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

Show Worked Solution

a.    \(\text{Form: Linear}\)

\(\text{Direction: Negative}\)
 

b.    \(\text{Slope}=-0.7\)

\(\text{This means that for each added minute of watching television per day, a participant, on average,}\)

\(\text{will exercise for 0.7 minutes less.}\)

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

\(\text{If someone watches no television, the LSRL predicts they will exercise for 64.3 minutes per day.}\)
 

c.    \(\text{At} \ \ x=42:\)

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

\(\therefore \ \text{Jo is expected to exercise for 34.9 minutes}\)
 

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

\(\text{The model predicts a negative value for time spent exercising, which is not possible.}\)

Statistics, STD2 S4 2025 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.    \(\text{Form: Linear. Direction: Negative}\)

Show Worked Solution

a.    \(\text{Form: Linear}\)

\(\text{Direction: Negative}\)
 

b.    \(\text{Slope}=-0.7\)

\(\text{This means that for each added minute of watching television per day, a participant, on average,}\)

\(\text{will exercise for 0.7 minutes less.}\)

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

\(\text{If someone watches no television, the LSRL predicts they will exercise for 64.3 minutes per day.}\)
 

c.    \(\text{At} \ \ x=42:\)

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

\(\therefore \ \text{Jo is expected to exercise for 34.9 minutes}\)
 

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

\(\text{The model predicts a negative value for time spent exercising, which is not possible.}\)

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)

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`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 2014* HSC 30b

The scatterplot shows the relationship between expenditure per primary school student, as a percentage of a country’s Gross Domestic Product (GDP), and the life expectancy in years for 15 countries.
 

 
 

  1. For the given data, the correlation coefficient,  `r`, is 0.83. What does this indicate about the relationship between expenditure per primary school student and life expectancy for the 15 countries?   (1 mark)

  2. For the data representing expenditure per primary school student,  `Q_L`  is 8.4 and  `Q_U`  is 22.5.

     

    What is the interquartile range?   (1 mark)

  3. Another country has an expenditure per primary school student of 47.6% of its GDP.

     

    Would this country be an outlier for this set of data? Justify your answer with calculations.   (2 marks)

  4. On the scatterplot, draw the least-squares line of best fit  `y = 1.29x + 49.9`.    (2 marks)

  5. Using this line, or otherwise, estimate the life expectancy in a country which has an expenditure per primary school student of 18% of its GDP.   (1 mark)

  6. Why is this line NOT useful for predicting life expectancy in a country which has expenditure per primary school student of 60% of its GDP?   (1 mark)

Show Answers Only
  1. `text(It indicates there is a strong positive)`

     

    `text(correlation between the two variables.)`

  2. `14.1`
  3. `text(Yes, because it’s > 43.65%)`
  4.  
  5. `73.1\ text(years)`
  6. `text(At 60% GDP, the line predicts a life expectancy)`

  7.  

    `text(of 127.3. This line of best fit is only accurate)`

  8.  

    `text(in a lower range of GDP expediture.)`

Show Worked Solution
i. `text(It indicates there is a strong positive)`
  `text(correlation between the two variables)`

 

ii. `text(IQR)` `= Q_U\ – Q_L`
    `= 22.5\ – 8.4`
    `= 14.1`

 

♦ Mean mark 35% 

iii.  `text(An outlier on the upper side must be more than)` 

`Q_u\ +1.5xxIQR`

`=22.5+(1.5xx14.1)`

`=\ text(43.65%)`

`:.\ text(A country with an expenditure of 47.6% is an outlier).`

 

iv.  

v.  `text(Life expectancy) ~~ 73.1\ text{years (see dotted line)}`

♦♦ Mean mark 39%

 

`text(Alternative Solution)`

`text(When)\ x=18`

`y=1.29(18)+49.9=73.12\ \ text(years)`

  

♦♦♦ Mean mark 0%. The toughest question on the 2014 paper.
COMMENT: Examiners regularly ask students to identify and comment on outliers where linear relationships break down.
vi.   `text(At 60% GDP, the line predicts a life)`
  `text(expectancy of 127.3. This line of best)`
  `text(fit is only predictive in a lower range)`
  `text(of GDP expenditure.)`

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
  1. `text(Gradient = 6)`
  2. `text(Males should grow 6 cm per)`

     

    `text(year between the ages 11-16.)`

  3. `h = 6a + 80`
  4. `text(182 cm)`
  5. `text(People slow and eventually stop growing)`

  6.  

    `text(after they become adults.)`

Show Worked Solution

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

ii.   `text{Males should grow 6cm per year between the}`

`text{ages 11–16.}`
 

♦♦ Mean marks of 38%, 26% and 25% respectively for parts (i)-(iii).
MARKER’S COMMENT: Interpreting gradients has been consistently examined in recent history and almost always poorly answered. 

iii.   `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`

 

iv.   `text{Substitue}\ \ a=17\ \ \text{into equation from part (iii):}`

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

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

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