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I have seen that there exists tests which see if k samples share the same variance such as the Levene's test.

I was wondering if there is an equivalent test when means are wanting to be examined.

An non-parametric test would be needed here.

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  • $\begingroup$ Most non-parametric tests are based on a stronger or slightly different null hypothesis, such none of the distributions is stochastically dominant. See en.wikipedia.org/wiki/…. $\endgroup$
    – whuber
    Feb 9, 2021 at 17:41
  • $\begingroup$ @whuber Does stochastic domaine imply difference in mean? I have seen people use the Mann Whitney U test for a difference in medians so does this not apply to the Kruskal-Wallis test? $\endgroup$
    – math111
    Feb 9, 2021 at 18:24
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    $\begingroup$ Yes, stochastic dominance implies a difference in mean--but the implication does not go the other way. $\endgroup$
    – whuber
    Feb 9, 2021 at 18:57
  • $\begingroup$ @whuber If I dont reject the null, do I conclude the samples are from the same distribution? The reason I ask is I also performed the Levene test and rejected the null implying a difference in variance. So how can the KW test tell us the distributors are the same. $\endgroup$
    – math111
    Feb 9, 2021 at 19:09
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    $\begingroup$ The null in the KW test does not assert all distributions are identical. $\endgroup$
    – whuber
    Feb 9, 2021 at 19:14

1 Answer 1

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Valid conclusions from a Kruskal-Wallis test can be difficult when populations have markedly different shapes. Here we use three samples of size $n=100$ from $\mathsf{Beta}(.1,.1)$ with mean $\mu_1 = 1/2,$ $\mathsf{Beta}(5,1)$ with mean $\mu_2 = 5/6,$ and $\mathsf{Beta}(3,3)$ with mean $\mu_3 = 1/2.$

Densities of the three distributions are shown below:

enter image description here

Sampling in R:

set.seed(2021)
x1 = rbeta(100, .1, .1)
x2 = rbeta(100, 5, 1)
x3 = rbeta(100, 3, 3)
x = c(x1,x2,x3);  gp=rep(1:3, each=100)

The three sample means are as follows:

mean(x1); mean(x2); mean(x3)
[1] 0.5259448
[1] 0.8361308
[1] 0.498663

hdr="Boxplots of Samples from BETA(.1,.1) [red], BETA(5.1) [green], BETA(3,3)"
boxplot(x ~ gp, col=c("red", "green", "blue"), main=hdr)

enter image description here

A Kruskal-Wallis test (not ordinarily recommended for comparing distributions of different shapes) shows a highly significant difference among the three samples.

kruskal.test(x ~ gp)

        Kruskal-Wallis rank sum test

data:  x by gp
Kruskal-Wallis chi-squared = 59.016, df = 2, p-value = 1.531e-13

Sometimes empirical CDF (ECDF) plots are used to judge visually whether one sample stochastically dominates another. A dominating sample will tend to plot to the right (hence below) samples it dominates. Here it seems clear that Sample 2 (green) dominates Sample 3 (blue), but not so clear whether Sample 2 dominates Sample 1 (red).

hdr="ECDFs of Samples from BETA(.1,.1) [red], 
     BETA(5.1) [green], BETA(3,3)"
plot(ecdf(x1), col="red", main=hdr)
 lines(ecdf(x2), col="green2")
 lines(ecdf(x3), col="blue")

enter image description here

Sometimes two-sample Wilcoxon rank sum tests are used ad hoc to determine differences is 'location' or 'domination'. Here are the P-values from two such tests, but I will leave the interpretation up to you. Ordinarily, one would want to see a P-value below about 1% or 2% (according to the Bonferroni method of avoiding 'false discovery') in order to declare differences.

wilcox.test(x2,x1)$p.value
[1] 0.06075133
wilcox.test(x2,x3)$p.value
[1] 2.750473e-24

Roughly estimated probabilities $P(X_2 > X_1)\approx 0.58$ and $P(X_2 > X_3)\approx 0.92.$

quantile(replicate(10^6, mean(sample(x2) > x1)), c(.025,.975))
 2.5% 97.5% 
 0.54  0.61 
quantile(replicate(10^6, mean(sample(x2) > x3)), c(.025,.975))
 2.5% 97.5% 
 0.88  0.95 
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  • $\begingroup$ Thank you for your answer. To conclude, there are no tests that can compare means without the need of the same shape? $\endgroup$
    – math111
    Feb 10, 2021 at 10:33
  • $\begingroup$ No simple answer. Need to specify exact situation. Sometimes 'same shape' means same var also. Welch t test will compare means of 2 normal distributions with different variances. // Kolmogorov-Smirnov will compare samples to see if their populations have same shape. Might distinguish btw large enough samples from $\mathsf{Norm}(\mu=\sigma=1)$ and $\mathsf{Exp}(1).$ which has unit mean and SD: ks.test(rnorm(10,2,1),rexp(10))$p.val may not reject, but ks.test(rnorm(100,1,1),rexp(100))$p.val often does. // Can find LR test to distinguish btw 2 exp'l distn's w/ $\ne$ means. $\endgroup$
    – BruceET
    Feb 10, 2021 at 11:50

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