Resource assessment exercises: finite population correction

This article is part of the Resource assessment exercises. See the category page for a (chronological) table of contents.

Finite population correction

If we observe all values \(y_{i\in U}\) we talk about a census. The mean is calculated, not estimated, i.e.,

SRSwoR <- sample(trees$dbh, size=N)
mean(SRSwoR)

## [1] 21.05

mean(trees$dbh)

## [1] 21.05

As noted before, SRSwoR stands for simple random sampling without replacement (woR). However, if we take a sample with replacement (SRS; set replace = TRUE) we get a slightly different value. This would be an estimate.

SS <- sample(trees$dbh, size = N, replace = TRUE)
mean(SRS)

## [1] 20.96

If we are interested in a population parameter and take an SRSwoR of size \(n=N\), we get the true population value. There is no doubt about its value (assuming that measurement errors are absent). However, if we estimate the standard error for the \(n=N\) sample we get a positive value instead of a zero \(s_{\bar{y}}\).

sd(SRSwoR)/sqrt(N)

## [1] 0.07416

This cannot be! The estimator of the standard error holds for sampling with replacement. For sampling without replacement we have to correct for the fact that we took a relatively large sample. The finite population correction (fpc) for a relatively large sample is defined as,

$\text{fpc}=1-\frac{n}{N}.$

1

Obviously, if \(n=N\) the fpc becomes zero. Suppose we take a sample of size \(n=25,000\) from trees, then

S25k <- sample(trees$dbh, size = 25000)
sd(S25k)/sqrt(25000) # without fpc

## [1] 0.08099

fpc <- 1 - 15000/30000
sqrt(var(S25k)/25000 * fpc)

## [1] 0.03307
  
sd(S25k)/sqrt(25000) * sqrt(fpc)

## [1] 0.03307

For the parametric standard error the fpc becomes,

$\text{fpc}=\frac{N-n}{N-1}.$

2

As a rule of thumb, we apply the fpc when the sampling fraction

$f=\frac{n}{N}$

3

exceeds 0.05, i.e., 5 percent.

Related articles

• Previous article: Standard error and confidence intervals
• Next article: Required sample size determination