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 $\$ 3.06$. 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 $\sigma$. Interpret the results. (a) The confidence interval for the population variance is $(5.44,20.66)$. (Round to two decimal places as needed.) Interpret the results. Select the correct choice below and fill in the answer box(es) to complete your choice. (Round to two decimal places as needed.) A. With $90 \%$ confidence, you can say that the population variance is less than B. With $10 \%$ confidence, you can say that the population variance is between and C. With $90 \%$ confidence, you can say that the population variance is between and D. With $10 \%$ confidence, you can say that the population variance is greater than
Solution
Step 1 ：The problem provides us with a sample standard deviation of $3.06$ and a sample size of $14$. We are asked to construct a $90\%$ confidence interval for the population variance and standard deviation.
Step 2 ：The formula for the confidence interval for the population variance is given by: \[\left( \frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}} \right)\] where $n$ is the sample size, $s^2$ is the sample variance, and $\chi^2_{\alpha/2, n-1}$ and $\chi^2_{1-\alpha/2, n-1}$ are the critical values for the chi-square distribution with $n-1$ degrees of freedom.
Step 3 ：Substituting the given values into the formula, we get the confidence interval for the population variance as $(5.44, 20.66)$.
Step 4 ：The confidence interval for the population standard deviation is the square root of the confidence interval for the population variance. So, the confidence interval for the population standard deviation is $(2.33, 4.55)$.
Step 5 ：Interpreting the results, with $90 \%$ confidence, we can say that the population variance is between $5.44$ and $20.66$ and the population standard deviation is between $2.33$ and $4.55$.
Step 6 ：\(\boxed{C. \text{With } 90 \% \text{ confidence, you can say that the population variance is between } 5.44 \text{ and } 20.66}\)
