smc-265-70-Linearise

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 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

0.009434
0.01000
1.020
2.698
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

3
511
772
957
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%.

CORE, FUR1 2020 VCAA 14 MC

In a study, the association between the number of tasks completed on a test and the time allowed for the test, in hours, was found to be non-linear.

The data can be linearised using a log₁₀ transformation applied to the variable number of tasks.

The equation of the least squares line for the transformed data is

log₁₀ (number of tasks) = 1.160 + 0.03617 × time

This equation predicts that the number of tasks completed when the time allowed for the test is three hours is closest to

Show Answers Only

`C`

Show Worked Solution

`log_10 text{(number of tasks)}`	`= 1.160 + 0.03617 xx 3`
	`= 1.26851`

`:.\ text(Number of tasks)`	`= 10^(1.26851)`
	`= 18.557…`
	`~~ 19`

`=> C`

Data Analysis, GEN1 2019 NHT 11 MC

The Human Development Index (HDI) and the mean number of children per woman for 13 countries are related.

This relationship is non-linear.

To linearise the data, a `log_10` transformation is applied to the response variable children.

A least squares line is then fitted to the linearised data.

The equation of this least squares line is

`log_10 (text(children)) = 1.06 - 0.00674 xx HDI`

Using this equation, the mean number of children per woman for a country with a HDI of 95 is predicted to be closest to

Show Answers Only

`C`

Show Worked Solution

`log_10(text(children))`	`= 1.06 – 0.00674 xx 95`
	`= 0.4197`

`:.\ text(children)`	`= 10^(0.4197)`
	`= 2.62…`

`=>\ C`

CORE, FUR1 2018 VCAA 12 MC

A `log_10(y)` transformation was used to linearise a set of non-linear bivariate data.

A least squares line was then fitted to the transformed data.

The equation of this least squares line is

`log_10(y) = 3.1 - 2.3x`

This equation is used to predict the value of `y` when `x = 1.1`

The value of `y` is closest to

–0.24
0.57
0.91
1.6
3.7

Show Answers Only

`E`

Show Worked Solution

`log_10y`	`= 3.1 – 2.3(1.1)`
	`= 0.57`
`:.y`	`= 10^(0.57)`
	`= 3.71…`

`=> E`

CORE, FUR2 2017 VCAA 4

The eggs laid by the female moths hatch and become caterpillars.

The following time series plot shows the total area, in hectares, of forest eaten by the caterpillars in a rural area during the period 1900 to 1980.

The data used to generate this plot is also given.

The association between area of forest eaten by the caterpillars and year is non-linear.

A log₁₀ transformation can be applied to the variable area to linearise the data.

When the equation of the least squares line that can be used to predict log₁₀ (area) from year is determined, the slope of this line is approximately 0.0085385
Round this value to three significant figures. (1 mark)
Perform the log₁₀ transformation to the variable area and determine the equation of the least squares line that can be used to predict log₁₀ (area) from year.
Write the values of the intercept and slope of this least squares line in the appropriate boxes provided below.
Round your answers to three significant figures. (2 marks)

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The least squares line predicts that the log₁₀ (area) of forest eaten by the caterpillars by the year 2020 will be approximately 2.85

Using this value of 2.85, calculate the expected area of forest that will be eaten by the caterpillars by the year 2020.
i. Round your answer to the nearest hectare. (1 mark)

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ii. Give a reason why this prediction may have limited reliability. (1 mark)

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a. `0.00854\ (text(3 sig fig))`

b. `log_10(text(area)) = −14.4 + 0.000854 xx text(year)`

c.i. `708\ text(hectares)`

c.ii. `text(This prediction extrapolates significantly from the given)`
`text(data range and as a result, its reliability decreases.)`

Show Worked Solution

a. `0.0085385 = 0.00854\ (text(3 sig fig))`

♦ Mean marks of part (a) and (b) 44%.

`log_10(text(area))`

`= −14.4 + 0.000854 xx text(year)`

♦♦ Mean mark part (c)(i) 29%.
COMMENT: When the question specifies using the value 2.85, use it!

c.i.	`log_10(text(Area))`	`= 2.85`
	`:.\ text(Area)`	`= 10^2.85`
		`= 707.94…`
		`= 708\ text(hectares)`

c.ii. `text(This prediction extrapolates significantly from the given)`

`text(data range and as a result, its reliability decreases.)`

CORE, FUR1 2016 VCAA 11-12 MC

The table below gives the Human Development Index (HDI) and the mean number of children per woman (children) for 14 countries in 2007.

A scatterplot of the data is also shown.

Part 1

The scatterplot is non-linear.

A log transformation applied to the variable children can be used to linearise the scatterplot.

With HDI as the explanatory variable, the equation of the least squares line fitted to the linearised data is closest to

log(children) = 1.1 – 0.0095 × HDI
children = 1.1 – 0.0095 × log(HDI)
log(children) = 8.0 – 0.77 × HDI
children = 8.0 – 0.77 × log(HDI)
log(children) = 21 – 10 × HDI

Part 2

There is a strong positive association between a country’s Human Development Index and its carbon dioxide emissions.

From this information, it can be concluded that

increasing a country’s carbon dioxide emissions will increase the Human Development Index of the country.
decreasing a country’s carbon dioxide emissions will increase the Human Development Index of the country.
this association must be a chance occurrence and can be safely ignored.
countries that have higher human development indices tend to have higher levels of carbon dioxide emissions.
countries that have higher human development indices tend to have lower levels of carbon dioxide emissions.

Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ D`

Show Worked Solution

`text(Part 1)`

`=>A`

`text(Part 2)`

`text(A strong positive association does not mean)`

`text(an increase in one variable causes an increase)`

`text(in the other.)`

`=> D`

CORE, FUR2 2011 VCAA 4

The average age of women at first marriage in years (average age) and average yearly income in dollars per person (income) were recorded for a group of 17 countries.

The results are displayed in Table 2. A scatterplot of the data is also shown.

The relationship between average age and income is nonlinear.

A log transformation can be applied to the variable income and used to linearise the scatterplot.

Apply this log transformation to the data and determine the equation of the least squares regression line that allows average age to be predicted from log (income).
Write the coefficients for this equation, correct to two decimal places, in the spaces provided. (2 marks)

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Use this equation to predict the average age of women at first marriage in a country with an average yearly income of $20 000 per person.
Write your answer correct to one decimal place. (1 mark)
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Show Answers Only

`text(Average age) = 2.39 + 5.89 xx log (text(income))`
`text(27.7 years)`

Show Worked Solution

a. `text(By calculator after applying the log transformation,)`

`text(Average age) = 2.39 + 5.89 xx log (text(income))`

♦♦ Parts (i) and (ii) had a combined mean mark 0f 34%.
MARKER’S COMMENT: Students struggled with the log transformation, incorrect data entry and choosing the correct dependent variable.

b. `text(The average age at first marriage with an)`

`text(average yearly income of $20 000)`

`=2.39+5.89 xx log(20\ 000)`

`=27.7\ text{years (1 d.p.)}`

CORE, FUR2 2014 VCAA 3

The scatterplot and table below show the population, in thousands, and the area, in square kilometres, for a sample of 21 outer suburbs of the same city.

In the outer suburbs, the relationship between population and area is non-linear.

A log transformation can be applied to the variable area to linearise the scatterplot.

Apply the log transformation to the data and determine the equation of the least squares regression line that allows the population of an outer suburb to be predicted from the logarithm of its area.
Write the slope and intercept of this regression line in the boxes provided below.
Write your answers, correct to one decimal place. (1 mark)

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Use this regression equation to predict the population of an outer suburb with an area of 90 km².
Write your answer, correct to the nearest one thousand people. (1 mark)

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Show Answers Only

`text(Population)\ = 7.7 + 7.7 xx log_(10)(text(area))`
`23\ 000`

Show Worked Solution

a. `text(By Calculator,)`

♦ Mean marks of 43% (part (a)) and 35% (part (b)).
MARKER’S COMMENT: Many students confused the base 10 logarithm with the natural logarithm.

`text(Population)\ = 7.7 + 7.7 xx log_(10)(text(area))`

b. `text(By Calculator,)`

`text(Population)`	`= 7.7 + 7.7 xx log_(10) 90`
	`=22.747…`
	`= 23\ 000\ text{(nearest thousand)}`

CORE, FUR1 2015 VCAA 11 MC

A log transformation is used to linearise the relationship between the weight of a mouse, in grams, and its age, in weeks.

When a least squares regression line is fitted to the transformed data, its equation is

`text(weight) = – 7 + 30 log_10(text(age))`

This equation predicts that a mouse aged five weeks has a weight, in grams, that is closest to

A. `14`

B. `21`

C. `23`

D. `41`

E. `143`

Show Answers Only

`A`

Show Worked Solution

`text(Weight)`	`= – 7 + 30log_10 5`
	`= 13.969…`

`=> A`

CORE, FUR1 2007 VCAA 9 MC

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

To linearise the scatterplot, she applies a log y transformation; that is, a log transformation is applied to the `y`-axis scale.

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

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

A. `log y= –217 + 88.0\ x`

B. `log y = –3.8 + 4.4\ x`

C. `log y = 3.1 + 0.008\ x`

D. `log y = 0.88 + 0.23\ x`

E. `log y = 1.58 + 0.002\ x`

Show Answers Only

`D`

Show Worked Solution

`text(Two transformed data points will be)`

`(5, log 98), (9, log 869)`

`:.\ text(Gradient)`	`=(log 869 – log 98)/(9-5)`
	`= 0.2369…`

`rArr D`

CORE, FUR1 2009 VCAA 12 MC

The mathematics achievement level (TIMSS score) for grade 8 students and the general rate of Internet use (%) for 10 countries are displayed in the scatterplot below.

To linearise the data, it would be best to plot

A. mathematics achievement against Internet use.

B. log (mathematics achievement) against Internet use.

C. mathematics achievement against log (Internet use).

D. mathematics achievement against (Internet use)².

E. ` 1/text(mathematics achievement)` against Internet use.

Show Answers Only

`C`

Show Worked Solution

`text(The shape of the data is logarithmic.)`

`:.\ text(To linearise the data, it would be best to)`

`text{plot mathematics achievement against}`

`text{log (Internet use). }`

`=> C`

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