The number of hours of reserve capacity of 10 randomly selected automotive batteries is shown to the right. $\begin{array}{llllll}1.79 & 1.85 & 1.54 & 1.68 & 1.71 \\ 1.98 & 1.31 & 1.53 & 1.44 & 2.06\end{array}$ Assume the sample is taken from a normally distributed population. Construct $90 \%$ confidence intervals for (a) the population variance $\sigma^{2}$ and (b) the population standard deviation $\sigma$. (Round to three decimal places a]s needed.) Interpret the results. Select the correct choice below and fill in the answer box(es) to complete your choice. (Round to three decimal places as needed.) A. With $10 \%$ confidence, you can say that the population standard deviation is between and hours of reserve capacity. C. With $90 \%$ confidence, you can say that the population standard deviation is less than hours of reserve capacity. B. With $90 \%$ confidence, you can say that the population standard deviation is between and hours of reserve capacity. D. With $10 \%$ confidence, you can say that the population standard deviation is greater than hours of reserve capacity.

Step 1 ：Given the data of the number of hours of reserve capacity of 10 randomly selected automotive batteries: \(1.79, 1.85, 1.54, 1.68, 1.71, 1.98, 1.31, 1.53, 1.44, 2.06\).

Step 2 ：We are asked to construct a 90% confidence interval for the population variance \(\sigma^{2}\) and the population standard deviation \(\sigma\).

Step 3 ：First, we calculate the sample variance \(s^{2}\) and the sample standard deviation \(s\). The sample variance is calculated as \(s^{2} = 0.056898888888888895\) and the sample standard deviation is \(s = 0.23853487981611599\).

Step 4 ：Next, we use the chi-square distribution to construct the confidence interval for the population variance and standard deviation. The degrees of freedom for the chi-square distribution is \(n-1\), where \(n\) is the sample size. In this case, \(n = 10\).

Step 5 ：The formula for the confidence interval for the variance is given by: \[\left(\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2}}, \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2}}\right)\] where \(s^{2}\) is the sample variance, \(n\) is the sample size, and \(\chi^{2}_{\alpha/2}\) and \(\chi^{2}_{1-\alpha/2}\) are the chi-square values for the upper and lower tails of the chi-square distribution.

Step 6 ：Using the chi-square distribution table, we find that \(\chi^{2}_{\alpha/2} = 3.325112843066815\) and \(\chi^{2}_{1-\alpha/2} = 16.918977604620448\).

Step 7 ：Substituting these values into the formula, we find the confidence interval for the variance to be \([0.030267195333372392, 0.15400680342856868]\).

Step 8 ：The confidence interval for the standard deviation is the square root of the confidence interval for the variance. Therefore, the confidence interval for the standard deviation is \([0.1739746973941107, 0.39243700568189116]\).

Step 9 ：Rounding to three decimal places, we find that with 90% confidence, the population standard deviation is between 0.174 and 0.392 hours of reserve capacity. Therefore, the final answer is \(\boxed{[0.174, 0.392]}\).