Question

in a random sample of five mobile devices, the mean repair cost was $65.00 and the standard deviation was $12.50. 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 \(n = 5\), so \(df=5-1 = 4\).
For a \(99\%\) confidence interval, \(\alpha=1 - 0.99=0.01\), and \(\frac{\alpha}{2}=0.005\).
Using the t - distribution table or a calculator, \(t_{\frac{\alpha}{2},df}=t_{0.005,4}=5.598\)

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

The formula for the margin of error for a t - distribution is \(E = t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\)
Given \(s = 12.50\), \(n = 5\), and \(t_{\frac{\alpha}{2}}=5.598\)
\(E=5.598\times\frac{12.50}{\sqrt{5}}\)
\(E=5.598\times\frac{12.50}{2.236}\)
\(E=5.598\times5.59\)
\(E\approx31.30\)

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x} + E\)
Given \(\bar{x}=65\)
\(65-31.30<\mu<65 + 31.30\)
\(33.70<\mu<96.30\)

Answer:

The \(99\%\) confidence interval for the population mean \(\mu\) is \((33.70,96.30)\)