Dr. Hackney STA Solutions pg 155

# Dr. Hackney STA Solutions pg 155 - Second Edition 9-13 9.46...

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 9-13 9.46 The proof is similar to that of Theorem 9.3.5: P θ ( θ ∈ C * ( X )) = P θ ( X ∈ A * ( θ )) ≤ P θ ( X ∈ A ( θ )) = P θ ( θ ∈ C ( X )) , where A and C are any competitors. The inequality follows directly from Definition 8.3.11. 9.47 Referring to (9.3.2), we want to show that for the upper confidence bound, P θ ( θ ∈ C ) ≤ 1- α if θ ≥ θ . We have P θ ( θ ∈ C ) = P θ ( θ ≤ ¯ X + z α σ/ √ n ) . Subtract θ from both sides and rearrange to get P θ ( θ ∈ C ) = P θ θ- θ σ/ √ n ≤ ¯ X- θ σ/ √ n + z α = P Z ≥ θ- θ σ/ √ n- z α , which is less than 1- α as long as θ ≥ θ . The solution for the lower confidence interval is similar. 9.48 a. Start with the hypothesis test H : θ ≥ θ versus H 1 : θ < θ . Arguing as in Example 8.2.4 and Exercise 8.47, we find that the LRT rejects H if ( ¯ X- θ ) / ( S/ √ n ) <- t n- 1 ,α . So the acceptance region is { x : (¯ x- θ ) / ( s/ √ n ) ≥ - t n- 1 ,α...
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