Simpson’s Diversity Index
Simpson’s Diversity Index is a measure both of species richness (i.e. the number of different species present) and species evenness (i.e. how evenly distributed each species is).
\(D = \frac{(N\;\times\;(N\;-\;1))}{(Σn\;\times\;(n\;-\;1))}\)- \(D\) = Simpson’s Diversity Index
- \(n\) = the number of individuals of each species
- \(N\) = the total number of individuals
Mann Whitney U test
Mann Whitney U is a statistical test that is used either to test whether there is a significant difference between the medians of two sets of data.
The Mann Whitney U test can only be used if there are at least 6 pairs of data. It does not require a normal distribution.
There are 3 steps to take when using the Mann Whitney U test
Step 1. State the null hypothesis
There is no significant difference between _______ and _______
Step 2. Calculate the Mann Whitney U statistic
[llatex]U_1= n_1 \times n_2 + 0.5 n_2 (n_2 + 1)\;- ∑ R_2[/latex]
[llatex]U_2 = n_1 \times n_2 + 0.5 n_1 (n_1 + 1)\;- ∑ R_1[/latex]
- `\(n_1\) is the number of values of \(x_1\)
- \(n_2\) is the number of values of \(x_2\)
- \(R_1\) is the ranks given to \(x_1\)
- \(R_2\) is the ranks given to \(x_2[latex]
Step 3. Test the significance of the result
Compare the value of U against the critical value for U at a confidence level of 95% / significance value of P = 0.05.
If U is equal to or smaller than the critical value (p=0.05) the REJECT the null hypothesis. There is a SIGNIFICANT difference between the 2 data sets.
If U is greater than the critical value, then ACCEPT the null hypothesis. There is NOT a significant difference between the 2 data sets.
Worked example
A biologist is investigating whether there is a difference in woodland flora between two contrasting areas of woodland, Site A and Site B. Eight randomly placed frame quadrat samples in each of two contrasting areas of woodland produced the following % cover of dog’s mercury, a common woodland plant. Here are the results.
Quadrat | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
Site A | 26 | 14 | 8 | 6 | 26 | 20 | 11 | 13 |
Site B | 28 | 16 | 26 | 25 | 25 | 24 | 16 | 28 |
Step 1. State the null hypothesis
There is no significant difference in species richness between Site A and Site B.
Step 2. Calculate Mann Whitney U statistic
(a) Give each result a rank. Calculate the sum of the ranks for the two columns.
Now arrange the data values in order, and give each value a rank in order from smallest to highest
Site A | 6 | 8 | 11 | 13 | 14 | 20 | 26 | 26 |
rank | 1 | 2 | 3 | 4 | 5 | 8 | 13 | 13 |
Site B | 16 | 16 | 24 | 25 | 25 | 26 | 28 | 28 |
rank | 6.5 | 6.5 | 9 | 10.5 | 10.5 | 13 | 15.5 | 15.5 |
(b) Calculate [latex]∑R_1\) and \(∑R_2\)
∑R_1
is the sum of the ranks in the first column (Site A) = 49
\(∑R_2\) is the sum of the ranks in the first column (Site B) = 87
\(n_1\) = 8 and \(n_2 = 8\)
(c) Calculate \(U_1\) and \(U_2\)
\(U_1 = 8\times 8 + 0.5 \times 8 \times (8 + 1) – 87 = 13\) \(U_2 = 8 \times8 + 0.5 \times 8 \times (8 + 1) – 49 = 51\)Step 3. Test the significance of the result
In this example, \(U_1\) = 13 and \(U_2\) = 51
Select the smaller or the values. In this case \(U_1\) is the smaller of the two values, so U=13
The critical value at p=0.05 significance level for \(n_1=8\) and \(n_2=8\) is 13. Since our calculated value of 13 = 13, the null hypothesis can be rejected.
In conclusion, there is a significant difference in species richness between Site A and Site B.

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