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 $H_1:~µ_1\gt µ_2$
$t_0=\frac{(x ̅_1-x ̅_2)-(µ_1-µ_2)}{\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}=\frac{(43.1-41.0)-0}{\sqrt {\frac{4.5^2}{23}+\frac{5.1^2}{13}}}=1.237$
$n=13$ (use the smaller value of $n$), so:
$d.f.=n-1=12$
Right-tailed test:
$t_α=t_{0.05}=1.782$
(According to Table VI, for d.f. = 12 and area in right tail = 0.05)
Since $t_0\lt t_α$, we do not reject the null hypothesis.
