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 (□,□). (round to two decimal places as needed.)

Explanation:

Step1: Determine the degrees of freedom and t - value

The degrees of freedom \(df=n - 1\), where \(n = 6\), so \(df=6-1 = 5\).
For a \(99\%\) confidence interval, the significance level \(\alpha=1 - 0.99=0.01\), and \(\frac{\alpha}{2}=0.005\).
Using the t - distribution table or a calculator, the t - value \(t_{\frac{\alpha}{2},df}=t_{0.005,5}=4.032\).

Step2: Calculate the margin of error \(E\)

The formula for the margin of error when using the t - distribution is \(E = t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\), where \(s = 14\) and \(n = 6\).
Substitute the values: \(E=4.032\times\frac{14}{\sqrt{6}}\).
First, calculate \(\frac{14}{\sqrt{6}}\approx5.715\).
Then \(E = 4.032\times5.715\approx23.05\).

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x}+E\), where \(\bar{x}=75\).
\(\bar{x}-E=75 - 23.05=51.95\) and \(\bar{x}+E=75 + 23.05 = 98.05\).

Answer:

\((51.95,98.05)\)