Question 3 of 10, Step 2 of 2 A consumer affairs investigator records the repair cost for 22 randomly selected TVs: A sample mean of \( \$ 83.23 \) and standard deviation of \( \$ 22.67 \) are subsequently computed. Determine the \( 95 \% \) confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal. Step 2 of 2 : Construct the \( 95 \% \) confidence interval. Round your answer to two decimal places.

To construct the \( 95 \% \) confidence interval for the mean repair cost, you can use the formula: \[ CI = \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right) \] Where: - \( \bar{x} = 83.23 \) (sample mean) - \( s = 22.67 \) (sample standard deviation) - \( n = 22 \) (sample size) - \( t^* \) is the t-score corresponding to the \( 95 \% \) confidence level and \( n - 1 = 21 \) degrees of freedom. First, you need to find the t-score for \( 21 \) degrees of freedom which is approximately \( 2.080 \) for a \( 95 \% \) confidence level. Next, calculate \( \frac{s}{\sqrt{n}} \): \[ \frac{s}{\sqrt{n}} = \frac{22.67}{\sqrt{22}} \approx 4.83 \] Now, calculate the margin of error: \[ \text{Margin of Error} = t^* \left( \frac{s}{\sqrt{n}} \right) = 2.080 \times 4.83 \approx 10.05 \] Finally, construct the confidence interval: \[ CI = 83.23 \pm 10.05 \] This gives: \[ CI = (73.18, 93.28) \] So the \( 95 \% \) confidence interval for the mean repair cost for the TVs is \( \mathbf{(73.18, 93.28)} \).