Chapter 25, Problem 17E

To determine

To create a $90\%$ confidence interval for how many more push-ups boys can do than girls, on average, at that high school.

Answer to Problem 17E

We are $90\%$ confident that the mean difference in push-ups is between $(1.6148,13.0519)$ .

Explanation of Solution

The table of the push-ups of the boys and girls are given in the question. Thus, the mean of the boys and girls are calculated as:

  $\begin{array}{l}{\overline{x}}_1=\frac{17+27+31+17+25+\mathrm{....}+11+34}{12}\\ =23.8333\\ {\overline{x}}_2=\frac{24+7+14+16+\mathrm{....}+31+8}{12}\\ =16.5\end{array}$

And the standard deviation of the two will be as:

  $\begin{array}{l}{s}_1=\sqrt{\frac{{(17-23.8333)}^2+{(27-23.8333)}^2+{(31-23.8333)}^2+{(17-23.8333)}^2+\mathrm{....}+{(34-23.8333)}^2}{12-1}}\\ =7.2090\\ s_2=\sqrt{\frac{{(24-16.5)}^2+{(7-16.5)}^2+{(14-16.5)}^2+\mathrm{....}+{(8-16.5)}^2}{12-1}}\\ =8.9392\end{array}$

Thus, the degree of freedom will be calculated as:

  $\begin{array}{l}df=\frac{{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2})}^2}{\frac{\raisebox{1ex}{$s_1^2$}\!\left/ \!\raisebox{-1ex}{$n_1$}\right.}{n_1-1}+\frac{\raisebox{1ex}{$s_2^2$}\!\left/ \!\raisebox{-1ex}{$n_2$}\right.}{n_2-1}}\\ =21\end{array}$

Now let us find out the critical value using the table T in appendix F with

  $\begin{array}{l}df=n-1\\ =21-1\\ =20\end{array}$

Then it is as:

  $t=1.725$

Thus, the confidence interval will be as:

  $\begin{array}{l}({\overline{x}}_1-{\overline{x}}_2)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(23.8333-16.5)-1.725\times \sqrt{\frac{{7.2090}^2}{12}+\frac{{8.9392}^2}{12}}\\ =1.6148\\ ({\overline{x}}_1-{\overline{x}}_2)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(23.8333-16.5)+1.725\times \sqrt{\frac{{7.2090}^2}{12}+\frac{{8.9392}^2}{12}}\\ =13.0519\end{array}$

Thus, we conclude that we are $90\%$ confident that the mean difference in push-ups is between $(1.6148,13.0519)$ .