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How to interpret the following result from Mauchly's test conducted in R?

Mauchly's test of sphericity Contrasts orthogonal to ~1

data: SSD matrix from lm(formula = Controls3 ~ 1)

W = 0, p-value = 1

From my understanding, a p-value of 1 should indicate that there is no difference in variances and that we can move forward with this assumption in further statistical analyses. However, from what I've seen, a test statistic of 0 is usually associated with an undefined p-value. So I'm confused as to what to make of this result. Any suggestions would be greatly appreciated. Thanks.

sf1
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  • I Just noticed this is basically an identical question to one you previously asked. Please edit your first question instead of posting a second one. – RAND Aug 05 '19 at 16:55
  • Sorry, I didn't get responses for that question so I rephrased it. I deleted the other question. – sf1 Aug 05 '19 at 18:41

1 Answers1

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The test statistic is distributed as a chi-square under the null hypothesis. The support of the chi-squared distribution included=s 0, as the smallest possible value, thus the probability of getting something as or more extreme is 1

For your situation is this a real data set or an example because a test statistic of 0 and p-value of 1 are hard to come by except for design. Also sphericity as an assumptions is typically violated, you might just want to adjust the degrees of freedom for the test just in case and instead of doing a hypothesis test to check your assumptions and then deciding whether or not to adjust the degrees of freedom.

RAND
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  • Thanks, this is is a real data set – sf1 Aug 05 '19 at 18:42
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    Like I mentioned I would probably not use Mauchly's test and instead just adjust the degrees of freedom for sphericity. I would also double check all calculations let led to a test statistic of 0 when testing sphericity with a real data set. – RAND Aug 05 '19 at 20:06