Business Statistics-S2(2012)-3009L3

8/16/2019 Business Statistics-S2(2012)-3009L3

1/20

Model AnswersSeries 2 2012 (3009)

For furtherinformationcontact us:

Tel. +44 (0) 8707 202909Email. [email protected]

LCCI International Qualifications

Business Statistics

Level 3


2/20

ASE3009/2/12/MA Page 1 of 18

Business Statistics Level 3Series 2 2012

How to use this booklet

Model Answers have been developed by EDI to offer additional information and guidance to Centres,teachers and candidates as they prepare for LCCI International Qualifications. The contents of thisbooklet 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, plusa fully worked example or sample answer (where applicable)

(3) Helpful Hints – where appropriate, additional guidance relating to individual

questions 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 standard required.The general standard of model answers is one that would achieve a Distinction grade. EDI accepts thatcandidates may offer other answers that could be equally valid.

© Education Development International plc 2012

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.


3/20

ASE3009/2/12/MA Page 2 of 18

QUESTION 1

A company wishes to estimate a regression line for the relationship between sales and advertising.To arrive at its decision regarding the best model to use the company calculates three correlationcoefficients.

(a) (i) The sales in any month depend on that month’s advertising with r = 0.15

(ii) The sales in any month depend on 50% of the previous month’s advertisingand 50% of that month’s advertising with r = 0.66

(iii) The sales in any month depend on the previous month’s advertising with r = 0.94

Interpret each of the above correlation coefficients and state which of the suggested models you wouldchose as the basis for predicting sales. Justify your answer.

(5 marks)

The advertising and sales figures for the last twelve months is shown in the following table.

Month Mar Apr May Jun Jul Aug Sep Oct Nov Dec Jan Feb

Advertising£00

20 22 36 45 26 35 80 65 36 72 29 35

Sales£000

33 45 47 65 89 37 55 117 85 49 105 38

(b) Using your conclusions from part (a) estimate the least squares regression line forthe above data.

(10 marks)

(c) Estimate the level of sales in May 2012 when advertising in April 2012 is £85,000 and May 2012£ 90,000. Comment on the likely accuracy of your answer.

(5 marks)

(Total 20 marks)


4/20

ASE3009/2/12/MA Page 3 of 18

MODEL ANSWER TO QUESTION 1Syllabus Topic 3.1: Correlation and regression

(a) (i) This is low positive correlation(ii) This is reasonably strong positive correlation(iii) This is strong positive correlation coefficient.

The best model to use is the sales in any month depend on the previous

month’s advertising as the coefficient of correlation is 0.94

(b) Σx = 466 Σy = 732 Σx2 = 24052 Σxy = 36477

22 x x n

y x xynb =

24662405211

7324663647711

= 1.268 (1.27)

n

x b

n

ya

=11

466268.1

11

732 = 12.818 (12.82)

y = 12.818 +1.268x

(c) y = 12.818 + 85x1.268 = 120.619 (£000)

Given the value of the coefficient of correlation the estimate could be consideredto be accurate, however the level of advertising is outside the range of the data onwhich the regression equation is based (extrapolation) which reduces the likelihoodof accuracy.

3 x 1

2

4 x 1

1m 1 ft 1 cao

1m 1cao

1 cao1m 1ft 1cao

2


5/20

ASE3009/2/12/MA Page 4 of 18

QUESTION 2

The average weight of a paperback book is 650 grams with a standard deviation of 50 grams. Aninternet bookshop usually receives orders for three books at a time. Assume the weight of the books isnormally distributed.

(a) Find the probability that an order for three books weighs:

(i) under 1900 grams(ii) over 1975 grams(iii) between 1900 and 1975 grams.

(13 marks)

The weight of the packaging for three books is normally distributed with mean 200 grams and standarddeviation of 20 grams.

(b) Find the probability that an order for three books and its packaging weighs more than2250 grams.

(4 marks)

The company’s delivery contract charges £2.75 for all deliveries which weigh 2250 grams and below and

£3.50 for deliveries which weigh more than 2250 grams.

(c) Estimate the expected cost for the delivery of 3 books.(3 marks)

(Total 20 marks)

MODEL ANSWER TO QUESTION 2Syllabus Topic 4.2: Normal distribution

(a) Joint mean = 3 x x = 3 x 650 = 1950 grams

Joint Standard deviation =2

3 sd =2

503 = 7500 = 86.6

(i) Under 1900 grams z =sd

x x =

6.86

19501900 = 0.58 (0.6), table p = 0.726

Ans 1-0.726 = 0.274

(ii) Over 1975 grams z =sd

x x =

6.86

19501975 = 0.29 (0.3), table p = 0.618

Answer = 1 – 0.618 = 0.382

(iii) 1900 to 1950 = 0.726 – 0.5 = 0.2261950 to 1975 = 0.618 - 0.5 = 0.118

Between 1900 and 1975 grams = 0.344

(b) Joint mean =21

x x = 1950+ 200 = 2150

Joint standard deviation =2

2

2

1 sd sd =

22206.86 = 88.9

z =sd

x x =

9.88

21502250 = 1.1, table p = 0.864 answer = 1 - 0.864 = 0.136

(c) Expected cost = 2.75 x 0.864 +3.50 x 0.136 = £2.85

1m 1cao

1m 1ft 1cao

1m 1ft 1 cao

1m

1cao

1m 1ft

1cao

1cao

1cao

1ft 1cao

1m 1ft 1 cao


6/20

ASE3009/2/12/MA Page 5 of 18

QUESTION 3

(a) Explain what is meant by the standard error of the mean.(4 marks)

A company advertises its job vacancies in the newspapers and through an internet recruitment site. A random sample of vacancies showed the response rate from the two media.

Job A B C D E F G H

Newspapers 37 45 36 57 29 21 46 89

Internet site 43 42 45 61 36 26 45 102

(b) Test whether the mean is higher from the internet site than the newspapers.(12 marks)

(c) State the alternative hypothesis if you had carried out a two tail test and the conclusionsyou would come to.

(4 marks)

(Total 20 marks)


7/20

ASE3009/2/12/MA Page 6 of 18

MODEL ANSWER TO QUESTION 3Syllabus Topic 2.1: Measures of location, Syllabus Topic 2.2: Measures of location,Syllabus Topic 5.1: Procedure for significance, Syllabus Topic 5.2: Type I and Type II errorsSyllabus Topic 5.3: Significance tests n=>30

(a) The standard error of the mean is the standard deviation of the distribution ofthe means when samples of a given size are taken from a population.

It equals nse

Where σ is the population standar d deviation and n is the sample size.

(b) Null hypothesis: There is no difference in the mean response of the two media. Alternative hypothesis: The internet site provides a greater mean response.

Degrees of freedom = n – 1 = 8-1 = 7Critical t value one tail = 1.90/3.00

d = 40/8 = 5 sdd =

1

2

n

d d =

7

186 = 5.16

z =

n

sd

d

d

0 =

8

16.5

05 = -2.74

Conclusions: Reject the null hypothesis at the 0.05 significance level. The internet siteprovides a greater response rate. Accept the null hypothesis at the 0.01 significance level.There is no difference in the response rate between the two media.

(c) The alternative hypothesis would be H1: There is a difference in the meanresponse between the two media.

A two tail test has critical t values of 2.37 and 3.50. The conclusion would be thesame as the absolute calculated t value >2.37 and < 3.50

Job A B C D E F G H Total

Newspapers 37 45 36 57 29 21 46 89

Internet site 43 42 45 61 36 26 45 102Difference d -6 3 -9 -4 -7 -5 1 -13 -40

d

36 9 81 16 49 25 1 169 386

2d d 1 64 16 1 4 0 36 64 186

2 x 2

1cao1cao

1 cao1cao

caod 1 Σd

2 or

Σ 2

d d 1cao

1m 1cao

1m 1cao

2ft

2 x 2


8/20

ASE3009/2/12/MA Page 7 of 18

QUESTION 4

A doctor’s surgery records the number of patients who attend per quarter over a three year period.

Patients (000)

Year/Quarter 1 2 3 4

2009 36 28 48 682010 44 36 56 76

2011 44 36 64 76

(a) Graph the data and comment on the pattern shown.(4 marks)

(b) Find the four quarter centred moving average and the average quarterly variationsfor the time series.

(10 marks)

(c) (i) Estimate the number of patients in quarter 1 and quarter 2 of 2012.

(ii) If a basic payment of £10.50 is made for each attendance at the surgery whatis the total expected income of the surgery for the estimated number of patientsover the first two quarters of 2012?

(6 marks)

(Total 20 marks)


9/20

ASE3009/2/12/MA Page 8 of 18

MODEL ANSWER TO QUESTION 4Syllabus Topic 3.2: Time series

(a)

Times Series

0

10

20

30

40

50

60

70

80

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

Time per iods

N u m b e r o f P a t i e n t s

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

2009 2010 2011

Comment: The data shows a strong quarterly variation with an upwards trend.

Patients Movingtotal 1

Movingtotal 2

Movingaverage/Trend

AdditiveDifferences

ORMultiplicative

Factors

36

28

48 180 368 46 2 1.04

68 188 384 48 20 1.42

44 196 400 50 -6 0.88

36 204 416 52 -16 0.69

56 212 424 53 3 1.06

76 212 424 53 23 1.43

44 212 432 54 -10 0.81

36 220 440 55 -19 0.65

64 220

76

1 axes,

scale labels2 plots

1 comment

MA/Trend = 6

Differences = 2

1m cao 1m 1ft 1m 1ft 1m 1ft or 1m 1ft


10/20

ASE3009/2/12/MA Page 9 of 18

MODEL ANSWER TOQUESTION 4 CONTINUED

(b)

Quarter 1 Quarter 2 Quarter 3 Quarter 4

1 2 3 4

Year 2008 2 20

2009 -6 -16 3 23

2010 -10 -19

Total quarterly variation -16 -35 5 43

Average quarterly variation -8 -17.5 2.5 21.5

(c) (i) Growth rate per quarter = Tn – T1 = 55 – 46 = 1.29n – 1 8 – 1

Predicted patients quarter 1 2012 = (55 + 1.29 x 3) – 8 = 58.87 – 8 = 50.87Predicted patients quarter 2 2012 = (55 + 1.29 x 4) – 17.5 = 60.16 – 17.5 = 42.66

(ii) (50.87 + 42.66) x £10.50 = 93.74 x £10.50 = £982.07 (000)

ORMultiplicative Factors

Quarter 1 Quarter 2 Quarter 3 Quarter 4

1 2 3 4

Year 2008 1.04 1.42

2009 0.88 0.69 1.06 1.43

2010 0.81 0.65

Total quarterly variation 1.69 1.35 2.10 2.85

Average quarterly variation 0.85 0.67 1.05 1.43

(c) (i) Growth rate per quarter = Tn – T1 = 55 – 46 = 1.29n – 1 8 – 1

Predicted patients quarter 1 2012 = (55 + 1.29 x 3) – 8 = 58.87 x 0.85 = 50.0Predicted patients quarter 2 2012 = (55 + 1.29 x 4) – 17.5 = 60.16 x 0.67 = 40.3

(ii) (50 + 40.3) x £10.50 = 90.3 x £10.50 £948.15 (000)

1m 1cao

1m 1cao

1m 1cao

1m 1cao

OR

1m 1cao

1m 1cao

1m 1cao

1m 1cao


11/20

ASE3009/2/12/MA Page 10 of 18

QUESTION 5

A company discusses with its marketing agency how it might collect information from its customers.The agency suggests three possibilities:

(i) personal interview(ii) telephone interview(iii) email interview.

(a) Describe two advantages and two disadvantages of each method.(12 marks)

(b) A previous sample of 4000 customers showed that 37% of the customers held apositive view of the company. How large a sample should be taken to be 95% certainthat the estimated results are within 3% of the views held by all customers?

(4 marks)

A survey using the sample size calculated in part (b) produces a result that 33% of customers hold apositive view of the company.

(c) By pooling the results of the two samples what percentage of customers hold a positive

view of the company?(4 marks)

(Total 20 marks)


12/20

ASE3009/2/12/MA Page 11 of 18

MODEL ANSWER TO QUESTION 5Syllabus Topic 1.2: Survey methodology

(a) Personal Interview. The interviewer and respondent meet face to face to conductthe interview.

Advantages: Social clues e.g. body language, can be obtained. Questions can beexplained. Further information can be developed from the respondent. No time delay

between question and answer.Disadvantages:

Interviewer bias or interference in the process, time taken totranscribe the interview, cost especially if respondents are widely spread. Need toagree a mutually agreeable time and place for the interview.

Telephone interview. The interview is carried out using the telephone.

Advantages: A wider range of respondent can be reached as there is no need fora face to face meeting. The cost will be lower. Able to react to the respondentsreplies.Disadvantages: No social clues available, no awareness of the circumstances inwhich the interview takes place which may affect the responses. Easier for therespondent to break-off the interview. Unable to standardise the process.

Email interview. A questionnaire is drawn up in the usual way and distributedby email to the respondents.

Advantages: A wider range of respondent can be reached as there is no need fora face to face meeting. The cost will be lower. No need for an agreed time for aninterview. The respondent can answer at their own pace. The responses can bedownloaded directly to the computer. The respondent may be prepared to giveanswers which might not usually be acceptable. The respondent has time to thinkabout the answer.Disadvantages: Complete lack of social clues. Unable to extend/develop a question.The interviewee may cost a lot of time for the interviewer. Due to the asynchronouscommunication of time, the interviewee might wait sometimes for days or weeks

before he/she answers the questions. The interviewee may fail to respond.Spontaneous answers may be reduced.

(b) z = 96.1 d = 0.03

n = z2/(d

2x(p(1-p)) 1.96

2 /(0.03

2x (0.37x0.63)) = 995

(1036 if z =2 used)

(c) Pooled probability = 4000 x 0.37 + 995 x 0.33 = 1480 + 328.35

= 1808.35/4995 = 0.362

2+2

2+2

2+2

More than oneword answersfor full credit

1.96 1cao1m 1ft 1cao

1m 1ft

1m 1cao


13/20

ASE3009/2/12/MA Page 12 of 18

QUESTION 6

A company sources its products from a number of countries and records the number of items sold byprice and country.

Origin Under £8 £8 and under £12 £12 and under £20 £20 and over

France 70 50 30 30

Spain 110 80 40 50

USA 60 60 30 20

Australia 70 40 20 20

(a) Test whether there is an association between price and country of origin.(12 marks)

Ten years ago the proportions of products bought by the company from each country were.

France Spain USA Australia

24% 33% 24% 19%

(b) Test whether the origin of the products has changed over the ten year period.(8 marks)

(Total 20 marks)


14/20

ASE3009/2/12/MA Page 13 of 18

MODEL ANSWER TOQUESTION 6 Syllabus Topic 1.1: Graphical presentationSyllabus Topic 5.5: Chi-square tests

(a) Null hypothesis: There is no association between the number of bottles soldin different price categories and the countries of origin.

Alternative hypothesis: There is association between the number of bottles

sold in different price categories and the countries of origin.

Degrees of freedom = (R-1)(C-1) = (4 -1)(4 – 1) = 9

Critical 2 = 16.92/21.67

Expected Values Contributions to 2

71.5 53.1 27.7 27.7

111.3 82.6 43.1 43.1

67.6 50.1 26.2 26.2

59.6 44.2 23.1 23.1

2 = 9.86

Conclusions: The calculated value of 2 is less than the critical values. There is no

association between the number of bottles sold in different price categories and thecountries of origin.

(b) Null hypothesis: The origin of products by country has not changed. Alternative hypothesis: The origin of products by country has changed.

Degrees of freedom = n-1 = 3

Critical 2 = 7.81/11.34

180 280 170 150 Observed

187.2 257.4 187.2 148.2 Expected

0.28 1.98 1.58 0.02 Contributions to 2

2 = 3.86

Conclusions: The calculated value of 2 is less than the critical values. The origin of

products by country has not changed.

1

11

Expectedvalues 1m2caoContributions

to 2 1m 2cao

1cao

2 ft

1

1

1cao1cao1m1cao

2ft

0.03 0.18 0.19 0.19

0.01 0.08 0.22 1.11

0.85 1.94 0.57 1.45

1.81 0.40 0.41 0.41


15/20

ASE3009/2/12/MA Page 14 of 18

QUESTION 7

The table below gives the revenues and volume of sales for four types of printer sold by a computersupplies company in 2005 and 2010

2005 2010

Printer Revenue (£000) Quantity (00) Revenue (£000) Quantity (00)

Axk 47 700 14 799 17

Mxk16 1848 22 3135 33

Uxb 900 15 624 13

Imf 1520 19 910 14

(a) Calculate:

(i) the Laspeyres base year weighted price index number for 2010 based on 2005(ii) the Paasche current year weighted price index number for 2010 based on 2005(iii) Comment on the difference in your answers in (i) and (ii).

(10 marks)

(b) Identify and describe five factors involved in creating a general price index suchas the Retail Price Index.

(10 marks)

(Total 20 marks)


16/20

ASE3009/2/12/MA Page 15 of 18

MODEL ANSWER TO QUESTION 7Syllabus Topic 2.5: Index numbers

(a) The prices for year 2005 = 700/14 = 50, 84, 60, 80The prices for year 2010 = 799/17 = 47, 95, 48, 65

00q p =4968 01q p =4703 11q p =5468 10q p =5522

Laspeyres = 100

00

01

q p

q p = = 94.7

Paasche =

10

11

q p

q p = = 99.0

Comment: Both indices show a reduction in price. The Paasche by less than theLaspeyres due to the increase in weight for item Mxk16

(b) Prices. How the prices are collected and which prices should be collected?

Choice of goods and services. Items need to be chosen which are representativeof the whole of the consumption for the population

Base year. A starting year for the index has to be chosen which allows for reliablecomparisons in later years. The base year value = 100

Weights. How is the relative importance of the items in the index determined?This is usually based on the proportion of annual expenditure by consumers on therepresentative items.

Choice of index number. How is the price index going to be constructed? Usually

this is based on

W WI

where W = the weights of the subgroups identified and I is

the value of the index for each subgroup.

1m 2cao

1m 2 cao

1m

1m

2 x 1

5 x 2

4703 x 100

4968

5468 x 100

5522


17/20

ASE3009/2/12/MA Page 16 of 18

QUESTION 8

A partnership takes a random sample of the businesses it audits and records the size of the fee itcharges.

Fee charged £000 Frequency

100 and under 200 23





600 and under 1000 10

(a) Estimate the arithmetic mean and the standard deviation of fee charged.(9 marks)

The median for the above data is £370.55.

(b) Estimate the coefficient of skew.(3 marks)

Three years previously a sample of 150 audit fees showed the arithmetic mean was 340.1 (£000) with astandard deviation of 159.8 (£000).

(c) Test whether the mean fee has changed over the three year period.(8 marks)

(Total 20 marks)


18/20

ASE3009/2/12/MA Page 17 of 18

MODEL ANSWER TO QUESTION 8Syllabus Topic 2.1: Measures of locationSyllabus Topic 2.2: Measures of locationSyllabus Topic 2.3: Measures of skewnessSyllabus Topic 5.1: Procedures for significance

(a)

Σf Σfx

Σf 2 x x Σfx

2

200 76100 4598950 33555000

f

fx x = 76100/200 = 380.5

f

x x f sd

2

=200

4598950 = 151.64

(b) Coefficient of Skew = (3(Mean – median)/standard deviation)

= 3(380.5(£000)-370.55)/151.64(£000) = 0.197

(c) Null hypothesis: There has been no change in the fee charged over the three yearperiod Alternative hypothesis: There has been a change in the fee charged over the threeyear period.

Two tail test for large samples Critical z values = 1.96/2.58

2

2

2

1

2

`1

21

n

s

n

s

x x z

=

150

8.159

200

64.151

1.3405.380

22

= 40.4/16.89 = 2.39 (2.4)

Conclusions: Reject the null hypothesis at the 0.05 level. There has been a change in thefee charged over the three year period.

Accept the null hypothesis at the 0.01 level. There has not been a change in the fee chargedover the three year period.

Σf =1cao

Mp 1caoΣfx 1m

Σf 2 x x / Σfx21m

1m 1cao

1m 1ft 1cao

1m

1cao1 for correct unit1cao

1cao

1

1m 1ft 1cao

2ft


20/20

EDI

International House

Siskin Parkway East

Middlemarch Business Park

Coventry CV3 4PE

UK

Tel. +44 (0) 8707 202909Fax. +44 (0) 2476 516505

Email. [email protected]

www.ediplc.com

Business Statistics-S2(2012)-3009L3

Documents

Transcript of Business Statistics-S2(2012)-3009L3