Edexcel IAL Maths

S1.4 Correlation and regression

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What this page covers

Overview

Scatter diagrams and the language of correlation: positive, negative, strong, weak, none

Calculating Sₓₓ, S_yy and S_xy from raw data or from summary statistics

The product moment correlation coefficient r: calculating it, interpreting it in context, and its limitations

The least-squares regression line y=a+bx: what "least squares" means and how to find a and b

Explanatory and response variables, and interpreting a and b in context

Interpolation and extrapolation, and how reliable an estimate is

Coding: the effect on r, on the summary statistics and on the regression line

In one minute

Key points

1. Correlation and regression measure and use a straight-line relationship between two variables, and three summary numbers, Sₓₓ, S_yy and S_xy, carry the whole topic.

2. Sₓₓ=Σx²-(Σx)²/n, S_yy=Σy²-(Σy)²/n, S_xy=Σxy-(Σx)(Σy)/n.

3. r=S_xy/√(SₓₓS_yy), always between −1 and 1. It measures linear association only, and correlation does not prove cause.

4. Regression line of y on x: b=S_xy/Sₓₓ, a=ȳ-bx̄. It passes through (x̄,ȳ) and minimises the sum of the squared vertical distances.

5. b is the change in y for each extra unit of x; a is the value of y when x=0 (often outside the data, so treat it with care).

6. Interpolation is reliable when r is close to ±1; extrapolation is not. Coding u=(x-p)/q does not change r (for q>0).

What does a scatter diagram show?

Section 1 of 14

Bivariate data. Pairs of values of two variables measured on each member of a sample, for example the height and the arm span of each student. A scatter diagram plots each pair as one point (x,y).

With positive correlation, as x increases y tends to increase: the points rise from left to right. With negative correlation they fall from left to right. With no correlation there is no straight-line pattern. The closer the points lie to a straight line, the stronger the (linear) correlation. In context, say what happens to one variable as the other increases: "days with more sunshine tend to generate more energy".

What does a scatter diagram show? (continued)

Section 1 of 14

Four sets of paired data with the product moment correlation coefficient of each, calculated from the plotted points. The U-shaped points follow a curve exactly, yet r=0: correlation only measures how close the points are to a straight line.

Common error

Writing just "positive correlation" when asked to interpret a diagram in context. Name the variables: "as the hours of sunshine increase, the energy generated tends to increase".

How do you calculate Sxx, Syy and Sxy?

Section 2 of 14

Summary statistics for paired data

(n = number of pairs): Sₓₓ=Σ(x-x̄)²=Σx²-(Σx)²/n, S_yy=Σ(y-ȳ)²=Σy²-(Σy)²/n S_xy=Σ(x-x̄)(y-ȳ)=Σxy-(Σx)(Σy)/n

Sₓₓ and S_yy are never negative. S_xy is positive for positive correlation and negative for negative correlation; it carries the sign of everything that follows. Why? For a point above-right of the mean point, both (x-x̄) and (y-ȳ) are positive, so its product is positive; points below-left also give positive products. A rising pattern therefore makes S_xy positive.

How do you calculate Sxx, Syy and Sxy? (continued)

Section 2 of 14

The solar-panel data split by lines through the mean point. Every point lies in a shaded quadrant, where (x-x̄)(y-ȳ)>0, so S_xy is positive.

In most questions you are given Σx, Σy, Σx², Σy² and Σxy: substitute into the right-hand forms. From raw data, use your calculator's statistics mode, but still write the formula with the numbers in it, and keep full calculator values of Sₓₓ, S_yy and S_xy: rounding early can change the third significant figure of r.

The solar-panel data used in the next three sections: hours of sunshine on 8 days and the energy a domestic solar panel generated on each of those days.

How do you calculate Sxx, Syy and Sxy? (continued)

Section 2 of 14

hours of sunshine, x	2.5	3.0	4.5	5.0	6.5	7.0	8.5	9.5
energy, y (kWh)	3.1	4.4	5.2	6.9	7.8	9.6	10.3	12.1

How do you calculate Sxx, Syy and Sxy? (continued)

Section 2 of 14 · Worked example 1 · Easy · Calculate

WORKED EXAMPLE · QUESTION

For the solar-panel data, Σx=46.5, Σy=59.4, Σx²=314.25, Σy²=509.12 and Σxy=399.25. Calculate Sₓₓ, S_yy and S_xy.

SOLUTION

Sₓₓ=314.25-46.5²/8=314.25-270.28125= 43.96875

S_yy=509.12-59.4²/8=509.12-441.045= 68.075

S_xy=399.25-(46.5×59.4)/8=399.25-345.2625= 53.9875, positive, matching the rising pattern of the points.

What does the product moment correlation coefficient measure?

Section 3 of 14

Product moment correlation coefficient.

A measure of the strength and direction of the linear relationship between two variables: r=S_xy/√(SₓₓS_yy), -1≤r≤1. r has no units, and swapping x and y gives the same value. A value outside -1≤r≤1 means an arithmetic error.

Value of r	Meaning
r=1	perfect positive linear correlation: every point on a rising straight line
close to 1 (e.g. 0.9)	strong positive linear correlation
around 0.4 to 0.6	moderate / weak positive linear correlation
close to 0	little or no linear correlation (there may still be a curved relationship)
close to −1	strong negative linear correlation

What does the product moment correlation coefficient measure? (continued)

Section 3 of 14

Value of r	Meaning
r=-1	perfect negative linear correlation: every point on a falling straight line

What does the product moment correlation coefficient measure? (continued)

Section 3 of 14 · Worked example 2 · Easy · Calculate, interpret

WORKED EXAMPLE · QUESTION

Using the results of Worked example 1, calculate the product moment correlation coefficient between hours of sunshine and energy generated, and interpret it in context.

SOLUTION

r=53.9875/(√(43.96875×68.075))=0.98680…= 0.987

Strong positive linear correlation: days with more hours of sunshine tend to generate more energy.

What does the product moment correlation coefficient measure? (continued)

Section 3 of 14

Exam tip — three parts to an interpretation

An interpretation of r needs the direction (positive/negative), the strength (strong/weak) and the context (both variables named). "Strong" on its own, or a bare number, is not an interpretation.

How do you find the least-squares regression line?

Section 4 of 14

Residuals for the solar-panel data. The largest is 0.72 kWh at x=7. No other straight line gives a smaller total of squared residuals than this one (1.79); the positive and negative residuals add up to exactly 0.

For each point, the residual is its vertical distance from the line: observed y − value of y on the line. The least-squares regression line of y on x is the straight line that makes the sum of the squares of the residuals as small as possible. The distances are measured parallel to the y-axis because y is the variable being predicted.

Regression line of y on x.

How do you find the least-squares regression line? (continued)

Section 4 of 14

y=a+bx, b=S_xy/Sₓₓ, a=ȳ-bx̄ Because a=ȳ-bx̄, the line always passes through the mean point (x̄,ȳ), a quick check on a sketch.

How do you find the least-squares regression line? (continued)

Section 4 of 14 · Worked example 3 · Medium · Find

The solar-panel data with the regression line y=0.288+1.23x. The line passes through the mean point (5.8125, 7.425), marked with a cross.

WORKED EXAMPLE · QUESTION

Using the results of Worked example 1, find the equation of the regression line of y on x for the solar-panel data, giving a and b to 3 significant figures.

SOLUTION

Gradient: b=(53.9875/43.96875)=1.2279…

Means: x̄=(46.5/8)=5.8125, ȳ=(59.4/8)=7.425.

Intercept, using the unrounded b: a=7.425-1.2279…×5.8125=0.2881

y=0.288+1.23x

How do you find the least-squares regression line? (continued)

Section 4 of 14

Common error

Rounding b to 1.23 before finding a: 7.425-1.23×5.8125=0.276, not 0.288. Keep the full value of b in your calculator for the next step.

Which variable is which, and what do a and b mean?

Section 5 of 14

Explanatory (independent) variable: the variable that is set, controlled or changes on its own; it goes on the horizontal axis as x. Response (dependent) variable: the variable whose value depends on, or is predicted from, the explanatory variable; it goes on the vertical axis as y.

Ask which variable could affect the other: hours of sunshine can affect the energy generated, not the other way round. The regression line of y on x is used to estimate y for a given x; it is not the right line for estimating x from y.

Interpreting b: "for each extra (unit of x), (the response) increases/decreases by about b (units of y)".

Interpreting a: the value of y when x=0, meaningful only if x=0 makes sense and is not far outside the data.

The phone data: hours of screen-on time x since a full charge, and battery left y (%), for one phone on 8 occasions.

Which variable is which, and what do a and b mean? (continued)

Section 5 of 14

x (hours)	1	1.5	2.5	3	4	4.5	5.5	6
y (%)	89	82	74	63	58	49	41	32

Which variable is which, and what do a and b mean? (continued)

Section 5 of 14 · Worked example 4 · Medium · State, find, interpret

WORKED EXAMPLE · QUESTION

For the phone data, n=8, Σx=28, Σy=488, Σx²=121, Σy²=32 560 and Σxy=1456. (a) State, with a reason, which is the explanatory variable. (b) Find the regression line of y on x. (c) Interpret the gradient and the intercept in context.

SOLUTION

(a) Screen-on time: using the screen drains the battery, so the battery level depends on the time, not the reverse.

(b) Sₓₓ=121-28²/8=23, S_xy=1456-(28×488)/8=-252. b=(-252)/23=-10.957…, a=61-(-10.957…)(3.5)=99.35: y=99.3-11.0x.

(c) b: the battery level falls by about 11 percentage points for each hour of screen-on time. a: with no screen-on time the battery is at about 99%, close to the full charge it starts from.

When can you trust an estimate from the line?

Section 6 of 14

The phone data cover 1 to 6 hours (shaded). The line y=99.3-11.0x reaches 0% at x=9.07, and beyond that it predicts impossible negative values: a straight-line model cannot hold for ever.

Interpolation: estimating y for an x inside the range of the data. Extrapolation: estimating y for an x outside the range of the data.

An interpolated estimate is reliable when the correlation is strong (r close to ±1). An extrapolated estimate is unreliable: nothing in the data shows that the straight-line pattern carries on beyond it. A weak correlation makes every estimate unreliable, inside the range or not.

When can you trust an estimate from the line? (continued)

Section 6 of 14 · Worked example 5 · Medium · Estimate, comment

WORKED EXAMPLE · QUESTION

Use the regression line from Worked example 4 to estimate the battery left after (a) 2 hours and (b) 10 hours of screen-on time. Comment on the reliability of each estimate.

SOLUTION

(a) y=99.348-10.957×2= 77.4%. Reliable: 2 hours is inside the data (interpolation), and r=-0.994 (using S_yy=2792 from the sums in Worked example 4) shows strong negative linear correlation.

(b) y=99.348-10.957×10= −10.2%. Unreliable: 10 hours is outside the data (extrapolation), and a negative battery level is impossible.

When can you trust an estimate from the line? (continued)

Section 6 of 14

Common error

Rearranging y=a+bx to estimate x from a given y. The line of y on x minimises vertical distances only; to estimate x you would need the regression line of x on y, a different line unless r=±1.

How does coding affect correlation and regression?

Section 7 of 14

Data are often coded to make the numbers smaller: u=(x-p)/q and v=(y-s)/t. Adding or subtracting a constant shifts the points; dividing by a positive constant rescales an axis. Neither changes the shape of the scatter diagram, so r is unchanged. Dividing by a negative constant reflects the diagram: r changes sign but keeps its size. To turn a coded regression line into one for x and y, substitute the coding into it and rearrange.

Coding and the summary statistics

(q,t>0): Sᵤᵤ=Sₓₓ/q², Sᵥᵥ=S_yy/t², Sᵤᵥ=S_xy/qt, rᵤᵥ=r_xy The constants p and s that are subtracted have no effect on any of the S values.

How does coding affect correlation and regression? (continued)

Section 7 of 14 · Worked example 6 · Hard · Find, hence, estimate

Hardness of the 8 bolts against furnace temperature, with the decoded line y=-81.2+0.362x.

WORKED EXAMPLE · QUESTION

A batch of 8 steel bolts is heat-treated at different furnace temperatures x °C and the hardness y of each is measured. The data are coded using u=(x-800)/5 and v=y-200, giving Σu=93, Σv=236, Σu²=1383, Σv²=7964, Σuv=3290. (a) Find the product moment correlation coefficient between x and y. (b) Find the regression line of v on u. (c) Hence find the regression line of y on x. (d) Estimate the hardness of a bolt treated at 860 °C.

How does coding affect correlation and regression? (continued)

Section 7 of 14 · Worked example 6 · Hard · Find, hence, estimate (continued)

The same 8 bolts plotted with the coded variables. The pattern is identical, only the axis labels change, so r=0.994 for u and v as well as for x and y.

SOLUTION

(a) Sᵤᵤ=1383-93²/8=301.875, Sᵥᵥ=7964-236²/8=1002, Sᵤᵥ=3290-(93×236)/8=546.5. r=546.5/(√(301.875×1002))=0.994; coding does not change r, so r=0.994.

(b) Gradient =(546.5/301.875)=1.8104…; ū=11.625, v̄=29.5; intercept =29.5-1.8104…×11.625=8.4547: v=8.45+1.81u.

(c) Substitute: y-200=8.4547+1.8104×(x-800)/5, so y=208.45+0.36207x-289.66: y=-81.2+0.362x.

How does coding affect correlation and regression? (continued)

Section 7 of 14 · Worked example 6 · Hard · Find, hence, estimate (continued)

(d) y=-81.202+0.36207×860= 230.

Check: Sₓₓ=5²×Sᵤᵤ=7546.875 and S_xy=5×Sᵤᵥ=2732.5, so b=2732.5÷7546.875=0.362. The intercept −81.2 would be the hardness at 0 °C, far outside the data, so it has no practical meaning.

What are the limitations of the correlation coefficient?

Section 8 of 14

Linear only. r measures how close the points are to a straight line. A strong curved relationship can have r near 0, so always look at the scatter diagram.

Correlation is not causation. Both variables may depend on a third: ice-cream sales and sunburn cases rise together because both increase in hot, sunny weather.

Outliers. One unusual point, often a recording error, can change r a lot, in either direction.

Small samples. With only a few points, a large r can happen by chance (testing for this is beyond S1).

The range of the data. r describes the relationship only within the values measured.

What are the limitations of the correlation coefficient? (continued)

Section 8 of 14 · Worked example 7 · Medium · Find, comment

One mistyped height (171 cm entered as 117 cm) drops r from 0.988 to 0.494 and drags the regression line down (dashed). Removing the error gives a much better model (solid).

WORKED EXAMPLE · QUESTION

The arm span x cm and height y cm of 10 students give Σx=1708, Σy=1654, Σx²=292 876, Σy²=277 202, Σxy=283 512, and r=0.494. One student's pair was recorded as (172, 117); it should be removed. (a) Find the new summary statistics for the remaining 9 students. (b) Find the new value of r and comment.

SOLUTION

What are the limitations of the correlation coefficient? (continued)

Section 8 of 14 · Worked example 7 · Medium · Find, comment (continued)

(a) Subtract the incorrect pair from every sum: Σx=1536, Σy=1537, Σx²=292 876-172²=263 292, Σy²=277 202-117²=263 513, Σxy=283 512-172×117=263 388.

(b) With n=9: Sₓₓ=263 292-1536²/9=1148, S_yy=263 513-1537²/9=1027.56, S_xy=263 388-(1536×1537)/9=1073.33. r=1073.33/(√(1148×1027.56))= 0.988. The correlation is now strong and positive: taller students tend to have longer arm spans; the single error had hidden this.

What are the limitations of the correlation coefficient? (continued)

Section 8 of 14

Common error

Removing a point by subtracting its values from Σx and Σy but leaving Σx², Σy², Σxy and n unchanged. Every sum loses that point's contribution, and n goes down by 1.

How do the pieces fit together in a full question?

Section 9 of 14

A full question usually runs: summary statistics → r and its meaning → regression line → interpret the gradient → estimate and judge reliability → coding or a change of units. Keep full calculator values throughout.

How do the pieces fit together in a full question? (continued)

Section 9 of 14 · Worked example 8 · Hard · Find, interpret, explain

WORKED EXAMPLE · QUESTION

The table gives the altitude x (hundreds of metres) and the mean July temperature y (°C) for 8 weather stations in one mountain region.

x	2	5	7	9	12	14	16	19
y	24.6	23.5	21.9	21.4	19.1	18.6	16.8	15.7

You may use Σx=84, Σy=161.6, Σx²=1116, Σy²=3334.48, Σxy=1569.3. (a) Find Sₓₓ, S_yy and S_xy. (b) Calculate r and interpret it in context. (c) Find the regression line of y on x. (d) Interpret the gradient in context. (e) Estimate the mean July temperature at a station 600 m above sea level, and comment on its reliability. (f) Explain why the line should not be used for a summit at 3500 m. (g) The altitudes are re-recorded in metres, w=100x. State the correlation coefficient between w and y, and the gradient of the regression line of y on w.

How do the pieces fit together in a full question? (continued)

Section 9 of 14 · Worked example 8 · Hard · Find, interpret, explain (continued)

Mean July temperature against altitude for the 8 stations, with the regression line through the mean point (10.5, 20.2) and the estimate for an altitude of 600 m.

SOLUTION

(a) Sₓₓ=1116-84²/8= 234; S_yy=3334.48-161.6²/8= 70.16; S_xy=1569.3-(84×161.6)/8= −127.5.

(b) r=(-127.5)/(√(234×70.16))= −0.995: strong negative linear correlation; higher stations tend to have lower mean July temperatures.

(c) b=(-127.5)/234=-0.54487…; x̄=10.5, ȳ=20.2; a=20.2-(-0.54487…)(10.5)=25.921: y=25.9-0.545x.

How do the pieces fit together in a full question? (continued)

Section 9 of 14 · Worked example 8 · Hard · Find, interpret, explain (continued)

(d) For each extra 100 m of altitude, the mean July temperature falls by about 0.545 °C.

(e) 600 m means x=6: y=25.921-0.54487×6= 22.7 °C; reliable, because x=6 is inside the data (interpolation) and the correlation is strong.

(f) 3500 m is x=35, far outside the range 2 to 19: extrapolation, so the linear pattern may not continue.

(g) Multiplying x by a positive constant does not change r; it divides the gradient by 100: r=-0.995, gradient −0.00545 °C per metre.

Exam tips for S1.4

Section 10 of 14

Exam tip 1 — write the formula with the numbers in

Show S_xy=1569.3-(84×161.6)/8 before the answer. If a calculator slip gives a wrong value, the method is still visible and later parts can follow through from it.

Exam tip 2 — keep unrounded values

Store Sₓₓ, S_yy, S_xy and b in your calculator memory. Give r, a and b to 3 significant figures at the end, unless the question asks otherwise.

Exam tip 3 — context in every interpretation

"For each extra hour of sunshine, about 1.23 kWh more energy is generated" is an interpretation; "b is the gradient" is not. Use the names and units of both variables.

Exam tips for S1.4 (continued)

Section 10 of 14

Exam tip 4 — check sign and size

r, S_xy and b must all have the same sign as the trend in the scatter diagram, and -1≤r≤1. The line must pass through (x̄,ȳ).

Exam tip 5 — reliability needs two reasons

To judge an estimate, say whether it is interpolation or extrapolation and what the value of r shows about the strength of the correlation.

Common misconceptions

Section 11 of 14

Misconception	Why it is wrong	Correct idea
r=0 means the variables are unrelated.	r only measures linear association; a curve can give r=0.	r=0 means no linear correlation; look at the scatter diagram.
A strong correlation proves that x causes y.	Both may depend on a third variable, or it may be coincidence.	Correlation describes association, not cause.
r=-0.9 is weaker than r=0.5.	The sign gives the direction, the size gives the strength.	−0.9 is a strong (negative) correlation.
Coding changes the correlation coefficient.	Shifting and positive rescaling do not change the shape of the scatter diagram.	r is unchanged by u=(x-p)/q with q>0.
The regression line of y on x can be rearranged to predict x.	It minimises vertical distances only.	Use it only to estimate y from x.
An estimate is reliable as long as r is close to 1.	The pattern is only known inside the data.	Extrapolated estimates are unreliable however strong r is.

How this topic connects

Section 12 of 14

S1.2 Measures of location and dispersion

Sₓₓ/n is the variance of x, so the coding rules here are the same ones used for means and standard deviations.

P1 Straight-line graphs

The regression line y=a+bx is a straight line: gradient and intercept are read and interpreted exactly as for y=mx+c.

P2/P3 Exponentials and logarithms

A curved relationship such as y=kxⁿ becomes linear when you plot log y against log x; correlation and regression then apply to the logged data.

Key formulae

Section 13 of 14

Formula	Use	In the formulae list?
Sₓₓ=Σx²-(Σx)²/n	spread of x; also S_yy for y	Yes
S_xy=Σxy-(Σx)(Σy)/n	how x and y vary together	Yes
r=S_xy/√(SₓₓS_yy)	product moment correlation coefficient	Yes
b=S_xy/Sₓₓ, a=ȳ-bx̄	regression line of y on x: y=a+bx	Yes
ȳ=a+bx̄	the line passes through the mean point	No – derive it
Sᵤᵤ=Sₓₓ/q², Sᵤᵥ=S_xy/(qt), rᵤᵥ=r_xy	coding u=(x-p)/q, v=(y-s)/t	No – learn it

Check your understanding

Section 14 of 14

1. For a set of data, Sₓₓ=40, S_yy=90 and S_xy=-48. Find r.

2. The regression line of y on x is y=12.4+0.75x, and x̄=20. Find ȳ.

3. Two variables have r=0.03. Does this show they are not related?

4. Data are coded with u=(x-50)/10 and v=(y-3)/2. The coded data have r=0.62. Find r for x and y.

5. Phone sales in a country and the number of people over 80 have both risen steadily, with r=0.97. Does selling phones make people live longer?

Check your understanding · answers

Section 14 of 14

1. r=-48÷√(40×90)=-48÷60=-0.8: fairly strong negative linear correlation.

2. The line passes through (x̄,ȳ), so ȳ=12.4+0.75×20=27.4.

3. No. It shows there is almost no linear correlation. The variables could still have a strong curved relationship, which a scatter diagram would show.

4. 0.62. Subtracting constants and dividing by positive constants does not change the correlation coefficient.

5. No. Correlation does not imply causation: both quantities increase over time (time, wealth and population growth are third variables), so they are correlated without one causing the other.

Check understanding

Practice questions

The answer and each step appear one click at a time.

Question 1 [5 marks]

An engineer records the wind speed x m s⁻¹ and the power output y kW of a wind turbine at 20 different times. The wind speeds recorded range from 4 m s⁻¹ to 12 m s⁻¹. The product moment correlation coefficient is 0.962 and the regression line of y on x is y = −62.4 + 31.5x.

(a) State, with a reason, which variable is the explanatory variable. [1]

(b) Interpret the value 31.5 in context. [1]

(c) Estimate the power output when the wind speed is 9 m s⁻¹, and comment on the reliability of your estimate. [2]

(d) Use the line to find the power output predicted for a wind speed of 1 m s⁻¹, and explain why this value should not be trusted. [1]

Mark scheme click to reveal each step

the wind speed, because the wind drives the turbine: the power output depends on the wind speed, not the reverse

for each increase of 1 m s⁻¹ in wind speed, the power output increases by about 31.5 kW

−62.4 + 31.5 × 9 = 221.1 kW

reliable: 9 is within the range of the data (interpolation) and r = 0.962 shows strong positive correlation

−30.9 kW, which is impossible (negative power); 1 m s⁻¹ is outside the data, so this is extrapolation

Question 2 [7 marks]

For 15 farms in one region, a researcher records the rainfall x mm during the growing season and the wheat yield y tonnes per hectare. The rainfall ranged from 38 mm to 91 mm. Sₓₓ = 1600, S_(yy) = 4.84, x̄ = 62.5, ȳ = 4.2, and the product moment correlation coefficient is exactly 0.6.

(a) Show that S_(xy) = 52.8. [2]

(b) Find the equation of the regression line of y on x. [3]

(c) Use your line to estimate the yield for a farm with 70 mm of rainfall, and comment on the reliability of the estimate. [2]

Mark scheme click to reveal each step

0.6 = ^(Sxy)⁄_(√(1600 × 4.84))

√(1600 × 4.84) = 40 × 2.2 = 88, so S_(xy) = 0.6 × 88 = 52.8

b = ^(52.8)⁄₁₆₀₀ = 0.033

a = 4.2 − 0.033 × 62.5 = 2.1375

y = 2.14 + 0.033x

2.1375 + 0.033 × 70 = 4.45 tonnes per hectare

70 mm is inside the data (interpolation), but r = 0.6 is only moderate correlation, so the estimate is not very reliable

Keep practising

S1.4 Correlation and regression

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