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Statistics, STD2 2026 HSC 22

Researchers were interested in whether the critics' scores for movies could be used to predict the general public's scores for movies.

A random sample of movies was selected, and the average score from the critics and the average score from the public were recorded.

A scatterplot of the dataset and the least-squares regression line are shown.
 

  1. What was the lowest score given by the public for a movie, correct to 1 decimal place?   (1 mark)
  2. The gradient of the least-squares regression line for the dataset is 0.54.
  3. Interpret the gradient in the context of the dataset.   (2 marks)
  4. A new data point is included in the dataset. The critics' score is 7.2 and the public's score is 6.
  5. Identify ONE effect this would have on the least-squares regression line.   (1 mark)
Show Answers Only

a.    \(6.1\)

 
b.    \(\text{Interpretation of gradient = 0.54:}\)

\(\text{For each 1-point increase in the critics’ score, the predicted public’s}\)

\(\text{score increases by 0.54 points.}\)
 

c.    \(\text{The new point lies below the LSRL.}\)

\(\text{Possible effects could include:}\)

\(\text{- lowering the LSRL at each critic’s score.}\)

\(\text{- increasing the LSRL gradient (its LHS will be relatively lower than its RHS).}\)

Show Worked Solution
a.     \(\text{Lowest public score}\) \(=\ \text{$y$-value of the lowest plotted point}\)
    \(=6.1\)

 
b.    \(\text{Interpretation of gradient = 0.54:}\)

\(\text{For each 1-point increase in the critics’ score, the predicted public’s}\)

\(\text{score increases by 0.54 points.}\)
 

c.    \(\text{The new point lies below the LSRL.}\)

\(\text{Possible effects could include:}\)

\(\text{- lowering the LSRL at each critic’s score.}\)

\(\text{- increasing the LSRL gradient (its LHS will be relatively lower than its RHS).}\)

Statistics, STD2 S4 2025 HSC 25

In a research study, participants were asked to record the number of minutes they spent watching television and the number of minutes they spent exercising each day over a period of 3 months. The averages for each participant were recorded and graphed.
 

  1. Describe the bivariate dataset in terms of its form and direction.   (2 marks)
  2. Form:  ..................................................................
  3. Direction:  ............................................................

The equation of the least-squares regression line for this dataset is

\(y=64.3-0.7 x\)

  1. Interpret the values of the slope and \(y\)-intercept of the regression line in the context of this dataset.   (2 marks)

  2. Jo spends an average of 42 minutes per day watching television.
  3. Use the equation of the regression line to determine how many minutes on average Jo is expected to exercise each day.   (1 mark)

  4. Explain why it is NOT appropriate to extrapolate the regression line to predict the average number of minutes of exercise per day for someone who watches an average of 2 hours of television per day.   (1 mark)

Show Answers Only

a.    \(\text{Form: Linear. Direction: Negative}\)

b.    \(\text{Slope}=-0.7\)

This means that for each added minute of watching television per day, a participant, on average, will exercise for 0.7 minutes less.

\(y \text{-intercept}=64.3\)

If someone watches no television, the LSRL predicts they will exercise for 64.3 minutes per day.

c.    Jo is expected to exercise for 34.9 minutes.

d.    \(\text{At} \ \  x=120\ \text{(2 hours),} \ \ y=64.3-0.7 \times 120=-19.7\)

The model predicts a negative value for time spent exercising, which is not possible.

Show Worked Solution

a.    Form: Linear

Direction: Negative

Mean mark (a) 51%.

b.    \(\text{Slope}=-0.7\)

This means that for each added minute of watching television per day, a participant, on average, will exercise for 0.7 minutes less.

\(y \text{-intercept}=64.3\)

If someone watches no television, the LSRL predicts they will exercise for 64.3 minutes per day.

♦♦♦ Mean mark (b) 20%.

c.    \(\text{At} \ \ x=42:\)

\(y=64.3-0.7 \times 42=34.9\)

\(\therefore\) Jo is expected to exercise for 34.9 minutes.
 

d.    \(\text{At} \ \  x=120\ \text{(2 hours),} \ \ y=64.3-0.7 \times 120=-19.7\)

The model predicts a negative value for time spent exercising, which is not possible.

♦ Mean mark (d) 45%.

Statistics, STD2 S4 2024 HSC 30

A researcher is studying anacondas (a type of snake).

A dataset recording the age (in years) and length (in cm) of female and male anacondas is displayed on the graph.

Anacondas reach maturity at about 4 years of age.
 

Write THREE observations about anacondas that may be made from the scatterplot. (Note: No calculations are required.)   (3 marks)

Show Answers Only

\(\text{Answers could include three of the following:}\)

\(\rightarrow\ \text{Female anacondas are longer than males of the equivalent age.}\)

\(\rightarrow\ \text{Female anacondas grow more quickly than male anacondas from birth until}\)

\(\text{maturity which can be seen by the steeper gradient of the LOBF for each dataset}\)

\(\text{over this period.}\)

\(\rightarrow\ \text{Female anacondas continue to grow to at least 10 years of age, well past their}\)

\(\text{age of maturity at 4 years of age.}\)

\(\rightarrow\ \text{Male anacondas’ growth slows noticeably and flattens out once they hit their}\)

\(\text{age of maturity at 4 years old.}\)

Show Worked Solution

\(\text{Answers could include three of the following:}\)

\(\rightarrow\ \text{Female anacondas are longer than males of the equivalent age.}\)

\(\rightarrow\ \text{Female anacondas grow more quickly than male anacondas from birth until}\)

\(\text{maturity which can be seen by the steeper gradient of the LOBF for each dataset}\)

\(\text{over this period.}\)

\(\rightarrow\ \text{Female anacondas continue to grow to at least 10 years of age, well past their}\)

\(\text{age of maturity at 4 years of age.}\)

\(\rightarrow\ \text{Male anacondas’ growth slows noticeably and flattens out once they hit their}\)

\(\text{age of maturity at 4 years old.}\)

Statistics, STD2 S4 2022 HSC 35

Jo is researching the relationship between the ages of teenage characters in television series and the ages of actors playing these characters.

After collecting the data, Jo finds that the correlation coefficient is 0.4564.

A scatterplot showing the data is drawn. The line of best fit with equation  `y=-7.51+1.85 x`, is also drawn.
 


 

Describe and interpret the data and other information provided, with reference to the context given.   (4 marks)

Show Answers Only

`text{Correlation coefficient}\ (r) = 0.4564`

      • `text{Moderate and positive correlation}`

`text{Gradient of LOBF}\ = 1.85`

      • `text{On average, each extra year of a character’s age results}`
        `text{in the actor being 1.85 years older.}`

`text{Mode of data set = 15 years}`
  

`text{Limitations}`

      • `text{Data set is very restricted with just a 4 year range of}`
        `text{character ages.}`
      • `text{LOBF not useful when extrapolated to the left as it drops}`
        `text{below zero (on y-axis).}`
      • `text{Relationship describes correlation only, not causation.}`
Show Worked Solution

`text{Correlation coefficient}\ (r) = 0.4564`

      • `text{Moderate and positive correlation}`

`text{Gradient of LOBF}\ = 1.85`

      • `text{On average, each extra year of a character’s age results}`
        `text{in the actor being 1.85 years older.}`

`text{Mode of data set = 15 years}`
  

`text{Limitations}`

      • `text{Data set is very restricted with just a 4 year range of}`
        `text{character ages.}`
      • `text{LOBF not useful when extrapolated to the left as it drops}`
        `text{below zero (on y-axis).}`
      • `text{Relationship describes correlation only, not causation.}`

♦♦ Mean mark 30%.

Statistics, STD2 S4 2022 HSC 12 MC

For a particular course, the recorded data show a relationship between the number of hours of study per week and the marks achieved out of 100 .

A least-squares regression line is fitted to this dataset. The equation of this line is given by

`M=20+3 H,`

where `M` is the predicted mark and `H` is the number of hours of study per week.

Based on this regression equation, which of the following is correct regarding the predicted mark of a student?

  1. It will be 3 for zero hours of study per week.
  2. It will be 20 for zero hours of study per week.
  3. It will increase by 20 for every additional hour of study per week.
  4. It will increase by 1 for every 3 additional hours of study per week.
Show Answers Only

`B`

Show Worked Solution

`text{Consider Option}\ B:`

`text{If zero hours of study are done per week}\ \ → \ \ H=0`

`:. M=20+(3 xx 0) = 20`

`=>B`

Statistics, STD2 S4 2021 HSC 33

For a sample of 17 inland towns in Australia, the height above sea level, `x` (metres), and the average maximum daily temperature, `y` (°C), were recorded.

The graph shows the data as well as a regression line.
 

     
 

The equation of the regression line is  `y = 29.2-0.011x`.

The correlation coefficient is  `r =-0.494`.

  1. i.  By using the equation of the regression line, predict the average maximum daily temperature, in degrees Celsius, for a town that is 540 m above sea level. Give your answer correct to one decimal place.   (1 mark)

  2. ii. The gradient of the regression line is −0.011. Interpret the value of this gradient in the given context.   (2 marks)

  3. The graph below shows the relationship between the latitude, `x` (degrees south), and the average maximum daily temperature, `y` (°C), for the same 17 towns, as well as a regression line.
     
     
         
     
    The equation of the regression line is  `y = 45.6-0.683x`.
  4. The correlation coefficient is  `r =-0.897`.
  5. Another inland town in Australia is 540 m above sea level. Its latitude is 28 degrees south.
  6. Which measurement, height above sea level or latitude, would be better to use to predict this town’s average maximum daily temperature? Give a reason for your answer.   (1 mark)

Show Answers Only

a.    i.  `23.3°text(C)`

ii. `text(See Worked Solutions)`

b.    `text(Latitude. Correlation coefficient shows a stronger relationship.)`

Show Worked Solution
a.i.     `y` `=29.2-0.011(540)`
    `=23.26`
    `=23.3°text{C  (1 d.p.)`
♦♦ Mean mark (a.ii.) 28%.

 
a.ii. `text(On average, the average maximum daily temperature of)`

`text(inland towns drops by 0.011 of a degree for every metre)`

`text(above sea level the town is situated.)`
 

♦♦♦ Mean mark (b) 18%.

b.    `text(The correlation co-efficient of the regression line using)`

`text(latitude is significantly stronger than the equivalent)`

`text(co-efficient for the regression line using height above sea)`

`text(level.)`

`:.\ text(The equation using latitude is preferred.)`

Statistics, STD2 S4 2021 HSC 28

A salesperson is interested in the relationship between the number of bottles of lemonade sold per day and the number of hours of sunshine on the day.

The diagram shows the dataset used in the investigation and the least-squares regression line.


 

  1. Find the equation of the least-squares regression line relating to the dataset.   (2 marks)

  2. Suppose a sixth data point was collected on a day which had 10 hours of sunshine. On that day 45 bottles of lemonade were sold.
  3. What would happen to the gradient found in part (a)?   (1 mark)

Show Answers Only

a.    `y=3.2x+2`

b.    `text{Gradient would increase (steepen).}`

Show Worked Solution

a.    `text(Method 1)`

♦ Mean mark part (a) 37%.

`text{Input data points (in Stats Mode “Ax + B”):}`

`(2,8), (3, 11), (5, 19), (6, 22), (9, 30)`

`=>\ y=3.2x + 2`
 

`text(Method 2)`
 

`text{Find gradient using (0, 2) and (5, 18)}:`

`m=(18-2)/(5-0) = 3.2,\ \ ytext(-intercept)\ = 2`

`:.\ text(Equation:)\ y=3.2x + 2`

Mean mark part (b) 53%.

 
 b.    `text(Method 1)`

`text{Add (10, 45) to the data set in Stats Mode above:}`

`text(Gradient increases to 4.1.)`
 

`text(Method 2)`

`text{Data point (10, 45) lies above the regression line.`

`:.\ text{Gradient would increase (steepen).}`

Statistics, 2ADV S2 EQ-Bank 22

The arm spans (in cm) and heights (in cm) for a group of 13 boys have been measured. The results are displayed in the table below.
 

CORE, FUR2 2008 VCAA 4

The aim is to find a linear equation that allows arm span to be predicted from height.

  1. What will be the independent variable in the equation?   (1 mark)

  2. Assuming a linear association, determine the equation of the least squares regression line that enables arm span to be predicted from height. Write this equation in terms of the variables arm span and height. Give the coefficients correct to two decimal places.   (2 marks)

  3. Using the equation that you have determined in part b., interpret the slope of the least squares regression line in terms of the variables height and arm span.   (1 mark)

Show Answers Only

a.    `text(Height)`

b.    `text(Arm span)\ = 1.09 xx text(height)-15.63`

c.    `text(On average, arm span increases by 1.09 cm)`

`text(for each 1 cm increase in height.)`

Show Worked Solution

a.    `text(Height)`

COMMENT: Calculator skills for finding the least squares regression line were required in NESA sample exam – know this critical skill well!

 

b.    `text(By calculator,)`

`text(Arm span)\ = 1.09 xx text(height)-15.63`
  

c.    `text(On average, arm span increases by 1.09 cm)`

`text(for each 1 cm increase in height.)`

Algebra, STD2 A2 2016 HSC 29e

The graph shows the life expectancy of people born between 1900 and 2000.
 


  1. According to the graph, what is the life expectancy of a person born in 1932?   (1 mark)

  2. With reference to the value of the gradient, explain the meaning of the gradient in this context.   (2 marks)

Show Answers Only

a.    `text(68 years)`

b.    `text(After 1900, life expectancy increases 0.25 years for each later year someone is born.)`

Show Worked Solution

a.    \(\text{68 years}\)

b.    \(\text{Using (1900,60), (1980,80):}\)

\(\text{Gradient}\) \(= \dfrac{y_2-y_1}{x_2-x_1}\)
  \(= \dfrac{80-60}{1980-1900}\)
  \(= 0.25\)

 
\(\text{After 1900, life expectancy increases by 0.25 years for}\)

\(\text{each year later that someone is born.}\)

♦♦ Mean mark (b) 33%.

Statistics, STD2 S4 2013 HSC 28b

Ahmed collected data on the age (`a`) and height (`h`) of males aged 11 to 16 years.

He created a scatterplot of the data and constructed a line of best fit to model the relationship between the age and height of males.
 

  1. Determine the gradient of the line of best fit shown on the graph.   (1 mark)

  2. Explain the meaning of the gradient in the context of the data.   (1 mark)

  3. Determine the equation of the line of best fit shown on the graph.   (2 marks)

  4. Use the line of best fit to predict the height of a typical 17-year-old male.   (1 mark)

  5. Why would this model not be useful for predicting the height of a typical 45-year-old male?   (1 mark)

Show Answers Only

a.    `text(Gradient = 6)`

b.    `text(Males should grow 6 cm per year between the ages 11-16.)`

c.    `h = 6a + 80`

d.    `text(182 cm)`

e.    `text(People slow and eventually stop growing after they become adults.)`

Show Worked Solution

a.    `text{Gradient}\ =(176-146)/(16-11)=30/5=6`
 

b.    `text{Males should grow 6cm per year between the ages 11–16.}`
 

♦♦ Mean marks of 38%, 26% and 25% respectively for parts (a)-(c).

c.    `text{Gradient = 6,  Passes through (11, 146)}`

`y-y_1` `=m(x-x_1)`
`h-146` `=6(a-11)`
`h` `=6a-66+146`
  `=6a + 80`

 

d.   `text{Substitute}\ \ a=17\ \ \text{into equation from part (c):}`

`h=(6 xx 17) +80=182`

`:.\ text{A typical 17 year old is expected to be 182cm.}`
 

e.    `text(People slow and eventually stop growing after they become adults.)`