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I have a sample size of 30 subjects, divided into three groups of 10 each. all the subjects have performed a repetition test, and each participant has a score out of 100. My main aim is to compare performance differences across the three groups using this final repetition score. To decide whether I should use parametric or non-parametric analysis, I need to check the normality of distribution of the score.

When I conducted shapiro-wilk test to test the distribution for each group, and I got(p>0.05) for all the groups. I also conducted a one-sample ks.test() for each group and the results were (p<0.01). I am confused, which one should I follow? which one of the results is more valid? taking into consideration my small sample size. below are the histograms and qqnorm plots of the three groups Histrograms and qqplots

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    $\begingroup$ It's literally impossible for this score to be normally distributed because its support is bounded. Testing this fact is pointless. $\endgroup$
    – Chris Haug
    Jul 22, 2017 at 14:30
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    $\begingroup$ See here, in particular Harvey's answer. $\endgroup$
    – Glen_b
    Jul 23, 2017 at 6:05
  • $\begingroup$ The size of your data sample is far too small for any meaningful conclusions to be drawn -- it doesn't matter which statistical tests you use to analyze it, any conclusions you draw will not be generalizable. $\endgroup$
    – Chill2Macht
    Jul 23, 2017 at 10:12

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Many people are highly skeptical of the K-S test these days, and put more stock in the S-W test. However, I am a bit confused by what you are saying: the K-S test typically has lower power to detect deviations from normality than the S-W test (which would one lead to expect the opposite result from what you write). Yet, in your case you might just examine whether results are robust either way, i.e. regardless of whether you use a parametric or non-parametric procedure to examine between-group differences.

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  • $\begingroup$ the KS is not lower power than SW against every alternative, nor is it the case when the SW has better power that will the p-value be lower on every possible sample you can get with that alternative. $\endgroup$
    – Glen_b
    Jul 23, 2017 at 6:05

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