turn on suggestions

Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type.

Showing results for

Find a Community

- Home
- /
- Analytics
- /
- Stat Procs
- /
- Wilcoxon signed rank test (exact distributions)

Topic Options

- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic for Current User
- Bookmark
- Subscribe
- Printer Friendly Page

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

05-30-2012 03:31 PM

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!!

Accepted Solutions

Solution

05-31-2012
02:01 PM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

05-31-2012 02:01 PM

For details of exact test calculation methods, you might have to look in the references within :

PG

PG

All Replies

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

05-30-2012 07:14 PM

This SUGI 1994 paper contains a macro that uses the DATA step to reproduce the WSR test, so you ought to be able to see exactly what happens for tied values.

http://www.sascommunity.org/sugi/SUGI94/Sugi-94-172%20Tian.pdf

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

05-31-2012 12:27 PM

Thanks for the paper Rick, but it doesn't really tell how SAS calculates the exact p-values, just shows an attached file in the macro code (WSRtable.dat). Thanks for your help though!! Have a great day!

Solution

05-31-2012
02:01 PM

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

05-31-2012 02:01 PM

For details of exact test calculation methods, you might have to look in the references within :

PG

PG