Summary
I'm trying to calculate the confidence interval for speedup of a program when I have many measurements for the old and new execution times.
Details
Speedup is calculated as: $\frac{time_{new}}{time_{old}}$, which is essentially a ratio. When looking for information about calculating the confidence interval of a ratio I looked at the question How to compute the confidence interval of the ratio of two normal means,which suggested using Fieller's theorem.
The Wikipedia article includes information about Approximate formulae and I believe my scenario would fall in case 1 because my standard error is relatively small compared to the measured execution time. So based on what Wikipedia says I should be able to use:
$$ Var \left(\frac{a}{b}\right) = \left(\frac{a}{b}\right)^2 \left(\frac{Var(a)}{a^2} + \frac{Var(b)}{b^2} \right) $$
This part makes sense to me, but where I got lost is when Wikipedia says:
From this a 95% confidence interval can be constructed in the usual way (degrees of freedom for $t^*$ is equal to the total number of values in the numerator and denominator minus 2).
I am familiar with using the steps described on Wikipedia to compute a confidence interval for a population, but I am not sure how to apply them when given this variance formula. The Wikipedia article on Fieller's theorem includes additional detail with an equation to use for "logged data" (I'm not certain what the means). But I don't think that equation is applicable for my problem.
Unfortunately I'm not able to make sense of how I would translate this to code (I would like implement my calculations in Ruby, but as long as I can figure out something like pseduocode for the equation the coding should not be a problem).
The question that I referenced earlier contains a like to an online calculator for computing the confidence interval of a quotient. I don't know how complex the code for this is, but if I could find something like it that would probably answer my question. I need to be able to calculate the CI in my own script rather than using an online calculator.