Would the following strategy work?
numtests <- 20
# Create a data frame: test1 results from trial 1
# test2 results from trial 2
# agree indicagtor if trial1= trial2 (value =1) or
# trial1<>trial2 (value =0)
data <- data.frame(test1 <-rbinom(numtests,1,0.5),
test2<-rbinom(numtests,1,0.5),agree<-test1*test2)
cat("Fraction of times test1=test2",sum(data$test2)/numtests,"\n")
# Choose one of the following tests:
prop.test(sum(data$agree),20)
binom.test(sum(data$agree),20)
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
>>> csrabak <cesar.ra...@gmail.com> 9/7/2011 8:31 PM >>>
Em 7/9/2011 16:53, array chip escreveu:
> Hi all, thanks very much for sharing your thoughts. and sorry for my
> describing the problem not clearly, my fault.
>
> My data is paired, that is 2 different diagnostic tests were performed on the
> same individuals. Each individual will have a test results from each of the 2
> tests. Then in the end, 2 accuracy rates were calculated for the 2 tests. And
> I want to test if there is a significant difference in the accuracy
> (proportion) between the 2 tests. My understanding is that prop.test() is
> appropriate for 2 independent proportions, whereas in my situation, the 2
> proportions are not independent calculated from "paired" data, right?
>
> the data would look like:
>
> pid test1 test2
> p1 1 0
> p2 1 1
> p3 0 1
> :
> :
>
> 1=test is correct; 0=not correct
>
> from the data above, we can calculate accuracy for test1 and test2, then to
> compare....
>
>
> So mcnemar.test() is good for that, right?
>
> Thanks
>
John,
From above clarifying I suggest you consider the use of kappa test. For
a list of possible ways of doing it in R try:
RSiteSearch("kappa",restrict="functions")
HTH
--
Cesar Rabak
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