18.4: Randomization Association
- Page ID
- 2197
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- Compute a randomization test for Pearson's \(r\)
A significance test for Pearson's \(r\) is described in the section inferential statistics for \(b\) and \(r\). The significance test described in that section assumes normality. This section describes a method for testing the significance of \(r\) that makes no distributional assumptions.
| X | 1.0 | 2.4 | 3.8 | 4.0 | 11.0 |
|---|---|---|---|---|---|
| Y | 1.0 | 2.0 | 2.3 | 3.7 | 2.5 |
The approach is to consider the \(X\) variable fixed and compare the correlation obtained in the actual data to the correlations that could be obtained by rearranging the \(Y\) variable. For the data shown in Table \(\PageIndex{1}\), the correlation between \(X\) and \(Y\) is \(0.385\). There is only one arrangement of \(Y\) that would produce a higher correlation. This arrangement is shown in Table \(\PageIndex{2}\) and the \(r\) is \(0.945\). Therefore, there are two arrangements of \(Y\) that lead to correlations as high or higher than the actual data.
| X | Y |
|---|---|
| 1.0 | 1.0 |
| 2.4 | 2.0 |
| 3.8 | 2.3 |
| 4.0 | 2.5 |
| 11.0 | 3.7 |
The next step is to calculate the number of possible arrangements of \(Y\). The number is simply \(N!\), where \(N\) is the number of pairs of scores. Here, the number of arrangements is \(5! = 120\). Therefore, the probability value is \(2/120 = 0.017\). Note that this is a one-tailed probability since it is the proportion of arrangements that give an \(r\) as large or larger. For the two-tailed probability, you would also count arrangements for which the value of \(r\) were less than or equal to \(-0.385\). In randomization tests, the two-tailed probability is not necessarily double the one-tailed probability.