This document describes how to perform a Mann-Whitney U test to compare two independent samples. It provides an example comparing the percentage improvement of patients using a new ointment versus a standard ointment. The steps are to: 1) state the null and alternative hypotheses, 2) rank all data values from smallest to largest, 3) calculate the totals for each sample, 4) determine the test statistic, 5) find the critical value, and 6) make a conclusion based on whether the test statistic is greater than or less than the critical value. In this example, the test statistic is greater than the critical value, so the null hypothesis that there is no difference between the samples is accepted.

1. M A T T B R A Y L E Y
Mann-Whitney U Test
A Mann-Whitney U Test is used to test the hypothesis that two independent samples
come from identical populations. It can be used where there are two sets of data which
are not paired.
You can tell that data is not paired when there are two sets of data in a table and the
number of data values for the first set of data is different to the second set. You can also
tell it’s not paired if there isn’t an additional row with labels for paired data values. This
would be an example of a paired set of data:
So a Mann-Whitney U Test would not be an appropriate test to use for this.
Person 1 2 3 4 5 6 7 8
A 2 7 3 9 1 4 2 5
B 4 20 7 2 27 9 3 6

2. Outpatients attending a clinic for a specific skin disease were asked to take part in a trail of a new ointment.
Of these patients, 22 were selected at random and listed in alphabetical order.
The first 12 were treated with the new ointment and the remaining 10 were treated with the standard
ointment.
The percentage improvement observed after six weeks of treatment was as follows:
Example Question
New
Ointment
34 29 18 25 36 21 42 50 38 26 31 41
Standard
Ointment
18 20 12 40 27 22 35 15 28 51
Carry out a Mann-Whitney U test, using the 5% significance level, to investigate whether patients who used
the new ointment achieved a greater percentage improvement.

3. 1. The first step is to write your hypotheses:
H0 is always: Samples are taken from identical populations
H1 is always: Samples are not taken from identical populations, the two
averages are different (two tailed test) OR one average is greater
than the other (one tailed test)
2. Rank the table of numbers in order from smallest to largest, like this:
Steps
14 12 3.5 8 16 6 20 21 17 9 13 19
New
Ointment
34 29 18 25 36 21 42 50 38 26 31 41
Standard
Ointment
18 20 12 40 27 22 35 15 28 51
3.5 5 1 18 10 7 15 2 11 22

4. 3. Add the ranks of the New Ointment and the ranks of the Standard
Ointment:
TN = 158.5 TS = 94.5
4. Calculate the test statistic for each of your totals. The formula is:
This is given to you in the formula booklet, so there’s no need to remember it!
Steps
14 12 3.5 8 16 6 20 21 17 9 13 19
New
Ointment
34 29 18 25 36 21 42 50 38 26 31 41
Standard
Ointment
18 20 12 40 27 22 35 15 28 51
3.5 5 1 18 10 7 15 2 11 22
‘T’ means total; in this question, I
have used ‘N’ to represent the New
Ointment and ‘S’ to represent the
Standard Ointment. You should set
it out in the same way.

5. The Test Statistic is the smaller of these two values:
39.5.
5. Use Page 33 in the Formula Booklet and
determine which table you need to use. This is
the second table for this question because we
are doing a two tailed test at the 5% significance
level. Use the values you’ve written down for m
and n to find your Critical Value. This is 30, for
our example.
Steps
TN = 158.5 TS = 94.5
U1 = 158.5 – 12(13) ÷ 2 = 80.5
U2 = 94.5 – 10(11) ÷ 2 = 39.5

6. Conclusion
If the Test Statistic is ≤ the
Critical Value, then you
Reject H0.
6. You can now conclude in context.
 The Test Statistic for this example is 39.5
 The Critical Value for this example is 30
As 39.5 is > 30, accept H0. There is significant evidence at the 5% level that there is a difference
between the percentage improvement and the use of the new ointment.
Notice that the same wording has been used to write the conclusion.