I don't know of an existing R function to do this. However, it
should not be too hard, especially if I had only one with the numbers
you gave. I'd compute the observed chi-square, then construct a series
of 4 nested "for" loops to generate all 5969040 = 22!/(15! 0! 3! 4!)
possible outcomes that sum to 22, compute the chi-square for each, and
count how many have a chi-square at least as extreme as what you
observed. If I wanted a general algorithm, that would take more work.
If you'd like more help than this, PLEASE do read the posting guide!
"http://www.R-project.org/posting-guide.html";, show us your code and
where you got stuck.
spencer graves
Christine Adrion wrote:
> Hello,
>
> I have a question concerning the R-function chisq.test.
>
> For example, I have some count data which can be categorized as follows
> class1: 15 observations
> class2: 0 observations
> class3: 3 observations
> class4: 4 observations
>
> I would like to test the hypothesis whether the population probabilities are
> all equal (=> Test for discrete uniform distribution)
> If you have a small sample size and therefore a sparse (1xr)-table, then
> assumptions for chisquare-goodness-of-fit test are violated (the numbers
> expected are less than 5 in more than 75% of the entries.)
>
> ####### R-Program: Chisquare-Test :#########
>
> mydata <- c(15,0,3,4)
> chisq.test(mydata, correct=TRUE, rescale.p = TRUE, simulate.p.value = TRUE, B
> = 2000)
>
>
> As you cannot ignore the small sample size, I use 'simulate.p.value' is
> 'TRUE' and therefore the p-value is computed by Monte Carlo simulation with
> 'B' replicates.
> But is it also the possible to use an EXACT version of a chisquare
> goodness-of-fit test without a Monte-Carlo-simulation? How can I calculate
> this in R?
>
>
>
> Any hint would be appreciated,
> Regards,
> Christine Adrion
> [[alternative HTML version deleted]]
>
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Spencer Graves, PhD
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