VCE General Maths Data Analysis 2018 Mini Test 2

This is the full VCE General Maths Exam with worked solutions. You can also try Mini-Tests, which are official VCAA exams split into short tests you can do anytime.

Number of marks: 10

Reading time: 4 minutes

Writing time: 22 minutes

Instructions
• Answer all questions in pencil on your Multiple-Choice Answer Sheet.
• Choose the response that is correct for the question.
• A correct answer scores 1; an incorrect answer scores 0.
• Marks will not be deducted for incorrect answers.
• No marks will be given if more than one answer is completed for any question.
• Unless otherwise indicated, the diagrams in this book are not drawn to scale.


Data analysis - 2018 (Part 2)

Question 1 [2018 Exam 1 Q9]

The coefficient of determination is 0.8339
The correlation coefficient \(r\) is closest to

  • A. –0.913
  • B. –0.834
  • C. –0.695
  • D. 0.834
  • E. 0.913
Correct Answer: A
Click here for full solution
Question 2 [2018 Exam 1 Q10]

In a study of the association between a person’s height, in centimetres, and body surface area, in square metres, the following least squares line was obtained.

body surface area = –1.1 + 0.019 × height

Which one of the following is a conclusion that can be made from this least squares line?

  • A. An increase of 1 m² in body surface area is associated with an increase of 0.019 cm in height.
  • B. An increase of 1 cm in height is associated with an increase of 0.019 m² in body surface area.
  • C. The correlation coefficient is 0.019
  • D. A person’s body surface area, in square metres, can be determined by adding 1.1 cm to their height.
  • E. A person’s height, in centimetres, can be determined by subtracting 1.1 from their body surface area, in square metres.
Correct Answer: B
Click here for full solution
Question 3 [2018 Exam 1 Q11]

Freya uses the following data to generate the scatterplot below.

x 1 2 3 4 5 6 7 8 9 10
y 105 48 35 23 18 16 12 12 9 9
Scatterplot of y vs x

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

  • A. \(\frac{1}{y} = -0.0039 + 0.012x\)
  • B. \(\frac{1}{y} = -0.025 + 1.1x\)
  • C. \(\frac{1}{y} = 7.8 - 0.082x\)
  • D. \(y = 45.3 + 59.7 \times \frac{1}{x}\)
  • E. \(y = 59.7 + 45.3 \times \frac{1}{x}\)
Correct Answer: A
Click here for full solution
Question 4 [2018 Exam 1 Q12]

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

  • A. –0.24
  • B. 0.57
  • C. 0.91
  • D. 1.6
  • E. 3.7
Correct Answer: E
Click here for full solution
Question 5 [2018 Exam 1 Q13]

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

Mean \(\bar{x} = 27.8\) \(\bar{y} = 33.4\)
Standard deviation \(s_x = 2.33\) \(s_y = 3.24\)
Equation of the least squares line \(y = -2.84 + 1.31x\)

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

  • A. 0.88
  • B. 0.89
  • C. 0.92
  • D. 0.94
  • E. 0.97
Correct Answer: D
Click here for full solution
Question 6 [2018 Exam 1 Q14]

A least squares line is fitted to a set of bivariate data.
Another least squares line is fitted with response and explanatory variables reversed.
Which one of the following statistics will not change in value?

  • A. the residual values
  • B. the predicted values
  • C. the correlation coefficient \(r\)
  • D. the slope of the least squares line
  • E. the intercept of the least squares line
Correct Answer: C
Click here for full solution
Question 7 [2018 Exam 1 Q15]

The table below shows the monthly profit, in dollars, of a new coffee shop for the first nine months of 2018.

Month Jan. Feb. Mar. Apr. May June July Aug. Sept.
Profit ($) 2890 1978 2402 2456 4651 3456 2823 2678 2345

Using four-mean smoothing with centring, the smoothed profit for May is closest to

  • A. $2502
  • B. $3294
  • C. $3503
  • D. $3804
  • E. $4651
Correct Answer: B
Click here for full solution
Question 8 [2018 Exam 1 Q16]

The quarterly sales figures for a large suburban garden centre, in millions of dollars, for 2016 and 2017 are displayed in the table below.

Year Quarter 1 Quarter 2 Quarter 3 Quarter 4
2016 1.73 2.87 3.34 1.23
2017 1.03 2.45 2.05 0.78

Using these sales figures, the seasonal index for Quarter 3 is closest to

  • A. 1.28
  • B. 1.30
  • C. 1.38
  • D. 1.46
  • E. 1.48
Correct Answer: C
Click here for full solution

End of Multiple-Choice Question Book

VCE is a registered trademark of the VCAA. The VCAA does not endorse or make any warranties regarding this study resource. Past VCE exams and related content can be accessed directly at www.vcaa.vic.edu.au

>