A machine that produces ball bearings is set so that the average diameter is 0.5

Question

A machine that produces ball bearings is set so that the average diameter is 0.5 inch. Assume the diameter of ball bearings is normally distributes. A sample of 10 ball bearings was measured, with sample mean of 0.493 inch and sample standard deviation of 0.05 inch. Can we conclude at 10% significance level that the mean diameter is not 0.5 inch? 7. a. Use the critical value approach and interpret the test result. b. Use the p-value approach and interpret the test result. c. Use the confidence interval approach and interpret the test result.

Explanation / Answer

a. Here we have the H0: mu=0.5, H1: mu not equal to 0.5. Thus for the random sample given we have the Z statistic as Z= (0.493-0.5)/(0.05/sqrt(10)) = -0.007/0.015 = -0.47. Thus this result is not signifcant and so we accept the null hypothesis that the mean = 0.5.

b. The p value approach considers the probability of a value greater than the Z stat obtained. So in the above case the p value will be very large as the Z stat is very small. This will mean that the p value will be >0.05. So at the 5% level of significance we accept the null hypothesis that mu=00.5.

c. Based on the data we can construct the confidence interval as follows 0.5 +/- 1.96*(0.05/sqrt(10)) = 0.53 and 0.47. Thus from the sample we have the mean as 0.493. This lies within the confidence interval above and so we conclude that the mean is 0.5.

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