Dear Michael,
Thanks very much for your answers! The purpose of my analysis is to test whether the contingency table x is different from the contingency table y. Or, to put it differently, whether there is a significant difference between the joint distribution A&B and A&C. Based on your answer I'm wondering whether the best way to do this is really a chisq.test? Or is there probably a different function or package I should use altogether? Thanks, Michael -----Original Message----- From: Meyners, Michael [mailto:meyner...@pg.com] Sent: Dienstag, 27. September 2011 17:00 To: Michael Haenlein; r-help@r-project.org Subject: RE: [R] Pearson chi-square test Just for completeness: the manual calculation you'd want is most likely sum((x-y)^2 / (x+y)) (that's one you can find on the Wikipedia link you provided). To get the same from chisq.test, try something like chisq.test(data.frame(x,y)[,c(3,6)]) (there are surely smarter ways, but at least it works here). Note that something like chisq.test(as.vector(x), as.vector(y)) will give a different test, i.e. based on a contingency table of x cross y). M. > -----Original Message----- > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- > project.org] On Behalf Of Meyners, Michael > Sent: Tuesday, September 27, 2011 13:28 > To: Michael Haenlein; r-help@r-project.org > Subject: Re: [R] Pearson chi-square test > > Not sure what you want to test here with two matrices, but reading the > manual helps here as well: > > y a vector; ignored if x is a matrix. > > x and y are matrices in your example, so it comes as no surprise that > you get different results. On top of that, your manual calculation is > not correct if you want to test whether two samples come from the same > distribution (so don't be surprised if R still gives a different > value...). > > HTH, Michael > > > -----Original Message----- > > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- > > project.org] On Behalf Of Michael Haenlein > > Sent: Tuesday, September 27, 2011 12:45 > > To: r-help@r-project.org > > Subject: [R] Pearson chi-square test > > > > Dear all, > > > > I have some trouble understanding the chisq.test function. > > Take the following example: > > > > set.seed(1) > > A <- cut(runif(100),c(0.0, 0.35, 0.50, 0.65, 1.00), labels=FALSE) > > B <- cut(runif(100),c(0.0, 0.25, 0.40, 0.75, 1.00), labels=FALSE) > > C <- cut(runif(100),c(0.0, 0.25, 0.50, 0.80, 1.00), labels=FALSE) > > x <- table(A,B) > > y <- table(A,C) > > > > When I calculate the test statistic by hand I get a value of > > approximately > > 75.9: > > http://en.wikipedia.org/wiki/Pearson's_chi- > > square_test#Calculating_the_test-statistic > > sum((x-y)^2/y) > > > > But when I do chisq.test(x,y) I get a value of 12.2 while > > chisq.test(y,x) > > gives a value of 10.3. > > > > I understand that I must be doing something wrong here, but I'm not > > sure > > what. > > > > Thanks, > > > > Michael [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
