@benso8 You say: "Using the mean for missing values is not ALWAYS a bad thing." so can you give a realistic example where this approach would be a neutral or a good thing?

@Aksakal Perhaps we're reading things differently, but from the question linked to: "The second [interpretation] is that the AUC of a classifier is equal to the probability that the classifier will rank a randomly chosen positive example higher than a randomly chosen negative example, i.e. P(score(x+)>score(x−))." which is what I was (hoping I was) saying above. "90% of the time" has a standard frequentist interpretation here. The answer that says "AUC is proportional to the number of correctly ordered pairs" is saying the same thing. The proportion of correctly ordered pairs is the AUC.

@Aksakal A proof, more complex than was asked for, of this relationship is given here

@RobertLong I'd also worry that fitting separate models inflates Type I error rates (c.f. normality testing for deciding between parametric and non-parametric tests). We might want a variable to be random rather than fixed but feel that with only two or three levels this presents issues, but even there I wouldn't look at both as equally valid choices.

@RobertLong The question of whether "test" should be a fixed or a random effect here starts with the question of whether this is the entire population of tests of interest (which is plausible given the information above, c.f. comparing two countries) or whether this is a sample of possible tests (which is also plausible, c.f. a sample of countries). This isn't a heuristic here any more than it would be when asking whether ethnicity or country, etc. would a random sample of levels or not. The question of whether there are sufficient levels is only relevant under the second interpretation.

I feel that a step might be missing here, namely are the tests a sample from a population of potential tests or are they the entire population of tests of interest? This point wasn't clear in the question for me. An answer in favour of the second option here would make any question around the number of levels required for random factors moot as test would then be a fixed factor. Personally, I wouldn't recommend looking at this with test as both a random and as a fixed factor as these start from different views of the research question.

Transformations will change the question, e.g. a log transformation will make this a comparison of geometric means. If the departures from normality and homoscedasticity of residuals are concerning and you really want to explore differences in (arithmetic) means, bootstrapping would be the obvious suggestion to me assuming you want the trinity of effect size, 95% CI, and p-value for interpretation and reporting.

If your hypothesis is definitely about differences in (arithmetic) means, that would rule out some obvious suggestions here, such as Kruskal-Wallis (which for the similar shapes across the groups you note would be an approximate test for differences in medians, with Dunn's test for the post-hoc comparisons--since you're using Stata [note the capitalisation, it's not an abbreviation], "net search dunntest" would find a user-written program for this).

This will depend on what you want (point estimates, well-behaved p-values and CIs) and how flexible you are willing to be about exactly what you are comparing (arithmetic means, geometric means, medians, location more generally).

If you regressed the distance people could throw a heavy object based on the ambient temperature, the slopes using "metres per °C", "inches per °F", "miles per K", and similar will all have the same meaning (and the same p-value will be calculated for each slope). In other words, you're correct that the (linear) scaling of any variable doesn't affect the equation in terms of its meaning. I can't think of any reason for such scaling beyond numerical stability or interpretation.

You can, if this makes sense to you, simply scale the variable to tens or hundreds of workers. This is the same as changing from millimeters to centimeters or from centimeters to meters respectively. Using "per hundred workers", 5 becomes 0.05, 5000 becomes 50.This will change an OR of exactly 1.01 per worker to 1.01^100=2.70 per 100 workers, for example.