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

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

where

$$T=M\text{'}\widehat{\sum }M.$$

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

$$C=-\left(n-r\right)\log \left(W\right)D,$$

where

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

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.