Answer
$t_0\gt t_α$: null hypothesis is rejected.
There is enough evidence to conclude that $µ_1\gt µ_2$.
Work Step by Step
$H_0:~µ_1=µ_2$ versus $H_1: µ_1\gt µ_2$
$t_0=\frac{(x ̅_1-x ̅_2)-(µ_1-µ_2)}{\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}=\frac{(50.2-42.0)-0}{\sqrt {\frac{6.4^2}{25}+\frac{9.9^2}{18}}}=3.081$
$n=18$ (use the smaller value of $n$), so:
$d.f.=n-1=17$
Right-tailed test:
$t_α=t_{0.10}=1.333$
(According to Table VI, for d.f. = 17 and area in right tail = 0.10)
Since $t_0\gt t_α$, we reject the null hypothesis.
