3 ms·

Interesting point. So the original paper (2012) didn't give a table or anything, but said: "Customers who signed at the beginning on average revealed higher use

Interesting point. So the original paper (2012) didn't give a table or anything, but said: "Customers who signed at the beginning on average revealed higher use (M = 26,098.4, SD = 12,253.4) than those who signed at the end [M = 23,670.6, SD = 12,621.4; F(1, 13,485) = 128.63, P < 0.001]." Was M = 26,098.4, SD = 12,253.4 enough to infer a uniform distribution?

ImaCake 5y ago
> Was M = 26,098.4, SD = 12,253.4 enough to infer a uniform distribution? No unfortunately it is not. It's perfectly plausible to have a normal distribution with mean = 26k and sd = 12k. Although those numbers do look kinda weird, but nothing you could verify. I am not sure what the F(*) means here, maybe an F statistic? But that seems wrong if you are comparing two normally distributed samples, you might expect a T statistic. To verify the distribution you would need a histogram or you could get fancy with a qqplot. You could also try a statistical test for a normal distribution but these fail on large sample sizes so the visual plots are your best bet.