Hi everybody,
One of the predictors (A) of my dependent variable clearly follows a quadratic function (A and A squared both significant). I want to see how this predictor interacts with another one (B, also significant). Yet, I do not know how to introduce it into the OLS regression model. So, if in a regular interaction I will introduce the interaction term as A*B, what should I do this time? A * A squared * B?? is it possible at all to test it?
Hello Isabel,
In general, your new model would be: Y' = b0 + b1*A + b2*A^2 + b3*B + b4A*B + b5*(A^2)*B
Good luck with your work.
Hello Isabel,
In general, your new model would be: Y' = b0 + b1*A + b2*A^2 + b3*B + b4A*B + b5*(A^2)*B
Good luck with your work.
As David Morse says, the key is to make two interaction variables, B*A and B*A-squared, and add these to the variables already in the model, namely B, A and A-squared. If one or both of the interaction variables are significant, you have an interaction effect :-)
Isabel -
Other comments about your model:
1) You could compare various models' fits using graphical residual analyses, and try to avoid overfitting using a cross-validation in each case.
2) If you know that when all independent variables are zero, y must be zero, then drop the intercept term.
3) Don't forget heteroscedasticity. (See https://www.researchgate.net/project/OLS-Regression-Should-Not-Be-a-Default-for-WLS-Regression, and updates.)
4) Does the model you select make sense from a subject matter perspective? If not, it could be a peculiarity of your sample. Hopefully cross-validation would find that, but there could be a shortcoming in your data collection process. - You said "I want to see how this predictor interacts with another one...." Did you expect this? - It is important to use the best set of independent variables, not too many, and not too few, and the best one or ones.
5) Miscellaneous: There's principal components, there are data quality issues, different types of models - various considerations. There are pros and cons to consider, but I think that if all other considerations are about 'equal,' it might be good to lean toward simplicity.
Cheers - Jim
Two points.
1. If you are using R, remember to use the I function (for identity). That is an error that I often see.
2. (and most important). You say " my dependent variable clearly follows a quadratic function (A and A squared both significant) ." Of course finding A and A^2 both being significant for tons of reasons does not imply that the relationship is quadratic. You can say it is not linear. Do you have any theoretical reason to suppose that it will be U or inverse U shaped?
Dear all, thank you very much for your answers! You gave a lot to think and to test.
Cheers!
You could also fit a non-linear smooth and an interaction which is a lot more flexible than a quadratic
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