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 (39.26, 90.74) (round to two decimal places as needed.)
the margin of error is $□. (round to two decimal places as needed.)

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\), where \(n = 5\). So \(df=5-1 = 4\).

Step2: Find the critical value \(t_{\alpha/2}\)

For a \(99\%\) confidence interval, \(\alpha=1 - 0.99=0.01\), and \(\alpha/2=0.005\). Using the t - distribution table or a calculator, for \(df = 4\) and \(\alpha/2=0.005\), \(t_{\alpha/2}=5.598\).

Step3: Calculate the margin of error \(E\)

The formula for the margin of error when using the t - distribution is \(E=t_{\alpha/2}\times\frac{s}{\sqrt{n}}\), where \(s = 12.50\) and \(n = 5\).

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Answer:

The margin of error is \(\$31.30\)