Answer
$t_0\gt -t_α$: null hypothesis is not rejected.
There is not enough evidence to conclude that $µ_1\lt µ_2$.
Work Step by Step
$H_0:~µ_1=µ_2$ versus $H_1:~µ_1\lt µ_2$
$t_0=\frac{(x ̅_1-x ̅_2)-(µ_1-µ_2)}{\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}=\frac{(96.6-98.3)-0}{\sqrt {\frac{3.2^2}{13}+\frac{2.5^2}{8}}}=-1.357$
$n=8$ (use the smaller value of $n$), so:
$d.f.=n-1=7$
Left-tailed test:
$t_α=t_{0.05}=1.895$
(According to Table VI, for d.f. = 7 and area in right tail = 0.05)
So, $-t_{\frac{α}{2}}=-1.895$
Since $t_0\gt -t_α$, we do not reject the null hypothesis.
