This section shows how you can use the Chi-squared test to compare two parts of the dunes.

## 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 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 \(n\), \(n x (n-1)\), \(N\) and \(D\) for each site

MOBILE DUNE | MOBILE DUNE | FIXED DUNE | FIXED DUNE | |
---|---|---|---|---|

Species | \(n\) | \(n x (n-1)\) | \(n\) | \(n x (n-1)\) |

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.

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.

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 \(U_2\)

\(U_1 = 10\; \times\; 10\; +\; 0.5\; \times\; 10\; \times\; (10\; +\; 1)\; – \;63 = 92\) \(U_2 = 10\; \times\; 10\; +\; 0.5\; \times\; 10\; \times\; (10\; +\; 1)\; – \;147 = 8\)### Step 3. Test the significance of the result

In this example, \(U_1 = 92\) and \(U_2 = 8\)

We select the smaller of the values. Here, \(U_2\) is the smaller of the two values, so \(U=8\)

Use a set of Mann-Whitney U critical value tables (example here) to find the critical value at the 95% confidence level.

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.