ASE3009209MA

Model Answers Series 2 2009 (3009)

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LCCI International Qualifications

Business Statistics Level 3

Page 1 of 19

Business Statistics Level 3 Series 2 2009

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 2009 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.

3009/2/09/MA Page 2 of 19

QUESTION 1 An oil exploration company records the life of its drill bits against a measure of rock hardness. For the last 12 jobs the records show: Job a b c d e f g h i j k l Hardness 20 14 26 31 35 23 29 31 14 27 29 22 Life (Hours) 122 170 100 83 58 97 67 60 157 94 83 98 (a) Plot the data on a scatter graph.

(4 marks) (b) Calculate the equation of the regression line for the life of drill bits based on the rock hardness.

(10 marks) The correlation coefficient given by the above data is -0.954. (c) Test whether the correlation coefficient differs significantly from zero.

(6 marks)

(Total 20 marks)

3009/2/09/MA Page 3 of 19

MODEL ANSWER TO QUESTION 1 (a)

Life v Hardness

0

20

40

60

80

100

120

140

160

180

0 5 10 15 20 25 30 35 40

Hardness

Life

Hou

rs

(b)

Hardness (x) Life Hours (y) X2 XY 20 122 400 2440 14 170 196 2380 26 100 676 2600 31 83 961 2573 35 58 1225 2030 23 97 529 2231 29 67 841 1943 31 60 961 1860 14 157 196 2198 27 94 729 2538 29 83 841 2407 22 98 484 2156 301 1189 8039 27356

3009/2/09/MA Page 4 of 19

MODEL ANSWER TO 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 = 12 – 2 =10, critical t0.05/0.01= 2.23/3.17

2r1

2nrt

−

−=

20.9541

2120.954t

−

−−= = -10.06

Conclusions: Reject the null hypothesis. The result is highly significant. The correlation coefficient differs from zero.

b = nΣXY - ΣXΣY nΣX2- Σ(X)2 12 x 27356 – 301 x 1189 12 x 8039 - 3012 328272 - 357889 96468 - 90601 -5.048 (-5.05)

a = ΣY - bΣX

n n

1189 – -5.048 x 301 12 12 99.08 – -126.62 225.7

y = 225.7 – 5.048x

3009/2/09/MA Page 5 of 19

QUESTION 2 An extract from a company’s personnel records shows the age and grades of staff:

AGE Grade Under 25 25 and under 50 50 and over Total

A 70 120 110 300 B 40 50 30 120 C 10 30 40 80

Total 120 200 180 500 (a) Test whether there is an association between age of a member of staff and grade of work.

(12 marks)

(b) Five years ago, the company was re-organised with the proportion of jobs in each grade set as: Has the company’s grade structure changed since this grade structure was introduced?

(8 marks)

(Total 20 marks)

Grade A B C 53% 26% 21%

3009/2/09/MA Page 6 of 19

MODEL ANSWER TO QUESTION 2 (a) Null hypothesis: there is no relationship between age and grade. Alternative hypothesis: there is a relationship between age and grade. Degrees of freedom (3 - 1) x (3 - 1) = 4 Critical Chi square = 9.49 (5%) = 13.28 (1%)

Conclusion: Reject the null hypothesis: there is a highly significant difference. There the distribution of grades has changed over time.

Observed Grade Under 25 30 and under 50 50 and

over TOTAL

A 70 120 110 300 B 40 50 30 120 C 10 30 40 80

Total 120 200 180 500

Expected 72 120 108 300 28.8 48 43.2 120 19.2 32 28.8 80

0.05555556 0 0.037037 4.35555556 0.083333 4.033333 4.40833333 0.125 4.355556

χ2= 17.4537

Conclusion: reject the null hypothesis at both the 5% and 1% levels. There is a strong relationship between age and grade.

(b) Null hypothesis: the distribution of grades has not changed over time. Alternative hypothesis: the distribution of grades has changed over time. Degrees of freedom c – 1 = 3 – 1 = 2 Critical value for Chi square = 5.99/9.21

Observed 300 120 80Expected 275 125 100

4.62 0.77 5.95

χ2= 11.34

3009/2/09/MA Page 7 of 19

QUESTION 3 (a) Explain what is meant by:

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

(4 marks) (b) The production process for manufacturing computer casings involves three independent stages;

cutting, forming and painting. The probability of a fault in the: cutting stage = 0.02

forming stage = 0.03

painting stage = 0.05

Using a tree diagram or otherwise, calculate the probability that in the manufacturing of a

computer casing:

(i) faults are found at all three stages of production (ii) no fault is found (iii) a fault is found at exactly one stage of production.

(12 marks) (c) An item is found to have one fault. What is the probability that the fault was in the forming stage?

(4 marks)

(Total 20 marks)

3009/2/09/MA Page 8 of 19

MODEL ANSWER TO QUESTION 3 (a) Mutually exclusive events: the events cannot occur simultaneously. Independent events: the occurrence of one does not affect the occurrence of the other(s). (b) (i) 3 faults 0.02 x 0.03 x 0.05 = 0.00003 (ii) 0 faults

1 – 0.02 = 0.98, 1 – 0.03 = 0.97 1 – 0.05 = 0.95

0.98 x 0.97 x 0.95 = 0.90307 (iii) 1 fault 0.02 x 0.97 x 0.95 = 0.01843 + 0.03 x 0.98 x 0 .95 = 0.02793 + 0.05 x 0.98 x 0.97 = 0.04753 = 0.09389 (c) If one fault a forming fault Pr(forming fault) = 0.02793 Pr(One fault) 0.09389 = 0.2975 (0.3)

3009/2/09/MA Page 9 of 19

QUESTION 4 (a) Explain the meaning of the term ‘sampling distribution of the mean’.

(4 marks)

(b) A random sample of the executives in two regional sales offices earn the following annual amounts.

Eastern

Region £000 Western

Region £000 17 26 23 14 33 18 24 27 17 25 10 30 25 16 19 17 24 22 20

Test whether the annual mean earnings vary between the regions.

(14 marks) (c) In the conclusion you drew in part (b), might a type 1 or type 2 error have occurred?

(2 marks)

(Total 20 marks)

3009/2/09/MA Page 10 of 19

MODEL ANSWER TO QUESTION 4 (a) Sampling distribution of the mean is the distribution of sample means when a large number of

samples of the same size are taken. When n, the sample size, is large, this will follow a normal distribution. If n, the sample size, is small, the sampling distribution will follow the ‘t’ distribution.

(b) Null hypothesis: There is no difference in the mean earnings paid in the different regions. Alt hypothesis: There is a significant difference in the mean earnings paid in the different regions. Degrees of freedom = n1 + n2 – 2 = 9 + 10 – 2 = 17 t0.05/0.01 = 2.11/2.90

192 215 Sample totals 21.33333 21.5 Sample means 42.25 28.5 Sample variances Joint Var =

34.97059 Se = 5.913594 Standard error

21.33333 – 21.5 . t = 5.913594√(1/9+1/10) 21.33333 – 21.5 5.913594√0.211 -0.16667 2.7164 -0.061 (-0.06134)

Conclusions: The calculated value t < critical t therefore accept the null hypothesis. There is no difference between the mean earnings paid in the two regions. (c) A type 2 error might have occurred as the null hypothesis was accepted.

3009/2/09/MA Page 11 of 19

QUESTION 5

(a) Why would you calculate a weighted index number?

(4 marks) (b) A small engineering firm purchases 4 components from overseas suppliers. It has found that

prices of, and expenditure on, these components have changed in the following manner between 2005 and 2008:

2005 2005 2008 2008

COMPONENT PRICE £ EXPENDITURE £ PRICE £ EXPENDITURE £ 121X 6 2400 8 4400 7 P 7 1960 10 2500 32P 14 1680 15 1800 4T 10 1840 13 1820

(i) Calculate the Laspeyres base year weighted index for prices between 2005 and 2008

(6 marks)

(ii) Calculate the Paasche current year weighted index for prices between 2005 and 2008. (6 marks)

(c) (i) Give one advantage and one disadvantage of using a current year weighted index (ii) Give one advantage and one disadvantage of using a base year weighted index.

(4 marks)

(Total 20 marks)

3009/2/09/MA Page 12 of 19

MODEL ANSWER TO QUESTION 5 (a) A weighted index reflects the quantities/prices involved in the measure and therefore

gives a more accurate result. (b)

(c) (i) Reflects the current situation as the weights are updated, need to recalculate both

numerator and denominator for each year, intermediate years cannot be compared. (ii) simpler to calculate, less cost involved in collecting weights each year, seems to

give more logical results over a number of years, weights stay the same. Base weights become out of date.

2005 2005 2008 2008COMPONENT PRICE £ EXPENDITURE £ PRICE £ EXPENDITURE £

121X 6 2400 8 44007 P 7 1960 10 250032P 14 1680 15 18004T 10 1840 13 1820

Base total 7880 Current total 10520

2005 quantities 2008 quantities 400 550 280 250 120 120 184 140

(i) base year index 400 x 8 + 280 x 10 + 120 x 15 + 184 x 13 x 100 7880

10192 x100 7880 1.2934 x100 129.3

(ii) current year index 10520 x 100 550 x 6 + 250 x 7 + 120 x 14 + 140 x 10

10520 x100 8130 1.2940 x100 129.4

3009/2/09/MA Page 13 of 19

QUESTION 6 (a) Explain the circumstances in which a normal distribution based z test is used in preference to a t test, when testing the mean of one sample.

(4 marks) The mean weight of a random sample of 12 bags of cement is 49.88 kg. The population standard deviation is known to be 0.5 kg. (b) Test, at the 0.01 significance level, whether the mean weight of the sample is significantly

different from the specified weight of the bags of cement given as 50.0 kg. (8 marks)

(c) A company introduces a new production technique which reduces the standard deviation to 0.15

kg. Assuming the sample size remains 12, what weight should the company set as the target weight so that 99% of the time the sample mean weight of the bags exceeds 50 kg?

(6 marks)

(d) Explain the impact of a smaller sample size upon the value of the standard error. (2 marks)

(Total 20 marks)

3009/2/09/MA Page 14 of 19

MODEL ANSWER TO QUESTION 6 (a) A normal distribution based z test is used when there is a large sample with known population

standard deviation; a ‘t’ test is used when the sample size is small (less than 30) and the population standard deviation is unknown.

(b) Null hypothesis: the mean weight of cement in a bag does not differ from the specified weight. Alternative hypothesis: the mean weight of cement in a bag does differ from the specified weight. Critical z value = 1.96/2.58

n

σµx

z−

=

12

0.55049.88 −

= = 0.144

0.12 = 0.83

Conclusion: The calculated z value is less than the critical z value at the 0.05 and 0.01 levels; the mean weight does not differ significantly from the specified weight. (c) 99% z value = +2.33

n

σµx

z−

=

12

0.1550x

2.33−

=+

+2.33 x 0.043 = x – 50 = 50 + 0.101

x = 50.101kg (d) A smaller sample size will, other things being equal, increase the standard error.

3009/2/09/MA Page 15 of 19

QUESTION 7 (a) Explain the terms: (i) stratified random sampling. (ii) quota sampling.

Give one advantage and one disadvantage of each sampling method. (8 marks)

(b) 300 employees in a company have identity numbers in the range 4000 to 4299. Explain how you

might take a random sample from these employees. Illustrate your answer by using the random number table below, starting at the top left number 123, select a random sample of 10 employees.

(6 marks)

Random Number Table

123 091 073 454 413 443 814 826 685 431 008 919 646 350 002 184 572 335 592 661 620 344 368 139 662 153 591 002 184 152 549 089 673 968 748 027 014 361 431 662 276 454 704 897 620 780 908 348 545 577 578 306 427 536 432 005 542 278 073 727 991 326 895 700 355 021 245 202 550 405 169 348 374 830 258 630 286 520 070 432 925 621 772 709 801 606 688 134 539 290

(c) Give 6 characteristics of a well-designed questionnaire. (6 marks)

(Total 20 marks)

3009/2/09/MA Page 16 of 19

MODEL ANSWER TO QUESTION 7 (a) (i) Stratified random sample: divides the population into groups with the same

characteristics and samples randomly from each group. Advantages: Reduces the sampling error. More representative of the population.

Provides greater information. Disadvantages: Cost, Obtaining the information to form the strata. (ii) Quota Sampling: Divides the population into groups with the same characteristics

and samples non-randomly from each group. Advantages: No need for a sampling frame, saving time and cost. Disadvantages: Not possible to calculate the standard error, selection of respondents

might be biased. (b) Allocation of random numbers. In practice the digit 4 can be ignored and values between

000 and 299 taken. This would give if read by row: 123, 91, 73, 8, 2, 184, 139, 153, 2, 184, 152, 89 Reject the second 2 and second 184. By column: 123, 8, 276, 169, 231, 91, 89, 73, 139, 2. Selection of 10 values within the range from the random number tables. 6 marks for 10 correct value drawn. Need to quote the figs as 4123 for full 6, 5 max if 123 etc given. (c) Clear questions

Lack of ambiguity eg avoidance of negatives Logical sequence of questions ie those on the same topic grouped together Only relevant questions Introductory letter to describe the purpose of the questionnaire Clear layout of questionnaire Avoid reliance on memory for answers Provide answers if possible eg yes/no Provide tick boxes for responses Include answer coding to reduce workload Provide 5 point scale for answers based on opinions

3009/2/09/MA Page 17 of 19

QUESTION 8 (a) Explain the difference between a one tail test and a two tail test.

(4 marks)

A company carried out a random sample of staff in its regional headquarters in Birmingham and Newcastle. It discovered that 48 out of 310 employees in Birmingham were dissatisfied with working conditions, whilst 38 out of 220 employees in Newcastle were dissatisfied with working conditions.

(b) Test whether more workers were dissatisfied with working conditions in Newcastle compared with

Birmingham. (12 marks)

(c) Explain what is meant by the term ‘the 95% confidence interval for the proportion’. (4 marks)

(Total 20 marks)

3009/2/09/MA Page 18 of 19

MODEL ANSWER TO QUESTION 8 (a) A one tail test is used to test if there is direction to the difference in mean or proportion,

a two tail test is used to test if there is a difference without regard for direction. (b) Null hypothesis: There is no difference in the proportion of staff dissatisfied with working

conditions in Birmingham and Newcastle.

Alternative hypothesis: More staff in are dissatisfied in Newcastle than in Birmingham. Critical z 0.05/0.01 = 1.64/2.33 p1 = 48/310 = 0.155, p2 = 38/220 = 0.173

Polled value of 220310

3848p

+

+= =

530

86 = 0.162

( )

=

+−

−

220

1

310

10.16210.162

0.155 0.173z = ( )0.0077710.136

0.018 = 0.55

Conclusions: There is insufficient evidence to reject the null hypothesis at the 5% significance level. There is no difference in the proportion dissatisfied with the working conditions in Newcastle and Birmingham.

(c) The 95% confidence interval means that if samples of the same size are taken, the

population proportion will lie within the stated range 95 times out of 100.

3009/2/09/MA Page 19 of 19 © Education Development International plc 2009

LEVEL 3

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