Contrast Analysis with R: Tutorial for obtaining contrast estimates in a 2-way factorial design

In this post, I want to show how contrast estimates can be obtained with R. In particular, I want to show how we can replicate, with R, a contrast analysis of an interaction contrast in a 2 x 4 between subjects design.

Our example is from Haans (2018; see also this post). It considers the effect of students' seating distance from the teacher and the educational performance of the students: the closer to the teacher the student is seated, the higher the performance. A "theory "explaining the effect is that the effect is mainly caused by the teacher having decreased levels of eye contact with the students sitting farther to the back in the lecture hall.

To test that theory, a experiment was conducted with N = 72 participants attending a lecture. The lecture was given to two independent groups of 36 participants. The first group attended the lecture while the teacher was wearing dark sunglasses, the second group attented the lecture while the teacher was not wearing sunglasses. All participants were randomly assigned to 1 of 4 possible rows, with row 1 being closest to the teacher and row 4 the furthest from the teacher The dependent variable was the score on a 10-item questionnaire about the contents of the lecture. So, we have a 2 by 4 factorial design, with n = 9 participants in each combination of the factor levels. 

Here we focus on obtaining an interaction contrast: we will estimate the extent to which the difference between the mean retention score of the participants on the first row and those on the other rows differs between the conditions with and without sunglasses. 

The interaction contrast with SPSS

Custom contrasts for the one-way repeated measures design using Lmer

Here is some code for doing one-way repeated measures analysis with lme4 and custom contrasts. We will use a repeated measures design with three conditions of the factor Treat and 20 participants. The contrasts are Helmert contrasts, but they differ from the built-in Helmert contrasts in that the sum of the absolute values of the contrasts weights equals 2 for each contrast.
The standard error of each contrast equals the square root of the product of the sum of the squared contrast weight w and the residual variance divided by the number of participants n.

$$\sigma_{\hat{\psi}} =  \sqrt{\sum{w_i}\sigma^2_e/n}$$

The residual variance equals the within treatment variance times 1 minus the correlation between conditions. (Which equals the within treatment variance minus 1 times the covariance $\rho\sigma^2_{within}$) .

$$\sigma^2_e = \sigma^2_{within}(1 - \rho)$$

In the example below, the within treatment variance equals 1 and the covariance 0.5 (so the value of the correlation is .50 as well). The residual variance is therefore equal to .50.

For the first contrasts, the weights are equal to {-1, 1, 0}, so the value of the standard error of the contrasts should be equal to the square root of 2*.50/20 = 0.2236. 

library(MASS)

library(lme4)

# setting up treatment and participants factors
nTreat = 3
nPP = 20
Treat <- factor(rep(1:nTreat, each=nPP))
PP <- factor(rep(1:nPP, nTreat))

# generate some random 
# specify means

means = c(0, .20, .50)

# create variance-covariance matrix
# var = 1, cov = .5
Sigma = matrix(rep(.5, 9), 3, 3)
diag(Sigma) <- 1

# generate the data; using empirical = TRUE
# so that variance and covariance are known
# set to FALSE for "real" random data

sco = as.vector(mvrnorm(nPP, means, Sigma, empirical = TRUE))

#setting up custom contrasts for Treatment factor 
myContrasts <- rbind(c(-1, 1, 0), c(-.5, -.5, 1))
contrasts(Treat) <- ginv(myContrasts)

#fit linear mixed effects model: 
myModel <- lmer(sco ~ Treat + (1|PP))

summary(myModel)

## Linear mixed model fit by REML ['lmerMod']
## Formula: sco ~ Treat + (1 | PP)
## 
## REML criterion at convergence: 157.6
## 
## Scaled residuals: 
##     Min      1Q  Median      3Q     Max 
## -2.6676 -0.4869  0.1056  0.6053  1.9529 
## 
## Random effects:
##  Groups   Name        Variance Std.Dev.
##  PP       (Intercept) 0.5      0.7071  
##  Residual             0.5      0.7071  
## Number of obs: 60, groups:  PP, 20
## 
## Fixed effects:
##             Estimate Std. Error t value
## (Intercept)   0.2333     0.1826   1.278
## Treat1        0.2000     0.2236   0.894
## Treat2        0.4000     0.1936   2.066
## 
## Correlation of Fixed Effects:
##        (Intr) Treat1
## Treat1 0.000        
## Treat2 0.000  0.000

Planning for a precise contrast estimate: the mixed model case

In a previous post (here), we saw how we can determine sample size for obtaining, with assurance, a precise interaction contrast estimate. In that post we considered a 2 x 2 factorial design. In this post, I will extend the discussion to the mixed model case. That is, we will consider sample size planning for a precise interaction estimate in case of a design with 2 fixed factors and two random factors: participant and stimulus (item). (A pdf version of this post can be found here: view pdf. )

In order to keep things relatively simple, we will focus on a design where both participants and items are nested under condition. So, each treatment condition has a unique sample of participants and items. We will call this design the both-within-condition design  (see, for instance, Westfall et al. 2014, for detailed descriptions of this design). We will analyse the 2 x 2 factorial design as a single factor design (the factor has a = 4 levels) and formulate an interaction contrast.

Planning for Precision: A confidence interval for the contrast estimate

In a previous post, which can be found here, I described how the relative error variance of a treatment mean can be obtained by combining variance components.  I concluded that post by mentioning how this relative error variance for the treatment mean can be used to obtain the variance of a contrast estimate. In this post, I will discuss a little more how this latter variance can be used to obtain a confidence interval for the contrast estimate, but we take a few steps back and consider a relatively simple study.

The plan of this post is as follows. We will have a look at the analysis of a factorial design and focus on estimating an interaction effect. We will consider both the NHST approach and an estimation approach. We will use both 'hand calculations' and SPSS.

An important didactic aspect of this post is to show the connection between the ANOVA source table and estimates of the standard error of a contrast estimate. Understanding that connections helps in understanding one of my planned posts on how obtaining these estimates work in the case of mixed model ANOVA.  See the final section of this post.

Planning for Precision: Introduction to variance components

Both theory underlying the Precision application and the use of the app in practice rely for a large part on specifying variance components. In this post, I will give you some more details about what these components are, and how they relate to the analysis of variance model underlying the app.

What is variance?

Let's start with a relatively simple conceptual explanation of variance. The key ideas are expected value and error.  Suppose you randomly select a single score from a population of possible values. Let's suppose furthermore that the population of values can be described with a normal distribution. There is actually no need to suppose a normal distribution, but it makes the explanation relatively easy to follow.

As you probably know, the normal distribution is centered around its mean value, which is (equal to) the parameter μ. We call this parameter the population mean.

Now, we select a single random value from the population. Let's call this value X.  Because we know something about the probability distribution of the population values, we are also in the position to specify an expected value for the score X. Let's use the symbol E(X) for this expected value. The value of E(X) proves (and can be proven) to be equal to the parameter μ. (Conceptually, the expectation of a variable can be considered as its long run average).

Of course, the actual value obtained will in general not be equal to the expected value, at least not if we sample from continuous distributions like the normal distribution. Let's call the difference between the value X and it's expectation E(X) = μ. an error, deviation or residual: e = X - E(X) = X -  μ.

We would like to have some indication of the extent to which X differs from its expectation, especially when E(X) is estimated on the basis of  a statistical model.  Thus, we would like to have something like E(X - E(X)) = E(X -  μ). The variance gives us such an indication, but does so in squared units, because working with the expected error itself always leads to the value 0  E(X -  μ) = E(X) - E( μ) =  μ -  μ = 0. (This simply says that on average the error is zero; the standard explanation is that negative and positive errors cancel out in the long run).

The variance is the expected squared deviation (mean squared error) between X and its expectation: E((X - E(X))²) = E(X -  μ)²), and the symbol for the population value is σ².

Some examples of variances (remember we are talking conceptually here):
- the variance of the mean, is the expected squared deviation between a sample mean and its expectation the population mean.
- the variance of the difference between two means:  the expected squared deviation between the sample difference and the population difference between two means.
- the variance of a contrast: the expected squared deviation between the sample value of the contrast and the population value of the contrast.

It's really not that complicated, I believe.