Model Answers Series 3 2010 (3009)

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LCCI International Qualifications

Business Statistics Level 3

3009/3/10/MA Page 1 of 20

Business Statistics Level 3 Series 3 2010

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 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 2010 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.

3009/3/10/MA Page 2 of 20

QUESTION 1 A local authority conducts a random sample of the number of planning applications examined per day by its planning officers over a period of five years. The data are shown below.

Number of planning applications examined

Number of days

40 and under 80 103

80 and under 120 68

120 and under 160 38

160 and under 200 22

200 and under 300 15

(a) Calculate the mean and standard deviation for the number of planning applications examined

per day. (9 marks)

(b) Calculate a 95% confidence interval estimate for the mean daily number of planning

applications examined. (5 marks)

The government has set a standard of 120 planning applications to be examined per day. (c) Test whether the local authority reaches the government standard.

(6 marks)

(Total 20 marks)

3009/3/10/MA Page 3 of 20

MODEL ANSWER TO QUESTION 1 (a)

Number of planning applications examined

Number of days

x fx

fx2 f

2

xx

40 and under 80 103 59.5 6128.5 364645.8 215114.5

80 and under 120 68 99.5 6766.0 673217.0 2209.3

120 and under 160 38 139.5 5301.0 739489.5 44706.6

160 and under 200 22 179.5 3949.0 708845.5 121450.8

200 and under 300 15 249.5 ..3742.5 ..933753.8 312337.4

246 25887.0 3419951.6 695818.6

f fx 2fx f

2

xx

Arithmetic mean = f

fxx

= 25887 = 105.2 applications per day

246

Standard deviation =

22

f

fx

f

fxs

2

246

25887

246

6.3419951s = 7.110732.13902 = 5.2828

or = 53.18 per day

f

xxf2

=

f

xxf2

f

xxf2

f

xxf2

f

xxf2

==

165

.1743409121

246

6.695818

3009/3/10/MA Page 4 of 20

QUESTION 1 CONTINUED

(b) 95% confidence interval z = 1.96

nxci 96.1

246

18.5396.12.105

= 105.2 6.65= 98.58 to 111.88 (c) Null Hypothesis: The local authority reaches the government standard. Alternative hypothesis: The local authority does not reach the government standard. One tail test z = 1.64

n

xz =

246

18.53

1202.105

= -14.27 = -4.21 3.39 Conclusion: There is evidence to reject the null hypothesis; the local authority has not reached the government standard.

3009/3/10MA Page 5 of 20

QUESTION 2 A random sample of 12 households records the value of their house and the annual expenditure on repairs to the house.

Value of house (£000)

Annual expenditure on repairs (£)

215 1850

323 2550

455 3150

329 1850

235 1750

425 1700

327 2600

385 3300

237 2750

155 1300

196 1600

176 1750

(a) Calculate the least squares regression line of expenditure on repairs based on the value of a

house. (10 marks)

(b) Using the regression equation found in (a) estimate the annual expenditure on repairs to a

house valued at £250,000. (2 marks)

The value of the coefficient of determination is 0.403 or 40.3%

(c) Explain what the coefficient of determination measures and comment on the accuracy of your answer in (b) above.

(4 marks)

(d) What factors, other than the value of the house, may affect the expenditure on repairs? (4 marks)

(Total 20 marks)

3009/3/10MA Page 6 of 20

MODEL ANSWER TO QUESTION 2 (a)

Value of house £000

Annual expenditure on repairs £ x

2 xy

215 1850 46225 397750

323 2550 104329 823650

455 3150 207025 1433250

329 1850 108241 608650

235 1750 55225 411250

425 1700 180625 722500

327 2600 106929 850200

385 3300 148225 1270500

237 2750 56169 651750

155 1300 24025 201500

196 1600 38416 313600

176 1750 30976 308000

3458 26150 1106410 7992600

x y x2 xy

22 xxn

yxxynb

23458110641012

261503458799260012b

1195776413276920

9042670095911200

b= 5484500 = 4.16 1319156

n

xb

n

ya

12

345816.4

12

26150

a = 981.1 y = 981.1 + 4.16x (b) Estimated costs of repair and maintenance = 981.1 + 4.16 x 250

= 981.1 + 1040. = 2021.1 (£) (c) The coefficient of determination measures the change in repairs expenditure due to the change in house value. Therefore, 40.3% of the change in repair expenditure is

due to the change in house value. The estimated repairs expenditure is not very accurate.

(d) Age of house, size of family, attitude to house repair, size and value may not be comparable.

3009/3/10MA Page 7 of 20

QUESTION 3 (a) Explain the circumstances in which a paired t test would be used in preference to a two sample

mean test for small independent samples. (4 marks)

A consumer protection magazine wishes to compare the cost of repairs between two companies. They used a sample of 9 businesses to request estimates for the cost of a repair by both companies.

Business Company X

Estimated cost of repairs £

Company Y Estimated cost of

repairs £

A 2400 2560

B 2600 2760

C 5696 5952

D 7936 7933

E 4000 4200

F 7040 7245

G 3930 3830

H 6000 5964

I 3576 3620

(b) Test whether the estimated cost of repairs differs between the two companies.

(12 marks) (c) What practical reasons may the two companies have for giving different estimates of repair

costs? (4 marks)

(Total 20 marks)

3009/3/10MA Page 8 of 20

MODEL ANSWER TO QUESTION 3 (a) The paired t test is used when a single sample is subject to two “treatments” compared with two samples being compared. (b) Null hypothesis: The estimated cost of repair does not differ between the two companies. Alternative hypothesis: The estimated cost of repair does differ between the two companies. Degrees of Freedom = n - 1 = 9 - 1 = 8 Critical t value 0.05 = 2.31: 0.01 = 3.36

Company X Estimated cost of repairs £

Company Y Estimated cost of repairs £

Difference d

)( dd 2)( dd d

2

2400 2560 -160 -61.556 3789.1 25600

2600 2760 -160 -61.556 3789.1 25600

5696 5952 -256 -157.556 24823.8 65536

7936 7933 3 101.444 10291.0 9

4000 4200 -200 -101.556 10313.5 40000

7040 7245 -205 -106.56 11354.1 42025

3930 3830 100 198.444 39380.2 10000

6000 5964 36 134.444 18075.3 1296

3576 3620 ..44 54.444 2964.2 …1936

-886 124780 212002

Σd 2)( dd Σd

2

n

dd =

sd =

22

n

d

n

d =

2

9

886

9

212002 = 117.7

or

n

ddsd

2

= 9

124780 = 117.7

1

0

ns

dt

d

=

19

7.117

44.98 = -2.36

-886 = -98.44 9

3009/3/10MA Page 9 of 20

QUESTION 3 CONTINUED Conclusion: The calculated value of t is greater than the critical value of t at the 0.05 level; reject the null hypothesis accept the alternative hypothesis, the estimated cost of repair does differ between the two companies. The results is not significant at the 0.01 level accept the null hypothesis, the estimates do not differ between the two companies. (c) The companies may have different cost structures due to eg labour costs, error in calculating costs, use of manufacturer’s or generic parts, quality of the work carried out.

3009/3/10MA Page 10 of 20

QUESTION 4

A random sample of workers in the Marketing department of a large company were interviewed and the degree of job satisfaction and their job status were recorded.

Job Status

Job Satisfaction

High Medium Low

Management 105 30 165

Skilled 69 32 89

Unskilled 56 38 96

(a) Test whether there is an association between job satisfaction and job status.

(12 marks)

(b) National data show that marketing workers express the following levels of job satisfaction.

Job Satisfaction

High Medium Low

30% 16% 54%

By combining the different job status of marketing workers’ test whether the views on job satisfaction of workers in the Marketing department differ from the national situation.

(8 marks)

(Total 20 marks)

3009/3/10MA Page 11 of 20

MODEL ANSWER TO QUESTION 4 (a) Null hypothesis: there is no association between job satisfaction and job status. Alternative hypothesis: there is association between job satisfaction and job status. Degrees of freedom (r - 1)(c - 1) = (3 - 1)(3 - 1) = 2 x 2 = 4

2 0.05 = 9.49; 0.01 = 13.28

Observed frequencies

Job Status

Job Satisfaction

High Medium Low

105 30 165

69 32 89

56 38 96

Expected frequencies

101.47 44.12 154.41

64.26 27.94 97.79

64.26 27.94 97.79

Contributions to

2

0.1228 4.5176 0.7261

0.3489 0.5896 0.7908

1.0629 3.6212 0.0329

chi sq= 11.8127 Conclusion: the calculated Chi-squared is greater than the critical value of Chi-squared at the 0.05 but not at the 0.01 significance level, there is some association between job satisfaction and job status. (b) Null hypothesis: marketing workers have the same levels of job satisfaction as marketing executives nationally. Alternative hypothesis: marketing workers do not have the same levels of job satisfaction as marketing executives nationally. Degrees of freedom (3 - 1) = 2

Critical value of 2 0.05 = 5.99 0.01 = 9.21

High Medium Low

Observed 230 100 350 Expected 204 108.8 367.2

3.314 0.712 0.806 Chi squared = 4.8313

Conclusion: the calculated value of

2 is less than the critical value of

2 there is insufficient

evidence to reject the null hypothesis that marketing executives have the same level of job satisfaction as marketing executives.

3009/3/10MA Page 12 of 20

QUESTION 5

A, B and C are candidates for a role as design manager. The Managing Director recommends a candidate and the company board must ratify the decision. The probability the Managing Director recommends A is 55%, B 25% and C 20%. The probability that the board ratifies A is 60%, B 30%

and C 10% and is independent of the recommendation made by the Managing Director. (a) Find the probability that

(i) B is a successful candidate (ii) No candidate is successful (iii) Given that one candidate is successful, it is candidate B.

(10 marks)

A container ship terminal can unload standard containers at the rate of 2000 per 8 hour day with a standard deviation of 200 containers. Assume the rate of unloading is normally distributed.

(b) Find the probability that more than 2360 containers are unloaded in a day.

(3 marks) (c) A ship carries 6500 containers what is the probability it will be unloaded in less than 3 days.

(7 marks)

(Total 20 marks)

3009/3/10/MA Page 13 of 20

MODEL ANSWER TO QUESTION 5 (a) (i) B successful = 0.25 x 0.3 = 0.075 (ii) No candidate is successful A = 0.55 x 0.4 = 0.22 B = 0.25 x 0.7 = 0.175 C = 0.20 x 0.9 = 0.18.. 0.575 (iii) Probability a candidate successful = 1 - 0.575 = 0.425 Probability B successful = 0.075 = 0.176 Probability success 0.425 (b) More than 2360 in a day

sd

xz

200

23602000z = 360/200 = 1.8

Table probability = 0.964 required probability = 1 - 0.964 = 0.036 (c) A joint normal distribution approach is needed.

Joint mean 321 xxx = 2000 + 2000 + 2000 = 6000

Joint standard deviation = 2

3

2

2

2

1 sdsdsd = 222 200200200 =346

sd

xz

346

60006500z = 500/346 = 1.4 (1.445)

Table probability = 0.919 required probability = 1- 0.919 = 0.081 (0.075)

3009/3/10/MA Page 14 of 20

QUESTION 6 (a) Explain why, when the sample size increases, the sample proportions cluster more closely

about the population proportion. (4 marks)

A company is concerned about the difference in sick leave between the US factories and the UK

factories. A random sample of 70 US staff showed 11 took more than 10 days sick leave absent, whilst a sample of 90 UK staff showed 18 took more than 10 days sick leave. (b) (i) Test whether the proportion of staff taking more than 10 days sick leave is higher in the

UK than the US (12 marks)

(ii) Explain what is meant by a Type 1 error and whether a Type 1 error may have been

committed in your conclusions in part (b) (i). (4 marks)

(Total 20 marks)

3009/3/10/MA Page 15 of 20

MODEL ANSWER TO QUESTION 6 (a) When samples of a given size are taken, the distribution of the sample proportions is referred to as the sampling distribution of the proportion .

The formula is se = n

pp 1( as n increases for any given proportion the value of se

will decrease. (b) (i) Null hypothesis: the proportion of staff with more than 10 days sick leave in the US does not differ from the proportion of staff with more than 10 days sick leave in the UK.

Alternative hypothesis: the proportion of staff with more than 10 days sick leave in the US is less than the proportion of staff with more than 10 days sick leave in the UK.

Critical z value for 0.05 significance level = -1.64 p1 = 11 = 0.1571; p2 = 18 = 0.20 70 90

21

2211

nn

pnpnp

= 0.1571 x 70 + 0.20 x 90 = 11 + 18 = 0.181 70 + 90 160

21

21

111

nnpp

ppz

90

1

70

1819.0181.0

20.01571.0

= -0.0429 = - 0.698 0.0614 Conclusions: the calculated z value is less than the critical value of z. There is insufficient evidence to reject the null hypothesis. The proportion of staff with more than 10 days sick leave in the US does not differ from the proportion of staff with more than 10 days sick leave in the UK. (ii) A type 1 error is when a true null hypothesis is rejected when it should be accepted. As the null hypothesis was accepted a type 1 error cannot have been made.

3009/3/10/MA Page 16 of 20

QUESTION 7 An investigation was carried out into the number of errors per shift made in dispatching internet orders from warehouses in London and in Hong Kong. Random samples were taken in both London and Hong Kong. A summary of the results is given below.

London Hong Kong

Arithmetic mean 225 211

Standard Deviation 31 23

Median 218 217

Sample size 55 47

(a) Calculate a measure of skewness for each of the cities.

(4 marks)

(b) Write a brief memo to the Sales Manager of the company highlighting the main features in the error statistics between the two cities.

(8 marks)

(c) Test whether the average number of errors in Hong Kong is less than the average number of errors in London.

(8 marks)

(Total 20 marks)

3009/3/10/MA Page 17 of 20

MODEL ANSWER TO QUESTION 7 (a) The measure of skew = 3(mean – median) Standard deviation London = 3(225 – 218) Hong Kong= 3(211 – 217) 31 23 = 0.677 = -0.783 (b) Memo layout: Title, Date, From, To and Subject A range of content is possible. The following is suggested The arithmetic mean and standard deviation are both larger for London than Hong Kong. Coefficient of variations 0.138 and -0.109 London is more variable than Hong Kong. The London Median is greater than the Hong Kong Median. The London Mean is greater than the Hong Kong Mean. The London mean must have more extreme high values whilst Hong Kong must have more extreme low values. The Hong Kong data is negatively skewed and London data is positively skewed but to approximately the same degree. (c) Null hypothesis: there is no difference in the number of errors made in dispatching internet orders from warehouses in London and in Hong Kong. Alternative hypothesis: there is a greater number of errors made in dispatching internet orders from warehouses in London than in Hong Kong. Critical value of z for 0.05 significance level = 1.64/2.33

1

2

2

1

2

1

21

n

s

n

s

xxz =

47

23

55

31

21122522

26.1147.17

14z =

23.5

14 = 2.62

Conclusions: The calculated value of z is more than the critical value of z. There is evidence to reject the null hypothesis. There is a greater number of errors made in dispatching internet orders from warehouses in London than in Hong Kong.

3009/3/10/MA Page 18 of 20

QUESTION 8 (a) Explain the impact of a business setting its quality control limits too wide.

(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. The average weight of pre-cut tuna is set at 175 grams with a standard deviation of 5 grams. Samples of 9 items at a time are taken from the production line to check the accuracy of the manufacturing process. (b) (i) Draw a control chart to monitor the process

(7 marks)

(ii) 7 samples taken from the production line had the following mean weight (grams):

Weight per piece (grams)

178.6 180.8 170.5 168.4 172.5 175.8 175.2

Plot these data on your control chart and comment appropriately

(5 marks)

(iii) If the sample mean had been wrongly set at 177 cm, assuming the standard deviation remains the same and the sample size is 9, what is the probability that a sample mean lies outside the upper action limit?

(4 marks)

(Total 20 marks)

3009/3/10/MA Page 19 of 20 © Education Development International plc 2010

MODEL ANSWER TO QUESTION 8 (a) By setting the quality control limits too wide the number of samples rejected

will decrease and costs of resetting the process will be reduced and form a lower level of rejected items. The business may gain a reputation for poor quality which may damage its potential sales. (b)

(i) Warning limits n

x 96.1 9

596.1175 = 175 ± 3.26 = 171.7 to 178.3

(171.67 to 178.33 for z =2)

Action limits n

x 09.3 9

509.3175 = 175 ± 5.15 = 169.85 to 180.15

(170 to 180 for z = 3) (ii)

168

170

172

174

176

178

180

1 2 3 4 5 6 7

Weig

ht p

er

pie

ce g

ram

s

Sample number

Quality Control Chart

Comment: Samples 2 and 4 lie outside the action limits and the process is unstable. It needs to be stopped and adjusted.

(iii)

n

xz

9

5

17715.180z

= 3.15 = 1.9 table proportion = 0.971 therefore answer = 0.029

1.67

3009/3/10/MA Page 20 of 20 © Education Development International plc 2010

LEVEL 3

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