Solved – How to analyse multiple between subjects factors in a mixed ANOVA in SPSS

anovarepeated measuresspss

I am performing a Mixed between subjects Anova where I have multiple time points that subjects have completed a measure at.

The within subjects factor is time.
The between subjects factors include gender, ethnicity, refugee status and English language.

My two questions are

Combined or separate ANOVAs: Is it possible in SPSS to do this as one analysis or do I need to do multiple Anovas one for each of my between subjects factors? If it is possible to do it in one model can you explain how specifically in SPSS?
Fixed versus random factors: My advisor told me to be sure to know whether I am using the models as fixed effects or random effects but she herself doesnt really understand this. I understand the difference ie fixed effects are like gender versus random effects that are random sample of the total # of levels in the population however I am not sure how to deal with this in SPSS or even tell what the default is. I performed my analyses using Analyze, General Linear Model, Repeated Measures.

Best Answer

Combined or separate ANOVAs

You probably want to include all the predictors to develop an overall model. By default SPSS will introduce all possible interactions between your predictors. I doubt you would be interested in testing whether there is a gender by ethnicity by refugee status interaction. As a starting point I'd click on the "Model" button and select only the effects of interest (e.g., main effects, perhaps interactions between between subjects factors and time; perhaps others). * Also by default SPSS REPEATED MEASURES will treat time as an unordered factor, whereas often it is better treating time as a numeric variable where you explore linear, quadratic or other functional effects of time. Polynomial contrasts are one simple option to explore in SPSS REPEATED MEASURES.
If you want to get more sophisticated then you might want to explore multilevel modelling options which provide more flexibility.

Fixed versus random

All your between subjects factors are presumably fixed factors in that you have measured all levels of the factor of interest, or at least you have not sampled your factors from a domain of possible factors.

Related Solutions

Repeated Measures ANOVA – Using LME/LMER in R for Two Within-Subject Factors

What you're fitting with aov is called a strip plot, and it's tricky to fit with lme because the subject:A and subject:B random effects are crossed.

Your first attempt is equivalent to aov(Y ~ A*B + Error(subject), data=d), which doesn't include all the random effects; your second attempt is the right idea, but the syntax for crossed random effects using lme is very tricky.

Using lme from the nlme package, the code would be

lme(Y ~ A*B, random=list(subject=pdBlocked(list(~1, pdIdent(~A-1), pdIdent(~B-1)))), data=d)

Using lmer from the lme4 package, the code would be something like

lmer(Y ~ A*B + (1|subject) + (1|A:subject) + (1|B:subject), data=d)

These threads from R-help may be helpful (and to give credit, that's where I got the nlme code from).

http://www.biostat.wustl.edu/archives/html/s-news/2005-01/msg00091.html

http://permalink.gmane.org/gmane.comp.lang.r.lme4.devel/3328

http://www.mail-archive.com/r-help@stat.math.ethz.ch/msg10843.html

This last link refers to p.165 of Pinheiro/Bates; that may be helpful too.

EDIT: Also note that in the data set you have, some of variance components are negative, which is not allowed using random effects with lme, so the results differ. A data set with all positive variance components can be created using a seed of 8. The results then agree. See this answer for details.

Also note that lme from nlme does not compute the denominator degrees of freedom correctly, so the F-statistics agree but not the p-values, and lmer from lme4 doesn't try too because it's very tricky in the presence of unbalanced crossed random effects, and may not even be a sensible thing to do. But that's more than I want to get into here.

Solved – Different F-ratios for within subjects effects when using SPSS and R’s aov

This may be because your between-groups variable, NativeLanguage, is unbalanced (12 English, 9 Other), in which case the type of Sums-of-Squares employed is going to affect the F values. By default, aov() uses Type 1 sums of squares, which isn't recommended with unbalanced designs. Instead, use the ezANOVA() function from the ez package:

my_anova = ezANOVA(
    data = C1
    , dv = .(RT)
    , wid = .(Subject)
    , within = .(Class)
    , between = .(NativeLanguage)
    , type = 3 #SPSS uses type 3 Sums-of-Squares
    , observed = .(NativeLanguage) #ensures appropriate effect size is computed
)
#note warning about data imbalance
print(my_anova)

This yields the results table:

$ANOVA
                Effect DFn DFd         F           p p<.05          ges
2       NativeLanguage   1  19  6.297322 0.021311506     * 0.2451177279
3                Class   1  19  1.976662 0.175885891       0.0009118545
4 NativeLanguage:Class   1  19 14.187926 0.001305329     * 0.0065510116

Which strangely still has a different report for the Class effect than what you're getting from SPSS/Statistica. Adding a detailed=TRUE argument to the ezANOVA() call above gives us a slightly more detailed results table (including the intercept and sums of squares):

$ANOVA
                Effect DFn DFd          SSn        SSd          F            p p<.05          ges
1          (Intercept)   1  19 7717585.5514 245636.967 596.954632 8.136647e-16     * 0.9587389575
2       NativeLanguage   1  19   81413.4195 245636.967   6.297322 2.131151e-02     * 0.2451177279
3                Class   1  19     303.1399   2913.831   1.976662 1.758859e-01       0.0009118545
4 NativeLanguage:Class   1  19    2175.8535   2913.831  14.187926 1.305329e-03     * 0.0065510116

This shows that the mismatch lies in the SSn for the class effect; ezANOVA (which uses car::Anova()) obtains an SSn of 303ish whereas SPSS/Statistica obtain an SSn of 569.