This is probably one of the more straight forward questions on here but here it is: I want to use a random number generator to sample X number of charts to look for the # occurrences of Y event. So let's say I randomly sample 100 charts, and 18% of them contain Y (or alternatively Y occurs X# of times within a sample of 100). How do I determine how confident I can be that this is the true estimate of the population of events (I may or may not have an exact population N.

Additionally, or equally important, how can I estimate the apriori number of cases which need to be randomly sampled to have a sufficient confidence interval ?

I know G power can be used for this but I just need some clarification on the process.

Many thanks

  • $\begingroup$ This question has been cross-posted on Math.SE as well. @SeanMichael, maybe you should delete one of these posts? Please do include a link when cross-posting, and do wait for an answer before doing so. $\endgroup$
    – PKG
    Commented Nov 2, 2014 at 19:40

1 Answer 1


It depends on what you mean by the population. If you want an estimate of the prevalence of a disease in the population as a whole, really this is more a question of study design. So, for instance:

  1. How are patients referred to this clinic? The prevalence of a disorder at a tertiary care center in a specialized clinic will be much higher than that found in family practice.
  2. Are there barriers to accessing care (e.g. insurance) that may cause the prevalence to be underestimated?
  3. What time window are you using to sample the cases? If you want the prevalence at a particular point (e.g. over one year) you need to make sure your random sample only includes charts from that time period.

There are many good texts on clinical epidemiology that can help you with some of these considerations:



If you're looking specifically at the prevalence in your particular environment, some of these concerns may be less important.

The equation for the sample size required to estimate a proportion (i.e. prevalence) in one sample is given by:

$$ n = \frac{z^{2}_{1-a/2}p(1-p)}{d^2} $$


  • d = the desired precision
  • p = the estimated proportion
  • z = the number of standard errors away from the mean (1.96 for 95% confidence)

If you don't have a good estimate of the proportion, choosing p =0.5 will maximize the sample size estimate.

  • $\begingroup$ That is quite helpful. especially the last part. To follow-up on the "d" term ("Desired precision") that would be, say, .95 (the "power")? $\endgroup$ Commented Nov 12, 2014 at 13:39
  • $\begingroup$ It's the margin of error (think the results of a poll). Here's a worked example for clarity: for 95% confidence, an anticipated proportion of 20%, and a precision of 5 percentage points you would have (1.96)^2 x 0.2(1-0.2)/(0.05)^2 = 245.8624 for the sample size. [Thanks to Refik Saskin who provided this example and equation to me originally.] $\endgroup$
    – Red_Star
    Commented Nov 15, 2014 at 23:42
  • $\begingroup$ Thanks redstar-- that is much appreciated! you are certainly good at this and I appreciate your help!! $\endgroup$ Commented Jan 16, 2015 at 15:45

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