OT: A test with dependent samples.
On Tue, 2009-02-10 at 21:03 -0500, Murray Cooper wrote:
Hi R-masters Well, I think this a complex problem because haven't a control group OR a not randomized study. But i think the solution is a Bayesian approach. I don't know the probability of vomiting in a cat but isn't 0, so i think de priori is a beta (1[0+1],73[72+1]). The likelihood is oblivious beta(13[12+1],61[60+1]) So the posteriori is beta(1,73)*beta(13,61)=beta(14,134) The expected valeu of posteriori is 0.1 in 72 cats is same 7.2 or 7 CATS is almost a half of numbers of study.
Bernardo Rangel Tura, M.D,MPH,Ph.D
National Institute of Cardiology
Brazil
> 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
> >>