As part of a survey, a marketing representative asks a random sample of 30 business owners how much they would be willing to pay for a website for their company. She finds that the sample standard deviation is $\$ 3429$. Assume the sample is taken from a normally distributed population. Construct $99 \%$ confidence intervals for (a) the population variance $\sigma^{2}$ and (b) the population standard deviation $\sigma$. Interpret the results.
(a) The confidence interval for the population variance is (Round to the nearest integer as needed.)

Step 1 ：The problem involves constructing a 99% confidence interval for the population variance and standard deviation. The sample size is 30 and the sample standard deviation is $3429.

Step 2 ：We use the chi-square distribution to construct the confidence interval for the population variance. The formula for the confidence interval is given by: \[\left( \frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}} \right)\] 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 degrees of freedom (n-1) and significance level \(\alpha\) (in this case, \(\alpha = 0.01\) for a 99% confidence interval).

Step 3 ：Substituting the given values into the formula, we get the chi-square values as 13.12114888796041 and 52.335617785933614.

Step 4 ：Using these chi-square values, we calculate the lower and upper bounds of the confidence interval for the population variance as 6515317.931178544 and 25987296.68503925 respectively.

Step 5 ：Rounding these values to the nearest integer, we get the final answer. The 99% confidence interval for the population variance is approximately \(\boxed{(\$6,515,318, \$25,987,297)}\). This means that we are 99% confident that the true population variance lies within this interval.