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exact goodness-of-fit test
2 messages · Christine A., Spencer Graves
2 days later
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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