Interval Estimation for a Change in a Binomial Proportion I'm not sure how to estimate the confidence interval (CI) for a change in a small sample size binomial proportion using the same sample set both times.
I have two methods that I would like to compare (A and B). I have tested both methods on the same sample ($n=28$) from a large population.
Method A gave the correct result 11 times but method B gave the correct result 17 times. I think that this indicates that method B is 17/11-1 = 55% better than method A. As well as this point estimate for the difference between the methods, I would like to understand the uncertainty caused by my small number of samples. How can I construct a 95% CI for the 55% improvement in performance please?

*

*In 11 cases, both test A and test B worked.


*In no cases, test A did work but test B didn't work.


*In 6  cases, test A didn't work but test B did work.


*In 11 cases, both test A and test B didn't work.
These proportions all relate to the same sample ($n=28$). They are not independent of each other.
Is there a way to calculate CIs that doesn't assume independence please? I would be happy with confidence intervals or with credible intervals and would also be interested in arguments as to why such measures of uncertainty were not appropriate.
 A: Comment: Thanks for additional info. Maybe start with $2\times 2$ table:
              A
          ---------
B         OK    Not       Tot
-----------------------------
OK        11      6        17      
Not        0     11        11
-----------------------------
T0t       11     17        28

Then a chi-squared test on this table has a very
small P-value, supporting the anticipated lack of
independence in the paired design. [A simulated P-value
is necessary because of the small counts.]
chisq.test(rbind(c(11,6),c(0,11)), sim=T, B=5000)

    Pearson's Chi-squared test 
    with simulated p-value 
    (based on 5000 replicates)

data:  rbind(c(11, 6), c(0, 11))
X-squared = 11.723, df = NA, p-value = 0.0009998

In such cases, where the chi-squared test requires
simulation, the simulated P-value is often about
the same as for Fisher's exact test.
fisher.test(rbind(c(11,6),c(0,11)))

         Fisher's Exact Test for Count Data

data:  rbind(c(11, 6), c(0, 11))
p-value = 0.0009334
alternative hypothesis: 
  true odds ratio is not equal to 1
95 percent confidence interval:
 2.983321      Inf
sample estimates:
odds ratio 
       Inf 

It seems clear that B does better than A, but with only 28 patients and six differences in responses, I do not want to speculate how to find a useful confidence interval for a difference in binomial proportions $p_B - p_A.$ Maybe someone else can answer that.
