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?

  • $\begingroup$ The output gives you a p-value (see the first sentence of the linked section). You compare it with your significance level, rejecting the null if $p\leq \alpha$. Note that it's only a comparison of means if you add additional assumptions $\endgroup$
    – Glen_b
    Feb 14 '17 at 23:17

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

    wilcox.test(raj_reps.a, raj_reps.b, exact=TRUE)

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.