# How to interpret linear regression where the dependent variable has been transformed by being sqaured?

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,

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.
• 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. Apr 22, 2018 at 17:53