What are wrong units?

The gradient has units of (y units) per (x unit). Include them explicitly.

§-Syllabus dot point

NSWMaths Standard 2Syllabus dot point

How is the least-squares regression line calculated, and how is it used to model a linear relationship between two variables?

Find and use the equation of the least-squares regression line to model a linear relationship between two variables

A focused answer to the HSC Maths Standard 2 dot point on the least-squares regression line. The equation $y = mx + b$ , finding the gradient and intercept on a calculator, predicting and reading values off the line, interpreting the gradient and intercept, interpolation versus extrapolation, and worked Australian examples.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-20

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

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What this dot point is asking

NESA wants you to do four things. Find the equation of the least-squares regression line using your calculator's statistics functions. Write it in $y = mx + b$ form. Use it to predict $y$ from $x$ . Interpret the gradient and intercept in the context of the worded problem. This is where the scatterplot and the correlation coefficient pay off: once the data looks linear and $r$ is reasonably strong, the regression line turns the relationship into a formula you can predict with.

The answer

The least-squares regression line

Bivariate data is data where each item has two measurements, an $x$ and a $y$ . For such data, the least-squares regression line is the straight line that makes the sum of the squared vertical distances from the data points to the line as small as possible. Those vertical distances are called residuals: how far each actual point sits above or below the line. Squaring them before adding gives big misses extra weight, so the line cannot drift away from the points. The result is the single straight line that fits the cloud best overall. It is the standard "line of best fit" for linear data.

The equation has the form:

y = m x + b

where $m$ is the gradient and $b$ is the $y$ -intercept. The whole point of finding it is that it lets you predict: feed in an $x$ , get the model's best estimate of $y$ .

Finding the line on a calculator

You will not be asked to compute the gradient and intercept by hand. The procedure on a NESA-approved scientific calculator:

Enter statistics mode (typically MODE STAT, then a 2-variable or $A + Bx$ option).
Enter the $(x, y)$ pairs into the statistical lists.
Read off $m$ (sometimes labelled $a$ or $B$ ) and $b$ (sometimes labelled $A$ or $a$ ) from the regression-result menu.
Read off $r$ (the correlation coefficient) at the same time, to check the fit is worth using.

Different calculator models use different labels, and some even swap the roles of $A$ and $B$ . So check which one is the gradient on your model. Practise on the exact calculator you will use in the exam, and clear any previous data first.

Build and use the line, stage by stage

Here is the full routine on one dataset: daily ice-cream van sales (in hundreds of dollars) against the daily maximum temperature in degrees Celsius over $15$ summer days. Watch it go from a cloud of dots to a usable prediction.

Stage 1, plot the bivariate data. Put the independent variable (temperature) on the horizontal axis and the response (sales) on the vertical axis, then plot each pair. The cloud rises to the right, so a positive linear model looks reasonable, which is the green light to fit a line.

Stage 2, find and draw the least-squares line. Enter the pairs into the calculator and read off the gradient and intercept; here $m = 0.33$ and $b = -3.4$ , so the line is $y = 0.33x - 3.4$ . Draw it straight through the middle of the cloud. It is the best linear fit, not a line forced through any two particular points.

Stage 3, read a prediction off the line. To predict sales on a $30$ degree day, you can either substitute into the equation, $y = 0.33 \times 30 - 3.4 = 6.5$ (so about $650), or read it straight off the graph: go up from $30$ on the temperature axis to the line, then across to the sales axis, landing at $6.5$ . Both routes agree, and on a graphical question you show the dashed read-off lines.

Stage 4, predict only inside the data range. The line can be extended forever, but you can only trust it across the temperatures the data actually covers (about $18$ to $39$ degrees), shaded below. A prediction inside that band is called interpolation and is safe. A prediction outside it, say at $45$ degrees, is called extrapolation and may be wrong. Nothing guarantees the trend keeps going: sales might level off or even fall once it is too hot to queue outside.

Predicting $y$ from $x$

Substitute the $x$ value into the line equation. The result is the model's predicted $y$ at that $x$ . The actual value will usually differ a little, because the line is the best fit over the whole dataset, not a guarantee for any single case. Quote the prediction with its units and round sensibly for the context.

Interpreting the gradient

The gradient $m$ has units of (y units) per (x unit). For every $1$ -unit increase in $x$ , the model says $y$ changes by $m$ units on average.

Always include the word "average" or "on average" and the units. The "on average" matters because the line is a best fit across many points, not an exact rule for any one observation, and markers reward it explicitly.

Interpreting the $y$ -intercept

The intercept $b$ is the predicted $y$ value when $x = 0$ . In context this is sometimes meaningful (a base salary at zero years of experience, say) and sometimes pure extrapolation (predicted food spending at zero income, which no real household has).

If $x = 0$ lies well outside the range of the data, say so: the intercept is an extrapolation and may carry no real-world meaning. State it as a number, then add the caveat.

Interpolation versus extrapolation

This is the single most common follow-up, so make it a habit:

Interpolation: predicting for an $x$ inside the range of the data. Generally reliable, because the line is fitted there.
Extrapolation: predicting for an $x$ outside the range of the data. Treat with caution; the linear trend may not continue, and the further out you go, the worse it can be.

Whenever you predict, glance at whether the $x$ sits inside the data range. If it does not, name the prediction as extrapolation and flag that it may be unreliable. (The separate interpolation and extrapolation dot point goes deeper on this.)

When to use the line

The least-squares line is appropriate when:

the scatterplot suggests an approximately linear relationship,
the correlation coefficient $|r|$ is moderately strong or stronger,
there are no extreme outliers distorting the fit.

If the scatterplot is clearly non-linear, the line will be a poor model even if $r$ is not zero, and a prediction from it can be badly off.

How exam questions ask about the regression line

"Write the equation of the least-squares regression line." Put the calculator's $m$ and $b$ into $y = mx + b$ explicitly.
"Use the line to predict $y$ when $x = \dots$ " Substitute and compute, then state the result with units; note if it is extrapolation.
"Interpret the gradient." Average change in $y$ per $1$ -unit increase in $x$ , in context, with units, including "on average".
"Interpret the $y$ -intercept." Predicted $y$ at $x = 0$ , in context, with a note if $x = 0$ is outside the data (extrapolation).
"Is this prediction reliable?" Reliable if interpolating; flag it if extrapolating or if the fit ( $r$ ) is weak.

Worked example

Regression line for used-car resale value

A dataset of car age and resale value for $15$ used Toyotas gives, from a calculator:

$m = -2500$ (with $y$ in dollars, $x$ in years)
$b = 38000$

Write the equation of the regression line: Substitute the calculator values directly into $y = mx + b$ :

V = -2500x + 38000

The negative gradient tells us immediately that older cars have lower predicted resale values.

Predict values by substituting into the equation: For a car that is $5$ years old, substitute $x = 5$ :

V = -2500 \times 5 + 38000 = -12500 + 38000 = 25\,500 \text{ dollars}

For a car that is $10$ years old:

V = -2500 \times 10 + 38000 = 13\,000 \text{ dollars}

Interpret the gradient in context: The gradient is $-2500$ . For every extra year of age, resale value drops by an average of $2500. The word "average" is important because the line is the best fit over the whole dataset, not an exact rule for every individual car.
Interpret the intercept, checking for extrapolation: The intercept is $ $38\,000$ . This is the predicted resale value at $x = 0$ , meaning a brand-new car. If the dataset includes cars near zero years old this is a reasonable baseline; if the youngest car in the sample is several years old, then $x = 0$ lies outside the data range and the intercept is an extrapolation that should not be trusted literally.
Answer: the regression line is $V = -2500x + 38000$ ; predicted values are $ $25\,500$ at $5$ years and $ $13\,000$ at $10$ years; the gradient means resale value falls by an average of $2500 per year.

Interpret a weekly entertainment-spending model

A regression of weekly entertainment spending on household income for ABS-style data gives:

y = 0.045x + 35

with both variables in dollars per week.

Interpret the gradient: The gradient is $0.045$ . For every extra dollar of weekly income, the household spends an average of $0.045 (about $4.5$ cents) more on entertainment. This reflects that entertainment spending scales slowly as a fraction of income.
Interpret the intercept, noting extrapolation: The intercept is $35$ . A household earning $0 per week is predicted to spend $35 on entertainment. No household in the dataset has zero income, so $x = 0$ is well outside the range of the data. The intercept is purely a model artefact and should not be interpreted literally.
Decide whether a prediction is interpolation or extrapolation: Predicting entertainment spending for a household earning $1500 per week, when the data ranged from $600 to $2500, is interpolation and is reasonable. Predicting for a household earning $8000 per week is extrapolation and should be flagged as unreliable.
Answer: gradient $0.045$ means an extra $4.5$ cents of entertainment spending per dollar of income (on average); the intercept of $35$ is an extrapolation and carries no practical meaning here; predictions are trustworthy only for incomes within the sampled range.

Edge cases worth knowing

A prediction that comes out negative or impossible. A model can predict a negative resale value or negative sales at extreme $x$ . That is a sign you have extrapolated past where the line makes sense; report the value but note the model has broken down.
A weak fit. If $|r|$ is small, the line still exists but its predictions are poor. Mention the weak correlation when judging reliability.
Outliers tilting the line. Because least squares squares the residuals, one far-off point can swing the gradient and intercept noticeably. Identify outliers from the scatterplot first.
Reversing the variables. The regression line of $y$ on $x$ is not the same as $x$ on $y$ . Fit the line with the response variable as $y$ , matching the question.

Exam-style practice questions

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

2022 HSC-style4 marksFor a dataset of

50

pairs, calculator output gives gradient

m = 2.5

and intercept

b = -8

for the least-squares regression line. Write the equation, predict

y

when

x = 12

, and interpret the gradient.

Show worked answer →

Equation: $y = 2.5 x - 8$ .

At $x = 12$ : $y = 2.5 \times 12 - 8 = 30 - 8 = 22$ .

Interpretation: for every increase of $1$ in $x$ , $y$ increases by $2.5$ (on average, according to the model).

Markers reward the equation, the substitution, and an interpretation of the gradient that uses the word "average" or "on average" to acknowledge the model is a best fit, not exact.

2023 HSC-style3 marksA linear model of weekly food spending (

y

, in $) on weekly income (

x

, in $) for

80

Sydney households gives the line

y = 0.18 x + 95

. Interpret the gradient and the

y

-intercept in this context.

Show worked answer →

Gradient $0.18$ : for every extra dollar of weekly income, the household spends, on average, an additional $0.18 on food. In other words, about $18\%$ of each additional dollar of income goes to food.

$y$ -intercept $95$ : a household with zero income is predicted to spend $95 per week on food. This is the model's baseline. In practice, $0 income is well outside the dataset, so the intercept may not be reliable (extrapolation).

Markers reward the gradient interpretation in context with units, the intercept interpretation in context, and a brief caveat about extrapolation for the intercept.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksA least-squares regression line is given by

y = 4x + 12

. (a) Write down the gradient and the

y

-intercept. (b) Use the line to predict

y

when

x = 8

Show worked solution →

Read off the gradient and intercept. The line is in the form $y = mx + b$ , so the gradient is $m = 4$ and the $y$ -intercept is $b = 12$ .

Predict by substituting. Put $x = 8$ into the equation:

y = 4 \times 8 + 12 = 32 + 12 = 44.

Check. The gradient is positive, so a larger $x$ should give a larger $y$ ; the predicted $44$ is above the intercept of $12$ , as expected. So the gradient is $4$ , the intercept is $12$ , and $y = 44$ when $x = 8$ .

foundation3 marksA phone plan's monthly bill is modelled by the least-squares line

y = 0.15x + 20

, where

x

is the number of call minutes used and

y

is the bill in dollars. (a) Predict the bill for a month with

300

call minutes. (b) Interpret the gradient in this context.

Show worked solution →

Predict the bill at $x = 300$ . Substitute $x = 300$ into the line:

y = 0.15 \times 300 + 20 = 45 + 20 = 65.

So the predicted bill is $65.

Interpret the gradient. The gradient is $0.15$ . For each extra call minute, the bill rises by an average of $0.15 (that is, $15$ cents per minute). The word "average" matters because the line is a best fit, not an exact rule for every month.

Check. The intercept $20 is the fixed monthly charge at $0$ minutes; adding $300$ minutes at $15$ cents each is $300 \times 0.15 = 45$ dollars, and $20 + 45 = 65$ , confirming the prediction.

foundation3 marksFor a group of students, the least-squares line relating exam mark

y

to weekly study hours

x

y = 6.2x + 58

. (a) Predict the mark for a student who studies

5

hours per week. (b) Interpret the gradient in context.

Show worked solution →

Predict the mark at $x = 5$ . Substitute $x = 5$ into the line:

y = 6.2 \times 5 + 58 = 31 + 58 = 89.

So the predicted exam mark is $89$ .

Interpret the gradient. The gradient is $6.2$ . For each extra hour of weekly study, the model predicts an exam mark that is, on average, $6.2$ marks higher. Including "on average" acknowledges that the line is a best fit across the whole group, not a guarantee for any one student.

Check. The gradient is positive, so more study should mean a higher mark; the predicted $89$ sits above the intercept of $58$ , which is consistent. The predicted mark is $89$ .

foundation3 marksAn electrician's charge is modelled by the least-squares line

y = 85x + 70

, where

x

is the number of hours worked and

y

is the total charge in dollars. (a) Predict the charge for a

3

hour job. (b) Interpret the

y

-intercept in this context.

Show worked solution →

Predict the charge at $x = 3$ . Substitute $x = 3$ into the line:

y = 85 \times 3 + 70 = 255 + 70 = 325.

So the predicted charge is $325.

Interpret the intercept. The $y$ -intercept is $70$ . It is the predicted charge when $x = 0$ hours, so it represents a fixed call-out fee of $70 that applies before any time is billed. Here $x = 0$ is meaningful, so the intercept has a sensible real-world reading.

Check. The call-out fee of $70 plus $3$ hours at $85 each is $70 + 3 \times 85 = 70 + 255 = 325$ , matching the prediction. The charge is $325.

core4 marksFive students recorded weekly study hours

x

and exam mark

y

as the pairs

(1, 52)

(2, 55)

(3, 62)

(4, 68)

(5, 73)

. A calculator gives the least-squares line as

y = 5.5x + 45.5

. (a) State the gradient and intercept. (b) Predict the mark for a student who studies

3.5

hours. (c) Interpret the gradient.

Show worked solution →

State the gradient and intercept. From the calculator line $y = 5.5x + 45.5$ , the gradient is $5.5$ and the $y$ -intercept is $45.5$ .

Predict the mark at $x = 3.5$ . Substitute $x = 3.5$ :

y = 5.5 \times 3.5 + 45.5 = 19.25 + 45.5 = 64.75.

So the predicted mark is about $65$ (to the nearest whole mark).

Interpret the gradient. The gradient $5.5$ means that for each extra hour of weekly study, the exam mark increases by an average of $5.5$ marks.

Check. The line should pass through the mean point. The mean of the $x$ values is $3$ and the mean of the $y$ values is $62$ ; substituting $x = 3$ gives $5.5 \times 3 + 45.5 = 16.5 + 45.5 = 62$ , which matches $\bar{y}$ , so the line is correct.

core4 marksOver five days a kiosk recorded the maximum temperature

x

(in degrees Celsius) and cold-drink sales

y

(in hundreds of dollars):

(15, 22)

(18, 32)

(21, 37)

(24, 42)

(27, 47)

. A calculator gives the least-squares line

y = 2x - 6

. (a) Predict sales when the temperature is

20

degrees. (b) Predict sales at

26

degrees. (c) Interpret the gradient in context.

Show worked solution →

Predict sales at $x = 20$ . Substitute $x = 20$ into the line:

y = 2 \times 20 - 6 = 40 - 6 = 34,

so predicted sales are $34$ hundreds of dollars, that is $3400.

Predict sales at $x = 26$ . Substitute $x = 26$ :

y = 2 \times 26 - 6 = 52 - 6 = 46,

so predicted sales are $46$ hundreds of dollars, that is $4600.

Interpret the gradient. The gradient is $2$ . For each extra degree of maximum temperature, sales rise by an average of $2$ hundred dollars, that is $200 per degree.

Check. The mean temperature is $21$ and the mean sales figure is $36$ ; substituting $x = 21$ gives $2 \times 21 - 6 = 36$ , matching $\bar{y}$ , so the line passes through the mean point as it should.

core4 marksA used-car dealer records the age

x

(in years) and value

y

(in thousands of dollars) of six cars of one model:

(1, 24)

(2, 24)

(3, 21)

(4, 18)

(5, 15)

(6, 9)

. A calculator gives the least-squares line

y = -3x + 29

. (a) Predict the value of a car that is

4.5

years old. (b) Interpret the gradient. (c) Interpret the

y

-intercept, commenting on its reliability.

Show worked solution →

Predict the value at $x = 4.5$ . Substitute $x = 4.5$ into the line:

y = -3 \times 4.5 + 29 = -13.5 + 29 = 15.5,

so the predicted value is $15.5$ thousand dollars, that is $15,500.

Interpret the gradient: The gradient is $-3$ . For each extra year of age, the value drops by an average of $3$ thousand dollars, that is $3000 per year. The negative sign shows value falls as age rises.
Interpret the intercept: The intercept is $29$ , the predicted value at $x = 0$ , a brand-new car: about $29,000. The youngest car in the data is $1$ year old, so $x = 0$ is just outside the data range; the intercept is a mild extrapolation and should be treated as an estimate.
Check: The mean age is $3.5$ and the mean value is $18.5$ ; substituting $x = 3.5$ gives $-3 \times 3.5 + 29 = -10.5 + 29 = 18.5$ , matching $\bar{y}$ , so the line is correct.

core4 marksFor a set of bivariate data the mean of the

x

values is

12

and the mean of the

y

values is

48

. The least-squares regression line has gradient

2.5

. (a) Using the fact that the line passes through the point of means

(12, 48)

, find the

y

-intercept. (b) Write the equation of the line. (c) Predict

y

when

x = 20

Show worked solution →

Use the mean point to find the intercept. The least-squares line always passes through $(\bar{x}, \bar{y}) = (12, 48)$ . With $y = 2.5x + b$ , substitute the mean point:

48 = 2.5 \times 12 + b = 30 + b,

so $b = 48 - 30 = 18$ .

Write the equation. With gradient $2.5$ and intercept $18$ :

y = 2.5x + 18.

Predict at $x = 20$ . Substitute $x = 20$ :

y = 2.5 \times 20 + 18 = 50 + 18 = 68.

Check. Substituting the mean $x = 12$ back into the line gives $2.5 \times 12 + 18 = 30 + 18 = 48 = \bar{y}$ , confirming the intercept is correct. So $y = 68$ when $x = 20$ .

exam6 marksA researcher records the daily maximum temperature

x

(in degrees Celsius) and the number of visitors

y

(in hundreds) at a pool on six days:

(10, 38)

(20, 51)

(30, 61)

(40, 72)

(50, 79)

(60, 89)

. A calculator gives the least-squares line

y = x + 30

. (a) Predict the number of visitors when the temperature is

35

degrees. (b) Predict the number of visitors at

80

degrees. (c) State which of your two predictions is more reliable, and explain why.

Show worked solution →

Predict at $x = 35$ . Substitute $x = 35$ into the line:

y = 35 + 30 = 65,

so about $65$ hundred visitors, that is $6500$ visitors.

Predict at $x = 80$ . Substitute $x = 80$ :

y = 80 + 30 = 110,

so about $110$ hundred visitors, that is $11\,000$ visitors.

Compare reliability. The temperatures in the data run from $10$ to $60$ degrees. The prediction at $x = 35$ lies inside this range, so it is interpolation and is reliable. The prediction at $x = 80$ lies outside the range, so it is extrapolation; the linear trend may not continue (a temperature of $80$ degrees is also physically unrealistic), so that prediction is unreliable.

Check. The mean temperature is $35$ and the mean visitor figure is $65$ ; substituting $x = 35$ gives $35 + 30 = 65 = \bar{y}$ , so the line passes through the mean point and the $x = 35$ prediction is sound.

exam5 marksA least-squares regression line relating a worker's productivity score

y

to years of experience

x

y = 1.9x + 12.4

. (a) Predict the productivity score for a worker with

18

years of experience. (b) Use the line to estimate the experience needed for a productivity score of

50

, giving your answer to one decimal place. (c) The data covered

2

20

years of experience. Comment on whether part (a) is interpolation or extrapolation.

Show worked solution →

Predict the score at $x = 18$ . Substitute $x = 18$ into the line:

y = 1.9 \times 18 + 12.4 = 34.2 + 12.4 = 46.6,

so the predicted productivity score is about $46.6$ .

Solve for the experience giving $y = 50$ . Set $y = 50$ and solve for $x$ :

50 = 1.9x + 12.4 \implies 1.9x = 37.6 \implies x = \frac{37.6}{1.9} \approx 19.8.

So about $19.8$ years of experience is needed (to one decimal place).

Classify part (a). The data covered $2$ to $20$ years. Since $x = 18$ lies inside this range, part (a) is interpolation and is reliable.

Check. Substituting $x = 19.8$ back gives $1.9 \times 19.8 + 12.4 = 37.62 + 12.4 = 50.02 \approx 50$ , confirming the estimate. So the score at $18$ years is about $46.6$ , and $50$ needs about $19.8$ years.

exam6 marksA gym records each member's number of weekly sessions

x

and a fitness score

y

for seven members:

(2, 14)

(4, 19)

(6, 21)

(8, 26)

(10, 30)

(12, 33)

(14, 39)

. A calculator gives the least-squares line

y = 2x + 10

. (a) State the gradient and intercept. (b) Interpret the gradient in context. (c) Predict the fitness score for a member doing

9

sessions a week. (d) Predict the score for

25

sessions a week and explain why this prediction should be treated with caution.

Show worked solution →

State the gradient and intercept: From $y = 2x + 10$ , the gradient is $2$ and the $y$ -intercept is $10$ .
Interpret the gradient: For each extra weekly session, the fitness score rises by an average of $2$ points.
Predict at $x = 9$: Substitute $x = 9$ :

y = 2 \times 9 + 10 = 18 + 10 = 28,

so the predicted fitness score is $28$ .

Predict at $x = 25$ and caution. Substitute $x = 25$ :

y = 2 \times 25 + 10 = 50 + 10 = 60.

The data covered $2$ to $14$ sessions, so $x = 25$ is well outside the range. This is extrapolation: the linear trend may not continue (fitness gains often level off), so the predicted $60$ is unreliable.

Check. The mean number of sessions is $8$ and the mean score is $26$ ; substituting $x = 8$ gives $2 \times 8 + 10 = 26 = \bar{y}$ , so the line passes through the mean point and the in-range prediction is sound.

exam7 marksAn analyst records weekly income

x

(in hundreds of dollars) and weekly savings

y

(in dollars) for six people:

(6, 35)

(8, 50)

(10, 65)

(12, 77)

(14, 86)

(16, 95)

. A calculator gives the least-squares line

y = 6x + 2

. (a) Interpret the gradient, taking care with the units. (b) Predict the weekly savings of a person with an income of $1100 per week. (c) Predict the savings for an income of $2000 per week, and state whether this is interpolation or extrapolation. (d) Interpret the

y

-intercept and comment on its reliability.

Show worked solution →

Interpret the gradient. Here $x$ is income in hundreds of dollars, so $1$ unit of $x$ is $100 of weekly income. The gradient $6$ means that for each extra $100 of weekly income, weekly savings rise by an average of $6.

Predict at an income of $1100. An income of $1100 is $x = 11$ (hundreds). Substitute $x = 11$ :

y = 6 \times 11 + 2 = 66 + 2 = 68,

so the predicted weekly savings are $68.

Predict at an income of $2000. An income of $2000 is $x = 20$ . Substitute $x = 20$ :

y = 6 \times 20 + 2 = 120 + 2 = 122,

so about $122. The data covered incomes from $x = 6$ to $x = 16$ (that is $600 to $1600), so $x = 20$ is outside the range: this is extrapolation.

Interpret the intercept. The intercept is $2$ , the predicted savings at $x = 0$ (zero income): about $2. No one in the data has zero income, so $x = 0$ is well outside the range; the intercept is an extrapolation and should not be read literally.

Check. The mean income is $x = 11$ and the mean savings figure is $68$ ; substituting $x = 11$ gives $6 \times 11 + 2 = 68 = \bar{y}$ , so the line passes through the mean point and the $1100 prediction is sound.

exam6 marksFor

14

summer days, a researcher records the daily maximum temperature

x

(in degrees Celsius) and the region's peak electricity demand

y

(in hundreds of megawatts). No raw points are given, only these summary statistics:

\bar{x} = 20

\bar{y} = 60

s_x = 5

s_y = 15

, and the correlation coefficient

r = 0.8

. The temperatures in the data range from

12

30

degrees. (a) Using

m = r \times \dfrac{s_y}{s_x}

and the fact that the line passes through the point of means, find the equation of the least-squares regression line. (b) Interpret the gradient and the

y

-intercept in context. (c) Predict the peak demand on a

28

degree day and state, with a reason, whether the prediction is reliable.

Show worked solution →

Find the gradient from the correlation and spreads. The gradient of the least-squares line is the correlation coefficient scaled by the ratio of the standard deviations:

m = r \times \frac{s_y}{s_x} = 0.8 \times \frac{15}{5} = 0.8 \times 3 = 2.4.

Find the intercept using the point of means. The least-squares line always passes through $(\bar{x}, \bar{y}) = (20, 60)$ . With $y = 2.4x + b$ , substitute the mean point and solve for $b$ :

b = \bar{y} - m\bar{x} = 60 - 2.4 \times 20 = 60 - 48 = 12.

So the equation is $y = 2.4x + 12$ .

Interpret the gradient: The gradient is $2.4$ . For each extra degree of maximum temperature, peak demand rises by an average of $2.4$ hundred megawatts, that is $240$ MW per degree. The word "average" matters because the line is a best fit, not an exact rule for any single day.
Interpret the intercept: The intercept is $12$ , the predicted demand at $x = 0$ degrees: $12$ hundred megawatts, that is $1200$ MW. Since the data only covers $12$ to $30$ degrees, $x = 0$ is far outside that range, so this baseline is an extrapolation and should not be read literally.
Predict at $x = 28$ and judge reliability: Substitute $x = 28$ into the line:

y = 2.4 \times 28 + 12 = 67.2 + 12 = 79.2,

so about $79.2$ hundred megawatts, that is roughly $7920$ MW. Because $28$ lies inside the data range $12$ to $30$ , this is interpolation, and with a strong correlation of $r = 0.8$ the prediction is reliable.

Answer: the line is $y = 2.4x + 12$ ; demand rises by an average of $240$ MW per degree and the intercept of $1200$ MW at $0$ degrees is an unreliable extrapolation; the predicted demand at $28$ degrees is about $7920$ MW, and this interpolation is reliable.

exam6 marksA surf club studies

15

junior members. For each it records

x

, the number of training sessions attended per week, and

y

, a rescue-skills score out of

100

. The data give

\bar{x} = 8

\bar{y} = 55

s_x = 3

s_y = 12.5

, and

r = 0.84

, with attendance ranging from

2

14

sessions. (a) Comment on the strength and direction of the linear relationship. (b) Find the equation of the least-squares regression line using

m = r \times \dfrac{s_y}{s_x}

. (c) Predict the score of a member who trains

10

times a week. (d) Predict the score for

20

sessions a week and explain, referring to both the data range and the value of

r

, why this prediction is the least trustworthy of the two.

Show worked solution →

Comment on the relationship. Since $r = 0.84$ is positive and close to $1$ , there is a strong positive linear relationship: members who train more often tend to score higher. As a guide, $r^2 = 0.84^2 \approx 0.71$ , so about $71\%$ of the variation in scores is explained by the linear model.

Find the gradient and intercept. The gradient is

m = r \times \frac{s_y}{s_x} = 0.84 \times \frac{12.5}{3} = 0.84 \times 4.1\overline{6} = 3.5.

The line passes through $(\bar{x}, \bar{y}) = (8, 55)$ , so

b = \bar{y} - m\bar{x} = 55 - 3.5 \times 8 = 55 - 28 = 27,

giving the equation $y = 3.5x + 27$ .

Predict at $x = 10$ . Substitute $x = 10$ :

y = 3.5 \times 10 + 27 = 35 + 27 = 62,

so the predicted score is $62$ out of $100$ .

Predict at $x = 20$ and judge it. Substitute $x = 20$ :

y = 3.5 \times 20 + 27 = 70 + 27 = 97.

The attendance data runs only from $2$ to $14$ sessions, so $x = 20$ is well outside the range: this is extrapolation. The linear trend may not continue (scores cannot exceed $100$ and improvement tends to level off), and even where the model does apply the correlation is only $0.84$ , not perfect, so predictions carry error. The $x = 20$ prediction is therefore the least trustworthy, whereas $x = 10$ is interpolation inside the data and is reliable.

Answer: the relationship is strong and positive ( $r = 0.84$ ); the line is $y = 3.5x + 27$ ; the predicted score at $10$ sessions is $62$ , while the predicted $97$ at $20$ sessions is an unreliable extrapolation beyond the data range.

exam5 marksSix Australian households were surveyed. For each,

x

is the number of residents and

y

is the average daily electricity use in kilowatt-hours. The number of residents took the values

2

3

4

5

6

across five of the homes (the sixth is omitted here) and the corresponding electricity figures were

33

41

49

56

66

. These five pairs have

s_x = 1.5

s_y = 12

and

r = 0.9

. (a) Find the mean number of residents and the mean daily usage from the raw values. (b) Hence find the equation of the least-squares regression line, using

m = r \times \dfrac{s_y}{s_x}

and the point of means. (c) Predict the daily usage of a

3

person household and of a

10

person household, classifying each prediction as interpolation or extrapolation.

Show worked solution →

Find the two means from the raw values. Add the residents and divide by $5$ , then do the same for the usage figures:

\bar{x} = \frac{2 + 3 + 4 + 5 + 6}{5} = \frac{20}{5} = 4,

\bar{y} = \frac{33 + 41 + 49 + 56 + 66}{5} = \frac{245}{5} = 49.

Find the gradient and intercept. The gradient is

m = r \times \frac{s_y}{s_x} = 0.9 \times \frac{12}{1.5} = 0.9 \times 8 = 7.2.

The line passes through $(\bar{x}, \bar{y}) = (4, 49)$ , so

b = \bar{y} - m\bar{x} = 49 - 7.2 \times 4 = 49 - 28.8 = 20.2,

giving the equation $y = 7.2x + 20.2$ .

Predict at $x = 3$ . Substitute $x = 3$ :

y = 7.2 \times 3 + 20.2 = 21.6 + 20.2 = 41.8,

so about $41.8$ kWh per day. Since $3$ lies inside the data range $2$ to $6$ residents, this is interpolation.

Predict at $x = 10$ . Substitute $x = 10$ :

y = 7.2 \times 10 + 20.2 = 72 + 20.2 = 92.2,

so about $92.2$ kWh per day. Since $10$ is outside the range $2$ to $6$ , this is extrapolation and should be treated with caution.

Answer: the means are $\bar{x} = 4$ residents and $\bar{y} = 49$ kWh; the line is $y = 7.2x + 20.2$ ; the predicted usage is about $41.8$ kWh for $3$ residents (interpolation, reliable) and about $92.2$ kWh for $10$ residents (extrapolation, unreliable).

What this dot point is asking

The answer

The least-squares regression line

Finding the line on a calculator

Build and use the line, stage by stage

Predicting yyy from xxx

Interpreting the gradient

Interpreting the yyy-intercept

Interpolation versus extrapolation

When to use the line

How exam questions ask about the regression line

Edge cases worth knowing

Exam-style practice questions

Practice questions

Predicting $y$ from $x$

Interpreting the $y$ -intercept