This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Second Edition 819 8.45 The verification of size is the same computation as in Exercise 8.37a. Example 8.3.3 shows that the power function m () for each of these tests is an increasing function. So for > 0 , m () > m (0 ) = . Hence, the tests are all unbiased. 8.47 a. This is very similar to the argument for Exercise 8.41. + b. By an argument similar to part (a), this LRT rejects H0 if T+ = X Y 
2 Sp 1 n + 1 m tn+m2, . +  c. Because H0 is the union of H0 and H0 , by the IUT method of Theorem 8.3.23 the test that rejects H0 if the tests in parts (a) and (b) both reject is a level test of H0 . That is, the test rejects H0 if T + tn+m2, and T  tn+m2, . d. Use Theorem 8.3.24. Consider parameter points with X  Y = and 0. For any , P (T + tn+m2, ) = . The power of the T  test is computed from the noncentral t distribution with noncentrality parameter x  Y  ()/[(1/n + 1/m)] = 2/[(1/n + 1/m)] which converges to as 0. Thus, P (T  tn+m2, ) 1 as 0. By Theorem 8.3.24, this IUT is a size test of H0 . 8.49 a. The pvalue is P = 7 or more successes out of 10 Bernoulli trials 10 1 7 2 = .171875.
7 = 1 2
8 1 2 3 + 10 8 1 2 1 2 2 + 10 9 1 2 9 1 2 1 + 10 10 1 2 10 1 2 0 b. Pvalue = P {X 3  = 1} = 1  P (X < 3  = 1) e1 12 e1 11 e1 10 + + .0803. = 1 2! 1! 0! c. Pvalue = P{
i Xi 9  3 = 3} = 1  P (Y < 9  3 = 3) 38 37 36 35 31 30 + + + ++ + 8! 7! 6! 5! 1! 0! .0038, =
3 1  e3 where Y = i=1 Xi Poisson(3). 8.50 From Exercise 7.26, (x) = where (x) = x a. For K > 0, P ( > Kx, a) = where Z n(0, 1). n 2 2 2 na n n( (x))2 /(22 ) e , 2 2 and we use the "+" if > 0 and the "" if < 0. en(+ (x))
K
2 /(2 2 ) d = P Z> n [K+ (x)] , ...
View
Full
Document
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.
 Spring '12
 Dr.Hackney
 Statistics

Click to edit the document details