Step 1

Here \(\displaystyle\sigma{2}={12.6}\) and \(\displaystyle{c}={90}\%\) which is 0.9.

Therefore \(\displaystyle{\frac{{{1}-{c}}}{{{2}}}}={\frac{{{1}-{0.9}}}{{{2}}}}={0.05}\) and \(\displaystyle{1}-{\left({\frac{{{1}-{c}}}{{{2}}}}\right)}={0.95}\)

Part A

For \(\displaystyle{n}={20}\) the degrees of freedom are:

\(\displaystyle{d}{f}={20}-{1}={19}\)

So the \(\displaystyle\chi\) squared table values are:

\(\displaystyle{X}_{{{0.95}}}={10.117}\) and \(\displaystyle{X}_{{{0.05}}}={30.144}\)

So the upper and lower bounds are given by:

\(\displaystyle{U}{B}={\left({\frac{{{19}}}{{{10.117}}}}\right)}{\left({12.6}\right)}\)

\(\displaystyle{U}{B}={23.66}\)

\(\displaystyle{L}{B}={\left({\frac{{{19}}}{{{30.144}}}}\right)}{\left({12.6}\right)}\)

\(\displaystyle{L}{B}={7.94}\)

Part B

For \(\displaystyle{n}={30}\) the degrees of freedom are:

\(\displaystyle{d}{f}={30}-{1}={29}\)

So the \(\displaystyle\chi\) squared table values are:

\(\displaystyle{X}_{{{0.95}}}={17.708}\) and \(\displaystyle{X}_{{{0.05}}}={42.557}\)

So the upper and lower bounds are given by:

\(\displaystyle{U}{B}={\left({\frac{{{29}}}{{{17.708}}}}\right)}{\left({12.6}\right)}\)

\(\displaystyle{U}{B}={20.63}\)

\(\displaystyle{L}{B}={\left({\frac{{{29}}}{{{42.557}}}}\right)}{\left({12.6}\right)}\)

\(\displaystyle{L}{B}={8.59}\)