Answer
$P$-value $\lt α$: null hypothesis is rejected.
There is enough evidence to conclude that the mean step pulse of men is less than the mean step pulse of women.
Work Step by Step
$t_0=\frac{(x ̅_1-x ̅_2)-(µ_1-µ_2)}{\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}=\frac{(112.3-118.3)-0}{\sqrt {\frac{11.3^2}{51}+\frac{14.2^2}{70}}}=-2.586$
Left-tailed test:
$n=51$ (use the smaller value of $n$), so:
$d.f.=n-1=50$
$P$-value $=P(t\lt t_0)=P(t\lt-2.586)=P(t\gt2.586)$
For $d.f.=50$ and the area to area in right tail equals to 0.01: $t=2.403$
For $d.f.=50$ and the area to area in right tail equals to 0.005: $t=2.678$
So, $0.005\lt P$-value $\lt0.01$.
Since $P$-value $\lt α$, we reject the null hypothesis.
