Past data indicate that the variance of measurements made on sheet metal stampings by experienced quality control inspectors is 0.18 (inch)2. Such measurements made by an inexperienced inspector could have too large a variance (perhaps because of inability to read instruments properly) or too small a variance (perhaps because unusually high or low measurements are discarded). If a new inspector measures 101 stampings with variance of 0.13 (inch)2, test at the 0.05 level of significance whether the inspector is making satisfactory measurements. Assume data is normally distributed.

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Past data indicate that the variance of measurements made on sheet metal stampings by experienced quality control inspectors is 0.18 (inch)2. Such measurements made by an  inexperienced inspector could have too large a variance (perhaps because of inability to read instruments properly) or too small a variance (perhaps because unusually high or low measurements are discarded). If a new inspector measures 101 stampings with variance of 0.13 (inch)2, test at the 0.05 level of significance whether the inspector is making satisfactory measurements. Assume data is normally distributed.

Karen Robbins · Accepted Answer

Step 1  Hypothesis:  H0:σ2=0.18  H1:σ2≠0.18  Test Statistic (X2)=(n−1)s2σ2  =(101−1)(0.13)(0.18)  =72.2222  Thus, the test statistic is 72.2222.  Step 2  P-value is calculated using chi-square test statistic by p-value calculator.  P−value=0.0328096  =0.0328  Since P-value is less than 0.05 level of significance, thus there is sufficient evidence to reject the null hypothesis. So σ2≠0.18.

Past data indicate that the variance of measurements made on sheet met

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