R help - Sep 2011 - 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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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 at r-project.org [mailto:r-help-bounces at r-
> project.org] On Behalf Of Michael Haenlein
> Sent: Tuesday, September 27, 2011 12:45
> To: r-help at 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 at 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.

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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