I performed a polynomial regression using the following formula:

lm(deviance ~ poly(myDF$distance,3,raw=T))

However, the summary output states that only the third term is significant:

                                     Estimate Std. Error t value Pr(>|t|)  
(Intercept)                         -0.014825   0.095987  -0.154   0.8774  
poly(myDF$distance, 3, raw = T)1     0.031286   0.143283   0.218   0.8273  
poly(myDF$distance, 3, raw = T)2    -0.080363   0.065591  -1.225   0.2215  
poly(myDF$distance, 3, raw = T)3     0.021517   0.009377   2.295   0.0224 * 

How is this to be interpreted? My first guess is that only the full third-degree model (including the lower degree terms) fits the data significantly better than the null hypothesis. Is this correct? Put simply: Does the non-significance of the first and second degree terms impair the goodness of the model?

  • $\begingroup$ The threads at stats.stackexchange.com/questions/39087, stats.stackexchange.com/questions/108657, and stats.stackexchange.com/questions/64192 (among many others) may shed light on these questions. $\endgroup$ – whuber Aug 20 '14 at 13:36
  • $\begingroup$ Unless the first- and second- order terms were coded in a way that is especially meaningful for these data, they are automatically significant when the entire regression is significant. The p-values for the individual terms have little meaning. That is because these three terms (plus the "intercept") collectively form a basis for a four-dimensional space but none of them individually occupy a privileged direction in that space. Whether any of them appear to be "significant" is merely an artifact of the particular choice of the basis (as made by the internals of the poly function). $\endgroup$ – whuber Aug 20 '14 at 13:42
  • 1
    $\begingroup$ @whuber, why not turn that into an 'official' answer (& maybe develop it a little, perhaps w/ some plots). It is similar to information that exists elsewhere (as you note), but I don't recall having seen it in quite that form / so clear & concise. It's a nice summary of the issue. It would be a shame if it remains buried in a comment. $\endgroup$ – gung - Reinstate Monica Jan 29 '15 at 22:39

Your "first guess" can't be evaluated just from this output. You'd have to look at the models without the cubic term to decide.

What you can say is that, in the model with the cubic term, the linear and quadratic terms are not significant. Still, they should almost surely be included due to the "hierarchy principle".

The non-significance of the lower order terms is not, in itself, a bad thing.

As to how to interpret the model - since you have only one independent variable, you can easily make a graph with the IV on the x axis and the DV on the y axis. I think this happens automatically as one of the plots of plot(model) but, if not, you can do it with

m1 <- lm(deviance ~ poly(myDF$distance,3,raw=T))
    plot(x = myDF$distance, y = m1$fitted.values)

You can also get the fitted value for any distance by plugging it into the regression equation (or using predict with newdata).

  • $\begingroup$ In the quadratic model both the linear and the quadratic term are significant but the whole model has a lower R² (I know, this is not the most appropriate criteria for model selection). Most importantly, a 2nd order polynomial is not really plausible in my case. A linear regression is also significant but the QQ-Plot tells me that it's a bad representation of the data. $\endgroup$ – Hav0k Aug 20 '14 at 11:44
  • $\begingroup$ I agree with you. Substantive plausibility is the most important thing. $\endgroup$ – Peter Flom - Reinstate Monica Aug 20 '14 at 12:43

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