If I had a response variable that was square-root transformed, and an explanatory variable that is log transformed, and I wished to back transform the model using the summary statistics below, such that Y ~ (X)^2, how would I interpret the meaning of the relationship between X and Y using the estimated Beta coefficient?
I thought it was interpreted as: "If there is a 1% increase in X, there is approximately a change of sqrt(2.1014))/100 units increase in Y."
Residuals:
Min 1Q Median 3Q Max
-37.051 -12.096 -4.908 9.701 68.071
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -3.0147 2.0827 -1.448 0.148
Dose.Back 2.1014 0.1679 12.514 <2e-16 ***
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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 16.28 on 1154 degrees of freedom
Multiple R-squared: 0.1195, Adjusted R-squared: 0.1187
F-statistic: 156.6 on 1 and 1154 DF, p-value: < 2.2e-16
Best Answer
Your model and its estimates posit that
$$\sqrt{Y} = 2.1014 D - 3.0147 + \varepsilon$$
where $D$ is
Dose.Back(or its logarithm) and $\varepsilon$ is a random variable of zero expectation whose standard deviation is approximately $16.28.$ Squaring both sides gives$$Y = (2.1014 D - 3.0147 + \varepsilon)^2.$$
Adding $0.01$ to $D$ yields the value
$$(2.1014 (D + 0.01) - 3.0147 + \varepsilon')^2.$$
The difference is
$$2(2.1014 D - 3.0147 + \varepsilon)(\varepsilon' - \varepsilon + (0.01)(2.1014)) + (\varepsilon' - \varepsilon + (0.01)(2.1014))^2.$$
This expression, as well as its expectation, are complicated. Let us therefore focus on the simpler question of how the expectation of $Y$ varies with $D$. Note that
$$\eqalign{ \mathbb{E}(Y) &= \mathbb{E}\left(2.1014 D - 3.0147 + \varepsilon\right)^2 \\ &= (2.1014D - 3.0147)^2 + 2(2.1014D - 3.0147) \mathbb{E}(\varepsilon) + \mathbb{E}(\varepsilon^2) \\ &=(2.1014D - 3.0147)^2 + 0 + (16.28)^2. }$$
(This result is of considerable interest in its own right because it reveals the role played by the mean squared error in interpreting the relationship between $D$ and $Y$.)
When $0.01$ is added to $D$ the value of $\mathbb{E}(Y)$ increases by
$$2(2.1014)(2.1014D - 3.0147)(0.01) + 2.1014(0.01)^2.$$
The last term $2.1014(0.01)^2 \approx 0.0002$ is so small compared to the squared errors (with their typical value of $16.28$) that we may neglect it. In this case, to a good approximation, this fitted model associates an (additive) increase in $D$ of $0.01$ with an increase in $Y$ of
$$2(2.1014)(2.1014D - 3.0147)(0.01) = 0.0883176 D - 0.126.$$
When $D$ is the natural logarithm of some quantity $d$, a 1% multiplicative increase in $d$ causes a value of approximately $0.01$ to be added to $D$, because
$$\log(1.01 d) = \log(1.01) + \log(d) = \left(0.01 - (0.01)^2/2 + \cdots\right) + D \approx 0.01 + D.$$
If you used a logarithm to another base $b$, entailing $D = \log_b(d) = \log(d)/\log(b),$ then a 1% multiplicative increase in $d$ causes a value of approximately $(0.01)/\log(b)$ to be added to $D$, so everywhere "$0.01$" occurs in the preceding formulas you must use $(0.01/\log(b))$ instead.