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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
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