Used for:

- A significant association between 2 sets of measurements from the same sample

5-30 pairs of measurements**Method**

- Create a table with the headings

Variable 1 |
Rank 1 |
Variable 2 |
Rank 2 |
Difference (d) |
d^{2} |

If two or more variables are the same, add the rank positions the number would have if they were the next ones in the sequence.

For example, if you got 2 ‘4’s in a row

Variable 1 |
Rank 1 |

5 |
2 |

4 |
1.5 (1*) |

4 |
1.5 (2*) |

For their rank, add 1+2 (*their standard rank) = 3

Then divide 2 by how many there are, which is also 2

So 3 ÷ 2 = 1.5

- if they are however the next numbers in the sequence, then divide by the number of positions used
- calculate the difference (d) between the two ranks
- square the difference of each
- Add up all the values of d
^{2 }to get the ∑ - Multiply the ∑ by 6
- To find n, it is the amount of data (the highest rank)
- Cube the value of n
- Subtract the value of n
^{3 }by the original value of n

The value will be between -1 (perfect negative correlation) and + 1 (perfect positive correlation). The closer the value is to 0, the weaker the correlation

e.g. “Some students carried out a belt transect across some sand dunes. They counted the number of banded snails they found as they moved away from the shoreline

Distance from Shore (m) | Number of banded snails |

0 | 10 |

5 | 2 |

10 | 8 |

15 | 7 |

20 | 5 |

25 | 8 |

30 | 2 |

results

Variable 1 |
Rank 1 |
Variable 2 |
Rank 2 |
Difference (d) |
d^{2} |

0 | 1 | 10 | 7 | -6 | 36 |

5 | 2 | 2 | 1.5 | 0.5 | 0.25 |

10 | 3 | 8 | 5.5 | -2.5 | 6.25 |

15 | 4 | 7 | 4 | 0 | 0 |

20 | 5 | 5 | 3 | 2 | 4 |

25 | 6 | 8 | 5.5 | 0.5 | 0.25 |

30 | 7 | 2 | 1.5 | 5.5 | 30.25 |

Total ∑ | 77 |

∑ = 77

= -0.375

As n = 7 (as there are 7 variables) and we use the 0.05 grid, the result is:

Critical value: 0.6786

The correlation is not significant as there is a 5% chance the results are due to chance, therefore we agree with the null hypothesis

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