96119633 Business Statistics Series 2 2011 Code3009

Model Answers Series 2 2011 (3009)

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LCCI International Qualifications

Business Statistics Level 3

3009/2/11/MA Page 1 of 18

Business Statistics Level 3 Series 2 2010

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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 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 to

see 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 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 that candidates may offer other answers that could be equally valid.

© Education Development International plc 2011 All rights reserved; no part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without prior written permission of the Publisher. The book may not be lent, resold, hired out or otherwise disposed of by way of trade in any form of binding or cover, other than that in which it is published, without the prior consent of the Publisher.

3009/2/11/MA Page 2 of 18

QUESTION 1 The Marketing Manager of an internet trading company has been investigating a possible link between the different parts of the world and customer expenditure. A random sample of 1250 customers showed the following distribution.

Expenditure

World Region

Europe Asia North America

Rest of the World

Under £200 60 90 115 35

£200 and under £750

120 260 280 40

£750 and over

70 50 105 25

(a) Test whether there is a significant association between different parts of the world and customer expenditure.

(12 marks)

Before trading on the internet, the company operated a mail order business, and the percentage distribution of customer expenditure was as follows:

(b) Has the distribution of customer expenditure changed since the company changed trading from mail order to the internet?

(8 marks)

(Total 20 marks)

Europe Asia North America

Rest of the World

19% 33% 40% 8%

3009/2/11/MA Page 3 of 18

MODEL ANSWER TO QUESTION 1 (a) Null hypothesis: There is no association between the different parts of the world and customer expenditure. Alternative hypothesis: There is association between the different parts of the world and customer expenditure. Degrees of freedom = (4-1) (3-1) = 6 Critical Chi-squared χ2 =12.59/16.81 D/f

expected 60 96 120 24 freq 140 224 280 56

50 80 100 20

contribution 0.00 0.38 0.21 5.04 to chi-squared χ2

2.86 5.79 0.00 4.57

8.00 11.25 0.25 1.25 chi-squared (χ2) = 39.59

Conclusions: Reject the null hypothesis at the 1% significance level. There is strong evidence of association between the different parts of the world and customer expenditure. (b) Null hypothesis: The distribution of customer expenditure has not changed. Alternative hypothesis: The distribution of customer expenditure has changed. Degrees of freedom = 4-1= 3 Critical chi-squared χ2 =7.81/11.34 Observed 250 400 500 100 Expected 237.5 412.5 500 100 Contribution to χ2

0.658 0.379 000.0 0.000 chi-squared χ2 = 1.037 Conclusion: Do not reject the null hypothesis. The distribution of customer expenditure has not changed.

NH/AH 1 1 cao Chi 1 1m 2 cao cao 1m 2cao 1 cao 2 ft 1 NH/AH 1cao 1cao 1 cao 1m + 1 χ2 2 ft

3009/2/11/MA Page 4 of 18

QUESTION 2 (a) Explain, and give a business example of, what is meant in probability by:

(i) mutually exclusive events (ii) dependent events.

(8 marks) An industrial chemical company has three new products (A, B and C) in development. It estimates the probability of successful development of the products as 0.5, 0.6 and 0.7 respectively. (b) Find the probability that:

(i) all three products are successful (ii) no product is successful (iii) one product is successful.

(8 marks)

(c) Given exactly one product is successful what is the probability it is product B? (4 marks)

(Total 20 marks)

3009/2/11/MA Page 5 of 18

MODEL ANSWER TO QUESTION 2 (a) (i) Mutually exclusive events are events where the occurrence

of one event prevents the occurrence of another event. Example, a product launch being both successful and unsuccessful.

(ii) Dependent events are events where the outcome of one event is related to the outcome of another event(s). Example, success of a product launch is dependent on market conditions.

(b) Find the probability that:

(i) all are successfully developed 0.5 x 0.6 x 0.7 = 0.21

(ii) no product is successful 0.5 x 0.4 x 0.3 = 0.06

(iii) one product is successful 0.5 x 0.4 x 0.3 + 0.5 x 0.6 x 0.3 + 0.5 x 0.4 x 0.7 0.06 + 0.09 + 0.14 = 0.29 (c) Probability B successfully developed = 0.09 = 0.31 Probability one successful 0.29 .

2 x 2 2 x 2 1m 1cao 1m 1cao 1m 2 cao 1cao 1m 2cao 1cao

3009/2/11/MA Page 6 of 18

QUESTION 3 (a) Explain what the correlation coefficient measures.

(4 marks) The table below shows the length of service for a random sample of 10 employees, and the value of their weekly output.

Employee Service (years) Value of weekly output

£000

a 5 35.3

b 7 46.3

c 5 34.9

d 8 35.2

e 10 45.9

f 7 35.2

g 14 46.1

h 9 37.9

i 5 35.8

j 4 36.9

(b) Calculate the product-moment correlation coefficient between length of service and the value of weekly output and comment on your answer.

(10 marks)

(c) Test if the correlation coefficient differs significantly from zero. (6 marks)

(Total 20 marks)

3009/2/11/MA Page 7 of 18

MODEL ANSWER TO QUESTION 3 (a) The correlation coefficient measures the degree of linear association between two, or more, variables. It can vary between perfect association r = +/-1 to 0 no association. (b)

ΣX ΣY ΣX2 ΣY2 ΣXY

( )( )22 5.38955.15397107463010

5.389742.297510

−×−×

×−×=r

( )( )25.2265824

929=r r = 0.68

Comment: Some positive correlation.

(c) Null hypothesis: The correlation coefficient does not differ from zero. Alternative hypothesis: The correlation coefficient does differ from zero. Degrees of freedom = n-2 = 10 – 2 = 8 Critical t value 2.306/3.355

21

2

r

nrt

−

−= =

268.01

21068.0

−−

= 4.0 Conclusions: Reject the null hypothesis. The correlation coefficient does differ from zero.

(Total 20 marks)

2 x 2 5 x 1 cao 1m 1ft 1 for 2 correct values 1cao 1 ft 1cao 1cao 1m 1cao 2ft

74 389.5 630 15397.55 2975.2

3009/2/11/MA Page 8 of 18

QUESTION 4 In 2010 a random sample of 360 small businesses revealed the following distribution for annual sales.

Sales £000 Number of businesses

Over 40 and up to 60 65





(a) Estimate the mean and the standard deviation of sales per business.

(8 marks) In the previous year mean sales was £93,650.

(b) Test whether there has been a significant increase in the average sales for small businesses.

(7 marks)

(c) Calculate a 95% confidence interval for the mean sales in 2010 (5 marks)

(Total 20 marks)

3009/2/11/MA Page 9 of 18

MODEL ANSWER TO QUESTION 4

∑ f ∑ fx

2

∑ fx

( )2xxf −∑

∑∑=

f

fxx

= 36035120

= 97.56 (£000)

22

−=∑∑

∑∑

f

fx

f

fxs

2

36035120

3603978200

−== 39.16 (£000)

or

( )n

xxs ∑ −

=2

= 360552049

= 39.16 (£000)

(b) Null Hypothesis: There has not been an increase in the average sales compared with the previous year. Alternative hypothesis: There has been an increase in the average sales compared with the previous year. Critical Z value 1.64/2.33

n

xz

σµ−=

36016.39

65.9356.97 −=

z = 1.89 Conclusions: There is evidence to suggest that the sales have increase at the 0.05 but not at the 1% significance level. There is some evidence there has been an increase in sales. (c) The 95% confidence interval = mean ± 1.96st dev/√n = 97.56±1.96 x 39.16/√360 = 93.51 to 101.6 or £93,510 to £101,600

Annual Sales £000 Mid point (x)

Number of businesses f

fx fx2 f ( )2xx −

Over 40 and up to 60 50 65 3250 162500 147000

Over 60 and up to 80 70 77 5390 377300 58466.8

Over 80 and up to 100 90 86 7740 696600 4909.43

Over 100 and up to 140 120 74 8880 1065600 37277.7

Over 140 and up to 200 170 58 9860 1676200 304395

360 35120 3978200 552049

mid pt 1 cao

∑ fx 1m

2

∑ fx or

( )2xxf −∑

1m 1m 1 cao Lose 1 for no £000 1m 1ft 1cao 1 cao 1 cao 1m 1 ft 1cao 2 ft 1 m 1 for1.96, 1 ft 2 cao inc £000

3009/2/11/MA Page 10 of 18

QUESTION 5 (a) What are the benefits of a good quality control system to a business?

(4 marks) Quality control procedures are used, which set the warning limits at the 0.025 probability point and action limits at the 0.001 probability point. This means, for example, that the upper action limit is set so that the probability of the means exceeding the limit is 0.001. In a manufacturing process, the internal diameter of a boring is set at 200 mm, with a standard deviation of 2 mm. Samples of 8 items at a time are taken from the production line to check the accuracy of the manufacturing process. (b) (i) Calculate the values of the action and warning limits and construct a quality control chart to monitor the manufacturing process.

(8 marks) (i) The results for 5 samples are given below. Plot these on your quality control chart and comment on the results.

(4 marks) sample number 1 2 3 4 5 sample mean mm 201.2 197.5 201.9 202.3 198.7 (c) The process mean changed to 200.1 mm and the standard deviation remained at 2 mm. Calculate the probability that the mean of a random sample of 8 components would be below the lower warning limit.

(4 marks)

(Total 20 marks)

3009/2/11/MA Page 11 of 18

MODEL ANSWER TO QUESTION 5 (a) It reduces wastage and costs, improves product reputation.

(b) Warning Limits nx

σ96.1±

8

296.1200 ±

= 200 ± 1.96 x 0.707 = 200 ± 1.39 = 198.61 to 201.39 (198.6/201.4)

Action Limits nx

σ09.3±

200 ± 3.09 x 0.707 = 200 ± 2.18 = 197.82 to 202.18 (197.8/202.2)

195

196

197

198

199

200

201

202

203

1 2 3 4 5 6

Dia

met

er m

m

Sample number

Quality Control Chart

Comment: the process is not in control. The mean values of the samples swing from below the LAL to above the UAL.

(c) n

xz

σµ−= =

707.01.20061.198 −

= 2.1

Probability = 0.982 Answer = 1-0.982 = 0.018

2 x2 1n 1.96/2 1 cao 1 cao 3.09/3: 1 cao 1 cao Plot of limits 1 cao, Scale 1, Labels and title 1, plot of means 2 2 ft 1m, 1ft, z 1 cao, p 1cao

UAL UWL Mean LWL LAL

3009/2/11/MA Page 12 of 18

QUESTION 6 (a) In what circumstances might a significance test based on the normal distribution be used in preference to a significance test based on the ‘t’ distribution?

(4 marks) A company manufacturing light bulbs develops a new process which it claims increases the life of the bulbs. 10 bulbs from the old process are tested to destruction using the old process and a further 10 bulbs from the new process are tested to destruction.

Life using the old process (hours)

2,300 2,290 2,360 2,340 2,330 2,370 2,240 2,320 2,330 2,350

Life using the new process (hours)

2,290 2,300 2,370 2,380 2,360 2,380 2,260 2,330 2,340 2,370

(b) Test whether the new process has increased the life of the light bulbs.

(12 marks)

(c) What is meant by Type 1 error? From your conclusions in (b) might a type 1 error have been committed?

(4 marks)

(Total 20 marks)

3009/2/11/MA Page 13 of 18

MODEL ANSWER TO QUESTION 6 (a) The normal distribution will be used when the sample size is large (greater than 30) and/or the population standard deviation is known. Two means test (b) Null hypothesis: The new process has not increased the life of the light bulbs. Alternative Hypothesis: The new process has increased the life of the light bulbs. Degrees of Freedom = (n-1) + (m–1) = (10-1) + (10-1) = 18 Critical t value = 1.73/2.55

X Y

2)( xx −

2)( yy − 2300 2290 529 2304

2290 2300 1089 1444

2360 2370 1369 1024

2340 2380 289 1764

2330 2360 49 484

2370 2380 2209 1764

2240 2260 6889 6084

2320 2330 9 64

2330 2340 49 4

2350 2370 729 1024

Σx 23230 Σy 23380 =−∑ 2)( xx

13210 =−∑ 2)( yy 15960

x= 2323 y =2338

+−∑ 2)( xx =−∑ 2)( yy 29170

s = 2)()( 22

−+−∑+−∑

mnyyxx

= 21010

1596013210

−++

= 18

29170

= 40.26

t = 101

101 +

−

s

yx

= 10

1

10

126.40

23382323

+×

−

= -0.83 Conclusion: The calculated value of t is less than the critical value of t at the 0.05 significance level, therefore do not reject the null hypothesis. There is no evidence to support the claim that the new process has increased the life of light bulbs. (c) A type 1 error is when the null hypothesis is rejected when it should be accepted. A type 1 error cannot have occurred at the 0.05 level as the null hypothesis was not rejected.

2 x 2 2 cao 1cao Σx/ Σy 1cao

2)( xx −∑ /

2)( yy −∑

1cao

x / y

1cao 1m 1cao 1m 1cao 2 ft 2 x 2

3009/2/11/MA Page 14 of 18

QUESTION 7 (a) Describe the four components of a time series.

(4 marks) The table below shows the quarterly sales for a company in 000 units.

Year Quarter 1 Quarter 2 Quarter 3 Quarter 4

2008 117 133 150 188

2009 143 152 177 252

2010 159 198 209 325

(b) Calculate the centered trend values and the average quarterly variations for the time series.

(10 marks)

(c) Estimate the sales for quarter 2 of 2011 and comment on the likely accuracy of the estimates.

(6 marks)

(Total 20 marks)

3009/2/11/MA Page 15 of 18

MODEL ANSWER TO QUESTION 7 (a) The four components of a time series are:

(1) Trend the long term direction, (2) Cyclical variation due, for example, to the economic cycle, (3) Seasonal variation due, for example, to festive events (4) Residual or Random variation without any apparent explanation.

(b)

Quarter Sales Moving Total 1

Moving Total 2 Trend

Additive Difference

Multiplicative Difference

1 117

2 133

3 150 588 1202 150.25 -0.25 0.998

4 188 614 1247 155.875 32.125 1.206

5 143 633 1293 161.625 -18.625 0.885

6 152 660 1384 173 -21 0.879

7 177 724 1464 183 -6 0.967

8 252 740 1526 190.75 61.25 1.321

9 159 786 1604 200.5 -41.5 0.793

10 198 818 1709 213.625 -15.625 0.927

11 209 891

12 325 Average Seasonal Variation Additive Method Quarter 1 Quarter 2 Quarter 3 Quarter 4 -0.25 32.125 -18.625 -21 -6 61.25 -41.5 -15.625 Average -30.0625 -18.3125 -3.125 46.6875 Average Seasonal Variation Multiplicative Method Quarter 1 Quarter 2 Quarter 3 Quarter 4 0.998 1.206 0.885 0.879 0.967 1.321 0.793 0.927 Average 0.839 0.903 0.983 1.264 Estimated Trend Growth = T8-T1 = 213.625 – 150.25 = 63.375 = 9.05 n-1 8 – 1 7 Estimated trend for quarter 2, 2011 = 213.625 + (4 x 9.05) = 249.8 Estimated Sales Additive Method

2011 Quarter

Estimated trend

Additive Seasonal Variation

Q2

Estimated Sales

Additive Method

Multiplicative Seasonal Variation

Q2

Estimated Sales

Multiplicative Method

2 249.8 -18.3 231.5 0.903 225.6 Comment: The forecast is only two periods ahead, therefore relative stability can be expected but it is an extrapolation

4 x 1 4 x 0.5 for naming factors MT1 1m 1cao MT2 1m 1cao Trend 1m 1 cao Differences 1m 1cao 1 transfer 1 cao 1m 1cao 1m 1cao 2 comment

3009/2/11/MA Page 16 of 18

QUESTION 8 (a) Explain how:

(i) a stratified random sample would be taken (ii) a quota sample would be taken.

In each case give one advantage and one disadvantage of the method.

(8 marks) (b) Explain the term sampling distribution of the mean.

(4 marks) The data below was obtained from counting the number of customers visiting each of two branches of a company over a number of weeks during 2010.

Branch X Branch Y

Mean number of customers 1516 1634

Standard deviation 158 186

Sample size (no. weeks) 32 41

(c) Test whether there is a significant difference between the mean number of customers visiting the two branches.

(8 marks)

(Total 20 marks)

3009/2/11/MA Page 17 of 18

MODEL ANSWER TO QUESTION 8 (a) A stratified random sample would be taken by allotting all the relevant members of the sampling frame to different strata depending on characteristics. Select a random sample in each strata. Advantages: the sampling error can be calculated, the result can be used in further statistical calculation and it reduces the sampling error. Disadvantages: a sampling frame is necessary, the relevant characteristics need to be identified and the sample may be drawn from a widely dispersed geographical area. Quota sampling involves taking a sample with a given number of people. Often this is done by stopping people in the street. Interviewers are given quota controls ie the characteristics that the respondents should have. These often relate to age, gender and income. Advantages: the sample is relatively cheap, there is no need for a sampling frame and no need for call-backs. Disadvantages: there may be bias in the choice of respondents, theoretically the standard error cannot be calculated so significance tests are not valid. (b) The sampling distribution of the mean refers to the variability which occurs in the value of the means as different samples of the same size are taken. The standard error of the means equals σ/√n. (c) Null hypothesis: There is no difference in the number of customers visiting Branch X and Branch Y Alternative hypothesis: there is a difference in the number of customers visiting Branch X and Branch Y

Critical z value 1.96/2.58

z = 2

22

1

21

21

n

s

n

s

xx

+

−

= 41

186

32

158

1634151622

+

−

= 118.0 = -2.93 40.30 Reject the null hypothesis, there is strong evidence there is a difference in the

number of customers visiting Branch X and Branch Y.

2 for explanation 1 adv 1 dis adv 2 for explanation (1 given number 1 quota control) 1 adv 1 dis adv 2 x 2 1 NH 1 AH 1 cao 1m 1 ft 1cao 2ft

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96119633 Business Statistics Series 2 2011 Code3009

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