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Unformatted text preview: j e 7m g 1f m m g 1f h lkji A )e lkh g ji j A e Sb u V6 'U 6 cY a2 e d 1fRP b u V6 'U 6 cY a2 !P 5` e e d QiW7I4 31)('&7G 6 20 $ " 1 u 5 C 4 QiF d $ 0 xw QV'3%yQ( P g 4 e v u t sg r p EA h'&$ 8ia q C 5`ihf Ie d b cC P a X `Y
denote a if . distribution. The pvalue is W7I4 V1)(U&%G 6 20 $ " CT CG1FS1B 1RA 5 @#QB G1 @ I HE EA G1F@E DB5@9 C 9 C P 9 C 9 C A 8754 31)('&%#! 6 20 $ " Chapter 11, Problem 1. Let . We proved earlier that . Homework 9: Partial Solutions be the cdf of under . Hence, . The pvalue is Insufficient evidence to reject. so function is (c) Let (b) Chapter 11, Problem 3. (a) The rejection region is (d) The pvalue is 0 since all Chapter 11, Problem 2. Straightforward. must be less than or equal to 1/2 under n Reject . 1 o . The power . Chapter 11, Problem 4. and the pvalue is Q 6 B 4 m Qi n P Qq P 4 f Q74 Q 4 1 4 F . The Wald test statistic is evidence against the null . The results are equivocal. I'm not convinced. Chapter 11, Problem 5. (a) The Wald statistic is The pvalue is
' & %" # ! against the null hypothesis of no difference. The confidence interval is . Going only on the test, one would conclude that these are two different authors since the means are different. But the confidence interval shows that, while the difference is statistically significant, the difference is small. Taken together I would not conclude the authors are different though you might have come to a different conclusion. (5b) A permutation test on the absolute difference of means yields a pvalue of .024. (This depends on your simulation so everyone will have a slightly different number.) The evidence against the null seems a bit weaker. Chapter 11, Problem 6. (a) Under . Hence, implies and so . (b) Under , and 2 C 6 P $ C cW754 5 4 6 and . Hence, ` e C ! h F P g 86754 4H E 6 I k i81F HTd 4 E 6 E D F9 6 0 6 G @ CC B C C 6 A @ P P 8754 8f 7! 6 9 C 6 6 4 6 4 e n 1414 ) P $Q ) P 4 (& ' # # #" ` %$" e ! T 0 ) 11 with estimated standard error which is moderate . A 95 per cent confidence interval is 6 4 174 ##"
) 3 21 which is strong evidence do this. The grader will accept any answer for this. Some posibilities include: a F e Ie e P e 4 d g iE ' P I G e e g g iE iE I G ' G ' e P I e e A e g g iE iE ' ' G G P e % h g e e A g e % g e iE e g i E e e ' ' e P e % h G eDA e % g G g g iE g iE P e % h g G fA g ' G e e % h e g iE G !!$ % g ' g iE G P % h A g g iE P h 5e g
G % W754 6 e e B . t D@IC 6 d t 6 EG P t E 5C 6 P P H D 6 r Chapter 11, Problem 6. Under 6 G I r f I rf g h k (c) and hence , 4 twosample binomial tests using a Bonferroni correction. I E As test, a likelihood ratio test (from the appendix). We could also do 4 different Chapter 11, Problem 9. (a) For each comparison we can use test g Chapter 11, Problem 8. As I said in class, we did not really cover how to , and since , 3 g and hence . So . d (We could also use a permutation test.) The results of the Wald test are Drug Chlorpromazine Dimenhydrinate Pentobarbital (100 mg) Pentobarbital (150 mg) test statistic 2.76 .64 .48 1.65 odd ratio 2.42 .82 1.18 1.67 pvalue .006 .520 .627 .100 The only significant result is Chlorpromazine with an odds ratio of 2.42. (b) There were 4 test so for Bonferroni we should use a significance level of . The first result is still significant. giving the estimated standard error when 4 e 6 e 6 o Wald test is to reject where e ! i ! Chapter 11, Problem 10. The mle is D # 9 # # g e u ' D 9 % % % e g iE P G 2 u ! T g 1c74 14 i versus using the Wald test and the Fisher information is . The ...
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 Fall '02
 LarryWasserman

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