Dr. Hackney STA Solutions pg 129

Dr. Hackney STA Solutions pg 129 - 8-8 Solutions Manual for...

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Unformatted text preview: 8-8 Solutions Manual for Statistical Inference d. If either X1 1 or X2 1, we should reject H0 , because if = 0, P (Xi < 1) = 1. Thus, consider the rejection region given by {(x1 , x2 ) : x1 + x2 > C} {(x1 , x2 ) : x1 > 1} {(x1 , x2 ) : x2 > 1}. The first set is the rejection region for 2 . The test with this rejection region has the same size as 2 because the last two sets both have probability 0 if = 0. But for 0 < < C - 1, The power function of this test is strictly larger than 2 (). If C - 1 , this test and 2 have the same power. 8.14 The CLT tells us that Z = ( i Xi - np)/ np(1 - p) is approximately n(0, 1). For a test that rejects H0 when i Xi > c, we need to find c and n to satisfy P We thus want Z> c-n(.49) n(.49)(.51) c-n(.49) n(.49)(.51) = .01 and P Z> c-n(.51) n(.51)(.49) = .99. = 2.33 and c-n(.51) n(.51)(.49) = -2.33. Solving these equations gives n = 13,567 and c = 6,783.5. 8.15 From the Neyman-Pearson lemma the UMP test rejects H0 if 2 f (x | 1 ) (21 ) e-i xi /(21 ) = = 2 2 -n/2 e-i x2 /(20 ) f (x | 0 ) i (20 ) -n/2 2 2 0 1 n exp 1 2 x2 i i 1 1 2 - 2 0 1 >k for some k 0. After some algebra, this is equivalent to rejecting if x2 > i i 2log (k (1 /0 ) ) 1 2 0 n - 1 2 1 =c because 1 1 2 - 2 > 0 . 0 1 This is the UMP test of size , where = P0 ( 2 2 , use the fact that i Xi /0 2 . Thus n = P0 i i 2 Xi > c). To determine c to obtain a specified 2 2 2 Xi /0 > c/0 2 = P 2 > c/0 , n 2 2 so we must have c/0 = 2 , which means c = 0 2 . n, n, 8.16 a. Size = P (reject H0 | H0 is true) = 1 Power = P (reject H0 | HA is true) = 1 b. Size = P (reject H0 | H0 is true) = 0 Power = P (reject H0 | HA is true) = 0 8.17 a. The likelihood function is -1 Type I error = 1. Type II error = 0. Type I error = 0. Type II error = 1. n j -1 yj . L(, |x, y) = n i xi ...
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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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