Thanks for the clarification, Jim. The terminology "previous"
was not self-explanatory!
The following implements (in a somewhat crude way, but explicit)
a solution to your question:
M <- matrix(c(1, 4, 23, 30), byrow=TRUE, ncol=2)
M
# [,1] [,2]
# [1,] 1 4
# [2,] 23 30
fisher.test(M)$p.value
# [1] 0.3917553
mina <- (-1) ; maxa <- 4
As <- (mina:maxa)
Ps <- numeric(length(As))
for(i in (1:length(As))){
a <- As[i]
dM <- matrix(c(a,-a,-a,a),nrow=2)
Ps[i] <- fisher.test(M+dM)$p.value
}
Ps
# [1] 0.070002593 0.391755250 1.000000000 0.639558667
# [5] 0.148849134 0.009276064
So, amongst the P-values, the one below the attained one (0.3917553)
which is greatest is Ps[5] = 0.148849134. This could be identified
by the expression
max(Ps[Ps < fisher.test(M)$p.value])
# [1] 0.1488491
If fisher.test(M)$p.value were already the smallest possible value,
then this expression would return -Inf.
Ted.
On 14-Apr-11 14:55:36, Jim Silverton wrote:
> What Ted and Peter did were Fisher's exact test, To get the previous
> attainable p-value, what you do is the the fisher exact test p-values
> of
> ALL the possible tables with margins fixed and choose the p-value that
> is
> just below the one for fisher's exact test of the original table.
>
> n Thu, Apr 14, 2011 at 3:01 AM, peter dalgaard <pda...@gmail.com>
> wrote:
>
>>
>> On Apr 14, 2011, at 01:29 , (Ted Harding) wrote:
>>
>> > On 13-Apr-11 17:40:53, Jim Silverton wrote:
>> >> I have a matrix say,
>> >>
>> >> 1 4
>> >> 23 30
>> >>
>> >> and I want to find the previously attainable fisher's exact test
>> >> p-value. Is there a way to do this in R?
>> >> --
>> >> Thanks,
>> >> Jim.
>> >
>> > I do not understand what you mean by "previously attainable".
>> >
>> > As far as that particular matrix is concerned, the fisher.test()
>> > function will yield its exact Fisher P-value:
>> >
>> > M <- matrix(c(1, 4, 23, 30), byrow=TRUE, nrow=2)
>> > M
>> > # [,1] [,2]
>> > # [1,] 1 4
>> > # [2,] 23 30
>> > fisher.test(M)
>> > # Fisher's Exact Test for Count Data
>> > # data: M
>> > # p-value = 0.3918
>> > # alternative hypothesis: true odds ratio is not equal to 1
>> > # 95 percent confidence interval:
>> > # 0.006355278 3.653391412
>> > # sample estimates:
>> > # odds ratio
>> > # 0.3316483
>> >
>> > So the P-value is 0.3918 (as attained now, and as attainable
>> > at any time previously if you had done the above ... !).
>> >
>>
>> What Ted said, plus
>>
>> f <- fisher.test(M)
>> f$p.value
>> # [1] 0.3917553
>>
>>
>> --
>> Peter Dalgaard
>> Center for Statistics, Copenhagen Business School
>> Solbjerg Plads 3, 2000 Frederiksberg, Denmark
>> Phone: (+45)38153501
>> Email: pd....@cbs.dk Priv: pda...@gmail.com
>>
>>
>
>
> --
> Thanks,
> Jim.
--------------------------------------------------------------------
E-Mail: (Ted Harding) <ted.hard...@wlandres.net>
Fax-to-email: +44 (0)870 094 0861
Date: 14-Apr-11 Time: 16:37:44
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