Correlation and Regression (AA SL)
When you have two sets of paired data, correlation tells you how strongly they move together and regression gives you a line to predict one from the other. This page covers reading and interpreting Pearson's \(r\), finding a regression line, and the traps examiners set around causation and extrapolation. It's part of the broader Descriptive Statistics & Correlation topic.
18 questions on this sub-topic.
Correlation and regression, in brief
Covered under IB syllabus reference SL4.4: linear correlation of bivariate data, Pearson's product-moment correlation coefficient \(r\), scatter diagrams and lines of best fit, and the equation of the regression line of \(y\) on \(x\), found using technology.
Pearson's \(r\)
Tells you the strength and direction of linear correlation between two variables, from \(-1\) (perfect negative) to \(+1\) (perfect positive).
Not in the formula booklet - you find it directly from your GDC once the data is entered, along with the regression line itself.
Correlation and causation
A strong correlation coefficient shows two variables move together, but it never proves that one causes the other - a third factor could be driving both.
Using a regression line far outside the data range (extrapolation) makes any estimate unreliable, since the linear trend isn't guaranteed to continue.
Need the full syllabus wording and formula-booklet reference table? See Descriptive Statistics & Correlation.
Worked examples
Two quantities have Pearson's product-moment correlation coefficient \(r=-0.85.\)
(a) Describe the linear relationship.
(b) State what \(r=0\) would indicate.
Worked solution
(a) Strong A1
negative linear correlation. A1
(b) No linear correlation between the variables. A1
The following data shows the study hours \(x\) and test score \(y\) for 8 students.
\(\bar{x} = 4.5,\quad \bar{y} = 62,\quad S_{xx} = 18,\quad S_{yy} = 720,\quad S_{xy} = 90\)
(a) Find the regression line of \(x\) on \(y\) in the form \(x = ay + b\).
(b) Use the line to estimate the number of hours studied if a student scored 75.
Worked solution
(a) \(a = \dfrac{S_{xy}}{S_{yy}} = \dfrac{90}{720} = 0.125\) M1
\(b = \bar{x} - a\bar{y} = 4.5 - 0.125(62) = 4.5 - 7.75 = -3.25\) M1
\(x = 0.125y - 3.25\) A1
(b) \(x = 0.125(75) - 3.25 = 9.375 - 3.25 = 6.125 \approx 6.13\) hours A1
Common mistakes
- Claiming correlation proves causation. A strong \(r\) value shows two variables move together; it never establishes that one causes the other, since a third factor could explain both.
- Extrapolating a regression line beyond the data range. Using \(y=2.4x+1.5\) to estimate \(y\) at \(x=50\) when the data only covers \(x=1\) to \(9\) produces an estimate with no guarantee of validity - the linear pattern might not continue.
- Mixing up \(y\) on \(x\) with \(x\) on \(y\). The two regression lines are not the same line rearranged - if a question asks you to predict \(x\) from a given \(y\), you need the regression line of \(x\) on \(y\), not the line of \(y\) on \(x\) solved for \(x\).
Ready to practise properly?
18 correlation and regression questions, marked instantly like the real exam.
Quick answers
What does Pearson's correlation coefficient r measure?
\(r\) measures the strength and direction of the linear relationship between two variables, ranging from \(-1\) (perfect negative) to \(+1\) (perfect positive). A value near \(0\) means little or no linear correlation.
Why is it unsafe to use a regression line to predict values far outside the data range?
This is called extrapolation. The regression line is only fitted to the data you have, so there's no guarantee the same linear trend continues beyond it - using it far outside the given range can produce unreliable estimates. For GDC steps to fit the line itself, see the parent topic's GDC guidance.