David Winsemius
On Feb 10, 2009, at 4:50 PM, David Winsemius wrote:
> 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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>
> ______________________________________________
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