## 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.