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Stratification and allocation — Examples

1 · Prerequisites

2 · Summary

These worked finite populations compare proportional and disproportionate allocations, show why the positive-variance hypothesis matters for the continuous Neyman formula, and calculate how nearest-integer rounding can violate a stratum's minimum sample size. The valid designs are checked with the design-weighted estimator and its exact variance formula. A corrected integer allocation must still respect each stratum's sample-size bounds; the continuous optimum alone does not solve that constrained problem.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Proportional and disproportionate stratified allocations

Example

Let $U_{1}$ have fixed values $7, 9, 11, 13$ and let $U_{2}$ have fixed values $18, 22$ . Compare independent stratified SRSWOR allocations $(n_{1}, n_{2}) = (2, 1)$ and $(1, 2)$ , both with total sample size three. The first allocation is proportional to stratum sizes; the second oversamples the smaller stratum.

Facts & Assumptions

Given: The six fixed values above, independent within-stratum SRSWOR, and the two stated allocations.

[F1]

Stratified sampling designs and the stratified total estimator: The estimator is ${\hat{T}}_{s t} = N_{1} {\overset{ˉ}{y}}_{1, R} + N_{2} {\overset{ˉ}{y}}_{2, R}$ ; certainty strata have zero variance contribution.

[F2]

Unbiasedness and variance of the stratified total estimator: The estimator is design-unbiased, and its variance is the sum of the stratum contributions $N_{h}^{2} (1 - n_{h} / N_{h}) S_{h}^{2} / n_{h}$ over noncertainty strata.

Verification

1.1givenchoosealgebra

The fixed means are ${\overset{ˉ}{y}}_{1} = 10$ and ${\overset{ˉ}{y}}_{2} = 20$ . The population variances are $S_{1}^{2} = ((- 3)^{2} + (- 1)^{2} + 1^{2} + 3^{2}) / 3 = 20 / 3$ and $S_{2}^{2} = ((- 2)^{2} + 2^{2}) / 1 = 8$ , so the total is $4 (10) + 2 (20) = 80$ .

2.1F1F2step 1.1

Both allocations satisfy $1 \leq n_{h} \leq N_{h}$ . By [F2] and step 1.1, each corresponding weighted estimator has expectation $80$ .

2.2F2step 1.1algebra

Under allocation $(2, 1)$ , [F2] and step 1.1 give variance $16 (1 - 2 / 4) (20 / 3) / 2 + 4 (1 - 1 / 2) 8 / 1 = 80 / 3 + 16 = 128 / 3$ .

2.3F1F2step 1.1algebra

Under allocation $(1, 2)$ , the second stratum is a census, so [F1], [F2], and step 1.1 give variance $16 (1 - 1 / 4) (20 / 3) / 1 = 80$ .

3.1step 2.1step 2.2step 2.3algebra∎

Step 2.1 gives unbiasedness for both designs, and steps 2.2–2.3 give $128 / 3 < 80$ for these fixed values.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A zero-variance stratum is outside the positive-variance allocation formula

Example

Let the first stratum have values $(5, 5)$ and the second have values $(0, 3, 6)$ . For total sample size three, the unqualified weights $N_{h} S_{h}$ assign no sample units to the first stratum, even though the stratified design requires at least one observation per stratum.

Facts & Assumptions

Given: Two fixed strata with those values, independent within-stratum SRSWOR, and total sample size $n = 3$ .

[F1]

Stratified sampling designs and the stratified total estimator: Each stratum sample size satisfies $1 \leq n_{h} \leq N_{h}$ ; $S_{h}^{2}$ is defined for noncertainty strata with $N_{h} > 1$ .

[F2]

Continuous fixed-size Neyman allocation minimizes stratified variance: The continuous Neyman formula requires positive variances in every variable noncertainty stratum.

[F3]

Unbiasedness and variance of the stratified total estimator: For a valid allocation, the total estimator is unbiased and its design variance is the sum of the noncertainty stratum contributions.

Verification

1.1givenchoosealgebra

The first stratum has $N_{1} = 2$ , ${\overset{ˉ}{y}}_{1} = 5$ , and $S_{1}^{2} = 0$ . The second has $N_{2} = 3$ , ${\overset{ˉ}{y}}_{2} = 3$ , and $S_{2}^{2} = ((- 3)^{2} + 0^{2} + 3^{2}) / 2 = 9$ , so $S_{2} = 3$ and the total is $19$ .

2.1F1F2step 1.1algebra

The unqualified proportionality weights would be $N_{1} S_{1} = 2 (0) = 0$ and $N_{2} S_{2} = 3 (3) = 9$ , giving formal continuous allocation $(x_{1}, x_{2}) = (0, 3)$ . But $S_{1} = 0$ violates [F2]'s positive-variance hypothesis, and $x_{1} = 0$ violates [F1]'s $n_{1} \geq 1$ condition.

3.1F1F3step 1.1step 2.1algebra∎

The valid allocation $(n_{1}, n_{2}) = (1, 2)$ has variance $2^{2} (1 - 1 / 2) 0 / 1 = 0$ in the first stratum and $3^{2} (1 - 2 / 3) 9 / 2 = 27 / 2$ in the second by [F3]. It is a valid design, whereas the formal zero allocation is not.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

Continuous Neyman allocation may require constrained rounding

Example

Take two strata of sizes $2$ and $10$ . In the first, let the values be $(- 1 / \sqrt{2}, 1 / \sqrt{2})$ ; in the second, take five copies each of $- 9 / \sqrt{10}$ and $9 / \sqrt{10}$ . With total sample size six, the continuous Neyman allocation rounds to zero units in the first stratum under nearest- integer rounding, which violates the stratified design's positive sample-size requirement.

Facts & Assumptions

Given: The fixed strata above, independent within-stratum SRSWOR, total sample size six, and componentwise rounding to the nearest integer.

[F1]

Stratified sampling designs and the stratified total estimator: A stratified design requires integer sample sizes $1 \leq n_{h} \leq N_{h}$ ; for a noncertainty stratum, $S_{h}^{2} = (N_{h} - 1)^{- 1} \sum_{i \in U_{h}} (y_{i} - {\overset{ˉ}{y}}_{h})^{2}$ .

[F2]

Continuous fixed-size Neyman allocation minimizes stratified variance: With positive noncertainty stratum variances and a fixed positive total sample size, the uncapped continuous optimum is proportional to $N_{h} S_{h}$ ; this relaxation imposes no integer, capacity, lower-bound, or minimum-sample constraints.

[F3]

Unbiasedness and variance of the stratified total estimator: For an admissible allocation, the stratified total's variance is the sum of the stratum terms $N_{h}^{2} (1 - n_{h} / N_{h}) S_{h}^{2} / n_{h}$ over noncertainty strata.

Verification

1.1givenchoosealgebra

The first stratum has mean zero and $S_{1}^{2} = ((- 1 / \sqrt{2})^{2} + (1 / \sqrt{2})^{2}) / (2 - 1) = 1$ . The second has mean zero and $S_{2}^{2} = 10 (9 / \sqrt{10})^{2} / (10 - 1) = 9$ , so $S_{2} = 3$ .

2.1F1F2step 1.1algebra

The Neyman weights are $N_{1} S_{1} = 2$ and $N_{2} S_{2} = 30$ . By [F2] and step 1.1, $x_{1} = 6 (2) / (2 + 30) = 3 / 8$ and $x_{2} = 6 (30) / (2 + 30) = 45 / 8$ . Nearest-integer rounding gives $(0, 6)$ , which violates [F1]'s $n_{1} \geq 1$ condition.

3.1F1F3step 1.1step 2.1algebra∎

Allocation $(n_{1}, n_{2}) = (1, 5)$ is admissible. By [F3] and step 1.1 its variance is $2^{2} (1 - 1 / 2) 1 / 1 + 10^{2} (1 - 5 / 10) 9 / 5 = 2 + 90 = 92$ ; this feasible correction is not claimed to solve every constrained integer problem.

Stratification and allocation — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Proportional and disproportionate stratified allocations

Example

Facts & Assumptions

Verification

A zero-variance stratum is outside the positive-variance allocation formula

Example

Facts & Assumptions

Verification

Continuous Neyman allocation may require constrained rounding

Example

Facts & Assumptions

Verification

Sources