[R] OT: A test with dependent samples.

[David Winsemius]

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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