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Biological and technical replication — Examples

1 · Prerequisites

2 · Summary

The example compares five technical assays on one animal with one assay on each of five independently sampled animals. Both designs use the same number of readings, but the second averages biological variation across five units. The variance comparison follows from the stated random-intercept model and does not establish that any particular sample or assay process satisfies it.

3 · Logical flowchart

4 · Definitions, theorems and proofs

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5 · Examples, counterexamples and false statements

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

Five assays on one animal versus one assay on five animals

Example

Suppose animals are independently sampled from a target population and all receive the same fixed treatment. Under Biological and technical replication model, let the common mean be $μ + τ$ , the biological variance be $σ_{b}^{2} = 4$ , and the technical measurement variance be $σ_{e}^{2} = 1$ . Compare two designs with five assays in total: assay one animal five times, or assay five animals once each. In both designs the reported estimator is the average of all five readings, and the model target is the population mean $μ + τ$ .

Facts & Assumptions

Given: The balanced model with common fixed treatment contribution, five total readings, $σ_{b}^{2} = 4$ , and $σ_{e}^{2} = 1$ .

[F1]

Biological and technical replication model: each reading contains a biological effect shared within its unit and an independent technical error; biological and technical variables have zero means and the stated variances.

[F2]

Mean and variance under balanced biological and technical replication: with fixed treatment contributions, the all-reading mean has expectation $μ + \overset{ˉ}{τ}$ and variance $σ_{b}^{2} / n + σ_{e}^{2} / (n m)$ .

Verification

Given: The balanced model, common fixed treatment, five total readings, $σ_{b}^{2} = 4$ , and $σ_{e}^{2} = 1$ .

1.1givenF2algebra

With one animal and five assays, $n = 1, m = 5$ and $\overset{ˉ}{τ} = τ$ . By [F2], the average has expectation $μ + τ$ and variance $4 / 1 + 1 / (1 \cdot 5) = 21 / 5$ .

1.2givenF2algebra

With five independently sampled animals and one assay per animal, $n = 5, m = 1$ . By [F2], the average again has expectation $μ + τ$ and variance $4 / 5 + 1 / (5 \cdot 1) = 1$ .

2.1givenF1step 1.1step 1.2algebra

Both designs use five readings, but the first variance exceeds the second by $21 / 5 - 1 = 16 / 5 > 0$ . The comparison isolates the biological component: averaging the same animal five times leaves its shared $B_{1}$ contribution unaveraged, while the five-animal design averages five independent biological effects.

3.1F2step 2.1algebra∎

If instead $σ_{b}^{2} = 0$ , both designs have variance $σ_{e}^{2} / 5$ ; the strict advantage in step 2.1 comes from positive biological variation.

Remarks

The calculation is conditional on the stated model and equal assay count. It does not show that technical replication is useless for measuring assay error, or that this variance model is adequate for any particular study. The randomization unit and population sampling unit must be identified from the actual design.