Pearson correlation and coefficient of determination | Year 12

01 / LEARN

Understand the idea.

Pearson correlation measures linear association

Pearson's correlation satisfies $- 1 \leq r \leq 1$ when both numerical variables vary. The sign describes linear direction and $∣ r ∣$ describes linear strength. Values near zero rule out a strong straight-line tendency, but do not rule out a curved association.

The centred sums compare deviations from each variable's own mean: $S_{x x} = \sum {(x_{i} - \bar{x})}^{2}$ , $S_{y y} = \sum {(y_{i} - \bar{y})}^{2}$ and $S_{x y} = \sum (x_{i} - \bar{x}) (y_{i} - \bar{y})$ . Keep corresponding observations paired when calculating or entering them into technology.

r = \frac{S_{x y}}{\sqrt{S_{x x} S_{y y}}}

, where the sums use deviations from the respective means.

Interpret r-squared in the fitted linear model

For ordinary least-squares regression with an intercept, $r^{2}$ is the coefficient of determination. The percentage $100 r^{2} %$ describes the observed response variation accounted for by that fitted linear relationship. It is a measure of sample fit, not a guarantee for new predictions.

Squaring removes the sign, so $r^{2}$ cannot tell you whether the association rises or falls. It also does not measure causation, a percentage of the response amount, or the proportion of individual predictions that are correct.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Calculate from centred sums

A data set has Sxx=20, Syy=45 and Sxy=-24. Calculate r and r².

Worked solution

1
Find the denominator from the two sums of squares.
$\sqrt{S_{x x} S_{y y}} = \sqrt{20 \times 45} = 30$
Both sums are positive, so the correlation denominator is nonzero.
2
Calculate the signed correlation.
$r = \frac{- 24}{30} = - 0.8$
The negative cross-product sum gives a negative linear association.
3
Square the correlation.
$r^{2} = {(- 0.8)}^{2} = 0.64$
The coefficient of determination is nonnegative even when the correlation is negative.
4
Interpret the result in the fitted model.
$100 r^{2} % = 64 %$
For a least-squares line with an intercept, 64% of the observed response variation is accounted for by the fitted linear relationship.

Answer

r=-0.8 and r²=0.64; negative linear association.

EXAMPLE 02

Use paired measurements

Calculate the correlation for the supplied data.

Paired measurements
x	y
1	3
2	4
3	4
4	6
5	8

Worked solution

1
Calculate the two means.
$\begin{aligned} \bar{x} & = \frac{1 + 2 + 3 + 4 + 5}{5} = 3 \\ \bar{y} & = \frac{3 + 4 + 4 + 6 + 8}{5} = 5 \end{aligned}$
Each mean uses all five observations of its own variable.
2
List the paired deviations.
$\begin{aligned} x_{i} - \bar{x} : & - 2, - 1, 0, 1, 2 \\ y_{i} - \bar{y} : & - 2, - 1, - 1, 1, 3 \end{aligned}$
Keeping the order ensures that each cross-product uses measurements from the same row.
3
Form the centred sums.
$\begin{aligned} S_{x x} & = 4 + 1 + 0 + 1 + 4 = 10 \\ S_{y y} & = 4 + 1 + 1 + 1 + 9 = 16 \\ S_{x y} & = 4 + 1 + 0 + 1 + 6 = 12 \end{aligned}$
Square deviations for the two variation sums and multiply paired deviations for the cross-product sum.
4
Calculate correlation and its square.
$\begin{aligned} r & = \frac{12}{\sqrt{10 \times 16}} \approx 0.9487 \\ r^{2} & = \frac{12^{2}}{10 \times 16} = 0.9 \end{aligned}$
The positive value indicates a strong positive linear association. Squaring the unrounded value avoids rounding error in the 90% determination coefficient.

Answer

Positive linear correlation; r≈0.9487, r²≈0.9.

03 / YOUR TURN

Try it yourself.

Check your understanding

For r=-0.6, find the coefficient of determination.

Show answer & reasoning

0.36, or 36%.

Square the signed correlation: $r^{2} = {(- 0.6)}^{2} = 0.36$ . Thus the coefficient of determination is $0.36$ , or $36 %$ ; the original negative direction is no longer visible in the square.

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

Given $S_{x x} = 16, S_{y y} = 25$ and $S_{x y} = 18$ , calculate $r$ and $r^{2}$ .
3 marks
A fit has $r = - 0.75$ . State its direction and percentage of response variation explained.
3 marks
A student reports $r = 1.12$ . Explain the problem.
3 marks
A fitted model has $r^{2} = 0.49$ and a negative slope. Find $r$ .
3 marks
If every $x$ -value is converted from metres to centimetres, what happens to $r$ ?
3 marks
All five response values are 7. Why is Pearson $r$ undefined?
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Given $S_{x x} = 16, S_{y y} = 25$ and $S_{x y} = 18$ , calculate $r$ and $r^{2}$ .
3 marks
The denominator is $\sqrt{16 \times 25} = 20$ .
Divide 18 by 20 and square the result.
$r = 0.9$ ; $r^{2} = 0.81$ .
A fit has $r = - 0.75$ . State its direction and percentage of response variation explained.
3 marks
The sign gives negative association.
$100 \times {(- 0.75)}^{2} = 56.25$ .
Negative direction; 56.25% explained.
A student reports $r = 1.12$ . Explain the problem.
3 marks
The theoretical range is $- 1 \leq r \leq 1$ .
Check data pairing, entry and the calculation.
A Pearson correlation cannot exceed 1.
A fitted model has $r^{2} = 0.49$ and a negative slope. Find $r$ .
3 marks
Take the square root, giving magnitude 0.7.
Use the slope direction to choose the negative sign.
$r = - 0.7$ .
If every $x$ -value is converted from metres to centimetres, what happens to $r$ ?
3 marks
Multiplication by a positive constant changes scale, not linear correlation.
It stays the same.
All five response values are 7. Why is Pearson $r$ undefined?
3 marks
A standardised measure of shared variation cannot be calculated.
The response has zero variation, so $S_{y y} = 0$ and the denominator is zero.

Checking saved progress…

Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 General Mathematics: statistical models. Original lessons and assessment practice; not an official SACE assessment.

Official curriculum source ↗

In this lesson

Understand the idea.

Pearson correlation measures linear association

Interpret r-squared in the fitted linear model

See the method in action.

Calculate from centred sums

Find the denominator from the two sums of squares.

Calculate the signed correlation.

Square the correlation.

Interpret the result in the fitted model.

Use paired measurements

Calculate the two means.

List the paired deviations.

Form the centred sums.

Calculate correlation and its square.

Try it yourself.

Check your understanding

Keep practising