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An interlaboratory consensus value is a value derived from results supplied by a specified group of laboratories to represent their common location for a stated measurand. It can serve as an assigned value in proficiency testing, but consensus and independent reference assignment are different ways of establishing that benchmark. The procedure must specify the participating results, estimator and associated measurement uncertainty.^[ref]

For an original example, three laboratories report $9.90$, $10.00$ and $10.10$ in the same units. Their arithmetic mean is $10.00$ and their sample standard deviation is $0.10$. Under a model of independent, equally distributed, unbiased laboratory errors, the estimated standard uncertainty of this mean is $0.10/\sqrt{3}\approx 0.0577$. This calculation does not include an unrecognized systematic error common to all three laboratories.

Choosing a location estimator is a substantive decision. For stipulated results $100.0$, $100.2$, $100.4$ and $105.0$, the arithmetic mean is $101.4$, whereas the median is $100.3$. Robust statistics can reduce the influence of an isolated extreme result, but cannot decide from these four numbers whether that result is erroneous or represents a different measurement method or material condition. The guide discusses robust approaches while distinguishing consensus assessment from fitness-for-purpose criteria.^[ref]

When laboratories supply valid, mutually uncorrelated standard uncertainties $u_i$ for unbiased estimates of the same quantity, an inverse-variance weighted mean usesThe last equality belongs to that model. Shared references, method-specific effects or underestimated $u_i$ can invalidate its interpretation. A laboratory reporting a small uncertainty should not receive overwhelming influence merely because that number was submitted.

An interlaboratory consensus value therefore summarizes a specified collection of evidence. It is not automatically the true value, a certified reference material value, or an independent test of those same laboratories. Shared bias in an interlaboratory consensus and the influence of participants on their own benchmark must be considered before interpreting a small deviation as evidence of accuracy.

A leave-one-out interlaboratory consensus estimates a benchmark for laboratory $i$ from the other participants. For a fixed weighted mean with positive coefficients $w_j$ and at least two results, defineThe omitted result does not enter its own benchmark. This removes the direct participant influence on a consensus assigned value for the stated fixed-weight estimator. It does not establish independence when laboratories share a systematic error. The proficiency-testing guide distinguishes participant consensus from expert or independent reference assignment; the leave-one-out construction here is a mathematical comparison, not a claim that the guide requires this scoring procedure.^[ref]

With equal weights and original results $(9,10,11)$, the omitted-result arithmetic means are $(10.5,10,9.5)$, giving residuals $(-1.5,0,1.5)$. Residuals from the full arithmetic mean are $(-1,0,1)$. In general,The larger magnitudes reflect a changed comparator rather than a changed laboratory result.

The joint dependence of leave-one-out consensus residuals matters when interpreting the collection of scores. Suppose the $X_i$ are independent random variables with a common mean and finite common variance ${\sigma }^2$. A result is independent of the arithmetic mean of the other results, soFor different participants, however, their leave-one-out benchmarks overlap. Linearity of covariance givesThus independent laboratory observations do not give independent residual scores. The zero-sum identity makes their covariance matrix a singular matrix. For $n=4$ and $\sigma =2$ units, each residual has standard deviation $\sqrt{16/3}\approx 2.31$ units, and each distinct pair has correlation coefficient $-1/3$. No normal distribution assumption is needed for these moment identities.

A leave-one-out inverse-variance weighted consensus applies an inverse-variance weighted mean to the remaining participants when their standard uncertainties are known. Under independent unbiased observations with weights $w_j=u_j^{-2}$, put $W={\sum }_j{w}_j$. ThenThese are uncertainties under that observation model; the resulting quantity should not automatically be called a proficiency-test z-score with a fixed assessment scale. Computing $({\sum }_j{w}_j{x}_j-w_i{x}_i)/(W-w_i)$ is algebraically convenient, but direct summation of the remaining terms can avoid cancellation when one participant dominates the total weight.

If a robust estimator has weights depending on the full data set, merely deleting one weighted numerator term need not reproduce the estimator fitted without that participant. The estimator must be recomputed according to its specified rule. Finally, a common additive calibration error remains undetectable in these residuals: replacing every $X_i$ by $X_i+B$ leaves all $D_i$ unchanged. Shared bias in an interlaboratory consensus therefore remains a separate question from self-influence and the variance of the residuals.

A leave-one-out inverse-variance weighted consensus applies an inverse-variance weighted mean to all laboratories except the one being assessed. Suppose their results $X_j$ are independent random variables with common expected value $\theta$ and known positive variances $u_j^2$. For laboratory $i$, defineUnder this model it is the best linear unbiased estimator of $\theta$ from the remaining results. The omitted observation does not determine its own leave-one-out interlaboratory consensus value.

The minimum-variance property follows directly from the Cauchy-Schwarz inequality. For any fixed coefficients $c_j$ satisfying ${\sum }_{j\ne i}{c}_j=1$, the linear unbiased estimator $T={\sum }_{j\ne i}{c}_j{X}_j$ hasEquality holds precisely for $c_j=u_j^{-2}/W_{-i}$. ThereforeThis is a comparison among linear unbiased estimators under the stated independent-error model; it requires no normal distribution assumption.

For an original example, assess a laboratory reporting $x_i=12$ with standard uncertainty $u_i=0.5$. The other laboratories report $(10,12,8)$ with respective standard uncertainties $(1,2,4)$, all in the same units. Their total precision is $W_{-i}=1+1/4+1/16=21/16$, givingThe held-out difference is $D_i=12/7\simeq 1.714$, withThe denominator includes the benchmark's uncertainty even though $X_i$ was excluded. This uncertainty-normalized difference and a proficiency-test z-score based on a predetermined assessment scale answer different questions.

Shared bias in an interlaboratory consensus changes the uncertainty model. Suppose instead $X_j=\theta +B+{\epsilon }_j$, where the common reference contribution $B$ has variance $u_B^2$ and the mutually independent residuals, independent also of $B$, have variances $s_j^2$. Coefficients summing to one preserve $B$ in the benchmark. Its variance is $u_B^2+\sum c_j^2{s}_j^2$, so the minimizing coefficients use $s_j^{-2}$, andUsing the full marginal uncertainties $\sqrt{u_B^2+s_j^2}$ as if they were independent changes the coefficients and loses the shared-reference structure. The propagation of uncertainty retains the covariance of dependent inputs.^[ref]

If the held-out laboratory shares the same $B$, that contribution cancels in $D_i$, while it remains in the benchmark's absolute measurement uncertainty. Precision of a difference and knowledge of an absolute value are distinct properties. Estimated, result-dependent uncertainties or additional correlations require analysis of the actual estimator; the fixed known-variance proof does not establish optimality for those cases.

A meta-analysis combines statistical evidence from several studies addressing a common estimand. It may assume a shared effect or use a random-effects meta-analysis to represent between-study variation. Inverse-variance weighted means combine independent effect estimates when their variances are known or estimated; differences in study validity and effect modifiers require substantive assessment, not just weighting.

Let ${\hat{\delta }}_D$ and ${\hat{\delta }}_I$ be independent, approximately unbiased estimates of one consistent treatment effect, with variances $v_D,v_I>0$. The inverse-variance weighted mean givesIndeed, weights $w,1-w$ give variance $w^2{v}_D+(1-w{)}^2{v}_I$; differentiating gives $w=v_I/(v_D+v_I)$, the weight on the direct estimate. Here the three displayed trials give ${\hat{\delta }}_D=-1.33$, $v_D=0.04$, and hence $\hat{\delta }\simeq -1.3203$ with variance $0.03448$. This numerical pooling uses the displayed evidence, not the unavailable data from all twelve trials. A shared study or shared control would introduce covariance, requiring the corresponding correlated-estimator formula; inconsistency would invalidate the common-effect interpretation.

Unbiasedness requires ${\sum }_i{\alpha }_i=1$. Let $S={\sum }_i{\sigma }_i^{-2}$ and minimize $V(\alpha )={\sum }_i{\sigma }_i^2{\alpha }_i^2$. The Lagrange multiplier equations for $V-\lambda ({\sum }_i{\alpha }_i-1)$ are $2{\sigma }_i^2{\alpha }_i=\lambda$, so normalization gives $\lambda =2/S$. HenceThe gradient at this point is $(2/S)1$, whose inner product with every feasible displacement $h$ satisfying $1^{T}h=0$ is zero. The Hessian matrix is $2\mathrm{diag}({\sigma }_1^2,\dots ,{\sigma }_N^2)$, and $h^T{\nabla }^{2}Vh=2{\sum }_i{\sigma }_i^2{h}_i^2>0$ for every nonzero feasible $h$. These are the first- and second-order constrained minimum conditions. Strict convexity also proves global uniqueness of this inverse-variance weighted mean; normal errors are not needed.

Independence makes the variance of an estimator equal toWith ${\alpha }_2=1-{\alpha }_1$, differentiate to obtain $2{\alpha }_1{\sigma }_1^2-2(1-{\alpha }_1){\sigma }_2^2=0$. The second derivative is $2({\sigma }_1^2+{\sigma }_2^2)>0$, so the unique minimum isEquivalently, the inverse-variance weighted mean uses ${\alpha }_i=K{\sigma }_i^{\gamma }$ withWe assume positive measurement variances. A zero-variance observation already supplies the true value exactly.

Equal variances make every weight $1/N$, so the inverse-variance weighted mean reduces to the arithmetic mean. Independence givesThus the variance scales as $N^{-1}$, whereas the uncertainty expressed as a standard deviation scales as $N^{-1/2}$.

Assuming independent Gaussian measurement errors with the quoted standard deviations, use the inverse-variance weighted mean:Here ${\sum }_i{w}_i=0.654514$ and ${\sum }_i{w}_i{x}_i=65.805556$, givingThe consistency statistic is ${\chi }^2=4.45$ for four degrees of freedom, so there is no evident excess scatter requiring error rescaling. A shared systematic error would require a covariance matrix and would not be reduced by this averaging rule.

For independent study estimates $y_i$ with within-study variances $v_i$, a usual approximate model is $y_i\mid {\delta }_i\sim N({\delta }_i,v_i)$ and ${\delta }_i\sim N(\mu ,{\tau }^2)$. Its marginal distribution is $N(\mu ,v_i+{\tau }^2)$, so for fixed heterogeneity ${\tau }^2$ the inverse-variance weighted mean uses weights $(v_i+{\tau }^2{)}^{-1}$. Here $\mu$ describes the mean effect across comparable studies; ${\tau }^2$ represents between-study variation. Estimation and uncertainty for heterogeneity are essential, especially with few studies.