EWORK 5
Question 17, 6.2.20-T
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In a random sample of four mobile devices, the mean repair cost was $\$ 90.00$ 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 $90 \%$ confidence interval for the population mean. Interpret the results.
The $90 \%$ confidence interval for the population mean $\mu$ is
(Round to two decimal places as needed.)

Step 1 ：First, we need to find the t-score. For a 90% confidence interval and degrees of freedom (df) = n - 1 = 4 - 1 = 3, the t-score (which can be found in a t-distribution table) is approximately 1.638.

Step 2 ：Next, we substitute the given values into the formula for the margin of error in a t-distribution, which is: \(E = t * (s/\sqrt{n})\). Substituting the values, we get \(E = 1.638 * (14/\sqrt{4}) = 1.638 * 7\).

Step 3 ：Calculating the above expression, we get \(E = 11.47\). So, the margin of error is \(\boxed{11.47}\).

Step 4 ：To construct the 90% confidence interval for the population mean, we subtract and add the margin of error from/to the sample mean. The lower limit is \($90.00 - $11.47 = $78.53\) and the upper limit is \($90.00 + $11.47 = $101.47\).

Step 5 ：So, the 90% confidence interval for the population mean is \(\boxed{($78.53, $101.47)}\).

Step 6 ：Interpretation: We are 90% confident that the true population mean repair cost for mobile devices is between $78.53 and $101.47.