This morning, in our mathematical statistics course, we’ve been discussing the ‘proportion test‘, i.e. given a sample of Bernoulli trials, with , we want to test
against
A natural test (which can be related to the maximum likelihood ratio test) is based on the statistic
The test function is here
To get the bounds of the acceptance region, we need the distribution of , under . Consider here a numerical application
n=20
p=.5
set.seed(1)
echantillon=sample(0:1,size=n,
prob=c(1-p,p),
replace=TRUE)
- the asymptotic distribution
The first (and standard idea) is to use the central limit theorem, since
So, under ,
Then while . The acceptance region is then between the two red lines, below,
T=sqrt(n)*(mean(echantillon)-.5)/
sqrt(mean(echantillon)*
(1-mean(echantillon)))
u=seq(-3,3,by=.01)
v=dnorm(u)
plot(u,v,type="l",lwd=2)
abline(v=qnorm(.025),col="red")
abline(v=qnorm(.975),col="red")
abline(v=T,col="blue")
Here we use the fact that
Using transformation of the ‘density’, we can (at least numerically) compute the (exact) distribution of
u=seq(-3,3,by=.01)
v=sqrt(.5*(1-.5))*n*dbinom(round(
(sqrt(.5*(1-.5))*u/sqrt(n)+.5)*n),
size=n,prob=.5)/sqrt(n)
Here I used a round value, it guess it would be better with a floor function, but here the graph looks symmetric (which is something I like)
abline(v=sqrt(n)*(qbinom(.025,size=n,prob=.5)/n-.5)/sqrt(.5*(1-.5)),col="red")
abline(v=sqrt(n)*(qbinom(.975,size=n,prob=.5)/n-.5)/sqrt(.5*(1-.5)),col="red")
lines(u,v,type="s")
- distribution based on Monte Carlo simulations
Probably more interesting, here we do not use the fact that we might know the distribution of the mean. We just generate random samples, under , and then compute ,
T=rep(NA,1000)
for(i in 1:1000){
x=sample(0:1,size=n,
prob=c(1-.5,.5),
replace=TRUE)
m=mean(x)
T[i]=(m-.5)/sqrt(m*(1-m))*sqrt(n)}
lines(density(T),lwd=2)
abline(v=quantile(T,.025),col="red")
abline(v=quantile(T,.975),col="red")