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I am running a linear regression model where the dependent variable (Y) is log-transformed. I am struggling on how to interpret the adjusted R-squared of this log-transformed model that is meaningful. Any insight is very much appreciated! Thank you.

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  • $\begingroup$ After the fit, you will need to de-transform the dependent variable and model predictions, and then manually calculate. $\endgroup$ – James Phillips Oct 4 '18 at 20:15
  • $\begingroup$ @JamesPhillips Thanks for your answer. Could you please elaborate a bit more what do you mean by manual calculation and how to perform it? $\endgroup$ – curiousmind Oct 4 '18 at 20:33
  • $\begingroup$ I calculate R-squared (R2) as "R2 = 1.0 - (absolute_error_variance / dependent_data_variance)" and the R-squared adjusted (R2adj) as "R2adj = 1.0 - (1.0 - R2)*((number_of_data_points - 1.0)/(number_of_data_points-number_of_parameters))" $\endgroup$ – James Phillips Oct 4 '18 at 22:36
  • $\begingroup$ @JamesPhillips If the relationship is assumed to be linear when the outcome is logged, why should OP back-transform the predictions to calculate an $R^2$? $\endgroup$ – Heteroskedastic Jim Oct 5 '18 at 1:08
  • $\begingroup$ @HeteroskedasticJim one of the reasons is that the R-squared calculation I wrote in my comment requires variance of the dependent variable, not variance of the log(dependent variable). $\endgroup$ – James Phillips Oct 5 '18 at 10:55
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Is the model of log-transformed Y the only model you are considering? Then you can just interpret the (unadjusted) R-squared in the usual way. For example, if the R-squared is 70%, then 70% of the variability in the log-transformed values of Y is accounted for by the predictor variables included in the model.

If you are considering several competing models for the log-transformed Y, then it makes sense to compare their explanatory power via the adjusted R-squared. In that case, the model with the highest value for the adjusted R-squared would be preferred - as explained, for example, here: http://blog.minitab.com/blog/adventures-in-statistics-2/multiple-regession-analysis-use-adjusted-r-squared-and-predicted-r-squared-to-include-the-correct-number-of-variables.

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  • $\begingroup$ How then can the model's explanatory power for the variability in the non-log-transformed values of Y be calculated? $\endgroup$ – James Phillips Oct 5 '18 at 11:06
  • $\begingroup$ @JamesPhillips R-squared is a model fit statistic. By modeling the log transformed Y, you are automatically claiming that it makes sense to think of this outcome on the log scale. In fact, predictors have a better linear relationship to the outcome on this scale. To all intents and purposes, the log transformed Y is your outcome. You just happened to measure its non log transformed counterpart. $\endgroup$ – Heteroskedastic Jim Oct 5 '18 at 13:11
  • $\begingroup$ @HeteroskedasticJim My understanding of the OP's question regards the modeling of Y. $\endgroup$ – James Phillips Oct 5 '18 at 14:27
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    $\begingroup$ @Isabella Thank you for your answer. Yes, I understand that I can interpret the R-squared in the usual way except for the fact that it accounts for the variability of log-transformed Y instead of Y. However, this interpretation is a bit difficult to grasp. I was wondering if it can be interpreted which is more intuitive? For example, for this model, I am interpreting the exponent of the coefficients for better understanding. Thank you. $\endgroup$ – curiousmind Oct 5 '18 at 14:37
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    $\begingroup$ @IsabellaGhement Thank you for the suggestion. I will try that. $\endgroup$ – curiousmind Oct 9 '18 at 16:03

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