# Standar Deviation of a ratio of means

First, my question.
I have 20 data points of a baseline ($$b$$) and 20 data points of a faster version ($$f$$).

I currently present $$\frac{\bar{b}}{\bar{f}}$$ in a bar graph in which I'd like to add error bars yet I'm unsure how to produce the standard error/standard deviation to show.

Second:
I tried looking around for answers but found them to be referring to slightly different questions than mine. Two posts [1, 2] here suggest Fieller's theorem, though the response to the first question suggest that there's a "simple" way to just calculate $$sd(\frac{\bar{b}}{\bar{f}})$$. The wikipedia page on Fieller's theorem suggests that it's necessary because the values may be in different units, suggesting to me at least that there's another, simpler way if they're both in the same units (which they are: ms).

My supervisor suggested pairing the times to calculate speedups and then taking the standard deviation of those but that feels wrong because there's no connection through which to justify the pairings.

• The quick answer is to use bootstrap, but: 1) what happens if $\bar{f}=0$? 2) $\bar{b}/\bar{f}$ is just one number, so what are you putting in your bar graph? If there is no natural pairing, then your supervisor is dead wrong about pairing values. – Dave Mar 3 '20 at 23:30
• 1) $\bar{f}$ will always be a time greater than zero. 2) I have multiple $\bar{f}$, I was just trying to simplify for the sake of the question. Do you have a link for bootstrap? – Braaedy Mar 3 '20 at 23:44
• If you have multiple $\bar{f}$ then I’m very concerned about what you’re doing. What are your data? What are your goals? – Dave Mar 4 '20 at 0:10
• if you are calculating a ratio this suggests a logarithmic transform might be appropriate anyway, and $\log(b)$ - $\log(f)$ is easier to calculate a standard error for. you could then back-transform a confidence interval if you wanted the ratio estimate on the natural scale. – George Savva Mar 4 '20 at 0:45
• @Dave Speedup. I have a baseline process completed in $\bar{b}$ milliseconds, and then four improved versions of the process completed in $\bar{f}_1, \bar{f}_2, \bar{f}_3, \bar{f}_4$ ms. ${speedup}_i=\frac{\bar{b}}{\bar{f}_i}$. I plot ${speedup}_i$ and am looking to add error bars corresponding to $sd({speedup}_i)$ – Braaedy Mar 4 '20 at 21:41