Scenario: A comparison on the extent of student engagement is to be conducted to check whether interventions would be needed to entice students to participate in school activities. The team randomly sampled students (proportional allocation) from the four Academic Programs offered by the school and met the desired sample size computed via Modified Cochran.
library(xlsx)
data <- read.xlsx("StudentEngagement.xlsx", sheetIndex = "Sheet1")
head(data)
## NA. Engagement Acad
## 1 1 3.53 Nursing
## 2 2 4.53 Nursing
## 3 3 3.85 Nursing
## 4 4 3.91 Nursing
## 5 5 4.35 Nursing
## 6 6 3.46 Nursing
tail(data)
## NA. Engagement Acad
## 73 73 3.77 Civil Engineering
## 74 74 3.36 Civil Engineering
## 75 75 2.87 Civil Engineering
## 76 76 3.81 Civil Engineering
## 77 77 3.56 Civil Engineering
## 78 78 3.38 Civil Engineering
str(data)
## 'data.frame': 78 obs. of 3 variables:
## $ NA. : chr "1" "2" "3" "4" ...
## $ Engagement: num 3.53 4.53 3.85 3.91 4.35 3.46 4.3 5.05 4.6 3.51 ...
## $ Acad : chr "Nursing" "Nursing" "Nursing" "Nursing" ...
summary(data)
## NA. Engagement Acad
## Length:78 Min. :2.160 Length:78
## Class :character 1st Qu.:3.270 Class :character
## Mode :character Median :3.635 Mode :character
## Mean :3.647
## 3rd Qu.:4.018
## Max. :5.210
data <- data[,-1]
unique(data$Acad)
## [1] "Nursing" "Business Administration"
## [3] "Computer Science" "Civil Engineering"
Nsg <- data[data$Acad == "Nursing",]$Engagement
BA <- data[data$Acad == "Business Administration",]$Engagement
CS <- data[data$Acad == "Computer Science",]$Engagement
CE <- data[data$Acad == "Civil Engineering",]$Engagement
library(nortest)
ad.test(Nsg)
##
## Anderson-Darling normality test
##
## data: Nsg
## A = 0.29386, p-value = 0.5579
ad.test(BA)
##
## Anderson-Darling normality test
##
## data: BA
## A = 0.33658, p-value = 0.4768
ad.test(CS)
##
## Anderson-Darling normality test
##
## data: CS
## A = 0.15365, p-value = 0.9488
ad.test(CE)
##
## Anderson-Darling normality test
##
## data: CE
## A = 0.47463, p-value = 0.2096
#Data sets tend to assume Normal Distribution
par(mfrow=c(2,2))
plot(density(Nsg), col = "darkorchid")
plot(density(BA), col = "darkorchid")
plot(density(CE), col = "darkorchid")
plot(density(CS), col = "darkorchid")
par(mfrow=c(2,2))
boxplot(Nsg, col = "darkorchid")
boxplot(BA, col = "darkorchid")
boxplot(CE, col = "darkorchid")
boxplot(CS, col = "darkorchid")
par(mfrow=c(2,2))
qqnorm(data[data$Acad == "Business Administration",]$Engagement); qqline(data[data$Acad == "Business Administration",]$Engagement)
qqnorm(data[data$Acad == "Civil Engineering",]$Engagement); qqline(data[data$Acad == "Civil Engineering",]$Engagement)
qqnorm(data[data$Acad == "Computer Science",]$Engagement); qqline(data[data$Acad == "Computer Science",]$Engagement)
qqnorm(data[data$Acad == "Nursing",]$Engagement); qqline(data[data$Acad == "Nursing",]$Engagement)
# Although statistics show that the distribution for each variable is approximately normal, the Boxplot and QQ-plot visualizations shows otherwise.
library(psych)
Desc <- describeBy(data$Engagement, data$Acad)
Desc <- rbind(Desc$`Business Administration`, Desc$`Civil Engineering`, Desc$`Computer Science`, Desc$`Nursing`)
Desc <- round(Desc[, c(2,5,7,8,9)],2)
Group <- c("Business Administration", "Civil Engineering", "Computer Science", "Nursing")
Group <- data.frame(Group)
Desc <- cbind(Group, Desc)
Desc
## Group n median mad min max
## X1 Business Administration 24 3.34 0.52 2.56 4.54
## X11 Civil Engineering 17 3.56 0.40 2.32 4.56
## X12 Computer Science 20 3.58 0.58 2.16 4.97
## X13 Nursing 17 4.07 0.64 3.46 5.21
#Median is the measure for central tendency given the result of the normality test
library(broom)
res <- tidy(kruskal.test(data$Engagement~data$Acad))
res <- res[,c(4,3,1,2)]
res
## # A tibble: 1 x 4
## method parameter statistic p.value
## <chr> <int> <dbl> <dbl>
## 1 Kruskal-Wallis rank sum test 3 15.0 0.00179
#Median is the measure for central tendency given the result of the normality test
library(FSA)
ph <- dunnTest(Engagement ~ Acad, data=data, method="bonferroni")
## Warning: Acad was coerced to a factor.
ph
## Comparison Z P.unadj P.adj
## 1 Business Administration - Civil Engineering -0.2439727 0.807251947 1.00000000
## 2 Business Administration - Computer Science -0.8174954 0.413645357 1.00000000
## 3 Civil Engineering - Computer Science -0.5158455 0.605962352 1.00000000
## 4 Business Administration - Nursing -3.6015833 0.000316285 0.00189771
## 5 Civil Engineering - Nursing -3.1031419 0.001914778 0.01148867
## 6 Computer Science - Nursing -2.7106479 0.006715190 0.04029114
#Median is the measure for central tendency given the result of the normality test