91Ó°ÊÓ

To calculate a$90\%$confidence interval for the difference between the mean right-leg strengths of males and females.

As the given data is normally distributed and the given population standard deviations are more or less equal, the pooled t-test can be used as follows:

$H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2$

$H_{\mathrm{Î\pm }}:{\mathrm{μ}}_1>{\mathrm{μ}}_2$

Here the pooled standard deviation needs to be calculated as follows:

Now, determine $t_{\frac{\mathrm{Î\pm }}{2}}$with

$$\begin{gathered} df=n_1+n_2-2 \\ =13+14-2 \end{gathered}$$

The confidence interval's endpoints can be calculated as:

$$\begin{gathered} \left({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2\right)-t_{\frac{a}{2}}×s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} \\ =(2127-1843)-1.708×479.33\sqrt{\frac{1}{13}+\frac{1}{14}} \end{gathered}$$

$≈31.333$

$$\begin{gathered} \left({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2\right)+t_{\frac{a}{2}}×s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} \\ =(2127-1843)+1.708×479.33\sqrt{\frac{1}{13}+\frac{1}{14}} \end{gathered}$$

$≈599.333$

It can be said with $90\%$confidence that the mean right leg strength of males is in the range of $31.333$less and $599.33$more than the mean right leg strength of females.