Statistical comparison of 2 small data sets for 2X increase in the population mean
Yes, in principle, resampling can help answer this question. 

incumbent &lt;- c(7.3, 8.4, 8.4, 8.5, 8.7, 9.1, 9.8, 11.0, 11.1, 11.9)
proposed  &lt;- c(17.3, 17.9, 19.2, 20.3, 20.5, 20.6, 21.1, 21.2, 21.3, 21.7)

set.seed(42)

M  &lt;- 2000
rs &lt;- double(M)

for (i in 1:M) {
    rs[i] &lt;- mean(sample(proposed, replace=T)) - 2 * mean(sample(incumbent, replace=T))
}


To make the assessment, you should choose one (not both) of the following:  

A. The (two-tailed) 90% confidence interval for the difference in the (weighted) means using Hall's method is:

ci.hall &lt;- 2 * (mean(proposed)-2*mean(incumbent)) - rev(quantile(rs,prob=c(0.05, 0.95)))
names(ci.hall) &lt;- rev(names(ci.hall))
ci.hall

   5%   95% 
-0.29  2.95 


This is appropriate if you have any concern about missing the possibility that mean(proposed) might actually be less than 2 * mean(incumbent).

B. The proportion of resample means >= 0 provides the (one-tailed) estimate that mean(proposed) is at least twice mean(incumbent):

sum(rs&gt;=0)/M

[1] 0.8915


The problem is that the samples are really rather small and resampling estimates
can be unstable for small n. The same issue applies if you want to assess normality and go with parametric comparisons. 

If you can get to, say, n >= 30, the approach described here should be fine.