# k-sample test for equality of means

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.

• 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/….
– whuber
Feb 9, 2021 at 17:41
• @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? Feb 9, 2021 at 18:24
• Yes, stochastic dominance implies a difference in mean--but the implication does not go the other way.
– whuber
Feb 9, 2021 at 18:57
• @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. Feb 9, 2021 at 19:09
• The null in the KW test does not assert all distributions are identical.
– whuber
Feb 9, 2021 at 19:14

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: 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)
 0.5259448
 0.8361308
 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) 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") 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)$$p.value  0.06075133 wilcox.test(x2,x3)$$p.value
 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

• Thank you for your answer. To conclude, there are no tests that can compare means without the need of the same shape? Feb 10, 2021 at 10:33
• 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. Feb 10, 2021 at 11:50