Answer
$F_0\gt F_{α,n_1-1,n_2-1}$: null hypothesis is rejected.
There is enough evidence to conclude that the mean CBCL scores are different for the three treatment groups.
Work Step by Step
$x ̅=\frac{11.7+9.0+12.4}{3}=11.0333$
$SST=n_1(x ̅_1-x ̅)^2+n_2(x ̅_2-x ̅)^2+n_3(x ̅_3-x ̅)^2=70(11.7-11.0333)^2+70(9.0-11.0333)^2+70(12.4-11.0333)^2=451.2636$
$MST=\frac{SSE}{k-1}=\frac{451.2636}{3-1}=225.63$
$SSE=(n_1-1)s_1^2+(n_2-1)s_2^2+(n_3-1)s_3^2=(70-1)21.6+(70-1)13.0+(70-1)8.4=2967$
$MSE=\frac{SSE}{n-k}=\frac{2967}{210-3}=14.33$
$F_0=\frac{MST}{MSE}=\frac{225.63}{14.33}=15.745$
Two-tailed test:
$F_{α,n_1-1,n_2-1}=$F_{0.05,2,207}$=3.04$
(According to table VIII, for $d.f._1=2$, $d.f._2=207$ and area in the right tail = 0.05)
Since $F_0\gt F_{α,n_1-1,n_2-1}$, we reject the null hypothesis.
