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).
- \(D\) = 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 \(n\), \(n\times(n-1)\), \(N\) and \(D\) for each site
| Mobile dune | Mobile dune | Fixed dune | Fixed dune | |
|---|---|---|---|---|
| Species | \(n\) | \(n\times(n-1)\) | \(n\) | \(n\times(n-1)\) |
| 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
\(U_1= n_1 \times n_2 + 0.5 n_2 (n_2 + 1)\;- ∑ R_2\)\(U_2 = n_1 \times n_2 + 0.5 n_1 (n_1 + 1)\;- ∑ R_1\)- \(n_1\) 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 1 (Grazed fixed dunes) | Site 2 (Ungrazed 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 \(U_2\)
\(U_1= 10 \times 10 + 0.5 \times 10 \times (10 + 1)\;- 63 = 92\)\(U_2= 10 \times 10 + 0.5 \times 10 \times (10 + 1)\;- 147 = 8\)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.
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