The commute distances at a huge corporation vary from employee to employee. It is known that the population of all these employee commute distances is approximately normally distributed. The corporation claims that the standard deviation of this population is $8.44 \mathrm{~km}$. You are a recruiter who wants to test this claim with a random sample of 32 employees. Based on your sample, follow the steps below to construct a $90 \%$ confidence interval for the population standard deviation of all the employee commute distances. Then state whether the confidence interval you construct contradicts the corporation's claim. (If necessary, consult a list of formulas.) (a) Click on "Take Sample" to see the results from the random sample. Take Sample \begin{tabular}{|c|c|c|c|} \hline Number of employees & Sample mean & \begin{tabular}{c} Sample standard \\ deviation \end{tabular} & \begin{tabular}{c} Sample variance \\ \hline 32 \end{tabular} \\ \hline 17.95 & 5.79 & 33.5241 \\ \hline \end{tabular} To find the confidence interval for the population standard deviation, first find the confidence interval for the population variance. Enter the values of the point estimate of the population variance, the sample size, the left critical value, and the right critical value you need for your $90 \%$ confidence interval for the population variance. (Choose the correct critical values from the table of critical values provided.) When you are done, select "Compute".
Solution
Step 1 ：The problem is asking for a 90% confidence interval for the population variance and then to use that to find the confidence interval for the population standard deviation. The sample size is 32, the sample variance is 33.5241.
Step 2 ：We can use a chi-squared distribution to find the critical values for the confidence interval for the variance. The degrees of freedom will be n-1, which is 31.
Step 3 ：The critical values for a 90% confidence interval with 31 degrees of freedom are 20.723 and 43.773 (from the chi-squared distribution table).
Step 4 ：We can then use these values to calculate the confidence interval for the variance, and then take the square root of those values to find the confidence interval for the standard deviation.
Step 5 ：The 90% confidence interval for the population variance is approximately (23.10, 53.90) and for the population standard deviation is approximately (4.81, 7.34).
Step 6 ：The corporation's claim that the standard deviation is 8.44 km is not within this interval, so the sample data contradicts the corporation's claim.
Step 7 ：Therefore, the final answer is \(\boxed{(4.81, 7.34)}\).
