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

by OC2447826

Department

Economics for Management StudiesCourse Code

MGEB12H3Professor

ATA Study Guide

MidtermThis

**preview**shows pages 1-3. to view the full**28 pages of the document.**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!

Last

Name:

First

Name:

ID

FOR MARKERS ONLY:

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.

MC

Q#1

Q#2

Total

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.

Solution:

0: = 4.6

1: > 4.6

50 =5.2 −4.6

2.49

√51

= 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?

Solution

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 rejecting “recognition 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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