Question

A sample of 100 of a certain machine part found an average
width of 0.755 in with a standard deviation of 0.012 in. Find the
99% confidence interval of the true mean of the width of the parts.
(What distrubution should be used?)

Answer #1

Solution :

Since, the population standard deviation is unknown therefore, the t-distribution should be used.

The 99% confidence interval for population mean is given as follows :

Where, x̅ is sample mean, s is sample standard deviation, n is sample size and t(0.01/2, n - 1) is critical t value to construct 99% confidence interval.

We have, n = 100, x̅ = 0.755, s = 0.012

Using t-table we get, t(0.01/2, 100 - 1) = 2.6264

Hence, 99% confidence interval for the true mean of the width of the parts is,

**The 99% confidence interval of the true mean of the
width of the parts is, (0.752, 0.758).**

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