Question

Grip and Leg Strength. Refer to Problem 6. Determine a $90\%$ confidence interval for the difference between the mean right-leg strengths of males and females. Interpret your result.

Short answer

$-31.33\text{to}599.33$

Step-by-step solution

Given Information

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

Explanation

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:

$H_0:{\mu }_1={\mu }_2$

$H_{\alpha }:{\mu }_1>{\mu }_2$

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

$$\begin{gathered} s\sqrt{\frac{{\left(n_1-1\right)}_{s_1}{}^2+{{\left(n_2-1\right)}_{s_2}}^2}{2}} \\ =\sqrt{\frac{(13-1){513}^2+(14-1){446}^2}{13+14-2}} \\ \approx 479.33 \end{gathered}$$

Now, determine $t_{\frac{\alpha }{2}}$with

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

$t_{0.05}=1.708$

The confidence interval's endpoints can be calculated as:

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

$\approx 31.333$

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

$\approx 599.333$

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.