R devel - Jul 2003 - The two chisq.test p values differ when the contingency table is transposed! (PR#3486)

Full_Name: Tao Shi
Version: 1.7.0
OS: Windows XP Professional
Submission from: (NULL) (149.142.163.65)

> x
     [,1] [,2]
[1,]  149  151
[2,]    1    8
> c2x<-chisq.test(x, simulate.p.value=T, B=100000)$p.value
> for(i in (1:20)){c2x<-c(c2x,chisq.test(x,
simulate.p.value=T,B=100000)$p.value)}
> c2tx<-chisq.test(t(x), simulate.p.value=T, B=100000)$p.value
> for(i in (1:20)){c2tx<-c(c2tx,chisq.test(t(x), simulate.p.value=T,
+ B=100000)$p.value)}
> cbind(c2x,c2tx)
          c2x    c2tx
 [1,] 0.03711 0.01683
 [2,] 0.03717 0.01713
 [3,] 0.03709 0.01609
 [4,] 0.03833 0.01657
 [5,] 0.03696 0.01668
 [6,] 0.03698 0.01660
 [7,] 0.03704 0.01805
 [8,] 0.03731 0.01699
 [9,] 0.03746 0.01683
[10,] 0.03671 0.01676
[11,] 0.03589 0.01689
[12,] 0.03684 0.01670
[13,] 0.03678 0.01709
[14,] 0.03742 0.01658
[15,] 0.03734 0.01664
[16,] 0.03723 0.01778
[17,] 0.03690 0.01615
[18,] 0.03650 0.01621
[19,] 0.03759 0.01740
[20,] 0.03712 0.01653
[21,] 0.03788 0.01702

>Date: Wed, 16 Jul 2003 01:27:25 +0200 (MET DST)
>From: shitao@ucla.edu
>> x
>     [,1] [,2]
>[1,]  149  151
>[2,]    1    8
>> c2x<-chisq.test(x, simulate.p.value=T, B=100000)$p.value
>> for(i in (1:20)){c2x<-c(c2x,chisq.test(x,
>simulate.p.value=T,B=100000)$p.value)}
>> c2tx<-chisq.test(t(x), simulate.p.value=T, B=100000)$p.value
>> for(i in (1:20)){c2tx<-c(c2tx,chisq.test(t(x), simulate.p.value=T,
>+ B=100000)$p.value)}
>> cbind(c2x,c2tx)
>          c2x    c2tx
> [1,] 0.03711 0.01683
> [2,] 0.03717 0.01713
The problem is in ctest/R/chisq.test.R, where the p-value is
calculated as 

            STATISTIC <- sum((x - E) ^ 2 / E)
            PARAMETER <- NA
            PVAL <- sum(tmp$results >= STATISTIC) / B

Here tmp$results is a collection of simulated chisquare values, but
because of different rounding, the statistics corresponding to tables
equal to the observed table are slightly smaller than the value
calculated in STATISTIC, and effectively the p-value is calcuated as

             PVAL <- sum(tmp$results > STATISTIC) / B

instead.

What's the appropriate fix here? 

PVAL <- sum(tmp$results > STATISTIC - .Machine$double.eps^0.5) / B

works on this example, but is there something better?

Duncan Murdoch