Xjyuk

I have data which is independent, but non-normal, with unequal variance. There are more than two groups, all with the same sample size. Which non-parametric test can I use?

Comment:

Simulated gamma data Here are simulated gamma data to illustrate some of the points in BenBolker's Answer (+1). Although
none of the traditional tests is completely satisfactory, the
Kruskal-Wallis test shows differences in groups for
my fake data (in which one difference between groups is rather large).

set.seed(123)
x1 = round(rgamma(20, 5, .2), 3)
x2 = round(rgamma(20, 5, .25), 3)
x3 = round(rgamma(20, 5, .35), 3)
sd(x1); sd(x2); sd(x3)
[1] 10.30572
[1] 6.218724
[1] 6.483086

Sample standard deviations differ, and box plots show different
dispersions along with different locations.

x = c(x1, x2, x3)
g = as.factor(rep(1:3, each=20))
boxplot(x ~ g, notch=T, col="skyblue2")

Notches in the sides of the boxes are nonparametric
confidence intervals calibrated so that, comparing
two groups, CIs that overlap suggest no significant
difference. So the first and last groups may differ significantly.

Kruskal-Wallis test, with ad hoc Wilcoxon comparisons. A Kruskal-Wallis test detects differences among the
three groups.

kruskal.test(x ~ g)

 Kruskal-Wallis rank sum test

data: x by g
Kruskal-Wallis chi-squared = 13.269, 
 df = 2, p-value = 0.001314

Ad hoc tests with two-sample Wilcoxon tests show significant differences between groups 1 and 3. P-values for the other two comparisons are not small enough to satisfy the Bonferroni criterion
against false discovery.

wilcox.test(x1,x2)$p.val
[1] 0.01674239
wilcox.test(x2,x3)$p.val
[1] 0.06343245
wilcox.test(x1,x3)$p.val
[1] 0.000667562

Welch one-factor ANOVA with ad hoc Welch t comparisons. Here we have moderately large samples and only moderate
skewness. A Welch one-way ANOVA, which does not
assume equal variances may be useful. The overall test gives a highly significant result; you could use Welch two-sample t tests for ad hoc comparisons.

oneway.test(x ~ g)

 One-way analysis of means 
 (not assuming equal variances)

data: x and g
F = 6.905, num df = 2.000, denom df = 36.733, 
 p-value = 0.002847

Permutation test focused on differences among group means. Finally, a permutation test with the standard deviation
of the three (permuted) group means as 'metric', shows a significant difference. This metric focuses on differences among group means. The test is nonparametric. The P-value 0.0008 is similar to that of the
Kruskal-Wallis test.

d.obs = sd(c(mean(x1),mean(x2),mean(x3)))
set.seed(915)
m = 10^5; d.prm = numeric(m)
for(i in 1:m) 
 x.prm = sample(x) # scrambe obs among gps
 a1=mean(x.prm[1:20]); a2=mean(x.prm[21:40]); a3=mean(x.prm[41:60])
 d.prm[i] = sd(c(a1,a1,a3)) 
mean(d.prm >= d.obs)
[1] 0.00075
length(unique(d.prm))
[1] 77417

Of the 100,000 iterations there were over 77,000
uniquely different values of d.prm. Their histogram
is shown below along with the value 5.235 of d.obs.

The Kruskal-Wallis test is a rank-based analogue of 1-way ANOVA, so it would be a reasonable approach to nonparametric testing of differences in location for >=2 groups.

HOWEVER: the "unequal variance" thing really messes you up. This answer discusses why unequal variances are problematic for Mann-Whitney tests (the 2-sample version of K-W/non-parametric version of the t-test), and the same problem applies to K-W, as discussed on Wikipedia (linked above):

If the researcher can make the assumptions of an identically shaped and scaled distribution for all groups, except for any difference in medians, then the null hypothesis is that the medians of all groups are equal, and the alternative hypothesis is that at least one population median of one group is different from the population median of at least one other group.

Loosely speaking, from my answer to the Mann-Whitney question:

If you are satisfied with showing that the distribution of the groups differs in some way from that of men, then you don't need the extra assumption [of equality of distribution except for location].