Suppose we have outcome variable Y and two predictors X1 and X2, where X1 and X2 sort of come from a type of category. For example, X1 is the relationship score with friends and X2 relationship score with family. Not the same, but they reflect some thing together (social relationship, in this case).

I want to set up a model to test the hypothesis that the diversity/balance of predictors can influence the outcome. Some thing like this: Suppose there are two persons who both have scores of 10 on X1 and X2 total. I suspect that the one with score of 5 on X1 and 5 on X2 will have better outcome than the one with 10 on X1 and 0 on X2.

I think a normal model (Y = X1 + X2) or an interaction term (Y = X1 + X2 + X1*X2) seem not to do the job. Maybe there needs to create new variable(s) but I'm not sure which to create.

  • $\begingroup$ Your predictors $X_1,X_2$ are ordinal, but how is your outcome measured? $\endgroup$ – abaumann May 9 '14 at 10:26
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    $\begingroup$ Why do you think an ordinary interaction would not do the job? It would certainly distinguish the cases in your "suspicion" (end of 2nd paragraph). $\endgroup$ – Peter Flom May 9 '14 at 10:48
  • $\begingroup$ @abaumann I have the outcome being dichotomous. Does it make some difference as compared with other measurement scales? $\endgroup$ – NonSleeper May 10 '14 at 5:04
  • $\begingroup$ @PeterFlom I think the interaction will tell us how the effect of one predictor depends on values of the other. It looks like that does not exactly give an impression, or a specific way of interpretation, as how the "balanced contribution" of predictors is an independent and important factor to consider in addition to the total scores they have. Not sure if I'm clear enough about this idea though. $\endgroup$ – NonSleeper May 10 '14 at 5:06
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    $\begingroup$ I don't know what you mean by "balanced contribution of predictors". Interactions can be measured many ways - the most common is the product of the two variables, but you can do things like product of a variable and the square of the other etc. $\endgroup$ – Peter Flom May 10 '14 at 10:39

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