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.

Question

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, [latex]C_A[/latex], and its 95% confidence interval.

Accepted Answer

The equation for a normal calibration curve using external standards is

[latex]S_{meas} = b_0 + b_1 × C_A[/latex]

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

[latex]X=\frac{\bar{Y}_X-b_0 }{b_1} [/latex] (5.20)

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

To calculate the standard deviation for the analyte’s concentration, we must determine the values for [latex]\bar y and \sum (x_i-\bar{x} )^2 [/latex]. 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 [latex]\sum (x_i-\bar{x} )^2 [/latex] 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,

[latex]\sum (x_i-\bar{x} )^2 =s^2(n – 1)[/latex]

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

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

Substituting known values into equation 5.21 gives

[latex]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}[/latex] 5.21

[latex]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[/latex]

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

[latex]μ_A = C_A ± ts_A = 0.241 ± (2.78)(0.0024) = 0.241 ± 0.007[/latex]

Table 5.1 Data for Hypothetical Multiple-Point External Standardization
[latex]C_S[/latex]	[latex]S_{meas}[/latex]
0.000	0.00
0.100	12.36
0.200	24.83
0.300	35.91
0.400	48.79
0.500	60.42

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

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