Answer
$t_0\lt -t_{\frac{α}{2}}$: null hypothesis is rejected.
There is enough evidence to conclude that there is a difference in test scores.
Work Step by Step
$x ̅_1,n_1~and~s_1$ refer to group 1 and $x ̅_2,n_2~and~s_2$ refer to group 2.
$x ̅_1=\frac{∑x_{1_i}}{n_1}=62.850$
$s_1=\sqrt {\frac{∑(x_{1_i}-x ̅_1)^2}{n_1}}=12.031$
$x ̅_2=\frac{∑x_{2_i}}{n_2}=54.352$
$s_2=\sqrt {\frac{∑(x_{2_i}-x ̅_2)^2}{n_2}}=9.224$
$H_0:~µ_1=µ_2$ versus $H_1:~µ_1\ne µ_2$
$t_0=\frac{(x ̅_1-x ̅_2)-(µ_1-µ_2)}{\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}=\frac{(62.850-54.352)-0}{\sqrt {\frac{12.031^2}{18}+\frac{9.224^2}{18}}}=2.378$
$n=18$, so:
$d.f.=n-1=17$
Two-tailed test:
$t_{\frac{α}{2}}=t_{0.025}=2.110$
(According to Table VI, for d.f. = 17 and area in right tail = 0.025)
Also, $-t_{\frac{α}{2}}=-2.110$
Since $t_0\lt -t_{\frac{α}{2}}$, we reject the null hypothesis.
