Hi 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) and chisq.test(tab) 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) x [1] "c" "a" "c" "c" "a" "c" "a" "c" "a" "a" y<-sample(1:5, 20, replace=T) > y [1] 2 5 1 1 2 5 2 3 1 5 5 5 1 5 5 3 2 2 5 1 > chisq.test(x,y) Error in chisq.test(x, y) : 'x' and 'y' must have the same length x<-sample(letters[1:3], 20, replace=T) > chisq.test(x,y) Pearson's Chi-squared test data: x and y X-squared = 4.7937, df = 6, p-value = 0.5705 Warning message: In chisq.test(x, y) : Chi-squared approximation may be incorrect > Regards Petr [EMAIL PROTECTED] napsal dne 06.12.2007 23:09:24: > > The chi-square does not need your two categorical variables to have equal > 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. You > can modify it for your data and should give you no error. > > data bloodtype; > input bloodtype$ state$ count@@; > datalines; > 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; > quit; > > > Best > Ramin > Gainesville > > > > 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 > > > >>temp<-chisq.test(x,y) > > > > but got error "the lengths of the two vectors don't match". any help > > will be appreciated > > -- > > Regards, > > Rana Shoaaib Mehmood > > > > ______________________________________________ > > 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. > > > > > > -- > View this message in context: http://www.nabble.com/testing-independence-of- > categorical-variables-tf4855773.html#a14202348 > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > 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. ______________________________________________ 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.