Business Statistics Level 3

Model Answers Series 2 2008 (Code 3009)

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Business Statistics Level 3 Series 2 2008

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 2008 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 2008 QUESTION 1 When it is under control a process produces components whose weight is normally distributed with mean 505mm and standard deviation 2 mm. Random samples of 5 components are selected at intervals and the mean length of each sample is measured. 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. (a) (i) Construct a control chart to monitor the mean length of these components. (8 marks) (ii) The means of 8 samples were: 506.2, 504.9, 504.7, 504.4, 504.8, 504.2, 504.6, and 503.7 Plot these values on your control chart and comment.

(4 marks) (b) If the process mean changed to 506 mm and the standard deviation remained at 2 mm, calculate

the probability that the mean of a randomly selected sample of 5 items would lie outside the warning limits.

(6 marks) (c) Give two business advantages of a good quality control scheme. (2 marks)

(Total 20 marks)

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499.00

500.00

501.00

502.00

503.00

504.00

505.00

506.00

507.00

508.00

509.00

1 2 3 4 5 6 7 8

MODEL ANSWER TO QUESTION 1

(a) Warning Limits n

σ1.96±x

5

21.96505 ± = 505 ± 1.96 x 0.89 = 505 ± 1.75 = 506.75 to 503.25

Action Limits n

σx 3.09±

505 ± 3.09 x 0.89 = 505 ± 2.76 = 507.76 to 502.24 Comment: All the points lie within the upper and lower control limits and therefore the process appears to be in control. However, the data shows a downwards trend and action should be taken to prevent the process moving out of control.

(b) n

xz

σ

µ−= =

0.89

506506.75 − =

0.89

0.75 = 0.84

Probability = 0.201

nσ

µxz

−= =

0.89

506503.25 − =

0.89

2.75− = -3.09

Probability = 0.001 Answer = 0.201 + 0.001 = 0.202 (c) Reduction of waste and therefore costs. Improved ability to market the product and greater sales.

UAL UWL

LWL

LAL

3009/2/08/MA Page 5 of 20

QUESTION 2 The manufacturer of an electronic component has made certain claims about the life of the component. (a) A random sample of 12 of these components was selected and the components were tested to

destruction. The lifetimes (in hours) were: 534 399 464 415 743 596 669 964 767 606 885 518 Set up a 95% confidence interval for the mean life of this component.

(12 marks) (b) The manufacturer claims the mean lifetime of the component is 700 hours. State the null

hypothesis and the alternative hypothesis and use the confidence interval from part (a) to test whether there is evidence that the mean life of the components differs from 700 hours.

(4 marks) (c) Explain what is meant by a type 2 error. State whether a type2 error may have occurred in the

conclusion you drew in part (b). (4 marks)

(Total 20 marks)

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MODEL ANSWER TO QUESTION 2 (a)

n

xx

∑= =

12

7560 = 630 hours

St deviation = 1

( )−

−∑

n

xx 2

= 112

360274

− or

22

−∑∑

=n

x

n

xsd

180.97 or 181 hours

( )2630

12

5123074−=sd

For a 95% Confidence interval v n-1 = 11, t = ± 2.20

n

sdxci 2.20±= =

12

1812.20630 ± = 630 ± 115.0

ci = 515 to 745 hours (b) Null hypothesis: the mean life of the component does not differ from 700 hours. Alternative hypothesis: the mean life of the component does differ from 700 hours. Using the 0.05 significance level the interval is 515 to 745.

Conclusion: The claimed value for the mean of 700 hours lies within the interval 515 to 745 hours. Accept the null hypothesis the mean life of the component does not differ from 700 hours.

(c) A type 2 error is when the null hypothesis is accepted when it should have been rejected. A type 2 error might have been made as the null hypothesis was not rejected.

x x2 _

x - x _

(x-x)2 534 285156 -96 9216 399 159201 -231 53361 464 215296 -166 27556 415 172225 -215 46225 743 552049 113 12769 596 355216 -34 1156 669 447561 39 1521 964 929296 334 111556 767 588289 137 18769 606 367236 -24 576 885 783225 255 65025 518 268324 -112 12544

Σx=7560 Σx2 = 5123074

( )2xx −∑ = 360274

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QUESTION 3 (a) Outline the main uses of index numbers from the point of view of a business person. (4 marks) The following data was collected from the wages department of a company.

Year 1 Year 4 Department Number of

employees Total wage cost

(£000) Number of employees

Total wage cost (£000)

Electrical 50 800 60 1200 Assembly 240 3408 120 1800 Testing 90 837 100 1000 Despatch 200 1400 420 3780

(b) (i) Calculate a Laspeyres index number for the increase in wages from year 1 to year 4. (ii) Calculate a Paasche index number for the increase in wages from year 1 to year 4.

(10 marks) (c) Contrast the use of Laspeyres and Paasche index numbers. (6 marks)

(Total 20 marks)

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MODEL ANSWER TO QUESTION 3 (a) Index numbers which are likely to be of most use to a business person are for the costs of raw

materials, which indicate how costs will have risen, wages to compare the position of the firm’s workers with others, and a general price index such as the Retail Price Index which would indicate possible future wage pressures.

(b) Year 1 Year 4 Department Number of

employees Total

wage cost (£000)

Number of employees

Total wage cost

(£000)

Yr1 Wage

w1

Yr 4 Wage

w4 Electrical 50 800 60 1200 16.0 20.0 Assembly 240 3408 120 1800 14.2 15.0 Testing 90 837 100 1000 9.3 10.0 Despatch 200 1400 420 3780 7.0 9.0 6445 7780

Σ p1q1 Σ p4q4

p4q1 p1q4 1000 960

36000 1704 900 930 1800 2940 7300 6534

Σ p4q 1 Σ p1q4

10011

14 xqp

qpL

∑

∑= = 100x

6445

7300 = 113.3

100x 41

44

qp

qpP

∑∑

= = 100x 6534

7780 = 119.1

(c) The Paasche index requires the actual weights to be found for each year of the series.

This can be expensive compared with the Laspeyres index which requires only the base year weights to be found.

The denominator of the formula for the Paasche index needs to be recalculated for each period. The Laspeyres denominator remains the same throughout the life of the index series.

The different years of a Laspeyres index number can be directly compared with each other whereas with the Paasche index the current year can only be compared with the base year.

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QUESTION 4 A company uses 2 procedures for assessing the suitability of staff for promotion: a test and an interview by a panel of three managers. The following data show the results of the promotion test and the scores the three managers gave each member of staff:

Candidate Test Manager X Manager Y Manager Z A 74 67 75 75 B 92 75 68 91 C 58 69 64 77 D 71 75 64 59 E 86 88 74 94 F 65 68 72 75 G 78 71 73 81 H 72 72 68 62 I 63 68 65 65 J 60 72 58 66

(a) Total the scores given by the three managers and find the correlation coefficient between the test

results and the managers’ total scores. Comment on your answer. (12 marks)

(b) Test whether the correlation coefficient differs from zero. (6 marks) (c) The correlation coefficients between the test results and the scores of the individual managers

were 0.61, 0.57 and 0.68 respectively. What is the implication of these results for the promotion procedure?

(2 marks)

(Total 20 marks)

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MODEL ANSWER TO QUESTION 4 (a)

( )( )

( )( ) ( )( )∑ ∑∑ ∑∑ ∑∑

−−

−=

2222 yynxxn

yxxynr

( )( )22 215110x46595971910x52803

719x215110x156143

−−

−=r

( )( )3278911069

14861=r = 0.78

Comment: There is strong positive association between the two variables. (b) The correlation coefficient does not differ significantly from zero. Alternative hypothesis: The correlation coefficient does differ significantly from zero. Degrees of freedom = n-2 = 10-2 = 8 Critical value of t0.05/0.01 = 2.31/3.36

21

2

r

nrt

−

−=

2(0.78)1

2100.78

−

−=t = 3.53

Conclusions: The calculated value of t is greater than the critical t value reject the null hypothesis. The correlation coefficient differs from zero.

(c) It is better to take the combined scores of the managers into account rather than those of an

individual manager.

Candidate Test x

Total scorey

x2

y2

xy

A 74 217 5476 47089 16058 B 92 234 8464 54756 21528 C 58 210 3364 44100 12180 D 71 198 5041 39204 14058 E 86 256 7396 65536 22016 F 65 215 4225 46225 13975 G 78 225 6084 50625 17550 H 72 202 5184 40804 14544 I 63 198 3969 39204 12474 J 60 196 3600 38416 11760

Totals 719 2151 52803 465959 156143

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QUESTION 5 (a) Explain how probabilities are determined using the classical and empirical approaches. Give a business example of each approach.

(6 marks) (b) An investment advisor considers the probability of a rise in the price of three shares in one

month’s time. The following table shows the probability of each share rising – where each share acts independently of each other.

Share Probability of a rise in

one month A 0.4 B 0.7 C 0.6

Find the probability that in one month’s time: (i) All three share prices rise. (ii) None of the three share prices rise. (iii) Exactly one of the three share prices rises.

(10 marks) (c) If one of the share prices has risen what is the probability it is share B? (4 marks)

(Total 20 marks)

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MODEL ANSWER TO QUESTION 5 (a) Classical approach to probability is where there are a set known number of outcomes which must

occur with given probabilities. A casino where the probability of certain cards appearing is known. This allows the owners to set the odds offered for winning combination.

Empirical approach to probability occurs when a process or number of events take place which can have different outcomes. For example, an oil exploration company boring test wells and from 100 wells bored 87 are dry.

(b) (i) Probability all rise 0.4 x 0.7 x 0.6 = 0.168 (ii) None increase 0.6 x 0.3 x 0.4 = 0.072 (iii) 0.4 x 0.3 x 0.4 = 0.048 0.6 x 0.7 x 0.4 = 0.168 0.6 x 0.3 x 0.6 = 0.108 0.324 (c) Probability B has risen 0.168 = 0.519 Probability 1 share has risen 0.324

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QUESTION 6 (a) Explain the difference between a one tail and a two tail test. (4 marks) The Accounts Department of a manufacturing company keeps a record of the debts of its customers. In 2000, from a random sample of 200 customers, 48 owed more than £1000. In 2007, from a random sample of 300 customers, 93 customers owed more than £1000. (b) Test whether the proportion owing more than £1000 has changed between 2000 and 2007.

(12 marks)

(c) Would your conclusion in (b) have been different if a one tail test had been used? Justify your

answer. (4 marks)

(Total 20 marks)

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MODEL ANSWER TO QUESTION 6 (a) A one tail test tests the direction of change either greater than or less than. A two tail test tests there is change both greater and less than. (b) Null hypothesis: there is no difference in the proportion of customers owing more than £1000.

Alternative hypothesis: there is a difference in the proportion of customers owing more than £1000.

p1 = 48/200 =0.24, p2 = 93/300 = 0.31 Critical z value for 0.05 significance level = ± 1.96

( )

+−

−=

21

21

111

nnpp

ppz where

21

2211

nn

pnpnp

+

+=

Pooled value of p = (48+93)/(200+300) = 0.282

( )

+−

−

300

1

200

10.28210.282

0.310.24 =

0.008330.202476

0.07

× = 1.704

Conclusions: There is insufficient evidence to reject the null hypothesis. There is no difference in the proportion owing more than £1000. (c) A one tailed test would use a Z value of +1.64. The results would still have been to reject the null

hypothesis and accept the alternative hypothesis, there is a difference in the proportion of customers owing more than £1000.

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QUESTION 7 (a) Explain the circumstances in which a chi-squared test can be used. (4 marks) (b) A car manufacturer sells its product in 3 sales regions, Europe, USA and Asia. A random sample

of 500 vehicles sold, showed the following distribution of sales region and engine capacity.

Region Engine Capacity Europe USA Asia Under 1000cc 40 20 60 1000cc and under 2000cc 100 120 30 Over 2000cc 30 90 10

Test whether there is any association between sales region and engine capacity. (12 marks)

When comparing the current proportion of vehicles in each engine category with 5 years ago a statistical test gives a chi-squared value of 6.83. Test whether the proportion of vehicles in each category has changed over the past five years.

(4 marks)

(Total 20 marks)

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MODEL ANSWER TO QUESTION 7 (a) A chi-squared test can be used to measure association where a contingency table is presented.

It can also be used to test for randomness or goodness of fit to a specified distribution. (b) Null hypothesis: There is no association between Sales Region and Engine Capacity. Alternative hypothesis: There is association between Sales Region and Engine Capacity Degrees of freedom (R-1)(C-1) = (3-1)(3-1) = 4 Critical chi-squared = 9.49/13.28 Observed Frequencies Expected frequencies

Engine Capacity

Europe USA Asia

Under 1000cc 40 20 60 40.8 55.2 24

1000cc and under 2000cc

100 120 30 85 115 50

Over 2000cc 30 90 10 44.2 59.8 26

Contributions to chi-squared

0.015686 22.44638 54 2.647059 0.217391 8 4.561991 15.25151 9.8461538

ΣX2 116.99

Conclusions: The calculated value of chi-squared is greater than the critical value of chi-squared at both the 0.05 and the 0.01 significance level. Reject the null hypothesis there is strong evidence that Region and Engine Capacity are associated.

(c) Null hypothesis: there has been no change in the proportion of vehicles in each engine category.

Alternative hypothesis: there has been a change in the proportion of vehicles in each engine category.

Critical X2 = Degrees of freedom = n-1 = 3-1 = 2 Critical X2 = 5.99/9.21

Conclusions: The calculated value of X2 is greater than the critical value at the 0.05 significance level, therefore there is evidence to show there has been a change in the proportion of vehicles in each engine capacity. However the calculated value of X2 is less than the critical value of X2 at the 0.01 level of significance therefore there is insufficient evidence to reject the null hypothesis

that there has been no change.

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QUESTION 8 (a) Briefly describe the components of a time series. (4 marks) The following table shows the daily sales over a three week period, in £, of a small store.

Monday Tuesday Wednesday Thursday Friday Saturday SundayWeek 1 476 694 767 775 651 437 495 Week 2 526 734 812 829 702 569 589 Week 3 559 797 851 895 745 618 754

(b) Use the method of moving averages to find the trend and the additive model to find the seasonal

variations. (10 marks)

(c) Predict the level of sales on Monday and Tuesday of week 4 and comment upon the likely

accuracy of your answer. (6 marks)

(Total 20 marks)

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MODEL ANSWER TO QUESTION 8 (a) Trend: the long term direction of change. Cyclical: a regular up and down pattern that repeats over a number of years. Seasonal variation: a regular up and down pattern that repeats over a year. Residual/random variation odd patterns of the data that have no demonstrable cause. (b)

Sales Moving Total Differences

476

Moving Average/

Trend 694 767 775 4295 613.6 161.4 651 4345 620.7 30.3 437 4385 626.4 -189.4 495 4430 632.9 -137.9 526 4484 640.6 -114.6 734 4535 647.9 86.1 812 4667 666.7 145.3 829 4761 680.1 148.9 702 4794 684.9 17.1 569 4857 693.9 -124.9 589 4896 699.4 -110.4 559 4962 708.9 -149.9 797 5005 715.0 82.0 851 5054 722.0 129.0 895 5219 745.6 149.4 745 618 754

Monday Tuesday Wednesday Thursday Friday Saturday Sunday 161.4 30.3 -189.4 -137.9

-114.6 86.1 145.3 148.9 17.1 -124.9 -110.4

-149.9 82 129 149.4 Total -264.5 168.1 274.3 459.7 47.4 -314.3 -248.3 Average S V -132.25 84.05 137.15 153.233 23.7 -157.15 -124.15

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MODEL ANSWER TO QUESTION 8 CONTINUED (c) Forecast Increase per day Tn-T1 = 745.6-613.6 = 9.43 n-1 15-1 Trend ASV Forecast Forecast

Monday 745.6 + (4 x 9.43) = 783.31 - 132.25 = 651.06 Tuesday 745.6 + (5 x 9.43) = 792.74 + 84.05 = 876.79

This is based on extrapolation and therefore the forecasts can be unreliable. However, as the forecast is only for one period ahead the results are likely to be reasonably accurate.

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