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.

 Site 1Site 2
 \(n\)\(n\)
Common reed220
Saw sedge140
Alder100
Sheep’s fescue07
Common bent018
Creeping buttercup08
Soft rush010

Calculate \(n\), \(n\;\times\;(n\;-\;1)\), \(N\) and \(D\) for each site

 Site 1Site 1Site 2Site 2
 \(n\)\(n\;\times\;(n\;-\;1)\)\(n\)\(n\;\times\;(n\;-\;1)\)
Common reed2246200
Saw sedge1418200
Alder109000
Sheep’s fescue00742
Common bent0018306
Creeping buttercup00856
Soft rush001090
TOTAL4673443494
 D = 46 (45) / 734D = 43 (42) / 494
 D = 2.82D = 3.66

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.

Distance from open water (m)Soil pin depth (cm)
013.0
314.5
612.8
915.5
1214.4
1511.0
188.4
217.0
248.5
276.5
308.5
336.5
365.6
394.0
425.2

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

Distance from open water (m)Distance from open water (m)Soil pin depth (cm)Soil pin depth (cm)
DataRankDataRank
0113.012
3214.514
6312.811
9415.515
12514.413
15611.010
1878.47
2187.06
2498.58.5
27106.54.5
30118.58.5
33126.54.5
36135.63
39144.01
42155.22

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

Distance from open water (m)Soil pin depth (cm)
DataRankDataRank\(D\)\(D^2\)
0113.01211121
3214.51412144
6312.811864
9415.51511121
12514.413864
15611.010416
1878.4700
2187.0624
2498.58.50.50.25
27106.54.55.530.25
30118.58.52.56.25
33126.54.57.556.25
36135.6310100
39144.0113169
42155.2213169

(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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