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1、Solutions Manual for Statistical Inference, Second EditionGeorge Casella University of FloridaRoger L. Berger North Carolina State University Damaris Santana University of Florida0-2Solutions Manual for Statistical Inference“When I hear you give your reasons,” I remarked, “the thing always appears t
2、o me to be so ridiculously simple that I could easily do it myself, though at each successive instance of yourreasoning I am baffled until you explain your process.” Dr. Watson to Sherlock Holmes A Scandal in Bohemia0.1 DescriptionThis solutions manual contains solutions for all odd numbered problem
3、s plus a large number of solutions for even numbered problems. Of the 624 exercises in Statistical Inference, Second Edition, this manual gives solutions for 484 (78%) of them. There is an obtuse pattern as to which solutionswere included in this manual. We assembled all of the solutions that we had
4、 from the first edition,and filled in so that all odd-numbered problems were done. In the passage from the first to thesecond edition, problems were shuffled with no attention paid to numbering (hence no attentionpaid to minimize the new effort), but rather we tried to put the problems in logical or
5、der.A major change from the first edition is the use of the computer, both symbolically through Mathematicatmand numerically using R. Some solutions are given as code in either of these lan- guages. Mathematicatmcan be purchased from Wolfram Research, and R is a free download from http:/www.r-projec
6、t.org/. Here is a detailed listing of the solutions included.ChapterNumber of ExercisesNumber of SolutionsMissing 1555126,30,36,42 2403734,38,40 350424,6,10,20,30,32,34,36 465528,14,22,28,36,40 48,50,52,56,58,60,62 569462,4,12,14,26,28 all even problems from 36 68 643358,16,26,28,34,36,38,42 766524,
7、14,16,28,30,32,34, 36,42,54,58,60,62,64 8585136,40,46,48,52,56,58 958412,8,10,20,22,24,26,28,30 32,38,40,42,44,50,54,56 104826all even problems except 4 and 32 1141354,20,22,24,26,40 123116all even problems0.2 AcknowledgementMany people contributed to the assembly of this solutions manual. We again
8、thank all of thosewho contributed solutions to the first edition many problems have carried over into the second edition. Moreover, throughout the years a number of people have been in constant touch with us, contributing to both the presentations and solutions. We apologize in advance for those we
9、forget to mention, and we especially thank Jay Beder, Yong Sung Joo, Michael Perlman, Rob Strawderman, and Tom Wehrly. Thank you all for your help.And, as we said the first time around, although we have benefited greatly from the assistance andACKNOWLEDGEMENT0-3comments of others in the assembly of
10、this manual, we are responsible for its ultimate correctness. To this end, we have tried our best but, as a wise man once said, “You pays your money and you takes your chances.”George Casella Roger L. Berger Damaris SantanaDecember, 2001Chapter 1Probability Theory“If any little problem comes your wa
11、y, I shall be happy, if I can, to give you a hint or two as to its solution.” Sherlock Holmes The Adventure of the Three Students1.1 a. Each sample point describes the result of the toss (H or T) for each of the four tosses. So, for example THTT denotes T on 1st, H on 2nd, T on 3rd and T on 4th. The
12、re are 24= 16 such sample points.b. The number of damaged leaves is a nonnegative integer. So we might use S = 0,1,2,.c. We might observe fractions of an hour. So we might use S = t : t 0, that is, the halfinfinite interval 0,).d. Suppose we weigh the rats in ounces. The weight must be greater than
13、zero so we might use S = (0,). If we know no 10-day-old rat weighs more than 100 oz., we could use S = (0,100.e. If n is the number of items in the shipment, then S = 0/n,1/n,.,1.1.2 For each of these equalities, you must show containment in both directions.a. x AB x A and x / B x A and x / A B x A(
14、A B). Also, x A and x / B x A and x Bc x A Bc.b. Suppose x B. Then either x A or x Ac. If x A, then x B A, and, hence x (B A)(B Ac). Thus B (B A)(B Ac). Now suppose x (B A)(B Ac). Then either x (B A) or x (B Ac). If x (B A), then x B. If x (B Ac), then x B. Thus (B A) (B Ac) B. Since the containment
15、 goes both ways, we have B = (B A) (B Ac). (Note, a more straightforward argument for this part simply uses the Distributive Law to state that (B A) (B Ac) = B (A Ac) = B S = B.)c. Similar to part a).d. From part b). A B = A (B A) (B Ac) = A (B A) A (B Ac) = A A (B Ac) = A (B Ac).1.3 a. x A B x A or x B x B A x A B x A and x B x B A.b. x A (B C) x A or x B C x A B or x C x (A B) C. (It can similarly be shown that A (B C) = (A C) B.) x A (B C) x A and x B and x C x (A B) C.c. x (A B)cx / A or x / B x Acand x Bc x Ac
