This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ology. 207 APPENDIX 3: SPEARMAN TEST FOR CORRELATION
The Spearman’s correlation coefficient is frequently used as alternative to the
standard (Pearson) correlation coefficient when only ordinal data are available.
Spearman’s rank order correlation coefficient is mathematically equivalent to the
Pearson correlation coefficient computed on ranks instead of scores. For the Spearman test, the scores on each variable must first be rank ordered with the lowest
score being assigned a rank of 1. The Spearman’s rank order test assumes both
variables are continuous and ordinal and that the data are independent. The procedure
usually followed is:
1. Select the significance level, α, and decide whether a one or two tailed test is
required. 2. Rank-order the scores on each variables separately and subtract each ranking on
the response variable from its associated ranking on the explanatory variable.
Compute D, the sum of the squares of the resulting differences. Find n, the
number of scores on either variable.
rs=1 - 6ΣD2
4. If n is larger than 30, use the normal approximation.
5. From the table in Appendix 5, find the relevant critical value(s) of r s and make
decision as specified at the top of the table in Appendix 5 Adapted from Leach, 1979. 208 APPENDIX 4 – THE KRUSKAL WALLIS AND WILCOXON RANK SUM TESTS
1. KRUSKAL WALLIS TEST
The Kruskal Wallis test is a direct generalisation of the Wilcoxon Rank Sum Test to
three, or more independent samples. The test attempts to decide whether the samples
of scores come from the same population (the null hypothesis) or from several
populations that differ in location (the alternative hypothesis). It assumes that the data
are independent and that the scores on the response variable consist of continuous
ordinal data. The procedure usually adopted is: 1. Select the significance level, α 2. Find n, the total number of scores, and t i, the number of scores in the ith sample.
Check that Σ ti=n. 3. Rank order all n scores
4. Find the sum of ranks in each sample. Denote the rank-sum of the ith sample R i.
Check that Σ Ri=n(n+1)/2 5. Calculate:
K=-3(n+1)+ 12 Σ Ri2
6. If more than a quarter of the response scores are involved in ties, go to 9
7. If more than three samples are being compared or if any of the sample sizes is
larger than 5, use the chi-square approximation in 10.
8. Find the critical value in the table in Appendix 6. Reject the null hypothesis if the
obtained K is larger than or equal to the critical value. If the null hypothesis is
rejected, use multiple comparisons to locate the effects. Otherwise, stop.
9. For extensive ties, find ui, the number of scores at each particular value of the
response variable and divide the value of K obtained in 5 by, 1 - Σ ui(ui-1)( ui+1)
10. Chi-square approximation. Where there are more than two samples, a distribution
known as the chi-square distribution frequently plays a role similar to that of the
normal distribution in two sampled cases in providing an approxim...
View Full Document
- Summer '14
- The Land