[R] OT: A test with dependent samples.
David Winsemius
dwinsemius at comcast.net
Tue Feb 10 22:50:34 CET 2009
In the biomedical arena, at least as I learned from Rosner's
introductory text, the usual approach to analyzing paired 2 x 2 tables
is McNemar's test.
?mcnemar.test
> mcnemar.test(matrix(c(73,0,61,12),2,2))
McNemar's Chi-squared test with continuity correction
data: matrix(c(73, 0, 61, 12), 2, 2)
McNemar's chi-squared = 59.0164, df = 1, p-value = 1.564e-14
The help page has citation to Agresti.
--
David winsemius
On Feb 10, 2009, at 4:33 PM, Rolf Turner wrote:
>
> I am appealing to the general collective wisdom of this
> list in respect of a statistics (rather than R) question. This
> question
> comes to me from a friend who is a veterinary oncologist. In a
> study that
> she is writing up there were 73 cats who were treated with a drug
> called
> piroxicam. None of the cats were observed to be subject to vomiting
> prior
> to treatment; 12 of the cats were subject to vomiting after treatment
> commenced. She wants to be able to say that the treatment had a
> ``significant''
> impact with respect to this unwanted side-effect.
>
> Initially she did a chi-squared test. (Presumably on the matrix
> matrix(c(73,0,61,12),2,2) --- she didn't give details and I didn't
> pursue
> this.) I pointed out to her that because of the dependence --- same 73
> cats pre- and post- treatment --- the chi-squared test is
> inappropriate.
>
> So what *is* appropriate? There is a dependence structure of some
> sort,
> but it seems to me to be impossible to estimate.
>
> After mulling it over for a long while (I'm slow!) I decided that a
> non-parametric approach, along the following lines, makes sense:
>
> We have 73 independent pairs of outcomes (a,b) where a or b is 0
> if the cat didn't barf, and is 1 if it did barf.
>
> We actually observe 61 (0,0) pairs and 12 (0,1) pairs.
>
> If there is no effect from the piroxicam, then (0,1) and (1,0) are
> equally likely. So given that the outcome is in {(0,1),(1,0)} the
> probability of each is 1/2.
>
> Thus we have a sequence of 12 (0,1)-s where (under the null
> hypothesis)
> the probability of each entry is 1/2. Hence the probability of this
> sequence is (1/2)^12 = 0.00024. So the p-value of the (one-sided)
> test
> is 0.00024. Hence the result is ``significant'' at the usual levels,
> and my vet friend is happy.
>
> I would very much appreciate comments on my reasoning. Have I made
> any
> goof-ups, missed any obvious pit-falls? Gone down a wrong garden
> path?
>
> Is there a better approach?
>
> Most importantly (!!!): Is there any literature in which this
> approach is
> spelled out? (The journal in which she wishes to publish will
> almost surely
> demand a citation. They *won't* want to see the reasoning spelled
> out in
> the paper.)
>
> I would conjecture that this sort of scenario must arise reasonably
> often
> in medical statistics and the suggested approach (if it is indeed
> valid
> and sensible) would be ``standard''. It might even have a name!
> But I
> have no idea where to start looking, so I thought I'd ask this
> wonderfully
> learned list.
>
> Thanks for any input.
>
> cheers,
>
> Rolf Turner
>
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