SQA AH style: regression-style practice: Using S_xx=40, S_xy=54, x=5 and y=12, find the least-squares regression line of y on x and predict y when x=7.

Gradient: b=S_xyS_xx=5440=1.35 (1 mark). Intercept from a=y-bx=12-1.35× 5=12-6.75=5.25 (1 mark). Line: y=5.25+1.35x (1 mark). Prediction at x=7: y=5.25+1.35× 7=5.25+9.45=14.7 (1 mark). Markers reward the gradient, the intercept, the equation and a correct prediction within the data range.

ScotlandStatisticsSyllabus dot point

How do you measure and model the linear relationship between two variables?

Analyse bivariate data using scatter plots, the sums of squares and products, the product-moment correlation coefficient, and the least-squares regression line, and assess the model with residual plots and the limitations of extrapolation.

A focused answer to the SQA Advanced Higher Statistics bivariate data content: scatter plots, the sums of squares Sxx, Syy and Sxy, the product-moment correlation coefficient, the least-squares regression line, prediction, residual plots and the dangers of extrapolation.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The building blocks are the sums of squares and products: $S_{xx}=\sum x^2-\dfrac{(\sum x)^2}{n}$ , $S_{yy}=\sum y^2-\dfrac{(\sum y)^2}{n}$ and $S_{xy}=\sum xy-\dfrac{\sum x\sum y}{n}$ . The product-moment correlation coefficient is $r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}$ , lying between $-1$ and $1$ . The least-squares regression line of $y$ on $x$ is $\hat{y}=a+bx$ with $b=\dfrac{S_{xy}}{S_{xx}}$ and $a=\bar{y}-b\bar{x}$ . A residual is $y-\hat{y}$ ; a residual plot showing random scatter supports the linear model, while a pattern signals it is wrong. Extrapolation beyond the data is unreliable.

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What this dot point is asking
Scatter plots and the sums of squares
The product-moment correlation coefficient
The least-squares regression line
Residual plots and limitations
Try this

What this dot point is asking

Bivariate analysis studies how two variables move together. The SQA wants you to picture the relationship with a scatter plot, to quantify its strength with the product-moment correlation coefficient, to fit a least-squares regression line, to use it for prediction, and crucially to judge the fit using residual plots and to recognise where prediction is unsafe.

Scatter plots and the sums of squares

A scatter plot shows the form (linear or not), direction (positive or negative) and strength of a relationship at a glance, and it should always come first.

These three quantities are the raw material for both correlation and regression, so computing them carefully is the first calculation in any bivariate question.

The product-moment correlation coefficient

The correlation coefficient $r$ measures the strength and direction of a linear relationship.

Calculate r from summary statistics

Given the three summary statistics $S_{xx}$ , $S_{yy}$ and $S_{xy}$ , we substitute directly into the formula for $r$ . The sign of $r$ tells us the direction of the relationship and its magnitude tells us the strength.

Step 1: Write the formula and identify the given values

The product-moment correlation coefficient is $r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}$ . Here $S_{xx}=25$ , $S_{yy}=64$ and $S_{xy}=-30$ .

Step 2: Compute the denominator

Multiply $S_{xx}$ by $S_{yy}$ and take the square root to obtain the denominator.

\sqrt{S_{xx}S_{yy}}=\sqrt{25\times 64}=\sqrt{1600}=40

Step 3: Evaluate $r$ and interpret

Divide $S_{xy}$ by the denominator, then interpret the result in context.

r=\dfrac{-30}{40}=-0.75

Final answer: $r=-0.75$ , indicating a fairly strong negative linear relationship: as $x$ increases, $y$ tends to decrease.

The least-squares regression line

The regression line of $y$ on $x$ is the straight line that minimises the sum of the squared vertical residuals.

Fit a regression line and predict

To fit the least-squares regression line $\hat{y}=a+bx$ we compute the gradient $b$ first (using the sums of squares), then use the fact that the line passes through the mean point $(\bar{x},\bar{y})$ to find the intercept $a$ .

Step 1: Compute the gradient

The gradient is the ratio of the sum of cross-products to the sum of squares of $x$ . It measures how much $y$ changes on average for each unit increase in $x$ .

b=\dfrac{S_{xy}}{S_{xx}}=\dfrac{-30}{25}=-1.2

Step 2: Compute the intercept

The regression line always passes through the sample mean point $(\bar{x},\bar{y})$ . Substituting into $\hat{y}=a+bx$ and solving for $a$ gives:

a=\bar{y}-b\bar{x}=20-(-1.2)(4)=20+4.8=24.8

Step 3: Write the equation and make a prediction

State the fitted line, then substitute the required $x$ -value. Prediction is only reliable within the observed range of $x$ .

\hat{y}=24.8-1.2x

At $x=5$ : $\hat{y}=24.8-1.2(5)=24.8-6=18.8$ , a reliable prediction because $x=5$ lies within the data range.

Final answer: The regression line is $\hat{y}=24.8-1.2x$ ; at $x=5$ , $\hat{y}=18.8$ .

Residual plots and limitations

A residual is the vertical gap between an observed point and the line, $e_i=y_i-\hat{y}_i$ . Plotting residuals against $x$ (or against $\hat{y}$ ) checks whether the straight-line model is appropriate.

Try this

Q1. Given $S_{xx}=50$ , $S_{yy}=72$ , $S_{xy}=48$ , find $r$ . [2 marks]

Cue. $r=\dfrac{48}{\sqrt{50\times 72}}=\dfrac{48}{60}=0.8$ , a strong positive linear relationship.

Q2. A residual plot of a fitted line shows a clear U-shape. State what this tells you. [1 mark]

Cue. The relationship is not linear, so a straight-line model is inappropriate and a curved model should be considered.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: correlation4 marksFor a sample of

n=6

pairs,

S_{xx}=40

S_{yy}=90

and

S_{xy}=54

. Calculate the product-moment correlation coefficient and describe the relationship.

Show worked answer →

Product-moment correlation: $r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\dfrac{54}{\sqrt{40\times 90}}$ (1 mark).

$\sqrt{40\times 90}=\sqrt{3600}=60$ (1 mark), so $r=\dfrac{54}{60}=0.9$ (1 mark).

Since $r=0.9$ is close to $1$ , there is a strong positive linear relationship: as $x$ increases, $y$ tends to increase (1 mark). Markers reward the formula, the computed denominator, the value of $r$ and a correct interpretation.

AH style: regression4 marksUsing

S_{xx}=40

S_{xy}=54

\bar{x}=5

and

\bar{y}=12

, find the least-squares regression line of

y

x

and predict

y

when

x=7

Show worked answer →

Gradient: $b=\dfrac{S_{xy}}{S_{xx}}=\dfrac{54}{40}=1.35$ (1 mark).

Intercept from $a=\bar{y}-b\bar{x}=12-1.35\times 5=12-6.75=5.25$ (1 mark).

Line: $\hat{y}=5.25+1.35x$ (1 mark).

Prediction at $x=7$ : $\hat{y}=5.25+1.35\times 7=5.25+9.45=14.7$ (1 mark). Markers reward the gradient, the intercept, the equation and a correct prediction within the data range.

Related dot points

Sources & how we know this

SQA Advanced Higher Statistics Course Specification (C803 77) — SQA (2023)
SQA Advanced Higher Statistics Data Booklet — SQA (2019)