## Mauchly’s Test of Sphericity

The regular p-value calculations in the repeated measures anova (`ranova`) are accurate if the theoretical distribution of the response variables have compound symmetry. This means that all response variables have the same variance, and each pair of response variables share a common correlation. That is,

`$\Sigma ={\sigma }^{2}\left(\begin{array}{cccc}1& \rho & \cdots & \rho \\ \rho & 1& \cdots & \rho \\ ⋮& ⋮& \ddots & ⋮\\ \rho & \rho & \cdots & 1\end{array}\right).$`

If the compound symmetry assumption is false, then the degrees of freedom for the repeated measures anova test must be adjusted by a factor ε, and the p-value must be computed using the adjusted values.

Compound symmetry implies sphericity.

For a repeated measures model with responses y1, y2, ..., sphericity means that all pair-wise differences y1 – y2, y1 – y3, ... have the same theoretical variance. Mauchly’s test is the most accepted test for sphericity.

Mauchly’s W statistic is

`$W=\frac{|T|}{{\left(trace\left(T\right)/p\right)}^{d}},$`

where

`$T=M\text{'}\sum ^{^}M.$`

M is a p-by-d orthogonal contrast matrix, Σ is the covariance matrix, p is the number of variables, and d = p – 1.

A chi-square test statistic assesses the significance of W. If n is the number of rows in the design matrix, and r is the rank of the design matrix, then the chi-square statistic is

`$C=-\left(n-r\right)\mathrm{log}\left(W\right)D,$`

where

`$D=1-\frac{2{d}^{2}+d+2}{6d\left(n-r\right)}.$`

The C test statistic has a chi-square distribution with (p(p – 1)/2) – 1 degrees of freedom. A small p-value for the Mauchly’s test indicates that the sphericity assumption does not hold.

The `rmanova` method computes the p-values for the repeated measures anova based on the results of the Mauchly’s test and each epsilon value.

## References

[1] Mauchly, J. W. “Significance Test for Sphericity of a Normal n-Variate Distribution. The Annals of Mathematical Statistics. Vol. 11, 1940, pp. 204–209.