# Mann-Whitney U Test

This Mann-Whitney U Test is used to find out if there is a significant difference between the medians of two sets of data.

There should be between 5 and 20 readings or values (e.g. quadrats or sweeps) in each data set. The data does not need to be normally distributed (i.e. it is a non-parametric test).

Before carrying out the statistical test, a null hypothesis needs to be formulated. The null hypothesis is different to the main study hypothesis and always states there will be no significant difference between the two data sets.

**Example**

Two ponds were surveyed to see if there was a difference in the number of invertebrate species found in each pond. Pond A was shaded by trees, Pond B was unshaded.

Null hypothesis:** There is no significant difference in the number of species found in the two ponds.**

**Results** (number of species per sweep, n is the number of sweeps)

Pond A | 6 | 2 | 11 | 5 | 7 | 3 | 6 | 5 | 6 | 4 | _{n}A = 10 |

Pond B |
11 |
5 |
8 |
4 |
13 |
6 |
15 |
7 |
12 |
10 |
_{n}B = 10 |

1. First put the data in order as one set of numbers (distinguish between the two data sets – Pond A is in normal type, Pond B is in blue) and ‘rank’ them all between 1 and, in this data set, 20.

Data | 2 | 3 | 4 |
4 | 5 |
5 | 5 | 6 |
6 | 6 | 6 | 7 |
7 | 8 |
10 |
11 |
11 | 12 |
13 |
15 |

Rank | 1 | 2 | 3.5 |
3.5 | 6 |
6 | 6 | 9.5 |
9.5 | 9.5 | 9.5 | 12.5 |
12.5 | 14 |
15 |
16.5 |
16.5 | 18 |
19 |
20 |

Note: If there is more than one reading with the same value they ‘share a rank’. For example, the value 4 would have the ranks 3 and 4 but because there are two readings with that value they have the same rank which is 3.5 (3+4/2). There are 3 readings with the value 5 so the rank given to each of them is 6 (5+6+7)/3 = 6.

2. Once you have ranked the data, you need to add up the ranks for each data set

ΣrA = Pond A = 1 + 2 + 3.5 + 6 + 6+ 9.5 + 9.5 + 9.5 + 12.5 + 16.5 = 76

**ΣrB = Pond B = 3.5 + 6 + 9.5 + 12.5 + 14 + 15 + 16.5 + 18 + 19 + 20 = 134**

3. You can then calculate the value of U for each pond

nA = number of samples from Pond A, nB = number of samples from Pond B

UA = nAnB + (nA(nA+1))/2 – ∑rA = 100 + 55 – 76 = 79

**UB = nAnB + (nB(nB+1))/2 – ∑rB = 100 + 55 – 134 = 21**

4. Take the smallest value of U (21 is smaller than 79). This is your calculated test statistic.

5. Look up the critical value of U from the Mann-Whitney U Test statistical table

6. If the calculated value is LESS THAN or EQUAL TO the critical value there is only a 5% probability that the null hypothesis is true and can, therefore, be rejected. Therefore, there is a statistically significant difference between the two data sets.

In our example, the **critical value is 23**.

Our **calculated value of 21** is LESS THAN the critical value and so, we reject the null hypothesis.

Your statistics results should go in the results section of your report.

e.g. “The Mann-Whitney U test was used to see if there was a statistically significant difference in the number of species in the two ponds because the measurements were of the ordinal level. The null hypothesis stated that there was no significant difference in the number of invertebrate species in the two ponds and was rejected because the calculated U value was less than the critical U value at the 5% significance level (n=10, Ucalc = 21, Ucrit = 23).”

Now consider the biological significance of the result… and link it back to the other data you have collected.