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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!)
Table continued on p. 12
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Table continued on p. 13
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