Question 5.10: Using the data from Table 5.1, determine the relationship be...

Equations 5.13 and 5.14 are written in terms of the general variables x and y. As you work through this example, remember that x represents the concentration of analyte in the standards (C_S), and that y corresponds to the signal (S_{meas}). We begin by setting up a table to help in the calculation of the summation terms \sum x_i, \sum y_i, \sum x_i²,  and  \sum x_iy_i which are needed for the calculation of b_0  \text{and}  b_1

b_1=\frac{n\sum{x_iy_i}-\sum x_i\sum y_i }{n\sum x_i^2-(\sum x_i)^2}    (5.13)

b_0=\frac{\sum y_i-b_1\sum x_i }{n}     (5.14)

Adding the values in each column gives

Substituting these values into equations 5.12 and 5.13 gives the estimated slope

y=\beta _0 + \beta _1 x         (5.12)

b_1=\frac{(6)(66.701)-(1.500)(182.31)}{(6)(0.550)-(1.500)^2} =120.706

and the estimated y-intercept

b_0=\frac{182.31-(120.706)(1.500)}{6}=0.209

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

S_{meas} = 120.70 ×  C_S + 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.