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When testing for normality and homogeneity of variance in SPSS, it showed this:

enter image description here

If I go by Kolmogorov-Smirnov, than the 'M' data is not normal, but if I go by Shapiro-Wilk, they all are normally distributed. However, the test of homogeneity of variance shows that based on the Mean and based on the trimmed mean, equal variance is not assumed. I know this would change which Post-hoc test I use for BG Anova but I'm not sure how it would affect my data if I use a Kruskal-Wallis Anova?

Should I use a Kruskal-Wallis ANOVA or a 1-way Between Groups ANOVA?

Thank you!

h77
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  • Failure to reject does not imply that the null is true (you cannot reasonably conclude "they all are normally distributed"; you don't have 100% power at every alternative). 2. Please see the following thread: https://stats.stackexchange.com/questions/2492/is-normality-testing-essentially-useless -- several good answers there; Harvey Motulsky's is a good starting point.
  • – Glen_b Apr 03 '20 at 02:09
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    Please do not make such huge changes in your post that they render other people's posts in this thread incomprehensible. – whuber Apr 20 '20 at 14:49