability of getting at least one with Group AB blood. Is the probability high enough so that the researcher can be reasonably sure of getting someone with Group AB blood?

S O L U T I O N

Step 1: Let A = at least 1 of the 20 subjects has Group AB blood.

Step 2: Identify the event that is the complement of A.

A = not getting at least 1 subject with Group AB blood among 20 = all 20 subjects have blood that is not Group AB

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4-3 Complements, Conditional Probability, and Bayes’ Theorem 145

PA RT 2 Conditional Probability We now consider the principle that the probability of an event is often affected by knowledge that some other event has occurred. For example, the probability of a golfer making a hole in one is 1>12,000 (based on past results), but if you have the additional knowledge that the selected person is a professional golfer, the probability changes to 1>2375 (based on data from USA Today).

Step 3: Find the probability of the complement by evaluating P1A2 If there is a 0.0526 probability of a person having Group AB blood, it follows that there is a 0.9474 probability of a person not having Group AB blood, and we get the following:

P1A2 = P1all 20 subjects have blood that is not Group AB2 = 0.9474 # 0.9474 # g # 0.9474 = 0.947420 = 0.339365436

Step 4: Find P(A) by evaluating 1 - P1A2. P1A2 = 1 - P1A2 = 1 - 0.339365436 = 0.661 1rounded2

I N T E R P R E TAT I O N

For a group of 20 subjects, there is a 0.661 probability of getting at least 1 person with Group AB blood. This probability is not very high, so if the researcher needs a person with Group AB blood, more than 20 subjects should be used.

DEFINITION

A conditional probability of an event is a probability obtained with the additional information that some other event has already occurred.

Notation

P1B � A2 denotes the conditional probability of event B occurring, given that event A has already occurred.

INTUITIVE APPROACH FOR FINDING P1B ∣ A2 The conditional probability of B occurring given that A has occurred can be found by assuming that event A has occurred and then calculating the probability that event B will occur, as illustrated in Example 2.

FORMAL APPROACH FOR FINDING P1B ∣ A2 The probability P1B � A2 can be found by dividing the probability of events A and B both occurring by the probability of event A:

P1B � A2 = P1A and B2 P1A2

Prosecutor’s Fallacy

The prosecu-

tor’s fallacy is

misunderstand-

ing or confusion

of two different

conditional

probabilities:

(1) the probability

that a defendant is innocent, giv-

en that forensic evidence shows

a match; (2) the probability that

forensics shows a match, given

that a person is innocent. The

prosecutor’s fallacy has led to

wrong convictions and imprison-

ment of some innocent people.

Lucia de Berk was a nurse

who was convicted of murder

and sentenced to prison in the

Netherlands. Hospital administra-

tors observed suspicious deaths

that occurred in hospital wards

where de Berk had been present.

An expert testified that there was

only 1 chance in 342 million that

her presence was a coincidence.

However, mathematician Richard

Gill calculated the probability to

be closer to 1>150, or possibly as low as 1>5. The court used the probability that the suspicious

deaths could have occurred with

de Berk present, given that she

was innocent. The court should

have considered the probability

that de Berk is innocent, given

that the suspicious deaths oc-

curred when she was present.

This error of the prosecutor’s

fallacy is subtle and can be

very difficult to understand and

recognize, yet it can lead to the

imprisonment of innocent people.

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146 CHAPTER 4 Probability

The preceding formula is a formal expression of conditional probability, but blind use of formulas is not recommended. Instead, we recommend the intuitive approach, as illustrated in Example 2.

TABLE 4-1 Results from Drug Tests of Job Applicants

Positive Test Result (Test shows drug use.)

Negative Test Result (Test shows no drug use.)

Subject Uses Drugs 45 (True Positive)

5 (False Negative)

Subject Does Not Use Drugs 25 (False Positive)

480 (True Negative)

EXAMPLE 2 Pre-Employment Drug Screening

Refer to Table 4-1 to find the following:

a. If 1 of the 555 test subjects is randomly selected, find the probability that the subject had a positive test result, given that the subject actually uses drugs. That is, find P(positive test result � subject uses drugs).

b. If 1 of the 555 test subjects is randomly selected, find the probability that the subject actually uses drugs, given that he or she had a positive test result. That is, find P(subject uses drugs � positive test result).

S O L U T I O N

a. Intuitive Approach: We want P(positive test result � subject uses drugs), the probability of getting someone with a positive test result, given that the se- lected subject uses drugs. Here is the key point: If we assume that the selected subject actually uses drugs, we are dealing only with the 50 subjects in the first row of Table 4-1. Among those 50 subjects, 45 had positive test results, so we get this result:

P1positive test result � subject uses drugs2 = 45 50

= 0.900

Formal Approach: The same result can be found by using the formula for P1B � A2given with the formal approach. We use the following notation.

P1B � A2 = P1positive test result � subject uses drugs2 where B = positive test result and A = subject uses drugs.

In the following calculation, we use P(subject uses drugs and had a posi- tive test result) = 45>555 and P(subject uses drugs) = 50>555 to get the following results:

P1B � A2 = P1A and B2 P1A2

becomes

P1positive test result � subject uses drugs2

= P1subject uses drugs and had a positive test result2

P1subject uses drugs2

= 45>555 50>555 = 0.900

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4-3 Complements, Conditional Probability, and Bayes’ Theorem 147

Confusion of the Inverse Note that in Example 2, P(positive test result � subject uses drugs) ≠ P(subject uses drugs � positive test result). This example proves that in general, P1B � A2 ≠ P1A � B2. (There could be individual cases where P1A � B2 and P1B � A2 are equal, but they are generally not equal.) To incorrectly think that P1B � A2 and P1A � B2 are equal or to incorrectly use one value in place of the other is called confusion of the inverse.

By comparing the intuitive approach to the formal approach, it should be clear that the intuitive approach is much easier to use, and it is also less likely to result in errors. The intuitive approach is based on an understanding of condi- tional probability, instead of manipulation of a formula, and understanding is so much better.

b. Here we want P(subject uses drugs � positive test result). If we assume that the subject had a positive test result, we are dealing with the 70 subjects in the first column of Table 4-1. Among those 70 subjects, 45 use drugs, so

P1subject uses drugs � positive test result2 = 45 70

= 0.643

Again, the same result can be found by applying the formula for conditional probability, but we will leave that for those with a special fondness for ma- nipulations with formulas.

I N T E R P R E TAT I O N

The first result of P(positive test result � subject uses drugs) = 0.900 indicates that a subject who uses drugs has a 0.900 probability of getting a positive test result. The second result of P(subject uses drugs � positive test result) = 0.643 indicates that for a subject who gets a positive test result, there is a 0.643 probability that this subject actually uses drugs. Note that P1positive test result � subject uses drugs2 P1subject uses drugs � positive test result2. See “Confusion of the Inverse” that follows.

EXAMPLE 3 Confusion of the Inverse

Consider these events:

D: It is dark outdoors.

M: It is midnight.

In the following, we conveniently ignore the Alaskan winter and other such anomalies.

P1D � M2 = 1 1It is certain to be dark given that it is midnight.2 P1M � D2 = 0 1The probability that it is exactly midnight given

that it is dark is almost zero.2 Here, P1D � M2 ≠ P1M � D2. Confusion of the inverse occurs when we incorrectly switch those probability values or think that they are equal.

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148 CHAPTER 4 Probability

PA RT 3 Bayes’ Theorem In this section we extend the discussion of conditional probability to include applica- tions of Bayes’ theorem (or Bayes’ rule), which we use for revising a probability value based on additional information that is later obtained.

Let’s consider a study showing that physicians often give very misleading in- formation when they experience confusion of the inverse. They tended to confuse P(cancer � positive test result) with P(positive test result � cancer). About 95% of physi- cians estimated P(cancer � positive test result) to be about 10 times too high, with the result that patients were given diagnoses that were very misleading, and patients were unnecessarily distressed by the incorrect information. Let’s take a closer look at this classic example, and let’s hope that we can give physicians information in a better format that is easy to understand.

EXAMPLE 4 Interpreting Medical Test Results

Assume cancer has a 1% prevalence rate, meaning that 1% of the population has cancer. Denoting the event of having a cancer by C, we have P1C2 = 0.01 for a subject randomly selected from the population. This result is included with the fol- lowing performance characteristics of the test for cancer (based on Probabilistic Reasoning in Clinical Medicine, by David Eddy, Cambridge University Press).

■ There is a 1% prevalence rate of the cancer. That is, P1C2 = 0.01. ■ The false positive rate is 10%. That is, P(positive test result given that cancer is not present) = 0.10.

■ The true positive rate is 80%. That is, P(positive test result given that cancer is present) = 0.80.

Find P1C � positive test result2. That is, find the probability that a subject actually has cancer given that he or she has a positive test result.

S O L U T I O N

Using the given information, we can construct a hypothetical population with the above characteristics. We can find the entries in Table 4-2 on the next page, as follows.

■ Assume that we have 1000 subjects. With a 1% prevalence rate, 10 of the sub- jects are expected to have cancer. The sum of the entries in the first row of val- ues is therefore 10.

■ The other 990 subjects do not have cancer. The sum of the entries in the second row of values is therefore 990.

■ Among the 990 subjects without cancer, 10% get positive test results, so 10% of the 990 cancer-free subjects in the second row get positive test results. See the entry of 99 in the second row.

■ For the 990 subjects in the second row, 99 test positive, so the other 891 must test negative. See the entry of 891 in the second row.

■ Among the 10 subjects with cancer in the first row, 80% of the test results are positive, so 80% of the 10 subjects in the first row test positive. See the entry of 8 in the first row.

■ The other 2 subjects in the first row test negative. See the entry of 2 in the first row.

In this section we extend the discussion of conditional probability to include applica tions of

Group Testing

During World

War II, the U.S.

Army tested

for syphilis

by giving

each soldier

an individual

blood test that was analyzed

separately. One researcher sug-

gested mixing pairs of blood

samples. After the mixed pairs

were tested, those with syphilis

could be identified by retest-

ing the few blood samples that

were in the pairs that tested

positive. Since the total number

of analyses was reduced by pair-

ing blood specimens, why not

combine them in groups of three

or four or more? This technique

of combining samples in groups

and retesting only those groups

that test positive is known as

group testing or pooled testing, or

composite testing. University of

Nebraska statistician Christopher

Bilder wrote an article about this

topic in Chance magazine, and

he cited some real applications.

He noted that the American

Red Cross uses group testing

to screen for specific diseases,

such as hepatitis, and group

testing is used by veterinarians

when cattle are tested for the

bovine viral diarrhea virus.

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4-3 Complements, Conditional Probability, and Bayes’ Theorem 149

The solution in Example 4 is not very difficult. Another approach is to compute the probability using this formula commonly given with Bayes’ theorem:

P1A � B2 = P1A2 # P1B � A2

3P1A2 # P1B � A24 + 3P1A2 # P1B � A24 If we replace A with C and replace B with “positive,” we get this solution for Example 4:

P1C � positive2 = P1C2 # P1positive � C2

P1C2 # P1positive � C2 + P1C2 # P1Positive � C2 =

0.01 # 0.80 10.01 # 0.802 + 10.99 # 0.102 = 0.0748

Study Results Here is a truly fascinating fact: When 100 physicians were given the information in Example 4, 95 of them estimated P(C � positive) to be around 0.70 to 0.80, so they were wrong by a factor of 10. Physicians are extremely intel- ligent, but here they likely suffered from confusion of the inverse. The given rate of 80% for positive test results among those who are true positives implies that P1positive � C2 = 0.80, but this is very different from P1C � positive2. The physi- cians would have done much better if they had seen the given information in the form of a table like Table 4-2.

The importance and usefulness of Bayes’ theorem is that it can be used with se- quential events, whereby new additional information is obtained for a subsequent event, and that new information is used to revise the probability of the initial event. In this context, the terms prior probability and posterior probability are commonly used.

To find P1C � positive test result2, see that the first column of values includes the positive test results. In that first column, the probability of randomly selecting a subject with cancer is 8>107 or 0.0748, so P1C � positive test result2 = 0.0748.

I N T E R P R E TAT I O N

For the data given in this example, a randomly selected subject has a 1% chance of cancer, but for a randomly selected subject given a test with a positive result, the chance of cancer increases to 7.48%. Based on the data given in this example, a positive test result should not be devastating news, because there is still a good chance that the test is wrong.

TABLE 4-2 Test Results

Positive Test Result (Test shows cancer.)

Negative Test Result (Test shows no cancer.)

Total

Cancer 8 (True Positive)

2 (False Negative)

10

No Cancer 99 (False Positive)

891 (True Negative)

990

DEFINITIONS

A prior probability is an initial probability value originally obtained before any ad- ditional information is obtained.

A posterior probability is a probability value that has been revised by using addi- tional information that is later obtained.

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150 CHAPTER 4 Probability

Relative to Example 4, P1C2 = 0.01, which is the probability that a randomly se- lected subject has cancer. P(C) is an example of a prior probability. Using the ad- ditional information that the subject has received a positive test result, we found that P1C � positive test result2 = 0.0748, and this is a posterior probability because it uses that additional information of the positive test result.

Statistical Literacy and Critical Thinking

1.  Language: Complement of “At Least One” Let A = the event of getting at least one defective pacemaker battery when 3 batteries are randomly selected with replacement from a batch. Write a statement describing event A.

2. Probability of At Least One Let A = the event of getting at least 1 defective pacemaker battery when 3 batteries are randomly selected with replacement from a batch. If 5% of the batteries in a batch are defective and the other 95% are all good, which of the following are correct?

a. P1A2 = 10.95210.95210.952 = 0.857 b. P1A2 = 1 - 10.95210.95210.952 = 0.143 c. P1A2 = 10.05210.05210.052 = 0.000125 3. Notation Let event G = subject has glaucoma (disorder of the eye) and let event Y = test indicates that “yes,” the subject has glaucoma. Use your own words to translate the notation P1Y � G2 into a verbal statement. 4. Confusion of the Inverse Using the same events G and Y described in Exercise 3, de- scribe confusion of the inverse.

At Least One. In Exercises 5–12, find the probability.

5. Three Girls Find the probability that when a couple has three children, at least one of them is a girl. (Assume that boys and girls are equally likely.)

6. Probability of a Girl Assuming that boys and girls are equally likely, find the probability of a couple having a boy when their third child is born, given that the first two children were both girls.

7. Births in the United States In the United States, the true probability of a baby being a boy is 0.512 (based on the data available at this writing). Among the next six randomly selected births in the United States, what is the probability that at least one of them is a girl?

8. Births in China In China, where many couples were allowed to have only one child, the probability of a baby being a boy was 0.545. Among six randomly selected births in China, what is the probability that at least one of them is a girl? Could this system continue to work indefinitely? (Phasing out of this policy was begun in 2015.)

9.  Phone Survey Subjects for the California Health Interview Survey are contacted using telephone numbers in which the last four digits are randomly selected (with replacement). Find the probability that for one such phone number, the last four digits include at least one 0.

10.  At Least One Correct Answer If you make random guesses for 10 multiple-choice MCAT test questions (each with five possible answers), what is the probability of getting at least 1 correct? If these questions are part of a practice test and an instructor says that you must get at least one correct answer before continuing, is there a good chance you will continue?

11. At Least One Defective Ultrasound Transducer A study showed that 39.8% of ultra- sound transducers are defective (based on data from “High Incidence of Defective Ultrasound

4-3 Basic Skills and Concepts

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4-3 Complements, Conditional Probability, and Bayes’ Theorem 151

Transducers in Use in Routine Clinical Practice,” by Martensson et al., European Journal of Echocardiography, Vol. 10, No. 1093.) An engineer needs at least one defective ultrasound transducer so she can try to identify the problem. If she randomly selects 10 ultrasound trans- ducers from a very large batch, what is the probability that she will get at least one that is de- fective? Is that probability high enough so that she can be reasonably sure of getting a defective transducer for her work?

12. Fruit Flies An experiment with fruit flies involves one parent with normal wings and one parent with vestigial wings. When these parents have an offspring, there is a 3>4 probability that the offspring has normal wings and a 1>4 probability of vestigial wings. If the parents give birth to five offspring, what is the probability that at least one of the offspring has vestigial wings? If researchers need at least one offspring with vestigial wings, can they be quite confi- dent of getting one?

Identical and Fraternal Twins. In Exercises 13–16, use the data in the following table. Instead of summarizing observed results, the entries reflect the actual probabilities based on births of twins (based on data from the Northern California Twin Registry and the ar- ticle “Bayesians, Frequentists, and Scientists,” by Bradley Efron, Journal of the American Statistical Association, Vol. 100, No. 469). Identical twins come from a single egg that splits into two embryos, and fraternal twins are from separate fertilized eggs. The table entries reflect the principle that among sets of twins, 1 ,3 are identical and 2 ,3 are fraternal. Also, identical twins must be of the same gender and the genders are equally likely (approxi- mately), and genders of fraternal twins are equally likely.

Boy>boy Boy>girl Girl>boy Girl>girl Identical Twins 5 0 0 5

Fraternal Twins 5 5 5 5

13. Identical Twins

a. After having a sonogram, a pregnant woman learns that she will have twins. What is the probability that she will have identical twins?

b. After studying the sonogram more closely, the physician tells the pregnant woman that she will give birth to twin boys. What is the probability that she will have identical twins? That is, find the probability of identical twins given that the twins consist of two boys.

14. Fraternal Twins

a. After having a sonogram, a pregnant woman learns that she will have twins. What is the probability that she will have fraternal twins?

b. After studying the sonogram more closely, the physician tells the pregnant woman that she will give birth to twins consisting of one boy and one girl. What is the probability that she will have fraternal twins?

15. Fraternal Twins If a pregnant woman is told that she will give birth to fraternal twins, what is the probability that she will have one child of each gender?

16. Fraternal Twins If a pregnant woman is told that she will give birth to fraternal twins, what is the probability that she will give birth to two girls?

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152 CHAPTER 4 Probability

In Exercises 17–20, refer to the accompanying table showing results from a Chembio test for hepatitis C among HIV-infected patients (based on data from a variety of sources).

Positive Test Result Negative Test Result

Hepatitis C 335 10

No Hepatitis C 2 1153

17. False Positive Find the probability of selecting a subject with a positive test result, given that the subject does not have hepatitis C. Why is this case problematic for test subjects?

18.  False Negative Find the probability of selecting a subject with a negative test result, given that the subject has hepatitis C. What would be an unfavorable consequence of this error?

19. Positive Predictive Value Find the positive predictive value for the test. That is, find the probability that a subject has hepatitis C, given that the test yields a positive result. Does the result make the test appear to be effective?

20. Negative Predictive Value Find the negative predictive value for the test. That is, find the probability that a subject does not have hepatitis C, given that the test yields a negative re- sult. Does the result make the test appear to be effective?

21. Redundancy in Computer Hard Drives Assume that there is a 3% rate of disk drive failures in a year (based on data from various sources including lifehacker.com).

a. If all of your computer data are stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that during a year, you can avoid catastrophe with at least one working drive? Express the result with four decimal places.

b. If copies of all of your computer data are stored on three independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive? Express the result with six decimal places. What is wrong with using the usual round-off rule for probabilities in this case?

22. Redundancy in Hospital Generators Assume that emergency backup generators fail 22% of the times when they are needed (based on data from Arshad Mansoor, senior vice presi- dent with the Electric Power Research Institute). A hospital has three backup generators so that power is available if at least one of them works in a power failure. Find the probability of having at least one of the backup generators working, given that a power failure has occurred. Does the result appear to be adequate for the hospital’s needs?

23. Composite Drug Test Based on the data in Table 4-1 on page 146, assume that the prob- ability of a randomly selected person testing positive for drug use is 0.126. If drug screening samples are collected from 5 random subjects and combined, find the probability that the com- bined sample will reveal a positive result. Is that probability low enough so that further testing of the individual samples is rarely necessary?

24.  Composite Water Samples The Fairfield County Department of Public Health tests water for the presence of E. coli (Escherichia coli) bacteria. To reduce laboratory costs, water samples from 10 public swimming areas are combined for one test, and further testing is done only if the combined sample tests positive. Based on past results, there is a 0.005 probability of finding E. coli bacteria in a public swimming area. Find the probability that a combined sample from 10 public swimming areas will reveal the presence of E. coli bacteria. Is that probability low enough so that further testing of the individual samples is rarely necessary?

25. Shared Birthdays Find the probability that of 25 randomly selected people, at least 2 share the same birthday.

4-3 Beyond the Basics

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4-4 Risks and Odds 153

Key Concept This section introduces absolute risk reduction, relative risk, and odds ratio as measures helpful for comparing probability values and measuring risk. This section also introduces “number needed to treat” as a measure of the number of sub- jects that must be treated in order to prevent the single occurrence of some event, such as a disease.

One simple way to measure risk is to use a probability value. For example, in one of the largest medical experiments ever conducted, it was found that among 200,745 children injected with the Salk vaccine, 33 developed paralytic polio (poliomyelitis). It follows that for this treatment group, P1polio2 = 33>200,745 = 0.000164. How- ever, that single measure does not give us any information about the rate of polio for those children who were injected with a placebo. The risk of polio for children treated with the Salk vaccine should be somehow compared to the risk of polio for those chil- dren given a placebo. Let’s consider the data summarized in Table 4-3.

4-4 Risks and Odds

TABLE 4-3 Prospective Study of Polio and the Salk Vaccine

Polio No polio Total

Salk Vaccine 33 200,712 200,745

Placebo 115 201,114 201,229

Based on the data in Table 4-3, we can identify the following probabilities:

Polio rate for treatment group: P1polio � Salk vaccine2 = 33 200,745

= 0.000164

Polio rate for placebo group: P1polio � placebo2 = 115 201,229

= 0.00571

Informal comparison of the preceding two probabilities likely suggests that there is a substantial difference between the two polio rates. Later chapters will use more effec- tive methods for determining whether the apparent difference is actually significant, but in this section we introduce some simple measures for comparing the two rates.

The preceding table can be generalized with the following format:

TABLE 4-4 Generalized Table Summarizing Results of a Prospective Study

Disease No Disease

Treatment a b

Placebo c d

We noted above that this section introduces some simple measures for comparing two rates, such as the polio rate for the Salk vaccine treatment group and the polio rate for the placebo group, as summarized in Table 4-3. We begin with the absolute risk reduction.

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154 CHAPTER 4 Probability

Absolute Risk Reduction

DEFINITION

When comparing two probabilities or rates, the absolute risk reduction is simply the absolute value of the following difference.

Absolute risk reduction = � P1event occurring in treatment group2 - P1event occurring in control group2 �

If the data are in the generalized format of Table 4−4, we can express the absolute risk reduction as follows:

Absolute risk reduction = � P1event occurring in treatment group2 - P1event occurring in control group2 �

= ` a a + b

- c

c + d `

(In the above expression, “treatment” might be replaced by the “presence of some condition” or some other equivalent description.)

CAUTION: The above definition of absolute risk reduction always results in the positive difference between a probability in the treatment group and a probability in the control group. Consider this when interpreting the effectiveness of the treatment in terms of being helpful or harmful. See Exercises 13−16, where those in an atorvastatin treatment group have a higher rate of infections than those in the placebo group. Be careful to interpret the results correctly.

EXAMPLE 1 Finding Absolute Risk Reduction

Using the data summarized in Table 4-3, find the absolute risk reduction, which can be used to measure the effectiveness of the Salk vaccine.

S O L U T I O N

Based on the data in Table 4-3, we have already found that P(polio � Salk vaccine) = 0.000164 and P(polio � placebo) = 0.000571. It follows that:

Absolute risk reduction = � P1polio � Salk vaccine2 - P1polio � placebo2 � = � 0.000164 - 0.000571 � = 0.000407

For a subject treated with the Salk vaccine, there is an absolute risk reduction of 0.000407 when compared to a subject given a placebo. That is, there are 0.0407% fewer events of polio for subjects treated with the Salk vaccine than for subjects given a placebo. This doesn’t seem like much of a reduction, but when considered in the con- text of the number of polio events in a large population, it is a significant reduction.

Relative Risk

Section 1-3 included definitions of retrospective and prospective studies:

■ Retrospective Study: Data are collected from a past time period by going back in time (through examinations of records, interviews, etc.).

■ Prospective Study: Data are collected in the future from groups or “cohorts” that share common factors.

Absolute Risk Reduction Monkey Typists

A classical

claim is that

a monkey

randomly

hitting a key-

board would

eventually

produce the complete works of

Shakespeare, assuming that it

continues to type century after

century. The multiplication rule

for probability has been used to

find such estimates. One

result of 1,000,000,000,000,000,

000,000,000,000,000,000,000

years is considered by some to

be too short. In the same spirit,

Sir Arthur Eddington wrote this

poem: “There once was a brainy

baboon, who always breathed

down a bassoon. For he said, ‘It

appears that in billions of years, I

shall certainly hit on a tune.’”

M04_TRIO9015_02_SE_C04_118-179.indd 154 30/07/16 11:35 AM

4-4 Risks and Odds 155

In a prospective study, a commonly used measure for comparing risk is relative risk. We first introduce the following notation, and then we define relative risk.

Notation

pt = proportion (or incidence rate) of some characteristic in a treatment group pc = proportion (or incidence rate) of some characteristic in a control group

DEFINITION

In a prospective study, the relative risk (or risk ratio or RR) of a characteristic is the ratio pt>pc, where pt is the proportion of the characteristic in the treatment (or exposed) group and pc is the proportion in the control group (or group not ex- posed). If the data are in the same format as the generalized Table 4-4, then the relative risk is found by evaluating

pt pc

=

a a + b

c c + d

Interpreting Relative Risk A relative risk value of 1 shows that the risk is the same for the treatment group and the control (or placebo) group. A relative risk value much greater than 1 shows that there is a much greater risk for the treatment group. The fol- lowing example illustrates how the relative risk of 0.287 shows that the risk of polio in the treatment group is much less than the risk of polio in the placebo group.

EXAMPLE 2 Computing Relative Risk

Using the data in Table 4-3, find the relative risk.

S O L U T I O N

For the sample data in Table 4-3, we will consider the treatment group to be the group of children given the Salk vaccine, and the control group is the group of chil- dren given a placebo. Using the preceding notation, we have

pt = proportion of polio in treatment group = 33

33 + 200,712 = 0.000164

pc = proportion of polio in control 1placebo2 group = 115

115 + 201,114 = 0.000571

Using the above values, we can now find the relative risk as follows.

Relative risk = pt pc

= 0.000164 0.000571

= 0.287

I N T E R P R E TAT I O N

We can interpret this result as follows: The polio rate for children given the Salk vaccine is 0.287 of the polio rate for children given a placebo. (A relative risk less than 1 indicates that the treatment results in a reduced risk.) If we were to consider the reciprocal value of 0.000571>0.000164 = 3.48, we see that children in the placebo group are 3.48 times more likely to get polio.

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156 CHAPTER 4 Probability

Number Needed to Treat

One problem with relative risk is that it may be misleading by suggesting that a treat- ment is superior or inferior, even when the absolute difference between rates is not very large. For example, if 3 out of 10,000 aspirin users were to experience an imme- diate cure of a cold compared to only 1 out of 10,000 placebo users, the relative risk of 3.00 correctly indicates that the incidence of immediate cold cures is three times as high for aspirin users, but the cure rates of 0.0003 and 0.0001 are so close that, for all practical purposes, aspirin should not be considered as a factor affecting the immedi- ate cure of a cold. With cure rates of 0.0003 and 0.0001, the absolute risk reduction is 0.0002. Because the absolute risk reduction is so small, the effectiveness of the aspirin treatment would be negligible. In such a situation, the number needed to treat would be a more effective measure that is not so misleading.

CAUTION: When interpreting relative risk, consider the incidence rates. If the probability of disease in an exposed group is 5>1,000,000 and the probability of disease in an unexposed group is 1>1,000,000, the relative risk is 5.0, which sounds really bad, but the very low incidence rates suggest that there isn’t much risk in either group.

DEFINITION

The number needed to treat (NNT) is the number of subjects that must be treated in order to prevent one event, such as a disease or adverse reaction. It is calcu- lated by dividing 1 by the absolute risk reduction.

number needed to treat = 1

absolute risk reduction

Round-Off Rule If the calculated value of the number needed to treat is not a whole number, round it up to the next larger whole number.

If the sample data are in the format of the generalized Table 4-4, then:

Number needed to treat = 1

` a a + b

- c

c + d ` 1rounded up to the next larger whole number2

If 3 out of 10,000 aspirin users were to experience an immediate cure of a cold compared to only 1 out of 10,000 placebo users, the absolute risk reduction is � 0.0003 - 0.0001 � = 0.0002, and the number needed to treat is 1>0.0002 = 5000. This means that we would need to treat 5000 subjects with colds to get one person who experiences an immediate cure.

EXAMPLE 3 Computing the Number Needed to Treat

Using the polio data in Table 4-3, find the number needed to treat, then interpret the result.

S O L U T I O N

In Example 1 we found that the absolute risk reduction is 0.000407. It is now easy to find the number needed to treat.

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4-4 Risks and Odds 157

Odds

So far in this chapter, we have used probability values to express likelihood of various events. Probability values are numbers between 0 and 1 inclusive. However, expres- sions of likelihood are often given as odds, such as 50:1 (or “50 to 1”).

Number needed to treat = 1

absolute risk reduction =

1 0.000407

= 2457.002457

= 2458 1rounded up2

I N T E R P R E TAT I O N

The result of 2458 can be interpreted as follows: We would need to vaccinate 2458 children with the Salk vaccine (instead of a placebo) to prevent one of the children from getting polio. Given the extremely serious consequences of polio, the Salk vaccine has been found to be very effective and important.

DEFINITIONS

The actual odds against event A occurring are the ratio P1A2>P1A2, usually ex- pressed in the form of a:b (or “a to b”), where a and b are integers. (Reduce using the largest common factor; if a = 16 and b = 4, express the odds as 4:1 instead of 16:4.)

The actual odds in favor of event A occurring are the ratio P1A2>P1A2, which is the reciprocal of the actual odds against that event. If the odds against an event are a:b, then the odds in favor are b:a.

Note that in the two preceding definitions, the actual odds against and the actual odds in favor describe the actual likelihood of some event. (Gambling situations typically use payoff odds, which describe the amount of profit relative to the amount of a bet. For example, if you bet on the number 7 in roulette, the actual odds against winning are 37:1, but the payoff odds are 35:1. Racetracks and casinos are in business to make a profit, so the payoff odds will usually differ from the actual odds.)

TABLE 4-5 Retrospective Study of Newborn Discharge and Rehospitalization

Rehospitalized within a week

Not rehospitalized within a week

Total

Early discharge (*30 hours)

457 3199 3656

Late discharge (30 + hours)

260 2860 3120

EXAMPLE 4 Rehospitalization and Discharge

Consider the data in Table 4-5 (based on results from “The Safety of Newborn Early Discharge,” by Liu, Clemens, Shay, Davis, and Novack, Journal of the American Medical Association, Vol. 278, No. 4).

a. For those babies discharged early, find the probability of being rehospitalized within a week.

b. For those babies discharged early, find the odds in favor of being rehospital- ized early.

continued

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158 CHAPTER 4 Probability

Odds Ratio

For the data in Table 4-5, how does the likelihood of rehospitalization differ between the early discharge group and the late discharge group? One way to address that question is to use the odds ratio.

S O L U T I O N

a. There were 3656 babies discharged early, and 457 of them were rehospital- ized within a week, so

P1rehospitalized2 = 457 3656

= 1 8

b. Because P(rehospitalized) = 1>8, it follows that P1rehospitalized2 = 1 - 1>8= 7>8. We can now find the odds in favor of rehospitalization as follows:

Odds in favor of rehospitalization for the early discharge group

= P1rehospitalized2 P1rehospitalized2 =

1>8 7>8 =

1 7

This result is often expressed as 1:7. With odds of 1:7 in favor of rehospitalization for babies discharged early, it

follows that the odds against rehospitalization for early discharge are 7:1.

DEFINITION

In a retrospective or prospective study, the odds ratio (OR or relative odds) is a measure of risk found by evaluating the ratio of the odds in favor of the treatment group (or case group exposed to the risk factor) to the odds in favor of the control group, evaluated as follows:

Odds ratio = odds in favor of treatment 1or exposed2 group

odds in favor of control group

If the data are in the format of the generalized Table 4-4 on page 153, then the odds ratio can be computed as follows:

Odds ratio = ad bc

EXAMPLE 5 Computing Odds Ratio

Using the data in Table 4-5, find the odds ratio for rehospitalization.

S O L U T I O N

For this example, we consider the case group to be the babies discharged early, and we consider the control group to be the babies discharged late. The preceding exam- ple showed that for the early discharge group, the odds in favor of rehospitalization are 1:7. Using similar calculations for the late discharge group, we get odds in favor of rehospitalization of 1>11 or 1:11. We can now find the odds ratio.

Odds ratio = odds in favor of rehospitalization in early discharge group

odds in favor of rehospitalization in late discharge group =

1>7 1>11

= 11 7

or 1.571

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4-4 Risks and Odds 159

Why Not Use Relative Risk for Retrospective Studies?

Relative risk makes sense only if the involved probabilities are good estimates of the actual incidence rates, as in a prospective study. Using relative risk for a retro- spective study could incorrectly involve a situation in which researchers can choose disease cases that are very different from actual incidence rates, with the result that the relative risk can be very wrong. That is the reason that relative risk is defined for prospective studies only. The odds ratio is defined for prospective and retrospective studies.

Relative Risk: Prospective study

Odds Ratio: Prospective study or retrospective study

Table 4–6 includes the results from a prospective study of 1000 randomly selected sub- jects, which is conducted to investigate the risk of lung cancer from smoking. Table 4-6 contains entries that are realistic based on current incidence rates. However, Table 4-7 below is based on a retrospective study in which the researcher went back in time to find 985 subjects with lung cancer and 985 subjects without lung cancer, so Table 4-7 does not reflect actual incidence rates. See the following.

From Table 4-6: P1lung cancer � smoker2 = 13>180 = 0.0722 From Table 4-7: P1lung cancer � smoker2 = 854>1021 = 0.836

The above probabilities are very different, so both of them cannot be good estimates of the likelihood of getting lung cancer from smoking. The above probability of 0.0722 is a good estimate because it is based on a prospective study with realistic incidence rates, but the probability of 0.836 is a poor estimate because it is based on the retro- spective study designed to include an equal number of subjects with lung cancer and subjects without lung cancer.

Now compare the relative risk values and the odds ratio values from Tables 4-6 and 4-7. See that the odds ratio values are approximately the same, but the relative risk values are dramatically different. The relative risk value of 29.6 from the prospective study is a good measure, but the relative risk of 6.1 from the retrospective study is a poor measure.

If we take advantage of the fact that Table 4-5 does correspond to the generalized Table 4-4 on page 153, then the odds ratio can also be calculated as follows:

Odds ratio = ad bc

= 14572128602 13199212602 = 1.571

I N T E R P R E TAT I O N

This result indicates that the odds in favor of rehospitalization are 1.571 times higher for babies discharged early when compared to those discharged late. This suggests that newborns discharged early are at substantially increased risk of rehos- pitalization.

TABLE 4-6 Prospective Study

RR = 29.6; OR = 31.8

Lung Cancer No Lung Cancer

Smoker 13 167

Nonsmoker 2 818

TABLE 4-7 Retrospective Study

RR = 6.1; OR = 31.9

Lung Cancer No Lung Cancer

Smoker 854 167

Nonsmoker 131 818

Total 985 985

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160 CHAPTER 4 Probability

SUMMARY OF KEY POINTS

Disease No Disease

Treatment a b

Placebo c d

Statistical Literacy and Critical Thinking

1. Notation The relative risk of a characteristic is the ratio pt>pc. What do pt and pc represent? 2.  Relative Risk Identify an important disadvantage of relative risk used with a relatively small difference between the rates in the treatment and control groups.

3.  Number Needed to Treat A measure of the effectiveness of an influenza vaccine is the number needed to treat, which is 37 (under certain conditions). Interpret that number. Does the result apply to every particular group of 37 subjects?

4.  Retrospective ,Prospective The odds ratio is a measure used in retrospective or pro- spective studies. Describe retrospective and prospective studies.

Headaches and Viagra In Exercises 5−12, use the data in the accompanying table (based on data from Pfizer, Inc.). That table describes results from a clinical trial of the drug Viagra. Some subjects were treated with Viagra while others were given a placebo; then headache events were recorded.

Headache No Headache

Viagra Treatment 117 617

Placebo 29 696

5. Type of Study Is the study retrospective or prospective?

4-4 Basic Skills and Concepts

■ Absolute risk reduction = ` a a + b

- c

c + d `

■ Number Needed to Treat (NNT) = 1

Absolute Risk Reduction

■ Actual odds against event A = P1A2 P1A2

■ Actual odds in favor of event A = P1A2 P1A2

Relative risk 1RR2 = pt pc

=

a a + b

c c + d

Odds ratio 1OR2 = ad bc

(for prospective only)

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4-4 Risks and Odds 161

6.  Probability For those in the Viagra treatment group, find the probability that the subject experienced a headache.

7. Comparing Probabilities Compare P(headache � Viagra treatment) and P(headache � placebo).

8. Absolute Risk Reduction Find the value of the absolute risk reduction for headaches in the treatment and placebo groups.

9. Number Needed to Treat Find the number of Viagra users that would need to stop using Viagra in order to prevent a single headache.

10. Odds For those in the Viagra treatment group, find the odds in favor of a headache, then find the odds against a headache.

11. Relative Risk Find the relative risk of a headache for those in the treatment group com- pared to those in the placebo group. Interpret the result.

12.  Odds Ratio Find the odds ratio for headaches in the treatment group compared to the placebo group, then interpret the result. Should Viagra users be concerned about headaches as an adverse reaction?

Clinical Trial of Atorvastatin (Lipitor). In Exercises 13−16, use the data in the accom- panying table that summarizes results from a clinical trial of atorvastatin (based on data from Parke-Davis).

Infection No Infection

Atorvastatin (10 mg) 89 774

Placebo 27 243

13. Absolute Risk Reduction

a. What is the probability of infection in the atorvastatin treatment group?

b. What is the probability of infection in the placebo group?

c. Find the value of the absolute risk reduction for infection in the placebo group and the atorv- astatin treatment group. Write a brief statement interpreting the result.

14. Number Needed to Treat Calculate the number needed to treat and interpret the result.

15. Odds For those who were treated with atorvastatin, find the odds in favor of an infection. Also find the odds in favor of an infection for those given a placebo. Is there much of a differ- ence between these two results?

16. Odds Ratio and Relative Risk Find the odds ratio and relative risk for an infection in the group treated with atorvastatin compared to the placebo group. Based on this result, does atorvastatin appear to increase the risk of an infection? Why or why not?

17. Odds Ratio and Relative Risk In a clinical trial of 2103 subjects treated with Nasonex (mometasone), 26 reported headaches. In a control group of 1671 subjects given a placebo, 22 reported headaches. Find the relative risk and odds ratio for the headache data. What do the results suggest about the risk of a headache from the Nasonex treatment?

18. Design of Experiments You would like to conduct a study to determine the effective- ness of seat belts in saving lives in car crashes.

a. What would be wrong with randomly selecting 2000 drivers, then randomly assigning half of them to a group that uses seat belts and another group that does not wear seat belts?

b. If 2000 drivers are randomly selected and separated into two groups according to whether they use seat belts, what is a practical obstacle in conducting a prospective study of the effec- tiveness of seat belts in car crashes?

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162 CHAPTER 4 Probability

In the biological and health sciences, rates are often used to describe the likelihood of an event. Rates are used by researchers and health professionals to monitor the health status of a community or population. Although any specific time interval could be used, we assume a time interval of one year throughout this section.

4-5 Rates of Mortality, Fertility, and Morbidity

DEFINITION

A rate describes the frequency of occurrence of some event. It is the relative fre- quency of an event, multiplied by some number, typically a value such as 1000 or 100,000. A rate can be expressed as

aa b bk

where

a = frequency count of the number of people for whom the event occurred

b = total number of people exposed to the risk of the event occurring

k = multiplier number, such as 1000 or 100,000

The above general definition is commonly applied to measures of mortality, fertility, and morbidity. For the following rates, mortality refers to deaths, fertility refers to births, and morbidity refers to diseases. Here are additional terms and their meanings:

■ Infants: Babies who were born alive

■ Neonates: Infants under the age of 28 days

■ Fetal Death: Occurs when a fetus is delivered without life after 20 weeks of gestation

■ Neonatal Death: Occurs when an infant dies under 28 days of age

Mortality Rates

Crude 1or unadjusted2 mortality rate = a deaths population size

bk

Infant mortality rate = adeaths of infants under 1 year of age number of live births

bk

Neonatal mortality rate = adeaths of infants under 28 days of age number of live births

bk Fetal mortality rate =

a fetuses delivered without life after 20 weeks of gestation number of live births + fetuses delivered without life after 20 weeks of gestation

bk

Perinatal mortality rate = afetal deaths + neonatal deaths live births + fetal deaths

bk

Fertility Rates

Crude birthrate = a live births population size

bk

General fertility rate = a live births number of women aged 15 - 44

bk

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4-5 Rates of Mortality, Fertility, and Morbidity 163

Morbidity (Disease) Rates

Incidence rate = areported new cases of disease population size

bk

Prevalence rate = anumber of people with disease at a given time population size at the given point in time

bk

EXAMPLE 1 Crude Mortality Rate

For a recent year in the United States, there were 2,515,458 deaths in a population of 312,799,495 people. Use those values with a multiplier of 1000 to find the crude mortality rate.

S O L U T I O N

With 2,515,458 people who died, with 312,799,495 people in the population, and letting k = 1000, we compute the crude mortality rate as follows:

Crude mortality rate = a deaths population size

bk = a 2,515,458 312,799,495

b1000

= 8.0 1rounded2

I N T E R P R E TAT I O N

For this particular year, the death rate is 8.0 people for each 1000 people in the population. Using the relative frequency definition of probability given in Section 4-1, we might also say that for a randomly selected person, the probability of death in this year is 2,515,458>312,799,495 = 0.00804. One important advantage of the mortality rate of 8.0 people (per 1000 people in the population) is that it results in a value that uses fewer decimal places and is generally easier to use and understand.

EXAMPLE 2 Infant Mortality Rate

The infant mortality rate is a very important measure of the health of a region. (According to the United Nations, the worldwide infant mortality rate is 49.4 per 1000 live births.) For a recent year, there were 3,953,590 live births in the United States, and there were 23,910 deaths of infants under 1 year of age. Using a mul- tiplying factor of k = 1000, find the infant mortality rate of the United States and compare it to the rate of 2.1 for Japan.

S O L U T I O N

The infant mortality rate is computed as shown below.

Infant mortality rate = adeaths of infants under 1 year of age number of live births

bk

= a 23,910 3,953,590

b1000

= 6.0 1rounded2

continued

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164 CHAPTER 4 Probability

A crude rate, as defined, is a single value based on crude totals. When compar- ing two different regions, such as Florida and Colorado, a comparison of rates can be misleading because of differences in factors such as age that might affect the rates. In a recent year, the crude mortality rates (per 1000 population) were 9.1 in Florida and 6.4 in Colorado. This is not too surprising, considering that in Florida, roughly 18% of the population is over the age of 65, compared to only 11% for Colorado. The higher mortality rate for Florida does not mean that Florida is less healthy; in this case, it appears that Florida has a higher death rate largely because it has a higher proportion of older residents. Instead of using crude rates, we might use either specific rates or adjusted rates.

Specific rates are rates specific for some particular group, such as people aged 18–24, or rates specific for some particular cause of death, such as deaths due to myo- cardial infarction.

Adjusted rates involve calculations that can be quite complicated, but they basi- cally make adjustments for important factors, such as age, gender, or race.

Because age is the characteristic that typically affects mortality the most, it is the most common factor used as the basis for adjustment. Calculations of adjusted rates involve the creation of a theoretical standardized population that is used for the re- gions being compared. A population of 1,000,000 people with the same composition as the United States is often used as the standardized population. Adjusted rates are valuable for comparing different regions, but they do not necessarily reflect the true crude death rates. Adjusted rates should not be used as death rates and they should not be compared to crude rates.

When assessing the accuracy of rates, we should consider the source. Mortality rates found in a source such as the Statistical Abstract of the United States (compiled by the U.S. Bureau of the Census) are likely to be quite accurate, because each state now has a mandatory death reporting system, although official government reports seem to take years to produce. However, morbidity rates are likely to be less accurate, because some diseases are known to be underreported or not reported at all. Some morbidity rates might be the result of very questionable surveys. However, some sur- veys, such as the annual National Health Survey, involve large samples of people who are very carefully chosen, so that results are likely to be very accurate.

Statistical Literacy and Critical Thinking

1. Birth Rate The birth rate in China is 12.3 per 1000. What exactly does that mean?

2. Rates Exercise 1 describes the birth rate in China as 12.3 per 1000. Another way to describe the birth rate is to give the rate as a proportion or probability of 0.0123. What advantage does the rate of “12.3 per 1000” have over the rate expressed as 0.0123?

3.  Expected Births Given that China has a birth rate of 12.3 per 1000 and a population of 1,360,762,587, about how many births are expected in a year?

4. Incidence and Prevalence What is the difference between a disease incidence rate and a disease prevalence rate?

4-5 Basic Skills and Concepts

I N T E R P R E TAT I O N

The infant mortality rate of 6.0 deaths per 1000 infants under 1 year of age is sub- stantially greater than the infant mortality rate of 2.1 in Japan.

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4-5 Rates of Mortality, Fertility, and Morbidity 165

Finding Rates. In Exercises 5−12, use the data in the accompanying table (based on data for a recent year from various sources, including the U.S. Census Bureau and the National Institutes of Health) to find the indicated rates. Round results to one decimal place, and use a multiplying factor of k = 1000 unless indicated otherwise.

Vital Statistics for the United States in One Year

Population: 312,799,495 Deaths: 2,515,458

Women aged 15–44: 61,488,227 Motor vehicle deaths: 33,783

Live births: 3,953,590 Fetuses delivered without life after 20 weeks of gestation: 26,148

Deaths of infants under 1 year of age: 23,910 Deaths of infants under 28 days of age: 15,973

HIV-infected persons: 1,155,792 Deaths from HIV infections: 7683

5. Find the neonatal mortality rate.

6. Find the fetal mortality rate.

7. Find the perinatal mortality rate.

8. Find the crude birth rate.

9. Find the general fertility rate.

10. Using a multiplier of k = 100,000, find the motor vehicle death incidence rate.

11. Find the HIV infection prevalence rate.

12. Find the HIV infection mortality rate for HIV-infected persons.

13. Finding Probability An example in this section involved the crude mortality rate, which was found to be 8.0 persons per 1000 population. Find the probability of randomly selecting someone and getting a person who died within the year. What advantage does the crude mortal- ity rate have over the probability value?

14. Finding Probability The crude death rate for China was recently 7.4, and that rate was computed using a multiplier of k = 1000.

a. Find the probability that a randomly selected Chinese person died within the year.

b. If two Chinese people are randomly selected, find the probability that they both died within the year, and express the result using three significant digits.

c. If two Chinese people are randomly selected, find the probability that neither of them died within the year, and express the result using three significant digits.

15. Finding Probability The crude death rate for Spain was recently 8.3, and that rate was computed using a multiplier of k = 1000.

a. Find the probability that a randomly selected Spaniard died within the year.

b. If two Spaniards are randomly selected, find the probability that they both died within the year, and express the result using three significant digits.

c. If two Spaniards are randomly selected, find the probability that at least one of them sur- vived the year, and express the result using six decimal places. What would be wrong with expressing the answer using three significant digits?

continued

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166 CHAPTER 4 Probability

16. Finding Probability In a recent year in the United States, there were 787,650 deaths due to cardiovascular disease, and the population was 312,799,495.

a. Find the crude mortality rate for cardiovascular disease. (This result is sometimes called the cause-specific death rate.)

b. Find the probability that a randomly selected person died of cardiovascular disease and express the result using three significant digits.

c. Find the probability that when three people are randomly selected, none of them died because of cardiovascular disease.

17.  Cause-of-Death Ratio In a recent year in the United States, there were 2,515,458 deaths, and 787,650 of them were due to cardiovascular disease. The cause-of-death ratio is expressed as follows:

adeaths due to specific disease total number of deaths

bk where k = 100

a. Find the cause-of-death ratio for cardiovascular disease.

b. If three of the deaths are randomly selected, find the probability that none of them are due to cardiovascular disease.

18. Crude Mortality Rates The table below lists numbers of deaths and population sizes for different age groups for Florida and the United States for a recent year.

a. Find the crude mortality rate for Florida and the crude mortality rate for the United States. Al- though we should not compare crude mortality rates, what does the comparison suggest in this case?

b. Using only the age group of 65 and older, find the mortality rates for Florida and the United States. Compare the results.

c. What percentage of the Florida population is made up of people aged 65 and older? What is the percentage for the United States? What do the results suggest about the crude mortality for Florida compared to the United States?

Age

0–24 25–64 65 and older

Florida deaths 3625 39,820 129,395

Florida population 5,716,861 9,842,031 3,375,303

U.S. deaths 63,208 619,982 1,832,268

U.S. population 103,542,603 163,960,163 45,296,729

19. Number of Deaths The number of deaths in the United States has been steadily increas- ing each year. Does this mean that the health of the nation is declining? Why or why not?

20. Comparing Rates In a recent year, the crude mortality rate of the United States was 8.0 (per 1000 population), and the corresponding crude mortality rate for China was 7.4. What is a major problem with comparing the crude mortality rates of the United States and China?

21. Adjusted Mortality Rate Refer to the data listed in Exercise 18. Change the Florida popula- tion sizes for the three age categories so that they fit the same age distribution as the U.S. popu- lation. Next, adjust the corresponding numbers of deaths proportionately. (Use the same Florida mortality rates for the individual age categories, but apply those rates to the adjusted population sizes.) Finally, compute the Florida mortality rate using the adjusted values. The result is a mortal- ity rate adjusted for the variable of age. (Better results could be obtained by using more age catego- ries.) How does this adjusted mortality rate for Florida compare to the mortality rate for the United States? (Note: There are other methods for computing adjusted rates than the one used here.)

4-5 Beyond the Basics

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4-6 Counting 167

MULTIPLICATION COUNTING RULE: For a sequence of events in which the first event can occur n1 ways, the second event can occur n2 ways, the third event can occur n3 ways, and so on, the total number of possibilities is n1 # n2 # n3 . . ..

EXAMPLE 1 Multiplication Counting Rule: DNA

In a linear triplet of three DNA nucleotides, each of the nucleotides can be any one of these four bases (with repetition allowed): A (adenine); C (cytosine); G (guanine); T (thymine). Two different examples of triplets are CTA and TTG. What is the total number of different possible triplets? Given that the four nucleotides are equally likely, what is the probability of getting the triplet of AAA?

S O L U T I O N

There are 4 different possibilities for each of the three nucleotides, so the total num- ber of different possible triplets is n1 # n2 # n3 = 4 # 4 # 4 = 64.

If the four nucleotides are equally likely, the probability of getting the triplet of AAA is 1>64 or 0.0156.

2. Factorial Rule

The factorial rule is used to find the total number of ways that n different items can be rearranged with different arrangements of the same items counted separately. The factorial rule uses the following notation.

NOTATION

The factorial symbol (!) denotes the product of decreasing positive whole num- bers. For example, 4! = 4 # 3 # 2 # 1 = 24. By special definition, 0! = 1.

FACTORIAL RULE The number of different arrangements (order matters) of n different items when all n of them are selected is n!.

The factorial rule is based on the principle that the first item may be selected n differ- ent ways, the second item may be selected n - 1 ways, and so on.

Routing problems often involve applications of the factorial rule, as in the follow- ing example.

Key Concept Probability problems typically require that we know the total number of simple events, but finding that number often requires one of the five rules presented in this section. In Section 4-2, with the addition rule, multiplication rule, and conditional probability, we encouraged intuitive rules based on understanding and we discour- aged blind use of formulas, but this section requires much greater use of formulas as we consider five different methods for counting the number of possible outcomes in a variety of situations. Not all counting problems can be solved with these five methods, but they do provide a strong foundation for the most common real applications.

1. Multiplication Counting Rule

The multiplication counting rule is used to find the total number of possibilities from some sequence of events.

4-6 Counting

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168 CHAPTER 4 Probability

Permutations and Combinations: Does Order Count?

When using different counting methods, it is essential to know whether different ar- rangements of the same items are counted only once or are counted separately. The terms permutations and combinations are standard in this context, and they are de- fined as follows:

EXAMPLE 2 Factorial Rule: Travel Itinerary

Quest Diagnostics collects blood specimens from different laboratories. A driver is dispatched to make collections at 5 different locations. How many different routes are possible?

S O L U T I O N

For those 5 different locations, the number of different routes is 5! = 5 # 4 # 3 # 2 # 1 = 120.

Note that this solution could have been done by applying the multiplica- tion counting rule. The first stop can be any one of the 5 locations, the second stop can be any one of the 4 remaining locations, and so on. The result is again 5 # 4 # 3 # 2 # 1 = 120. Use of the factorial rule has the advantage of including the factorial symbol, which is sure to impress.

DEFINITIONS

Permutations of items are arrangements in which different sequences of the same items are counted separately. (The letter arrangements of abc, acb, bac, bca, cab, and cba are all counted separately as six different permutations.)

Combinations of items are arrangements in which different sequences of the same items are counted as being the same. (The letter arrangements of abc, acb, bac, bca, cab, and cba are all considered to be the same single combination.)

Mnemonics for Permutations and Combinations

■ Remember “Permutations Position,” where the alliteration reminds us that with permutations, the positions of the items makes a difference.

■ Remember “Combinations Committee,” which reminds us that with members of a committee, rearrangements of the same members result in the same committee, so order does not count.

3. Permutations Rule (When All of the Items Are Different)

The permutations rule is used when there are n different items available for selection, we must select r of them without replacement, and the sequence of the items matters. The result is the total number of arrangements (or permutations) that are possible. (Re- member: Rearrangements of the same items are counted as different permutations.)

PERMUTATIONS RULE: When n different items are available and r of them are selected without replacement, the number of different permutations (order counts) is given by

nPr = n!

1n - r2!

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4-6 Counting 169

4. Permutations Rule (When Some Items Are Identical to Others)

When n items are all selected without replacement, but some items are identical, the number of possible permutations (order matters) is found by using the following rule.

EXAMPLE 3 Permutations Rule (with Different Items): Clinical Trial of New Drug

When testing a new drug, Phase I requires only 5 volunteers, and the objective is to assess the drug’s safety. To be very cautious, we plan to treat the 5 subjects in sequence, so that any particularly adverse effect can allow us to stop the treatments before any other subjects are treated. If 8 volunteers are available, how many differ- ent sequences of 5 subjects are possible?

S O L U T I O N

We need to select r = 5 subjects from n = 8 volunteers that are available. The num- ber of different sequences of arrangements is found as shown:

nPr = n!

1n - r2! = 8!

18 - 52! = 6720

There are 6720 different possible arrangements of 5 subjects selected from the 8 that are available.

PERMUTATIONS RULE (WHEN SOME ITEMS ARE IDENTICAL TO OTHERS)

The number of different permutations (order counts) when n items are available and all n of them are selected without replacement, but some of the items are iden- tical to others, is found as follows:

n! n1!n2! . . . nk!

where n1 are alike, n2 are alike,…, and nk are alike.

EXAMPLE 4 Permutations Rule (with Some Identical Items): Designing Surveys

When designing surveys, pollsters sometimes repeat a question to see if a subject is thoughtlessly providing answers just to finish quickly. For one particular survey with 10 questions, 2 of the questions are identical to each other, and 3 other ques- tions are also identical to each other. For this survey, how many different arrange- ments are possible? Is it practical to survey enough subjects so that every different possible arrangement is used?

S O L U T I O N

We have 10 questions with 2 that are identical to each other and 3 others that are also identical to each other, and we want the number of permutations. Using the rule for permutations with some items identical to others, we get

n! n1!n2! . . . nk!

= 10! 2!3!

= 3,628,800

2 # 6 = 302,400

continued

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170 CHAPTER 4 Probability

5. Combinations Rule

The combinations rule is used when there are n different items available for selection, only r of them are selected without replacement, and order does not matter. The result is the total number of combinations that are possible. (Remember: Rearrangements of the same items are considered to be the same combination.)

I N T E R P R E TAT I O N

There are 302,400 different possible arrangements of the 10 questions. It is not practical to accommodate every possible permutation. For typical surveys, the num- ber of respondents is somewhere around 1000.

COMBINATIONS RULE:

When n different items are available, but only r of them are selected without replace- ment, the number of different combinations (order does not matter) is found as follows:

nCr = n!

1n - r2!r!

EXAMPLE 5 Combinations Rule: Phase I of a Clinical Trial

When testing a new drug on humans, a clinical test is normally done in three phases. Phase I is conducted with a relatively small number of healthy volunteers. Assume that we want to treat 20 healthy humans with a new drug, and we have 30 suitable volunteers available. If 20 subjects are selected from the 30 that are available, and the 20 selected subjects are all treated at the same time, how many different treatment groups are possible?

S O L U T I O N

Because all subjects are treated at the same time, order is irrelevant, so we need to find the number of different possible combinations. With n = 30 subjects available and with r = 20 subjects selected, the number of combinations is found as follows.

nCr = n!

1n - r2!r! = 30!

130 - 202!20! = 30!

10! # 20! = 30,045,015

I N T E R P R E TAT I O N

There are 30,045,015 different possible combinations.

Permutations or Combinations? Because choosing between permutations and com- binations can often be tricky, we provide the following example that emphasizes the difference between them.

EXAMPLE 6 Permutations and Combinations: Officers and Committees

The Portland Medical Center must appoint three corporate officers: chief executive officer (CEO), executive chairperson, and chief operating officer (COO). It must also appoint a planning committee with three different members. There are eight qualified candidates, and officers can also serve on the planning committee.

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4-6 Counting 171

a. How many di�erent ways can the o¸cers be appointed?

b. How many di�erent ways can the committee be appointed?

S O L U T I O N

Note that in part (a), order is important because the officers have very different functions. However, in part (b), the order of selection is irrelevant because the com- mittee members all serve the same function.

a. Because order does count, we want the number of permutations of r = 3 people selected from the n = 8 available people. We get

nPr = n!

1n - r2! = 8!

18 - 32! = 336

b. Because order does not count, we want the number of combinations of r = 3 people selected from the n = 8 available people. We get

nCr = n!

1n - r2!r! = 8!

18 - 32!3! = 56

With order taken into account, there are 336 different ways that the officers can be appointed, but without order taken into account, there are 56 different possible com- mittees.

Statistical Literacy and Critical Thinking

1.  Notation What does the symbol ! represent? Six different patients can be scheduled for X-ray films 6! different ways, so what is the actual number of ways that six people can be scheduled for X-ray films?

2.  Permutations ,Combinations What is the basic difference between permutations and combinations?

3. Notation Evaluate 9C4. What does the result represent?

4. Notation Evaluate 9P4. What does the result represent?

In Exercises 5–30, express all probabilities as fractions.

5. Pin Numbers The Kinsale Medical Supply Company issues pin numbers to its employees so that they can access an online database. A hacker must randomly guess the correct pin code for the Information Technology supervisor, and that pin code consists of four digits (each 0 through 9) that must be entered in the correct order. Repetition of digits is allowed. What is the probability of a correct guess on the first try?

6. Social Security Numbers A Social Security number consists of nine digits in a particular order, and repetition of digits is allowed. After seeing the last four digits printed on a receipt, if you randomly select the other digits, what is the probability of getting the correct Social Secu- rity number of the person who was given the receipt?

7.  Assigning Shifts The staff supervisor at the Wellington Medical Center must assign a team of two physicians to work the emergency room on Saturday night. If there are 19 physi- cians available and two of them are randomly selected, what is the probability of getting the two youngest physicians?

4-6 Basic Skills and Concepts

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172 CHAPTER 4 Probability

8. Review Board The supervisor at the Wellington Medical Center must select three nurses from 11 who are available for a review board. How many different ways can that be done?

9. Blood Test Quest Diagnostics has just received 8 different blood samples. If they are tested in random order, what is the probability that they are tested in the alphabetical order of the sub- jects who provided the samples?

10. Radio Station Call Letters If radio station call letters must begin with either K or W and must contain a total of either three or four letters, how many different possibilities are there?

11. Scheduling Routes A new director of the Veterans Health Administration plans to visit one hospital in each of five different states. If the five states are randomly selected from all 50 states without replacement and the order is also random, what is the probability that she visits Idaho, Oregon, Alaska, New Jersey, and Ohio, in that order?

12. Survey Reliability A health survey with 12 questions is designed so that 3 of the ques- tions are identical and 4 other questions are identical (except for minor changes in wording). How many different ways can the 12 questions be arranged?

13. Safety with Numbers A safe “combination” consists of four numbers between 0 and 99, and the safe is designed so that numbers can be repeated. If someone tries to gain access to the safe, what is the probability that he or she will get the correct combination on the first attempt? Assume that the numbers are randomly selected. Given the number of possibilities, does it seem feasible to try opening the safe by making random guesses for the combination?

14. Electricity The control panel for an MRI device uses five color-coded wires. If we trou- bleshoot by testing two wires at a time, how many different tests are required for every possible pairing of two wires?

15.  Clinical Trial In a clinical trial of the drug atorvastatin (Lipitor), one group of subjects was given placebos, a second group was given treatments of 10 mg, a third group was given treatments of 20 mg, a fourth group was given treatments of 40 mg, and a fifth group was given treatments of 80 mg. If the Phase I trial involved 15 subjects randomly assigned to the five groups with three in each group, how many different ways can the groups be formed?

16.  Emergency Room Instead of treating emergency room patients in the order that they arrive, it is common to treat those with more serious problems first. If an emergency room has seven different patients, how many ways can they be arranged in sequence?

17. ZIP Code If you randomly select five digits, each between 0 and 9, with repetition allowed, what is the probability you will get the ZIP code of the Secretary of Health and Human Services?

18. FedEx Deliveries With a short time remaining in the day, a FedEx driver has time to make deliveries at 6 locations among the 9 locations remaining. How many different routes are possible?

19. Phone Numbers Current rules for telephone area codes allow the use of digits 2–9 for the first digit and 0–9 for the second and third digits. How many different area codes are pos- sible with these rules? That same rule applies to the exchange numbers, which are the three digits immediately preceding the last four digits of a phone number. Given both of those rules, how many ten-digit phone numbers are possible? Given that these rules apply to the United States and Canada and a few islands, are there enough possible phone numbers? (Assume that the combined population is about 400,000,000.)

20. Classic Counting Problem A classic counting problem is to determine the number of different ways that the letters of “Mississippi” can be arranged. Find that number.

21.  Corporate Officers and Committees The Newport Medical Supply Company must appoint a president, chief executive officer (CEO), chief operating officer (COO), and chief financial officer (CFO). It must also appoint a strategic planning committee with four different members. There are 10 qualified candidates, and officers can also serve on the committee.

a. How many different ways can the four officers be appointed?

b. How many different ways can a committee of four be appointed? continued

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4-6 Counting 173

c. What is the probability of randomly selecting the committee members and getting the four youngest of the qualified candidates?

22. Card Access You have an identification card used for access to a secure area of the Wel- lington Medical Center. It’s dark and you can’t see your card when you insert it. The card must be inserted with the front side up and the printing configured so that the beginning of your name enters first.

a. What is the probability of selecting a random position and inserting the card with the result that the card is inserted correctly?

b. What is the probability of randomly selecting the card’s position and finding that it is incor- rectly inserted on the first attempt, but it is correctly inserted on the second attempt?

c. How many random selections are required to be absolutely sure that the card works because it is inserted correctly?

23. Amino Acids With 8 different amino acids available, 5 are to be selected to form a chain (called a polypeptide chain) in which order counts. How many different chains are possible?

24. Identity Theft with Credit Cards Credit card numbers typically have 16 digits, but not all of them are random.

a. What is the probability of randomly generating 16 digits and getting your MasterCard number?

b. Receipts often show the last four digits of a credit card number. If only those last four digits are known, what is the probability of randomly generating the other digits of your MasterCard number?

c. Discover cards begin with the digits 6011. If you know that the first four digits are 6011 and you also know the last four digits of a Discover card, what is the probability of randomly gener- ating the other digits and getting all of them correct? Is this something to worry about?

25. What a Word! One of the longest words in standard statistics terminology is “homosce- dasticity.” How many ways can the letters in that word be arranged?

26. Phase I of a Clinical Trial A clinical test on humans of a new drug is normally done in three phases. Phase I is conducted with a relatively small number of healthy volunteers. For example, a Phase I test of bexarotene involved only 14 subjects. Assume that we want to treat 14 healthy humans with this new drug and we have 16 suitable volunteers available.

a. If the subjects are selected and treated one at a time in sequence, how many different sequen- tial arrangements are possible if 14 people are selected from the 16 that are available?

b. If 14 subjects are selected from the 16 that are available, and the 14 selected subjects are all treated at the same time, how many different treatment groups are possible?

c. If 14 subjects are randomly selected and treated at the same time, what is the probability of selecting the 14 youngest subjects?

27. Lightning and Lottery As of this writing, the Mega Millions lottery is run in 44 states. Winning the jackpot requires that you select the correct five different numbers between 1 and 75 and, in a separate drawing, you must also select the correct single number between 1 and 15. Find the probability of winning the jackpot if you buy one ticket. How does the result compare to the probability of being struck by lightning in a year, which the National Weather Service estimates to be 1>960,000? 28. Designing Experiment Clinical trials of Nasonex involved a group given placebos and another group given treatments of Nasonex. Assume that a preliminary Phase I trial is to be conducted with 12 subjects, including 6 men and 6 women. If 6 of the 12 subjects are randomly selected for the treatment group, find the probability of getting 6 subjects of the same gender. Would there be a problem with having members of the treatment group all of the same gender?

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174 CHAPTER 4 Probability

29. Morse Codes The International Morse code is a way of transmitting coded text by using sequences of on>off tones. Each character is 1 or 2 or 3 or 4 or 5 segments long, and each seg- ment is either a dot or a dash. For example, the letter G is transmitted as two dashes followed by a dot, as in — — •. How many different characters are possible with this scheme? Are there enough characters for the alphabet and numbers?

30.  Mendel’s Peas Mendel conducted some his famous experiments with peas that were either smooth yellow plants or wrinkly green plants. If four peas are randomly selected from a batch consisting of four smooth yellow plants and four wrinkly green plants, find the probabil- ity that the four selected peas are of the same type.

31. Computer Variable Names A common computer programming rule is that names of variables must be between one and eight characters long. The first character can be any of the 26 letters, while successive characters can be any of the 26 letters or any of the 10 digits. For example, allowable variable names include A, BBB, and M3477K. How many different vari- able names are possible? (Ignore the difference between uppercase and lowercase letters.)

32. Handshakes

a. Five physicians gather for a meeting about a patient. If each physician shakes hands with each other physician exactly once, what is the total number of handshakes?

b. If n physicians shake hands with each other exactly once, what is the total number of hand- shakes?

c. How many different ways can five physicians be seated at a round table? (Assume that if everyone moves to the right, the seating arrangement is the same.)

d. How many different ways can n physicians be seated at a round table?

4-6 Beyond the Basics

1. Standard Tests Standard tests, such as the MCAT, tend to make extensive use of multiple- choice questions because they are easy to grade using software. If one such multiple-choice question has possible correct answers of a, b, c, d, e, what is the probability of a wrong answer if the answer is a random guess?

2.  Likelihood of Disease After obtaining a patient’s positive test result, a physician con- cludes that there is a 30% chance that the subject has a disease. What is the probability that the subject does not have the disease?

3. Months If a month is randomly selected after mixing the pages from a calendar, what is the probability that it is a month containing the letter y?

4.  Sigmoidoscopy ,Colonoscopy Based on data from the Centers for Disease Control, 67.7% of males over the age of 50 have had a sigmoidoscopy or colonoscopy. If two males over the age of 60 are randomly selected, what is the probability that they both have had a sigmoid- oscopy or colonoscopy?

5. Subjective Probability Estimate the probability that the next time you get a cut, it requires stitches.

Chapter Quick Quiz

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In Exercises 6–10, use the following results from tests of an experiment to test the effective- ness of an experimental vaccine for children (based on data from USA Today). Express all probabilities in decimal form.

Developed Flu Did Not Develop Flu

Vaccine Treatment 14 1056

Placebo 95 437

6. If 1 of the 1602 subjects is randomly selected, find the probability of getting 1 that devel- oped flu.

7. If 1 of the 1602 subjects is randomly selected, find the probability of getting 1 who had the vaccine treatment or developed flu.

8. If 1 of the 1602 subjects is randomly selected, find the probability of getting 1 who had the vaccine treatment and developed flu.

9. Find the probability of randomly selecting 2 subjects without replacement and finding that they both developed flu.

10. Find the probability of randomly selecting 1 of the subjects and getting 1 who developed flu, given that the subject was given the vaccine treatment.

In Exercises 1–10, use the data in the accompanying table and express all results in decimal form. (The results are based on “Splinting vs Surgery in the Treatment of Carpal Tunnel Syn- drome,” by Gerritsen et al., Journal of the American Medical Association, Vol. 288, No. 10.)

Treatment for Carpal Tunnel Syndrome

Successful Treatment Unsuccessful Treatment

Splint Treatment 60 23

Surgery Treatment 67 6

1. Success If 1 of the patients is randomly selected, find the probability of selecting someone with a successful treatment.

2. Success Find the probability of randomly selecting a patient and getting one with a suc- cessful treatment, given that the patient was treated with splinting.

3. Success Find the probability of randomly selecting a patient and getting one with a suc- cessful treatment, given that the patient was treated with surgery.

4. Success or Surgery If 1 of the patients is randomly selected, find the probability of get- ting a patient who had a successful treatment or was treated with surgery.

5. No Success or Splint If 1 of the patients is randomly selected, find the probability of get- ting someone who had an unsuccessful treatment or was treated with a splint.

6. Both Successful If 2 patients are randomly selected without replacement, find the prob- ability that they both had successful treatments.

7. Both Successful If 2 patients are randomly selected with replacement, find the probability that they both had successful treatments.

8. Complement If A represents the event of randomly selecting one patient included in the table and getting someone who was treated with surgery, what does A represent? Find the value of P1A2.

Review Exercises

CHAPTER 4 Review Exercises 175

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176 CHAPTER 4 Probability

9. Complement If A represents the event of randomly selecting one patient included in the table and getting someone who had a successful treatment, what does A represent? Find the value of P1A2. 10. All Three Successful If 3 patients are randomly selected without replacement, find the probability that all three had successful treatments.

11. Vision Correction About 75% of the U.S. population uses some type of vision correction (such as glasses or contact lenses).

a. If someone is randomly selected, what is the probability that he or she does not use vision correction?

b. If four different people are randomly selected, what is the probability that they all use vision correction?

c. Would it be unlikely to randomly select four people and find that they all use vision correc- tion? Why or why not?

12. National Statistics Day

a. If a person is randomly selected, find the probability that his or her birthday is October 18, which is National Statistics Day in Japan. Ignore leap years.

b. If a person is randomly selected, find the probability that his or her birthday is in October. Ignore leap years.

c. Estimate a subjective probability for the event of randomly selecting an adult American and getting someone who knows that October 18 is National Statistics Day in Japan.

d. Is it unlikely to randomly select an adult American and get someone who knows that Octo- ber 18 is National Statistics Day in Japan?

13. Composite Sampling for Diabetes Currently, the rate for new cases of diabetes in a year is 3.4 per 1000 (based on data from the Centers for Disease Control and Prevention). When testing for the presence of diabetes, the Portland Diagnostics Laboratory saves money by com- bining blood samples for tests. The combined sample tests positive if at least one person has diabetes. If the combined sample tests positive, then the individual blood tests are performed. In a test for diabetes, blood samples from 10 randomly selected subjects are combined. Find the probability that the combined sample tests positive with at least 1 of the 10 people having diabetes. Is it likely that such combined samples test positive?

14. Redundancy Using battery-powered alarm clocks, it is estimated that the probability of failure on any given day is 1>1000. a. What is the probability that the alarm clock works for an important event?

b. When using two alarm clocks for an important event, what is the probability that at least one of them works?

Cumulative Review Exercises

1. Fatal Drunk Driving Listed below are the blood alcohol concentrations (g>dL) of drivers convicted of drunk driving in fatal car crashes (based on data from the National Highway Traf- fic Safety Administration).

0.09 0.11 0.11 0.13 0.14 0.15 0.17 0.17 0.18 0.18 0.23 0.35

Find the value of the following statistics and include appropriate units.

a. mean b. median c. midrange d. range

e. standard deviation f. variance

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2. Fatal Drunk Driving Use the same data given in Exercise 1.

a. Identify the 5-number summary and also identify any values that appear to be outliers.

b. Construct a boxplot. c. Construct a stemplot.

3. Organ Donors USA Today provided information about a survey (conducted for Donate Life America) of 5100 adult Internet users. Of the respondents, 2346 said they are willing to donate organs after death. In this survey, 100 adults were surveyed in each state and the District of Co- lumbia, and results were weighted to account for the different state population sizes.

a. What percentage of respondents said that they are willing to donate organs after death?

b. Based on the poll results, what is the probability of randomly selecting an adult who is will- ing to donate organs after death?

c. What term is used to describe the sampling method of randomly selecting 100 adults from each state and the District of Columbia?

4. Sampling Eye Color Based on a study by Dr. P. Sorita Soni, eye colors in the United States are as follows: 40% brown, 35% blue, 12% green, 7% gray, 6% hazel.

a. A statistics instructor collects eye color data from her students. What is the name for this type of sample?

b. Identify one factor that might make the sample from part (a) biased and not representative of the general population of people in the United States.

c. What is the probability that a randomly selected person will have brown or blue eyes?

d. If two people are randomly selected, what is the probability that at least one of them has brown eyes?

5. Blood Pressure and Platelets Given below are the systolic blood pressure measurements (mm Hg) and blood platelet counts (1000 cells>mL) of the first few subjects included in Data Set 1 “Body Data” in Appendix B. Construct a graph suitable for exploring an association be- tween systolic blood pressure and blood platelet count. What does the graph suggest about that association?

Systolic 100 112 134 126 114 134 118 138 114 124

Platelet 319 187 297 170 140 192 191 286 263 193

CHAPTER 4 Technology Project 177

Simulations Calculating probabilities are sometimes painfully difficult, but simulations pro- vide us with a very practical alternative to calculations based on formal rules. A simulation of a procedure is a process that behaves the same way as the procedure so that similar results are produced. Instead of calculating the probability of getting exactly 5 boys in 10 births, you could repeatedly toss 10 coins and count the number of times that exactly 5 heads (or simulated “boys”) occur. Better yet, you could do the simulation with a random number generator on a computer or calculator to randomly generate 1s (or simulated “boys”) and 0s (or simulated “girls”). Let’s consider this probability exercise:

Find the probability that among 50 randomly selected people, at least 3 have the same birthday.

For the above problem, a simulation begins by representing birthdays by integers from 1 through 365, where 1 represents a birthday of January 1, and 2 represents January 2, and so on. We can simulate 50 birthdays by using a calculator or computer to generate 50

Technology Project

continued

M04_TRIO9015_02_SE_C04_118-179.indd 177 30/07/16 11:35 AM

178 CHAPTER 4 Probability

random numbers (with repetition allowed) between 1 and 365. Those numbers can then be sorted, so it becomes easy to examine the list to determine whether any 3 of the simulated birth dates are the same. (After sorting, equal numbers are adjacent.) We can repeat the process as many times as we wish, until we are satisfied that we have a good estimate of the probability. Use technology to simulate 20 different groups of 50 birthdays. Use the results to estimate the probability that among 50 randomly selected people, at least 3 have the same birthday.

Summary of Simulation Functions:

Statdisk: Select Data from the top menu, select Uniform Generator from the dropdown menu.

Excel: Click Insert Function fx, select Math & Trig, select RANDBETWEEN. Copy to additional cells.

TI-83 ,84 Plus: Press L, select PROB from the top menu, select randInt from the menu. StatCrunch: Select Data from the top menu, select Simulate from the dropdown

menu, select Discrete Uniform from the submenu.

Minitab: Select Calc from the top menu, select Random Data from the drop- down menu, select Integer from the submenu.

FROM DATA TO DECISION

Critical Thinking: Interpreting results from a test for smoking It is estimated that roughly half of patients who smoke lie when asked if they smoke. Pulse CO-oximeters may be a way to get information about smoking without relying on pa- tients’ statements. Pulse CO-oximeters use light that shines through a fingernail, and it measures carboxyhemoglobin (carbon monoxide in blood). These devices are used by fire-

men and emergency departments to detect carbon monoxide poisoning, but they can also be used to identify smokers. The accompanying table lists results from people aged 18–44 when the pulse CO-oximeter is set to detect a 6% or higher level of carboxyhemoglobin (based on data from “Carbon Monoxide Test Can Be Used to Identify Smoker,” by Patrice Wendling, Internal Medicine News, Vol. 40., No. 1, and Centers for Disease Control and Prevention).

CO-Oximetry Test for Smoking

Positive Test Result Negative Test Result

Smoker 49 57

Nonsmoker 24 370

Analyzing the Results

1. False Positive Based on the results in the table, find the probability that a subject is not a smoker, given that the test result is positive.

2. True Positive Based on the results in the table, find the probability that a subject smokes, given that the test result is positive.

3. False Negative Based on the results in the table, find the probability that a subject smokes, given that the test result is negative.

4. True Negative Based on the results in the table, find the probability that a subject does not smoke, given that the test result is negative.

5. Sensitivity Find the sensitivity of the test by finding the probability of a true positive, given that the subject actually smokes.

6. Specificity Find the specificity of the test by finding the probability of a true negative, given that the subject does not smoke.

7. Positive Predictive Value Find the positive predictive value of the test by finding the probability that the subject smokes, given that the test yields a positive result.

8. Negative Predictive Value Find the negative predictive value of the test by finding the probability that the subject does not smoke, given that the test yields a negative result.

9. Confusion of the Inverse Find the following values, then compare them. In this case, what is confusion of the inverse?

• P(smoker � positive test result) • P(positive test result � smoker)

M04_TRIO9015_02_SE_C04_118-179.indd 178 13/09/16 5:45 PM

Cooperative Group Activities

1. In-class activity Divide into groups of three or four and use coin flipping to develop a simula- tion that emulates the kingdom that abides by this decree: After a mother gives birth to a son, she will not have any other children. If this decree is followed, does the proportion of girls increase?

2. In-class activity Divide into groups of three or four and use actual thumbtacks or Hershey’s Kisses candies, or paper cups, to estimate the probability that when dropped, they will land with the point (or open side) up. How many trials are necessary to get a result that appears to be reasonably accurate when rounded to the first decimal place?

3. Out-of-class activity Marine biologists often use the capture-recapture method as a way to estimate the size of a population, such as the number of fish in a lake. This method involves capturing a sample from the population, tagging each member in the sample, and then return- ing it to the population. A second sample is later captured, and the tagged members are counted along with the total size of this second sample. The results can be used to estimate the size of the population.

Instead of capturing real fish, simulate the procedure using some uniform collection of items such as colored beads, M&Ms, or index cards. Start with a large collection of at least 200 of such items. Collect a sample of 50 and use a marker to “tag” each one. Replace the tagged items, mix the whole population, then select a second sample and proceed to estimate the population size. Compare the result to the actual population size obtained by counting all of the items.

4. Out-of-class activity In Cumulative Review Exercise 4, it was noted that eye colors in the United States are distributed as follows: 40% brown, 35% blue, 12% green, 7% gray, 6% hazel. That distribution can form the basis for probabilities. Conduct a survey by asking fellow students to identify the color of their eyes. Does the probability of 0.4 for brown eyes appear to be consistent with your results? Why would a large sample be required to confirm that P(hazel eyes) = 0.06?

CHAPTER 4 Cooperative Group Activities 179

M04_TRIO9015_02_SE_C04_118-179.indd 179 30/07/16 11:35 AM

180

Probability Distributions

Binomial Probability Distributions

Poisson Probability Distributions

5-1

5-2

5-3

Is the XSORT Gender Selection Method Effective?CHAPTER PROBLEM

Discrete Probability Distributions

We live in a time with incredible advances in technology, med-

icine, and health care. Cloning is no longer science fiction. We

have iPads, iPhones, virtual-reality headsets, and self-driving

cars. We carry calculators that can instantly execute many

complex statistical calculations. Heart pacemakers have defi-

brillators capable of shocking and restarting stopped hearts.

Couples use procedures that are claimed to greatly increase

the chance of having a baby with a desired gender.

Some people argue that gender selection methods should

be banned, regardless of the reason, while others enthusiasti-

cally support the use of such methods. Lisa Belkin asked in the

New York Times Magazine, “If we allow parents to choose the

sex of their child today, how long will it be before they order up

eye color, hair color, personality traits, and IQ?” There are some

convincing arguments in favor of at least limited use of gender

selection. One such argument involves couples carrying

5

M05_TRIO9015_02_SE_C05_180-215.indd 180 04/08/16 4:03 PM

X-linked recessive genes. For some of these couples, any male

children have a 50% chance of inheriting a disorder, but none

of the female children will inherit the disorder. These couples

may want to use gender selection as a way to ensure that they

have baby girls, thereby guaranteeing that a disorder will not

be inherited by any of their children.

The Genetics & IVF Institute in Fairfax, Virginia, devel-

oped a technique called MicroSort and claimed that it in-

creases the chances of a couple having a baby with a desired

gender. (Clinical trials of MicroSort have been discontinued.)

The MicroSort XSORT method is claimed to increase the

chances of a couple having a baby girl, and the MicroSort

YSORT method is claimed to increase the chances of a

baby boy. The latest results for the XSORT method consist

of 945 couples who wanted to have baby girls. After using

the XSORT technique, 879 of those couples had baby girls.

(See Figure 5-1 for a bar graph illustrating these results.)

We usually expect that in 945 births, the number of girls

should be somewhere around 472 or 473. Given that 879 out

of 945 couples had girls, can we conclude that the XSORT

technique is effective, or might we explain the outcome as

just a chance sample result? In answering that question, we

will use principles of probability to determine whether the

observed birth results differ significantly from results that we

would expect from random chance. This is a common goal

of inferential statistics: Determine whether results can be

reasonably explained by random chance or whether random

chance doesn’t appear to be a feasible explanation, so that

other factors are influencing results. In this chapter we pres-

ent methods that allow us to find the probabilities we need

for determining whether the XSORT results are significant,

suggesting that the method is effective.

Figure 5-2 on the next page provides a visual illustration of what this chapter accom- plishes. When investigating the numbers of heads in two coin tosses, we can use the following two different approaches:

• Use real sample data to find actual results: The approach of Chapters 2 and 3 is to collect sample data from actual coin tosses, then summarize the results in a table representing the frequency distribution, and then find statistics, such as the sample mean x and the sample standard deviation s.

• Use probabilities to find expected results: Using principles of probability from Chapter 4, we can find the probability for each possible number of heads in two tosses. Then we could summarize the results in a table representing a probability distribution.

In this chapter we merge the above two approaches as we create a table de- scribing what we expect to happen (instead of what did happen), then we find the population mean m and population standard deviation s. The table at the extreme right in Figure 5-2 is a probability distribution, because it describes the distribution using probabilities instead of frequency counts. The remainder of this book and the core of inferential statistics are based on some knowledge of probability distributions. In this chapter we focus on discrete probability distributions.

Chapter Objectives 181

CHAPTER OBJECTIVES

> > >

FIGURE 5-1 Results from the XSORT Method of Gender Selection

M05_TRIO9015_02_SE_C05_180-215.indd 181 31/08/16 3:32 PM

182 CHAPTER 5 Discrete Probability Distributions

Key Concept This section introduces the concept of a random variable and the con- cept of a probability distribution. We illustrate how a probability histogram is a graph that visually depicts a probability distribution. We show how to find the important parameters of mean, standard deviation, and variance for a probability distribution. Most importantly, we describe how to determine whether outcomes are significant (significantly low or significantly high). We begin with the related concepts of random variable and probability distribution.

5-1 Probability Distributions

Count numbers of heads in tosses of two coins.

Collect sample data from two coin tosses, then find statistics and create graphs.

Number of heads

x 0 1 2

f 27 56 17

Create a theoretical model of the expected results, then find parameters.

Find the probability for each possible number of heads in two coin tosses.

P(0) 5 0.25 P(1) 5 0.50 P(2) 5 0.25

Chapters 2 and 3

Chapter 4

Chapters 2 and 3 Chapter 5

Chapter 4

x 5 0.9 s 5 0.7 Number of

heads in two coin tosses

x 0 1 2

P(x) 0.25 0.50 0.25

m 5 1.0 s 5 0.7

FIGURE 5-2

Here are the chapter objectives:

Probability Distributions

• Define random variable and probability distribution. • Determine when a potential probability distribution actually satisfies the necessary

requirements.

• Given a probability distribution, compute the mean and standard deviation, then use those results to determine whether results are significantly low or significantly high.

Binomial Probability Distributions

• Describe a binomial probability distribution and find probability values for a binomial distribution.

• Compute the mean and standard deviation for a binomial distribution, then use those results to determine whether results are significantly low or significantly high.

Poisson Probability Distributions

• Describe a Poisson probability distribution and find probability values for a Poisson distribution.

5-1

5-2

5-3

Collect sample data from two coin tosses, then find statistics and create graphs.

Number of heads

x 0 1 2

f 27 56 17

Create a theoretical model of the expected results, then find parameters.

Find the probability for each possible number of heads in two coin tosses.

P(0) 5 0.25 P(1) 5 0.50 P(2) 5 0.25

Chapters 2 and 3 Chapter 5

Chapter 4

x 5 0.9 s 5 0.7 Number of

heads in two coin tosses

x 0 1 2

P(x)x)x 0.25 0.50 0.25

m 5 1.0 s 5 0.7

Here are the chapter objectives:

Probability Distributions

• Define random variable and probability distribution. • Determine when a potential probability distribution actually satisfies the necessary

requirements.

• Given a probability distribution, compute the mean and standard deviation, then use those results to determine whether results are significantly low or significantly high.

Binomial Probability Distributions

• Describe a binomial probability distribution and find probability values for a binomial distribution.

• Compute the mean and standard deviation for a binomial distribution, then use those results to determine whether results are significantly low or significantly low or significantly low significantly high.

Poisson Probability Distributions

• Describe a Poisson probability distribution and find probability values for a Poisson distribution.

M05_TRIO9015_02_SE_C05_180-215.indd 182 31/08/16 3:32 PM

5-1 Probability Distributions 183

PA RT 1 Basic Concepts of a Probability Distribution

DEFINITIONS

A random variable is a variable (typically represented by x) that has a single nu- merical value, determined by chance, for each outcome of a procedure.

A probability distribution is a description that gives the probability for each value of the random variable. It is often expressed in the format of a table, formula, or graph.

In Section 1-2 we made a distinction between discrete and continuous data. Random variables may also be discrete or continuous, and the following two defini- tions are consistent with those given in Section 1-2.

DEFINITIONS

A discrete random variable has a collection of values that is finite or countable. (If there are infinitely many values, the number of values is countable if it is pos- sible to count them individually, such as the number of tosses of a coin before getting heads.)

A continuous random variable has infinitely many values, and the collec- tion of values is not countable. (That is, it is impossible to count the individual items because at least some of them are on a continuous scale, such as body temperatures.)

This chapter deals exclusively with discrete random variables, but the following chapters deal with continuous random variables.

Probability Distribution: Requirements

Every probability distribution must satisfy each of the following three requirements.

1. There is a numerical (not categorical) random variable x, and its number values are associated with corresponding probabilities.

2. ΣP(x) = 1 where x assumes all possible values. (The sum of all probabilities must be 1, but sums such as 0.999 or 1.001 are acceptable because they result from rounding errors.)

3. 0 … P(x) … 1 for every individual value of the random variable x. (That is, each probability value must be between 0 and 1 inclusive.)

The second requirement comes from the simple fact that the random variable x represents all possible events in the entire sample space, so we are certain (with prob- ability 1) that one of the events will occur. The third requirement comes from the basic principle that any probability value must be 0 or 1 or a value between 0 and 1.

EXAMPLE 1 Genetics

Although the Chapter Problem involves 945 births, let’s consider a simpler example that involves only two births with the following random variable:

x = number of girls in two births

The above x is a random variable because its numerical values depend on chance. With two births, the number of girls can be 0, 1, or 2, and Table 5-1 is a probability

continued

M05_TRIO9015_02_SE_C05_180-215.indd 183 04/08/16 4:03 PM

184 CHAPTER 5 Discrete Probability Distributions

Notation for 0+

In tables such as Table 5-1 or the binomial probabilities listed in Table A-1 in Appendix A, we sometimes use 0 + to represent a probability value that is positive but very small, such as 0.000000123. When rounding a probability value for inclu- sion in such a table, rounding to 0 would be misleading because it would incor- rectly suggest that the event is impossible.

Probability Histogram: Graph of a Probability Distribution

There are various ways to graph a probability distribution, but for now we will con- sider only the probability histogram. Figure 5-3 is a probability histogram corre- sponding to Table 5-1. Notice that it is similar to a relative frequency histogram (de- scribed in Section 2-2), but the vertical scale shows probabilities instead of relative frequencies based on actual sample results.

distribution because it gives the probability for each value of the random variable x and it satisfies the three requirements listed earlier:

1. The variable x is a numerical random variable and its values are associated with probabilities, as in Table 5-1.

2. ΣP(x) = 0.25 + 0.50 + 0.25 = 1 3. Each value of P(x) is between 0 and 1. (Specifically, 0.25 and 0.50 and 0.25

are each between 0 and 1 inclusive.)

The random variable x in Table 5-1 is a discrete random variable, because it has three possible values (0, 1, 2), and 3 is a finite number, so this satisfies the require- ment of being finite or countable.

TABLE 5-1 Probability Distribution for the Number of Girls in Two Births

x: Number of Girls

P(x)

0 0.25

1 0.50

2 0.25

FIGURE 5-3 Probability Histogram for Number of Girls in Two Births

In Figure 5-3, we see that the values of 0, 1, 2 along the horizontal axis are lo- cated at the centers of the rectangles. This implies that the rectangles are each 1 unit wide, so the areas of the rectangles are 0.25, 0.50, and 0.25. The areas of these rectan- gles are the same as the probabilities in Table 5-1. We will see in Chapter 6 and future chapters that such a correspondence between areas and probabilities is very useful.

Probability Formula Example 1 involves a table, but a probability distribution

could also be in the form of a formula. Consider the formula P(x) = 1

2(2 - x)! x!

M05_TRIO9015_02_SE_C05_180-215.indd 184 31/08/16 3:32 PM

5-1 Probability Distributions 185

(where x can be 0, 1, or 2). Using that formula, we find that P102 = 0.25, P112 = 0.50, and P122 = 0.25. The probabilities found using this formula are the same as those in Table 5-1. This formula does describe a probability distribution because the three re- quirements are satisfied, as shown in Example 1.

TABLE 5-2 Hospital Job Interview Mistakes

x P(x)

Inappropriate attire 0.50

Being late 0.44

Lack of eye contact 0.33

Checking phone or texting 0.30

Total 1.57

EXAMPLE 2 Hospital Job Interview Mistakes

Hiring managers were asked to identify the biggest mistakes that job applicants make during an interview, and Table 5-2 is based on their responses (based on data from an Adecco survey). Does Table 5-2 describe a probability distribution?

S O L U T I O N

Table 5-2 violates the first requirement because x is not a numerical random vari- able. Instead, the “values” of x are categorical data, not numbers. Table 5-2 also violates the second requirement because the sum of the probabilities is 1.57, but that sum should be 1. Because the three requirements are not all satisfied, we conclude that Table 5-2 does not describe a probability distribution.

Parameters of a Probability Distribution Remember that with a probability distribution, we have a description of a population instead of a sample, so the values of the mean, standard deviation, and variance are parameters, not statistics. The mean, variance, and standard deviation of a discrete probability distribution can be found with the following formulas:

FORMULA 5-1 Mean M for a probability distribution

m = Σ3x # P1x24

FORMULA 5-2 Variance S2 for a probability distribution

s2 = Σ31x - m22 # P1x24 (This format is easier to understand.)

FORMULA 5-3 Variance S2 for a probability distribution

s2 = Σ3x2 # P1x24 - m2 (This format is easier for manual calculations.)

FORMULA 5-4 Standard deviation S for a probability distribution

s = 2Σ3x2 # P1x24 - m2

M05_TRIO9015_02_SE_C05_180-215.indd 185 04/08/16 4:03 PM

186 CHAPTER 5 Discrete Probability Distributions

When applying Formulas 5-1 through 5-4, use the following rule for rounding results.

Round-Off Rule For m, s, And s2 From A Probability Distribution

Round results by carrying one more decimal place than the number of decimal places used for the random variable x. If the values of x are integers, round m, s, and s2 to one decimal place.

Exceptions to Round-Off Rule In some special cases, the above round-off rule re- sults in values that are misleading or inappropriate. For example, with four-engine jets the mean number of jet engines working successfully throughout a flight is 3.999714286, which becomes 4.0 when rounded, but that is misleading because it suggests that all jet engines always work successfully. Here we need more precision to correctly reflect the true mean, such as the precision in 3.999714.

Expected Value The mean of a discrete random variable x is the theoretical mean outcome for infinitely many trials. We can think of that mean as the expected value in the sense that it is the average value that we would expect to get if the trials could continue indefinitely.

DEFINITION

The expected value of a discrete random variable x is denoted by E, and it is the mean value of the outcomes, so E = m and E can also be found by evaluating Σ3x # P1x24, as in Formula 5-1.

CAUTION An expected value need not be a whole number, even if the different possible values of x might all be whole numbers. The expected number of girls in five births is 2.5, even though five particular births can never result in 2.5 girls. If we were to survey many couples with five children, we expect that the mean number of girls will be 2.5.

EXAMPLE 3 Finding the Mean, Variance, and Standard Deviation

Table 5-1 on page 184 describes the probability distribution for the number of girls in two births (assuming that boys and girls are equally likely). Find the mean, vari- ance, and standard deviation for the probability distribution described in Table 5-1 from Example 1.

S O L U T I O N

In Table 5-3, the two columns at the left describe the probability distribution given earlier in Table 5-1. We create the two columns at the right for the purposes of the calculations required.

Using Formulas 5-1 and 5-2 and the table results, we get

Mean: m = Σ3x # P1x24 = 1.0 Variance: s2 = Σ31x - m22 # P1x24 = 0.5

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5-1 Probability Distributions 187

I N T E R P R E TAT I O N

Assuming that boys and girls are equally likely in two births, the mean number of girls is 1.0, the variance is 0.50 girls2, and the standard deviation is 0.7 girl. Also, the expected value for the number of girls in two births is 1.0 girl, which is the same value as the mean. If we were to collect data on a large number of trials with two births in each trial, we expect to get a mean of 1.0 girl.

The standard deviation is the square root of the variance, so

Standard deviation: s = 20.5 = 0.707107 = 0.7 1rounded2 Rounding: In Table 5-3, we use m = 1.0. If m had been the value of 1.23456, we might round m to 1.2, but we should use its unrounded value of 1.23456 in Table 5-3 calculations. Rounding in the middle of calculations can lead to results with errors that are too large.

TABLE 5-3 Calculating m and s for a Probability Distribution

x P 1x2 x # P 1x2 1x − m 22 # P1x2 0 0.25 0 # 0.25 = 0.00 10 - 122 # 0.25 = 0.25 1 0.50 1 # 0.50 = 0.50 11 - 122 # 0.50 = 0.00 2 0.25 2 # 0.25 = 0.50 12 - 122 # 0.25 = 0.25

Total 1.00 c

m = Σ3x # P1x24 0.50

c s2 = Σ31x - m22 # P1x24

Making Sense of Results: Significant Values We present the following two different approaches for determining whether a value of a random variable x is significantly low or high.

Identifying Significant Results with the Range Rule of Thumb

The range rule of thumb (introduced in Section 3-2) may be helpful in interpreting the value of a standard deviation. According to the range rule of thumb, the vast major- ity of values should lie within 2 standard deviations of the mean, so we can consider a value to be significant if it is at least 2 standard deviations away from the mean. We can therefore identify “significant” values as follows:

Range Rule of Thumb for Identifying Significant Values

Significantly low values are 1m - 2s2 or lower. Significantly high values are 1m + 2s2 or higher. Values not significant: Between 1m - 2s2 and 1m + 2s2

Figure 3-3 from Section 3-2 illustrates the above criteria:

Values not significant Significantly low values

Significantly high values

mm − 2s m + 2s

HINT Know that the use of the number 2 in the range rule of thumb is somewhat arbitrary, and this is a guideline, not an absolutely rigid rule.

M05_TRIO9015_02_SE_C05_180-215.indd 187 04/08/16 4:03 PM

188 CHAPTER 5 Discrete Probability Distributions

Identifying Significant Results with Probabilities:

■ Significantly high number of successes: x successes among n trials is a signifi- cantly high number of successes if the probability of x or more successes is 0.05 or less. That is, x is a significantly high number of successes if P(x or more) … 0.05.*

■ Significantly low number of successes: x successes among n trials is a significantly low number of successes if the probability of x or fewer successes is 0.05 or less. That is, x is a significantly low number of successes if P(x or fewer) … 0.05.*

EXAMPLE 4 Identifying Significant Results with the Range Rule of Thumb

In Example 3 we found that with two births, the mean number of girls is m = 1.0 girl and the standard deviation is s = 0.7 girl. Use those results and the range rule of thumb to determine whether 2 girls is a significantly high number of girls.

S O L U T I O N

Using the range rule of thumb, the value of 2 girls is significantly high if it is greater than or equal to m + 2s. With m = 1.0 girl and s = 0.7 girl, we get

m + 2s = 1 + 210.72 = 2.4 girls Significantly high numbers of girls are 2.4 and above.

I N T E R P R E TAT I O N

Based on these results, we conclude that 2 girls is not a significantly high number of girls (because 2 is not greater than or equal to 2.4).

*The value 0.05 is not absolutely rigid. Other values, such as 0.01, could be used to distinguish between results that are significant and those that are not significant.

Identification of significantly low or significantly high numbers of successes is some- times used for the purpose of rejecting assumptions, as stated in the following rare event rule.

The Rare Event Rule for Inferential Statistics

If, under a given assumption, the probability of a particular outcome is very small and the outcome occurs significantly less than or significantly greater than what we expect with that assumption, we conclude that the assumption is probably not correct.

For example, if testing the assumption that boys and girls are equally likely, the out- come of 20 girls in 100 births is significantly low and would be a basis for rejecting that assumption.

EXAMPLE 5 Identifying Significant Results with Probabilities

Is 879 girls in 945 births a significantly high number of girls?

What does the result suggest about the Chapter Problem, which includes results from the XSORT method of gender selection? (Among 945 births from parents us- ing the XSORT method, there were 879 girls. Is 879 girls in those 945 births significantly high?)

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5-1 Probability Distributions 189

Not Exactly, but “At Least as Extreme”

It should be obvious that among 945 births, 879 girls is significantly high, whereas 475 girls is not significantly high. What makes 879 girls significant while 475 girls is not significant? It is not probabilities of exactly 879 girls and 475 girls (they are both less than 0.026). It is the fact that the probability of 879 or more girls is very low (0.0000), but the probability of 475 or more girls is not low (0.448).

PA RT 2 Expected Value and Rationale for Formulas Expected Value

In Part 1 of this section we noted that the expected value of a random variable x is equal to the mean m. We can therefore find the expected value by computing Σ3x # P1x24, just as we do for finding the value of m.

S O L U T I O N

A result of 879 girls in 945 births is greater than we expect with random chance, but we need to determine whether 879 girls is significantly high. Here, the relevant probability is the probability of getting 879 or more girls in 945 births. Using methods covered later in Section 5-2, we can find that P(879 or more girls in 945 births) = 0.0000 (rounded). Because the probability of getting 879 or more girls is less than or equal to 0.05, we conclude that 879 girls in 945 births is a significantly high number of girls. See Figure 5-4, which is a probability histogram showing the probability for the different numbers of girls.

FIGURE 5-4 Probability Histogram of Girls in 945 Births

EXAMPLE 6 Births

Assuming that boys and girls are equally likely, find the expected number of girls in 945 births. Instead of using Formula 5-1, just think about the number of girls expected in 945 births.

S O L U T I O N

The expected number of girls in 945 births is 472.5 girls. continued

I N T E R P R E TAT I O N

It is unlikely that we would get 879 or more girls in 945 births by chance. It follows that 879 girls in 945 births is significantly high, so the XSORT method appears to be effective (but this does not prove that the XSORT method is responsible for the large number of girls).

0.0000 (rounded). Because the probability of getting 879 or more girls is

Do Boys or Girls Run in the Family?

One of the

authors of

this book, his

siblings, and

his siblings’

children consist

of 11 males and

only 1 female.

Is this an example of a phenom-

enon whereby one particular

gender runs in a family? This

issue was studied by examin-

ing a random sample of 8770

households in the United States.

The results were reported in the

Chance magazine article “Does

Having Boys or Girls Run in the

Family?” by Joseph Rodgers

and Debby Doughty. Part of

their analysis involves use of the

binomial probability distribution

discussed in this section. Their

conclusion is that “We found no

compelling evidence that sex

bias runs in the family.”

M05_TRIO9015_02_SE_C05_180-215.indd 189 04/08/16 4:03 PM

190 CHAPTER 5 Discrete Probability Distributions

Rationale for Formulas 5-1 Through 5-4

Instead of blindly accepting and using formulas, it is much better to have some un- derstanding of why they work. When computing the mean from a frequency distribu- tion, f represents class frequency and N represents population size. In the expression below, we rewrite the formula for the mean of a frequency table so that it applies to a population. In the fraction f>N, the value of f is the frequency with which the value x occurs and N is the population size, so f>N is the probability for the value of x. When we replace f>N with P(x), we make the transition from relative frequency based on a limited number of observations to probability based on infinitely many trials. This result shows why Formula 5-1 is as given earlier in this section.

m = Σ1f # x2

N = Σ c f

# x N d = Σ cx # f

N d = Σ3x # P1x24

Similar reasoning enables us to take the variance formula from Chapter 3 and apply it to a random variable for a probability distribution; the result is Formula 5-2. Formula 5-3 is a shortcut version that will always produce the same result as Formula 5-2. Although Formula 5-3 is usually easier to work with, Formula 5-2 is easier to under- stand directly. Based on Formula 5-2, we can express the standard deviation as

s = 2Σ31x - m22 # P1x24 or as the equivalent form given in Formula 5-4.

I N T E R P R E TAT I O N

In any specific sample of 945 births, we can never get 472.5 girls, but 472.5 girls is the expected value in the sense that it would be the mean from many samples of 945 births.

Statistical Literacy and Critical Thinking

1. Random Variable The accompanying table lists probabilities for the corresponding num- bers of girls in four births. What is the random variable, what are its possible values, and are its values numerical?

5-1 Basic Skills and Concepts

Number of Girls in Four Births

Number of Girls x

P(x)

0 0.063

1 0.250

2 0.375

3 0.250

4 0.063

2. Discrete or Continuous? Is the random variable given in the accompanying table discrete or continuous? Explain.

3.  Probability Distribution For the accompanying table, is the sum of the values of P(x) equal to 1, as required for a probability distribution? Does the table describe a probability distribution?

4. Significant For 100 births, P(exactly 56 girls) = 0.0390 and P(56 or more girls) = 0.136. Is 56 girls in 100 births a significantly high number of girls? Which probability is relevant to answering that question?

Identifying Discrete and Continuous Random Variables. In Exercises 5 and 6, refer to the given values, then identify which of the following is most appropriate: discrete ran- dom variable, continuous random variable, or not a random variable.

5. a. Exact weights of the next 100 babies born in the United States

b. Responses to the survey question “Which health plan do you have?”

continued

M05_TRIO9015_02_SE_C05_180-215.indd 190 04/08/16 4:03 PM

5-1 Probability Distributions 191

c. Numbers of families that must be surveyed before finding one with 10 children

d. Exact foot lengths of humans

e. Shoe sizes (such as 8 or 8½) of humans

6. a. Grades (A, B, C, D, F) earned in biostatistics classes

b. Heights of students in biostatistics classes

c. Numbers of students in biostatistics classes

d. Eye colors of biostatistics students

e. Numbers of times biostatistics students must toss a coin before getting heads

Identifying Probability Distributions. In Exercises 7–14, determine whether a prob- ability distribution is given. If a probability distribution is given, find its mean and standard deviation. If a probability distribution is not given, identify the requirements that are not satisfied.

7. Genetic Disorder Five males with an X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the X-linked genetic disorder.

x P(x)

0 0.031

1 0.156

2 0.313

3 0.313

4 0.156

5 0.031

8.  Male Color Blindness When conducting research on color blindness in males, a researcher forms random groups with five males in each group. The random variable x is the number of males in the group who have a form of color blindness (based on data from the National Institutes of Health).

x P(x)

0 0.659

1 0.287

2 0.050

3 0.004

4 0.001

5 0 +

9.  Genetics Experiment A genetics experiment involves off- spring peas in groups of four. A researcher reports that for one group, the number of peas with white flowers has a probability dis- tribution as given in the accompanying table.

x P(x)

0 0.04

1 0.26

2 0.36

3 0.20

4 0.08

10. Mortality Study For a group of four men, the probability dis- tribution for the number x who live through the next year is as given in the accompanying table.

x P(x)

0 0.0000

1 0.0001

2 0.0006

3 0.0387

4 0.9606

M05_TRIO9015_02_SE_C05_180-215.indd 191 04/08/16 4:03 PM

192 CHAPTER 5 Discrete Probability Distributions

11. Genetic Disorder Three males with an X-linked genetic dis- order have one child each. The random variable x is the number of children among the three who inherit the X-linked genetic disorder.

x P(x)

0 0.4219

1 0.4219

2 0.1406

3 0.0156

12. Diseased Seedlings An experiment involves groups of four seedlings grown under controlled conditions. The random variable x is the number of seedlings in a group that meet specific criteria for being classified as “diseased.”

x P(x)

0 0.805

1 0.113

2 0.057

3 0.009

4 0.002

Genetics. In Exercises 13–18, refer to the accompanying table, which describes results from groups of 8 births from 8 different sets of parents. The random variable x represents the number of girls among 8 children.

Number of Girls x

P(x)

0 0.004

1 0.031

2 0.109

3 0.219

4 0.273

5 0.219

6 0.109

7 0.031

8 0.004

13. Mean and Standard Deviation Find the mean and standard deviation for the numbers of girls in 8 births.

14.  Range Rule of Thumb for Significant Events Use the range rule of thumb to determine whether 1 girl in 8 births is a sig- nificantly low number of girls.

15.  Range Rule of Thumb for Significant Events Use the range rule of thumb to determine whether 6 girls in 8 births is a sig- nificantly high number of girls.

16. Using Probabilities for Significant Events

a. Find the probability of getting exactly 7 girls in 8 births.

b. Find the probability of getting 7 or more girls in 8 births.

c. Which probability is relevant for determining whether 7 is a significantly high number of girls in 10 births: the result from part (a) or part (b)?

d. Is 7 a significantly high number of girls in 8 births? Why or why not?

17. Using Probabilities for Significant Events

a. Find the probability of getting exactly 6 girls in 8 births.

b. Find the probability of getting 6 or more girls in 8 births.

c. Which probability is relevant for determining whether 6 is a significantly high number of girls in 8 births: the result from part (a) or part (b)?

d. Is 6 a significantly high number of girls in 8 births? Why or why not?

18. Using Probabilities for Significant Events

a. Find the probability of getting exactly 1 girl in 8 births.

b. Find the probability of getting 1 or fewer girls in 8 births.

c. Which probability is relevant for determining whether 1 is a significantly low number of girls in 8 births: the result from part (a) or part (b)?

d. Is 1 a significantly low number of girls in 8 births? Why or why not?

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5-2 Binomial Probability Distributions 193

Sleepwalking. In Exercises 19–23, refer to the accompanying table, which describes the numbers of adults in groups of five who reported sleepwalking (based on data from “Prevalence and Comorbidity of Nocturnal Wandering In the U.S. Adult General Population,” by Ohayon et al., Neurology, Vol. 78, No. 20).

19. Mean and Standard Deviation Find the mean and standard deviation for the numbers of sleepwalkers in groups of five.

x P(x)

0 0.172

1 0.363

2 0.306

3 0.129

4 0.027

5 0.002 20.  Range Rule of Thumb for Significant Events Use the range rule of thumb to determine whether 4 is a significantly high number of sleepwalkers in a group of 5 adults.

21.  Range Rule of Thumb for Significant Events Use the range rule of thumb to determine whether 3 is a significantly high number of sleepwalkers in a group of 5 adults.

Key Concept Section 5-1 introduced the important concept of a discrete proba- bility distribution. Among the various discrete probability distributions that exist, the focus of this section is the binomial probability distribution. Part 1 of this sec- tion introduces the binomial probability distribution along with methods for find- ing probabilities. Part 2 presents easy methods for finding the mean and standard deviation of a binomial distribution. As in other sections, we stress the importance of interpreting probability values to determine whether events are significantly low or significantly high.

PA RT 1 Basics of Binomial Probability Distribution Binomial probability distributions allow us to deal with circumstances in which the outcomes belong to two categories, such as cured>not cured or acceptable>defective or survived>died.

5-2 Binomial Probability Distributions

22. Using Probabilities for Identifying Significant Events

a. Find the probability of getting exactly 4 sleepwalkers among 5 adults.

b. Find the probability of getting 4 or more sleepwalkers among 5 adults.

c. Which probability is relevant for determining whether 4 is a significantly high number of sleepwalkers among 5 adults: the result from part (a) or part (b)?

d. Is 4 a significantly high number of sleepwalkers among 5 adults? Why or why not?

23. Using Probabilities for Identifying Significant Events

a. Find the probability of getting exactly 1 sleepwalker among 5 adults.

b. Find the probability of getting 1 or fewer sleepwalkers among 5 adults.

c. Which probability is relevant for determining whether 1 is a significantly low number of sleepwalkers among 5 adults: the result from part (a) or part (b)?

d. Is 1 a significantly low number of sleepwalkers among 5 adults? Why or why not?

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194 CHAPTER 5 Discrete Probability Distributions

Notation For Binomial Probability Distributions

S and F (success and failure) denote the two possible categories of all outcomes.

P1S2 = p 1p = probability of a success2 P1F2 = 1 - p = q 1q = probability of a failure2 n the fixed number of trials

x a specific number of successes in n trials, so x can be any whole number between 0 and n, inclusive

p probability of success in one of the n trials

q probability of failure in one of the n trials

P(x) probability of getting exactly x successes among the n trials

The word success as used here is arbitrary and does not necessarily represent something good. Either of the two possible categories may be called the success S as long as its probability is identified as p. (The value of q can always be found from q = 1 - p. If p = 0.95, then q = 1 - 0.95 = 0.05.)

DEFINITION

A binomial probability distribution results from a procedure that meets these four requirements:

1. The procedure has a fixed number of trials. (A trial is a single observation.)

2. The trials must be independent, meaning that the outcome of any individual trial doesn’t a�ect the probabilities in the other trials.

3. Each trial must have all outcomes classified into exactly two categories, com- monly referred to as success and failure.

4. The probability of a success remains the same in all trials.

CAUTION When using a binomial probability distribution, always be sure that x and p are consistent in the sense that they both refer to the same category being called a success.

EXAMPLE 1 Hybridization Experiments

When Gregor Mendel conducted his famous hybridization experiments, he used peas with green pods and peas with yellow pods. Because green is dominant and yellow is recessive, when crossing two parents with the green>yellow pair of genes, we expect that 3>4 of the offspring peas should have green pods. That is, P(green pod) = 3>4. Assume that all parents have the green>yellow combination of genes, and we want to find the probability that exactly three of five offspring peas have green pods.

a. Does this procedure result in a binomial distribution?

b. If this procedure does result in a binomial distribution, identify the values of n, x, p, and q.

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5-2 Binomial Probability Distributions 195

Treating Dependent Events as Independent

When selecting a sample (as in a survey), we usually sample without replacement. Sampling without replacement results in dependent events, which violates a require- ment of a binomial distribution. However, we can often treat the events as if they were independent by applying the following 5% guideline introduced in Section 4-2:

5% Guideline for Cumbersome Calculations

When sampling without replacement and the sample size is no more than 5% of the size of the population, treat the selections as being independent (even though they are actually dependent).

Methods for Finding Binomial Probabilities We now proceed with three methods for finding the probabilities corresponding to the random variable x in a binomial distribution. The first method involves calculations using the binomial probability formula and is the basis for the other two methods. The second method involves the use of software or a calculator, and the third method involves the use of the Appendix Table A-1. (With technology so widespread, such tables are becoming obsolete.) If using technology that automatically produces bino- mial probabilities, we recommend that you solve one or two exercises using Method 1 to better understand the basis for the calculations.

S O L U T I O N

a. This procedure does satisfy the requirements for a binomial distribution, as shown below.

1. The number of trials (5) is fixed.

2. The 5 trials are independent because the probability of any o§spring pea having a green pod is not a§ected by the outcome of any other o§spring pea.

3. Each of the 5 trials has two categories of outcomes: The pea has a green pod or it does not.

4. For each o§spring pea, the probability that it has a green pod is 3>4 or 0.75, and that probability remains the same for each of the 5 peas.

b. Having concluded that the given procedure does result in a binomial distribu- tion, we now proceed to identify the values of n, x, p, and q.

1. With 5 o§spring peas, we have n = 5.

2. We want the probability of exactly 3 peas with green pods, so x = 3.

3. The probability of success (getting a pea with a green pod) for one selec- tion is 0.75, so p = 0.75.

4. The probability of failure (not getting a green pod) is 0.25, so q = 0.25.

Again, it is very important to be sure that x and p both refer to the same concept of “success.” In this example, we use x to count the number of peas with green pods, so p must be the probability that a pea has a green pod. Therefore, x and p do use the same concept of success (green pod) here.

Not at Home

Pollsters cannot

simply ignore

those who were

not at home

when they

were called

the first time.

One solution

is to make repeated callback

attempts until the person can

be reached. Alfred Politz and

Willard Simmons describe a way

to compensate for those missed

calls without making repeated

callbacks. They suggest weight-

ing results based on how often

people are not at home. For

example, a person at home only

two days out of six will have a

2>6 or 1>3 probability of being at home when called the first time.

When such a person is reached

the first time, his or her results

are weighted to count three

times as much as someone who

is always home. This weighting

is a compensation for the other

similar people who are home two

days out of six and were not at

home when called the first time.

This clever solution was first

presented in 1949.

M05_TRIO9015_02_SE_C05_180-215.indd 195 04/08/16 4:03 PM

196 CHAPTER 5 Discrete Probability Distributions

FORMULA 5-5 Binomial Probability Formula

P1x2 = n!1n - x2!x! # px # qn - x for x = 0, 1, 2, c, n

where

n = number of trials x = number of successes among n trials p = probability of success in any one trial q = probability of failure in any one trial 1q = 1 - p2

Formula 5-5 can also be expressed as P1x2 = nCx # px # qn - x. With x items identi- cal to themselves, and n - x other items identical to themselves, the number of permutations is nCx = n!31n - x2!x!4, so the two sides of this equation are inter- changeable. The factorial symbol !, introduced in Section 4-6, denotes the product of decreasing factors. Two examples of factorials are 3! = 3 # 2 # 1 = 6 and 0! = 1 (by definition).

EXAMPLE 2 Hybridization Experiment

Assuming that the probability of a pea having a green pod is 0.75 (as in Example 1), use the binomial probability formula to find the probability of getting exactly 3 peas with green pods when 5 offspring peas are generated. That is, find P(3) given that n = 5, x = 3, p = 0.75, and q = 0.25.

S O L U T I O N

Using the given values of n, x, p, and q in the binomial probability formula (Formula 5-5), we get

P132 = 5!15 - 32!3! # 0.753 # 0.255 - 3

= 5!

2!3! # 0.421875 # 0.0625

= 110210.421875210.06252 = 0.263671875 The probability of getting exactly 3 peas with green pods among 5 offspring peas is 0.264 (rounded to three significant digits).

Calculation hint: When computing a probability with the binomial probability for- mula, it’s helpful to get a single number for n!>31n - x2!x!4 or nCx, a single num- ber for px, and a single number for qn - x, and then simply multiply the three factors together as shown in the third line of the calculation in the preceding example. Don’t round when you find those three factors; round only at the end, and round to three significant digits.

Method 1: Using the Binomial Probability Formula In a binomial probability distribution, probabilities can be calculated by using Formula 5-5.

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5-2 Binomial Probability Distributions 197

Method 2: Using Technology Technology can be used to find binomial probabili- ties. The screen displays listing binomial probabilities for n = 5 and p = 0.75, as in Example 2, are given. Notice that in each display, the probability distribution is given as a table.

Statdisk

Minitab

Excel

TI-83>84 Plus

Method 3: Using Table A-1 in Appendix A This method can be skipped if tech- nology is available. Table A-1 in Appendix A lists binomial probabilities for select values of n and p. It cannot be used if n > 8 or if the probability p is not one of the 13 values included in the table.

To use the table of binomial probabilities, we must first locate n and the desired corresponding value of x. At this stage, one row of numbers should be isolated. Now align that row with the desired probability of p by using the column across the top. The isolated number represents the desired probability. A very small probability, such as 0.000064, is indicated by 0 + .

EXAMPLE 3 Births

Assuming that boys and girls are equally likely, find the probability of getting ex- actly 5 boys in 8 randomly selected births.

continued

M05_TRIO9015_02_SE_C05_180-215.indd 197 04/08/16 4:04 PM

198 CHAPTER 5 Discrete Probability Distributions

PA RT 2 Using Mean and Standard Deviation for Critical Thinking

Section 5-1 included formulas for finding the mean, variance, and standard deviation from any discrete probability distribution. A binomial distribution is a particular type of discrete probability distribution, so we could use those same formulas, but if we know the values of n and p, it is much easier to use the following:

S O L U T I O N

Because boys and girls are assumed to be equally likely, we have p = 0.5. Because there are 8 births we have n = 8. Because we want the probability of exactly 5 boys, we have x = 5.

Refer to Table A-1with n = 8, x = 0.5, and p = 0.5. Locate n = 8 at the left, then find the row of probabilities for x = 5. Next, look across the top row to find the column of values under p = 0.50. Table A-1 shows that P(5 boys) = 0.219.

For Binomial Distributions

Formula 5-6 Mean: m = np

Formula 5-7 Variance: s2 = npq

Formula 5-8 Standard Deviation: s = 1npq

As in earlier sections, finding values for m and s can be great fun, but it is especially important to interpret and understand those values, so the range rule of thumb and the rare event rule for inferential statistics can be very helpful. Here is a brief summary of the range rule of thumb: Values are significantly low or high if they differ from the mean by more than 2 standard deviations, as described by the following:

Range Rule of Thumb

Significantly low values … 1m - 2s2 Significantly high values Ú 1m + 2s2 Values not significant: Between 1m - 2s2 and 1m + 2s2

EXAMPLE 4 Hybridization Experiment

Use Formulas 5-6 and 5-8 to find the mean and standard deviation for the numbers of peas with green pods when groups of 5 offspring peas are generated. Assume that there is a 0.75 probability that an offspring pea has a green pod.

S O L U T I O N

Using the values n = 5, p = 0.75, and q = 0.25, Formulas 5-6 and 5-8 can be applied as follows:

m = np = 15210.752 = 3.8 s = 1npq = 215210.75210.252 = 1.0 1rounded2

M05_TRIO9015_02_SE_C05_180-215.indd 198 31/08/16 3:32 PM

5-2 Binomial Probability Distributions 199

Formula 5-6 for the mean makes sense intuitively. If 75% of peas have green pods and five offspring peas are generated, we expect to get around 5 # 0.75 = 3.8 peas with green pods. This result can be generalized as m = np. The variance and standard deviation are not so easily justified, and we omit the complicated algebraic manipula- tions that lead to Formulas 5-7 and 5-8. Instead, refer again to the preceding example and Table 5-3 on page 187 to verify that for a binomial distribution, Formulas 5-6, 5-7, and 5-8 will produce the same results as Formulas 5-1, 5-3, and 5-4.

EXAMPLE 5 Genetics

In an actual experiment, Mendel generated 580 offspring peas. He claimed that 75%, or 435, of them would have green pods. The actual experiment resulted in 428 peas with green pods.

a. Assuming that groups of 580 o§spring peas are generated, find the mean and standard deviation for the numbers of peas with green pods.

b. Use the range rule of thumb to find the numbers of peas with green pods that separate significantly low values and significantly high values from values that are not significant. Based on those numbers, can we conclude that Mendel’s actual result of 428 peas with green pods is significantly low or significantly high? Does this suggest that Mendel’s value of 75% is wrong?

S O L U T I O N

a. With n = 580 o§spring peas, with p = 0.75, and q = 0.25, we can find the mean and standard deviation for the numbers of peas with green pods as follows:

m = np = 1580210.752 = 435.0 s = 1npq = 21580210.75210.252 = 10.4

For groups of 580 o§spring peas, the mean number of peas with green pods is 435.0 and the standard deviation is 10.4.

b. We must now interpret the results to determine whether Mendel’s actual result of 428 peas is a result that could easily occur by chance, or whether that result is so unlikely that the assumed rate of 75% is wrong. We will use the range rule of thumb as follows:

Significantly low values … 1m - 2s2 = 435.0 - 2110.42 = 414.2 Significantly high values Ú 1m + 2s2 = 435.0 + 2110.42 = 455.8

I N T E R P R E TAT I O N

Based on these results, significantly low values are 414.2 or lower, and significantly high values are 455.8 or higher. That is, if Mendel generated many groups of 580 offspring peas and if his 75% rate is correct, the numbers of peas with green pods should usually fall between 414.2 and 455.8. Mendel actually got 428 peas with green pods, and that value is neither significantly low nor significantly high, so the experimental results are consistent with the 75% rate. The results do not suggest that Mendel’s claimed rate of 75% is wrong.

M05_TRIO9015_02_SE_C05_180-215.indd 199 04/08/16 4:04 PM

200 CHAPTER 5 Discrete Probability Distributions

Variation in Statistics Example 5 is a good illustration of the importance of variation in statistics. In a traditional algebra course, we might conclude that 428 is not 75% of 580 simply because 428 does not equal 435 (which is 75% of 580). However, in statistics we recognize that sample results vary. We don’t expect to get exactly 75% of the peas with green pods. We recognize that as long as the results don’t vary too far away from the claimed rate of 75%, they are consistent with that claimed rate of 75%.

In this section we presented easy procedures for finding values of the mean m and standard deviation s from a binomial probability distribution. However, it is really important to be able to interpret those values by using such tools as the range rule of thumb for distinguishing values that are significantly low or significantly high from values that are not significant.

Instead of the range rule of thumb, we could also use probabilities to determine when values are significantly high or significantly low.

Using Probabilities to Determine When Results Are Significantly High or Low

■ Significantly high number of successes: x successes among n trials is a significantly high number of successes if the probability of x or more successes is 0.05 or less. That is, x is a significantly high number of successes if P(x or more) … 0.05.*

■ Significantly low number of successes: x successes among n trials is a significantly low number of successes if the probability of x or fewer successes is 0.05 or less. That is, x is a significantly low number of successes if P(x or fewer) … 0.05.*

*The value 0.05 is not absolutely rigid. Other values, such as 0.01, could be used to distinguish between results that are significant and those that are not significant.

Rationale for the Binomial Probability Formula

The binomial probability formula is the basis for all three methods presented in this section. Instead of accepting and using that formula blindly, let’s see why it works.

In Example 2, we used the binomial probability formula to find the probability of getting exactly 3 peas with green pods when 5 offspring peas are generated. With P(green pod) = 0.75, we can use the multiplication rule from Section 4-2 to find the probability that the first 3 peas have green pods while the last 2 peas do not have green pods. We get the following result:

P13 peas with green pods followed by 2 peas with pods that are not green2 = 0.75 # 0.75 # 0.75 # 0.25 # 0.25 = 0.753 # 0.252 = 0.0264

This result gives a probability of generating five offspring in which three have green pods. However, it does not give the probability of getting exactly three peas with green pods be- cause it assumes a particular arrangement for three offspring peas with green pods. Other arrangements for generating three offspring peas with green pods are possible.

In Section 4-6 we saw that with three subjects identical to each other (such as peas with green pods) and two other subjects identical to each other (such as peas without green pods), the total number of arrangements, or permutations, is 5!>315 - 32!3!4, or 10. Each of those 10 different arrangements has a probability of 0.753 # 0.252, so the total probability is as follows:

P13 peas with green pods among 52 = 5!15 - 32!3! # 0.753 # 0.252

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5-2 Binomial Probability Distributions 201

This particular result can be generalized as the binomial probability formula (Formula 5-5). That is, the binomial probability formula is a combination of the multiplication rule of probability and the counting rule for the number of arrange- ments of n items when x of them are identical to each other and the other n - x are identical to each other.

P (x) = n!

(n - x)!x! # px # qn - x

The number of outcomes with exactly x successes among n trials

The probability of x successes among n trials for any one particular order

2 2

Binomial Distributions Access tech instructions, videos, and data sets at www.TriolaStats.com

TECH CENTER

Statistical Literacy and Critical Thinking

1.  Hybridization Assume that 75% of offspring peas have green pods. Suppose we want to find the probability that when five offspring peas are randomly selected, exactly two of them are green. What is wrong with using the multiplication rule to find the prob- ability of getting two peas with green pods followed by three peas with yellow pods: (0.75)(0.75)(0.25)(0.25)(0.25) = 0.00879?

2. Variation and Notation Assume that we want to find the probability that among five off- spring peas, exactly two of them have green pods. Also assume that 75% of offspring peas have green pods (and the others have yellow pods).

a. Identify the values of n, x, p, and q.

b. For groups of 5 randomly selected offspring peas, find the mean, standard deviation, and variance for the numbers of peas among five that have green pods. Include appropriate units.

3.  Independent Events Based on a KRC Research survey, when 1020 adults were asked about hand hygiene, 44% said that they wash their hands after using public transportation. Con- sider the probability that among 30 different adults randomly selected from the 1020 who were surveyed, there are at least 10 who wash their hands after using public transportation. Given that these subjects were selected without replacement, are the 30 selections independent? Can they be treated as being independent? Can the probability be found using the binomial prob- ability formula?

4.  Notation of 0+ Using the same survey from Exercise 3, the probability of randomly selecting 30 of the 1020 adults and getting exactly 24 who wash hands after using public trans- portation is represented as 0+. What does 0+ indicate? Does 0+ indicate that it is impossible to get exactly 24 adults who wash their hands after using public transportation?

5-2 Basic Skills and Concepts

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202 CHAPTER 5 Discrete Probability Distributions

Identifying Binomial Distributions. In Exercises 5–12, determine whether the given procedure results in a binomial distribution (or a distribution that can be treated as bino- mial). For those that are not binomial, identify at least one requirement that is not satisfied.

5.  Clinical Trial of YSORT The YSORT method of gender selection, developed by the Genetics & IVF Institute, was designed to increase the likelihood that a baby will be a boy. When 291 couples used the YSORT method and gave birth to 291 babies, the weights of the babies were recorded.

6.  Clinical Trial of YSORT The YSORT method of gender selection, developed by the Genetics & IVF Institute, was designed to increase the likelihood that a baby will be a boy. When 291 couples use the YSORT method and give birth to 291 babies, the genders of the babies are recorded.

7.  Clinical Trial of Lipitor Treating 863 subjects with Lipitor (atorvastatin) and recording whether there is a “yes” response when they are each asked if they experienced a headache (based on data from Pfizer, Inc.).

8. Clinical Trial of Lipitor Treating 863 subjects with Lipitor (atorvastatin) and asking each subject how their head feels (based on data from Pfizer, Inc.).

9.  Nicorette Treating 50 smokers with Nicorette and asking them how their mouth and throat feel.

10. Nicorette Treating 50 smokers with Nicorette and recording whether there is a “yes” re- sponse when they are asked if they experience any mouth or throat soreness.

11. Defibrillators Determining whether each of 500 defibrillators is acceptable or defective.

12. Defibrillators Counting the numbers of defects in each of 500 defibrillators.

Binomial Probability Formula. In Exercises 13 and 14, answer the questions designed to help understand the rationale for the binomial probability formula.

13. Guessing Answers Standard tests, such as the SAT, ACT, or Medical College Admis- sion Test (MCAT) typically use multiple choice questions, each with five possible answers (a, b, c, d, e), one of which is correct. Assume that you guess the answers to the first three ques- tions.

a. Use the multiplication rule to find the probability that the first two guesses are wrong and the third is correct. That is, find P(WWC), where W denotes a wrong answer and C denotes a correct answer.

b. Beginning with WWC, make a complete list of the different possible arrangements of two wrong answers and one correct answer; then find the probability for each entry in the list.

c. Based on the preceding results, what is the probability of getting exactly one correct answer when three guesses are made?

14.  Vision Correction 53% of adults use eyeglasses for vision correction (based on data from a Vision Council survey). Four adults are randomly selected.

a. Use the multiplication rule to find the probability that the first three use eyeglasses and the fourth does not use eyeglasses. That is, find P(EEEN), where E denotes an adult who uses eye- glasses and N denotes an adult who does not use eyeglasses.

b. Beginning with EEEN, make a complete list of the different possible arrangements of three adults who use eyeglasses and one who does not use eyeglasses; then find the probability for each entry in the list.

c. Based on the preceding results, what is the probability of getting exactly three adults who use eyeglasses and one who does not?

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5-2 Binomial Probability Distributions 203

MCAT Test. In Exercises 15–20, assume that random guesses are made for eight multiple choice questions on an MCAT test, so that there are n = 8 trials, each with probability of success (correct) given by p = 0.20. Find the indicated probability for the number of correct answers.

15. Find the probability that the number x of correct answers is exactly 7.

16. Find the probability that the number x of correct answers is at least 4.

17. Find the probability that the number x of correct answers is fewer than 3.

18. Find the probability that the number x of correct answers is no more than 2.

19. Find the probability of no correct answers.

20. Find the probability that at least one answer is correct.

In Exercises 21–24, assume that when adults are randomly selected, 21% do not require vision correction (based on data from a Vision Council survey).

21. If 8 adults are randomly selected, find the probability that exactly 2 of them do not require vision correction.

22. If 20 adults are randomly selected, find the probability that exactly 5 of them do not require vision correction.

23. If 10 adults are randomly selected, find the probability that at least 3 of them do not require vision correction.

24.  If 12 adults are randomly selected, find the probability that fewer than 3 of them do not require vision correction.

Significance with Range Rule of Thumb. In Exercises 25 and 26, assume that different groups of couples use the XSORT method of gender selection and each couple gives birth to one baby. The XSORT method is designed to increase the likelihood that a baby will be a girl, but assume that the method has no effect, so the probability of a girl is 0.5.

25. Gender Selection Assume that the groups consist of 36 couples.

a. Find the mean and standard deviation for the numbers of girls in groups of 36 births.

b. Use the range rule of thumb to find the values separating results that are significantly low or significantly high.

c. Is the result of 26 girls a result that is significantly high? What does it suggest about the ef- fectiveness of the XSORT method?

26. Gender Selection Assume that the groups consist of 16 couples.

a. Find the mean and standard deviation for the numbers of girls in groups of 16 births.

b. Use the range rule of thumb to find the values separating results that are significantly low or significantly high.

c. Is the result of 11 girls a result that is significantly high? What does it suggest about the ef- fectiveness of the XSORT method?

Significance with Range Rule of Thumb. In Exercises 27 and 28, assume that hybrid- ization experiments are conducted with peas having the property that for offspring, there is a 0.75 probability that a pea has green pods (as in one of Mendel’s famous experiments).

27. Hybrids Assume that offspring peas are randomly selected in groups of 10.

a. Find the mean and standard deviation for the numbers of peas with green pods in the groups of 10.

continued

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204 CHAPTER 5 Discrete Probability Distributions

b. Use the range rule of thumb to find the values separating results that are significantly low or significantly high.

c. Is the result of 9 peas with green pods a result that is significantly high? Why or why not?

28. Hybrids Assume that offspring peas are randomly selected in groups of 16.

a. Find the mean and standard deviation for the numbers of peas with green pods in the groups of 16.

b. Use the range rule of thumb to find the values separating results that are significantly low or significantly high.

c. Is a result of 7 peas with green pods a result that is significantly low? Why or why not?

Composite Sampling. Exercises 29 and 30 involve the method of composite sampling, whereby a medical testing laboratory saves time and money by combining blood samples for tests so that only one test is conducted for several people. A combined sample tests positive if at least one person has the disease. If a combined sample tests positive, then individual blood tests are used to identify the individual with the disease.

29. HIV It is estimated that worldwide, 1% of those aged 15–49 are infected with the human immunodeficiency virus (HIV) (based on data from the National Institutes of Health). In tests for HIV, blood samples from 36 people are combined. What is the probability that the com- bined sample tests positive for HIV? Is it unlikely for such a combined sample to test positive?

30.  Blood Donor Testing The American Red Cross tests every unit of donated blood for several infectious diseases, including hepatitis B, hepatitis C, HIV, syphilis, and West Nile virus infection. Blood samples from 16 donors are combined and tested, and all 16 individual samples are approved only if the combined sample passes all tests. If there is 1.4% chance that a random individual fails any of the tests, find the probability that the combined sample is not approved.

Acceptance Sampling. Exercises 31 and 32 involve the method of acceptance sampling, whereby a shipment of a large number of items is accepted based on test results from a sample of the items.

31. Aspirin The MedAssist Pharmaceutical Company receives large shipments of aspirin tab- lets and uses this acceptance sampling plan: Randomly select and test 40 tablets, and then ac- cept the whole batch if there is only one or none that doesn’t meet the required specifications. If one shipment of 5000 aspirin tablets actually has a 3% rate of defects, what is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected?

32.  AAA Batteries AAA batteries are made by companies including Duracell, Energizer, Eveready, and Panasonic, and they are used to power Prestige Medical Xenon pocket otoscopes (those things that physicians use to look in your ears). When purchasing bulk orders of AAA batteries, a manufacturer of otoscopes uses this acceptance sampling plan: Randomly select 50 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 2 batteries do not meet specifications. A shipment contains 2000 AAA batteries, and 2% of them do not meet specifications. What is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected?

Ultimate Binomial Exercises! Exercises 33−36 involve finding binomial probabilities, finding parameters, and determining whether values are significantly high or low by using the range rule of thumb and probabilities.

33.  Gender Selection At an early stage of clinical trials of the XSORT method of gender selection, 14 couples using that method gave birth to 13 girls and 1 boy.

a. Assuming that the XSORT method has no effect and boys and girls are equally likely, use the range rule of thumb to identify the limits separating values that are significantly low and

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5-2 Binomial Probability Distributions 205

those that are significantly high (for the number of girls in 14 births). Based on the results, is the result of 13 girls significantly high?

b. Find the probability of exactly 13 girls in 14 births, assuming that the XSORT method has no effect.

c. Find the probability of 13 or more girls in 14 births, assuming that the XSORT method has no effect.

d. Which probability is relevant for determining whether 13 girls is significantly high: the probability from part (b) or part (c)? Based on the relevant probability, is the result of 13 girls significantly high?

e. What do the results suggest about the effectiveness of the XSORT method?

34. Clinical Trial A treatment for hypertension has been found to be successful in 60% of the patient population. In a test of a new treatment, 40 subjects are treated for hypertension and 29 of these subjects experience success with the new treatment.

a. Assuming that the old success rate of 60% still applies, use the range rule of thumb to iden- tify the limits separating numbers of successes that are significantly low or significantly high. Based on the results, is 29 successes among the 40 subjects significantly high?

b. Find the probability that exactly 29 of the 40 cases are successes, assuming that the general success rate is 60%.

c. Find the probability that 29 or more of the cases are successes, assuming that the general success rate is 60%.

d. Which probability is relevant for determining whether 29 successes is significantly high: the probability from part (b) or part (c)? Based on the relevant probability, is the result of 29 suc- cesses significantly high?

e. What do the results suggest about the effectiveness of the new treatment?

35. Hybrids One of Mendel’s famous experiments with peas included 47 offspring, and 34 of them had long stems. Mendel claimed that under the same conditions, 75% of offspring peas would have long stems. Assume that Mendel’s claim of 75% is true, and assume that a sample consists of 47 offspring peas.

a. Use the range rule of thumb to identify the limits separating values that are significantly low and those that are significantly high. Based on the results, is the result of 34 peas with long stems either significantly low or significantly high?

b. Find the probability of exactly 34 peas with long stems.

c. Find the probability of 34 or fewer peas with long stems.

d. Which probability is relevant for determining whether 34 peas with long stems is signifi- cantly low: the probability from part (b) or part (c)? Based on the relevant probability, is the result of 34 peas with long stems significantly low?

e. What do the results suggest about Mendel’s claim of 75%?

36. Vaccine For a specific group of subjects, there is a 5% chance of influenza (“flu”). When 80 subjects were treated with a vaccine, only one of them presented with influenza.

a. Use the range rule of thumb to identify the limits separating values that are significantly low and those that are significantly high. Based on the results, is the result of one subject getting influenza either significantly low or significantly high?

b. Find the probability of exactly one subject experiencing influenza, assuming that the vaccine has no effect.

c. Find the probability of one or fewer subjects experiencing influenza, assuming that the vaccine has no effect. continued

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206 CHAPTER 5 Discrete Probability Distributions

d. Which probability is relevant for determining whether one subject experiencing influenza is significantly low: the probability from part (b) or part (c)? Based on the relevant probability, is the result of one subject experiencing influenza significantly low?

e. What do the results suggest about the effectiveness of the vaccine?

37. Geometric Distribution If a procedure meets all the conditions of a binomial distribution except that the number of trials is not fixed, then the geometric distribution can be used. The probability of getting the first success on the xth trial is given by P1x2 = p11 - p2x - 1, where p is the probability of success on any one trial. Subjects are randomly selected for the National Health and Nutrition Examination Survey conducted by the National Center for Health Statis- tics, Centers for Disease Control and Prevention. The probability that someone is a universal donor (with group O and type Rh negative blood) is 0.06. Find the probability that the first subject to be a universal blood donor is the fifth person selected.

38. Multinomial Distribution The binomial distribution applies only to cases involving two types of outcomes, whereas the multinomial distribution involves more than two categories. Suppose we have three types of mutually exclusive outcomes denoted by A, B, and C. Let P1A2 = p1, P1B2 = p2, and P1C2 = p3. In n independent trials, the probability of x1 out- comes of type A, x2 outcomes of type B, and x3 outcomes of type C is given by

n! 1x12!1x22!1x32!

# px11 # px22 # px33 Data Set 8 “IQ and Lead” in Appendix B includes 78 subjects from a low lead exposure group, 22 subjects from a medium lead exposure group, and 21 subjects from a high lead exposure group. Find the probability of randomly selecting 10 subjects for a follow-up study and getting 5 from the low lead group, 2 from the medium lead group, and 3 from the high lead group. Assume that the selections are made with replacement. Can we use the above expression for finding the probability if the sampling is done without replacement?

39. Hypergeometric Distribution If we sample from a small finite population without re- placement, the binomial distribution should not be used because the events are not indepen- dent. If sampling is done without replacement and the outcomes belong to one of two types, we can use the hypergeometric distribution. If a population has A objects of one type, while the remaining B objects are of the other type, and if n objects are sampled without replacement, then the probability of getting x objects of type A and n - x objects of type B is

P1x2 = A!1A - x2!x! # B! 1B - n + x2!1n - x2! ,

1A + B2! 1A + B - n2!n!

In a medical research project, there are 20 subjects available and 4 of them are infected with HIV, while the other 16 are not infected. If 8 of the subjects are randomly selected without re- placement, what is the probability that 3 of the subjects are infected with HIV, while the other 5 are not infected? What is the probability if the sampling is done with replacement?

5-2 Beyond the Basics

Key Concept In Section 5-1 we introduced general discrete probability distributions and in Section 5-2 we considered binomial probability distributions, which is one particular category of discrete probability distributions. In this section we introduce Poisson probability distributions, which are another category of discrete probability distributions.

5-3 Poisson Probability Distributions

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5-3 Poisson Probability Distributions 207

The following definition states that Poisson distributions are used with occur- rences of an event over a specified interval, and here are some applications:

■ Number of Internet users logging onto WebMD in one day

■ Number of patients arriving at an emergency room in one hour

■ Number of Atlantic hurricanes in one year

DEFINITION

A Poisson probability distribution is a discrete probability distribution that ap- plies to occurrences of some event over a specified interval. The random variable x is the number of occurrences of the event in an interval. The interval can be time, distance, area, volume, or some similar unit. The probability of the event occurring x times over an interval is given by Formula 5-9.

FORMULA 5-9 Poisson Probability Distribution

P1x2 = m x # e-m x!

where

e ≈ 2.71828

m = mean number of occurrences of the event in the intervals

Requirements for the Poisson Probability Distribution

1. The random variable x is the number of occurrences of an event in some interval.

2. The occurrences must be random.

3. The occurrences must be independent of each other.

4. The occurrences must be uniformly distributed over the interval being used.

Parameters of the Poisson Probability Distribution

■ The mean is m. ■ The standard deviation is s = 1m.

Properties of the Poisson Probability Distribution

1. A particular Poisson distribution is determined only by the mean μ.

2. A Poisson distribution has possible x values of 0, 1, 2, . . . with no upper limit.

EXAMPLE 1 Hospital Births

In a recent year, there were 4229 births at NYU Langone Medical Center (based on data from the NYU Langone website). Assume that the number of births each day is about the same, and assume that the Poisson distribution is a suitable model.

a. Find m, the mean number of births per day.

b. Find the probability that on a randomly selected day, there are exactly 10 births. That is, find P(10), where P(x) is the probability of x births in a day.

continued

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208 CHAPTER 5 Discrete Probability Distributions

Poisson Distribution as Approximation to Binomial The Poisson distribution is sometimes used to approximate the binomial distribution when n is large and p is small. One rule of thumb is to use such an approximation when the following two requirements are both satisfied.

Requirements for Using Poisson as an Approximation to Binomial

1. n Ú 100 2. np … 10

If both requirements are satisfied and we want to use the Poisson distribution as an approximation to the binomial distribution, we need a value for m. That value can be calculated by using Formula 5-6 (from Section 5-2):

S O L U T I O N

a. The Poisson distribution applies because we are dealing with the occurrences of an event (births) over some interval (a day). The mean number of births per day is

m = Number of births Number of days

= 4229 365

= 11.5863

b. Using Formula 5-9, the probability of x = 10 births in a day is found as shown here (with x = 10, m = 11.5863, and e = 2.71828):

P1102 = m x # e-m x!

= 11.586310 # 2.71828-11.5863

10! = 0.112

The probability of exactly 10 births in a day is 0.112.

FORMULA 5-6 Mean for Poisson as an Approximation to Binomial

m = np

EXAMPLE 2 Influenza

In one year, the rate of influenza is 5%. If 120 people are randomly selected, find the probability of getting at least one who contracts influenza.

S O L U T I O N

The time interval is a year. With n = 120 and p = 0.05, the conditions n Ú 100 and np … 10 are both satisfied, so we can use the Poisson distribution as an approx- imation to the binomial distribution. We first need the value of m, which is found as follows:

m = np = 1120210.052 = 6 Having found the value of m, we can proceed to find the probability for specific values of x. Because we want the probability that x is “at least 1,” we will use the clever strategy of first finding P(0), the probability of no subjects getting influenza.

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5-3 Poisson Probability Distributions 209

The probability of at least one subject getting influenza can then be found by sub- tracting that result from 1. We find P(0) by using x = 0, m = 6, and e = 2.71828, as shown here:

P102 = m x # e-m x !

= 60 # 2.71828-6

0! =

1 # 0.00248 1

= 0.00248

Using the Poisson distribution as an approximation to the binomial distribution, we find that there is a 0.00248 probability of no subjects with influenza, so the prob- ability of at least one subject with influenza is 1 - 0.00248 = 0.998. If we use the binomial distribution, we again get a probability of 0.998, so the Poisson approxi- mation is quite good here.

Poisson Distributions Access tech instructions, videos, and data sets at www.TriolaStats.com

TECH CENTER

Statistical Literacy and Critical Thinking

1.  Notation In analyzing patient admissions at NYU Langone Medical Center, we find that 31,645 patients were admitted in a recent year (based on data from the NYU Langone website). Assume that we want to find the probability of exactly 85 patient admissions in a randomly selected day. In applying Formula 5-9, identify the values of m, x, and e. Also, briefly describe what each of those symbols represents.

2. Patient Admissions Use the same patient admission data given in Exercise 1. Let the ran- dom variable x represent the number of patient admissions in one day, and assume that it has a Poisson distribution. What is the standard deviation for the values of the random variable x? What is the variance?

3.  Poisson Probability Distribution The random variable x represents the number of pa- tient admissions in a day, as described in Exercise 1. Assume that the random variable x has a Poisson distribution. What are the possible values of x? Is a value of x = 90.3 possible? Is x a discrete random variable or a continuous random variable?

4. Probability if 0 For Formula 5-9, what does P(0) represent? Simplify Formula 5-9 for the case in which x = 0.

Births. In Exercises 5–8, assume that the Poisson distribution applies, assume that the mean number of births at the NYU Langone Medical Center is 11.5863 per day, and proceed to find the probability that in a randomly selected day, the number of births is the value given.

5. Births Find the probability that in a day, there will be exactly 12 births.

6. Births Find the probability that in a day, there will be exactly 9 births.

7. Births Find the probability that in a day, there will be at least 1 birth.

8. Births Find the probability that in a day, there will be at least 2 births.

9. Murders In a recent year, there were 333 murders in New York City. Find the mean number of murders per day; then use that result to find the probability that in a day, there are no mur- ders. Does it appear that there are expected to be many days with no murders?

5-3 Basic Skills and Concepts

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210 CHAPTER 5 Discrete Probability Distributions

10. Deaths from Horse Kicks A classical example of the Poisson distribution involves the number of deaths caused by horse kicks to men in the Prussian Army between 1875 and 1894. Data for 14 corps were combined for the 20-year period, and the 280 corps-years included a to- tal of 196 deaths. After finding the mean number of deaths per corps-year, find the probability that a randomly selected corps-year has the following numbers of deaths: (a) 0, (b) 1, (c) 2, (d) 3, (e) 4. The actual results consisted of these frequencies: 0 deaths (in 144 corps-years); 1 death (in 91 corps-years); 2 deaths (in 32 corps-years); 3 deaths (in 11 corps-years); 4 deaths (in 2 corps-years). Compare the actual results to those expected by using the Poisson probabilities. Does the Poisson distribution serve as a good tool for predicting the actual results?

11.  World War II Bombs In analyzing hits by V-1 buzz bombs in World War II, South London was partitioned into 576 regions, each with an area of 0.25 km2. A total of 535 bombs hit the combined area of 576 regions.

a. Find the probability that a randomly selected region had exactly 2 hits.

b. Among the 576 regions, find the expected number of regions with exactly 2 hits.

c. How does the result from part (b) compare to this actual result: There were 93 regions that had exactly 2 hits?

12. Disease Cluster Neuroblastoma, a rare form of cancer, occurs in 11 children in a million, so its probability is 0.000011. Four cases of neuroblastoma occurred in Oak Park, Illinois, which had 12,429 children.

a. Assuming that neuroblastoma occurs as usual, find the mean number of cases in groups of 12,429 children.

b. Using the unrounded mean from part (a), find the probability that the number of neuroblas- toma cases in a group of 12,429 children is 0 or 1.

c. What is the probability of more than one case of neuroblastoma?

d. Does the cluster of four cases appear to be attributable to random chance? Why or why not?

13. Car Fatalities The recent rate of car fatalities was 33,561 fatalities for 2969 billion miles traveled (based on data from the National Highway Traffic Safety Administration). Find the probability that for the next billion miles traveled, there will be at least one fatality. What does the result indicate about the likelihood of at least one fatality?

14. Dandelions Dandelions are studied for their effects on crop production and lawn growth. In one region, the mean number of dandelions per square meter was found to be 7.0 (based on data from Manitoba Agriculture and Food).

a. Find the probability of no dandelions in an area of 1 m2.

b. Find the probability of at least one dandelion in an area of 1 m2.

c. Find the probability of at most two dandelions in an area of 1 m2.

15. Rubella During the last 13 years in the United States, there were 138 cases of rubella.

a. Find the mean number of cases of rubella per year. Round the result to four decimal places.

b. Find the probability of no cases of rubella in a year.

c. Find the probability of exactly 9 cases of rubella in a year. How does it compare to the 2 years among 13 that had exactly 9 cases of rubella?

16.  Diphtheria During the past 34 years, there were 56 cases of diphtheria in the United States.

a. Find the mean number of cases of diphtheria per year. Express the result with five decimal places.

b. Find the probability of no cases of diphtheria in a year.

c. Find the probability that the number of diphtheria cases in a year is 5 or fewer. If a year has more than 5 cases of diphtheria, is that a significantly high number?

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6. Drinking Does the table describe a probability distribution? Why or why not?

7.  Drinking Find the mean of the number of heavy drinkers in groups of five randomly se- lected adult males.

8. Drinking Based on the table, the standard deviation is 0.5 male. What is the variance? In- clude appropriate units.

9. Drinking What does the probability of 0 + indicate? Does it indicate that among five ran- domly selected adult males, it is impossible for all of them to be heavy drinkers?

10. Drinking What is the probability that fewer than three of the five adult males are heavy drinkers? If we were to find that among 5 randomly selected adult males, there are 4 heavy drinkers, is 4 significantly high?

17. Probability Histogram for a Poisson Distribution Construct the probability histogram for Exercise 16. Is the Poisson probability distribution a normal distribution or is it skewed?

5-3 Beyond the Basics

In Exercises 1–10, use the following: Based on data from the National Center for Health Statistics, 20.6% of adult males smoke. A random sample of 64 male adults is obtained.

1. Smoking Find the mean number of smokers in groups of 64 randomly selected adult males.

2. Smoking Find the standard deviation of the number of smokers in groups of 64 randomly selected adult males.

3. Smoking Are the results from Exercises 1 and 2 statistics or parameters?

4. Smoking For a random sample of 64 males, find the numbers separating the outcomes that are significantly high or significantly low.

5.  Smoking Find the probability that the first 8 randomly selected males include exactly 3 who smoke.

In Exercises 6–10, use the following: Five male adults are randomly selected, and the table in the margin lists the probabilities for the number that are heavy drinkers (based on data from the National Center for Health Statistics). Males are considered to be heavy drinkers if they have at least 14 drinks per week “on average.”

Chapter Quick Quiz

x P(x)

0 0.762

1 0.213

2 0.024

3 0.001

4 0 +

5 0 +

In Exercises 1–5, assume that 28% of randomly selected adults have high cholesterol (with a level of at least 240 mg ,dL or are taking medicine to reduce cholesterol), based on results from the National Center for Health Statistics. Assume that a group of five adults is ran- domly selected.

1. Cholesterol Find the probability that exactly two of the five adults have high cholesterol.

2.  Cholesterol Find the probability that at least one of the five adults has high cholesterol. Does the result apply to five adults from the same family? Why or why not?

3. Cholesterol Find the mean and standard deviation for the numbers of adults in groups of five who have high cholesterol.

4. Cholesterol If all five of the adults have high cholesterol, is five significantly high? Why or why not?

Review Exercises

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212 CHAPTER 5 Discrete Probability Distributions

5.  Cholesterol If the group of five adults includes exactly 1 with high cholesterol, is that value of 1 significantly low?

6.  Security Survey In a USA Today poll, subjects were asked if passwords should be re- placed with biometric security, such as fingerprints. The results from that poll have been used to create the accompanying table. Does this table describe a probability distribution? Why or why not?

Response P(x)

Yes 0.53

No 0.17

Not sure 0.30

7. Condom Failure Rate According to the Department of Health and Human Services, the failure rate for male condoms is 18%. The accompanying table is based on the failure rate of 18%, where x represents the number of condoms that fail when six are tested.

a. Does the table describe a probability distribution? Why or why not?

b. Assuming that the table does describe a probability distribution, find its mean.

c. Assuming that the table does describe a probability distribution, find its standard deviation.

d. If 6 condoms are tested and 5 of them fail, is 5 a significantly high number of failures? Why or why not?

e. What does the symbol 0 + represent?

x P(x)

0 0.304

1 0.400

2 0.220

3 0.064

4 0.011

5 0.001

6 0 +

8.  Poisson: Deaths Currently, an average of 143 residents of Madison, CT (population 17,858), die each year (based on data from the U.S. National Center for Health Statistics).

a. Find the mean number of deaths per day.

b. Find the probability that on a given day, there are no deaths.

c. Find the probability that on a given day, there are more than two deaths.

d. Based on the preceding results, should Madison have a contingency plan to handle more than two deaths per day? Why or why not?

Cumulative Review Exercises

1.  Manatee Deaths Listed below are the annual numbers of manatee deaths from boats in Florida for each of the past 10 years, listed in chronological order.

69 79 92 73 90 97 83 88 81 72

a. Find the mean.

b. Find the median.

c. Find the range.

d. Find the standard deviation.

e. Find the variance.

f. Describe an important characteristic of the data that is not addressed by the statistics found in parts (a) through (e).

g. Use the range rule of thumb to identify the values separating significant values from those that are not significant.

h. Based on the result from part (f), do any of the years have a number of manatee deaths that is significantly low or significantly high?

i. What is the level of measurement of the data: nominal, ordinal, interval, or ratio?

j. Are the data discrete or continuous?

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2. Analysis of Last Digits The accompanying table lists the last or rightmost digits of weights of the females listed in Data Set 1 “Body Data” in Appendix B. The last digits of a data set can sometimes be used to determine whether the data have been measured or simply reported. The presence of disproportionately more 0s and 5s is often a sure sign that the data have been re- ported instead of measured.

a. Using the table, find the mean and standard deviation of those last digits. Are the results statistics or parameters?

b. Examine the given table to determine if there is anything about the sample data (such as disproportionately more 0s and 5s) suggesting that the given last digits are not random? Or do they appear to be random?

c. Does the table describe a probability distribution? Why or why not?

x f

0 11

1 11

2 16

3 14

4 16

5 24

6 13

7 14

8 19

9 9

3. Government Health Plan Fox News broadcast a graph similar to the one shown here. The graph is intended to compare the number of people actually enrolled in a government health plan (left bar) and the goal for the number of enrollees (right bar). Does the graph depict the data correctly or is it somehow misleading? Explain.

4. Diabetes Among adults in the United States, 11.5% have diabetes (based on data from the National Center for Health Statistics).

a. Find the probability that a randomly selected adult does not have diabetes.

b. Find the probability that two randomly selected adults both have diabetes.

c. Find the probability that among three randomly selected adults, at least one has diabetes.

d. For groups of 40 randomly selected adults, find the mean and standard deviation for the numbers of adults having diabetes. Are these results statistics or parameters?

e. If 40 adults are randomly selected and 10 of them have diabetes, is 10 a result that is signifi- cantly low or significantly high? Why?

CHAPTER 5 Cumulative Review Exercises 213

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214 CHAPTER 5 Discrete Probability Distributions

FROM DATA TO DECISION

Critical Thinking: Determining criteria for concluding that a gender selection method is effective

You are responsible for analyzing results from a clinical trial of the effectiveness of a new method of gender selection. Assume that the sample size of n = 75 couples has already been established, and each couple will have one child. Fur- ther assume that each of the couples will be subjected to a treatment that supposedly increases the likelihood that the child will be a girl. Assume that with no treatment, the prob- ability of a baby being a girl is 0.488, which is currently the correct value in the United States.

There is a danger in obtaining results first, then making conclusions about the results. If the results are close to show- ing the effectiveness of a treatment, it might be tempting to

conclude that there is an effect when, in reality, there is no ef- fect. It is better to establish criteria before obtaining results.

a. Using the methods of this chapter, identify the criteria that should be used for concluding that the treatment is effective in in- creasing the likelihood of a girl. Among the 75 births, how many girls would you require in order to conclude that the gender selec- tion procedure is effective? Explain how you arrived at this result.

b. If 60% of the 75 babies are girls, is that result high enough to conclude that the gender selection method is effective? Why or why not?

c. If 64% of the 75 babies are girls, is that result high enough to conclude that the gender selection method is effective? Why or why not?

Cooperative Group Activities

1. In-class activity Win $1,000,000! The James Randi Educational Foundation offers a $1,000,000 prize to anyone who can show “under proper observing conditions, evidence of any paranormal, supernatural, or occult power or event.” Divide into groups of three. Select one person who will be tested for extrasensory perception (ESP) by trying to correctly identify a digit (0–9) randomly selected by another member of the group. Conduct at least 20 trials. Another group member should record the randomly selected digit, the digit guessed by the subject, and whether the guess was correct or wrong. Construct the table for the probability distribution of randomly generated digits, construct the relative frequency table for the ran- dom digits that were actually obtained, and construct a relative frequency table for the guesses that were made. After comparing the three tables, what do you conclude? What proportion of guesses is correct? Does it seem that the subject has the ability to select the correct digit signifi- cantly more often than would be expected by chance?

Mendel’s Hybrid Experiments One of Mendel’s famous experiments with peas included 1064 offspring, and 787 of them had long stems. Mendel claimed that under the same condi- tions, 75% of offspring peas would have long stems. Assume that Mendel’s claim of 75% is true, and assume that a sample consists of 1064 offspring peas.

• Use the range rule of thumb to identify the limits separating values that are significantly low and those that are significantly high. Based on the results, is the result of 787 peas with long stems either significantly low or significantly high?

• Find the probability of exactly 787 peas with long stems.

• Find the probability of 787 or fewer peas with long stems.

• Which probability is relevant for determining whether 787 peas with long stems is signifi- cantly low: the probability from part (b) or part (c)? Based on the relevant probability, is the result of 787 peas with long stems significantly low?

• What do the results suggest about Mendel’s claim of 75%?

Technology Project

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2.  In-class activity See the preceding activity and design an experiment that would be ef- fective in testing someone’s claim that he or she has the ability to identify the color of a card selected from a standard deck of playing cards. Describe the experiment with great detail. Be- cause the prize of $1,000,000 is at stake, we want to be careful to avoid the serious mistake of concluding that the person has a paranormal power when that power is not actually present. There will likely be some chance that the subject could make random guesses and be correct every time, so identify a probability that is reasonable for the event of the subject passing the test with random guesses. Be sure that the test is designed so that this probability is equal to or less than the probability value selected.

3.  In-class activity Suppose we want to identify the probability distribution for the number of children in families with at least one child. For each student in the class, find the number of brothers and sisters and record the total number of children (including the student) in each family. Construct the relative frequency table for the result obtained. (The values of the random variable x will be 1, 2, 3, . . . .) What is wrong with using this relative frequency table as an es- timate of the probability distribution for the number of children in randomly selected families?

4. Out-of-class activity The analysis of the last digits of data can sometimes reveal whether the data have been collected through actual measurements or reported by the subjects. Refer to an almanac or the Internet and find a collection of data (such as lengths of rivers in the world), then analyze the distribution of last digits to determine whether the values were ob- tained through actual measurements.

5. Out-of-class activity The photos shown below depict famous statisticians with first names of David and John, not necessarily in the order shown. Conduct a survey by asking this ques- tion: “Which man is named David and which man is named John?” Do the respondents ap- pear to give results significantly different from what is expected with random guesses? (See “Who Do You Look Like? Evidence of Facial Stereotypes for Male Names” by Lea, Thomas, Lamkin, and Bell, Psychonomic Bulletin & Review, Vol. 14, Issue 5.)

CHAPTER 5 Cooperative Group Activities 215

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216

The Standard Normal Distribution

Real Applications of Normal Distributions

Sampling Distributions and Estimators

The Central Limit Theorem

Assessing Normality

Normal as Approximation to Binomial

6-1

6-2

6-3

6-4

6-5

6-6

What Is a Normal Pulse Rate?CHAPTER PROBLEM

Normal Probability Distributions

Exploring the pulse rates of adult males and females in Data

Set 1 “Body Data” from Appendix B reveals the following:

• Adult males have pulse rates with a mean of 69.6 bpm

(beats per minute), a standard deviation of 11.3 bpm, and a

distribution that is approximately normal.

• Adult females have pulse rates with a mean of 74.0 bpm,

a standard deviation of 12.5 bpm, and a distribution that is

approximately normal.

• There appears to be a significant difference between pulse

rates of males and females.

For the purposes of this chapter, we will use the above results as

reasonable estimates of population parameters. See the following:

m s Distribution

Male Adult Pulse Rates (bpm) 69.6 11.3 Normal

Female Adult Pulse Rates (bpm) 74.0 12.5 Normal

6

M06_TRIO9015_02_SE_C06_216-281.indd 216 11/08/16 4:59 PM

Physicians routinely measure pulse rates of patients, and the

normal range is generally considered to be between 60 bpm and

100 bpm. Here are conditions for pulse rates outside that range:

Tachycardia: Pulse rate greater than 100 bpm.

Bradycardia: Pulse rate less than 60 bpm

An excessively high pulse rate (tachycardia) is generally more

of a problem than an excessively low pulse rate (bradycar-

dia). An excessively high pulse rate can indicate a high risk of

stroke, heart disease, or can even cause death. An excessively

low pulse can occur with an athlete in peak physical condition,

and some drugs such as beta blockers can also cause an

excessively low pulse rate.

Here are some questions that can be addressed with the

methods of this chapter:

• What is the proportion of adult males who are expected to

have pulse rates greater than 100 bpm?

• What is the proportion of adult males who are expected to

have pulse rates less than 60 bpm.

• For males, if we introduce a criterion whereby the highest

1% of pulse rates are considered to be significantly high,

what is the cutoff?

• For males, if we introduce a criterion whereby the lowest 1%

of pulse rates are considered to be significantly low, what is

the cutoff?

These same questions can be posed for adult females, and we will

address such questions using the methods in this chapter.

Chapter 5 introduced discrete probability distributions, but in this chapter we introduce continuous probability distributions, and most of this chapter focuses on normal distri- butions. Here are the chapter objectives:

The Standard Normal Distribution

• Describe the characteristics of a standard normal distribution. • Find the probability of some range of z values in a standard normal distribution. • Find z scores corresponding to regions under the curve representing a standard

normal distribution.

Real Applications of Normal Distributions

• Develop the ability to describe a normal distribution (not necessarily a standard normal distribution).

• Find the probability of some range of values in a normal distribution. • Find x scores corresponding to regions under the curve representing a normal

distribution.

Sampling Distributions and Estimators

• Develop the ability to describe a sampling distribution of a statistic. • Determine whether a statistic serves as a good estimator of the corresponding

population parameter.

The Central Limit Theorem

• Describe what the central limit theorem states. • Apply the central limit theorem by finding the probability that a sample mean falls

within some specified range of values.

• Identify conditions for which it is appropriate to use a normal distribution for the distribution of sample means.

6-1

6-2

6-3

6-4

Chapter Objectives 217

CHAPTER OBJECTIVES

> > >

continued

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218 CHAPTER 6 Normal Probability Distributions

Key Concept In this section we present the standard normal distribution, which is a specific normal distribution having the following three properties:

1. Bell-shaped: The graph of the standard normal distribution is bell-shaped (as in Figure 6-1).

2. m = 0: The standard normal distribution has a mean equal to 0.

3. s = 1: The standard normal distribution has a standard deviation equal to 1.

In this section we develop the skill to find areas (or probabilities or relative frequen- cies) corresponding to various regions under the graph of the standard normal distri- bution. In addition, we find z scores that correspond to areas under the graph. These skills become important in the next section as we study nonstandard normal distribu- tions and the real and important applications that they involve.

Normal Distributions There are infinitely many different normal distributions, depending on the values used for the mean and standard deviation. We begin with a brief introduction to this general family of normal distributions.

6-1 The Standard Normal Distribution

Assessing Normality

• Develop the ability to examine histograms, outliers, and normal quantile plots to determine whether sample data appear to be from a population having a distribution that is approximately normal.

Normal as Approximation to Binomial

• Identify conditions for which it is appropriate to use a normal distribution as an ap- proximation to a binomial probability distribution.

• Use the normal distribution for approximating probabilities for a binomial distribution.

6-5

6-6

Assessing Normality

• Develop the ability to examine histograms, outliers, and normal quantile plots to determine whether sample data appear to be from a population having a distribution that is approximately normal.

Normal as Approximation to Binomial

• Identify conditions for which it is appropriate to use a normal distribution as an ap-Identify conditions for which it is appropriate to use a normal distribution as an ap-Identify conditions f proximation to a binomial probability distribution.

• Use the normal distribution for approximating probabilities for a binomial distribution.

DEFINITION

If a continuous random variable has a distribution with a graph that is symmetric and bell-shaped, as in Figure 6-1, and it can be described by the equation given as Formula 6-1, we say that it has a normal distribution.

FORMULA 6-1

y = e-

1 21x -ms 22

s22p

In this book, we won’t actually use Formula 6-1, but examining the right side of the equation reveals that any particular normal distribution is determined by two parame- ters: the population mean, m, and population standard deviation, s. (In Formula 6-1, x is a variable that can change, p = 3.14159, and e = 2.71828.) Once specific values are selected for m and s, Formula 6-1 is an equation relating x and y, and we can graph

m

Value

Curve is bell-shaped and symmetric

FIGURE 6-1 The Normal Distribution

M06_TRIO9015_02_SE_C06_216-281.indd 218 11/08/16 4:59 PM

6-1 The Standard Normal Distribution 219

that equation to get a result that will look like Figure 6-1. And that’s about all we need to know about Formula 6-1!

Uniform Distributions The major focus of this chapter is the concept of a normal probability distribution, but we begin with a uniform distribution so that we can see the following two very impor- tant properties:

1. The area under the graph of a continuous probability distribution is equal to 1.

2. There is a correspondence between area and probability, so probabilities can be found by identifying the corresponding areas in the graph using this formula for the area of a rectangle:

Area = height * width

EXAMPLE 1 Waiting Times for Emergency Room Check-In

During certain time periods at a hospital in New York City, patients arriving at the emergency room have waiting times that are uniformly distributed between 0 minutes and 5 minutes, as illustrated in Figure 6-2.

Refer to Figure 6-2 to see these properties:

■ All of the different possible waiting times are equally likely.

■ Waiting times can be any value between 0 min and 5 min, so it is possible to have a waiting time of 1.234567 min.

■ By assigning the probability of 0.2 to the height of the vertical line in Figure 6-2, the enclosed area is exactly 1. (In general, we should make the height of the ver- tical line in a uniform distribution equal to 1>range.)

0

0.2

1 2

Area 5 1

3 4 5

x (waiting time in minutes)

P (x

)

FIGURE 6-2 Uniform Distribution of Waiting Time

DEFINITION

A continuous random variable has a uniform distribution if its values are spread evenly over the range of possibilities. The graph of a uniform distribution results in a rectangular shape.

Density Curve The graph of any continuous probability distribution is called a density curve, and any density curve must satisfy the requirement that the total area under the curve is exactly 1. This requirement that the area must equal 1 simplifies probability problems, so the following statement is really important:

Because the total area under any density curve is equal to 1, there is a correspondence between area and probability.

M06_TRIO9015_02_SE_C06_216-281.indd 219 11/08/16 4:59 PM

220 CHAPTER 6 Normal Probability Distributions

Standard Normal Distribution The density curve of a uniform distribution is a horizontal straight line, so we can find the area of any rectangular region by applying this formula:

Area = height * width

Because the density curve of a normal distribution has a more complicated bell shape as shown in Figure 6-1, it is more difficult to find areas. However, the basic principle is the same: There is a correspondence between area and probability. In Figure 6-4 we show that for a standard normal distribution, the area under the density curve is equal to 1. In Figure 6-4, we use “z Score” as a label for the horizontal axis, and this is common for the standard normal distribution, defined as follows.

EXAMPLE 2 Waiting Times at an Emergency Room

Given the uniform distribution illustrated in Figure 6-2, find the probability that a randomly selected patient has a waiting time of at least 2 minutes.

S O L U T I O N

The shaded area in Figure 6-3 represents waiting times of at least 2 minutes. Because the total area under the density curve is equal to 1, there is a correspondence between area and probability. We can easily find the desired probability by using areas as follows:

P1wait time of at least 2 min2 = height * width of shaded area in Figure 6@3 = 0.2 * 3 = 0.6

0

0.2

1 2

Area 5 0.2 3 3 5 0.6

3 4 5

x (waiting time in minutes)

P (x

)

FIGURE 6-3 Using Area to Find Probability

I N T E R P R E TAT I O N

The probability of randomly selecting a patient with a waiting time of at least 2 minutes is 0.6.

1 2 30

z Score

Area 5 1

212223

FIGURE 6-4 Standard Normal Distribution

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6-1 The Standard Normal Distribution 221

Finding Probabilities When Given z Scores It is not easy to manually find areas in Figure 6-4, but we can find areas (or prob- abilities) for many different regions in Figure 6-4 by using technology, or we can also use Table A-2 (in Appendix A and the Formulas and Tables insert card). Key features of the different methods are summarized in Table 6-1, which follows. (StatCrunch provides options for a cumulative left region, a cumulative right region, or the region between two boundaries.) Because calculators and software generally give more ac- curate results than Table A-2, we strongly recommend using technology. (When there are discrepancies, answers in Appendix D will generally include results based on tech- nology as well as answers based on Table A-2.)

TABLE 6-1 Formats Used for Finding Normal Distribution Areas

Cumulative Area from the Left The following provide the cumulative area from the left up to a vertical line above a specific value of z:

• Table A-2 • Statdisk • Minitab • Excel • StatCrunch

z

Cumulative Left Region

Area Between Two Boundaries The following provide the area bounded on the left and bounded on the right by vertical lines above specific values.

• TI-83 ,84 Plus calculator • StatCrunch

UpperLower

Area Between Two Boundaries

DEFINITION

The standard normal distribution is a normal distribution with the parameters of m = 0 and s = 1. The total area under its density curve is equal to 1 (as in Figure 6-4).

Table A-2: If using Table A-2, it is essential to understand these points:

1. Table A-2 is designed only for the standard normal distribution, which is a nor- mal distribution with a mean of 0 and a standard deviation of 1.

2. Table A-2 is on two pages, with the left page for negative z scores and the right page for positive z scores.

3. Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z score.

continued

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222 CHAPTER 6 Normal Probability Distributions

The following examples illustrate procedures that can be used with real and im- portant applications introduced in the following sections.

CAUTION When working with a normal distribution, be careful to avoid confusion between z scores and areas.

z 5 1.270

Area 5 0.8980 (from Table A-2)

FIGURE 6-5 Finding Area to the Left of z = 1.27

EXAMPLE 3 Bone Density Test

A bone mineral density test can be helpful in identifying the presence or likelihood of osteoporosis, a disease causing bones to become more fragile and more likely to break. The result of a bone density test is commonly measured as a z score. The population of z scores is normally distributed with a mean of 0 and a standard de- viation of 1, so these test results meet the requirements of a standard normal distri- bution, and the graph of the bone density test scores is as shown in Figure 6-5.

A randomly selected adult undergoes a bone density test. Find the probability that this person has a bone density test score less than 1.27.

S O L U T I O N

Note that the following are the same (because of the aforementioned correspon- dence between probability and area):

■ Probability that the bone density test score is less than 1.27

■ Shaded area shown in Figure 6-5

So we need to find the area in Figure 6-5 below z = 1.27. If using Table A-2, begin with the z score of 1.27 by locating 1.2 in the left column; next find the value in the adjoining row of probabilities that is directly below 0.07, as shown in the excerpt on the top of the following page. Table A-2 shows that there is an area of 0.8980 corresponding to z = 1.27. We want the area below 1.27, and Table A-2 gives the cumulative area from the left, so the desired area is 0.8980. Because of the correspondence between area and probability, we know that the probability of a z score below 1.27 is 0.8980.

4. When working with a graph, avoid confusion between z scores and areas.

z score: Distance along the horizontal scale of the standard normal dis- tribution (corresponding to the number of standard deviations above or below the mean); refer to the leftmost column and top row of Table A-2.

Area: Region under the curve; refer to the values in the body of Table A-2.

5. The part of the z score denoting hundredths is found across the top row of Table A-2.

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6-1 The Standard Normal Distribution 223

TABLE A-2 (continued) Cumulative Area from the LEFT

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359

0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753

0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141

1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621

1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830

1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015

1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177

1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319

I N T E R P R E TAT I O N

The probability that a randomly selected person has a bone density test result below 1.27 is 0.8980, shown as the shaded region in Figure 6-5. Another way to interpret this result is to conclude that 89.80% of people have bone density levels below 1.27.

EXAMPLE 4 Bone Density Test: Finding the Area to the Right of a Value

Using the same bone density test from Example 3, find the probability that a ran- domly selected person has a result above - 1.00. A value above - 1.00 is considered to be in the “normal” range of bone density readings.

S O L U T I O N

We again find the desired probability by finding a corresponding area. We are look- ing for the area of the region to the right of z = - 1.00 that is shaded in Figure 6-6. The Statdisk display on the top of the following page shows that the area to the right of z = - 1.00 is 0.841345.

If we use Table A-2, we should know that it is designed to apply only to cumu- lative areas from the left. Referring to the page with negative z scores, we find that the cumulative area from the left up to z = - 1.00 is 0.1587, as shown in Figure 6-6. Because the total area under the curve is 1, we can find the shaded area by subtracting 0.1587 from 1. The result is 0.8413. Even though Table A-2 is designed only for cumulative areas from the left, we can use it to find cumulative areas from the right, as shown in Figure 6-6.

0.1587

z 5 –1.00

1. Use z 5 21.00 in Table A-2 to find this area. 2. Because

the total area is 1, this area is 1 2 0.1587 5 0.8413

FIGURE 6-6 Finding the Area to the Right of z = − 1 continued

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224 CHAPTER 6 Normal Probability Distributions

Example 4 illustrates a way that Table A-2 can be used indirectly to find a cumu- lative area from the right. The following example illustrates another way that we can find an area indirectly by using Table A-2.

I N T E R P R E TAT I O N

Because of the correspondence between probability and area, we conclude that the probability of randomly selecting someone with a bone density reading above - 1 is 0.8413 (which is the area to the right of z = - 1.00). We could also say that 84.13% of people have bone density levels above - 1.00.

z 5 21.00

0.1587

This area 0.1587

z 5 22.50

0.0062

minus this area 2 0.0062

22.50 21.00

0.1525

equals this area 5 0.1525

FIGURE 6-7 Finding the Area Between Two z Scores

EXAMPLE 5 Bone Density Test: Finding the Area Between Two Values

A bone density test reading between - 1.00 and - 2.50 indicates that the subject has osteopenia, which is some bone loss. Find the probability that a randomly selected subject has a reading between - 1.00 and - 2.50.

S O L U T I O N

We are again dealing with normally distributed values having a mean of 0 and a standard deviation of 1. The values between - 1.00 and - 2.50 correspond to the shaded region in the third graph included in Figure 6-7. Table A-2 cannot be used to find that area directly, but we can use this table to find the following:

1. The area to the left of z = - 1.00 is 0.1587. 2. The area to the left of z = - 2.50 is 0.0062. 3. The area between z = - 2.50 and z = - 1.00 (the shaded area at the far right in

Figure 6-7) is the di¢erence between the areas found in the preceding two steps:

Statdisk

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6-1 The Standard Normal Distribution 225

Example 5 can be generalized as the following rule:

The area corresponding to the region between two z scores can be found by finding the di�erence between the two areas found in Table A-2.

Figure 6-8 illustrates this general rule. The shaded region B can be found by calculat- ing the difference between two areas found from Table A-2.

I N T E R P R E TAT I O N

Using the correspondence between probability and area, we conclude that there is a probability of 0.1525 that a randomly selected subject has a bone density read- ing between - 1.00 and - 2.50. Another way to interpret this result is to state that 15.25% of people have osteopenia, with bone density readings between - 1.00 and - 2.50.

HINT Don’t try to memorize a rule or formula for this case. Focus on understanding by using a graph. Draw a graph, shade the desired area, and then get creative to think of a way to find the desired area by working with cumulative areas from the left.

0z Left z Right

A

B

Shaded area B 5 (areas A and B combined) — (area A)

FIGURE 6-8 Finding the Area Between Two z Scores

Probabilities such as those in the preceding examples can also be expressed with the following notation.

Notation

P1a 6 z 6 b2 denotes the probability that the z score is between a and b. P1z 7 a2 denotes the probability that the z score is greater than a. P1z 6 a2 denotes the probability that the z score is less than a. With this notation, P1- 2.50 6 z 6 - 1.002 = 0.1525 states in symbols that the

probability of a z score falling between - 2.50 and - 1.00 is 0.1525 (as in Example 5).

Finding z Scores from Known Areas Examples 3, 4, and 5 all involved the standard normal distribution, and they were all examples with this same format: Given z scores, find areas (or probabilities). In many cases, we need a method for reversing the format: Given a known area (or probability), find the corresponding z score. In such cases, it is really important to avoid confusion between z scores and areas. Remember, z scores are distances along the horizontal

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226 CHAPTER 6 Normal Probability Distributions

scale, but areas (or probabilities) are regions under the density curve. (Table A-2 lists z-scores in the left column and across the top row, but areas are found in the body of the table.) We should also remember that z scores positioned in the left half of the curve are always negative. If we already know a probability and want to find the cor- responding z score, we use the following procedure.

Procedure for Finding a z Score from a Known Area

1. Draw a bell-shaped curve and identify the region under the curve that corre- sponds to the given probability. If that region is not a cumulative region from the left, work instead with a known region that is a cumulative region from the left.

2. Use technology or Table A-2 to find the z score. With Table A-2, use the cu- mulative area from the left, locate the closest probability in the body of the table, and identify the corresponding z score.

Special Cases In the solution to Example 6 that follows, Table A-2 leads to a z score of 1.645, which is midway between 1.64 and 1.65. When using Table A-2, we can usually avoid interpolation by simply selecting the closest value. The accompanying table lists special cases that are often used in a wide variety of applications. (For one of those special cases, the value of z = 2.576 gives an area slightly closer to the area of 0.9950, but z = 2.575 has the advantage of being the value exactly midway be- tween z = 2.57 and z = 2.58.) Except in these special cases, we can usually select the closest value in the table. (If a desired value is midway between two table values, select the larger value.) For z scores above 3.49, we can use 0.9999 as an approxima- tion of the cumulative area from the left; for z scores below - 3.49, we can use 0.0001 as an approximation of the cumulative area from the left.

Special Cases in Table A-2

z Score

Cumulative Area from the Left

1.645 0.9500

-1.645 0.0500

2.575 0.9950

-2.575 0.0050

Above 3.49 0.9999

Below -3.49 0.0001

0

Area 5 0.95

z 5 ?

FIGURE 6-9 Finding the 95th Percentile

EXAMPLE 6 Bone Density Test: Finding a Test Score

Use the same bone density test scores used in earlier examples. Those scores are normally distributed with a mean of 0 and a standard deviation of 1, so they meet the requirements of a standard normal distribution. Find the bone density score cor- responding to P95, the 95th percentile. That is, find the bone density score that sepa- rates the bottom 95% from the top 5%. See Figure 6-9.

S O L U T I O N

Figure 6-9 shows the z score that is the 95th percentile, with 95% of the area (or 0.95) below it.

Technology: We could find the z score using technology. The following Excel display shows that the z score with an area of 0.95 to its left is z = 1.644853627, or 1.645 when rounded.

M06_TRIO9015_02_SE_C06_216-281.indd 226 11/08/16 4:59 PM

6-1 The Standard Normal Distribution 227

Excel

Table A-2: If using Table A-2, search for the area of 0.95 in the body of the table and then find the corresponding z score. In Table A-2 we find the areas of 0.9495 and 0.9505, but there’s an asterisk with a special note indicating that 0.9500 corresponds to a z score of 1.645. We can now conclude that the z score in Figure 6-9 is 1.645, so the 95th percentile is z = 1.645.

I N T E R P R E TAT I O N

For bone density test scores, 95% of the scores are less than or equal to 1.645, and 5% of them are greater than or equal to 1.645.

I N T E R P R E TAT I O N

For the population of bone density test scores, 2.5% of the scores are equal to or less than - 1.96 and 2.5% of the scores are equal to or greater than 1.96. Another in- terpretation is that 95% of all bone density test scores are between - 1.96 and 1.96.

0 z 5 1.96z 5 21.96

Area 5 0.025Area 5 0.025

To find this z score, locate the cumulative area to the left in Table A–2. Locate 0.975 in the body of Table A–2.

FIGURE 6-10 Finding z Scores

EXAMPLE 7 Bone Density Test

Using the same bone density test described in Example 3, we have a standard normal distribution with a mean of 0 and a standard deviation of 1. Find the bone density test score that separates the bottom 2.5% and find the score that separates the top 2.5%.

S O L U T I O N

The required z scores are shown in Figure 6-10. Those z scores can be found using technology. If using Table A-2 to find the z score located to the left, we search the body of the table for an area of 0.025. The result is z = - 1.96. To find the z score located to the right, we search the body of Table A-2 for an area of 0.975. (Remember that Table A-2 always gives cumulative areas from the left.) The result is z = 1.96. The values of z = - 1.96 and z = 1.96 separate the bottom 2.5% and the top 2.5%, as shown in Figure 6-10.

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228 CHAPTER 6 Normal Probability Distributions

Critical Values For a normal distribution, a critical value is a z score on the bor- derline separating those z scores that are significantly low or significantly high. Com- mon critical values are z = - 1.96 and z = 1.96, and they are obtained as shown in Example 7. In Example 7, values of z = - 1.96 or lower are significantly low because only 2.5% of the population have scores at or below - 1.96, and the values at or above z = 1.96 are significantly high because only 2.5% of the population have scores at or above 1.96. Critical values will become extremely important in subsequent chapters. The following notation is used for critical z values found by using the standard normal distribution.

DEFINITION

For the standard normal distribution, a critical value is a z score on the borderline separating those z scores that are significantly low or significantly high.

CAUTION When finding a value of za for a particular value of a, note that a is the area to the right of za, but Table A-2 and some technologies give cumulative areas to the left of a given z score. To find the value of za, resolve that conflict by using the value of 1 - a. For example, to find z0.1, refer to the z score with an area of 0.9 to its left.

Notation

The expression za denotes the z score with an area of a to its right. (a is the Greek letter alpha.)

EXAMPLE 8 Finding the Critical Value zA

Find the value of z0.025. (Let a = 0.025 in the expression za.)

S O L U T I O N

The notation of z0.025 is used to represent the z score with an area of 0.025 to its right. Refer to Figure 6-10 and note that the value of z = 1.96 has an area of 0.025 to its right, so z0.025 = 1.96. Note that z0.025 corresponds to a cumulative left area of 0.975.

Examples 3 through 7 in this section are based on the real application of the bone density test, with scores that are normally distributed with a mean of 0 and standard deviation of 1, so that these scores have a standard normal distribution. Apart from the bone density test scores, it is rare to find such convenient parameters, because typical normal distributions have means different from 0 and standard deviations dif- ferent from 1. In the next section we present methods for working with such normal distributions.

Finding z Scores>Areas (Standard Normal) Access tech instructions, videos, and data sets at www.TriolaStats.com

TECH CENTER

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6-1 The Standard Normal Distribution 229

5. Greater than 3.00 minutes. 6. Less than 4.00 minutes.

7. Between 2 minutes and 3 minutes. 8. Between 2.5 minutes and 4.5 minutes.

Standard Normal Distribution. In Exercises 9–12, find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.

9. 

z 5 0.44

10. 

z 5 21.04

11. 

z 5 20.84 z 5 1.28

12. 

z 5 21.07 z 5 0.67

Statistical Literacy and Critical Thinking

1. Normal Distribution What’s wrong with the following statement? “Because the digits 0, 1, 2, . . . , 9 are the normal results from lottery drawings, such randomly selected numbers have a normal distribution.”

2.  Normal Distribution A normal distribution is informally described as a probability dis- tribution that is “bell-shaped” when graphed. Draw a rough sketch of a curve having the bell shape that is characteristic of a normal distribution.

3. Standard Normal Distribution Identify the two requirements necessary for a normal dis- tribution to be a standard normal distribution.

4. Notation What does the notation za indicate?

Continuous Uniform Distribution. In Exercises 5–8, refer to the continuous uniform distribution depicted in Figure 6-2 and described in Example 1. Assume that a patient is randomly selected, and find the probability that the waiting time is within the given range.

6-1 Basic Skills and Concepts

0

0.2

1 2

Area 5 1

3 4 5

x (waiting time in minutes)

P (x

)

FIGURE 6-2 Uniform Distribution of Waiting Time

M06_TRIO9015_02_SE_C06_216-281.indd 229 11/08/16 4:59 PM

230 CHAPTER 6 Normal Probability Distributions

Standard Normal Distribution. In Exercises 13–16, find the indicated z score. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.

13. 

0

0.8907

z

14. 

z 0

0.3050

15. 

z 0

0.9265

16. 

z0

0.2061

Standard Normal Distribution. In Exercises 17–36, assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers to four decimal places.

17. Less than - 1.23. 18. Less than - 1.96.

19. Less than 1.28. 20. Less than 2.56.

21. Greater than 0.25. 22. Greater than 0.18.

23. Greater than - 2.00. 24. Greater than –3.05.

25. Between 2.00 and 3.00. 26. Between 1.50 and 2.50.

27. Between and - 2.55 and - 2.00. 28. Between - 2.75 and –0.75.

29. Between - 2.00 and 2.00. 30. Between - 3.00 and 3.00.

31. Between - 1.00 and 5.00. 32. Between - 4.27 and 2.34.

33. Less than 4.55. 34. Greater than - 3.75.

35. Greater than 0. 36. Less than 0.

Finding Bone Density Scores. In Exercises 37–40 assume that a randomly selected sub- ject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, and then find the bone den- sity test score corresponding to the given information. Round results to two decimal places.

37.  Find P99, the 99th percentile. This is the bone density score separating the bottom 99% from the top 1%.

38.  Find P10, the 10th percentile. This is the bone density score separating the bottom 10% from the top 90%.

39. If bone density scores in the bottom 2% and the top 2% are used as cutoff points for levels that are too low or too high, find the two readings that are cutoff values.

40. Find the bone density scores that can be used as cutoff values separating the lowest 3% and highest 3%.

M06_TRIO9015_02_SE_C06_216-281.indd 230 11/08/16 4:59 PM

6-2 Real Applications of Normal Distributions 231

Critical Values. In Exercises 41–44, find the indicated critical value. Round results to two decimal places.

41. z0.10 42. z0.02 43. z0.04 44. z0.15

Basis for the Range Rule of Thumb and the Empirical Rule. In Exercises 45–48, find the indicated area under the curve of the standard normal distribution; then convert it to a percentage and fill in the blank. The results form the basis for the range rule of thumb and the empirical rule introduced in Section 3-2.

45. About About % of the area is between z = - 1 and z = 1 (or within 1 standard devia- tion of the mean).

46. About % of the area is between z = - 2 and z = 2 (or within 2 standard deviations of the mean).

47. About % About % of the area is between z = - 3 and z = 3 (or within 3 standard devia- tions of the mean).

48. About %   of the area is between z = - 3.5 and z = 3.5 (or within 3.5 standard deviations of the mean).

49. Significance For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of scores that are

a. significantly high (or at least 2 standard deviations above the mean).

b. significantly low (or at least 2 standard deviations below the mean).

c. not significant (or less than 2 standard deviations away from the mean).

50. Distributions In a continuous uniform distribution,

m = minimum + maximum

2 and s =

range

212 a. Find the mean and standard deviation for the distribution of the waiting times represented in Figure 6-2, which accompanies Exercises 5–8.

b. For a continuous uniform distribution with m = 0 and s = 1, the minimum is -23 and the maximum is 23. For this continuous uniform distribution, find the probability of randomly selecting a value between - 1 and 1, and compare it to the value that would be obtained by in- correctly treating the distribution as a standard normal distribution. Does the distribution affect the results very much?

6-1 Beyond the Basics

Key Concept Now we really get real as we extend the methods of the previous section so that we can work with any nonstandard normal distribution (with a mean different from 0 and>or a standard deviation different from 1). The key is a simple conversion (Formula 6-2) that allows us to “standardize” any normal distribution so that x values can be transformed to z scores; then the methods of the preceding section can be used.

6-2 Real Applications of Normal Distributions

M06_TRIO9015_02_SE_C06_216-281.indd 231 11/08/16 4:59 PM

232 CHAPTER 6 Normal Probability Distributions

FORMULA 6-2

z = x - m s

(round z scores to 2 decimal places)

Figure 6-11 illustrates the conversion from a nonstandard to a standard normal distribution. The area in any normal distribution bounded by some score x (as in Figure 6-11a) is the same as the area bounded by the corresponding z score in the standard normal distribution (as in Figure 6-11b).

xm

m

s z 5

x 2

z

PP

0 (b) Standard Normal Distribution

(a) Nonstandard Normal Distribution

FIGURE 6-11 Converting Distributions

Some calculators and software do not require the use of Formula 6-2 to convert to z scores because probabilities can be found directly. However, if using Table A-2, we must first convert values to standard z scores.

When finding areas with a nonstandard normal distribution, use the following procedure.

Procedure for Finding Areas with a Nonstandard Normal Distribution

1. Sketch a normal curve, label the mean and any specific x values, and then shade the region representing the desired probability.

2. For each relevant value x that is a boundary for the shaded region, use Formula 6-2 to convert that value to the equivalent z score. (With many technologies, this step can be skipped.)

3. Use technology (software or a calculator) or Table A-2 to find the area of the shaded region. This area is the desired probability.

The following example illustrates the above procedure.

EXAMPLE 1 What Is the Proportion of Adult Males with Pulse Rates Greater Than 100 Bpm?

In the Chapter Problem we noted that pulse rates of adult males are normally dis- tributed with a mean of 69.6 bpm and a standard deviation of 11.3 bpm. Find the proportion of adult males with a pulse rate greater than 100 bpm. These males are considered to be at a high risk of stroke, heart disease, or cardiac death.

S O L U T I O N

Step 1: See Figure 6-12, which incorporates this information: Men have pulse rates that are normally distributed with a mean of 69.6 bpm and a standard deviation of 11.3 bpm. The shaded region represents the men with pulse rates above 100 bpm.

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6-2 Real Applications of Normal Distributions 233

0.0036

100m 5 69.6 x (pulse rate)

z 5 0 z 5 2.69 z scale

FIGURE 6-12 Pulse Rates of Men

Step 2: We can convert a pulse rate of 100 bpm to the z score of 2.69 by using Formula 6-2 as follows:

z = x - m s

= 100 - 69.6

11.3 = 2.69 1rounded to two decimal places2

Step 3: Technology: Technology can be used to find that the area to the right of 100 bpm in Figure 6-12 is 0.0036. (With many technologies, Step 2 can be skipped.)

Table A-2: Use Table A-2 to find that the cumulative area to the left of z = 2.69 is 0.9964. (Remember, Table A-2 is designed so that all areas are cumulative areas from the left.) Because the total area under the curve is 1, it follows that the shaded area in Figure 6-12 is 1 - 0.9964 = 0.0036.

I N T E R P R E TAT I O N

The proportion of men with pulse rates above 100 bpm is 0.0036, which is roughly 4 men in a thousand. This is a very rare event, so an adult male presenting with a pulse rate above 100 bpm is extremely rare, or there is some medical condition that is causing the high pulse rate.

EXAMPLE 2 Normal Pulse Rates

Normal pulse rates are generally considered to be between 60 bpm and 100 bpm. Given that pulse rates of adult males are normally distributed with a mean of 69.6 bpm and a standard deviation of 11.3 bpm, find the percentage of males with normal pulse rates.

S O L U T I O N

Figure 6-13 shows the shaded region representing men with pulse rates between 60 bpm and 100 bpm.

Step 1: See Figure 6-13 on the next page, which incorporates this information: Men have pulse rates that are normally distributed with a mean of 69.6 bpm and a standard deviation of 11.3 bpm. The shaded region represents the men with pulse rates between 60 bpm and 100 bpm.

continued

High Cost of Low Quality

The Federal

Drug Adminis-

tration recently

reached an

agreement

whereby a

pharmaceutical

company, the Schering-Plough

Corporation, would pay a record

$500 million for failure to cor-

rect problems in manufactur-

ing drugs. According to a New

York Times article by Melody

Petersen, “Some of the problems

relate to the lack of controls that

would identify faulty medicines,

while others stem from outdated

equipment. They involve some

200 medicines, including Claritin,

the allergy medicine that is

Schering’s top-selling product.”

company, the Schering-Plough

M06_TRIO9015_02_SE_C06_216-281.indd 233 11/08/16 4:59 PM

234 CHAPTER 6 Normal Probability Distributions

Step 2: With some technologies, the shaded area in Figure 6-13 can be found directly and it is not necessary to convert the x scores of 60 bpm and 100 bpm to z scores. (See Step 3.)

If using Table A-2, we cannot find the shaded area directly, but we can find it indirectly by using the same procedures from Section 6-1, as follows: (1) Find the cumulative area from the left up to 100 bpm (or z = 2.69); (2) find the cumulative area from the left up to 60 bpm (or z = - 0.85); (3) find the difference between those two areas. The pulse rates of 100 bpm and 60 bpm are converted to z scores by using Formula 6-2 as follows:

For x = 100 bpm: z = x - m s

= 100 - 69.6

11.3 = 2.69

1z = 2.69 yields an area of 0.9964.2

For x = 60 bpm: z = x - m 11.3

= 60 - 69.6

11.3 = - 0.85

1z = - 0.85 yields an area of 0.1977.2 Step 3: Technology: Technology will show that the shaded area in Figure 6-13 is 0.7986.

Table A-2: Refer to Table A-2 with z = 2.69 and find that the cumulative area to the left of z = 2.69 is 0.9964. (Remember, Table A-2 is designed so that all areas are cumulative areas from the left.) Table A-2 also shows that z = - 0.85 corre- sponds to an area of 0.1977. Because the areas of 0.9964 and 0.1977 are cumulative areas from the left, we find the shaded area in Figure 6-13 as follows:

Shaded area in Figure 6@13 = 0.9964 - 0.1977 = 0.7987

There is a small discrepancy between the area of 0.7986 found from technology and the area of 0.7987 found from Table A-2. The area obtained from technology is more accurate because it is based on unrounded z scores, whereas Table A-2 requires z scores rounded to two decimal places.

I N T E R P R E TAT I O N

Expressing the result as a percentage, we conclude that about 80% of men have pulse rates between 60 bpm and 100 bpm.

0.7986

10060 m 5 69.6 x (pulse rate)

z 5 0 z 5 2.69z 5 20.85 z scale

FIGURE 6-13 Pulse Rates of Men

M06_TRIO9015_02_SE_C06_216-281.indd 234 11/08/16 4:59 PM

6-2 Real Applications of Normal Distributions 235

Finding Values from Known Areas Here are helpful hints for those cases in which the area (or probability or percentage) is known and we must find the relevant value(s):

1. Graphs are extremely helpful in visualizing, understanding, and successfully working with normal probability distributions, so they should always be used.

2. Don’t confuse z scores and areas. Remember, z scores are distances along the horizontal scale, but areas are regions under the normal curve. Table A-2 lists z scores in the left columns and across the top row, but areas are found in the body of the table.

3. Choose the correct (right>left) side of the graph. A value separating the top 10% from the others will be located on the right side of the graph, but a value separating the bottom 10% will be located on the left side of the graph.

4. A z score must be negative whenever it is located in the left half of the normal distribution.

5. Areas (or probabilities) are always between 0 and 1, and they are never negative.

Procedure for Finding Values from Known Areas or Probabilities

1. Sketch a normal distribution curve, write the given probability or percent- age in the appropriate region of the graph, and identify the x value(s) being sought.

2. Use technology. If technology is not available, use Table A-2 by referring to the body of Table A-2 to find the area to the left of x; then identify the z score corresponding to that area.

3. If you know z and must convert to the equivalent x value, use Formula 6-2 by entering the values for m, s, and the z score found in Step 2; then solve for x. Based on Formula 6-2, we can solve for x as follows:

x = m + 1z # s2 1another form of Formula 6@22 c

(If z is located to the left of the mean, be sure that it is a negative number.)

4. Refer to the sketch of the curve to verify that the solution makes sense in the context of the graph and in the context of the problem.

The following example uses this procedure for finding a value from a known area.

EXAMPLE 3 Pulse Rates

Given that pulse rates of adult males are normally distributed with a mean of 69.6 bpm and a standard deviation of 11.3 bpm, find the pulse rate that separates the highest 1% from the lowest 99%. That is, find P99.

S O L U T I O N

Step 1: Figure 6-14 on the next page shows the normal distribution with the pulse rate x that we want to identify. The shaded area represents the lowest 99% of the pulse rates.

continued

M06_TRIO9015_02_SE_C06_216-281.indd 235 31/08/16 3:48 PM

236 CHAPTER 6 Normal Probability Distributions

Table A-2: If using Table A-2, search for an area of 0.9900 in the body of the table. The area of 0.9900 corresponds to z = 2.33.

Step 3: With z = 2.33, m = 69.6 bpm, and s = 11.3 bpm, we can solve for x by us- ing Formula 6-2:

z = x - m s becomes 2.33 =

x - 69.6 11.3

The result of x = 95.929 bpm can be found directly or by using the following ver- sion of Formula 6-2:

x = m + 1z # s2 = 69.6 + 12.33 # 11.32 = 95.929 bpm Step 4: The solution of x = 95.9 bpm (rounded) in Figure 6-14 is reasonable be- cause it is greater than the mean of 69.6 bpm.

I N T E R P R E TAT I O N

The male pulse rate of 95.9 (rounded) separates the top 1% from the bottom 99%.

0.99

x 5 ?m 5 69.6 x (pulse rate)

z 5 0 z 5 2.83 z scale

FIGURE 6-14 Finding the 99th Percentile

Step 2: Technology: Technology will provide the value of x in Figure 6-14. For example, see the accompanying Excel display showing that x = 95.88773098 bpm, or 95.9 bpm when rounded.

Excel

Significance

In Chapter 4 we saw that probabilities can be used to determine whether values are significantly high or significantly low. Chapter 4 referred to x successes among n tri- als, but we can adapt those criteria to apply to continuous variables as follows:

Significantly high: The value x is significantly high if P(x or greater) … 0.05.* Significantly low: The value x is significantly low if P(x or less) … 0.05.*

*The value of 0.05 is not absolutely rigid, and other values such as 0.01 could be used instead.

M06_TRIO9015_02_SE_C06_216-281.indd 236 11/08/16 4:59 PM

6-2 Real Applications of Normal Distributions 237

Step 2: Technology: Technology will show that the values of x in Figure 6-15 are 53.4 beats per minute and 94.6 beats per minute when rounded.

Table A-2: If using Table A-2, we must work with cumulative areas from the left. For the leftmost value of x, the cumulative area from the left is 0.05, so search for an area of 0.05 in the body of the table to get z = - 1.645 (identified by the aster- isk between 0.0505 and 0.0495). For the rightmost value of x, the cumulative area from the left is 0.95, so search for an area of 0.9500 in the body of the table to get z = 1.645 (identified by the asterisk between 0.9495 and 0.9505). Having found the two z scores, we now proceed to convert them to pulse rates.

Step 3: We now solve for the two values of x by using Formula 6-2 directly or by using the following version of Formula 6-2:

Leftmost value of x: x = m + 1z # s2 = 74.0 + 1- 1.645 # 12.52 = 53.4 Rightmost value of x: x = m + 1z # s2 = 74.0 + 11.645 # 12.52 = 94.6

Step 4: Referring to Figure 6-15, we see that the leftmost value of x = 53.4 is rea- sonable because it is less than the mean of 74.0. Also, the rightmost value of 94.6 is reasonable because it is above the mean of 74.0.

I N T E R P R E TAT I O N

Here are the pulse rates of women that are significant:

■ Significantly low: 53.4 beats per minute or lower

■ Significantly high: 94.6 beats per minute or higher

Physicians could use these results to investigate health issues that could cause pulse rates to be significantly low or significantly high.

0.05

x 5 ?x 5 ? m 5 74.0

0.05

FIGURE 6-15 Pulse Rates of Women

EXAMPLE 4 Significantly Low or Significantly High Female Pulse Rates

Use the preceding criteria to identify pulse rates of women that are significantly low or significantly high. Based on Data Set 1 “Body Data” in Appendix B, assume that women have normally distributed pulse rates with a mean of 74.0 beats per minute and a standard deviation of 12.5 beats per minute.

S O L U T I O N

Step 1: We begin with the graph shown in Figure 6-15. We have entered the mean of 74.0, and we have identified the x values separating the lowest 5% and the highest 5%.

M06_TRIO9015_02_SE_C06_216-281.indd 237 11/08/16 4:59 PM

238 CHAPTER 6 Normal Probability Distributions

Finding x Values>Areas Access tech instructions, videos, and data sets at www.TriolaStats.com

TECH CENTER

Statistical Literacy and Critical Thinking

1.  Birth Weights Based on Data Set 3 “Births” in Appendix B, birth weights are normally distributed with a mean of 3152.0 g and a standard deviation of 693.4 g.

a. What are the values of the mean and standard deviation after converting all birth weights to z scores using z = 1x - m2>s? b. The original birth weights are in grams. What are the units of the corresponding z scores?

2.  Birth Weights Based on Data Set 3 “Births” in Appendix B, birth weights are normally distributed with a mean of 3152.0 g and a standard deviation of 693.4 g.

a. For the bell-shaped graph, what is the area under the curve?

b. What is the value of the median?

c. What is the value of the mode?

d. What is the value of the variance?

3. Normal Distributions What is the difference between a standard normal distribution and a nonstandard normal distribution?

4. Random Digits Computers are commonly used to randomly generate digits of telephone numbers to be called when conducting the California Health Survey. Can the methods of this section be used to find the probability that when one digit is randomly generated, it is less than 3? Why or why not? What is the probability of getting a digit less than 3?

IQ Scores. In Exercises 5–8, find the area of the shaded region. The graphs depict IQ scores of adults, and those scores are normally distributed with a mean of 100 and a stan- dard deviation of 15 (as on the Wechsler IQ test).

5. 

118

6. 

91

7. 

79 133

8. 

124112

6-2 Basic Skills and Concepts

91

M06_TRIO9015_02_SE_C06_216-281.indd 238 11/08/16 4:59 PM

6-2 Real Applications of Normal Distributions 239

IQ Scores. In Exercises 9–12, find the indicated IQ score and round to the nearest whole number. The graphs depict IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler IQ test).

9. 

x

0.9918

10. 

x

0.1587

11. 

x

0.9798

12. 

x

0.9099

Female Pulse Rates. In Exercises 13–20, assume that an adult female is randomly selected. Females have pulse rates that are normally distributed with a mean of 74.0 beats per minute and a standard deviation of 12.5 beats per minute (based on Data Set 1 “Body Data” in Appendix B). (Hint: Draw a graph in each case.)

13. Find the probability of a pulse rate less than 100 beats per minute.

14. Find the probability of a pulse rate greater than 80 beats per minute.

15. Find the probability of a pulse rate between 60 beats per minute and 70 beats per minute.

16. Find the probability of a pulse rate between 70 beats per minute and 90 beats per minute.

17. Find P90, which is the pulse rate separating the bottom 90% from the top 10%.

18.  Find the first quartile Q1, which is the pulse rate separating the bottom 25% from the top 75%.

19.  Significance Instead of using 0.05 for identifying significant values, use the criteria that a value x is significantly high if P(x or greater) … 0.01 and a value is significantly low if P(x or less) … 0.01. Find the pulse rates separating significant values from those that are not significant. Using these criteria, is a pulse rate of 102 beats per minute significantly high?

20.  Significance Instead of using 0.05 for identifying significant values, use the criteria that a value x is significantly high if P(x or greater) … 0.025 and a value is significantly low if P(x or less) … 0.025. Find the pulse rates separating significant values from those that are not significant. Using these criteria, is a pulse rate of 48 beats per minute significantly low?

21. Eye Contact In a study of facial behavior, people in a control group are timed for eye con- tact in a 5-minute period. Their times are normally distributed with a mean of 184.0 seconds and a standard deviation of 55.0 seconds (based on data from “Ethological Study of Facial Behavior in Nonparanoid and Paranoid Schizophrenic Patients,” by Pittman, Olk, Orr, and Singh, Psychiatry, Vol. 144, No. 1). For a randomly selected person from the control group, find the probability that the eye contact time is greater than 230.0 seconds, which is the mean for paranoid schizophrenics. Based on personal experience, does the result appear to be the proportion of people who are paranoid schizophrenics?

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240 CHAPTER 6 Normal Probability Distributions

22.  Body Temperatures Based on sample results in Data Set 2 “Body Temperatures” in Appendix B, assume that human body temperatures are normally distributed with a mean of 98.20°F and a standard deviation of 0.62°F.

a. According to emedicinehealth.com, a body temperature of 100.4oF or above is considered to be a fever. What percentage of normal and healthy persons would be considered to have a fever? Does this percentage suggest that a cutoff of 100.4oF is appropriate?

b. Physicians want to select a minimum temperature for requiring further medical tests. What should that temperature be if we want only 2.0% of healthy people to exceed it? (Such a result is a false positive, meaning that the test result is positive, but the subject is not really sick.)

23.  Low Birth Weight The University of Maryland Medical Center considers “low birth weights” to be those less than 5.5 lb or 2495 g. Birth weights are normally distributed with a mean of 3152.0 g and a standard deviation of 693.4 g (based on Data Set 3 “Births” in Appendix B).

a. If a birth weight is randomly selected, what is the probability that it is a “low birth weight”?

b. Find the weights considered to be significantly low using the criterion of a probability of 0.05 or less. How do these results compare to the criterion of 2495 g?

c. Compare the results from parts (a) and (b).

24. Durations of Pregnancies The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.

a. In a letter to “Dear Abby,” a wife claimed to have given birth 308 days after a brief visit from her husband, who was working in another country. Find the probability of a pregnancy lasting 308 days or longer. What does the result suggest?

b. If we stipulate that a baby is premature if the duration of pregnancy is in the lowest 3%, find the duration that separates premature babies from those who are not premature. Premature babies often require special care, and this result could be helpful to hospital administrators in planning for that care.

Large Data Sets. In Exercises 25 and 26, refer to the data sets in Appendix B and use software or a calculator.

25.  Diastolic Blood Pressure of Males Refer to Data Set 1 in Appendix B and use the diastolic blood pressures of males.

a. Find the mean and standard deviation, and verify that the data have a distribution that is roughly normal. Round the results using three decimal places.

b. Treating the unrounded values of the mean and standard deviation as parameters, and assum- ing that male diastolic blood pressures are normally distributed, find diastolic blood pressures separating the lowest 2.5% and the highest 2.5%. These values could be helpful when physi- cians try to determine whether diastolic blood pressures are significantly low or significantly high.

26.  Diastolic Blood Pressure of Females Repeat the preceding exercise using females instead of males.

27. Outliers For the purposes of constructing modified boxplots as described in Section 3-3, outliers are defined as data values that are above Q3 by an amount greater than 1.5 * IQR or below Q1 by an amount greater than 1.5 * IQR, where IQR is the interquartile range. Using this definition of outliers, find the probability that when a value is randomly selected from a normal distribution, it is an outlier.

6-2 Beyond the Basics

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6-3 Sampling Distributions and Estimators 241

A Short Story Among the population of all adults, exactly 40% have brown eyes (the authors just know this). In a survey of 1000 adults, 42% of the subjects were observed to have brown eyes. Being so intrigued by this, 50,000 people became so enthusias- tic that they each conducted their own individual survey of 1000 randomly selected adults. Each of these 50,000 new surveyors reported the percentage that they found, with results such as 38%, 39%, and 43%. The authors obtained each of the 50,000

Key Concept We now consider the concept of a sampling distribution of a statistic. Instead of working with values from the original population, we want to focus on the values of statistics (such as sample proportions or sample means) obtained from the population. Figure 6-16 shows the key points that we need to know, so try really, really hard to understand the story that Figure 6-16 tells.

6-3 Sampling Distributions and Estimators

Proportions

Sample 1

Sample 2

Sample 3

(Population Proportion is p)

Distribution of Sample Proportions

Sample Proportions

Sample

Sample proportions tend to have a normal distribution

p̂ p̂ p̂ p̂ p̂

p̂ p̂ p̂ p̂ p̂

p̂ p̂ p̂

p̂ p̂ p̂ p̂ p̂ p̂

p̂

p ˆ

• • •

1 p̂

2 p̂

3 p̂

Sampling Procedure: Randomly select n values and find the proportion p for each sample.

ˆ

Means

Sample 1

Sample 2

Sample 3

(Population Mean is m)

Distribution of Sample Means

Sample MeansSample

Sample means tend to have a normal distribution

• • •

Sampling Procedure: Randomly select n values and find the mean x for each sample.

x x x x x x x x x x x x x x x x xX1

X2

X3

x x x x

x

Variances

Sample 1

Sample 2

Sample 3

(Population Variance is s2)

Distribution of Sample Variances

Sample Variances

Sample

Sample variances tend to have a skewed distribution

• • •

Sampling Procedure: Randomly select n values and find the variance s2 for each sample.

s2 1

s2 2

s2 3

s2 s2 s2

s2 s2 s2

s2 s2 s2 s2 s2 s2 s2 s2 s2 s2 s2 s2 s2 s2 s2 s2

s2 s2 s2 s2 s2

FIGURE 6-16 General Behavior of Sampling Distributions

continued

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242 CHAPTER 6 Normal Probability Distributions

sample percentages, changed them to proportions, and then they constructed the his- togram shown in Figure 6-17. Notice anything about the shape of the histogram? It’s normal. Notice anything about the mean of the sample proportions? They are centered about the value of 0.40, which happens to be the population proportion. Moral: When samples of the same size are taken from the same population, the following two prop- erties apply:

1. Sample proportions tend to be normally distributed.

2. The mean of sample proportions is the same as the population mean.The implications of the preceding properties will be extensive in the chapters that follow.

FIGURE 6-17 Histogram of 50,000 Sample Proportions

Let’s formally define sampling distribution, the main character in the preceding short story.

DEFINITION

The sampling distribution of a statistic (such as a sample proportion or sample mean) is the distribution of all values of the statistic when all possible samples of the same size n are taken from the same population. (The sampling distribution of a statistic is typically represented as a probability distribution in the format of a prob- ability histogram, formula, or table.)

Sampling Distribution of Sample Proportion The preceding general definition of a sampling distribution of a statistic can now be restated for the specific case of a sample proportion:

DEFINITION

The sampling distribution of the sample proportion is the distribution of sample proportions (or the distribution of the variable pn ), with all samples having the same sample size n taken from the same population. (The sampling distribution of the sample proportion is typically represented as a probability distribution in the format of a probability histogram, formula, or table.)

We need to distinguish between a population proportion p and a sample proportion, and the following notation is common and will be used throughout the remainder of this book, so it’s very important.

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6-3 Sampling Distributions and Estimators 243

Notation for Proportions

p = population proportion pn = sample proportion

HINT pn is pronounced “p-hat.” When symbols are used above a letter, as in x and pn , they represent statistics, not parameters.

Behavior of Sample Proportions

1. The distribution of sample proportions tends to approximate a normal distribution.

2. Sample proportions target the value of the population proportion in the sense that the mean of all of the sample proportions pn is equal to the popu- lation proportion p; the expected value of the sample proportion is equal to the population proportion.

EXAMPLE 1 Sampling Distribution of the Sample Proportion

Consider repeating this process: Roll a die 5 times and find the proportion of odd numbers (1 or 3 or 5). What do we know about the behavior of all sample propor- tions that are generated as this process continues indefinitely?

S O L U T I O N

Figure 6-18 illustrates a process of rolling a die 5 times and finding the proportion of odd numbers. (Figure 6-18 shows results from repeating this process 10,000 times, but the true sampling distribution of the sample proportion involves repeating the process indefinitely.) Figure 6-18 shows that the sample proportions are approxi- mately normally distributed. (Because the values of 1, 2, 3, 4, 5, 6 are all equally likely, the proportion of odd numbers in the population is 0.5, and Figure 6-18 shows that the sample proportions have a mean of 0.50.)

Proportions

Sample 1

Sample 2

Sample 3

(Population Proportion is p 5 0.5)

Distribution of Sample Proportions

Sample Proportions

Sample

p 5 0.5 Sample proportions are approximately normal

• • •

0.2

0.4

0.8

Sampling Procedure: Roll a die 5 times and find the proportion p of odd numbers for each sample.

ˆ

FIGURE 6-18 Sample Proportions from 10,000 Trials

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244 CHAPTER 6 Normal Probability Distributions

Behavior of Sample Means

1. The distribution of sample means tends to be a normal distribution. (This will be discussed further in the following section, but the distribution tends to be- come closer to a normal distribution as the sample size increases.)

2. The sample means target the value of the population mean. (That is, the mean of the sample means is the population mean. The expected value of the sample mean is equal to the population mean.)

DEFINITION

The sampling distribution of the sample mean is the distribution of all possible sample means (or the distribution of the variable x), with all samples having the same sample size n taken from the same population. (The sampling distribution of the sample mean is typically represented as a probability distribution in the format of a probability histogram, formula, or table.)

EXAMPLE 2 Sampling Distribution of the Sample Mean

A pediatrician has three patients with measles and they are ages 4, 5, and 9. Consider the population of {4, 5, 9}. If two ages are randomly selected with replace- ment from the population {4, 5, 9}, identify the sampling distribution of the sample mean by creating a table representing the probability distribution of the sample mean. Do the values of the sample mean target the value of the population mean?

S O L U T I O N

If two values are randomly selected with replacement from the population {4, 5, 9}, the leftmost column of Table 6-2 lists the nine different possible samples. The second column lists the corresponding sample means. The nine samples are equally likely with a probability of 1>9. We saw in Section 5-1 that a probability distribution gives the probability for each value of a random variable, as in the second and third col- umns of Table 6-2. The second and third columns of Table 6-2 represent the sampling distribution of the sample mean. In Table 6-2, some of the sample mean values are repeated, so we combined them in Table 6-3.

TABLE 6-2 Sampling Distribution of Mean

Sample Sample Mean x Probability

4, 4 4.0 1>9 4, 5 4.5 1>9 4, 9 6.5 1>9 5, 4 4.5 1>9 5, 5 5.0 1>9 5, 9 7.0 1>9 9, 4 6.5 1>9 9, 5 7.0 1>9 9, 9 9.0 1>9

TABLE 6-3 Sampling Distribution of Mean (Condensed)

Sample Mean x Probability

4.0 1>9 4.5 2>9 5.0 1>9 6.5 2>9 7.0 2>9 9.0 1>9

Sampling Distribution of the Sample Mean We now consider sample means.

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6-3 Sampling Distributions and Estimators 245

If we were to create a probability histogram from Table 6-2, it would not have the bell shape that is characteristic of a normal distribution, but that’s because we are working with such small samples. If the population of {4, 5, 9} were much larger and if we were selecting samples much larger than n = 2, as in this example, we would get a probability histogram that is much closer to being bell-shaped, indicat- ing a normal distribution, as in Example 3.

I N T E R P R E TAT I O N

Because Table 6-3 lists the possible values of the sample mean along with their cor- responding probabilities, Table 6-3 is an example of a sampling distribution of a sample mean.

The value of the mean of the population {4, 5, 9} is m = 6.0. Using either Table 6-2 or 6-3, we could calculate the mean of the sample values and we get 6.0. Because the mean of the sample means (6.0) is equal to the mean of the population (6.0), we conclude that the values of the sample mean do target the value of the population mean. It’s unfortunate that this sounds so much like doublespeak, but this illustrates that the mean of the sample means is equal to the population mean m.

HINT Read the last sentence of the above paragraph a few times until it makes sense.

EXAMPLE 3 Sampling Distribution of the Sample Mean

Consider repeating this process: Roll a die 5 times to randomly select 5 values from the population {1, 2, 3, 4, 5, 6}, then find the mean x of the results. What do we know about the behavior of all sample means that are generated as this process con- tinues indefinitely?

S O L U T I O N

Figure 6-19 illustrates a process of rolling a die 5 times and finding the mean of the results. Figure 6-19 shows results from repeating this process 10,000 times, but the true sampling distribution of the mean involves repeating the process indefinitely. Because the values of 1, 2, 3, 4, 5, 6 are all equally likely, the population has a mean of m = 3.5. The 10,000 sample means included in Figure 6-19 have a mean of 3.5. If the process is continued indefinitely, the mean of the sample means will be 3.5. Also, Figure 6-19 shows that the distribution of the sample means is approximately a normal distribution.

Means

Sample 1

Sample 2

Sample 3

(Population Mean is m 5 3.5)

Distribution of Sample Means

Sample Means

Sample

m 5 3.5 Sample means are

approximately normal

• • •

3.4

4.4

2.8

Sampling Procedure: Roll a die 5 times and find the mean x for each sample.

FIGURE 6-19 Sample Means from 10,000 Trials

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246 CHAPTER 6 Normal Probability Distributions

Sampling Distribution of the Sample Variance Let’s now consider the sampling distribution of sample variances.

DEFINITION

The sampling distribution of the sample variance is the distribution of sample variances (the variable s2), with all samples having the same sample size n taken from the same population. (The sampling distribution of the sample variance is typi- cally represented as a probability distribution in the format of a table, probability histogram, or formula.)

CAUTION When working with population standard deviations or variances, be sure to evaluate them correctly. In Section 3-2 we saw that the computations for population standard deviations or variances involve division by the population size N instead of n - 1, as shown here.

Population standard deviation: s = B Σ1x - m22

N

Population variance: s2 = Σ1x - m22

N

Because the calculations are typically performed with software or calculators, be care- ful to correctly distinguish between the variance of a sample and the variance of a population.

Behavior of Sample Variances

1. The distribution of sample variances tends to be a distribution skewed to the right.

2. The sample variances target the value of the population variance. (That is, the mean of the sample variances is the population variance. The expected value of the sample variance is equal to the population variance.)

EXAMPLE 4 Sampling Distribution of the Sample Variance

Consider repeating this process: Roll a die 5 times and find the variance s2 of the results. What do we know about the behavior of all sample variances that are gener- ated as this process continues indefinitely?

S O L U T I O N

Figure 6-20 illustrates a process of rolling a die 5 times and finding the variance of the results. Figure 6-20 shows results from repeating this process 10,000 times, but the true sampling distribution of the sample variance involves repeating the process indefinitely. Because the values of 1, 2, 3, 4, 5, 6 are all equally likely, the popu- lation has a variance of s2 = 2.9, and the 10,000 sample variances included in Figure 6-20 have a mean of 2.9. If the process is continued indefinitely, the mean of the sample variances will be 2.9. Also, Figure 6-20 shows that the distribution of the sample variances is a skewed distribution, not a normal distribution with its characteristic bell shape.

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6-3 Sampling Distributions and Estimators 247

Estimators: Unbiased and Biased The preceding examples show that sample proportions, means, and variances tend to target the corresponding population parameters. More formally, we say that sample proportions, means, and variances are unbiased estimators. See the following two definitions.

DEFINITIONS

An estimator is a statistic used to infer (or estimate) the value of a population parameter.

An unbiased estimator is a statistic that targets the value of the corresponding population parameter in the sense that the sampling distribution of the statistic has a mean that is equal to the corresponding population parameter.

Unbiased Estimators These statistics are unbiased estimators. That is, they each target the value of the corresponding population parameter (with a sampling distribu- tion having a mean equal to the population parameter):

■ Proportion pn

■ Mean x

■ Variance s2

Biased Estimators These statistics are biased estimators. That is, they do not target the value of the corresponding population parameter:

■ Median

■ Range

■ Standard deviation s

Important Note: The sample standard deviations do not target the population standard deviation s, but the bias is relatively small in large samples, so s is often used to estimate S even though s is a biased estimator of s.

Variances

Sample 1

Sample 2

Sample 3

(Population Variance is s2 5 2.9)

Distribution of Sample Variances

Sample Variances

Sample

s2 5 2.9 Sample variances tend to have a skewed distribution

• • •

1.8

2.3

2.2

Sampling Procedure: Roll a die 5 times and find the variance s2

for each sample.

FIGURE 6-20 Sample Variances from 10,000 Trials

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248 CHAPTER 6 Normal Probability Distributions

b. The last two columns of Table 6-4 list the values of the range along with the corresponding probabilities, so the last two columns constitute a table summa- rizing the probability distribution. Table 6-4 therefore describes the sampling distribution of the sample range.

c. The mean of the sample ranges in Table 6-4 is 20>9, or 2.2. The population of {4, 5, 9} has a range of 9 - 4 = 5. Because the mean of the sample ranges (2.2) is not equal to the population range (5), the sample ranges do not target the value of the population range.

d. Because the sample ranges do not target the population range, the sample range is a biased estimator of the population range.

I N T E R P R E TAT I O N

Because the sample range is a biased estimator of the population range, a sample range should generally not be used to estimate the value of the population range.

EXAMPLE 5 Sampling Distribution of the Sample Range

As in Example 2, consider samples of size n = 2 randomly selected from the population {4, 5, 9}.

a. List the di¢erent possible samples along with the probability of each sample, then find the range for each sample.

b. Describe the sampling distribution of the sample range in the format of a table summarizing the probability distribution.

c. Based on the results, do the sample ranges target the population range, which is 9 - 4 = 5?

d. What do these results indicate about the sample range as an estimator of the population range?

S O L U T I O N

a. In Table 6-4 we list the nine di¢erent possible samples of size n = 2 selected with replacement from the population {4, 5, 9}. The nine samples are equally likely, so each has probability 1>9. Table 6-4 also shows the range for each of the nine samples.

TABLE 6-4 Sampling Distribution of Range

Sample Sample Range Probability

4, 4 0 1>9 4, 5 1 1>9 4, 9 5 1>9 5, 4 1 1>9 5, 5 0 1>9 5, 9 4 1>9 9, 4 5 1>9 9, 5 4 1>9 9, 9 0 1>9

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6-3 Sampling Distributions and Estimators 249

Why Sample with Replacement? All of the examples in this section involved sam- pling with replacement. Sampling without replacement would have the very practi- cal advantage of avoiding wasteful duplication whenever the same item is selected more than once. Many of the statistical procedures discussed in the following chapters are based on the assumption that sampling is conducted with replacement because of these two very important reasons:

1. When selecting a relatively small sample from a large population, it makes no significant difference whether we sample with replacement or without replace- ment.

2. Sampling with replacement results in independent events that are unaf- fected by previous outcomes, and independent events are easier to analyze and result in simpler calculations and formulas.

Statistical Literacy and Critical Thinking

1. Births There are about 11,000 births each day in the United States, and the proportion of boys born in the United States is 0.512. Assume that each day, 100 births are randomly selected and the proportion of boys is recorded.

a. What do you know about the mean of the sample proportions?

b. What do you know about the shape of the distribution of the sample proportions?

2.  Sampling with Replacement The Orangetown Medical Research Center randomly se- lects 100 births in the United States each day, and the proportion of boys is recorded for each sample.

a. Do you think the births are randomly selected with replacement or without replacement?

b. Give two reasons why statistical methods tend to be based on the assumption that sampling is conducted with replacement, instead of without replacement.

3.  Unbiased Estimators Data Set 3 “Births” in Appendix B includes birth weights of 400 babies. If we compute the values of sample statistics from that sample, which of the following statistics are unbiased estimators of the corresponding population parameters: sample mean; sample median; sample range; sample variance; sample standard deviation; sample proportion?

4.  Sampling Distribution Data Set 3 “Births” in Appendix B includes a sample of birth weights. If we explore this sample of 400 birth weights by constructing a histogram and find- ing the mean and standard deviation, do those results describe the sampling distribution of the mean? Why or why not?

5.  Good Sample? A geneticist is investigating the proportion of boys born in the world population. Because she is based in China, she obtains sample data from that country. Is the resulting sample proportion a good estimator of the population proportion of boys born world- wide? Why or why not?

6. Physicians There are about 900,000 active physicians in the United States, and they have annual incomes with a distribution that is skewed instead of being normal. Many different sam- ples of 40 physicians are randomly selected, and the mean annual income is computed for each sample.

a. What is the approximate shape of the distribution of the sample means (uniform, normal, skewed, other)?

b. What value do the sample means target? That is, what is the mean of all such sample means?

6-3 Basic Skills and Concepts

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250 CHAPTER 6 Normal Probability Distributions

In Exercises 7–10, use the same population of {4, 5, 9} that was used in Examples 2 and 5. As in Examples 2 and 5, assume that samples of size n = 2 are randomly selected with replacement.

7. Sampling Distribution of the Sample Variance

a. Find the value of the population variance s2.

b. Table 6-2 describes the sampling distribution of the sample mean. Construct a similar table representing the sampling distribution of the sample variance s2. Then combine values of s2 that are the same, as in Table 6-3 (Hint: See Example 2 on page 244 for Tables 6-2 and 6-3, which describe the sampling distribution of the sample mean.)

c. Find the mean of the sampling distribution of the sample variance.

d. Based on the preceding results, is the sample variance an unbiased estimator of the popula- tion variance? Why or why not?

8. Sampling Distribution of the Sample Standard Deviation For the following, round results to three decimal places.

a. Find the value of the population standard deviation s.

b. Table 6-2 describes the sampling distribution of the sample mean. Construct a similar table representing the sampling distribution of the sample standard deviation s. Then combine values of s that are the same, as in Table 6-3. (Hint: See Example 2 on page 244 for Tables 6-2 and 6-3, which describe the sampling distribution of the sample mean.)

c. Find the mean of the sampling distribution of the sample standard deviation.

d. Based on the preceding results, is the sample standard deviation an unbiased estimator of the population standard deviation? Why or why not?

9. Sampling Distribution of the Sample Median

a. Find the value of the population median.

b. Table 6-2 describes the sampling distribution of the sample mean. Construct a similar table representing the sampling distribution of the sample median. Then combine values of the me- dian that are the same, as in Table 6-3. (Hint: See Example 2 on page 244 for Tables 6-2 and 6-3, which describe the sampling distribution of the sample mean.)

c. Find the mean of the sampling distribution of the sample median.

d. Based on the preceding results, is the sample median an unbiased estimator of the popula- tion median? Why or why not?

10. Sampling Distribution of the Sample Proportion

a. For the population, find the proportion of odd numbers.

b. Table 6-2 describes the sampling distribution of the sample mean. Construct a similar table representing the sampling distribution of the sample proportion of odd numbers. Then combine values of the sample proportion that are the same, as in Table 6-3. (Hint: See Example 2 on page 244 for Tables 6-2 and 6-3, which describe the sampling distribution of the sample mean.)

c. Find the mean of the sampling distribution of the sample proportion of odd numbers.

d. Based on the preceding results, is the sample proportion an unbiased estimator of the popu- lation proportion? Why or why not?

In Exercises 11–14, use the population of {34, 36, 41, 51} of the amounts of caffeine (mg ,12 oz) in Coca-Cola Zero, Diet Pepsi, Dr Pepper, and Mellow Yello Zero. Assume that random samples of size n = 2 are selected with replacement.

11. Sampling Distribution of the Sample Mean

a. After identifying the 16 different possible samples, find the mean of each sample, then construct a table representing the sampling distribution of the sample mean. In the table,

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6-3 Sampling Distributions and Estimators 251

combine values of the sample mean that are the same. (Hint: See Table 6-3 in Example 2 on page 244.)

b. Compare the mean of the population {34, 36, 41, 51} to the mean of the sampling distribu- tion of the sample mean.

c. Do the sample means target the value of the population mean? In general, do sample means make good estimators of population means? Why or why not?

12. Sampling Distribution of the Median Repeat Exercise 11 using medians instead of means.

13. Sampling Distribution of the Range Repeat Exercise 11 using ranges instead of means.

14.  Sampling Distribution of the Variance Repeat Exercise 11 using variances instead of means.

15.  Births: Sampling Distribution of Sample Proportion When two births are ran- domly selected, the sample space for genders is bb, bg, gb, and gg (where b = boy and g = girl). Assume that those four outcomes are equally likely. Construct a table that describes the sampling distribution of the sample proportion of girls from two births. Does the mean of the sample proportions equal the proportion of girls in two births? Does the result suggest that a sample proportion is an unbiased estimator of a population proportion?

16. Births: Sampling Distribution of Sample Proportion For three births, assume that the genders are equally likely. Construct a table that describes the sampling distribution of the sample proportion of girls from three births. Does the mean of the sample proportions equal the proportion of girls in three births? (Hint: See Exercise 15 for two births.)

17. MCAT Tests Because they enable efficient procedures for evaluating answers, multiple choice questions are commonly used on standardized tests, such as the MCAT or the GRE Biology test. Such questions typically have five choices, one of which is correct. Assume that you must make random guesses for two such questions. Assume that both questions have correct answers of “a.”

a. After listing the 25 different possible samples, find the proportion of correct answers in each sample; then construct a table that describes the sampling distribution of the sample propor- tions of correct responses.

b. Find the mean of the sampling distribution of the sample proportion.

c. Is the mean of the sampling distribution [from part (b)] equal to the population proportion of correct responses? Does the mean of the sampling distribution of proportions always equal the population proportion?

18. Hybridization A hybridization experiment begins with four peas having yellow pods and one pea having a green pod. Two of the peas are randomly selected with replacement from this population.

a. After identifying the 25 different possible samples, find the proportion of peas with yellow pods in each of them; then construct a table to describe the sampling distribution of the propor- tions of peas with yellow pods.

b. Find the mean of the sampling distribution.

c. Is the mean of the sampling distribution [from part (b)] equal to the population proportion of peas with yellow pods? Does the mean of the sampling distribution of proportions always equal the population proportion?

19. Using a Formula to Describe a Sampling Distribution Exercise 15 “Births” re- quires the construction of a table that describes the sampling distribution of the proportions of girls from two births. Consider the formula shown here, and evaluate that formula using sample

6-3 Beyond the Basics

continued

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252 CHAPTER 6 Normal Probability Distributions

proportions (represented by x) of 0, 0.5, and 1. Based on the results, does the formula describe the sampling distribution? Why or why not?

P1x2 = 1 212 - 2x2!12x2! where x = 0, 0.5, 1

20. Mean Absolute Deviation Is the mean absolute deviation of a sample a good statistic for estimating the mean absolute deviation of the population? Why or why not? (Hint: See Example 5.)

Key Concept In the preceding section we saw that the sampling distribution of sample means tends to be a normal distribution as the sample size increases. In this section we introduce and apply the central limit theorem. The central limit theorem allows us to use a normal distribution for some very meaningful and important applications.

6-4 The Central Limit Theorem

CENTRAL LIMIT THEOREM

For all samples of the same size n with n 7 30, the sampling distribution of x can be approximated by a normal distribution with mean m and standard deviation s>1n.

Given any population with any distribution (uniform, skewed, whatever), the dis- tribution of sample means x can be approximated by a normal distribution when the samples are large enough with n 7 30. (There are some special cases of very non- normal distributions for which the requirement of n 7 30 isn’t quite enough, so the number 30 should be higher in those cases, but those cases are relatively rare.)

EXAMPLE 1 HDL Cholesterol of Females

Figures 6-21 and 6-22 illustrate the central limit theorem.

■ Original data: Figure 6-21 is a histogram of the high-density lipoprotein (HDL) cholesterol measures (mg>dL) of the 147 females listed in Data Set 1 “Body Data” in Appendix B, and those measures have a distribution that is skewed to the right instead of being normal.

■ Sample means: Figure 6-22 is a histogram of 100 sample means. Each sample includes 100 HDL cholesterol measures of females, and this histogram shows that the sample means have a distribution that is very close to being normal.

FIGURE 6-21 Nonnormal Distribution: HDL Cholesterol from 147 Women

FIGURE 6-22 Approximately Normal Distribution: Means from Samples of Size n = 100 of HDL Cholesterol from Females

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6-4 The Central Limit Theorem 253

A Universal Truth Example 1 and the central limit theorem are truly remarkable be- cause they describe a rule of nature that works throughout the universe. If we could send a spaceship to a distant planet “in a galaxy far, far away,” and if we collect samples of rocks (all of the same large sample size) and weigh them, the sample means would have a distribution that is approximately normal. Think about the significance of that!

The following key points form the foundation for estimating population param- eters and hypothesis testing—topics discussed at length in the following chapters.

I N T E R P R E TAT I O N

The original HDL cholesterol measurements depicted in Figure 6-21 have a skewed distribution, but when we collect samples and compute their means, those sample means tend to have a distribution that is normal.

The Central Limit Theorem and the Sampling Distribution of x

Given

1. Population (with any distribution) has mean m and standard deviation s.

2. Simple random samples all of the same size n are selected from the population.

Practical Rules for Real Applications Involving a Sample Mean x

Requirements: Population has a normal distribution or n + 30:

Mean of all values of x: mx = m

Standard deviation of all values of x: sx = s

1n z score conversion of x: z =

x - m s

1n Original population is not normally distributed and n " 30: The distribution of x might not be approximated well by a normal distribution, and the methods of this section might not apply. Use other methods, such as nonparametric methods or bootstrapping methods (Section 7-4).

Considerations for Practical Problem Solving

1. Check Requirements: When working with the mean from a sample, verify that the normal distribution can be used by confirming that the original population has a normal distribution or the sample size is n 7 30.

2. Individual Value or Mean from a Sample? Determine whether you are using a normal distribution with a single value x or the mean x from a sample of n values. See the following.

• Individual value: When working with an individual value from a normally distributed population, use the methods of

Section 6-2 with z = x - m s

.

• Mean from a sample of values: When working with a mean for some sample of n values, be sure to use the value of

s>1n for the standard deviation of the sample means, so use z = x - m s

1n .

KEY ELEMENTS

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254 CHAPTER 6 Normal Probability Distributions

The following new notation is used for the mean and standard deviation of the distribution of x.

NOTATION FOR THE SAMPLING DISTRIBUTION OF x

If all possible simple random samples of size n are selected from a population with mean m and standard deviation s, the mean of all sample means is denoted by mx and the standard deviation of all sample means is denoted by sx.

Mean of all values of x: mx = m

Standard deviation of all values of x: sx = s

1n Note: sx is called the standard error of the mean and is sometimes denoted as SEM.

Applying the Central Limit Theorem Many practical problems can be solved with the central limit theorem. Example 2 is a good illustration of the central limit theorem because we can see the difference between working with an individual value in part (a) and working with the mean for a sample in part (b). Study Example 2 carefully to understand the fundamental differ- ence between the procedures used in parts (a) and (b). In particular, note that when

working with an individual value, we use z = x - m s

, but when working with the

mean x for a collection of sample values, we use z = x - m s>1n.

EXAMPLE 2 Pulse Rates of Women

In the Chapter Problem it was noted that women have normally distributed pulse rates with a mean of 74.0 bpm and a standard deviation of 12.5 bpm.

a. Find the probability that 1 randomly selected woman has a pulse rate greater than 80 bpm.

b. Find the probability that a sample of 16 randomly selected women have a mean pulse rate greater than 80 bpm.

c. Given that part (b) involves a sample size that is not larger than 30, why can the central limit theorem be used?

S O L U T I O N

a. Approach Used for an Individual Value: Use the methods presented in Section 6-2 because we are dealing with an individual value from a nor- mally distributed population. We seek the area of the green-shaded region in Figure 6-23(a).

Technology: If using technology (as described at the end of Section 6-2), we find that the green-shaded area in the graph at the left is 0.3156.

Table A-2: If using Table A-2, we convert the pulse rate of 80 bpm to the cor- responding z score, as shown here:

z = x - m s

= 80 - 74.0

12.5 = 0.48

We refer to Table A-2 to find that the cumulative area to the left of z = 0.48 is 0.6844, so the green-shaded area in Figure 6-23(a) is 1 - 0.6844 = 0.3156.

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6-4 The Central Limit Theorem 255

b. Approach Used for the Mean of Sample Values: Use the central limit theo- rem because we are dealing with the mean of a sample of 16 women, not an individual woman.

Requirement check for part b We can use the normal distribution if the original population is normally distributed or n 7 30. The sample size is not greater than 30, but the original population of pulse rates of women has a normal distribution, so samples of any size will yield means that are normally distributed.

Because we are now dealing with a distribution of sample means, we must use the parameters mx and sx, which are evaluated as follows:

mx = m = 74.0

sx = s

1n = 12.5

116 = 3.125

We want to find the green-shaded area shown in Figure 6-23(b).

Technology: If using technology, the green-shaded area in Figure 6-23(b) is 0.0274.

Table A-2: If using Table A-2, we convert the value of x = 80 bpm to the corresponding z score of z = 1.92, as shown here:

z = x - mx sx

= 80 - 74.0

12.5

116 =

6 3.125

= 1.92

From Table A-2 we find that the cumulative area to the left of z = 1.92 is 0.9726, so the green-shaded area of Figure 6-23(b) is 1 - 0.9726 = 0.0274.

c. Even though the sample size is not greater than 30, we can use the central limit theorem because the population of pulse rates of women is normally

continued

The Fuzzy Central Limit Theorem

In The Cartoon

Guide to Statis-

tics, by Gonick

and Smith,

the authors

describe the

Fuzzy Central

Limit Theorem as follows: “Data

that are influenced by many small

and unrelated random effects

are approximately normally

distributed. This explains why

the normal is everywhere: stock

market fluctuations, student

weights, yearly temperature

averages, SAT scores: All are the

result of many different effects.”

People’s heights, for example,

are the results of hereditary

factors, environmental factors,

nutrition, health care, geographic

region, and other influences,

which, when combined, produce

normally distributed values.

Limit Theorem as follows: “Data

x 5 80

(s 5 12.5)

m 5 74.0

Individual women pulse rates

Means of pulse rates from samples of women (16 in each sample)

x 5 80 mx 5 74.0

(sx 5 3.125)

FIGURE 6-23 Female Pulse Rates

(a) (b)

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256 CHAPTER 6 Normal Probability Distributions

Example 2 shows that we can use the same basic procedures from Section 6-2, but we must remember to correctly adjust the standard deviation when working with a sample mean instead of an individual sample value.

Introduction to Hypothesis Testing Carefully examine the conclusions that are reached in the next example illustrating the type of thinking that is the basis for the important procedure of hypothesis testing (formally introduced in Chapter 8). Example 3 uses the rare event rule for inferential statistics, first presented in Section 4-1:

Identifying Significant Results with Probabilities: The Rare Event Rule for Inferential Statistics

If, under a given assumption, the probability of a particular observed event is very small and the observed event occurs significantly less than or significantly greater than what we typically expect with that assumption, we conclude that the assumption is probably not correct.

The following example illustrates the above rare event rule.

distributed. As noted in the requirement check for part (b), samples of any size will yield means that are normally distributed.

I N T E R P R E TAT I O N

There is a 0.3156 probability that an individual woman will have a pulse rate greater than 80 bpm, and there is a 0.0274 probability that 16 randomly selected women will have pulse rates with a mean greater than 80 bpm.

EXAMPLE 3 Body Temperatures

Assume that the population of human body temperatures has a mean of 98.6°F, as is commonly believed. Also assume that the population standard deviation is 0.62°F (based on data from University of Maryland researchers). If a sample of size n = 106 is randomly selected, find the probability of getting a mean of 98.2°F or lower. (The value of 98.2°F was actually obtained from researchers; see the mid- night temperatures for Day 2 in Data Set 2 “Body Temperatures” in Appendix B.)

S O L U T I O N

We work under the assumption that the population of human body temperatures has a mean of 98.6°F. We weren’t given the distribution of the population, but because the sample size n = 106 exceeds 30, we use the central limit theorem and conclude that the distribution of sample means is a normal distribution with these parameters:

mx = m = 98.6 1by assumption2

sx = s

1n = 0.62

1106 = 0.0602197

Figure 6-24 shows the shaded area (see the tiny left tail of the graph) correspond- ing to the probability we seek. Having already found the parameters that apply to the distribution shown in Figure 6-24, we can now find the shaded area by using the same procedures developed in Section 6-2.

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6-4 The Central Limit Theorem 257

I N T E R P R E TAT I O N

The result shows that if the mean of our body temperatures is really 98.6°F, as we as- sumed, then there is an extremely small probability of getting a sample mean of 98.2°F or lower when 106 subjects are randomly selected. University of Maryland researchers did obtain such a sample mean, and after confirming that the sample is sound, there are two feasible explanations: (1) The population mean really is 98.6°F and their sample represents a chance event that is extremely rare; (2) the population mean is actually lower than the assumed value of 98.6°F and so their sample is typical. Because the probability is so low, it is more reasonable to conclude that the population mean is lower than 98.6°F. In reality it appears that the true mean body temperature is closer to 98.2°F!

This is the type of reasoning used in hypothesis testing, to be introduced in Chapter 8. For now, we should focus on the use of the central limit theorem for finding the probability of 0.0001, but we should also observe that this theorem will be used later in applying some very important concepts in statistics.

Technology: If we use technology to find the shaded area in Figure 6-24, we get 0.0000000000155, which can be expressed as 0+.

Table A-2: If we use Table A-2 to find the shaded area in Figure 6-24, we must first convert the score of x = 98.20°F to the corresponding z score:

z = x - mx sx

= 98.20 - 98.6

0.0602197 = - 6.64

Referring to Table A-2 we find that z = - 6.64 is off the chart, but for values of z below - 3.49, we use an area of 0.0001 for the cumulative left area up to z = - 3.49. We therefore conclude that the shaded region in Figure 6-24 is 0.0001.

0 z

mx 5 98.6

26.64

x 5 98.2

0.0001

FIGURE 6-24 Means of Body Temperatures from Samples of Size n = 106

Correction for a Finite Population In applying the central limit theorem, our use of sx = s>1n assumes that the popula- tion has infinitely many members. When we sample with replacement, the population is effectively infinite. When sampling without replacement from a finite population, we may need to adjust sx. Here is a common rule of thumb:

When sampling without replacement and the sample size n is greater than 5% of the finite population size N (that is, n + 0.05N), adjust the standard deviation of sample means Sx by multiplying it by this finite population correction factor:

A N - n N - 1

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258 CHAPTER 6 Normal Probability Distributions

Except for Exercise 21 “Correcting for a Finite Population,” the examples and exer- cises in this section assume that the finite population correction factor does not apply, because we are sampling with replacement, or the population is infinite, or the sample size doesn’t exceed 5% of the population size.

Statistical Literacy and Critical Thinking

1.  Requirements A researcher collects a simple random sample of grade-point averages of biostatistics students and she calculates the mean of this sample. Under what conditions can that sample mean be treated as a value from a population having a normal distribution?

2. Small Sample Weights of adult human brains are normally distributed. Samples of weights of adult human brains, each of size n = 15, are randomly collected and the sample means are found. Is it correct to conclude that the sample means cannot be treated as being from a normal distribution because the sample size is too small? Explain.

3. Notation In general, what do the symbols mx and sx represent? What are the values of mx and sx for samples of size 64 randomly selected from the population of IQ scores with popula- tion mean of 100 and standard deviation of 15?

4.  Annual Incomes Annual incomes of physicians are known to have a distribution that is skewed to the right instead of being normally distributed. Assume that we collect a large 1n 7 302 random sample of annual incomes of physicians. Can the distribution of those incomes in that sample be approximated by a normal distribution because the sample is large? Why or why not?

Using the Central Limit Theorem. In Exercises 5–8, assume that females have pulse rates that are normally distributed with a mean of 74.0 beats per minute and a standard deviation of 12.5 beats per minute (based on Data Set 1 “Body Data” in Appendix B).

5. a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 80 beats per minute.

b. If 16 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 80 beats per minute.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

6. a. If 1 adult female is randomly selected, find the probability that her pulse rate is greater than 70 beats per minute.

b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean greater than 70 beats per minute.

c. Why can the normal distribution be used in part (b), even though n 6 30?

7. a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 72 beats per minute and 76 beats per minute.

b. If 4 adult females are randomly selected, find the probability that they have pulse rates with a mean between 72 beats per minute and 76 beats per minute.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

8. a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 78 beats per minute and 90 beats per minute.

b. If 16 adult females are randomly selected, find the probability that they have pulse rates with a mean between 78 beats per minute and 90 beats per minute.

c. Why can the normal distribution be used in part (b), even though n 6 30?

6-4 Basic Skills and Concepts

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6-4 The Central Limit Theorem 259

9.  Hemoglobin in Men Hemoglobin levels in adult males are normally distributed with a mean of 14.7 g>dL and a standard deviation of 1.3 g>dL (based on data from the National Health and Nutrition Examination Survey).

a. The normal hemoglobin range for men is 13.6 g>dL to 17.7 g>dL. What percentage of men have hemoglobin levels in the normal range?

b. If we randomly collect samples of men with 9 in each sample, what percentage of those samples have a mean hemoglobin level that is within the normal range?

10.  Hemoglobin in Women Hemoglobin levels in adult females are normally distributed with a mean of 13.0 g>dL and a standard deviation of 1.3 g>dL (based on data from the Na- tional Health and Nutrition Examination Survey).

a. The normal hemoglobin range for women is 12.1 g>dL to 15.1 g>dL. What percentage of women have hemoglobin levels in the normal range?

b. If we randomly collect samples of women with 9 in each sample, what percentage of those samples have a mean hemoglobin level that is within the normal range?

11.  Diastolic BP in Women Diastolic blood pressure is a measure of the pressure when arteries rest between heartbeats. Diastolic blood pressure levels in women are normally distrib- uted with a mean of 70.2 mm Hg and a standard deviation of 11.2 mm Hg (based on Data Set 1 “Body Data” in Appendix B).

a. A diastolic blood pressure level above 90 mm Hg is considered to be hypertension. What percentage of women have hypertension?

b. If we randomly collect samples of women with 4 in each sample, what percentage of those samples have a mean above 90 mm Hg?

12. Diastolic BP in Men Diastolic blood pressure is a measure of the pressure when arteries rest between heartbeats. Diastolic blood pressure levels in men are normally distributed with a mean of 71.3 mm Hg and a standard deviation of 12.0 mm Hg (based on Data Set 1 “Body Data” in Appendix B).

a. A diastolic blood pressure level above 90 mm Hg is considered to be hypertension. What percentage of men have hypertension?

b. If we randomly collect samples of men with 4 in each sample, what percentage of those samples have a mean above 90 mm Hg?

13.  Mensa Membership in Mensa requires a score in the top 2% on a standard intelligence test. The Wechsler IQ test is designed for a mean of 100 and a standard deviation of 15, and scores are normally distributed.

a. Find the minimum Wechsler IQ test score that satisfies the Mensa requirement.

b. If 4 randomly selected adults take the Wechsler IQ test, find the probability that their mean score is at least 131.

c. If 4 subjects take the Wechsler IQ test and they have a mean of 132, but the individual scores are lost, can we conclude that all 4 of them are eligible for Mensa?

14.  Sleep The amounts of times that adults sleep are normally distributed with a mean of 6.8 hours and a standard deviation of 1.4 hours (based on data from multiple sources, includ- ing a Gallup poll and the American Journal of Epidemiology). A common recommendation is that we get between 7 and 9 hours of sleep each night.

a. For someone randomly selected, find the probability that they get between 7 and 9 hours of sleep in a night.

b. If we randomly collect a sample of 5 adults, what is the probability that the sample mean is between 7 hours and 9 hours?

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260 CHAPTER 6 Normal Probability Distributions

Ergonomics. Exercises 15–20 involve applications of ergonomics, which is a discipline focused on the design of tools and equipment so that they can be used safely, comfortably, and efficiently.

15. Water Taxi Safety Passengers died when a water taxi sank in Baltimore’s Inner Harbor. Men are typically heavier than women and children, so when loading a water taxi, assume a worst-case scenario in which all passengers are men. Assume that weights of men are normally distributed with a mean of 189 lb and a standard deviation of 39 lb (based on Data Set 1 “Body Data” in Appendix B). The water taxi that sank had a stated capacity of 25 passengers, and the boat was rated for a load limit of 3500 lb.

a. Given that the water taxi that sank was rated for a load limit of 3500 lb, what is the maxi- mum mean weight of the passengers if the boat is filled to the stated capacity of 25 passengers?

b. If the water taxi is filled with 25 randomly selected men, what is the probability that their mean weight exceeds the value from part (a)?

c. After the water taxi sank, the weight assumptions were revised so that the new capacity be- came 20 passengers. If the water taxi is filled with 20 randomly selected men, what is the prob- ability that their mean weight exceeds 175 lb, which is the maximum mean weight that does not cause the total load to exceed 3500 lb?

d. Is the new capacity of 20 passengers safe?

16. Designing Manholes According to the website www.torchmate.com, “manhole covers must be a minimum of 22 in. in diameter, but can be as much as 60 in. in diameter.” Assume that a manhole is constructed to have a circular opening with a diameter of 22 in. Men have shoulder breadths that are normally distributed with a mean of 18.2 in. and a standard deviation of 1.0 in. (based on data from the National Health and Nutrition Examination Survey).

a. What percentage of men will fit into the manhole?

b. Assume that Connecticut’s Eversource company employs 36 men who work in manholes. If 36 men are randomly selected, what is the probability that their mean shoulder breadth is less than 18.5 in.? Does this result suggest that money can be saved by making smaller manholes with a diameter of 18.5 in.? Why or why not?

17. Southwest Airlines Seats Southwest Airlines currently has a seat width of 17 in. Men have hip breadths that are normally distributed with a mean of 14.4 in. and a standard deviation of 1.0 in. (based on anthropometric survey data from Gordon, Churchill, et al.).

a. Find the probability that if an individual man is randomly selected, his hip breadth will be greater than 17 in.

b. Southwest Airlines uses a Boeing 737 for some of its flights, and that aircraft seats 122 pas- sengers. If the plane is full with 122 randomly selected men, find the probability that these men have a mean hip breadth greater than 17 in.

c. Which result should be considered for any changes in seat design: the result from part (a) or part (b)?

18.  Redesign of Ejection Seats When women were finally allowed to become pilots of fighter jets, engineers needed to redesign the ejection seats because they had been originally designed for men only. The ACES-II ejection seats were designed for men weighing between 140 lb and 211 lb. Weights of women are now normally distributed with a mean of 171 lb and a standard deviation of 46 lb (based on Data Set 1 “Body Data” in Appendix B).

a. If 1 woman is randomly selected, find the probability that her weight is between 140 lb and 211 lb.

b. If 25 different women are randomly selected, find the probability that their mean weight is between 140 lb and 211 lb.

c. When redesigning the fighter jet ejection seats to better accommodate women, which probability is more relevant: the result from part (a) or the result from part (b)? Why?

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6-5 Assessing Normality 261

19. Doorway Height The Boeing 757-200 ER airliner carries 200 passengers and has doors with a height of 72 in. Heights of men are normally distributed with a mean of 68.6 in. and a standard deviation of 2.8 in. (based on Data Set 1 “Body Data” in Appendix B).

a. If a male passenger is randomly selected, find the probability that he can fit through the doorway without bending.

b. If half of the 200 passengers are men, find the probability that the mean height of the 100 men is less than 72 in.

c. When considering the comfort and safety of passengers, which result is more relevant: the probability from part (a) or the probability from part (b)? Why?

d. When considering the comfort and safety of passengers, why are women ignored in this case?

20.  Loading Aircraft Before every flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. The Bombardier Dash 8 aircraft can carry 37 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6200 lb. The pilot sees that the plane is full and all passengers are men. The aircraft will be overloaded if the mean weight of the passengers is greater than 6200 lb>37 = 167.6 lb. What is the probability that the aircraft is overloaded? Should the pilot take any action to correct for an overloaded aircraft? Assume that weights of men are normally distributed with a mean of 189 lb and a standard deviation of 39 lb (based on Data Set 1 “Body Data” in Appendix B).

21. Correcting for a Finite Population In a study of babies born with very low birth weights, 275 children were given IQ tests at age 8, and their scores approximated a normal distribution with m = 95.5 and s = 16.0 (based on data from “Neurobehavioral Outcomes of School-Age Children Born Extremely Low Birth Weight or Very Preterm,” by Anderson et al., Journal of the American Medical Association, Vol. 289, No. 24). Fifty of those children are to be ran- domly selected without replacement for a follow-up study.

a. When considering the distribution of the mean IQ scores for samples of 50 children, should sx be corrected by using the finite population correction factor? Why or why not? What is the value of sx?

b. Find the probability that the mean IQ score of the follow-up sample is between 95 and 105.

6-4 Beyond the Basics

Key Concept The following chapters include important statistical methods requiring that sample data are from a population having a normal distribution. In this section we present criteria for determining whether the requirement of a normal distribution is satisfied. The criteria involve (1) visual inspection of a histogram to see if it is roughly bell-shaped; (2) identifying any outliers; and (3) constructing a normal quantile plot.

PA RT 1 Basic Concepts of Assessing Normality When trying to determine whether a collection of data has a distribution that is approximately normal, we can visually inspect a histogram to see if it is approxi- mately bell-shaped (as discussed in Section 2-2), we can identify outliers, and we can also use a normal quantile plot (discussed briefly in Section 2-2).

6-5 Assessing Normality

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262 CHAPTER 6 Normal Probability Distributions

Procedure for Determining Whether It Is Reasonable to Assume That Sample Data Are from a Population Having a Normal Distribution

1. Histogram: Construct a histogram. If the histogram departs dramatically from a bell shape, conclude that the data do not have a normal distribution.

2. Outliers: Identify outliers. If there is more than one outlier present, conclude that the data might not have a normal distribution. (Just one outlier could be an error or the result of chance variation, but be careful, because even a single out- lier can have a dramatic effect on results.)

3. Normal quantile plot: If the histogram is basically symmetric and the num- ber of outliers is 0 or 1, use technology to generate a normal quantile plot. Apply the following criteria to determine whether the distribution is nor- mal. (These criteria can be used loosely for small samples, but they should be used more strictly for large samples.)

Normal Distribution: The population distribution is normal if the pattern of the points is reasonably close to a straight line and the points do not show some systematic pattern that is not a straight-line pattern.

Not a Normal Distribution: The population distribution is not normal if either or both of these two conditions apply:

• The points do not lie reasonably close to a straight line.

• The points show some systematic pattern that is not a straight-line pattern.

Histograms and Normal Quantile Plots

In Part 2 of this section we describe the process of constructing a normal quantile plot, but for now we focus on interpreting a normal quantile plot. The following displays show histograms of data along with the corresponding normal quantile plots.

Normal: The first case shows a histogram of IQ scores that is close to being bell- shaped, so the histogram suggests that the IQ scores are from a normal distribution. The corresponding normal quantile plot shows points that are reasonably close to a straight-line pattern, and the points do not show any other systematic pattern that is not a straight line. It is safe to assume that these IQ scores are from a population that has a normal distribution.

DEFINITION

A normal quantile plot (or normal probability plot) is a graph of points (x, y) where each x value is from the original set of sample data, and each y value is the corresponding z score that is expected from the standard normal distribution.

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6-5 Assessing Normality 263

Uniform: The second case shows a histogram of data having a uniform (flat) distribu- tion. The corresponding normal quantile plot suggests that the data are not normally distributed. Although the pattern of points is reasonably close to a straight-line pattern, there is another systematic pattern that is not a straight-line pattern. We conclude that these sample values are from a population having a distribution that is not normal.

Skewed: The third case shows a histogram of the HDL cholesterol measurements. The shape of the histogram is skewed to the right. The corresponding normal quantile plot shows points that are not close to a straight-line pattern. These HDL cholesterol measurements are from a population having a distribution that is not normal.

Tools for Determining Normality ■ Histogram ,Outliers: If the requirement of a normal distribution is not too strict, simply look at a histogram and find the number of outliers. If the histo- gram is roughly bell-shaped and the number of outliers is 0 or 1, treat the popula- tion as if it has a normal distribution.

■ Normal Quantile Plot: Normal quantile plots can be difficult to construct on your own, but they can be generated with suitable technology.

■ Advanced Methods: In addition to the procedures discussed in this section, there are other more advanced procedures for assessing normality, such as the chi-square goodness-of-fit test, the Lilliefors test, the Anderson-Darling test, the Jarque-Bera test, and the Ryan-Joiner test (discussed briefly in Part 2).

PA RT 2 Manual Construction of Normal Quantile Plots

The following is a relatively simple procedure for manually constructing a normal quantile plot, and it is the same procedure used by Statdisk and the TI-83>84 Plus cal- culator. Some statistical packages use various other approaches, but the interpretation of the graph is essentially the same.

: The second case shows a histogram of data having a uniform (flat) distribu- tion. The corresponding normal quantile plot suggests that the data are not normally distributed. Although the pattern of points is reasonably close to a straight-line pattern,

. We conclude that

The Placebo Effect

It has long been

believed that

placebos actu-

ally help some

patients. In fact,

some formal

studies have

shown that when given a placebo

(a treatment with no medicinal

value), many test subjects show

some improvement. Estimates of

improvement rates have typically

ranged between one-third and

two-thirds of patients. However,

a more recent study suggests

that placebos have no real ef-

fect. An article in New England

Journal of Medicine (Vol. 334,

No. 21) was based on research

of 114 medical studies over

50 years. The authors of the

article concluded that placebos

appear to have some effect only

for relieving pain, but not for

other physical conditions. They

concluded that apart from clinical

trials, the use of placebos “can-

not be recommended.”

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264 CHAPTER 6 Normal Probability Distributions

Manual Construction of a Normal Quantile Plot

Step 1: First sort the data by arranging the values in order from lowest to highest.

Step 2: With a sample of size n, each value represents a proportion of 1>n of the sample. Using the known sample size n, find the values of 12n,

3 2n,

5 2n, and so

on, until you get n values. These values are the cumulative areas to the left of the corresponding sample values.

Step 3: Use the standard normal distribution (software or a calculator or Table A-2) to find the z scores corresponding to the cumulative left areas found in Step 2. (These are the z scores that are expected from a normally distributed sample.)

Step 4: Match the original sorted data values with their corresponding z scores found in Step 3; then plot the points (x, y), where each x is an original sam- ple value and y is the corresponding z score.

Step 5: Examine the normal quantile plot and use the criteria given in Part 1. Con- clude that the population has a normal distribution if the pattern of the points is reasonably close to a straight line and the points do not show some systematic pattern that is not a straight-line pattern.

EXAMPLE 1 Platelet Counts

Consider this sample of five patient platelet counts (1000 cells>mL): 125, 229, 236, 257, 234. With only five values, a histogram will not be very helpful in revealing the distribution of the data. Instead, construct a normal quantile plot for these five val- ues and determine whether they appear to come from a population that is normally distributed.

S O L U T I O N

The following steps correspond to those listed in the procedure above for construct- ing a normal quantile plot.

Step 1: First, sort the data by arranging them in order. We get 125, 229, 234, 236, 257.

Step 2: With a sample of size n = 5, each value represents a proportion of 1>5 of the sample, so we proceed to identify the cumulative areas to the left of the corresponding sample values. The cumulative left areas, which are expressed in general as 12n,

3 2n,

5 2n, and so on, become these specific areas for this example with

n = 5: 110, 3 10,

5 10,

7 10,

9 10. These cumulative left areas expressed in decimal form are 0.1,

0.3, 0.5, 0.7, and 0.9.

Step 3: We now use technology (or Table A-2) with the cumulative left areas of 0.1000, 0.3000, 0.5000, 0.7000, and 0.9000 to find these corresponding z scores: - 1.28, - 0.52, 0, 0.52, and 1.28. (For example, the z score of - 1.28 has an area of 0.1000 to its left.)

Step 4: We now pair the original sorted platelet counts with their corresponding z scores. We get these (x, y) coordinates, which are plotted in the following Statdisk display:

1125, - 1.282, 1229, - 0.522, 1234, 02, 1236, 0.522, 1257, 1.282

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6-5 Assessing Normality 265

Ryan-Joiner Test The Ryan-Joiner test is one of several formal tests of normality, each having its own advantages and disadvantages. Statdisk has a feature of Normal- ity Assessment that displays a histogram, normal quantile plot, the number of poten- tial outliers, and results from the Ryan-Joiner test.

Statdisk

I N T E R P R E TAT I O N

We examine the normal quantile plot in the Statdisk display. The points do not ap- pear to lie reasonably close to the straight line, so we conclude that the sample of five platelet counts does not appear to be from a normally distributed population.

EXAMPLE 2 Platelet Counts

Example 1 used a sample of five platelet counts. We can use the Normality Assess- ment feature of Statdisk with a different sample of the 300 platelet counts listed in Data Set 1 “Body Data” in Appendix B.

Statdisk

continued

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266 CHAPTER 6 Normal Probability Distributions

Data Transformations Many data sets have a distribution that is not normal, but we can transform the data so that the modified values have a normal distribution. One common transformation is to transform each value of x by taking its logarithm. (You can use natural logarithms or logarithms with base 10. If any original values are 0, take logarithms of values of x + 1). If the distribution of the logarithms of the values is a normal distribution, the distribution of the original values is called a lognormal distribution. (See Exercises 19 “Transformations” and 20 “Lognormal Distribu- tion”.) In addition to transformations with logarithms, there are other transformations, such as replacing each x value with 1x, or 1>x, or x2. In addition to getting a required normal distribution when the original data values are not normally distributed, such transformations can be used to correct deficiencies, such as a requirement (found in later chapters) that different data sets have the same variance.

Let’s use the display with the three criteria for assessing normality.

1. Histogram: We can see that the histogram is skewed to the left instead of being bell-shaped.

2. Outliers: The display shows that there are 20 possible outliers. If we examine a sorted list of the 300 platelet counts, there are platelet counts that appear to be outliers.

3. Normal quantile plot: The points in the normal quantile plot do not fit a straight-line pattern very well. We conclude that the 300 platelet counts do not appear to be from a population with a normal distribution.

Normal Quantile Plots Access tech instructions, videos, and data sets at www.TriolaStats.com

TECH CENTER

Statistical Literacy and Critical Thinking

1. Normal Quantile Plot Data Set 1 “Body Data” in Appendix B includes the heights of 147 randomly selected women, and heights of women are normally distributed. If you were to con- struct a histogram of the 147 heights of women in Data Set 1, what shape do you expect the histogram to have? If you were to construct a normal quantile plot of those same heights, what pattern would you expect to see in the graph?

2.  Normal Quantile Plot After constructing a histogram of the ages of the 147 women in- cluded in Data Set 1 “Body Data” in Appendix B, you see that the histogram is far from being bell-shaped. What do you now know about the pattern of points in the normal quantile plot?

3. Small Sample An article includes elapsed times (hours) to lumbar puncture for 19 patients who entered emergency rooms with sudden and severe “thunderclap” headaches (based on data from “Thunderclap Headache and Normal Computed Tomographic Results: Value of Cerebro- spinal Fluid Analysis,” by DuPont et al., Mayo Clinic Proceedings, Vol. 83, No. 12). Given that the sample size is less than 30, what requirement must be met in order to treat the sample mean as a value from a normally distributed population? Identify three tools for verifying that requirement.

6-5 Basic Skills and Concepts

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6-5 Assessing Normality 267

4. Assessing Normality The accompanying histogram is constructed from the diastolic blood pressure measurements of the 147 women included in Data Set 1 “Body Data” in Appendix B. If you plan to conduct further statistical tests and there is a loose requirement of a normally distrib- uted population, what do you conclude about the population distribution based on this histogram?

Minitab

Interpreting Normal Quantile Plots. In Exercises 5–8, examine the normal quantile plot and determine whether the sample data appear to be from a population with a normal distribution.

5.  Head Lengths of Bears The normal quantile plot represents the head lengths (in.) of bears listed in Data Set 11 “Bear Measurements.”

6. Diet Pepsi The normal quantile plot represents weights (pounds) of the contents of cans of Diet Pepsi.

7.  Patient Service Times The normal quantile plot represents service times (minutes) of randomly selected patients.

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