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 [EMAIL PROTECTED] 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, [EMAIL PROTECTED]
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
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