Dr. Hackney STA Solutions pg 129

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

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8-8 Solutions Manual for Statistical Inference d. If either X 1 1 or X 2 1, we should reject H 0 , because if θ = 0, P ( X i < 1) = 1. Thus, consider the rejection region given by { ( x 1 , x 2 ): x 1 + x 2 > C } { ( x 1 , x 2 ): x 1 > 1 } { ( x 1 , x 2 ): x 2 > 1 } . The first set is the rejection region for φ 2 . The test with this rejection region has the same size as φ 2 because the last two sets both have probability 0 if θ = 0. But for 0 < θ < C - 1, The power function of this test is strictly larger than β 2 ( θ ). If C - 1 θ , this test and φ 2 have the same power. 8.14 The CLT tells us that Z = ( i X i - np ) / np (1 - p ) is approximately n(0 , 1). For a test that rejects H 0 when i X i > c , we need to find c and n to satisfy P Z > c - n ( . 49) n ( . 49)( . 51) = . 01 and P Z > c - n ( . 51) n ( . 51)( . 49) = . 99 . We thus want c - n ( . 49) n ( . 49)( . 51) = 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 H 0 if f ( x | σ 1 ) f ( x | σ 0 ) = (2 πσ 2 1 ) - n/ 2 e - Σ i x 2 i / (2 σ 2 1 ) (2 πσ 2 0 ) - n/ 2 e - Σ i x 2 i / (2 σ 2 0 ) = σ 0 σ 1 n exp 1 2 i x 2 i 1 σ 2 0 - 1 σ 2 1 > k for some k 0. After some algebra, this is equivalent to rejecting if
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