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Michael Haenlein

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]]

______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Meyners, Michael

Re: 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: [hidden email] [mailto:r-help-bounces@r-
> project.org] On Behalf Of Michael Haenlein
> Sent: Tuesday, September 27, 2011 12:45
> To: [hidden email]
> 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]]
>
> ______________________________________________
> [hidden email] 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.

______________________________________________
[hidden email] 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.

Meyners, Michael

Re: 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: [hidden email] [mailto:r-help-bounces@r-
> project.org] On Behalf Of Meyners, Michael
> Sent: Tuesday, September 27, 2011 13:28
> To: Michael Haenlein; [hidden email]
> 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: [hidden email] [mailto:r-help-bounces@r-
> > project.org] On Behalf Of Michael Haenlein
> > Sent: Tuesday, September 27, 2011 12:45
> > To: [hidden email]
> > 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]]
> >
> > ______________________________________________
> > [hidden email] 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.
>
> ______________________________________________
> [hidden email] 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.

Michael Haenlein

Re: Pearson chi-square test

In reply to this post by Michael Haenlein

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:[hidden email]]
Sent: Dienstag, 27. September 2011 17:00
To: Michael Haenlein; [hidden email]
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: [hidden email] [mailto:r-help-bounces@r-

> project.org] On Behalf Of Meyners, Michael

> Sent: Tuesday, September 27, 2011 13:28

> To: Michael Haenlein; [hidden email]

> 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: [hidden email] [mailto:r-help-bounces@r-

> > project.org] On Behalf Of Michael Haenlein

> > Sent: Tuesday, September 27, 2011 12:45

> > To: [hidden email]

> > 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]]

______________________________________________
[hidden email] 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.

Meyners, Michael

Re: Pearson chi-square test

I suspect that the chisquare-test might not be appropriate, as you have constraints (same number of observations for A in both contingency tables). I further suspect that there is no test readily available for that, but I might be wrong. Maybe randomization tests could help here, but it would require a bit of thinking AND programming to accomplish that. chisq.test might give you an approximate solution, but I can't say how good this will be (and it might also depend on the data, btw).
Best, Michael

From: Michael Haenlein
Sent: Tuesday, September 27, 2011 17:05
To: [hidden email]
Cc: Meyners, Michael
Subject: RE: [R] Pearson chi-square test

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
Sent: Dienstag, 27. September 2011 17:00
To: Michael Haenlein; [hidden email]
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.