Thanks Ted (and other) for your suggestion. Here I have implemented
following:
Tab <- matrix(c(8, 10, 12, 6), nr = 2)
Simu_Number <- 50000
Tab_Simulate <- vector("list", length = Simu_Number)
for (i in 1:Simu_Number) {
Tab_Simulate[[i]] <- matrix(rmultinom(1, sum(Tab), rep(0.25,
4)), nrow = 2) ## All Cells have equal probability
}
Sample_ChiSq <- sapply(Tab_Simulate, function(x) {
Statistic <-
sum((chisq.test(as.table(x))$observed -
chisq.test(as.table(x))$expected)^2/chisq.test(as.table(x))$expected)
return(Statistic)
})
length(Sample_ChiSq[Sample_ChiSq < qchisq(0.95, 1)])/Simu_Number
hist(Sample_ChiSq, freq = FALSE)
lines(dchisq(seq(min(Sample_ChiSq), max(Sample_ChiSq), by = 0.5), 1))
However I think I am making some serious mistake as histogram did not
match the density curve.
Can somebody help me where I am making mistake?
Thanks and regards,
On 01-12-2012 21:45, (Ted Harding) wrote:
You will need to be clear about whether you are conditioning on
the marginal totals as well as on the overall total. As stated,
you are only asking for the overall total (36) to be fixed.
In that case, one possible (and by no means unique) approach
would be to:
[A]: Choose any four "random integers" a,b,c,d that add up to 36
(even there, you still have to make a choice about what distribution
to adopt for the "random integers").
[B]: Place the results into the 2x2 matrix; then evaluate chi-squared.
[C]: Repeat until you have enough cases.
Example (using equiprobable multinomial to generate 4 "random integers")
Tab <- matrix(rmultinom(1,36,c(1,1,1,1)/4), nrow=2)
A more specific choice would be to fix the row and column probablilties
(but not the sample row and column totals), e.g.:
P.row1 <- 0.25 ; P.row2 <- 1 - P.row1
P.col1 <- 0.50 ; P.col2 <- 1 - P.col1
Then, adopting the hypothesis of independence between rows and columns:
P11 <- P.row1*P.col1
P21 <- P.row2*P.col1
P12 <- P.row1*P.col2
P22 <- P.row2*P.col2
Tab <- matrix(rmultinom(1,36,c(P11,P21,P12,P22)), nrow=2)
On the other hand, if you want to also fix the sample row margins
and column margins, then your sampled table needs to be generated
using a hypergeometric distribution, (which is the basis of the
fisher.test mentioned by Bert Gunter). Since explaining how to
do this is a bit mjore complicated than the above, please first
confirm what constraints (e.g. only total count; row-margins
only; col-margins only; both row- & col-margins) you wich to impose.
Hoping this helps,
Ted.
On 01-Dec-2012 14:28:24 Christofer Bogaso wrote:
Thanks Bert for your reply.
I am trying to understand/visualize the Sample Chi-Squared Statistic's
Null distribution. Therefore I need all possible Contingency tables
under Independence case.
What could be better way to visualize that?
Thanks and regards,
On 01 December 2012 20:03:00, Bert Gunter wrote:
Christopher:
Don't do this!
If I understand you correctly, you want FIsher's exact test. This is
already available in R, using far smarter algorithms then you would. See:
?fisher.test
-- Bert
On Sat, Dec 1, 2012 at 5:48 AM, Christofer Bogaso
<bogaso.christo...@gmail.com <mailto:bogaso.christo...@gmail.com>> wrote:
Thanks John for your reply. However still not clear how I should
proceed.
My goal is to generate all possible contingency tables. Basically
I want to see the distribution of Chi-squared Statistic under
independence (NULL).
So I was thinking if I can generate all possible permutation of
integer numbers having sum equal to (8 + 10 + 12 + 6) = 36. Is
there any R function to do that?
Thanks and regards,
On 01-12-2012 18:39, John Kane wrote:
Are you basically asking for all possible permutations of the
table? If so see ?permn in the combinat package.
John Kane
Kingston ON Canada
-----Original Message-----
From:bogaso.christo...@gmail.com
<mailto:bogaso.christo...@gmail.com>
Sent: Sat, 01 Dec 2012 18:10:15 +0545
To:r-help@r-project.org <mailto:r-help@r-project.org>
Subject: [R] Getting all possible contingency tables
Hello all,
Let say I have 2-way contingency table:
Tab <- matrix(c(8, 10, 12, 6), nr = 2)
and the Chi-squared test could not reject the independence:
> chisq.test(Tab)
Pearson's Chi-squared test with Yates'
continuity correction
data: Tab
X-squared = 1.0125, df = 1, p-value = 0.3143
However I want to get all possible contingency tables
under this
independence scenario (one of them would obviously be the
given table
as, we could not reject the independence), and for each
such table I
want to calculate the Ch-sq statistic.
Can somebody help me how to generate all such tables?
Thanks and regards,
-------------------------------------------------
E-Mail: (Ted Harding)<ted.hard...@wlandres.net>
Date: 01-Dec-2012 Time: 16:00:16
This message was sent by XFMail
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