I wonder if someone can help me with an ANCOVA covariate question? Perhaps you may know the answer to my problem.

Subjects were assigned to a treatment or control group, and I measured a (continuous) dependent variable. There is also a covariate, and it is continuous. But the covariate is not independent for each subject.


Subjects are assigned to two different kinds of teaching method (treatment and control). In each condition, they do 15 tasks e.g. homework problems, and they get grades for each (continuous dependent variable). But the homework problems have varying levels of difficulty (continuous covariate) and we wish to control for that (also study it).

Is it ok to just treat problem-difficulty as a covariate?

Put another way, each record in the data is a single subject on a single problem. But the covariate is not indepedent for each subject on each problem. Instead, it is a function only of the problem; all subjects have the same value for problem-difficulty for a given problem. Is it ok to just treat problem-difficulty as a covariate?

Thank you very much for any insights.


1 Answer 1


I don't think you can do this with (M)ANCOVA in the conventional way, because that lets you have only one covariate. In a sense you don't have one outcome variable, you have 15 of them. (In another sense, you have one).

Your solution is to use a method allows you to have one covariate and one outcome - that is, you make the data 'long' instead of 'wide'. Then you can fit a multilevel model. (Alternatively, you can keep the data wide and fit a structural equation model - these are equivalent).

  • 1
    $\begingroup$ Hi. A multilevel model is quite right. Thank you. Until today I had only been able to find models in which the "level 2" predictor is the group identity, i.e. Yij = B0j + eij; B0j = C00 + u0j. But now I found that I can also use a level-2 continuous predictor, i.e. Yij = B0j + eij; B0j = C00 + C1*groupX + u0j. So, thank you... $\endgroup$
    – david
    Commented Oct 3, 2013 at 8:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.