D. Harvey - Modern Analytical Chemistry (794078), страница 38

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The standard deviation of theslope and intercept are given ass b1 =s b0 =ns r2=n ∑ x i2 – (∑ x i )2s r2 ∑ x i2=n ∑ x i2 – (∑ x i )2s r2∑ (x i – x )25.16s r2 ∑ x i2n ∑ (x i – x )25.17These standard deviations can be used to establish confidence intervals for the trueslope and the true y-interceptβ1 = b1 ± tsb15.18βo = bo ± tsb05.19where t is selected for a significance level of α and for n – 2 degrees of freedom.Note that the terms tsb1 and tsb0 do not contain a factor of ( n ) –1 because the confidence interval is based on a single regression line.

Again, many calculators, spreadsheets, and computer software packages can handle the calculation of sb0 and sb1 andthe corresponding confidence intervals for β0 and β1. Example 5.11 illustrates thecalculations.EXAMPLE 5.11Calculate the 95% confidence intervals for the slope and y-interceptdetermined in Example 5.10.SOLUTIONAgain, as you work through this example, remember that x represents theconcentration of analyte in the standards (CS), and y corresponds to the signal(Smeas). To begin with, it is necessary to calculate the standard deviation aboutthe regression. This requires that we first calculate the predicted signals, ŷi,using the slope and y-intercept determined in Example 5.10.

Taking the firststandard as an example, the predicted signal isŷi = b0 + b1x = 0.209 + (120.706)(0.100) = 12.280121standard deviation about the regressionThe uncertainty in a regression analysisdue to indeterminate error (sr).1400-CH05 9/8/99 3:59 PM Page 122122Modern Analytical ChemistryThe results for all six solutions are shown in the following table.xiyi0.0000.1000.2000.3000.4000.5000.0012.3624.8335.9148.7960.42ŷi0.20912.28024.35036.42148.49160.562(yi – ŷi)20.04370.00640.23040.26110.08940.0202Adding together the data in the last column gives the numerator of equation5.15, Σ(y i – ŷ i ) 2 , as 0.6512. The standard deviation about the regression,therefore, issr =0.6512= 0.40356–2Next we calculate sb1 and sb0 using equations 5.16 and 5.17.

Values for thesummation terms Σx2i and Σxi are found in Example 5.10.s b1 =ns r2=n ∑ x i2 – (∑ x i )2(6)(0.4035)2= 0.965(6)(0.550) – (1.500)2s b0 =s r2 ∑ x i2=n ∑ x i2 – (∑ x i )2(0.4035)2 (0.550)= 0.292(6)(0.550) – (1.500)2Finally, the 95% confidence intervals (α = 0.05, 4 degrees of freedom) for theslope and y-intercept areβ1 = b1 ± tsb1 = 120.706 ± (2.78)(0.965) = 120.7 ± 2.7β0 = b0 ± tsb0 = 0.209 ± (2.78)(0.292) = 0.2 ± 0.8The standard deviation about the regression, sr, suggests that the measuredsignals are precise to only the first decimal place. For this reason, we report theslope and intercept to only a single decimal place.To minimize the uncertainty in the predicted slope and y-intercept, calibrationcurves are best prepared by selecting standards that are evenly spaced over a widerange of concentrations or amounts of analyte. The reason for this can be rationalized by examining equations 5.16 and 5.17.

For example, both sb0 and sb1 can be– 2, which is present in the deminimized by increasing the value of the term Σ(xi – x)nominators of both equations. Thus, increasing the range of concentrations used inpreparing standards decreases the uncertainty in the slope and the y-intercept. Furthermore, to minimize the uncertainty in the y-intercept, it also is necessary to decrease the value of the term Σx2i in equation 5.17. This is accomplished by spreadingthe calibration standards evenly over their range.Using the Regression Equation Once the regression equation is known, we can useit to determine the concentration of analyte in a sample.

When using a normal calibration curve with external standards or an internal standards calibration curve, we–measure an average signal for our sample, YX, and use it to calculate the value of X1400-CH05 9/8/99 3:59 PM Page 123Chapter 5 Calibrations, Standardizations, and Blank CorrectionsX =YX – b0b15.20The standard deviation for the calculated value of X is given by the followingequation1/ 21(YX – y )2 sr  1sX =5.21 + +b1  m n b 12 ∑ (x i – x )2 –where m is the number of replicate samples used to establish YX, n is the number of–calibration standards, y is the average signal for the standards, and xi and –x are theindividual and mean concentrations of the standards.8 Once sX is known the confidence interval for the analyte’s concentration can be calculated asµX= X ± tsXwhere µX is the expected value of X in the absence of determinate errors, and thevalue of t is determined by the desired level of confidence and for n – 2 degrees offreedom.

The following example illustrates the use of these equations for an analysisusing a normal calibration curve with external standards.EXAMPLE 5.12Three replicate determinations are made of the signal for a samplecontaining an unknown concentration of analyte, yielding values of 29.32,29.16, and 29.51. Using the regression line from Examples 5.10 and 5.11,determine the analyte’s concentration, CA, and its 95% confidence interval.SOLUTIONThe equation for a normal calibration curve using external standards isSmeas = b0 + b1 × CA–thus, Y X is the average signal of 29.33, and X is the analyte’s concentra–tion.

Substituting the value of YX into equation 5.20 along with the estimatedslope and the y-intercept for the regression line gives the analyte’sconcentration asCA = X =YX – b029.33 – 0.209== 0.241b1120.706To calculate the standard deviation for the analyte’s concentration, we mustdetermine the values for –y and Σ(xi – –x)2. The former is just the average signalfor the standards used to construct the calibration curve. From thedata in Table 5.1, we easily calculate that –y is 30.385. Calculating Σ(xi – –x)2looks formidable, but we can simplify the calculation by recognizing that thissum of squares term is simply the numerator in a standard deviation equation;thus,– 2 = s2(n – 1)Σ(xi – x)where s is the standard deviation for the concentration of analyte in thestandards used to construct the calibration curve. Using the data in Table 5.1,we find that s is 0.1871 and– 2 = (0.1871)2(6 – 1) = 0.175Σ(x – x)i1231400-CH05 9/8/99 3:59 PM Page 124124Modern Analytical ChemistrySubstituting known values into equation 5.21 givessA = sX0.4035=120.7061/ 2 1 1 (29.33 – 30.385)2  + + 3 6 (120.706)2 (0.175) = 0.0024Finally, the 95% confidence interval for 4 degrees of freedom isµA = CA ± tsA = 0.241 ± (2.78)(0.0024) = 0.241 ± 0.007In a standard addition the analyte’s concentration is determined by extrapolating the calibration curve to find the x-intercept.

In this case the value of X isResidual errorX = x -intercept =and the standard deviation in X is1/ 20sXxiResidual error(a)0s= rb1 1(y )2 + 2 n b1 ∑ (x i – x )2 where n is the number of standards used in preparing the standard additions calibration curve (including the sample with no added standard), and –y is the averagesignal for the n standards. Because the analyte’s concentration is determined by extrapolation, rather than by interpolation, sX for the method of standard additionsgenerally is larger than for a normal calibration curve.A linear regression analysis should not be accepted without evaluating thevalidity of the model on which the calculations were based.

Perhaps the simplestway to evaluate a regression analysis is to calculate and plot the residual errorfor each value of x. The residual error for a single calibration standard, ri, is given asri = yi – ŷixi(b)Residual error–b0b10xi(c)Figure 5.11Plot of the residual error in y as a functionof x. The distribution of the residuals in(a) indicates that the regression modelwas appropriate for the data, and thedistributions in (b) and (c) indicate that themodel does not provide a good fit for thedata.If the regression model is valid, then the residual errors should be randomly distributed about an average residual error of 0, with no apparent trend toward eithersmaller or larger residual errors (Figure 5.11a).

Trends such as those shown in Figures 5.11b and 5.11c provide evidence that at least one of the assumptions on whichthe regression model is based are incorrect. For example, the trend toward largerresidual errors in Figure 5.11b suggests that the indeterminate errors affecting y arenot independent of the value of x. In Figure 5.11c the residual errors are not randomly distributed, suggesting that the data cannot be modeled with a straight-linerelationship. Regression methods for these two cases are discussed in the followingsections.5C.3 Weighted Linear Regression with Errors in yEquations 5.13 for the slope, b1, and 5.14 for the y-intercept, b0, assume thatindeterminate errors equally affect each value of y.

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