11_02_09 - k since * = min ( , 1 ) H : = vs. H 1 : Reject H...

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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.
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H 0 : θ Θ 0 vs. H 1 : θ Θ 1 Θ 0 Θ 1 = . Reject H 0 if Λ = ( 29 ( 29 θ L max θ L max 1 0 θ θ Θ Θ k Λ * = ( 29 ( 29 θ L max θ L max 1 0 0 θ θ Θ Θ Θ
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Unformatted text preview: k since * = min ( , 1 ) H : = vs. H 1 : Reject H if = ( 29 ( 29 L 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 / . That is, f X ( x ) = x e -, x > 0. H : = vs. H 1 : L ( ) = =-n i x i e 1 = - = n i i n x 1 exp . = X 1 . = ( 29 ( 29 L L = - - = = n i i n n i i n x x x x 1 1 1 1 exp exp = ( 29 { } exp x n x e n n n-. k { } exp x x- c Reject H if { } exp x x- c ....
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11_02_09 - k since * = min ( , 1 ) H : = vs. H 1 : Reject H...

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