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:meyners.m@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-bounces@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-bounces@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

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