Answer
$-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$: null hypothesis is not rejected.
There is not enough evidence to conclude that $M_x\ne M_y$.
Work Step by Step
Two-tailed test.
Large-sample case:
$T=S-\frac{n_1(n_1+1)}{2}=590-\frac{22(22+1)}{2}=590-253=337$
$z_0=\frac{T-\frac{n_1n_2}{2}}{\sqrt {\frac{n_1n_2(n_1+n_2+1)}{12}}}=\frac{337-\frac{22\times25}{2}}{\sqrt {\frac{22\times25(22+25+1)}{12}}}=1.32$
$z_{\frac{α}{2}}=z_{0.025}$
If the area of the standard normal curve to the right of $z_{0.025}$ is 0.025, then the area of the standard normal curve to the left of $z_{0.025}$ is $1−0.025=0.975$
According to Table V, the z-score which gives the closest value to 0.975 is 1.96.
Also, $-z_{\frac{α}{2}}=-z_{0.025}=-1.96$
Since $-z_{\frac{α}{2}}\lt z_0\lt z_{\frac{α}{2}}$, we do not reject the null hypothesis.
