Step 1
Solution:
We need to construct the 99% confidence interval for the population variance. We have been provided with the following information about the sample variance and sample size:
\(\displaystyle{s}^{{{2}}}={1369}\)

\(\displaystyle{n}={20}\) Step 2 The critical values for \(\displaystyle\alpha={0.01}\) and \(\displaystyle{d}{f}={19}\) degrees of freedom are \(\displaystyle{{X}_{{{L}}}^{{{2}}}}={{X}_{{{1}-\frac{\alpha}{{2}},{n}-{1}}}^{{{2}}}}={6.844}\) and \(\displaystyle{{X}_{{{U}}}^{{{2}}}}={{X}_{{{1}-\frac{\alpha}{{2}},{n}-{1}}}^{{{2}}}}={38.5823}\). The corresponding confidence interval is computed as shown below: \(\displaystyle{C}{I}{\left({V}{a}{r}{i}{a}{n}{c}{e}\right)}={\left({\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{{{X}_{{\frac{\alpha}{{2}},{n}-{1}}}^{{{2}}}}}}},{\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{{{X}_{{{1}-\frac{\alpha}{{2}},{n}-{1}}}^{{{2}}}}}}}\right)}\)

\(\displaystyle{\left({\frac{{{\left({20}-{1}\right)}\times{1369}}}{{{38.5823}}}},{\frac{{{\left({20}-{1}\right)}\times{1369}}}{{{6.844}}}}\right.}\)

\(\displaystyle={\left({674.17},{3800.5711}\right)}\) Now that we have the limits for the confidence interval, the limits for the 99% confidence interval for the population standard deviation are obtained by simply taking the squared root of the limits of the confidence interval for the variance, so then: CI(Standard Deviation) \(\displaystyle={\left(\sqrt{{{674.17}}},\sqrt{{{3800.5711}}}\right)}={\left({25.9648},{61.6488}\right)}\) Therefore, based on the data provided, the 99% confidence interval for the population variance is \(\displaystyle{674.17}{<}\sigma^{{{2}}}{<}{3800.5711}\)</span>, and the 99% confidence interval for the population standard deviation is \(\displaystyle{25.9648}{<}\sigma{<}{61.6488}\)</span>. Step 3 Ci for population variance (674.17 , 3800.57) Ci for population standard deviation ( 25.96 , 61.65)

\(\displaystyle{n}={20}\) Step 2 The critical values for \(\displaystyle\alpha={0.01}\) and \(\displaystyle{d}{f}={19}\) degrees of freedom are \(\displaystyle{{X}_{{{L}}}^{{{2}}}}={{X}_{{{1}-\frac{\alpha}{{2}},{n}-{1}}}^{{{2}}}}={6.844}\) and \(\displaystyle{{X}_{{{U}}}^{{{2}}}}={{X}_{{{1}-\frac{\alpha}{{2}},{n}-{1}}}^{{{2}}}}={38.5823}\). The corresponding confidence interval is computed as shown below: \(\displaystyle{C}{I}{\left({V}{a}{r}{i}{a}{n}{c}{e}\right)}={\left({\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{{{X}_{{\frac{\alpha}{{2}},{n}-{1}}}^{{{2}}}}}}},{\frac{{{\left({n}-{1}\right)}{s}^{{{2}}}}}{{{{X}_{{{1}-\frac{\alpha}{{2}},{n}-{1}}}^{{{2}}}}}}}\right)}\)

\(\displaystyle{\left({\frac{{{\left({20}-{1}\right)}\times{1369}}}{{{38.5823}}}},{\frac{{{\left({20}-{1}\right)}\times{1369}}}{{{6.844}}}}\right.}\)

\(\displaystyle={\left({674.17},{3800.5711}\right)}\) Now that we have the limits for the confidence interval, the limits for the 99% confidence interval for the population standard deviation are obtained by simply taking the squared root of the limits of the confidence interval for the variance, so then: CI(Standard Deviation) \(\displaystyle={\left(\sqrt{{{674.17}}},\sqrt{{{3800.5711}}}\right)}={\left({25.9648},{61.6488}\right)}\) Therefore, based on the data provided, the 99% confidence interval for the population variance is \(\displaystyle{674.17}{<}\sigma^{{{2}}}{<}{3800.5711}\)</span>, and the 99% confidence interval for the population standard deviation is \(\displaystyle{25.9648}{<}\sigma{<}{61.6488}\)</span>. Step 3 Ci for population variance (674.17 , 3800.57) Ci for population standard deviation ( 25.96 , 61.65)