Say that I have a population of $N=500$ individuals, and I'm sampling $n=60$ of them for a certain characteristic which is expected to be quite rare (say green eyes). In my sample I find out there are 3 persons (5% of the 60, $p=0.05$) with green eyes, and now I want to calculate the margin of error of my estimate for an interval of confidence of 95%.

I start by calculating the standard error $SE$ using the Finite Population Correction (FPC) since the sample and population size have a similar order of magnitude:

$SE = \sqrt{\frac{p(1-p)}{n} \cdot \frac{N-n}{N-1}} = \sqrt{\frac{0.05(1-0.05)}{60} \cdot \frac{500-60}{500-1}} = 0.0264$

Then the margin of error $MOE$ of this estimate equals the critical value of the t distribution $z(0.05)$ for the 95% interval of confidence (0.05 significance level) times the standard error $SE$:

$MOE = z(0.05) \cdot SE = 1.96 \cdot 0.0264 = 0.0517$

I have some trouble interpreting this margin of error. As far as I know, an interval of confidence of 95% with a MOE of 5% means that I can be 95% sure that the actual percentage of people with green eyes in my population is within plus or minus 5% of the percentage I measured via sampling. I think a more specific interpretation is that if I had to repeat my random sampling many times, 95% of the time I would sample a value which is plus or minus 5% of the value I sampled.

Given the calculations above, I would say I can be 95% sure that the actual percentage of people with green eyes in my population is within plus or minus 5% of the percentage I measured via sampling, i.e, within (5-5.17)% (whatever that means... 0% I guess?) and (5+5.17) = 10%. Alternatively, if I had to repeat the sampling, 95% of the time I would measure a percentage of individuals with green eyes between (5-5.17)% (again, 0%?) and (5+5.17) = 10.17%.

While intuitive, I am afraid there's some issue with this interpretation. I don't like very much the margin of error being larger than my sampled percentage, and hence their difference going below zero. For sure we know that in the 500 individuals there's at least 3 with green eyes (I sampled them!), so I feel like the lower-bound for the percentage of people with green eyes should be at least 3/500 = 0.6% rather than the 0% I "calculate" from (5-5.17)%.

I feel like I am missing a big piece of the puzzle here. Ideas?

  • 1
    $\begingroup$ You seem be trying to use an MoE as if it were equivalent to confidence limits, but this is a situation where that's not the case. To construct Binomial confidence limits, please see stats.stackexchange.com/search?q=clopper+pearson. For hypergeometric limits--namely, where the small population size matters--I'm not sure whether a good answer has yet been posted, but you might search for keywords like "correction" or "hypergeometric." $\endgroup$ – whuber Mar 7 at 16:37
  • $\begingroup$ What makes this a situation where the MoE can't be used to define confidence limits? Is it because the sampled proportion is small, and you can do this only when it is 50%? Thanks for the hint on the Clopper-Pearson intervals, but I see there's a huge literature in the space and no universally used method (Agresti-Coull, Jeffreys, Wilson...), let alone for small populations (with the hypergeometric correction). More than "textbook" stuff I see research papers published quite recently, proposing a plethora of methods. Is it such an advanced field? This problem happens often in practice I think. $\endgroup$ – st1led Mar 8 at 10:53
  • 1
    $\begingroup$ It's that there is no simple, sample-size and population-parameter invariant relationship between the MoE and a working confidence interval. That's partly, but not entirely, a function of a small sample. I wouldn't say the literature is "huge:" as with any other statistical problem, there are a few procedures that have been developed, studied, tested, and effectively applied. Part of the art of statistics is knowing how to evaluate and select an appropriate procedure for a particular problem. The results are pretty clear: you usually won't go wrong with the C-P method. $\endgroup$ – whuber Mar 8 at 12:02

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