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To correct for a left-skewed distribution I have squared my dependent variable in my linear regression. I was wondering how this affects how I can interpret it?

I was also wondering how I would interpret a regression where the dependent variable had been cubed?

Many thanks,

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If you decide to model the transformation of a variable, then you can simply undo the transformation once you have the model.

For example, if you model $z = y^2$ as $$z = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \cdots + \beta_k x_k + \epsilon$$ then you can replace the $y$ variable and solve for it: $$\begin{align}y^2 & = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \cdots + \beta_k x_k + \epsilon\\ y & = \sqrt{\beta_0 + \beta_1 x_1 + \beta_2 x_2 + \cdots + \beta_k x_k + \epsilon} \end{align}$$ Of course, this often requires a number of restrictions (like the radicand can't be negative or the transformation must be one-to-one). However, if all works well, you will just end up with a non-linear function relating the independent variables to the dependent variable.

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  • $\begingroup$ Hi, Thank you very much for your help! I was wondering whether you would be able to help me understand how to interpret each coefficient in this case? For example, after running the regression, if the coefficient of one independent variable is 10, another is 15 and the third is 20, how do I understand what the influence of each of these has on the dependent variable? Should I use predicted probabilities? $\endgroup$
    – I12345
    Apr 22, 2018 at 17:48
  • $\begingroup$ Because the model is no longer linear, the discussion of partial slopes becomes challenging. You could convert the initial model (using $y^2$ to standardized values so the resulting coefficients are at least in standard units. This at least will give you a general idea of which value has the most impact on the dependent variable the further those values are from that variable's mean. $\endgroup$
    – Gregg H
    Apr 22, 2018 at 17:53

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