The most commonly used form of linear regression is based on three assump- tions: (1) that any difference between the experimental data and the calculated regression line is due to indeterminate errors affecting the values of y, (2) that these indeterminate errors are normally distributed, and (3) that the indetermi- nate errors in ydo not depend on the value of x.Because we assume that indeter- minate errors are the same for all standards, each standard contributes equally in estimating the slope and y-intercept. For this reason the result is considered an unweighted linear regression.
The second assumption is generally true because of the central limit theorem outlined in Chapter 4. The validity of the two remaining assumptions is less cer- tain and should be evaluated before accepting the results of a linear regression.
In particular, the first assumption is always suspect since there will certainly be some indeterminate errors affecting the values of x. In preparing a calibration curve, however, it is not unusual for the relative standard deviation of the mea- sured signal (y) to be significantly larger than that for the concentration of ana- lyte in the standards (x). In such circumstances, the first assumption is usually reasonable.
Finding the Estimated Slope and y-Intercept The derivation of equations for cal- culating the estimated slope and y-intercept can be found in standard statistical texts7and is not developed here. The resulting equation for the slope is given as
5.13
and the equation for the y-intercept is
5.14 Although equations 5.13 and 5.14 appear formidable, it is only necessary to evaluate four summation terms. In addition, many calculators, spreadsheets, and other com- puter software packages are capable of performing a linear regression analysis based on this model. To save time and to avoid tedious calculations, learn how to use one of these tools. For illustrative purposes, the necessary calculations are shown in de- tail in the following example.
EXAMPLE
5.10
Using the data from Table 5.1, determine the relationship between Smeasand CS
by an unweighted linear regression.
SOLUTION
Equations 5.13 and 5.14 are written in terms of the general variables xand y.
As you work through this example, remember that x represents the concentration of analyte in the standards (CS), and that ycorresponds to the signal (Smeas). We begin by setting up a table to help in the calculation of the summation terms Σxi, Σyi, Σx2i, and Σxiyiwhich are needed for the calculation of b0and b1
b y b x
n
i i
0 = ∑ – 1∑
b n x y x y
n x x
i i i i
i i
1 = ∑ ∑ ∑2
∑
∑ –
– ( )
2
Figure 5.10
Normal calibration curve for the hypothetical data in Table 5.1, showing the regression line.
xi yi xi2 xiyi
0.000 0.00 0.000 0.000
0.100 12.36 0.010 1.236
0.200 24.83 0.040 4.966
0.300 35.91 0.090 10.773
0.400 48.79 0.160 19.516
0.500 60.42 0.250 30.210
Adding the values in each column gives
Σxi= 1.500 Σyi = 182.31 Σx2i= 0.550 Σxiyi= 66.701 Substituting these values into equations 5.12 and 5.13 gives the estimated slope
and the estimated y-intercept
The relationship between the signal and the analyte, therefore, is Smeas= 120.70×CS+ 0.21
Note that for now we keep enough significant figures to match the number of decimal places to which the signal was measured. The resulting calibration curve is shown in Figure 5.10.
b0 182 31 120 706 1 500
6 0 209
= . – ( . )( . ) = . b1
2
6 66 701 1 500 182 31
6 0 550 1 500 120 706
= ( )( . ) – ( . )( . ) =
( )( . ) – ( . ) .
120
Modern Analytical Chemistry80
60
40
20
0 Smeas
CA
0.0 0.1 0.2 0.3 0.4 0.5 0.6
Uncertainty in the Regression Analysis As shown in Figure 5.10, the regression line need not pass through the data points (this is the consequence of indeter- minate errors affecting the signal). The cumulative deviation of the data from the regression line is used to calculate the uncertainty in the regression due to
Chapter 5 Calibrations, Standardizations, and Blank Corrections
121
standard deviation about the regression The uncertainty in a regression analysis due to indeterminate error (sr).
indeterminate error. This is called the standard deviation about the regression, sr, and is given as
5.15 where yiis the ithexperimental value, and ˆyiis the corresponding value predicted by the regression line
yˆi=b0+b1xi
There is an obvious similarity between equation 5.15 and the standard deviation in- troduced in Chapter 4, except that the sum of squares term for sris determined rela- tive to ˆyiinstead of y,– and the denominator is n– 2 instead of n– 1; n– 2 indicates that the linear regression analysis has only n– 2 degrees of freedom since two pa- rameters, the slope and the intercept, are used to calculate the values of ˆyi.
A more useful representation of uncertainty is to consider the effect of indeter- minate errors on the predicted slope and intercept. The standard deviation of the slope and intercept are given as
5.16
5.17 These standard deviations can be used to establish confidence intervals for the true slope and the true y-intercept
β1=b1±tsb1 5.18 βo=bo±tsb0 5.19 where t is selected for a significance level of αand for n– 2 degrees of freedom.
Note that the terms tsb1and tsb0do not contain a factor of because the con- fidence interval is based on a single regression line. Again, many calculators, spread- sheets, and computer software packages can handle the calculation of sb0and sb1and the corresponding confidence intervals for β0 and β1. Example 5.11 illustrates the calculations.
EXAMPLE
5.11
Calculate the 95% confidence intervals for the slope and y-intercept determined in Example 5.10.
SOLUTION
Again, as you work through this example, remember that xrepresents the concentration of analyte in the standards (CS), and ycorresponds to the signal (Smeas). To begin with, it is necessary to calculate the standard deviation about the regression. This requires that we first calculate the predicted signals, ˆyi, using the slope and y-intercept determined in Example 5.10. Taking the first standard as an example, the predicted signal is
yˆi=b0+b1x= 0.209 + (120.706)(0.100) = 12.280 ( n)–1
s s x
n x x
s x
n x x
b i
i i
i i
0 = ∑ 2 2
∑
∑ = ∑
∑
r2 2 2
r2 2
– ( ) ( – )
s ns
n x x
s x x
b
i i i
1
r2 2
r2
= ∑ ∑ =
∑
– ( )2 ( – )2
s y – y
n
i i
in
r = ∑= ( ˆ )
–
1 2
2
The results for all six solutions are shown in the following table.
xi yi yˆi (yi – ˆyi)2
0.000 0.00 0.209 0.0437
0.100 12.36 12.280 0.0064
0.200 24.83 24.350 0.2304
0.300 35.91 36.421 0.2611
0.400 48.79 48.491 0.0894
0.500 60.42 60.562 0.0202
Adding together the data in the last column gives the numerator of equation 5.15, Σ(yi– ˆyi)2, as 0.6512. The standard deviation about the regression, therefore, is
Next we calculate sb1 and sb0using equations 5.16 and 5.17. Values for the summation terms Σx2iand Σxiare found in Example 5.10.
Finally, the 95% confidence intervals (α= 0.05, 4 degrees of freedom) for the slope and y-intercept are
β1=b1±tsb1= 120.706 ± (2.78)(0.965) = 120.7 ± 2.7 β0=b0±tsb0= 0.209 ± (2.78)(0.292) = 0.2 ± 0.8
The standard deviation about the regression, sr, suggests that the measured signals are precise to only the first decimal place. For this reason, we report the slope and intercept to only a single decimal place.
To minimize the uncertainty in the predicted slope and y-intercept, calibration curves are best prepared by selecting standards that are evenly spaced over a wide range of concentrations or amounts of analyte. The reason for this can be rational- ized by examining equations 5.16 and 5.17. For example, both sb0and sb1can be minimized by increasing the value of the term Σ(xi– –x)2, which is present in the de- nominators of both equations. Thus, increasing the range of concentrations used in preparing standards decreases the uncertainty in the slope and the y-intercept. Fur- thermore, to minimize the uncertainty in the y-intercept, it also is necessary to de- crease the value of the term Σx2iin equation 5.17. This is accomplished by spreading the calibration standards evenly over their range.
Using the Regression Equation Once the regression equation is known, we can use it to determine the concentration of analyte in a sample. When using a normal cali- bration 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 X
s s x
n x x
b i
i i
0 2
2
2
0 4035 0 550
6 0 550 1 500 0 292
= ∑
∑
∑ r22 2 = =
– ( )
( . ) ( . ) ( )( . ) – ( . ) .
s ns
n x x
b
i i
1
r2
= 2
∑
∑ = =
– ( )
( )( . )
( )( . ) – ( . ) .
2
2 2
6 0 4035
6 0 550 1 500 0 965 sr = 0 6512 =
6. 2 0 4035
– .
122
Modern Analytical ChemistryChapter 5 Calibrations, Standardizations, and Blank Corrections
123
5.20 The standard deviation for the calculated value of X is given by the following equation
5.21 where mis the number of replicate samples used to establish –
YX, nis the number of calibration standards, –yis the average signal for the standards, and xiand –xare the individual and mean concentrations of the standards.8Once sXis known the confi- dence interval for the analyte’s concentration can be calculated as
µX=X±tsX
where µXis the expected value of Xin the absence of determinate errors, and the value of tis determined by the desired level of confidence and for n– 2 degrees of freedom. The following example illustrates the use of these equations for an analysis using a normal calibration curve with external standards.
EXAMPLE
5.12
Three replicate determinations are made of the signal for a sample containing 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.
SOLUTION
The equation for a normal calibration curve using external standards is Smeas=b0+b1×CA
thus, –
YXis the average signal of 29.33, and X is the analyte’s concentra- tion. Substituting the value of –
YXinto equation 5.20 along with the estimated slope and the y-intercept for the regression line gives the analyte’s concentration as
To calculate the standard deviation for the analyte’s concentration, we must determine the values for –yand Σ(xi– –x)2. The former is just the average signal for the standards used to construct the calibration curve. From the data in Table 5.1, we easily calculate that –y is 30.385. Calculating Σ(xi– –x)2 looks formidable, but we can simplify the calculation by recognizing that this sum of squares term is simply the numerator in a standard deviation equation;
thus,
Σ
(xi– –x)2=s2(n– 1)where s is the standard deviation for the concentration of analyte in the standards used to construct the calibration curve. Using the data in Table 5.1, we find that sis 0.1871 and
Σ
(xi– –x)2= (0.1871)2(6 – 1) = 0.175C X Y – b
A = = Xb 0 = =
1
29 33 0 209
120 706 0 241 . – .
. .
s s
b m n
Y y
b x – x
X X
i
= + +
∑
r 1
2
12 2
1 2
1 1 ( – )
( )
/
X Y b
b
= X – 0 1
Figure 5.11
Plot of the residual error in yas a function of x.The distribution of the residuals in (a) indicates that the regression model was appropriate for the data, and the distributions in (b) and (c) indicate that the model does not provide a good fit for the data.
Substituting known values into equation 5.21 gives
Finally, the 95% confidence interval for 4 degrees of freedom is µA=CA±tsA= 0.241±(2.78)(0.0024) = 0.241 ± 0.007
In a standard addition the analyte’s concentration is determined by extrapolat- ing the calibration curve to find the x-intercept. In this case the value of Xis
and the standard deviation in Xis
where nis the number of standards used in preparing the standard additions cali- bration curve (including the sample with no added standard), and –yis the average signal for the nstandards. Because the analyte’s concentration is determined by ex- trapolation, rather than by interpolation, sXfor the method of standard additions generally is larger than for a normal calibration curve.
A linear regression analysis should not be accepted without evaluating the validity of the model on which the calculations were based. Perhaps the simplest way to evaluate a regression analysis is to calculate and plot the residual error for each value of x.The residual error for a single calibration standard, ri, is given as
ri=yi– ˆyi
If the regression model is valid, then the residual errors should be randomly distrib- uted about an average residual error of 0, with no apparent trend toward either smaller or larger residual errors (Figure 5.11a). Trends such as those shown in Fig- ures 5.11b and 5.11c provide evidence that at least one of the assumptions on which the regression model is based are incorrect. For example, the trend toward larger residual errors in Figure 5.11b suggests that the indeterminate errors affecting yare not independent of the value of x.In Figure 5.11c the residual errors are not ran- domly distributed, suggesting that the data cannot be modeled with a straight-line relationship. Regression methods for these two cases are discussed in the following sections.