a)

Computation of \(80\%\) confidence interval is obtained below:

It is assumed that the population is approximately normal.

Here, \(\displaystyle\alpha\ {i}{s}\ {0.20}{\left(={1}-{0.80}\right)}\).

Degrees of freedom is \(\displaystyle{24}={\left({n}-{1}\right)}.\)

Using EXCEL formula \(\displaystyle“={T}.{I}{N}{V}{.2}{T}{\left({0.2},{24}\right)}\text{"}\), the critical value is obtained as 1.3178. That is, \(\displaystyle{t}_{{\frac{\alpha}{{2}}}}={1.3178}.\)

The \(80\%\) confidence interval is:

\(\displaystyle{C}{I}={16.3}\pm{\left({1.3178}\right)}\frac{6}{\sqrt{{25}}}\)

\(\displaystyle={\left(\begin{matrix}{16.3}-{1.5814}\\{16.3}+{1.5814}\end{matrix}\right)}\)

\(\displaystyle={\left({14.7186},{17.8814}\right)}\)

Step 2

Computation of \(90\%\) confidence interval is obtained below:

It is assumed that the population is approximately normal.

Here, \(\displaystyle\alpha\ {i}{s}\ {0.10}{\left(={1}-{0.90}\right)}.\)

Degrees of freedom is \(\displaystyle{24}{\left(={n}-{1}\right)}.\)

Using EXCEL formula \(\displaystyle“={T}.{I}{N}{V}{.2}{T}{\left({0.10},{24}\right)}\text{"}\), the critical value is obtained as 1.7108. That is, \(\displaystyle{t}_{{\frac{\alpha}{{2}}}}={1.7108}.\)

The \(90\%\) confidence interval is:

\(\displaystyle{C}{I}={16.3}\pm{\left({1.7108}\right)}\frac{6}{\sqrt{{25}}}\)

\(\displaystyle={\left(\begin{matrix}{16.3}-{1.7108}\\{16.3}+{1.7108}\end{matrix}\right)}\)

\(\displaystyle={\left({14.589},{18.011}\right)}\)

Step 3

Computation of \(99\%\) confidence interval is obtained below:

It is assumed that the population is approximately normal.

Here, \(\displaystyle\alpha\ {i}{s}\ {0.01}{\left(={1}-{0.99}\right)}.\)

Degrees of freedom is \(\displaystyle{24}{\left(={n}-{1}\right)}\)

Using EXCEL formula \(\displaystyle“={T}.{I}{N}{V}{.2}{T}{\left({0.01},{24}\right)}\text{"}\), the critical value is obtained as 2.7969. That is, \(\displaystyle{t}_{{\frac{\alpha}{{2}}}}={2.7969}.\)

The \(99\%\) confidence interval is:

\(\displaystyle{C}{I}={16.3}\pm{\left({1.7108}\right)}\frac{6}{\sqrt{{25}}}\)

\(\displaystyle={\left(\begin{matrix}{16.3}-{2.053}\\{16.3}+{2.053}\end{matrix}\right)}\)

\(\displaystyle={\left({14.247},{18.353}\right)}\)

Step 4

The confidence level is increased, and then the length of the confidence interval gets wider.

Therefore, the confidence interval gets wider.