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in a growth curve like model, I try to test whether a trajectory (over time) is better described by a linear or quadratic trend. I fitted a linear mixed model (time nested in participants) and called orthogonal polynomial contrasts using the poly() function in R:

mod <- lmerTest::lmer(outcome ~ poly(time, 2) + (poly(time, 2) | id), data=dat.long)
parameters::parameters(mod) returns the following output:

Parameter Coefficient SE 95% CI t(1215) p
(Intercept) 6.75 0.15 [ 6.46, 7.04] 45.35 < .001
Time [1st degree] -58.01 4.08 [-66.01, -50.02] -14.22 < .001
Time [2nd degree] -9.00 2.56 [-14.02, -3.98] -3.51 < .001

I now wonder which of the following interpretations is correct:
(1) The test statistics refer to the "absolute" fit of the corresponding trends.
Thus, the linear model fits the data better (|t| lin > |t| qua) than the quadratic model.
(2) The quadratic term describes the increment of the quadratic model (vs. linear term only), thus suggesting that the quadratic model fits the data better (t(1215) = -3.51, p < .001).

(I suspect it is (2), but I would very much appreciate any comments/advice.)

Thank you so much for your help!

Marcel

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There seems to be some confusion.

You have fitted one model that has a fixed intercept, a linear term and a quadratic term, plus some random effects.

(1) The test statistics refer to the "absolute" fit of the corresponding trends. Thus, the linear model fits the data better (|t| lin > |t| qua) than the quadratic model.

This does not make sense because you have not fitted two models.

(2) The quadratic term describes the increment of the quadratic model (vs. linear term only), thus suggesting that the quadratic model fits the data better (t(1215) = -3.51, p < .001).

Again this does not make sense for the same reason.

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  • $\begingroup$ Does this answer your question ? If so please consider marking it as the accepted answer. If not, please let us know why. Also, if you haven't already, please consider upvoting it. $\endgroup$
    – Robert Long
    Commented Jul 24, 2021 at 12:05
  • $\begingroup$ Thanks a lot for your helpful answer! To make sure I interpret the results correctly: (1) Can the second row (i.e., 1st degree) be interpreted as a negative linear trend? (2) Does the significance of the quadratic term mean that we should not drop the quadratic component of the model? I remember being tought that the first polynomial does not reflect a linear trend but the "velocity" at x = 0. Here, however, where orthogonal contrasts are used, the linear component remains the same, no matter whether this is the only component in the model or whether I add a quadratic term (see above). $\endgroup$
    – mw21
    Commented Aug 6, 2021 at 6:55

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