Step 1

From the given information, \(\displaystyle{S}={1.22}\) milligrams and \(\displaystyle{n}={27}\).

The level of confidence is 0.90.

\(\displaystyle{1}-\alpha={0.90}\)

\(\displaystyle\alpha={1}-{0.90}\)

\(\displaystyle\alpha={0.10}\)

Step 2

Then, a \(90\%\) interval for the population variance is

\(\displaystyle{C}{I}={\left(\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{{1}-\frac{a}{{n}},{n}-{1}}}^{{2}}}}<\sigma^{2}<\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{\frac{a}{{n}},{n}-{1}}}^{{2}}}}\right)}\)</span>

\(\displaystyle={\left(\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{{X}_{{{1}-\frac{0.10}{{2}},{27}-{1}}}^{{2}}}}<\sigma^{2}<\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{{X}_{{\frac{0.10}{{2}},{27}-{1}}}^{{2}}}}\right)}\)</span>

\(\displaystyle={\left(\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.95},{26}}}^{{2}}}}<\sigma^{2}<\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.5},{26}}}^{{2}}}}\right)}\)</span>

\(\displaystyle={\left(\frac{{{26}\times{1.22}^{2}}}{{38.855}}<\sigma^{2}<\frac{{{26}\times{1.22}^{2}}}{{15.379}}\right)}\)</span>

\(\displaystyle={\left({0.9952}<\sigma^{2}<{2.5163}\right)}\)</span>

Interpretation:

There is \(90\%\) confidence that the population variance lies between 0.9952 and 2.5163.

Step 3

The \(90\%\) confidence interval for the population standard deviation is

\(\displaystyle{C}{I}={\left(\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{{1}-\frac{a}{{n}},{n}-{1}}}^{{2}}}}<\sigma<\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{\frac{a}{{n}},{n}-{1}}}^{{2}}}}\right)}\)</span>

\(\displaystyle={\left(\sqrt{{\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{X}_{{{1}-\frac{0.10}{{2}},{27}-{1}}}}}}<\sigma<\sqrt{{\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{{X}_{{\frac{0.10}{{2}},{27}-{1}}}^{{2}}}}}}\right)}\)</span>

\(\displaystyle={\left(\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.95},{26}}}^{{2}}}}}}<\sigma<\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.05},{26}}}^{{2}}}}}}\right)}\)</span>

\(\displaystyle{\left(\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{38.885}}}}<\sigma<\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{15.379}}}}\right)}\)</span> [(Using excel function), (=CHISQ.INV(0.95,26)), (Using excel function), (=CHISQ.INV(0.05,26))]

\(\displaystyle={\left({0.9976}<\sigma<{1.5863}\right)}\)</span>

Interpretation:

There is \(90\%\) confidence that the population standard deviation lies between 0.9976 and 1.5863.

From the given information, \(\displaystyle{S}={1.22}\) milligrams and \(\displaystyle{n}={27}\).

The level of confidence is 0.90.

\(\displaystyle{1}-\alpha={0.90}\)

\(\displaystyle\alpha={1}-{0.90}\)

\(\displaystyle\alpha={0.10}\)

Step 2

Then, a \(90\%\) interval for the population variance is

\(\displaystyle{C}{I}={\left(\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{{1}-\frac{a}{{n}},{n}-{1}}}^{{2}}}}<\sigma^{2}<\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{\frac{a}{{n}},{n}-{1}}}^{{2}}}}\right)}\)</span>

\(\displaystyle={\left(\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{{X}_{{{1}-\frac{0.10}{{2}},{27}-{1}}}^{{2}}}}<\sigma^{2}<\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{{X}_{{\frac{0.10}{{2}},{27}-{1}}}^{{2}}}}\right)}\)</span>

\(\displaystyle={\left(\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.95},{26}}}^{{2}}}}<\sigma^{2}<\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.5},{26}}}^{{2}}}}\right)}\)</span>

\(\displaystyle={\left(\frac{{{26}\times{1.22}^{2}}}{{38.855}}<\sigma^{2}<\frac{{{26}\times{1.22}^{2}}}{{15.379}}\right)}\)</span>

\(\displaystyle={\left({0.9952}<\sigma^{2}<{2.5163}\right)}\)</span>

Interpretation:

There is \(90\%\) confidence that the population variance lies between 0.9952 and 2.5163.

Step 3

The \(90\%\) confidence interval for the population standard deviation is

\(\displaystyle{C}{I}={\left(\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{{1}-\frac{a}{{n}},{n}-{1}}}^{{2}}}}<\sigma<\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{\frac{a}{{n}},{n}-{1}}}^{{2}}}}\right)}\)</span>

\(\displaystyle={\left(\sqrt{{\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{X}_{{{1}-\frac{0.10}{{2}},{27}-{1}}}}}}<\sigma<\sqrt{{\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{{X}_{{\frac{0.10}{{2}},{27}-{1}}}^{{2}}}}}}\right)}\)</span>

\(\displaystyle={\left(\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.95},{26}}}^{{2}}}}}}<\sigma<\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.05},{26}}}^{{2}}}}}}\right)}\)</span>

\(\displaystyle{\left(\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{38.885}}}}<\sigma<\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{15.379}}}}\right)}\)</span> [(Using excel function), (=CHISQ.INV(0.95,26)), (Using excel function), (=CHISQ.INV(0.05,26))]

\(\displaystyle={\left({0.9976}<\sigma<{1.5863}\right)}\)</span>

Interpretation:

There is \(90\%\) confidence that the population standard deviation lies between 0.9976 and 1.5863.