10_26_11ans - STAT 409 Fall 2011 Examples for 10/26/2011...

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Examples for 10/26/2011 Fall 2011 Consider a family of probability distributions with a p.d.f. of the form f ( x ; θ ) = θ x θ – 1 , 0 < x < 1, zero elsewhere, θ 1. To test the simple hypothesis H 0 : θ = 1 against the alternative simple hypothesis H 1 : θ = θ 1 , where θ 1 > 1, we will use a random sample X 1 , X 2 of size n = 2. Intuition: n = 1 f 0 ( x ) = < < o.w. 0 1 0 1 x f 1 ( x ) = < < o.w. 0 1 0 2 x x H 0 : X has p.d.f. f 0 ( x ) vs. H 1 : X has p.d.f. f 1 ( x ) . θ = 1 θ = 2 ( ) ( ) ( ) x x x x 1 0 2 1 ; ; H L H L λ = = . ( ) , ... , , 2 1 λ n x x x k x c . If α = significance level, then c = 1 – α . Then Power = 1 – c 2 = 2 α α 2 . a) Use the likelihood ratio to show that the best ( most powerful ) rejection region is given by C = { ( x 1 , x 2 ) : x 1 x 2 c }. ( ) ( ) ( ) 2 1 1 2 1 2 1 , ; , ; 1 , θ L L λ x x x x x x = = 1 2 1 1 2 1 1 1 θ θ θ 1 - - x x = ( ) 1 2 1 2 1 1 θ θ 1 - x x . Best
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10_26_11ans - STAT 409 Fall 2011 Examples for 10/26/2011...

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