Discussion for Freshwater Succession

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 1	Site 2
	\(n\)	\(n\)
Common reed	22	0
Saw sedge	14	0
Alder	10	0
Sheep’s fescue	0	7
Common bent	0	18
Creeping buttercup	0	8
Soft rush	0	10

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

	Site 1	Site 1	Site 2	Site 2
	\(n\)	\(n\;\times\;(n\;-\;1)\)	\(n\)	\(n\;\times\;(n\;-\;1)\)
Common reed	22	462	0	0
Saw sedge	14	182	0	0
Alder	10	90	0	0
Sheep’s fescue	0	0	7	42
Common bent	0	0	18	306
Creeping buttercup	0	0	8	56
Soft rush	0	0	10	90
TOTAL	46	734	43	494
	D = 46 (45) / 734	D = 43 (42) / 494
	D = 2.82	D = 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)
0	13.0
3	14.5
6	12.8
9	15.5
12	14.4
15	11.0
18	8.4
21	7.0
24	8.5
27	6.5
30	8.5
33	6.5
36	5.6
39	4.0
42	5.2

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)
Data	Rank	Data	Rank
0	1	13.0	12
3	2	14.5	14
6	3	12.8	11
9	4	15.5	15
12	5	14.4	13
15	6	11.0	10
18	7	8.4	7
21	8	7.0	6
24	9	8.5	8.5
27	10	6.5	4.5
30	11	8.5	8.5
33	12	6.5	4.5
36	13	5.6	3
39	14	4.0	1
42	15	5.2	2

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

Distance from open water (m)	Soil pin depth (cm)
Data	Rank	Data	Rank	\(D\)	\(D^2\)
0	1	13.0	12	11	121
3	2	14.5	14	12	144
6	3	12.8	11	8	64
9	4	15.5	15	11	121
12	5	14.4	13	8	64
15	6	11.0	10	4	16
18	7	8.4	7	0	0
21	8	7.0	6	2	4
24	9	8.5	8.5	0.5	0.25
27	10	6.5	4.5	5.5	30.25
30	11	8.5	8.5	2.5	6.25
33	12	6.5	4.5	7.5	56.25
36	13	5.6	3	10	100
39	14	4.0	1	13	169
42	15	5.2	2	13	169

(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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