Calculation of residual standard deviation and r-squared Related questions here and here (but not answered that satisfactorily in my view).
From Gelman and Hill, Q3.2:

Suppose that, for a certain population, we can predict log earnings
from log height as follows:

*

*A person who is 66 inches tall is predicted to have earnings of    $30,000.

*Every increase of 1% in height corresponds to a predicted    increase of 0.8% in earnings.

*The earnings of approximately 95% of people fall within a factor of 1.1 of predicted values.

(a) Give the equation of the regression line and the residual standard
deviation of the regression.
(b) Suppose the standard deviation of log heights is 5% in this
population. What, then, is the R2 of the regression model described
here?

I can calculate the equation of the regression line easily enough using the data given and: $$\text{log}(y) = \alpha + \beta \, \text{log}(x)$$
where $y$ is earnings and $x$ is height.
I then took: $$\text{log}(1.1)/1.96 \approx 0.0486$$ to be the residual standard deviation.
I think this then gives $$R^2 = 1 - \frac{0.0486^2}{0.05^2} \approx 0.0541$$
which I think is wrong?
Questions: what are the correct residual standard deviation and $R^2$?
NB: the question in the book for part (b) says log heights but I think that's a typo and they meant log earnings.
 A: I think that making the error term $\epsilon$ explicit would help. In log terms we would have:
$$\log(y) = \alpha + \beta \log(x) + \epsilon.$$
Taking exponents, the actual model would look like the following:
$$ y = e^{\alpha} x^\beta e^{\epsilon}.$$
Note that whenever the question talks about a log variable we should refer to the first model and otherwise to the second. Hence, the third point:

The earnings of approximately 95% of people fall within a factor of
1.1 of predicted values.

refers to the second model, where, as you can see, the error term $e^\epsilon$ is no longer gaussian due to exponentiation, making the 95%-1.96$\sigma$ rule not valid. Instead, what the sentence is saying is that
95% percent of the values of $e^\epsilon$ are below 1.1, or equivalently that the CDF value at 0.95 is 1.1. If $\epsilon$ is normal, $e^\epsilon$ is log-normal with parameters 0 and $\sigma_\epsilon$, the latter being the one you can infer from the information about the CDF. Unfortunately, it's not straightforward to do this analytically (due to the form that the aforementioned CDF takes) which makes me doubt whether all of this is correct (either the question or my intrepretation/answer).
