Answer
$t_0\gt t_α$: null hypothesis is rejected.
There is enough evidence to conclude that the experimental group experienced a larger mean improvement than the control group.
Work Step by Step
$x ̅_1,n_1~and~s_1$ refer to experimental group and $x ̅_2,n_2~and~s_2$ refer to control group.
$H_0:~µ_1=µ_2$ versus $H_1:~µ_1\gt µ_2$
$t_0=\frac{(x ̅_1-x ̅_2)-(µ_1-µ_2)}{\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}=\frac{(14.8-8.1)-0}{\sqrt {\frac{12.5^2}{55}+\frac{12.7^2}{60}}}=2.849$
$n=55$ (use the smaller value of $n$), so:
$d.f.=n-1=54$
Right-tailed test:
$t_α=t_{0.01}=2.403$
(According to Table VI, for d.f. = 50, the closest value to 54, and area in right tail = 0.01)
Since $t_0\gt t_α$, we reject the null hypothesis.
