高手你好，
     我大概看了下这篇文献，在这篇文献的第八页，给出了各种方法的样本数范围，比如说Skewness-Kurtosis需要大于9，Shapiro-Wilk小于2000，Shapiro-Francia小于5000。Kolmogorov-Smirnov没有，就是说范围不限制了吧？在stata中help normal还可以找到Kolmogorov-Smirnov检验的公式。可在这篇文献中又有这样的一句话：
Stata .ksmirnov command is not used for testing normality.
所以我想问问您是怎么理解的？Kolmogorov-Smirnov应该还是可以进行正态性检验的吧？还是说它属于类似正态性检验？

thanks

Hi,

Kolmogorov-Smirnov test was designed to compare two experimentally-determined distributions. When testing for normality, you compare an experimental distribution against a hypothetical ideal (Suppose the true normal distribution with mean mu and variance sigma, but both parameter mu and sigma you don't know), if you use Stata .ksmirnov, you may test it against normal like the example commands in (help ksmirnov),
webuse ksxmpl
summarize x
ksmirnov x = normal((x-r(mean))/r(sd))
which is using the sample mean r(mean) and the sample variance r(sd)^2 as the true parameters. By doing that the result of KS test will be conservative in the sense that it will be less likely to reject the null hypothesis, then you need to apply the Lilliefors correction which is not being done in .ksmirnov... In conclusion, Stata .ksmirnov command is not ready and so useful for testing normality. I recommend that you skip using the Kolmogorov-Smirnov test, and switch instead to the other alternative tests. But first rethink whether the normality tests are providing you with useful information.
      

Thanks， then if N>5000, which normality tests is better, ksmirnov(for one sample or two sample) or sktest? And could the test of ksmirnov apply to many more samples?