I carried out a wilcoxon test (using wilcox.test
in R) in 2 sample groups of size 135 and 48.
My results in r where:
wilcox.test(raj_reps.a, raj_reps.b)
Wilcoxon rank sum test with continuity correction
data: raj_reps.a and raj_reps.b
W = 2359.5, p-value = 0.005217
alternative hypothesis: true location shift is not equal to 0
My H0 was that the groups are equal in means, the alternative hypothesis was that the values in group a are different (lower) than in group b.
Can I accept H0? as the p-value is lower than 0.05?
What exactly means the W value? I read about it and found out its the number of ranks and used as critical value which I can look up in tables. Unfortunately most of the tables are not covering such huge sample sizes.
Can anyone tell me if this high W means, that I can accept H0 or do I have to reject it?
Best Answer
The Wilcoxon test does not test for equality of means, rather it tests
$$H_0: P(X_a > X_b) = 0.5$$
namely that a randomly drawn observation of group a has 50% chance of being larger than a randomly drawn observation from group b. Only if you see location-shift (i.e. distributions in both groups have the same shape but different mean (location)) than you can formulate your conclusion in terms of means. With the P-value you have you can reject $H_0$ on the 5% significance level. The $W$ is the Wilcoxon test statistic and is, as the name says, the sum of the ranks in one of both groups. You could enumerate the exact distribution of $W$ under $H_0$ by typing
This is very computer-intensive, so usually for such large sample sizes the asymptotic normality of $W$ is used for calculating the p-value, see
?wilcox.test