Dan Bolser wrote:
I have the following contingency table
dat <- matrix(c(1,506,13714,878702),nr=2)
And I want to test if their is an association between events
A:{a,not(a)} and B:{b,not(b)}
| b | not(b) | --------+-----+--------+ a | 1 | 13714 | --------+-----+--------+ not(a) | 506 | 878702 | --------+-----+--------+
I am worried that prop.test and chisq.test are not valid given the low
counts and low probabilites associated with 'sucess' in each category.
> test <- matrix( c(1,506, 13714, 878702), 2,2) > test [,1] [,2] [1,] 1 13714 [2,] 506 878702 > chisq.test(test)
Pearson's Chi-squared test with Yates' continuity correction
data: test X-squared = 5.1584, df = 1, p-value = 0.02313
> chisq.test(test,sim=TRUE)
Pearson's Chi-squared test with simulated p-value (based on 2000
replicates)data: test X-squared = 6.0115, df = NA, p-value = 0.02099
> chisq.test(test,sim=TRUE, B=200000)
# To rule out simulation uncertainty
Pearson's Chi-squared test with simulated p-value (based on 200000
replicates)data: test X-squared = 6.0115, df = NA, p-value = 0.01634
> N <- sum(test)
> rows <- rowSums(test)
> cols <- colSums(test)
> E <- rows %o% cols/N
> E
[,1] [,2]
[1,] 7.787351 13707.21
[2,] 499.212649 878708.79
>None of the expected'eds are lesser than 5, an often used rule of thumb (which might even be
to conservative. Just to check on the distribution:
rows <- round(rows) cols <- round(cols)
> pvals <- sapply(r2dtable(100000, rows, cols), function(x) chisq.test(x)$p.value)
> hist(pvals)
# not very good approximation to uniform histogram, but:
> sum(pvals < 0.05)/100000
[1] 0.03068
> sum(pvals < 0.01)/100000
[1] 0.00669
So the true levels are not very far from the calculated by te chisq approximation, and it seems safe to use it.
All of this to show that with R you are not anymore dependent on old "rules os thumb",
you can investigate for yourself.
Kjetil
Is it safe to use them, and what is the alternative? (given that fisher.test can't handle this data... hold the phone...
I just found fisher.test can handle this data if the test is one-tailed and not two-tailed.
I don't understand the difference between chisq.test, prop.test and fisher.test when the hybrid=1 option is used for the fisher.test.
I was using the binomial distribution to test the 'extremity' of the observed data, but now I think I know why that is inapropriate, however, with the binomial (and its approximation) at least I know what I am doing. And I can do it in perl easily...
Generally, how should I calculate fisher.test in perl (i.e. what are its principles). When is it safe to approximate fisher to chisq?
I cannot get insight into this problem...
How come if I do...
dat <- matrix(c(50,60,100,100),nr=2)
prop.test(dat)$p.value chisq.test(dat)$p.value fisher.test(dat)$p.value
I get
[1] 0.5173269 [1] 0.5173269 [1] 0.4771358
When I looked at the binomial distribution and the normal approximation thereof with similar counts I never had a p-value difference > 0.004
I am so fed up with this problem :(
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--
Kjetil Halvorsen.
Peace is the most effective weapon of mass construction.
-- Mahdi Elmandjra______________________________________________ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
