One-Way ANOVA
One-Way ANOVA Hypothesis
In the paired t-test topic note, and the two-sample t-test topic note we discussed how to compare the means between two groups. What if you'd like to compare the means of two or more groups? This problem can be solved using a one-way ANOVA. The hypothesis of the One-Way ANOVA states:
- \(H_0\) : there is no difference between the means of each group
- \(H_a\) : One or more of the group means is different from the other group means
Between Group Sum of Squares
To determine this difference we first calculate the between group sum of squares:
Explanation of Terms
- \(SS_between\) : between group sum of squares
- \(a\) : group a
- \(z\) : last group
- \(x_a\) : mean of group a
- \(n_a\) : sample size of group a
- \(\overline{X}\) : mean of all groups
So here we see that for each group we will subtract the total mean from the group mean and multiply that value by sample size of that group. We then take all those values for each group and add them together to get the sum of squares.
Within Group Sum of Squares
Now we will need to calculate the within group sum of squares:
Explanation of Terms
- \(SS_{within}\) : within group sum of squares
- \(a\) : group a
- \(z\) : last group
- \(x_ia\) : value i in group a
- \(\overline{x_a}\) : mean of group a
Here we note that for each group we sum the difference from the mean for each value and add them together. Then when we are done adding these differences for each group we add those values together to get the within group sum of squares.
One-Way ANOVA F-Statistic
Now to get our F-statistic we need to calculate the between group mean square value and the within group mean square value:
Explanation of Terms
- \(F\) : F-statistic
- \(MS_{between}\) : between group mean square value
- \(MS_{within}\) : within group mean square value
- \(k\) : number of groups
- \(t\) : total number of observations between all groups
- \(SS_{between}\) : between group sum of squares
Visualizing our data
As you can see this is pretty laborious. Luckily, R can do these calculations for us and we will use R to determine if there is any significant difference in age between the patient's country of origin. Here we will filter out countries that have less than 5 observations. First let's try a visual inspection of our data:
# one-way ANOVA
library(tidyverse)
# load meta data
meta <- read.table("./data/gbm_cptac_2021/data_clinical_patient.txt",
header = T,
sep="\t")
# isolate countries and ages
countries_ages <- meta %>%
filter(COUNTRY_OF_ORIGIN %in% c("China","Poland","Russia","United States"))
# plot our data
ggplot(countries_ages,aes(x=COUNTRY_OF_ORIGIN,y=AGE,fill=COUNTRY_OF_ORIGIN)) +
geom_boxplot()+
theme_bw()+
labs(
x="",
y="AGE",
fill="Country of Origin"
)
Here we can visually see that the mean age of patients from China is lower than the other groups.
One-Way ANOVA in R
Let's now apply a One-Way ANOVA in R and output the summary of results:
# calculate our one-way ANOVA
anova_res <- aov(AGE ~ COUNTRY_OF_ORIGIN, data = countries_ages)
# print out a summary of results
summary(anova_res)
Df Sum Sq Mean Sq F value Pr(>F)
COUNTRY_OF_ORIGIN 3 1583 527.6 3.824 0.0128 *
Residuals 84 11591 138.0
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Explanation of Results
- Here we see that our between sum of squares is
1583and our within sum of squares is11591 - We also note that our between group mean square value is
527.6and our within mean square valueis138.0 - By dividing our between group mean square value by the within mean square value we get an F-statistic of
3.824 - We notice a p-value of
0.0128, below 0.05, giving us enough evidence to reject the null:- that there is not a difference between group means
Assumptions
Like any statistical test, we make assumptions. The assumptions made by the one-way ANOVA are:
- The residuls are normally distributed
- The variances of each group are equal (Homoscedasticity)
Note
The residuals for a One-Way ANOVA are calculated by taking each value and subtracting the mean of the group that value belongs to
To test the normality of the residuals we will use the Shapiro-Wilk test:
# use the shapiro-wilk test to determine the normality of the residuals
shapiro.test(x = residuals(anova_res))
Shapiro-Wilk normality test
data: residuals(anova_res)
W = 0.98732, p-value = 0.5527
Given our p-value is above 0.05 we do not have enough evidence to reject the null hypothesis of the Shapiro-Wilk Test; that the data are normally distributed. In other words, if the p-value is above 0.05 your data are normally distributed. To determine the homoscedasticity of our data we will use the Levene test:
# apply the levene test to our data
library(car)
leveneTest(AGE ~ COUNTRY_OF_ORIGIN, data = countries_ages)
Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 3 0.1514 0.9285
84
Here we note that the p-value is above 0.05, indicating we don't have enough evidence to say that there is a significant difference in the variances between groups.
Non-Parametric Alternative
A non-parametric test is often used when either the assumptions about the distribution are not met. Additionally, these tests do not depend on the parameter they are assessing. Here, if the assumptions above are not met we can use the non-parametric equivalent, the Kruskal-Wallis rank sum test:
# run the non-parametric alternative to the one-way ANOVA
# the Kruskal-Wallis rank sum test
kruskal.test(AGE ~ COUNTRY_OF_ORIGIN,
data = countries_ages)
Kruskal-Wallis rank sum test
data: AGE by COUNTRY_OF_ORIGIN
Kruskal-Wallis chi-squared = 9.1676, df = 3, p-value = 0.02714