Biology Fieldwork

A Level

Discussion

Fieldwork in sand dunes

4. Discussion

This section shows how you can use the Chi-squared test to compare two parts of the dunes. 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 = (Nxx(N-1)) / (Σ nxx(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 Traeth-y-goes sand dunes. Raw data from point quadrats is used. The number means the total hits per species per site.

Species Mobile dune Fixed dune
Marram grass 10 4
Sea holly 3 0
Sand fescue 1 11
Saltwort 2 0
Dandelion 0 8

Calculate `n`, `nxx(n-1)`, `N` and `D` for each site

  Mobile dune Fixed dune
Species `n` `nxx(n-1)` `n` `nxx(n-1)`
Marram grass 10 90 4 12
Sea holly 3 6 0 0
Sand fescue 1 2 11 110
Saltwort 2 2 0 0
Dandelion 0 0 8 56
TOTAL 17 100 23 178
  D = 17(16) / 100 D = 23(22) / 178
  D = 2.72 D = 2.84

The larger the value of D, the higher the species diversity. A low value of D could be due to low overall species richness (like at the strand line) or to the dominance of one species (as in dune scrub). You could present your results as a graph

Mann Whitney U test

Mann Whitney U is a statistical test that is used either to test whether there is a significant difference between the medians of two sets of data.

The Mann Whitney U test can only be used if there are at least 6 pairs of data. It does not require a normal distribution.

There are 3 steps to take when using the Mann Whitney U test

Step 1. State the null hypothesis

There is no significant difference between _______ and _______

Step 2. Calculate the Mann Whitney U statistic

`U_1= n_1 xx n_2 + 0.5 n_2 (n_2 + 1) - ∑ R_2`

`U_2 = n_1 xx n_2 + 0.5 n_1 (n_1 + 1) - ∑ R_1`

  • `n_1` is the number of values of `x_1`
  • `n_2` is the number of values of `x_2`
  • `R_1` is the ranks given to `x_1`
  • `R_2` is the ranks given to `x_2`

Step 3. Test the significance of the result

Compare the value of U against the critical value for U at a confidence level of 95% / significance value of P = 0.05.

If U is equal to or smaller than the critical value (p=0.05) the REJECT the null hypothesis. There is a SIGNIFICANT difference between the 2 data sets.

If U is greater than the critical value, then ACCEPT the null hypothesis. There is NOT a significant difference between the 2 data sets.

Worked example

A biologist is investigating whether managed areas of the sand dunes have greater species richness than unmanaged areas. A frame quadrat was used to record number of species at 10 randomly chosen points in each of two sites: one in a grazed section of the fixed dunes and one in an ungrazed section of the fixed dunes.

Here are the results.

Number of times out of 10 the pin hits vegetation
Site 1 (Grazed fixed dunes) Site 2 (Ungrazed fixed dunes)
2 2
3 3
0 6
1 4
2 5
1 3
2 3
2 2
1 4
0 3

Step 1. State the null hypothesis

There is no significant difference in species richness between the grazed and the ungrazed parts of the fixed dunes.

Step 2. Calculate Mann Whitney U statistic

(a) Give each result a rank. Calculate the sum of the ranks for the two columns.

Number of times out of 10 the pin hits vegetation
Site 1 (Grazed fixed dunes) Site 2 (Ungrazed fixed dunes)
Number Rank Number Rank
2 8.5 2 8.5
3 14 3 14
0 1.5 6 20
1 4 4 17.5
2 8.5 5 19
1 4 3 14
2 8.5 3 14
2 8.5 2 8.5
1 4 4 17.5
0 1.5 3 14

(b) Calculate `∑R_1`and `∑R_2`

`∑R_1` is the sum of the ranks in the first column (deciduous woodland) = `63`

`∑R_2` is the sum of the ranks in the first column (evergreen woodland) = `147`

`n_1 = 10` and `n_2 = 10`

(c) Calculate `U_1` and `U_2`

`U_1 = 10 xx 10 + 0.5 xx 10 xx (10 + 1) – 63 = 92`

`U_2 = 10 xx 10 + 0.5 xx 10 xx (10 + 1) – 147 = 8`

Step 3. Test the significance of the result

In this example, `U_1 = 92` and `U_2 = 8`

`U` is the smaller of the two values, so `U=8`

The critical value at `p=0.05` significance level for `n_1=10` and `n_2=10` is `23`. Since our calculated value of `8 < 23`, the null hypothesis can be rejected.

In conclusion, there is a significant difference in species richness between the grazed and the ungrazed parts of the fixed dunes.