Mann-Whitney U Test

1. Overview

Mann-Whitney U Test is equivalent to Wilcoxon Rank Sum Test.

As in the Wilcoxon Rank Sum Test, we mix the samples of two independent populations with sample sizes $n_1,n_2$, rank the values, and share the rank for tied values, giving the rank sums $W_1$ and $W_2$ for each population.

2. Test Statistic

The test statistics $U_1,U_2$ are defined as

$$\begin{array}{rl}{U}_1& =n_1{n}_2+\frac{n_1(n_1+1)}{2}-W_1,\\ U_2& =n_1{n}_2+\frac{n_2(n_2+1)}{2}-W_2.\end{array}$$

Define the test statistic $U$ to be the smaller of the two values:

$$U=\min (U_1,U_2).$$

The rejection region is

$$\mathcal{RR}=\left\{U\le u_{\alpha /2}\right\},$$

where $u_{\alpha /2}$ should be determined from the Mann-Whitney U Table. Since we have chosen the smaller of the two values, the rejection region is always on the left side. The one-tailed test is $\mathcal{RR}=\left\{U\le u_{\alpha }\right\}$, only changing the significance level.

For large sample tests, the asymptotic normality of $U$ is the same as in the Wilcoxon Rank Sum Test.

3. Example

Example (Salary comparison). Consider the salary of two groups of employees, where the sample sizes are $n_1=12$ and $n_2=11$. Test if the median salary of two groups are equal under the significance level $\alpha =0.05$.

Solution

1. Rank the salaries of two groups and calculate the rank sum $W_1,W_2$. Here the results are $W_1=149,W_2=127$.
2. Calculate the test statistic $U_1=61,U_2=71$, and thus $U=61$.
3. Find the rejection region from the Mann-Whitney U Table: $\mathcal{RR}=\{U\le u_{\alpha /2}^L\}=\{U\le 33\}$.
4. Since $U_0=\min (U_1,U_2)=U_1=61\notin \mathcal{RR}$, we fail to reject the null hypothesis. Thus, there is no significant difference in the median salary of two groups.

Related Notes

• Wilcoxon Rank Sum Test
• Nonparametric Tests Overview