# How to Conduct a Two-Way ANOVA in R This tutorial explains how to conduct a two-way ANOVA in R.

## What is a Two-Way ANOVA?

two-way ANOVA(“analysis of variance”) is used to determine whether or not there is a statistically significant difference between the means of three or more independent groups that have been split on two factors.

The purpose of a two-way ANOVA is to determine how two factors impact a response variable, and to determine whether or not there is an interaction between the two factors on the response variable.

This type of test is called a two-way ANOVA because we are analyzing how two factors impacts our response variable.

## How to Conduct a Two-Way ANOVA in R

The following example illustrates how to conduct a two-way ANOVA in R.

### Background

Suppose we want to determine if exercise (no exercise, light exercise, intense exercise) and gender (male, female) impact weight loss. In this case, the two factors we’re studying are exercise and gender and the response variable is weight loss, measured in pounds. We can conduct a two-way ANOVA to determine if exercise and gender impact weight loss and to determine if there is an interaction between exercise and gender on weight loss.

We recruit 30 men and 30 women to participate in an experiment in which we randomly assign 10 of each to follow a program of either no exercise, light exercise, or intense exercise for one month.

The following code creates the data frame we’ll be working with:

```#make this example reproducible
set.seed(10)

#create data frame
data <- data.frame(gender = rep(c("Male", "Female"), each = 30),
exercise = rep(c("None", "Light", "Intense"), each = 10, times = 2),
weight_loss = c(runif(10, -3, 3), runif(10, 0, 5), runif(10, 5, 9),
runif(10, -4, 2), runif(10, 0, 3), runif(10, 3, 8)))

#view first six rows of data frame

#  gender exercise weight_loss
#1   Male     None  0.04486922
#2   Male     None -1.15938896
#3   Male     None -0.43855400
#4   Male     None  1.15861249
#5   Male     None -2.48918419
#6   Male     None -1.64738030

#see how many participants are in each group
table(data\$gender, data\$exercise)

#         Intense Light None
#  Female      10    10   10
#  Male        10    10   10
```

### Exploring the Data

Before we even fit the two-way ANOVA model, we can gain a better understanding of the data by finding the mean and standard deviation of weight loss for each of the six treatment groups using the dplyr package:

```#load dplyr package
library(dplyr)

#find mean and standard deviation of weight loss for each treatment group
data %>%
group_by(gender, exercise) %>%
summarise(mean = mean(weight_loss),
sd = sd(weight_loss))

# A tibble: 6 x 4
# Groups:   gender 
#  gender exercise   mean    sd
#
#1 Female Intense   5.31  1.02
#2 Female Light     0.920 0.835
#3 Female None     -0.501 1.77
#4 Male   Intense   7.37  0.928
#5 Male   Light     2.13  1.22
#6 Male   None     -0.698 1.12
```

We can also create a boxplot for each of the six treatment groups to visualize the distribution of weight loss for each group:

```#set margins so that axis labels on boxplot don't get cut off
par(mar=c(8, 4.1, 4.1, 2.1))

#create boxplots
boxplot(weight_loss ~ gender:exercise,
data = data,
main = "Weight Loss Distribution by Group",
xlab = "Group",
ylab = "Weight Loss",
col = "steelblue",
border = "black",
las = 2 #make x-axis labels perpendicular
)``` Right away we can see that the two groups who participated in intense exercise appear to have greater weight loss values. We can also see that males tend to have higher weight loss values for the intense and light exercise groups compared to females.

Next, we’ll fit the two-way ANOVA model to our data to see if these visual differences are actually statistically significant.

### Fitting the Two-Way ANOVA Model

The general syntax to fit a two-way ANOVA model in R is as follows:

aov(response variable ~ predictor_variable1 * predictor_variable2, data = dataset)

Note that the * between the two predictor variables indicates that we also want to test for an interaction effect between the two predictor variables.

In our example, we can use the following code to fit the two-way ANOVA model, using weight_loss as the response variable and gender and exercise as our two predictor variables. We can then use the summary() function to view the output of our model:

```#fit the two-way ANOVA model
model <- aov(weight_loss ~ gender * exercise, data = data)

#view the model output
summary(model)

#                Df Sum Sq Mean Sq F value Pr(>F)
#gender           1   15.8   15.80  11.197 0.0015 **
#exercise         2  505.6  252.78 179.087 <2e-16 ***
#gender:exercise  2   13.0    6.51   4.615 0.0141 *
#Residuals       54   76.2    1.41
#---
#Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
```

From the model output we can see that genderexercise, and the interaction between the two variables are all statistically significant at the .05 significance level.

### Checking the Model Assumptions

Before we go any further, we should check to see that the assumptions of our model are met so that the our results from the model are reliable. In particular, a two-way ANOVA assumes:

1. Independence – the observations in each group need to be independent of each other. Since we used a randomized design, this assumption should be met so we don’t need to worry too much about this.

2. Normality – the dependent variable should be approximately normally distributed for each combination of the groups of the two factors.

One way to check this assumption is to create a histogram of the model residuals. If the residuals are roughly normally distributed, this assumption should be met.

```#define model residuals
resid <- model\$residuals

#create histogram of residuals
hist(resid, main = "Histogram of Residuals", xlab = "Residuals", col = "steelblue")``` The residuals are roughly normally distributed, so we can assume the normality assumption is met.

3. Equal Variance – the variances for each group are equal or approximately equal.

One way to check this assumption is to conduct a Levene’s Test for equality of variances using the car package:

```#load car package
library(car)

#conduct Levene's Test for equality of variances
leveneTest(weight_loss ~ gender * exercise, data = data)

#Levene's Test for Homogeneity of Variance (center = median)
#      Df F value Pr(>F)
#group  5  1.8547 0.1177
#      54
```

Since the p-value of the test is greater than our significance level of 0.05, we can assume that our assumption of equality of variances among groups is met.

### Analyzing Treatment Differences

Once we have verified that the model assumptions are met, we can then conduct a post hoc test to determine exactly which treatment groups differ from one another.

For our post hoc test, we will use the function TukeyHSD() to conduct Tukey’s Test for multiple comparisons:

```#perform Tukey's Test for multiple comparisons
TukeyHSD(model, conf.level=.95)

#  Tukey multiple comparisons of means
#    95% family-wise confidence level
#
#Fit: aov(formula = weight_loss ~ gender * exercise, data = data)
#
#\$gender
#                diff       lwr      upr     p adj
#Male-Female 1.026456 0.4114451 1.641467 0.0014967
#
#\$exercise
#                   diff       lwr       upr   p adj
#Light-Intense -4.813064 -5.718493 -3.907635 0.0e+00
#None-Intense  -6.938966 -7.844395 -6.033537 0.0e+00
#None-Light    -2.125902 -3.031331 -1.220473 1.8e-06
#
#\$`gender:exercise`
#                                  diff        lwr         upr     p adj
#Male:Intense-Female:Intense  2.0628297  0.4930588  3.63260067 0.0036746
#Female:Light-Female:Intense -4.3883563 -5.9581272 -2.81858535 0.0000000
#Male:Light-Female:Intense   -3.1749419 -4.7447128 -1.60517092 0.0000027
#Female:None-Female:Intense  -5.8091131 -7.3788841 -4.23934219 0.0000000
#Male:None-Female:Intense    -6.0059891 -7.5757600 -4.43621813 0.0000000
#Female:Light-Male:Intense   -6.4511860 -8.0209570 -4.88141508 0.0000000
#Male:Light-Male:Intense     -5.2377716 -6.8075425 -3.66800066 0.0000000
#Female:None-Male:Intense    -7.8719429 -9.4417138 -6.30217192 0.0000000
#Male:None-Male:Intense      -8.0688188 -9.6385897 -6.49904786 0.0000000
#Male:Light-Female:Light      1.2134144 -0.3563565  2.78318536 0.2185439
#Female:None-Female:Light    -1.4207568 -2.9905278  0.14901410 0.0974193
#Male:None-Female:Light      -1.6176328 -3.1874037 -0.04786184 0.0398106
#Female:None-Male:Light      -2.6341713 -4.2039422 -1.06440032 0.0001050
#Male:None-Male:Light        -2.8310472 -4.4008181 -1.26127627 0.0000284
#Male:None-Female:None       -0.1968759 -1.7666469  1.37289500 0.9990364```

The p-value indicates whether or not there is a statistically significant difference between each group. For example, in the last row above we see that the male group with no exercise did not experience a statistically significant difference in weight loss compared to the female group with no exercise (p-value: 0.990364).

We can also visualize the 95% confidence intervals that result from the Tukey Test by using the plot() function in R:

```#set axis margins so labels don't get cut off
par(mar=c(4.1, 13, 4.1, 2.1))

#create confidence interval for each comparison
plot(TukeyHSD(model, conf.level=.95), las = 2)
``` ### Reporting the Results of the Two-Way ANOVA

Lastly, we can report the results of the two-way ANOVA in such a way that summarizes the findings:

A two-way ANOVA was conducted to examine the effects of gender (male, female) and exercise regimen (none, light, intense) on weight loss (measure in lbs). There was a statistically significant interaction between the effects of gender and exercise on weight loss (F(2, 54) = 4.615, p = 0.0141). Tukey’s HSD post hoc tests were carried out.

For males, an intense exercise regimen lead to significantly higher weight loss compared to both a light regimen (p < .0001) and no exercise regimen (p < .0001). In addition for males, a light regimen lead to significantly higher weight loss compared to no exercise regimen (p < .0001).

For females, an intense exercise regimen lead to significantly higher weight loss compared to both a light regimen (p < .0001) and no exercise regimen (p < .0001).

Normality checks and Levene’s test were conducted to verify that the ANOVA assumptions were met.