Dr. Hackney STA Solutions pg 128

Dr. Hackney STA Solutions pg 128 - Second Edition 8-7 We...

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Unformatted text preview: Second Edition 8-7 We have that, for ˆ σ 2 > 1, ∂ ∂ (ˆ σ 2 ) log λ ( x ) = n 2 1 ˆ σ 2- 1 < . So λ ( x ) is decreasing in ˆ σ 2 , and rejecting H for small values of λ ( x ) is equivalent to rejecting for large values of ˆ σ 2 , that is, large values of S 2 . The LRT accepts H if and only if S 2 < k , where k is a constant. We can pick the prior parameters so that the acceptance regions match in this way. First, pick α large enough that M/ ( n- 1) > k . Then, as β varies between 0 and ∞ , ( M- 2 /β ) / ( n- 1) varies between-∞ and M/ ( n- 1). So, for some choice of β , ( M- 2 /β ) / ( n- 1) = k and the acceptance regions match. 8.12 a. For H : μ ≤ 0 vs. H 1 : μ > 0 the LRT is to reject H if ¯ x > cσ/ √ n (Example 8.3.3). For α = . 05 take c = 1 . 645. The power function is β ( μ ) = P ¯ X- μ σ/ √ n > 1 . 645- μ σ/ √ n = P Z > 1 . 645- √ nμ σ ....
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