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176      5 Non-Parametric Tests of Hypotheses


           Example 5.3
           Q: According to Mendel’s Heredity Theory, a cross breeding of yellow and green
           peas should produce them in a proportion of three times more yellow peas than
           green peas. A  cross  breeding of  yellow  and green peas  was performed  and
           produced 176  yellow peas and  48 green  peas. Are these experimental results
           explainable by the Theory?
           A: Given the theoretically expected values of the proportion of yellow peas, the
           test is formalised as:

              H 0:   P(ω =1) = ¾ ;
              H 1:   P(ω =1) ≠ ¾.

              In  order to apply the binomial test to this example, using SPSS, we start by
           filling in a datasheet as shown in Table 5.2.
              Next, in order to specify that category 1 of pea-type occurs 176 times and the
           category 0 occurs 48 times, we use the “weight cases” option of SPSS, as shown in
           Commands 5.2. In  the  Weight Cases   window  we specify that  the  weight
           variable is n.
              Finally, with the binomial   command of SPSS, we obtain the results shown in
           Table 5.3, using  0.75 (¾)  as the tested proportion.  Note the “Based on Z
           Approximation” foot message displayed by SPSS. The two-tailed significance is
           0.248, so therefore, we do not reject the null hypothesis P(ω =1) = 0.75.

           Table 5.2. Datasheet for Example 5.3.

                    group                 pea-type                 n
                      1                      1                    176
                      2                      0                    48

           Table 5.3. Binomial test results obtained with SPSS for the Example 5.3.
                                                 Observed           Asymp. Sig.
                                 Category   n             Test Prop.
                                                  Prop.              (1-tailed)
                                                                           a
            PEA_TYPE    Group 1     1      176     0.79     0.75      0.124
                        Group 2     0       48     0.21
                         Total             224     1.00
           a  Based on Z approximation.


              Let us  now carry  out this test using the  values  of the  standardised  normal
           distribution. The important values to be computed are:

              np = 224×0.75 = 168;
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