This section shows how you can use the Spearman’s Rank Correlation Coefficient to investigate changes with distance from open water. Information about other statistical tests can be found here.

## 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 at Woodmere. Raw data from point quadrats is used. The number means the total hits per species per site.

Calculate $$n$$, $$n\;\times\;(n\;-\;1)$$, $$N$$ 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 or to the dominance of one species.

## Spearman’s Rank Correlation Test

Spearman’s Rank Correlation is a statistical test to test whether there is a significant relationship between two sets of data.

The Spearman’s Rank Correlation test can only be used if there are at least 10 (ideally at least 15-15) pairs of data.

There are 3 steps to take when using the Spearman’s Rank Correlation Test

### Step 1. State the null hypothesis

There is no significant relationship between _______ and _______

### Step 2. Calculate the Spearman’s Rank Correlation Coefficient

$$r_s = 1-\frac{(6∑D^2)}{n(n^2-1)}$$
• $$r_s$$ = Spearman’s Rank correlation coefficient
• $$D$$ = differences between ranks
• $$n$$ = number of pairs of measurements

Step 3. Test the significance of the result

Compare the value of $$r_s$$ that you have calculated against the critical value for $$r_s$$ at a confidence level of 95% / significance value of p = 0.05.

If $$r_s$$ is equal to or above the critical value (p=0.05) the REJECT the null hypothesis. There is a SIGNIFICANT relationship between the 2 variables.

A positive sign for $$r_s$$ indicates a significant positive relationship and a negative sign indicates a significant negative relationship.

If $$r_s$$ (ignoring any sign) is less than the critical value, ACCEPT the null hypothesis. There is NO SIGNIFICANT relationship between the 2 variables.

### Worked example

A biologist is investigating whether soil moisture decreases with distance from open water at Woodmere. A systematic transect is taken with readings every 4m from the edge of the open water. A soil pin is used to measure soil moisture levels.

### Step 1. State the null hypothesis

There is no significant relationship between soil moisture and distance from open water at Woodmere

### Step 2. Calculate the Spearman’s Rank Correlation Coefficient

(a) Rank the measurements

(b) Calculate $$D$$ and $$D^2$$

(c) Calculate $$∑D^2$$i.e. the sum of the $$D^2$$column= 1065

(d) Calculate r_s

$$r_s = 1-\frac{(6∑D^2)}{n(n^2-1)}$$ $$r_s = 1\;-\;\frac{(6\;\times\;1065)}{ (15\times(225\;-\;1))}$$ $$r_s = -0.90$$

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

The critical value at $$p=0.05$$ significance level for $$15$$ pairs of measurements is $$0.521$$

Since our calculated value of $$0.90<0.521$$ (ignore the minus sign), the null hypothesis is rejected.

In conclusion, there is a significant relationship between soil moisture and distance from open water at Woodmere.

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