nonlinear relationship - linear and quadratic term both positive, quadratic term insignificant

[Felicia Hollingsworth]

Dear all,

I've been trying to model nonlinear relationships in a multilevel model (mixed).

I'm just not really sure how to interpret my findings. In the model with both the linear and the quadratic term

mixed wag c.SM c.SM#c.SM (other independent variables) || ID:

the linear term is not significiant and the regression coefficient is positive whereas the quadratic term is significant and also positive.
I checked the joint significance via

test c.SM c.SM#c.SM

which was significant. So I calculated the vertex (-b/2a) which lies in the range of values. Can I assume that I have a U-shaped relationship even though the linear term is not significant and both terms have a positive regression coefficient? I've only found cases in which the regression coefficients were positive/negative or negative/positive for the linear and quadratic term.

Thanks a lot in advance!

[Richard Williams]

Once you add squared (or interaction) terms to a model the interpretation of the linear term changes and you generally don't worry whether it is significant or not. And you don't drop lower-level terms when you have higher level terms in the model. For more, see

https://www3.nd.edu/~rwilliam/stats2/l53.pdf

https://www3.nd.edu/~rwilliam/stats2/l55.pdf

[Felicia Hollingsworth]

Thank you for your answer! So the non-significance doesn't seem to be a problem, that's good! But can I assume a U-shaped relationship even though the linear and the quadratic term are both positive?

[Felicia Hollingsworth]

Oh, and I just realized, I made a mistake in the title, the quadratic term is significant, the linear term isn't.

[Richard Williams]

See

https://www3.nd.edu/~rwilliam/stats2/l61.pdf

especially p. 4. Yes, you should get a U shaped relationship, but not necessarily with the observed (or even possible) range of values.

Easy way to see this: suppose the linear effect is 0 so all you have is the squared term. Between the range of -5 to 5 you will get a U shape, e.g. -5^2 = 5^2, -4^2 = 4^2, 0^2 = 0.

[Nick Cox]

If the coefficient of the square is not zero, a quadratic will always be U-shaped in your sense, although parabolic is a better word.

As you're aware there is still the question of whether the turning-point is within the range of the data.

Both coefficients being positive are perfectly consistent with a U too. For example, see the results of

twoway function 1 + 2*x + 0.3*x^2, ra(-10 10)

I would use twoway function to see what the curve looks like in your case.

[Felicia Hollingsworth]

Thank you both for your answers.

I used a mean-centered variable for the linear and the quadratic term so that's why the results confused me a bit. After recalculating everything with the uncentered variable everything made more sense. I used marginsplot to plot the predictive margins (for both the uncentered and centered variable) and it showed me the parabolic (thanks for the tip!) relationship I expected with the vertex lying within the range of data.