Answer
$w_{\frac{α}{2}}\lt T\lt w_{1-\frac{α}{2}}$: null hypothesis is not rejected.
There is not enough evidence to conclude that the grades administered in each class are different.
Work Step by Step
$H_0:M_B=M_A$ versus $H_1:M_B\ne M_A$
Small-sample case:
$S=3+3+3+10+10+10+10+10+10+19.5+19.5+19.5+19.5+26.5+29.5=203$
$T=S-\frac{n_1(n_1+1)}{2}=203-\frac{15(15+1)}{2}=203-120=83$
Critical values:
$w_{\frac{α}{2}}=w_{0.025}=65$
(According to table XIII, for $n_1=15$, $n_2=15$ and $\frac{α}{2}=0.025$)
$w_{1-\frac{α}{2}}=n_1n_2-w_{\frac{α}{2}}$
$w_{0.975}=15\times15-65=160$
Since $w_{\frac{α}{2}}\lt T\lt w_{1-\frac{α}{2}}$, we do not reject the null hypothesis.
