Sand Dune Succession at Studland Beach

July 2010 Environmental News Worksheet

Sand dune succession at Studland Beach

The following data were collected by students on a field trip to Studland Beach on the English south coast. The table shows a summary of slope angle, percentage cover of marram grass and percentage cover of heather. The students took three transects through the dunes. The averaged-results are shown in the table.

Distance from the sea / m / Average slope angle / ˚ / Percentage cover of marram grass / % / Percentage cover of heather / %
0 / 5 / 0 / 0
10 / 15 / 37 / 0
20 / 5 / 63 / 0
30 / 5 / 48 / 0
40 / 3 / 8 / 34
50 / -7 / 0 / 43
60 / 3 / 0 / 100
70 / -5 / 0 / 100
80 / -2 / 10 / 35
90 / 0 / 0 / 68

Questions

(i)Using the data in Table A, and the grid provided, plot the dune profile and the kite diagrams for the percentage cover of marram and the percentage cover of heather. [9 marks]

(ii)Compare the distribution of marram with that of heather. [4 marks]

(iii)Comment on the usefulness of kite diagrams for showing distribution of vegetation. [3 marks]

(iv)Work out a Spearman’s Rank Correlation Coefficient (see Environmental Systems and Societies pp. 340–343) for the relationship between marram grass and heather i.e. is there a correlation between the distribution of marram grass and the distribution of heather. [6 marks]

Answers

(i)As on the artwork below – 3 marks for each graph. [9 max]

(ii)Marram is located closer to the sea – this is because it is better able to cope with the windy, salty, arid conditions associated with the shoreline. In contrast, heather dominates away from the shoreline. There is one anomaly at 80m where the proportion of heather falls and there is a slight increase in marram. [4 max]

(iii)Kite diagrams are very useful for showing the relative location/distribution of a large number of species, although only two are shown here. There could be 8–10 different species shown on a single kite diagram (see Environmental Systems and Societies, pg 348). One disadvantage is that many students find the split scale (100% = 50% above the line and 50% below the line) difficult to comprehend. [3 max]

(iv)The formula for Spearman’s rank Correlation Coefficient (Rs) is Rs = 1 – (6Σd2/n3-n) Where d is the difference in ranks and n is the number of observations. Thus:

Marram / % / Heather / % / Rank marram / Rank heather / Difference in ranks / Difference2
0 / 0 / =8 / =8.5 / 0.5 / 0.25
37 / 0 / 3 / =8.5 / 5.5 / 30.25
63 / 0 / 1 / =8.5 / 7.5 / 56.25
48 / 0 / 2 / =8.5 / 6.5 / 42.25
8 / 34 / 5 / 6 / 1 / 1
0 / 43 / =8 / 4 / 4 / 16
0 / 100 / =8 / =1.5 / 6.5 / 42.25
0 / 100 / =8 / =1.5 / 6.5 / 42.25
10 / 35 / 4 / 5 / 1 / 1
0 / 68 / =8 / 3 / 5 / 25

Σd2 = 256.5

Thus, Rs = 1-(6x 256.5/103 – 10) = 1 – (1539/990) = 1-1.55 = -0.55[4 marks]

The result tells us that there is a negative correlation, as one species increases the other decreases. When we check the Tables of Statistical Significance we find that the 95% level of significance is 0.56 and the 99% level of significance is 0.75. So, with our result of 0.55, we cannot claim that the results are statistically significant i.e. there is a correlation but it is not statistically significant.

[2 marks]

TOK

1539/990 (as in the computation above) gives an answer of 1.554545 recurring. If you round up the number 5, you get an answer of 1.56 to 2 d.p.

1-1.56 gives an answer of -0.56, which is statistically significant!