Confidence interval for a mean (finite population)

Explanation

When the total population has a known size \(N\) and is sampled without replacement, the observations are not strictly independent. The finite population correction factor (FPC) reduces the standard error in proportion to the sampled fraction \(f = n/N\).

The result is a narrower CI than the standard one: the larger \(f\) is, the greater the gain in precision. If \(N \to \infty\) (or \(f < 5\,\%\)), FPC ≈ 1 and both CIs coincide.

Formula for the CI with FPC

\( \text{FPC} = \sqrt{\frac{N-n}{N-1}} \)

\( \text{SE}_{\text{FPC}} = \frac{s}{\sqrt{n}} \cdot \text{FPC} \)

\( \text{CI} = \bar{x} \pm t_{\alpha/2,\;n-1} \cdot \text{SE}_{\text{FPC}} \)

• \(\bar{x}\): sample mean.
• \(s\): sample standard deviation (with divisor n−1).
• FPC: finite population correction factor.
• \(t_{\alpha/2,n-1}\): Student's t quantile with \(n-1\) degrees of freedom. For large n it converges to the normal quantile z.

When the FPC matters

• \(f = n/N > 5\,\%\): the correction noticeably narrows the CI.
• \(f > 20\,\%\): the reduction in the CI is substantial.
• \(f < 5\,\%\): the difference between the corrected and standard CI is under 3% and can be ignored.

Worked example

A company with N = 1,200 employees takes a sample of n = 127. The average time for a task is \(\bar{x} = 72\) minutes with sample standard deviation \(s = 12\) minutes. We want the 95% CI.

The sampling fraction is \(f = 127/1{,}200 \approx 10.6\,\%\), above the 5% threshold, so the correction is relevant.

\(\text{FPC} = \sqrt{(1200-127)/(1200-1)} = \sqrt{1073/1199} \approx 0.9463\).

The standard error without correction is \(12/\sqrt{127} \approx 1.065\) min, and with FPC it is \(1.065 \times 0.9463 \approx 1.008\) min.

The critical t value with 126 df at 95% is \(t_{0.025,126} \approx 1.979\).

CI with FPC: \(72 \pm 1.979 \times 1.008 = [70.0,\; 74.0]\) minutes.

Without correction the CI would be \([69.9,\; 74.1]\). The correction reduces the width from 4.2 to 4.0 minutes, a 5.4% reduction in this case.

Model assumptions

• Simple random sampling without replacement from a population of exactly known size N.
• The variable follows an approximately normal distribution, or n is large enough (CLT).
• N is fixed and known before sampling.
• The t distribution is used (σ unknown). If σ is known, replace t with z.

How to interpret the result

The interval \([L, U]\) is the plausible range for the population mean \(\mu\) given the chosen confidence level. The frequentist interpretation is the same as for any CI: if you repeated the sampling and built the CI with the same procedure many times, approximately \(C \times 100\,\%\) of those intervals would contain the true value of \(\mu\). The CI corrected by FPC is always equal to or narrower than the standard one: this is not a bias, but the correct reflection of the fact that sampling \(n/N\,\%\) of the population without replacement provides more information than sampling with replacement.

The difference between the two CIs depends directly on the sampling fraction. If \(n/N < 5\,\%\), the FPC is close to 1 and both intervals practically coincide. If \(n/N\) exceeds 20%, the width reduction can be substantial, as the calculator shows by comparing the two CIs side by side in the chart. When the entire population is sampled (\(n = N\)), the CI collapses to a point: the uncertainty disappears because every element is known.

• Connection with hypothesis testing: if a reference value \(\mu_0\) falls outside the CI with FPC, the data reject it at significance level \(\alpha = 1 - C\) in the equivalent two-sided test for finite populations.
• Which interval to use: always use the CI with FPC when \(n/N > 5\,\%\) and the design is simple random sampling without replacement. Reporting only the standard CI when the fraction is large overstates the real uncertainty.

References

• Cochran, W. G. (1977). Sampling Techniques (3rd ed.). Wiley.
• Lohr, S. L. (2019). Sampling: Design and Analysis (3rd ed.). Chapman & Hall.