Dr. Hackney STA Solutions pg 141

Dr. Hackney STA Solutions pg 141 - 8-20 Solutions Manual...

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Unformatted text preview: 8-20 Solutions Manual for Statistical Inference b. As a , + (x) x so P ( > K) P Z > n x (K-) . c. For K = 0, the answer in part (b) is 1 - (p-value) for H0 : 0. 8.51 If < p(x), sup P (W (X) c ) = < p(x) = sup P (W (X) W (x)). 0 0 Thus W (x) < c and we could not reject H0 at level having observed x. On the other hand, if p(x), sup P (W (X) c ) = p(x) = sup P (W (X) W (x)). 0 0 Either W (x) c in which case we could reject H0 at level having observed x or W (x) < c . But, in the latter case we could use c = W (x) and have {x : W (x ) c } define a size rejection region. Then we could reject H0 at level having observed x. 8.53 a. P (- < < ) = 1 1 1 + 2 2 2 2 e- - 2 /(2 2 ) d = 1 1 + = 1. 2 2 b. First calculate the posterior density. Because n -n(-)2 /(22 ) e x , f (|) = x 2 we can calculate the marginal density as m () x = = 2 2 1 1 1 f (|0) + x f (|) x e- /(2 ) d 2 2 - 2 2 2 1 n -n2 /(22 ) 1 1 x2 e x + e- /[2(( /n)+ )] 2 2 2 2 ( 2 /n)+ 2 1 (see Exercise 7.22). Then P ( = 0|) = 2 f (|0)/m (). x x x c. P |X| > x = 0 = 1 - P |X| x = 0 = 1 - P - X x = 0 x = 2 1- x/(/ n) , where is the standard normal cdf. d. For 2 = 2 = 1 and n = 9 we have a p-value of 2 (1 - (3)) and x P ( = 0| x) = 1+ 1 812 /20 e x 10 -1 . The p-value of x is usually smaller than the Bayes posterior probability except when x is very close to the value specified by H0 . The following table illustrates this. Some p-values and posterior probabilities (n = 9) x 0 .1 .15 .2 .5 .6533 .7 1 .7642 .6528 .5486 .1336 .05 .0358 .7597 .7523 .7427 .7290 .5347 .3595 .3030 p-value of x posterior P ( = 0|) x 1 .0026 .0522 2 0 0 ...
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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