Analytical Chem istry - DePauw University

Chapter 5 Standardizing Analytical Methods179Example 5.11Three replicate analyses for a sample containing an unknown concentrationof analyte, yield values for S samp of 29.32, 29.16 and 29.51. Usingthe results from Example 5.9 and Example 5.10, determine the analyte’sconcentration, C A , and its 95% confidence interval.So l u t i o nThe average signal, S samp, is 29.33, which, using equation 5.24 and theslope and the y-intercept from Example 5.9, gives the analyte’s concentrationasCAS=samp− b029. 33−0.209= = 0.241b 120.7061To calculate the standard deviation for the analyte’s concentration we must∑( )determine the values for S stdand C −Cstdistd. The former is just theaverage signal for the calibration standards, which, using the data in Table∑( )5.1, is 30.385. Calculating C −Cstdistdlooks formidable, but we can simplifyits calculation by recognizing that this sum of squares term is thenumerator in a standard deviation equation; thus,∑22( C −C s nstd std ) = ( C ) × −1i2std2( )where s Cstdis the standard deviation for the concentration of analyte inthe calibration standards. Using the data in Table 5.1 we find that s Cstdis0.1871 and22∑( C −Cstd std ) = ( 0. 1871) × ( 6− 1) = 0.175iSubstituting known values into equation 5.25 gives20.4035 1 1 ( 29. 33−30.385)s CA= + += 0.00242120.706 3 6 120. 706 0.175( ) ×Finally, the 95% confidence interval for 4 degrees of freedom is= ± = 0. 241± ( 278 . × 0. 0024)= 0. 241±0.007µ CC tsA A C AFigure 5.12 shows the calibration curve with curves showing the 95%confidence interval for C A .You can find values for t in Appendix 4.