Step 1
From the provided information,
Sample size \(\displaystyle{\left({n}\right)}={244}\)
Sample mean \(\displaystyle{\left(\overline{{{x}}}\right)}={28.3}\) hg
Sample standard deviation \(\displaystyle{\left({s}\right)}={7.1}\) hg
Step 2
Since, the population standard deviation is unknown, therefore, the t distribution will be used.
Confidence level = 98%
Level of significance \(\displaystyle{\left(\alpha\right)}={1}-{0.98}={0.02}\)
The degree of freedom \(\displaystyle={n}–{1}={244}–{1}={243}\)
The critical value of t at 243 degree of freedom with 0.01 level of significance from the t value table is 2.34.
Step 3
The required 98% confidence interval can be obtained as:
\(\displaystyle{C}{I}=\overline{{{x}}}\pm{t}_{{\frac{\alpha}{{2}},{n}-{1}}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

\(\displaystyle={28.3}\pm{\left({2.34}\right)}{\frac{{{7.1}}}{{\sqrt{{{224}}}}}}\)

\(\displaystyle={28.3}\pm{1.1}\)

\(\displaystyle={\left({27.2},{29.4}\right)}\) Thus, the confidence interval is \(\displaystyle{27.2}{<}\mu{<}{29.4}\)</span>. No, the results between the two confidence intervals are not very different. The confidence interval limits are almost similar.

\(\displaystyle={28.3}\pm{\left({2.34}\right)}{\frac{{{7.1}}}{{\sqrt{{{224}}}}}}\)

\(\displaystyle={28.3}\pm{1.1}\)

\(\displaystyle={\left({27.2},{29.4}\right)}\) Thus, the confidence interval is \(\displaystyle{27.2}{<}\mu{<}{29.4}\)</span>. No, the results between the two confidence intervals are not very different. The confidence interval limits are almost similar.