= Batch-specific blank allocation
\b[Batch-specific blank allocation] assigns background controls to the specimen groups whose preparation conditions they actually represent. In an experiment on <blank-corrected carbon recovery in polymer tests>, a batch may share inoculum, medium, collection time or reagent preparation. A precisely measured blank from another batch can still be the wrong correction. OECD's test framework uses parallel inoculum controls and warns that endogenous activity with a test substance need not equal the control's activity.<Test No. 301: Ready Biodegradability by OECD>[General principle §4; method 301 B §§1–3]{ref}
For an original independent-batch model, estimate a weighted net mean by
$$
\widehat\mu=\sum_b w_b(\overline T_b-\overline B_b),\qquad
\operatorname{Var}(\widehat\mu)=\sum_b w_b^2\left(\frac{\sigma_{T,b}^2}{n_{T,b}}+\frac{\sigma_{B,b}^2}{n_{B,b}}\right).
$$
The weights specify the scientific target; they are not automatically the observed fractions of specimens. Within a batch, reusing its blank estimate creates the <shared-blank covariance in carbon recovery>. Common corrections across batches require additional <covariance> terms rather than the independence assumed here.<Guide to the expression of uncertainty in measurement by JCGM>[§5.2.2, equation 13; §5.2.4, common corrections]{ref}
If specimen counts are fixed and each blank costs the same, minimizing the remaining <variance> under a total blank-count budget gives $n_{B,b}\propto|w_b|\sigma_{B,b}$. Consider two batches with weights $0.8,0.2$, ten specimen readings each, specimen standard deviation 1 mg, blank standard deviation 2 mg and twelve blank readings available. Ten blanks in the first batch and two in the second give <variance> $0.64(0.1+4/10)+0.04(0.1+4/2)=0.404\,\mathrm{mg^2}$. Splitting blanks six and six gives approximately $0.5213\,\mathrm{mg^2}$. These stipulated numbers illustrate allocation; integer and minimum-replication constraints still apply.
<Batch-matched blank weighting> also controls bias. If batch backgrounds are 1 and 5 mg, their scientifically weighted background is $0.8(1)+0.2(5)=1.8$ mg. Subtracting an equally pooled blank mean of 3 mg instead biases that weighted net result downward by 1.2 mg. Extra replication of the pooled estimate does not remove this mismatch.
Record batch membership and independent preparations before deciding which blanks to pool. Allocation can reduce sampling <variance>; it cannot prove that a background correction is chemically additive.
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