Used for:
- A significant association between 2 sets of measurements from the same sample
5-30 pairs of measurementsMethod
- Create a table with the headings
| Variable 1 | Rank 1 | Variable 2 | Rank 2 | Difference (d) | d2 |
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 d2 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 n3 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) | d2 |
| 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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