Question

in a random sample of five mobile devices, the mean repair cost was $70.00 and the standard deviation was $11.50. assume the population is normally distributed and use a t - distribution to find the margin of error and construct a 90% confidence interval for the population mean. interpret the results. the 90% confidence interval for the population mean μ is (□,□). (round to two decimal places as needed.)

Explanation:

Step1: Find the degrees of freedom and t - value

The degrees of freedom \(df=n - 1\), where \(n = 5\), so \(df=5-1 = 4\).
For a 90% confidence interval, \(\alpha=1 - 0.90=0.10\), and \(\frac{\alpha}{2}=0.05\).
Looking up in the t - distribution table, \(t_{\frac{\alpha}{2},df}=t_{0.05,4}=2.132\).

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

The formula for the margin of error when using the t - distribution is \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\).
Given \(s = 11.50\), \(n = 5\), and \(t_{\frac{\alpha}{2}}=2.132\).
\(E=2.132\times\frac{11.50}{\sqrt{5}}\)
\(E=2.132\times\frac{11.50}{2.236}\)
\(E=2.132\times5.143\)
\(E\approx10.97\)

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x}+E\).
Given \(\bar{x}=70.00\) and \(E = 10.97\).
\(\bar{x}-E=70.00 - 10.97=59.03\)
\(\bar{x}+E=70.00+10.97 = 80.97\)

Answer:

\((59.03,80.97)\)