Correlation & Regression — SL 4.2 Correlation and linear regression; interpreting slope/intercept and residuals (technology-supported) - IB Mathematics Applications and Interpretation

SECTION 1

Historical Context & Motivation

Humans have always looked for patterns in data, but for most of history we relied on intuition rather than mathematics. Is there a connection between how many hours you study and the grade you earn? Does temperature affect ice-cream sales? These questions feel simple, yet answering them rigorously required centuries of mathematical development. The story of correlation and regression begins in the 1800s, when scientists first tried to quantify how two measurements move together.

1805 • Legendre's Method of Least Squares

French mathematician Adrien-Marie Legendre published the method of least squares, a technique for fitting the best straight line through a set of data points by minimizing the sum of squared differences.

1886 • Galton Coins 'Regression'

Sir Francis Galton studied the heights of parents and children, discovering that children's heights tended to regress toward the mean. He called this phenomenon 'regression,' giving the technique its modern name.

1896 • Pearson's Correlation Coefficient

Karl Pearson formalized the Pearson product-moment correlation coefficient (r), providing a single number between −1 and 1 that measures the strength and direction of a linear relationship.

1970s–Today • Technology-Supported Analysis

With graphing calculators and software like GDCs, spreadsheets, and Desmos, students and researchers can compute regression equations and correlation coefficients in seconds, shifting the focus from calculation to interpretation.

The central question this topic addresses is deceptively straightforward: When two variables seem related, how do we measure the strength of that relationship and use it to make predictions? In IB Mathematics: Applications and Interpretation SL 4.2, you will learn how to answer this question using scatter plots, the correlation coefficient r, the regression line ŷ = ax + b, and residual analysis — all supported by technology.

SECTION 2

Core Principles & Definitions

Before diving into calculations, it is essential to build a clear vocabulary. The ideas below form the foundation of everything you will do with bivariate data — data sets that pair up two variables for each individual or observation.

Bivariate Data & Scatter Plots

Bivariate data consists of paired observations (x, y). When plotted on coordinate axes, each pair becomes a dot, and the collection of dots is called a scatter plot. The independent (explanatory) variable goes on the x-axis and the dependent (response) variable on the y-axis.

Correlation Coefficient (r)

The Pearson correlation coefficient (r) is a number from −1 to 1 that describes both the direction and strength of a linear relationship. Values close to 1 or −1 indicate a strong linear pattern, while values near 0 suggest little or no linear association.

Line of Best Fit (Regression Line)

The regression line ŷ = ax + b is the unique straight line that minimizes the sum of the squared residuals. It passes through the point (x̄, ȳ) and is calculated using the least-squares method.

Residuals

A residual is the vertical distance between an observed y-value and the y-value predicted by the regression line: residual = y − ŷ. Positive residuals sit above the line; negative residuals sit below it.

Interpolation vs. Extrapolation

Interpolation is predicting within the range of the given data and is generally reliable. Extrapolation is predicting outside that range and can be unreliable because the linear trend may not continue.

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KEY TAKEAWAY

Think of correlation like a friendship rating. If two variables are strongly positively correlated (r close to 1), they are 'best friends' — when one goes up, the other goes up too. If they are strongly negatively correlated (r close to −1), they are 'opposites' — when one goes up, the other goes down. An r near 0 means they are strangers who act independently of each other. The regression line is your best prediction tool: it draws the tightest-fitting line through the data so you can estimate one variable from the other.

SECTION 3

Visual Explanation — The Scatter Plot & Regression Line

A scatter plot is the starting point for every correlation and regression analysis. The diagram below shows a bivariate data set with a regression line drawn through it. Notice how the data points cluster around the line, with some sitting above it (positive residuals) and others below (negative residuals). The vertical dashed segments illustrate the residuals themselves.

Each cyan dot represents an (hours, score) pair. The violet line is the least-squares regression line ŷ = 6.1x + 26. The pink and green dashed segments show residuals — the vertical distances between observed points and the line. Points above the line have positive residuals; points below have negative residuals.

In the diagram, notice that the data points generally trend upward from left to right. This indicates a positive correlation: as hours of study increase, test scores tend to increase as well. The regression line captures this upward trend and provides a formula you can use for predictions. The residuals show you where the model is imperfect — no real-world data falls perfectly on a line. Examining whether residuals are randomly scattered (good) or show a curved pattern (bad) tells you whether a linear model is appropriate.

SECTION 4

Mathematical Framework

In IB Applications and Interpretation SL, you are expected to use your GDC (graphing display calculator) or approved technology to compute the values of r, a, and b. However, understanding the formulas helps you know what your calculator is doing and why the outputs make sense.

REGRESSION LINE (LEAST-SQUARES)

ŷ = ax + b

ŷ = predicted value of the dependent variable; a = slope (gradient) of the line; b = y-intercept (the value of ŷ when x = 0); x = value of the independent variable.

SLOPE OF REGRESSION LINE

a = Sₓᵧ / Sₓₓ

Sₓᵧ = Σ(xᵢ − x̄)(yᵢ − ȳ) is the sum of the products of deviations; Sₓₓ = Σ(xᵢ − x̄)² is the sum of squared deviations of x. Your GDC computes these automatically.

PEARSON CORRELATION COEFFICIENT

r = Sₓᵧ / √(Sₓₓ × Sᵧᵧ)

Sᵧᵧ = Σ(yᵢ − ȳ)². The value of r always satisfies −1 ≤ r ≤ 1. A positive r means a positive linear trend; a negative r means a negative linear trend.

RESIDUAL

eᵢ = yᵢ − ŷᵢ

yᵢ = observed (actual) value; ŷᵢ = value predicted by the regression line at the same x. A residual of zero means the point sits exactly on the line.

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IB Exam Tip

On the IB exam, you will not be asked to compute r or the regression equation by hand. You will use your GDC. The exam focuses on interpreting the slope, intercept, correlation coefficient, and residuals in context. Always write your interpretation in the context of the problem (e.g., 'For every additional hour of study, the predicted test score increases by 6.1 points').

SECTION 5

Interpreting the Correlation Coefficient

One of the most common exam tasks is to describe the correlation between two variables. You need to state both the direction (positive or negative) and the strength (weak, moderate, or strong) of the correlation. The spectrum bar below gives you a visual reference for how to classify values of r.

Strength of Linear Correlation (r)

Perfect negative Perfect positive

−1 (0%)−0.7 (15%)−0.4 (25%)0 (50%)0.4 (65%)0.7 (75%)1 (100%)

Four scatter plots illustrate how different values of r look visually. As |r| approaches 1, data points cluster more tightly around the regression line. When r ≈ 0, data points appear as a shapeless cloud. The information boxes below summarize how to interpret the slope and intercept in context.

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Correlation ≠ Causation

A strong correlation does not prove that one variable causes changes in the other. For example, ice-cream sales and drowning rates are strongly positively correlated — but buying ice cream does not cause drowning. Both increase because of a lurking variable: hot weather. Always be cautious about claiming cause and effect based on correlation alone.

SECTION 6

Worked Example

A teacher records the number of absences (x) and the final exam score (y) for eight students. The data are shown below. Let's walk through a complete analysis using technology.

Student	Absences (x)	Exam Score (y)
A	2	88
B	5	74
C	3	82
D	8	60
E	1	92
F	6	68
G	4	78
H	7	65

Absences vs. Exam Score data

Finding and Interpreting the Regression Line

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Step 1 — Enter Data into Your GDC

Enter the absences (x) into List 1 and the exam scores (y) into List 2 on your GDC. Make sure both lists have the same number of entries (8 in this case).
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Step 2 — Run Linear Regression

Use the statistics menu to perform linear regression (LinReg or equivalent). The GDC outputs: a = −4.37 (3 s.f.), b = 96.5 (3 s.f.), and r = −0.987 (3 s.f.).

ŷ = −4.37x + 96.5, r = −0.987
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Step 3 — Interpret the Slope

The slope a = −4.37 means that for each additional absence, the predicted exam score decreases by approximately 4.37 points. The negative slope reflects the negative correlation between absences and scores.
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Step 4 — Interpret the y-Intercept

The y-intercept b = 96.5 means that a student with zero absences would be predicted to score about 96.5 on the exam. This is within a reasonable range, so the intercept has a sensible real-world interpretation here.
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Step 5 — Interpret the Correlation Coefficient

Since r = −0.987, there is a very strong negative linear correlation between absences and exam scores. The value is very close to −1, which tells us the data points lie almost exactly along the regression line.
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Step 6 — Make a Prediction and Calculate the Residual

Predict the score for a student with 3 absences: ŷ = −4.37(3) + 96.5 = −13.11 + 96.5 = 83.4 (3 s.f.). The actual score for Student C (x = 3) is 82. The residual is: e = 82 − 83.4 = −1.4. The negative residual tells us the student scored slightly below the prediction.

Predicted score = 83.4; Residual = −1.4
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Step 7 — Assess Reliability of a Prediction

If someone asks you to predict the score for a student with 15 absences, be cautious! The data only goes up to x = 8. Predicting at x = 15 would be extrapolation and the result (ŷ = −4.37 × 15 + 96.5 = 31) may be unreliable because the linear trend might not hold that far.

SECTION 7

Strengths & Limitations of Linear Regression

Linear regression is one of the most widely used statistical tools in the world, but it is not perfect. Understanding when it works well and when it fails is just as important as knowing how to compute the regression line.

Strengths	Limitations
Simple to compute and interpret — a single equation summarizes the relationship.	Only models linear relationships; curved patterns will be poorly described.
The correlation coefficient r gives a quick measure of strength and direction.	Sensitive to outliers — a single extreme point can dramatically shift the line.
Enables predictions (interpolation) within the data range with reasonable accuracy.	Extrapolation beyond the data range is unreliable and can produce nonsensical predictions.
Residual analysis helps diagnose whether the model is appropriate.	Correlation does not imply causation — lurking variables may create misleading associations.
Technology makes the computation nearly instantaneous on a GDC or spreadsheet.	Requires bivariate quantitative data — cannot be applied to categorical variables directly.

Strengths vs. Limitations of Linear Regression

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KEY TAKEAWAY

Think of the regression line as a GPS route. It gives you the best estimated path from A to B based on the roads (data) it knows about. If you stay on roads it has mapped (interpolation), the directions are reliable. But if you drive off into unmapped territory (extrapolation), the GPS might send you into a lake. Also, just because two cities are connected by a road doesn't mean one city caused the other to be built — correlation is not causation.

SECTION 8

Connection to Advanced Theory

The linear regression you learn in SL 4.2 is the foundation for more powerful techniques you may encounter in HL Mathematics or university statistics courses. The table below shows how the concepts you have learned relate to their advanced counterparts.

SL 4.2 Concept	Advanced Extension
Simple linear regression (one x variable)	Multiple regression: ŷ = a₁x₁ + a₂x₂ + … + b, using several predictor variables simultaneously
Pearson's r for linear correlation	Coefficient of determination r² (the proportion of variance explained by the model); Spearman's rank correlation for non-linear monotonic relationships
Residuals (y − ŷ) checked visually	Formal residual analysis including tests for normality, homoscedasticity (constant variance), and independence
Line of best fit (linear model)	Non-linear regression models: quadratic, exponential, logarithmic, logistic, and polynomial curves fitted to data

SL 4.2 vs. Advanced Statistics

The key idea connecting SL to higher-level work is that r² (the coefficient of determination) tells you what fraction of the variability in y is 'explained' by the model. For instance, if r = −0.987, then r² ≈ 0.974, meaning about 97.4% of the variation in exam scores can be explained by the number of absences. This powerful interpretation extends directly into university-level statistics and data science.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A researcher finds that the correlation between daily temperature and the number of hot drinks sold at a café is r = −0.82. Describe the strength and direction of this correlation in context, and explain whether this means that high temperatures cause fewer hot drinks to be sold.

Show answer

The value r = −0.82 indicates a strong negative linear correlation between daily temperature and hot drink sales. This means that as daily temperature increases, the number of hot drinks sold tends to decrease, and the relationship is quite consistent. However, correlation does not prove causation. While it is plausible that higher temperatures directly reduce demand for hot drinks, there could be other lurking variables (such as seasonal menu changes or tourism patterns) contributing to the relationship. To establish causation, a controlled experiment would be needed.

Question 2: BASIC CALCULATION

The regression line for the relationship between hours of exercise per week (x) and resting heart rate in beats per minute (y) is ŷ = −1.8x + 78. (a) Interpret the slope in context. (b) Interpret the y-intercept in context. (c) Predict the resting heart rate for someone who exercises 5 hours per week.

Show answer

(a) The slope of −1.8 means that for each additional hour of exercise per week, the predicted resting heart rate decreases by 1.8 beats per minute. (b) The y-intercept of 78 means that a person who does 0 hours of exercise per week is predicted to have a resting heart rate of 78 bpm; this is a reasonable baseline value. (c) Substituting x = 5: ŷ = −1.8(5) + 78 = −9 + 78 = 69 bpm. A person who exercises 5 hours per week is predicted to have a resting heart rate of 69 beats per minute.

Question 3: INTERMEDIATE

Using the regression equation ŷ = −1.8x + 78 from Problem 2, a person who exercises 4 hours per week actually has a resting heart rate of 74 bpm. (a) Calculate the residual. (b) Explain what the sign of the residual tells you. (c) If the data set included exercise values from x = 1 to x = 10, would it be appropriate to predict the heart rate for someone exercising 20 hours per week? Justify your answer.

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(a) First find the predicted value: ŷ = −1.8(4) + 78 = −7.2 + 78 = 70.8 bpm. The residual is e = y − ŷ = 74 − 70.8 = 3.2 bpm. (b) The positive residual means the actual heart rate (74) is higher than the model predicted (70.8). In context, this person's heart rate is about 3.2 bpm above what the regression model would expect for someone who exercises 4 hours per week. (c) No, predicting at x = 20 would be extrapolation since the data only ranges from x = 1 to x = 10. The linear relationship may not continue beyond the observed data, and predicting at x = 20 could yield an unreliable or even negative heart rate, which is physically impossible.

Question 4: APPLIED

A car dealership records the age (in years) and selling price (in thousands of dollars) of 10 used cars of the same model. Using a GDC, the regression line is found to be ŷ = −2.15x + 28.3 with r = −0.946. (a) Describe the correlation. (b) Interpret the slope and y-intercept in context. (c) Estimate the price of a 6-year-old car. (d) A 4-year-old car sold for $22,000. Calculate the residual and explain what it suggests about this car.

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(a) The correlation is strong and negative (r = −0.946), meaning that as the age of a car increases, its selling price tends to decrease, and the relationship is nearly linear. (b) The slope of −2.15 means each additional year of age reduces the predicted selling price by approximately $2,150. The y-intercept of 28.3 suggests a brand-new car (age 0) of this model would have a predicted price of $28,300 — a reasonable starting value. (c) For x = 6: ŷ = −2.15(6) + 28.3 = −12.9 + 28.3 = 15.4, so the predicted price is approximately $15,400. (d) Predicted price at x = 4: ŷ = −2.15(4) + 28.3 = −8.6 + 28.3 = 19.7, i.e., $19,700. Residual = 22 − 19.7 = 2.3, or $2,300. The positive residual indicates this car sold for $2,300 more than the model predicted, perhaps because it had low mileage or was in exceptional condition.

Question 5: CRITICAL THINKING

Two students analyse the same data set. Student A plots 'advertising spend' on the x-axis and 'revenue' on the y-axis, obtaining ŷ = 3.2x + 50 with r = 0.91. Student B plots 'revenue' on the x-axis and 'advertising spend' on the y-axis, obtaining a different regression equation. (a) Will Student B get the same value of r? Explain. (b) Will Student B get the same regression line? Explain why or why not. (c) Discuss which student's model is more appropriate for predicting revenue from advertising spend, and what assumption must be checked about the residuals before trusting either model.

Show answer

(a) Yes, Student B will get the same value of r = 0.91. The Pearson correlation coefficient measures the mutual linear association between two variables and does not depend on which variable is treated as x or y. (b) No, the regression lines will be different. The regression of y on x minimizes vertical distances to the line, while the regression of x on y minimizes horizontal distances. These are two distinct lines (unless r = ±1, in which case they coincide). The slope and intercept will change because the minimization criterion is different. (c) Student A's model is more appropriate for predicting revenue from advertising spend, since revenue (the outcome to be predicted) is the dependent variable on the y-axis. Before trusting either model, the students should check that the residuals are randomly scattered with no obvious pattern (such as a curve or a fan shape). A pattern in the residuals would suggest a linear model is not appropriate and a different type of model might be needed.

SUMMARY

Lesson Summary

Summary

In SL 4.2, you learned to analyse bivariate data using scatter plots to visualize relationships. The Pearson correlation coefficient (r) quantifies the direction and strength of a linear association on a scale from −1 to 1. Using technology (your GDC), you can find the least-squares regression line ŷ = ax + b, where the slope (a) tells you how much ŷ changes for each one-unit increase in x, and the y-intercept (b) gives the predicted value when x = 0.

Residuals (y − ŷ) measure how far each data point falls from the regression line and should be randomly scattered if a linear model is appropriate. Always interpret your results in context. Use interpolation for reliable predictions within the data range, and be cautious with extrapolation outside it. Finally, remember that correlation does not imply causation — lurking variables may create associations that do not reflect a direct cause-and-effect relationship.