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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.

Thank you in advance for your replies!

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  • $\begingroup$ 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. $\endgroup$ – whuber Dec 4 '18 at 17:40
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    $\begingroup$ 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". $\endgroup$ – Glen_b Dec 5 '18 at 12:23
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It is a lower bound. For example, 2 minutes ago, I noticed that we can test normality with lillie.test() function in R software. Then, I used my normal distributed data to get K-S and Shapiro-Wilk test results. I had the statistics from SPSS. Shapiro-Wilk stats were same. But K-S test results were different. R said that p value of K-S was 0.178 and SPSS said that p value of K-S was 0.200 (and This is a lower bound of true significance). I got that my data is normal. It's still normal but SPSS gave us a lower bound.

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