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I have the following multiple linear regression model: Log(y) = B0 + B1X1 + B2X2 + B3x3 + e. X1 is a dummy that can take 0 = male and 1 = female and X2 and X3 are continuous variables.

I am not entirely sure on how to interpret the coefficients for the variables. The coefficient for the dummy variable is 0,20. Does that mean, that changing from male to female (male is baseline) the Y will increase by an average of 20%. Is it directly translated into percentage?

And for the continuous variables, the coefficient for X2 is 0,1. Does that mean that increasing X2 with 1 unit increases Y with an average of 10%? Again, is it directly translated into percentage?


marked as duplicate by COOLSerdash, Richard Hardy, Siong Thye Goh, kjetil b halvorsen, Michael Chernick Apr 27 at 20:49

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Positive coefficients somehow indicate a positive effect, but they don't simply turn into percentages. There is a transformation. Let's say your model is $\log y = b_0+b_1x_1$; this means $y=e^{b_0+b_1x_1}=A_0e^{b_1x_1}$. So, dummy or not, if $x_1$ increases by $1$ unit, $y$ increases by $e^{b_1}$, i.e. if $b_1=0.2$, $y$ increases by $e^{0.2}\approx 1.22$, i.e. $22\%$. The case is similar for your continuous variable.

  • $\begingroup$ I have read that if the natural log is used it is approximately translated into a percentage change, if the change in x is small. Is that correct? $\endgroup$ – maS Apr 27 at 15:45
  • 1
    $\begingroup$ Yes, but approximate is the key term here, because Taylor expansion of $e^x$ is: $e^x=1+x+x^2/2!,...\approx 1+x$ when $x$ is small. Which means $x\%$ increase in something. Here, to execute this idea, your coefficients should be small. $\endgroup$ – gunes Apr 27 at 15:47
  • $\begingroup$ Thanks gunes. Would you recommend using natural log then as my log transformation or what base should i use? cant seem to find a solid explanation of the choice $\endgroup$ – maS Apr 27 at 16:08
  • $\begingroup$ Also: inference on transformed variables $\ne$ inference on un-transformed variables. $\endgroup$ – Alexis Apr 27 at 16:57
  • $\begingroup$ @maS using base N results in model $y=e^{b_0\log N+b_1\log N x_1}$, where you solve for $a_i=b_i\log N$. It's one to one. $\endgroup$ – gunes Apr 27 at 17:36

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