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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.

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.

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. w < w0.05, n=12 = 13 v Null and Alternative Hypothesis b. H, u = 0.265 h. v Test Statistic: Hu<0.265 d. v Reject the Null Hypothesis if: v Conclusion: C.r = min (R+,R-) d. Fail to reject the null hypothesis. w = 9 is not less than or equal to the critical value w %23 = 5 0.05,n 9 e. Mean ball diameter f. Hu = 0.265 %3D Hu + 0.265 g. w = min (w+,w-) h.w S wo.05, n=9 = i. HoH = 0.265 %3D HH > 0.265 1. Reject the null hypothesis. The mean ball diameter claim is not 0.265 mm.
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. w < w0.05, n=12 = 13 v Null and Alternative Hypothesis b. H, u = 0.265 h. v Test Statistic: Hu<0.265 d. v Reject the Null Hypothesis if: v Conclusion: C.r = min (R+,R-) d. Fail to reject the null hypothesis. w = 9 is not less than or equal to the critical value w %23 = 5 0.05,n 9 e. Mean ball diameter f. Hu = 0.265 %3D Hu + 0.265 g. w = min (w+,w-) h.w S wo.05, n=9 = i. HoH = 0.265 %3D HH > 0.265 1. Reject the null hypothesis. The mean ball diameter claim is not 0.265 mm.
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