Finite Population Correction

Why a Finite Population Changes Standard Error

When sampling from a large population, increasing the sample size usually reduces the standard error. But if the population itself is finite and sampling is without replacement, the sample also removes members from the pool as it is drawn. A sample of 80 distinct units from a population of 100 leaves much less uncertainty about the population than 80 independent draws that could repeatedly select the same units.

The adjustment for this reduction in variability is called the finite population correction, or FPC. It matters most when the sample is a substantial fraction of the population. It is especially important near a census, when almost every population member is observed.

Definition: Suppose a simple random sample of size $n$ is drawn without replacement from a finite population of size $N$. The finite population correction for the standard error is $$ \sqrt{\frac{N-n}{N-1}}. $$ It multiplies the standard error formula that would apply to independent sampling with replacement from the same finite population. The correction is defined for $1\leq n\leq N$.

A simple random sample without replacement (SRSWOR) means that every possible subset of $n$ distinct population members has the same chance of selection. The FPC formulas below rely on this sampling design and on the finite population being the target of interest. They do not automatically apply to a convenience sample, a sample with unequal selection probabilities, or a design with clustering.

Standard Errors With the Correction

For a quantitative variable, let $\mu$ be the mean of the $N$ population values and let $\sigma^2$ denote their variance, defined here with divisor $N$: $$ \sigma^2=\frac{1}{N}\sum_{i=1}^{N}(y_i-\mu)^2. $$ For the sample mean $\bar{x}$, the standard error under SRSWOR is

SE(\bar{x})=\frac{\sigma}{\sqrt{n}}\sqrt{\frac{N-n}{N-1}}.

The first factor, $\sigma/\sqrt{n}$, is the standard error for independent sampling with replacement from this population. The second factor adjusts it for sampling without replacement. If a problem instead defines a finite-population variance using divisor $N-1$, the equivalent formula is $SE(\bar{x})=\sqrt{(1-n/N)S^2/n}$, where $S^2=\sum (y_i-\mu)^2/(N-1)$. The two forms agree because $S^2=N\sigma^2/(N-1)$.

For a categorical characteristic with population proportion $p$, the sample proportion $\hat{p}$ has standard error

SE(\hat{p})=\sqrt{\frac{p(1-p)}{n}}\sqrt{\frac{N-n}{N-1}}.

This expression uses the same FPC. When the finite population contains a proportion $p$ of units with the characteristic, the variance of the corresponding zero-or-one values, using divisor $N$, is $p(1-p)$. The formulas describe the standard deviation of the statistic across all possible SRSWOR samples of size $n$, not the variability among individual population members.

Worked Example: Adjusting a Proportion Standard Error

Suppose a hypothetical organization has $N=1{,}000$ members, of whom 35% have completed a particular training course. A simple random sample without replacement of $n=100$ members is selected. Find the standard error of the sample proportion who completed the course.

Check the conditions. The population is a specified finite set of 1,000 members, the sampling method is SRSWOR, and the target proportion is for those members. Thus the finite-population formula applies. Here $p=0.35$ and $1-p=0.65$.

First calculate the standard error without the correction, then multiply by the FPC:

SE(\hat{p}) =\sqrt{\frac{0.35(0.65)}{100}}\sqrt{\frac{1{,}000-100}{1{,}000-1}} =\sqrt{0.002275}\sqrt{\frac{900}{999}} \approx 0.04527.

Without the FPC, the standard error would be $\sqrt{0.002275}\approx 0.04770$. The sample is 10% of the population, and the correction lowers the standard error by a modest amount. In context, across repeated SRSWOR samples of 100 members, the sample proportion typically varies around the population proportion 0.35 with a standard deviation of about 0.04527, or 4.527 percentage points.

Worked Example: Standard Error for a Finite-Population Mean

Imagine a hypothetical population of $N=40$ machine parts. Their measured lifetimes have population mean $\mu=500$ hours and population standard deviation $\sigma=12$ hours, with $\sigma$ calculated using divisor $N$. A technician takes an SRSWOR of $n=10$ parts. Find the standard error of the sample mean lifetime.

Check the conditions. The target is the mean lifetime of these 40 parts, and the sample is a simple random sample without replacement. The stated population standard deviation uses the convention for the mean formula above.

The sample mean’s standard error is

SE(\bar{x}) =\frac{12}{\sqrt{10}}\sqrt{\frac{40-10}{40-1}} =\sqrt{\frac{144}{10}\cdot\frac{30}{39}} =\sqrt{11.076923\ldots} \approx 3.328\text{ hours}.

If the parts had instead been selected independently with replacement, the standard error would be $12/\sqrt{10}\approx 3.795$ hours. Without replacement, observing one part slightly reduces the uncertainty about the remaining parts, so the standard error is lower. The population mean of 500 hours does not enter this calculation because the standard error depends on the population spread, not its location.

Why the Correction Has This Form

The FPC can be justified by tracking how inclusion in a fixed-size sample connects the observations. Let the finite population values be $y_1,\ldots,y_N$, and write $z_i=y_i-\mu$, so $\sum z_i=0$. Let $I_i$ equal 1 when population member $i$ is selected and 0 otherwise. Under SRSWOR, $P(I_i=1)=n/N$, so $\operatorname{Var}(I_i)=n(N-n)/N^2$. For distinct members $i$ and $j$, both are selected with probability $n(n-1)/(N(N-1))$, giving

\operatorname{Cov}(I_i,I_j) =\frac{n(n-1)}{N(N-1)}-\left(\frac{n}{N}\right)^2 =-\frac{n(N-n)}{N^2(N-1)}.

The sample mean minus the population mean is $\bar{x}-\mu=(1/n)\sum I_i z_i$. Expanding its variance gives a sum of the individual variances and the pairwise covariances. Since $\sum_{i\ne j}z_i z_j=-\sum_i z_i^2$, substitution yields

\begin{aligned} \operatorname{Var}(\bar{x}) &=\frac{1}{n^2}\left[ \frac{n(N-n)}{N^2}\sum_i z_i^2 +\frac{n(N-n)}{N^2(N-1)}\sum_i z_i^2 \right]\\ &=\frac{N-n}{n(N-1)}\left(\frac{1}{N}\sum_i z_i^2\right) =\frac{\sigma^2}{n}\frac{N-n}{N-1}. \end{aligned}

Taking the square root gives the standard error formula. The negative covariance is the key: under a fixed sample size, selecting one member makes the inclusion of another slightly less likely. That dependence reduces the variance compared with independent draws.

Samples Near a Census

The correction depends on the sampling fraction $n/N$. If $n$ is small relative to $N$, then $(N-n)/(N-1)$ is close to 1, and the correction changes the standard error little. A common approximation is to ignore the FPC when the sample is no more than about 5% of the population. This is a practical rule, not a mathematical boundary: the exact FPC formula remains valid whenever the stated sampling conditions hold.

As $n$ approaches $N$, the correction decreases sharply. At a census, $n=N$, and its value is zero. This is appropriate: if every member of the finite population is measured, the sample mean and sample proportion equal their population values, so there is no sampling error for those quantities.

Worked Example: A Sample Close to a Census

A hypothetical list contains $N=24$ households, and 9 of them use a particular heating system. An SRSWOR of $n=20$ households is selected. Find the standard error of the sample proportion of households using that system.

Check the conditions. The target is the proportion among the 24 listed households, and 20 distinct households are selected by simple random sampling without replacement. The known population proportion is $p=9/24=0.375$, so $1-p=0.625$.

Applying the formula,

SE(\hat{p}) =\sqrt{\frac{0.375(0.625)}{20}}\sqrt{\frac{24-20}{24-1}} =\sqrt{0.01171875\cdot\frac{4}{23}} \approx 0.04514.

Without the correction, the standard error would be $\sqrt{0.01171875}\approx 0.1083$. The sample includes five-sixths of the households, so its proportion is much less variable than a proportion based on 20 independent draws with replacement. If all 24 households were surveyed, the standard error for estimating this finite population’s proportion by the sample proportion would be zero.

Common Mistakes and Practical Checks

The FPC is easy to apply once the population and sampling design are identified, but several errors can lead to an incorrect standard error or interpretation.

Using the correction for sampling with replacement. Repeated independent draws do not deplete the population, so this without-replacement correction does not apply to that design.
Using the wrong population size. $N$ must count the units in the finite population that defines the target, not a broader group that was not sampled.
Applying the correction to the standard error twice. Multiply the usual independent-sampling standard error by the FPC once; do not multiply its variance by the FPC rather than by the squared FPC.
Assuming a large sample always needs the correction. The relevant quantity is the sample’s fraction of the population. A sample of 500 from 1,000 is a large fraction; a sample of 500 from 10 million is not.
Interpreting the correction as a change in the population parameter. It reduces the sampling variability of the estimator. It does not change $p$, $\mu$, or the expected value of the sample statistic.

Before using an FPC, identify the finite population, verify that the sample is an SRS without replacement, and confirm that the parameter of interest describes that population. Then calculate the uncorrected standard error and multiply by $\sqrt{(N-n)/(N-1)}$. Keeping these steps separate helps prevent arithmetic and interpretation errors.

Key takeaway: For an SRSWOR from a finite population, the FPC is $\sqrt{(N-n)/(N-1)}$. It reduces standard errors when the sample is a substantial fraction of the population and makes them zero for a census.

Check Your Understanding

Use the conditions and formulas in the tutorial to answer these questions.

A sample of 30 records is selected without replacement from a finite list of 600. What is the FPC, and would it change the standard error substantially?
A population has $N=50$, population standard deviation $\sigma=8$, and a simple random sample without replacement has $n=25$. Write the standard error formula for the sample mean using the FPC.
Why does sampling without replacement produce a smaller standard error than independent sampling with replacement?
A statistician samples every member of a finite population. What is the FPC, and what does it imply about sampling error for the population mean?
A sample is drawn with replacement from a finite list. Should the FPC for sampling without replacement be applied? Explain.

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