Analytical Chem istry - DePauw University

Chapter 5 Standardizing Analytical Methods181sCAsr1= +b n1( S )std2 21 ∑( stdistd )i( ) −b C C2where n is the number of standard additions (including the sample with noadded standard), and S stdis the average signal for the n standards. Becausewe determine the analyte’s concentration by extrapolation, rather than byinterpolation, s CAfor the method of standard additions generally is largerthan for a normal calibration curve.Ev a l u a t i n g a Li n e a r Re g r e s s i o n Mod e lYou should never accept the result of a linear regression analysis withoutevaluating the validity of the your model. Perhaps the simplest way to evaluatea regression analysis is to examine the residual errors. As we saw earlier,the residual error for a single calibration standard, r i , isr = ( y − yˆ )i i iIf your regression model is valid, then the residual errors should be randomlydistributed about an average residual error of zero, with no apparenttrend toward either smaller or larger residual errors (Figure 5.13a). Trendssuch as those shown in Figure 5.13b and Figure 5.13c provide evidence thatat least one of the model’s assumptions is incorrect. For example, a trendtoward larger residual errors at higher concentrations, as shown in Figure5.13b, suggests that the indeterminate errors affecting the signal are notindependent of the analyte’s concentration. In Figure 5.13c, the residual(a) (b) (c)residual errorresidual errorresidual error0.0 0.1 0.2 0.3 0.4 0.5C std0.0 0.1 0.2 0.3 0.4 0.5C std0.0 0.1 0.2 0.3 0.4 0.5C stdFigure 5.13 Plot of the residual error in the signal, S std , as a function of the concentration of analyte, C std for anunweighted straight-line regression model. The red line shows a residual error of zero. The distribution of the residualerror in (a) indicates that the unweighted linear regression model is appropriate. The increase in the residual errors in(b) for higher concentrations of analyte, suggest that a weighted straight-line regression is more appropriate. For (c), thecurved pattern to the residuals suggests that a straight-line model is inappropriate; linear regression using a quadraticmodel might produce a better fit.