16. Non-parametric Tests

16.7 Spearman Rank Correlation Coefficient

This is a rank alternative to the Pearson correlation coefficient that may be used when the assumption of normality is not met for hypothesis testing. It is defined by

    \[ r_{s} = 1 - \frac{6 \sum_{i=1}^{n} d_{i}^{2}}{n(n^{2}-1)} \]

where

    \begin{eqnarray*} n & = & \mbox{sample size}\\ \\ d_{i} & = & \mbox{difference in ranks for data point } i \\ & = & r_{x_{i}} - r_{y_{i}} \end{eqnarray*}

where r_{x_{i}} = x rank of point i and r_{y_{i}} = y rank of point ii.

To test H_{0}: r_{s} = 0 versus H_{1}: r_{s} \neq 0 use r_{s} itself as the test statistic and r_{s,\mbox{crit}} from the Rank Correlation Coefficient Critical Values Table as the critical statistic. (Note that the Rank Correlation Coefficient Critical Values Table requires n < 30.) Reject H_{0} if r_{s} > r_{s,\mbox{crit}}.

 

Example 16.9 : Determine if the Spearman correlation between two textbook ratings, data given below, is significant.

Book rating 1 (x) rating 2 (y) rank x rank y d d^{2}
A 4 4 2 1 1 1
B 10 6 5 2 3 9
C 18 20 7 8 -1 1
D 20 14 8 6 2 4
E 12 16 6 7 -1 1
F 2 8 1 4 -3 9
G 5 11 3 5 -2 4
H 9 7 4 3 1 1
n=8 \sum d^{2} = 30

Note the preliminary data reduction (ranking and rank differences, d) done to the right side of the table.

1. Hypothesis.

H_{0}: r_{s} = 0
H_{1}: r_{s} \neq 0

(Note that population values are inferred in the hypotheses statement.)

2. Critical statistic.

From the Rank Correlation Coefficient Critical Values Table with \alpha = 0.05 and n = 8 find

    \[ r_{s,\mbox{crit}} = 0.738 \]

3. Test statistic.

    \[ r_{s} = 1 - \frac{6 \sum_{i=1}^{n} d_{i}^{2}}{n(n^{2}-1)} = 1 - \frac{(6) (30)}{8(8^{2}-1)} = 0.643 \]

4. Decide.

    \[ (r_{s} = 0.643) < (r_{s,\mbox{crit}} = 0.738) \]

so do not reject H_{0}.

5. Interpretation.

There is no significant correlation between the ratings.

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