I'm about to apply Kruskal-Wallis test (non-parametric ANOVA) on three groups of unequal length. I was taught/advised to apply Krukal-Wallis only if:

  • dependent variable is at least at ordinal level of measurement
  • group's $ n > 5 $ (otherwise H statistic is not $ \chi ^2 $ distributed, so exact p-value cannot be calculated, etc.)
  • Wikipedia page says that it requires an identically-shaped and scaled distribution for each group

Apart from that, what are general considerations when applying Kruskal-Wallis test? Should I check homoscedascity with Fligner-Killeen's test, and what should I do in case of small $ n $, or unequal group sizes?

There are exactRankTests::wilcox.exact and PASWR::wilcoxE.test functions in R, but I can't seem to find an analogous one for Kruskal-Wallis...

Advice, anyone?

  • 1
    $\begingroup$ A permutation test might be appropriate here. $\endgroup$
    – chl
    Mar 27 '11 at 19:52
  • $\begingroup$ I added a little something: "unequal groups". $\endgroup$
    – aL3xa
    Mar 27 '11 at 21:12
  • $\begingroup$ Can you tell us what your three $n$s are? $\endgroup$
    – onestop
    Mar 27 '11 at 22:09
  • $\begingroup$ Well, if I do that, then this question can easily be marked as an assignment/homework. =) But, OK, say: 10, 50, 100. $\endgroup$
    – aL3xa
    Mar 27 '11 at 22:37

With small, and possibly unequal group sizes, I'd go with chl's and onestop's suggestion and do a Monte-Carlo permutation test. For the permutation test to be valid, you need exchangeability under $H_{0}$. If all distributions have the same shape (and are therefore identical under $H_{0}$), this is true.

Here's a first try at looking at the case of 3 groups and no ties. First, let's compare the asymptotic $\chi^{2}$ distribution function against a MC-permutation one for given group sizes (this implementation will break for larger group sizes).

P  <- 3                     # number of groups
Nj <- c(4, 8, 6)            # group sizes
N  <- sum(Nj)               # total number of subjects
IV <- factor(rep(1:P, Nj))  # grouping factor
alpha <- 0.05               # alpha-level

# there are N! permutations of ranks within the total sample, but we only want 5000
nPerms <- min(factorial(N), 5000)

# random sample of all N! permutations
# sample(1:factorial(N), nPerms) doesn't work for N! >= .Machine$integer.max
permIdx <- unique(round(runif(nPerms) * (factorial(N)-1)))
nPerms  <- length(permIdx)
H       <- numeric(nPerms)  # vector to later contain the test statistics

# function to calculate test statistic from a given rank permutation
getH <- function(ranks) {
    Rj <- tapply(ranks, IV, sum)
    (12 / (N*(N+1))) * sum((1/Nj) * (Rj-(Nj*(N+1) / 2))^2)

# all test statistics for the random sample of rank permutations (breaks for larger N)
# numperm() internally orders all N! permutations and returns the one with a desired index
library(sna)                # for numperm()
for(i in seq(along=permIdx)) { H[i] <- getH(numperm(N, permIdx[i]-1)) }

# cumulative relative frequencies of test statistic from random permutations
pKWH   <- cumsum(table(round(H, 4)) / nPerms)
qPerm  <- quantile(H, probs=1-alpha)  # critical value for level alpha from permutations
qAsymp <- qchisq(1-alpha, P-1)        # critical value for level alpha from chi^2

# illustration of cumRelFreq vs. chi^2 distribution function and resp. critical values
plot(names(pKWH), pKWH, main="Kruskal-Wallis: permutation vs. asymptotic",
     type="n", xlab="h", ylab="P(H <= h)", cex.lab=1.4)
points(names(pKWH), pKWH, pch=16, col="red")
curve(pchisq(x, P-1), lwd=2, n=200, add=TRUE)
abline(h=0.95, col="blue")                         # level alpha
abline(v=c(qPerm, qAsymp), col=c("red", "black"))  # critical values
legend(x="bottomright", legend=c("permutation", "asymptotic"),
       pch=c(16, NA), col=c("red", "black"), lty=c(NA, 1), lwd=c(NA, 2))

enter image description here

Now for an actual MC-permutation test. This compares the asymptotic $\chi^{2}$-derived p-value with the result from coin's oneway_test() and the cumulative relative frequency distribution from the MC-permutation sample above.

> DV1 <- round(rnorm(Nj[1], 100, 15), 2)  # data group 1
> DV2 <- round(rnorm(Nj[2], 110, 15), 2)  # data group 2
> DV3 <- round(rnorm(Nj[3], 120, 15), 2)  # data group 3
> DV  <- c(DV1, DV2, DV3)                 # all data
> kruskal.test(DV ~ IV)                   # asymptotic p-value
Kruskal-Wallis rank sum test
data:  DV by IV 
Kruskal-Wallis chi-squared = 7.6506, df = 2, p-value = 0.02181

> library(coin)                           # for oneway_test()
> oneway_test(DV ~ IV, distribution=approximate(B=9999))
Approximative K-Sample Permutation Test
data:  DV by IV (1, 2, 3) 
maxT = 2.5463, p-value = 0.0191

> Hobs <- getH(rank(DV))                  # observed test statistic

# proportion of test statistics at least as extreme as observed one (+1)
> (pPerm <- (sum(H >= Hobs) + 1) / (length(H) + 1))
[1] 0.0139972
  • $\begingroup$ Thank you so much for this thorough post (and accent on R). $\endgroup$
    – aL3xa
    Mar 28 '11 at 12:38
  • You need not check homoscedasticity. Kruskal and Wallis stated in their original paper that the “test may be fairly insensitive to differences in variability”.
  • If there is no exact test available, you can use bootstrap.
  • $\begingroup$ Yeah, right... bootstrap is always an option. Thanks! $\endgroup$
    – aL3xa
    Mar 27 '11 at 19:41
  • 1
    $\begingroup$ @GaBorgulya About your first point, it seems to me that non-parametric tests make the assumption that there's a shift in central tendency, but that overall the distributions that are compared have the same shape. $\endgroup$
    – chl
    Mar 27 '11 at 19:54
  • $\begingroup$ So what should I use to test basic assumptions? Does that mean that I need to apply a homoscedascity test? $\endgroup$
    – aL3xa
    Mar 27 '11 at 20:05
  • 2
    $\begingroup$ Unfortunatey boostrapping doesn't work well with very small samples either. @chl's suggestion of a permutation test is a much better idea. Monte Carlo permutation tests are a good option when exact tests haven't been programmed or would take unfeasibly long. $\endgroup$
    – onestop
    Mar 27 '11 at 22:07
  • $\begingroup$ @chl The $H_0$ of KW is that the multiple samples come from the same distribution of any shape. Looking at the test statistic I agree that the test is more powerful in case of shifts only. But although the original paper(permalink) also states that the $H_A$ "is that the samples come from populations of approximately the same form, but shifted", I don't see where the test statistic relies on that assumption. $\endgroup$
    – GaBorgulya
    Mar 27 '11 at 22:28

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