Mann-Whitney U Test Calculator

Mann-Whitney U Test (Wilcoxon Rank-Sum Test)

The Mann-Whitney U test compares two independent groups to determine if one tends to have larger values than the other. It is a non-parametric alternative to the independent samples t-test, with no normality assumption needed.

When to use it:

• Two independent groups of data
• Data is not normally distributed
• Ordinal data (ranked categories)
• Comparing medians of two groups

The U statistic:

U1 = n1n2 + n1(n1+1)/2 - R1 U2 = n1*n2 - U1

Where n1, n2 are group sizes and R1 is the sum of ranks assigned to Group 1.

Normal approximation (the standard route once the groups get large):

mu_U = n1 * n2 / 2 sigma_U = sqrt( n1n2(n1+n2+1) / 12 ) z = (U - mu_U) / sigma_U

Interpretation:

Null hypothesis: Both groups come from the same distribution. A significant result means one group tends to produce higher values than the other.

Effect size (rank-biserial correlation): r = 1 - (2 * U1) / (n1 * n2) Values near +1 or -1 indicate a large effect; near 0 indicates no effect. The sign carries the direction: positive means Group 1 tends to be higher, negative means Group 2 does. This is why the formula uses U1 rather than the smaller of the two U values. Feeding it min(U1, U2) throws the sign away and makes every result look positive, which hides exactly the thing you ran the test to learn.

Tie correction: When tied ranks exist, the standard deviation is adjusted: sigma_U = sqrt( n1n2/12 * (N+1 - sum(t^3-t) / (N(N-1))) ) where N = n1+n2 and t = size of each tied rank group.

Exact versus approximate p-values: The normal approximation above is a convenience for larger samples. With small groups it is noticeably off, and both groups being under about 8 is where it strays most. For small samples with no ties this calculator counts the null distribution directly instead: it works out how many of the possible rank splits give a U at least as extreme as yours, and divides. That is the exact p-value, and it is what published Mann-Whitney tables contain. The page tells you which one it used. Ties make the exact count invalid, so with ties present it falls back to the tie-corrected normal approximation and says so.