smc-265-71-Linearise – Squared/Inverse

Data Analysis, GEN1 2024 NHT 11-12 MC

The table below shows the birth rate, in number of births per 1000 people, and the average annual income, in dollars per person, for a sample of 12 countries.

A scatterplot displaying the data is also shown.

Question 11

A squared transformation applied to the variable birth rate can be used to linearise the scatterplot.

The equation of the least squares line is

$(\textit{birth rate})^2=953-0.0333 \times \textit{income}$

Using this equation, the predicted birth rate, in number of births per 1000 people, for a country with an average annual income of $18 500 is closest to

17.7
18.0
18.4
337
113 535

Question 12

Coefficients of determination were calculated for

birth rate vs income (coefficient 1)
(birth rate)$^2$ vs income (coefficient 2 )
birth rate vs (income)$^2$ (coefficient 3 )

These coefficients were ranked in order from largest to smallest.

The order would be

coefficient 1 , coefficient 2 , coefficient 3
coefficient 2 , coefficient 3 , coefficient 1
coefficient 2 , coefficient 1 , coefficient 3
coefficient 3, coefficient 1, coefficient 2
coefficient 3, coefficient 2, coefficient 1

Show Answers Only

$\text{Question 11:}\ C$

$\text{Question 12:}\ E$

Show Worked Solution

$\text{Question 11}$

$\textit{(birth rate)}^2 = 953-0.0333 \times 18\,500=336.95 $

$\textit{birth rate}\ = \sqrt{336.95} = 18.356 $

$\Rightarrow C $

$\text{Question 12}$

$\text{Calculate the coefficient of determination for each pair:}$

$\text{Coefficient 1 = 0.845, coefficient 2 = 0.915, coefficient 3 = 0.948}$

$\Rightarrow E$

Data Analysis, GEN1 2024 VCAA 11-12 MC

The number of breeding pairs of a small parrot species has been declining over recent years.

The table below shows the number of breeding pairs counted, pairs, and the year number, year, for the last 12 years. A scatterplot of this data is also provided.

The association between pairs and year is non-linear.

Part 1

The scatterplot can be linearised using a logarithmic (base 10) transformation applied to the explanatory variable.

The least squares equation calculated from the transformed data is closest to

$\log _{10}(pairs)=2.44-0.0257 \times year$
$\log _{10}(pairs)=151-303 \times year$
$pairs =274-12.3 \times \log _{10}(year)$
$pairs =303-151 \times \log _{10}( year)$

Part 2

A reciprocal transformation applied to the variable $pairs$ can also be used to linearise the scatterplot.

When a least squares line is fitted to the plot of $\dfrac{1}{pairs}$ versus $year$, the largest difference between the actual value and the predicted value occurs at $year$

Show Answers Only

Part 1: $D$

Part 2: $A$

Show Worked Solution

Part 1

$\text{Apply reciprocal transformation and find regression line:}$

$\text{Least squares equation:}\ \ pairs=303-151\times\log_{10}{(year)}$

$\Rightarrow D$

Part 2

$\text{Largest difference occurs at year 1}$

$\Rightarrow A$

♦ Mean mark (Part 1) 48%.

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

$\textit{height}\ =-13.04 + 40.22 \times \dfrac{1}{\textit{age}}$
$\textit{height}\ =-10.74+8.30 \times \dfrac{1}{\textit{age}}$
$\textit{height}\ =2.14 + 0.63 \times \dfrac{1}{\textit{age}}$
$\textit{height}\ =13.04-40.22 \times \dfrac{1}{\textit{age}}$
$\textit{height}\ =16.56-22.47 \times \dfrac{1}{\textit{age}}$

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

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} $

$\Rightarrow D$

CORE, FUR2 2021 VCAA 5

A method for predicting future time differences in the 100 m freestyle swim is to use the formula

difference = winning time women – winning time men

A resulting data and time series plot are shown below. The plot is clearly non-linear.

Apply a reciprocal transformation to the variable difference to linearise the data. Fit a least squares line to the transformed data and write its equation below.
Round the values of the intercept and the slope to four significant figures. (2 marks)

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Use the equation from part a. to predict, in seconds, the difference the women's and men's winning times in the year 2032.
Round your answer to one decimal place. (1 mark)

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Show Answers Only

`1/text{difference} = -2.234 + 0.001209 xx text{year}`
`4.5 \ text{seconds}`

Show Worked Solution

a. `text{By CAS}: \ x text{-variable = year}, \ \ y text{-variable} = 1/text{difference}`

`:. \ text{Equation of least squares line:}`

`1/text{difference} = -2.234 + 0.001209 xx text{year}`

b. `1/text{difference}\ = -2.234 + 0.001209 xx 2032= 0.222688`

`:. \ text{difference}\ = 4.490 …= 4.5 \ text{seconds (to 1 d.p.)}`

CORE, FUR2 2020 VCAA 6

The table below shows the mean age, in years, and the mean height, in centimetres, of 648 women from seven different age groups.

What was the difference, in centimetres, between the mean height of the women in their twenties and the mean height of the women in their eighties? (1 mark)

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A scatterplot displaying this data shows an association between the mean height and the mean age of these women. In an initial analysis of the data, a line is fitted to the data by eye, as shown.

Describe this association in terms of strength and direction. (1 mark)

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The line on the scatterplot passes through the points (20,168) and (85,157).

Using these two points, determine the equation of this line. Write the values of the intercept and the slope in the appropriate boxes below.

Round your answers to three significant figures. (1 mark)

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mean height =

× mean age

In a further analysis of the data, a least squares line was fitted.

The associated residual plot that was generated is shown below.

The residual plot indicates that the association between the mean height and the mean age of women is non-linear.

The data presented in the table in part a is repeated below. It can be linearised by applying an appropriate transformation to the variable mean age.

Apply an appropriate transformation to the variable mean age to linearise the data. Fit a least squares line to the transformed data and write its equation below.

Round the values of the intercept and the slope to four significant figures. (2 marks)

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Show Answers Only

`10.4\ text(cm)`
`text(Strong and negative)`
`text(mean height) = 171 – 0.169 xx text(mean age)`
`text(mean height) = 167.9 – 0.001621 xx text{(mean age)}^2`

Show Worked Solution

a.	`text(Difference)`	`= 167.1 – 156.7`
		`= 10.4\ text(cm)`

Mean mark part b. 51%.

b. `text(Strong and negative.)`

♦♦ Mean mark part c. 23%.

c. `text(Slope) = (157 – 168)/(85 – 20) = -0.169`

`text(Equation of line)`

`y – 168`	`= -0.1692 (x – 20)`
`y`	`= -0.169x + 171`

`:.\ text(mean height) = 171 – 0.169 xx text(mean age)`

D. `text(By CAS)`

`text(mean height) = 167.9 – 0.001621 xx text{(mean age)}^2`

Data Analysis, GEN1 2019 NHT 10 MC

A student uses the data in the table below to construct the scatterplot shown below.

A squared transformation is applied to the variable `x` to linearise the data.

A least squares line is fitted to this linearised data with `x^2` as the explanatory variable.

The equation of this least squares line is closest to

`y = 94.1 - 12.3x^2`
`y = 95.8 - 1.57x^2`
`y = 8.76 - 0.0768x^2`
`y = 107 - 111x^2`
`y = 107 - 0.0768x^2`

Show Answers Only

`B`

Show Worked Solution

`text(Input all data points into CAS:)`

`y = 95.8 – 1.57x^2`

`=> B`

CORE, FUR1 2019 VCAA 12 MC

The table below shows the values of two variables `x` and `y`.

The associated scatterplot is also shown.

The explanatory variable is `x`.

The scatterplot is non-linear.

A squared transformation applied to the variable `x` can be used to linearise the scatterplot.

The equation of the least squares line fitted to the linearised data is closest to

`y = – 1.34 + 0.546x`
`y = – 1.34 + 0.546x^2`
`y = 3.93 - 0.00864x^2`
`y = 34.6 - 10.5x`
`y = 34.6 - 10.5x^2`

Show Answers Only

`B`

Show Worked Solution

`y = -1.34 + 0.546x^2\ \ \ text{(by CAS)}`

`=> B`

CORE, FUR1 2018 VCAA 11 MC

Freya uses the following data to generate the scatterplot below.

The scatterplot shows that the data is non-linear.

To linearise the data, Freya applies a reciprocal transformation to the variable `y`.

She then fits a least squares line to the transformed data.

With `x` as the explanatory variable, the equation of this least squares line is closest to

`1/y = −0.0039 + 0.012x`
`1/y = −0.025 + 1.1x`
`1/y = 7.8 - 0.082x`
`y = 45.3 + 59.7 xx 1/x`
`y = 59.7 + 45.3 xx 1/x`

Show Answers Only

`A`

Show Worked Solution

`text(By trial and error:)`

COMMENT: Carefully choosing a data point to test all options can sometimes eliminate all incorrect answers.

`text(Consider Option)\ A,`

`text{Using (9,9) to test .. }`

`1/9`	`~~ -0.0039 + 0.012x`
`0.111…`	`~~0.1041\ \ text{(a close approximation)}`

`text(All other options are not close.)`

`=> A`

CORE, FUR2 2008 VCAA 5

The number of hours spent doing homework each week (homework hours) and the number of hours spent watching television each week (television hours) were recorded for a group of 20 Year 12 students.

The results are displayed in the table below and a scatterplot constructed as shown.

The relationship between homework hours per week and television hours per week is clearly nonlinear.

A reciprocal transformation applied to the variable, television hours, can be used to linearise the scatterplot.

Apply this reciprocal transformation to the data and determine the equation of the least squares regression line that allows homework hours to be predicted from the reciprocal of television hours.
Write the coefficients correct to two decimal places. (2 marks)

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If a student spends 12 hours per week watching television, use the least squares regression line to predict the number of hours that the student spends doing homework. Give your answer correct to one decimal place. (1 mark)

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Show Answers Only

`text(Homework hours) = 5.12 + 102.90 xx 1/(text{TV hours})`
`13.7`

Show Worked Solution

a. `text(By calculator,)`

♦♦♦ Avg mean mark for both parts 16%.
MARKER’S COMMENT: A majority of students didn’t understand the term “reciprocal”.

`text(Homework hours) = 5.12 + 102.90 xx 1/(text{TV hours})`

b.	`text(Homework hours)`	`= 5.12 + 102.90 xx 1/12`
		`= 13.695`
		`=13.7\ \ text{(to 1 d.p.)}`

CORE, FUR2 2012 VCAA 4

The wind speeds (in km/h) that were recorded at the weather station at 9.00 am and 3.00 pm respectively on 18 days in November are given in the table below. A scatterplot has been constructed from this data set.

Let the wind speed at 9.00 am be represented by the variable ws9.00am and the wind speed at 3.00 pm be represented by the variable ws3.00pm.

The relationship between ws9.00am and ws3.00pm shown in the scatterplot above is nonlinear.

A squared transformation can be applied to the variable ws3.00pm to linearise the data in the scatterplot.

Apply the squared transformation to the variable ws3.00pm and determine the equation of the least squares regression line that allows (ws3.00pm)² to be predicted from ws9.00am.

In the boxes provided, write the coefficients for this equation, correct to one decimal place. (2 marks)

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Use this equation to predict the wind speed at 3.00 pm on a day when the wind speed at 9.00 am is 24 km/h.

Write your answer, correct to the nearest whole number. (1 mark)

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Show Answers Only

`(ws3.00p m)^2 = 3.4 + 6.6 xx ws9.00am`
`13`

Show Worked Solution

a. `text(By Calculator,)`

MARKER’S COMMENT: Many students were unable to deal with the squared transformation.

`(ws3.00p m)^2 = 3.4 + 6.6 xx ws9.00am`

b. `text(Substitute)\ \ ws9.00am = 24\ \ text(into the equation:)`

`(ws3.00p m)^2=3.4+6.6 xx 24=161.8`

`:. ws3.00p m≈ 12.720… =13\ \ text{(nearest whole)}`

CORE, FUR1 2010 VCAA 11 MC

A student uses the following data to construct the scatterplot shown below.

A reciprocal transformation is applied to the `x`-axis and is used to linearise the scatterplot.

With `y` as the dependent variable, the slope of the least squares regression line that is fitted to the linearised plot is closest to

A. `–249`

B. `–25`

C. `0.004`

D. `25`

E. `249`

Show Answers Only

`E`

Show Worked Solution

`text(By calculator,)`

♦♦ Mean mark 35%.

`text(Gradient of)\ y\ text(versus)\ 1/x\ text(is closest to 249.)`

`=> E`

CORE, FUR1 2006 VCAA 9 MC

A student uses the following data to construct the scatterplot shown below.

To linearise the scatterplot, she applies an `x`-squared transformation.

She then fits a least squares regression line to the transformed data with `y` as the dependent variable.

The equation of this least squares regression line is closest to

A. `y = 7.1 + 2.9x^2`

B. `y = –29.5 + 26.8x^2`

C. `y = 26.8 - 29.5x^2`

D. `y = 1.3 + 0.04x^2`

E. `y = –2.2 + 0.3x^2`

Show Answers Only

`A`

Show Worked Solution

`text(By calculator,)`

♦ Mean mark 42%.

`y=7.147+2.9387 x^2`

`rArr A`

CORE, FUR1 2013 VCAA 10 MC

The data in the scatterplot below shows the width, in cm, and the surface area, in cm², of leaves sampled from 10 different trees. The scatterplot is non-linear.

To linearise the scatterplot, (width)² is plotted against area and a least squares regression line is then fitted to the linearised plot.

The equation of this least squares regression line is

(width)² = 1.8 + 0.8 × area

Using this equation, a leaf with a surface area of 120 cm² is predicted to have a width, in cm, closest to

A. 9.2

B. 9.9

C. 10.6

D. 84.6

E. 97.8

Show Answers Only

`B`

Show Worked Solution

`text(Substituting)`

`text(Width)^2`	`=1.8+0.8 xx area`
	`=1.8+(0.8×120)`
	`=97.8`
`:.\ text(Width)`	`=\sqrt97.8`
	`=9.889…\ text(cm)`

`=> B`

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