Suppose the data for the first sample (Samp1) are 22, 24, 25, 29, 30 and the data for the second sample (Samp2) are 16, 21, 22, 23. The output from the Mann-Whitney test is:

Method
η₁: median of C1
η₂: median of C2
Difference: η₁ - η₂

Descriptive Statistics
Sample N Median
C1 5 25.0
C2 4 21.5

Estimation for Difference
CI for Achieved
Difference Difference Confidence
6 (-0.0000000, 13) 96.27%

Test
Null hypothesis H₀: η₁ - η₂ = 0
Alternative hypothesis H₁: η₁ - η₂ ≠ 0

Method W-Value P-Value
Not adjusted for ties 33.50 0.050
Adjusted for ties 33.50 0.049

The point estimate for η_{1} – η_{2} is the median of all possible pairwise differences between the two samples.

For this example, there are 5*4 = 20 pairwise differences. The possible pairwise differences for this example are: 22-16 = 6, 22-21 = 1, 22-22= 0, 22-23= −1, 8, 3, 2, 1, 9, 4, 3, 2, 13, 8, 7, 6, 14, 9, 8, 7.

You can get all the pairwise differences between 2 columns in Minitab by choosing

.The median of these differences is 6.

W = (number of positive differences) + 0.5(number of differences that equal 0) + 0.5(n_{1}(n_{1}+1)) where n_{1} = number of observations in the first sample.

For this example, W = 18 + 0.5(1) + 0.5*5*6 = 18 + 0.5 + 15 = 33.5.

The p-value is based on the test statistic for W. The test statistic, Z, (which is not part of the output) is a normal approximation using the mean and variance of W.

Mean of W = 0.5(n_{1} (n_{1} + n_{2} + 1)) variance of W = n_{1}*n_{2}(n_{1}+n_{2}+1)/12 where n_{1} and n_{2} are the number of observations in the first and second sample, respectively.

Z = (|W - mean of W| - .5)/square root of the variance of W.

Subtracting the .5 from the numerator is the continuity correction factor.

The p-value for H_{a}: η_{1} < η_{2} is CDF(Z). The p-value for H_{a}: η_{1} > η_{2} is (1 - CDF(Z)). The p-value for H_{a}: η_{1} ≠ η_{2} is 2*(1 - CDF(Z)). Where CDF is the cumulative probability of a standard normal distribution.

For this example:

- mean of W = 0.5*5(5+4+1) = 2.5*10 = 25
- variance of W = 5*4(5+4+1)/12 = 20*10/12 = 200/12 = 16.6667

Z = (|33.5 - 25| - .5)/sqrt(16.6667) = 1.9596

The p-value for H_{a}: η_{1} ≠ η_{2} is 2*(1 - 0.974979.) = 0.05.

You can get the cumulative probabilities in Minitab by choosing

.