Discussion for Saltmarsh Succession

This section shows how you can use the Spearman’s Rank Correlation Coefficient to investigate changes with distance inland from the seashore. 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 Abernant Saltmarsh. Raw data from point quadrats is used. The number means the total hits per species per site.

	Site 1	Site 2
	\(n\)	\(n\)
Green algae	4	0
Glasswort	9	0
Cordgrass	6	0
Saltmarsh grass	18	68
Seablite	10	1
Sea purslane	10	29
Sea spurrey	0	6
Scurvy grass	0	10
Thrift	0	6
Sea lavender	0	2
Sea aster	0	8
Sea milkwort	0	1

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

	Site 1	Site 1	Site 2	Site 2
	\(n\)	\(n\;\times\;(n\;-\;1)\)	\(n\)	\(n\;\times\;(n\;-\;1)\)
Green algae	4	12	0	0
Glasswort	9	72	0	0
Cordgrass	6	30	0	0
Saltmarsh grass	18	306	68	4556
Seablite	10	90	1	0
Sea purslane	10	90	29	812
Sea spurrey	0	0	6	30
Scurvy grass	0	0	10	90
Thrift	0	0	6	30
Sea lavender	0	0	2	2
Sea aster	0	0	8	56
Sea milkwort	0	0	1	0
TOTAL	73	840	142	5686
	D = 73 (72) / 840	D = 142 (141) / 5686
	D = 6.26	D = 3.52

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 biodiversity increases with height above sea level at Abernant Saltmarsh. A systematic transect is taken with readings every 0.25m above sea level. Raw data from point quadrats is used to calculate the Simpson’s Diversity Index.

Height above sea level (m)	Simpson’s Diversity Index
0.00	0.00
0.25	2.92
0.50	6.26
0.75	4.28
1.00	3.68
1.25	3.52
1.50	4.54
1.75	4.64
2.00	3.56
2.25	2.23

Step 1. State the null hypothesis

There is no significant relationship between height above sea level and Simpson’s Diversity Index at Abernant Saltmarsh.

Step 2. Calculate the Spearman’s Rank Correlation Coefficient

(a) Rank the measurements

Height above sea level (m)	Simpson’s Diversity Index
Data	Rank	Data	Rank
0.00	1	0.00	1
0.25	2	2.92	3
0.50	3	6.26	10
0.75	4	4.28	7
1.00	5	3.68	6
1.25	6	3.52	4
1.50	7	4.54	8
1.75	8	4.64	9
2.00	9	3.56	5
2.25	10	2.23	2

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

Distance along spit (m)	Mean pebble size (cm)
Data	Rank	Data	Rank	\(D\)	\(D^2\)
0.00	1	0.00	1	0	0
0.25	2	2.92	3	1	1
0.50	3	6.26	10	7	49
0.75	4	4.28	7	3	9
1.00	5	3.68	6	1	1
1.25	6	3.52	4	2	4
1.50	7	4.54	8	1	1
1.75	8	4.64	9	1	1
2.00	9	3.56	5	4	16
2.25	10	2.23	2	8	64

(c) Calculate \(∑D^2\) i.e. the sum of the \(D^2\) column`= 146`

(d) Calculate \(r_s\)

\(r_s = 1-\frac{(6∑D^2)}{n(n^2-1)}\)\(r_s = 1-/frac{(6\;\times\;146)}{(10\;\times\;(100\;-\;1))}\)

\(r_s\) = +0.12

Step 3. Test the significance of the result

The critical value at \(p=0.05\) significance level for \(10\) pairs of measurements is \(0.648\)

Since our calculated value of \(0.12<0.648\) (ignore the minus sign), the null hypothesis cannot be rejected.

In conclusion, there is no significant relationship between height above sea level and Simpson’s Diversity Index at Abernant Saltmarsh.

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