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testing independence of categorical variables

2 messages · Ramin Shamshiri, PIKAL Petr

Original
Ramin Shamshiri
# 
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:
?^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:

  
    

      
      
    

  


  


    

    

      
      

        


  

        

      PIKAL Petr
    

    

      
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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

            

      
      
      
    

        
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)

            

      
      
      
    

        
[1] 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

Warning message:
In chisq.test(x, y) : Chi-squared approximation may be incorrect

            

      
      
      
    

        
Regards
Petr


r-help-bounces at r-project.org napsal dne 06.12.2007 23:09:24:

            

      
      
      
    

        
equal

            

      
      
      
    

        
You

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        
http://www.nabble.com/testing-independence-of-

            

      
      
      
    

        

      
      
    

        
http://www.R-project.org/posting-guide.html