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 $\mathrm{\$}3.45$. Assume the sample is taken from a normally distributed population. Construct $99\mathrm{\%}$ 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 $99\mathrm{\%}$ 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 $99\mathrm{\%}$ 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: $(\frac{(n-1)s^2}{{\chi }_{1-\alpha /2}^2},\frac{(n-1)s^2}{{\chi }_{\alpha /2}^2})$, where $\alpha =1-0.99=0.01$.

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

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: $(\sqrt{1.66},\sqrt{17.02})$ = $(1.29,4.12)$.

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