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# 10_26_09ans - STAT 410 Examples for H0 = 0 Fall 2009 H 1 =...

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STAT 410 Examples for 10/26/2009 Fall 2009 H 0 : θ = θ 0 vs. H 1 : θ = θ 1 . Likelihood Ratio: ( 29 ( ( 29 ,..., , ; ,..., , ; ,..., , 2 1 1 2 1 0 2 1 L L λ n n n x x x x x x x x x θ θ = . Neyman-Pearson Theorem : { ( x 1 , x 2 , … , x n ) : ( k x x x n ,..., , 2 1 λ } ( Reject H 0 if ( k x x x n ,..., , 2 1 λ ) is the best (most powerful) rejection region. ½ . 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 0 : λ = 5 vs. H 1 : λ = 2 . a) Use the likelihood ratio to find the best rejection region. ( 29 ( ( 29 x x x e e e x x x 3 2 5 1 0 5 . 2 2 5 ; ; L L λ - - - = = = θ θ . ( k x < λ x > c . Reject H 0 if X > c . b) If the rejection region is “Reject H 0 if X > 0.50”, find the significance level α of the test. α = P ( Reject H 0 | H 0 is true ) = P ( X > 0.50 | λ = 5 ) = - 5 . 0 5 5 dx x e = e 2.5 0.082 .

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c) If the rejection region is “Reject H 0 if X > 0.50”, find the power of the test at λ = 2. Power = P ( Reject H 0 | H 0 is NOT true ) = P ( X > 0.50 | λ = 2 ) = - 5 . 0 2 2 dx x e = e 1 0.3679 . d) Find the rejection region with the significance level α = 0.05. 0.05 = α = P ( Reject H 0 | H 0 is true ) = P ( X > c | λ = 5 ) = - c dx x e 5 5 = e 5 c . c = ln 0.05 / 5 0.599 . e) Find the power of the test from part (d) at λ = 2. Power = P ( Reject H 0 | H 0 is NOT true ) = P ( X > c | λ = 2 ) = - c dx x e 2 2 = e 2 c 0.3017 .
¾ . 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 0 : λ = 2 vs. H 1 : λ = 5 . a) Use the likelihood ratio to find the best rejection region.

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