Toby Lipman posted on evidence-based-health:
> Now it seems to me that these calculations relate to populations assumed
> to be infinitely large. If the populations from which the samples are
> taken are smaller, then the CIs must also be smaller. For example it
> would plainly be crazy to give a 95% CI of 27% to 73% in the first
> example if the population from which the sample is taken is only 25,
> since the maximum variability would be from 40% to 60%.
Good observation. In fact, as the sampling fraction approaches 100% of the
population size, sampling variability (and thus standard error) approaches
zero. At 100%, the population parameter is measured exactly so there is no
meaning to a confidence interval.
> How can this be allowed for, and at what size of population does it
> seriously affect CIs?
The solution is application of the so-called Finite Population Correction
Factor. FPC generally can be ignored when sampling less than 8% of a
population, but somewhere around a sampling fraction of 5-10% the correction
becomes large enough to matter. It can be applied to either correct the
variance (and thus width of confidence intervals) or the sample size needed
for a specified level of precision. To correct the variance, multiply it by
the quantity (N-n)/(N-1) where N represents size of the population and n
represents size of the sample. To obtain a corrected sample size (n')
adjusted for sampling fraction (n/N), calculate the uncorrected sample size
required (n) in the usual manner and then compute n' from n' = n/(1 + n/N).
Using your example of sample size n=20 from a population size N=25, for a
given level of precision one would need only n/(1+0.8) or just 11 (not 20)
observations. A description of these calculations is found in many books on
statistics or sampling techniques.
David Birnbaum, PhD, MPH
Adjunct Professor
Dept. of Health Care & Epidemiology
University of British Columbia, Canada
|