Comparing two proportions using a discrete "testing" prior.
-------------------------------------------------
Example (from Berry):  Three Duke students were interested in whether basketball
players are more effective when under less pressure. They considered the 
three-point shots attempted by Duke's 1992-3 basketball team in the First 
half vs Second half of games, thinking there would be more pressure in the 
second half. Regard the following as random samples from larger populations: 
Of the 211 three-point shots attempted in the first half 71, or 33.6%, were 
successful; of the 255 attempted in the second half, 90 (35.3%) were successful.

We illustrate the discrete approach.  Suppose each proportion can be .3, 
.35, .4, .45, or .5.  We believe that the two proportions are equal with 
probability .5.  The following testing prior, constructed by the program 
'pp_disct', places equal probs on the diagonal and equal probs on the 
nondiagonal values such that P(diagonal)=.5.

MTB > exec 'pp_disct'

 FOR EACH P DISTRIBUTION:
 ------------------------
 INPUT LO AND HI VALUES:
DATA> .3 .5
 
 INPUT NUMBER OF MODELS:
DATA> 5
 
 INPUT PROBABILITY THAT P1=P2:
DATA> .5
 
INPUT OBSERVED NUMBER OF SUCCESSES AND FAILURES IN FIRST SAMPLE:
DATA> 0 0
 
INPUT OBSERVED NUMBER OF SUCCESSES AND FAILURES IN SECOND SAMPLE:
DATA> 0 0
 
Posterior distribution of P1 and P2:

 ROWS: PER_1     COLUMNS: PER_2
 
          30       35       40       45       50
 30 0.100000 0.025000 0.025000 0.025000 0.025000
 35 0.025000 0.100000 0.025000 0.025000 0.025000
 40 0.025000 0.025000 0.100000 0.025000 0.025000
 45 0.025000 0.025000 0.025000 0.100000 0.025000
 50 0.025000 0.025000 0.025000 0.025000 0.100000
 

TYPE 'Y' AND RETURN TO SEE A TABLE OF THE POSTERIOR DISTRIBUTION
OF THE DIFFERENCE IN PROBABILITIES P2-P1:
y

 Row   DIFF  P_DIFF

   1  -0.20   0.025
   2  -0.15   0.050
   3  -0.10   0.075
   4  -0.05   0.100
   5   0.00   0.500
   6   0.05   0.100
   7   0.10   0.075
   8   0.15   0.050
   9   0.20   0.025

The program 'pp_disct' is used to find the posterior density. The marginal probs for d=p2-p1 is given.  Note that P(d=0)=.76.  There is support for the hypothesis of equal proportions for this data.

MTB > exec 'pp_disct'

 FOR EACH P DISTRIBUTION:
 ------------------------
 INPUT LO AND HI VALUES:
DATA> .3 .5
 
 INPUT NUMBER OF MODELS:
DATA> 5
 
 INPUT PROBABILITY THAT P1=P2:
DATA> .5
 
INPUT OBSERVED NUMBER OF SUCCESSES AND FAILURES IN FIRST SAMPLE:
DATA> 71 140
 
INPUT OBSERVED NUMBER OF SUCCESSES AND FAILURES IN SECOND SAMPLE:
DATA> 90 165
 
Posterior distribution of P1 and P2:

 ROWS: PER_1     COLUMNS: PER_2
 
          30       35       40       45       50
 30  0.07132  0.09245  0.02815  0.00066  0.00000
 35  0.03152  0.65363  0.04975  0.00116  0.00000
 40  0.00561  0.02911  0.03545  0.00021  0.00000
 45  0.00012  0.00064  0.00019  0.00002  0.00000
 50  0.00000  0.00000  0.00000  0.00000  0.00000
 
 Row   DIFF     P_DIFF

   1  -0.20   0.000000
   2  -0.15   0.000125
   3  -0.10   0.006254
   4  -0.05   0.060819
   5   0.00   0.760422
   6   0.05   0.142409
   7   0.10   0.029311
   8   0.15   0.000660
   9   0.20   0.000001