Question

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 $3.54. 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 (\\( \square \\), \\( \square \\)). (round to two decimal places as needed.)

Explanation:

Step1: Determine the degrees of freedom and critical values

The degrees of freedom \(df=n - 1\), where \(n = 14\), so \(df=14 - 1=13\).
For a \(99\%\) confidence interval, \(\alpha=1 - 0.99 = 0.01\). The critical values are \(\chi_{L}^{2}=\chi_{1-\frac{\alpha}{2},df}^{2}=\chi_{0.995,13}^{2}=3.565\) and \(\chi_{R}^{2}=\chi_{\frac{\alpha}{2},df}^{2}=\chi_{0.005,13}^{2}=29.819\) (from the chi - square distribution table).
The sample standard deviation \(s = 3.54\), so the sample variance \(s^{2}=3.54^{2}=12.5316\).

Step2: Calculate the confidence interval for the population variance \(\sigma^{2}\)

The formula for the confidence interval for the population variance is \(\frac{(n - 1)s^{2}}{\chi_{R}^{2}}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{L}^{2}}\).
Substitute the values: \(\frac{(14 - 1)\times12.5316}{29.819}\leq\sigma^{2}\leq\frac{(14 - 1)\times12.5316}{3.565}\).
First, calculate \(\frac{13\times12.5316}{29.819}=\frac{162.9108}{29.819}\approx5.46\).
Second, calculate \(\frac{13\times12.5316}{3.565}=\frac{162.9108}{3.565}\approx45.70\).

Answer:

The confidence interval for the population variance is \((5.46,45.70)\)