GIFSQ Tools · Module 15 — Metrology, Validation & Laboratory Accreditation · Tool 283 of 500
Fit a least-squares straight line to calibration standards, check the coefficient of determination, and read an unknown sample concentration from its response.
Try sample values — fills the form with the worked example below.
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Slope
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Intercept
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R²
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Sample concentration
Equation: Slope b = (nΣxy − ΣxΣy) ÷ (nΣx² − (Σx)²); intercept a = (Σy − bΣx) ÷ n; R² = 1 − Σ(y − ŷ)² ÷ Σ(y − ȳ)²; sample concentration = (y_sample − a) ÷ b.
Method: Ordinary least-squares linear regression per ISO 8466-1 (calibration and evaluation of analytical methods) and Eurachem/CITAC CG4. A straight-line fit is the usual model for instrumental calibration over a validated range.
Assumptions: the response is linear over the calibration range; the x values are known without error; the scatter of y is similar at every level; the sample response lies inside the calibrated range.
Limitations: least squares assumes the error is all in y; a curved response needs a weighted or quadratic fit; extrapolating beyond the highest standard is not valid; a high R² alone does not prove linearity — inspect the residuals.
Rounding: slope and intercept to 4 decimal places; R² to 4 decimal places; sample concentration to 4 decimal places; intermediates unrounded.
This is a calculation aid, not a laboratory report — confirm against your accredited method before citing a value.
Method: Ordinary least-squares linear regression per ISO 8466-1 (calibration and evaluation of analytical methods) and Eurachem/CITAC CG4. A straight-line fit is the usual model for instrumental calibration over a validated range.
Assumptions: the response is linear over the calibration range; the x values are known without error; the scatter of y is similar at every level; the sample response lies inside the calibrated range.
Limitations: least squares assumes the error is all in y; a curved response needs a weighted or quadratic fit; extrapolating beyond the highest standard is not valid; a high R² alone does not prove linearity — inspect the residuals.
Rounding: slope and intercept to 4 decimal places; R² to 4 decimal places; sample concentration to 4 decimal places; intermediates unrounded.
This is a calculation aid, not a laboratory report — confirm against your accredited method before citing a value.
Worked example: Standards 0, 1, 2, 3, 4 mg/L with responses 0.01, 0.21, 0.39, 0.61, 0.79: slope = 0.1960, intercept = 0.0100, R² = 0.9992; a sample response of 0.50 gives (0.50 − 0.01) ÷ 0.196 = 2.5000 mg/L.
Enter values above to see results.
Accept the curve only when R² meets your method criterion (commonly 0.99 or better) and the sample response falls between the lowest and highest standards. Tool 283 of 500 — GIFSQ
