A manager records the repair cost for 4 randomly selected TVs. A sample mean of \( \$ 88.46 \) and standard deviation of \( \$ 17.20 \) are subsequently computed, Determine the \( 99 \% \) confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal. Step 2 of 2 : Construct the \( 99 \% \) confidence interval. Round your answer to two decimal places.

To construct a \( 99\% \) confidence interval for the mean repair cost of TVs, we'll use the **t-distribution** because the sample size is small (\( n = 4 \)) and the population standard deviation is unknown. ### Given: - **Sample mean (\(\bar{x}\))**: \$88.46 - **Sample standard deviation (\(s\))**: \$17.20 - **Sample size (\(n\))**: 4 - **Confidence level**: 99% ### Steps to Calculate the Confidence Interval: 1. **Determine the degrees of freedom (df):** \[ df = n - 1 = 4 - 1 = 3 \] 2. **Find the critical t-value (\(t_{\alpha/2}\)) for 99% confidence and \(df = 3\):** For a 99% confidence interval, the significance level (\(\alpha\)) is 0.01, and \(\alpha/2 = 0.005\). Consulting a t-table or using a calculator, the critical t-value is approximately: \[ t_{0.005, 3} \approx 4.541 \] 3. **Calculate the standard error (SE):** \[ SE = \frac{s}{\sqrt{n}} = \frac{17.20}{\sqrt{4}} = \frac{17.20}{2} = 8.60 \] 4. **Compute the margin of error (ME):** \[ ME = t_{\alpha/2} \times SE = 4.541 \times 8.60 \approx 39.05 \] 5. **Determine the confidence interval:** \[ \text{Lower limit} = \bar{x} - ME = 88.46 - 39.05 = 49.41 \] \[ \text{Upper limit} = \bar{x} + ME = 88.46 + 39.05 = 127.51 \] ### **\(99\%\) Confidence Interval:** \[ (\$49.41, \$127.51) \] This means we are 99% confident that the true mean repair cost for the TVs lies between \$49.41 and \$127.51. **Final Answer:** The 99 % confidence interval for the mean repair cost is ($49.41, $127.51).