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I have analyzed some data on vegetation change as a function of change in soil parameters. I compared a dataset from 2001 with a dataset from 2018 (fully balanced).

To investigate the change in vegetation (expressed as Bray-Curtis dissimilarity indice) as a function of change in soil parameters (just 2018 minus 2001 value) i ran several models. When comparing using AIC I ended up with a model of 9 soil parameters explaning vegetation change.

For visualizing the effect of different soil parameters i wanted to keep it simple and use a forest plot, but it turned out to be paradoxal. I saw in several ecological papers how researchers presented the results of their models using forest plots and found it an elegant and seemingly simple way to present results. I needed to scale my covariates however, because as R stated: "the predictor variables are on very different scales". The model looks fine, but how do interpret the results of my model with scaled variables?

Below is the result of the linear mixed model using a forest plot to visualize results.

Model: Vegetation change expressed as dissimilarity commposition between 2001 and 2018~change in total N + change in total P + change in PO4 (etc.) + (1|block/plot) +(1|Year)

Forest plot of scaled coefficients from Linear mixed model

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  • $\begingroup$ An exact answer would depend on how you rescaled your variables (did you standardize i.e. subtract the mean and divide by standard deviation?). Could you add that information. Also, this answer may be useful background for you: stats.stackexchange.com/questions/29781/… $\endgroup$
    – mkt
    Commented May 13, 2019 at 8:36
  • $\begingroup$ I used Z-score (scale function in R) to normalize the dataset. Thanks for the link! $\endgroup$
    – Stevestingray
    Commented May 13, 2019 at 9:11
  • $\begingroup$ My dependant variable is vegetation change. My independant variables are the covariates (as i was in the believe these are the same in this case). $\endgroup$
    – Stevestingray
    Commented May 13, 2019 at 10:30

1 Answer 1

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Noting for readers who might have missed it that you standardized (i.e. rescaled by z-score) only the predictors and not your response variable.

The linear model coefficients can be interpreted as the change in the response (i.e. dependent variable) for a 1 standard deviation increase in the predictor (i.e. independent variable). In your case, for example: a 1 standard deviation increase in Total N in soil is associated with a decrease (because of the negative coefficient value) in the vegetation change index by ~0.03 units.

Suppose instead that you had standardized all your data i.e. both predictors and the response variable. In that case, the coefficients could be interpreted as the change in the response variable (in standard deviations) for a 1 standard deviation change in the predictor.

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    $\begingroup$ Thanks so much for the explanation! I defined my model better in the top. Unfortunately, I did not normalize the dependant variable (vegetation change). I will run a new model with the normalized vegetation scores. $\endgroup$
    – Stevestingray
    Commented May 13, 2019 at 10:28
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    $\begingroup$ @TomvanHeusden No need to do that! I've edited my answer to explain things based on your comment here. $\endgroup$
    – mkt
    Commented May 13, 2019 at 10:34
  • $\begingroup$ Thanks a lot for your explanation! One last question if thats OK: I can't really wrap my head around the interpretation. I know what it means now, but i can't imagine what the effect is in terms of standard deviations. If i understand correctly: A one standard deviation change is not an in- or decrease in the value of the predictor? How does a 'one standard deviation' relate to the original values? What does it mean ín the field?'. Sorry if it's is a simple question, but i cant figure it out. $\endgroup$
    – Stevestingray
    Commented May 13, 2019 at 11:32
  • $\begingroup$ @TomvanHeusden No problem, I should have been clearer: instead of change, I should have said increase. I've edited the answer to correct that. $\endgroup$
    – mkt
    Commented May 13, 2019 at 11:53
  • $\begingroup$ @TomvanHeusden How much that is really depends on the values in your dataset, since the standard deviation is calculated from the data; this is unlike unstandardized coefficients which are more easy to apply outside the context of your data. But you can also transform the values back onto the original scale. $\endgroup$
    – mkt
    Commented May 13, 2019 at 11:54

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