The annual earnings of 13 randomly selected computer software engineers have a sample standard deviation of $\$ 3634$. Assume the sample is from a normally distributed population. Construct a confidence interval for the population variance $\sigma^{2}$ and the population standard deviation $\sigma$. Use a $95 \%$ level of confidence. Interpret the results. What is the confidence interval for the population variance $\sigma^{2}$ ? ( 6790665,35985259 ) (Round to the nearest integer as needed.) Interpret the results. Select the correct choice below and fill in the answer box(es) to complete your choice. (Round to the nearest integer as needed.) A. With $5 \%$ confidence, you can say that the population variance is between and . C. With $95 \%$ confidence, you can say that the population variance is between 6790665 and 35985259 . B. With $95 \%$ confidence, you can say that the population variance is greater than D. With $5 \%$ confidence, you can say that the population variance is less than a. What is the confidence interval for the population standard deviation $\sigma$ ? ( $\square, \square$ ) (Round to the nearest integer as needed.)

Step 1 ：Given that the sample standard deviation is $3634 and the sample size is 13, we can use the chi-square distribution to construct the confidence interval for the variance.

Step 2 ：The degrees of freedom will be the sample size minus 1, which is 12.

Step 3 ：The formula for the confidence interval for the 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 n is the sample size, s is the sample standard deviation, and \(\chi^2_{\alpha/2, n-1}\) and \(\chi^2_{1-\alpha/2, n-1}\) are the chi-square values for the given confidence level and degrees of freedom.

Step 4 ：Substituting the given values into the formula, we get the confidence interval for the variance as \(\left(\frac{(13-1)3634^2}{\chi^2_{0.025, 12}}, \frac{(13-1)3634^2}{\chi^2_{0.975, 12}}\right)\), which simplifies to \(\boxed{(6790665, 35985259)}\).

Step 5 ：For the standard deviation, we simply take the square root of the variance interval. This gives us the confidence interval for the standard deviation as \(\boxed{(2606, 5999)}\).

Step 6 ：This means that we are 95% confident that the true population variance lies between 6790665 and 35985259, and the true population standard deviation lies between 2606 and 5999.