Dr. Hackney STA Solutions pg 128

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

## This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

Ask a homework question - tutors are online