# Understanding warning message “Ties are present” in Kruskal-Wallis post hoc

I'm running post-hoc comparisons after a Kruskal-Wallis test. I'm using the PMCMR package.

> posthoc.kruskal.nemenyi.test( preference ~ instrument)

Pairwise comparisons using Tukey and Kramer (Nemenyi) test
with Tukey-Dist approximation for independent samples

data:  preference by instrument

Cello Drums Guitar
Drums  0.157 -     -
Guitar 0.400 0.953 -
Harp   0.013 0.783 0.458

Warning message:
In posthoc.kruskal.nemenyi.test.default(c(50L, 50L, 50L, 50L, 49L,  :
Ties are present, p-values are not corrected.


I'm confused by the warning message. Can anyone explain what it means and how I can correct it?

• I had the same warning occuring when running posthoc.kruskal.nemenyi.test() . So I adjusted the ties by running data$x <- rank(data$x, ties.method="average") before running posthoc.kruskal.nemenyi.test(data$x ~ data$y) again but I still got the same warning regarding "Ties are present". Is that a general problem or does this problem derive from my data? – Constanze Sep 2 '15 at 11:27

A tie means that you have several observations share the same value (hence the same rank). For example, a sample consists of observations: $1, 3, 3, 5, 10, 10, 10$. "$3$" and "$10$" are two ties, where $3$ has replicates of $2$ and $10$ has replicates of $3$. Such a sample corresponds to the rank statistics: $1, 2, 2, 4, 5, 5, 5$.
Two tie-breaking methods are common, one is "breaking ties by random". Namely, we regenerate distinct ranks randomly among the ties. Continuing the above example, to the tie "$2, 2$", we may draw two numbers without replacement from the set $\{2, 3\}$, then assign them to the second and third positions, for example, "3, 2". Similarly, we can do that for the tie $10$. A possible adjusted rank statistics can be $1, 3, 2, 4, 6, 5, 7$, hence the ties got broken. The disadvantage of this method is that you may get different test statistics among different analysis, since the tie-breaking is by random.
The second method is "averaging". That is, average assigns each tied element the "average" rank. Using this method, the original rank statistics becomes: $1, 2.5, 2.5, 4, 6, 6, 6$. This method essentially adjusts the ties instead of breaking them.