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Detection and quantification limits — Examples

1 · Prerequisites

2 · Summary

Worked examples substitute the five-percent false-positive and false-negative targets and ten-percent CV criterion into the known-blank normal formulas. They then change the model by adding estimated-blank variance or concentration-dependent variance. The final example treats a below-threshold report as a censored event and computes its probability under the declared response law.

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 →

Known-blank normal limits at five-percent error and ten-percent CV

Example

Assume AC and σ0>0. In the known-blank model X∼N(c,σ02), take α=β=0.05 and kQ=0.10. Then LC=z0.95σ0,LD=2z0.95σ0,LQ=10σ0. The NIST SP 958 table reports the corresponding known-blank values rounded to 1.64σ0, 3.29σ0, and 10σ0. These are conditional on the equal-known-variance normal law and the stated error and CV targets.

Facts & Assumptions

Given: AC; σ0>0; the known-blank experiment and definitions in Exact limits in the known-blank equal-variance normal model and Critical, detection, and quantification levels, with α=β=0.05 and kQ=0.10.

[A1]
[F1]

Under the known-blank equal-variance normal model, LC=z1−ασ0, LD=(z1−α+z1−β)σ0, and LQ=σ0/kQ. Exact limits in the known-blank equal-variance normal model

[F2]

The standard normal q-quantile is the unique zq with Φ(zq)=q, and z1−q=−zq. The standard normal CDF has unique symmetric quantiles

Verification

Given: AC; σ0>0; α=β=0.05=1/20; kQ=0.10=1/10; the known-blank model; and the suppliers [F1]–[F2].

1.1A1F1F2algebra

Since 1−α=1−β=0.95, [A1] and [F1] give LC=z0.95σ0 and LD=(z0.95+z0.95)σ0=2z0.95σ0, with z0.95 the unique quantile from [F2].

2.1F1step 1.1algebra∎

By [F1], the stated CV criterion gives LQ=σ0/kQ=σ0/(1/10)=10σ0. Together with step 1.1, this gives all three exact limits.

Source notes

The exact expressions follow from [F1]. The NIST table reports the known-blank row as 1.64, 3.29, and 10 times the blank standard deviation for the same targets; those table entries are rounded values, not exact replacements for z0.95.

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

An estimated blank increases the effective normal scale

Example

Assume AC, σ0>0, and τ>0. Suppose the blank-corrected response is explicitly modeled by Dc∼N(c,σ02+τ2) for every c≥0, where τ2>0 is the declared additional variance contribution from estimating the blank. Put σeff=σ02+τ2. The exact normal formulas use σeff in place of σ0: LC=z1−ασeff,LD=(z1−α+z1−β)σeff,LQ=σeffkQ. This is a stated comparison model, not a claim that every blank-estimation procedure produces an independent normal error with this variance. The blank-correction uncertainty must be included in the response law used to calculate the limits.

Facts & Assumptions

Given: AC; σ0>0, τ>0, 0<α,β<1/2, kQ>0, and the corrected-response law in the Example.

[A1]

AC is assumed by the known-normal limit theorem used below. The Axiom of Choice Exact limits in the known-blank equal-variance normal model

[F1]

For a known-blank equal-variance normal family with scale σ>0, the three limits are z1−ασ, (z1−α+z1−β)σ, and σ/kQ. Exact limits in the known-blank equal-variance normal model

Verification

Given: AC; σ0>0, τ>0, 0<α,β<1/2, kQ>0; and the corrected-response model in the Example.

1.1givenA1F1algebra

Set σeff=σ02+τ2. Since σ0>0 and τ>0, σeff>σ0>0. Thus the corrected-response family has exactly the form required by [F1], with σ=σeff.

2.1A1F1step 1.1algebra∎

Substitution into [F1] yields the three displayed limits. Since 0<α,β<1/2, both z1−α and z1−β are positive; since kQ>0, each multiplier of σeff is positive. Together with σeff>σ0, each corresponding limit is strictly larger than its known-blank value.

Source notes

Eurachem §5.3.2 and Example 4 show that routine averaging and blank correction change the standard deviation used for LOD/LOQ; §5.3.5 notes that day-to-day blank variation can require intermediate-precision data. The example's exact normal law is an explicit assumption, and the limit formulas are proved locally from [F1].

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Detection-limit assumptions and left-censored readings

Example

Consider the heteroscedastic response law Yc∼N(c,σ02(1+ρc)2) for c≥0, with σ0>0 and ρ≥0. The blank variance is σ02, but the standard deviation increases with concentration when ρ>0. For 0<α,β<1/2, write a=z1−β. The critical level is LC=z1−ασ0. A finite detection limit for false-negative target β exists exactly when aρσ0<1, and then LD=LC+aσ01−aρσ0. For kQ>0, the coefficient-of-variation criterion has a finite quantification limit exactly when kQ>ρσ0, and then LQ=σ0kQ−ρσ0. When a reporting procedure records only the category Y≤LC, its likelihood contribution at concentration c is the probability Pc(Y≤LC)=Φ ⁣(LC−cσ0(1+ρc)), not a density value at LC.

Facts & Assumptions

Given: AC; the heteroscedastic family and parameter restrictions in the Example; and the positive-decision rule is Y>LC.

[A1]

AC is required by the standard normal probability-law construction. The Axiom of Choice Standard normal and normal laws

[F1]

The standard normal CDF is strictly increasing, has no atoms, and its quantiles obey zq=−z1−q for every q∈(0,1). The standard normal CDF has unique symmetric quantiles

[F2]

A normal response with mean m and scale σ has expectation m and variance σ2. Location-scale normal laws have their stated mean and variance

[F3]

N(c,σ2) is the law of c+σZ for standard normal Z. Standard normal and normal laws

[F4]

A real random variable's CDF is F(x)=P(Y≤x). Cumulative distribution function of a real random variable

[F5]

The exact critical level and the false-negative target are defined relative to the specified blank and signal laws. Critical, detection, and quantification levels

[F6]

For a probability measure, P(Ac)=1−P(A). Basic identities for a probability measure

Verification

Given: The response model, positive-decision rule, and targets in the Example.

1.1A1F1F3F5F6givenalgebra

At c=0, the model is N(0,σ02), so [F3] represents Y0=σ0Z in law. Set LC=z1−ασ0; by [F1], [F3], and complement identity [F6], P0(Y0>LC)=1−Φ(z1−α)=α. Also [F1] gives z1/2=−z1/2, hence z1/2=0 and Φ(0)=1/2. Since 1−α>1/2, strict increase gives z1−α>0, and [F5] makes this an exact critical level.

1.2F2givenalgebra

By [F2], CV⁡(c)=σ(c)/c=σ0/c+ρσ0 for c>0. If kQ>ρσ0, the inequality CV⁡(c)≤kQ is equivalent to c≥σ0/(kQ−ρσ0), giving the displayed least value. If kQ≤ρσ0, then CV⁡(c)>ρσ0≥kQ for every finite c>0, so no finite quantification limit exists.

2.1A1F1F3F5givenstep 1.1algebra

For c>0, the scale is σ(c)=σ0(1+ρc)>0. By [F3], Pc(Y≤LC)=Φ((LC−c)/σ(c)). Since a=z1−β>0, strict increase in [F1] makes this equal to β exactly when c−LC=aσ0(1+ρc),a=z1−β>0. Rearranging gives c(1−aρσ0)=LC+aσ0, whose right-hand side is positive. If aρσ0<1, the unique solution is the positive displayed LD. If aρσ0=1, the equation would equate zero to a positive number; if aρσ0>1, its only algebraic solution is negative. Thus a nonnegative detection limit exists exactly under the stated condition.

3.1A1F1F3F4step 2.1algebra∎

If the recorded observation is the category Y≤LC, its probability under the stated law is the CDF at LC, namely the displayed Φ((LC−c)/σ(c)) by [F4]. The affine normal law has no atom at LC by [F1] and [F3], while the CDF gives the probability of the reported half-line. Substituting LC as an observed point is therefore not the likelihood of this censored report.

Source notes

Eurachem §5.3.2 explains that blank correction changes the standard deviation used for LOD/LOQ; §5.3.5 discusses changing blanks across days. The heteroscedastic law and censoring rule here are explicit model assumptions. Their consequences are derived above rather than imported as universal measurement claims.

Sources