The margin of error formula is given by
[tex]T_c\times\frac{s}{\sqrt[]{n}}[/tex]where T_c is the critical T-value for n=21 degrees of freedom and s is the standard deviation. Then, for n=21 and 98% confidence level, we have that
[tex]T_c=2.5176[/tex]Therefore, by substituting these values into the margin of error formula, we have
[tex]T_c\times\frac{s}{\sqrt[]{n}}=2.5176\times\frac{6}{\sqrt[]{21}}[/tex]which gives
[tex]T_c\times\frac{s}{\sqrt[]{n}}=3.2963[/tex]Then, the margin of error is 3.2963 and the confidence interval for the given mean is:
[tex]43.7037\le\bar{x}\le50.2963[/tex]