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Statistics, 2ADV S2 2025 HSC 14

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.    \(\text{Form: Linear}\)

\(\text{Direction: Negative}\)
 

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.    \(\text{34.9 minutes}\)
 

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

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

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, 2ADV S2 2023 HSC 18

A university uses gas to heat its buildings. Over a period of 10 weekdays during winter, the gas used each day was measured in megawatts (MW) and the average outside temperature each day was recorded in degrees Celsius (°C).

Using `x` as the average daily outside temperature and `y` as the total daily gas usage, the equation of the least-squares regression line was found.

The equation of the regression line predicts that when the temperature is 0°C, the daily gas usage is 236 MW.

The ten temperatures measured were: 0°, 0°, 0°, 2°, 5°, 7°, 8°, 9°, 9°, 10°,

The total gas usage for the ten weekdays was 1840 MW.

In any bivariate dataset, the least-squares regression line passes through the point `(bar x,bar y)`, where `bar x` is the sample mean of the `x`-values and `bary` is the sample mean of the `y`-values.

  1. Using the information provided, plot the point `(bar x,bar y)` and the `y`-intercept of the least-squares regression line on the grid.  (3 marks)
     

 

  1. What is the equation of the regression line?  (2 marks)

  2. In the context of the dataset, identify ONE problem with using the regression line to predict gas usage when the average outside temperature is 23°C.  (1 mark)

Show Answers Only

a.    
       

b.    `y=-10.4x+236`

c.    `text{Answers could include one of the following:}`

`text{→ 23°C is outside the range of the dataset and requires the trend}`

`text{to be extrapolated.}`

`text{→ At 23°C, the equation predicts negative daily gas usage.}`

Show Worked Solution

a.    `barx=(0+0+0+2+5+7+8+9+9+10)/10=5^@text{C}`

`bary=1840/10=184`

`text{Regression line passes through:}\ (0,236) and (5,184)`
 

 

b.    `m=(y_2-y_1)/(x_2-x_1)=(184-236)/(5-0)=-10.4`

`text{Equation of line}\ m=-10.4\ text{passing through}\ (0,236):`

`(y-y_1)` `=m(x-x_1)`  
`y-236` `=-10.4(x-0)`  
`y` `=-10.4x+236`  

 
c.    `text{Answers could include one of the following:}`

`text{→ 23°C is outside the range of the dataset and requires the trend}`

`text{to be extrapolated.}`

`text{→ At 23°C, the equation predicts negative daily gas usage.}`

Statistics, 2ADV S2 2021 HSC 17

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
  1. i.  `23.3°text(C)`
  2. ii. `text(See Worked Solutions)`
  3. `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.)`
 

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.)`
 

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, 2ADV S2 2020 HSC 27

A cricket is an insect. The male cricket produces a chirping sound.

A scientist wants to explore the relationship between the temperature in degrees Celsius and the number of cricket chirps heard in a 15-second time interval.

Once a day for 20 days, the scientist collects data. Based on the 20 data points, the scientist provides the information below.

  • A box-plot of the temperature data is shown.
     
       
  • The mean temperature in the dataset is 0.525°C below the median temperature in the dataset.
  • A total of 684 chirps was counted when collecting the 20 data points.

The scientist fits a least-squares regression line using the data `(x, y)`, where `x` is the temperature in degrees Celsius and `y` is the number of chirps heard in a 15-second time interval. The equation of this line is

`y = −10.6063 + bx`,

where `b` is the slope of the regression,

The least-squares regression line passes through the point  `(barx, bary)`, where  `barx`  is the sample mean of the temperature data and  `bary`  is the sample mean of the chirp data.

Calculate the number of chirps expected in a 15-second interval when the temperature is 19° Celsius. Give your answer correct to the nearest whole number.  (5 marks)

Show Answers Only

`29\ text(chirps)`

Show Worked Solution

`y = −10.6063 + bx`

♦ Mean mark 46%.

`text(Find)\ b:`

`text(Line passes through)\ \ (barx, bary)`

`barx` `= 22 – 0.525`
  `= 21.475`

 

`bary` `= text(total chirps)/text(number of data points)`
  `= 684/20`
  `= 34.2`

 

`34.2` `= −10.6063 + b(21.475)`
`:.b` `= 44.8063/21.475`
  `~~ 2.0864`

 

`text(If)\ \ x = 19,`

`y` `= −10.6063 + 2.0864 xx 19`
  `= 29.03`
  `= 29\ text(chirps)`

Statistics, STD2 S4 EQ-Bank 4

Ten high school students have their height and the length of their right foot measured.

The results are recorded in the table below.
 


 

  1. Using technology, calculate Pearson's correlation coefficient for the data. Give your answer to 3 decimal places.  (1 mark)

  2. Describe the strength of the association between height and length of right foot for these students.  (1 mark)

  3. Using technology, determine the least squares regression line that allows height to be predicted from right foot length.  (1 mark)

Show Answers Only
  1. `0.941\ \ (text(to 3 d.p.))`
  2. `text(The association is positive and strong.)`
  3. `text(Height) =47.4 + 4.7 xx text(foot length)`
Show Worked Solution

i.   `text(By calculator,)`

COMMENT: Issues here? YouTube has short and excellent help videos – search your calculator model and topic – eg. “fx-82 correlation” .

`r` `= 0.94095…`
  `= 0.941\ \ (text(to 3 d.p.))`

 

ii.   `text(The association is positive and strong.)`

 

iii.   `x\ text(value ⇒ foot length (independent variables))`

`y\ text(value ⇒ height.)`

`text(By calculator:)`

`text(Height) = 47.4 + 4.7 xx text(foot length)`

Statistics, 2ADV S2 EQ-Bank 3

The table below lists the average life span (in years) and average sleeping time (in hours/day) of 9 animal species.
 


 

  1. Using sleeping time as the independent variable, calculate the least squares regression line. (1 mark)

  2. A wallaby species sleeps for 4.5 hours, on average, each day.

     

    Use your equation from part i to predict its expected life span, to the nearest year.   (1 mark)

Show Answers Only
  1. `text(life span) = 42.89 – 2.85 xx text(sleeping time)`
  2. `30\ text(years)`
Show Worked Solution

i.    `text(By calculator:)`

COMMENT: Issues here? YouTube has short and excellent help videos – search your calculator model and topic – eg. “fx-82 regression line” .

`text(life span) = 42.89 – 2.85 xx text(sleeping time)`
 

ii.   `text(Predicted life span of wallaby)`

`= 42.89 – 2.85 xx 4.5` 

`= 30.06…`

`= 30\ text(years)`

Statistics, 2ADV S2 EQ-Bank 2

The table below lists the average body weight (in kilograms) and average brain weight (in grams) of nine animal species.
 


 

A least squares regression line is fitted to the data using body weight as the independent variable.

  1. Calculate the equation of the least squares regression line. (1 mark)

  2. If dingos have an average body weight of 22.3 kilograms, calculate the predicted average brain weight of a dingo using your answer to part i.   (1 mark)

Show Answers Only
  1. `text(brain weight) = 49.4 + 2.68 xx text(body weight)`
  2. `109\ text(grams)`
Show Worked Solution

i.   `text(By calculator:)`

COMMENT: Know this critical calculator skill!.

`text(brain weight) = 49.4 + 2.68 xx text(body weight)`

 

ii.   `text(Predicted brain weight of a dingo)`

`= 49.4 + 2.68 xx 22.3` 

`=109.164`

`= 109\ text(grams)`

Statistics, 2ADV S2 EQ-Bank 1

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
  1. `text(Height)`
  2. `text(Arm span)\ = 1.09 xx text(height) – 15.63`
  3. `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.)`