# Binomial mid-p value

I've been under the impression that the mid-$p$ values generally control the Type I error, and consequently confidence intervals based on mid-$p$ values control the coverage. However I have checked that this is not true with the binom.midp() function of the binomSamSize package (see below).

Is it a problem with binom.midp()?

library(binomSamSize)

coverage <- function(p, n, conf){
bounds <- binom.midp(0:n,n,conf)
low <- bounds$lower up <- bounds$upper
covered <- which(low<p & p <up)-1
sum(dbinom(covered, n, p))
}

n <- 100
conf <- 95/100
p <- seq(0.01,0.99,length.out=100)

cover <- sapply(p, function(p){
coverage(p,n,conf)
}
)

plot(p,cover,type="l")
abline(h=conf, lty=2)


EDIT: Now I have checked the significance level of the hypothesis test based on the mid-p value (very easy to implement), and the results are in agreement:

### mid-p value for H0:{p>p0}
pv <- function(x,n,p0){
pbinom(x, n, p0)-dbinom(x,n,p0)/2
}
### typeI error
typeI <- function(p,n,alpha){
pvs <- pv(0:n,n,p)
rejected <- which(pvs<alpha)-1
sum(dbinom(rejected,n,p))
}

### test
n <- 100
alpha <- 5/100
p <- seq(0.01,0.99,length.out=100)

probs <- sapply(p, function(p){
typeI(p,n,alpha)
}
)

plot(p,probs,type="l")
abline(h=alpha,lty=2)


-
Hi Stephane. You might want to look at L. D. Brown, T. Cai, and A. DasGupta (2001), Interval estimation for a binomial proportion, Statistical Science, vol. 16, no. 2, 101-133. There, they point out a connection between the Clopper-Pearson (guaranteed coverage), mid-P, and Jeffreys (no guaranteed coverage, especially near end) intervals. See Section 3.2 and Section 4.3. I have not gone through all the details, but I believe the effect you are seeing is real. –  cardinal Apr 8 '12 at 20:18
Thanks cardinal. You posted your comment and I posted my edit at the same time. Nevertheless I remain sure than mid-p values control the Type I for certain models. But not the binomial one, as we saw. –  Stéphane Laurent Apr 9 '12 at 6:42
... but I would not be surprised that the binomial mid-p value unconditionally achieve a good Type I error in the context of this question stats.stackexchange.com/questions/25972/… –  Stéphane Laurent Apr 9 '12 at 6:44