angelo.arcadi at virgilio.it
2011-Dec-08 21:58 UTC
[R] prop.test() and the simultaneous confidence interval for multiple proportions in R
Dear list members,
I want to perform in R the analysis "simultaneous confidence interval for
multiple proportions", as illustrated in the article of Agresti et al.
(2008) "Simultaneous confidence intervals for comparing binomial
parameter", Biometrics 64, 1270-1275.
If I am not wrong the R function implementing the Agresti et al. method is
prop.test(). I ask an help because I have some difficulties in reading the
output of that function.
As a case study, I need to apply such analysis on the following simple prolbem:
I did an experiment in which 12 participants had to choose between 3 conditions
when provided with 3 stimuli.
Stimulus Condition1 Condition2 Condition 3
A 9 1 2
B 10 2 0
C 8 2 2
My goal is to prove that it is not by chance that Condition 1 is preferred
rather than the other two conditions.
So, I apply the function prop.test(), summing the values of Conditions 2 and 3):
table<-matrix(c(9,3,10,2,8,4),ncol=2,byrow=T)
rownames(table)<-c("stimulusA","stimulusB","stimulusC")
colnames(table)<-c("Condition1","Conditions2and3")
> table
Condition1 Conditions2and3
stimulusA 9 3
stimulusB 10 2
stimulusC 8 4
prop.test(table)
> prop.test(table)
3-sample test for equality of proportions without continuity correction
data: table
X-squared = 0.8889, df = 2, p-value = 0.6412
alternative hypothesis: two.sided
sample estimates:
prop 1 prop 2 prop 3
0.7500000 0.8333333 0.6666667
Warning message:
In prop.test(table) : Chi-squared approximation may be incorrect
I don't understand where I can deduct that Condition1 is more preferred than
Conditions 2 and 3.
Should I simply look at the p-value?
The fact is that tried with a more extreme example, but the p-value results
still above 0.05:
This is the table I used:
> table2
Condition1 Condition2
stimulusA 12 0
stimulusB 10 2
stimulusC 11 1
> table2<-matrix(c(12,0,10,2,11,1),ncol=2,byrow=T)
>
rownames(table2)<-c("stimulusA","stimulusB","stimulusC")
> colnames(table2)<-c("Condition1","Condition2")
> prop.test(table2)
3-sample test for equality of proportions without continuity correction
data: table2
X-squared = 2.1818, df = 2, p-value = 0.3359
alternative hypothesis: two.sided
sample estimates:
prop 1 prop 2 prop 3
1.0000000 0.8333333 0.9166667
Warning message:
In prop.test(table2) : Chi-squared approximation may be incorrect
Could you please enlighten me?
Thanks in advance
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Jean V Adams
2011-Dec-12 13:34 UTC
[R] prop.test() and the simultaneous confidence interval for multiple proportions in R
?prop.test
The very first line in the help file on the function prop.test():
"prop.test can be used for testing the null that the proportions
(probabilities of success) in several groups are the same ..."
How to interpret the results of the prop.test() you ran? The proportion
preferring Condition 1 with stimulus A was 75%, that with B was 83%, and
that with C was 67%. Are these three proportions the same? No, X2_2 =
0.89, P = 0.64.
A search on-line yielded some "R code to find simultaneous confidence
intervals for binomial proportions" based on Agresti et al. (2008).
Perhaps you would find that useful.
http://www.stat.ufl.edu/~aa/cda/R/multcomp/ryu-simultaneous.pdf
Jean
angelo.arcadi@virgilio.it wrote on 12/08/2011 03:58:38 PM:
> Dear list members,
> I want to perform in R the analysis "simultaneous confidence
> interval for multiple proportions", as illustrated in the article of
> Agresti et al. (2008) "Simultaneous confidence intervals for
> comparing binomial parameter", Biometrics 64, 1270-1275.
>
> If I am not wrong the R function implementing the Agresti et al.
> method is prop.test(). I ask an help because I have some
> difficulties in reading the output of that function.
>
> As a case study, I need to apply such analysis on the following
> simple prolbem:
>
> I did an experiment in which 12 participants had to choose between 3
> conditions when provided with 3 stimuli.
>
> Stimulus Condition1 Condition2 Condition 3
> A 9 1 2
> B 10 2 0
> C 8 2 2
>
> My goal is to prove that it is not by chance that Condition 1 is
> preferred rather than the other two conditions.
>
> So, I apply the function prop.test(), summing the values of
> Conditions 2 and 3):
>
> table<-matrix(c(9,3,10,2,8,4),ncol=2,byrow=T)
>
rownames(table)<-c("stimulusA","stimulusB","stimulusC")
> colnames(table)<-c("Condition1","Conditions2and3")
>
> > table
> Condition1 Conditions2and3
> stimulusA 9 3
> stimulusB 10 2
> stimulusC 8 4
>
>
> prop.test(table)
>
> > prop.test(table)
>
> 3-sample test for equality of proportions without continuity
correction>
> data: table
> X-squared = 0.8889, df = 2, p-value = 0.6412
> alternative hypothesis: two.sided
> sample estimates:
> prop 1 prop 2 prop 3
> 0.7500000 0.8333333 0.6666667
>
> Warning message:
> In prop.test(table) : Chi-squared approximation may be incorrect
>
> I don't understand where I can deduct that Condition1 is more
> preferred than Conditions 2 and 3.
> Should I simply look at the p-value?
>
> The fact is that tried with a more extreme example, but the p-value
> results still above 0.05:
> This is the table I used:
>
> > table2
> Condition1 Condition2
> stimulusA 12 0
> stimulusB 10 2
> stimulusC 11 1
>
> > table2<-matrix(c(12,0,10,2,11,1),ncol=2,byrow=T)
> >
rownames(table2)<-c("stimulusA","stimulusB","stimulusC")
> >
colnames(table2)<-c("Condition1","Condition2")
> > prop.test(table2)
>
> 3-sample test for equality of proportions without continuity
correction>
> data: table2
> X-squared = 2.1818, df = 2, p-value = 0.3359
> alternative hypothesis: two.sided
> sample estimates:
> prop 1 prop 2 prop 3
> 1.0000000 0.8333333 0.9166667
>
> Warning message:
> In prop.test(table2) : Chi-squared approximation may be incorrect
>
>
>
> Could you please enlighten me?
>
>
> Thanks in advance
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