SQL Analysis

Study Abroad Affordability Analysis

Identifying the most affordable countries and cities for international students

Problem Statement

Which countries and cities offer the most affordable overall study-abroad options when you consider both tuition and cost of living, and how do they compare?

Dataset

Source: Our World in Data - International Education Costs dataset

Records: 907 programs across 70+ countries

SQL Commands Used

SELECT, FROM

Basic data retrieval

GROUP BY

Aggregating by country, city, and degree level

AVG()

Calculating average tuition, rent per location

COUNT()

Counting programs per country/level

ORDER BY

Sorting from cheapest to most expensive

Calculated columns

Total cost formula (tuition + living + fees)

Analysis

International students can save $40,000-$80,000 over the duration of their degree by choosing European or Latin American destinations versus USA or Australia, making higher education accessible to broader demographics.

SQL Queries

Which country has the lowest tuition fee

SELECT country, AVG(tuition_USD) AS avg_tuition
FROM international_education_costs
GROUP BY country
ORDER BY avg_tuition;

Which country has the most programs

SELECT country, COUNT(program) AS number_of_programs
FROM international_education_costs
GROUP BY country
ORDER BY number_of_programs DESC;

Which city has the cheapest rent

SELECT country, city, AVG(rent_USD) AS avg_rent
FROM international_education_costs
GROUP BY city, country
ORDER BY avg_rent;

How many programs by degree level

SELECT level, COUNT(program) AS number_of_programs
FROM international_education_costs
GROUP BY level
ORDER BY number_of_programs;

Total Annual Cost

SELECT country, city, university, program, level,
  (tuition_USD + (rent_USD * 12 * duration_years) +
   visafee_USD + (insurance_USD * duration_years)) AS total_cost
FROM international_education_costs
ORDER BY total_cost

Tools Used

SQLPostgreSQLData AnalysisAggregation Functions