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Dr. Hackney STA Solutions pg 141

# Dr. Hackney STA Solutions pg 141 - 8-20 Solutions Manual...

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8-20 Solutions Manual for Statistical Inference b. As a → ∞ , δ + ( x ) ¯ x so P ( θ > K ) P Z > n σ ( K - ¯ x ) . c. For K = 0, the answer in part (b) is 1 - (p-value) for H 0 : θ 0. 8.51 If α < p ( x ), sup θ Θ 0 P ( W ( X ) c α ) = α < p ( x ) = sup θ Θ 0 P ( W ( X ) W ( x )) . Thus W ( x ) < c α and we could not reject H 0 at level α having observed x . On the other hand, if α p ( x ), sup θ Θ 0 P ( W ( X ) c α ) = α p ( x ) = sup θ Θ 0 P ( W ( X ) W ( x )) . Either W ( x ) c α in which case we could reject H 0 at level α having observed x or W ( x ) < c α . But, in the latter case we could use c α = W ( x ) and have { x : W ( x ) c α } define a size α rejection region. Then we could reject H 0 at level α having observed x . 8.53 a. P ( -∞ < θ < ) = 1 2 + 1 2 1 2 πτ 2 -∞ e - θ 2 / (2 τ 2 ) = 1 2 + 1 2 = 1 . b. First calculate the posterior density. Because f x | θ ) = n 2 πσ e - n x - θ ) 2 / (2 σ 2 ) , we can calculate the marginal density as m π x ) = 1 2 f x | 0) + 1 2 -∞ f x | θ ) 1 2 πτ e - θ 2 / (2 τ 2 ) = 1 2 n 2 πσ e - n ¯ x 2 / (2 σ 2 ) + 1 2 1 2 π ( σ 2 /n )+ τ 2 e - ¯ x 2 / [2(( σ 2 /n )+ τ 2 )] (see Exercise 7.22). Then P ( θ = 0 | ¯ x ) = 1 2 f x | 0) /m π x ). c. P ( | ¯ X | > ¯ x θ = 0 ) = 1 - P ( | ¯ X | ≤
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