Spearman Correlation - Example

Example of Statistical Test using the Spearman Rank-Difference Correlation Coefficient

Is there a significant positive correlation between the rankings of 10 children on a reading test and their teacher's ranking of their reading ability? In this problem we are relating a set of scores (interval level of measurement) with the teacher's ranking of the children in reading (ordinal level of measurement). To do this we first convert the reading test scores to ranks by assigning the highest score a rank of 1, the next highest a rank of 2, etc. Now we are looking at rankings on two variables and can use the Spearman Rank-Difference Correlation Coefficient to test the significance of the relationship. The two set of ranks, as well as the difference between the pairs of ranks (D) and the differences squared (D²), are shown in the following table.

Worksheet to Calculate the Correlation between Students Ranks on a Reading Test and Teacher's Ranking of the Students on Reading

Reading Test Score Rank	Teacher's Ranking on Reading	D	D2
1	3	-2	4
2	2	0	0
3	1	2	4
4	4	0	0
5	5	0	0
6	6	0	0
7	8	-1	1
8	7	1	1
9	10	-1	1
10	9	1	1
Total			12

From the table we can see that:

df = N - 2 = 10 - 2 = 8

We now have the information we need to complete the six step process for testing statistical hypotheses for our research problem.

1. State the null hypothesis and the alternative hypothesis based on your research question.

   H₀: r_S = 0

   H₁: r_S > 0

   Note: Our null hypothesis states that there is no significant relationship between the two variables. The alternative hypothesis states that there is a significant positive correlation between the two variables.

2. Set the alpha level.
   
   Note: As usual we will set our alpha level at .05, we have 5 chances in 100 of making a type I error.
3. Calculate the value of the appropriate statistic. Also indicate the degrees of freedom for the statistical test if necessary.

   r_S = .93

   df = N - 2 = 10 - 2 = 8

4. Write the decision rule for rejecting the null hypothesis.

   Reject H₀ if r_S >= .549

   Note: To write the decision rule we had to know the critical value for r_S, with an alpha level of .05, and 8 degrees of freedom. We can do this by looking at Appendix Table B, this is the same table we used for the Pearson r, and noting the tabled value for the column for the .10 level and the row for 8 df (.549).

   Note: We used the .10 column because we are doing a one-tailed test with an alpha of .05 As noted in our problem above the the Pearson r, in the table of critical values for r, the .10 column is used for alpha = .10 (two-tailed test) and for alpha = .05 (one-tailed test).

5. Write a summary statement based on the decision.
   Reject H₀, p < .05, one-tailed
   Note: Since our calculated value of r_S (.93) is greater than .549, we reject the null hypothesis and accept the alternative hypothesis.
6. Write a statement of results in standard English.
   There is a significant positive correlation between the children's ranks on a reading test and their teacher's ranking of them on reading.