If the difference is zero and we know that, then I do not see any interest in doing so. In a factor levels we search this but here I think such contrasts are not useful.

Though there won't be statistical difference in the effect, your interest may be realized in SAS by using PROC MIXED as if there will be. If there won't be,  the confidence interval estimate includes Zero. If there will be,  CI does not. The question is that you do not tell me, what is be in the second level or other levels.

If I understand the question .... This is a classic situation for using a computer intensive method. This assumes that you have a large number of replicates (16 replicates for a permutation test, 20 for all others). If you have less data, then this is pointless.

In SAS take your data set and randomly select a subpopulation of size n. Run proc Mixed, save the results. Randomly select another subpopulation, and run proc mixed. Do this 10,000 to 50,000 times. If you really want to understand your data, try doing this again for subpopulations of different sizes. You can do this by modifying the program that is published in the appendix of this manuscript:

Randomization Tests: Example Using Morphological Differences in Aphis gossypii (Homoptera: Aphididae) T. A. EBERT, W. S. FARGO, B. CARTWRIGHT, F. R. HALL

Annals of the Entomological Society of America 10/1998; 91(6):761-770

The PDF is on my ResearchGate account.

I highly recommend that you spend some time playing with your data, or start by playing with the data that are also included in the cited manuscript, or use data generated using a random number generator. Think of things this way: You run an experiment, gather data, and calculate a mean and standard deviation. Yet you know that if you were to do exactly the same experiment you would get a slightly different answer. If someone else would do exactly the same experiment they too would get a different answer. We hope that the difference is not too large, but that is hope. The point is that you calculated a mean and standard deviation, but by repeating the experiment you will be able to calculate a mean and standard deviation for your original mean, and a mean and standard deviation for your original standard deviation. To start you used your experimental design to test your system. Now you are testing your experimental design.

How does changing subpopulation size influence the outcome?

How does changing the number of randomizations change the outcome? What happens if I use 10 randomizations (run 1000 sets of 10 randomizations)? What if I use 100, 1000, 10,000 randomizations? You can now use this outcome to quantitatively justify a specific number of randomizations that should be used in your system.

I believe the question is simpler than that. It refers to coding of the contrast/estimate statments in SAS.I may have explained incorrectly or superficially. The experiment has four classes, and each class represents a combinatory arrangement of group (2 levels [control and treatment]) and dose (2 levels [low and high]). Let me explain better:

A: it is the combinatory arrangement of control group and low dose;

B: it is the combinatory arrangement of control group and high dose;

C: it is the combinatory arrangement of treatment group and low dose;

D: it is the combinatory arrangement of treatment group and high dose.

Initially, I performed the contrast/estimate I was interested between group, dose, and their interactions. But now, I want to do contrast/estimate between (1) control group - low dose (A) and control group - low dose (A); and (2) control group - high dose (B) and control group - high dose (B) (i.e., -A +A +B -B, or in other words, -1 +1 +1 -1). I cannot include (at least I do not know how to do it) in the contrast statement a term informing SAS that I want to contrast class A against class A, and class B against class B. As I said before, this contrast is important for me (and will be very useful) because I want to know the confidence interval estimates (lower and upper) even though the contratast estimation will be zero (0) indicating no statistical significance. If there is a simpler way to obtain the information I wish, I will be glad to know.

let me know If my question is still a  little bit confusing.

Ok, I'll try again, though maybe still a bit confused.

So you want to contrast high dose versus low dose for the control group, and you want to contrast low versus high for the treatment.

If you have a variable named "treatment" where 0=control and 1=treatment then the simple way is to type the following SAS code anywhere between your input statement and the start of the analysis statements.

I would probably have a program like:

Data Name1;

Infile (file name);

Input (variables);

If treatment=0 then output;

Run;

Data Name1; Set Name1;

(Your analysis code)

Run the program. Save your answer.

Change the SAS statement to this: 

If treatment=1 then output;

Rerun the program.

Does this accomplish what you want?

Hello Dr. Ebert. Yes, this accomplish what I wanted. Thank you! Also, thank you all for the asnwers.

I think it has DATE NOT WHAT YOU WANT TO DO= NS  not difference.

NEXT: Beside the fixed effect column, create a new column named place of division number to which data to compare within that effect the rest of the effects that you have no interest in second place, ahead of proc mixed data = mydata; by subdivision; keep the same model, it will compare to the first subdivision in the fixed effect specified, IN CONTRAST place fixed effect. Eg.

input subd ....effect fixed.....

cards;

1 A

1 A

1 A

2 A

2 A

2 A

0 B 

0 B 

0 B 

0 B 

0 B 

0 B 

proc mixed data = mydata; by subdivision;

randomeffect effectfixed class subdivision;

model y = effectfixed subdivision;

random randomeffect

contrast 'decomposing the fixed effects' my ger coefficients;

estimate '' '' '' 'My ger coefficients;

run;