[R] OT: A test with dependent samples.
David Winsemius
dwinsemius at comcast.net
Wed Feb 11 04:16:15 CET 2009
Respectfully, I must disagree. (And it's not my cats, but those of
Turner's colleague.) I particularly disagree with using a Fisher's
exact test as did Turner, as it would double the sample size
improperly (even though the FET is known to be conservative.)
Your strategy appears very much in the Bayesian tradition with a very
informative prior. The strategy of calculating CI's and seeing if they
overlap appears pretty non-standard, I must say.
It appears to me that the exact version of the McNemar test offered by
Rosner is equivalent to his explication of the Wilcoxon signed rank
test which also tests for the null of symmetry of deviation on either
side of the signs of difference in rankings.
-- David Winsemius
On Feb 10, 2009, at 9:03 PM, Murray Cooper wrote:
> David,
>
> If you really want to do a test on this data, I would suggest
> a Fisher's Exact test, but you want to use hypergeometric
> probabilities. You would probably want to try the CMH
> test, if the function allows a single table and actually uses
> hypergeometric probabilities.
>
> My suggestion, would be to calculate the frequency of
> vomiting, for animals that didn't vomit before, calculate
> the CIs and then use some historical data on the vomiting
> rate, for non-treated cats and see whether it falls inside the
> CIs for your treated animals. If it does, then you might
> conclude that the vomiting rate, for treated cats, is
> similar to non-treated cats.
>
> Murray M Cooper, Ph.D.
> Richland Statistics
> 9800 N 24th St
> Richland, MI, USA 49083
> Mail: richstat at earthlink.net
>
> ----- Original Message ----- From: "David Winsemius" <dwinsemius at comcast.net
> >
> To: "Rolf Turner" <r.turner at auckland.ac.nz>
> Cc: "R-help Forum" <r-help at r-project.org>
> Sent: Tuesday, February 10, 2009 4:50 PM
> Subject: Re: [R] OT: A test with dependent samples.
>
>
>> 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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>>> {{dropped: 9}}
>>>
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>>
>> ______________________________________________
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>
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