Comparing the means of more than two groups
There are mainly two techniques are used to compare the one-sample mean to a standard known mean. These two techniques are:
- Analysis of Variance (ANOVA)
- One way ANOVA
- Two way ANOVA
- MANOVA Test
- Kruskal–Wallis Test
One way ANOVA
The one-way analysis of variance (ANOVA), also known as one-factor ANOVA, is an extension of independent two-samples t-test for comparing means in a situation where there are more than two groups. In one-way ANOVA, the data is organized into several groups base on one single grouping variable.
Implementation in R:
For performing the one-way analysis of variance (ANOVA) in R, use the function aov(). The function summary.aov() is used to summarize the analysis of the variance model. The syntax for the function is given below.
Syntax: aov(formula, data = NULL)
Parameters:
- formula: A formula specifying the model.
- data: A data frame in which the variables specified in the formula will be found
Example:
One way ANOVA test is performed using mtcars dataset which comes preinstalled with dplyr package between disp attribute, a continuous attribute, and gear attribute, a categorical attribute.
R
# R program to illustrate# one way ANOVA test# Loading the package library(dplyr) # Calculate test statistics using aov function mtcars_aov <- aov(mtcars $ disp ~ factor(mtcars $ gear)) print(summary(mtcars_aov)) |
Output:
Df Sum Sq Mean Sq F value Pr(>F)
factor(mtcars$gear) 2 280221 140110 20.73 2.56e-06 ***
Residuals 29 195964 6757
—
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
The summary shows that the gear attribute is very significant to displacement(Three stars denoting it). Also, P value less than 0.05, so it proves that gear is significant to displacement i.e related to each other, and we reject the Null Hypothesis.
Two way ANOVA
Two-way ANOVA test is used to evaluate simultaneously the effect of two grouping variables (A and B) on a response variable. It takes two categorical groups into consideration.
Implementation in R:
For performing the two-way analysis of variance (ANOVA) in R, also use the function aov(). The function summary.aov() is used to summarize the analysis of variance model. The syntax for the function is given below.
Syntax: aov(formula, data = NULL)
Parameters:
- formula: A formula specifying the model.
- data: A data frame in which the variables specified in the formula will be found
Example: Two way ANOVA test is performed using mtcars dataset which comes preinstalled with dplyr package between disp attribute, a continuous attribute and gear attribute, a categorical attribute, am attribute, a categorical attribute.
R
# R program to illustrate# two way ANOVA test# Loading the package library(dplyr) # Calculate test statistics using aov function mtcars_aov2 = aov(mtcars $ disp ~ factor(mtcars $ gear) * factor(mtcars $ am)) print(summary(mtcars_aov2)) |
Output:
Df Sum Sq Mean Sq F value Pr(>F)
factor(mtcars$gear) 2 280221 140110 20.695 3.03e-06 ***
factor(mtcars$am) 1 6399 6399 0.945 0.339
Residuals 28 189565 6770
—
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
The summary shows that gear attribute is very significant to displacement(Three stars denoting it) and am attribute is not much significant to displacement. P-value of gear is less than 0.05, so it proves that gear is significant to displacement i.e related to each other. P-value of am is greater than 0.05, am is not significant to displacement i.e not related to each other.
MANOVA Test
Multivariate analysis of variance (MANOVA) is simply an ANOVA (Analysis of variance) with several dependent variables. It is a continuation of the ANOVA. In an ANOVA, we test for statistical differences on one continuous dependent variable by an independent grouping variable. The MANOVA continues this analysis by taking multiple continuous dependent variables and bundles them collectively into a weighted linear composite variable. The MANOVA compares whether or not the newly created combination varies by the different levels, or groups, of the independent variable.
Implementation in R:
R provides a method manova() to perform the MANOVA test. The class “manova” differs from class “aov” in selecting a different summary method. The function manova() calls aov and then add class “manova” to the result object for each stratum.
Syntax: manova(formula, data = NULL, projections = FALSE, qr = TRUE, contrasts = NULL, …)
Parameters:
- formula: A formula specifying the model.
- data: A data frame in which the variables specified in the formula will be found. If missing, the variables are searched for in the standard way.
- projections: Logical flag
- qr: Logical flag
- contrasts: A list of contrasts to be used for some of the factors in the formula.
…: Arguments to be passed to lm, such as subset or na.action
Example: To perform the MANOVA test in R let’s take iris data set.
R
# R program to illustrate # MANOVA test # Import required library library(dplyr) # Taking iris data set myData = iris # Show a random sample set.seed(1234) dplyr::sample_n(myData, 10) |
Output:
Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.5 2.5 4.0 1.3 versicolor 2 5.6 2.5 3.9 1.1 versicolor 3 6.0 2.9 4.5 1.5 versicolor 4 6.4 3.2 5.3 2.3 virginica 5 4.3 3.0 1.1 0.1 setosa 6 7.2 3.2 6.0 1.8 virginica 7 5.9 3.0 4.2 1.5 versicolor 8 4.6 3.1 1.5 0.2 setosa 9 7.9 3.8 6.4 2.0 virginica 10 5.1 3.4 1.5 0.2 setosa
To know if there is any important difference, in sepal and petal length, between the different species then perform MANOVA test. Hence, the function manova() can be used as follows.
R
# Taking two dependent variable sepal = iris$Sepal.Length petal = iris$Petal.Length # MANOVA test result = manova(cbind(Sepal.Length, Petal.Length) ~ Species, data = iris) summary(result) |
Output:
Df Pillai approx F num Df den Df Pr(>F)
Species 2 0.9885 71.829 4 294 < 2.2e-16 ***
Residuals 147
—
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
From the output above, it can be seen that the two variables are highly significantly different among Species.
Kruskal–Wallis test
The Kruskal–Wallis test is a rank-based test that is similar to the Mann–Whitney U test but can be applied to one-way data with more than two groups. It is a non-parametric alternative to the one-way ANOVA test, which extends the two-samples Wilcoxon test. A group of data samples is independent if they come from unrelated populations and the samples do not affect each other. Using the Kruskal-Wallis Test, it can be decided whether the population distributions are similar without assuming them to follow the normal distribution.
Implementation in R:
R provides a method kruskal.test() which is available in the stats package to perform a Kruskal-Wallis rank-sum test.
Syntax: kruskal.test(x, g, formula, data, subset, na.action, …)
Parameters:
- x: a numeric vector of data values, or a list of numeric data vectors.
- g: a vector or factor object giving the group for the corresponding elements of x
- formula: a formula of the form response ~ group where response gives the data values and group a vector or factor of the corresponding groups.
- data: an optional matrix or data frame containing the variables in the formula .
- subset: an optional vector specifying a subset of observations to be used.
- na.action: a function which indicates what should happen when the data contain NA
- …: further arguments to be passed to or from methods.
Example: Let’s use the built-in R data set named PlantGrowth. It contains the weight of plants obtained under control and two different treatment conditions.
R
# Preparing the data set # to perform Kruskal-Wallis Test # Taking the PlantGrowth data set myData = PlantGrowth print(myData) # Show the group levels print(levels(myData$group)) |
Output:
weight group 1 4.17 ctrl 2 5.58 ctrl 3 5.18 ctrl 4 6.11 ctrl 5 4.50 ctrl 6 4.61 ctrl 7 5.17 ctrl 8 4.53 ctrl 9 5.33 ctrl 10 5.14 ctrl 11 4.81 trt1 12 4.17 trt1 13 4.41 trt1 14 3.59 trt1 15 5.87 trt1 16 3.83 trt1 17 6.03 trt1 18 4.89 trt1 19 4.32 trt1 20 4.69 trt1 21 6.31 trt2 22 5.12 trt2 23 5.54 trt2 24 5.50 trt2 25 5.37 trt2 26 5.29 trt2 27 4.92 trt2 28 6.15 trt2 29 5.80 trt2 30 5.26 trt2 [1] "ctrl" "trt1" "trt2"
Here the column “group” is called factor and the different categories (“ctr”, “trt1”, “trt2”) are named factor levels. The levels are ordered alphabetically. The problem statement is we want to know if there is any significant difference between the average weights of plants in the 3 experimental conditions. And the test can be performed using the function kruskal.test() as given below.
R
# R program to illustrate # Kruskal-Wallis Test # Taking the PlantGrowth data set myData = PlantGrowth # Performing Kruskal-Wallis test result = kruskal.test(weight ~ group, data = myData) print(result) |
Output:
Kruskal-Wallis rank sum test
data: weight by group
Kruskal-Wallis chi-squared = 7.9882, df = 2, p-value = 0.01842
As the p-value is less than the significance level 0.05, it can be concluded that there are significant differences between the treatment groups.
Comparing Means in R Programming
There are many cases in data analysis where you’ll want to compare means for two populations or samples and which technique you should use depends on what type of data you have and how that data is grouped together. The comparison of means tests helps to determine if your groups have similar means. So this article contains statistical tests to use for comparing means in R programming. These tests include:
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