Answer
$t_0\gt t_{\frac{α}{2}}$: null hypothesis is rejected.
There is enough evidence to conclude that there is a difference in mean CBCL scores between the control group and the RAP group.
Work Step by Step
$H_0:~µ_{CG}=µ_{RAP}$ versus $H_1:~µ_{CG}\neµ_{RAP}$
$t_0=\frac{(x ̅_1-x ̅_2)-(µ_1-µ_2)}{\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}=\frac{(11.7-9.0)-0}{\sqrt {\frac{21.6}{70}+\frac{13.0}{70}}}=3.840$
$n=70$, so:
$d.f.=n-1=69$
Two-tailed test:
$t_{\frac{α}{2}}=t_{0.025}=1.994$
(According to Table VI, for d.f. = 70, the closest value to 69, and area in right tail = 0.025)
Since $t_0\gt t_{\frac{α}{2}}$, we reject the null hypothesis.
