[R] exact goodness-of-fit test

[Spencer Graves]

	  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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