# 18.8: Rank Randomization for Association

Skills to Develop

• Compute Spearman's $$ρ$$
• Test Spearman's $$ρ$$ for significance

The rank randomization test for association is equivalent to the randomization test for Pearson's r except that the numbers are converted to ranks before the analysis is done. Table $$\PageIndex{1}$$ shows $$5$$ values of $$X$$ and $$Y$$. Table $$\PageIndex{2}$$ shows these same data converted to ranks (separately for $$X$$ and $$Y$$).

Table $$\PageIndex{1}$$: Example data

X Y
1.0 1.0
2.4 2.0
3.8 2.3
4.0 3.7
11.0 2.5

Table $$\PageIndex{2}$$: Ranked data

X Y
1 1
2 2
3 3
4 5
5 4

The approach is to consider the $$X$$ variable fixed and compare the correlation obtained in the actual ranked data to the correlations that could be obtained by rearranging the $$Y$$ variable ranks. For the ranked data shown in Table $$\PageIndex{2}$$, the correlation between $$X$$ and $$Y$$ is $$0.90$$. The correlation of ranks is called "Spearman's $$ρ$$."

Table $$\PageIndex{3}$$: Ranked data with correlation of $$1.0$$

X Y
1 1
2 2
3 3
4 4
5 5

There is only one arrangement of $$Y$$ that produces a higher correlation than $$0.90$$: A correlation of $$1.0$$ results if the fourth and fifth observations' $$Y$$ values are switched (see Table $$\PageIndex{3}$$). There are also three other arrangements that produce an $$r$$ of $$0.90$$ (see Tables $$\PageIndex{4}$$, $$\PageIndex{5}$$, and $$\PageIndex{6}$$). Therefore, there are five arrangements of $$Y$$ that lead to correlations as high or higher than the actual ranked data (Tables $$\PageIndex{2}$$ through $$\PageIndex{6}$$).

Table $$\PageIndex{4}$$: Ranked data with correlation of $$0.90$$

X Y
1 1
2 2
3 4
4 3
5 5

Table $$\PageIndex{5}$$: Ranked data with correlation of $$0.90$$

X Y
1 1
2 3
3 2
4 4
5 5

Table $$\PageIndex{6}$$: Ranked data with correlation of $$0.90$$

X Y
1 2
2 1
3 3
4 4
5 5

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 $$5/120 = 0.042$$. Note that this is a one-tailed probability since it is the proportion of arrangements that give a correlation as large or larger. The two-tailed probability is $$0.084$$.

Since it is hard to count up all the possibilities when the sample size is even moderately large, it is convenient to have a table of critical values.

Table of critical values for Spearman's $$\rho$$

From the table linked to above, you can see that the critical value for a one-tailed test with $$5$$ observations at the $$0.05$$ level is $$0.90$$. Since the correlation for the sample data is $$0.90$$, the association is significant at the $$0.05$$ level (one-tailed). As shown above, the probability value is $$0.042$$. Since the critical value for a two-tailed test is $$1.0$$, Spearman's $$ρ$$ is not significant in a two-tailed test.

### Contributor

• Online Statistics Education: A Multimedia Course of Study (http://onlinestatbook.com/). Project Leader: David M. Lane, Rice University.