I would like to calculate the 95% CI for my binomial data. The set up is that I have 20 devices, each measuring the same standard three times. The data are binomial and come out as Found/Not Found (1,0). For the sake of argument lets say that each run comes up with 50% positive. Using the 95% CI formula on some practice data (see below):
$ \hat p \pm z*\sqrt{ \hat p*(1- \hat p)/n }$
We would have a 95% CI of 0.5 +/- 0.11 (or spanning 28.1% thru 71.9%) and R1=R2=R3.
I could simply use n=60 and p=.5 and get a 95% CI of 0.5 +/- .12 (37.3%-62.6%).
However since each device is used 3 times measuring the same thing, I seems appropriate to account for my repeated measures? Notice I have made the test data to be widely divergent, I do not think this is the case for my data.
I've used Cochrane's Q Test in the past to test for differences in paired data but in this case I want to find the global CI so I am not sure it is the right way of going.
How does one calculate the CI in this situation?
Test Data
Device R1 R2 R3
1 1 1 0
2 1 0 0
3 1 1 0
4 1 0 0
5 1 1 0
6 1 0 0
7 1 1 0
8 1 0 0
9 1 1 0
10 1 0 0
11 0 1 1
12 0 0 1
13 0 1 1
14 0 0 1
15 0 1 1
16 0 0 1
17 0 1 1
18 0 0 1
19 0 1 1
20 0 0 1