This section shows how you can use the Spearman’s Rank Correlation Coefficient to investigate changes with distance inland from the seashore. 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).

\(D = \frac{(N\;\times\;(N\;-\;1))}{(Σn\;\times\;(n\;-\;1))}\)- \(D\) = 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 at Abernant Saltmarsh. Raw data from point quadrats is used. The number means the total hits per species per site.

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

\(n\) | \(n\) | |

Green algae | 4 | 0 |

Glasswort | 9 | 0 |

Cordgrass | 6 | 0 |

Saltmarsh grass | 18 | 68 |

Seablite | 10 | 1 |

Sea purslane | 10 | 29 |

Sea spurrey | 0 | 6 |

Scurvy grass | 0 | 10 |

Thrift | 0 | 6 |

Sea lavender | 0 | 2 |

Sea aster | 0 | 8 |

Sea milkwort | 0 | 1 |

Calculate \(n\), \(n\;\times\;(n\;-\;1)\), \(N\) and \(D\) for each site

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

\(n\) | \(n\;\times\;(n\;-\;1)\) | \(n\) | \(n\;\times\;(n\;-\;1)\) | |

Green algae | 4 | 12 | 0 | 0 |

Glasswort | 9 | 72 | 0 | 0 |

Cordgrass | 6 | 30 | 0 | 0 |

Saltmarsh grass | 18 | 306 | 68 | 4556 |

Seablite | 10 | 90 | 1 | 0 |

Sea purslane | 10 | 90 | 29 | 812 |

Sea spurrey | 0 | 0 | 6 | 30 |

Scurvy grass | 0 | 0 | 10 | 90 |

Thrift | 0 | 0 | 6 | 30 |

Sea lavender | 0 | 0 | 2 | 2 |

Sea aster | 0 | 0 | 8 | 56 |

Sea milkwort | 0 | 0 | 1 | 0 |

TOTAL | 73 | 840 | 142 | 5686 |

D = 73 (72) / 840 | D = 142 (141) / 5686 | |||

D = 6.26 | D = 3.52 |

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

\(r_s = 1-\frac{(6∑D^2)}{n(n^2-1)}\)- \(r_s\) = 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 biodiversity increases with height above sea level at Abernant Saltmarsh. A systematic transect is taken with readings every 0.25m above sea level. Raw data from point quadrats is used to calculate the Simpson’s Diversity Index.

Height above sea level (m) | Simpson’s Diversity Index |
---|---|

0.00 | 0.00 |

0.25 | 2.92 |

0.50 | 6.26 |

0.75 | 4.28 |

1.00 | 3.68 |

1.25 | 3.52 |

1.50 | 4.54 |

1.75 | 4.64 |

2.00 | 3.56 |

2.25 | 2.23 |

### Step 1. State the null hypothesis

There is no significant relationship between height above sea level and Simpson’s Diversity Index at Abernant Saltmarsh.

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

(a) Rank the measurements

Height above sea level (m) | Simpson’s Diversity Index | ||
---|---|---|---|

Data | Rank | Data | Rank |

0.00 | 1 | 0.00 | 1 |

0.25 | 2 | 2.92 | 3 |

0.50 | 3 | 6.26 | 10 |

0.75 | 4 | 4.28 | 7 |

1.00 | 5 | 3.68 | 6 |

1.25 | 6 | 3.52 | 4 |

1.50 | 7 | 4.54 | 8 |

1.75 | 8 | 4.64 | 9 |

2.00 | 9 | 3.56 | 5 |

2.25 | 10 | 2.23 | 2 |

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

Distance along spit (m) | Mean pebble size (cm) | ||||
---|---|---|---|---|---|

Data | Rank | Data | Rank | \(D\) | \(D^2\) |

0.00 | 1 | 0.00 | 1 | 0 | 0 |

0.25 | 2 | 2.92 | 3 | 1 | 1 |

0.50 | 3 | 6.26 | 10 | 7 | 49 |

0.75 | 4 | 4.28 | 7 | 3 | 9 |

1.00 | 5 | 3.68 | 6 | 1 | 1 |

1.25 | 6 | 3.52 | 4 | 2 | 4 |

1.50 | 7 | 4.54 | 8 | 1 | 1 |

1.75 | 8 | 4.64 | 9 | 1 | 1 |

2.00 | 9 | 3.56 | 5 | 4 | 16 |

2.25 | 10 | 2.23 | 2 | 8 | 64 |

(c) Calculate \(∑D^2\) i.e. the sum of the \(D^2\) column`= 146`

(d) Calculate \(r_s\)

\(r_s = 1-\frac{(6∑D^2)}{n(n^2-1)}\)\(r_s = 1-/frac{(6\;\times\;146)}{(10\;\times\;(100\;-\;1))}\)\(r_s\) = +0.12

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

The critical value at \(p=0.05\) significance level for \(10\) pairs of measurements is \(0.648\)

Since our calculated value of \(0.12<0.648\) (ignore the minus sign), the null hypothesis cannot be rejected.

In conclusion, there is no significant relationship between height above sea level and Simpson’s Diversity Index at Abernant Saltmarsh.