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10_28_11 - STAT 409 Fall 2011 Examples for Example 1 1 2 3...

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STAT 409 Examples for 10/28/2011 Fall 2011 Example 1: 1 2 3 4 5 p 0 ( x ) 0.25 0.15 0.20 0.05 0.35 p 1 ( x ) 0.20 0.20 0.20 0.20 0.20 H 0 : X has p.m.f. p 0 ( x ) vs. H 1 : X has p.m.f. p 1 ( x ) . 1 2 3 4 5 λ ( x ) 1.25 0.75 1 0.25 1.75 Reject H 0 if x = 4 α = 0.05, Power = 0.20. Reject H 0 if x = 4 or 2 α = 0.20, Power = 0.40. Reject H 0 if x = 4 Reject H 0 with probability 3 1 if x = 2 α = 0.10, Power = 30 8 .
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Example 2: 1 2 3 4 5 p 0 ( x ) 0.25 0.15 0.20 0.05 0.35 p 1 ( x ) 0.05 0.30 0.40 0.20 0.05 H 0 : X has p.m.f. p 0 ( x ) vs. H 1 : X has p.m.f. p 1 ( x ) . 1 2 3 4 5 λ ( x ) 5 0.5 0.5 0.25 7 Reject H 0 if x = 4 α = 0.05, Power = 0.20. Reject H 0 if x = 4 Reject H 0 with probability 3 1 if x = 2 α = 0.10, Power = 0.30. Reject H 0 if x = 4 Reject H 0 with probability 4 1 if x = 3 α = 0.10, Power = 0.30. Reject H 0 if x = 4 Reject H 0 with probability 6 1 if x = 2 Reject H 0 with probability 8 1 if x = 3 α = 0.10, Power = 0.30.
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H 0 : θ Θ 0 vs. H 1 : θ Θ 1 Θ 0 Θ 1 = . Reject H 0 if Λ = ( ) ( ) θ L max θ L max 1 0 θ θ Θ Θ k Λ * = ( ) ( ) θ L max θ L max 1 0 0 θ θ Θ Θ Θ k since Λ * = min ( Λ , 1 ) H 0 : θ = θ 0 vs. H 1 : θ θ 0 Reject H 0 if Λ = ( ) ( ) θ ˆ L θ L 0 k Example 3: Let X 1 , X 2 , … , X n be a random sample of size n from an Exponential
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