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mont4e_sm_ch15_supplemental

# mont4e_sm_ch15_supplemental - Supplemental Exercises 15-33...

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Supplemental Exercises 15-33 a) 1. Parameter of interest is median surface finish 0 . 10 ~ : 3 0 . 10 ~ : . 2 1 0 = μ μ H H 4. α = 0.05 5. The test statistic is the observed number of plus differences or r + = 5. 6. We reject H 0 if the P-value corresponding to r + = 5 is less than or equal to α = 0.05. 7. Using the binomial distribution with n = 10 and p = 0.5, P-value = 2P( R * 5| p = 0.5) =1.0 8. Conclusion, we cannot reject H 0 . We cannot reject the claim that the median is 10 μ in. b) 1. Parameter of interest is median of surface finish 0 . 10 ~ : 3 0 . 10 ~ : . 2 1 0 = μ μ H H 4. α =0.05 5. Test statistic is n n r z 5 . 0 5 . 0 0 = + 6. We reject H 0 if the | Z 0 | > Z 0.025 = 1.96 7. Computation: 0 10 5 . 0 ) 10 ( 5 . 0 5 0 = = z 8. Conclusion, cannot reject H 0 . There is not sufficient evidence that the median surface finish differs from 10 microinches. The P-value = 2[1- Φ (0)] = 1 15-34 a) 1. The parameter of interest is the median fluoride emissions. 6 ~ : . 3 6 ~ : . 2 1 0 < = μ μ H H 4. α =0.05 5. The test statistic is the observed number of plus differences or r + = 4. 6. We reject H 0 if the P-value corresponding to r + = 4 is less than or equal to α = 0.05. 7. Using the binomial distribution with n = 15 and p = 0.5, P-value = P(R + 4 | p = 0.5) = 0.1334 8. Conclusion, do not reject H 0 . The data does not support the claim that the median fluoride impurity level is less than 6.

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mont4e_sm_ch15_supplemental - Supplemental Exercises 15-33...

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