Question

in a random sample of six people, the mean driving distance to work was 23.9 miles and the standard deviation was 7.2 miles. assuming the population is normally distributed and using the t - distribution, a 99% confidence interval for the population mean μ is (12.0, 35.8) (and the margin of error is 11.9). through research, it has been found that the population standard deviation of driving distances to work is 6.3. using the standard normal distribution with the appropriate calculations for a standard deviation that is known, find the margin of error and construct a 99% confidence interval for the population mean μ. interpret and compare the results.

identify the margin of error.
6.6 miles
(round to one decimal place as needed.)
construct a 99% confidence interval for the population mean.
(□,□)
(round to one decimal place as needed.)

Explanation:

Step1: Recall the formula for confidence interval

The formula for a confidence interval when population standard deviation \(\sigma\) is known is \(\bar{x}\pm E\), where \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\) (margin of error), \(\bar{x}\) is the sample mean, \(z_{\alpha/2}\) is the z - score, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. For a \(99\%\) confidence interval, \(\alpha=1 - 0.99=0.01\), and \(\alpha/2=0.005\). From the standard normal table, \(z_{0.005} = 2.576\). We are given \(\bar{x}=23.9\), \(\sigma = 6.3\), \(n = 6\).

Step2: Calculate the confidence interval

The lower limit of the confidence interval is \(\bar{x}-E\) and the upper limit is \(\bar{x}+E\). We know \(E = 6.6\) (already calculated as the margin of error).
The lower limit \(=23.9-6.6=17.3\)
The upper limit \(=23.9 + 6.6=30.5\)

Answer:

\((17.3,30.5)\)