Homework Assignment №1 at MVA
Author: Anna Kuznetsova
2025-01-13
1. DATA IMPORT AND DESCRIPTION
# Importing a dataset from Kaggle.com, of Student Performance Prediction:
mydata <- read.table("~/Library/Mobile Documents/com~apple~CloudDocs/Documents/MA_HW1.csv",
header = TRUE, sep = ",", dec = ".")
head(mydata, 10)## ID Gender Study.Hours Sleep.Hours Attendance Grades.Italian Grades.Spanish
## 1 1 1 3.4 8.2 60 79 89
## 2 2 1 3.2 5.9 76 77 78
## 3 3 2 3.2 9.3 41 32 33
## 4 4 1 3.2 8.2 47 34 45
## 5 5 2 3.8 10.0 75 76 81
## 6 6 2 3.4 9.0 47 50 55
## 7 7 2 7.9 8.1 63 80 74
## 8 8 1 1.4 8.0 47 55 56
## 9 9 2 5.4 8.8 67 73 68
## 10 10 1 1.4 9.6 42 59 56
## Unit.Pass
## 1 1
## 2 1
## 3 2
## 4 2
## 5 1
## 6 2
## 7 1
## 8 2
## 9 1
## 10 2EXPLANATION OF THE DATA:
- Unit of observation: An individual student.
- Sample size: 165 units of observation.
DESCRIPTION OF THE VARIABLES:
- Student.ID: Randomly assigned number of each student (1-65).
- Gender: The gender of each student, either male (1) or female (2).
- Study.Hours: Average daily time spent studying (in hours).
- Sleep.Hours: Average daily time spent sleeping (in hours).
- Attendance: The percentage of classes attended by a student (in percentages, %).
- Grades.Italian: Final performance score of a student in Italian language exam (1-100 points).
- Grades.Spanish: Final performance score of a student in Spanish language exam (1-100 points).
- Unit.Pass: The final decision whether the language unit is passed or no, based on the grades obtained in both exams, with a condition of obtaining a grade higher than 60 on each exam (1 - passed, 2 - failed).
DATA SOURCE: The dataset is retrieved from Keggle. This is a synthetic representation of student performance, designed to mimic real-world scenarios by considering key factors such as study habits, sleep patterns, and class attendance. Each row represents a hypothetical student, and the dataset includes both input features and the calculated target variable (course grades and passing the unit).
# Basic description of the data frame structure. Shows the number of units and variables, and types of variables (num = numeric, int = integer / full number, etc):
str(mydata)## 'data.frame': 165 obs. of 8 variables:
## $ ID : int 1 2 3 4 5 6 7 8 9 10 ...
## $ Gender : int 1 1 2 1 2 2 2 1 2 1 ...
## $ Study.Hours : num 3.4 3.2 3.2 3.2 3.8 3.4 7.9 1.4 5.4 1.4 ...
## $ Sleep.Hours : num 8.2 5.9 9.3 8.2 10 9 8.1 8 8.8 9.6 ...
## $ Attendance : num 60 76 41 47 75 47 63 47 67 42 ...
## $ Grades.Italian: num 79 77 32 34 76 50 80 55 73 59 ...
## $ Grades.Spanish: num 89 78 33 45 81 55 74 56 68 56 ...
## $ Unit.Pass : int 1 1 2 2 1 2 1 2 1 2 ...2. DATA MANIPULATION
# Renaming some columns titles in order to make the data frame neater:
colnames(mydata)[3] <- ("Study") #Its previous name was Study.Hours
colnames(mydata)[4] <- ("Sleep") #Its previous name was Sleep.Hours
colnames(mydata)[6] <- ("Italian") #Its previous name was Grades.Italian
colnames(mydata)[7] <- ("Spanish") #Its previous name was Grades.Spanish
colnames(mydata)[8] <- ("Unit") #Its previous name was Unit.Pass
head(mydata, 5)## ID Gender Study Sleep Attendance Italian Spanish Unit
## 1 1 1 3.4 8.2 60 79 89 1
## 2 2 1 3.2 5.9 76 77 78 1
## 3 3 2 3.2 9.3 41 32 33 2
## 4 4 1 3.2 8.2 47 34 45 2
## 5 5 2 3.8 10.0 75 76 81 1# Changing the depiction of Gender from "1" and "2" to words "Male" and "Female":
mydata$Gender <- factor(mydata$Gender,
levels = c("1", "2"),
labels = c("Male", "Female"))
# Changing the depiction of Unit from "1" and "2" to words "Pass" and "Fail":
mydata$Unit <- factor(mydata$Unit,
levels = c("1", "2"),
labels = c("Pass", "Fail"))
head(mydata, 5)## ID Gender Study Sleep Attendance Italian Spanish Unit
## 1 1 Male 3.4 8.2 60 79 89 Pass
## 2 2 Male 3.2 5.9 76 77 78 Pass
## 3 3 Female 3.2 9.3 41 32 33 Fail
## 4 4 Male 3.2 8.2 47 34 45 Fail
## 5 5 Female 3.8 10.0 75 76 81 PassDESCRIPTIVE STATISTICS: Presenting some descriptive statistics and explaining few estimates of parameters.
summary(mydata [ , -1]) #Removing ID because it doesn't make sense to analyse it.## Gender Study Sleep Attendance Italian
## Male :87 Min. :1.10 Min. : 5.100 Min. : 40.00 Min. : 32.00
## Female:78 1st Qu.:3.20 1st Qu.: 7.100 1st Qu.: 53.00 1st Qu.: 60.00
## Median :3.70 Median : 8.200 Median : 67.00 Median : 76.00
## Mean :4.48 Mean : 7.978 Mean : 68.04 Mean : 73.47
## 3rd Qu.:5.40 3rd Qu.: 9.000 3rd Qu.: 80.00 3rd Qu.: 89.00
## Max. :9.80 Max. :10.000 Max. :100.00 Max. :100.00
## Spanish Unit
## Min. : 33.00 Pass:125
## 1st Qu.: 68.00 Fail: 40
## Median : 78.00
## Mean : 75.04
## 3rd Qu.: 88.00
## Max. :100.00mean(mydata$Attendance)## [1] 68.03636- The average percentage of class participation of all students is approximately 68%.
median(mydata$Spanish)## [1] 78- Half of students passed Spanish exam with a grade lower than 78 points, and other half with a grade higher than 78 points.
#install.packages("pastecs")
library(pastecs)
round(stat.desc(mydata[ , -c(1,2,8)]), 1) #Round data to one decimal point.## Study Sleep Attendance Italian Spanish
## nbr.val 165.0 165.0 165.0 165.0 165.0
## nbr.null 0.0 0.0 0.0 0.0 0.0
## nbr.na 0.0 0.0 0.0 0.0 0.0
## min 1.1 5.1 40.0 32.0 33.0
## max 9.8 10.0 100.0 100.0 100.0
## range 8.7 4.9 60.0 68.0 67.0
## sum 739.2 1316.4 11226.0 12123.0 12381.0
## median 3.7 8.2 67.0 76.0 78.0
## mean 4.5 8.0 68.0 73.5 75.0
## SE.mean 0.2 0.1 1.3 1.4 1.4
## CI.mean.0.95 0.3 0.2 2.6 2.8 2.7
## var 4.9 1.8 292.7 337.6 305.4
## std.dev 2.2 1.3 17.1 18.4 17.5
## coef.var 0.5 0.2 0.3 0.3 0.2Minimum and Maximum of Study: the minimum average hours spent studying is 1.1, and the maximum is 9.8.
Comparing Mean and Median of Sleep: mean = 8.0, median = 8.2. Since the numbers differ, and mean is less than (to the left of) the median, then the data for Sleep should be skewed to the left. But the difference is not very significant, so it is not likely that there are outliers.
Range of Attendance: the maximum difference between the percent of attendance of two students is 60.
Third quantile of Italian: approximately 75 percent of students got less than 89.0 points on the exam, and approximately 25 percent of students earned more than 89.0 points.
SE.mean of Spanish: standard error of mean of 1.4 tells how much the sample mean could vary if the samples of 165 students are repeatedly taken. Since sample mean is high (75.0), the SE.mean suggests the sample mean is quite a good estimate of population mean.
Coef.Var of Study: coefficient of variation (CV) compares relative variability or the dispersion of data across different datasets, regardless their units; low coefficient of variation implies that data is clustered around the mean, as in this case, where CV is 0.5.
3.1. HYPOTHESIS TESTING
One unit of observation (a student) is measured twice — for Italian exam and for Spanish exam. Therefore, the variables are dependent or paired, and we need to analyse two dependent samples. The hypothesis test methods are Paired T-Test for parametric test, and either the Sign test or the Wilcoxon Signed Rank Test for non-parametric test.
RESEARCH QUESTION: Is the average grade obtained for the Italian exam different from the average grade obtained for the Spanish exam, in the observed sample of 165 students?
3.1.1. PARAMETRIC TEST - Paired T-Test:
HYPOTHESIS:
H0: Mean value of grades for Italian course is equal to mean value of grades for Spanish course.
H0: 𝜇Italian = 𝜇Spanish, or: 𝜇Difference = 0
H1: Mean value of grades for Italian course is not equal to mean value of grades for Spanish course.
H1: 𝜇Italian ≠ 𝜇Spanish, or: 𝜇Difference ≠ 0
CONDITIONS for Paired T-Test:
- Analyzed variable is numeric.
- The differences between variables are normal.
- Variables are comparable (same scale).
COMPARING DATA WITH STATISTICS:
# Showing only relevant data for hypothesis testing:
head(mydata [ , c(6, 7)], 5)## Italian Spanish
## 1 79 89
## 2 77 78
## 3 32 33
## 4 34 45
## 5 76 81# Using descriptive statistics to compare the means:
library(pastecs)
round(stat.desc(mydata[ , c(6,7)]), 1)## Italian Spanish
## nbr.val 165.0 165.0
## nbr.null 0.0 0.0
## nbr.na 0.0 0.0
## min 32.0 33.0
## max 100.0 100.0
## range 68.0 67.0
## sum 12123.0 12381.0
## median 76.0 78.0
## mean 73.5 75.0
## SE.mean 1.4 1.4
## CI.mean.0.95 2.8 2.7
## var 337.6 305.4
## std.dev 18.4 17.5
## coef.var 0.3 0.2- Mean grade of Italian (73.5) and mean grade of Spanish (75.0) are very close numbers.
# Creating a new variable that shows a difference between each grade in each course.
mydata$Difference <- mydata$Italian - mydata$Spanish# Building a histogram chart:
#install.packages("ggplot2")
library(ggplot2)
ggplot(mydata, aes(x = Difference)) +
geom_histogram(binwidth = 3, fill = "slategray2", colour = "skyblue4") +
ylab ("Frequency") + xlab ("Difference")
- The histogram of differences looks normal.
SHAPIRO-WILK NORMALITY TEST:
Checking the condition of normality.
H0: Differences between grades of Italian and Spanish are normally distributed.
H1: Differences between grades of Italian and Spanish are not normally distributed.
# Shapiro-Wilk normality test for checking whether differences between variables are normally distributed:
shapiro.test(mydata$Difference)##
## Shapiro-Wilk normality test
##
## data: mydata$Difference
## W = 0.98702, p-value = 0.1301# Showing a visualization of distribution of differences, expecting to see all spots directly on the line.
#install.packages("ggpubr")
library(ggpubr)
ggqqplot(mydata$Difference,
color = "steelblue")
- P-value is higher than L (0.05). We cannot reject H0. We assume that variables are normally distributed (normality assumption is met), therefore we can perform a parametric test - Paired T-Test.
PAIRED T-TEST:
# Performing a Paired T-Test to check whether two means are equal.
t.test(mydata$Italian, mydata$Spanish,
paired = TRUE,
alternative = "two.sided")##
## Paired t-test
##
## data: mydata$Italian and mydata$Spanish
## t = -2.3268, df = 164, p-value = 0.0212
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
## -2.8905626 -0.2367101
## sample estimates:
## mean difference
## -1.563636H0: 𝜇Italian = 𝜇Spanish, or: 𝜇Difference = 0
H1: 𝜇Italian ≠ 𝜇Spanish, or: 𝜇Difference ≠ 0
We reject H0 at p<0.05.
# Checking the effect size, or how large the difference between tested variables is:
#install.packages("effectsize")
library(effectsize)
effectsize::cohens_d(mydata$Difference)## Cohen's d | 95% CI
## --------------------------
## -0.18 | [-0.33, -0.03]# Interpreting the effect size.
interpret_cohens_d(0.18, rules = "sawilowsky2009")## [1] "very small"
## (Rules: sawilowsky2009)The sample is large, so even a small difference is significant.
CONCLUSION: Based on the sample data, we reject the null hypothesis about the equality of average grades for Italian and Spanish exams (p<0.05), and assume a conclusion that the average grades are different. We found a very small effect size, expressed in Cohen’s D statistic by Sawilowsky rules of -0.18. Average grade for Italian exam is smaller than average grade for Spanish exam.
3.1.2. NON-PARAMERTIC TEST - Wilcoxon Signed Rank Test:
The variables are interval, and we use Wilcoxon Signed Rank Test.
HYPOTHESIS:
H0: Location distribution of grades for Italian course is equal to location distribution of grades for Spanish course.
H1: Location distribution of grades for Italian course is not equal to location distribution of grades for Spanish course.
# Performing a Wilcoxon Signed Rank Test:
wilcox.test(mydata$Italian, mydata$Spanish,
paired = TRUE,
correct = FALSE,
exact = FALSE,
alternative = "two.sided")##
## Wilcoxon signed rank test
##
## data: mydata$Italian and mydata$Spanish
## V = 4921.5, p-value = 0.03335
## alternative hypothesis: true location shift is not equal to 0effectsize(wilcox.test(mydata$Italian, mydata$Spanish,
paired = TRUE,
correct = FALSE,
exact = FALSE,
alternative = "two.sided"))## r (rank biserial) | 95% CI
## ----------------------------------
## -0.20 | [-0.36, -0.02]interpret_rank_biserial(0.20)## [1] "medium"
## (Rules: funder2019)Reject H0 at p<0.05.
Location distributions for grades for Italian and Spanish exams are not equal. We found a medium effect size of -0.2 by Funder statistic.
3.1.4. CONCLUSION (the most appropriate test):
The assumptions for parametric test are met, and parametric tests hold more statistical power, so we prefer the results from the Paired T-Test: we reject the null hypothesis about the equality of average grades for Italian and Spanish exams (p<0.05), and assume a conclusion that there is a very small effect sizeof -0.18, meaning that the average grade for Italian exam is smaller than average grade for Spanish exam.
3.2. CORRELATION ANALYSIS
For numeric variables, we use parametric test Pearson correlation, which checks the degree of relationship (interdependence) of variables.
RESEARCH QUESTION: Is there a relationship between the percentage of Attention and grades for Italian exam?
HYPOTHESIS:
H0: There is no relationship between Attendance and grades for Italian exam.
H0: 𝜌Attendance, Italian = 0
H1: There is a relationship between Attendance and grades for Italian exam.
H1: 𝜌Attendance, Italian ≠ 0
# Showing a scatter plot matrix with a correlation between Attendance and grades for Italian exam.
#install.packages("car")
library(car)## Loading required package: carDatascatterplotMatrix(mydata[ , c(5,6)], smooth = FALSE)
- We can see a clearly defined linear positive relationship without significant outliers.
# Showing scatter plots and corresponding correlation coefficients.
#install.packages("GGally")
library(GGally)## Registered S3 method overwritten by 'GGally':
## method from
## +.gg ggplot2ggpairs(mydata[ , c(5,6)])
- The number 0.880 implies a linear relationship between Attendance and grades for Italian exam that is positive and strong (L=0.001).
# Showing a correlation matrix:
#install.packages("Hmisc")
library(Hmisc)##
## Attaching package: 'Hmisc'## The following objects are masked from 'package:base':
##
## format.pval, unitsrcorr(as.matrix(mydata[ , c(5,6)]),
type = "pearson")## Attendance Italian
## Attendance 1.00 0.88
## Italian 0.88 1.00
##
## n= 165
##
##
## P
## Attendance Italian
## Attendance 0
## Italian 0- CONCLUSION: Based on the sample data, we reject the null hypothesis about the absence of a relationship between the percentage of Attention and grades for Italian exam (p=0), and assume a conclusion that there is a linear positive strong correlation of 0.88, based on the scatterplot and Pearson test performed through the correlation matrix. This conclusion implies that the attendance of classes and higher performance on Italian exam is highly correlated.
3.3. ASSOCIATION OF CATEGORICAL VARIABLES
Association between two categorical variables is checked with the analysis of frequencies performed with a parametric test Pearson Chi-Square Test or non-parametric test Fisher’s Exact Probability Test.
RESEARCH QUESTION: Does the chance of passing the unit vary depending on the gender of a student?
HYPOTHESIS:
H0: There is no association between passing the unit and a student’s gender.
H1: There is an association between passing the unit and a student’s gender.
ASSUMPTIONS::
Observations are independent of each other.
All expected frequencies are greater than 5.
# Performing Chi Square Test:
chi_square <- chisq.test(mydata$Unit, mydata$Gender,
correct = TRUE) #Implement Yates' correction to 2x2 table.
chi_square##
## Pearson's Chi-squared test with Yates' continuity correction
##
## data: mydata$Unit and mydata$Gender
## X-squared = 0.76838, df = 1, p-value = 0.3807- P-value is higher than 0.05. We do not reject H0, and assume there is not enough evidence to find an association.
# Using a function that gives the sums of expected or theoretical frequencies:
addmargins(round(chi_square$expected, 2))## mydata$Gender
## mydata$Unit Male Female Sum
## Pass 65.91 59.09 125
## Fail 21.09 18.91 40
## Sum 87.00 78.00 165- All expected frequencies are higher than 5 (assumption met).
- If we assume no association between passing the unit and the gender, we expect 65.91 males to pass the unit, and 21.09 to fail the unit.
- If we assume no association between passing the unit and the gender, we expect 59.09 females to pass the unit, and 18.91 to fail the unit.
- Overall, we expect 125 students to pass, and 40 students to fail the unit.
# Using a function that gives the sums of observed or empirical frequencies:
addmargins(chi_square$observed)## mydata$Gender
## mydata$Unit Male Female Sum
## Pass 63 62 125
## Fail 24 16 40
## Sum 87 78 165- From all observed 87 males, 63 passed the unit, and 24 didn’t.
- From all observed 78 females, 62 passed the unit, and 16 didn’t.
- Overall, from the observed sample of 165 students, 125 students passed the unit, and 40 didn’t.
- The overall observed passing rate is 63/87 or 0.72 for males and 62/78 or 0.79 for females, which implies that the rates are nearly equal.
- COMPARISON OF EMPIRICAL AND EXPECTED: We can conclude that there are only minor differences between expected and observed frequencies.
# Using a function that shows standard residuals.
round(chi_square$residuals, 2)## mydata$Gender
## mydata$Unit Male Female
## Pass -0.36 0.38
## Fail 0.63 -0.67- All standardized residuals observed are lower than 1.96 in absolute terms, and therefore there are no statistically significant differences. Since the residuals’ values are close to zero, the observed values align closely with the expected values, and there is no significant deviation.
# Using a function that gives proportions or the structure of observed frequencies.
addmargins(round(prop.table(chi_square$observed, 1), 3), 2)## mydata$Gender
## mydata$Unit Male Female Sum
## Pass 0.504 0.496 1.000
## Fail 0.600 0.400 1.000- Out of those who passed, 50.4 were males and 49.6 were females.
- Out of those who failed, 60 percent were males, and 40 percent were females.
addmargins(round(prop.table(chi_square$observed, 2), 3), 1)## mydata$Gender
## mydata$Unit Male Female
## Pass 0.724 0.795
## Fail 0.276 0.205
## Sum 1.000 1.000- Out of all males, there are 72.4 percent of males who passed the unit, and 27.6 percent who failed.
- Out of all females, there are 79.5 percent of females who passed the unit, and 20.5 percent who failed.
# Checking the effect size by Cramer's V statistics:
library(effectsize)
effectsize::cramers_v(mydata$Unit, mydata$Gender)## Cramer's V (adj.) | 95% CI
## --------------------------------
## 0.03 | [0.00, 1.00]
##
## - One-sided CIs: upper bound fixed at [1.00].interpret_cramers_v(0.03)## [1] "tiny"
## (Rules: funder2019)- The effect size is tiny (r<0.05) by Funder and Ozer statistic.
# Checking the effect size by the usage of Odds ratio (only for 2x2 table):
oddsratio(mydata$Unit, mydata$Gender)## Odds ratio | 95% CI
## -------------------------
## 0.68 | [0.33, 1.40]- Odds ratio should be larger than 1, so we calculate it manually as 1/0.68, and it’s equal to 1.47.
interpret_oddsratio(1.47)## [1] "small"
## (Rules: cohen1988)The effect size by the Odds ratio is small by Cohen’s D statistic. By Chen statistic, the value of 1.47 is interpreted as “very small”.
CONCLUSION: Based on the sample data, we do not reject the null hypothesis about the absence of association between passing the unit and a gender of a student, and assume a conclusion that there is no association, based on the findings from Pearson Chi-Square Test and Standardazed Residuals testing. We found a tiny or very small effect size, expressed Cramer’s V statistic of 0.03, and in Cohen’s D statistic of 1.47, and the effect is not statistically significant.