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                                           TABLE 24.1
                                           Ten Measurements of Lead Concentration (µg/L)
                                           on Identical Specimens from Five Laboratories
                                           Lab 1    Lab 2    Lab 3   Lab 4   Lab 5
                                              3.4    4.5      5.3     3.2     3.3
                                              3.0    3.7      4.7     3.4     2.4
                                              3.4    3.8      3.6     3.1     2.7
                                              5.0    3.9      5.0     3.0     3.2
                                              5.1    4.3      3.6     3.9     3.3
                                              5.5    3.9      4.5     2.0     2.9
                                              5.4    4.1      4.6     1.9     4.4
                                              4.2    4.0      5.3     2.7     3.4
                                              3.8    3.0      3.9     3.8     4.8
                                              4.2    4.5      4.1     4.2     3.0
                                              = 4.30  3.97   4.46     3.12    3.34
                                            y i
                                            2
                                           s i  = 0.82  0.19  0.41    0.58    0.54
                        The within-treatment sum of squares is calculated from the residuals of the observations within a
                       treatment and the average for that treatment. The variance within each treatment is:

                                                            n t  (  y t )
                                                               y ti –
                                                        s t =  ∑ --------------------
                                                        2
                                                               n t –  1
                                                            i=1
                       where y ti  are the n t  observations under each treatment.
                        Assuming that all treatments have the same population variance, we can pool the k sample variances
                                                        2
                       to estimate the within-treatment variance  (s w ):
                                                              (
                                                             k
                                                       s w =  ∑ t=1 n t –  1)s t 2
                                                       2
                                                           ---------------------------------
                                                               (
                                                             k
                                                            ∑ t=1 n t –  1)
                                               ()
                                                2
                       The between-treatment variance  s b   is calculated using the treatment averages   and the grand average, y:
                                                                                 y t
                                                            k  (   y)  2
                                                           ∑ t=1 n t y t –
                                                      s b =  ----------------------------------
                                                       2
                                                              k –  1
                                                                                     2     2
                       If there are an equal number of observations in each treatment the equations for s w   and s b   simplify to:
                                                           ( n t –  k  2
                                                       s w =  ---------------------------------
                                                       2
                                                               1)∑ t=1 s t
                                                              N –  k
                       and
                                                               (
                                                              k
                                                           n t ∑ t=1 y t – y)  2
                                                      s b =  ----------------------------------
                                                       2
                                                              k –  1
                                                        2     2
                       The logic of the comparison between for  s w   and  s b   goes like this:
                                                            2
                          1. The pooled variance within treatments (s w )  is based on N − k degrees of freedom. It will be
                             unaffected by real differences between the means of the k treatments. Assuming no hidden
                             factors are affecting the results,   estimates the pure measurement error variance  σ 2 .
                                                      2
                                                      s w
                          2.  If there are no real differences between the treatment averages other than what would be
                                                                              2
                             expected to occur by chance, the variance between treatments (s b )   also reflects only random
                                                                                     2
                             measurement error. As such, it would be nearly the same magnitude as s w   and would give a
                             second estimate of  σ 2 .
                       © 2002 By CRC Press LLC
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