# How to calculate a confidence interval for estimating a proportion when the population is very small?

I plan to execute a little study on whether dentists tend to grant privately insured patients (PRIV) a sooner appointment compared with publicly insured patients (PUB). I want to restrict this study to the dentists of a single town in Germany.

In this town there are about 500 dental practices and I am willing to call maximum 100 practices (twice, once as a PUB and once as PRIV). (A sample without replacement of size up to 100)

So the result of this examination would be a list of up to 100 practices with an associated value of 1 - if the appointment for the PRIV was more than two days sooner - and a 0 - if not. The proportion of 1s is what I want to estimate.

Now the sources I have been reading always assumed a much larger population with a finite population correction close to 1. This is why I doubt that blindly using a conservative or approximate confidence interval would lead to a statistically meaningful result.

1. So my first question is, of course, whether it is reasonable at all trying to estimate the proportion using a confidence interval in this case?

2. And if yes, then what statistical technology/method (leading to a CI) would be a good choice for this use case?

One idea that comes to mind is using a sample size of 25 f.x., in that case the FPC would be 0.95 - "close enough" to 1. But it seems stupid trying to improve the quality of an estimation by reducing the information used for the estimation.

If you sample 100 dentists from the 500 in town, & observe $x$ who give privately-insured patients earlier appointments, $X$ is distributed according to the hypergeometric distribution

$$\Pr(X=x)=\frac{\binom{\nu}{x}\binom{500-\nu}{100-x}}{\binom{500}{100}}$$

where $\nu$ is the number of dentists in town who give privately-insured patients earlier appointments. The usual exact $\alpha$ confidence interval for $\nu$ is given by

$$\left\{\nu : \min\left[\Pr\nolimits_{\nu}(X \leq x), \Pr\nolimits_{\nu}(X\geq x)\right] > \frac{\alpha}{2}\right\}$$

though Blaker (2000), "Confidence curves and improved exact confidence intervals for discrete distributions" The Canadian Journal of Statistics, 28, 4 shows how Spjøtvoll's notion of acceptability can be used to give often shorter intervals with the correct coverage & sensible nesting properties.

A couple of oddities in your analysis:

1. The waiting time difference between appointments could be considered as a continuous variable—why discretize it & lose information?
2. You're considering this as a sample from the finite number of dentists, whereas it might be more natural to consider it as a sample from the infinite number of potential bookings; no individual dentist is likely to be completely consistent in his scheduling practices.
• Hey Scortchi, thanks for your input! This is the combinatorial solution which bootstraps using x and calculates the precise probabilities - I will use this for my little study. (1) I will also have a look at the actual measured differences - don't worry! But for answering the given question this is probably the better approach b/c the observed difference will be depending on attributes like location of the dentist practice - also a lot of practices simply refuse publicly insured patients at time (Diff = inf) Commented Sep 16, 2013 at 15:21
• (2) isn't a necessary feature of a meaningful sample that you can associate with every element of the population a probability of showing up in the sample? I wouldn't know how to calculate that for a population of hypothetical bookings. But your point is valid - but I assume that on average dentists are considerably consistent. Commented Sep 16, 2013 at 15:28

This problem sounds very closely related to statistical power analysis, which considers how to design experiments that will, with some level of confidence, be able to observe effects of some specified size. Since you are only willing to call 100 dentists, the detectable effect size at some fixed power and level is in turn restricted to be larger than you would be able to detect if you called, say, all 500 dentists. This webpage might be a helpful starting point. http://www.statsoft.com/textbook/power-analysis/

• +1 because you introduce a new cue (power analysis) but as you are not actually answering the question - of course I cannot mark it as an answer. but still - very much appreciated! Nonetheless - your statement with "1000 dentists" confuses me a bit, as I state that there are no more than 500 in that town. Commented Sep 12, 2013 at 13:16
• Naturally; I should have noted that this was really a long comment.
– Sycorax
Commented Sep 12, 2013 at 13:33