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Stage 4: Data analysis

Stage 4: Data presentation | Statistics

Data presentation

The kite diagram below shows the results of a transect conducted at a saltmarsh in south Wales.

Statistics

Spearman's Rank Correlation Coefficient

You can use the Spearman's Rank Correlation Coefficient is used to investigate how vegetation changes along an environmental gradient. Need more information about this test?

Worked example

A point quadrat was used to sample vegetation along an interrupted belt transect running from the sea-shore inland. 100 hits were made at each station on the transect. The results are shown below. The Simpson's Diversity Index has been calculated for each of the 12 sites.

Site	1	2	3	4	5	6	7	8	9	10	11	12
Bladder wrack	0	0	13	1	0	0	0	0	0	0	0	0
bramble	0	0	0	0	0	0	0	0	0	1	2	23
Cocksfoot grass	0	0	0	0	0	0	13	23	12	0	5	19
Common scurvy grass	0	0	0	0	0	1	0	0	0	0	0	0
Cord grass	0	7	53	41	17	0	0	0	0	0	0	0
Glasswort	0	16	13	0	0	0	0	0	0	0	0	0
Green algae	4	22	27	5	18	19	0	0	0	0	0	0
Meadow grass	0	0	0	0	0	1	21	39	25	39	61	71
Moss	0	0	0	0	0	0	0	0	0	0	0	3
Nettle	0	0	0	0	0	0	0	0	0	0	0	1
Parsley water-dropwort	0	0	0	0	0	0	0	0	0	7	0	0
Red fescue Grass	0	0	0	0	0	0	76	80	81	0	0	4
Saltmarsh grass	0	0	13	0	0	5	0	0	0	0	0	0
Sea beet	0	0	0	0	0	0	9	0	0	1	0	0
Sea lavender	0	0	0	0	0	0	2	0	0	0	0	0
Sea plantain	0	0	0	0	0	0	0	0	0	0	0	0
Sea purslane	0	0	0	60	86	88	8	10	0	11	0	0
Sea wormwood	0	0	0	0	0	0	0	0	0	0	13	7
Thistle	0	0	0	0	0	0	0	0	1	8	3	0
Vetch	0	0	0	0	0	0	0	0	0	2	0	0
Yarrow	0	0	0	0	0	0	0	0	0	5	0	0
Yorkshire fog grass	0	0	0	0	0	0	0	15	7	80	75	42
Wracks	13	0	0	0	0	1	0	0	0	0	0	0
Sea aster	0	0	0	0	5	0	0	0	0	0	0	0
Moss	0	0	0	0	0	0	0	0	0	0	0	6
Cinqufoil	0	0	0	0	0	0	0	0	0	0	0	3
Sea carrot	0	0	0	0	0	0	0	0	0	10	11	0
Bare ground	97	69	9	7	0	0	0	0	0	0	0	0
Simpson Yule Species Diversity Index	0.360	0.610	0.714	0.536	0.494	0.385	0.607	0.685	0.535	0.692	0.665	0.756

A bar chart of the species diversity results is shown below.

Is there a change in species diversity with distance inland in this saltmarsh? You can use the Spearman's Rank correlation coefficient.

The null hypothesis is that there is no significant correlation between diatance inland and Simpson's Diversity Index.

Rank the data

Site	rank	Simpson's Index	rank
1	1	0.360	1
2	2	0.610	7
3	3	0.714	11
4	4	0.536	5
5	5	0.494	3
6	6	0.385	2
7	7	0.607	6
8	8	0.685	9
9	9	0.535	4
10	10	0.692	10
11	11	0.665	8
12	12	0.756	12

Calculate the difference (d) and d2.

Site	rank	Simpson's Index	rank	d	d2
1	1	0.360	1	1	1
2	2	0.610	7	5	25
3	3	0.714	11	8	64
4	4	0.536	5	1	1
5	5	0.494	3	2	4
6	6	0.385	2	4	16
7	7	0.607	6	1	1
8	8	0.685	9	1	1
9	9	0.535	4	5	25
10	10	0.692	10	0	0
11	11	0.665	8	3	9
12	12	0.756	12	0	0
Total =					150

rs = 1 - (6 x 150) / 12 x 143

rs = 1 - 900/1716

rs = 0.52

In our example, at the p=0.05 probability level where degrees of freedom = 11, the critical value of rs = 0.618

Since 0.52 < 0.168, the null hypothesis is not rejected at the p=0.05 level.

Therefore, we are 95% certain that there is no correlation between distance inland and Simpson's Diversity Index.

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