Computing ANOVA in R

Question

Edit: I think the data was poorly given, varables coke, pepsi, and sprite were just given in 2 columns when I got it, but it has nothing to do with before and after treatment?

here's the original questions since someone asked For the study, 2 males and 2 females are picked randomly from NYC, and a random sample of 2 males and 2 females are picked from LA. Each person is given 6 drinks (2 from each brand), where the order of the drinks for each person is randomized. Each person rates the drink’s taste, Y, on a scale of 60 to 100, where 100 is the best tasted.

Please let me know if I can provide any other info. Pretty much I am asked to fit ANOVA. End of Edit#

I would like to understand ANOVA in R, I am working in this data set.

dat <- structure(list(city = structure(c(2L, 2L, 2L, 2L, 1L, 1L, 1L, 
1L), .Label = c("LA", "NY"), class = "factor"), sex = structure(c(2L, 
2L, 1L, 1L, 2L, 2L, 1L, 1L), .Label = c("F", "M"), class = "factor"), 
    rater = 1:8, frit0 = c(77L, 72L, 78L, 84L, 77L, 78L, 83L, 
    81L), frit1 = c(76L, 71L, 78L, 84L, 76L, 77L, 86L, 84L), 
    utz0 = c(78L, 73L, 78L, 84L, 74L, 81L, 85L, 83L), utz1 = c(80L, 
    75L, 80L, 86L, 72L, 73L, 81L, 79L), weiss0 = c(78L, 76L, 
    84L, 90L, 78L, 79L, 88L, 86L), weiss1 = c(81L, 73L, 81L, 
    87L, 81L, 82L, 92L, 90L)), .Names = c("city", "sex", "rater", 
"coke0", "coke1", "pepsi0", "pepsi1", "sprite0", "sprite1"), 
 class = "data.frame", row.names = c(NA, -8L))

I want to do an analysis of variance, check assumptions, and interpret results. I am sure this is done many times by the experts, so, what is the correct step to do it?

I have tried

am1 <- aov(coke1 ~ coke0 + sex, data=dat2)
summary(am1)

Is this correct? Or even relevant? thanks in advance!

Answer 1

If you think in terms of linear models instead of going directly to ANOVA you may see something very important in the data. Consider the following plot of the data for looking at the response coke1 as a function of coke0 and sex. You'll notice that the responses for Males are perfectly linear making the use of a linear model moot as we already know the 'line of best fit' since the data is a line.

library(ggplot2)
ggplot(dat) +
  aes(x = coke0, y = coke1, color = sex) + 
  geom_point() + 
  stat_smooth(method = 'lm')

You may continue with an ANOVA, I would do so as:

g <- lm(coke1 ~ sex + coke0, data = dat)
plot(g) # graphics to check some model assumptions
anova(g)

The ANOVA results should look like:

Analysis of Variance Table

Response: coke1
          Df  Sum Sq Mean Sq F value    Pr(>F)     
sex        1 128.000 128.000  72.804 0.0003639 ***
coke0      1  49.209  49.209  27.989 0.0032170 ** 
Residuals  5   8.791   1.758                      
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Answer 2

so, from describing your data, it seems your dataset is in the wrong format. Generaly I would recomend having one set of data per row, so each rating of a drink is one row, with an indication of their sex, city, the drink type (and if you knew it the order of the tasting). Luckily Hadley Wickham and the amazing reshape2 package comes quickly to your aid

library(reshape2)
library(ggplot2)

melteddata<-melt(data=dat, id.vars=c("city", "sex", "rater"))

Now, you can examine your data to see what you can see

qplot(data=melteddata, y=value, x=variable, colour=sex, shape=city)

Generaly, Males score lower than Women, and maybe LA scores higher than NY. Finaly, you can do your anova, investigating the variability when trying to predict response based on all the other variables.

aovobj<-aov(value~city*sex*rater*variable, data=melteddata)

summary(aovobj)