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I read some questions about this subject, but I couldn't find an answer.

I'm having trouble interpreting the practical effect of the polynomial predictor variable on the response variable.

My model is:

y ~ poly(x,3) + z

My result is:

            Estimate     SE     Z    P

(Intercept) -2.851  0.234   -12.173  < 0.0001
poly(x)1    -0.784  1.036   -0.758   0.449
poly(x)2     1.937  0.845    2.293   0.022 *
poly(x)3     2.754  0.768    3.587   0.0003 **
z            0.342  0.105    3.268   0.001 *

I think it extremely complicated to describe the curvature just by looking at the parameter estimates. When I plot the model considering only “poly(x,3)”, I observe clear evidence of a curvilinear relationship, like this:

enter image description here

Note that I don't have only one independent variable in my model, so my question is: Can I show the table with all predictor variables together (e.g. y ~ poly(x,3)) and, after that, interpret them separately using graphs?

By the way, my hypothesis is that the responsible variable “y” increases until an optimum level of a gradient of the predictor variable “x”, but decreases after that (I am really interested in this hump-shaped relationship), while presents a positive response to the variable “z”.

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  • $\begingroup$ I'm not quite sure what your graphs are showing. Can you be more specific? What do the lines and the dots represent? $\endgroup$ – Tim Mak Apr 7 '20 at 3:38
  • $\begingroup$ The dots in both graphs represent my real sample data. The lines represent the model using just variable x or y. Then, in graphic (a) the line represents the model "y ~ poly(x,3)" and in graphic (b) the line represents the model "y ~ z". However, my model considers both predictor variables: "y ~ poly(x,3) + z". I made the graphs separately because I am not able to understand the effect of the polynomial variable just by viewing the results table, since the second and third-order terms are positive. $\endgroup$ – Barbara Quirino Apr 7 '20 at 3:58
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Let me try to rephrase the question. You are asking how your model y ~ poly(x,3) + z can be interpreted to provide evidence for whether there exists a change point in the domain of x such that y increases with x before the change point and decreases afterwards.

However, I would question the choice of the model in the first place. Why poly(x,3)? I think more generally, your problem falls under the domain of change point detection, which is a difficult problem beyond the scope of the usual linear regression model.

However, since your dataset seems small, I think it is wise not to go beyond simple methods, and regression does provide an easy solution -- simply fit y ~ poly(x,2) + z, i.e. a quadratic on x instead of a cubic, and examine the significance of the quadratic term. If significant, it implies that the relationship is not linear, and from graphical examination, you can argue that it goes up and then down again.

As for how to present the data graphically, you can plot residuals(y ~ z) ~ x, or alternatively residuals(y ~ z) ~ residuals(x ~ z). Both will give you evidence of the relationship between x and y after accounting for z.

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  • $\begingroup$ Yes, that's exactly what I meant. Thank you so much for all the suggestions. I'll try to plot the residuals as you suggested. Regarding the model, I chose poly(x,3) because the model with "poly (x, 2)" was not significant, while the model with only the lower term was. I read that a third-order term easily allows the model to bend down after the peak. Also, the value of R2 doubled from the first to the third-order model, and a likelihood test showed that the more complex model fits better. $\endgroup$ – Barbara Quirino Apr 7 '20 at 14:13
  • $\begingroup$ Actually, I forgot to mention that I'm not interested in the exact numbers, but in the response trend (it is an ecological study). $\endgroup$ – Barbara Quirino Apr 7 '20 at 14:13
  • $\begingroup$ It's not a good idea to choose model based on whether a term is significant, unless you are solely interested in prediction. $\endgroup$ – Tim Mak Apr 8 '20 at 1:55
  • $\begingroup$ Actually I suppose it's ok to use poly(x,3) if your graphical analysis suggests it's a good fit. If the model is significantly better than the linear one, you can argue that there's a non-linear relationship. $\endgroup$ – Tim Mak Apr 9 '20 at 2:21

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