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1400-CH04  9/8/99  3:55 PM  Page 91






                                                                               Chapter 4 Evaluating Analytical Data  91

                     SOLUTION
                     We begin by summarizing the mean and standard deviation for the data
                     reported by each analyst. These values are
                                                 –
                                                 X A = 86.83%
                                                 s A = 0.32
                                                 –
                                                 X B = 82.71%
                                                 s B = 2.16
                     A two-tailed F-test of the following null and alternative hypotheses
                                                            2
                                              2
                                         H 0 :  s A = s B 2  H A :  s A ¹s B 2
                     is used to determine whether a pooled standard deviation can be calculated.
                     The test statistic is

                                               s B 2  (. ) 2
                                                       6
                                                     21
                                         F exp =   =       =45 .6
                                               s 2  (. ) 2
                                                     03
                                                       2
                                                A
                     Since F exp is larger than the critical value of 7.15 for F(0.05, 5, 5), the null
                     hypothesis is rejected and the alternative hypothesis that the variances are
                     significantly different is accepted. As a result, a pooled standard deviation
                     cannot be calculated.
                        The mean values obtained by the two analysts are compared using a two-
                     tailed t-test. The null and alternative hypotheses are
                                             –   –          –   –
                                        H 0 : X A  = X B  H A: X A  ≠ X B
                     Since a pooled standard deviation could not be calculated, the test statistic, t exp ,
                     is calculated using equation 4.19

                                   X -  X B            86 .83  -82 .71
                                     A
                         t exp =                =                      =  . 462
                                  2       2             2          2
                                                          6
                                                               21
                                                            +
                                                       2
                                                                     6
                                                    03
                                   n ) +
                                                                 6
                                 s (/ A  ( s n )   [( . ) / ] [( . ) / ]
                                           / B
                                  A       B
                     and the degrees of freedom are calculated using equation 4.22
                                                   2
                                          2
                                     [( . 032 6  +( . 216 6  2
                                                    / )]
                                           / )
                          n=                                     -2  = . 5 3  » 5
                                                     2
                                   2
                                                              1
                                                      / ) /(6
                              {( . 032 6  2  + )}  +{( . 216 6  2  +)}
                                    / ) /(6
                                           1
                     The critical value for t(0.05, 5) is 2.57. Since the calculated value of t exp is
                     greater than t(0.05, 5) we reject the null hypothesis and accept the alternative
                     hypothesis that the mean values for %w/w Na 2 CO 3 reported by the two
                     analysts are significantly different at the chosen significance level.
                 Paired Data In some situations the variation within the data sets being compared
                 is more significant than the difference between the means of the two data sets. This
                 is commonly encountered in clinical and environmental studies, where the data
                 being compared usually consist of a set of samples drawn from several populations.
                 For example, a study designed to investigate two procedures for monitoring the
                 concentration of glucose in blood might involve blood samples drawn from ten pa-
                 tients. If the variation in the blood glucose levels among the patients is significantly
                 larger than the anticipated variation between the methods, then an analysis in which
                 the data are treated as unpaired will fail to find a significant difference between the
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