Finite Population Correction

When a simple random sample is drawn without replacement from a finite population, each draw reveals information about the remaining units. The sample mean keeps the same expectation, but its sampling variance becomes smaller than the independent-draw formula suggests.

Suppose a company has N=200 eligible candidates and interviews n=20 without replacement. The sample is 10% of the whole population, so one selection noticeably changes who remains available.

If Yi is candidate i's score when selected, the estimator is the sample mean Ybar=(1/n) sum Yi. Under simple random sampling, Ybar is unbiased for the finite-population mean mu.

After one unit is selected, it cannot be selected again. For distinct draw positions, knowing the first outcome changes the distribution of the second; the selections are dependent even though the sampling design is fair.

Under simple random sampling without replacement, two distinct draws have negative covariance: Cov(Yi,Yj)=-sigma^2/(N-1). If one draw is above the population mean, the remaining pool is slightly lower on average.

For two different population units, the joint inclusion probability is n(n-1)/[N(N-1)], which is smaller than the product (n/N)^2. Their inclusion indicators therefore have negative covariance.

For dependent draws, Var(sum Yi) equals the sum of individual variances plus the sum over all ordered covariance terms. There are n(n-1) ordered pairs, so their combined contribution matters.

Because Ybar=(1/n) sum Yi, its variance is 1/n^2 times the variance of the sum. The scaling applies to both the n variance terms and the n(n-1) covariance terms.

Substituting Var(Yi)=sigma^2 and Cov(Yi,Yj)=-sigma^2/(N-1) gives Var(Ybar)=(1/n^2)[n sigma^2-n(n-1)sigma^2/(N-1)].

Factoring n sigma^2 and simplifying the remaining fraction gives Var(Ybar)=(sigma^2/n)[(N-n)/(N-1)]. The independent-draw variance is multiplied by one finite-population correction.

The variance finite population correction is (N-n)/(N-1). It lies from 0 to 1 for 1<=n<=N, equals 1 when n=1, and reaches 0 when the entire population is observed.

Variance is multiplied by (N-n)/(N-1), while standard deviation and standard error are multiplied by its square root. Confusing these two factors over-corrects or under-corrects uncertainty.

In an imaginary infinite population, removing one unit barely changes the next draw. In a finite population, every observed unit removes uncertainty from the remaining pool, so repeated sample means vary less.

It appears when the probability model matches simple random sampling without replacement. With replacement, draws are independent and the correction disappears; the sampling design, not a preference, determines the formula.

N must count the actual finite sampling frame for this selection process and time window. A department, hiring cycle, or shortlist can define different populations, so N must match the claim being made.

As the sampling fraction n/N grows, the correction shrinks and uncertainty falls. When n is tiny relative to N, the factor is close to one and the correction has little practical effect.

Imagine repeatedly drawing n=20 candidates from the same N=200 without replacement. The distribution of sample means is narrower than it would be under independent draws, even though the center stays at the population mean.

A value such as 78 summarizes the observed sample, but reliability comes from its standard error or confidence interval. The correction changes that uncertainty statement, not the point estimate itself.

A smaller standard error creates a narrower interval around the same estimate. If a decision compares a population mean with a benchmark, the corrected interval can provide more precise evidence while leaving the estimate unchanged.

Confirm a finite population of size N, simple random sampling without replacement, sample size n, and a mean or total whose design variance is being computed. Then use the variance factor or its square root at the correct stage.