2006 Series 4 (3709) Singapore

7/31/2019 2006 Series 4 (3709) Singapore

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Business

StatisticsLevel 3

Model AnswersSeries 4 2006 Singapore (Code 3709)


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Education Development International plc 2006 Company Registration No: 3914767All rights reserved. This publication in its entirety is the copyright of Education Development International plc.

Reproduction either in whole or in part is forbidden without written permission from Education Development International plc.

International House Siskin Parkway East Middlemarch Business Park Coventry CV3 4PETelephone: +44 (0) 8707 202909 Facsimile: + 44 (0) 24 7651 6566

Email: [email protected]

Vision StatementOur vision is to contribute to the achievements of learners around

the world by providing integrated assessment and learning services,adapted to meet both local market and wider occupational needs

and delivered to international standards.


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1

Business Statistics Level 3Series 4 2006

How to use this booklet

Model Answers have been developed by Education Development International plc (EDI) to offeradditional information and guidance to Centres, teachers and candidates as they prepare for LCCIInternational Qualifications. The contents of this booklet are divided into 3 elements:

(1) Questions reproduced from the printed examination paper

(2) Model Answers summary of the main points that the Chief Examiner expected tosee in the answers to each question in the examination paper,plus a fully worked example or sample answer (where applicable)

(3) Helpful Hints where appropriate, additional guidance relating to individualquestions or to examination technique

Teachers and candidates should find this booklet an invaluable teaching tool and an aid to success.

EDI provides Model Answers to help candidates gain a general understanding of the standardrequired. The general standard of model answers is one that would achieve a Distinction grade. EDIaccepts that candidates may offer other answers that could be equally valid.

Education Development International plc 2006

All rights reserved; no part of this publication may be reproduced, stored in a retrieval system ortransmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwisewithout prior written permission of the Publisher. The book may not be lent, resold, hired out orotherwise disposed of by way of trade in any form of binding or cover, other than that in which it ispublished, without the prior consent of the Publisher.


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3709/4/06/MA 1

Business Statistics Level 3Series 4 2006

QUESTION 1

(a) Explain how you would decide whether to use the additive model or themultiplicative model to analyse a time series.

(4 marks)

The quarterly sales of a retail outlet over a three year period are given in the table below.

Quarterly Sales ($000s)

1 2 3 4

Year 1

Year 2

Year 3

20

25

30

40

50

60

60

70

80

40

50

60

(b) Plot the time series on a diagram and comment on what the diagram shows.(4 marks)

(c) Using the moving average method find the trend within the data.(4 marks)

Further analysis of a larger set of data has revealed the following quarterly variation factors.(These factors have been calculated using the multiplicative model)

Quarter 1 0.5

Quarter 2 1.0

Quarter 3 1.5

Quarter 4 1.0

(d) Using the above information forecast the sales figures for Quarters 1 and 2 of year 4.(4 marks)

(e) Calculate the seasonally adjusted sales figures for year 3 and comment.(4 marks)

(Total 20 marks)


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3709/4/06/MA CONTINUED ON NEXT PAGE2

MODEL ANSWER TO QUESTION 1

(a)

Examine the difference between the data and estimated trend values.

Data - Trend constant indicates additive model to be used.

Data - Trend increasing or decreasing indicates multiplicative model to be

used.Examine the relationship between the trend and the seasonal factors, if

independent - additive, if dependent - multiplicative.

If diagram used the above points could be seen.

(b)

Graph of Sales

0

20

4060

80

0 1 2 3 4 5 6 7 8 9 10 11 12 13

Quarters

Sales($0

00)

Comment: Upward trend in sales. Regular annual pattern present.

Year Quarter Sales Sum 4s Sum 8s Moving

Average

Trend

1 1 20

2 40

160

3 60 325.000 40.625

165

4 40 340.000 42.500

175

2 1 25 360.000 45.000

1852 50 380.000 47.500

195

3 70 395.000 49.375

200

4 50 410.000 51.250

210


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3709/4/06/MA 3

MODEL ANSWER TO QUESTION 1 CONTINUED

3 1 30 430.000 53.750

220

2 60 450.000 56.250

230

3 80

4 60

(d)

Average change in trend = Last trend value - First trend value

Number of changes in trend

= 56.25 - 40.625 = 2.232

7

Forecast = Predicted trend adjusted by the average quarterly variation

Year 4 Quarter 1 Forecast = [56.25 + 3(2.232)] x 0.5 = 31.473 ($31,473)

Year 4 Quarter 2 Forecast = [56.25 + 4(2.232)] x 1.0 = 65.178 ($65,178)

(e)

Seasonally adjusted figure = Actual data adjusted by the averageseasonal variation

Sales ASV S.A.F.

($000)

Year 3 Qt 1 30 0.50 60.000

Qt 2 60 1.00 60.000

Qt 3 80 1.50 53.333

Qt 4 60 1.00 60.000

Comment: The seasonally adjusted sales figures for year 3 are the same

for quarters 1,2 and 4. There is a fall in the figure for quarter 3.


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3709/4/06/MA 4

QUESTION 2

The data below relates to the monthly manufacturing cost of a company in thousands of dollars, andthe monthly output in thousands of units. The data has been collected over a ten month period.

(a) Determine the least squares regression equation of monthly manufacturing cost on monthlyoutput and calculate the correlation coefficient between the two variables.

(12 marks)

(b) Forecast the manufacturing cost for November given that the expected output for the month is7500 units.

(2 marks)

(c) Carry out a suitable test to ascertain whether the correlation coefficient is significantly differentfrom zero.

(6 marks)

(Total 20 marks)

Month MonthlyManufacturing

Cost ($thousands)

Output(thousands units)

JanFebMarAprMayJunJul

AugSepOct

40504020303025304050

9856374357


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(a)

y x xy y2 x2

40.00 9.00 360.00 1600.00 81.00

50.00 8.00 400.00 2500.00 64.00

40.00 5.00 200.00 1600.00 25.0020.00 6.00 120.00 400.00 36.00

30.00 3.00 90.00 900.00 9.00

30.00 7.00 210.00 900.00 49.00

25.00 4.00 100.00 625.00 16.00

30.00 3.00 90.00 900.00 9.00

40.00 5.00 200.00 1600.00 25.00

50.00 7.00 350.00 2500.00 49.00

355.00 57.00 2120.00 13525.00 363.00

y = a + bx

b = 10(2120) - (57)(355) b = 2.5330

10(363) - (57)2

a = 355 - (2.533)(57) a = 21.063

10 10

y =21.063 + 2.533x (Round values acceptable)

CorrelationCoefficient

r = 965

1874.76

r = 0.5147

(b)

x = 7.5 Cost = 21.063 + 2.533(7.5)

Cost = 40.06

Cost = 40,060

( )r

n xy x y

n x x n y y=

( ( ) )( ( ) )2 2 2 2

bn xy x y

n x x=

2 2( )


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3709/4/06/MA 6


(c)

H0: = 0, H1: 0

Critical t value (8 df, 5% & 1%) = 2.31 & 3.36

t = 1.4558 = 1.698

0.8574

The calculated t value is less than both critical values.

There is no evidence to reject H0. The correlation coefficientis not significantly different from zero.

tr n

r=

2

12

2649.01

2105147.0t

=


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3709/4/06/MA 7

QUESTION 3

(a) Briefly explain the following and give an appropriate example for each.

(i) Mutually exclusive events(ii) Independent events

(4 marks)

A company has three independent sales teams. The number of sales persons in each team and thedistribution of males and females within the teams is shown below.

SalesTeam

Male Female

A

B

C

32

20

16

16

20

40

(b) Calculate the probability that a sales person selected at random

(i) belongs to sales team A(ii) belongs to sales team A given that she is a female(iii) is male given that he is from sales team C(iv) is male and belongs to sales team B

(10 marks)

The company is about to launch a new product that will be sold by all three sales teams. The demandfor the new product is uncertain and two pricing strategies (X and Y) are to be considered. The profitearned per unit of product for each strategy is shown below along with the probability distr ibution forweekly demand.

Pricing Strategy

X (profit per unit = $30) Y (profit per unit = $25)

Weeklydemand(units)

Probability Weeklydemand(units)

Probability

300

400

500

0.3

0.5

0.2

300

400

500

600

0.2

0.3

0.4

0.1

(c) Calculate the expected weekly profit under each pricing strategy and advise the company onwhich strategy should be chosen.

(6 marks)

(Total 20 marks)


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3709/4/06/MA 8


(a)

Mutually Exclusive - events are mutually exclusive if they cannot occur at the same time.

e.g. tossing a coin - one cannot see botha head and a tail showing face up.

Independent events - events are independent if the outcome of one

has no effect upon the outcome of the other.

e.g. rolling a fair die and scoring a six does not

affect the outcome of the second and subsequent

rolls of the die.

(b)

Sales Male Female

Team

A 32 16 48

B 20 20 40

C 16 40 56

68 76 144

(i) 48 = 0.3333

144

(ii) 16 = 0.2105

76

(iii) 16 = 0.2857

56

(iv) 20 = 0.1389

144

other methods acceptable

(c) Expected weekly demand for strategy X

= (300 x 0.3) + (400 x 0.5) + (500 x 0.2) = 390 units

Expected weekly profit for strategy X = 390 x $30 = $ 11,700

Expected weekly demand for strategy Y

= (300 x 0.2) + (400 x 0.3) + (500 x 0.4) + (600 x 0.1) = 440 units

Expected weekly profit for strategy Y = 440 x $25 = $ 11,000

Strategy X is preferred


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3709/4/06/MA 9

QUESTION 4

A product passes through three production processes A, B and C before it is ready for packing andshipment. The times taken for each production process are independently normally distributed withdetails given below.

ProductionProcess Mean Time(minutes) StandardDeviation(minutes)

A

B

C

200

100

100

15

10

20

(a) Find the probability that a product selected at random passes through process A

(i) in less than 230 minutes(ii) in less than 185 minutes

(6 marks)

(b) Find the probability that a product selected at random passes through processes A and B in lessthan 250 minutes.

(5 marks)

(c) Find the probability that a product selected at random passes through all three processes within450 minutes.

(5 marks)

(d) In a period when 2000 products were produced, estimate how many were ready for packing andshipment within 6 hours.

(4 marks)

(Total 20 marks)


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(a)

(i) Z = x - Z = 230 - 200 = 2 15

area = 0.977area of interest = 0.977

p(x < 230 minutes) = 0.977

(ii) Z = 185 - 200 = -1

15

area = 0.841

area of interest = 1 - 0.841 = 0.159

p(x < 185 minutes) = 0.159

(b)mean = 200+100 = 300 mins

st dev = 152 + 102 = 18.0278

250 - 300 = -2.7738 (say -2.8)

18.0278

area = 0.997

area of interest = 1 - 0.997 = 0.003

p(x


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3709/4/06/MA 11

MODEL ANSWER TO QUESITON 4 CONTINUED

(d)

6 hours = 360 minutes

mean = 400 mins st dev = 26.9258 mins

360 - 400 = -1.4856 (say -1.5) area = 0.933

26.9258

p(product passes through all processes within 360 mins) = 1 - 0.933 = 0.067

Of 2000 units the number ready within 6 hours =

2000 x 0.067 = 134


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3709/4/06/MA 12

QUESTION 5

In 2005 the average weekly output of Factory A production operatives was 62 units per operative. In2006 a random sample of 400 operatives working in the same factory was taken and the outputrecorded.

Weekly Output(units) Number of Operatives

50 - 5455 - 5960 - 6465 - 6970 - 74

40801606060

(a) Carry out a suitable test to determine whether the 2006 average weekly output is greaterthan the 2005 level.

(8 marks)

Another random sample, this time involving 300 production operatives from Factory B, was taken in2006. The proportion of operatives with output levels within the range 60 units to 64 units was found tobe 0.45.

(b) Carry out a suitable test to determine whether there is a significant difference in theproportion of operatives, from the two factories, producing within the range 60 units to 64 units.

(8 marks)

(c) Explain the terms type I and type II errors and state which error may have been madein parts a) and b).

(4 marks)

(Total 20 marks)


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(a)

x f fx fx2

52 40 2080 108160

57 80 4560 259920

62 160 9920 61504067 60 4020 269340

72 60 4320 311040

400 24900 1563500

mean = 24900 62.25 units400

std.dev = 5.8 units

H0: = 62 units

H1: > 62 unitsSignificance level 5% or 1%

Critical Z value 1.64 or 2.33

Calculated Z value

Z = 0.86

The calculated Z value is greater than both critical values

there is strong evidence to reject the null hypothesis.

The average weekly output in 2006 is greater than in 2005.

n

sxZ =

400

8.5

62-25.62=Z

400

25100-

400

1563500..

2

=devstd

22

f

fx

f

fxs

=


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(b)

PA = 0.4 PB = 0.45

Null hypothesis A = B

Alternative hypothesis A BSig level

5% or1%

Critical Value Z = +/-1.96 or +/-2.58

Calculated Value

where

p = 0.4214

= -0.05

0.0377

= -1.3258

The calculated Z value is greater than the critical valuesthere is no evidence to reject the null hypothesis. There is no

difference in the pproportion of workers producing within the range

60 units to 65 units.

(c)

Type I error (Reject H0 when it is actually true.)

Type II error (Accept H0 when it is actually false.)

part a) type I

part b) type II

n

1+

n

1)p-1(p

p-p=Z

BA

BA

pn p n p

n n

A A B B

A B

=+

+

300400

)45.0(300)40.0(400p

+

+=

+

=

300

1

400

1)4214.0-1(4214.0

45.0-40.0Z


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3709/4/06/MA 15

QUESTION 6

(a) Briefly describe how you would decide whether to use a paired comparison t test or anindependent sample t test.

(4 marks)

A company is considering the introduction of a training scheme to help improve performance. An initial

trial amongst a sample of the workforce has taken place. The output per hour of a randomly selectedgroup of 10 employees who have taken part in the trial was recorded before and after attending thetraining program. The results are shown in the table below.

Employee 1 2 3 4 5 6 7 8 9 10

Output per hour(Before training) 10 12 10 11 12 14 14 14 12 12

Output per hour(After training) 12 12 12 12 14 18 18 16 10 12

(b) Carry out a test to determine whether there is any evidence that the training improves theperformance of the workforce and advise the company on whether the training schemeshould be extended to include all production workers.

(16 marks)

(Total 20 marks)


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3709/4/06/MA 16


(a)

Paired comparison indication of pairing, same subject used usually

t test same size samples used

before and after

e.g. sales per representative before and after anadvertising campaign.

Independent samples no evidence of pairing, often different sized samples

test used

e.g. sales comparison between two separate and

independent sales offices.

(b)

Before After d d2 d - d (d - d)2

Training Training

10 12 2 4 0.5 0.25

12 12 0 0 -1.5 2.25

10 12 2 4 0.5 0.25

11 12 1 1 -0.5 0.25

12 14 2 4 0.5 0.25

14 18 4 16 2.5 6.25

14 18 4 16 2.5 6.25

14 16 2 4 0.5 0.25

12 10 -2 4 -3.5 12.25

12 12 0 0 -1.5 2.25

15 53 30.5

d = 15 = 1.5

10

sd = = = 1.8409

H0: d = 0

H1: d > 0

Significance level5% or1%

Degrees of Freedom n - 1 = 9

Critical t values 1.83 & 2.82

Calculated t value

t = 2.58

The calculated t value lies between the two critical values. There is some

evidence to reject the null hypothesis.

The result of the test is not conclusive, it might be advisable to assess the

improvement again by carrying out a further test.

( )d dn

2

1 9

5.30

td

s

n

= 0

10

8409.1

05.1t =


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3709/4/06/MA 17

QUESTION 7

An internet company has carried out a survey investigating the amount of time spent surfing theinternet by different age groups. The table below summarizes the results of the survey.

Time Spent Surfing the Internet per week (hours)Age (years)

Less than 4 4 to 10 More than 10

Less than 3025 35 65

30 to 5535 35 35

Over 55

40 20 10

(a) Carry out a suitable test to ascertain whether there is a link between age and the amount oftime spent surfing the internet. Comment upon why the internet company might find thisinformation useful.

(12 marks)

A similar survey conducted two years ago suggested that 40% of the population spent less than 4hours per week surfing the internet, 35% spent between 4 and 10 hours per week on the internetand the remainder spent more than 10 hours on the internet.

(b) Carry out a test to determine whether the amount of time spent on the internet haschanged over the past two years.(8 marks)

(Total 20 marks)


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3709/4/06/MA 18


(a)

Null Hypothesis: There is no link between age and the amountof time spent on the internet.

Alternative Hyp. There is a link between age and the amount

of time spent on the internet.

Observed 25 35 65 125

35 35 35 105

40 20 10 70

100 90 110 300

Expected 41.67 37.50 45.83 125.00

35.00 31.50 38.50 105.00

23.33 21.00 25.67 70.00

100.00 90.00 110.00 300.00

Contribution to 2

6.67 0.17 8.02 14.850.00 0.39 0.32 0.71

11.90 0.05 9.56 21.52

37.07

2 = 37.07

Significance Level 5% or 1%Degrees ofFreedom (3-1)(3-1) = 4

Critical 2 Value 9.49 or 13.28

Calculated 2 Value = 37.07

The calculated 2value is greater than both critical values, there is verystrong evidence to reject the null hypothesis. This information would be

useful in targeting customers during advertising and marketing campaigns.(b)

observed 100 90 110

expected 120 105 75

Contribution to 2 3.3333 2.1429 16.3333Calculated 2 Value = 21.8095

Null Hypothesis: No change in amount of time spent on internet

Alternative Hypothesis: There is a change

Significance Level 5% or1%Degrees ofFreedom n - 1 = 3 - 1 = 2

Critical 2 Value 5.99 or 9.21

The calculated 2value is greater than both critical values, there isstrong evidence to reject the null hypothesis. There appears to have been a

change in the use of the internet.


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3709/4/06/MA 19

QUESTION 8

(a) Briefly explain the use of a mean chart in quality control processes.(4 marks)

A company in its quality control procedures sets the warning limit at the 0.025 probability pointand the action limit at the 0.001 probability point. This means, for example that the upper action

line is set so that the probability of the mean exceeding the line is 0.001.

The thickness of a component is normally distributed with a mean thickness of 12.5 mm and astandard deviation of 1.5 mm. At intervals, random samples of 9 components are selected fromthe production line and inspected.

(b) Construct a control chart to help monitor the production process.(8 marks)

(c) Plot the following sample mean thicknesses on your graph and comment.

12.3 12.5 12.8 12.8 12.9 13.2 13.6 13.8(4 marks)

(d) Assuming that the mean thickness has been correctly set at 12.5 mm, find the probability that asingle component selected at random has a thickness greater than the upper warning limit butless than the upper action limit.

(4 marks)

(Total 20 marks)


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3709/4/06/MA Education Development International plc 200620


(a)

The mean chart provides a way to assess whether the mean dimension

remains within set limits. These limits are called warning limits and action

limits. There is only a small chance that the mean dimension will lie beyond

these limits and so if this happens the process will need to be checked or

even stopped. By monitoring samples in this way it is hoped that thequantity of rejects is kept low. Problems with the process can be resolved

at an early stage.

(b) & (c)

Warning LimitsUsing

x +/- Z n

12.5 +/- 2 1.5 11.5 mm to 13.5 mm

9

Action Limits

12.5 +/- 3 1.5 11.0 mm to 14.0 mm9

The sample results are increasing in thickness. Samples 7 and 8 are both

outside the upper warning limit. This pattern suggests that the process is

out of control. The process should be stopped and corrections made.

(d)

= 1 (area = 0.841)

= 0.67 (area = 0.758)

area of interest = 0.841 - 0.758 = 0.083

p(13.5 mm < x < 14.0 mm) = 0.083

Mean Chart

10

11

12

13

14

15

0 1 2 3 4 5 6 7 8 9

Sample

Th

ickness(mm) UAL

UWL

LWL

LAL

=x

Z5.1

5.120.14Z

=

=x

Z5.1

5.125.13Z

=


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2006 Series 4 (3709) Singapore

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