[R] comparing proportions
array chip
arrayprofile at yahoo.com
Thu Feb 10 23:17:29 CET 2011
Robert, thank you!
I tried all 3 models you suggested. Since each subject only has one line of data
in my dataset, would including Subject as a factor in glm() or lm() lead to 0 df
for resaiduals?
Attached is my dataset and here is my version of the 3 models:
test<-read.table("test.txt",sep='\t',header=T)
library(lme4)
## model 1: generalized mixed model
fit1<-glmer(cbind(success,failure)~group+(1|subject),data=test,family=binomial)
## model 2: generalized model
fit2<-glm(cbind(success,failure)~group,data=test,family=binomial)
## model 3: linear model
fit3<-lm(prop.success~group,weights=success+failure,data=test)
> fit1
Generalized linear mixed model fit by the Laplace approximation
Formula: cbind(success, failure) ~ group + (1 | subject)
Data: test
AIC BIC logLik deviance
54.75 57.89 -24.38 48.75
Random effects:
Groups Name Variance Std.Dev.
subject (Intercept) 1.8256 1.3511
Number of obs: 21, groups: subject, 21
Fixed effects:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -0.6785 0.4950 -1.371 0.170
groupgroup2 -0.7030 0.6974 -1.008 0.313
> summary(fit2)
Call:
glm(formula = cbind(success, failure) ~ group, family = binomial,
data = test)
Deviance Residuals:
Min 1Q Median 3Q Max
-2.8204 -2.0789 -0.5407 1.0403 4.0539
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -0.4400 0.2080 -2.115 0.0344 *
groupgroup2 -0.6587 0.3108 -2.119 0.0341 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 78.723 on 20 degrees of freedom
Residual deviance: 74.152 on 19 degrees of freedom
> summary(fit3)
Call:
lm(formula = prop.success ~ group, data = test, weights = success +
failure)
Residuals:
Min 1Q Median 3Q Max
-1.1080 -0.7071 -0.2261 0.4754 1.9157
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.39175 0.08809 4.447 0.000276 ***
groupgroup2 -0.14175 0.12364 -1.146 0.265838
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.8676 on 19 degrees of freedom
Multiple R-squared: 0.0647, Adjusted R-squared: 0.01548
F-statistic: 1.314 on 1 and 19 DF, p-value: 0.2658
As you can see, all 3 models gave quite different results, the model 2
(generalized linear model) gave significant p value while model 1 (generalized
mixed model) and model 3 (regular linear model) did not. Also as you can from
the data, prop.success has some value outside the range 0.15 to 0.85, so maybe
regular linear model may not be appropriate?
Thank you,
John
________________________________
From: Robert A LaBudde <ral at lcfltd.com>
To: array chip <arrayprofile at yahoo.com>
Cc: Bert Gunter <gunter.berton at gene.com>; R-help at stat.math.ethz.ch
Sent: Thu, February 10, 2011 12:54:44 PM
Subject: Re: [R] comparing proportions
1. If you use a random effects model, you should make Subject the random factor.
I.e., a random intercepts model with 1|Subject. Group is a fixed effect: You
have only 2 groups. Even if you had more than 2 groups, treating Group as random
would return a standard deviation, not a P-value as you wanted. Finally, I doubt
you believe the groups used are meaningless, and only the population of groups
is of interest. Instead you consider them special, so Group is a fixed effect.
2. The number of observations for each Subject is the number of trials, which
you previously indicated were 7 to 10 in the cases listed.
3. If you have no interest in the Subject effect, you can use a fixed Subject
factor instead with glm() instead of glmer() or other mixed model function. This
is a good idea so long as the number of subjects is, say, less than 10.
Otherwise a mixed model would be a better idea.
I suggest you fit all three models to learn about what you're doing: 1) glmer()
or equivalent, with cbind(successes, failures) ~ 1|Subject + Group; 2) glm()
with cbind(successes, failures) ~ Subject + Group; and 3) lm(p ~ Subject +
Group), where p is the proportion success for a particular subject and group.
Then compare the results. They will probably all 3 give the same conclusion to
the hypothesis question about Group. I would guess the glmer() P-value will be
larger, then the glm() and finally the lm(), but the last two may reverse. The
lm() model may actually perform fairly well, as the Edgeworth series converges
rapidly to normal for binomial distributions with p within 0.15 to 0.85 and 10+
replicates, as I stated before.
I'd be interested in seeing the results of these 3 fits myself just for
curiosity.
At 01:21 PM 2/10/2011, array chip wrote:
> Robert and Bert, thank you both very much for the response, really appreciated.
>I agree that using regular ANOVA (or regular t test) may not be wise during the
>normality issue. So I am debating between generalized linear model using glm(..,
>family=binomial) or generalized linear mixed effect model using glmer(...,
>family=binomial). I will forward to Robert an offline list email I sent to Bert
>about whether using (1|subject) versus (1|group) in mixed model specification.
>If using (1|group), both models will give me the same testing for fixed effects,
>which is what I am mainly interested in. So do I really need a mixed model here?
>
> Thanks again
>
> John
>
>
> From: Bert Gunter <gunter.berton at gene.com>
> To: Robert A LaBudde <ral at lcfltd.com>
> Cc: array chip <arrayprofile at yahoo.com>
> Sent: Thu, February 10, 2011 10:04:06 AM
> Subject: Re: [R] comparing proportions
>
> Robert:
>
> Yes, exactly. In an offlist email exchange, he clarified this for me,
> and I suggested exactly what you did, also with the cautions that his
> initial ad hoc suggestions were unwise. His subsequent post to R-help
> and the sig-mixed-models lists were the result, although he appears to
> have specified the model incorrectly in his glmer function (as
> (1|Group) instead of (1|subject).
>
> Cheers,
> Bert
>
> On Thu, Feb 10, 2011 at 9:55 AM, Robert A LaBudde
><<mailto:ral at lcfltd.com>ral at lcfltd.com> wrote:
> > prop.test() is applicable to a binomial experiment in each of two classes.
> >
> > Your experiment is binomial only at the subject level. You then have
> > multiple subjects in each of your groups.
> >
> > You have a random factor "Subjects" that must be accounted for.
> >
> > The best way to analyze is a generalized linear mixed model with a binomial
> > distribution family and a logit or probit link. You will probably have to
> > investigate overdispersion. If you have a small number of subjects, and
> > don't care about the among-subject effect, you can model them as fixed
> > effects and use glm() instead.
> >
> > Your original question, I believe, related to doing an ANOVA assuming
> > normality. In order for this to work with this kind of proportion problem,
> > you generally won't get good results unless the number of replicates per
> > subject is 12 or more, and the proportions involved are within 0.15 to 0.85.
> > Otherwise you will have biased confidence intervals and significance tests.
> >
> >
> >
> > At 07:51 PM 2/9/2011, array chip wrote:
> >>
> >> Content-type: text/plain
> >> Content-disposition: inline
> >> Content-length: 2969
> >>
> >> Hi Bert,
> >>
> >> Thanks for your reply. If I understand correctly, prop.test() is not
> >> suitable to
> >> my situation. The input to prop.test() is 2 numbers for each group (# of
> >> success
> >> and # of trials, for example, groups 1 has 5 success out of 10 trials;
> >> group 2
> >> has 3 success out of 7 trials; etc. prop.test() tests whether the
> >> probability of
> >> success is the same across groups.
> >>
> >> In my case, each group has several subjects and each subject has 2 numbers
> >> (#
> >> success and # trials). So
> >>
> >> for group 1:
> >> subject 1: 5 success, 10 trials
> >> subject 2: 3 success, 8 trials
> >> :
> >> :
> >>
> >> for group 2:
> >> subject a: 7 success, 9 trials
> >> subject b: 6 success, 7 trials
> >> :
> >> :
> >>
> >> I want to test whether the probability of success in group 1 is the same
> >> as in
> >> group 2. It's like comparing 2 groups of samples using t test, what I am
> >> uncertain about is that whether regular t test (or non-pamametric test) is
> >> still
> >> appropriate here when the response variable is actually proportions.
> >>
> >> I guess prop.test() can not be used with my dataset, or I may be wrong?
> >>
> >> Thanks
> >>
> >> John
> >>
> >>
> >>
> >>
> >>
> >>
> >>
> >> ________________________________
> >> From: Bert Gunter <<mailto:gunter.berton at gene.com>gunter.berton at gene.com>
> >>
> >> Sent: Wed, February 9, 2011 3:58:05 PM
> >> Subject: Re: [R] comparing proportions
> >>
> >> 1. Is this a homework problem?
> >>
> >> 2. ?prop.test
> >>
> >> 3. If you haven't done so already, get and consult a basic statistical
> >> methods book to help you with questions such as this.
> >>
> >> -- Bert
> >>
> >>
> >> > Hi, I have a dataset that has 2 groups of samples. For each sample, then
> >> > response measured is the number of success (no.success) obatined with
> >> > the
> >> >number
> >> > of trials (no.trials). So a porportion of success (prpop.success) can be
> >> > computed as no.success/no.trials. Now the objective is to test if there
> >> > is a
> >> > statistical significant difference in the proportion of success between
> >> > the 2
> >> > groups of samples (say n1=20, n2=30).
> >> >
> >> > I can think of 2 ways to do the test:
> >> >
> >> > 1. regular t test based on the variable prop.success
> >> > 2. Mann-Whitney test based on the variable prop.success
> >> > 2. do a binomial regression as:
> >> > fit<-glm(cbind(no.success,no.trials-no.success) ~ group, data=data,
> >> > family=binomial)
> >> > anova(fit, test='Chisq')
> >> >
> >> > My questions is:
> >> > 1. Is t test appropriate for comparing 2 groups of proportions?
> >> > 2. how about Mann-Whitney non-parametric test?
> >> > 3. Among the 3, which technique is more appropriate?
> >> > 4. any other technique you can suggest?
> >> >
> >> > Thank you,
> >> >
> >> > John
> >> >
> >> >
> >> >
> >> > [[alternative HTML version deleted]]
> >> >
> >> >
> >> > ______________________________________________
> >> > <mailto:R-help at r-project.org>R-help at r-project.org mailing list
> >> >
><https://stat.ethz.ch/mailman/listinfo/r-help>https://stat.ethz.ch/mailman/listinfo/r-help
>
> >> > PLEASE do read the posting guide
> >> >
><http://www.R-project.org/posting-guide.html>http://www.R-project.org/posting-guide.html
>
> >> > and provide commented, minimal, self-contained, reproducible code.
> >> >
> >> >
> >>
> >>
> >>
> >> --
> >> Bert Gunter
> >> Genentech Nonclinical Biostatistics
> >>
> >>
> >>
> >>
> >> [[alternative HTML version deleted]]
> >>
> >>
> >> ______________________________________________
> >> <mailto:R-help at r-project.org>R-help at r-project.org mailing list
> >>
><https://stat.ethz.ch/mailman/listinfo/r-help>https://stat.ethz.ch/mailman/listinfo/r-help
>
> >> PLEASE do read the posting guide
> >>
><http://www.R-project.org/posting-guide.html>http://www.R-project.org/posting-guide.html
>
> >> and provide commented, minimal, self-contained, reproducible code.
> >
> > ================================================================
> > Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail:
><mailto:ral at lcfltd.com>ral at lcfltd.com
> > Least Cost Formulations, Ltd. URL:
><http://lcfltd.com/>http://lcfltd.com/
> > 824 Timberlake Drive Tel: 757-467-0954
> > Virginia Beach, VA 23464-3239 Fax: 757-467-2947
> >
> > "Vere scire est per causas scire"
> > ================================================================
> >
> >
>
>
>
> --
> Bert Gunter
> Genentech Nonclinical Biostatistics
> 467-7374
><http://devo.gene.com/groups/devo/depts/ncb/home.shtml>http://devo.gene.com/groups/devo/depts/ncb/home.shtml
>l
================================================================
Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: ral at lcfltd.com
Least Cost Formulations, Ltd. URL: http://lcfltd.com/
824 Timberlake Drive Tel: 757-467-0954
Virginia Beach, VA 23464-3239 Fax: 757-467-2947
"Vere scire est per causas scire"
================================================================
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