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Dr. Hackney STA Solutions pg 131

# Dr. Hackney STA Solutions pg 131 - 8-10 Solutions Manual...

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8-10 Solutions Manual for Statistical Inference b. The size is . 05 = β ( θ 0 ) = 1 + Φ( - c ) - Φ( c ) which implies c = 1 . 96. The power (1 - type II error) is . 75 β ( θ 0 + σ ) = 1 + Φ( - c - n ) - Φ( c - n ) = 1 + Φ( - 1 . 96 - n ) 0 - Φ(1 . 96 - n ) . Φ( - . 675) . 25 implies 1 . 96 - n = - . 675 implies n = 6 . 943 7. 8.19 The pdf of Y is f ( y | θ ) = 1 θ y (1 ) - 1 e - y 1 , y > 0 . By the Neyman-Pearson Lemma, the UMP test will reject if 1 2 y - 1 / 2 e y - y 1 / 2 = f ( y | 2) f ( y | 1) > k. To see the form of this rejection region, we compute d dy 1 2 y - 1 / 2 e y - y 1 / 2 = 1 2 y - 3 / 2 e y - y 1 / 2 y - y 1 / 2 2 - 1 2 which is negative for y < 1 and positive for y > 1. Thus f ( y | 2) /f ( y | 1) is decreasing for y 1 and increasing for y 1. Hence, rejecting for f ( y | 2) /f ( y | 1) > k is equivalent to rejecting for y c 0 or y c 1 . To obtain a size α test, the constants c 0 and c 1 must satisfy α = P ( Y c 0 | θ = 1) + P ( Y c 1 | θ = 1) = 1 - e - c 0 + e - c 1 and f ( c 0 | 2) f ( c 0 | 1) = f ( c 1 | 2) f ( c 1 | 1) . Solving these two equations numerically, for α = . 10, yields c 0 = . 076546 and c 1 = 3 . 637798.
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