data(iris)
head(iris)## Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## 1 5.1 3.5 1.4 0.2 setosa
## 2 4.9 3.0 1.4 0.2 setosa
## 3 4.7 3.2 1.3 0.2 setosa
## 4 4.6 3.1 1.5 0.2 setosa
## 5 5.0 3.6 1.4 0.2 setosa
## 6 5.4 3.9 1.7 0.4 setosasummary(iris)## Sepal.Length Sepal.Width Petal.Length Petal.Width
## Min. :4.300 Min. :2.000 Min. :1.000 Min. :0.100
## 1st Qu.:5.100 1st Qu.:2.800 1st Qu.:1.600 1st Qu.:0.300
## Median :5.800 Median :3.000 Median :4.350 Median :1.300
## Mean :5.843 Mean :3.057 Mean :3.758 Mean :1.199
## 3rd Qu.:6.400 3rd Qu.:3.300 3rd Qu.:5.100 3rd Qu.:1.800
## Max. :7.900 Max. :4.400 Max. :6.900 Max. :2.500
## Species
## setosa :50
## versicolor:50
## virginica :50
##
##
## DESCRIPTION: The data was taken from the R studio dataset. It has 150 observations, 3 numerical and 1 categorical variable, which are:
Sepal.Length: The length of the sepal, in centimeters.
Sepal.Width: The width of the sepal, in centimeters.
Petal.Length: The length of the petal, in centimeters.
Petal.Width: The width of the petal, in centimeters.
Species: The species of the iris flower, which can be setosa, versicolor, or virginica.
#1. INDEPENDENT T-TEST/WILCOXON RUNK SUM TEST
NORMALITY CHECK:
Ho: Distribution of variables Sepal.Lenght is normal for both species of Iris (versicolor & virginica).
H1: Distribution of variables Sepal.Lenght is not normal for both species of Iris (Versicolor & virginica).
shapiro.test(iris$Sepal.Length[iris$Species == "versicolor"])##
## Shapiro-Wilk normality test
##
## data: iris$Sepal.Length[iris$Species == "versicolor"]
## W = 0.97784, p-value = 0.4647shapiro.test(iris$Sepal.Length[iris$Species == "virginica"])##
## Shapiro-Wilk normality test
##
## data: iris$Sepal.Length[iris$Species == "virginica"]
## W = 0.97118, p-value = 0.2583Given the results of the shapiro test we cannot reject the null hypothesis (since p-values are above the 0.05 threshold).
VARIANCES CHECK:
Ho: Variances are the same.
H1: Variances are not the same.
library(car)## Loading required package: carDataleveneTest(iris$Sepal.Length ~ iris$Species)## Levene's Test for Homogeneity of Variance (center = median)
## Df F value Pr(>F)
## group 2 6.3527 0.002259 **
## 147
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Given the results of the Levene’s test we reject the null hypothesis at p=0.002.
ASSUMPTIONS ARE VIOLATED=> NON-PARAMETRIC TEST:
Ho: True location shift is equal to 0.
H1: True location shift is not equal to 0.
wilcox.test(iris$Sepal.Length[iris$Species == "versicolor"] , iris$Sepal.Length[iris$Species == "virginica"] )##
## Wilcoxon rank sum test with continuity correction
##
## data: iris$Sepal.Length[iris$Species == "versicolor"] and iris$Sepal.Length[iris$Species == "virginica"]
## W = 526, p-value = 5.869e-07
## alternative hypothesis: true location shift is not equal to 0Given the Wilcoxon rank sum test we can reject the null hypthesis at p=0.001.
#2. ONE-WAY ANOVA
iris$SpeciesF <- factor(iris$Species,
labels= c("setosa", "versicolor", "virginica"),
levels= c("setosa", "versicolor", "virginica"))NORMALITY CHECK:
Ho: Distribution of Petal length is normally distributed.
H1: Distribution of Petal length is not normaly distributed.
library(dplyr)##
## Attaching package: 'dplyr'## The following object is masked from 'package:car':
##
## recode## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionlibrary(rstatix)##
## Attaching package: 'rstatix'## The following objects are masked from 'package:effectsize':
##
## cohens_d, eta_squared## The following object is masked from 'package:stats':
##
## filteriris %>%
group_by(SpeciesF) %>%
shapiro_test(Petal.Length)## # A tibble: 3 × 4
## SpeciesF variable statistic p
## <fct> <chr> <dbl> <dbl>
## 1 setosa Petal.Length 0.955 0.0548
## 2 versicolor Petal.Length 0.966 0.158
## 3 virginica Petal.Length 0.962 0.110Given the Shapiro-Wilk test, we cannot reject the null hypothesis in all variables.
Ho:Variances are the same.
H1: Variances are not the same.
library(car)
leveneTest(iris$Petal.Length, group= iris$SpeciesF)## Levene's Test for Homogeneity of Variance (center = median)
## Df F value Pr(>F)
## group 2 19.48 3.129e-08 ***
## 147
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1P value is lower than 0.05 therefore we reject the null hypothesis, variances are not the same.
#install.packages("onewaytests")
library(onewaytests)
welch.test(Petal.Length ~ SpeciesF, data=iris)##
## Welch's Heteroscedastic F Test (alpha = 0.05)
## -------------------------------------------------------------
## data : Petal.Length and SpeciesF
##
## statistic : 1828.092
## num df : 2
## denom df : 78.07296
## p.value : 2.693327e-66
##
## Result : Difference is statistically significant.
## -------------------------------------------------------------pairwise.t.test(x=iris$Petal.Length, g= iris$SpeciesF,
p.adjust.method = "bonferroni")##
## Pairwise comparisons using t tests with pooled SD
##
## data: iris$Petal.Length and iris$SpeciesF
##
## setosa versicolor
## versicolor <2e-16 -
## virginica <2e-16 <2e-16
##
## P value adjustment method: bonferroniEvery variable is different, since the p value is less than 0.05.
iris$Size <- ifelse(iris$Sepal.Length < 5, "Small",
ifelse(iris$Sepal.Length >= 5 & iris$Sepal.Length < 7, "Medium", "Large"))iris$SizeF <- factor(iris$Size,
labels= c("Small", "Medium", "Large"),
levels= c("Small", "Medium", "Large"))#3. CHI-SQUARE TEST
Ho: There is no association between size and species.
H1: There is association between two categorical variables.
results <- chisq.test(iris$SpeciesF, iris$SizeF,
correct=TRUE)## Warning in chisq.test(iris$SpeciesF, iris$SizeF, correct = TRUE): Chi-squared
## approximation may be incorrectresults##
## Pearson's Chi-squared test
##
## data: iris$SpeciesF and iris$SizeF
## X-squared = 57.575, df = 4, p-value = 9.369e-12P value is less than 0.001, therefore we can reject the null hypothesis.
addmargins(results$observed)## iris$SizeF
## iris$SpeciesF Small Medium Large Sum
## setosa 20 30 0 50
## versicolor 1 48 1 50
## virginica 1 37 12 50
## Sum 22 115 13 150round(results$expected, 2)## iris$SizeF
## iris$SpeciesF Small Medium Large
## setosa 7.33 38.33 4.33
## versicolor 7.33 38.33 4.33
## virginica 7.33 38.33 4.33The assumptions are violated.
round (results$res, 2)## iris$SizeF
## iris$SpeciesF Small Medium Large
## setosa 4.68 -1.35 -2.08
## versicolor -2.34 1.56 -1.60
## virginica -2.34 -0.22 3.68addmargins(round(prop.table(results$observed) , 3))## iris$SizeF
## iris$SpeciesF Small Medium Large Sum
## setosa 0.133 0.200 0.000 0.333
## versicolor 0.007 0.320 0.007 0.334
## virginica 0.007 0.247 0.080 0.334
## Sum 0.147 0.767 0.087 1.001addmargins(round(prop.table(results$observed, 1), 3), 2)## iris$SizeF
## iris$SpeciesF Small Medium Large Sum
## setosa 0.40 0.60 0.00 1.00
## versicolor 0.02 0.96 0.02 1.00
## virginica 0.02 0.74 0.24 1.00addmargins(round(prop.table(results$observed, 2), 3), 1)## iris$SizeF
## iris$SpeciesF Small Medium Large
## setosa 0.909 0.261 0.000
## versicolor 0.045 0.417 0.077
## virginica 0.045 0.322 0.923
## Sum 0.999 1.000 1.000fisher.test(iris$SizeF, iris$SpeciesF)##
## Fisher's Exact Test for Count Data
##
## data: iris$SizeF and iris$SpeciesF
## p-value = 3.538e-12
## alternative hypothesis: two.sidedWe can reject the null hypothesis at p=0.001.