Finite-Population Sample Means and the Correction Factor

Why the Sampling Fraction Matters

When observations are sampled without replacement from a fixed finite population, each selected unit changes which units remain available. Consequently, the sample values are dependent. The dependence matters for the variance of a sample total or mean: a sample that covers a substantial share of the population is less variable than one that covers only a small share.

The exact adjustment is the finite-population correction. We derive it for simple random sampling without replacement (SRSWOR), in which every subset of \(n\) distinct population units has the same probability of being selected. The population values themselves are fixed; the randomness comes only from the selection of the sample.

Population and Sample Notation

Let the finite population consist of fixed values \(y_1,\ldots,y_N\), with \(N\geq 2\). Write \(s\) for the random set of selected units, where \(s\) is chosen by SRSWOR and \(1\leq n\leq N\). Define the population total and mean by

T_N=\sum_{i=1}^N y_i, \qquad \bar{Y}_N=\frac{1}{N}\sum_{i=1}^N y_i.

We use the finite-population variance with denominator \(N-1\), and also name the version with denominator \(N\), since the distinction is important in boundary cases.

Definition: The finite-population variance and the population mean squared deviation are, respectively,

S^2=\frac{1}{N-1}\sum_{i=1}^N (y_i-\bar{Y}_N)^2, \qquad V_N=\frac{1}{N}\sum_{i=1}^N (y_i-\bar{Y}_N)^2 =\frac{N-1}{N}S^2.

The sample total and sample mean are \(T_s=\sum_{i\in s}y_i\) and \(\bar{y}_s=T_s/n\). Expectation and variance below are taken over the random choice of \(s\), with the population values held fixed.

Inclusion Indicators and the Exact Results

For each unit \(i\), let \(I_i=1\) if unit \(i\) is selected and \(I_i=0\) otherwise. Under SRSWOR, \(P(I_i=1)=n/N\). For two distinct units \(i\) and \(j\), the probability that both are selected is \(n(n-1)/(N(N-1))\). These indicator probabilities give the exact design expectation and variance.

Theorem: For an SRSWOR of size \(n\) from a fixed population of size \(N\geq2\),

E[T_s]=n\bar{Y}_N,\qquad E[\bar{y}_s]=\bar{Y}_N,

and

\operatorname{Var}(T_s)=n\left(1-\frac{n}{N}\right)S^2, \qquad \operatorname{Var}(\bar{y}_s)=\frac{1-n/N}{n}S^2.

Here \(1-n/N\) is the finite-population correction to the variance formula written using \(S^2\).

Derivation

The sample total can be written as \(T_s=\sum_{i=1}^N I_i y_i\). Since \(E[I_i]=n/N\), linearity of expectation yields

E[T_s]=\sum_{i=1}^N y_i E[I_i] =\frac{n}{N}\sum_{i=1}^N y_i =n\bar{Y}_N.

Dividing by \(n\) proves \(E[\bar{y}_s]=\bar{Y}_N\). Thus the sample mean is design-unbiased for the finite-population mean.

For the variance, put \(z_i=y_i-\bar{Y}_N\), so that \(\sum_i z_i=0\). Subtracting the nonrandom quantity \(n\bar{Y}_N\) from \(T_s\) gives

T_s-n\bar{Y}_N=\sum_{i=1}^N I_i z_i.

Write \(f=n/N\). The indicator variance is \(\operatorname{Var}(I_i)=f(1-f)\). For distinct \(i,j\), the joint inclusion probability gives

\begin{aligned} \operatorname{Cov}(I_i,I_j) &=E[I_iI_j]-E[I_i]E[I_j]\\ &=\frac{n(n-1)}{N(N-1)}-\frac{n^2}{N^2}\\ &=-\frac{f(1-f)}{N-1}. \end{aligned}

The covariance is negative: including one unit makes the joint inclusion of another less likely than it would be under independent selection. Expanding the variance of the centered total and using these indicator variances and covariances gives

\begin{aligned} \operatorname{Var}(T_s) &=f(1-f)\sum_{i=1}^N z_i^2 -\frac{2f(1-f)}{N-1}\sum_{i<j}z_i z_j. \end{aligned}

Because \(\sum_i z_i=0\), squaring this equality shows that \(2\sum_{i<j}z_i z_j=-\sum_i z_i^2\). Substitution therefore yields

\begin{aligned} \operatorname{Var}(T_s) &=f(1-f)\left(1+\frac{1}{N-1}\right)\sum_{i=1}^N z_i^2\\ &=f(1-f)\frac{N}{N-1}(N-1)S^2\\ &=n(1-f)S^2. \end{aligned}

Finally, \(T_s=n\bar{y}_s\), so scaling variance by \(1/n^2\) gives \(\operatorname{Var}(\bar{y}_s)=(1-f)S^2/n\). This proves the theorem. \(\square\)

Worked Examples

Worked Example: Checking the Formula by Enumerating Samples

Take the four fixed population values \(2,4,6,8\), and select \(n=2\) units by SRSWOR. Find the design mean and variance of the sample mean, then verify the variance by listing all possible samples.

The population mean is \(\bar{Y}_N=(2+4+6+8)/4=5\). The squared deviations from 5 sum to \(9+1+1+9=20\), so \(S^2=20/(4-1)=20/3\). Here \(f=2/4=1/2\). The formulas give

E[\bar{y}_s]=5, \qquad \operatorname{Var}(\bar{y}_s) =\frac{1-1/2}{2}\cdot\frac{20}{3} =\frac{5}{3}.

There are six equally likely two-unit samples. Their means are \(3,4,5,5,6,7\). Their average is \(30/6=5\), and their variance is

\frac{(3-5)^2+(4-5)^2+(5-5)^2+(5-5)^2+(6-5)^2+(7-5)^2}{6} =\frac{10}{6}=\frac{5}{3}.

The enumeration agrees with both the exact mean and variance.

Worked Example: One Draw and the Two Population Variance Conventions

Use the same four values, \(2,4,6,8\), but now select just one unit. A single selected value is uniformly distributed over those four fixed values. Its mean is 5, and its variance is

V_N=\frac{9+1+1+9}{4}=5.

The formula using \(S^2\) gives the same result:

\operatorname{Var}(\bar{y}_s) =\frac{1-1/4}{1}\cdot\frac{20}{3} =\frac{3}{4}\cdot\frac{20}{3} =5.

Since \(n=1\), the sample mean is the selected value itself. Its variance is \(V_N=(N-1)S^2/N\), not \(S^2\). This check shows why the two denominator conventions must not be interchanged.

Worked Example: A Sample Covering Most of the Population

Suppose a population has five fixed values \(3,5,5,7,10\), and four are selected by SRSWOR. Calculate the expected sample total and mean and their variances.

The population mean is \(\bar{Y}_N=30/5=6\). The squared deviations sum to \(9+1+1+1+16=28\), so \(S^2=28/4=7\). The sampling fraction is \(f=4/5\). Therefore,

E[T_s]=4(6)=24,\qquad E[\bar{y}_s]=6,

and

\operatorname{Var}(T_s)=4\left(1-\frac45\right)7=\frac{28}{5}=5.6, \qquad \operatorname{Var}(\bar{y}_s)=\frac{1/5}{4}\,7=\frac{7}{20}=0.35.

The correction is small because the sample contains four of the five units. If all five were selected, the sample mean would equal 6 with certainty and its variance would be zero.

Interpreting the Finite-Population Correction

The correction factor \(1-n/N\) multiplies the variance when it is expressed using \(S^2\). Accordingly, the standard deviation of the sample mean is

\sqrt{\operatorname{Var}(\bar{y}_s)} =\frac{S}{\sqrt{n}}\sqrt{1-\frac{n}{N}}.

The factor \(\sqrt{1-n/N}\) is called the finite-population correction for the standard error. As the sampling fraction increases, the correction decreases. At a full census, \(n=N\), it is zero because the sample mean then equals the population mean exactly. For a small sampling fraction, the correction is near one, so the without-replacement variance is close to \(S^2/n\).

An equivalent expression using the denominator-\(N\) variance \(V_N\) is

\operatorname{Var}(\bar{y}_s) =\frac{N-n}{n(N-1)}V_N.

For \(n=1\), this expression reduces to \(V_N\), as it must. Keeping track of which population variance convention is being used prevents a common boundary-case error.

Common Mistakes

Using the wrong population variance convention. With \(S^2\) defined using denominator \(N-1\), the formula is \((1-n/N)S^2/n\). For one draw, this equals \((N-1)S^2/N=V_N\), not \(S^2\).
Applying the correction to the wrong quantity. The factor \(1-n/N\) multiplies the variance. The corresponding standard-error multiplier is its square root.
Forgetting the census boundary case. When \(n=N\), the total and mean are fixed at \(T_N\) and \(\bar{Y}_N\), so both variances must be zero.
Treating the population values as random. These are design expectations and variances: the values \(y_i\) are fixed, and only the sample selection varies.

Key takeaway: Under SRSWOR, the sample mean is unbiased for the fixed population mean, and its exact design variance is \((1-n/N)S^2/n\). The finite-population correction accounts for the reduction in uncertainty as the sample covers more of the population.

Check Your Understanding

Use the definitions and formulas above to answer these questions.

For a population with \(N=20\), a sample of size \(n=5\) has population variance \(S^2=12\). What are the design expectation and variance of its sample mean?
Why is the covariance between two distinct inclusion indicators negative under SRSWOR?
If \(n=1\), which population variance convention gives the variance of the selected value directly: \(S^2\) or \(V_N\)?
What must the variance of the sample mean be when \(n=N\), and how does the formula show this?
How does the finite-population correction to the standard error differ from the correction to the variance?