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MGEB12H3 Study Guide - Midterm Guide: Academic Dishonesty, Scantron Corporation, Test StatisticExam

Economics for Management Studies
Course Code
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MGEB12 Quantitative Methods in Economics-II
Winter 2019
Test-1 (Solutions)
Instructor: Ata Mazaheri
Instructions: This is a closed book test. A formula sheet is attached. You are allowed the use
of a calculator.
Write your full name and your student ID # on both this test book as well as the
Scantron paper.
You have 90 Minutes.
Good Luck!
The University of Toronto's Code of Behaviour on Academic Matters applies to all
University of Toronto Scarborough students. The Code prohibits all forms of
academic dishonesty including, but not limited to, cheating, plagiarism, and the use of
unauthorized aids. Students violating the Code may be subject to penalties up to and
including suspension or expulsion from the University.
Marks Earned
Maximum Marks
Possible 68 12 15 95

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Page 2 of 28
Part A (Long Questions)
Question-1 (12 Points) Our counseling center on campus is concerned that most students
requiring therapy do not take advantage of their services. Right now students attend only 4.6
sessions in a given year! Administrators are considering having patients sign a contract stating
they will attend at least 10 sessions in an academic year.
We had 51 patients sign the contract. For this sample, the average number of times they attended
therapy sessions was 5.2 and the standard deviation was 2.49.
Question: Does signing the contract actually increase participation/attendance?
a) [6 Points] State the null and alternative hypotheses. Using the p-value approach, test the
hypothesis at the 1% level of significance.
0: = 4.6
1: > 4.6
50 =5.2 4.6
= 1.785
P-Value: Since t statistics is between t0.05,50= 1.676 t0.025,50= 2.009, the p-value will be
between 2.5% and 5%.
Decision: Since the p-value is more than 1%. We cannot reject the null. Therefore, we can
not conclude the alternative.

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Page 3 of 28
b) [6 Points] Ignore part (a). It has been shown many times that on a certain memory test,
recognition is substantially better than recall. However, the P-Value for the data from your
sample was .12, so you were unable to reject the null hypothesis that recall and recognition
produce the same results.
b1: What would be the null and the alternative in this case? What constitutes a Type I error in this
problem? What is a Type II error?
b2: What type of error did you make?
H0: Recognition is not better than recall (they are equal)
H1: Recognition is better than recall
Type-I error: Rejecting "recognition is no better than recall" and conclude that it is better when in
fact it is not (Type I error)
Type-II error: Not rejectingrecognition is not better than recall” and as a result not to conclude
that recognition is better than recall when in reality it is.
(2 Points) In this case we have committed Type-II error (P-value is large so we do not reject the
null when the alternative is correct).
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