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 *** --- 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 `

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#### 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

$$eqalign{ mathbb{E}(Y) &= mathbb{E}left(2.1014 D – 3.0147 + varepsilonright)^2 \ &= (2.1014D – 3.0147)^2 + 2(2.1014D – 3.0147) mathbb{E}(varepsilon) + mathbb{E}(varepsilon^2) \ &=(2.1014D – 3.0147)^2 + 0 + (16.28)^2. }$$

(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

$$log(1.01 d) = log(1.01) + log(d) = left(0.01 – (0.01)^2/2 + cdotsright) + D approx 0.01 + D.$$

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.

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