mont4e_sm_ch09_sec05 - Section 9-5 9-76 a) 1) The parameter...

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9-34 Section 9-5 9-76 a) 1) The parameter of interest is the true fraction of satisfied customers. 2) H 0 : p = 0.9 3) H 1 : p 0.9 4) α = 0.05 5) () 0 0 0 0 1 p np np x z = or n p p p p z 0 0 0 0 1 ˆ = ; Either approach will yield the same conclusion 6) Reject H 0 if z 0 < z α /2 where z α /2 = z 0.025 = 1.96 or z 0 > z α /2 where z α /2 = z 0.025 = 1.96 7) x = 850 n = 1000 85 . 0 1000 850 ˆ = = p 27 . 5 ) 1 . 0 )( 9 . 0 ( 1000 ) 9 . 0 ( 1000 850 1 0 0 0 0 = = = p np np x z 8) Because -5.27<-1.96 reject the null hypothesis and conclude the true fraction of satisfied customers is significantly different from 0.9 at α = 0.05. The P-value: 2(1- Φ (5.27)) 2(1-1) 0 b) The 95% confidence interval for the fraction of surveyed customers is: 87 . 0 827 . 0 1000 ) 15 . 0 ( 85 . 0 96 . 1 85 . 1000 ) 15 . 0 ( 85 . 0 96 . 1 85 . ) ˆ 1 ( ˆ ˆ ) ˆ 1 ( ˆ ˆ 2 / 2 / + + p p n p p z p p n p p z p α Because 0.9 is not included in the confidence interval, we reject the null hypothesis at α = 0.05. 9-77 a) 1) The parameter of interest is the true fraction of rejected parts 2) H 0 : p = 0.03 3) H 1 : p < 0.03 4) α = 0.05 5) 0 0 0 0 1 p np np x z = or n p p p p z 0 0 0 0 1 ˆ = ; Either approach will yield the same conclusion 6) Reject H 0 if z 0 < z α where z α = z 0.05 = 1.65 7) x = 10 n = 500 02 . 0 500 10 ˆ = = p 31 . 1 ) 97 . 0 )( 03 . 0 ( 500 ) 03 . 0 ( 500 10 1 0 0 0 0 = = = p np np x z
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9-35 8) Because 1.31 > 1.65, do not reject the null hypothesis. There is not enough evidence to conclude that the true fraction of rejected parts is significantly less than 0.03 at α = 0.05. P-value = Φ (-1.31) = 0.095 b) The upper one-sided 95% confidence interval for the fraction of rejected parts is: 0303 . 0 500 ) 98 . 0 ( 02 . 0 65 . 1 02 . ) ˆ 1 ( ˆ ˆ + p p n p p z p p α Because 0.03<0.0303 the we fail to reject the null hypothesis 9-78 a) 1) The parameter of interest is the true fraction defective integrated circuits 2) H 0 : p = 0.05 3) H 1 : p 0.05 4) α = 0.05 5) () 0 0 0 0 1 p np np x z = or n p p p p z 0 0 0 0 1 ˆ = ; Either approach will yield the same conclusion 6) Reject H 0 if z 0 < z α /2 where z α /2 = z 0.025 = 1.96 or z 0 > z α /2 where z α /2 = z 0.025 = 1.96 7) x = 13 n = 300 ± . p == 13 300 0 043 53 . 0
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This note was uploaded on 10/12/2009 for the course IND E 315 taught by Professor Kailashkapur during the Spring '09 term at University of Washington.

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mont4e_sm_ch09_sec05 - Section 9-5 9-76 a) 1) The parameter...

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