Question 5.12: Three replicate determinations are made of the signal for a ...

The equation for a normal calibration curve using external standards is

S_{meas} = b_0 + b_1 × C_A

thus, Ȳ_X is the average signal of 29.33, and X is the analyte’s concentration. Substituting the value of Ȳ_X into equation 5.20 along with the estimated slope and the y-intercept for the regression line gives the analyte’s concentration as

X=\frac{\bar{Y}_X-b_0 }{b_1}        (5.20)

C_A=X=\frac{\bar{Y}_X-b_0 }{b_1} =\frac{29.33-0.209}{120.706} =0.241

To calculate the standard deviation for the analyte’s concentration, we must determine the values for \bar y  and  \sum (x_i-\bar{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 ȳ is 30.385. Calculating \sum (x_i-\bar{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,

\sum (x_i-\bar{x} )^2 =s^2(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 s is 0.1871 and

\sum (x_i-\bar{x} )^2 =(0.1871)^2(6 – 1) = 0.175

Substituting known values into equation 5.21 gives

s_X=\frac{s_r}{b_1}\left\{\frac{1}{m}+\frac{1}{n}+\frac{(\bar{Y}_X-\bar{y})^2}{b_1^2\sum{(x_i-\bar{x} )^2} } \right\}^{1/2}                       5.21

s_A=s_X=\frac{0.4035}{120.706} \left\{\frac{1}{3}+\frac{1}{6}+\frac{(29.33-30.385)^2}{(120.706)^2(0.175)} \right\} ^{1/2}=0.0024

Finally, the 95% confidence interval for 4 degrees of freedom is

μ_A = C_A ± ts_A = 0.241 ± (2.78)(0.0024) = 0.241 ± 0.007