1. 14 Elements of Nonparametric Statistics
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14.5
Rank Correlation
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3. Rank Correlation
Charles Spearman developed the rank correlation coefficient in the early 1900s. It is a nonparametric alternative to the linear correlation coefficient (Pearson’s product moment, r)
The Spearman rank correlation coefficient, rs, is found by using this formula:

4. Rank Correlation
Where di is the difference in the paired rankings and n is the number of pairs of data. The value of rs will range from –1 to 1 and will be used in much the same manner as Pearson’s linear correlation coefficient, r, was used.
The Spearman rank coefficient is defined by using formula (3.1), with data rankings substituted for quantitative x and y values.

5. Rank Correlation
The original data may be rankings, or if the data are quantitative, each variable must be ranked separately; then the rankings are used as pairs.
If there are no ties in the rankings, formula (14.11) is equivalent to formula (3.1). Formula (14.11) provides us with an easier procedure to use for calculating the rs statistic.
Assumptions for inferences about rank correlation The n ordered pairs of data form a random sample, and the variables are ordinal or numerical.

6. Rank Correlation
The null hypothesis that we will be testing is: “There is no correlation between the two rankings.” The alternative hypothesis may be either two-tailed—there is correlation—or one-tailed if we anticipate either positive or negative correlation.
The critical region will be on the side (s) corresponding to the specific alternative that is expected. For example, if we suspect negative correlation, then the critical region will be in the left-hand tail.

7. Example 13 – Calculating the Spearman Rank Correlation Coefficient
Let’s consider a hypothetical situation in which four judges rank five contestants in a contest.
Let’s identify the judges as A, B, C, and D and the contestants as a, b, c, d, and e. Table 14.8 lists the awarded rankings.
Rankings for Five Contestants
Table 14.8

8. Example 13 – Calculating the Spearman Rank Correlation Coefficient
cont’d
When we compare judges A and B, we see that they ranked the contestants in exactly the opposite order: perfect disagreement (see Table 14.9).
From our previous work with correlation, we expect the calculated value for rs to be exactly –1 for these data.
Rankings of A and B
Table 14.9

9. Example 13 – Calculating the Spearman Rank Correlation Coefficient
cont’d
We have:
When judges A and C are compared, we see that their rankings of the contestants are identical (see Table 14.10).
Rankings of A and C
Table 14.10

10. Example 13 – Calculating the Spearman Rank Correlation Coefficient
cont’d
We would expect to find a calculated correlation coefficient of +1 for these data:

11. Example 13 – Calculating the Spearman Rank Correlation Coefficient
cont’d
By comparing the rankings of judge A with those of judge B and then with those of judge C, we have seen the extremes: total agreement and total disagreement.
Now let’s compare the rankings of judge A with those of judge D (see Table 14.11). There seems to be no real agreement or disagreement here.
Rankings of A and D
Table 14.11

12. Example 13 – Calculating the Spearman Rank Correlation Coefficient
cont’d
Let’s compute rs:
The result is fairly close to zero, which is what we should have suspected, since there was no real agreement or disagreement.

13. Rank Correlation
The test of significance will result in a failure to reject the null hypothesis when rs is close to zero; the test will result in a rejection of the null hypothesis when rs is found to be close to +1 or –1.
The critical values in Table 15 in Appendix B are the positive critical values only.
Since the null hypothesis is “The population correlation coefficient is zero (i.e., s = 0),” we have a symmetrical test statistic.

14. Rank Correlation
Hence we need only add a plus or minus sign to the value found in the table, as appropriate. The sign is determined by the specific alternative that we have in mind.
When there are only a few ties, it is common practice to use formula (14.11).
Even though the resulting value of rs is not exactly equal to the value that would occur if formula (3.1) were used, it is generally considered to be an acceptable estimate.

15. Rank Correlation
Example 14 shows the procedure for handling ties and uses formula (14.11) for the calculation of rs.
When ties occur in either set of the ordered pairs of rankings, assign each tied observation the mean of the ranks that would have been assigned had there been no ties, as was done for the Mann–Whitney U test.

16. Example 14 – One-tailed Hypothesis Test
Students who finish exams more quickly than the rest of the class are often thought to be smarter. Table presents the scores and order of finish for 12 students on a recent 1-hour exam.
At the 0.01 level, do these data support the alternative hypothesis that the first students to complete an exam have higher grades?
Data on Exam Scores [TA14-12]
Table 14.12

17. Example 14 – Solution
Step 1 a. Parameter of interest: The rank correlation coefficient between score and order of finish, s
b. Statement of hypotheses:
Ho: Order of finish has no relationship to exam score.
Ha: The first to finish tend to have higher grades.
Step 2 a. Assumptions: The 12 ordered pairs of data form a random sample; order of finish is an ordinal variable and test score is numerical.

18. Example 14 – Solution
cont’d
b. Test statistic: The Spearman rank correlation coefficient, rs
c. Level of significance:  = 0.01 for a one-tailed test
Step 3 a. Sample information: The data are given in Table
Data on Exam Scores [TA14-12]
Table 14.12

19. Example 14 – Solution
cont’d
b. Test statistic: Rank the scores from highest to lowest, assigning the highest score the rank number 1, as shown. (Order of finish is already ranked.)
The rankings and preliminary calculations are shown in Table
Rankings of Test Scores and Differences
Table 14.13

20. Example 14 – Solution Using formula (14.11), we obtain Thus, = 0.503
cont’d
Using formula (14.11), we obtain
Thus, = 0.503

21. Example 14 – Solution Step 4 Probability Distribution: p-Value:
cont’d
Step 4 Probability Distribution:
p-Value:
a. Since the concern is for “positive” values, the p-value is the area to the right: P = P (rs  for n = 12)

22. Example 14 – Solution To find the p-value, you have two options:
cont’d
To find the p-value, you have two options:
1. Use Table 15 (Appendix B) to place bounds on the p- value. Table 15 lists only two-tailed  (this test is one- tailed, so divide the column heading by 2): < P < 0.05.
2. Use a computer or calculator to find the p-value: P =
b. The p-value is not smaller than  .

23. Example 14 – Solution Classical:
cont’d
Classical:
The critical region is one-tailed because Ha expresses concern for values related to “positive.” Since the table is for two-tailed, the critical value is located at the intersection of the  = 0.02 column ( = 0.01 in each tail) and the n = 12 row of table 15 :
b is not in the critical region, as shown in the figure.

24. Example 14 – Solution Step 5 a. Decision: Fail to reject Ho. cont’d
b. Conclusion: These sample data do not provide sufficient evidence to enable us to conclude that the first students to finish have higher grades, at the 0.01 level of significance.

25. Rank Correlation
Calculating the p-Value for the Spearman Rank Correlation Test
Method 1: Use Table 15 in Appendix B to place bounds on the p-value. By inspecting the n = 12 row of Table 15, you can determine an interval within which the p-value lies.
Locate the value of rs along the n = 12 row and read the bounds from the top of the table. Table 15 lists only two-tailed values (therefore, you must divide by 2 for a one-tailed test). We find < P < 0.05.

26. Rank Correlation
Method 2: If you are doing the hypothesis test with the aid of a computer or graphing calculator, most likely it will calculate the p-value for you.