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              90     Modern Analytical Chemistry


                                                         4
                                                  EXAMPLE  .19
                                                  Tables 4.1 and 4.8 show results for two separate experiments to determine the
                                                  mass of a circulating U.S. penny. Determine whether there is a difference in the
                                                  means of these analyses at a= 0.05.
                                                  SOLUTION
                                                  To begin with, we must determine whether the variances for the two analyses
                                                  are significantly different. This is done using an F-test as outlined in Example
                                                  4.18. Since no significant difference was found, a pooled standard deviation
                                                  with 10 degrees of freedom is calculated

                                                                              1
                                                                         ( n A - ) s 2 A  +( n B -) s 2 B
                                                                                         1
                                                                 s pool =
                                                                             n A + n B -2
                                                                               0
                                                                         (7  - )( .00259 )  +(5  -)( .00138 )
                                                                             1
                                                                                            1
                                                                                              0
                                                                     =
                                                                                   7  +5  -2
                                                                       0
                                                                     = .0459
                                                  where the subscript A indicates the data in Table 4.1, and the subscript B
                                                  indicates the data in Table 4.8. The comparison of the means for the two
                                                  analyses is based on the null hypothesis
                                                                                 –   –
                                                                            H 0 :  X A = X B
                                                  and a two-tailed alternative hypothesis
                                                                                 –   –
                                                                            H A :  X A ¹X B
                                                  Since the standard deviations can be pooled, the test statistic is calculated using
                                                  equation 4.20
                                                                   X -  X B         . 3 117  -3 .081
                                                                    A
                                                        t exp =                =                  =  . 134
                                                                    1
                                                                     n +1
                                                                                        ( / +1
                                                                                               5
                                                              s pool  (/ A  n / B )  . 0 0459 1 7  / )
                                                  The critical value for t(0.05, 10), from Appendix 1B, is 2.23. Since t exp is less than
                                                  t(0.05, 10) the null hypothesis is retained, and there is no evidence that the two
                                                  sets of pennies are significantly different at the chosen significance level.

                                                         4
                                                  EXAMPLE  .20
                                                  The %w/w Na 2 CO 3 in soda ash can be determined by an acid–base titration.
                                                  The results obtained by two analysts are shown here. Determine whether the
                                                  difference in their mean values is significant at a= 0.05.
                                                                   Analyst A          Analyst B
                                                                     86.82              81.01
                                                                     87.04              86.15
                                                                     86.93              81.73
                                                                     87.01              83.19
                                                                     86.20              80.27
                                                                     87.00              83.94
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