Question

in a random sample of four mobile devices, the mean repair cost was $80.00 and the standard deviation was $12.00. 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 (44.95, 115.05) (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 and t - value

The degrees of freedom \(df=n - 1\), where \(n = 4\), so \(df=4-1 = 3\).
For a \(99\%\) confidence interval, using the t - distribution table, \(t_{\alpha/2}=5.841\) (with \(df = 3\) and \(\alpha=1 - 0.99=0.01\), \(\alpha/2=0.005\)).

Step2: Calculate the margin of error formula

The formula for the margin of error \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\), where \(s = 12\) and \(n = 4\).
Substitute the values: \(E=5.841\times\frac{12}{\sqrt{4}}\).
First, \(\sqrt{4}=2\), then \(\frac{12}{2}=6\).
So \(E=5.841\times6\).
\(E = 35.05\) (rounded to two decimal places).

Answer:

The margin of error is \(\$35.05\)