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Data Analysis, GEN2 2024 VCAA 3

The Olympic gold medal-winning height for the women's high jump, \(\textit{Wgold}\), is often lower than the best height achieved in other international women's high jump competitions in that same year.

The table below lists the Olympic year, \(\textit{year}\), the gold medal-winning height, \(\textit{Wgold}\), in metres, and the best height achieved in all international women's high jump competitions in that same year, \(\textit{Wbest}\), in metres, for each Olympic year from 1972 to 2020.

A scatterplot of \(\textit{Wbest}\) versus \(\textit{Wgold}\) for this data is also provided.
  

year 1972 1976 1980 1984 1988 1992 1996 2000 2004 2008 2012 2016 2020

Wgold
(m)

 1.92 1.93 1.97 2.02 2.03 2.02 2.05 2.01 2.06 2.05 2.05 1.97 2.04

Wbest
(m)

 1.94 1.96 1.98 2.07 2.07 2.05 2.05 2.02 2.06 2.06 2.05 2.01 2.05

 

When a least squares line is fitted to the scatterplot, the equation is found to be:

\(Wbest =0.300+0.860 \times Wgold\)

The correlation coefficient is 0.9318

  1. Name the response variable in this equation.   (1 mark)

  2. Draw the least squares line on the scatterplot above.  (1 mark)

  3. Determine the value of the coefficient of determination as a percentage.  (1 mark)
  4. Round your answer to one decimal place.
  5. Describe the association between \(\textit{Wbest}\) and \(\textit{Wgold}\) in terms of strength and direction.  (1 mark)

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\text { strength } \rule[-1ex]{0pt}{0pt} & \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
\hline
\rule{0pt}{2.5ex}\text { direction } \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}

  1. Referring to the equation of the least squares line, interpret the value of the slope in terms of the variables \(\textit{Wbest}\) and \(\textit{Wgold}\).  (1 mark)

  2. In 1984, the \(\textit{Wbest}\) value was 2.07 m for a \(\textit{Wgold}\) value of 2.02 m .
  3. Show that when this least squares line is fitted to the scatterplot, the residual value for this point is 0.0328.  (2 marks)

  4. The residual plot obtained when the least squares line was fitted to the data is shown below. The residual value from part f is missing from the residual plot.
     

    1. Complete the residual plot by adding the residual value from part f, drawn as a cross ( X ), to the residual plot above.   (1 mark)

    2. In part b, a least squares line was fitted to the scatterplot. Does the residual plot from part g justify this? Briefly explain your answer.  (1 mark)

  1. In 1964, the gold medal-winning height, \(\textit{Wgold}\), was 1.90m . When the least squares line is used to predict \(\textit{Wbest}\), it is found to be 1.934 m .
  2. Explain why this prediction is not likely to be reliable.  (1 mark)

Show Answers Only

a.    \(Wbest\)

b.    

c.    \(86.8\%\)

d.    \(\text{Strong, positive}\)

e.    \(Wbest\ \text{will increase, on average by 0.86 metres for every metre of increase in}\ Wgold.\)

f.      \(Wbest\) \(=0.300 +0.86\times 2.02\)
    \(=2.0372\)

\(\therefore\ \text{Residual}\ =2.07-2.0372=0.0328\)

g.i.

g.ii.  \(\text{Yes, it is justified as there is no clear pattern, linear or otherwise.}\)

h.    \(\text{This prediction is outside the data range (1972 – 2020 → extrapolation)}\)

\(\text{and therefore cannot be relied upon.}\)

Show Worked Solution

a.    \(Wbest\)

b.    \(\text{Using points:}\ (1.90, 1.934)\ \text{and}\ (2.00, 2.02)\)

c.    \(r=0.9318\)

\(r^2=0.9318^2=0.8682\dots\)

\(\therefore\ \text{Coefficient of determination} \approx 86.8\%\)

d.    \(\text{Strong, positive}\)

e.    \(Wbest\ \text{will increase, on average by 0.86 metres for every metre of increase in}\ Wgold.\)   

f.      \(Wbest\) \(=0.300 +0.86\times 2.02\)
    \(=2.0372\)

\(\therefore\ \text{Residual}\ =2.07-2.0372=0.0328\)

g.i.

g.ii.  \(\text{Yes, it is justified as there is no clear pattern, linear or otherwise.}\)

h.    \(\text{This prediction is outside the data range (1972 – 2020 → extrapolation)}\)

\(\text{and therefore cannot be relied upon.}\)

Data Analysis, GEN2 2023 VCAA 3

The scatterplot below plots the average monthly ice cream consumption, in litres/person, against average monthly temperature, in °C. The data for the graph was recorded in the Northern Hemisphere.
 

When a least squares line is fitted to the scatterplot, the equation is found to be:

consumption = 0.1404 + 0.0024 × temperature

The coefficient of determination is 0.7212

  1. Draw the least squares line on the scatterplot graph above.  (1 mark)

  2. Determine the value of the correlation coefficient \(r\).
  3. Round your answer to three decimal places.  (1 mark)

  4. Describe the association between average monthly ice cream consumption and average monthly temperature in terms of strength, direction and form.  (1 mark)

    \begin{array} {|l|c|}
    \hline
    \rule{0pt}{2.5ex} \textbf{strength} \rule[-1ex]{0pt}{0pt} & \quad \quad \quad \quad \quad \quad \quad \quad \\
    \hline
    \rule{0pt}{2.5ex} \textbf{direction} \rule[-1ex]{0pt}{0pt} & \\
    \hline
    \rule{0pt}{2.5ex} \textbf{form} \rule[-1ex]{0pt}{0pt} & \\
    \hline
    \end{array}

  5. Referring to the equation of the least squares line, interpret the value of the intercept in terms of the variables consumption and temperature.  (1 mark)

  6. Use the equation of the least squares line to predict the average monthly ice cream consumption, in litres per person, when the monthly average temperature is –6°C.  (1 mark)

  7. Write down whether this prediction is an interpolation or an extrapolation.  (1 mark)

Show Answers Only

a.    
         

b.    \(r = \sqrt{0.7212} = 0.849\ \text{(3 d.p.)}\)

c.   

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{strength} \rule[-1ex]{0pt}{0pt} & \text{strong} \\
\hline
\rule{0pt}{2.5ex} \textbf{direction} \rule[-1ex]{0pt}{0pt} & \text{positive} \\
\hline
\rule{0pt}{2.5ex} \textbf{form} \rule[-1ex]{0pt}{0pt} & \text{linear} \\
\hline
\end{array}

 
d.    \(\text{At 0°C, the predicted average consumption is:}\)

\(\textit{consumption} = 0.1404 + 0.002 \times 0 = 0.1404\ \text{L/person}\)

 
e.    \(\text{Find consumption} (c)\ \text{when temperature} (t) = -6:\)

\(c=0.1404 + 0.002 \times -6 = 0.1284\ \text{L/person} \)

 
f.     \(\text{Extrapolation (even though the axes extend to –6°C, the data set} \)

\(\text{range finishes with a lower limit around –4.5°C.)}\)

Show Worked Solution

a.    
         

b.    \(r = \sqrt{0.7212} = 0.849\ \text{(3 d.p.)}\)

c.   

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{strength} \rule[-1ex]{0pt}{0pt} & \text{strong} \\
\hline
\rule{0pt}{2.5ex} \textbf{direction} \rule[-1ex]{0pt}{0pt} & \text{positive} \\
\hline
\rule{0pt}{2.5ex} \textbf{form} \rule[-1ex]{0pt}{0pt} & \text{linear} \\
\hline
\end{array}

 
d.    \(\text{At 0°C, the predicted average consumption is:}\)

\(\textit{consumption} = 0.1404 + 0.002 \times 0 = 0.1404\ \text{L/person}\)

 
e.    \(\text{Find consumption} (c)\ \text{when temperature} (t) = -6:\)

\(c=0.1404 + 0.002 \times -6 = 0.1284\ \text{L/person} \)

 
f.     \(\text{Extrapolation (even though the axes extend to –6°C, the data set} \)

\(\text{range finishes with a lower limit around –4.5°C.)}\)

Data Analysis, GEN1 2022 VCAA 7-8 MC

The association between the weight of a seal's spleen, spleen weight, in grams, and its age, in months, for a sample of seals is non-linear.

This association can be linearised by applying a \(\log _{10}\) transformation to the variable spleen weight.
 

The equation of the least squares line for this scatterplot is

\(\log _{10}\) (spleen weight) = 2.698 + 0.009434 × age

 
Question 7

The equation of the least squares line predicts that, on average, for each one-month increase in the age of the seals, the increase in the value of \(\log _{10}\) (spleen weight) is

  1. 0.009434
  2. 0.01000
  3. 1.020
  4. 2.698
  5. 5.213

 
Question 8

Using the equation of the least squares line, the predicted spleen weight of a 30-month-old seal, in grams, is

  1. 3
  2. 511
  3. 772
  4. 957
  5. 1192
Show Answers Only

\(\text{Question 7:} \ A\)

\(\text{Question 8:} \ D\)

Show Worked Solution

\(\text{Question 7}\)

\(\text{Graph passes through (0, 2.7) and (130, 3.93)} \)

\(\text{Gradient}\ \approx \dfrac{3.93-2.7}{130} \approx 0.0946 \)

\(\Rightarrow A\)
 

\(\text{Question 8}\)

\(\text{Let the predicted spleen weight be}\ w:\)

\(\log_{10} w\) \(=2.698 + 0.009434 \times 30\)  
\(\log_{10} w\) \(=2.98102\)  
\(w\) \(= 10^{2.98102}\)  
  \(=957.2381527\)  

 
\(\Rightarrow D\)

Mean mark (Q8) 56%.

Data Analysis, GEN2 2023 VCAA 1

Data was collected to investigate the use of electronic images to automate the sizing of oysters for sale. The variables in this study were:

    • ID: identity number of the oyster
    • weight: weight of the oyster in grams (g)
    • volume: volume of the oyster in cubic centimetres (cm³)
    • image size: oyster size determined from its electronic image (in megapixels)
    • size: oyster size when offered for sale: small, medium or large

The data collected for a sample of 15 oysters is displayed in the table.
 

  1. Write down the number of categorical variables in the table.   (1 mark)
  2. Determine, in grams:
    1. the mean weight of all the oysters in this sample.   (1 mark)
    2. the median weight of the large oysters in this sample.   (1 mark)
  3. When a least squares line is used to model the association between oyster weight and volume, the equation is:

        \(\textit{volume} = 0.780 + 0.953 \times \textit{weight} \)

    1. Name the response variable in this equation.
    2. Complete the following sentence by filling in the blank space provided.
    3. This equation predicts that, on average, each 10 g increase in the weight of an oyster is associated with a ________________ cm³ increase in its volume.
  2. A least squares line can also be used to model the association between an oyster's volume, in cm³, and its electronic image size, in megapixels. In this model, image size is the explanatory variable.
  3. Using data from the table, determine the equation of this least squares line. Use the template below to write your answer. Round the values of the intercept and slope to four significant figures.
  4. The number of megapixels needed to construct an accurate electronic image of an oyster is approximately normally distributed.
  5. Measurements made on recently harvested oysters showed that:
    • 97.5% of the electronic images contain less than 4.6 megapixels
    • 84% of the electronic images contain more than 4.3 megapixels.
  1. Use the 68-95-99.7% rule to determine, in megapixels, the mean and standard deviation of this normal distribution.
Show Answers Only

a.    \(\text{Categorical variables = 2 (ID and size)}\)

b.i.  \(\text{Mean weight}\ = \dfrac{\text{sum of oyster weights}}{15} = \dfrac{171.3}{15} = 11.42 \)

b.ii.  \(\text{Median}\ = 11.4 \)

c.i.   \(\text{Volume}\)

c.ii.  \(\text{Increase}\ = 0.953 \times 10 = 9.53\ \text{cm}^{3} \)

d.    \(\text{Volume}\ = 0.002857 + 2.571 \times \text{image size} \)

e.    \(s_x = 0.1 \)

\(\bar x = 4.4\) 

Show Worked Solution

a.    \(\text{Categorical variables = 2 (ID and size)}\)
 

b.i.  \(\text{Mean weight}\ = \dfrac{\text{sum of oyster weights}}{15} = \dfrac{171.3}{15} = 11.42 \)
 

b.ii.  \(\text{15 data points}\ \ \Rightarrow \ \ \text{Median = 8th data point (in order)}\)

 \(\text{Median}\ = 11.4 \)
 

c.i.   \(\text{Volume}\)
 

c.ii.  \(\text{Increase}\ = 0.953 \times 10 = 9.53\ \text{cm}^{3} \)
 

d.    \(\text{Input the image size column values}\ (x)\ \text{and volume} \)

\(\text{values}\ (y)\ \text{into the calculator:}\)

\(\textit{Volume}\ = 0.002857 + 2.571 \times \textit{image size} \)
 

e.    \(z\text{-score (4.6)}\ = 2\ \ \Rightarrow \bar x + 2 \times s_x = 4.6\ …\ (1)\)

\(z\text{-score (4.3)}\ = -1\ \ \Rightarrow \bar x-s_x = 4.3\ …\ (2)\)

\( (1)-(2) \)

\(3 s_x = 0.3 \ \ \Rightarrow\ \ s_x = 0.1 \)

\(\bar x = 4.4\) 

Data Analysis, GEN1 2023 VCAA 11-12 MC

The table below shows the height, in metres, and the age, in years, for 11 plantation trees. A scatterplot displaying this data is also shown.
 

Question 11

A reciprocal transformation applied to the variable age can be used to linearise the scatterplot.

With \(\dfrac{1}{\textit{age}}\) as the explanatory variable, the equation of the least squares line fitted to the linearised data is closest to

  1. \(\textit{height}\ =-13.04 + 40.22 \times \dfrac{1}{\textit{age}}\)
  2. \(\textit{height}\ =-10.74+8.30 \times \dfrac{1}{\textit{age}}\)
  3. \(\textit{height}\ =2.14 + 0.63 \times \dfrac{1}{\textit{age}}\)
  4. \(\textit{height}\ =13.04-40.22 \times \dfrac{1}{\textit{age}}\)
  5. \(\textit{height}\ =16.56-22.47 \times \dfrac{1}{\textit{age}}\)

 
Question 12

The scatterplot can also be linearised using a logarithm (base 10) transformation applied to the variable age.

The equation of the least squares line is

\(\textit{height }=-3.8+12.6 \times \log _{10}(\textit{age}) \)

Using this equation, the age, in years, of a tree with a height of 8.52 m is closest to

  1. 7.9
  2. 8.9
  3. 9.1
  4. 9.5
  5. 9.9
Show Answers Only

\(\text{Question 11:}\ D\)

\(\text{Question 12:}\ D\)

Show Worked Solution

\(\text{Question 11}\)

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & 4 & 5 & 6 & 7 & 8 \\
\hline
\rule{0pt}{2.5ex} \dfrac{1}{x} \rule[-1ex]{0pt}{0pt} & 0.25 & 0.2 & 0.167 & 0.143 & 0.125 \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 3.5 & 4.0 & 6.0 & 7.8 & 8.0 \\
\hline
\end{array}

\(\text{By inspection of the table: as}\ \dfrac{1}{x} ↓, \ y ↑\ \text{(gradient is negative)} \)

\(\text{Eliminate options A, B and C.}\)

\(\text{Option D is clearly a better LSRL by considering linearised inputs.}\)

\(\text{Consider}\ \dfrac{1}{x} = 0.2:\)

\(\text{Option D …}\ 13.04-40.22 \times 0.2 = 4.99\ \text{(vs actual 6.0)}\)

\(\text{Option E …}\ 16.56-22.47 \times 0.2 = 12.06\ \text{(vs actual 6.0)}\)

\(\Rightarrow D\)
 

\(\text{Question 12}\)

\(8.52\) \(=-3.8+12.6 \times \log{10}(\textit{age}) \)  
\(\log{10}(\textit{age}) \) \(= \dfrac{8.52+3.8}{12.6} \)  
  \(=0.977…\)  
\(\textit{age}\) \(=10^{0.977…} \)  
  \(=9.50\ \text{years} \)  

 
\(\Rightarrow D\)

Data Analysis, GEN1 2023 VCAA 7-8 MC

A teacher analysed the class marks of 15 students who sat two tests.

The test 1 mark and test 2 mark, all whole number values, are shown in the scatterplot below.

A least squares line has been fitted to the scatterplot.
 

Question 7

The equation of the least squares line is closest to

  1. test 2 mark = – 6.83 + 1.55 × test 1 mark
  2. test 2 mark = 15.05 + 0.645 × test 1 mark
  3. test 2 mark = – 6.78 + 0.645 × test 1 mark
  4. test 2 mark = 1.36 + 1.55 × test 1 mark
  5. test 2 mark = 6.83 + 1.55 × test 1 mark

 
Question 8

The least squares line shows the predicted test 2 mark for each student based on their test 1 mark.

The number of students whose actual test 2 mark was within two marks of that predicted by the line is

  1. 3
  2. 4
  3. 5
  4. 6
  5. 7
Show Answers Only

\(\text{Question 7:}\ A\)

\(\text{Question 8:}\ C\)

Show Worked Solution

\(\text{Question 7}\)

\(\text{By inspection, gradient is greater than 1 (eliminate B and C)}\)

\(\text{LSRL passes through (16, 18):}\)

\(\text{Option A:}\ -6.83 + 1.55 \times 16 = 18.0\ \checkmark \)

\(\Rightarrow A\)
 

\(\text{Question 8}\)

\(\text{5 values are within 1 grid height (measured vertically), or 2 marks,}\)

\(\text{from the LRSR.}\)

\(\Rightarrow C\)

CORE, FUR2 2021 VCAA 4

The time series plot below shows that the winning time for both men and women in the 100 m freestyle swim in the Olympic Games has been decreasing during the period 1912 to 2016.
 

Least squares lines are used to model the trend for both men and women.

The least squares line for the men's winning time has been drawn on the time series plot above.

The equation of the least squares line for men is

winning time men = 356.9 – 0.1544 × year

The equation of the least squares line for women is

winning time women = 538.9 – 0.2430 × year

  1. Draw the least squares line for winning time women on the time series plot above.  (1 mark)
  2. The difference between the women's predicted winning time and the men's predicted winning time can be calculated using the formula.
  3.       difference = winning time women – winning time men
  4. Use the equation of the least squares lines and the formula above to calculate the difference predicted for the 2024 Olympic Games.
  5. Round your answer to one decimal place.  (2 marks)
  6. The Olympic Games are held every four years. The next Olympic Games will be held in 2024, then 2028, 2032 and so on.
  7. In which Olympic year do the two least squares lines predict that the wining time for women will first be faster than the winning time for men in the 100 m freesytle?  (2 marks)
Show Answers Only
  1. `text{See Worked Solutions}`
  2. `2.7 \ text{seconds}`
  3. `2056`
Show Worked Solution

a.   

`text{Find end points for the women’s graph.}`

`1908 \ text{data point} = 538.9 – 0.2430 xx 1908 = 75.256`

`2020 \ text{data point} = 538.9 – 0.2430 xx 2020 = 48.04`

`=> \ text{Line passes through} \ (1908, 75.3) \ text{and} \ (2020, 48.0)`
 

b.   `text{difference}` `= (538.9 – 0.2430 xx 2024) – (356.9 – 0.1544 xx 2024)`
    `= 2.673 …`
    `= 2.7 \ text{seconds (to 1 d.p.)}`

 
c.    `text{Times will be equal when}`

`538.9 – 0.2430 xx text{year} = 356.9 – 0.1544 xx text{year}`

`text{year} = 2054.17 …`

`text{1st Olympic year after} \ 2054.17 = 2056`

CORE, FUR2 2021 VCAA 3

The time series plot below shows the winning time, in seconds, for the women's 100 m freestyle swim plotted against year, for each year that the Olympic Games were held during the period 1956 to 2016.

A least squares line has been fitted to the plot to model the decreasing trend in the winning time over this period.
 

The equation of the least squares line is

winning time = 357.1 – 0.1515 × year

The coefficient of determination is 0.8794

  1. Name the explanatory variable in this time series plot.  (1 mark)
  2. Determine the value of the correlation coefficient (`r`).
  3. Round your answer to three decimal places.  (1 mark)
  4. Write down the average decrease in winning time, in seconds per year, during the period 1956 to 2016.  (1 mark)
  5. The predicted winning time for the women's 100 m freestyle in 2000 was 54.10 seconds.
  6. The actual winning time for the women's 100 m freestyle in 2000 was 53.83 seconds.
  7. Determine the residual value in seconds.  (1 mark)
  8. The following equation can be used to predict the winning time for the women's 100 m freestyle in the future.
  9.      winning time =  357.1 – 0.1515 × year
  10.  i. Show that the predicted winning time for the women's 100 m freestyle in 2032 is 49.252 seconds.  (1 mark)
  11. ii. What assumption is being made when this equation is used to predict the winning time for the women's 100 m freestyle in 2032?   (1 mark)
Show Answers Only
  1. `text{year}`
  2. `- 0.938`
  3. `0.1515 \ text{seconds}`
  4. `-0.27`
  5.  i. `49.252 \ text{seconds}`
  6. ii. `text{The same trend continues when the graph is extended beyond 2016.}`
Show Worked Solution

a.      `text{year}`
 

b. `r^2` `= 0.8794 \ text{(given)}`
  `r` `= ± sqrt{0.8794}`
    `= ± 0.938 \ text{(to 3 d.p.)}`

 
`text{By inspection of graph, correlation is negative}`

`:. \ r = -0.938`
 

c.    `text{Average decrease in winning time = 0.1515 seconds}`

`text{(this is given by the slope of the line.)}`
 

d.    `text{Residual Value}` `= text{actual} – text{predicted}`
    `= 53.83 – 54.10`
    `= -0.27`

 

e.i.      `text{winning time (2032)}` `= 357.1 – 0.1515 xx 2032`
      `=49.252 \ text{seconds}`

 

e.ii.  `text{The assumption is that the graph is accurate when it is extended}`

  `text{beyond 2016 (i.e decreasing trend continues).}`

CORE, FUR2 2020 VCAA 5

The scatterplot below shows body density, in kilograms per litre, plotted against waist measurement, in centimetres, for 250 men.
 


 

When a least squares line is fitted to the scatterplot, the equation of this line is

body density = 1.195 – 0.001512 × waist measurement

  1. Draw the graph of this least squares line on the scatterplot above.  (1 mark)

(Answer on the scatterplot above.)

  1. Use the equation of this least squares line to predict the body density of a man whose waist measurement is 65 cm.

     

    Round your answer to two decimal places.  (1 mark)

  2. When using the equation of this least squares line to make the prediction in part b., are you extrapolating or interpolating?  (1 mark)
  3. Interpret the slope of this least squares line in terms of a man’s body density and waist measurement.  (1 mark)
  4. In this study, the body density of the man with a waist measurement of 122 cm was 0.995 kg/litre.

     

    Show that, when this least squares line is fitted to the scatterplot, the residual, rounded to two decimal places, is –0.02  (1 mark)

  5. The coefficient of determination for this data is 0.6783

     

    Write down the value of the correlation coefficient `r`.

     

    Round your answer to three decimal places.  (1 mark)

  6. The residual plot associated with fitting a least squares line to this data is shown below.
     

     

       
     

     

    Does this residual plot support the assumption of linearity that was made when fitting this line to this data? Briefly explain your answer.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `1.10\ text(kg/litre)`
  3. `text(extrapolating)`
  4. `text(See Worked Solutions)`
  5. `-0.02`
  6. `-0.824`
  7. `text(See Worked Solutions)`
Show Worked Solution

a.   `text(LSRL passes through)\ (60, 1.1043) and (130, 0.998)`

♦ Mean mark part a. 41%.

b.   `text(body density)` `= 1.195 – 0.001512 xx 65`
    `= 1.09672`
    `= 1.10\ text{kg/litre (to 2 d.p.)}`
♦ Mean mark part c. 38%.
c.   `text(A waist of 65 cm is outside the)`
  `text(range of the existing data set.)`

 
`:.\ text(Extrapolating)`

 

♦ Mean mark part d. 44%.
d.   `text(Body density decreases by 0.001512 kg/litre)`
  `text(for each increase in waist size of 1 cm.)`

 

e.   `text{Body density (predicted)}`

`= 1.195 – 0.001512 xx 122`

`~~ 1.0105\ text(kg/litre)`
 

`text(Residual)` `= text(Actual – predicted)`
  `~~ 0.995 – 1.0105`
  `~~ -0.0155`
  `~~ -0.02\ text{(to 2 d.p.)}`

 

♦♦ Mean mark part f. 25%.

f.   `r` `= -sqrt(0.6783)`
    `= – 0.8235…`
    `= -0.824\ text{(to 3 d.p.)}`

 

g.  `text(The residual plot has no pattern and is)`

`text(centred around zero.)`

`:.\ text(It supports the assumption of linearity)`

`text(of the LSRL.)`

CORE, FUR1 2020 VCAA 13 MC

A least squares line of the form  `y = a + bx`  is fitted to a scatterplot.

Which one of the following is always true?

  1. As many of the data points in the scatterplot as possible will lie on the line.
  2. The data points in the scatterplot will be divided so that there are as many data points above the line as there are below the line.
  3. The sum of the squares of the shortest distances from the line to each data point will be a minimum.
  4. The sum of the squares of the horizontal distances from the line to each data point will be a minimum.
  5. The sum of the squares of the vertical distances from the line to each data point will be a minimum.
Show Answers Only

`E`

Show Worked Solution

`text(The sum of the squares of the vertical distances from)`

♦♦ Mean mark 34%.

`text(the line to each data point will be a minimum.)`

`=>  E`

CORE, FUR1-NHT 2019 VCAA 9 MC

A least squares line of the form  `y = a + bx`  is fitted to a set of bivariate data for the variables `x` and `y`.

For this set of bivariate data,  `barx = 5.50`,  `bary = 5.60`,  `s_x = 3.03`,  `s_y =1.78`  and  `a = 3.1`

The slope of the least squares line, `b`, is closest to

  1.  0.44
  2.  0.45
  3.  0.58
  4.  0.59
  5.  0.76
Show Answers Only

`B`

Show Worked Solution
`bar y` `=a+b barx`
`5.60` `= 3.1 – b(5.50)`
`b` `= (5.60 – 3.1)/5.50`
  `= 04545…`

 
`=> B`

CORE, FUR2 2019 VCAA 5

The scatterplot below shows the atmospheric pressure, in hectopascals (hPa), at 3 pm (pressure 3 pm) plotted against the atmospheric pressure, in hectopascals, at 9 am (pressure 9 am) for 23 days in November 2017 at a particular weather station.
 


 

A least squares line has been fitted to the scatterplot as shown.

The equation of this line is

pressure 3 pm = 111.4 + 0.8894 × pressure 9 am

  1. Interpret the slope of this least squares line in terms of the atmospheric pressure at this weather station at 9 am and at 3 pm.  (1 mark)
  2. Use the equation of the least squares line to predict the atmospheric pressure at 3 pm when the atmospheric pressure at 9 am is 1025 hPa.

     

    Round your answer to the nearest whole number.  (1 mark)

  3. Is the prediction made in part b. an example of extrapolation or interpolation?  (1 mark)
  4. Determine the residual when the atmospheric pressure at 9 am is 1013 hPa.

     

    Round your answer to the nearest whole number.  (1 mark)

  5. The mean and the standard deviation of pressure 9 am and pressure 3 pm for these 23 days are shown in Table 4 below.
     


 

    1. Use the equation of the least squares line and the information in Table 4 to show that the correlation coefficient for this data, rounded to three decimal places, is  `r` = 0.966  (1 mark)
    2. What percentage of the variation in pressure 3 pm is explained by the variation in pressure 9 am?

       

      Round your answer to one decimal place.  (1 mark)

  1. The residual plot associated with the least squares line is shown below.
     


 

    1. The residual plot above can be used to test one of the assumptions about the nature of the association between the atmospheric pressure at 3 pm and the atmospheric pressure at 9 am.

       

      What is this assumption?  (1 mark)

    2. The residual plot above does not support this assumption.

       

      Explain why.  (1 mark)

Show Answers Only
  1. `text(An increase in 1hPa of pressure at 9 am is associated)`
    `text(with an increase of 0.8894 hPa of pressure at 3 pm.)`
  2. `1023\ text(hPa)`
  3. `text(Interpolation)`
  4. `3\ text(hPa)`
    1. `0.966`
    2. `93.3%`
    1. `text(The assumption is that a linear relationship)`
      `text(exists between the pressure at 9 am and the)`
      `text(pressure at 3 pm.)`
    2. `text(The residual plot does not appear to be random.)`
Show Worked Solution

a.    `text(An increase in 1hPa of pressure at 9 am is associated)`

`text(with an increase of 0.8894 hPa of pressure at 3 pm.)`

 

b.   `text(pressure 3 pm)` `= 111.4 + 0.8894 xx 1025`
    `= 1023\ text(hPa)`

 

c.  `text{Interpolation (1025 is within the given data range)}`

 

d.   `text(Residual)` `= text(actual) – text(predicted)`
    `= 1015 – (111.4 + 0.8894 xx 1013)`
    `= 1015 – 1012.36`
    `= 2.63…`
    `~~ 3\ text(hPa)`

 

e.i.   `r= b (s_x)/(s_y)`

    `= 0.8894 xx 4.5477/4.1884`
    `= 0.96569…`
    `= 0.966`

 

e.ii.   `r` `= 0.966`
  `r^2` `= 0.9331`
    `= 93.3%`

 

f.i.   `text(The assumption is that a linear relationship)`
 

`text(exists between the pressure at 9 am and the)`

`text(pressure at 3 pm.)`

 

f.ii.  `text(The residual plot does not appear to be random.)`

CORE, FUR1 2019 VCAA 11 MC

A study was conducted to investigate the effect of drinking coffee on sleep.

In this study, the amount of sleep, in hours, and the amount of coffee drunk, in cups, on a given day were recorded for a group of adults.

The following summary statistics were generated.

On average, for each additional cup of coffee drunk, the amount of sleep

  1. decreased by 0.55 hours.
  2. decreased by 0.77 hours.
  3. decreased by 1.1 hours.
  4. increased by 1.1 hours.
  5. increased by 2.3 hours.
Show Answers Only

`A`

Show Worked Solution
`b` `= r xx (s_y)/(s_x)`
  `= -0.770 xx 1.12/1.56`
  `= -0.55\ text(hours)`

 
`:.\ text(Sleep decreased by 0.55 hours for each)`

`text(additional cup of coffee.)`

`=>  A`

CORE, FUR1 2018 VCAA 13 MC

The statistical analysis of a set of bivariate data involving variables `x` and `y` resulted in the information displayed in the table below.

Using this information, the value of the correlation coefficient `r` for this set of bivariate data is closest to

  1. 0.88
  2. 0.89
  3. 0.92
  4. 0.94
  5. 0.97
Show Answers Only

`D`

Show Worked Solution

♦ Mean mark 49%.

`text(Gradient)` `= r(s_y)/(s_x)`
`1.31` `= r xx 3.24/2.33`
`:. r` `= 1.31 xx 2.33/3.24`
  `= 0.942…`

`=> D`

CORE, FUR2 2013 VCAA 3

The development index and the average pay rate for workers, in dollars per hour, for a selection of 25 countries are displayed in the scatterplot below.
 

CORE, FUR2 2013 VCAA 31  
 

The table below contains the values of some statistics that have been calculated for this data.
 

     CORE, FUR2 2013 VCAA 32
 

  1. Determine the standardised value of the development index (`z` score) for a country with a development index of 91.

     

    Write your answer, correct to one decimal place.  (1 mark)

  2. Use the information in the table to show that the equation of the least squares regression line for a country’s development index, `y`, in terms of its average pay rate, `x`, is given by
     
         `y = 81.3 + 0.272x`  (2 marks)
  3. The country with an average pay rate of $14.30 per hour has a development index of 83.

     

    Determine the residual value when the least squares regression line given in part (b) is used to predict this country’s development index.

     

    Write your answer, correct to one decimal place.  (2 marks)

Show Answers Only
  1. `1.8`
  2. `text(See Worked Solutions)`
  3. `−2.2`
Show Worked Solution
MARKER’S COMMENT: Many students made a careless error by using the Average pay rate (`x`-variable).
a.    `z` `= (y – bar y)/s`
    `= (91 – 85.6)/2.99`
    `=1.806…`
    `= 1.8\ \ text{(to 1 d.p.)}`

 

♦ Parts (b) and (c) had a combined mean mark of 48%.
b.    `b` `= r xx s_y/s_x`
    `= 0.488 xx 2.99/5.37`
    `=0.2717…`

 

`a` `= bary – b barx`
  `= 85.6 − 0.272 xx 15.7`
  `= 81.329…`

 
`:.\ text(The least squares line is)`

`y = 81.3 + 0.272x`

 

c.    `text(Predicted)` `= 81.3 + 0.272 xx 14.30`
    `= 85.1896…`

 

`:.\ text(Residual)` `= 83 – 85.1896…`
  `= − 2.1896…`
  `= – 2.2`

CORE, FUR1 2010 VCAA 10 MC

For a set of bivariate data that involves the variables `x` and `y`, with `y` as the response variable

`r = – 0.644, \ \ barx = 5.30, \ \ bary = 5.60, \ \ s_x = 3.06, \ \ s_y = 3.20`

The equation of the least squares regression line is closest to

A.   `y = 9.2 - 0.7x`

B.   `y = 9.2 + 0.7x`

C.   `y = 2.0 - 0.6x`

D.   `y = 2.0 - 0.7x`

E.   `y = 2.0 + 0.7x`

Show Answers Only

`A`

Show Worked Solution

`y = a + bx\ \ text(where)`

`b` `= r(s_y)/(s_x)`
  `=– 0.644 xx 3.20/3.06`
  `= – 0.6734…`

 

`a` `= bary – b barx`
  `=5.30 – (– 0.6734… xx 5.60)`
  `= 9.07…`

 
`:. y ~~ 9.2 – 0.7x`

`=> A`

CORE, FUR1 2015 VCAA 10 MC

For a set of bivariate data that involves the variables `x` and `y`:

`r = –0.47`,   `barx = 1.8`,   `s_x = 1.2`,   `bary = 7.2`,   `s_y = 0.85`

Given the information above, the least squares regression line predicting `y` from `x` is closest to

A.   `y = 8.4 - 0.66x`

B.   `y = 8.4 + 0.66x`

C.   `y = 7.8 - 0.33x`

D.   `y = 7.8 + 0.33x`

E.   `y = 1.8 + 5.4x`

Show Answers Only

`C`

Show Worked Solution

`text(Regression line has the form,)`

`y = a + bx\ \ \ text(where)`

`b` `=r s_y/s_x`
  `= – 0.47 xx (0.85)/1.2`
  `= -0.332…`

 

`a` `=bar y – b bar x`
  `= 7.2 – (-0.332… xx 1.8)`
  `= 7.799…`

`=> C` 

CORE, FUR1 2006 VCAA 7 MC

For a set of bivariate data, involving the variables `x` and `y`,

`r =– 0.5675, \ bar x = 4.56, \ s_x = 2.61, \ bar y = 23.93 \ and\  s_y = 6.98`

The equation of the least squares regression line  `y = a + bx`  is closest to

A.   `y= 30.9 - 1.52x`

B.   `y = 17.0 - 1.52x`

C.   `y = – 17.0 + 1.52x`

D.   `y = 30.9 - 0.2x`

E.   `y = 24.9 - 0.2x`

Show Answers Only

`A`

Show Worked Solution
`b` `=r xx s_y/s_x`
  `=-0.5675 xx 6.98/2.61`
  `=-1.517…`
`a` `=bar y – b barx`
  `=23.93 – (- 1.517… xx 4.56)`
  `=30.85…`

`rArr A`