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07_31ans - STAT 410 Examples for Summer 2009 H θ = θ vs H...

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Unformatted text preview: STAT 410 Examples for 07/31/2009 Summer 2009 H : θ = θ vs. H 1 : θ = θ 1 . Likelihood Ratio: ( 29 ( 29 ( 29 ,..., , ; ,..., , ; ,..., , 2 1 1 2 1 2 1 L L λ n n n x x x x x x x x x θ θ = . Neyman-Pearson Theorem : C = { ( x 1 , x 2 , … , x n ) : ( 29 k x x x n ≤ ,..., , 2 1 λ }. ( “ Reject H if ( 29 k x x x n ≤ ,..., , 2 1 λ ” ) is the best (most powerful) rejection region. 4. Let X 1 , X 2 , … , X n be a random sample of size n from the distribution with probability density function ( 29 ( 29 ( 29 θ 2 X X ln 1 θ θ ; x x x f x f ⋅- = = , x > 1, θ > 1. Find the form of the uniformly most powerful rejection region for testing H : θ = 2 vs. H 1 : θ > 2. Let θ > 2. ( 29 ( 29 ( 29 ( 29 ∏ ∏ = = ⋅- = = n i i i n i i i n n n x x x x x x x x x x x x x 1 θ 2 1 2 2 1 1 2 1 2 1 ln 1 θ ln ,..., , ; ,..., , ; ,..., , H L H L λ . ( 29 ( 29 n n i i n x x x x 2 2 θ 1 2 1 1 θ ,..., , λ- - = ∏ = . ( 29 k x x x n ≤ ,..., , 2 1 λ ⇔ c x n i i ≤ ∏ = 1 ( since θ > 2 ). 4½. Let X have an Exponential distribution with mean 1 / λ . That is, f X ( x ) = x e λ λ- , x > 0. Consider the test H : λ = 5 vs. H 1 : λ = 2 . a) Use the likelihood ratio to find the best rejection region. ( 29 ( 29 ( 29 x x x e e e x x x 3 2 5 1 5 . 2 2 5 ; ; L L λ--- ⋅ = = = θ θ . ( 29 k x < λ ⇔ x > c . Reject H if X > c . b) If the rejection region is “Reject H if X > 0.50”, find the significance level α of the test. α = P ( Reject H | H is true ) = P ( X > 0.50 | λ = 5 ) = ∫ ∞- 5 . 5 5 dx x e = e – 2.5 ≈ 0.082 . c) If the rejection region is “Reject H if X > 0.50”, find the power of the test at λ = 2. Power = P ( Reject H | H is NOT true ) = P ( X > 0.50 | λ = 2 ) = ∫ ∞- 5 . 2 2 dx x e = e – 1 ≈ 0.3679 . d) Find the rejection region with the significance level α = 0.05....
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07_31ans - STAT 410 Examples for Summer 2009 H θ = θ vs H...

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