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Unformatted text preview: Second Edition 913 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...
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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.
 Spring '12
 Dr.Hackney
 Statistics

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