Rostyslav_Mykhalchuk_HW1
Homework Assignment 1
- Name: Rostyslav Mykhalchuk
- StudentID: 12321015
- Course number: 5708
Task 1-3: Finding, importing and displaying the data
mydata <- read.table("./loan.csv",
header = TRUE,
sep = ",",
quote = "\"") #I used quote = "\"" because R has not understood apostrophe correctly
head(mydata)## age gender occupation education_level marital_status income credit_score loan_status
## 1 32 Male Engineer Bachelor's Married 85000 720 Approved
## 2 45 Female Teacher Master's Single 62000 680 Approved
## 3 28 Male Student High School Single 25000 590 Denied
## 4 51 Female Manager Bachelor's Married 105000 780 Approved
## 5 36 Male Accountant Bachelor's Married 75000 710 Approved
## 6 24 Female Nurse Associate's Single 48000 640 Denied
Task 4: Explaining my data
Unit of the observation:
- Individual person
Sample size:
- 61 observations
Description of the variables:
age- The age of the individual
- Type of variable: Numerical ratio
- Units: Years
- The age of the individual
gender- The gender of the individual, either ‘Male’ or ‘Female’.
- Type of variable: Categorical nominal
- The gender of the individual, either ‘Male’ or ‘Female’.
occupation- The occupation of the individual
- Type of variable: Categorical nominal
education_level- The highest level of education attained by the individual, such as ‘High School’, ‘Associate’s’, ‘Bachelor’s’, ‘Master’s’, or ‘Doctoral’.
- Type of variable: Categorical ordinal
marital_status- The marital status of the individual, either ‘Single’ or ‘Married’.
- Type of variable: Categorical nominal
income- The annual income of the individual
- Type of variable: Numerical ratio
- Units: Dollars
- The annual income of the individual
credit_score- The credit score of the individual, ranging from 300 to 850.
- Type of variable: Numerical interval
- The credit score of the individual, ranging from 300 to 850.
loan_status- The target variable, indicating whether the loan application was ‘Approved’ or ‘Denied’.
- Type of variable: Categorical nominal
Task 5: Source of data
Source: Mandala, S. K. (n.d.). Simple Loan Classification
Dataset. Kaggle. Retrieved 24 March 2025, from https://www.kaggle.com/datasets/sujithmandala/simple-loan-classification-dataset
Task 6: Data manipulation
6.1 Renaming variables and dropping all N/As
# install.packages("dplyr")
# install.packages("tidyr")
library(dplyr)
library(tidyr)
mydata <- mydata %>%
rename(Income_USD = income, Age = age, Gender = gender, Occupation = occupation, Education_level = education_level, Marital_status = marital_status, Credit_score = credit_score, Loan_status = loan_status) %>%
drop_na()6.2 Creating a new variable, showing an income in euros, and other data manipulation
# Here I will indicate an exchange rate and show an income in euros
USD_EUR <- 0.92
mydata$Income_EUR <- mydata$Income_USD * USD_EUR
# Adding a new variable called "ID"
mydata$ID <- c(1:61)
mydata <- mydata[, c(10, 1:9)] # Moves column 10 to the front
# Changing the "Male" and "Female" into "M" and "F" (Actually, I have a pretty clean data and don't need this data manipulation, so it is done just in order to show my knowledge)
mydata$Gender <- factor(mydata$Gender,
levels = c("Male", "Female"),
labels = c("M", "F"))
head(mydata)## ID Age Gender Occupation Education_level Marital_status Income_USD Credit_score Loan_status Income_EUR
## 1 1 32 M Engineer Bachelor's Married 85000 720 Approved 78200
## 2 2 45 F Teacher Master's Single 62000 680 Approved 57040
## 3 3 28 M Student High School Single 25000 590 Denied 23000
## 4 4 51 F Manager Bachelor's Married 105000 780 Approved 96600
## 5 5 36 M Accountant Bachelor's Married 75000 710 Approved 69000
## 6 6 24 F Nurse Associate's Single 48000 640 Denied 441606.3 Creating a new data frame, which includes only male
mydata_M <- mydata[mydata$Gender == "M", ]
head(mydata_M)## ID Age Gender Occupation Education_level Marital_status Income_USD Credit_score Loan_status Income_EUR
## 1 1 32 M Engineer Bachelor's Married 85000 720 Approved 78200
## 3 3 28 M Student High School Single 25000 590 Denied 23000
## 5 5 36 M Accountant Bachelor's Married 75000 710 Approved 69000
## 7 7 42 M Lawyer Doctoral Married 120000 790 Approved 110400
## 9 9 37 M IT Master's Married 92000 750 Approved 84640
## 11 11 55 M Consultant Master's Married 110000 770 Approved 1012006.4 Creating a new data frame, which includes only individuals aged younger than 30
mydata_under30 <- mydata[mydata$Age < 30, ]
head(mydata_under30)## ID Age Gender Occupation Education_level Marital_status Income_USD Credit_score Loan_status Income_EUR
## 3 3 28 M Student High School Single 25000 590 Denied 23000
## 6 6 24 F Nurse Associate's Single 48000 640 Denied 44160
## 8 8 29 F Artist Bachelor's Single 38000 620 Denied 34960
## 13 13 26 M Salesman High School Single 42000 610 Denied 38640
## 16 16 27 F Designer Bachelor's Single 52000 650 Denied 47840
## 22 22 25 F Receptionist High School Single 35000 580 Denied 32200Task 7: Descriptive statistics
summary(mydata[ , -c(1, 3, 4, 5, 6, 9)]) #Dropping several non-numerical variables in order to receive observable statistics as a summary## Age Income_USD Credit_score Income_EUR
## Min. :24.00 Min. : 25000 Min. :560.0 Min. : 23000
## 1st Qu.:30.00 1st Qu.: 52000 1st Qu.:650.0 1st Qu.: 47840
## Median :36.00 Median : 78000 Median :720.0 Median : 71760
## Mean :37.08 Mean : 78984 Mean :709.8 Mean : 72665
## 3rd Qu.:43.00 3rd Qu.: 98000 3rd Qu.:770.0 3rd Qu.: 90160
## Max. :55.00 Max. :180000 Max. :830.0 Max. :165600Based on the given data frame:
- The median age of the sample is 36 years, meaning that half
of the individuals are 36 years old or younger, and the other
half are older than 36.
- The maximum reported annual income in the sample is $180,000, equivalent to approximately €165,600 (possible outlier). An average income, however, equals to $78,984 (~€72,665)
- The first quartile of credit score distribution is 650, indicating that 25% of individuals in the sample have credit rating that is less than or equal to 650.
head(mydata[order(mydata$Age), -c(1, 3, 4, 5, 6, 9, 10)])## Age Income_USD Credit_score
## 6 24 48000 640
## 22 25 35000 580
## 58 25 32000 570
## 13 26 42000 610
## 38 26 28000 560
## 16 27 52000 650avg_income <- mean(mydata$Income_USD)
youngest_income <- mydata[6, "Income_USD"]
difference_youngest_and_avg <- (youngest_income/avg_income)-1
difference_youngest_and_avg <- round(difference_youngest_and_avg, 2)
print(difference_youngest_and_avg)## [1] -0.39- The youngest individual earns $48,000 annually, which is approximately 39% lower than the sample’s average income.
Descriptive statistics for the selected variables:
# install.packages("psych")
library(psych)
describeBy(mydata$Income_USD, group = mydata$Gender)##
## Descriptive statistics by group
## group: M
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 31 80677.42 26507.09 85000 81640 22239 25000 130000 105000 -0.34 -0.68 4760.81
## ------------------------------------------------------------------------------------------------------------------------------------------------------
## group: F
## vars n mean sd median trimmed mad min max range skew kurtosis se
## X1 1 30 77233.33 40331.97 66500 71958.33 34841.1 28000 180000 152000 0.97 0.19 7363.58Income comparison between male and female based on the given data shows that:
- On average, males in the sample earn slightly more
($80,677.42) than females ($77,233.33)
- The standard deviation (SD), which measures income variability, is higher among females ($40,331.97) than males ($26,507.09). This suggests that female incomes are more widely spread out in the sample
- The skewness values show that female income distribution is slightly positively skewed (0.97), meaning more females earn below the average, while male income is slightly negatively skewed (−0.34), indicating a few higher earners pulling the average up
library(modeest)## Registered S3 method overwritten by 'rmutil':
## method from
## plot.residuals psychmlv(mydata$Occupation) # Most likely value## [1] "Engineer"- Most frequent occupation: Engineer
Task 8: Presenting graphs of the distribution
8.1 Histogram
# Showing how annual incomes are distributed
hist(mydata$Income_EUR,
main = "Distribution of annual income in USD",
xlab = "Income",
ylab = "Frequency",
breaks = seq(from = 0, to = 200000, by = 10000),
col = "forestgreen")
# Showing how ages are spread across given sample
hist(mydata$Age,
main = "Distribution of ages",
xlab = "Age",
ylab = "Frequency",
breaks = seq(from = 10, to = 70, by = 5),
col = "bisque")
- The distribution of annual income is positively skewed and deviates from a normal distribution due to a few high-income outliers
- Assumptions of normality should be treated with caution
- In contrast, the age distribution looks more symmetrical and is similar to a normal distribution, with the majority of individuals in their 30s
8.2 Boxplot
# Showing how annual incomes are distributed
boxplot(mydata$Income_USD,
xlab = "Annual Income (in USD)")
- There are two visible outliers on the higher end, suggesting that a few individuals earn significantly more than the rest
- The interquartile range spans from about $50,000 to $100,000, showing where most incomes fall
- Overall, this confirms the income distribution is not symmetrical and is affected by outliers
Thus, there are 2 outliers that should be removed.
head(mydata[order(-mydata$Income_USD), ])## ID Age Gender Occupation Education_level Marital_status Income_USD Credit_score Loan_status Income_EUR
## 10 10 48 F Doctor Doctoral Married 180000 820 Approved 165600
## 48 48 48 F Doctor Doctoral Married 175000 830 Approved 161000
## 34 34 47 F Dentist Doctoral Married 140000 810 Approved 128800
## 45 45 42 M Lawyer Doctoral Married 130000 800 Approved 119600
## 18 18 41 F Pharmacist Doctoral Married 125000 800 Approved 115000
## 7 7 42 M Lawyer Doctoral Married 120000 790 Approved 110400mydata_new <- mydata[-c(10, 48), ]
head(mydata_new[order(-mydata_new$Income_USD), ])## ID Age Gender Occupation Education_level Marital_status Income_USD Credit_score Loan_status Income_EUR
## 34 34 47 F Dentist Doctoral Married 140000 810 Approved 128800
## 45 45 42 M Lawyer Doctoral Married 130000 800 Approved 119600
## 18 18 41 F Pharmacist Doctoral Married 125000 800 Approved 115000
## 7 7 42 M Lawyer Doctoral Married 120000 790 Approved 110400
## 56 56 50 F Professor Doctoral Married 120000 810 Approved 110400
## 23 23 46 M Banker Master's Married 115000 780 Approved 105800boxplot(mydata_new$Income_USD,
xlab = "Income (in USD)")
- The updated boxplot shows a more compact and symmetric distribution
of income after removing the two extreme outliers
boxplot(mydata$Age,
xlab = "Age (in years)")
- The median age is centered within the box, suggesting a fairly symmetric distribution
- There are no outliers, and the whiskers extend evenly, supporting
the idea of a roughly normal distribution
# install.packages("ggplot2")
library(ggplot2)##
## Attaching package: 'ggplot2'## The following objects are masked from 'package:psych':
##
## %+%, alphaggplot(mydata, aes(x=Loan_status, y=Income_USD, fill=Gender)) +
geom_boxplot() +
scale_fill_brewer(palette="Spectral") +
xlab("Loan status") +
labs(fill="Gender")
- Individuals with approved loans tend to have higher incomes than those with denied loans
- Among approved applicants, females generally have slightly higher incomes than males
- For denied loans, income levels are lower overall, with less variation, and males appear to earn slightly less than females on average (which could have been influenced by outliers)
8.3 Scatter plot:
# Used when we have 2 numerical variables and seek the relationship between them
# install.packages("car")
library(car)## Loading required package: carData##
## Attaching package: 'car'## The following object is masked from 'package:psych':
##
## logit## The following object is masked from 'package:dplyr':
##
## recodescatterplot(y = mydata$Age,
x = mydata$Income_USD,
ylab = "Age (in years)",
xlab = "Annual Income (in USD)",
smooth = FALSE)
- Scatter plot was used in order to show the relationship between age and income in the sample
- This graph shows a clear positive correlation between age and annual income
- Based on the given data, the older the individual, the higher annual
income he or she earns
scatterplot(Age ~ Income_USD | mydata$Gender,
ylab = "Age (in years)",
xlab = "Annual Income (in USD)",
smooth = FALSE,
data = mydata)
- Income and age are positively correlated for both genders
- However, for the same annual income, females in the sample are on
average older than males, as shown by the higher placement of the female
regression line
scatterplotMatrix(mydata[ , -c(1, 3, 4, 5, 6, 9, 10)],
smooth = FALSE)
- The scatter plot matrix was used in order to show the relationship between age, annual income and credit score
- On the main diagonal one can see the distributions of those 3 variables
- The scatter plot matrix displays positive linear relationships between age, income, and credit score
- The older the individual, the higher their income and the better their credit score tends to be
- Income is right-skewed, with a few high earners standing out, age is right-skewed, and credit score is left-skewed
- Stronger correlations are observed between income and credit score, followed by age and income