The following data were recorded during the preparation of a calibration curve, where Smeas and s are the mean and standard deviation, respectively, for three replicate measurements of the signal.

Question

The following data were recorded during the preparation of a calibration curve, where [latex]\bar S_{meas}[/latex] and s are the mean and standard deviation, respectively, for three replicate measurements of the signal.

[latex]C_A[/latex]	[latex]\bar S_{meas}[/latex]	s
0.000	0.00	0.02
0.100	12.36	0.02
0.200	24.83	0.07
0.300	35.91	0.13
0.400	48.79	0.22
0.500	60.42	0.33

Determine the relationship between [latex]\bar S_{meas} and C_A[/latex] using a weighted linear regression model.

Accepted Answer

Once again, as you work through this example, remember that x represents the concentration of analyte in the standards ([latex]C_S[/latex]), and y corresponds to the average signal ([latex]\bar S_{meas}[/latex]). We begin by setting up a table to aid in the calculation of the weighting factor.

[latex]C_A[/latex]	[latex]\bar S_{meas}[/latex]	s
0.000	0.00	0.02
0.100	12.36	0.02
0.200	24.83	0.07
0.300	35.91	0.13
0.400	48.79	0.22
0.500	60.42	0.33

Adding together the values in the forth column gives

[latex]Σs_i^{–2} = 5293.09[/latex]

which is used to calculate the weights in the last column. As a check on the calculation, the sum of the weights in the last column should equal the number of calibration standards, n. In this case

[latex]Σw_i = 6.0000[/latex]

After the individual weights have been calculated, a second table is used to aid in calculating the four summation terms in equations 5.22 and 5.23.

[latex]b_1=\frac{n\sum w_ix_iy_i-\sum w_ix_i \sum w_iy_i}{n\sum{w_ix_i^2-(\sum{w_ix_i} )^2}} [/latex] 5.22

[latex]b_0=\frac{\sum w_iy_i-b_1\sum w_ix_i}{n}[/latex] 5.23

[latex]x_i[/latex]	[latex]y_i[/latex]	[latex]w_i[/latex]	[latex]w_ix_i[/latex]	[latex]w_iy_i[/latex]	[latex]w_ix_i²[/latex]	[latex]w_ix_iy_i[/latex]
0.000	0.00	2.8339	0.0000	0.0000	0.0000	0.0000
0.100	12.36	2.8339	0.2834	35.0270	0.0283	3.5027
0.200	24.83	0.2313	0.0463	5.7432	0.0093	1.1486
0.300	35.91	0.0671	0.0201	2.4096	0.0060	0.7229
0.400	48.79	0.0234	0.0094	1.1417	0.0037	0.4567
0.500	60.42	0.0104	0.0052	0.6284	0.0026	0.3142

Adding the values in the last four columns gives

[latex] Σw_ix_i = 0.3644 Σw_iy_i = 44.9499 Σw_ix_i^2= 0.0499 Σw_ix_iy_i = 6.1451[/latex]

Substituting these values into the equations 5.22 and 5.23 gives the estimated slope

[latex]b_1=\frac{(6)(6.1451)-(0.3644)(44.9499)}{(6)(0.0499)-(0.3644)^2} =122.985[/latex]

and the estimated y-intercept

[latex]b_0=\frac{44.9499-(122.985)(0.3644)}{6} =0.0224[/latex]

The relationship between the signal and the concentration of the analyte, therefore, is

[latex]\bar S_{meas}= 122.98 × C_A + 0.02 [/latex]

with the calibration curve shown in Figure 5.12.

Question 5.13: The following data were recorded during the preparation of a...

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