The waiting times (in minutes) of a random sample of 22 people at a bank have a sample standard deviation of 4.7 minutes. Construct a confidence interval for the population variance $\sigma^{2}$ and the population standard deviation $\sigma$. Use a $95 \%$ level of confidence. Assume the sample is from a normally distributed population. $(13.1,45.1)$ (Round to one decimal place as needed) Interpret the results. Select the correct choice below and fill in the answer box(es) to complete your choice. (Round to one decimal place as needed.) A. With $95 \%$ confidence, you can say that the population variance is less than C. With $5 \%$ confidence, you can say that the population variance is greater than B. With $95 \%$ confidence, you can say that the population variance is between 13.1 and 45.1 D. With $5 \%$ confidence, you can say that the population variance is between and What is the confidence interval for the population standard deviation $\sigma$ ? (Round to one decimal place as needed.)

Step 1 ：Given that the sample size (n) is 22, the sample standard deviation (s) is 4.7, and the level of confidence is 95%, we can use the chi-square distribution to calculate the confidence interval for the population variance.

Step 2 ：The formula for the confidence interval for the population variance is \(\left(\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2, n-1}}, \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2, n-1}}\right)\), where \(\chi^{2}_{\alpha/2, n-1}\) and \(\chi^{2}_{1-\alpha/2, n-1}\) are the chi-square values for the given degrees of freedom and significance level.

Step 3 ：By substituting the given values into the formula, we get the chi-square values as \(\chi^{2}_{\alpha/2, n-1} = 10.282897782522863\) and \(\chi^{2}_{1-\alpha/2, n-1} = 35.478875905727264\).

Step 4 ：Substituting these chi-square values into the formula, we get the lower and upper limits of the variance as 13.07510421786265 and 45.11276974749688 respectively.

Step 5 ：Taking the square root of these limits, we get the lower and upper limits of the standard deviation as 3.615951357231268 and 6.716604033847528 respectively.

Step 6 ：Rounding these values to one decimal place, we get the $95\%$ confidence interval for the population variance as \(\boxed{(13.1, 45.1)}\) and for the population standard deviation as \(\boxed{(3.6, 6.7)}\).

Step 7 ：Therefore, with $95\%$ confidence, we can say that the population variance is between 13.1 and 45.1 and the population standard deviation is between 3.6 and 6.7.