# Biology Fieldwork

A Level

# Discussion

## 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.