[R] testing independence of categorical variables
petr.pikal at precheza.cz
Fri Dec 7 08:46:20 CET 2007
Well, R does exactly what it says. From help page.
"Otherwise, x and y must be vectors or factors of the same length"
I do not know SAS but I presume that
> tables bloodtype*state
gives you something like
tab <- table(bloodtype, state)
shall give you the expected result. You can also do directly
chisq.test(bloodtype, state). But what you cannot do is to test vectors
unequal **lengths**, and that is what he did. I beleve that you can not do
it in SAS either.
x<-sample(letters[1:3], 10, replace=T)
 "c" "a" "c" "c" "a" "c" "a" "c" "a" "a"
y<-sample(1:5, 20, replace=T)
 2 5 1 1 2 5 2 3 1 5 5 5 1 5 5 3 2 2 5 1
Error in chisq.test(x, y) : 'x' and 'y' must have the same length
x<-sample(letters[1:3], 20, replace=T)
Pearson's Chi-squared test
data: x and y
X-squared = 4.7937, df = 6, p-value = 0.5705
In chisq.test(x, y) : Chi-squared approximation may be incorrect
r-help-bounces at r-project.org napsal dne 06.12.2007 23:09:24:
> The chi-square does not need your two categorical variables to have
> levels, nor limitation for the number of levels.
> The Chi-square procedure is as follow:
> χ^2=∑_(All Cells)▒〖(Observed-Expected)〗^2/Expected
> Expected Cell= E_ij=n((i^th RowTotal)/n)((j^th RowTotal)/n)
> Degree of Freedom=df= (row-1)(Col-1)
> This way should not give you any errors if your calculations are all
> correct. I usually use SAS for calculations like this. Below is a sample
> code I wrote to test whether US_State and Blood type are independent.
> can modify it for your data and should give you no error.
> data bloodtype;
> input bloodtype$ state$ count@@;
> A FL 122 B FL 117
> AB FL 19 O FL 244
> A IA 1781 B IA 351
> AB IA 289 O IA 3301
> A MO 353 B MO 269
> AB MO 60 O MO 713
> proc freq data=bloodtype;
> tables bloodtype*state
> / cellchi2 chisq expected norow nocol nopercent;
> weight count;
> Shoaaib Mehmood wrote:
> > hi,
> > is there a way of calculating of measuring dependence between two
> > categorical variables. i tried using the chi square test to test for
> > independence but i got error saying that the lengths of the two
> > vectors don't match. Suppose X and Y are two factors. X has 5 levels
> > and Y has 7 levels. This is what i tried doing
> > but got error "the lengths of the two vectors don't match". any help
> > will be appreciated
> > --
> > Regards,
> > Rana Shoaaib Mehmood
> > ______________________________________________
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