In linear regression, how to test if doubling of x leads to a doubling of log-transformed y? In a simple linear regression with a log-transformed y variable, how do I test if a doubling of the x variable leads to a doubling of the y variable?
 A: The question asks to test whether the slope equals $1$.  This can most easily be conducted by regressing $\log(y) - \log(x)$ against $\log(x),$ but can also be done by post-processing the regression summary (when the original data might not be available, for instance).

Analysis
The model is
$$\mathbb{E}[\log(y)] = \beta_0 + \beta_1\log(x).$$
Doubling $x$ to $x'$ changes $\log(x)$ to $\log(2x) = \log(2) + \log(x)$.  Applying that change to the model yields
$$\mathbb{E}[\log(y')] = \beta_0 + \beta_1(\log(x) + \log(2)) = \mathbb{E}[\log(y)] + \beta_1\log(2).$$
You would like to test whether this is reasonably close to a doubling of $\mathbb{E}[\log(y)]$, which would add $\log(2)$ to it.  Comparing, it is evident the question comes down to whether $\log(2) = \beta_1\log(2)$; that is, we need to test whether $\beta_1=1$.
The usual output of a linear regression only tests whether the coefficients are zero, not $1.$  Here is sample output from such a regression:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   1.0817     0.2029   5.331 9.16e-06 ***
x.log         0.4125     0.2636   1.565    0.128   

The second line reports that the estimate of $\beta_1$ is $\hat{\beta}_1 = 0.4125$ with a standard error of $0.2636$, a t value of $1.565 = 0.4135/0.2536,$ and a p-value of $0.128$.  The relatively large p-value only tells us not to reject the null hypothesis that $\beta_1=0$: it has no direct bearing on whether $\beta_1=1$.
Solution
There are (at least) two ways to fix that.


*

*An easy solution is to regress $\log(y) - \log(x)$ against $\log(x)$.  That is because
$$\mathbb{E}[\log(y)-\log(x)] = \beta_0 + \beta_1\log(x) - \log(x) = \beta_0 + (\beta_1-1)\log(x).$$
The new slope is $\beta_1-1,$ so comparing that to $0$ is tantamount to comparing $\beta_1$ to $1$.  For example, here is the new output:
                Estimate Std. Error t value Pr(>|t|)    
(Intercept)   1.0817     0.2029   5.331 9.16e-06 ***
x.log        -0.5875     0.2636  -2.228   0.0335 * 

Indeed, $\hat{\beta}_1-1$ = $-0.5875 = 0.4125 - 1$ as expected.  The standard error does not change, but this time the t value is $(\hat{\beta}_1 - 1)/se(\hat{\beta}_1) = -2.228.$  The corresponding p value of $0.0335$ suggests there may be significant evidence that $\beta_1-1\ne 0$; that is, that $\beta_1 \ne 1.$ 

*You can post process the original results.  Returning to the first output, we may directly compute $(\hat{\beta}_1 - 1)/se(\hat{\beta}_1) = (0.4125 - 1)/0.2636 = -2.228.$  This is converted to a p-value using a two-sided area under the Student t curve with the appropriate degrees of freedom ($30$ in this case, because there were $32$ data values minus two estimated parameters).  The p-value is $0.03350966,$ agreeing with the previous output. 
