[R] fisher exact vs. simulated chi-square
Dirk Janssen
dirkj at rz.uni-leipzig.de
Tue Apr 22 14:07:40 CEST 2003
Dear All,
I have a problem understanding the difference between the outcome of a
fisher exact test and a chi-square test (with simulated p.value).
For some sample data (see below), fisher reports p=.02337. The normal
chi-square test complains about "approximation may be incorrect",
because there is a column with cells with very small values. I
therefore tried the chi-square with simulated p-values, but this still
gives p=.04037. I also simulated the p-value myself, using r2dtable,
getting the same result, p=0.04 (approx).
Why is this substantially higher than what the fisher exact says? Do
the two tests make different assumptions? I noticed that the
discrepancy gets smaller when I increase the number of observations
for column A3. Does this mean that the simulated chi-square is still
sensitive to cells with small counts, even though it does not give me
the warning?
Thanks in advance,
Dirk Janssen
------------------------------------------------------------------
> ta <- matrix(c(45,85,27,32,40,34,1,2,1),nc=3,
dimnames=list(c("A","B","C"),c("A1","A2","A3")))
> ta
A1 A2 A3
A 45 32 1
B 85 40 2
C 27 34 1
> fisher.test(ta)
Fisher's Exact Test for Count Data
data: ta
p-value = 0.02337
alternative hypothesis: two.sided
> chisq.test(ta, simulate=T, B=100000)
Pearson's Chi-squared test with simulated p-value (based on 1e+05
replicates)
data: ta
X-squared = 9.6976, df = NA, p-value = 0.04037
> chisq.test(ta)
Pearson's Chi-squared test
data: ta
X-squared = 9.6976, df = 4, p-value = 0.04584
Warning message:
Chi-squared approximation may be incorrect in: chisq.test(ta)
# simulate values by hand, based on r2dtable example
> expected <- outer(rowSums(ta), colSums(ta), "*") / sum(ta)
> meanSqResid <- function(x) mean((x - expected) ^ 2 / expected)
> sum(sapply(r2dtable(100000, rowSums(ta), colSums(ta)), meanSqResid)
>= meanSqResid(ta))/ 100000
[1] 0.03939
# is similar to
> sum(sapply(r2dtable(100000, rowSums(ta), colSums(ta)),
function(x) { chisq.test(x)$statistic })
>= 9.6976)/ 100000
[1] 0.04044
There were 50 or more warnings (use warnings() to see the first 50)
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