Model Answers Series 4 2010 (3009)

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LCCI International Qualifications

Business Statistics Level 3

3009/4/10/MA Page 1 of 17

Business Statistics Level 3 Series 4 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.

3009/4/10/MA Page 2 of 17

QUESTION 1 (a) When might a measure of rank correlation be used in preference to the product-moment

correlation coefficient? (4 marks)

A random sample of cars was used to measure the relationship between the official fuel consumption and the engine capacity of the car. The results are given below.

Engine capacity (litres)

Petrol Consumption

(Miles per gallon)

1.2 47.8

1.4 45.5

1.6 42.8

2.0 34.0

4.7 18.0

1.6 40.9

3.5 31.0

5.5 24.4

3.0 25.9

(Source Whatcar? September 2008) (b) Calculate the product-moment correlation coefficient and comment on your answer. (10 marks) (c) Test whether the correlation coefficient differs significantly from zero. (6 marks)

(Total 20 marks)

3009/4/10/MA Page 3 of 17

MODEL ANSWER TO QUESTION 1 (a) The rank correlation coefficient is used when only ordinal or

positional data is available; product-moment correlation is used

when interval data is available

(b)

Engine capacity litres

Petrol Consumption

Miles per gallon

x y x2 y

2 xy

1.2 47.8 1.44 2284.84 57.36

1.4 45.5 1.96 2070.25 63.70

1.6 42.8 2.56 1831.84 68.48

2.0 34.0 4.00 1156.00 68.00

4.7 18.0 22.09 324.00 84.60

1.6 40.9 2.56 1672.81 65.44

3.5 31.0 12.25 961.00 108.50

5.5 24.4 30.25 595.36 134.20

3.0 25.9 9.00 670.81 77.70

24.5 310.3 86.11 11566.91 727.98

Σx Σy Σx2 Σy

2 Σxy

2222 yynxxn

yxxynr

22 3.31091.1156695.2411.869

3.3105.2498.7279r

1.781674.174

53.1050r

8989.0r (-0.9)

Strong – negative correlation

3009/4/10/MA Page 4 of 17

QUESTION 1 CONTINUED

(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 = 9-2 = 7

Critical t value =2.37/3.50

21

2

r

nrt

2)8989.0(1

298989.0t = -5.43

Conclusion, The calculated t value is greater than the critical t value at 0.05

and 0.01 levels. Reject the null hypothesis there is a highly significant difference from 0

3009/4/10/MA Page 5 of 17

QUESTION 2 (a) Distinguish between the circumstances in which a paired t test is used in preference to a two-

sample independent t test. (4 marks)

A company took a random sample of the expenses claimed by senior managers in 2008 and 2009. (data to the nearest £ hundred)

2008 2009

£ £

42 45

44 46

45 60

48 52

55 45

56 52

46 49

42 53

45

(b) Test whether there has been a significant increase in expenses between the two years.

(14 marks) (c) If a two tail test had been carried out would the conclusion remain the same as that drawn in

part (b)? Justify your answer. (2 marks)

(Total 20 marks)

3009/4/10/MA Page 6 of 17

MODEL ANSWER TO QUESTION 2 (a) A paired t test will be used where a single sample is subject to two treatments or a “before and after” measurement. A two independent sample test is used when two different samples are compared. (b) Null hypothesis: There has not been an increase in the level of expenses between 2008 and 2009. Alternative hypothesis: There has been an increase in expenses between 2003 and 2004. Critical value of t at 0.05 significance level, 8+9-2 =15 degrees of freedom = 1.750.05/2.600.01 1 tail

2008 2009

£ £ x

2

y2

2

11 xx 2

yy

42 45 1764 2025 27.563 21.778

44 46 1936 2116 10.563 13.444

45 60 2025 3600 5.0625 106.78

48 52 2304 2704 0.5625 5.4444

55 45 3025 2025 60.063 21.778

56 52 3136 2704 76.563 5.4444

46 49 2116 2401 1.5625 0.4444

42 53 1764 2809 27.563 11.111

45 2025 21.778

378 447 18070 22409 209.5 208

Σx Σy Σ x2 Σ y

2 Σ

2

11 xx Σ 2

22 xx

n

xx

1

8

378 = 47.25

n

xy

2

9

447 = 49.67

22

1n

x

n

xs

2

8

378

8

18070 = 5.12

or n

xx2

= 8

5.209 = 5.12

22

2m

y

m

ys

2

9

447

9

22409 = 4.81

3009/4/10/MA Page 7 of 17

QUESTION 2 CONTINUED

or m

yy2

= 9

208 = 4.81

s = 2

2

2

2

1

mn

smsn or s =

2

22

mn

yyxx = 5.28

mns

yxt

11

1

9

1

8

128.5

67.4925.47=

56.2

42.2 = - 0.94

Conclusions: The calculated value of t is less than the critical value of t.0.05 There is no evidence to reject the null hypothesis. There has not been an increase in the level of expenses between 2008 and 2009. (c) The critical values of t0.05/0.01 would be 2.13/2.95. The answer would be the same at the 0.05 significance level where the null hypothesis would not be rejected. The answer would remain the same at the 0.01 significance level.

3009/4/10/MA Page 8 of 17

QUESTION 3 A company suspects there are regional differences within the UK in the form of transport used by its customers to visit its DIY superstores. It carries out a random sample of customers‟ transport use.

Region

North UK Midlands South UK

Form of Transport

Car 143 256 231

Foot 56 121 123

Public transport 51 73 146

(a) Test whether the company‟s suspicions are correct. (12 marks) A previous survey of general shopping trips had shown that the percentages made by different transport methods were.

Transport Method (%)

Car Foot Public transport

65% 20% 15%

(b) Test whether the transport pattern for DIY superstores shown in part (a) varies from the general

pattern shown in the previous survey. (8 marks)

(Total 20 marks)

3009/4/10/MA Page 9 of 17

MODEL ANSWER TO QUESTION 3 (a) Null hypothesis: There is no association between region and the form of transport

used by its customers to visit the stores.

Alternative hypothesis: There is an association between region and the form of transport used by its customers to visit the stores.

Degrees of freedom = (3-1) x (3-1) = 4

Critical 2 0.05 significance level = 9.49 (0.01, 13.28)

Expected values

Contributions to 2

1.05 1.65 3.78

0.68 0.64 0.03

0.49 7.88 9.98 Chi-squared =26.18

Conclusions: The calculated value of Chi-squared is greater than the critical value of

2 at both the 0.05 and 0.01 levels of significance. Reject the null hypothesis. There is

an association between the region and the form of transport used by customers. (b) Null hypothesis: The form of transport used has not changed over time. Alternative hypothesis: The form of transport used has changed over time.

Critical value of 2 , 2 degrees of freedom = 5.990.05 and 9.210.01

Actual 630 300 270 Expected 780 240 180

Contribution to Chi-squared

28.84615 15 45 Chi-squared = 88.85

Conclusions: The calculated value of Chi-squared is greater than the critical Chi-squared at the 0.01 level therefore reject the null hypothesis there is very strong evidence that the form of transport has changed over time.

131.25 236.25 262.50

62.50 112.50 125.00

56.25 101.25 112.50

3009/4/10/MA Page 10 of 17

QUESTION 4

(a) Identify and describe five factors which should be taken into account when designing a consumer price index.

(10 marks) (b) The government wishes to increase the indirect taxes on Tobacco and Alcohol drink but does

not wish the measure of price inflation to reflect the increase. The current breakdown of the RPI (Retail Price Index) is shown below.

Category Weight Index(1987 =100)

Food 111 188.4

Catering 47 267.7

Alcoholic drink 59 229.2

Tobacco 27 381.4

Housing 254 321.0

Fuel and light 33 283.9

Household goods 66 160.0

Household services 64 205.3

Clothing and footwear 42 86.8

Personal goods and services 41 223.0

Motoring expenditure 133 179.7

Fares and other travel costs 20 285.3

Leisure goods 38 84.7

Leisure services 65 294.3

(Based on the Retail Price Index, Monthly Digest of Statistics, February, 2009 Jan 1987 =100)

(i) Calculate the Retail Price Index including all items (5 marks)

(ii) By removing the two categories Tobacco and Alcoholic drink, calculate the adjusted value

of the index.

Explain the reason why the index value has changed as your answer suggests. (5 marks)

(Total 20 marks)

3009/4/10/MA Page 11 of 17

MODEL ANSWER TO QUESTION 4 (a) Constituents. As the index is representative of all goods and services a sample

of items has to be chosen. This raises the issue of how the sample is to be selected.

Prices. The same items might have different prices depending on the range of different shops/retail-outlets and different size towns and cities. The problem is how to create a representative price. The selection of items changes as people‟s purchasing patterns change e.g. the disappearance of black and white TV, the addition of DVD players

Weights. Each item and each group of items has to have a relative importance in the index. This is usually based on the proportion of income spent on the item or group of items in the previous year. The weights are usually up-dated annually.

Base year. To avoid exaggeration or diminution of changes in the index the base year has to be chosen so as to minimise these two effects.

Type of Index. A large number of different forms of indices can be used.

The retail price index is a value index based on changing weights and prices of items which is compared with the base year.

Other relevant points can gain credit. (b) (i)

Category Weight Index WxI

Food 111 188.4 20912.4

Catering 47 267.7 12581.9

Alcoholic drink 59 229.2 13522.8

Tobacco 27 381.4 10297.8

Housing 254 321 81534

Fuel and light 33 283.9 9368.7

Household goods 66 160 10560

Household services 64 205.3 13139.2

Clothing and footwear 42 86.8 3645.6

Personal goods and services 41 223 9143

Motoring expenditure 133 179.7 23900.1

Fares and other travel costs 20 285.3 5706

Leisure goods 38 84.7 3218.6

Leisure services 65 294.3 19129.5

Total 1000 236659.6

RPI = W

WI =

1000

6.236659 = 236.7

(ii) RPI excluding Alcohol and Tobacco ΣW = 1000-86 = 914 ΣWI = (236659.6 – (13522.8+10297.8))

W

WI =

914

212839 = 232.9

The answer is less because the excluded category tobacco has an index value substantially higher than the all items RPI value.

3009/4/10/MA Page 12 of 17

QUESTION 5

(a) Describe the characteristics of the Normal Distribution. (4 marks) (b) The weekly demand for a product is normally distributed with mean of 2400 units and standard

deviation of 400 units. Calculate the probability that weekly demand is

(i) less than 1800 units (ii) between 2200 units and 2800 units.

(8 marks) (c) Individual units of the product have a mean weight of 250 grams with a standard deviation of 10

grams, the weights are normally distributed. The product is packed in boxes of 12 units.

(i) Calculate the mean weight and standard deviation of the contents of a box of 12 units. (5 marks)

(ii) What weight should the company quote for the weight of the contents of a box if it wishes

it to be exceeded on 99% of occasions? (3 marks)

(Total 20 marks)

MODEL ANSWER TO QUESTION 5 (a) The characteristics of the Normal Distribution are: It is symmetrical, bell shaped,

the mean median and mode coincide. A standardised normal distribution has a standard deviation with a value of 1. The tails of the distribution never touch the „x‟ axis (the distribution is asymptotic to the „x‟ axis).

(b (i) xx

z = 400

24001800 = -1.5, p = 1-0.933 = 0.067

(ii) xx

z1 = 400

24002200 = -0.5, p = 0.692-0.5 = 0.192

xx

z2 = 400

24002800 = 1, p = 0.841-0.5 =0.341

between 2200 units and 2800 units = 0.192+0.341 =0.533 (c) (i) Mean weight of a box = 12 x 250 = 3000g

Standard deviation of a box of 12 = 21 10 x 2 = 1200 =34.64g

(ii) sdx 33.2 = 3000 – 80.71

= 2919.29g

3009/4/10/MA Page 13 of 17

QUESTION 6 (a) What is meant by a 95% confidence interval?` (4 marks) (b) A random sample of 320 postings to the sales ledger by the head office of a company showed 7

errors, whilst a random sample of 400 postings to the sales ledger by the branch office showed 12 errors.

Test whether the proportion of errors differs between the head office and the branch office.

(12 marks) (c) Explain the difference between a type 1 and type 2 error and identify which type might occur in

your answer to part (b). (4 marks)

(Total 20 marks)

MODEL ANSWER TO QUESTION 6 (a) The 95% confidence interval means that if 100 samples of a given size are taken the population parameter will lie within the stated interval on average 95 of occasions. (b) Null hypothesis: the proportion of errors does not differ between the two offices.

Alternative hypothesis: the proportion of errors does differ between the two offices.

Critical z value 0.05 significance level = ±1.96 (0.01, ±2.58) p1 = 7/320 = 0.021875, p2 = 12/400 =0.03

21

2211

nn

pnpnp

400320

40003.0320021875.0

720

19 = 0.02639

21

21

111

nnpp

ppz

400

1

320

102639.0102639.0

03.0021875.0

= -0.008125 = -0.68 0.01202 Conclusions: The calculated value of z is less than the critical value of z, There is insufficient evidence to reject the null hypothesis. The proportion of errors does not differ between the two offices. (c) A type one error is the rejection of a true null hypothesis. A type 2 error is the acceptance of false null hypothesis. A type 2 error might have occurred.

3009/4/10/MA Page 14 of 17

QUESTION 7 A company wishes to sample the opinion of its workforce towards new working practices it is planning to introduce. (a) What is meant by the sampling frame when a sample survey is carried out?

Give an example of what might be used as a sampling frame in the situation outlined above. (4 marks)

(b) Identify and describe two appropriate methods by which the sample might be taken and state

two advantages and two disadvantages of each of the methods you suggest. (12 marks)

(c) What sample size is needed to be 95% confident of being within 0.03 of the population

proportion when there is no current estimate of the population proportion? (4 marks)

(Total 20 marks)

MODEL ANSWER TO QUESTION 7 (a) The sampling frame is a list of the total population which is to be surveyed. In this case it will be a list of all the workforce (perhaps using payroll information). (b) A simple random sample whereby all members of the workforce have an equal chance of being chosen. Each member of the work force is allotted a number and a random sample, up to the required size is selected by using random number tables or a random number generator. Advantages The standard error of the sample statistic can be calculated. Each item of the population has an equal chance of being chosen. Disadvantages A sampling frame has to exist (which in this case is possible). The sample drawn may not be representative of the underlying population. A stratified random sample could be used where the workforce is divided into homogeneous sub groups which may be thought to have differing views on

the topic. Advantages The results will be more representative of the underlying population. Information will be gained about each of the sub groups Greater precision, the standard error of a stratified random sample is less than that of a simple random sample. Disadvantages A more complex sampling frame is needed. The relevant strata need to be identified. It is more complex and hence expensive than a simple random sample.

(Other appropriate methods may be suggested) (c) When there is no current estimate of the population proportion the value of p is taken to be 0.5. The value for z for 95% confidence = ±1.96

2

22

03.0

5.096.1n

0009.0

25.08416.3 = 1067.11 (1068 or 1111/1112 if z=2 used)

3009/4/10/MA Page 15 of 17

QUESTION 8 (a) Explain what is meant by the sampling distribution of the mean. (4 marks) (b) The table below shows the results of two random samples carried out into the level of

productivity of a major international manufacturing company, with two plants in the UK and Belgium.

Test whether the sample means differ between the UK and Belgium. (8 marks) (c) Pool the data for the means and standard deviation and hence estimate the 95% confidence

interval for the joint mean. (8 marks)

(Total 20 marks)

UK Belgium

Mean productivity (units per shift) 675 702

Standard Deviation 56 48

Sample Size 100 140

3009/4/10/MA Page 16 of 17

MODEL ANSWER TO QUESTION 8 (a) When a number of samples of a given size n are taken from a population with mean, μ and standard deviation, σ and the means of the samples calculated, the sample means will be distributed with a standard deviation

(standard error) n

and mean μ.

(b) Null hypothesis: There is no difference in the mean productivity between the UK and Belgium. Alternative hypothesis: There is a difference in the mean productivity between the UK and Belgium. The critical z value = 0.05 significance level =1.96 (0.01, 2,58)

= -3.9046 Conclusions: The calculated value of z is greater than the critical value of z at both the 0.05 and 0.01 critical values. Reject the null hypothesis, accept the alternative hypothesis that productivity does differ between the UK and Belgium. (c) Pooled mean

= 117780 = 690.75 240 Pooled standard deviation

Sd p =

21

2

22

2

11

nn

snsn =

140100

4814056100 22

240

636160 = 51.48

95% confidence interval = n

x 96.1 =240

48.5196.175.690

= 690.75 ± 6.51 = 684.24 to 697.26

1

2

2

1

2

1

21

n

s

n

s

xxz

140

48

100

56

702675

22

21

2211

nn

xnxnx

140100

140702100675

3009/4/10/MA Page 17 of 17 © Education Development International plc 2010

LEVEL 3

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