# [R] significance test interquartile ranges

peter dalgaard pdalgd at gmail.com
Sat Jul 14 14:08:23 CEST 2012

```On Jul 14, 2012, at 12:25 , Rui Barradas wrote:

> Hello,
>
> There's a test for iqr equality, of Westenberg (1948), that can be found on-line if one really looks. It starts creating a 1 sample pool from the two samples and computing the 1st and 3rd quartiles. Then a three column table where the rows correspond to the samples is built. The middle column is the counts between the quartiles and the side ones to the outsides. These columns are collapsed into one and a Fisher exact test is conducted on the 2x2 resulting table.

That's just wrong, is it not? Just because things were suggested by someone semi-famous, it doesn't mean that they actually work...

Take two normal distributions, equal in size,  with a sufficiently large difference between the means, so that there is no material overlap. The quartiles of the pooled sample will then be the medians of the original samples, and the test will be that one sample has the same number above its median as the other has below its median.

If it weren't for the pooling business, I'd say that it was a sane test for equality of quartiles, but not for the IQR.

>
> R code could be:
>
>
> iqr.test <- function(x, y){
> 	qq <- quantile(c(x, y), prob = c(0.25, 0.75))
> 	a <- sum(qq < x & x < qq)
> 	b <- length(x) - a
> 	c <- sum(qq < y & y < qq)
> 	d <- length(y) - b
> 	m <- matrix(c(a, c, b, d), ncol = 2)
> 	numer <- sum(lfactorial(c(margin.table(m, 1), margin.table(m, 2))))
> 	denom <- sum(lfactorial(c(a, b, c, d, sum(m))))
> 	p.value <- 2*exp(numer - denom)
> 	data.name <- deparse(substitute(x))
> 	data.name <- paste(data.name, ", ", deparse(substitute(y)), sep="")
> 	method <- "Westenberg-Mood test for IQR range equality"
> 	alternative <- "the IQRs are not equal"
> 	ht <- list(
> 		p.value = p.value,
> 		method = method,
> 		alternative = alternative,
> 		data.name = data.name
> 	)
> 	class(ht) <- "htest"
> 	ht
> }
>
> n <- 1e3
> pv <- numeric(n)
> set.seed(2319)
> for(i in 1:n){
> 	x <- rnorm(sample(20:30, 1), 4, 1)
> 	y <- rchisq(sample(20:40, 1), df=4)
> 	pv[i] <- iqr.test(x, y)\$p.value
> }
>
> sum(pv < 0.05)/n  # 0.8
>
>

To wit:

> iqr.test(rnorm(100), rnorm(100,10,3))

Westenberg-Mood test for IQR range equality

data:  rnorm(100), rnorm(100, 10, 3)
p-value = 0.2248
alternative hypothesis: the IQRs are not equal

> replicate(10,iqr.test(rnorm(100), rnorm(100,10,3))\$p.value)
 0.2248312 0.2248312 0.2248312 0.2248312 0.2248312 0.2248312 0.2248312
 0.2248312 0.2248312 0.2248312

--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com

```