Choosing between and properly interpreting Analysis of Variance vs Wilcoxon rank sum test (Newb on this, so please assume good faith but low skill)
I have this set of data where I'm interested in whether there is a significant and meaningful difference in consumption (measured in local currency) per person based on whether the person didread a piece of information.
> summary(df)
 didread    consumption      
 0:20295   Min.   :    0.00  
 1: 5518   1st Qu.:   40.19  
           Median :  108.42  
           Mean   :  194.36  
           3rd Qu.:  227.23  
           Max.   :13245.55
> sd(df$consumption)
[1] 330.9634

The distribution of consumption is highly skewed. The density chart of log10(consumption + 1), grouped by didread shows

Now, an analysis of variance shows
> aov.out <- aov(log10(consumption + 1) ~ didread, data=df)
> summary(aov.out)
               Df Sum Sq Mean Sq F value   Pr(>F)    
didread         1     19  19.375   31.99 1.56e-08 ***
Residuals   25811  15630   0.606                     
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> TukeyHSD(aov.out)
  Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = log10(consumption + 1) ~ didread, data = df)

$didread
           diff         lwr         upr p adj
1-0 -0.06682713 -0.08998316 -0.04367109     0

Also, trying a Wilcoxon rank sum test:
> with(df, wilcox.test(log10(consumption + 1) ~ didread, conf.int=TRUE))

    Wilcoxon rank sum test with continuity correction

data:  log10(consumption + 1) by didread
W = 60112000, p-value < 2.2e-16
alternative hypothesis: true location shift is not equal to 0
95 percent confidence interval:
 0.04743923 0.08262361
sample estimates:
difference in location 
            0.06515087

Now for my questions:


*

*Am I in the right here in interpreting this as that both of these tests say that the individuals in the didread == 1 group consumed 1 - exp(-0.066) = 6% less than didread == 0?

*Which of the tests (aov vs wilcox.test) is more appropriate given these distributions? If I would like to report confidence intervals, which should I use?

*Are there any post-hoc tests I should consult in whether these results are meaningful, or is that a business decision from this point on?

*Anything else I should consider in this, to not make a fool out of myself?


Edit 1: Venturing further into new territory; attempting to answer the "distribution of residuals" questions
Is this what @whuber is asking about?
df.lm <- with(subset(df, consumption > 0), lm(log10(consumption) ~ didread))
summary(df.lm)
data.stdres=rstandard(df.lm)
qqnorm(data.stdres)
qqline(data.stdres)


 A: The interpretation is a percent change, but not exactly what you gave. The tests give the difference between $didread_0$ and $didread_1$ so the model is similar to (pardon the notation please)
$$
log_{10}(consumption+1) = didread_0 + difference\\
consumption+1 = 10^{didread_0}10^{difference}
$$
The percent change of $(consumption+1)$ is 
$10^{difference}=10^{-0.06682713}=0.8573791$. 
This means that $didread_1$ is a 14.3% drop from $didread_0$. You can check this by comparing mean(df[df$didread==1,]$consumption) and mean(df[df$didread==0,]$consumption) * 10^(-0.06682713) which should be almost equal.
For the Wilcoxon test, it looks like the reference level was switched to $didread_1$. So there $didread_0$ has a $10^{0.06515087}=1.16$ or 16% increase from $didread_1$.
For the ANOVA test, the QQPlot is helpful and you can see some problems in one tail of the plot. There are also other residual plots that you can check for problems. For the Wilcoxon test, it assumes independence of samples and that the only difference in (consumption) distributions is a shift in the means. This looks to be the case except for near zero where $didread_0$ is now left of $didread_1$ instead of right of it. With slight problems in both tests' assumptions, I don't see a clear winner. Personally, I think the Wilcoxon test's assumptions are better met and would go with that one.
When computing the confidence intervals, don't forget to transform properly. The 95% confidence interval for ANOVA would be 
$
\Big[\big(1-10^{-0.08998316}\big),\big(1-10^{-0.04367109}\big)\Big] = [0.187, 0.096]
$
so $didread_1$ is 9.6% to 18.7% less consumption than $didread_0$. Again the Wilcoxon test is similar but in the reverse direction.
$
\Big[10^{0.04743923},10^{0.08262361}\Big] = [1.11, 1.21]
$
so $didread_0$ is 11% to 21% more consumption than $didread_1$
