# 12_04ans - STAT 410 Examples for 12/04/2009 Fall 2009 H :...

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Unformatted text preview: STAT 410 Examples for 12/04/2009 Fall 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 : { ( 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. 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 : 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. 0.05 = α = P ( Reject H | H 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 | H 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....
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## This note was uploaded on 04/15/2010 for the course STAT 410 taught by Professor Monrad during the Fall '08 term at University of Illinois, Urbana Champaign.

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12_04ans - STAT 410 Examples for 12/04/2009 Fall 2009 H :...

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