For n <= 20, proc univariate uses the exact distribution to compute the significance of S, but what exactly happens when there are tied ranks within the small dataset? I've read all 3 references (Iman, Conover, and Lehmann), but none of them really explain what happens when ties occur with a small sample (n <= 20) with the exact distribution. An example is below...
Consider the data:
Grp1 Grp2 Diff Rank
.32 .39 -0.07 3.5
.4 .47 -0.07 3.5
.11 .11 0.00 ---
.47 .43 0.04 1
.32 .42 -0.10 5
.35 .3 0.05 2
.32 .43 -0.11 6
.63 .98 -0.35 8
.5 .86 -0.36 9
.6 .79 -0.19 7
Sum of ranks for positive differences (ri+) = 3
Given that the ranks are {1, 2, 3.5, 3.5, 5, 6, 7, 8, 9}, the only ways to get a sum of ranks that is less than or equal to 3 is for the set of positive ranks to be one of:
Set Sum
{} = 0
{1} = 1
{2} = 2
{1,2} = 3
So, there are 4 configurations on the left-hand side extreme and 4 on the right. Thus, the p-value should be 8/2^9 = 8/512 = 0.0156. However, SAS reports p=10/512 = 0.0195.
Perhaps, SAS is saying that {3.5} is either {3} or {4} with ½ probability. Thus, this would be ½ more cases. Since there are two ranks of 3.5, each could be {3} with ½ probability and thus there would be a total of 5 configurations on the left extreme and 5 on the right?
Set Sum
{} = 0 with 100% probability = 1.0 case
{1} = 1 with 100% probability = 1.0 case
{2} = 2 with 100% probability = 1.0 case
{1,2} = 3 with 100% probability = 1.0 case
{3.5} = 3 with 50% probability = 0.5 case
{3.5} = 3 with 50% probability = 0.5 case
---------
5.0 cases
If you know how SAS is computing the exact p-value with ties (all possible combinations of the sum of ranks less than or equal to the sum of positive ranks) or know where a 'useful' reference might be, please let me know. Thanks in advance!!
