# Sufficient sample size for skewed populations

I have measured 50000 individuals from a certain population, and 800 (~1.6%) of them have a certain characteristic (say green eyes). I can be pretty sure that this is an effect that is present in a small fraction of the population.

Now I have another 1500 individuals from the same population that all have another characteristic (say brown hair). I'd like to sample from this second population to estimate how many of them also have green eyes.

How can I derive a sufficient sample size for the task?

In principle, I would just calculate the sample size for whatever confidence and margin of error I like. For a population of 1500 and a standard 95% confidence level with 5% error that would be a sample of 306 individuals -- assuming a 50-50 split on the response distribution (calculated using what I think is a standard calculator: http://www.raosoft.com/samplesize.html).

What throws me a bit off the book is that I know that only a small fraction of the population have green eyes. Is this sample size of 306 sufficient to precisely assess that? Should I sample more, because I am looking at an effect (green eyes) that is rare in the population? Conversely, can I get away with sampling less, because (assuming independence between having brown hair and green eyes), chances are that the vast majority of my samples won't have green eyes?

In the meantime, since sampling is expensive in my context, I ran a little pilot with a smaller sample size (67 out of the 1500), and found out that 2 of the 67 (3%) have green eyes. I suppose this means that I have a 3% response distribution on my sample, which should then mean that I can be relatively sure that my answer is correct -- at least more than if the results were closer to a 50-50 split (green eyes vs. no green eyes). Can I then claim that I have a 95% confidence level with a 4% margin of error, meaning that I can be 95% sure that 0-7% of the 1500 have green eyes? (always calculated in the same way, with standard sampling size calculators)

For the given sample size $$n$$ and population size $$N$$, the variance ($$Var$$) of estimated proportion is proportional to $$\pi(1-\pi)$$, where $$\pi$$ is proportion of green eyes in the population. So the variance achieve the largest when $$\pi = 0.5$$ (a 50-50 split). The 95% confidence interval is calculated by estimated proportion $$\pm 1.96\sqrt{(Var)}$$. It means when the sample size and population size are fixed, the width of 95% CI is largest when $$\pi = 0.5$$ (a 50-50 split). When $$\pi$$ goes to smaller, the width of 95% CI will decrease. It means the sample size calculated under the assumption of $$\pi = 0.5$$ is conservative approach, and it is enough for $$\pi \ne 0.5$$. If you want, you can change $$\pi$$ to the reasonable value to get smaller and more appropriate sample size.
• When the sample size = 67 and $\pi = 1.6$%, the width of 95 CI is about 2*1.96*sqrt(0.984*0.016/67) = 6%. So the margin error = half of width of CI = 3%. Dec 8, 2018 at 16:28