Answer
$t_0\lt t_α$: null hypothesis is not rejected.
There is not enough evidence to conclude that $µ_1\gt µ_2$.
Work Step by Step
$H_0:~µ_1=µ_2$ versus $H_1:~µ_1\gt µ_2$
$t_0=\frac{(x ̅_1-x ̅_2)-(µ_1-µ_2)}{\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}=\frac{(48.2-45.2)-0}{\sqrt {\frac{8.4^2}{45}+\frac{10.3^2}{41}}}=1.472$
$n=41$ (use the smaller value of $n$), so:
$d.f.=n-1=40$
Right-tailed test:
$t_α=t_{0.01}=2.423$
(According to Table VI, for d.f. = 40 and area in right tail = 0.01)
Since $t_0\lt t_α$, we do not reject the null hypothesis.
