library(tidyverse) # Tidy data processing
library(ggformula) # Formula based plots
library(mosaic) # Data inspection and Statistical Inference
library(broom) # Tidy outputs from Statistical Analyses
library(infer) # Statistical Inference, Permutation/Bootstrap
library(patchwork) # Arranging Plots
library(ggprism) # Interesting Categorical Axes
library(supernova) # Beginner-Friendly ANOVA Tables
Comparing Multiple Means with ANOVA
Setting up R Packages
Plot Theme
Show the Code
# https://stackoverflow.com/questions/74491138/ggplot-custom-fonts-not-working-in-quarto
# Chunk options
knitr::opts_chunk$set(
fig.width = 7,
fig.asp = 0.618, # Golden Ratio
# out.width = "80%",
fig.align = "center"
)
### Ggplot Theme
### https://rpubs.com/mclaire19/ggplot2-custom-themes
theme_custom <- function() {
font <- "Roboto Condensed" # assign font family up front
theme_classic(base_size = 14) %+replace% # replace elements we want to change
theme(
panel.grid.minor = element_blank(), # strip minor gridlines
text = element_text(family = font),
# text elements
plot.title = element_text( # title
family = font, # set font family
# size = 20, #set font size
face = "bold", # bold typeface
hjust = 0, # left align
# vjust = 2 #raise slightly
margin = margin(0, 0, 10, 0)
),
plot.subtitle = element_text( # subtitle
family = font, # font family
# size = 14, #font size
hjust = 0,
margin = margin(2, 0, 5, 0)
),
plot.caption = element_text( # caption
family = font, # font family
size = 8, # font size
hjust = 1
), # right align
axis.title = element_text( # axis titles
family = font, # font family
size = 10 # font size
),
axis.text = element_text( # axis text
family = font, # axis family
size = 8
) # font size
)
}
# Set graph theme
theme_set(new = theme_custom())
#
Introduction
Suppose we have three sales strategies on our website, to sell a certain product, say men’s shirts. We have observations of customer website interactions over several months. How do we know which strategy makes people buy the fastest ?
If there is a University course that is offered in parallel in three different classrooms, is there a difference between the average marks obtained by students in each of the classrooms?
In each case we have a set of Quant observations in each Qual category: Interaction Time
vs Sales Strategy
in the first example, and Student Marks
vs Classroom
in the second. We can take mean scores in each category and decide to compare them. How do we make the comparisons? One way would be to compare them pair-wise, doing as many t-tests
as there are pairs. But with this rapidly becomes intractable and also dangerous: with increasing number of groups
, the number of mean-comparisons becomes very large \(N\choose 2\) and with each comparison the possibility of some difference showing up, just by chance, increases! And we end up making the wrong inference and perhaps the wrong decision. The trick is of course to make comparisons all at once and ANOVA is the technique that allows us to do just that.
In this tutorial, we will compare the Hatching Time of frog spawn1, at three different lab temperatures.
In this tutorial, our research question is:
Based on the sample dataset at hand, how does frogspawn hatching time vary with different temperature settings?
Workflow: Read the Data
Download the data by clicking the button below.
Save the CSV in a subfolder titled “data” inside your R work folder.
frogs_orig <- read_csv("data/frogs.csv")
frogs_orig
Our response variable is the hatching Time
. Our explanatory variable is a factor, Temperature
, with 3 levels: 13°C, 18°C and 25°C. Different samples of spawn were subject to each of these temperatures respectively.
Workflow: Clean the Data
The data is in wide-format, with a separate column for each Temperature, and a common column for Sample ID. This is good for humans, but poor for a computer: there are NA entries since not all samples of spawn can be subject to all temperatures. (E.g. Sample ID #1 was maintained at 13°C, and there are NAs in the other two columns, which we don’t need).
We will first stack up the Temperature
columns into a single column, separate that into pieces and then retain just the number part (13, 18, 25), getting rid of the word Temperature
from the column titles. Then the remaining numerical column with temperatures (13, 18, 25) will be converted into a factor.
We will use pivot_longer()
and separate_wider_regex()
to achieve this. [See this animation for pivot_longer(): https://haswal.github.io/pivot/ ]
frogs_orig %>%
pivot_longer(
.,
cols = starts_with("Temperature"),
cols_vary = "fastest",
# new in pivot_longer
names_to = "Temp",
values_to = "Time"
) %>%
drop_na() %>%
##
separate_wider_regex(
cols = Temp,
# knock off the unnecessary "Temperature" word
# Just keep the digits thereafter
patterns = c("Temperature", TempFac = "\\d+"),
cols_remove = TRUE
) %>%
# Convert Temp into TempFac, a 3-level factor
mutate(TempFac = factor(
x = TempFac,
levels = c(13, 18, 25),
labels = c("13", "18", "25")
)) %>%
rename("Id" = `Frogspawn sample id`) -> frogs_long
frogs_long
##
frogs_long %>% count(TempFac)
So we have cleaned up our data and have 20
samples for Hatching Time
per TempFac
setting.
Workflow: EDA
Let us plot some histograms and boxplots of Hatching Time:
# Set graph theme
theme_set(new = theme_custom())
##
gf_histogram(~Time,
fill = ~TempFac,
data = frogs_long, alpha = 0.5
) %>%
gf_vline(xintercept = ~ mean(Time)) %>%
gf_labs(
title = "Histograms of Hatching Time Distributions vs Temperature",
x = "Hatching Time", y = "Count"
) %>%
gf_text(7 ~ (mean(Time) + 2),
label = "Overall Mean"
) %>%
gf_refine(guides(fill = guide_legend(title = "Temperature level (°C)")))
# Set graph theme
theme_set(new = theme_custom())
##
gf_boxplot(
data = frogs_long,
Time ~ TempFac,
fill = ~TempFac,
alpha = 0.5
) %>%
gf_vline(xintercept = ~ mean(Time)) %>%
gf_labs(
title = "Boxplots of Hatching Time Distributions vs Temperature",
x = "Temperature", y = "Hatching Time",
caption = "Using ggprism"
) %>%
gf_refine(
scale_x_discrete(guide = "prism_bracket"),
guides(fill = guide_legend(title = "Temperature level (°C)"))
)
The histograms look well separated and the box plots also show very little overlap. So we can reasonably hypothesize that Temperature has a significant effect on Hatching Time.
Let’s go ahead with our ANOVA test.
Workflow: ANOVA
We will first execute the ANOVA test with code and evaluate the results. Then we will do an intuitive walkthrough of the process and finally, hand-calculate entire analysis for clear understanding. For now, a little faith!
R offers a very simple command aov
to execute an ANOVA test: Note the familiar formula
of stating the variables:
frogs_anova <- aov(Time ~ TempFac, data = frogs_long)
This creates an ANOVA model object, called frogs_anova
. We can examine the ANOVA model object best with a package called supernova
2:
# library(supernova)
# Set graph theme
theme_set(new = theme_custom())
#
supernova::pairwise(frogs_anova,
correction = "Bonferroni", # Try "Tukey"
alpha = 0.05, # 95% CI calculation
var_equal = TRUE, # We'll see
plot = TRUE
)
group_1 group_2 diff pooled_se t df lower upper p_adj
<chr> <chr> <dbl> <dbl> <dbl> <int> <dbl> <dbl> <dbl>
1 18 13 -5.300 0.257 -20.608 57 -5.861 -4.739 .0000
2 25 13 -10.100 0.257 -39.272 57 -10.661 -9.539 .0000
3 25 18 -4.800 0.257 -18.664 57 -5.361 -4.239 .0000
This table + error-bar plot gives us a clear comparison between each pair of the three groups of observations defined by TempFac
. The differences in spawn hatching Time
between each pair of TempFac
settings are given by the diff
column. Also shown are the confidence intervals for each of these differences (none of which include \(0\)); the p-value
s for each of these differences is also negligible. Thus we can conclude that the effect of temperature on hatching time is significant.
To find which specific value of TempFac
has the most effect
will require pairwise comparison of the group means, using a standard t-test
. The confidence level for such repeated comparisons will need what is called Bonferroni correction3 to prevent us from detecting a significant (pair-wise) difference simply by chance. To do this we take \(\alpha = 0.05\), the confidence level used and divide it by \(K\), the number of pair-wise comparisons we intend to make. This new value is used to decide on the significance of the estimated parameter. So the pairwise comparisons in our current data will have to use \(\alpha/3 = 0.0166\) as the confidence level. The supernova::pairwise()
function did this for us very neatly!
There are also other ways, such as the “Tukey correction” for multiple tests.
All that is very well, but what is happening under the hood of the aov()
command?
Consider a data set with a single Quant and a single Qual variable. The Qual variable has two levels, the Quant data has 20 observations per Qual level.
All Data: In Fig A, the horizontal black line is the overall mean of quant
, denoted as \(\mu_{tot}\). The vertical black lines to the points show the departures of each point from this overall mean. The sum of squares of these vertical black lines in Fig A is called the Total Sum of Squares (SST).
\[ SST = \Sigma (y - \mu_{tot})^2 \tag{1}\]
Grouped Data: In Fig B, the horizontal green and red lines are the means of the individual groups, respectively \(\mu_A\) and \(\mu_B\). The green and red vertical lines are the departures, or errors, of each point from its own group-mean. The sum of the squares of the green and red lines is called the Total Error Sum of Squares (SSE).
\[ SSE = \Sigma [(y - \mu_A)^2] + \Sigma (y - \mu_B)^2] \tag{2}\]
Improvement: We take the difference in the squared error sums:
\[ SSA = SST - SSE \tag{3}\]
\(SSA\) is called the Treatment Sum of Squares, the “improvement” in going from believing in one mean to believing in two.
Improvement Ratio: \(SSA/SSE\) might now help us decide whether two means are better than one.
Let us compute these numbers for our toy dataset:
Analysis of Variance Table (Type III SS)
Model: quant ~ qual
SS df MS F PRE p
----- --------------- | -------- -- ------- ------- ----- -----
Model (error reduced) | 823.407 1 823.407 139.356 .7857 .0000
Error (from model) | 224.529 38 5.909
----- --------------- | -------- -- ------- ------- ----- -----
Total (empty model) | 1047.935 39 26.870
What do we see?
- All Data: \(SST = 1047.935\).
- Grouped Data: \(SSE = 224.529\).
- Improvement: \(SSA = SST-SSE\) = \(823.407\).
- Improvement Ratio: Before we set up this ratio, we must realize that each of these measures uses a different number of observations! So the comparison is done after scaling each of \(SSA\) and \(SSE\) by the number of observations influencing them. (a sort of per capita, or average, squared error, an idea we saw when we defined Standard Errors): \(F_{stat} = \frac{SSA / df_{SSA}}{SSE / df_{SSE}}\), where \(df_{SSA} = 1\) and \(df_{SSE} = 38\) are respectively the degrees of freedom in \(SSA\) and \(SSE\).
-
Large Enough Ratio?: The value of the
F-statistic
from the table above is \(\frac{823.407}{5.909} = 139.356\). Is this ratio big enough?F-statistic
is compared with a critical value of theF-critical
to help us decide. (Here, it is.) - Belief: So we now believe in the idea of two means.
-
Back to Mean Differences: Finally, in order to find which of the means is significantly different from others (if there are more than two!), we need to make a pair-wise comparison of the means, applying the
Bonferroni correction
as stated before. This means we divide the criticalp.value
we expect by the number of comparisons we make between levels of the Qual variable.supernova
did this for us in the error-bar plot above.
When divide each of \(SSA\) and \(SSE\) by their degrees of freedom, this gives us a ratio of variances, the F-statistic. And so we are in effect deciding if means are significantly different by analyzing (a ratio of) variances! Hence the name, AN-alysis O-f VA-riance, ANOVA.
So this may seem like a great Hero’s Journey, where we start with means and differences, go into sums of squares, differences and comparisons of error ratios, and return to the means where we started, only to know them properly now.
Now that we understand what aov()
is doing, let us hand-calculate the numbers for our frogs
dataset and check. Let us visualize our calculations first.
Let us get the ready table from supernova
first, and then systematically calculate all numbers with understanding:
supernova::supernova(frogs_anova)
Analysis of Variance Table (Type III SS)
Model: Time ~ TempFac
SS df MS F PRE p
----- --------------- | -------- -- ------- ------- ----- -----
Model (error reduced) | 1020.933 2 510.467 385.897 .9312 .0000
Error (from model) | 75.400 57 1.323
----- --------------- | -------- -- ------- ------- ----- -----
Total (empty model) | 1096.333 59 18.582
Here are the SST, SSE, and the SSA:
# Calculate overall sum squares SST
frogs_overall <- frogs_long %>%
summarise(
overall_mean_time = mean(Time),
# Overall mean across all readings
# The Black Line
SST = sum((Time - overall_mean_time)^2),
n = n()
) # Always do this with `summarise`
frogs_overall
##
SST <- frogs_overall$SST
SST
[1] 1096.333
# Calculate sums of square errors *within* each group
# with respect to individual group means
frogs_within_groups <- frogs_long %>%
group_by(TempFac) %>%
summarise(
grouped_mean_time = mean(Time), # The Coloured Lines
grouped_variance_time = var(Time),
group_error_squares = sum((Time - grouped_mean_time)^2),
n = n()
)
frogs_within_groups
##
frogs_SSE <- frogs_within_groups %>%
summarise(SSE = sum(group_error_squares))
##
SSE <- frogs_SSE$SSE
SSE
[1] 75.4
SST
SSE
SSA <- SST - SSE
SSA
[1] 1096.333
[1] 75.4
[1] 1020.933
We have \(SST = 1096\), \(SSE = 75.4\) and therefore \(SSA = 1020.9\).
In order to calculate the F-Statistic, we need to compute the variances, using these sum of squares. We obtain variances by dividing by their Degrees of Freedom:
\[ F_{stat} = \frac{SSA / df_{SSA}}{SSE / df_{SSE}} \]
where \(df_{SSA}\) and \(df_{SSE}\) are respectively the degrees of freedom in SSA and SSE.
Let us calculate these Degrees of Freedom.
With \(k = 3\) levels in the factor TempFac
, and \(n = 20\) points per level, \(SST\) clearly has degree of freedom \(kn-1 = 3*20~ -1 = 59\), since it uses all observations but loses one degree to calculate the global mean. (If each level did not have the same number of points \(n\), we simply take all observations less one as the degrees of freedom for \(SST\)).
\(SSE\) has \(k*(n-1) = 3 * (20 -1) = 57\) as degrees of freedom, since each of the \(k\) groups there are \(n\) observations and each group loses one degree to calculate its own group mean.
And therefore \(SSA\), being their difference, has \(kn-1 -k*(n-1) = k-1 = 2\) degrees of freedom.
These are, of course, as shown in the df column in the supernova tabel above. We can still calculate these in R, for the sake of method and clarity (and pedantry):
# Error Sum of Squares SSE
df_SSE <- frogs_long %>%
# Takes into account "unbalanced" situations
# Where groups are not equal in size
group_by(TempFac) %>%
summarise(per_group_df_SSE = n() - 1) %>%
summarise(df_SSE = sum(per_group_df_SSE)) %>%
as.numeric()
## Overall Sum of Squares SST
df_SST <- frogs_long %>%
summarise(df_SST = n() - 1) %>%
as.integer()
# Treatment Sum of Squares SSA
k <- length(unique(frogs_long$TempFac))
df_SSA <- k - 1
The degrees of freedom for the quantities are:
df_SST
df_SSE
df_SSA
[1] 59
[1] 57
[1] 2
Now we are ready to compute the F-statistic: dividing each sum-of-squares byt its degrees of freedom gives us variances which we will compare, using the F-statistic as a ratio:
# Finally F_Stat!
# Combine the sum-square_error for each level of the factor
# Weighted by degrees of freedom **per level**
# Which are of course equal here ;-D
MSE <- frogs_within_groups %>%
summarise(mean_square_error = sum(group_error_squares / df_SSE)) %>%
as.numeric()
MSE
[1] 1.322807
##
MSA <- SSA / df_SSA # This is OK
MSA
[1] 510.4667
##
F_stat <- MSA / MSE
F_stat
[1] 385.8966
The F-stat
is compared with a critical value of the F-statistic, F_crit
which is computed using the formula for the f-distribution in R. As with our hypothesis tests, we set the significance level to \(\alpha = 0.95\), but here with the Bonferroni correction, and quote the two relevant degrees of freedom as parameters to qf()
which computes the critical F value F_critical
as a quartile:
F_crit <-
qf(
p = (1 - 0.05 / 3), # Significance level is 5% + Bonferroni Correction
df1 = df_SSA, # Numerator degrees of freedom
df2 = df_SSE # Denominator degrees of freedom
)
F_crit
F_stat
[1] 4.403048
[1] 385.8966
The F_crit
value can also be seen in a plot4,5:
# Set graph theme
theme_set(new = theme_custom())
#
mosaic::xpf(
q = F_crit,
df1 = df_SSA, df2 = df_SSE, method = "gg",
log.p = FALSE, lower.tail = TRUE,
return = "plot"
) %>%
gf_vline(xintercept = F_crit) %>%
gf_label(0.75 ~ 5,
label = "F_critical",
inherit = F, show.legend = F
) %>%
gf_labs(
title = "F distribution for Frogs Data",
subtitle = "F_critical = 4.403"
)
Any value of F more than the F_crit
occurs with smaller probability than \(0.05/3 = 0.017\). Our F_stat is much higher than F_crit
, by orders of magnitude! And so we can say with confidence that Temperature has a significant effect on spawn Time.
And that is how ANOVA computes!
Stating the Model
And supernova
gives us a nice linear equation relating Hatching_Time
to TempFac
:
supernova::equation(frogs_anova)
Fitted equation:
Time = 26.3 + -5.3*TempFac18 + -10.1*TempFac25 + e
TempFac18
and TempFac25
are binary {0,1} coded variables, representing the test situation. e
is the remaining error. The equation models the means at each value of TempFac
.
Workflow: Checking ANOVA Assumptions
ANOVA makes 3 fundamental assumptions:
- Data (and errors) are normally distributed.
- Variances are equal.
- Observations are independent.
We can check these using checks and graphs.
Checks for Normality
The shapiro.wilk
test tests if a vector of numeric data is normally distributed and rejects the hypothesis of normality when the p-value is less than or equal to 0.05.
shapiro.test(x = frogs_long$Time)
Shapiro-Wilk normality test
data: frogs_long$Time
W = 0.92752, p-value = 0.001561
The p-value is very low and we cannot reject the (alternative) hypothesis that the overall data is not normal. How about normality at each level of the factor?
The shapiro.wilk
test makes a NULL Hypothesis that the data are normally distributed and estimates the probability that the given data could have happened by chance. Except for TempFac = 18
the p.value
s are less than 0.05 and we can reject the NULL hypothesis that each of these is normally distributed. Perhaps this is a sign that we need more than 20 samples per factor level. Let there be more frogs !!! இன்னும தவளைகள் வேண்டும்!! !!
We can also check the residuals post-model:
# Set graph theme
theme_set(new = theme_custom())
#
frogs_anova$residuals %>%
as_tibble() %>%
gf_dhistogram(~value, data = .) %>%
gf_fitdistr()
##
frogs_anova$residuals %>%
as_tibble() %>%
gf_qq(~value, data = .) %>%
gf_qqstep() %>%
gf_qqline()
##
shapiro.test(frogs_anova$residuals)
Unsurprisingly, the residuals are also not normally distributed either.
Check for Similar Variance
Response data with different variances at different levels of an explanatory variable are said to exhibit heteroscedasticity. This violates one of the assumptions of ANOVA.
To check if the Time
readings are similar in variance
across levels of TempFac
, we can use the Levene Test, or since our per-group observations are not normally distributed, a non-parametric rank-based Fligner-Killeen Test. The NULL hypothesis is that the data are with similar variances. The tests assess how probable this is with the given data assuming this NULL hypothesis:
frogs_long %>%
group_by(TempFac) %>%
summarise(variance = var(Time))
# Not too different...OK on with the test
DescTools::LeveneTest(Time ~ TempFac, data = frogs_long)
##
fligner.test(Time ~ TempFac, data = frogs_long)
Fligner-Killeen test of homogeneity of variances
data: Time by TempFac
Fligner-Killeen:med chi-squared = 0.53898, df = 2, p-value = 0.7638
It seems that there is no cause for concern here; the data do not have significantly different variances.
Independent Observations
This is an experiment design concern; the way the data is gathered must be specified such that data for each level of the factors ( factor combinations if there are more than one) should be independent.
Workflow: Effect Size
The simplest way to find the actual effect sizes
detected by an ANOVA test is something we have already done, with the supernova
package: Here is the table and plot again:
# Set graph theme
theme_set(new = theme_custom())
#
frogs_supernova <-
supernova::pairwise(frogs_anova,
plot = TRUE,
alpha = 0.05,
correction = "Bonferroni"
)
frogs_supernova
group_1 group_2 diff pooled_se t df lower upper p_adj
<chr> <chr> <dbl> <dbl> <dbl> <int> <dbl> <dbl> <dbl>
1 18 13 -5.300 0.257 -20.608 57 -5.861 -4.739 .0000
2 25 13 -10.100 0.257 -39.272 57 -10.661 -9.539 .0000
3 25 18 -4.800 0.257 -18.664 57 -5.361 -4.239 .0000
This table, the plot, and the equation we set up earlier all give us the sense of how the TempFac
affects Time
. The differences are given pair-wise between levels of the Qual factor, TempFac
, and the standard error
has been declared in pooled fashion (all groups together).
We can also use (paradoxically) the summary.lm()
command:
tidy_anova <-
frogs_anova %>%
summary.lm() %>%
broom::tidy()
tidy_anova
It may take a bit of effort to understand this. First the TempFac
is arranged in order of levels, and the mean
at the \(TempFac = 13\) is titled Intercept
. That is \(26.3\). The other two means for levels \(18\) and \(25\) are stated as differences from this intercept, \(-5.3\) and \(-10.1\) respectively. The p.value
for all these effect sizes is well below the desired confidence level of \(0.05\).
Observe that the std.error
for the intercept is \(0.257\) while that for TempFac18
and TempFac25
is \(0.257 \times \sqrt2 = 0.363\) since the latter are differences in means, while the former is a single mean. The Variance of a difference is the sum of the individual variances, which are equal here.
We can easily plot bar-chart with error bars for the effect size:
# Set graph theme
theme_set(new = theme_custom())
#
tidy_anova %>%
mutate(
hi = estimate + std.error,
lo = estimate - std.error
) %>%
gf_hline(
data = ., yintercept = 0,
colour = "grey",
linewidth = 2
) %>%
gf_col(estimate ~ term,
fill = "grey",
color = "black",
width = 0.15
) %>%
gf_errorbar(hi + lo ~ term,
color = "blue",
width = 0.2
) %>%
gf_point(estimate ~ term,
color = "red",
size = 3.5
) %>%
gf_refine(scale_x_discrete("Temp Values",
labels = c("13°C", "18°C", "25°C")
))
If we want an “absolute value” plot for effect size, it needs just a little bit of work:
# Merging group averages with `std.error`
# Set graph theme
theme_set(new = theme_custom())
#
frogs_long %>%
group_by(TempFac) %>%
summarise(mean = mean(Time)) %>%
cbind(std.error = tidy_anova$std.error) %>%
mutate(
hi = mean + std.error,
lo = mean - std.error
) %>%
gf_hline(
data = ., yintercept = 0,
colour = "grey",
linewidth = 2
) %>%
gf_col(mean ~ TempFac,
fill = "grey",
color = "black", width = 0.15
) %>%
gf_errorbar(hi + lo ~ TempFac,
color = "blue",
width = 0.2
) %>%
gf_point(mean ~ TempFac,
color = "red",
size = 3.5
) %>%
gf_refine(scale_x_discrete("Temp Values",
labels = c("13°C", "18°C", "25°C")
))
In both graphs, note the difference in the error-bar heights.
The ANOVA test does not tell us that the “treatments” (i.e. levels of TempFac
) are equally effective. We need to use a multiple comparison procedure to arrive at an answer to that question. We compute the pair-wise differences in effect-size:
Tukey multiple comparisons of means
95% family-wise confidence level
Fit: aov(formula = Time ~ TempFac, data = frogs_long)
$TempFac
diff lwr upr p adj
18-13 -5.3 -6.175224 -4.424776 0
25-13 -10.1 -10.975224 -9.224776 0
25-18 -4.8 -5.675224 -3.924776 0
We see that each of the pairwise differences in effect-size is significant, with p = 0
!
Workflow: ANOVA using Permutation Tests
We wish to establish the significance of the effect size due to each of the levels in TempFac
. From the normality tests conducted earlier we see that except at one level of TempFac
, the times are are not normally distributed. Hence we opt for a Permutation Test to check for significance of effect.
As remarked in Ernst8, the non-parametric permutation test can be both exact and also intuitively easier for students to grasp.
We proceed with a Permutation Test for TempFac
. We shuffle the levels (13, 18, 25) randomly between the Times and repeat the ANOVA test each time and calculate the F-statistic. The Null distribution is the distribution of the F-statistic over the many permutations and the p-value is given by the proportion of times the F-statistic equals or exceeds that observed.
We will use infer
to do this: We calculate the observed F-stat with infer
, which also has a very direct, if verbose, syntax for doing permutation tests:
observed_infer <-
frogs_long %>%
specify(Time ~ TempFac) %>%
hypothesise(null = "independence") %>%
calculate(stat = "F")
observed_infer
We see that the observed F-Statistic is of course \(385.8966\) as before. Now we use infer
to generate a NULL distribution using permutation of the factor TempFac
:
null_dist_infer <- frogs_long %>%
specify(Time ~ TempFac) %>%
hypothesise(null = "independence") %>%
generate(reps = 4999, type = "permute") %>%
calculate(stat = "F")
##
null_dist_infer
##
null_dist_infer %>%
visualise(method = "simulation") +
shade_p_value(obs_stat = observed_infer$stat, direction = "right") +
scale_x_continuous(trans = "log10", expand = c(0, 0)) +
coord_cartesian(xlim = c(0.2, 500), clip = "off") +
annotation_logticks(outside = FALSE) +
theme_custom()
As seen, the infer
based permutation test also shows that the permutationally generated F-statistics are nowhere near that which was observed. The effect of TempFac
is very strong.
Wait, But Why?
- In marketing, design, or business research, similar quantities may be measured across different locations, or stores, or categories of people, for instance.
- ANOVA is the tool to decide if the Quant variable has differences across the Qual categories.
- This approach can be extended to more than one Qual variable, and also if there is another Quant variable in the mix.
Conclusions
We have discussed ANOVA as a means of modelling the effects of a Categorical variable on a Continuous (Quant) variable. ANOVA can be carried out using the standard formula aov
when assumptions on distributions, variances, and independence are met. Permutation ANOVA tests can be carried out when these assumptions do not quite hold.
What if we have two Categorical variables as predictors?
We then need to perform a Two-Way ANOVA analysis, where we look at the predictors individually (main effects) and together (interaction effects). Here too, we need to verify if the number of observations are balanced across all combinations of factors of the two Qualitative predictors. There are three different classical approaches (Type1, Type2, and Type3 ANOVA) for testing hypotheses in ANOVA for unbalanced designs, as they are called. (Langsrud 2003).
Note that when we specified our research question, we had no specific hypothesis about the means, other than that they might be different. In many situations, we may have reason to believe in the relative “ordering” of the means for different levels of the Categorical variable. The one-sided t-test is the simplest example (e.g., \(\mu_1 >= 0\) and \(\mu_1 >= \mu_2\)); this readily extends to the multi-parameter setting, where more than one inequality constraint can be imposed on the parameters (e.g., \(\mu_1 <= \mu_2 <= \mu_3\).
It is possible to incorporate these beliefs into the ANOVA model, using what is called as informative hypothesis testing, which have certain advantages compared to unconstrained models. The R package called restriktor
has the capability to develop such models with beliefs.
Your Turn
Try the simple datasets at https://www.performingmusicresearch.com/datasets/
Can you try to ANOVA-analyse the datasets we dealt with in plotting Groups with Boxplots?
References
- The ANOVA tutorial at Our Coding Club
- Antoine Soetewey. How to: one-way ANOVA by hand. https://statsandr.com/blog/how-to-one-way-anova-by-hand/
- ANOVA in R - Stats and R https://statsandr.com/blog/anova-in-r/
- Michael Crawley.(2013) The R Book,second edition. Chapter 11.
- David C Howell, Permutation Tests for Factorial ANOVA Designs
- Marti Anderson, Permutation tests for univariate or multivariate analysis of variance and regression
- Judd, Charles M., Gary H. McClelland, and Carey S. Ryan.(2017). “Introduction to Data Analysis.” In, 1–9. Routledge. https://doi.org/10.4324/9781315744131-1.
- Patil, I. (2021). Visualizations with statistical details: The ‘ggstatsplot’ approach. Journal of Open Source Software, 6(61), 3167, doi:10.21105/joss.03167
- Langsrud, Øyvind. (2003). ANOVA for unbalanced data: Use type II instead of type III sums of squares. Statistics and Computing. 13. 163-167. https://doi.org/10.1023/A:1023260610025. https://www.researchgate.net/publication/220286726_ANOVA_for_unbalanced_data_Use_type_II_instead_of_type_III_sums_of_squares
- Kim TK. (2017). Understanding one-way ANOVA using conceptual figures. Korean J Anesthesiol. 2017 Feb;70(1):22-26. https://ekja.org/upload/pdf/kjae-70-22.pdf
- Anova – Type I/II/III SS explained.https://mcfromnz.wordpress.com/2011/03/02/anova-type-iiiiii-ss-explained/
R Package Citations
Footnotes
The ANOVA tutorial at Our Coding Club.↩︎
Pruim R, Kaplan DT, Horton NJ (2017). “The mosaic Package: Helping Students to ‘Think with Data’ Using R.” The R Journal, 9(1), 77–102. https://journal.r-project.org/archive/2017/RJ-2017-024/index.html.↩︎
mosaic::xpf()
gives both a graph and the probabilities.↩︎ggplot2 Based Plots with Statistical Details • ggstatsplot https://indrajeetpatil.github.io/ggstatsplot/↩︎
Ernst, Michael D. 2004. “Permutation Methods: A Basis for Exact Inference.” Statistical Science 19 (4): 676–85. doi:10.1214/088342304000000396.↩︎
Citation
@online{2023,
author = {},
title = {Comparing {Multiple} {Means} with {ANOVA}},
date = {2023-03-28},
url = {https://av-quarto.netlify.app/content/courses/Analytics/Inference/Modules/130-ThreeMeansOrMore/},
langid = {en},
abstract = {ANOVA to investigate how frogspawn hatching time varies
with temperature.}
}