Answer
$t_0\gt t_α$: null hypothesis is rejected.
There is enough evidence to conclude that the students perform better on the final exam in the fall semester with the online homework system.
Work Step by Step
$x ̅_1,n_1~and~s_1$ refer to fall semester and $x ̅_2,n_2~and~s_2$ refer to spring semester.
$H_0:~µ_1=µ_2$ versus $H_1:~µ_1\gt µ_2$
$t_0=\frac{(x ̅_1-x ̅_2)-(µ_1-µ_2)}{\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}=\frac{(73.6-67.9)-0}{\sqrt {\frac{10.3^2}{27}+\frac{12.4^2}{25}}}=1.795$
$n=25$ (use the smaller value of $n$), so:
$d.f.=n-1=24$
Right-tailed test:
$t_α=t_{0.05}=1.711$
(According to Table VI, for d.f. = 24 and area in right tail = 0.05)
Since $t_0\gt t_α$, we reject the null hypothesis.
