Thank you Stephanie. This is perfect!
On May 10, 12:34 pm, Stephanie Kovalchik wrote:
> JC,
>
> If each row are the counts for a 2 x 2 contingency table - so for the
> ith contingency table you have counts for row 1 c(Y08[i],Z08[i]) and
> row 2 (Y09[i],Z09[i]) then you could use apply:
>
> X
JC,
If each row are the counts for a 2 x 2 contingency table - so for the
ith contingency table you have counts for row 1 c(Y08[i],Z08[i]) and
row 2 (Y09[i],Z09[i]) then you could use apply:
X <- cbind(vdata$Y08,vdata$X08-vdata$Y08,vdata$Y09,vdata$X09-vdata$Y98)
f.chisq <- function(x){
m <
On May 9, 2009, at 4:53 PM, JC wrote:
I am very new to R. I have some data from a CVS stored in vdata with 4
columns labeled:
X08, Y08, X09, Y09.
I have created two new "columns" like so:
Z08 <- (vdata$X08-vdata$Y08)
Z09 <- (vdata$X09-vdata$Y09)
I would like to use chisq.test for each "row
I am very new to R. I have some data from a CVS stored in vdata with 4
columns labeled:
X08, Y08, X09, Y09.
I have created two new "columns" like so:
Z08 <- (vdata$X08-vdata$Y08)
Z09 <- (vdata$X09-vdata$Y09)
I would like to use chisq.test for each "row" and output the p-value
for each in a sto
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