Scatter Diagrams and Regression Lines

Exploring relationships between variables and making predictions using the line of best fit

SPEC 2.2 • DATA PRESENTATION AND INTERPRETATION

Bivariate Data

Definition

Bivariate data consists of pairs of values (x, y) where each pair represents two related measurements from the same individual or item.

When we collect data on two variables simultaneously, we can investigate whether there is a relationship between them.

Examples of Bivariate Data:

Height and weight of people

Hours studied and exam score

Temperature and ice cream sales

Age of car and its value

Questions We Can Ask:

Is there a relationship?

How strong is it?

Can we predict one from the other?

What is the equation of the relationship?

Key Concept

Bivariate data is visualised using a scatter diagram - each point represents one pair of values.

Scatter Diagrams

A scatter diagram (or scatter plot) displays bivariate data as points on a coordinate grid.

Scatter diagram showing positive linear relationship

Tip

The dashed line is the regression line (line of best fit) - it shows the general trend in the data.

Explanatory and Response Variables

Explanatory Variable (x)

The independent variable that we believe influences or explains changes in the other variable.

Plotted on the horizontal axis.

Response Variable (y)

The dependent variable that responds to or is affected by changes in the explanatory variable.

Plotted on the vertical axis.

ContextExplanatory (x)Response (y)

Study habitsHours studiedExam score

AdvertisingAd spending (£)Sales revenue (£)

Plant growthAmount of fertiliserHeight of plant

Car depreciationAge of car (years)Value of car (£)

Important

The choice of which variable is explanatory depends on the context and what you're trying to investigate. Ask: "Which variable do I think influences the other?"

Types of Correlation

Correlation describes the strength and direction of a linear relationship between two variables.

Strong Positive

r close to +1

Weak Positive

r between 0 and +0.5

No Correlation

r close to 0

Strong Negative

r close to -1

Weak Negative

r between -0.5 and 0

Non-linear

r misleading here!

Important

Linear regression is only appropriate for data showing a linear pattern. Always plot your data first!

Correlation vs Causation

Key Principle

Correlation does NOT imply causation

Just because two variables show correlation, it does not mean one causes the other.

Two variables have a causal relationship if:

A change in one variable directly causes a change in the other.

Example: Ice cream sales & drownings

Strong positive correlation exists, but ice cream doesn't cause drowning!

Third variable: Hot weather increases both ice cream sales AND swimming (leading to more drownings)

Reasons correlation ≠ causation:

Confounding variable - a third factor affects both

Reverse causation - the effect causes the cause

Coincidence - no real relationship

Sampling issues - biased or small sample

Exam Tip

When asked to comment on whether data shows a causal relationship, consider: context, whether there could be confounding variables, and whether the sample is appropriate (random, large enough).

01Worked Example - Plotting Data

The table shows the number of hours spent revising (x) and the exam score (y) for 6 students.

Hours (x)2457810

Score (y)354552606875

(a) State the explanatory and response variables.

(b) Describe the relationship shown.

a)

Explanatory (x): Hours spent revising - this is what we control/change

Response (y): Exam score - this is what we measure as a result

positive linear relationship

Worked Example - Identifying Variables02

A researcher investigates the relationship between the age of a car and its resale value. Which variable should be the explanatory variable? Justify your answer.

1

The age of the car should be the explanatory variable (x).

2

Justification: The age of a car influences its value, not the other way around. As time passes, cars depreciate in value. We cannot change a car's value to make it older or younger.

3

The resale value is the response variable (y) because it responds to (is affected by) the age.

Expected relationship: Negative correlation - as age increases, value decreases.

The Regression Line

Definition

The regression line of y on x is the line that minimises the sum of squared residuals:

\sum (y_{i} - \overset{y}{^}_{i})^{2}

Least Squares Method

Residual:

e_{i} = y_{i} - \overset{y}{^}_{i}

We minimise:

S = \sum e_{i}^{2} = \sum (y_{i} - \overset{y}{^}_{i})^{2}

Setting

\frac{\partial S}{\partial a} = 0

and

\frac{\partial S}{\partial b} = 0

gives:

b = \frac{S _{x y}}{S _{xx}}

a = \overset{y}{ˉ} - b \overset{x}{ˉ}

Key Point

The regression line always passes through

(\overset{x}{ˉ}, \overset{y}{ˉ})

. The formulas minimise the total squared vertical distance from points to the line.

Equation of the Regression Line

Formula

y = a + b x

b = gradient (slope) of the linea = y-intercept (value of y when x = 0)

Calculating the Coefficients

b = \frac{S _{x y}}{S _{xx}}

a = \overset{y}{ˉ} - b \overset{x}{ˉ}

Remember

The formulas for

S_{xx}

and

S_{x y}

are:

S_{xx} = \sum x^{2} - \frac{( \sum x ) ^{2}}{n}

S_{x y} = \sum x y - \frac{( \sum x ) ( \sum y )}{n}

Tip

Always calculate b first, then use it to find a. The line must pass through

(\overset{x}{ˉ}, \overset{y}{ˉ})

.

Worked Example - Calculating Sxy and Sxx03

For the data: n = 8, Σx = 40, Σy = 120, Σx² = 240, Σxy = 680. Calculate Sₓₓ and S_xy.

1

Calculate

S_{xx}

:

S_{xx} = \sum x^{2} - \frac{( \sum x ) ^{2}}{n} = 240 - \frac{4 0 ^{2}}{8} = 240 - \frac{1600}{8} = 240 - 200 = 40

2

Calculate

S_{x y}

:

S_{x y} = \sum x y - \frac{( \sum x ) ( \sum y )}{n} = 680 - \frac{40 \times 120}{8} = 680 - \frac{4800}{8} = 680 - 600 = 80

Answer: Sₓₓ = 40, S_xy = 80

Worked Example - Finding Regression Equation04

Using the values from Example 3 (n = 8, Σx = 40, Σy = 120, Sₓₓ = 40, S_xy = 80), find the equation of the regression line y on x.

1

Calculate the means:

\overset{x}{ˉ} = \frac{\sum x}{n} = \frac{40}{8} = 5

\overset{y}{ˉ} = \frac{\sum y}{n} = \frac{120}{8} = 15

2

Calculate b (gradient):

b = \frac{S _{x y}}{S _{xx}} = \frac{80}{40} = 2

3

Calculate a (y-intercept):

a = \overset{y}{ˉ} - b \overset{x}{ˉ} = 15 - 2 (5) = 15 - 10 = 5

Regression equation: y = 5 + 2x

Interpreting a and b

Gradient (b)

The gradient tells you how much y changes for each unit increase in x.

Interpretation: "For every 1 unit increase in [x variable], the [y variable] increases/decreases by b units on average."

Y-intercept (a)

The y-intercept is the predicted value of y when x = 0.

Interpretation: "When [x variable] is zero, the predicted [y variable] is a units."

Caution

The y-intercept may not always have a meaningful interpretation:

If x = 0 is outside the range of data, interpreting a may be extrapolation

Sometimes x = 0 is impossible (e.g., "0 hours of sunlight" for plant growth)

Example Interpretation

For y = 5 + 2x where x = hours studied and y = exam score:

b = 2: Each additional hour of study increases the predicted exam score by 2 marks

a = 5: A student who studies 0 hours would be predicted to score 5 marks

05Worked Example - Interpretation

A company finds the regression line for advertising spend (x, in £1000s) and weekly sales (y, in £1000s) is: y = 12 + 3.5x

Interpret the values of the gradient and y-intercept in context.

1

Gradient (b = 3.5):

For every additional £1000 spent on advertising, the weekly sales are predicted to increase by £3500 on average.

2

Y-intercept (a = 12):

When no money is spent on advertising (x = 0), the predicted weekly sales are £12,000. This represents the baseline sales without advertising.

3

Note: The y-intercept has a sensible interpretation here because a company could genuinely have zero advertising spend.

Using the Regression Line

The regression line can be used to predict values of y for given values of x.

Interpolation

Making predictions within the range of the original data.

Generally reliable

Extrapolation

Making predictions outside the range of the original data.

Potentially unreliable

Important

Extrapolation assumes the linear relationship continues beyond the data - this may not be true!

06Worked Example - Prediction

The regression line for temperature (x °C, range 10-30) and ice cream sales (y, in hundreds) is: y = 0.5 + 0.8x

(a) Predict sales when temperature is 25°C

(b) Would it be appropriate to use this equation to predict sales at 40°C?

a)

Substitute x = 25 into y = 0.5 + 0.8x:

y = 0.5 + 0.8 (25) = 0.5 + 20 = 20.5

Predicted sales: 2050 ice creams

This is interpolation (25°C is within range 10-30°C), so it's reliable.

b)

No, this would not be appropriate. 40°C is outside the data range (10-30°C), so this is extrapolation which is unreliable.

Limitations of Regression

When Regression May Be Inappropriate

Non-linear relationships: If the scatter diagram shows a curve, a linear model is inappropriate

Outliers: Extreme values can distort the regression line significantly

Extrapolation: Predictions outside the data range are unreliable

No causal relationship: Correlation does not imply causation

Remember

Correlation ≠ Causation. Just because two variables are correlated doesn't mean one causes the other. There could be a third variable affecting both, or it could be coincidence.

Calculator: Linear Regression

Setup & Data Entry

1. Press MODE → 2 (STAT) → 2 (A+BX)

2. Enter data pairs: [x] = [y] =

Example: For (2, 35): Press 2 = 35 =

Getting Results

1. Press SHIFT → 1 (STAT)

2. Select 5 (Reg) for regression menu

3. Choose: 1=A, 2=B, 3=r

Key Statistics

OptionSymbolMeaning1Ay-intercept (a)2BGradient (b)3rCorrelation coefficient

Casio fx-83GT CW

Exam Tip

Always show your formula substitutions even when using calculator results. Write: "Using calculator: a = ..., b = ..." then state your equation.

Exponential Models

When data shows a non-linear pattern, we can use logarithms to transform it into a linear relationship.

Power Model: y = axⁿ

Taking logs of both sides:log y = log a + n log xCode using: Y = log y and X = log x

Gives linear form: Y = log a + nX

Exponential Model: y = kbˣ

Taking logs of both sides:log y = log k + x log bCode using: Y = log y and X = x

Gives linear form: Y = log k + (log b)X

Finding Constants from Coded Data

For y = axⁿ:

Intercept of Y on X gives log a → a = 10^intercept

Gradient gives n directlyFor y = kbˣ:

Intercept gives log k → k = 10^intercept

Gradient gives log b → b = 10^gradient

How to Identify the Model

If coded with X = log x, Y = log y → Power model (y = axⁿ)

If coded with X = x, Y = log y → Exponential model (y = kbˣ)

07Worked Example - Exponential Model

Data are coded using X = t and Y = log g. The regression line of Y on X is found to be Y = -0.2215 + 0.0792X.

(a) State whether the relationship is of the form y = axⁿ or y = kbˣ.

(b) Find the values of the constants in the original model.

a)

Since X = t (not log t) and Y = log g, this is an exponential model: g = kb^t

b)

Finding k: The intercept is log k = -0.2215

k = 10^-0.2215 = 0.600 (3 s.f.)

•

Finding b: The gradient is log b = 0.0792

b = 10^0.0792 = 1.20 (3 s.f.)

Model: g = 0.600 × 1.20^t

08Worked Example - Power Model

The relationship between pressure P and volume V is modelled as P = aVⁿ. Data are coded using X = log V and Y = log P. The regression line of Y on X is: Y = 2.4 - 1.4X

(a) Explain why coding X = log V and Y = log P confirms this is a power model.

(b) Find the values of a and n.

a)

For P = aVⁿ, taking logs: log P = log a + n log V

With Y = log P and X = log V, this gives Y = log a + nX

This matches the linear form Y = c + mX, confirming a power model.

b)

Finding n: The gradient equals n

n = -1.4

•

Finding a: The intercept is log a = 2.4

a = 10^2.4 = 251 (3 s.f.)

Model: P = 251V^-1.4

Practice

Question 1: Given: n = 10, Σx = 50, Σy = 200, Σx² = 300, Σxy = 1200

(a) Calculate Sₓₓ and S_xy

(b) Find the equation of the regression line y on x

(c) Predict y when x = 7

a)

S_{xx} = 300 - \frac{5 0 ^{2}}{10} = 300 - 250 = 50

S_{x y} = 1200 - \frac{50 \times 200}{10} = 1200 - 1000 = 200

b)

\overset{x}{ˉ} = 5, \overset{y}{ˉ} = 20

b = \frac{200}{50} = 4

a = 20 - 4 (5) = 0

y = 4x

c)

y = 4 (7) = 28

Practice

Question 2: The regression line for engine size (x litres) and fuel consumption (y mpg) is: y = 55 - 8x

The data was collected from cars with engine sizes between 1.0 and 3.0 litres.

(a) Interpret the gradient in context.

(b) Interpret the y-intercept and comment on its validity.

(c) Would it be reliable to use this equation to predict the fuel consumption of a car with a 5.0 litre engine? Explain.

a)

For each additional litre of engine size, fuel consumption decreases by 8 mpg on average.

b)

When engine size is 0 litres, predicted consumption is 55 mpg. This is NOT valid as a car cannot have 0 litre engine - extrapolation beyond data range.

c)

No - this is extrapolation (5.0L is outside 1.0-3.0L range). The linear relationship may not hold for very large engines.

Practice

Question 3: A biologist collects data on the age (x months) and mass (y grams) of a species of fish. She calculates the regression line as y = 15 + 12x for fish aged 2-10 months.

(a) Predict the mass of a fish aged 6 months. Is this prediction reliable?

(b) A student uses the equation to predict that a fish aged 50 months would have mass 615g. Explain why this prediction is inappropriate.

a)

y = 15 + 12(6) = 15 + 72 = 87 grams

This is reliable as 6 months is within the data range (2-10 months) - interpolation.

b)

This is extrapolation (50 months far outside 2-10 month range).

Fish don't grow linearly forever - growth typically slows or stops at maturity. The linear model only applies to young, growing fish. 615g may be biologically impossible for this species.

Summary

Key Concepts

ConceptDescription

Explanatory variable (x)The independent variable that influences the response

Response variable (y)The dependent variable that responds to changes in x

Regression liney = a + bx, line of best fit minimising squared residuals

Gradient (b)Change in y for each unit increase in x

InterpolationPrediction within data range - reliable

ExtrapolationPrediction outside data range - unreliable

Key Formulas

b = \frac{S _{x y}}{S _{xx}}

and

a = \overset{y}{ˉ} - b \overset{x}{ˉ}

where

S_{xx} = \sum x^{2} - \frac{( \sum x ) ^{2}}{n}

and

S_{x y} = \sum x y - \frac{( \sum x ) ( \sum y )}{n}

Exam Skills

Always interpret coefficients in context

State whether prediction is interpolation or extrapolation

Explain why extrapolation is unreliable with context-specific reasons