Save \& Exit Certify Lesson: 8.3 Estumaung rupure Question 3 of 10 , Step 2 of 2 A researcher records the repair cost for 27 randomly selected refrigerators. A sample mean of \( \$ 97.89 \) and standard deviation of \( \$ 17.53 \) are subsequently computed. Determine the \( 90 \% \) confidence interval for the mean repair cost for the refrigerators. Assume the population is approximately normal. Step 2 of 2 : Construct the \( 90 \% \) confidence interval. Round your answer to two decimal places.

To construct a 90% confidence interval for the mean repair cost of the refrigerators, we can use the formula for the confidence interval: \[ \text{Confidence Interval} = \bar{x} \pm z \left(\frac{s}{\sqrt{n}}\right) \] Where: - \(\bar{x} = 97.89\) (sample mean) - \(s = 17.53\) (sample standard deviation) - \(n = 27\) (sample size) - \(z\) is the z-score corresponding to a 90% confidence level, which is approximately 1.645. Now, calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{17.53}{\sqrt{27}} \approx 3.39 \] Next, calculate the margin of error (ME): \[ ME = z \times SE = 1.645 \times 3.39 \approx 5.57 \] Finally, construct the confidence interval: \[ \text{Lower limit} = \bar{x} - ME = 97.89 - 5.57 \approx 92.32 \] \[ \text{Upper limit} = \bar{x} + ME = 97.89 + 5.57 \approx 103.46 \] Thus, the 90% confidence interval for the mean repair cost for the refrigerators is: \[ \text{Confidence Interval} \approx (92.32, 103.46) \] So, rounding to two decimal places, the final answer is \( (92.32, 103.46) \).