# Biology Fieldwork

A Level

# Discussion

## 4. Discussion

This section shows how you can use the Spearman's Rank Correlation Coefficient to investigate changes with distance from open water. 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 at Woodmere. Raw data from point quadrats is used. The number means the total hits per species per site.

Site 1 | Site 2 | |
---|---|---|

`n` | `n` | |

Common reed | 22 | 0 |

Saw sedge | 14 | 0 |

Alder | 10 | 0 |

Sheep's fescue | 0 | 7 |

Common bent | 0 | 18 |

Creeping buttercup | 0 | 8 |

Soft rush | 0 | 10 |

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

Site 1 | Site 2 | |||
---|---|---|---|---|

`n` | `nxx(n-1)` | `n` | `nxx(n-1)` | |

`n` | `n` | |||

Common reed | 22 | 462 | 0 | 0 |

Saw sedge | 14 | 182 | 0 | 0 |

Alder | 10 | 90 | 0 | 0 |

Sheep's fescue | 0 | 0 | 7 | 42 |

Common bent | 0 | 0 | 18 | 306 |

Creeping buttercup | 0 | 0 | 8 | 56 |

Soft rush | 0 | 0 | 10 | 90 |

TOTAL | 46 | 734 | 43 | 494 |

D = 46 (45) / 734 | D = 43 (42) / 494 | |||

D = 2.82 | D = 3.66 |

The larger the value of D, the higher the species diversity. A low value of D could be due to low overall species richness or to the dominance of one species.

## Spearman’s Rank Correlation Test

Spearman’s Rank Correlation is a statistical test to test whether there is a significant relationship between two sets of data.

The Spearman’s Rank Correlation test can only be used if there are at least 10 (ideally at least 15-15) pairs of data.

There are 3 steps to take when using the Spearman’s Rank Correlation Test

### Step 1. State the null hypothesis

There is no significant relationship between _______ and _______

### Step 2. Calculate the Spearman’s Rank Correlation Coefficient

`r_s = 1-(6∑D^2) / (n(n^2-1))`

- `r_s` = Spearman's Rank correlation coefficient
- `D` = differences between ranks
- `n` = number of pairs of measurements

Step 3. Test the significance of the result

Compare the value of `r_s` that you have calculated *against* the critical value for `r_s` at a confidence level of 95% / significance value of p = 0.05.

If `r_s` is equal to or above the critical value (p=0.05) the REJECT the null hypothesis. There is a SIGNIFICANT relationship between the 2 variables.

A positive sign for `r_s` indicates a significant positive relationship and a negative sign indicates a significant negative relationship.

If `r_s` (ignoring any sign) is less than the critical value, ACCEPT the null hypothesis. There is NO SIGNIFICANT relationship between the 2 variables.

### Worked example

A biologist is investigating whether soil moisture decreases with distance from open water at Woodmere. A systematic transect is taken with readings every 4m from the edge of the open water. A soil pin is used to measure soil moisture levels.

Distance from open water (m) | Soil pin depth (cm) |
---|---|

0 | 13.0 |

3 | 14.5 |

6 | 12.8 |

9 | 15.5 |

12 | 14.4 |

15 | 11.0 |

18 | 8.4 |

21 | 7.0 |

24 | 8.5 |

27 | 6.5 |

30 | 8.5 |

33 | 6.5 |

36 | 5.6 |

39 | 4.0 |

42 | 5.2 |

### Step 1. State the null hypothesis

There is no significant relationship between soil moisture and distance from open water at Woodmere

### Step 2. Calculate the Spearman’s Rank Correlation Coefficient

(a) Rank the measurements

Distance from open water (m) | Soil pin depth (cm) | ||
---|---|---|---|

Data | Rank | Data | Rank |

0 | 1 | 13.0 | 12 |

3 | 2 | 14.5 | 14 |

6 | 3 | 12.8 | 11 |

9 | 4 | 15.5 | 15 |

12 | 5 | 14.4 | 13 |

15 | 6 | 11.0 | 10 |

18 | 7 | 8.4 | 7 |

21 | 8 | 7.0 | 6 |

24 | 9 | 8.5 | 8.5 |

27 | 10 | 6.5 | 4.5 |

30 | 11 | 8.5 | 8.5 |

33 | 12 | 6.5 | 4.5 |

36 | 13 | 5.6 | 3 |

39 | 14 | 4.0 | 1 |

42 | 15 | 5.2 | 2 |

(b) Calculate `D` and `D^2`

Distance from open water (m) | Soil pin depth (cm) | ||||
---|---|---|---|---|---|

Data | Rank | Data | Rank | `D` | `D^2` |

0 | 1 | 13.0 | 12 | 11 | 121 |

3 | 2 | 14.5 | 14 | 12 | 144 |

6 | 3 | 12.8 | 11 | 8 | 64 |

9 | 4 | 15.5 | 15 | 11 | 121 |

12 | 5 | 14.4 | 13 | 8 | 64 |

15 | 6 | 11.0 | 10 | 4 | 16 |

18 | 7 | 8.4 | 7 | 0 | 0 |

21 | 8 | 7.0 | 6 | 2 | 4 |

24 | 9 | 8.5 | 8.5 | 0.5 | 0.25 |

27 | 10 | 6.5 | 4.5 | 5.5 | 30.25 |

30 | 11 | 8.5 | 8.5 | 2.5 | 6.25 |

33 | 12 | 6.5 | 4.5 | 7.5 | 56.25 |

36 | 13 | 5.6 | 3 | 10 | 100 |

39 | 14 | 4.0 | 1 | 13 | 169 |

42 | 15 | 5.2 | 2 | 13 | 169 |

(c) Calculate `∑D^2`i.e. the sum of the `D^2`column`= 1065`

(d) Calculate `r_s`

`r_s = 1-(6∑D^2) / (n(n^2-1))`

`r_s = 1-(6xx1065) / (15xx(225-1))`

`r_s = -0.90`

### Step 3. Test the significance of the result

The critical value at `p=0.05` significance level for `15` pairs of measurements is `0.521`

Since our calculated value of `0.90<0.521` (ignore the minus sign), the null hypothesis is rejected.

In conclusion, there is a significant relationship between soil moisture and distance from open water at Woodmere