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## Checking normality in repeated measures ANOVA

In regression analyses, of which repeated measures designs are but one example, we make assumptions about the residuals. Residuals are the bits of each observation which are not explained by a set of predictors. The radius test (Andrews et al, 1973) on residuals checks the appropriateness of a repeated measures design. It compares the sums of squared standardised residuals to a chi-square distribution by assuming that the observed chi-squares are a random (uniform) sample from that distribution.

Annotated code is given below for a response over 4 time points but can be adapted for particular examples as further illustrated with an example from a 3x2 design.

```GLM
a b c d
/WSFACTOR = time 4
/METHOD = SSTYPE(3)
/SAVE = ZRESID
/CRITERIA = ALPHA(.05)
/WSDESIGN = time .

* the sum of the squares of the residuals is
* chi-square. So obtain 1- p-values. The
* residuals are contained in columns zre_1 to
* zre_4.

compute sum = cdf.chisq(zre_1**2 + zre_2**2 + zre_3**2 + zre_4**2,4).

* then compare the p-values to a uniform
* distribution. If they follow a uniform
* distribution
* then the residuals are normally distributed as
* assumed by the model. This follows
* since the sum of squared residuals can be
* considered as randomly sampled from
* a chi-square distribution.

NPAR TESTS
/K-S(UNIFORM)= sum
/MISSING ANALYSIS.```

For example if fitting an ANOVA with two repeated measures factors one would fit the model with two main effects and an interaction, output the standardised residuals and compare the sum of their squares

```GLM
a b c d e f
/WSFACTOR = cond 3 time 2
/METHOD = SSTYPE(3)
/SAVE = ZRESID
/CRITERIA = ALPHA(.05)
/WSDESIGN = cond time cond*time.

* the sum of the squares of the residuals is
* chi-square. So obtain 1- p-values. The
* residuals are contained in columns zre_1 to
* zre_6.

compute sum = cdf.chisq(zre_1**2 + zre_2**2 + zre_3**2 + zre_4**2 + zre_5**2 + zre_6**2,6).

* then compare the p-values to a uniform
* distribution. If they follow a uniform
* distribution
* then the residuals are normally distributed as
* assumed by the model. This follows
* since the sum of squared residuals can be
* considered as randomly sampled from
* a chi-square distribution.

NPAR TESTS
/K-S(UNIFORM)= sum
/MISSING ANALYSIS.```

Note also that Schmider et al. (2010) performed simulations on a one-way between subjects ANOVA and concluded that, at least in this case, ANOVA is robust to violations in normality particularly for large n greater than 100. You can alternatively assess normality in repeated measures by looking at the normality of all the possible pairs of differences between the levels of the factors constituting the repeated measures e.g. if you have four levels of repeated measures there are six possible pairwise differences between pairs of these six levels that are assumed to be normally distributed in the analysis of variance.

Reference

Schmider, E., Ziegler, M., Danay, E., Beyer, L., & Bühner, M. (2010). Is it really robust? Reinvestigating the robustness of ANOVA against violations of the normal distribution assumption. Methodology: European Journal Of Research Methods For The Behavioral And Social Sciences, 6(4), 147-151.