Transcription
INSTRUCTOR’SSOLUTIONS MANUALPROBABILITY ANDSTATISTICAL INFERENCENINTH EDITIONROBERT V. HOGGUniversity of IowaElliot A. TanisHope CollegeDale L. ZimmermanUniversity of IowaBoston Columbus Indianapolis New York San Francisco Upper Saddle RiverAmsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal TorontoDelhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo
The author and publisher of this book have used their best efforts in preparing this book. These efforts include thedevelopment, research, and testing of the theories and programs to determine their effectiveness. The author andpublisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentationcontained in this book. The author and publisher shall not be liable in any event for incidental or consequentialdamages in connection with, or arising out of, the furnishing, performance, or use of these programs.Reproduced by Pearson from electronic files supplied by the author.Copyright 2015 Pearson Education, Inc.Publishing as Pearson, 75 Arlington Street, Boston, MA 02116.All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in anyform or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior writtenpermission of the publisher. Printed in the United States of America.ISBN-13: 978-0-321-91379-1ISBN-10: 0-321-91379-5www.pearsonhighered.com
iiiContentsPrefacev1 Probability1.1 Properties of Probability1.2 Methods of Enumeration1.3 Conditional Probability1.4 Independent Events . .1.5 Bayes’ Theorem . . . . .1123452 Discrete Distributions2.1 Random Variables of the Discrete Type2.2 Mathematical Expectation . . . . . . . .2.3 Special Mathematical Expectations . . .2.4 The Binomial Distribution . . . . . . . .2.5 The Negative Binomial Distribution . .2.6 The Poisson Distribution . . . . . . . .7710111416173 Continuous Distributions3.1 Random Variables of the Continuous Type . . . . . . . .3.2 The Exponential, Gamma, and Chi-Square Distributions3.3 The Normal Distribution . . . . . . . . . . . . . . . . . .3.4 Additional Models . . . . . . . . . . . . . . . . . . . . .19192528294 Bivariate Distributions4.1 Bivariate Distributions of the Discrete Type . .4.2 The Correlation Coefficient . . . . . . . . . . .4.3 Conditional Distributions . . . . . . . . . . . .4.4 Bivariate Distributions of the Continuous Type4.5 The Bivariate Normal Distribution . . . . . . .333334353641.43434448505255565859.5 Distributions of Functions of Random Variables5.1 Functions of One Random Variable . . . . . . . . . . . . .5.2 Transformations of Two Random Variables . . . . . . . .5.3 Several Random Variables . . . . . . . . . . . . . . . . . .5.4 The Moment-Generating Function Technique . . . . . . .5.5 Random Functions Associated with Normal Distributions5.6 The Central Limit Theorem . . . . . . . . . . . . . . . . .5.7 Approximations for Discrete Distributions . . . . . . . . .5.8 Chebyshev’s Inequality and Convergence in Probability .5.9 Limiting Moment-Generating Functions . . . . . . . . . .c 2015 Pearson Education, Inc.Copyright .
ivContents6 Point Estimation6.1 Descriptive Statistics . . . . . . . . . .6.2 Exploratory Data Analysis . . . . . . .6.3 Order Statistics . . . . . . . . . . . . .6.4 Maximum Likelihood Estimation . . .6.5 A Simple Regression Problem . . . . .6.6 Asymptotic Distributions of MaximumLikelihood Estimators . . . . . . . . .6.7 Sufficient Statistics . . . . . . . . . . .6.8 Bayesian Estimation . . . . . . . . . .6.9 More Bayesian Concepts . . . . . . . .616163687174.787981837 Interval Estimation7.1 Confidence Intervals for Means . . . . . . . . . . . .7.2 Confidence Intervals for the Difference of Two Means7.3 Confidence Intervals For Proportions . . . . . . . . .7.4 Sample Size . . . . . . . . . . . . . . . . . . . . . . .7.5 Distribution-Free Confidence Intervals for Percentiles7.6 More Regression . . . . . . . . . . . . . . . . . . . .7.7 Resampling Methods . . . . . . . . . . . . . . . . . .85858688899092988 Tests of Statistical Hypotheses8.1 Tests About One Mean . . . . . . .8.2 Tests of the Equality of Two Means8.3 Tests about Proportions . . . . . . .8.4 The Wilcoxon Tests . . . . . . . . .8.5 Power of a Statistical Test . . . . . .8.6 Best Critical Regions . . . . . . . . .8.7 Likelihood Ratio Tests . . . . . . . .1051051071101111151191219 More Tests9.1 Chi-Square Goodness-of-Fit Tests . . . . . . .9.2 Contingency Tables . . . . . . . . . . . . . . .9.3 One-Factor Analysis of Variance . . . . . . .9.4 Two-Way Analysis of Variance . . . . . . . .9.5 General Factorial and 2k Factorial Designs . .9.6 Tests Concerning Regression and Correlation9.7 Statistical Quality Control . . . . . . . . . . .125125128128132133134135.c 2015 Pearson Education, Inc.Copyright
PrefacevPrefaceThis solutions manual provides answers for the even-numbered exercises in Probability and StatisticalInference, 9th edition, by Robert V. Hogg, Elliot A. Tanis, and Dale L. Zimmerman. Completesolutions are given for most of these exercises. You, the instructor, may decide how many of thesesolutions and answers you want to make available to your students. Note that the answers for theodd-numbered exercises are given in the textbook.All of the figures in this manual were generated using Maple, a computer algebra system. Mostof the figures were generated and many of the solutions, especially those involving data, were solvedusing procedures that were written by Zaven Karian from Denison University. We thank him forproviding these. These procedures are available free of charge for your use. They are available fordown load at http://www.math.hope.edu/tanis/. Short descriptions of these procedures are provided on the “Maple Card.” Complete descriptions of these procedures are given in Probability andStatistics: Explorations with MAPLE, second edition, 1999, written by Zaven Karian and Elliot Tanis, published by Prentice Hall (ISBN 0-13-021536-8). You can download a copy of this manual ur hope is that this solutions manual will be helpful to each of you in your teaching.If you find an error or wish to make a suggestion, send these to Elliot Tanis, [email protected],and he will post corrections on his web page, c 2015 Pearson Education, Inc.Copyright
vic 2015 Pearson Education, Inc.Copyright
Chapter 1 Probability1Chapter 1Probability1.1Properties of Probability1.1-2 Sketch a figure and fill in the probabilities of each of the disjoint sets.Let A {insure more than one car}, P (A) 0.85.Let B {insure a sports car}, P (B) 0.23.Let C {insure exactly one car}, P (C) 0.15.It is also given that P (A B) 0.17. Since A C φ, P (A C) 0. It follows thatP (A B C 0 ) 0.17. Thus P (A0 B C 0 ) 0.06 and P (A0 B 0 C) 0.09.1.1-4 (a) S {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, TTHH,HTHT, THTH, THHT, HTTT, THTT, TTHT, TTTH, TTTT};(b) (i) 5/16, (ii) 0, (iii) 11/16, (iv) 4/16, (v) 4/16, (vi) 9/16, (vii) 4/16.1.1-6 (a) P (A B) 0.4 0.5 0.3 0.6;(b)A (A B 0 ) (A B)P (A) P (A B 0 ) P (A B)0.4 P (A B 0 ) 0.30P (A B ) 0.1;(c) P (A0 B 0 ) P [(A B)0 ] 1 P (A B) 1 0.3 0.7.1.1-8 Let A {lab work done}, B {referral to a specialist},P (A) 0.41, P (B) 0.53, P ([A B]0 ) 0.21.P (A B) P (A) P (B) P (A B)0.79 0.41 0.53 P (A B)P (A B) 0.41 0.53 0.79 0.15.1.1-10A B C P (A B C) A (B C)P (A) P (B C) P [A (B C)]P (A) P (B) P (C) P (B C) P [(A B) (A C)]P (A) P (B) P (C) P (B C) P (A B) P (A C) P (A B C).1.1-12 (a) 1/3; (b) 2/3; (c) 0; (d) 1/2.c 2015 Pearson Education, Inc.Copyright
2Section 1.2 Methods of Enumeration 2[r r( 3/2)]3 1 .1.1-14 P (A) 2r21.1-16 Note that the respective probabilities are p0 , p1 p0 /4, p2 p0 /42 , · · ·. Xp0 14kk 0p01 1/4p0 1 34151 .16161 p 0 p1 1 1.2Methods of Enumeration1.2-2 (a) (4)(5)(2) 40; (b) (2)(2)(2) 8.µ ¶61.2-4 (a) 4 80;3(b) 4(26 ) 256;(c)(4 1 3)! 20.(4 1)!3!1.2-6 S { DDD, DDFD, DFDD, FDDD, DDFFD, DFDFD, FDDFD, DFFDD,FDFDD, FFDDD, FFF, FFDF, FDFF, DFFF FFDDF, FDFDF,DFFDF, FDDFF, DFDFF, DDFFF } so there are 20 possibilities.1.2-8 3 · 3 · 212 36,864.µ¶ µ¶n 1n 11.2-10 rr 11.2-1202µn (n 1)!(n 1)! r!(n 1 r)! (r 1)!(n r)! n!(n r)(n 1)! r(n 1)! r!(n r)!r!(n r)!µ ¶n µ ¶nXXnn(1 1)n ( 1)r (1)n r ( 1)r.rrr 0r 0n(1 1) n µ ¶Xnr 0rr(1) (1)n r n µ ¶Xnr 0r.¶10 1 3645!1.2-14 886,163,135.3636!9!µ ¶µ¶19 52 19102,48636µ ¶1.2-16 (a) 0.2917;52351,3259µ ¶µ ¶µ ¶µ ¶µ ¶µ ¶µ ¶19 10 7 3 5 2 67,695321 0 1 0 2µ ¶ 0.00622. (b)521,236,6649c 2015 Pearson Education, Inc.Copyright µ ¶n.r
Chapter 1 Probability1.33Conditional Probability1.3-2 (a)1041;1456(b)392;633(c)649.823(d) The proportion of women who favor a gun law is greater than the proportion of menwho favor a gun law.1.3-4 (a) P (HH) (b) P (HC) 13 121· ;52 511713 1313· ;52 51204(c) P (Non-Ace Heart, Ace) P (Ace of Hearts, Non-Heart Ace) 12 41 3511· · .52 51 52 5152 · 51521.3-6 Let H {died from heart disease}; P {at least one parent had heart disease}.P (H P 0 ) N (H P 0 )110 .N (P 0 )64813 2 1·· ;20 19 181140µ ¶µ ¶3 171121µ ¶ ·(b) ;20173803µ ¶µ¶3179X1352 2k 2µ ¶·(c) 0.4605.2020 2k76k 12k1.3-8 (a)(d) Draw second. The probability of winning is 1 0.4605 0.5395.52 51 50 49 48 478,808,975····· 0.74141;52 52 52 52 52 5211,881,376(b) P (A0 ) 1 P (A) 0.25859.1.3-10 (a) P (A) 1.3-12 (a) It doesn’t matter because P (B1 ) (b) P (B) 111, P (B5 ) , P (B18 ) ;18181812 on each draw.1891.3-14 (a) P (A1 ) 30/100;(b) P (A3 B2 ) 9/100;(c) P (A2 B3 ) 41/100 28/100 9/100 60/100;c 2015 Pearson Education, Inc.Copyright
4Section 1.4 Independent Events(d) P (A1 B2 ) 11/41;(e) P (B1 A3 ) 13/29.1.3-161.43 5 2 423· · .5 8 5 840Independent Events1.4-2 (a) P (A B) P (A)P (B) (0.3)(0.6) 0.18;P (A B) P (A) P (B) P (A B) 0.3 0.6 0.18 0.72.(b) P (A B) 0P (A B) 0.P (B)0.61.4-4 Proof of (b): P (A0 B) P (B)P (A0 B)P (B)[1 P (A B)]P (B)[1 P (A)]P (B)P (A0 ).Proof of (c): P (A0 B 0 ) P [(A B)0 ]1 P (A B)1 P (A) P (B) P (A B)1 P (A) P (B) P (A)P (B)[1 P (A)][1 P (B)]P (A0 )P (B 0 ).1.4-6P [A (B C)] P [A B C] P (A)P (B)P (C) P (A)P (B C).P [A (B C)] P [A0 (B C 0 )]P [A0 B 0 C 0 ]1.4-8 P [(A B) (A C)]P (A B) P (A C) P (A B C)P (A)P (B) P (A)P (C) P (A)P (B)P (C)P (A)[P (B) P (C) P (B C)]P (A)P (B C).P (A0 C 0 B)P (B)[P (A0 C 0 ) B]P (B)[1 P (A C B)]P (B)[1 P (A C)]P (B)P [(A C)0 ]P (B)P (A0 C 0 )P (B)P (A0 )P (C 0 )P (A0 )P (B)P (C 0 )P (A0 )P (B C 0 ) P [(A B C)0 ] 1 P (A B C) 1 P (A) P (B) P (C) P (A)P (B) P (A)P (C) P (B)P (C) P A)P (B)P (C) [1 P (A)][1 P (B)][1 P (C)] P (A0 )P (B 0 )P (C 0 ).21 2 3 1 4 3 5 2 3· · · · · · .6 6 6 6 6 6 6 6 69c 2015 Pearson Education, Inc.Copyright
This solutions manual provides answers for the even-numbered exercises in Probability and Statistical Inference, 9th edition, by Robert V. Hogg, Elliot A. Tanis, and Dale L. Zimmerman. Complete solutions
