Return to Answer

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

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.

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

The linear model coefficients can be interpreted as the change in the independent variable for a 1 standard deviation increase in the predictor. 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 independent variable. In that case, the coefficients could be interpreted as the change in the independent variable (in standard deviations) for a 1 standard deviation change in the predictor.

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

Noting that you standardized (i.e. rescaling by z-score) only the predictors and not your independent variable.

The linear model coefficients can be interpreted as the change in the independent variable for a 1 standard deviation increase in the predictor. 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.

If you had standardized all your data i.e. both predictors and the independent variable: the coefficients could be interpreted as the change in the independent variable (in standard deviations) for a 1 standard deviation change in the predictor.

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

The linear model coefficients can be interpreted as the change in the independent variable for a 1 standard deviation increase in the predictor. 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 independent variable. In that case, the coefficients could be interpreted as the change in the independent variable (in standard deviations) for a 1 standard deviation change in the predictor.