Answer
$t_0\lt -t_α$: null hypothesis is rejected.
There is enough evidence to conclude that $µ_1\lt µ_2$
Work Step by Step
$H_0:~µ_1=µ_2$ versus $H_1: µ_1\lt µ_2$
$t_0=\frac{(x ̅_1-x ̅_2)-(µ_1-µ_2)}{\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}=\frac{(94.2-115.2)-0}{\sqrt {\frac{15.9^2}{40}+\frac{23.0^2}{32}}}=-4.393$
$n=32$ (use the smaller value of $n$), so:
$d.f.=n-1=31$
Left-tailed test:
$t_α=t_{0.05}=1.696$
(According to Table VI, for d.f. = 31 and area in right tail = 0.05)
So, $-t_{\frac{α}{2}}=-1.696$
Since $t_0\lt -t_α$, we reject the null hypothesis.
