Mann-Whitney Test

A non-parametric test comparing two groups by ranking all observations — valid without normality assumptions.

Hypothesis test — rejection region and test statistic

✓ Fail to reject H₀ — p-value ≈ 0.0719 > α=0.05

α = 0.05

z = 1.80

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Definition

The Mann-Whitney U test (also called the Wilcoxon rank-sum test) is a nonparametric test that compares two independent groups without assuming normality.

Null hypothesis: the two populations have the same distribution (or, in the location-shift version, the same median).

Procedure:

Pool both groups and rank all observations (smallest = 1)
Compute $U_1 = n_1 n_2 + \frac{n_1(n_1+1)}{2} - R_1$ , where $R_1$ is the sum of ranks for group 1
$U_2 = n_1 n_2 - U_1$ (they sum to $n_1 n_2$ )
The test statistic is $U = \min(U_1, U_2)$ ; compare to the Mann-Whitney distribution (or use normal approximation for large $n$ )

Key properties

Uses only the rank order of observations, never their actual numeric values
Valid for any continuous distribution — no normality assumption required
$U_1$ and $U_2$ always sum to exactly $n_1 n_2$
Nearly as powerful as the t-test under normality, and often more powerful when data is skewed or heavy-tailed

Common mistakes

Interpreting a significant result as "the medians differ": strictly, the test detects stochastic dominance — if the two distributions differ in shape (not just location), "equal medians" isn't quite the right null hypothesis being tested
Ignoring ties: many tied values require a correction factor in the normal approximation; ignoring it can distort the p-value

Comparing response times

Group A (new UI): $\{2.1, 3.4, 1.9, 4.0\}$ , Group B (old UI): $\{3.8, 5.2, 4.7, 6.1\}$ .

Pool and rank: 1.9(A), 2.1(A), 3.4(A), 3.8(B), 4.0(A), 4.7(B), 5.2(B), 6.1(B).

$R_A = 1+2+3+5 = 11$ , $U_A = 4\cdot4 + 10 - 11 = 15$ , $U_B = 16 - 15 = 1$ .

$U = \min(15, 1) = 1$ . Small $U$ suggests Group A has generally smaller values (faster).

Try it

Why might you use the Mann-Whitney test instead of a two-sample t-test when comparing incomes of two professional groups?

Solution

Income distributions are typically right-skewed with heavy tails (a few very high earners). The two-sample t-test assumes (approximately) normal distributions, which fails badly with heavy-tailed skewed data — the test statistic doesn't follow the t-distribution, and the Type I error rate is inflated.

The Mann-Whitney test uses only the ranks of the observations, not their actual values. It's insensitive to outliers and skewness. The test is valid for any continuous distribution — no normality required.

Additionally, income data may have practical outliers (billionaires in the sample) that massively influence the mean but barely affect the rank ordering.

Related concepts

Statistics· Statistical Tests

T-TestTesting whether a mean differs from a reference value or whether two group means differ, using the t-distribution when sample sizes are small.

Statistics· Statistical Tests

Kruskal-Wallis TestThe non-parametric analogue of one-way ANOVA — comparing three or more groups using ranks rather than raw values.

Statistics· Inference

Hypothesis TestingA formal framework for deciding whether data provides enough evidence to reject a default assumption.