August2011 - Questions

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SMM882 Quantitative Methods for Business

9th

August 2011 14:00-16:30

Instructions to students:

Please answer ALL FOUR questions and indicate clearly which answer belongs to which

question. Please motivate all your answers by using appropriate explanations; an answer (e.g.,

“the p-value is 0.13”), even if it is correct, will be awarded zero marks without an explanation of

how you obtained the answer or why you think it is correct.

This examination paper consists of 13 printed pages: the title page, the examination questions (6

pages), a formula sheet (Appendix 1; 1 page), the standard normal distribution table (Appendix

2; 2 pages), and the t distribution table (Appendix 3; 3 pages).

Materials:

Number of answer booklets to be provided: 1

Calculator types that are permitted: Casio FX-83 GT+, Casio FX-85 GT+, Casio FX-83 MS,Casio FX-83 ES, Casio FX-85 MS, and Casio FX-85 ES. Al l o ther calculators are st rict ly

prohibited.

Dictionaries are not permitted.

The examination paper may NOT be removed from the examination room.

Examiner : Dr Hans TW Frankort

MSc in Management



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Question 2 (25 marks)

A global construction firm is interested in understanding the age profile of its construction

employees visiting a General Practitioner for back injuries. The responsible health and safety

executive expects that the average age of such employees is at least 35 years.

a. Formulate the null and alternative hypotheses that can be used to test this

expectation.

(5 marks)

He subsequently draws a random sample of 50 employees visiting a General Practitioner for

back injuries and he obtains a sample mean age of 35.78 and a sample standard deviation of

4.6.

b. Using the p-value to assess significance against α = 0.05, should the null hypothesis

be rejected?

(5 marks)

c. If you drew a smaller sample, would you be more or less likely to reject the null

hypothesis?

(5 marks)

d. Using a 90% confidence interval to conduct the hypothesis test, should the null

hypothesis be rejected?

(5 marks)

e. If α became larger, would you be more or less likely to reject the null hypothesis?

(5 marks)



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Question 4 (30 marks)

For business schools, it is important to understand how study behavior affects students’ exam

marks. For example, Guess Business School regularly analyzes the relationships between exam

marks and a range of study behaviors. Recently, it gathered data on the marks obtained by 52

students in the resit exam of one of its core modules, and on two study behaviors plus students’

marks in the original exam. Table 1 shows the variables created to analyze these data.

Table 1. The four variables as used in the analysis

Variable Explanation

Resit_mark Resit mark of a student, on a scale from 0 to 100

Tutorials0 (student attended no tutorials) or1 (student attended at least one tutorial)

Email_office0 (student did not ask questions by email / in an office meeting) or1 (student asked questions by email / in an office meeting)

Original_mark Original mark of a student, on a scale from 0 to 49

Table 2 shows descriptive statistics for these four variables, plus the correlations among them.

Table 2. Means, standard deviations, and correlations

Variable MeanStandard

deviation1. 2. 3.

1. Resit_mark 61.96 15.50 1.002. Tutorials 0.52 0.50 0.43 1.00

3. Email_office 0.44 0.50 0.40 0.24 1.00

4. Original_mark 23.46 11.54 0.54 0.56 0.01

a. How many students attended at least one tutorial? And how many students asked

questions by email or in an office meeting?

(5 marks)

b. Students with a lower Original_mark might be more likely to attend tutorials and ask

questions by email or in an office meeting. Does Table 2 support this expectation?

(5 marks)

To explore in more depth the possible determinants of students’ Resit_mark, Guess Business

School estimated two regression equations using the least squares method as follows:

(1) Resit_mark = β0 + β1(Tutorials) + β2(Email_office)

(2) Resit_mark = γ0 + γ1(Tutorials) + γ2(Email_office) + γ3(Original_mark)

Table 3 shows the results of the OLS estimation of equation (1).



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Table 3. Estimated regression equation of Resit_mark versus Tutorials and Email_office

Adjusted R-square

0.25

Dependent variable: Resit_mark

F(2, 49) Sig.

9.39 0.000

β Std. Error

(constant) 52.01 2.96

Tutorials 10.89 3.84

Email_office 9.71 3.87

c. Using 95% confidence intervals, interpret β1 and β2 in this estimated regressionequation.

(5 marks)

Guess Business School would like to use the results in Table 3 to obtain the predicted

Resit_mark for four categories of students:

i. Students that neither attended tutorials nor asked questions by email / in an office meeting

ii. Students that attended tutorials, but did not ask questions by email / in an office meeting

iii. Students that did not attend tutorials, but did ask questions by email / in an office meeting

iv. Students that both attended tutorials and asked questions by email / in an office meeting

d. Using Table 3, give the predicted Resit_mark for all four student categories.

(5 marks)

Table 4 shows the results of the OLS estimation of equation (2).

Table 4. Estimated regression equation of Resit_mark versus Tutorials, Email_office, and

Original_mark

Adjusted R-square

0.41

Dependent variable: Resit_mark

F(3, 48) Sig.

12.98 0.000

γ Std. Error

(constant) 40.00 4.07

Tutorials 1.73 4.14

Email_office 11.66 3.45

Original_mark 0.68 0.18



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e. Based on the information in Table 4, what are the t-values for the intercept

(constant), Tutorials, Email_office, and Original_mark? Which of the four are

significant at or beyond α = 0.01?

(5 marks)

f. Again give the predicted Resit_mark for student categories i, ii, iii, and iv, but now

base your answer on the estimates as shown in Table 4. Assume that for all four

categories, the Original_mark equals the mean Original_mark.

(5 marks)

****** END OF EXAMINATION PAPER ******



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Appendix 1: Formu la Sheet

N n ≤

( ) ∑= )( x xp X E

( )∑ −= )()( 2 x p x X Var µ

13 QQ IQR −=

( )( )

1−

−−= ∑

n

y y x xs

ii

XY

( )( )

N

y x Y i X i

XY

µ µ σ

−−= ∑

Y X

XY XY

ss

sr =

Y X

XY XY

σ σ

σ ρ =

σ

−= X

z X

x z

σ

µ 0−=

X

pn

xt

σ

µ 01

−=

−−

ib

i

pn

bt

σ

β 01

−=

−−

( )

1

2

2

−

−=∑

n

x xs

i

2ss =

( )

N

xi∑ −=

2

2 µ

σ

2σ σ =

n

s

N

n N X

1−

−=σ

n N

n N X

σ σ

1−

−=

n

s X =σ

n X

σ σ =

X z x σ α 2/± X pnt x σ α 2/,1−−

±

1

1)1(1 22

−−

−−−= pn

n RadjR 21

1)( j

j R

X VIF −

=



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Appendix 2: Cumulat ive Probab il it ies fo r the Standard Normal Dis tr ibut ion (2 pages!)

Table continued on p. 10



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Appendix 3: t Distribution (3 pages!)




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