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Preparation for Test 2-solutions

# Preparation for Test 2-solutions - Preparation for Test II...

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Preparation for Test II Problem#1 a)H 0 : p 1 =p 2 =p 3 =1/3 H A : At least one p i differs from its hypothesized value N=434 E(n1)=n p 1 =1/3 434=144,67 E(n2)=n p 2 =1/3 434=144,67 E(n3)=n p 3 =1/3 434=144,67 735 , 87 67 , 144 ) 67 , 144 81 ( 67 , 144 ) 67 , 144 119 ( 67 , 144 ) 67 , 144 234 ( ) ( )) ( ( 2 2 2 2 2 = - + - + - = - = ni e ni E ni χ Rejection region: For α=0,01 and df=k-1=2 is 21 , 9 01 , 0 2 = χ Reject H 0 if 01 , 0 2 2 χ χ 87,735>9m21 Opinions are not evenly divided b) n p p p p f f f pf f ) 1 ( 96 , 1 96 , 1 - ± = ± σ Where f p =234/434=0,5391 =0,5391 0469 , 0 5391 , 0 434 5391 , 0 1 ( 5391 , 0 96 , 1 ± = - ± Problem #2 Ho: 1 p =0.32 2 p =0.50 3 p =0.18 α H : At least one of probabilities differs from their hypothesized values. a) E( 1 n )=85x0.32=27.2 E( 2 n )=85x0.50=42.5 E( 3 n )=85x0.18=15.3 [ ] ) ( ) ( 2 2 i i i n E n E n x - =

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( 29 ( 29 ( 29 3 . 15 3 . 15 17 5 . 42 5 . 42 32 2 . 27 5 . 27 36 2 2 2 2 - + - + - = x =5.630 df=n-1=3-1=2 05 . 0 = α 99 . 5 2 = α x , Rejection region ( 29 + ∞ ; 99 . 5 Since observed 2 x statistics does not fall into rejection region, we cannot reject Ho with 05 . 0 = α . b) ( 29 529 . 0 ; 317 . 0 106 . 0 423 . 0 85 ) 577 . 0 ( ) 423 . 0 ( 96 . 1 85 36 ˆ ˆ 2 = ± = ± = ± p z p δ α Problem #3 c) 2 1 σ - Population variance of errors made by Novice Inspector d) 2 2 σ
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