Solved – Wilcoxon Test – non normality, non equal variances, sample size not the same

hypothesis testingnonparametricnormality-assumptionwilcoxon-mann-whitney-test

I know there are already a lot of posts out there, but I couldn't find this exact combination in any of them.
Comparing two samples (Prices associated with men and with women), but I have neither the same sample size ($n = 790$ vs $n=795$) nor equal variance or normality.

My hypothesis is whether the prices for Women are greater than prices for Men. The Wilcoxon test is significant on a 4 % level.

Can I actually say anything helpful since so many assumptions are violated? Would another test be better?

EDIT Some additional infos:

Prices Women: Median 28.00, Mean 28.47,  Std. Dev. 17.17,  Skewness 0.91, Kurtosis 5.41
Prices Men:   Median 26.00, Mean, 29.08, Std. Dev. 22.43,  Skewness 2.39, Kurtosis 12.74

Best Answer

tl;dr if you want to interpret the rejection of the null hypothesis as evidence that prices for women are greater than those for men, then you do need the assumption of equal variance (in fact, equal distributions) between the two populations. If you are satisfied with showing that the distribution of prices for women differs in some way from that of men, then you don't need the extra assumption.

You don't need to worry about unequal sample size (this will affect the power of the test, but not its validity) or Normality.

For what it's worth, testing whether one group's values are larger on average than another group's when their variances also differ is a surprisingly deep question, even for Normally distributed data (where it's known as the Behrens-Fisher problem).

Referring to the Wikipedia page: the "very general formulation" says:

Under the null hypothesis H0, the distributions of both populations are equal.[3]

The alternative hypothesis H1 is that the distributions are not equal.

The next paragraph says:

Under more strict assumptions than the general formulation above, e.g., if the responses are assumed to be continuous and the alternative is restricted to a shift in location, i.e., $F_1(x) = F_2(x + δ)$, we can interpret a significant Mann–Whitney U test as showing a difference in medians ...

(emphasis added)

Note to technical readers: I think this is a reasonable summary, but if anyone wants to be more rigorous, feel free to comment or edit or post an alternative answer ...

Related Solutions

Solved – Normality assumption and sample size

Disputes about normality with large N are often to do with tests of normality, not normality per se. For larger sample sizes passing a test of normality, like Shapiro-Wilks is not required. Consider the following in R.

findNonNormal <- function(n = 5000){
    p <- 1
    while(p > 0.05) {
        y <- rnorm(n)
        p <- shapiro.test(y)$p.value
        }
    y
    }

y <- findNonNormal()
hist(y)
qqnorm(y)

The results show a remarkably normal distribution that the test says is not normal. That's because the power of the test is so high with that N that it finds non normal distributions with very small deviations. You could easily find similar results with the N's you mentioned.

Generally, passing an eyeball test of normality is all that's needed. This eyeball test needs to be adjusted with N. If you feel you cannot do the assessment just do some simulations with a similar N and see what typical data from a truly normal distribution look like.

If your data really are not normal don't do the parameteric tests. But, contrary to your belief, a large N with reasonably normal distributions is when the power of a parametric test becomes most valuable. It allows one to make estimates of the parameters in the population, and the better and larger the sample the more accurate those estimates will be.

Additionally, if you're looking at a t-test, for example, the distribution of the data is going to be bimodal with a strong effect. It's because there are two means in the distribution. So, the requirement is not that the data look normal but the residuals look normal. This is true for your ANOVA as well.

Solved – T-test equal variances, unequal sample sizes

Theoretically, if the assumption of equal variances is satisfied and the dependent variable is normally distributed you could run a t-test despite the unequal sample sizes between the two comparison groups. I suggest you read these two posts (1 & 2) which deal with similar issues. If the assumption of normality is not satisfied then a Mann-Whitney test is appropriate as you have already suggested. You don't provide more information regarding your research questions and data availability to actually recommend a better statistical test.

Best Answer

Related Solutions

Solved – Normality assumption and sample size

Solved – T-test equal variances, unequal sample sizes

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