Question

in a random sample of six mobile devices, the mean repair cost was $75.00 and the standard deviation was $14.00. assume the population is normally distributed and use a t - distribution to find the margin of error and construct a 99% confidence interval for the population mean. interpret the results. the 99% confidence interval for the population mean μ is (51.95, 98.05). (round to two decimal places as needed.) the margin of error is $\square$. (round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for margin of error

The formula for margin of error \(E\) using t - distribution is \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\). Also, the confidence interval is \(\bar{x}-E<\mu <\bar{x} + E\). Given \(\bar{x}-E = 51.95\) and \(\bar{x}=75\).

Step2: Solve for \(E\)

We know that \(\bar{x}-E = 51.95\). Substitute \(\bar{x}=75\) into the equation. Then \(E=\bar{x}-51.95\).

Answer:

\(23.05\)