[R] Getting all possible contingency tables

[Christofer Bogaso]

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.christofer at gmail.com  <mailto:bogaso.christofer at 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.christofer at gmail.com
>>>              <mailto:bogaso.christofer at gmail.com>
>>>              Sent: Sat, 01 Dec 2012 18:10:15 +0545
>>>              To:r-help at r-project.org  <mailto:r-help at 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.Harding at wlandres.net>
> Date: 01-Dec-2012  Time: 16:00:16
> This message was sent by XFMail
> -------------------------------------------------
>