This question: 1 point(s) possib Use technology to construct the confidence intervals for the population variance $\sigma^{2}$ and the population standard deviation $\sigma$. Assume the sample is taken from a normally distributed population. \[ c=0.95, s^{2}=20.25, n=29 \] The confidence interval for the population variance is $\square . \square$. (Round to two decimal places as needed.) The confidence interval for the population standard deviation is ( $\square . \square$ ). (Round to two decimal places as needed.)

Step 1 ：Given values are: confidence level \(c = 0.95\), sample variance \(s^{2} = 20.25\), and sample size \(n = 29\).

Step 2 ：Calculate the value of alpha, which is \(1 - c = 0.05\).

Step 3 ：Calculate the degrees of freedom, which is \(n - 1 = 28\).

Step 4 ：Calculate the chi-square values \(\chi^{2}_{\alpha/2, n-1}\) and \(\chi^{2}_{1-\alpha/2, n-1}\) using the chi-square distribution. The calculated values are \(\chi^{2}_{\alpha/2, n-1} = 44.46\) and \(\chi^{2}_{1-\alpha/2, n-1} = 15.31\).

Step 5 ：Substitute these values into the formulas to calculate the confidence intervals for the variance. The calculated values are \(variance_{lower} = 12.75\) and \(variance_{upper} = 37.04\).

Step 6 ：Substitute these values into the formulas to calculate the confidence intervals for the standard deviation. The calculated values are \(std\_dev_{lower} = 3.57\) and \(std\_dev_{upper} = 6.09\).

Step 7 ：Final Answer: The confidence interval for the population variance is \(\boxed{(12.75, 37.04)}\). The confidence interval for the population standard deviation is \(\boxed{(3.57, 6.09)}\).