significance test interquartile ranges
Hello, Em 14-07-2012 13:08, peter dalgaard escreveu:
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.
Right, thank you! It forced me to pay more attention to what I was reading. The "test is aimed at differences in scale only, presuming no difference in location" http://www.stat.ncsu.edu/information/library/mimeo.archive/ISMS_1986_1499.pdf The original can be found at http://www.dwc.knaw.nl/DL/publications/PU00018486.pdf If we subtract the median of each sample to each of them, the medians become zero but the IQRs remain as they were. In my simulation I had chosen samples from distributions with equal mean, and that point passed unnoticed. The code should then be slightly revised. I'll repost it because there was a typo in the 'method' member of the returned list iqr.test <- function(x, y){ data.name <- deparse(substitute(x)) data.name <- paste(data.name, ", ", deparse(substitute(y)), sep="") x <- x - median(x) y <- y - median(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) method <- "Westenberg-Mood test for IQR 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 } Rui Barradas
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