5 ms·

You don't even need the graphs, miles driven per year is a sum. Sums converge to gaussians yet here the deviation was half of mean which means they used a unifo

You don't even need the graphs, miles driven per year is a sum. Sums converge to gaussians yet here the deviation was half of mean which means they used a uniform variable generator for miles.

throwthere 5y ago
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.
deleted 5y ago
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voldacar 5y ago
Taleb pointed this out on twitter iirc
thaumasiotes 5y ago
> yet here the deviation was half of mean which means they used a uniform variable What am I missing? The mean of a normal distribution can be any value. A normal with mean 2 and deviation 1 has deviation equal to half of the mean. Is this related to the fact that miles driven can't be negative (unlike a sample from a genuine normal)?
- CuriouslyC 5y ago
  I think this is an overzealous statement of the central limit theorem. A log-normal distribution would probably be a better model in this case.

You don't even need the graphs, miles driven per year is a sum. Sums converge to gaussians yet here the deviation was half of mean which means they used a uniform variable generator for miles.

throwthere 5y ago

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.

deleted 5y ago

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voldacar 5y ago

Taleb pointed this out on twitter iirc

thaumasiotes 5y ago

> yet here the deviation was half of mean which means they used a uniform variable What am I missing? The mean of a normal distribution can be any value. A normal with mean 2 and deviation 1 has deviation equal to half of the mean. Is this related to the fact that miles driven can't be negative (unlike a sample from a genuine normal)?

CuriouslyC 5y ago
I think this is an overzealous statement of the central limit theorem. A log-normal distribution would probably be a better model in this case.