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Cluster and two-stage sampling — Examples

1 · Prerequisites

2 · Summary

The worked village-and-field example enumerates every outcome of a small two-stage SRSWOR design and computes its exact expansion estimate and design variance directly. The counterexample uses unequal PSU sizes and an explicit common within-PSU correlation model to show why the equal-size design-effect shortcut cannot use the arithmetic mean cluster size as a universal substitute.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-6-luna)Open item page →

A two-stage village-and-field sample

Example

Two villages are the PSUs, and the four fields in each village are its SSUs. Their fixed values are $y_{1 \cdot} = (2, 4, 6, 8), y_{2 \cdot} = (1, 3, 5, 7) .$ Select one village uniformly, then select two of its four fields uniformly without replacement. The two-stage expansion estimator has mean $36$ , the fixed total over all eight fields, and exact design variance $368 / 3$ .

Facts & Assumptions

Given: The displayed fixed field values; first-stage SRSWOR of one village from two; second-stage SRSWOR of two fields from the selected village.

[F1]

Two-stage sampling laws and the expansion estimator: The joint sample mass is the first-stage mass times the selected PSUs' second-stage masses, and $\hat{T} = \sum_{i \in S_{1}} {\hat{T}}_{i} / π_{i}^{(1)}$ with ${\hat{T}}_{i} = \sum_{j \in S_{i}} y_{i j} / π_{j ∣ i}^{(2)}$ .

[F2]

Moments, variance, and covariance on a probability space: Variance is the expectation of the centered square.

[A1]

Each stage uses the uniform SRSWOR law just stated; hence each of the six field pairs in a selected village has conditional probability $1 / 6$ and each of the twelve joint outcomes has probability $1 / 12$ .

Verification

1.1givenF1A1algebra

The village totals are $T_{1} = 20$ and $T_{2} = 16$ , $π_{1}^{(1)} = π_{2}^{(1)} = 1 / 2$ , and every second-stage field inclusion probability is $1 / 2$ ; hence a selected pair with value sum $s$ gives ${\hat{T}}_{i} = 2 s$ and $\hat{T} = 4 s$ .

2.1F1A1step 1.1algebra

Enumerating the six pairs in village 1 gives expansion estimates $(24, 32, 40, 40, 48, 56)$ and in village 2 $(16, 24, 32, 32, 40, 48)$ ; their combined sum is $432$ , so the mean of the twelve equally likely outcomes is $432 / 12 = 36 = T_{1} + T_{2}$ .

3.1F2A1step 2.1algebra∎

The twelve estimates in step 2.1 have squared deviations from $36$ summing to $1472$ ; by [F2] and their common probability $1 / 12$ , the exact design variance is $1472 / 12 = 368 / 3$ , by direct finite enumeration without an AC-dependent conditional-variance theorem.

Remarks

This finite enumeration is an independent check of one two-stage design; it does not replace the general variance proof on the A page. Holmes–Huber’s chapter supplies the experimental-design caution that repeated observations within one group do not create additional independent units; it does not prove the design-based expansion or variance formulas used here.

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Average cluster size does not determine the unequal-size design effect

Statement refuted

For unequal cluster sizes, with independent PSUs, equal marginal variance, common within-cluster correlation, and negligible FPC, the design effect is still $1 + (\overset{ˉ}{m} - 1) ρ$ after substituting the arithmetic mean size $\overset{ˉ}{m}$ for the equal cluster size. In the example below, $\overset{ˉ}{m} = 2$ and the within-pair correlation is $1$ , so this proposed formula predicts $2$ , while the exact model-averaged design-effect ratio tends to $5 / 2$ as $M \to \infty$ with fixed $k$ .

The two designs are evaluated separately on each fixed realization of the model values: $S_{C}$ is a uniform SRSWOR of $k$ PSUs and $S_{E}$ is a uniform SRSWOR of $2 k$ SSUs. Define ${\hat{μ}}_{C} = \frac{M}{k N} \sum_{i \in S_{C}} T_{i}, {\hat{μ}}_{E} = \frac{1}{2 k} \sum_{(i, j) \in S_{E}} y_{i j},$ where $N = \sum_{i} n_{i} = 2 M$ . The design effect here is $E_{B} [{Var}_{d} ({\hat{μ}}_{C} ∣ B)] / E_{B} [{Var}_{d} ({\hat{μ}}_{E} ∣ B)]$ .

Facts & Assumptions

Given: An even number $M \geq 4$ of PSUs, half of size $1$ and half of size $3$ ; independent random variables $B_{i}$ with mean $0$ and variance $σ^{2} > 0$ ; fixed SSU values $y_{i j} = B_{i}$ conditional on $(B_{i})$ ; and a cluster SRSWOR of $k$ PSUs, $1 \leq k < M$ , compared with an element SRSWOR of $2 k$ SSUs from the same finite population. This matches the expected SSU count in the cluster sample. Define the design effect as the ratio of the two design variances, each averaged over the model for $(B_{i})$ .

[F1]

Cluster sampling units, cluster totals, and one-stage SRSWOR: A cluster design selects PSUs and, in one-stage sampling, observes all SSUs in selected PSUs; the cluster total is $T_{i} = \sum_{j} y_{i j}$ .

[F2]

Unbiasedness and variance of the SRSWOR cluster-total estimator: Under SRSWOR of $k$ from $M$ PSUs, the total estimator has variance $M^{2} (1 - k / M) S_{T}^{2} / k$ .

[F3]

Moments, variance, and covariance on a probability space: Variance is the expectation of the centered square, and covariance is the expectation of the centered product.

[F4]

Covariance is symmetric and bilinear in finite linear combinations: Covariance is symmetric and bilinear in finite linear combinations of square-integrable variables.

[F5]

Variance of a finite sum as the sum of all variances and covariances: The variance of a finite sum of square-integrable real random variables is the sum of all pairwise covariances.

[A1]

Conditional on the model values, the finite population is fixed and each design is an ordinary uniform SRSWOR. The model variables $B_{i}$ are independent across PSUs and share the stated mean and variance.

Counterexample

1.1givenF1A1algebra

The cluster sizes are $1$ and $3$ in equal proportions, so $N = 2 M$ and $\overset{ˉ}{m} = 2$ ; in every size- $3$ cluster the values equal $B_{i}$ , giving correlation $ρ = 1$ for each within-cluster pair that exists (size- $1$ clusters have none), and every element has marginal variance $σ^{2}$ . The proposed average-size substitution therefore predicts $1 + (2 - 1) \cdot 1 = 2$ .

1.2givenA1F2F3F4F5algebra

For ${\overset{ˉ}{Y}}_{U} = N^{- 1} \sum_{i} n_{i} B_{i}$ , independence gives $E_{B} [{\overset{ˉ}{Y}}_{U}] = 0$ and ${Var}_{B} ({\overset{ˉ}{Y}}_{U}) = 5 σ^{2} / (4 M)$ ; each element has $E_{B} [y_{i j}^{2}] = σ^{2}$ , so the identity $\sum_{i, j} (y_{i j} - {\overset{ˉ}{Y}}_{U})^{2} = \sum_{i, j} y_{i j}^{2} - N {\overset{ˉ}{Y}}_{U}^{2}$ and $S_{U}^{2} = (N - 1)^{- 1} \sum_{i, j} (y_{i j} - {\overset{ˉ}{Y}}_{U})^{2}$ give $E_{B} [S_{U}^{2}] = (N σ^{2} - N {Var}_{B} ({\overset{ˉ}{Y}}_{U})) / (N - 1) = σ^{2} (2 M - 5 / 2) / (2 M - 1)$ . By [F2] on the singleton-PSU partition with sample size $2 k$ , the element mean estimator has variance $(1 - 2 k / N) S_{U}^{2} / (2 k) = (1 - k / M) S_{U}^{2} / (2 k)$ , so $E_{B} [{Var}_{d} ({\hat{μ}}_{E} ∣ B)] = [(1 - k / M) / (2 k)] σ^{2} (2 M - 5 / 2) / (2 M - 1)$ .

2.1givenA1F2F3F4F5step 1.1algebra

Let $n_{i} \in {1, 3}$ be the PSU sizes; then $T_{i} = n_{i} B_{i}$ , $\sum_{i} n_{i}^{2} = 5 M$ , and independence gives ${Var}_{B} (\overset{ˉ}{T}) = σ^{2} \sum_{i} n_{i}^{2} / M^{2}$ . Using $\sum_{i} (T_{i} - \overset{ˉ}{T})^{2} = \sum_{i} T_{i}^{2} - M {\overset{ˉ}{T}}^{2}$ and $E_{B} [T_{i}] = 0$ yields $E_{B} [\sum_{i} (T_{i} - \overset{ˉ}{T})^{2}] = \sum_{i} {Var}_{B} (T_{i}) - M {Var}_{B} (\overset{ˉ}{T}) = ((M - 1) / M) σ^{2} \sum_{i} n_{i}^{2}$ , hence $E_{B} [S_{T}^{2}] = (1 / (M - 1)) E_{B} [\sum_{i} (T_{i} - \overset{ˉ}{T})^{2}] = (σ^{2} / M) \sum_{i} n_{i}^{2} = 5 σ^{2}$ . By [F2], dividing the total-estimator variance by $N^{2} = (2 M)^{2}$ gives $E_{B} [{Var}_{d} ({\hat{μ}}_{C} ∣ B)] = (5 σ^{2} / (4 k)) (1 - k / M)$ .

3.1step 1.1step 1.2step 2.1algebra∎

The common positive factor $(1 - k / M) σ^{2} / (2 k)$ cancels, so $DEFF = (5 / 2) (2 M - 1) / (2 M - 5 / 2)$ ; at $M = 100$ this is $199 / 79$ , not $2$ , and it tends to $5 / 2$ as $M \to \infty$ , refuting the average-size substitution.

Remarks

This refutes only the arithmetic-mean-size substitution. It does not give a general design-effect formula for unequal cluster sizes, unequal weights, or nonexchangeable covariance structures; each requires its own stated estimator and variance calculation.

Cluster and two-stage sampling — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A two-stage village-and-field sample

Example

Facts & Assumptions

Verification

Remarks

Average cluster size does not determine the unequal-size design effect

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources