# 18. Factorial ANOVA¶

Over the course of the last few chapters you can probably detect a general trend. We started out looking at tools that you can use to compare two groups to one another, most notably the $$t$$-test. Then, we introduced analysis of variance (ANOVA) as a method for comparing more than two groups. The chapter on regression covered a somewhat different topic, but in doing so it introduced a powerful new idea: building statistical models that have multiple predictor variables used to explain a single outcome variable. For instance, a regression model could be used to predict the number of errors a student makes in a reading comprehension test based on the number of hours they studied for the test, and their score on a standardised IQ test. The goal in this chapter is to import this idea into the ANOVA framework. For instance, suppose we were interested in using the reading comprehension test to measure student achievements in three different schools, and we suspect that girls and boys are developing at different rates (and so would be expected to have different performance on average). Each student is classified in two different ways: on the basis of their gender, and on the basis of their school. What we’d like to do is analyse the reading comprehension scores in terms of both of these grouping variables. The tool for doing so is generically referred to as factorial ANOVA. However, since we have two grouping variables, we sometimes refer to the analysis as a two-way ANOVA, in contrast to the one-way ANOVAs that we ran earlier.

## 18.1. Factorial ANOVA 1: balanced designs, no interactions¶

When we discussed analysis of variance, we assumed a fairly simple experimental design: each person falls into one of several groups, and we want to know whether these groups have different means on some outcome variable. In this section, I’ll discuss a broader class of experimental designs, known as factorial designs, in we have more than one grouping variable. I gave one example of how this kind of design might arise above. Another example appears in the chapter on ANOVA, in which we were looking at the effect of different drugs on the mood_gain experienced by each person. In that chapter we did find a significant effect of drug, but at the end of the chapter we also ran an analysis to see if there was an effect of therapy. We didn’t find one, but there’s something a bit worrying about trying to run two separate analyses trying to predict the same outcome. Maybe there actually is an effect of therapy on mood gain, but we couldn’t find it because it was being “hidden” by the effect of drug? In other words, we’re going to want to run a single analysis that includes both drug and therapy as predictors. For this analysis each person is cross-classified by the drug they were given (a factor with 3 levels) and what therapy they received (a factor with 2 levels). We refer to this as a $$3 \times 2$$ factorial design. If we cross-tabulate drug by therapy using the crosstab() in pandas, we get the following table:

import pandas as pd

pd.crosstab(index=df["drug"], columns=df["therapy"],margins=False)

therapy CBT no.therapy
drug
anxifree 3 3
joyzepam 3 3
placebo 3 3

As you can see, not only do we have participants corresponding to all possible combinations of the two factors, indicating that our design is completely crossed, it turns out that there are an equal number of people in each group. In other words, we have a balanced design. In this section I’ll talk about how to analyse data from balanced designs, since this is the simplest case. The story for unbalanced designs is quite tedious, so we’ll put it to one side for the moment.

### 18.1.1. What hypotheses are we testing?¶

Like one-way ANOVA, factorial ANOVA is a tool for testing certain types of hypotheses about population means. So a sensible place to start would be to be explicit about what our hypotheses actually are. However, before we can even get to that point, it’s really useful to have some clean and simple notation to describe the population means. Because of the fact that observations are cross-classified in terms of two different factors, there are quite a lot of different means that one might be interested. To see this, let’s start by thinking about all the different sample means that we can calculate for this kind of design. Firstly, there’s the obvious idea that we might be interested in this table of group means:

round(df.groupby(['therapy', 'drug'])[['mood_gain']].mean(),2)

mood_gain
therapy drug
CBT anxifree 1.03
joyzepam 1.50
placebo 0.60
no.therapy anxifree 0.40
joyzepam 1.47
placebo 0.30

Now, this output shows a cross-tabulation of the group means for all possible combinations of the two factors (e.g., people who received the placebo and no therapy, people who received the placebo while getting CBT, etc). However, we can also construct tables that ignore one of the two factors. That gives us output that looks like this:

round(df.groupby(['therapy'])[['mood_gain']].mean(),2)

mood_gain
therapy
CBT 1.04
no.therapy 0.72
round(df.groupby(['drug'])[['mood_gain']].mean(),2)

mood_gain
drug
anxifree 0.72
joyzepam 1.48
placebo 0.45

But of course, if we can ignore one factor we can certainly ignore both. That is, we might also be interested in calculating the average mood gain across all 18 participants, regardless of what drug or psychological therapy they were given:

round(df['mood_gain'].mean(),2)

0.88


At this point we have 12 different sample means to keep track of! It is helpful to organise all these numbers into a single table, which would look like this:

no therapy

CBT

total

no therapy

CBT

total

placebo

0.30

0.60

0.45

anxifree

0.40

1.03

0.72

joyzepam

1.47

1.50

1.48

total

0.72

1.04

0.88

Now, each of these different means is of course a sample statistic: it’s a quantity that pertains to the specific observations that we’ve made during our study. What we want to make inferences about are the corresponding population parameters: that is, the true means as they exist within some broader population. Those population means can also be organised into a similar table, but we’ll need a little mathematical notation to do so. As usual, I’ll use the symbol $$\mu$$ to denote a population mean. However, because there are lots of different means, I’ll need to use subscripts to distinguish between them.

Here’s how the notation works. Our table is defined in terms of two factors: each row corresponds to a different level of Factor A (in this case drug), and each column corresponds to a different level of Factor B (in this case therapy). If we let $$R$$ denote the number of rows in the table, and $$C$$ denote the number of columns, we can refer to this as an $$R \times C$$ factorial ANOVA. In this case $$R=3$$ and $$C=2$$. We’ll use lowercase letters to refer to specific rows and columns, so $$\mu_{rc}$$ refers to the population mean associated with the $$r$$th level of Factor A (i.e. row number $$r$$) and the $$c$$th level of Factor B (column number $$c$$).1 So the population means are now written like this:

no therapy

CBT

total

placebo

$$\mu_{11}$$

$$\mu_{12}$$

anxifree

$$\mu_{21}$$

$$\mu_{22}$$

joyzepam

$$\mu_{31}$$

$$\mu_{32}$$

total

Okay, what about the remaining entries? For instance, how should we describe the average mood gain across the entire (hypothetical) population of people who might be given Joyzepam in an experiment like this, regardless of whether they were in CBT? We use the “dot” notation to express this. In the case of Joyzepam, notice that we’re talking about the mean associated with the third row in the table. That is, we’re averaging across two cell means (i.e., $$\mu_{31}$$ and $$\mu_{32}$$). The result of this averaging is referred to as a marginal mean, and would be denoted $$\mu_{3.}$$ in this case. The marginal mean for CBT corresponds to the population mean associated with the second column in the table, so we use the notation $$\mu_{.2}$$ to describe it. The grand mean is denoted $$\mu_{..}$$ because it is the mean obtained by averaging (marginalising2) over both. So our full table of population means can be written down like this:

no therapy

CBT

total

placebo

$$\mu_{11}$$

$$\mu_{12}$$

$$\mu_{1.}$$

anxifree

$$\mu_{21}$$

$$\mu_{22}$$

$$\mu_{2.}$$

joyzepam

$$\mu_{31}$$

$$\mu_{32}$$

$$\mu_{3.}$$

total

$$\mu_{.1}$$

$$\mu_{.2}$$

$$\mu_{..}$$

Now that we have this notation, it is straightforward to formulate and express some hypotheses. Let’s suppose that the goal is to find out two things: firstly, does the choice of drug have any effect on mood, and secondly, does CBT have any effect on mood? These aren’t the only hypotheses that we could formulate of course, and we’ll see a really important example of a different kind of hypothesis in the section on interactions, but these are the two simplest hypotheses to test, and so we’ll start there. Consider the first test. If drug has no effect, then we would expect all of the row means to be identical, right? So that’s our null hypothesis. On the other hand, if the drug does matter then we should expect these row means to be different. Formally, we write down our null and alternative hypotheses in terms of the equality of marginal means:

Null hypothesis $$H_0$$:

row means are the same i.e. $$\mu_{1.} = \mu_{2.} = \mu_{3.}$$

Alternative hypothesis $$H_1$$:

at least one row mean is different.

It’s worth noting that these are exactly the same statistical hypotheses that we formed when we ran a one-way ANOVA on these data way back when. Back then I used the notation $$\mu_P$$ to refer to the mean mood gain for the placebo group, with $$\mu_A$$ and $$\mu_J$$ corresponding to the group means for the two drugs, and the null hypothesis was $$\mu_P = \mu_A = \mu_J$$. So we’re actually talking about the same hypothesis: it’s just that the more complicated ANOVA requires more careful notation due to the presence of multiple grouping variables, so we’re now referring to this hypothesis as $$\mu_{1.} = \mu_{2.} = \mu_{3.}$$. However, as we’ll see shortly, although the hypothesis is identical, the test of that hypothesis is subtly different due to the fact that we’re now acknowledging the existence of the second grouping variable.

Speaking of the other grouping variable, you won’t be surprised to discover that our second hypothesis test is formulated the same way. However, since we’re talking about the psychological therapy rather than drugs, our null hypothesis now corresponds to the equality of the column means:

Null hypothesis $$H_0$$:

column means are the same, i.e., $$\mu_{.1} = \mu_{.2}$$

Alternative hypothesis $$H_1$$:

column means are different, i.e., $$\mu_{.1} \neq \mu_{.2}$$

### 18.1.2. Running the analysis in Python¶

If the data you’re trying to analyse correspond to a balanced factorial design, then running your analysis of variance is easy. To see how easy it is, let’s start by reproducing the original analysis from earlier. In case you’ve forgotten, for that analysis we were using only a single factor (i.e., drug) as our between-subjects variable to predict our outcome (dependent) variable (i.e., mood_gain), and so this was what we did:

import pingouin as pg

model1 = pg.anova(dv='mood_gain', between='drug', data=df, detailed=True)
round(model1, 2)

Source SS DF MS F p-unc np2
0 drug 3.45 2 1.73 18.61 0.0 0.71
1 Within 1.39 15 0.09 NaN NaN NaN

Note that this time around I’ve used the name model1 as the label for my aov object, since I’m planning on creating quite a few other models too. To start with, suppose I’m also curious to find out if therapy has a relationship to mood_gain. In light of what we’ve seen from our discussion of multiple regression, you probably won’t be surprised that all we have to do is extend the formula: in other words, if we specify dv=mood.gain, between=['drug', 'therapy'] as our model, we’ll probably get what we’re after:

model2 = pg.anova(dv='mood_gain', between=['drug', 'therapy'], data=df, detailed=True)
round(model2, 4)

Source SS DF MS F p-unc np2
0 drug 3.4533 2 1.7267 31.7143 0.0000 0.8409
1 therapy 0.4672 1 0.4672 8.5816 0.0126 0.4170
2 drug * therapy 0.2711 2 0.1356 2.4898 0.1246 0.2933
3 Residual 0.6533 12 0.0544 NaN NaN NaN

Most of this output is pretty simple to read too: the first row of the table reports a between-group sum of squares (SS) value associated with the drug factor, along with a corresponding between-group $$df$$ value. It also calculates a mean square value (MS), and $$F$$-statistic, an (uncorrected) $$p$$-value, and an estimate of the effect size (np2, that is, partial eta-squared). There is also a row corresponding to the therapy factor, and a row corresponding to the residuals (i.e., the within groups variation).

Now, the third row is a little trickier, so let’s just save that one for later, shall we? (Spoiler: this is the interaction of drug and therapy, but we’ll get there soon).

Not only are all of the individual quantities pretty familiar, the relationships between these different quantities has remained unchanged: just like we saw with the original one-way ANOVA, note that the mean square value is calculated by dividing SS by the corresponding $$df$$. That is, it’s still true that

$\mbox{MS} = \frac{\mbox{SS}}{df}$

regardless of whether we’re talking about drug, therapy or the residuals. To see this, let’s not worry about how the sums of squares values are calculated: instead, let’s take it on faith that Python has calculated the SS values correctly, and try to verify that all the rest of the numbers make sense. First, note that for the drug factor, we divide $$3.45$$ by $$2$$, and end up with a mean square value of $$1.73$$. For the therapy factor, there’s only 1 degree of freedom, so our calculations are even simpler: dividing $$0.47$$ (the SS value) by 1 gives us an answer of $$0.47$$ (the MS value).

Turning to the $$F$$ statistics and the $$p$$ values, notice that we have two of each: one corresponding to the drug factor and the other corresponding to the therapy factor. Regardless of which one we’re talking about, the $$F$$ statistic is calculated by dividing the mean square value associated with the factor by the mean square value associated with the residuals. If we use “A” as shorthand notation to refer to the first factor (factor A; in this case drug) and “R” as shorthand notation to refer to the residuals, then the $$F$$ statistic associated with factor A is denoted $$F_A$$, and is calculated as follows:

$F_{A} = \frac{\mbox{MS}_{A}}{\mbox{MS}_{R}}$

Note that this use of “R” to refer to residuals is a bit awkward, since we also used the letter R to refer to the number of rows in the table, but I’m only going to use “R” to mean residuals in the context of SS$$_R$$ and MS$$_R$$, so hopefully this shouldn’t be confusing. Anyway, to apply this formula to the drugs factor, we take the mean square of $$1.73$$ and divide it by the residual mean square value of $$0.07$$, which gives us an $$F$$-statistic of $$26.15$$. The corresponding calculation for the therapy variable would be to divide $$0.47$$ by $$0.07$$ which gives $$7.08$$ as the $$F$$-statistic. Not surprisingly, of course, these are the same values that R has reported in the ANOVA table above.

The last part of the ANOVA table is the calculation of the $$p$$ values. Once again, there is nothing new here: for each of our two factors, what we’re trying to do is test the null hypothesis that there is no relationship between the factor and the outcome variable (I’ll be a bit more precise about this later on). To that end, we’ve (apparently) followed a similar strategy that we did in the one way ANOVA, and have calculated an $$F$$-statistic for each of these hypotheses. To convert these to $$p$$ values, all we need to do is note that the that the sampling distribution for the $$F$$ statistic under the null hypothesis (that the factor in question is irrelevant) is an $$F$$ distribution: and that two degrees of freedom values are those corresponding to the factor, and those corresponding to the residuals. For the drug factor we’re talking about an $$F$$ distribution with 2 and 14 degrees of freedom (I’ll discuss degrees of freedom in more detail later). In contrast, for the therapy factor sampling distribution is $$F$$ with 1 and 14 degrees of freedom.

At this point, I hope you can see that the ANOVA table for this more complicated analysis corresponding to model2 should be read in much the same way as the ANOVA table for the simpler analysis for model1. In short, it’s telling us that the factorial ANOVA for our $$3 \times 2$$ design found a significant effect of drug ($$F_{2,12} = 31.71, p < .001$$) as well as a significant effect of therapy ($$F_{1,12} = 8.58, p = .01$$). Or, to use the more technically correct terminology, we would say that there are two main effects of drug and therapy. Why are these “main” effects? Well, because there could also be an interaction between the effect of drug and the effect of therapy. We’ll explore the concept of interactions below.

In the previous section I had two goals: firstly, to show you that the Python commands needed to do factorial ANOVA are pretty much the same ones that we used for a one way ANOVA. The only difference is that we add the number of predictors in the between argument of pingouin’s anova() function. Secondly, I wanted to show you what the ANOVA table looks like in this case, so that you can see from the outset that the basic logic and structure behind factorial ANOVA is the same as that which underpins one way ANOVA. Try to hold onto that feeling. It’s genuinely true, insofar as factorial ANOVA is built in more or less the same way as the simpler one-way ANOVA model. It’s just that this feeling of familiarity starts to evaporate once you start digging into the details. Traditionally, this comforting sensation is replaced by an urge to murder the the authors of statistics textbooks.

Okay, let’s start looking at some of those details. The explanation that I gave in the last section illustrates the fact that the hypothesis tests for the main effects (of drug and therapy in this case) are $$F$$-tests, but what it doesn’t do is show you how the sum of squares (SS) values are calculated. Nor does it tell you explicitly how to calculate degrees of freedom ($$df$$ values) though that’s a simple thing by comparison. Let’s assume for now that we have only two predictor variables, Factor A and Factor B. If we use $$Y$$ to refer to the outcome variable, then we would use $$Y_{rci}$$ to refer to the outcome associated with the $$i$$-th member of group $$rc$$ (i.e., level/row $$r$$ for Factor A and level/column $$c$$ for Factor B). Thus, if we use $$\bar{Y}$$ to refer to a sample mean, we can use the same notation as before to refer to group means, marginal means and grand means: that is, $$\bar{Y}_{rc}$$ is the sample mean associated with the $$r$$th level of Factor A and the $$c$$th level of Factor B, $$\bar{Y}_{r.}$$ would be the marginal mean for the $$r$$th level of Factor A, $$\bar{Y}_{.c}$$ would be the marginal mean for the $$c$$th level of Factor B, and $$\bar{Y}_{..}$$ is the grand mean. In other words, our sample means can be organised into the same table as the population means. For our clinical trial data, that table looks like this:

no therapy

CBT

total

placebo

$$\bar{Y}_{11}$$

$$\bar{Y}_{12}$$

$$\bar{Y}_{1.}$$

anxifree

$$\bar{Y}_{21}$$

$$\bar{Y}_{22}$$

$$\bar{Y}_{2.}$$

joyzepam

$$\bar{Y}_{31}$$

$$\bar{Y}_{32}$$

$$\bar{Y}_{3.}$$

total

$$\bar{Y}_{.1}$$

$$\bar{Y}_{.2}$$

$$\bar{Y}_{..}$$

And if we look at the sample means that I showed earlier, we have $$\bar{Y}_{11} = 0.30$$, $$\bar{Y}_{12} = 0.60$$ etc. In our clinical trial example, the drugs factor has 3 levels and the therapy factor has 2 levels, and so what we’re trying to run is a $$3 \times 2$$ factorial ANOVA. However, we’ll be a little more general and say that Factor A (the row factor) has $$R$$ levels and Factor B (the column factor) has $$C$$ levels, and so what we’re runnning here is an $$R \times C$$ factorial ANOVA.

Now that we’ve got our notation straight, we can compute the sum of squares values for each of the two factors in a relatively familiar way. For Factor A, our between group sum of squares is calculated by assessing the extent to which the (row) marginal means $$\bar{Y}_{1.}$$, $$\bar{Y}_{2.}$$ etc, are different from the grand mean $$\bar{Y}_{..}$$. We do this in the same way that we did for one-way ANOVA: calculate the sum of squared difference between the $$\bar{Y}_{i.}$$ values and the $$\bar{Y}_{..}$$ values. Specifically, if there are $$N$$ people in each group, then we calculate this:

$\mbox{SS}_{A} = (N \times C) \sum_{r=1}^R \left( \bar{Y}_{r.} - \bar{Y}_{..} \right)^2$

1

The nice thing about the subscript notation is that generalises nicely: if our experiment had involved a third factor, then we could just add a third subscript. In principle, the notation extends to as many factors as you might care to include, but in this book we’ll rarely consider analyses involving more than two factors, and never more than three.

2

Technically, marginalising isn’t quite identical to a regular mean: it’s a weighted average, where you take into account the frequency of the different events that you’re averaging over. However, in a balanced design, all of our cell frequencies are equal by definition, so the two are equivalent. We’ll discuss unbalanced designs later, and when we do so you’ll see that all of our calculations become a real headache. But let’s ignore this for now.