Linear models, R² and causation

Height depends on the number of days, so $d$ is the explanatory variable ($x$ list) and $h$ is the response ($y$ list). The calculator gives a slope of 0.6147 and an intercept of 5.7056:

$$h = 0.61d + 5.71 \approx 0.6d + 5.7$$

(A) has the right numbers with $d$ and $h$ swapped, which is a different line. (C) and (D) swap the slope and the intercept. 51.9% of students chose (B). (A) and (D) each drew about 20%.

a) Time is the explanatory variable, so $t$ goes in the $x$ list and $n$ in the $y$ list:

$$n = 42.6t + 55.4$$

1 mark · correctly determines the equation of the least-squares line

The guide says “Accept variables $n$ and $t$ only”, so the same numbers written as $y = 42.6x + 55.4$ did not earn the mark. The 2022 subject report used this question to show the equation stated in the given variables.

b)

$$\begin{aligned} n &= 42.6(21) + 55.4 \\ &= 950 \end{aligned}$$

1 mark · substitutes into equation from Question 16a)

1 mark · predicts number of sales

Month 21 is fifteen months past the data, which only covers months 1 to 6. That is extrapolation. This question did not ask for a comment, but an “evaluate” question would.

a) Define the variables first. That is a mark on its own.

Let $x$ = the number of years since 2014 and $y$ = the business’s annual profit in thousands of dollars.

1 mark · correctly defines the variables

So 2015 is $x = 1$ and 2020 is $x = 6$. The calculator gives:

$$y = 4.286x + 34.267$$

1 mark · correctly determines the equation of the least-squares line

The guide accepted other variable names such as $t$ and $P$, rounded parameters, and the line with $x$ as the year itself, $y = 4.286x - 8597.162$.

b) 2021 is $x = 7$:

$$\begin{aligned} y &= 4.286 \times 7 + 34.267 \\ &= 64.269 \end{aligned}$$

1 mark · correctly determines the $x$ value

64.269 thousand dollars is $64 269, which is $64 300 to the nearest hundred dollars.

1 mark · determines profit

If you use the year itself as $x$, keep the full slope in your calculator. With the rounded slope, $4.286 \times 2021 - 8597.162 = 64.844$, which is $64 800, because a slope error of about 0.0003 gets multiplied by 2021. With full precision both forms give 64.267.

a) This one is two points, not a calculator fit.

$$\begin{aligned} m &= \frac{40 - 20}{7 - 2} \\ &= 4 \end{aligned}$$

1 mark · correctly determines the slope

$$\begin{aligned} p - 20 &= 4(n - 2) \\ p &= 4n + 12 \end{aligned}$$

1 mark · determines equation of least-squares line

Again the guide accepted “variables $p$ and $n$ only”. Check the intercept on the graph: the line crosses the vertical axis at 12.

b)

$$\begin{aligned} p &= 4(15) + 12 \\ &= 72 \end{aligned}$$

1 mark · substitutes into equation from Question 23a)

1 mark · predicts hourly pay, including units

The unit is part of the mark, so write $72 per hour. The data stops before 10 years, so 15 years is an extrapolation.

$$\begin{aligned} m &= 0.926 \times \frac{41.7}{2.3} \\ &= 16.79 \end{aligned}$$

$$\begin{aligned} c &= 68.6 - 16.79 \times 5.2 \\ &= -18.70 \end{aligned}$$

(A) swaps the means in $c$, $5.2 - 16.79 \times 68.6$. (C) flips the standard deviations, $0.926 \times \frac{2.3}{41.7} = 0.05$. (D) does both. QCAA did not publish option percentages for 2020.

Step 1 · Notice what is missing. You have $\bar{y}$, $s_y$ and $r$, but not $\bar{x}$ or $s_x$. The $x$ values are the years of operation, 1 to 10, so enter them in your calculator.

$$\bar{x} = 5.5 \qquad s_x = 3.02765$$

1 mark · correctly determines $\bar{x}$ and $s_x$

The guide also accepted $x = 0, 1, \ldots, 9$. Then $\bar{x} = 4.5$, the intercept is 5192.05, the 11th year is $x = 10$, and the prediction is the same 15 120.83.

Step 2 · Slope. The guide writes the line as $y = a + bx$, so its $b$ is the slope and its $a$ is the intercept.

$$\begin{aligned} b &= r\frac{s_y}{s_x} \\ &= 0.9987 \times \frac{3010}{3.02765} \\ &= 992.878 \end{aligned}$$

1 mark · determines $b$

Step 3 · Intercept.

$$\begin{aligned} a &= \bar{y} - b\bar{x} \\ &= 9660 - 992.878 \times 5.5 \\ &= 4199.17 \end{aligned}$$

1 mark · determines $a$

Step 4 · Predict for the 11th year.

$$\begin{aligned} y &= 4199.17 + 992.878(11) \\ &= 15\,120.83 \end{aligned}$$

Predicted profit in the 11th year is $15 121.

1 mark · determines 11th year profit to the nearest dollar

1 mark · shows logical organisation communicating key steps

In a show-that the answer is printed in the question, so every mark is for the working. Put $\bar{x}$, $s_x$, the slope and the intercept on the page. Keep calculator precision: with $s_x = 3.03$ you get $15 117 and your working no longer shows the result. The 2022 report listed the same five steps as the effective responses.

This runs the line backwards. You cannot fit a line with two unknowns in the data, but you know one point the line must pass through: $(\bar{x}, \bar{y})$.

Step 1 · Read the slope and intercept. The guide calls them $b$ and $a$.

$$b = 2.1875 \qquad a = 0.0625$$

1 mark · correctly identifies the $a$ and $b$ values

Step 2 · The mean of $x$.

$$\bar{x} = \frac{4 + 3 + 8 + 4 + 6}{5} = 5$$

1 mark · correctly determines $\bar{x}$

Step 3 · The mean of $y$, from $a = \bar{y} - b\bar{x}$.

$$\begin{aligned} 0.0625 &= \bar{y} - 2.1875 \times 5 \\ \bar{y} &= 11 \end{aligned}$$

1 mark · determines $\bar{y}$

Step 4 · The missing total.

$$\begin{aligned} 11 &= \frac{4 + 8 + p + q + 16}{5} \\ 55 &= 28 + p + q \\ p + q &= 27 \end{aligned}$$

1 mark · determines sum of missing values

Step 5 · Use the difference of 3. If $q = p + 3$:

$$\begin{aligned} p + p + 3 &= 27 \\ p &= 12 \\ q &= 15 \end{aligned}$$

1 mark · determines values for $p$ and $q$

1 mark · shows logical organisation, communicating key steps

The guide also accepted $p = 15$ and $q = 12$, and trial and error for $p + q = 27$. You can check which pair fits the whole line: with $p = 12$ and $q = 15$ the calculator returns exactly $y = 2.1875x + 0.0625$ and $r = 0.875$.

(C) is the only plot with points scattered above and below the axis with no pattern.

(A) dips then rises, a curve. (B) has every residual above zero. (D) climbs steadily, a trend the line has missed. QCAA did not publish option percentages for 2020.

(A) scatters above and below zero with no pattern.

(B) curves upwards, (C) falls steadily, and (D) is a U shape. 43.3% of students chose (A) and 39.8% chose (C). (C) looks tidy because its points nearly form a line, but residuals that form a line mean the model has missed a trend.

The 2021 report called this “a reverse understanding of the use of a residual plot”. Usually you go from data to residuals. Here you go from residuals back to the line. The guide writes the line as $y = bx + a$ and uses $y_A$ for actual and $y_P$ for predicted.

Step 1 · The 31-year-old’s predicted value. The plot shows a residual of $-0.75$ at age 31.

$$\begin{aligned} y_A - y_P &= -0.75 \\ 119 - y_P &= -0.75 \\ y_P &= 119.75 \end{aligned}$$

1 mark · correctly determines the $y_P$ value

Step 2 · Slope.

$$\begin{aligned} b &= r\frac{s_y}{s_x} \\ &= 0.875 \times \frac{6}{4} \\ &= 1.3125 \end{aligned}$$

1 mark · correctly determines the $b$ value

Step 3 · Intercept. The predicted point (31, 119.75) is on the line:

$$\begin{aligned} 119.75 &= 1.3125 \times 31 + a \\ a &= 79.0625 \end{aligned}$$

1 mark · determines $a$ value

Step 4 · Predicted value at 40.

$$\begin{aligned} y &= 1.3125 \times 40 + 79.0625 \\ &= 131.5625 \end{aligned}$$

1 mark · determines predicted $y$ value for oldest person

Step 5 · Add the residual back. The residual at age 40 is about 1.4:

$$\begin{aligned} y &= 131.5625 + 1.4 \\ &= 132.9625 \end{aligned}$$

The oldest person in the sample has a systolic blood pressure of 133.

1 mark · determines actual systolic blood pressure as a whole number

1 mark · shows logical organisation communicating key steps

The guide accepted a residual from 1.3 to 1.5, and every value in that range gives 133. It also allowed follow-through from an earlier error.

a) Trend: the long-term trend is positive, because the amount of rainfall generally increases as time increases. Seasonality: the data is seasonal, with a high 4th quarter every year.

1 mark · appropriately describes the long-term trend

1 mark · appropriately describes the seasonality

Trend and seasonality are taught in Topic guide 4.

b) The intercept is the prediction at $x = 0$. Since $x = 1$ is the first quarter of 2016, $x = 0$ is the quarter before it, the 4th quarter of 2015.

$y$-intercept: the model predicts that 156.5 mm of rainfall was falling in the 4th quarter of 2015.

1 mark · appropriately interprets the $y$-intercept

Slope: on average, an additional 1.763 mm of rainfall fell each quarter.

1 mark · appropriately interprets the slope

The guide accepted other wording, such as “a rate of rainfall of 1.763 mm/quarter”. Both readings carry the unit and the time step. “The slope is 1.763” on its own interprets nothing.

a) $0.997 \times 100\% = 99.7\%$. The working for this part is in Topic guide 2.

1 mark · correctly states the percentage

b) slope = 49

1 mark · correctly identifies the slope value as 49

Australia’s average weekly earnings increased each year by $49.

1 mark · correctly interprets the slope as an increase per year

The 2025 report used this part to show the difference between stating the slope and interpreting it as a change in one variable for each unit of the other.

c) Substitute $x = 2035$:

$$\begin{aligned} y &= 49(2035) - 97\,140 \\ &= 2575 \end{aligned}$$

1 mark · correctly substitutes into the least-squares line equation

1 mark · predicts value

The predicted average weekly earnings in 2035 are $2575. The data runs from 2013 to 2023, so 2035 is twelve years past it. That is extrapolation. No mark asked for the comment here.

Step 1 · Use $R^2$ to place Leigh. 0.95 < 0.96, so age explains a higher percentage of the account balance variation for industry B. Leigh works in industry B, and Tony, aged 50, works in industry A.

1 mark · correctly identifies dataset for which age explains a higher percentage of the account balance variation

Step 2 · Two models. Let $x$ = age and $y$ = account balance in dollars. The calculator gives:

$$\begin{aligned} \text{Industry A: } y &= 7910x - 205\,520 \\ \text{Industry B: } y &= 9570x - 243\,440 \end{aligned}$$

1 mark · correctly determines linear model for age vs account balance for industry A data

1 mark · correctly determines linear model for age vs account balance for industry B data

Step 3 · Predict each balance.

$$\begin{aligned} \text{Leigh (B, } x = 40\text{): } y &= 9570(40) - 243\,440 = 139\,360 \\ \text{Tony (A, } x = 50\text{): } y &= 7910(50) - 205\,520 = 189\,980 \end{aligned}$$

1 mark · substitutes $x = 40$ into appropriate equation and calculates Leigh’s current account balance

1 mark · substitutes $x = 50$ into appropriate equation and calculates Tony’s current account balance

Step 4 · The difference.

$$189\,980 - 139\,360 = 50\,620$$

The difference in account balances for Leigh and Tony is predicted to be $50 620.

1 mark · calculates difference in current account balances for Leigh and Tony

1 mark · shows logical organisation communicating key steps

The 2023 report highlighted a response that stated each equation in defined variables before substituting the ages. Both ages sit inside the data (22 to 72), so these are interpolations.

$$\begin{aligned} t &= 26.781 + 12.974(5) \\ &= 91.651 \end{aligned}$$

91.651 rounds to 92. (C) cuts off the decimals instead of rounding. (A) and (B) come from $12.974 \times 5 = 64.87$ with the intercept left off. 80.6% of students chose (D).

The data runs from 2011 to 2019. 2018 is inside that range, so any estimate for it is interpolation. The points either side, 2017 and 2019, are both about 2.7 million.

66.0% of students chose (A). 29.7% chose (B), the right value with the wrong word. A missing point inside the range is still interpolation.

Substitute $x = 25$ into each line, then check where each data set stops.

Only (A) gives 20, and its data stops at about 20 km, so 25 km is an extrapolation.

39.7% of students chose (A). More, 46.4%, chose (B). Plot (B) predicts 25 high-rises at 25 km, and its data reaches 25 km, so it fails on both counts.

a) Go up from 9 days to the line and across: 11 cm.

1 mark · provides the correct value including units

The guide accepted 10.8 to 11.2 cm.

b) Interpolation. The data covers days 1 to 24, and day 9 is inside that.

1 mark · correctly classifies the prediction as interpolation

c) The least-squares line does suggest that at 29 days the seedling will be 32 cm high. However, the data values are levelling off at about 25 cm, so extrapolating to day 29 is unwise.

1 mark · identifies that the least-squares line supports the statement

1 mark · identifies potential dangers of extrapolation

The guide accepted one danger. Two moves earned both marks: agree with what the line says, then explain why you should not trust it there. The levelling off shows in the last week of data, where the points sit below the line.

Step 1 · The model. Latitude $x$ is the explanatory variable. The calculator gives a slope of $-0.227$ and an intercept of 16.7:

$$y = 16.7 - 0.227x$$

1 mark · correctly determines the values for the slope and vertical axis intercept

1 mark · determines least-squares line equation

Step 2 · Predict for Hobart.

$$\begin{aligned} y &= 16.7 - 0.227(43) \\ &= 6.9 \end{aligned}$$

1 mark · substitutes latitude into least-squares line equation

The predicted ultraviolet index is 7.

1 mark · predicts UV index

Step 3 · Compare with Table 2, then conclude. An ultraviolet index of 7 corresponds to high, so the notification is reasonable.

1 mark · provides appropriate statement of reasonableness linked to prior working

With full calculator precision the prediction is 6.93, which also gives 7. The guide accepted non-integer values and an alternative approach: substitute for $y$ and solve for $x$.

Hobart, at 43° S, is outside the data, which runs from 12° S to 38° S. That makes it an extrapolation, worth a sentence in your conclusion even though the guide did not require one.

The stem does not tell you to add the profits. “Merged into one company” is the cue: C’s profit is B1 plus B2.

Step 1 · Combine the businesses.

1 mark · correctly sums B1 and B2

The guide also accepted summing all the raw values, or summing two separate models.

Step 2 · Fit the line. Let $x$ = the number of years after the end of 2025 and $y$ = C’s annual profit in dollars, so 2022 is $x = -3$ and 2025 is $x = 0$. The least-squares line for the four sums is:

$$y = 200\,000x + 1\,140\,000$$

1 mark · determines a model

The marking guide’s sample uses the same slope but starts its line at the actual 2025 profit, $y = 200\,000x + 1\,200\,000$. The guide accepted alternative approaches, including modelling B1 and B2 separately, which gives the same prediction as the line above.

Step 3 · Predict for 2040. $2040 - 2025 = 15$, so substitute $x = 15$:

$$\begin{aligned} y &= 200\,000(15) + 1\,140\,000 \\ &= 4\,140\,000 \end{aligned}$$

1 mark · determines suitable substitution

1 mark · determines C’s predicted annual profit at end of 2040

With the guide’s line the prediction is $4 200 000. The guide also accepted 4200, in thousands of dollars.

Step 4 · Compare, then decide.

$$4\,140\,000 > 4\,000\,000$$

C’s predicted annual profit at the end of 2040 is more than $4 million, so the director’s prediction is reasonable.

1 mark · provides decision

1 mark · shows logical organisation

The decision mark “can only be awarded if prior working supports decision”, so write the comparison. Either model gives a value above $4 million, so the decision is the same.

A stronger answer adds the limitation: 2040 is fifteen years past four years of data, so this is a long extrapolation. The 2025 report described the effective response as “a justified decision and a suitable reasonableness statement, supported by a comparison between the student’s calculated result and the given prediction”.

Both quantities increase together, so the association is positive. Nothing in the observation shows one causes the other. Both are more likely driven by a third variable, hot weather.

(A) and (B) claim causation. (C) has the wrong direction. QCAA did not publish option percentages for 2020.

Causal means a change in the explanatory variable produces a change in the response variable, so the response depends on it. That is always true of a causal relationship.

(A) and (B) describe the non-causal explanation, a third variable. (C) fails because a causal relationship can be negative. 82.5% of students chose (D).

A confounding variable is another variable, other than the explanatory variable, that influences the response variable, and usually the explanatory variable as well.

(A) describes a discrete variable. (B) describes a controlled variable, one held constant. (C) is close to a description of the explanatory variable itself. Only 37.5% of students chose (D), and 28.1% chose (B).

For (A) to (C) there is an obvious mechanism: more children need more schools, phone use drains the battery, customers buy the bread. For (D) there is none. How tall drivers are cannot change how many cars are at a petrol station, so a strong association there would most likely be coincidence.

42.2% of students chose (D). Each of the other three options drew 18% to 20%.

a) Pearson’s correlation coefficient is 0.986, so it is a strong association.

1 mark · correctly identifies the value of Pearson’s correlation coefficient

1 mark · correctly describes the strength as strong

Reading strength from $r$ is a Topic 1 skill, worked in full in Topic guide 2.

b) A confounding variable is another variable that has a similar effect on the response variable.

1 mark · correctly defines confounding variable

Accepted equivalents included “another variable that affects both the explanatory (independent) and response (dependent) variables” and “another variable that suggests there is a causal relationship between the explanatory (independent) and response (dependent) variables when there is not”. The 2024 report printed several definitions that scored.

c) Air temperature, $x$, and number of vehicles, $y$, are both lower at earlier and later times of day, when train users’ vehicles are arriving at and leaving the station, and both higher in the middle of the day, when train users are at work.

1 mark · correctly explains why $t$ could be a confounding variable

The explanation links time of day to both variables. That link is what earns the mark.