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Dr. Hackney STA Solutions pg 132

# Dr. Hackney STA Solutions pg 132 - Second Edition 10 k 10-k...

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Second Edition 8-11 b. The size of the test is P ( Y 6 | p = 1 / 2) = 10 k =6 ( 10 k ) ( 1 2 ) k ( 1 2 ) 10 - k . 377. The power function is β ( θ ) = 10 k =6 ( 10 k ) θ k (1 - θ ) 10 - k c. There is a nonrandomized UMP test for all α levels corresponding to the probabilities P ( Y i | p = 1 / 2), where i is an integer. For n = 10, α can have any of the values 0, 1 1024 , 11 1024 , 56 1024 , 176 1024 , 386 1024 , 638 1024 , 848 1024 , 968 1024 , 1013 1024 , 1023 1024 , and 1. 8.23 a. The test is Reject H 0 if X > 1 / 2. So the power function is β ( θ ) = P θ ( X > 1 / 2) = 1 1 / 2 Γ( θ +1) Γ( θ )Γ(1) x θ - 1 (1 - x ) 1 - 1 dx = θ 1 θ x θ 1 1 / 2 = 1 - 1 2 θ . The size is sup θ H 0 β ( θ ) = sup θ 1 (1 - 1 / 2 θ ) = 1 - 1 / 2 = 1 / 2. b. By the Neyman-Pearson Lemma, the most powerful test of H 0 : θ = 1 vs. H 1 : θ = 2 is given by Reject H 0 if f ( x | 2) /f ( x | 1) > k for some k 0. Substituting the beta pdf gives f ( x | 2) f ( x | 1) = 1 β (2 , 1) x 2 - 1 (1 - x ) 1 - 1 1 β (1 , 1) x 1 - 1 (1 - x ) 1 - 1 = Γ(3) Γ(2)Γ(1) x = 2 x. Thus, the MP test is Reject H 0 if X > k/ 2. We now use the α level to determine k . We have α = sup θ Θ 0 β ( θ ) = β (1) = 1 k/ 2 f X ( x | 1) dx = 1 k/ 2 1 β (1 , 1) x 1 - 1 (1 - x ) 1 - 1 dx = 1 - k 2 .
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