This section shows how you can use the Chi-squared test to compare two parts of the dunes.

## 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

### Worked example

A biologist is comparing species diversity at two sites within Traeth-y-goes sand dunes. Raw data from point quadrats is used. The number means the total hits per species per site.

Calculate $$n$$, $$n x (n-1)$$, $$N$$ and $$D$$ for each site

The larger the value of D, the higher the species diversity. A low value of D could be due to low overall species richness (like at the strand line) or to the dominance of one species (as in dune scrub). You could present your results as a graph

## 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

$$U_1= n_1 \times n_2 + 0.5 n_2 (n_2 + 1)\;- ∑ R_2$$ $$U_2 = n_1 \times n_2 + 0.5 n_1 (n_1 + 1)\;- ∑ R_1$$
• $$n_1$$ 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 $$x_2$$

### 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 managed areas of the sand dunes have greater species richness than unmanaged areas. A frame quadrat was used to record number of species at 10 randomly chosen points in each of two sites: one in a grazed section of the fixed dunes and one in an ungrazed section of the fixed dunes.

Here are the results.

### Step 1. State the null hypothesis

There is no significant difference in species richness between the grazed and the ungrazed parts of the fixed dunes.

### Step 2. Calculate Mann Whitney U statistic

(a) Give each result a rank. Calculate the sum of the ranks for the two columns.

(b) Calculate $$∑R_1$$ and $$∑R_2$$

$$∑R_1$$ is the sum of the ranks in the first column (deciduous woodland) = 63

$$∑R_2$$ is the sum of the ranks in the first column (evergreen woodland) = 147

$$n_1 = 10$$ and $$n_2 = 10$$

(c) Calculate $$U_1$$ and $$U_2$$

$$U_1 = 10\; \times\; 10\; +\; 0.5\; \times\; 10\; \times\; (10\; +\; 1)\; – \;63 = 92$$ $$U_2 = 10\; \times\; 10\; +\; 0.5\; \times\; 10\; \times\; (10\; +\; 1)\; – \;147 = 8$$

### Step 3. Test the significance of the result

In this example, $$U_1 = 92$$ and $$U_2 = 8$$

We select the smaller of the values. Here, $$U_2$$ is the smaller of the two values, so $$U=8$$

Use a set of Mann-Whitney U critical value tables (example here) to find the critical value at the 95% confidence level.

The critical value at $$p=0.05$$ significance level for $$n_1=10$$ and $$n_2=10$$ is $$23$$. Since our calculated value of $$8 < 23$$, the null hypothesis can be rejected.

In conclusion, there is a significant difference in species richness between the grazed and the ungrazed parts of the fixed dunes.

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