One-way Repeated-Measures ANOVA

First let’s load in the sample data. It’s “stacked”" in the usual way, where each row is a single observation, and columns code variables—in this case treatment, which has four levels, and subject, which has 10 levels.

datafilename <- "http://www.gribblelab.org/stats2019/data/oneWayRepdata.csv"
mydata$treatment <- factor(mydata$treatment)
mydata$subject <- factor(mydata$subject)
mydata
##    dv treatment subject
## 1   8         1       1
## 2  10         2       1
## 3   7         3       1
## 4   5         4       1
## 5   9         1       2
## 6   9         2       2
## 7   8         3       2
## 8   6         4       2
## 9   7         1       3
## 10  5         2       3
## 11  8         3       3
## 12  4         4       3
## 13  9         1       4
## 14  6         2       4
## 15  5         3       4
## 16  7         4       4
## 17  8         1       5
## 18  7         2       5
## 19  7         3       5
## 20  6         4       5
## 21  5         1       6
## 22  4         2       6
## 23  4         3       6
## 24  3         4       6
## 25  7         1       7
## 26  6         2       7
## 27  5         3       7
## 28  4         4       7
## 29  8         1       8
## 30  8         2       8
## 31  6         3       8
## 32  6         4       8
## 33  9         1       9
## 34  8         2       9
## 35  6         3       9
## 36  5         4       9
## 37  7         1      10
## 38  7         2      10
## 39  4         3      10
## 40  5         4      10

demo: ignore subjects

As a demonstration, let’s run this as if it were simply a between-subjects ANOVA, in other words, ignoring the fact that observations from the four levels of treatment come from the same subjects:

am1 <- aov(dv ~ treatment, data=mydata)
summary(am1)
##             Df Sum Sq Mean Sq F value  Pr(>F)
## treatment    3   38.9  12.967   6.062 0.00189 **
## Residuals   36   77.0   2.139
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

In this case the F-test of the treatment effect is significant at $$p=0.001890$$.

demo: include subjects as if it were a two-factor between-subjects design

Now as another demonstration, let’s run this ANOVA again as if it were a two-factor design, in which we include subject as a factor. In this case, note that we cannot include both the main effects and the interaction effect—since there is only one observation from each subject in each condition, there are not enough degrees of freedom. So we will simply specify a model that has the two main effects and leaves out the interaction effect:

am2 <- aov(dv ~ treatment + subject, data=mydata)
summary(am2)
##             Df Sum Sq Mean Sq F value   Pr(>F)
## treatment    3   38.9  12.967  12.241 3.06e-05 ***
## subject      9   48.4   5.378   5.077 0.000471 ***
## Residuals   27   28.6   1.059
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Note how the SS, df, and MS for the treatment effect are identical to the previous case where the subject factor was not included in the model. What’s changed, is that the error term (the term denoted Residuals) is now smaller. In other words, in the first model (am1), the SS was 77.00 — this represents the “leftover”" variability in the dependent variable that is unaccounted for by the treatment factor. In the am2 model, the subject factor has a SS of 48.4, which is now sliced out of the error term, leaving only 28.6. In other words, the subject factor has accounted for a chunk of the variance in the dependent variable, leaving us with a smaller error term in the ANOVA. This is good, because it means that the F-ratio for the treatment effect is larger (and hence the p-value is smaller)—since it is equal to the MS for treatment divided by the MS for the error term.

Note that this decrease in MS error is not “free”“, in other words there is a cost: we give up degrees of freedom (was 36 in am1 but is now reduced to 27 in am2). Lower degrees of freedom in the error term means that (all other things being equal) the MS error term will be larger, which (all other things being equal) means a smaller F-ratio for our test of the treatment effect. Now all other things are not equal, because the inclusion of subjects has lowered the error term SS as well … so whether the inclusion of additional factors helps us, in the end, or not, ultimately depends on the usual question: is the reduction in error gained by the inclusion of the additional term in the model worth it, given that we have to”pay" a certain number of degrees of freedom?

Correct method (univariate approach)

Now the correct way of running a repeated-measures ANOVA is not to pretend that this is a 2-factor between-subjects design, minus the interaction term—(even though in the end, the computations of the ANOVA table are the same). Here is the correct way of doing it:

am3 <- aov(dv ~ treatment + Error(subject/treatment), data=mydata)
summary(am3)
##
## Error: subject
##           Df Sum Sq Mean Sq F value Pr(>F)
## Residuals  9   48.4   5.378
##
## Error: subject:treatment
##           Df Sum Sq Mean Sq F value   Pr(>F)
## treatment  3   38.9  12.967   12.24 3.06e-05 ***
## Residuals 27   28.6   1.059
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Note that the specification of the model has a new element we have not seen before, namely the addition of an error term: +Error(subject/treatment). This notation tells R to slice out of the model an additional error term corresponding to the variance accounted for by subjects. The subjects/treatment notation tells R that the treatment factor is the repeated-measures factor over which subjects applies.

As you can see the output is essentially the same, just reorganized a bit. The variance accounted for by subjects has not only been removed from the model’s overall error term, but has even been physically separated in the output. So now, the effect of treatment is significant at p=3.06e-05.

This is the correct method of running a single factor repeated-measures ANOVA, using what is called a univariate approach. That is to say, there is a single dependent variable (called dv in our data frame mydata). We have been doing univariate procedures all along in the course so far.

Multivariate approach

There is another general method of performing ANOVA when there are one or more repeated-measures factors, called the multivariate approach. Maxwell & Delaney devote a lot of material to going over the rationale. In short, although it is slightly more complex conceptually, one benefit of using a multivariate approach is that it provides ways of testing the sphericity assumption—the assumption of homogeneity of variance of differences between groups—as well as providing corrected versions of the F-test assuming the assumption has been violated. As you know from your reading, repeated-measures ANOVA is actually quite sensitive to the sphericity assumption, and what’s more, the sphericity assumption is very often violated.

Here is how to perform a repeated-measures ANOVA in R using a multivariate approach. First, we must reorganize the data into a format in which each row represents a single subject, and columns represent levels of the treatment factor. This may be familiar to those of you who have used SPSS—this is the way repeated-measures factors are organized in SPSS data tables as well.

response <- with(mydata,cbind(dv[treatment==1], dv[treatment==2],
dv[treatment==3], dv[treatment==4]))

Our data are now essentially in a matrix format:

response
##       [,1] [,2] [,3] [,4]
##  [1,]    8   10    7    5
##  [2,]    9    9    8    6
##  [3,]    7    5    8    4
##  [4,]    9    6    5    7
##  [5,]    8    7    7    6
##  [6,]    5    4    4    3
##  [7,]    7    6    5    4
##  [8,]    8    8    6    6
##  [9,]    9    8    6    5
## [10,]    7    7    4    5

We now form the multivariate model using the lm() function in R:

mlm1 <- lm(response ~ 1)
mlm1
##
## Call:
## lm(formula = response ~ 1)
##
## Coefficients:
##              [,1]  [,2]  [,3]  [,4]
## (Intercept)  7.7   7.0   6.0   5.1

The ~1 notation simply tells R that there are no between-subjects factors here … in other words, only fit the model using intercepts. If you type mlm1 to look at the model object you will see that four intercepts were fit—one representing the mean of each of the four levels of the dependent variable:

Now we must set up a variable that defines the design of our study, which is simple in this case, it’s a single factor with four levels:

rfactor <- factor(c("r1", "r2", "r3", "r4"))

Now we must load the car library into R (we only have to do this once in a given session), because we are going to make use of the Anova() function (note the capital A, this is different than the anova() function). The Anova() function calculates ANOVA tables for a number of different kinds of model objects including multivariate model objects.

library(car)
## Loading required package: carData

We now define a new anova model object mlm1.aov by a function call to Anova():

mlm1.aov <- Anova(mlm1, idata=data.frame(rfactor), idesign = ~rfactor, type="III")

The first argument, mlm1, is our multivariate model defined above. The second argument, idata=data.frame(rfactor) passes information about the within-subjects variable, in other words how many levels there are. The third argument, idesign=~rfactor, passes information about the within-subjects design, in other words that the variable that rfactor describes is the repeated-measures variable. The fourth argument, type="III", instructs Anova() to calculate the “Type-III” sums of squares when forming the ANOVA table. This only matters for so-called “unbalanced designs” in which there are different numbers of observations in different groups. Maxwell and Delaney do a good job of explaining how this is related to different ways of computing sums of squares. In the summary() command we specify multivariate=FALSE because we want to suppress the portion of the output of the Anova() function that is related to multivariate statistical tests—these are really only relevant when the experimental design truly is a multivariate design (in other words when there are multiple dependent variables).

Here is the output we get:

summary(mlm1.aov, multivariate=FALSE)
##
## Univariate Type III Repeated-Measures ANOVA Assuming Sphericity
##
##             Sum Sq num Df Error SS den Df F value    Pr(>F)
## (Intercept) 1664.1      1     48.4      9 309.440 2.808e-08 ***
## rfactor       38.9      3     28.6     27  12.241 3.060e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
##
## Mauchly Tests for Sphericity
##
##         Test statistic p-value
## rfactor        0.34613 0.14884
##
##
## Greenhouse-Geisser and Huynh-Feldt Corrections
##  for Departure from Sphericity
##
##         GG eps Pr(>F[GG])
## rfactor 0.7426  0.0002388 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
##            HF eps   Pr(>F[HF])
## rfactor 0.9981017 3.106252e-05

First we see that the test of the repeated measures factor (called rfactor here) is significant at p=3.06e-05. Then we see that a test of sphericity was performed (Mauchly’s test), which is not significant (p=0.14884). Nevertheless, we get two corrected versions of the F-test: one called Greenhouse-Geisser, and another, slightly less conservative correction, called Huynh-Feldt. In this case, both are also significant (the p-values for GG and HF are marked Pr(>F[GG]) and Pr(>F[HF]), respectively). See your Maxwell and Delaney readings for details about how these corrections are computed.

Comparisons between individual means

There are two legitimate ways of testing differences between individual means. The first is to use the same F-ratio that we have seen from the between-subjects ANOVA, for testing contrasts $$\psi$$:

$$$F_{comp} = \frac{\psi^{2}/\sum_{j=1}^{a}(c^{2}_{j}/n_{j})}{MS_{error}}$$$

where $$\psi$$ is the contrast of interest:

$$$\psi = \sum_{j=1}^{a}(c_{j}\bar{Y}_{j})$$$

So for example if we want to test the mean of group 1 minus the mean of group 2, the weights on the contrast would be $$c_{1}=(+1)$$ and $$c_{2}=(-1)$$. We would simply compute the $$MS_{error}$$ term based on the ANOVA output. Remember, $$MS_{error}=SS_{error}/df_{error}$$:

df2 <- 27 # df for error term
sserr <- 28.6 # ss error term
mserr <- sserr/df2 # compute mserr
n <- 10 # num subjects per group
a <- 4 # num groups
mresp1 <- mean(response[,1]) # mean of resp1
mresp2 <- mean(response[,2]) # mean of resp2
sscomp <- (mresp1-mresp2)^2 # ss comparison
dfcomp <- 1 # df comparison
Fcomp <- (n*sscomp/2)/(mserr) # Fobs for comparison
Fcomp
## [1] 2.312937

We can then compute a probability using the pf() function:

pcomp <- 1-pf(Fcomp, dfcomp, df2) # pobs for comparison
pcomp
## [1] 0.1399272

In this case, the mean of group 1 is not significantly different than the mean of group 2, $$p=0.1399$$. Note that this is a pairwise comparison that is uncorrected for Type-I error. We can, for example, perform this as a Tukey test, by transforming $$Fcomp$$ into a value of $$q$$ (studentized range statistic), and then computing a probability (see Maxwell & Delaney, pg. 550, equation 41)

qobs <- sqrt(2*Fcomp) # compute q value
pt <- 1 - ptukey(qobs, 4, (4-1)*(n-1)) # df for q are (a,(a-1)(n-1))
pt
## [1] 0.4394847

Again, mean 1 is not significantly different than mean 2, $$p=0.5605$$.

If you’re concerned about violating the sphericity assumption (heterogeneity of variances of differences between groups), then you might notice that the method above is based on using the $$MS_{error}$$ term from the ANOVA for the denominator of the F-test. Essentially this is a sort of pooled estimate of variance based on all groups. If sphericity is violated, then it may be a better approach to perform a pairwise test in a way that only uses variances of the groups of interest to perform the test. One way to do this is by forming a contrast, based on a set of weights (as above), but that is applied not to the means of each group, but to individual subject scores, to create composite scores $$\psi_{i}$$:

$$$\psi_{i} = \sum_{j=1}^{a}c_{j}Y_{ij}$$$

Then the values of the composite scores $$\psi_{i}$$ are evaluated using a t-test. The extent to which they are non-zero reflects the extent to which the contrast is significant. In other words the t-test tests the hypothesis that the composite scores were drawn from a zero-mean population. Here is an example, again comparing group 1 vs group 2:

# fancy matrix multiplication way of computing composite scores:
mycontrast <- c(+1, -1, 0, 0)
dscores <- response %*% mycontrast
# or you can simply do this:
dscores <- response[,1]-response[,2]
t.test(dscores)
##
##  One Sample t-test
##
## data:  dscores
## t = 1.655, df = 9, p-value = 0.1323
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  -0.2567852  1.6567852
## sample estimates:
## mean of x
##       0.7

So the difference between groups 1 and 2 is not significant ($$p=0.1323$$). Note that this test is uncorrected for Type-I error. You could always transform a value of the $$t$$ statistic into an F (when $$df_{num}=1$$, $$F=t^{2}$$), then into a $$q$$ statistic (as above) to perform it as a Tukey test.

A more straightforward way: pairwise.t.test()

There is a function called pairwise.t.test() that will do pairwise t tests given data in a format like we loaded in originally, namely dependent variable in one column, and a grouping variable coding factor levels in another column. We can also ask for some flavours of multiple comparisons corrections, like for example the Bonferroni-Holm correction. Here is how we would use it to do post-hoc pairwise t-tests on our repeated measures data, using Bonferroni-Holm corrections:

with(mydata, pairwise.t.test(x=dv, g=treatment, p.adjust.method="holm", paired=TRUE))
##
##  Pairwise comparisons using paired t tests
##
## data:  dv and treatment
##
##   1       2     3
## 2 0.320   -     -
## 3 0.021   0.320 -
## 4 5.5e-06 0.021 0.320
##
## P value adjustment method: holm

We’re given a table of p-values for our t-tests. in this case treatment levels 1 and 3 are significantly different (p=0.021), levels 1 and 4 are significantly different (p=5.5e-06), and levels 2 and 4 are significantly different (p=0.021).