How to interpret regression coefficients when outcome variable was transformed to $1 / \sqrt{Y}$? In OLS regression, what does a one unit change in X represent in terms of a beta-related change in Y when the outcome variable has been transformed to $1/\sqrt(Y)$? 
(Transformation was done to ensure linearity in the model. My predictor variables are 0-1 dummies.)
I've consulted the following useful posts, but I still don't know what is the answer:
https://stats.stackexchange.com/a/2148
Back-transformation of regression coefficients
How to interpret regression coefficients when response was transformed by the 4th root?
 A: The model is
$$\frac{1}{\sqrt{Y}} = \beta_0 + \beta_1 X_1 + \cdots + \beta_p X_p + \varepsilon$$
where $Y$ is the original outcome, the $X_i$ are the explanatory variables, the $\beta_i$ are the coefficients, and $\varepsilon$ are iid, mean-zero error terms.  Writing $b_i$ for the estimated value of $\beta_i$, we see that a one-unit change in $X_i$ adds $b_i$ to the right hand side.  Starting from any baseline set of values $(x_1, \ldots, x_p)$, this induces a change in predicted values from $\widehat{1/\sqrt{y}} = b_0 + b_1 x_1 + \cdots + b_p x_p$ to $\widehat{1/\sqrt{y'}} = b_0 + b_1 x_1 + \cdots + b_p x_p + b_i$.  Subtracting the first equation from the second gives
$$\frac{1}{\sqrt{\hat{y'}}} - \frac{1}{\sqrt{\hat{y}}} = b_i.$$
Solving for $\hat{y'}$ gives
$$\hat{y'} = \frac{\hat{y}}{1 + 2b_i\sqrt{\hat{y}} + b_i^2 \hat{y}}.$$
One may stop here, but often we seek simpler expressions: the behavior of this one might not be any easier to understand than the original model.  Simplification can be achieved provided  $b_i$ is very small.  If necessary, we can contemplate a tiny change in $X_i$, say by an amount $\delta$, which would replace $b_i$ in the preceding equation by $\delta b_i$.  Using a sufficiently small value of $\delta$ will assure the denominator is close to $1$.  When it is,
$$\frac{\hat{y}}{1 + 2\delta b_i\sqrt{\hat{y}} + \delta^2 b_i^2 \hat{y}} \approx \hat{y}(1 - 2\delta b_i\sqrt{\hat{y}} - \delta^2 b_i^2 \hat{y}),$$
whence the change in predicted values is
$$\hat{y'} - \hat{y} \approx -\delta (2b_i\sqrt{\hat{y}} + \delta b_i^2 \hat{y}).$$
Taking $\delta$ to be so small that $\delta b_i^2 \hat{y} \ll 2 b_i\sqrt{\hat{y}}$ allows us to drop the second term in the right hand side.  That is, for very tiny changes, the predicted outcome changes by $-(2b_i\sqrt{\hat{y}})$ times the amount of change in $x_i$.

Comments
The appearance of the negative sign indicates that an increase in $X_i$ will decrease $Y$ when $b_i$ is positive and increase $Y$ when $b_i$ is negative.  Normally, we avoid this (potentially confusing) sign reversal by using $-1/\sqrt{Y}$ instead of $1/\sqrt{Y}$ when making a reciprocal square root transformation (or any other transformation that reverses the order of numbers).
This solution method is always applicable no matter how $Y$ is re-expressed, but it can lead to complicated algebra for other transformations of $Y$.  Those who know the basics of differential calculus will recognize that all we're doing here is approximating the change in $\hat{y}$ to first order using its derivative with respect to $x_i$, so they will be able to avoid most of the algebraic manipulations.
