Sample size calculator for a mean in a finite population

Explanation

When you estimate a population mean and the total population size N is known and finite, the standard formula overestimates the sample needed. The finite population correction factor (FPC) takes advantage of the fact that sampling a large fraction of the population provides more information than the infinite formula predicts.

The correction is relevant when n₀/N > 5%. For example, for N = 1,000 and n₀ = 139, the sampling fraction is 13.9%, which justifies the correction. If N > 10,000 and n₀ is small, the savings are minimal and you can use the standard formula for a mean.

Sample size formula

\( n_0 = \left(\frac{Z\,\sigma}{E}\right)^2 \)

\( n = \frac{n_0}{1 + \dfrac{n_0-1}{N}} \)

• n₀: uncorrected size (standard formula for infinite population).
• n: size corrected for the finiteness of the population.
• σ: expected standard deviation of the variable.
• E: maximum tolerable absolute margin of error.
• N: known total population size.

Quick setup

• σ: obtain from historical data, a pilot study or the literature. If you don't have it, use range/4 as a conservative estimate.
• N: must be the exact, known count of the target population before sampling.
• E: define the error in the same units as σ — at what difference would your decision change?
• Confidence level: 95% is the standard; 99% for critical decisions.

Worked example

The HR department of an industrial plant wants to estimate the average time operators spend on a specific assembly task. The plant has N = 1,200 workers on staff, an exactly known number, which makes the finite population correction mandatory.

Previous ergonomic studies indicate that the standard deviation of task time is σ = 12 minutes. The methods engineer sets an acceptable margin of error of E = 2 minutes and works with a 95% confidence level (z = 1.96).

The first step is to calculate the sample size without correction, as if the population were infinite:

\( n_0 = \left(\dfrac{z \cdot \sigma}{E}\right)^2 = \left(\dfrac{1.96 \times 12}{2}\right)^2 = (11.76)^2 \approx 138.3 \rightarrow 139 \)

Next, the finite population correction is applied, which reduces the sample size when the sampling fraction \( f = n_0/N \) is not negligible:

\( n = \dfrac{n_0}{1 + \dfrac{n_0 - 1}{N}} = \dfrac{139}{1 + \dfrac{138}{1200}} = \dfrac{139}{1.115} \approx 124.7 \rightarrow \mathbf{125} \)

The savings amount to 14 interviews compared with the uncorrected design. The sampling fraction is \( f \approx 125/1200 \approx 10.4\,\% \), large enough for the correction to be relevant (rule of thumb: correct whenever f > 5%).

With a sample of 125 randomly selected operators, it can be stated with 95% confidence that the estimated mean time does not deviate by more than ±2 minutes from the true mean time of the entire workforce. This result makes it possible to size cycle times and detect bottlenecks with sufficient precision for decision-making.

Model assumptions

• Simple random sampling without replacement from a population of exactly known size N.
• The variable approximately follows a normal distribution, or n is large enough (CLT).
• N is fixed and does not change during sampling.

Common uses

• Average task time in a specific plant or department.
• Average spending of active customers in a loyalty program.
• Average consumption in a closed community (building, campus, residence).
• Inventory: average weight or value of items in a known warehouse.

How to interpret the result

The value \(n\) is the minimum sample size corrected for the finiteness of the population to estimate a mean with the specified margin of error \(E\) and confidence level. The finite population correction (FPC) reduces \(n\) relative to the infinite-population case: the larger the sampling fraction \(f = n/N\), the greater the savings. If \(f < 5\,\%\), the FPC is practically irrelevant (it reduces \(n\) by less than 5%) and the standard formula would be sufficient; if \(f > 20\,\%\), the savings are substantial and the correction is necessary to avoid oversizing the study.

It is essential that \(N\) represents the actual size of the target population, not the approximate sampling frame nor the general reference population. For example, if the study targets the employees of a specific company with 450 employees, \(N = 450\); using an inflated \(N\) (such as the workforce of every company in the sector) would eliminate the advantage of the correction. Run a sensitivity analysis testing \(N \pm 10\,\%\) to assess how much an imprecise estimate of the population size affects the resulting \(n\); with large populations the sensitivity is low, but with small populations it can be significant.

If you anticipate non-response or exclusion rates, calculate the recruitment \(n\) by dividing the corrected \(n\) by \((1 - \text{dropout rate})\), always verifying that the result does not exceed \(N\). If, after the adjustment, the required sample size exceeds 80% of the population, consider a census instead of sampling. Once the data has been collected, use the confidence interval calculator for the mean, also applying the FPC, to obtain the study's actual CI.

References and further reading

• Wikipedia (en): Finite population correction — derivation and when the correction is relevant.
• Wikipedia (en): Simple random sample — fundamentals of sampling without replacement.
• Cochran, W. G. (1977). Sampling Techniques (3rd ed.). Wiley. — classic reference on sampling in finite populations.

Frequently asked questions

• When is it worth applying the correction? When n₀/N > 5%. With smaller fractions, the savings are under 5% of observations.
• What if N isn't known exactly? Use the uncorrected formula; it will overestimate the sample but the result will always be valid.
• Can n be greater than N? No: the correction guarantees n < N. If n₀ ≥ N, a full census is necessary.
• Why not always use the correction? Because it requires knowing N exactly. If N is uncertain or variable, the correction can be misleading.