Chemistry · Lesson

Calibration, reference materials and traceability

Connect indications to quantity values with a stated calibration model. Inspect residuals, reference-material scope and uncertainty instead of treating a high fit statistic as complete validation.

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By the end, you can

Invert a stipulated linear response within its stated interval.
Identify systematic residual curvature despite a high R².
Propagate independent standard uncertainties through a dilution ratio.
Interpret a reference comparison without claiming universal accuracy.

Calibration connects references, indications and results

Calibration establishes relationships between reference quantity values and indications with their uncertainties, then supports deriving results from indications. It is different from adjusting an instrument or checking whether it meets a requirement. A calibrated response is still interpreted through a model and its applicable interval.

Reference values, preparation, matrix, stability, blank contribution and response variability matter. Calibration should use an appropriate model and assess deviations from it. A curve fitted to several points does not establish selectivity for the intended constituent, or that the procedure captures the whole sampling target.

Fit and invert one explicit response model

An invented dataset has concentrations C=0, 1, 2, 3, 4 mg L⁻¹ and indications R=1.04, 2.94, 5.01, 7.00, 9.01 in declared arbitrary response unit U. Ordinary unweighted least squares gives R̂=b+mC with intercept b=1.00 U and slope m=2.00 U L mg⁻¹. Its response residuals R−R̂ are 0.0400, −0.0600, 0.0100, 0, 0.0100 U.

For a new indication R=6.00 U within this calibration interval, the point estimate is C=(R−b)/m=2.50 mg L⁻¹. This inversion does not yet supply the estimate’s uncertainty. The ordinary regression model places independent constant-variance errors in response; significant reference-value uncertainty, changing variance or correlated readings require appropriate treatment.

Inspect the response residuals and independent checks. Applying a fit outside its tested interval is extrapolation, with additional risk of model failure. A large R² reports the fraction of variation explained by the chosen regression in the supplied dataset; it does not prove the model is correct.

A high R² can coexist with a wrong model

A separate invented noiseless response is R=1+10C+0.2C² U when C is numerically expressed in mg L⁻¹. At C=0, 1, 2, 3, 4, 5, the values are 1, 11.2, 21.8, 32.8, 44.2, 56 U. The corresponding dimensional coefficients are 1 U, 10 U L mg⁻¹ and 0.2 U L² mg⁻².

Fitting a straight line instead gives R̂=1/3+11C on the same numerical basis and R²≈0.999295. Yet its residuals are +2/3, −2/15, −8/15, −8/15, −2/15,+2/3 U: positive at both ends and negative in the middle. The exact data have constant second differences 0.400 U, exposing the quadratic generating model.

The figure shows this residual pattern directly. It is a demonstration of misspecification for a known noiseless toy, not a statistical lack-of-fit test on measured data. A few noisy observations would support a more limited conclusion and require more evidence before choosing a different calibration.

Systematic residual curvature despite R²≈0.999295

Response residual / U

Toy concentration / mg L⁻¹

Six quadratic-model residuals
— Known residual function

Underlying data

Toy concentration / mg L⁻¹	Response residual / U
0	0.6666666666666666 · Six quadratic-model residuals
1	-0.13333333333333341 · Six quadratic-model residuals
2	-0.5333333333333333 · Six quadratic-model residuals
3	-0.5333333333333331 · Six quadratic-model residuals
4	-0.1333333333333332 · Six quadratic-model residuals
5	0.6666666666666666 · Six quadratic-model residuals

Original noiseless quadratic response sampled at six chosen concentrations and deliberately fitted with a straight line. Residuals are observed toy response minus straight-line prediction in U. The dashed curve is the exact residual function 0.2C²−C+2/3 on the stated mg L⁻¹ numerical basis. This is a known-model teaching comparison, not experimental lack-of-fit testing.

Traceability and reference suitability need documentation

Metrological traceability belongs to a measurement result: a documented calibration chain links it to a reference, with uncertainty contributions. A label on an instrument or an asserted reference source alone does not document the whole result. Traceability also does not guarantee sufficiently small uncertainty or freedom from mistakes.

A certified reference material assigns specified property values with uncertainties and traceabilities in its documentation. Check chemical entity, matrix, lot, stability, intended use, minimum representative portion and whether a value is certified, indicative or informational. Certification of one property does not certify every possible measurement of the material.

Reference materials can assess calibration, trueness or other performance questions when their scope matches the intended result. A result from one suitable check does not establish the method’s entire range, matrices, selectivity or sampling behavior.

A derived reference value carries input uncertainty

For an optional deeper invented model, let C=C_stock V_a/V_f, with stock 100 mg L⁻¹, aliquot volume 2.00 mL and final mixture volume 20.0 mL. The resulting concentration is 10.0 mg L⁻¹. The denominator is final volume, not 20.0+2.00 mL. Assume no analyte loss or transformation.

Choose independent standard uncertainties u(C_stock)=2.00 mg L⁻¹, u(V_a)=0.0100 mL and u(V_f)=0.0200 mL. First-order derivative contributions are +0.200,+0.0500, −0.0100 mg L⁻¹. Their squared sum is 0.0426 (mg L⁻¹)², giving standard uncertainty u(C)≈0.206398 mg L⁻¹.

The minus sign in the denominator derivative does not make a negative variance contribution when inputs are independent. Correlations, common standards or other significant effects would need their own terms. These assigned values illustrate a measurement model rather than a preparation instruction.

Compatibility is a chosen check, not proof of every claim

Compare an invented result 9.70 mg L⁻¹ with standard uncertainty 0.200 to a reference 10.0 mg L⁻¹ with standard uncertainty 0.100, for the same quantity and matrix. Assume the uncertainties independent. Difference is −0.300 mg L⁻¹ and its standard uncertainty is √(0.200²+0.100²)≈0.223607 mg L⁻¹.

A chosen illustrative bound ±2u(d) is approximately ±0.447214 mg L⁻¹, so the difference lies within it. This is compatibility under one stated rule. It does not prove absence of bias, universal accuracy, documented traceability or acceptance for every intended use.

Worked examples

Invert the specified calibration

Use the invented fitted response R̂=1.00+2.00C and new indication 6.00 U.

Subtract the intercept: 6.00−1.00=5.00 U.
Divide by slope 2.00 U L mg⁻¹.
C=2.50 mg L⁻¹, within the supplied 0–4 interval.

A point estimate needing its own uncertainty and performance evidence.

Look beyond the fit statistic

Compare the six exact quadratic response values with their straight fit.

The straight fit has slope 11 and intercept 1/3 on the stated basis.
R²≈0.999295.
Positive end residuals and negative central residuals remain.
Second differences of the generated response are 0.400 U.

A large R² does not remove known systematic curvature.

Propagate the stipulated independent inputs

Use stock 100 mg L⁻¹, volume ratio 2/20 and the assigned standard uncertainties.

Concentration is 10.0 mg L⁻¹.
Derivative uncertainty contributions are 0.200, 0.0500, −0.0100 mg L⁻¹.
Variance sum is 0.0426 (mg L⁻¹)².
Standard uncertainty is approximately 0.206398 mg L⁻¹.

The derived concentration retains input uncertainty under the first-order model.

Evaluate one chosen difference bound

Use result 9.70±u(0.200) and reference 10.0±u(0.100), in mg L⁻¹ with independent standard uncertainties.

Difference is −0.300 mg L⁻¹.
u(d)≈0.223607 mg L⁻¹.
Chosen 2u(d) bound is approximately 0.447214 mg L⁻¹.
The difference lies within that bound.

Compatibility by the selected check, without a universal method-acceptance claim.

Try it yourself

Use the same invented linear fit for indication 8.00 U. Find the concentration point estimate.
Show the answer and reasoning
3.50 mg L⁻¹.
(8.00−1.00)/2.00=3.50, within the supplied interval.
For the quadratic toy at C=2.50 mg L⁻¹, find its residual from the stated straight fit.
Show the answer and reasoning
−7/12 U, approximately −0.583333 U.
Evaluate 0.2(2.5)²−2.5+2/3; this interpolated residual follows the known toy function.
Does certification of one reference-material property certify every analyte and quantity in it?
Show the answer and reasoning
No.
Assigned values, uncertainty, traceability and intended uses are property-specific.
Does documented traceability alone guarantee uncertainty small enough for the intended decision?
Show the answer and reasoning
No.
Traceability and fitness for an intended purpose are related but distinct requirements.

What to carry forward

State the response model and valid interval.
Inspect residuals as well as summary fit statistics.
Traceability belongs to a documented result.
Reference-material certification has a specific scope.
Derived reference values retain relevant input uncertainties.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.