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, we can use the formula for the confidence interval: \[ CI = \bar{x} \pm z \left( \frac{s}{\sqrt{n}} \right) \] Where: - \( \bar{x} \) is the sample mean - \( z \) is the z-score corresponding to the desired confidence level - \( s \) is the sample standard deviation - \( n \) is the sample size ### Step 1: Identify the known values - Sample mean (\( \bar{x} \)): \( 97.89 \) - Sample standard deviation (\( s \)): \( 17.53 \) - Sample size (\( n \)): \( 27 \) ### Step 2: Find the z-score for a \( 90\% \) confidence level For a \( 90\% \) confidence level, the z-score can be found using a standard normal distribution table or calculator. The critical value \( z \) for \( 90\% \) confidence is approximately \( 1.645 \). ### Step 3: Calculate the standard error (SE) The standard error (SE) is calculated as follows: \[ SE = \frac{s}{\sqrt{n}} = \frac{17.53}{\sqrt{27}} \] ### Step 4: Calculate the confidence interval Now we can calculate the confidence interval using the values we have: \[ CI = 97.89 \pm 1.645 \left( \frac{17.53}{\sqrt{27}} \right) \] Let's perform the calculations step by step. 1. Calculate the standard error (SE). 2. Calculate the margin of error (ME). 3. Construct the confidence interval. I'll perform these calculations now. Calculate the value by following steps: - step0: Calculate: \(\frac{17.53}{\sqrt{27}}\) - step1: Simplify the root: \(\frac{17.53}{3\sqrt{3}}\) - step2: Convert the expressions: \(\frac{\frac{1753}{100}}{3\sqrt{3}}\) - step3: Multiply by the reciprocal: \(\frac{1753}{100}\times \frac{1}{3\sqrt{3}}\) - step4: Multiply the fractions: \(\frac{1753}{100\times 3\sqrt{3}}\) - step5: Multiply: \(\frac{1753}{300\sqrt{3}}\) - step6: Multiply by the Conjugate: \(\frac{1753\sqrt{3}}{300\sqrt{3}\times \sqrt{3}}\) - step7: Multiply the numbers: \(\frac{1753\sqrt{3}}{900}\) Calculate or simplify the expression \( 1.645*(17.53/\sqrt(27)) \). Calculate the value by following steps: - step0: Calculate: \(1.645\left(\frac{17.53}{\sqrt{27}}\right)\) - step1: Simplify the root: \(1.645\left(\frac{17.53}{3\sqrt{3}}\right)\) - step2: Divide the terms: \(1.645\times \frac{1753}{300\sqrt{3}}\) - step3: Convert the expressions: \(\frac{329}{200}\times \frac{1753}{300\sqrt{3}}\) - step4: Multiply the fractions: \(\frac{329\times 1753}{200\times 300\sqrt{3}}\) - step5: Multiply the terms: \(\frac{576737}{200\times 300\sqrt{3}}\) - step6: Multiply the terms: \(\frac{576737}{60000\sqrt{3}}\) - step7: Multiply by the Conjugate: \(\frac{576737\sqrt{3}}{60000\sqrt{3}\times \sqrt{3}}\) - step8: Multiply the numbers: \(\frac{576737\sqrt{3}}{180000}\) ### Step 3: Calculate the Standard Error (SE) The standard error (SE) is calculated as follows: \[ SE \approx 3.37 \] ### Step 4: Calculate the Margin of Error (ME) Now, we calculate the margin of error (ME): \[ ME \approx 5.55 \] ### Step 5: Construct the Confidence Interval Now we can construct the confidence interval: \[ CI = 97.89 \pm 5.55 \] Calculating the lower and upper bounds: - Lower bound: \( 97.89 - 5.55 \) - Upper bound: \( 97.89 + 5.55 \) Let's perform these final calculations. Calculate the value by following steps: - step0: Calculate: \(97.89+5.55\) - step1: Add the numbers: \(103.44\) Calculate or simplify the expression \( 97.89 - 5.55 \). Calculate the value by following steps: - step0: Calculate: \(97.89-5.55\) - step1: Subtract the numbers: \(92.34\) ### Final Step: Calculate the Confidence Interval Bounds - Lower bound: \( 92.34 \) - Upper bound: \( 103.44 \) ### Conclusion The \( 90\% \) confidence interval for the mean repair cost for the refrigerators is: \[ (92.34, 103.44) \] Thus, rounding to two decimal places, the confidence interval is: \[ \boxed{(92.34, 103.44)} \]