[R] comparing matched proportions using glm

[Charles C. Berry]

On Thu, 4 Oct 2007, Corry Gellatly wrote:

>
> Dear R users,
>
> Is it possible to use a generalized linear model to do a binomial
> comparison of one list of proportions with a matched list of proportions
> to test for a difference?
>
> So, for example:
>
> list 1  		list 2
>
> a1  |  b1        	a2 |  b2
>
> 3   |  4          7  |  9
> 6   |  7          5  |  1
> 9   |  1          3  |  1
>
>
> I want to compare list 1 with list 2 and the samples are matched.


Meaning that

 	 3     4          7    9

are the _counts_ in one stratum of three in all?

And you have an hypothesis that claims the proportions are equal in each 
stratum??

The obvious candidate for that setup is a log-linear model for the counts 
in a 2 by 2 by 3 table.

See

 	?loglin

and

 	?loglm (in MASS)

and the refernces therein.

You can do this type of work in glm() if you understand surrogate Poisson 
models as outlined in

McCullagh P. and Nelder, J. A. (1989) Generalized Linear Models. London: 
Chapman and Hall.

HTH,

Chuck

> Obviously, I could add the columns and do a binomial test, i.e.
> prop.test(c(18,15),c(30,26)), however, I have a large dataset so this
> would reduce the power of my analysis. I could compare the ratios, i.e.
> a1/(a1+b1) compared to a2/(a2+b2) for the samples in each list, however,
> this does not account for the difference in sample sizes between samples
> in each list.
>
> I have tried a glm where I bind a2 and b2 as the y variable, i.e.
> y<-cbind(a2,b2) and also bind a1 and b1 as the x variable, i.e.
> y<-cbind(a1,b1) and run <-glm(y~x,binomial)
>
> I get this type of output:
>
> 	Call:
> 	glm(formula = y ~ x, family = binomial)
>
> 	Deviance Residuals:
> 	     Min        1Q    Median        3Q       Max
> 	-3.20426  -0.72686  -0.01822   0.68320   4.05035
>
> 	Coefficients:
> 	             Estimate Std. Error z value Pr(>|z|)
> 	(Intercept)  0.178369   0.186421   0.957    0.339
> 	xa1     	 0.008109   0.017430   0.465    0.642
> 	xb1		-0.026666   0.018153  -1.469    0.142
>
> 	(Dispersion parameter for binomial family taken to be 1)
>
> 	    Null deviance: 565.14  on 467  degrees of freedom
> 	Residual deviance: 559.69  on 465  degrees of freedom
> 	AIC: 1883.3
>
> 	Number of Fisher Scoring iterations: 3
>
>
> Is this output meaningful? It seems that y is not compared directly with
> x, but rather compared with a1 and b1, which is not intended?
>
> I wonder if this is a suitable approach to the problem? I'll be very
> grateful for any advice or suggestions.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

Charles C. Berry                            (858) 534-2098
                                             Dept of Family/Preventive Medicine
E mailto:cberry at tajo.ucsd.edu	            UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901