Business Statistics Level 3 Model Answers Series 2 2005 (Code 3009)

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Business Statistics Level 3 Series 2 2005

How to use this booklet

Model Answers have been developed by Education Development International plc (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 2005 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.

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Business Statistics Level 3 Series 2 2005 Question 1 (a) Describe the characteristics of the normal distribution. (4 marks) A company employs a team of electricians to maintain its machines. The time taken to repair a

machine follows a normal distribution with a mean time of 20 minutes and a standard deviation of 4 minutes. Whilst the machine is not working it is taken out of service.

(b) Calculate the probability that the time taken to repair a machine is: (i) less than 18 minutes (ii) between 18 minutes and 26 minutes. (6 marks) (c) The electricians wish to establish a benchmark time within which they can guarantee a 90% completion rate. What is that benchmark time? (4 marks) On some occasions when a machine needs repairing an electrician is not available. The mean

time taken for an electrician to become available is 10 minutes with a standard deviation of 5 minutes. The time taken for an electrician to become available is normally distributed. (d) If an electrician has to become available, what is the probability that finding an electrician and completing the repair takes more than 45 minutes? (6 marks) (Total 20 marks)

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Model Answer for Question 1 (a) Symmetrical bell shaped curve, asymptotic to the x axis, mean, median and mode are equal.

(b) 21

42

420–18z

sd–z ====µx

Proportion = 0.692 Answer = 1 – 0.692 = 0.308

5.146

420–26z

sd–z ====µx

Proportion = 0.933 therefore greater than 26 mins = 0.067 Between 18 minutes and 26 minutes. = 0.692 – 0.067 = 0.625 (c) 90% or 0.9 gives z = 1.3

1.3 = 4

20–x=

sd

µ–x

1.3 x 4 = x – 20 x = 5.2 + 20 = 25.2 (d) Joint mean = 2xx +1 = 20 + 10 = 30

Joint standard deviation = 22

21 sdsd +

= 4.64154 22 ==+

3246

15463045 .

..–– ==== z

sdxz µ

Proportion = 0.989 Probability more than 45 = 1 – 0.989 = 0.011

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Question 2 The table below shows information for a random sample of 10 workers. Their ages and salaries are given in the table below:

Worker Age (years) Salary £000

a 33 25.3 b 55 36.3 c 45 14.9 d 56 25.2 e 38 35.9 f 42 25.2 g 54 36.1 h 22 14.9 i 31 25.8 j 37 26.9

(a) Draw a scatter diagram. (3 marks) (b) Calculate the product-moment correlation coefficient and comment on your answer. (11 marks) (c) Test whether the correlation coefficient differs significantly from zero. (6 marks) (Total 20 marks)

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Model Answer for Question 2 (a) (b) x y x2 y2 xy

33 25.3 1089 640.09 834.955 36.3 3025 1317.69 1996.545 14.9 2025 222.01 670.556 25.2 3136 635.04 1411.238 35.9 1444 1288.81 1364.242 25.2 1764 635.04 1058.454 36.1 2916 1303.21 1949.422 14.9 484 222.01 327.831 25.8 961 665.64 799.837 26.9 1369 723.61 995.3413 266.5 18213 7653.15 11408

xΣ yΣ 2xΣ 2yΣ xyΣ

r = ))(–n)()(–n(

))((–n

22 22 yyxx

yxxy

ΣΣΣΣ

ΣΣΣ

( )( )22 5.266–15.653,7x10413–213,18x10

5.266x413–408,11x10r =

r = ( )( ) 25.509,5x561,11

5.015,425.022,71–5.531,76569,170–130,182

5.064,110–080,114=

r = 0.503 There is weak positive linear relationship between the variables.

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Model Answer for Question 2 continued Null hypothesis: the correlation coefficient does not differ from zero Alternative Hypothesis: the correlation coefficient does differ from zero Critical t value, 8 degrees of freedom = 2.31/3.36

2r–1

2–nr=t t =

2(0.503)–1

2–100.503

t = 25311.0–1

4231.1

= 8642.04231.1

= 1.647 Conclusion: The calculated value of t is less than the critical t value therefore do not reject the null hypothesis. The correlation does not differ from zero.

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Question 3 (a) Describe the four components of a time series decomposition. (4 marks) The table below shows the quarterly imports figures for a country in £billions:

Year Quarter 1 Quarter 2 Quarter 3 Quarter 4

2002 15 25 38 41

2003 20 38 51 58

2004 31 47 60 73

(b) Calculate the centred moving average for the quarterly imports and estimate the average seasonal variations. (10 marks) (c) Forecast the level of imports for the four quarters of 2005 and comment on the likely accuracy

ofyour answer. (6 marks) (Total 20 marks)

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Model Answer for Question 3 continued (a) The four components are the long term trend, the cyclical movement, the

seasonal variation and the residual or random element. (b)

Imports £B M Total M Avg 1 M Avg 2 Differences 15 or Trend 25 38 29.75 7.62541 31 8.37520 34.25 -15.87538 37.5 -1.62551 41.75 7.87558 44.5 12.37531 46.75 -16.87547 49 -3.87560

119 124 137 150 167 178 187 196 211

52.75

30.37532.62535.87539.62543.12545.62547.87550.875

73

1m 1cao 1m 1cao 1m 1cao 1m 1cao

Qtr 1 Qtr 2 Qtr 3 Qtr 4 -15.875 -1.625 7.625 8.375 -16.875 -3.875 7.875 12.375Average seasonal variation -16.375 -2.75 7.75 10.375

Quarterly growth rate = 1–valuestrendofNumber

trendFirst–trendLast

1–8

375.30–875.50

= 2.929 Forecast Quarter 1 = (50.875 + 2.929 x 3) -16.375 = 43 Quarter 2 = (50.875 + 2.929 x 4) -2.75 = 60 Quarter 3 = (50.875 + 2.929 x 5) + 7.75 = 73 Quarter 4 = (50.875 + 2.929 x 6) +10.375 = 79 The forecast should be fairly accurate as the data is reasonably stable and the length of forecast is not too great.

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Question 4 An estate agent took a random sample of house buyers to find out if there was a relationship between family income and house value.

Family Income House Value

Under £25,000

£25,000 and under £50,000

£50,000 and over

Under £100,000 30 50 10

£100,000 and under £200,000

110 120 40

£200,000 and over 40 30 70 (a) Test for association between family income and house value. (12 marks) (b) By combining the data for income levels test to see if the proportion of houses over £200,000 sold by the estate agent differs significantly from the national proportion of 0.26. (8 marks) (Total 20 marks)

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Model Answer for Question 4 (a) Null hypothesis: there is no association between the value of the house and family income. Alternative hypothesis: there is association between the value of the house and family income. Degrees of freedom = (3 – 1)(3 – 1) = 4 Critical 2χ 0.05/0.01 = 9.49/13.28

Observed 30 50 10

90

110 120 40 270

40 30 70 140

180 200 120 500 Expected 32.4 36 21.6 90 97.2 108 64.8 270 50.4 56 33.6 140 180 200 120 500 Contributions 0.178 5.444 6.229 To χ 2 1.685 1.333 9.491 2.146 12.071 39.433 χ 2 = 78.007

Conclusions: The value of 2χ is greater than the critical value. The difference is highly significant. Reject the null hypothesis. There is a highly significant association between the value of the house and family income.

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Model Answer for Question 4 continued (b) Null hypothesis: there is no significant difference between the national proportion of people buying houses over £200,000 and the estate agents sales’ proportion. Alternative hypothesis: there is a significant difference between the national proportion of people buying houses over £200,000 and the estate agents sales’ proportion. Critical z 0.05 = 1.96

5002601260

2602801 ).–(.

.–.π)–π(π– ==

n

pz

z = 0003848.0

02.0 = 1.0195 (1.02)

Conclusion: As the calculated value of z is less than the critical value of z do not reject the null hypothesis. There is no significant difference between the national proportion of people buying houses over £200,000 and the estate agents sales’ proportion.

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Question 5 (a) Explain the meaning of the term the sampling distribution of the mean illustrating the difference in

application when there are large and small sample sizes. (6 marks)

Random samples were taken of the price of three bedroomed apartments in urban and rural areas.

Urban area £000 Rural area £000

149 189

155 157

141 142

165 128

189 155

135 169

149 175

187 176

146 194

153 (b) Test to discover whether the price of three-bedroomed apartments is significantly cheaper in urban areas than rural areas. (14 marks) (Total 20 marks)

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Model Answer for Question 5 (a) As samples of a given size are taken the mean values will vary. The distribution of the mean values is referred to as the sampling distribution of the mean. When the sample size is large, 30 or greater, the sampling distribution is normally distributed with the mean of the sample means equal to the population mean. When the sample size is small and the data is normally distributed, less than 30, the sampling distribution follows the ‘t’ distribution. (b) Null Hypothesis: The price of a three bedroomed apartment in urban areas is not significantly cheaper than the price of a three bedroomed apartment in rural areas. Alternative Hypothesis: The price of a three bedroomed apartment in urban areas is significantly cheaper than the price of a three bedroomed apartment in rural areas. Degrees of freedom = n + m – 2 = 10 + 9 – 2 = 17 Critical ‘t’ = 2.11/2.90

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Model Answer for Question 5 continued x y 2)–( xx 2)–( yy

149 189 62.41 576155 157 3.61 64141 142 252.81 529165 128 65.61 1369189 155 1030.41 100135 169 479.61 16149 175 62.41 100187 176 906.01 121146 194 118.81 841

153 15.211569 1485 2996.9 3716

22 )()( y–yx–xyx ∑∑∑∑

156.9=10

1,569=x 165

9485,1

==y

Pooled sd = 2–mn

)()( 22

++ y–yx–x

= 2–89716,39.996,2

++ =

179.712,6 = 19.87

t =

m1

n1s +

y–x t =

91

10187.19

165–9.156

+

t = 12.9

1.8– = –0.887

Conclusion: The calculated value is less than the critical value, so there is insufficient evidence to reject the null hypothesis. The price of a three bedroomed apartment in urban areas is not significantly cheaper than the price of a three bedroomed apartment in rural areas.

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Question 6 (a) Explain the benefits to a business of using a quality control system. (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 mean exceeding the limit is 0.001. The length of a product is set at 300 cm with a standard deviation of 2 cm. 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. (8 marks) (ii) The first 7 samples taken had the following mean lengths (in cm):

Length cm 298.4 302.5 301.6 298.2 302.2 302 299.8 Plot these data on your control chart and comment appropriately. (4 marks) (iii) If the sample mean had been wrongly set at 299 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 lower warning limit?

(4 marks) (Total 20 marks)

CONTINUED ON THE NEXT PAGE 17

Model Answer for Question 6 (a) The most important benefit to a business of a good quality control system

is the better reputation it will have with its customers. It allows the company to detect faults in a system quickly and be able to

rectify them. This should reduce waste and cut costs. It should ensure that the output is of good quality. (b) (i)

Quality Control Chart

297.5

298.5

299.5

300.5

301.5

302.5

303.5

1 2 3 4 5 6 7

Sample Number

Leng

th c

m

UAC

UWL

MEAN

LWL

LAL:

(ii) Mean length = 300 cm, σ = 2 cm n = 9

Warning limits = n

96.1 σ±x

9296.1300 ±

Warning limits = )used2zif333.1(307.1300 =± 298.697 to 301.307 (298.666 to 301.333)

Action limits = 9

23.09±300=r

n

σ3.09±x

Action limits = 300 ± 2.06 297.94 to 302.06

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Model Answer for Question 6 continued (iii) The lower warning limit is 298.697

n

–z σµx

= 45.0–

92

299–697.298z ==

using z = 0.4 probability = 1 – 0.655 = 0.345 using z = 0.5 probability = 1 – 0.692 = 0.308

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Question 7 (a) Explain the difference between a Type I and Type II error. (4 marks) The records of a company contain the following data on order value for the past two months:

September October

Mean value per order £ 640 625

Standard deviation £ 40 35

Sample size 60 55 (b) Has there been a significant decrease in the mean value per order between September and October? (7 marks) (c) Pool the data for September and October to give a best estimate of the mean value and the

standard deviation per order. Using these data, estimate a 95% confidence interval for the mean value per order.

(9 marks) (Total 20 marks)

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Model Answer for Question 7 (a) A Type I error is the rejection of a true null hypothesis

A Type II error is the acceptance of a false null hypothesis (b) Null hypothesis: there is no significance difference in the mean order value between September and October. Alternative hypothesis: there is significant difference in the mean order value between September and October. Critical z value = ± 1.64, ± 2

z =

1

22

1

21

21

ns

ns

–

+

xx z =

5535

6040

625–64022

+

z = 996.615

94.4815

27.2267.2615

==+

z = 2.14 Conclusion: there is sufficient evidence to reject the null hypothesis at the 5%, the mean value of orders between September and October has decreased. There is insufficient evidence to reject the null hypothesis at the 1%, accept the alternative hypothesis the mean value of orders between September and October has decreased.

(c) Joint mean = n1 x 1 + n2 x 2 = 83.632115

775,72115

625x55640x60==

+

Joint Standard Deviation = 5560

35x5540x60nn

sdnsdn 22

21

222

211

++

=++

69.3765.420,1115

375,163==

z value for 95% significance level = ± 1.96

ci = ±x 1.96 nσ ci = 632.83 ± 1.96

11569.37

= 632.83 ± 6.89 = 625.94 to 639.72

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Question 8 (a) Explain what is meant by and give a business example of: (i) Independent events. (ii) Conditional probability. (4 marks) A company is planning the launch of a new product. It estimates that the probability of good market conditions is 60%. If market conditions are good the probability of a successful launch is 70%, if market conditions are poor the probability of a successful launch is 50%. (b) Find the probability that the launch is successful. (5 marks) (c) If the product launch is unsuccessful what is the probability that the market conditions were poor? (6 marks) The estimated returns from the new product launch are:

£ millions Market conditions are good and the product launch is successful 80 Market conditions are good and the product launch is unsuccessful –18 Market conditions are poor and the product launch is successful 50 Market conditions are poor and the product launch is unsuccessful –25

(d) What is the expected profit from the new product launch? (5 marks) (Total 20 marks)

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Model Answer for Question 8 (a) (i) An independent event is one where the occurrence of one event does not affect the probability of a second event eg a rise in costs and an increase in supply. (ii) The conditional probability of an event is the probability that one event occurs given another event has occurred eg given a customer buys a product today, what is the probability they saw last night’s TV advertisement? (b) Probability good market conditions and successful = 0.6 x 0.7 = 0.42 Probability bad market conditions and successful = 0.4 x 0.5 = 0.20 0.62 (c) Probability launch unsuccessful = 1 – 0.62 = 0.38 Probability unsuccessful and bad market conditions = 0.5 x 0.4 = 0.20

38.020.0

ulunsuccessflaunchobabilityPrconditionsmarketbadandulunsuccessflaunchobabilityPr

=

= 0.526 (d) Market condition and outcome Prob £m £m Market conditions are good and the product launch is successful 0.42 x 80 33.60 Market conditions are good and the product launch is unsuccessful 0.18 x–18 –3.24 Market conditions are poor and the product launch is successful 0.20 x 50 10.00 Market conditions are poor and the product launch is unsuccessful 0.20 x–25 –5.00 35.36

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