2 x 2 chisq.test (PR#8415)
This is the same as PR#8265, from a person also not giving his/her name but sharing your ISP. Please don't submit a repeat, as the FAQ asks. After last time this was raised, I checked Yates' original paper and Fisher's book and it seems that R's formula follows what they say. Do remember that 0 is an impossible value for a chisq-distributed variable, and so one would not expect the corrected distribution to take that value with positive probability.
On Tue, 20 Dec 2005 cig69410 at syd.odn.ne.jp wrote:
Full_Name: nobody
Do you expect us to take the word of `nobody' as to the correct definition of a statistic? We need your credentials and references.
Version: 2.2.0 OS: any Submission from: (NULL) (219.66.34.183) 2 x 2 table, such as
x
[,1] [,2] [1,] 10 12 [2,] 11 13
chisq.test(x)
Pearson's Chi-squared test with Yates'
continuity correction
data: x
X-squared = 0.0732, df = 1, p-value = 0.7868
but, X-squared = 0.0732 is over corrected.
when abs(a*d-b*c) <= sum(a,b,c,d), chisq.value must be 0!, and P-value must be
1!
code of chisq.test must be as follows
# if (correct && nrow(x) == 2 && ncol(x) == 2) {
# YATES <- 0.5
# METHOD <- paste(METHOD, "with Yates' continuity correction")
# }
# else YATES <- 0
# STATISTIC <- sum((abs(x - E) - YATES)^2/E)
## replace begin
if (correct && nrow(x) == 2 && ncol(x) == 2) {
STATISTIC <- if (abs(x[1,1]*x[2,2]-x[1,2]*x[2,1]) < sum(x)/2) 0
else sum((abs(x - E) - 0.5)^2/E)
METHOD <- paste(METHOD, "with Yates' continuity correction")
}
else STATISTIC <- sum((abs(x - E))^2/E)
## replace end
Brian D. Ripley, ripley at stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595