# "This is a lower bound of true significance." in Kolmogorov-Smirnov. WHat does it mean?

I' conducting a Normality Test for a numerical variable (level of a substance in blood) in a sample of 368 subjects.
As I will need to run analysis of covariance (ANCOVA) and partial correlations, I find myself a little bit in doubt with this result. According to the p value of the Kolmogorov-Smirnov (0.200), my variable has a normal distibution (while according for Shapiro-Wilk it hasn't). Anyway, I'm wondering what "This is a lower bound of true significance" means, referring to the p = 0.200. Should I care about it? To note, histogram, skewness and kurtosis suggest a normal distribution of the data. Outliers have been excluded with the z score method (>3). If I transform the values in log10, the distribution results non/normal.

• Part of this is a FAQ: ANOVA does not care how your response variable is distributed. Thus, although your question about "lower bound" has a (simple) answer, it might not be helpful to you.
– whuber
Dec 4, 2018 at 17:40
• In addition "According to the p value of the Kolmogorov-Smirnov (0.200), my variable has a normal distibution" is not a correct interpretation of a failure to reject the null". Dec 5, 2018 at 12:23
• Make sure that you understand the meaning of p-values. A non-significant p-value doesn't tell you that the null hypothesis is true. It tells you that you have failed to find evidence that it is false. It's still probably false. Aug 29, 2021 at 3:32

The Kolmogorov-Smirnov test is a test of the null hypothesis that your data come from some prespecified distribution, $$F$$. Coming from a Normal distribution is a much wider null hypothesis; there are lots of Normal distributions.
At one time it was common to find the best-fitting Normal distribution to your data (by estimating the mean and variance from the data), then pretend that was a prespecified distribution and do a Kolmogorov-Smirnov test. It's pretty clear that you will be less likely to reject the null that way -- the fitted normal distribution will fit your data better, because it has been fitted to your data. So, if you take the test statistic $$D$$ and your observed value $$d$$, and calculate the $$p$$-value $$P(D\geq d)$$ as if you had a prespecified distribution, the number you get will be too large. It will be larger than the true \$P(D>d) for the testing procedure you actually used.
Lilliefors's test fixes this up. It uses the same test statistic $$D$$ (the maximum vertical difference between cumulative distribution functions), but it calculates the null distribution of $$D$$ taking account that you have estimated the mean and variance from the same data. How does it do that? By brute-force simulation. Someone (Hubert Lilliefors, or perhaps his research assistants) generated lots of sets of data from Normal distributions, did the KS test on each set, and made a big table of the null sampling distribution. They did this in 1967 when it was a big deal; you could do the same thing in a few minutes now with R but we've gotten used to using the stored tables.