From the provided information,

Sample size \((n) = 244\)

Sample mean \((\overline{x}) = 28.3 hg\)

Sample standard deviation \((s) = 7.1 hg\)

Since, the population standard deviation is unknown, therefore, the t distribution will be used.

Confidence level \(= 98\%\)

Level of significance \((\alpha) = 1 – 0.98 = 0.02\)

The degree of freedom = 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.

The required \(98\%\) confidence interval can be obtained as:

\(CI = \overline{x} \pm t_{\alpha/2,n-1}\frac{S}{\sqrt{n}}\)

\(= 28.3 \pm (2.34) \frac{7.1}{\sqrt{244}}\)

\(= 28.3 \pm 1.1\)

\(= (27. 2, 29.4)\)

Thus, the confidence interval is \(27.2 < \mu < 29.4.\)

No, the results between the two confidence intervals are not very different.

The confidence interval limits are almost similar. Therefore, the correct option is B).

Sample size \((n) = 244\)

Sample mean \((\overline{x}) = 28.3 hg\)

Sample standard deviation \((s) = 7.1 hg\)

Since, the population standard deviation is unknown, therefore, the t distribution will be used.

Confidence level \(= 98\%\)

Level of significance \((\alpha) = 1 – 0.98 = 0.02\)

The degree of freedom = 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.

The required \(98\%\) confidence interval can be obtained as:

\(CI = \overline{x} \pm t_{\alpha/2,n-1}\frac{S}{\sqrt{n}}\)

\(= 28.3 \pm (2.34) \frac{7.1}{\sqrt{244}}\)

\(= 28.3 \pm 1.1\)

\(= (27. 2, 29.4)\)

Thus, the confidence interval is \(27.2 < \mu < 29.4.\)

No, the results between the two confidence intervals are not very different.

The confidence interval limits are almost similar. Therefore, the correct option is B).