Page 139 - Modern Analytical Chemistry
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1400-CH05  9/8/99  3:59 PM  Page 122





              122    Modern Analytical Chemistry


                                                  The results for all six solutions are shown in the following table.

                                                         x i           y i             ˆ y i      (y i –ˆy i ) 2
                                                        0.000         0.00           0.209        0.0437
                                                        0.100        12.36          12.280        0.0064
                                                        0.200        24.83          24.350        0.2304
                                                        0.300        35.91          36.421        0.2611
                                                        0.400        48.79          48.491        0.0894
                                                        0.500        60.42          60.562        0.0202
                                                  Adding together the data in the last column gives the numerator of equation
                                                  5.15, S(y i – ˆ y i ) , as 0.6512. The standard deviation about the regression,
                                                              2
                                                  therefore, is
                                                                             .
                                                                            0 6512
                                                                                     .
                                                                      s r =       =0 4035
                                                                            6 –  2
                                                                          using equations 5.16 and 5.17. Values for the
                                                  Next we calculate s b1  and s b0
                                                                  2
                                                  summation terms Sx and Sx i are found in Example 5.10.
                                                                   i
                                                                     2                      2
                                                                                    6
                                                                   ns r            ()(. 0 4035 )
                                                          =                 =                     = . 0 965
                                                       s b 1      2       2                    2
                                                                                6
                                                              nx –(   å  x i )  ()(. 0 550 ) – ( . 1 500 )
                                                               å
                                                                  i
                                                                  2   2                 2
                                                                                  0 4035) ( .
                                                                 s å  x i        (.       0 550)
                                                                  r
                                                                                                   0 292
                                                          =                 =                     = .
                                                       s b 0      2       2                    2
                                                                                6 0 550) – ( .
                                                             n å  x –( å x i )  ()(.      1 500)
                                                                  i
                                                  Finally, the 95% confidence intervals (a= 0.05, 4 degrees of freedom) for the
                                                  slope and y-intercept are
                                                                       = 120.706 ± (2.78)(0.965) = 120.7 ± 2.7
                                                            b 1 = b 1 ± ts b1
                                                                         = 0.209 ± (2.78)(0.292) = 0.2 ± 0.8
                                                              b 0 = b 0 ± ts b0
                                                  The standard deviation about the regression, s r , suggests that the measured
                                                  signals are precise to only the first decimal place. For this reason, we report the
                                                  slope and intercept to only a single decimal place.
                                                  To minimize the uncertainty in the predicted slope and y-intercept, calibration
                                              curves are best prepared by selecting standards that are evenly spaced over a wide
                                              range of concentrations or amounts of analyte. The reason for this can be rational-
                                                                                                            can be
                                              ized by examining equations 5.16 and 5.17. For example, both s b0  and s b1
                                              minimized by increasing the value of the term S(x i – x) , which is present in the de-
                                                                                         – 2
                                              nominators of both equations. Thus, increasing the range of concentrations used in
                                              preparing standards decreases the uncertainty in the slope and the y-intercept. Fur-
                                              thermore, to minimize the uncertainty in the y-intercept, it also is necessary to de-
                                                                       2
                                              crease the value of the term Sx in equation 5.17. This is accomplished by spreading
                                                                       i
                                              the calibration standards evenly over their range.
                                              Using the Regression Equation Once the regression equation is known, we can use
                                              it to determine the concentration of analyte in a sample. When using a normal cali-
                                              bration curve with external standards or an internal standards calibration curve, we
                                                                               –
                                              measure an average signal for our sample, Y X, and use it to calculate the value of X
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