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I have problem with interpreting the OLS regression result with the dependent variable square root transformed when doing difference-in-differences analysis.

Our regression model is:
$$ Y = β_0 + β_1 {\rm policy} + β_2 {\rm treated} + β_3 {\rm policy} \times {\rm treated} + \ldots + β_i X_i + \varepsilon $$ Where ${\rm policy}$ is a dummy variable indicate the policy change (0=pre-policy vs. 1=post-policy). ${\rm treated}$ is also a dummy variable indicate treatment or control group (0=control, 1=treatment)

To deal with the skewed distribution, $Y$ was square root transformed. Just wondering how can we interpret the result? Say, after the policy change, $Y$ was increased / decreased by ____?

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marked as duplicate by whuber May 20 '15 at 19:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Let $X = \sqrt{Y}$. I am assuming that $$ X = \beta_0 + \beta_{1} \text{policy} + \beta_{2} \text{treated} + \beta_{3} \text{policy} \times \text{treated} + \dots + \beta_{i}X_{i} + \varepsilon$$

Suppose after a policy change, the expected value of $X$ increased by $c$ adjusted for the other variables.

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    $\begingroup$ If the square root was taken to normalize the residuals, it wouldn't be the average that increased by $c^2$ but the median. Also note that if the relationship were rectilinear in the original space, it would be curving in the transformed space & vice versa. $\endgroup$ – gung - Reinstate Monica May 20 '15 at 19:09
  • $\begingroup$ A thing to watch out for is inference in untransforming the outcome. Because $\sqrt{\sigma^{2}_{\beta}} \ne \sigma^{2}_{\sqrt{\beta}}$, you cannot assume that predictors found to be significant in a model of $Y^{2}$ would likewise be significant in a model of $Y$ (i.e. we would expect different values of the various t statistics for the $\beta$s in $Y^{2} \sim \mathbf{BX}$ as compared to $Y \sim \mathbf{B}\sqrt{\mathbf{X}}$). $\endgroup$ – Alexis May 20 '15 at 19:12
  • $\begingroup$ This answer suddenly seems incomplete: it does not appear to say anything. $\endgroup$ – whuber May 20 '15 at 19:19
  • $\begingroup$ Thank @gung square root was used to deal with the skewed distribution of Y. In this case, would you guy please give more deals about the interpretation? $\endgroup$ – Xiaohui Wang May 21 '15 at 1:56
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    $\begingroup$ @XiaohuiWang, I'm not sure if I follow you. You should not transform Y to normalize its distribution, only the residuals (see here). Re: how to interpret your situation, that is covered in the linked duplicate thread. $\endgroup$ – gung - Reinstate Monica May 21 '15 at 2:01

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