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Re: kolmogorov_smirnov_test


From: Olli Saarela
Subject: Re: kolmogorov_smirnov_test
Date: Wed, 11 Jun 2008 19:33:33 +0300
User-agent: Thunderbird 2.0.0.14 (Windows/20080421)

But still I am very suspicious about this test.

When I run
p = kolmogorov_smirnov_test(rndnorm, "normal", 0.47977, 0.34159)
I get p=0 and it is ok with me.
But when I change mean and var to any random numbers I get exactly the
same result!!!

In statistical testing, the more data points you have, the smaller deviations are found statistically significant. Earlier you wrote the rndnorm vector has 1e5 values that are most likely normally distributed. This is such a large number, that even relatively small changes in the distribution are statistically significant.

Simulating your case:

> x = sqrt(0.34159) * randn(1e5, 1) + 0.47977;
> kolmogorov_smirnov_test(x, "normal", 0.47977, 0.34159);
pval: 0.958304
> kolmogorov_smirnov_test(x, "normal", 0.475, 0.34159);
pval: 0.0318814
> kolmogorov_smirnov_test(x, "normal", 0.47, 0.34159);
pval: 1.27959e-005

Unless your "random mean and var values" are both very close to the actual values in the data AND the distribution of the data is very close to normal, the answer p=0 is a good approximation.

Cheers,
  Olli



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