Suppose a pet supplies company uses three manufacturing centers for the production of its cat toys. The defect classification for item production contains three levels: no defects, minimal defects but still acceptable for use, and significant defects where the item does not meet quality standards. Connor, a quality assurance specialist at the pet supplies company, plans to run a chi-square test of homogeneity to determine if the proportion of cat toy items in each defect category is the same across the three manufacturing centers. He sets the significance level for the test at a = 0.05. He randomly selects 388 items from Center A, 406 items from Center B, 274 items from Center C, and records the defect classification of each item. The sample results are summarized in the contingency table. Center A Center B Center C Total Observed count Expected count Chi-square Observed count Expected count Chi-square Observed count Expected count Chi-square No defects p-value = 355 353.49 0.00648 378 369.89 0.17800 240 249.63 0.37130 973 Minimal defects (acceptable) 18 14.17 1.03610 9 14.83 2.28928 12 10.01 0.39753 39 Significant defects (discard) 15 20.34 1.40403 19 21.29 0.24599 22 14.37 4.05526 56 x² = 9.984 If you wish, you may download the data in your preferred format. CrunchIt! CSV Excel JMP Mac Text Minitab PC Text R SPSS TI Calc Total Select the accurate statement regarding Connor's hypothesis test decision and conclusion. 388 406 274 Use software to compute the p-value for the chi-square statistic, x², for Connor's hypothesis test. You may find this list of software manuals helpful. Provide your answer with precision to three decimal places. Avoid rounding until the final step. 1068 O Connor should reject the null hypothesis because there is sufficient evidence (p-value < 0.05) to reject the null. Thus, Connor should conclude that there is homogeneity in the defect classification proportions across the manufacturing centers. Connor should reject the null hypothesis because there is sufficient evidence (p-value < 0.05) to reject the null. Thus, Connor should conclude that the manufacturing center used is not independent of the defective rate. O Connor should fail to reject the null hypothesis because there is insufficient evidence (p-value > 0.05) to reject the null. Thus, Connor should conclude that defect classification proportions are uniform across the manufacturing centers. Connor should reject the null hypothesis because there is sufficient evidence (p-value < 0.05) to reject the null. Thus, Connor should conclude that at least one of the proportions for a defect classification is different among the three manufacturing centers. nnor should fail to reject the null hypothesis because there is insufficient evidence (p-value < 0.05) to reject the null. Thus, Connor should conclude that the frequency distribution for each defect classification is the same across the manufacturing centers.

Suppose a pet supplies company uses three manufacturing centers for the production of its cat toys. The defect classification for item production contains three levels: no defects, minimal defects but still acceptable for use, and significant defects where the item does not meet quality standards. Connor, a quality assurance specialist at the pet supplies company, plans to run a chi-square test of homogeneity to determine if the proportion of cat toy items in each defect category is the same across the three manufacturing centers. He sets the significance level for the test at a = 0.05. He randomly selects 388 items from Center A, 406 items from Center B, 274 items from Center C, and records the defect classification of each item. The sample results are summarized in the contingency table. Center A Center B Center C Total Observed count Expected count Chi-square Observed count Expected count Chi-square Observed count Expected count Chi-square No defects p-value = 355 353.49 0.00648 378 369.89 0.17800 240 249.63 0.37130 973 Minimal defects (acceptable) 18 14.17 1.03610 9 14.83 2.28928 12 10.01 0.39753 39 Significant defects (discard) 15 20.34 1.40403 19 21.29 0.24599 22 14.37 4.05526 56 x² = 9.984 If you wish, you may download the data in your preferred format. CrunchIt! CSV Excel JMP Mac Text Minitab PC Text R SPSS TI Calc Total Select the accurate statement regarding Connor's hypothesis test decision and conclusion. 388 406 274 Use software to compute the p-value for the chi-square statistic, x², for Connor's hypothesis test. You may find this list of software manuals helpful. Provide your answer with precision to three decimal places. Avoid rounding until the final step. 1068 O Connor should reject the null hypothesis because there is sufficient evidence (p-value < 0.05) to reject the null. Thus, Connor should conclude that there is homogeneity in the defect classification proportions across the manufacturing centers. Connor should reject the null hypothesis because there is sufficient evidence (p-value < 0.05) to reject the null. Thus, Connor should conclude that the manufacturing center used is not independent of the defective rate. O Connor should fail to reject the null hypothesis because there is insufficient evidence (p-value > 0.05) to reject the null. Thus, Connor should conclude that defect classification proportions are uniform across the manufacturing centers. Connor should reject the null hypothesis because there is sufficient evidence (p-value < 0.05) to reject the null. Thus, Connor should conclude that at least one of the proportions for a defect classification is different among the three manufacturing centers. nnor should fail to reject the null hypothesis because there is insufficient evidence (p-value < 0.05) to reject the null. Thus, Connor should conclude that the frequency distribution for each defect classification is the same across the manufacturing centers.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.3: Measures Of Spread

Problem 1GP

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