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A quality inspector measured the diameter of a ball bearing using a new type of caliper. The results were as follows (in mm): 0.265, 0.263, 0.266, 0.267, 0.267, 0.265, 0.267,0.267, 0.265, 0.268, 0.268, and 0.263. Use the Wilcoxon signed-rank test to evaluate the claim that the mean ball diameter is 0.265 mm. Use a = 0.05. Match the answers. Parameter of Interest: Null and Alternative Hypothesis Test Statistic:

A quality inspector measured the diameter of a ball bearing using a new type of caliper. The results were as follows (in mm): 0.265, 0.263, 0.266, 0.267, 0.267, 0.265, 0.267,0.267, 0.265, 0.268, 0.268, and 0.263. Use the Wilcoxon signed-rank test to evaluate the claim that the mean ball diameter is 0.265 mm. Use a = 0.05. Match the answers. Parameter of Interest: Null and Alternative Hypothesis Test Statistic:

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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A quality inspector measured the diameter of a ball bearing using a new type of caliper. The results were as follows (in mm): 0.265, 0.263, 0.266, 0.267, 0.267, 0.265, 0.267,0.267, 0.265, 0.268, 0.268, and 0.263. Use the Wilcoxon signed-rank test to evaluate the claim that the mean ball diameter is 0.265 mm. Use a = 0.05. Match the answers. - v Parameter of Interest: a. H:u = 0.265 v Null and Alternative Hypothesis H:µ > 0.265 v Test Statistic: b. w = min (w+,w-) v Reject the Null Hypothesis if: c. Fail to reject the null hypothesis. v Conclusion: w = 9 is not less than or equal to the critical value w 0.05.1=9= 5 d. Mean ball diameter e. Reject the null hypothesis. The mean ball diameter claim is not 0.265 mm. f. H:u = 0.265 H:µ + 0.265 g.r = min (R+,R-) h. Η: μ:0.265 H:u < 0.265 i. w < wo.05. J=12 = 13 j. w < W0.05. n=9
Transcribed Image Text:A quality inspector measured the diameter of a ball bearing using a new type of caliper. The results were as follows (in mm): 0.265, 0.263, 0.266, 0.267, 0.267, 0.265, 0.267,0.267, 0.265, 0.268, 0.268, and 0.263. Use the Wilcoxon signed-rank test to evaluate the claim that the mean ball diameter is 0.265 mm. Use a = 0.05. Match the answers. - v Parameter of Interest: a. H:u = 0.265 v Null and Alternative Hypothesis H:µ > 0.265 v Test Statistic: b. w = min (w+,w-) v Reject the Null Hypothesis if: c. Fail to reject the null hypothesis. v Conclusion: w = 9 is not less than or equal to the critical value w 0.05.1=9= 5 d. Mean ball diameter e. Reject the null hypothesis. The mean ball diameter claim is not 0.265 mm. f. H:u = 0.265 H:µ + 0.265 g.r = min (R+,R-) h. Η: μ:0.265 H:u < 0.265 i. w < wo.05. J=12 = 13 j. w < W0.05. n=9
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