1. Drop all subjects where both test results are not present. These
are uninformative on the difference.
2. Perform McNemar test on the remaining table. (Only the different
pairs are informative.)
This will give you a p-value on the data that actually contrasts the
two tests. (For your example, the p-value is
> X2<- data.frame(x1.m, x2.m)
> tX2<- table(X2)
> tX2
x2.m
x1.m 0 1
0 6 5
1 15 4
> mcnemar.test(tX2)
McNemar's Chi-squared test with continuity correction
data: tX2
McNemar's chi-squared = 4.05, df = 1, p-value = 0.04417
The low power is a consequence of the data not being informative.
You'll have to live with it.
If you want to squeeze any more out this data, I'd guess about the
only way to do it would be via a maximum likelihood approach with
marginals for the one test only data, an assumption of, e.g., a
binomial distribution for each test, and then use a likelihood ratio
test for p1=p2 vs. p1!=p2.
If you really feel up to it, you could program a randomization or
bootstrap test equivalent to the ML approach, maintaining the
marginal totals involved.
At 01:23 PM 7/24/2009, Tal Galili wrote:
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
--
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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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