Answer
$P$-value $\lt α$: null hypothesis is rejected.
There is enough evidence to conclude that women are more flexible than men.
Work Step by Step
$x ̅_1,n_1~and~s_1$ refer to women and $x ̅_2,n_2~and~s_2$ refer to men.
$t_0=\frac{(x ̅_1-x ̅_2)-(µ_1-µ_2)}{\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}=\frac{(20.99-18.64)-0}{\sqrt {\frac{2.07^2}{31}+\frac{3.29^2}{45}}}=3.818$
Left-tailed test:
$n=31$ (use the smaller value of $n$), so:
$d.f.=n-1=30$
$P$-value $=P(t\gt t_0)=P(t\gt3.818)$
For $d.f.=30$ and the area to area in right tail equals to 0.0005: $t=3.646$
So, $P$-value $\lt0.0005$.
Since $P$-value $\lt α$, we reject the null hypothesis.
