5/28/2018 AB103-2008-2009-Sem1-Exam-Soln

1/6

1

SOLUTIONS: AB103 Final Examination Semester 1, 2008-2009

Question 1

1. (A)2. (C)3. (B)4. (C)5. (D)6. (A)7. (D)8. (B)9. (E)10. (B)

Question 2

The following decision variables are defined:

Li = Amount invested in Lieman MiniBonds at the beginning of year i = 1, 2, 3, 4Mi = Amount invested in Merry and Link Bonds at the beginning of year i = 1, 2, 3, 4F2 = Amount invested in Far East Bonds at the beginning of year 2A4 = Amount invested in American Bonds at the beginning of year 4MMi = Amount invested in money market funds at the beginning of year year i = 1, 2,3, 4X = Amount accumulated at the beginning of Year 5

Maximize X

The following constraints are defined:

Beginning of Year 1L1 + M1 + MM1 = 150000

Beginning of Year 2L2 + M2 + F2 + MM2 = 1.02MM1

Beginning of Year 3

L3 + MM3 = 1.02MM2 + 1.15L1

Beginning of Year 4A4 + MM4 = 1.02MM3 + 1.15L2 + 1.25M1

Beginning of Year 5X = 1.02MM4 + 1.15L3 + 1.25M2 + 1.30F2 +

1.10A4

5/28/2018 AB103-2008-2009-Sem1-Exam-Soln

2/6

2

Note: Variables M3, L4 and M4 are not included as they will not impact the solution.

Question 3

a. Optimal decision is to choose FD. Expected returns 2500

No Research

0.40 0.60

EQ EQ 10000.00-

4000.00 1600.00

FD 2500.00 2500.00 2500.00 2500.00

EVwPI 10000.00 2500.00 5500.00

EVPI 3000.00

EVPI = $3000Since the research costs $500 < EVPI, a priori it might be worthwhile toconduct research.

Also see Decision Tree.

b.

Revised Probabilities

P(Ik|Sj) P(Ik Sj) P(Sj | Ik)

P(Sj) I1 I2 I1 I2 I1 I2

S1 0.40 0.60 0.40 0.24 0.16 0.67 0.25

S2 0.60 0.20 0.80 0.12 0.48 0.33 0.75

0.36 0.64

S1 S2

I1 0.67 0.33

EQ 10000.00-

4000.00 5380.00

FD 2500.00 2500.00 2500.00

I2 S1 S2

0.25 0.75

EQ 10000.00-

4000.00 -500.00

FD 2500.00 2500.00 2500.00

I1 I2

0.36 0.64

5380.00 2500.00 3536.80

5/28/2018 AB103-2008-2009-Sem1-Exam-Soln

3/6

3

Also See Decision Tree

EV with MR (after accounting for cost of research) = 3536.8-500 = 3036.8Since this is greater than the EV without MR, optimal strategy is to conduct research.Also see decision tree.

If I1 choose EQIf I2 choose FD

EVSI = EVwSI EV without SI = 3536.8 - 2500 = 1036.8.

Note:1. It is okay to work with alternate payoffs e.g., $10,000 + Returns or % returns.

Question 4

(a)

( )

( )

( ) ( )

( )

( ) ( )

22 2

2 22 2

2

2 2

0 1

02 2

1 1

6 25 16 8 69 14

6 25 16 8 69 14

16 1 14 1

i : against :

Under the test statistic where =

25 6923 26

11 7 14 9

Calculated 6 25

A A B B

A A B B

A B

A B A B

s / n s / nA B

s / n s / nA B

n n

A B

. / . /

. / . /

H H

X XH t ~ t

S S

n n

.

. .

t .

+

+

+

+

=

=

+

=

=

=

26 025 0

3 1828 69

16 14

As the calculated 3 182 2 056 we reject at =0.05

and conclude that the average profits generated by staff under the two

training programs are different.

,.

..

t . t . , H

= +

= < =

6 marks

( )

( )

2 2

26 025ii 95 confidence interval for :

6 25 8 6911.7-14.9 2 05616 14

5 268 1 132

A BA B A B ,.

A B

s s% x x t

n n

. ..

. , .

+

= +

=

2 marks

(iii) In hypothesis testing in (i) with two sided alternative hypothesis and CI in (ii)

both we used =0.05. Hence our conclusion in (i) must be consistent

5/28/2018 AB103-2008-2009-Sem1-Exam-Soln

4/6

4

with the results in (ii), i.e., reject H0as ( )0 5 268 1 132. , . 2 marks

(b)

( ) 0

1

i : The propotion of patients with congestive heart failure and alcohol consumption

are independent.

: The propotion of patients with congestive heart failure and alcohol consumption

are not indepen

H

H

( ) ( )

( )

2

2 2

0

2

2 2

2..01 0

dent.

( )Under the test statistic ~ with 2 df.

146 131.6132 292 273.8484Calculated 10.197

131.6132 273.8484

As the calculated ( 10.197) 9.210 we reject at =0.01 and

ij ij

ij

f eH

e

H

=

= + + =

= > =

K

coclude

that the propotion of patients with congestive heart failure and alcohol consumption arenot independent.

6 marks

( )

( ) ( )

( )

025

ii Let denotethe population proportion of moderate drinkers.

95 confidence interval for :

696 1913 1 696 19136961 96

1913 1913

0 342 0 385

m

m mm m .

p

p q% p p z

n

/ /.

. , .

=

=

3 marks

We are 95 % confident that the true proportion of moderate drinkers will be inthe interval (0.342, 0.385)

1 mark

Question 5

(a)ANOVA(b)

Model Sum of Squares df Mean Square F Sig.1 Regression 30781779.116 3 10260593.039 13.869 .000(a)

Residual 13317116.157 18 739839.786

Total 44098895.273 21

a Predictors: (Constant), x3, x1, x2b Dependent Variable: y

(b)

5/28/2018 AB103-2008-2009-Sem1-Exam-Soln

5/6

5

( ) ( )

( )

2

2

/ ( 1)Adjusted 1

/ 1

13317116.157 / (22 3 1)1 1 0.352 64.7%

44098895.273/ 21

a

i

SSE n k R

y y n

=

= = =

2 marks

This implies that the fitted model has explained 64.7% of the total sample variation in y values, afteradjusting for the sample size and the number of independent variables in the model.

1 mark

(c)

H0: 1= 2= 3= 0 against H1: At least one of 1, 2and 3not equal to 0.1 mark

, 1The test statistic ~ where =22 and =3.

Calculated 13.869

k n k

MSRF F n k

MSE

MSRF

MSE

=

= =

2 marks

The calculated F= 13.869 > F.05, 3, 18=3.16 and we conclude that at least one ofthe three variablesx1,x2 andx3 are useful for predictingy.

1 mark

(d)

H0: 1= 0 against H1: 10.

10 1

1

Under The test statistic ~ where =22 and =3.

( )

1 mark

Calculated t

n kH t t n kse

=

( )18

22.8611.25

18.293

1 mark

value 2 1.25 0.2

As the -value is very large, we fail to

P P t

p

= =

= > >

0reject for any

5/28/2018 AB103-2008-2009-Sem1-Exam-Soln

6/6

6

(iii) 95% CI for the average difference:

( )( )

( )

2 18 025 2

2689 2 101 471 345

1698 7042 3679 2958

,. t se

. .

. , .

=

=

2 marks