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. What is the confidence interval for the population variance $\sigma^{2}$ ? (Round to one decimal place as needed.)

Step 1 ：We are given a random sample of 22 people waiting at a bank, with a sample standard deviation of 4.7 minutes. We are asked to construct a confidence interval for the population variance \(\sigma^{2}\) and the population standard deviation \(\sigma\) at a 95% level of confidence. We assume the sample is from a normally distributed population.

Step 2 ：The formula for the confidence interval for the population variance \(\sigma^{2}\) 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, \(s^{2}\) is the sample variance, \(\alpha\) is the significance level, and \(\chi^{2}_{\alpha/2, n-1}\) and \(\chi^{2}_{1-\alpha/2, n-1}\) are the critical values of the Chi-Square distribution with \(n-1\) degrees of freedom.

Step 3 ：Substituting the given values into the formula, we have \(n=22\), \(s=4.7\), and \(\alpha=0.05\).

Step 4 ：The critical values of the Chi-Square distribution with 21 degrees of freedom are approximately 10.282897782522863 and 35.478875905727264 for \(\chi^{2}_{\alpha/2, 21}\) and \(\chi^{2}_{1-\alpha/2, 21}\) respectively.

Step 5 ：Substituting these values into the formula, we find the confidence interval for the population variance \(\sigma^{2}\) to be approximately (13.07510421786265, 45.11276974749688).

Step 6 ：Rounding to one decimal place, the confidence interval for the population variance \(\sigma^{2}\) is \(\boxed{(13.1, 45.1)}\).