Variable type

The dependent variable body mass is quantitative continuous, while both independent variables sex and species are qualitative variables (with at least 2 levels).

Therefore, this assumption is met.

If your dependent variable is quantitative discrete, this is count data, which, strictly speaking, should be analyzed using a generalized linear model, not an ANOVA.

Independence

Independence is usually checked based on the design of the experiment and how data have been collected.²

To keep it simple, observations are usually:

• independent if each experimental unit (here a penguin) has been measured only once and the observations are collected from a representative and randomly selected portion of the population, or
• dependent if each experimental unit has been measured at least twice (as it is often the case in the medical field for example, with two measurements on the same subjects; one before and one after the treatment).

In our case, body mass has been measured only once on each penguin, and on a representative and random sample of the population, so the independence assumption is met.

Note that if your data correspond to observations made several times on the same experimental units (for example, if one of the factors is a treatment (A or B) and the second factor is time (day 1, day 2 and day 3), and measurements are made at each time point on the same subjects), a two-way mixed ANOVA should be used.

Normality

We have a large sample in all subgroups (each combination of the levels of the two factors, called cell):

table(dat$species, dat$sex)

##            
##             female male
##   Adelie        73   73
##   Chinstrap     34   34
##   Gentoo        58   61

so normality does not need to be checked.

For completeness, we still show how to verify normality, as if we had a small samples.

There are several methods to test the normality assumption. The most common methods being:

• a QQ-plot by group or on the residuals, and/or
• a histogram by group or on the residuals, and/or
• a normality test (Shapiro-Wilk test for instance) by group or on the residuals.

The easiest/shortest way is to verify the normality with a QQ-plot on the residuals. To draw this plot, we first need to save the model:

# save model
mod <- aov(body_mass_g ~ sex * species,
  data = dat
)

This piece of code will be explained further.

Now we can draw the QQ-plot on the residuals. We show two ways to do so, first with the plot() function and second with the qqPlot() function from the {car} package:

# method 1
plot(mod, which = 2)

# method 2
library(car)

qqPlot(mod$residuals,
  id = FALSE # remove point identification
)

Code for method 1 is slightly shorter, but it misses the confidence interval around the reference line.

If points follow the straight line (called Henry’s line) and fall within the confidence band, we can assume normality. This is the case here.

If you prefer to verify the normality based on a histogram of the residuals, here is the code:

# histogram
hist(mod$residuals)

The histogram of the residuals show a gaussian distribution, which is in line with the conclusion from the QQ-plot.

Although the QQ-plot and histogram is largely enough to verify the normality, if you want to test it more formally with a statistical test, the Shapiro-Wilk test can be applied on the residuals as well:

# normality test
shapiro.test(mod$residuals)

## 
## 	Shapiro-Wilk normality test
## 
## data:  mod$residuals
## W = 0.99776, p-value = 0.9367

\(\Rightarrow\) We do not reject the null hypothesis that the residuals follow a normal distribution (\(p\)-value = 0.937).

From the QQ-plot, histogram and Shapiro-Wilk test, we conclude that we do not reject the null hypothesis of normality of the residuals.

The normality assumption is thus verified, we can now check the equality of the variances.

Note that if the normality assumption is not met, many transformations can be applied on the dependent variable to improve it, the most common ones being the logarithmic (log() function in R) and the Box-Cox transformations. If the normality assumption is still not met on the transformed data, the non-parametric version of the two-way ANOVA, the Scheirer–Ray–Hare test, can be used. Alternatively, a permutation test can also be used.

Homogeneity of variances

Equality of variances, also referred as homogeneity of variances or homoscedasticity, can be verified visually with the plot() function:

plot(mod, which = 3)

Since the spread of the residuals is constant, the red smooth line is horizontal and flat, so it looks like the constant variance assumption is satisfied here.

The diagnostic plot above is sufficient, but if you prefer it can also be tested more formally with the Levene’s test (also from the {car} package):³

leveneTest(mod)

## Levene's Test for Homogeneity of Variance (center = median)
##        Df F value Pr(>F)
## group   5  1.3908 0.2272
##       327

\(\Rightarrow\) We do not reject the null hypothesis that the variances are equal (\(p\)-value = 0.227).

Both the visual and formal approaches give the same conclusion; we do not reject the hypothesis of homogeneity of the variances.

Note that, as for the normality, the logarithmic and Box-Cox transformations may improve homogeneity of the residuals.

Outliers

The easiest and most common way to detect outliers is visually thanks to boxplots by groups.

For females and males:

library(ggplot2)

# boxplots by sex
ggplot(dat) +
  aes(x = sex, y = body_mass_g) +
  geom_boxplot()

For the three species:

# boxplots by species
ggplot(dat) +
  aes(x = species, y = body_mass_g) +
  geom_boxplot()

There are, as defined by the interquartile range criterion, two outliers for the species Chinstrap. These points are, nonetheless, not extreme enough to bias results.

Therefore, we consider that the assumption of no significant outliers is met.

Preliminary analyses

Before performing any statistical test, it is a good practice to make some descriptive statistics in order to have a first overview of the data, and perhaps, have a glimpse of the results to be expected.

This can be done via descriptive statistics or plots.

# mean by group
aggregate(body_mass_g ~ species + sex,
  data = dat,
  FUN = mean
)

##     species    sex body_mass_g
## 1    Adelie female    3368.836
## 2 Chinstrap female    3527.206
## 3    Gentoo female    4679.741
## 4    Adelie   male    4043.493
## 5 Chinstrap   male    3938.971
## 6    Gentoo   male    5484.836

# mean and sd by group
library(dplyr)

group_by(dat, sex, species) %>%
  summarise(
    mean = round(mean(body_mass_g, na.rm = TRUE)),
    sd = round(sd(body_mass_g, na.rm = TRUE))
  )

## # A tibble: 8 × 4
## # Groups:   sex [3]
##   sex    species    mean    sd
##   <fct>  <fct>     <dbl> <dbl>
## 1 female Adelie     3369   269
## 2 female Chinstrap  3527   285
## 3 female Gentoo     4680   282
## 4 male   Adelie     4043   347
## 5 male   Chinstrap  3939   362
## 6 male   Gentoo     5485   313
## 7 <NA>   Adelie     3540   477
## 8 <NA>   Gentoo     4588   338

# boxplot by group
library(ggplot2)

ggplot(dat) +
  aes(x = species, y = body_mass_g, fill = sex) +
  geom_boxplot()

dat %>%
  filter(!is.na(sex)) %>%
  ggplot() +
  aes(x = species, y = body_mass_g, fill = sex) +
  geom_boxplot()

dat %>%
  filter(!is.na(sex)) %>%
  ggplot() +
  aes(x = sex, y = body_mass_g, fill = species) +
  geom_boxplot()

Two-way ANOVA in R

As mentioned earlier, including an interaction effect in a two-way ANOVA is not compulsory. However, in order to avoid flawed conclusions, it is recommended to first check whether the interaction is significant or not, and depending on the results, include it or not.

If the interaction is not significant, it is safe to remove it from the final model. On the contrary, if the interaction is significant, it should be included in the final model which will be used to interpret results.

We thus start with a model which includes the two main effects (i.e., sex and species) and the interaction:

# Two-way ANOVA with interaction
# save model
mod <- aov(body_mass_g ~ sex * species,
  data = dat
)

# print results
summary(mod)

##              Df    Sum Sq  Mean Sq F value   Pr(>F)    
## sex           1  38878897 38878897 406.145  < 2e-16 ***
## species       2 143401584 71700792 749.016  < 2e-16 ***
## sex:species   2   1676557   838278   8.757 0.000197 ***
## Residuals   327  31302628    95727                     
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 11 observations deleted due to missingness

Similar to a one-way ANOVA, the principle of a two-way ANOVA is based on the total dispersion of the data, and its decomposition into four components:

1. the share attributable to the first factor
2. the share attributable to the second factor
3. the share attributable to the interaction of the 2 factors
4. the unexplained, or residual portion.

The sum of squares (column Sum Sq) shows these four components. The two-way ANOVA consists of using a statistical test to determine whether each of the dispersion component (attributable to the 2 factors studied and to their interaction) is significantly greater than the residual component. If this is the case, we conclude that the effect considered (factor A, factor B or the interaction) is significant.

We see that the species explain a large part of the variability of body mass. It is the most important factor in explaining this variability.

The \(p\)-values are displayed in the last column of the output above (Pr(>F)). From these \(p\)-values, we conclude that, at the 5% significance level:

• controlling for the species, body mass is significantly different between the two sexes,
• controlling for the sex, body mass is significantly different for at least one species, and
• the interaction between sex and species (displayed at the line sex:species in the output above) is significant.

So from the significant interaction effect, we have just seen that the relationship between body mass and species is different between males and females. Since it is significant, we have to keep it in the model and we should interpret results from that model.

If, on the contrary, the interaction was not significant (that is, if the \(p\)-value \(\ge\) 0.05) we would have removed this interaction effect from the model. For illustrative purposes, below the code for a two-way ANOVA without interaction, referred as an additive model:

# Two-way ANOVA without interaction
aov(body_mass_g ~ sex + species,
  data = dat
)

For the readers who are used to perform linear regressions in R, you will notice that the structure of the code for a two-way ANOVA is in fact similar:

• the formula is dependent variable ~ independent variables
• the + sign is used to include independent variables without an interaction⁴
• the * sign is used to include independent variables with an interaction

The resemblance with a linear regression is not a surprise because a two-way ANOVA, like all ANOVA, is actually a linear model.

Note that the following code works as well, and give the same results:⁵

# method 2
mod2 <- lm(body_mass_g ~ sex * species,
  data = dat
)

Anova(mod2)

## Anova Table (Type II tests)
## 
## Response: body_mass_g
##                Sum Sq  Df F value    Pr(>F)    
## sex          37090262   1 387.460 < 2.2e-16 ***
## species     143401584   2 749.016 < 2.2e-16 ***
## sex:species   1676557   2   8.757 0.0001973 ***
## Residuals    31302628 327                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Note that the aov() function assumes a balanced design, meaning that we have equal sample sizes within levels of our independent grouping variables. Moreover, aov() uses the type I sums of squares, so we can obtain different \(p\)-values when we write y ~ A * B and y ~ B * A.

For unbalanced design, that is, unequal numbers of subjects in each subgroup, the recommended methods are:

• the type II ANOVA when there is no significant interaction, which can be done in R with Anova(mod, type = "II") or Anova(mod, type = 2), where mod is the name of your saved model, and
• the type III ANOVA when there is a significant interaction, which can be done in R with Anova(mod, type = "III") or Anova(mod, type = 3).

This is beyond the scope of the post and we assume a balanced design here. For the interested reader, see this detailed discussion about type I, type II and type III ANOVA.

Pairwise comparisons

Through the two main effects being significant, we concluded that:

• controlling for the species, body mass is different between females and males, and
• controlling for the sex, body mass is different for at least one species.

If body mass is different between the two sexes, given that there are exactly two sexes, it must be because body mass is significantly different between females and males.

If one wants to know which sex has the highest body mass, it can be deduced from the means and/or boxplots by subgroup. Here, it is clear that males have a significantly higher body mass than females.

However, it is not so straightforward for the species. Let me explain why it is not as easy as for the sexes.

There are three species (Adelie, Chinstrap and Gentoo), so there are 3 pairs of species:

1. Adelie and Chinstrap
2. Adelie and Gentoo
3. Chinstrap and Gentoo

If body mass is significantly different for at least one species, it could be that:

• body mass is significantly different between Adelie and Chinstrap but not significantly different between Adelie and Gentoo, and not significantly different between Chinstrap and Gentoo, or
• body mass is significantly different between Adelie and Gentoo but not significantly different between Adelie and Chinstrap, and not significantly different between Chinstrap and Gentoo, or
• body mass is significantly different between Chinstrap and Gentoo but not significantly different between Adelie and Chinstrap, and not significantly different between Adelie and Gentoo.

Or, it could also be that:

• body mass is significantly different between Adelie and Chinstrap, and between Adelie and Gentoo, but not significantly different between Chinstrap and Gentoo, or
• body mass is significantly different between Adelie and Chinstrap, and between Chinstrap and Gentoo, but not significantly different between Adelie and Gentoo, or
• body mass is significantly different between Chinstrap and Gentoo, and between Adelie and Gentoo, but not significantly different between Adelie and Chinstrap.

Last, it could also be that body mass is significantly different between all species.

As for a one-way ANOVA, we cannot, at this stage, know precisely which species is different from which one in terms of body mass. To know this, we need to compare each species two by two thanks to post-hoc tests (also known as pairwise comparisons).

There are several post-hoc tests, the most common ones being the Tukey HSD which tests all possible pairs of groups, and the Dunett’s test which compares all groups to a reference group. As mentioned earlier, these tests should not be done on the sex variable because there are only two levels.

In this post, we show only the Tukey HSD test. For the interested reader, the Dunnett’s test is illustrated here.

As for the one-way ANOVA, the Tukey HSD test can be done in R as follows:

# method 1
TukeyHSD(mod,
  which = "species"
)

##   Tukey multiple comparisons of means
##     95% family-wise confidence level
## 
## Fit: aov(formula = body_mass_g ~ sex * species, data = dat)
## 
## $species
##                        diff       lwr       upr     p adj
## Chinstrap-Adelie   26.92385  -80.0258  133.8735 0.8241288
## Gentoo-Adelie    1377.65816 1287.6926 1467.6237 0.0000000
## Gentoo-Chinstrap 1350.73431 1239.9964 1461.4722 0.0000000

or using the {multcomp} package:

# method 2
library(multcomp)

res_tukey <- glht(
  aov(body_mass_g ~ sex + species,
    data = dat
  ),
  linfct = mcp(species = "Tukey")
)

summary(res_tukey)

## 
## 	 Simultaneous Tests for General Linear Hypotheses
## 
## Multiple Comparisons of Means: Tukey Contrasts
## 
## 
## Fit: aov(formula = body_mass_g ~ sex + species, data = dat)
## 
## Linear Hypotheses:
##                         Estimate Std. Error t value Pr(>|t|)    
## Chinstrap - Adelie == 0    26.92      46.48   0.579     0.83    
## Gentoo - Adelie == 0     1377.86      39.10  35.236   <1e-05 ***
## Gentoo - Chinstrap == 0  1350.93      48.13  28.067   <1e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## (Adjusted p values reported -- single-step method)

or using the pairwise.t.test() function using the \(p\)-value adjustment method of your choice:⁶

# method 3
pairwise.t.test(dat$body_mass_g, dat$species,
  p.adjust.method = "BH"
)

## 
## 	Pairwise comparisons using t tests with pooled SD 
## 
## data:  dat$body_mass_g and dat$species 
## 
##           Adelie Chinstrap
## Chinstrap 0.63   -        
## Gentoo    <2e-16 <2e-16   
## 
## P value adjustment method: BH

Note that when using the second method, it is the model without the interaction that needs to be specified into the glht() function, even if the interaction is significant. Moreover, do not forget to replace mod and species in my code with the name of your model and the name of your independent variable.

Both methods give the same results, that is:

• body mass is not significantly different between Chinstrap and Adelie (adjusted \(p\)-value = 0.83),
• body mass is significantly different between Gentoo and Adelie (adjusted \(p\)-value < 0.001), and
• body mass is significantly different between Gentoo and Chinstrap (adjusted \(p\)-value < 0.001).

Remember that it is the adjusted \(p\)-values that are reported, to prevent the issue of multiple testing which occurs when comparing several pairs of groups.

If you would like to compare all combinations of groups, it can be done with the TukeyHSD() function and specifying the interaction in the which argument:

# all combinations of sex and species
TukeyHSD(mod,
  which = "sex:species"
)

##   Tukey multiple comparisons of means
##     95% family-wise confidence level
## 
## Fit: aov(formula = body_mass_g ~ sex * species, data = dat)
## 
## $`sex:species`
##                                      diff       lwr       upr     p adj
## male:Adelie-female:Adelie        674.6575  527.8486  821.4664 0.0000000
## female:Chinstrap-female:Adelie   158.3703  -25.7874  342.5279 0.1376213
## male:Chinstrap-female:Adelie     570.1350  385.9773  754.2926 0.0000000
## female:Gentoo-female:Adelie     1310.9058 1154.8934 1466.9181 0.0000000
## male:Gentoo-female:Adelie       2116.0004 1962.1408 2269.8601 0.0000000
## female:Chinstrap-male:Adelie    -516.2873 -700.4449 -332.1296 0.0000000
## male:Chinstrap-male:Adelie      -104.5226 -288.6802   79.6351 0.5812048
## female:Gentoo-male:Adelie        636.2482  480.2359  792.2606 0.0000000
## male:Gentoo-male:Adelie         1441.3429 1287.4832 1595.2026 0.0000000
## male:Chinstrap-female:Chinstrap  411.7647  196.6479  626.8815 0.0000012
## female:Gentoo-female:Chinstrap  1152.5355  960.9603 1344.1107 0.0000000
## male:Gentoo-female:Chinstrap    1957.6302 1767.8040 2147.4564 0.0000000
## female:Gentoo-male:Chinstrap     740.7708  549.1956  932.3460 0.0000000
## male:Gentoo-male:Chinstrap      1545.8655 1356.0392 1735.6917 0.0000000
## male:Gentoo-female:Gentoo        805.0947  642.4300  967.7594 0.0000000

Or with the HSD.test() function from the {agricolae} package, which denotes subgroups that are not significantly different from each other with the same letter:

library(agricolae)

HSD.test(mod,
  trt = c("sex", "species"),
  console = TRUE # print results
)

## 
## Study: mod ~ c("sex", "species")
## 
## HSD Test for body_mass_g 
## 
## Mean Square Error:  95726.69 
## 
## sex:species,  means
## 
##                  body_mass_g      std  r  Min  Max
## female:Adelie       3368.836 269.3801 73 2850 3900
## female:Chinstrap    3527.206 285.3339 34 2700 4150
## female:Gentoo       4679.741 281.5783 58 3950 5200
## male:Adelie         4043.493 346.8116 73 3325 4775
## male:Chinstrap      3938.971 362.1376 34 3250 4800
## male:Gentoo         5484.836 313.1586 61 4750 6300
## 
## Alpha: 0.05 ; DF Error: 327 
## Critical Value of Studentized Range: 4.054126 
## 
## Groups according to probability of means differences and alpha level( 0.05 )
## 
## Treatments with the same letter are not significantly different.
## 
##                  body_mass_g groups
## male:Gentoo         5484.836      a
## female:Gentoo       4679.741      b
## male:Adelie         4043.493      c
## male:Chinstrap      3938.971      c
## female:Chinstrap    3527.206      d
## female:Adelie       3368.836      d

If you have many groups to compare, plotting them might be easier to interpret:

# set axis margins so labels do not get cut off
par(mar = c(4.1, 13.5, 4.1, 2.1))

# create confidence interval for each comparison
plot(TukeyHSD(mod, which = "sex:species"),
  las = 2 # rotate x-axis ticks
)

From the outputs and plot above, we conclude that all combinations of sex and species are significantly different, except between female Chinstrap and female Adelie (\(p\)-value = 0.138) and male Chinstrap and male Adelie (\(p\)-value = 0.581).

These results, which are by the way in line with the boxplots shown above and which will be confirmed with the visualizations below, concludes the two-way ANOVA in R.

Visualizations

If you would like to visualize results in a different way to what has already been presented in the preliminary analyses, below are some ideas of useful plots.

First, with the mean and standard error of the mean by subgroup using the allEffects() function from the {effects} package:

# method 1
library(effects)

plot(allEffects(mod))

Or using the {ggpubr} package:

# method 2
library(ggpubr)

ggline(subset(dat, !is.na(sex)), # remove NA level for sex
  x = "species",
  y = "body_mass_g",
  color = "sex",
  add = c("mean_se") # add mean and standard error
) +
  labs(y = "Mean of body mass (g)")

Alternatively, using {Rmisc} and {ggplot2}:

library(Rmisc)

# compute mean and standard error of the mean by subgroup
summary_stat <- summarySE(dat,
  measurevar = "body_mass_g",
  groupvars = c("species", "sex")
)

# plot mean and standard error of the mean
ggplot(
  subset(summary_stat, !is.na(sex)), # remove NA level for sex
  aes(x = species, y = body_mass_g, colour = sex)
) +
  geom_errorbar(aes(ymin = body_mass_g - se, ymax = body_mass_g + se), # add error bars
    width = 0.1 # width of error bars
  ) +
  geom_point() +
  labs(y = "Mean of body mass (g)")

Second, if you prefer to draw only the mean by subgroup:

with(
  dat,
  interaction.plot(species, sex, body_mass_g)
)

Last but not least, for those of you who are familiar with GraphPad, you are most likely familiar with plotting means and error bars as follows:

# plot mean and standard error of the mean as barplots
ggplot(
  subset(summary_stat, !is.na(sex)), # remove NA level for sex
  aes(x = species, y = body_mass_g, fill = sex)
) +
  geom_bar(position = position_dodge(), stat = "identity") +
  geom_errorbar(aes(ymin = body_mass_g - se, ymax = body_mass_g + se), # add error bars
    width = 0.25, # width of error bars
    position = position_dodge(.9)
  ) +
  labs(y = "Mean of body mass (g)")