ch15 - CHAPTER 15 Section 15-2 15-1 1 The parameter of...

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Unformatted text preview: CHAPTER 15 Section 15-2 15-1 1. The parameter of interest is median of pH. . 7 ~ : 3 . 7 ~ : . 2 1 ≠ = µ µ H H 4. α =0.05 5. The test statistic is the observed number of plus differences or r + = 8. 6. We reject H if the P-value corresponding to r + = 3 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 + ≥ 3 | p = 0.5) = 1 8. Conclusion: we cannot reject H . There is not enough evidence to reject the manufacturer’s claim that the median of the pH is 7.0 15-2 1. The parameter of interest is median titanium content. 5 . 8 ~ : 3 5 . 8 ~ : . 2 1 ≠ = µ µ H H 4. α =0.05 5. The test statistic is the observed number of plus differences or r + = 7. 6. We reject H if the P-value corresponding to r + = 7 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 * ≤ 7|p=0.5)=0.359 8. Conclusion: we cannot reject H . There is not enough evidence to reject the manufacturer’s claim that the median of the titanium content is 8.5. 15-3 1. Parameter of interest is the median impurity level. 5 . 2 ~ : . 3 5 . 2 ~ : . 2 1 < = µ µ H H 4. α =0.05 5. The test statistic is the observed number of plus differences or r + = 2. 6. We reject H if the P-value corresponding to r + = 2 is less than or equal to α = 0.05. 7. Using the binomial distribution with n = 22 and p = 0.5, P-value = P(R + ≤ 2 | p = 0.5) = 0.0002 8. Conclusion, reject H . The data supports the claim that the median is impurity level is less than 2.5. 15-4 1. The parameter of interest is median of pH. . 7 ~ : 3 . 7 ~ : . 2 1 ≠ = µ µ H H 4. α =0.05 5. The test statistic is n n r z 5 . 5 . * − = 6. We reject H if |Z |>1.96 for α =0.05. 7. r*=8 and 90 . 1 10 5 . ) 10 ( 5 . 8 5 . 5 . * = − = − = n n r z 8. Conclusion: we cannot reject H . There is not enough evidence to reject the manufacturer’s claim that the median of the pH is 7.0 P-value = 2[1 - P(|Z | < 1.90)] = 2(0.0287) = 0.0574 15-1 15-5 a) 1. Parameter of interest is the median compressive strength 2250 ~ : . 3 2250 ~ : . 2 1 > = µ µ H H 4. α = 0.05 5. The test statistic is the observed number of plus differences or r + = 7. 6. We reject H if the P-value corresponding to r + = 7 is less than or equal to α =0.05. 7. Using the binomial distribution with n = 12 and p = 0.5, P-value = P( R + ≥ 7 | p = 0.5) = 0.3872 8. Conclusion, cannot reject H . There is not enough evidence to conclude that the median compressive strength is greater than 2250. b) 1. Parameter of interest is the median compressive strength 2250 ~ : . 3 2250 ~ : . 2 1 > = µ µ H H 4. α =0.05 5. Test statistic is n n r z 5 . 5 . − = + 6. We reject H if the | Z | > Z 0.025 = 1.96 7. Computation: 577 ....
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ch15 - CHAPTER 15 Section 15-2 15-1 1 The parameter of...

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