Daniel's response about using a mixed model
"whopped me alongside the head", so now I'm
thinking more clearly. Here are a few more comments:
1. The LRT based on ML or the resampling
alternative is probably the most powerful among
reasonable algorithms to do a test on your
problem. You might even get a survival package to do it for you.
2. The McNemar test with casewise deletion should
work well enough, so long as you don't lose too
much data (e.g., less than 50%). Unfortunately,
this seems to be your situation.
3. The mixed model (or, more properly, the
generalized mixed model) approach will involve a
number of new assumptions not necessary in more direct methods.
4. A Bayesian approach would be better than 1)
above if you've got some useful prior information.
5. If you have a lot of missing data (> 50%), as
you appear to have from your example, you
generally will get more power than the McNemar
test from just ignoring the matching on subjects,
treating it as unpaired data (2 independent
samples). Not as good as method 1), but the high
missing rate would justify the independence
assumption, as it would reduce severely whatever
correlation is present on the same vs. different
subjects. The extra data may be worth more than
the pairing, particularly if there is a large
percentage of concordant pairs in the McNemar test of symmetry.
6. Finally, why are you even doing a test? Tests
don't provide physically useful information, only
that your study has enough subjects to
statistically detect an effect or not. Wouldn't
you be better off asking how to obtain a valid
confidence interval for the difference or ratio
of proportions for the two tests? The concordant pairs would then be useful.
At 03:17 PM 7/24/2009, Daniel Malter wrote:
Hi Tal, you can use the lme4 library and use a random effects model. I will
walk you through the example below (though, generally you should ask a
statistician at your school about this). At the very end, I will include a
loop for Monte-Carlo simulations that shows that the estimation of the fixed
effects is unbiased, which assures you that you are estimating the "correct"
coefficient. In addition, as Robert says, you should remove all subjects for
which both observations are missing.
##First, we simulate some data.
##Note how the data structure matches the data structure you have:
id=rep(1:100,each=2)
#the subject ID
test=rep(0:1,100)
# first or second measurement/test (your variable of interest)
randeff=rep(rnorm(100,0,1),each=2)
# a random effect for each subject
linear.predictor=randeff+2*test
#the linear predictor
#note the coefficient on the fixed effect of "test" is 2
probablty=exp(linear.predictor)/(1+exp(linear.predictor))
#compute the probability using the linear predictor
y=rbinom(200,1,probablty)
#draw 0/1 dependent variables
#with probability according to the variable probablty
miss=sample(1:200,20,replace=FALSE)
#some indices for missing variables
y[miss]=NA
#replace the dependent variable with NAs for the "miss"ing indices
##Some additional data preparation
id=factor(id)
#make sure id is coded as a factor/dummy
mydata=data.frame(y,test,id)
#bind the data you need for estimation in a dataframe
##Run the analysis
library(lme4)
reg=lmer(y~test+(1|id),binomial,data=mydata)
summary(reg)
##And here are the Monte-Carlo simulations
library(lme4)
bootstraps=200
estim=matrix(0,nrow=bootstraps,ncol=2)
for(i in 1:bootstraps){
id=rep(1:100,each=2)
test=rep(0:1,100)
randeff=rep(rnorm(100,0,1),each=2)
linear.predictor=randeff+2*test
probablty=exp(linear.predictor)/(1+exp(linear.predictor))
y=rbinom(200,1,probablty)
miss=sample(1:200,20,replace=FALSE)
y[miss]=NA
id=factor(id)
mydata=data.frame(y,test,id)
reg=lmer(y~test+(1|id),binomial,data=mydata)
estim[i,]=...@fixef
}
mean(estim[,1])
#mean estimate of the intercept from 200 MC-simulations
#should be close to zero (no intercept in the linear.predictor)
mean(estim[,2])
#mean estimate of the 'test' fixed effect from 200 MC-simulations
#should be close to 2 (as in the linear.predictor)
#plot estimated coefficients from 200 MC simulations
par(mfcol=c(1,2))
hist(estim[,1],main="Intercept")
hist(estim[,2],main="Effect of variable 'test'")
HTH,
Daniel
-------------------------
cuncta stricte discussurus
-------------------------
-----Ursprüngliche Nachricht-----
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von Tal Galili
Gesendet: Friday, July 24, 2009 1:23 PM
An: r-help@r-project.org
Betreff: [R] How to test frequency independence (in a 2 by 2 table) withmany
missing values
Hello dear R help group.
My question is statistical and not R specific, yet I hope some of you might
be willing to help.
*Experiment settings*: We have a list of subjects. each of them went
through two tests with the answer to each can be either 0 or 1.
*Goal*: We want to know if the two experiments yielded different results in
the subjects answers.
*Statistical test (and why it won't work)*: Naturally we would turn to
performing a mcnemar test. But here is the twist: we have missing values in
our data.
For our purpose, let's assume the missingnes is completely at random, and we
also have no data to use for imputation. Also, we have much more missing
data for experiment number 2 then in experiment number 1.
So the question is, under these settings, how do we test for experiment
effect on the outcome?
So far I have thought of two options:
1) To perform the test only for subjects that has both values. But they are
too scarce and will yield low power.
2) To treat the data as independent and do a pearson's chi square test
(well, an exact fisher test that is) on all the non-missing data that we
have. The problem with this is that our data is not fully independent (which
is a prerequisite to chi test, if I understood it correctly). So I am not
sure if that is a valid procedure or not.
Any insights will be warmly welcomed.
p.s: here is an example code producing this scenario.
set.seed(102)
x1 <- rbinom(100, 1, .5)
x2 <- rbinom(100, 1, .3)
X <- data.frame(x1,x2)
tX <- table(X)
margin.table(tX,1)
margin.table(tX,2)
mcnemar.test(tX)
put.missings <- function(x.vector, na.percent) { turn.na <-
rbinom(length(x.vector), 1, na.percent) x.vector[turn.na == 1] <- NA
return(x.vector)
}
x1.m <- put.missings(x1, .3)
x2.m <- put.missings(x2, .6)
tX.m <- rbind(table(na.omit(x1.m)), table(na.omit(x2.m)))
fisher.test(tX.m)
With regards,
Tal
--
----------------------------------------------
My contact information:
Tal Galili
Phone number: 972-50-3373767
FaceBook: Tal Galili
My Blogs:
http://www.r-statistics.com/
http://www.talgalili.com
http://www.biostatistics.co.il
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================================================================
Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: r...@lcfltd.com
Least Cost Formulations, Ltd. URL: http://lcfltd.com/
824 Timberlake Drive Tel: 757-467-0954
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