Assume that the hardness of iron bars produced by one plant is normally distribu

Question

Assume that the hardness of iron bars produced by one plant is normally distributed with unknown mean and variance. A random sample of n = 7 bars from the output of the plant is measured to determine whether the average hardness meets the desired figure mu 0 = 172. The following measurements: 167, 174, 179, 164, 163, 160, 168 were obtained. Test H_0: mu = 172 versus the alternative Ha: mu Notequalto 172 using an appropriate test with alpha = .01 Compute 99% confidence interval for mu. Compare to Q1. State conclusions.

Explanation / Answer

Sample size n=7

Degree of freedom=n-1=7-1=6

Two tailed Critical t=±3.707

From data we have xbar=167.86 and s=6.62

So test statistic t=(xbar-172)/(s/sqrt(n))

=(167.86-172)/(6.62/sqrt(7))

=-1.645

As calculated t>-3.707, we do not reject the null hypothesis and conclude that average hardness meets the desired figure of 172.

99% confidence interval =(xbar)±3.707*(s/sqrt(n))

=(167.86)±3.707*(6.62/sqrt(7))

=(158.58               177.14)

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