If you check the textbooks, confidence intervals are a little ambiguously defined. Where they are defined in terms of a recipe, they usually refer to samples from a normal population or for a large number of samples. Clearly the situation here is neither.
Looking closely at the assumptions in the standard recipes shows two things :- They assume the sample size is large. They use the sample statistic as an estimator of the population statistic at some point.
One can get a handle on this situation by inverting the problem as follows :- Assume that the probability of finding the mouth open is p, about which we know nothing, (ie. p is a random variable which has a flat distribution between [0,1]), what is the probability distribution of p GIVEN that in n observations, r of them showed the mouth was open.
One can solve this by:-
Write a computer program to :-
Here is a very simple C++ program to do steps 1) to 4)
#include <stdio.h>
#include <stdlib.h>
#include <iostream.h>
#include "abmath.h" // Provides drandom(), gives a double between 0
// and 1
main( int argc, char * argv[])
{
int n, r, rr;
const int N = 10000;
char * tailPtr;
double p;
n = strtol( argv[ 1], &tailPtr, 10);
r = strtol( argv[ 2], &tailPtr, 10);
cerr << "(n,r)=(" << n << ", " << r << ")\n";
for( int i=0; i < N;)
{
p = drandom();
rr = 0;
for( int j=0; j<n; j++)
if( drandom() < p)
rr++;
if( rr == r)
{
cout << p << '\n';
i++;
}
}
}