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SOLUTIONS Question 1 1. (D) 2. (D) 3. (D) 4. (A) 5. (B) 6. (D) 7. (C) 8. (A) 9. (D) 10. (B) Question 2 The decision variables are:

Br = slices of bread in the lunch Bu = Tablespoons of butter in the lunch Je = Tablespoons of jelly in the lunch C = number of cookies in the lunch M = cups of milk in the lunch Ju = cups of juice in the lunch Minimize C = 5.5Br + 4.5Bu + 3.0Je + 8.0C + 16M + 32Ju subject to 60Br + 100Bu + 45Je + 65C + 160M + 100Ju ≥ 450 calories 60Br + 100Bu + 45Je + 65C + 160M + 100Ju ≤ 650 calories 10Br + 65Bu + 25C + 80M ≤ 0.35(60Br + 100Bu + 45Je + 65C + 160M + 100Ju) 3Je + 2M + 50Ju ≥ 60mg Vitamin C 3.5Br + 4.5Bu + 1.5C + 8M + Ju ≥ 12mg Protein Br = 2 slices Bu ≥ 1.5 Je M + Ju ≥ 1 cup and Br ≥ 0, Bu ≥ 0, Je ≥ 0, C ≥ 0, M ≥ 0, Ju ≥ 0. One difficulty that may be encountered in implementing this model is that the model may recommend continuous quantities e.g., 1.2 cup juice and 1.39 cup milk which may pose practical difficulties. We may use our judgment to appropriately round off these continuous values.

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Question 3 No research Growth Contraction 0.6 0.4 Equity 15 -5 7 Bonds 2 5 3.2 MM 2.5 2.5 2.5

Revised Probabilities

Sj P(Ik | Sj) P( Ik ∩ Sj) P(Sj | Ik) FG FC FG FC FG FC Growth 0.6 0.85 0.15 0.51 0.09 0.927 0.200 Contraction 0.4 0.1 0.9 0.04 0.36 0.073 0.800 0.55 0.45

With Research Growth Contraction 0.927 0.073 FG Equity 15 -5 13.54 Bonds 2 5 2.219 MM 2.5 2.5 2.5 Growth Contraction 0.2 0.8 FC Equity 15 -5 -1 Bonds 2 5 4.4 MM 2.5 2.5 2.5 FG FC Research 0.55 0.45 13.54 4.4 9.427

Optimal Decision Conduct Market Research

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If Study forecasts growth (FG), invest in equity funds If Study forecasts contraction (FC), invest in bond funds The maximum you should be willing to pay for the economic forecast is 2.427% of the total investment $2m = $48540. Question 4 (a)

( )

( )2

0 1

i Let be papulation mean difference between the nicotine content of current and new filters. Assume that the differences in nicotine content are from a random

sample from

: 0 against :

d

d d

d

N , .

H H

μ

μ σ

μ =

( )

( )

0 12

8 05 0

02 marks

Under the test statistic where =9.

1 32Calculated 1 692 35 9

2 marksAs the calculated 1 69 1 86 we fail to reject at =0.05 and conclude that the new fi

d

d

dn d

d d

,.

XH t ~ t nS / n

.t .. /

t . t . , H

μ

α

−

>

=

= =

= < =

( )lter does not leave less nicotine than the current filter.

1 mark

( ) ( ) ( )

( ) ( )

( )

1 8

8 10 8 05

ii 1 69

From the tabulated values in Table 4, 1 397 and 1 860

Hence for the test in (i), 0.05 < < 0.10.2 marks

dn

,. ,.

P value P t Cal.t P t .

P t . P t . .

P value

−− = > = >

= =

−

4

( ) ( )( )

( )

( )

05iii The critical value in terms of the mean difference,

2 35 9 1 645

1 28861 2886 2So the power of the test at 2 002 35 90 9082

0 81813 marks

d d d .

d

x / n z

. / .

... P Z. /

P Z ..

σ

μ

=

=

=

−⎛ ⎞= = >⎜ ⎟⎝ ⎠

= > −

=

(b) ( )

( )

( )

( )

2 2 2 20 M W 1 M W

2 2

2

0 1 12

i : against :

1 mark

Given 21 16 27102436 21160000

Under the test statistic

1 mark27102436Calculated 1 28121160000

1 markAs the calculated F 0 7807

M W

M W M W

Mn ,n

W

H H

n ,n ,s ,s

sH F ~ Fs

F .

.

σ σ σ σ

− −

= ≠

= = = =

=

= =

=

( )

20 15 025 02 76 we fail to reject at =0.05 and conclude that there is no evidence of salary discrimination between menand women.

2 marks

, ,.F . , H α< =

5

( )

( )

( )( )

2 22 2M W 05 1 12 2

05 1 1

1ii 95 confidence interval for / :

2 marks

27102436 1 27102436 2 2021160000 2 33 21160000

0 55 2 82

3 marks

W M

M W

M M. ,n ,n

W . ,n ,n W

s s% , Fs F s

, ..

. , .

σ σ − −− −

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠=

Question 5 (a) ( ) 0 1 1 2 2 3 3

1

2

iwhere is the $ amount of merchandise purchased during the visit,

is the time spent (in minutes) at the site during the purchase visit,1 if MasterCard is used0 otherwise,

y x x xy

x

x

β β β β ε= + + + +

=

( )

3

2

1 if Visa is used 0 otherwise,

and is the error term with mean 0 and variance and for =1,2, , , they are independent.

4 marks

x

i nεε σ

⎧⎨⎩⎧

= ⎨⎩

K

( )

( )

0 1 1 2 2 3 3 4 1 2 3 1 3

1 2 3where the variables , and are explaind in ( ) above.2 marks

ii y x x x x x x xy , x x x i

β β β β β β ε= + + + + + + +

(b) (i) 1 10 064ˆ .β = has no meaningful interpretation as 2

1x is also in the model. (1 mark)

2 12 439ˆ .β = − gives difference in the mean maintenance time between the two types of machines, i.e. the mean maintenance time for machine B is 12.439 less than machine A.

(1 mark)

6

3 411ˆ .β = − reveals that the curve of the relationship between maintenance time and machine age is mound-shaped.

(1 mark) ( )

( )0 3 1 3ii : 0 : 0

1 mark

H against Hβ β= ≠

( ) ( )( )

30 3 4

3

18 025 0

iii Under : 0 the test statistc, where 22

1 mark0 411Cal. 2 585

0 159As the cal. (= -2.585) < 2 101 we reject and conclude that thequadratic term is useful for

nH t t nse

t

t t H,.

ˆ~ .ˆ

. ..

. ,

ββ

β−= = =

−= = −

= −

( )predicting the maintenace time.

2 marks ( )

( )

0

1

05 05

iv : There is no first - order autocorrelation : There is positive or negative first - order autocorrelation

1 markFrom the SPSS output, cal. d = 2.252Critical values : 1 05, 1 66

4L U

H against H

d d,. ,.. . ,= =

( )

05 05

05 05

0

2 34 4 - 4 1 05 2 95As 1 66 d = 2.252 < 4 - 2 34 we fail to reject

at = 0.10 and conclude that no first - order autocorrelation. 3 marks

U L

U U

d dd Cal. d

H

,. ,.

,. ,.

. , . .. . ,

α

− = = − =

= < =

( )

( )

2

2

v Adj. for the multiple regression model and simple lineasr regression

models are 0.936, 0.701, respectively. As the change in Adj. 0 936 0 701 0 235 23 50 it is clear that the multiple regressi

R

R . . . . % ,= − = =

( )on is better.

3 marks