Example 54.2: Multilevel Response

The PROBIT Procedure

Example 54.2: Multilevel Response

In this example, two preparations, a standard preparation and a test preparation, are each given at several dose levels to groups of insects. The symptoms are recorded for each insect within each group, and two multilevel probit models are fit. Because the natural sort order of the three levels is not the same as the response order, the ORDER=DATA option is specified in the PROC statement to get the desired order.

The following statements produce Output 54.2.1:

   data multi;
      input Prep $ Dose Symptoms $ N;
      LDose=log10(Dose);
      if Prep='test' then PrepDose=LDose;
      else PrepDose=0;
      datalines;
   stand     10      None       33
   stand     10      Mild        7
   stand     10      Severe     10
   stand     20      None       17
   stand     20      Mild       13
   stand     20      Severe     17
   stand     30      None       14
   stand     30      Mild        3
   stand     30      Severe     28
   stand     40      None        9
   stand     40      Mild        8
   stand     40      Severe     32
   test      10      None       44
   test      10      Mild        6
   test      10      Severe      0
   test      20      None       32
   test      20      Mild       10
   test      20      Severe     12
   test      30      None       23
   test      30      Mild        7
   test      30      Severe     21
   test      40      None       16
   test      40      Mild        6
   test      40      Severe     19
   ;

   proc probit order=data;
      class Prep Symptoms;
      nonpara: model Symptoms=Prep LDose PrepDose / lackfit;
      weight N;
      parallel: model Symptoms=Prep LDose / lackfit;
      weight N;
      title 'Probit Models for Symptom Severity';
   run;

The first model uses the PrepDose variable to allow for nonparallelism between the dose response curves for the two preparations. The results of this first model indicate that the parameter for the PrepDose variable is not significant, having a Wald chi-square of 0.73. Also, since the first model is a generalization of the second, a likelihood ratio test statistic for this same parameter can be obtained by multiplying the difference in log likelihoods between the two models by 2. The value obtained, 2 ×(-345.94 - (-346.31)), is 0.73. This is in close agreement with the Wald chi-square from the first model. The lack-of-fit test statistics for the two models do not indicate a problem with either fit.

Output 54.2.1: Multilevel Response: PROC PROBIT