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Examen

ECS3706 EXAM PACK 2025

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ECS3706 Latest exam pack questions and answers and summarized notes for exam preparation. Updated for 2025 exams . For assistance Whats-App.0.6.7..1.7.1..1.7.3.9 . All the best on your exams!!

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Publié le
13 octobre 2024
Nombre de pages
367
Écrit en
2024/2025
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Examen
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ECS3706
EXAM PACK




FOR ASSISTANCE WITH THIS MODULE +27 67 171 1739

,UNIVERSITY EXAMINATIONS




May/June 2024

ECS3706

ECONOMETRICS

100 marks
Duration: 4 hours
EXAMINERS:
FIRST: PROF S NHAMO
SECOND: DR M ZUKA
EXTERNAL: DR T QABHOBHO

CONFIDENTIAL

This paper consists of 10 pages, including a formulae sheet (page 7), three statistical tables
(pages 8 to 10), and the special front page.

Instructions:
(1) This paper consists of two sections.
Section A: Answer all three (3) questions, which together count for 60 marks:
(20 + 20 + 20) = 60.
(2) Section B: Answer any two (2) of the three (3) questions. Each question counts
for 20 marks: (2 x 20) = 40.
(3) Once you have completed your answers, submit your answers as a single document
in PDF format using the e-Assessment tool on the myExams portal. It is
preferable for you to type your answers (font: Arial 12) and then convert your
document to PDF for submission. However, if this is impossible, you may write your
answers down and scan them into a PDF file. Please write legibly.
(4) Emailed submissions WILL NOT be marked.
(5) Start with a cover page stating the module code (ECS3706) and your student
number.
(6) This should be followed by your answers to the questions.
(7) There is no need for a table of contents, an introduction, a conclusion, or a list of
references (as was required in your assignments). Simply answer the questions
asked.
(8) Ensure that each question and sub-question is clearly numbered.
(9) While you are not required to cite your sources, this does not mean that you can copy
information from any source. You need to answer the questions in your own words.
Plagiarism will not be tolerated and will result in disciplinary action if detected.
(10) Ensure that your PDF document is NOT encrypted to a “secured” mode and that it is
NOT password protected, as these files cannot be marked. Virus-infected files will
also not be marked.




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, Page 2 of 10 ECS3706
CONFIDENTIAL May/June 2024




ECS3706

Date: 22 May 2024




INSTRUCTIONS ON THE DAY OF ASSESSMENT: YOUR EXAM QR CODE & EXAM
ACCESS CODE
Ensure you are connected to the internet in order to log into the Invigilator
App and scan this QR code.

If you encounter difficulty in scanning the QR code, you can alternatively
enter the Exam Access Code below the QR code to start the invigilation.

If you finish your assessment before the app timer has run out you need to
press the ‘Finish Assessment’ button and follow instructions before you exit
or minimise the app.

Scan the QR code at the start time of the assessment, unless otherwise
specified by your institution, note that you can only scan this QR code once.
Exam Access Code: c042d3ea
The QR code is only scannable for a limited time and it should
therefore be scanned as soon as possible after the assessment
commencement time.

Keep the Invigilator App open on your smartphone for the full duration of
the assessment. You are not allowed to minimise or leave the app during
your assessment.

You must adhere to the assessment time limit communicated to you by your institution. The timer displayed in the Invigilator
may vary depending on the start time of invigilation.

FURTHER GUIDANCE
If you only have one device you may access your assessment in the application by pressing the ‘Access Browser’ button in
the top right corner of your app.

Once the QR code is scanned, ensure your media volume is turned up and place your smartphone next to you.

The Invigilator App will notify you with a notification beep when you are required to action a photo request, which you
should then perform as soon as possible.

We recommend that you keep your smartphone on charge for the duration of the assessment.

Ensure you are connected to the internet when commencing invigilation. You also need to be connected to the internet at the
end of the assessment in order to upload the app data.

If your assessment has multiple online sections, tests or attempts, you should NOT finish the invigilation until your entire
assessment has been completed.

Should you encounter any technical difficulty, please WhatsApp The Invigilator Helpdesk on +27 (0)73 505 8273
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, Page 3 of 10 ECS3706
CONFIDENTIAL May/June 2024



SECTION A (60 marks)

Answer ALL three (3) questions in section A.

Section A requires brief and to-the-point answers.

In most cases, list or briefly explain what is required.

Using statistical notations (mathematical symbols) to explain concepts may be advantageous – but
please ensure that you also explain their meaning.

You do not need to re-explain concepts that have been previously dealt with. If required, you may refer
to your previous answer/s.

In general, each mark represents one correct fact or correct interpretation.



QUESTION A1 (20 marks)

Consider the following estimated equation (Equation 1)

𝐶̂𝑡 = 20.5 − 0.09𝑃𝑡 + 0.82𝑌𝐷𝑡 − 0.52𝐷1𝑡 − 1.2𝐷2𝑡 − 1.6𝐷3𝑡

Where Ct = per capita kilograms of beef consumed in South Africa in quarter t
Pt = the price of 100Kgs of beef (in Rand) in quarter t
YDt = per capita disposable income (in Rand) in quarter t
D1t = Dummy equal to 1 in the first quarter (Jan. – March) of the year and 0
otherwise
D2t = Dummy equal to 1 in the second quarter of the year and 0 otherwise
D3t =Dummy equal to 1 in the third quarter of the year and 0 otherwise


(a) What is the meaning of the estimated coefficient of YD? (3)
(b) Specify the expected signs of each of the coefficients. Explain your reasoning. (5)
(c) Suppose we changed the definition of D3t so that it was equal to 1 in the fourth quarter
and 0 otherwise and re-estimated the equation with all other variables unchanged. Which
of the estimated coefficients would change? Would your answer to part (b) above
change? Explain your answer. (3)
(d) Using Equation 1, explain the meaning of a dummy variable trap. (4)
(e) Equation 1 is now modified into Equation 2 as follows:

𝐶̂𝑡 = 20.5 − 0.09𝑃𝑡 + 0.82𝑌𝐷𝑡 − 0.52𝐷1𝑡 − 1.2𝐷2𝑡 − 1.6𝐷3𝑡 + 0.6(𝑌𝐷𝑡 ∗ 𝐷1𝑡 )

What is the difference between Equation 1 and Equation 2? Interpret 0.6, the coefficient
of (𝑌𝐷𝑡 ∗ 𝐷1𝑡 ) (5)




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