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

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

Adding together the values in the forth column gives

Σs_i^{–2} = 5293.09

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

Σw_i = 6.0000

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.

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}}                  5.22

b_0=\frac{\sum w_iy_i-b_1\sum w_ix_i}{n}            5.23

Adding the values in the last four columns gives

Σw_ix_i = 0.3644           Σw_iy_i = 44.9499            Σw_ix_i^2= 0.0499            Σw_ix_iy_i = 6.1451

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

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

and the estimated y-intercept

b_0=\frac{44.9499-(122.985)(0.3644)}{6} =0.0224

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

\bar S_{meas}= 122.98 × C_A + 0.02

with the calibration curve shown in Figure 5.12.