I'm doing a two-way between-within ANOVA in SPSS. I have two groups with 9 subjects each (so total = 18), and 24 levels of one repeated measure.

I understand why Mauchly's test of Sphericity has no meaning when there are are only 2 levels of a repeated measures factor, but I notice (using General Linear Model.....repeated measures in SPSS) that Mauchly's test of Sphericity also appears to be undefined (or gives the useless output of Mauchly's W = '.0' , p = '.') when the number of levels of a repeated measure is equal to or greater than the number of cases (subjects). In these instances, even though Mauchly's statistic is not calculated, Greenhouse-Geisser, Huynh-Feldt, and Lower-Bound Epsilon values are calculated.

I would be really happy if someone could provide some insight on why Mauchly's statistic is not calculated in these cases and what should be done to assess sphericity in the absence of Mauchly's statistic.

  • $\begingroup$ Chi-square approximation of the statistic's distribution under H0 is valid only for quite large, and greater than the number of RM levels, sample size. $\endgroup$
    – ttnphns
    Commented Mar 27, 2013 at 10:16
  • $\begingroup$ Non-sphericity means that the response of each subject is modeled by a 24-dimensional multivariate normal distribution with unconstrained covariance matrix. It is not possible to estimate this covariance matrix with 18 subjects (does it work if you add a fictive third group of 9 individuals ?) $\endgroup$ Commented Mar 27, 2013 at 13:32
  • 1
    $\begingroup$ @Stephane, don't know exactly what you mean but you may be mixing up Bartlett's and Mauchly's sphericities. $\endgroup$
    – ttnphns
    Commented Mar 27, 2013 at 14:36
  • 1
    $\begingroup$ Nope. It's Bartlett's multivariate sphericity test. Mauchly's involves testing of variances of differences between RM-levels. $\endgroup$
    – ttnphns
    Commented Mar 27, 2013 at 16:45
  • 1
    $\begingroup$ unfortunately I have the same problem: I have one factor(time) with three levels(3 time points). in each time I have 3 cases (subjects). I dont have mauchly's tests significance value. but the sig with: sphericity assumed, Greenhouse-Geisser, Huynh-Feldt, and Lower-Bound Epsilon are totally different! How can I evaluate the difference between time points? Thanks $\endgroup$
    – user122698
    Commented Jul 8, 2016 at 23:29

2 Answers 2


In simple terms, one of the assumptions of a RM anova is that all the time points need to be correlated with each other to the same degree. The mauchly's tests this assumption, that all the times are related similarly.

When you have just 2 time points, you have only one correlation, between time 1 and 2. There is nothing else to compare this to, so the assumption is always met. Sphericity assumed, Greenhouse-Geisser, Huynh-Feldt, and Lower bound Epsilon values should all be the same in this case.

This is why you'd never use the Mauchly's test in a paired t-test, because they always have only 2 time points.

Greenhouse-Geisser, Huynh-Feldt, and Epsilon values should all be the same in this case.

  • $\begingroup$ This does not answer OP's question at all. $\endgroup$
    – amoeba
    Commented Oct 13, 2016 at 21:39

I had the same problem. Although this is an incomplete answer, hopefully it is helpful.

The p-value that shows up in the table for Mauchly's test is based on the chi-squared distribution. That calculation involves taking the log of Mauchly's W. Whenever W = 0, the log is undefined and the test is inconclusive.

As far as why W = 0 in these cases, I can't help you. The calculation for Mauchly's W involves more linear algebra than I know, but you can read more about it yourself here.

What to do?

Oberfeld and Franke (2012) conducted a simulation study of different repeated measures procedures. They note (without any citation or discussion) that in situations where you have a mixed model (with both within-subjects and between-subjects effects) that you simply must have more participants than observation periods.

If you have no between-subjects effects, they recommend using the Huynh-Feldt correction. It is valid when there are more observation periods than N.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.