Confidence interval of ratio of successes

Question

Suppose there are 100 observations of a binary outcome variable, and that 35 of them are successes. To calculate the 95% conf interval for the number of successes I have used prop.test(35,100) in R (equivalent to binom.test), yielding 25.9-45.3 for the number of successes. This is more or less clear to me.

The difficulty comes when plotting the ratio of successes/no-successes. The observed ratio would be 35/65, and I'd like to add error bars indicating the 95% conf interval. What about calculating the 95% conf interval of these ratios as: 25.9/(100-25.9) - 45.3/(100-45.3).

This makes some sense to me. However, as the number of not-successes and successes depend on each other, I am not sure that approach is appropriate: 25.9/(100-25.9) would correspond to the lower 95% conf interval of successes divided by the upper 95% conf interval of not successes. It would be too conservative. May be it would be better as 25.9/65 - 45.3/65.

Thanks for any hint to solve this probably simple question. Apologies if it's not clear enough - I do not speak (nor think) statisticsh fluently.

Fede

Answer 1

For notation we'll use $k$ for the number of successes.

Remember that the 95% confidence interval of $k$ is not the interval where the population average lies. The 95% confidence interval shows us that if you repeat your experiment many times we expect 95% of the sample means to lie in that interval.

If you want a confidence interval of $\frac{k}{100-k}$ you have to deal with one problem first. If you repeated the experiment many times there is a chance that you get $k=100$, this gives you an undefined value which you cannot say is inside or outside a confidence interval. To fix this just define $\frac{100}{0}=\infty$. Now we can say that it is outside of any finite interval.

You are lucky, the relationship between $k$ and $\frac{k}{100-k}$ is a special case where we have that one value increases only when the other increases. This means that your idea was correct. If you repeated the experiment many times you would find 95% of values for $k$ are in $(25.9,45.3)$, for the special case this implies that 95% of the values for $\frac{k}{100-k}$ are in $\left(\frac{25.9}{100-25.9}, \frac{45.3}{100-45.3}\right)$