Import your data into R

1. Prepare your data as specified here: Best practices for preparing your data set for R

2. Save your data in an external .txt tab or .csv files

3. Import your data into R as follow:

# If .txt tab file, use this
my_data <- read.delim(file.choose())
# Or, if .csv file, use this
my_data <- read.csv(file.choose())

Here, we’ll use the built-in R data set named ToothGrowth. It contains data from a study evaluating the effect of vitamin C on tooth growth in Guinea pigs. The experiment has been performed on 60 pigs, where each animal received one of three dose levels of vitamin C (0.5, 1, and 2 mg/day) by one of two delivery methods, (orange juice or ascorbic acid (a form of vitamin C and coded as VC). Tooth length was measured and a sample of the data is shown below.

# Store the data in the variable my_data
my_data <- ToothGrowth

Check your data

To get an idea of what the data look like, we display a random sample of the data using the function sample_n()[in dplyr package]. First, install dplyr if you don’t have it:

install.packages("dplyr")

# Show a random sample
set.seed(1234)
dplyr::sample_n(my_data, 10)

    len supp dose
38  9.4   OJ  0.5
36 10.0   OJ  0.5
37  8.2   OJ  0.5
50 27.3   OJ  1.0
59 29.4   OJ  2.0
1   4.2   VC  0.5
13 15.2   VC  1.0
56 30.9   OJ  2.0
27 26.7   VC  2.0
53 22.4   OJ  2.0

# Check the structure
str(my_data)

'data.frame':   60 obs. of  3 variables:
 $ len : num  4.2 11.5 7.3 5.8 6.4 10 11.2 11.2 5.2 7 ...
 $ supp: Factor w/ 2 levels "OJ","VC": 2 2 2 2 2 2 2 2 2 2 ...
 $ dose: num  0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 ...

From the output above, R considers “dose” as a numeric variable. We’ll convert it as a factor variable (i.e., grouping variable) as follow.

# Convert dose as a factor and recode the levels
# as "D0.5", "D1", "D2"
my_data$dose <- factor(my_data$dose, 
                  levels = c(0.5, 1, 2),
                  labels = c("D0.5", "D1", "D2"))
head(my_data)

   len supp dose
1  4.2   VC D0.5
2 11.5   VC D0.5
3  7.3   VC D0.5
4  5.8   VC D0.5
5  6.4   VC D0.5
6 10.0   VC D0.5

Question: We want to know if tooth length depends on supp and dose.

• Generate frequency tables:

table(my_data$supp, my_data$dose)

    
     D0.5 D1 D2
  OJ   10 10 10
  VC   10 10 10

We have 2X3 design cells with the factors being supp and dose and 10 subjects in each cell. Here, we have a balanced design. In the next sections I’ll describe how to analyse data from balanced designs, since this is the simplest case.

Visualize your data

Box plots and line plots can be used to visualize group differences:

• Box plot to plot the data grouped by the combinations of the levels of the two factors.

• Two-way interaction plot, which plots the mean (or other summary) of the response for two-way combinations of factors, thereby illustrating possible interactions.

• To use R base graphs read this: R base graphs. Here, we’ll use the ggpubr R package for an easy ggplot2-based data visualization.

• Install the latest version of ggpubr from GitHub as follow (recommended):

# Install
if(!require(devtools)) install.packages("devtools")
devtools::install_github("kassambara/ggpubr")

• Or, install from CRAN as follow:

install.packages("ggpubr")

• Visualize your data with ggpubr:

# Box plot with multiple groups
# +++++++++++++++++++++
# Plot tooth length ("len") by groups ("dose")
# Color box plot by a second group: "supp"
library("ggpubr")
ggboxplot(my_data, x = "dose", y = "len", color = "supp",
          palette = c("#00AFBB", "#E7B800"))

Two-Way ANOVA Test in R

# Line plots with multiple groups
# +++++++++++++++++++++++
# Plot tooth length ("len") by groups ("dose")
# Color box plot by a second group: "supp"
# Add error bars: mean_se
# (other values include: mean_sd, mean_ci, median_iqr, ....)
library("ggpubr")
ggline(my_data, x = "dose", y = "len", color = "supp",
       add = c("mean_se", "dotplot"),
       palette = c("#00AFBB", "#E7B800"))

Two-Way ANOVA Test in R

If you still want to use R base graphs, type the following scripts:

# Box plot with two factor variables
boxplot(len ~ supp * dose, data=my_data, frame = FALSE, 
        col = c("#00AFBB", "#E7B800"), ylab="Tooth Length")
# Two-way interaction plot
interaction.plot(x.factor = my_data$dose, trace.factor = my_data$supp, 
                 response = my_data$len, fun = mean, 
                 type = "b", legend = TRUE, 
                 xlab = "Dose", ylab="Tooth Length",
                 pch=c(1,19), col = c("#00AFBB", "#E7B800"))

Arguments used for the function interaction.plot():

Compute two-way ANOVA test

We want to know if tooth length depends on supp and dose.

The R function aov() can be used to answer this question. The function summary.aov() is used to summarize the analysis of variance model.

res.aov2 <- aov(len ~ supp + dose, data = my_data)
summary(res.aov2)

            Df Sum Sq Mean Sq F value   Pr(>F)    
supp         1  205.4   205.4   14.02 0.000429 ***
dose         2 2426.4  1213.2   82.81  < 2e-16 ***
Residuals   56  820.4    14.7                     
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The output includes the columns F value and Pr(>F) corresponding to the p-value of the test.

From the ANOVA table we can conclude that both supp and dose are statistically significant. dose is the most significant factor variable. These results would lead us to believe that changing delivery methods (supp) or the dose of vitamin C, will impact significantly the mean tooth length.

Not the above fitted model is called additive model. It makes an assumption that the two factor variables are independent. If you think that these two variables might interact to create an synergistic effect, replace the plus symbol (+) by an asterisk (*), as follow.

# Two-way ANOVA with interaction effect
# These two calls are equivalent
res.aov3 <- aov(len ~ supp * dose, data = my_data)
res.aov3 <- aov(len ~ supp + dose + supp:dose, data = my_data)
summary(res.aov3)

            Df Sum Sq Mean Sq F value   Pr(>F)    
supp         1  205.4   205.4  15.572 0.000231 ***
dose         2 2426.4  1213.2  92.000  < 2e-16 ***
supp:dose    2  108.3    54.2   4.107 0.021860 *  
Residuals   54  712.1    13.2                     
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

It can be seen that the two main effects (supp and dose) are statistically significant, as well as their interaction.

Note that, in the situation where the interaction is not significant you should use the additive model.

Interpret the results

From the ANOVA results, you can conclude the following, based on the p-values and a significance level of 0.05:

• the p-value of supp is 0.000429 (significant), which indicates that the levels of supp are associated with significant different tooth length.
• the p-value of dose is < 2e-16 (significant), which indicates that the levels of dose are associated with significant different tooth length.
• the p-value for the interaction between supp*dose is 0.02 (significant), which indicates that the relationships between dose and tooth length depends on the supp method.

Compute some summary statistics

• Compute mean and SD by groups using dplyr R package:

require("dplyr")
group_by(my_data, supp, dose) %>%
  summarise(
    count = n(),
    mean = mean(len, na.rm = TRUE),
    sd = sd(len, na.rm = TRUE)
  )

Source: local data frame [6 x 5]
Groups: supp [?]
    supp   dose count  mean       sd
  (fctr) (fctr) (int) (dbl)    (dbl)
1     OJ   D0.5    10 13.23 4.459709
2     OJ     D1    10 22.70 3.910953
3     OJ     D2    10 26.06 2.655058
4     VC   D0.5    10  7.98 2.746634
5     VC     D1    10 16.77 2.515309
6     VC     D2    10 26.14 4.797731

• It’s also possible to use the function model.tables() as follow:

model.tables(res.aov3, type="means", se = TRUE)

Multiple pairwise-comparison between the means of groups

In ANOVA test, a significant p-value indicates that some of the group means are different, but we don’t know which pairs of groups are different.

It’s possible to perform multiple pairwise-comparison, to determine if the mean difference between specific pairs of group are statistically significant.

TukeyHSD(res.aov3, which = "dose")

  Tukey multiple comparisons of means
    95% family-wise confidence level
Fit: aov(formula = len ~ supp + dose + supp:dose, data = my_data)
$dose
          diff       lwr       upr   p adj
D1-D0.5  9.130  6.362488 11.897512 0.0e+00
D2-D0.5 15.495 12.727488 18.262512 0.0e+00
D2-D1    6.365  3.597488  9.132512 2.7e-06

glht(model, lincft)

library(multcomp)
summary(glht(res.aov2, linfct = mcp(dose = "Tukey")))

     Simultaneous Tests for General Linear Hypotheses
Multiple Comparisons of Means: Tukey Contrasts
Fit: aov(formula = len ~ supp + dose, data = my_data)
Linear Hypotheses:
               Estimate Std. Error t value Pr(>|t|)    
D1 - D0.5 == 0    9.130      1.210   7.543   <1e-05 ***
D2 - D0.5 == 0   15.495      1.210  12.802   <1e-05 ***
D2 - D1 == 0      6.365      1.210   5.259   <1e-05 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Adjusted p values reported -- single-step method)

pairwise.t.test(my_data$len, my_data$dose,
                p.adjust.method = "BH")

    Pairwise comparisons using t tests with pooled SD 
data:  my_data$len and my_data$dose 
   D0.5    D1     
D1 1.0e-08 -      
D2 4.4e-16 1.4e-05
P value adjustment method: BH 

Check ANOVA assumptions: test validity?

ANOVA assumes that the data are normally distributed and the variance across groups are homogeneous. We can check that with some diagnostic plots.

Check the homogeneity of variance assumption

The residuals versus fits plot is used to check the homogeneity of variances. In the plot below, there is no evident relationships between residuals and fitted values (the mean of each groups), which is good. So, we can assume the homogeneity of variances.

# 1. Homogeneity of variances
plot(res.aov3, 1)

Two-Way ANOVA Test in R

Points 32 and 23 are detected as outliers, which can severely affect normality and homogeneity of variance. It can be useful to remove outliers to meet the test assumptions.

Use the Levene’s test to check the homogeneity of variances. The function leveneTest() [in car package] will be used:

library(car)
leveneTest(len ~ supp*dose, data = my_data)

Levene's Test for Homogeneity of Variance (center = median)
      Df F value Pr(>F)
group  5  1.7086 0.1484
      54               

From the output above we can see that the p-value is not less than the significance level of 0.05. This means that there is no evidence to suggest that the variance across groups is statistically significantly different. Therefore, we can assume the homogeneity of variances in the different treatment groups.

Check the normality assumpttion

Normality plot of the residuals. In the plot below, the quantiles of the residuals are plotted against the quantiles of the normal distribution. A 45-degree reference line is also plotted.

The normal probability plot of residuals is used to verify the assumption that the residuals are normally distributed.

The normal probability plot of the residuals should approximately follow a straight line.

# 2. Normality
plot(res.aov3, 2)

Two-Way ANOVA Test in R

As all the points fall approximately along this reference line, we can assume normality.

The conclusion above, is supported by the Shapiro-Wilk test on the ANOVA residuals (W = 0.98, p = 0.5) which finds no indication that normality is violated.

# Extract the residuals
aov_residuals <- residuals(object = res.aov3)
# Run Shapiro-Wilk test
shapiro.test(x = aov_residuals )

    Shapiro-Wilk normality test
data:  aov_residuals
W = 0.98499, p-value = 0.6694