The total size of the sample of the high school athletes is 233. The meanof the times they spend on the practicing is 1.6 h and the standard deviation is 0.5 h.
Obtain the z-value from the z-table of \(\displaystyle{99}\%\) confidence corresponding to the given mean and standard deviation is 2.576.
The confidence interval for the mean populationis calculated as,
\(\displaystyle{C}=\ \overline{{{x}}}\ \pm\ {z}\ \times\ {\frac{{{s}}}{{\sqrt{{{n}}}}}}\)

\(\displaystyle={1.6}\ \pm\ {\left({2.576}\right)}\ {\left({\frac{{{0.5}}}{{\sqrt{{{233}}}}}}\right)}\)

\(\displaystyle={1.6}\ \pm\ {0.0084}\)

\(\displaystyle={1.516},\ {1.684}\) Thus, the confidence interval for the mean population is \(\displaystyle{1.516}\ \leq\ \mu\ \leq\ {1684}.\)

\(\displaystyle={1.6}\ \pm\ {\left({2.576}\right)}\ {\left({\frac{{{0.5}}}{{\sqrt{{{233}}}}}}\right)}\)

\(\displaystyle={1.6}\ \pm\ {0.0084}\)

\(\displaystyle={1.516},\ {1.684}\) Thus, the confidence interval for the mean population is \(\displaystyle{1.516}\ \leq\ \mu\ \leq\ {1684}.\)