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# 11_02_09 - ≤ k since Λ = min Λ 1 H θ = θ vs H 1 θ...

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STAT 409 Examples for 11/02/2009 Fall 2009 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.
H 0 : θ Θ 0 vs. H 1 : θ Θ 1 Θ 0 Θ 1 = . Reject H 0 if Λ = ( ( 29 θ L max θ L max 1 0 θ θ Θ Θ k Λ * = ( ( 29 θ L max θ L max 1 0 0 θ θ Θ Θ Θ
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Unformatted text preview: ≤ k since Λ * = min ( Λ , 1 ) H : θ = θ vs. H 1 : θ ≠ θ Reject H if Λ = ( 29 ( 29 θ ˆ L θ L ≤ k Example 3 : Let X 1 , X 2 , … , X n be a random sample of size n from an Exponential distribution with mean 1 / λ . That is, f X ( x ) = x e λ λ-, x > 0. H : θ = θ vs. H 1 : θ ≠ θ L ( λ ) = ∏ =-n i x i e 1 λ λ = -∑ = n i i n x 1 λ λ exp . λ ˆ = X 1 . Λ = ( 29 ( 29 θ ˆ L θ L = - -∑ ∑ = = n i i n n i i n x x x x 1 1 1 1 exp exp λ λ = ( 29 { } λ λ exp x n x e n n n-. Λ ≤ k ⇔ { } λ exp x x-≤ c Reject H if { } λ exp x x-≤ c ....
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