Answer
$z_0\gt z_α$: null hypothesis is rejected.
There is enough evidence to conclude that $M_x\gt M_y$.
Work Step by Step
Right-tailed test.
Large-sample case:
$T=S-\frac{n_1(n_1+1)}{2}=1310-\frac{34(34+1)}{2}=1310-595=715$
$z_0=\frac{T-\frac{n_1n_2}{2}}{\sqrt {\frac{n_1n_2(n_1+n_2+1)}{12}}}=\frac{715-\frac{34\times30}{2}}{\sqrt {\frac{34\times30(34+30+1)}{12}}}=2.76$
$z_α=z_{0.05}$
If the area of the standard normal curve to the right of $z_{0.05}$ is 0.05, then the area of the standard normal curve to the left of $z_{0.05}$ is $1−0.05=0.95$
According to Table V, there are 2 z-scores which give the closest value to 0.95: 1.64 and 1.65. So, let's find the mean of these z-scores: $\frac{1.64+1.65}{2}=1.645$
Since $z_0\gt z_α$, we reject the null hypothesis.
