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Statistics, STD2 S4 2024 HSC 19

A teacher was exploring the relationship between students' marks for an assignment and their marks for a test. The data for five different students are shown on the graph.

The least-squares regression line is also shown.
 

  1. What is the equation of the least-squares regression line for this dataset?   (2 marks)

  2. Another student, whose marks are not on the graph, scored 5 for the assignment and 12 on the test.
  3. Did this student do better or worse on the test than the regression line predicts? Provide a reason for your answer.   (1 mark)

Show Answers Only

a.    \(\text {Data points: }(3,11),(5,14),(6,12),(7,15),(9,20)\)

\(y \,\text {-intercept}=6\)

\(m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{20-6}{10-0}=1.4\)

\(\text{LSRL}\ \ \Rightarrow \ y=1.4x+6\)
 

b.    \(\text {The student’s mark \((5,12)\) sits below the LSRL.}\)

 \(\text {Therefore the student did worse than expected.}\)

Show Worked Solution

a.    \(\text {Data points: }(3,11),(5,14),(6,12),(7,15),(9,20)\)

\(y\,\text {-intercept}=6\)

\(m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{20-6}{10-0}=1.4\)

\(\text{LSRL}\ \ \Rightarrow \ y=1.4x+6\)
  

b.    \(\text {The student’s mark \((5,12)\) sits below the LSRL.}\)

 \(\text {Therefore the student did worse than expected.}\)

Statistics, STD2 S4 2023 HSC 34

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)`
 

♦ Mean mark (a) 44%.

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`  
♦♦♦ Mean mark (b) 21%.

 
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.}`

♦♦♦ Mean mark 23%.

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
  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.)`
♦♦ Mean mark part 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 part 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
  1. `y=3.2x+2`
  2. `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, STD2 S4 2020 HSC 36

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 line.

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 18%.

`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.)`