∑Richard Hummel Math

Mann-Whitney Test

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

Mann-Whitney rank strip — pool both groups, rank all observations
1A2A3A4A5B6B7A8B9B10B11B
Group A rank sum
W_A = 17
Expected under H₀
E[W] = 30
Group B rank sum
W_B = 49

Under H₀ (groups identical), ranks are randomly distributed — each group's sum should be ≈ 30. Group B's higher rank sum suggests it tends to have larger values.

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:

  1. Pool both groups and rank all observations (smallest = 1)
  2. Compute U1=n1n2+n1(n1+1)2−R1U_1 = n_1 n_2 + \frac{n_1(n_1+1)}{2} - R_1, where R1R_1 is the sum of ranks for group 1
  3. U2=n1n2−U1U_2 = n_1 n_2 - U_1 (they sum to n1n2n_1 n_2)
  4. The test statistic is U=min⁡(U1,U2)U = \min(U_1, U_2); compare to the Mann-Whitney distribution (or use normal approximation for large nn)
Key properties
  • Uses only the rank order of observations, never their actual numeric values
  • Valid for any continuous distribution — no normality assumption required
  • U1U_1 and U2U_2 always sum to exactly n1n2n_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}\{2.1, 3.4, 1.9, 4.0\}, Group B (old UI): {3.8,5.2,4.7,6.1}\{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).

RA=1+2+3+5=11R_A = 1+2+3+5 = 11, UA=4⋅4+10−11=15U_A = 4\cdot4 + 10 - 11 = 15, UB=16−15=1U_B = 16 - 15 = 1.

U=min⁡(15,1)=1U = \min(15, 1) = 1. Small UU 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

View in full concept graph →