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As i am new to statistics and statistical terminology i am struggling a bit with interpreting the results of some of the things i do/have to do.

I am using RStudio as a tool for that. More precisely i am using a package nortest to perform normality distribution tests on my data. Running the Anderson-Darling normality test on my data returns A = 2946.8, p-value < 2.2e-16. A is i am guessing a statistical value of the test itself and the p-value is.. well p-value.

My question is: What does the p-value in this case signify? Is it the probability of my data sets distribution to differ from normal distribution? If yes, is it safe to assume that the distribution of the data is really-really similar to normal distribution?

Since the goal is outlier detection one of the methods proposed for this task (not by me) was a standard Z score cut off which is really sensitive to extremes in the data. The method i proposed is using the adjusted boxplot so it takes into consideration whether the data is skewed or not.

Note: The entire data set is 8702 records. Update: After all the research papers and materials i provided for my team leader, the argument against my suggestion and for using Z score is that by using the Z score it gives you something you can look at (the Z score of the observation) in comparison to the adj. boxplot where you can't observe any statistical indicator that describes the data in some way. Yea.. I know...

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    $\begingroup$ I am not familiar with RStudio. The Anderson-Darling test is a goodness of fit test for normality. The null hypothesis is that the data come from a normal distribution. The p-value indicates the probability that the Anderson-Darling test statistic is as extreme or more extreme than what would be expected under the null hypothesis.. A low p-value suggests that the data is not from a normal distribution. $\endgroup$ – Michael R. Chernick Dec 8 '16 at 12:55
  • $\begingroup$ There is some controversy about when it is appropriate to use p-values. The p-value is often misinterpreted in the literature. You can go to the website of the American Statistical Association to learn more about that. You go to www.amstat.org. $\endgroup$ – Michael R. Chernick Dec 8 '16 at 12:59
  • $\begingroup$ @MichaelChernick This is very interesting. Thank you for the link. The whole point of these tests i am doing is to convince people at my office that it is not a good idea to use Z score to flag outliers in our data as Z score performs good only on data that is normally distributed. And it seems that the only way to do it is to prove it to them somehow that the distribution is not normal. $\endgroup$ – Emil Filipov Dec 8 '16 at 13:02
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    $\begingroup$ You do not tell us your sample size but bear in mind that with a sufficiently large sample almost any hypothesis about the data would be rejected. $\endgroup$ – mdewey Dec 8 '16 at 13:21
  • $\begingroup$ @mdewey Hi there. I added that info in the question. Anyway the data i am running the test is 8702 records. Not 100 000 or more. Just around 9k :-) $\endgroup$ – Emil Filipov Dec 8 '16 at 13:29
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The Anderson-Darling test's null hypothesis is that the tested distribution is actually from the distribution specification it is tested against, here the null would be that your distribution is actually normal.

In general, a p-value so low suggests that you can safely reject the null hypothesis on all relevant significance levels. I.e. here it rejects the null of normality, your distribution is very probably not normal.

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    $\begingroup$ You're welcome. You may flag my answer as "accepted" to show other users that this topic has a satisfactory solution. $\endgroup$ – Benjamin Dec 8 '16 at 15:24
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    $\begingroup$ Beware of the use of a P-value to diagnose the properties of a dataset. The suggestion of @Bryan is worthy of consideration (I've extended it a little). $\endgroup$ – Michael Lew Dec 8 '16 at 20:30
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A good way to make a point about the distribution of data is to just plot it in comparison to plots of data that fit the conventional distributions. A visual display often yields better insights into the data than any test summary can.

With a sample of thousands of values you can easily plot a histogram of the values to get a nice visual display of the empirical distribution.

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  • $\begingroup$ True. The distribution of my data does not even closely resemble anything that remotely looks like normal, but that is not convincing enough that using Z score for outlier detection is a bad idea. $\endgroup$ – Emil Filipov Dec 9 '16 at 7:20
  • $\begingroup$ Alas, there are times when ignorance is invincible. I had an ex-wife who was convinced that, if you did not top off when filling your gas tank, you would reduce your gas mileage. No amount of evidence from mechanics or engineers could change her mind. $\endgroup$ – Bryan Dec 9 '16 at 15:16
  • $\begingroup$ I can't help but laugh on this one. Anyway, i spent the whole day mining through the data and i finally managed to come up with a ton of evidence that the data is not symmetric. Which is nice.. Now i only have to find a test that tells me what distribution i should assume for my data, so i can start doing something... :-D $\endgroup$ – Emil Filipov Dec 9 '16 at 16:02
  • $\begingroup$ For finding a distribution, I don't think there is any omnibus test. The best you have are tests against specific distributions, which you have to try distribution by distribution. $\endgroup$ – Bryan Feb 7 '17 at 13:09

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