The data for a sample of 80 steel parts, given in the accompanying table, show t

Question

The data for a sample of 80 steel parts, given in the accompanying table, show the reported difference, in inches, between the actual length of the steel part and the specified length of the steel part. For example, a value of -0.002 represents a steel part that is 0.002 inch shorter than the specified length. Complete parts (a) through (d). b. Construct a 95% confidence interval estimate of the population mean. -0.00445 lessthanorequalto mu lessthanorequalto .000520 (Round to six decimal places as needed.) c. Compare the conclusions reached in (a) and (b). Choose the correct answer below. A. The confidence interval and hypothesis test both show there is sufficient evidence that the mean difference is not equal to 0.0 inches. B. The confidence interval and hypothesis test both show that there is insufficient evidence that the mean difference is not equal to 0.0 inches. C. The confidence interval shows insufficient evidence while the hypothesis test shows sufficient evidence that the mean difference is not equal to 0.0 inches. D. The confidence interval shows sufficient evidence while the hypothesis test shows insufficient evidence that the mean difference is not equal to 0.0 inches. d. Because n = 80, do you have to be concerned about the normality assumption needed for the t test and t interval? A. Yes, because the population distribution must be normal for all sample sizes. B. Yes, because the normality assumption is needed if the hypothesis best shows sufficient evidence that the mean difference is not equal to 0.0 inches. C. No, because the normality assumption is needed only if the hypothesis test shows sufficient evidence that the mean difference is not equal to 0.0 inches. D. With the large sample size, the t test is still valid for approximately normal population distributions, unless the population is very skewed.

Explanation / Answer

Answer

As we know that the t-test is applied to test the mean value assumption when the population standard deviation is not given. Beside that we usually assume normality for a moderately or perfectly symmetric data set which has quite large number of sample values . In such cases we do the test using t- statistic.

c) So here as it is computed that the confidence interval is (-0.000445 , 0.000520) which includes 0 so with 95% confidence or 0.05 level of significance the confidence interval and the hypothesis both show that there is insufficient evidence that the mean difference is not equal to 0.

Hence option B is the right one

d) Because n=80 which is quitely large sample size the sample values follows normality where here also the t-test is also valid, since t- test is derived from normally distributed variables. Now if the population is very skewed then we cannot assume normality then we have to be concerned about this normality assumption.

Hence the right option is option D

Hence the answer...............

Thank you.............

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