Page 162 - Applied Statistics Using SPSS, STATISTICA, MATLAB and R
P. 162

142      4 Parametric Tests of Hypotheses


           the alternative hypothesis being that there is at least one pair with unequal means,
           µ i ≠ µ j.
                                                                      c
              We now assume that H 0 is assessed using two-means tests for all ( ) pairs of
                                                                      2
           the c  means. Moreover, we assume that every two-means test is performed at a
           95% confidence level, i.e., the probability of not rejecting the null hypothesis when
           true, for every two-means comparison, is 95%:

               ( P µ i  = µ j  |  H 0ij ) =  . 0  95 ,                     4.18

           where H 0ij is the null hypothesis for the  two-means test referring to the  i and  j
           samples.
              The probability of rejecting the null hypothesis 4.17 for the c means, when it is
           true, is expressed as follows in terms of the two-means tests:

              α =  ( P  reject  H 0  |  H 0 )                  .           4.19
                =  ( P µ ≠ µ 2  |  H  0  or µ ≠  µ 3  |  H 0  orK or µ c− 1  ≠ µ c  |  H 0 )
                                  1
                     1

              Assuming the two-means tests are independent, we rewrite 4.19 as:

              α = 1−  ( P µ =  µ 2  |  H 0 )P (µ =  µ 3  |  H 0  )K  ( P µ c− 1  =  µ c  |  H 0  ) .    4.20
                                     1
                       1

              Since H 0 is more restrictive than any H 0ij, as it implies conditions on more than
           two means, we have  P (µ ≠  µ j  |  H 0ij  )   ≥  P (µ ≠  µ j  |  H 0 ) , or, equivalently,
                                                     i
                                  i
             ( P µ =  µ j  |  H 0ij ) ≤  (µ =  µ j  |  H 0  ) .
                            P
               i
                                i
              Thus:

              α ≥ 1−  ( P µ 1  = µ 2  |  H 012 )P (µ 1  = µ 3  |  H 013 )K  ( P µ c− 1  = µ c  |  H 0c−  , 1 c ) .    4.21

              For instance,  for  c =  3, using 4.18 and 4.21, we obtain a Type  I Error
                     3
           α  ≥ 1−0.95 =  0.14.  For higher values of  c the Type  I Error degrades rapidly.
           Therefore, we need an approach that assesses the null hypothesis 4.17 in a “global”
           way, instead of assessing it using individual two-means tests.
              In the following sections we describe the  analysis of variance (ANOVA)
           approach, which provides a suitable methodology to test the “global” null
           hypothesis 4.17. We only describe the ANOVA approach for one or two grouping
           variables (effects or  factors). Moreover,  we  only consider the so-called “fixed
           factors” model, i.e., we only consider  making inferences on several fixed
           categories of a factor, observed in the dataset, and do not approach the problem of
           having to infer to more categories than the observed ones (the so called “random
           factors” model).
   157   158   159   160   161   162   163   164   165   166   167