Nonparametric ANOVA

A model-free alternative to ANOVA.

Kris Sankaran true
09-30-2021

Readings 3.11, Rmarkdown

  1. What can we do when the errors seem very far from normally distributed? Imagine all attempts to transform the data have failed. An alternative is to use nonparametric ANOVA. The figure below gives an example of highly skewed data, for which a standard ANOVA might not be appropriate.

  1. The intuition is that, if all the groups have the same means, then no group should be consistently ranked higher than another. Consider hte figure below. In the left case, there seems to be a difference in the groups, just by looking at the ranks. In the right, the groups seem more or less comparable.

  1. To implement this idea quantitatively,

    • Transform the data to their ranks. The smallest of the \(y_{ij}\) becomes 1, the next smallest becomes 2, etc. Denote these ranks by \(R_{ij}\).
    • Compute the test statistic \[ \frac{1}{S^2}\left[\sum_{i = 1}^{a} \frac{R_{i\cdot}^2}{n_{i}} - \frac{N\left(N + 1\right)^2}{4}\right] \] where we define \(R_{i\cdot}\) to be the sum of the ranks in the \(i^{th}\) group, and \[ S^2 = \frac{1}{N - 1} \left[\sum_{i, j} R_{ij}^2 - \frac{N\left(N + 1\right)^2}{4}\right]. \]
    • Compare the test statistic to a \(\chi^2_{a -1}\) reference distribution. If it seems too large to be plausible, reject the null hypothesis.

  2. Where did this test statistic come from? It’s possible to show that the statistic is equivalent to \[ \frac{\sum_{i} n_{i}\left(\bar{R}_{i} - \bar{R}\right)^2}{\frac{1}{N - 1}\sum_{i, j} \left(R_{ij} - \bar{R}\right)^2} \] which compares the average rank in group \(i\) to the average rank overall, and standardizes by the overall variance of the ranks. The first formula is the one presented in the book, though, and it’s easier to calculate by hand.

  3. Why not always use nonparametric ANOVA? If the data are actually normal, than this approach has less power than standard ANOVA. If you have doubts about validity, a safe approach is to try both. If the approaches approximately agree, default to standard ANOVA.

Code Example

  1. This test is implemented by the kruskal.test function. It expects input in the same form as lm in the earlier ANOVA examples. Below, we apply the test to the etch rate data. The \(p\)-value indicates that the groups have significantly different ranks, which is consistent with our previous findings.
library(readr)
etch_rate <- read_csv("https://uwmadison.box.com/shared/static/vw3ldbgvgn7rupt4tz3ditl1mpupw44h.csv")
etch_rate$power <- as.factor(etch_rate$power) # want to think of power as distinct groups
kruskal.test(rate ~ power, data = etch_rate)


    Kruskal-Wallis rank sum test

data:  rate by power
Kruskal-Wallis chi-squared = 16.907, df = 3, p-value =
0.0007386