A magazine includes a report on the energy costs per year for 32 -inch liquid crystal display (LCD) televisions. The article states that 14 randomly selected 32-inch LCD televisions have a sample standard deviation of $\$ 3.45$. 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 two decimal places as needed.)

Step 1 ：Given that the sample standard deviation, s = $3.45$ and the sample size, n = $14$.

Step 2 ：We are asked to find the $99\%$ confidence interval for the population variance and standard deviation. The degrees of freedom, df = n - 1 = $14 - 1 = 13$.

Step 3 ：The chi-square values for $99\%$ confidence interval with $13$ degrees of freedom are $\chi_{0.005}^{2} = 27.204$ and $\chi_{0.995}^{2} = 2.650$.

Step 4 ：The confidence interval for the population variance, $\sigma^{2}$, is given by the formula: $\left(\frac{(n-1)s^{2}}{\chi_{1-\alpha/2}^{2}}, \frac{(n-1)s^{2}}{\chi_{\alpha/2}^{2}}\right)$, where $\alpha = 1 - 0.99 = 0.01$.

Step 5 ：Substituting the given values into the formula, we get: $\left(\frac{(14-1)(3.45)^{2}}{27.204}, \frac{(14-1)(3.45)^{2}}{2.650}\right)$ = $\left(\frac{45.2025}{27.204}, \frac{45.2025}{2.650}\right)$ = $(1.66, 17.02)$.

Step 6 ：The confidence interval for the population standard deviation, $\sigma$, is given by the square root of the confidence interval for the population variance: $\left(\sqrt{1.66}, \sqrt{17.02}\right)$ = $(1.29, 4.12)$.

Step 7 ：Therefore, we are $99\%$ confident that the population variance is between $1.66$ and $17.02$, and the population standard deviation is between $1.29$ and $4.12$.