MODEL里面直接用*引入交互效应的变量

Since you do not provide data, I use simulated one to illustrate the key points.

Note: Continuous variavle=z, genno=x, sex=x2

By construction in my simulation, there is a interaction     

x * x2= {1 , 2 , 3,
              2, 4 , 6}

1) you need to test which model is better explaned by data.
  (y= x*x2 z or y= x x2 x*x2)   vs  y=x x2 z

This can be done likelihood ratio test. The model (y= x*x2 z or y= x x2 x*x2) is better because the data is generated by interaction of x*x2, so the simple linear approximation is rejected.

2) you may test if the cell of (x=1, x2=2) and the cell of (x=2, x2=1) is the same,  it should be the same by construction.

3) Then should be interpret your results, the interaction exists. That is to say each cell (x*x2) has differenct effect except for the cell of (x=1, x2=2) and the cell of (x=2, x2=1).

Run  the following program and check the results.

Hope this helps.


data t1;
   do x=1 to 3;
       do x2=1 to 2;
          do obs=1 to 10;
           z=rannor(123);
          y= 1+ x*x2 +1*z+ rannor(7789);
          output;
          end;
        end;
    end;
run;

proc mixed data=t1 method=ML;
class x x2;
model y= x*x2 z /solution;
run;


proc mixed data=t1 method=ML;
class x x2;
model y=x x2 z/solution;
run;

感谢crackman 版主和bobguy。不好意思，由于我水平差，bobguy提供的方法我没有完全看明白，但还是要感谢。我已经询问了我的老师，他给我做了解释，问题已经解决。其实他给我的解释非常简单，就是和版主说的一样，他给我的解释很详细，可以使我理解这样做的原因。
解释大概是这样的：在基因型中，我只关心其中一种表型和男性的交互作用，所以他让我把基因型分成TT（我关心的表型）和非TT型的，性别男性和女性，然后做交互作用，算出beta值，和or值。

Likelihood Ratio test (LR test) is typical and easy to understand in statistics. Many procedures in SAS have implemented such test. The calculation yourself is not necessary. For example, if you want to give a LR test on an interaction term, you can use

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