Answer
$-t_{\frac{α}{2}}\lt t_0\lt t_{\frac{α}{2}}$: null hypothesis is not rejected.
There is not enough evidence to conclude that individuals walk at different speeds depending on whether they are departing or arriving.
Work Step by Step
$x ̅_1,n_1~and~s_1$ refer to departure and $x ̅_2,n_2~and~s_2$ refer to arrival.
$H_0:~µ_1=µ_2$ versus $H_1:~µ_1\ne µ_2$
$t_0=\frac{(x ̅_1-x ̅_2)-(µ_1-µ_2)}{\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}=\frac{(260-269)-0}{\sqrt {\frac{53^2}{35}+\frac{34^2}{35}}}=-0.846$
$n=35$, so:
$d.f.=n-1=34$
Two-tailed test:
$t_{\frac{α}{2}}=t_{0.025}=2.032$
(According to Table VI, for d.f. = 34 and area in right tail = 0.025)
Also, $-t_{\frac{α}{2}}=-2.032$
Since $-t_{\frac{α}{2}}\lt t_0\lt t_{\frac{α}{2}}$, we do not reject the null hypothesis.
