Solution to Exercise 14.29.

  The row factor (first subscript; level index denoted i) is Male genotype and it has 6 levels (i ranges from 1 to 6). The column factor is Sex combination (second subscript; level index denoted j) and has 3 levels which we denote MM, MF, and FF. The parameters of interest are p_ij, the proportion of experimental units with Male genotype i and Sex combination j (I'm not exactly clear on what the experimental unit is here). The marginal frequencies are $p_{i \cdot}$, the proportion of units with Male genotype i irrespective of Sex combination, and $p_{\cdot j}$, the proportion of units with Sex combination j irrespective of Male genotype. The null hypothesis is

$$H_0 : p_{ij} = p_{i \cdot} \star p_{\cdot j}, \mbox{ for all $i$\space and $j$} ,$$

and the alternative is

$$H_1 : p_{ij} \ne p_{i \cdot} \star p_{\cdot j}, \mbox{ for some $i$\space and $j$} ,$$

i.e. the negation of the null hypothesis.

To do the analysis, I entered the data into minitab. In order to use the cross classified data choice under Stat $\rightarrow$ Tables, I had to unstack the columns and make appropriate columns with the levels of the factors using the Manip and Calc menus. The final data appears below:

Mgtype M/M M/F F/F Mgt2 SxCmb counts

1 35 80 39 1 1 35

2 41 84 45 2 1 41

3 33 87 31 3 1 33

4 8 26 8 4 1 8

5 5 11 6 5 1 5

6 30 65 20 6 1 30

1 2 80

2 2 84

3 2 87

4 2 26

5 2 11

6 2 65

1 3 39

2 3 45

3 3 31

4 3 8

5 3 6

6 3 20

Then I ran the minitab Tables with Frequencies in the ``counts'' column. The output appears below:

Tabulated Statistics

  
 
 
 
 Rows: MGeno2     Columns: SexComb
 
           1        2        3      All
  
 1     22.73    51.95    25.32   100.00
       23.03    22.66    26.17    23.55
        5.35    12.23     5.96    23.55
          35       80       39      154
  
 2     24.12    49.41    26.47   100.00
       26.97    23.80    30.20    25.99
        6.27    12.84     6.88    25.99
          41       84       45      170
  
 3     21.85    57.62    20.53   100.00
       21.71    24.65    20.81    23.09
        5.05    13.30     4.74    23.09
          33       87       31      151
  
 4     19.05    61.90    19.05   100.00
        5.26     7.37     5.37     6.42
        1.22     3.98     1.22     6.42
           8       26        8       42
  
 5     22.73    50.00    27.27   100.00
        3.29     3.12     4.03     3.36
        0.76     1.68     0.92     3.36
           5       11        6       22
  
 6     26.09    56.52    17.39   100.00
       19.74    18.41    13.42    17.58
        4.59     9.94     3.06    17.58
          30       65       20      115
  
 All   23.24    53.98    22.78   100.00
      100.00   100.00   100.00   100.00
       23.24    53.98    22.78   100.00
         152      353      149      654
 
Chi-Square = 6.463, DF = 10, P-Value = 0.775

 
  Cell Contents --
                  % of Row
                  % of Col
                  % of Tbl
                  Count

The P-value of 0.775 says that these data do not have a statistically significant departure from the hypothesis that the distribution of sex combinations is homogeneous with respect to the different genotypes.