Answer
The Mann-Whitney U Test and Wilcoxon rank-sum test both apply to the same situation and always lead to the same conclusions.
Work Step by Step
Using the Mann-Whitney U Test
$U={n_{1}}n_{2}+\frac{n_{1}(n_{1}+1)}{2} - R= 12\times11 +\frac{12(12+1)}{2}-123.5=86.5$
$z=\frac{U-\frac{n_{1}n_{2}}{2}}{\sqrt \frac{n_{1}n_{2}(n_{1}+n_{2+1)}}{12}}=1.26$
We are testing (with $\alpha = 0.05$) the hypothesis that the two populations have equal medians, so we have a two-tailed test with critical values $z = \pm1.96$. The test statistic found using The Mann-Whitney U test is $z =\pm 1.26$ does not fall within the critical region, so we fail to reject the null hypothesis that the populations have the same median.
The Mann-Whitney U Test is equivalent to the Wilcoxon rank-sum test for independent samples in the sense that they both apply to the same situation and always lead to the same conclusions.
