Comparing Means in R Programming

16 May, 2022
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:
T-test
Wilcoxon test
ANOVA test
Kruskal-Wallis test
So as we have discussed before various techniques are used depending on what type of data we have and how the data is grouped together. So let’ discuss one by one techniques depending on the different types of data.
There are mainly two techniques used to compare the one-sample mean to a standard known mean. These two techniques are:
One Sample T-test
One-Sample Wilcoxon Test
The One-Sample T-Test is used to test the statistical difference between a sample mean and a known or assumed/hypothesized value of the mean in the population.
Implementation in R:
For performing a one-sample t-test in R, use the function t.test(). The syntax for the function is given below:
Syntax: t.test(x, mu = 0)
Parameters:
x: the name of the variable of interest
mu: set equal to the mean specified by the null hypothesis
Example:
R
# R program to illustrate# One sample t-test set.seed(0)sweetSold <- c(rnorm(50, mean = 140, sd = 5)) # Ho: mu = 150# Using the t.test()result = t.test(sweetSold, mu = 150) # Print the resultprint(result)
 Output:
One Sample t-test

data:  sweetSold
t = -15.249, df = 49, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 150
95 percent confidence interval:
 138.8176 141.4217
sample estimates:
mean of x 
 140.1197 
The one-sample Wilcoxon signed-rank test is a non-parametric alternative to a one-sample t-test when the data cannot be assumed to be normally distributed. It’s used to determine whether the median of the sample is equal to a known standard value i.e. a theoretical value.
Implementation in R:
To perform a one-sample Wilcoxon-test, R provides a function wilcox.test() that can be used as follows:
Syntax: wilcox.test(x, mu = 0, alternative = “two.sided”)
Parameters:
x: a numeric vector containing your data values
mu: the theoretical mean/median value. Default is 0 but you can change it.
alternative: the alternative hypothesis. Allowed value is one of “two.sided” (default), “greater” or “less”.
Example: Here, let’s use an example data set containing the weight of 10 rabbits. Let’s know if the median weight of the rabbit differs from 25g?
R
# R program to illustrate# one-sample Wilcoxon signed-rank test # The data setset.seed(1234)myData = data.frame(name = paste0(rep("R_", 10), 1:10),weight = round(rnorm(10, 30, 2), 1)) # Print the dataprint(myData) # One-sample wilcoxon testresult = wilcox.test(myData$weight, mu = 25) # Printing the resultsprint(result)
Output:
    name weight
1   R_1   27.6
2   R_2   30.6
3   R_3   32.2
4   R_4   25.3
5   R_5   30.9
6   R_6   31.0
7   R_7   28.9
8   R_8   28.9
9   R_9   28.9
10 R_10   28.2

    Wilcoxon signed rank test with continuity correction

data:  myData$weight
V = 55, p-value = 0.005793
alternative hypothesis: true location is not equal to 25
In the above output, the p-value of the test is 0.005793, which is less than the significance level alpha = 0.05. So we can reject the null hypothesis and conclude that the average weight of the rabbit is significantly different from 25g with a p-value = 0.005793.
There are mainly two techniques are used to compare the means of paired samples. These two techniques are:
Paired sample T-test
Paired Samples Wilcoxon Test
This is a statistical procedure that is used to determine whether the mean difference between two sets of observations is zero. In a paired sample t-test, each subject is measured two times, resulting in pairs of observations.
Implementation in R:
For performing a one-sample t-test in R, use the function t.test(). The syntax for the function is given below.
Syntax: t.test(x, y, paired =TRUE)
Parameters:
x, y: numeric vectors
paired: a logical value specifying that we want to compute a paired t-test
Example:
R
# R program to illustrate# Paired sample t-test set.seed(0) # Taking two numeric vectorsshopOne <- rnorm(50, mean = 140, sd = 4.5)shopTwo <- rnorm(50, mean = 150, sd = 4) # Using t.tset()result = t.test(shopOne, shopTwo,                var.equal = TRUE) # Print the resultprint(result)
 Output:
Two Sample t-test
data:  shopOne and shopTwo
t = -13.158, df = 98, p-value < 2.2e-16
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -11.482807  -8.473061
sample estimates:
mean of x mean of y 
 140.1077  150.0856 
The paired samples Wilcoxon test is a non-parametric alternative to paired t-test used to compare paired data. It’s used when data are not normally distributed.
Implementation in R:
To perform Paired Samples Wilcoxon-test, the R provides a function wilcox.test() that can be used as follows:
Syntax: wilcox.test(x, y, paired = TRUE, alternative = “two.sided”)
Parameters:
x, y: numeric vectors
paired: a logical value specifying that we want to compute a paired Wilcoxon test
alternative: the alternative hypothesis. Allowed value is one of “two.sided” (default), “greater” or “less”.
Example: Here, let’s use an example data set, which contains the weight of 10 rabbits before and after the treatment. We want to know, if there is any significant difference in the median weights before and after treatment?
R
# R program to illustrate# Paired Samples Wilcoxon Test # The data set# Weight of the rabbit before treatmentbefore <-c(190.1, 190.9, 172.7, 213, 231.4,        196.9, 172.2, 285.5, 225.2, 113.7) # Weight of the rabbit after treatmentafter <-c(392.9, 313.2, 345.1, 393, 434,        227.9, 422, 383.9, 392.3, 352.2) # Create a data framemyData <- data.frame(group = rep(c("before", "after"), each = 10),weight = c(before, after)) # Print all dataprint(myData) # Paired Samples Wilcoxon Testresult = wilcox.test(before, after, paired = TRUE) # Printing the resultsprint(result)
Output:
   group weight
1  before  190.1
2  before  190.9
3  before  172.7
4  before  213.0
5  before  231.4
6  before  196.9
7  before  172.2
8  before  285.5
9  before  225.2
10 before  113.7
11  after  392.9
12  after  313.2
13  after  345.1
14  after  393.0
15  after  434.0
16  after  227.9
17  after  422.0
18  after  383.9
19  after  392.3
20  after  352.2

    Wilcoxon signed rank test

data:  before and after
V = 0, p-value = 0.001953
alternative hypothesis: true location shift is not equal to 0
In the above output, the p-value of the test is 0.001953, which is less than the significance level alpha = 0.05. We can conclude that the median weight of the mice before treatment is significantly different from the median weight after treatment with a p-value = 0.001953.
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 ANOVATwo way ANOVAMANOVA Test
One way ANOVA
Two way ANOVA
MANOVA Test
Kruskal–Wallis Test
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 packagelibrary(dplyr) # Calculate test statistics using aov functionmtcars_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 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 packagelibrary(dplyr) # Calculate test statistics using aov functionmtcars_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.
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 librarylibrary(dplyr) # Taking iris data setmyData = iris # Show a random sampleset.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 variablesepal = iris$Sepal.Lengthpetal = iris$Petal.Length # MANOVA testresult = 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.
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 setmyData = PlantGrowthprint(myData) # Show the group levelsprint(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 setmyData = PlantGrowth # Performing Kruskal-Wallis testresult = 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.
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