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I read in this link: enter link description here

It said: Outliers tend to increase the estimate of sample variance, thus decreasing the calculated F statistic for the ANOVA and lowering the chance of rejecting the null hypothesis.

I am very confused.

Null Hypothesis: Data is normally distributed. So, if a data has outliers, is not normally distributed, fails to reject the null hypothesis and increases the chances of rejecting the null hypothesis.

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    $\begingroup$ The null hypothesis you have stated is not the correct one for a one-way anova, which is what the cited article is talking about. $\endgroup$ – Sal Mangiafico Oct 23 '19 at 10:47
  • $\begingroup$ @SalMangiafico. Thanks. Sorry, is the null hypothesis for One-way ANOVA assumption is The null hypothesis is that data is normally distributed and the alternative hypothesis is that the data is not normally distributed.? Is very confusing $\endgroup$ – Kew Hsein Oct 23 '19 at 11:18
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    $\begingroup$ No, a one-way anova tests the null hypothesis that the means of all groups are equal. $\endgroup$ – Sal Mangiafico Oct 23 '19 at 11:40
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    $\begingroup$ The null hypothesis in ANOVA is that a bunch of means are equal. If you are quoting some source as saying otherwise, the source is incompetent and indeed likely to be (very) confusing. Nothing rules out a normal distribution producing outliers, as a great deal depends on what is an outlier, and a normal distribution in principle has infinite range and there is a positive probability of any value whatsoever occurring! $\endgroup$ – Nick Cox Oct 23 '19 at 11:41
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    $\begingroup$ Not correct. Absence of outliers no more implies normality than presence of outliers implies the opposite. There are many flavours of boxplots but assuming that you are talking about points more than 1.5 IQR from the nearer quartile then their existence is consistent with many distributions. $\endgroup$ – Nick Cox Oct 23 '19 at 12:53
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The idea your link is trying to get across is that normality is a necessary assumption for ANOVA to work. Departures from normality may cause outliers, which would hinder your ability to draw conclusions based on that ANOVA.

The presence of outliers isn't in itself conclusive to say whether your data is normally distributed or not. If you're unsure, make a Q-Q plot of your data.

Additionally, one or two outliers probably isn't enough to throw your normality assumption off. But if half of your points are outliers, then your data likely isn't normally distributed.

EDIT: As Nick Cox rightly points out, normality (or lacktherof) isn't the death of an ANOVA analysis (thanks to the ability to transform variables and such). Just be careful.

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    $\begingroup$ If that idea is what the link is trying to convey then the idea is exaggerated. First, everyone has to be clear that normality here is even ideally conditional on the systematic structure implied by the model. It's not about the marginal normality of the outcome variable. Second, there is much flexibility on many levels, say in applying ANOVA on some transformed scale. Otherwise put: your second and third paragraphs convey good points, but the first paragraph appears complicit in a common exaggeration. $\endgroup$ – Nick Cox Oct 23 '19 at 12:59
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    $\begingroup$ @GridAlien, Thanks. I look at this video youtube.com/watch?v=4JMLdDu1PJg at 5.40 minutes. It said that no outliners in boxplot (absence of outliers), meet the assumption of absence of outliers. $\endgroup$ – Kew Hsein Oct 23 '19 at 12:59
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    $\begingroup$ @Kew Hsein I guess English is not your first language, which is fine by me, but either way looking for outliers is not an assumption; it is just part of exploring data thoroughly that should be part of any analysis. $\endgroup$ – Nick Cox Oct 23 '19 at 13:18
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    $\begingroup$ Sorry, but life is short and I won't watch videos or read links to answer comments. The Laerd site does not receive high regard from experienced members here. Find a good textbook! $\endgroup$ – Nick Cox Oct 23 '19 at 13:25
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    $\begingroup$ @KewHsein If that is your takeaway message from the comments, you should really take up Nick's suggestion to read a good introductory book on statistics first. $\endgroup$ – Frans Rodenburg Oct 23 '19 at 13:46

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