Used for:

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

5-30 pairs of measurementsMethod 1. 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

1. if they are however the next numbers in the sequence, then divide by the number of positions used
2. calculate the difference (d) between the two ranks
3. square the difference of each
4. Add up all the values of dto get the ∑
5. Multiply the ∑ by 6
6. To find n, it is the amount of data (the highest rank)
7. Cube the value of n
8. Subtract the value of nby 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 *we use 0.05 as that is a 5% factor of error

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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