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Can I use Steel with unequal sample sizes?

mls1984

Community Trekker

Joined:

Oct 21, 2015

I have a treatment vs. control situation, with 9 observations for the control and 2 of 3 treatments, while there is only 6 observations for the last treatment. JMP allows me to perform the “Steel with control” analysis, but is this valid when the original paper states that it should be used for equal sample sizes?

1 ACCEPTED SOLUTION

Accepted Solutions
Solution

The documentation team (of which I am a member) recently added more detail to the documentation on the non-parametric multiple comparison tests, but, unfortunately, this will not appear until JMP 13. But to your question, the Steel with Control procedure in JMP does allow unequal treatment sizes.  Each pair of treatments is compared by ranking only the observations in the two treatments, calculating the mean rank for each treatment, and then computing the "Score Mean Difference", which is the difference in ranks with a continuity correction applied. The standardized statistic is asymptotically normal.  This is based on Hsu (1996).

Here are the details of the continuity correction that is applied to the Score Mean Difference:

  1. Denote the number of observations in the first treatment by n1 and the number in the second treatment by n2.
  2. Rank the observations within the sample consisting of these two treatments. Tied ranks are averaged.
  3. Denote the sum of the ranks for the first treatment by ScoreSum1 and for the second treatment by ScoreSum2.
  4. If the difference in mean scores is positive, then the Score Mean Difference is given as follows:
    Score Mean Difference = (ScoreSum1 - 0.5)/n1 - (ScoreSum2 + 0.5)/n2
  5. If the difference in mean scores is negative, then the Score Mean Difference is given as follows:
    Score Mean Difference = (ScoreSum1 + 0.5)/n1 - (ScoreSum2 -0.5)/n2

I hope this helps answer your question.

6 REPLIES
KarenC

Super User

Joined:

Feb 10, 2013

No comment on your particular situation as it often "depends".  What are you data assumptions, where did the data come from, why are your sample size unbalanced, why is normality not a reasonable assumption,on so on and so forth. (I am not looking for you to answer any of these questions, just posing the questions that quickly come to mind when thinking about such a questions).

However, I am interested in multiple comparisons and so "learning via JMP" I looked up the Steel with Control test in the JMP documentation and see no mention of balanced vs. unbalanced situations.  However, the documentation references Hsu (1996) and in Hsu he does discuss the unbalanced situations (both one and two sided) but comments that "software to compete....does not seem readily available".  So now almost 20 years since Hsu it appears that the software might actually be readily available.  Hopefully a JMP developer can "jump" in here and confirm that the Steel procedure implemented in JMP does indeed handle the unbalanced cases as one would expect.

kevin_c_anderso

Community Trekker

Joined:

Jun 5, 2014

Referencing Table 6.8 of Dmitrienko, A., et.al.(2007); Pharmaceutical Statistics using SAS: A Practical Guide; SAS Press pg 144:

Steel-Dwass methods are for equal group sample sizes, and the Dunn's methods are for unequal group sample sizes.

But, like Karen states, without a JMP developer confirmation, a leap of faith about what JMP is doing is required.

mls1984

Community Trekker

Joined:

Oct 21, 2015

Thank you both for your comments :-)

Solution

The documentation team (of which I am a member) recently added more detail to the documentation on the non-parametric multiple comparison tests, but, unfortunately, this will not appear until JMP 13. But to your question, the Steel with Control procedure in JMP does allow unequal treatment sizes.  Each pair of treatments is compared by ranking only the observations in the two treatments, calculating the mean rank for each treatment, and then computing the "Score Mean Difference", which is the difference in ranks with a continuity correction applied. The standardized statistic is asymptotically normal.  This is based on Hsu (1996).

Here are the details of the continuity correction that is applied to the Score Mean Difference:

  1. Denote the number of observations in the first treatment by n1 and the number in the second treatment by n2.
  2. Rank the observations within the sample consisting of these two treatments. Tied ranks are averaged.
  3. Denote the sum of the ranks for the first treatment by ScoreSum1 and for the second treatment by ScoreSum2.
  4. If the difference in mean scores is positive, then the Score Mean Difference is given as follows:
    Score Mean Difference = (ScoreSum1 - 0.5)/n1 - (ScoreSum2 + 0.5)/n2
  5. If the difference in mean scores is negative, then the Score Mean Difference is given as follows:
    Score Mean Difference = (ScoreSum1 + 0.5)/n1 - (ScoreSum2 -0.5)/n2

I hope this helps answer your question.

mls1984

Community Trekker

Joined:

Oct 21, 2015

Thanks Marie, for your reply and detailed information. Now I know that I can use this procedure without worrying about the validity of the results :-)

marie_gaudard0

Joined:

Jan 14, 2015

I'm so pleased that this helped! Good luck with your analysis!

--Marie