i got 30 binary variables and i want to find out if there is any combination of this variables which leads to a high value on a metric variable.

For example: i got the binary variables "loud", "hot", "new", "nice", "hard" and i got the dependent metric variable "liking" (1-100).

I want to find out which combination of "loud", "hot", "new", "nice" and "hard" leads to the highest ratings on "liking". If it was just 2 binary variables i would think a 2-factorial anova would do the job? But i got 30 binary variables and i have no idea how to do this.

i could calc 30 t-tests to finde out which binary has an effect alone. But if i do so, i wont finde out how they interact with each other. i could calc a regression with all the 30 variables and have the same problem.

Is there any method to finde out the best combination of binary-ratings for a high liking value?

And as a second theoretical question: If there is any combination of binary variables which leads to a high rating, does that imply that all the binary variables in this combination are correlated? If yes, i could look for correlations.

thanks a lot for all your help!


You can't check every possible combination. There are $2^{30} = 1,073,741,824$ of them, so unless you have a truly gigantic sample, you won't even be able to even find a case for each of them, let alone model them.

A more tenable approach is to check for every effect of a single binary variable and assume no interactions, or only allow all the two-way interactions (of which there are a mere ${30 \choose 2} = 435$). Whatever combinations you want to check, you can fit a linear regression model with whatever predictors you want, then look at what the model says is the combination of covariates with the greatest conditional mean.

  • $\begingroup$ a little late, but thanks for this helpful suggestions! $\endgroup$ – TinglTanglBob Aug 10 '18 at 11:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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