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[1] < x & x < qq[2])
> b <- length(x) - a
> c <- sum(qq[1] < y & y < qq[2])
> 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)
[1] 0.2248312 0.2248312 0.2248312 0.2248312 0.2248312 0.2248312 0.2248312
[8] 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....@cbs.dk Priv: pda...@gmail.com
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