Save \& Exit Certify Lesson: 8.3 Estimating rop Question 4 of 10 , Step 2 of 2 Correct A manager records the repair cost for 17 randomly selected stereos. A sample mean of \( \$ 82.37 \) and standard deviation of \( \$ 19.66 \) are subsequently computed. Determine the \( 80 \% \) confidence interval for the mean repair cost for the stereos. Assume the population is approximately normal. step 2 of 2 : construct the \( 80 \% \) confidence interval. Round your answer to two decimal places.

To construct the \( 80\% \) confidence interval for the mean repair cost, first, we need the standard error of the mean (SEM). The SEM is calculated using the formula: \[ SEM = \frac{s}{\sqrt{n}} \] where \( s \) is the standard deviation and \( n \) is the sample size. For your data, \( s = 19.66 \) and \( n = 17 \). Calculating SEM: \[ SEM = \frac{19.66}{\sqrt{17}} \approx 4.78 \] Next, we need to find the critical value associated with an \( 80\% \) confidence level. For a two-tailed test with \( \alpha = 0.20 \), we look for the critical value \( t \) from the t-distribution with \( n - 1 = 16 \) degrees of freedom. The critical value \( t \) can be approximated as \( 1.337 \). Now, we can calculate the margin of error (ME): \[ ME = t \times SEM \approx 1.337 \times 4.78 \approx 6.39 \] Finally, construct the confidence interval: \[ \text{Lower limit} = \bar{x} - ME = 82.37 - 6.39 \approx 75.98 \] \[ \text{Upper limit} = \bar{x} + ME = 82.37 + 6.39 \approx 88.76 \] Thus, the \( 80\% \) confidence interval for the mean repair cost for the stereos is approximately \( (75.98, 88.76) \).