Hi Everyone,
I have a question about Mixed-Design Anova in R. I want to obtain Mauchly�s test of Sphericity and the Greenhouse-Geisser correction. I have managed to do it in SPSS: GLM Measure1 Measure2 Measure3 Measure4 Measure5 Measure6 BY Grouping /WSFACTOR=Measure 6 Polynomial /METHOD=SSTYPE(3) /PLOT=PROFILE(Measure*Grouping) /CRITERIA=ALPHA(.05) /WSDESIGN=Measure /DESIGN=Grouping. I have tried to replicate this in R: library("dplyr") library("tidyr") library("ggplot2") library("ez") PatientID <- c(1:10) Measure1 <- c(3,5,7,4,NA,7,4,4,7,2) Measure2 <- c(1,2,5,6,8,9,5,NA,6,7) Measure3 <- c(3,3,5,7,NA,4,5,7,8,1) Measure4 <- c(1,2,5,NA,3,NA,6,7,3,6) Measure5 <- c(2,3,NA,8,3,5,6,3,6,4) Measure6 <- c(1,2,4,6,8,3,5,6,NA,4) Grouping <- c(1,0,1,1,1,0,0,1,1,0) dataframe <- data.frame(PatientID, Measure1, Measure2, Measure3, Measure4, Measure5, Measure6, Grouping) dataframe$Grouping <- as.factor(dataframe$Grouping) dataframe ezPrecis(dataframe) glimpse(dataframe) dataframe %>% count(PatientID) dataframe %>% count(PatientID, Grouping, Measure1, Measure2, Measure3, Measure4, Measure5, Measure6) %>% filter(PatientID %in% c(1:243)) %>% print(n = 10) # So, we have a mixed design with one between factor (Grouping) and 6 within factors (Measure 1 to 6). dat_means <- dataframe %>% group_by(Grouping, Measure1, Measure2, Measure3, Measure4, Measure5, Measure6) %>% summarise(mRT = mean(c(Measure1, Measure2, Measure3, Measure4, Measure5, Measure6))) %>% ungroup() View(dat_means) ggplot(dat_means, aes(c(Measure1, Measure2, Measure3, Measure4, Measure5, Measure6), mRT, colour = Grouping)) + geom_line(aes(group = Grouping)) + geom_point(aes(shape = Grouping), size = 3) + facet_wrap(~group) ANOVA <- ezANOVA(dat, x, PatientID, within = .( c(Measure1, Measure2, Measure3, Measure4, Measure5, Measure6)), between = Grouping, type = 3) print(ANOVA) However, this does not work. I know I am probably doing it completely wrong, but I do not know how to solve it. Besides that, I do not know what to fill in at the �x�. Can somebody help me? Thank you in advance. Lisa [[alternative HTML version deleted]]
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