Hi,

I found your question interesting and decided to take a closer look.

First I found this little excerpt here, which can probably help for some similar results with Mauchly's Test for Sphericity: http://study.sagepub.com/sites/default/files/spss_tip_15.1.pdf (I think it is from Andy Field's "Discovering Statistics Using IBM SPSS Statistics").

It states three reasons why SPSS could yield such results, which are

• only two repeated measures levels
• perfect sphericity (sphericity estimate of one)
• p-value cannot be computed because Chi-Square and df are zero

Since you have a df>0, that couldn't exactly be the problem here, but we are closing in.

Taking a closer look at the details of Mauchly's Test (e.g. here: https://www.mathworks.com/help/stats/mauchlys-test-of-sphericity.html), we can see that the Mauchly-W goes into a log function and since your Mauchly-W is zero and log(0) is undefined, I'd like to quote a popular German saying: "now we have the salad".

So this explains the missing Chi-Square and p-values: Chi-Square is undefined and p cannot be computed without it. You'd have to take a closer look at the computation of the Mauchly-W for your case data to find out why it turns out to be zero.

Now as mentioned by the others, Mauchly's Test for Sphericity is prone to miss sphericity violations for small sample sizes anyway (see e.g. https://statistics.laerd.com/statistical-guides/sphericity-statistical-guide.php).

Since sphericity is often violated in practice, I'd therefore also suggest using the Greenhouse-Geisser corrected values just in case.

I hope this helps narrowing down this part of your question and wish you best of luck.

Kind regards,

Martin

Hello Ajit, thank you for your reply. I have two factors and I have repeated measures. Therefore I choosed the two-way repeated ANOVA. So I dont fully understand your first part of the comment.

Regarding the Mauchly-Test I assume you mean, that I have not enough subjects?

Did I get this right?

hello Safa,

I already read this threat but I'm very grateful that you searched for a solution anyway. But unfortunately it doesnt solve my problem. My ANOVA is 3x17 and the sample size (31) is bigger than the number of repeats. therefore I still have my problem.... I assume its anyway somehow related to the number of repeats, but so far I didnt find anything.

With so many repeated measures, it is very hard to satisfy sphericity assumption. You can try forming a mixed model or do not bother with the significant result of Mauchly's test and go ahead with the results.

I'm not 100% why the test isn't reported but my intuition would be that the interaction is for a 3x17 design and hence in effect for differences between 51 sets of scores with every other set. That gives over 2500 parameters to estimate and I suspect this simply exhausts the df for the test.

In this case you can still estimate epsilon and can be certain that the uncorrected test is far too liberal. It would be safer to use the Greenhouse-Geisser correction in this case.

Thank you for the advice. I will be cautious and use Greenhouse-Geissler

Hi Lukas,

in n=31 you are just scratching normal distribution assumption, so Mauchly checking for sphericity (i.e. equal variance for 3+ samples) is most unlikely to give any useful outcome here.

If you are standing on solid statistical ground or know someone who is, probably better use an HLM (hierarical linear modelling) with measure points t0, t1,...tx (lvl 1) nested in participants (lvl 2), else just report the empirical data with descriptive statistics, probably with t-testing

http://www.methodenberatung.uzh.ch/de/datenanalyse/unterschiede/zentral/ttestabh.html

Best regards, viel Erfolg und Gruß aus Wachtberg

Andreas

Hi,

I found your question interesting and decided to take a closer look.

First I found this little excerpt here, which can probably help for some similar results with Mauchly's Test for Sphericity: http://study.sagepub.com/sites/default/files/spss_tip_15.1.pdf (I think it is from Andy Field's "Discovering Statistics Using IBM SPSS Statistics").

It states three reasons why SPSS could yield such results, which are

• only two repeated measures levels
• perfect sphericity (sphericity estimate of one)
• p-value cannot be computed because Chi-Square and df are zero

Since you have a df>0, that couldn't exactly be the problem here, but we are closing in.

Taking a closer look at the details of Mauchly's Test (e.g. here: https://www.mathworks.com/help/stats/mauchlys-test-of-sphericity.html), we can see that the Mauchly-W goes into a log function and since your Mauchly-W is zero and log(0) is undefined, I'd like to quote a popular German saying: "now we have the salad".

So this explains the missing Chi-Square and p-values: Chi-Square is undefined and p cannot be computed without it. You'd have to take a closer look at the computation of the Mauchly-W for your case data to find out why it turns out to be zero.

Now as mentioned by the others, Mauchly's Test for Sphericity is prone to miss sphericity violations for small sample sizes anyway (see e.g. https://statistics.laerd.com/statistical-guides/sphericity-statistical-guide.php).

Since sphericity is often violated in practice, I'd therefore also suggest using the Greenhouse-Geisser corrected values just in case.

I hope this helps narrowing down this part of your question and wish you best of luck.

Kind regards,

Martin

I would advise never using the Mauchley test (for reasons mentioned above among others). The epsilon estimates (which are not part of the test per se) are more useful. I would always recommend a correction if epsilon falls much below 1 (say .975 or lower).

Hello Lukas,

The reason for the null answer for Mauchley's test (note that this is for the interaction term) is that your sample size (N = 31) is smaller than the number of data points (3 x 17 = 51) required. You do get legitimate Mauchley results for both of the main effects (since 31 is larger than both 3 and 17, individually).

For your second question expressing concern over normality of scores, the Friedman two-way anova by ranks may suffice as a backup to the parametric test, assuming that the treatments were again repeated across all participants. If not, then the Kruskal-Wallis one-eay anova would be the choice.

Good luck with your work!