It seems to me that your two approaches are calculating CIs for different quantities. The bootstrap methods are calculating a CI for the difference in medians, while the Wilcoxon approach is calculating a CI for the median of the differences. If this were the mean, those would be the same, but not for the median:

A =c(619, 600, 490, 1076, 654, 955, 563, 955, 827, 873, 1253)
B =c(346, 507, 598, 228, 576, 338, 1153, 354, 560, 517, 381)

> median(A)-median(B)
[1] 320
> median(A-B)
[1] 273

--
Patrick Breheny
Assistant Professor
Department of Biostatistics
Department of Statistics
University of Kentucky


On 02/18/2012 05:05 AM, Vittorio Colagrande wrote:
Dear R-group,



I have run into a problem in estimating confidence intervals for the median 
difference.

I want to establish a confidence interval at (1- alpha) level for the 
difference between the

medians of two indipendent samples (size n and m), by using the Wilcoxon 
distribution

or with bootstrap methods.

First method, we consider the z matrix of d=n∙m differences of the first and 
second sample

data and we order these differences in a y vector. By the Wilcoxon 
distribuition (W) we

determine the q quantile such that Prob(W<q)= alpha/2, inf and sup of 
confidence interval are

respectively the q-th and (d-q+1)-th elements of the y vector, while we obtain 
the difference

median by the y distribution. This method is also used to establish the CI by 
wilcox.test.

Example. Two indipendent sample A and B (n=11, m=13)  of CD4 count cells 
(T-helper cells):

A =c(619, 600, 490, 1076, 654, 955, 563, 955, 827, 873, 1253)

B =c(346, 507, 598, 228, 576, 338, 1153, 354, 560, 517, 381, 415, 626)



1)      CI 95% by matrix z and y vector:

n=length(A)

m=length(B)

d=n*m

z=matrix(0,m,n)

for(j in seq_len(n))

z[, j]=A[j] - B

y=sort(as.vector(z))

q=qwilcox(0.05/2,n,m,lower.tail = TRUE, log.p = FALSE)

inf=y[q]

sup=y[d-q+1]

med=median(y)

results:  inf = 100, sup = 516 and med = 300.



2)      CI 95% by wilcox.test:

       I=wilcox.test(A,B,conf.lev=0.95,conf.int=TRUE,exact=F,correct=T)

      inf=I$conf.int[1]

      sup=I$conf.int[2].

results:  inf = 99.9, sup = 516.



Second method, bootstrap each sample separately, creating the sampling 
distribution for

each median. Then calculate the difference between the two medians, and create 
the

sampling distribution of those differences. This is the sampling distribution 
we care about.

Once we have that distribution we can establish a confidence interval. Some CI 
95%,

with reference to the CD4 example, one given below.



1)     First procedure (package boot):

library(boot)

       n=length(A)

       m=length(B)

       y=c(A,B)

      camp=data.frame(group=rep(c(1,2),c(n,m)),y)

      dif.median=function(data,i) {

      d=data[i,]

      n1=n+1

      m1=n+m

      median(d$y[1:n])-median(d$y[n1:m1])  }

      dif.boot=boot(camp,dif.median,R=10000, strata=camp$group)

      boot.ci(dif.boot, conf =0.95, type="bca")

results:  inf = 59, sup = 574.



2)      Second procedure (package pairwiseCI):

library(pairwiseCI)

MedDiff=Median.diff(A, B, conf.level=0.95, alternative="two.sided",R=10000)

MedDiff$conf.int

MedDiff$estimate

results:  inf = 56, sup = 574, median=320



3)      Third procedure (stratified bootstrap):

       dif<- numeric(10000)

       for(i in seq_len(10000))

       dif[i]<- median(sample(A, replace=TRUE)) - median(sample(B, 
replace=TRUE))

       quantile(dif,prob=c(0.5,(1-0.95)/2,(1-(1-0.95)/2)))

results: inf = 56, sup = 574, median = 313.



4)      Fourth procedure (package simpleboot)

  library(simpleboot)

  boot_diff<- two.boot(A, B, median, R = 10000)

  boot.ci(boot_diff,conf=0.95,type="bca")

results: inf = 59, sup = 574.



The bootstrap procedures do get the same results, but the confidence intervals 
are

significantly different from those obtained using the method that refers to the 
Wilcoxon

distribution.
Problem: does this difference depend on really "different" methods

or on incorrect implementation of the bootstrap technique?



I will greatly appreciate any clarification you could provide.

Best regards.

Vittorio Colagrande

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