Analytical Chem istry - DePauw University

Chapter 5 Standardizing Analytical Methods2050. 710− 29. 57× ( 2. 371×10 )= 0.00156−2b 0=The regression equation isS std = 29.57 × C std + 0.0015To calculate the 95% confidence intervals, we first need to determinethe standard deviation about the regression. The following table willhelp us organize the calculation.x i y iŷ i( y yˆ )− 2i i0.000 0.00 0.0015 2.250×10 –61.55×10 –3 0.050 0.0473 7.110×10 –63.16×10 –3 0.093 0.0949 3.768×10 –64.74×10 –3 0.143 0.1417 1.791×10 –66.34×10 –3 0.188 0.1890 9.483×10 –77.92×10 –3 0.236 0.2357 9.339×10 –8Adding together the data in the last column gives the numerator ofequation 5.19 as 1.596×10 –5 . The standard deviation about the regression,therefore, iss r=1.596×106−2−6= 1.997×10−3Next, we need to calculate the standard deviations for the slope and they-intercept using equation 5.20 and equation 5.21.−3 26× (. 1997×10 )s b1== 0.3007−4− 2 26× (. 1 378× 10 ) − ( 2.371×10 )−3 2 −4(. 1 997× 10 ) × ( 1. 378×10 )s b0=−4−6× (. 1 378×10 ) − ( 2. 371×10 )2 2= 1.441×10−3The 95% confidence intervals areβ 1 1 1= b ± ts b= 29. 57 ± ( 278 . × 0. 3007) = 29. 57 M -1 ± 0.85 M -1−3β 0 0 0= b ± ts b= 0. 0015 ± { 278 . × ( 1. 441× 10 } = 0. 0015± 0.0040With an average S samp of 0.114, the concentration of analyte, C A , isCAS=samp− b00. 114−0.0015= = 380 . × 10-1b 29.57 M1−3 M