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I'm struggling with interpreting the output from R for a two-way repeated measures ANOVA. I used m1.emax.aov <- aov(mean_emax_norm ~ (cond5*stride) + Error(dog_id/(cond5*stride)), df3.emg.m1) and the summary tables are throwing me.

I'm used to just seeing a summary table with the main effects and interaction listed once; each with their own error term. I don't know how to interpret the output below. I did not expect to see cond5 for Error:dog_id, stride for Error:dog_id:cond5, or cond5:stride for Error:dog_id:stride.

How do you interpret, for instance, stride is significant for Error:dog_id:cond5 but not for Error:dog_id:stride?

Any guidance would be greatly appreciated.

summary(m1.emax.aov)

Error: dog_id
      Df Sum Sq Mean Sq F value Pr(>F)  
cond5      2  706.5   353.3   12.69  0.011 *
Residuals  5  139.1    27.8                 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Error: dog_id:cond5
          Df Sum Sq Mean Sq F value Pr(>F)   
cond5      4  48.30   12.07   3.453 0.0216 * 
stride     1  36.07   36.07  10.316 0.0035 **
Residuals 26  90.91    3.50                  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Error: dog_id:stride
             Df Sum Sq Mean Sq F value Pr(>F)
stride        2   7.42   3.709   0.343  0.717
cond5:stride  3   9.99   3.331   0.308  0.819
Residuals    11 119.02  10.820               

Error: dog_id:cond5:stride
             Df Sum Sq Mean Sq F value Pr(>F)  
cond5:stride  8  103.8  12.975   2.498 0.0224 *
Residuals    52  270.1   5.194                 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
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    $\begingroup$ Does the output of with(df3.emg.m1, table(dog_id, cond5, stride)) result in every cell being equal in size? And if not, why? $\endgroup$
    – John
    Commented Sep 25, 2013 at 23:42
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    $\begingroup$ Cell sizes are not all equal. There's some missing data. Some of the trials had to be dropped for some of the subjects. $\endgroup$ Commented Sep 26, 2013 at 0:06

1 Answer 1

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If cond5 and and stride are both repeated measures and dog_id is what they're repeated on then you've specified your model correctly. The results are uninterpretable due to data issues.

When performing a repeated measures ANOVA you need each cell of the design to be equal in size. Each subject has to have an entry for each variable. If that does not occur then the ANOVA results can be what you've seen.

You're missing data for individual subjects. Therefore, you must drop those subjects if you want to stick with RM ANOVA for analysis. You'll then get a result back that is more like what you expect.

You could look up linear mixed effects analysis or multi level modeling and use that as a substitute for repeated measures ANOVA. That kind of analysis can be used to treat subjects more appropriately as random effects. For recent papers with both advice and tutorials for performing such analysis in R you might try Jaeger (2008) and Barr (2013). There are also lots of questions answered on the subject on this site.

Barr, D. J., Levy, R., Scheepers, C., & Tily, H. J. (2013). Random effects structure for confirma- tory hypothesis testing: Keep it maximal. Journal of Memory and Language, 68, 255-278. doi: dx.doi.org/10.1016/j.jml.2012.11.001

Jaeger, F. T. (2008). Categorical data analysis: Away from anovas (transformation or not) and towards logit mixed models. Journal of Memory and Language, 59, 434-446.

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    $\begingroup$ I feel like this does not quite answer the OP's question. This is a very good comment indeed--and I wholeheartedly agree with it--but the missing data have nothing to do with which error term to use in the denominator of the different F-tests of a repeated-measures ANOVA, which is what the question is about. $\endgroup$ Commented Sep 26, 2013 at 2:34
  • $\begingroup$ The questioner indicates from the question, and correctly specified model, that they actually know the right error terms. They're just confused by the output, which was not correct for the specified model. The output wasn't caused by confusion over knowing the right error terms but the structure of the data. $\endgroup$
    – John
    Commented Sep 26, 2013 at 3:27

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