Business Statistics Level 3

Model Answers Series 4 2007 (Code 3009)

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Business Statistics Level 3 Series 4 2007

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 2007 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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3009/4/07/MA Page 3 of 20

Business Statistics Level 3 Series 4 2007 QUESTION 1 The directors of a company decided to analyse its debt. A random sample of 1000 debts was selected and the following results were obtained: (a) Calculate the mean and standard deviation debt, giving your answers to the nearest whole £.

(8 marks)

The value of the median is £365. (b) Calculate the coefficient of skewness and comment on your answer.

(4 marks) (c) Last year the average debt was £432. Test whether there has been a significant increase in average debt.

(8 marks)

(Total 20 marks)

Range of Debts Number of debts

£ 0 and less than 200 220

200 and less than 400 340 400 and less than 600 190 600 and less than 1000 200

1000 and less than 2000 50

3009/4/07/MA Page 4 of 20

MODEL ANSWER TO QUESTION 1 (a)

Range of Debts Mid Pt Number of debts

£ fx fx2

0 and less than 200 100 220 22000 2200000200 and less than 400 300 340 102000 30600000400 and less than 600 500 190 95000 47500000600 and less than 1000 800 200 160000 128000000

1000 and less than 2000 1500 50 75000 112500000 1000 454000 320800000

∑f ∑fx ∑fx2

∑fx 454000 x = ∑f = 1000 = £454

Standard Deviation

−

=

1000454000

1000320800000

114684 = £339

(b) Coefficient of skew = ( )

SDMedianMean −3

=( )

3393654543 −

= 0.788

Comment: There is some positive skew (c) Null hypothesis: There has not been an increase in the average debt, compared with the previous year. Alternative hypothesis: There has been an increase in the average debt, compared with the previous year. Critical z = 1.64/2.33

nxzσ

µ−=

1000339432454 −

= 72.10

22= = 2.05

Conclusions: There is evidence to suggest that the average debt has increased at the 0.05 significance level. Reject the null hypothesis, accept the alternative hypothesis. There has been an increase in average debts. However, at the 0.01 level there is insufficient evidence to reject the null hypothesis. The company’s average debt has not increased.

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QUESTION 2 (a) Explain how you would decide whether to use the additive model or the multiplicative model to analyse a time series.

(4 marks) The following data show the quarterly production of a company for the past three years: Production (000 Units) Year Jan-Mar Apr-June Jul-Sept Oct-Dec 2004 48 56 70 96 2005 64 76 86 132 2006 72 100 108 188 (b) Plot the time series.

(3 marks)

(c) Using a suitable model, find using the method of moving averages, the trend and average seasonal variations. Plot the trend on your graph.

(13 marks)

(Total 20 marks)

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MODEL ANSWER TO QUESTION 2 (a) The choice of the additive or the multiplicative model in times series depends upon the nature of the data. If, for example, there is a strong trend in the data and the seasonal variations are increasing as the trend increases the multiplicative model is used. However, if the seasonal variations are not increasing absolutely as the trend increases the additive model can be used. Or, graphs can be used to illustrate these points. (b)

Production v Time

0

20

40

60

80

100

120

140

160

180

200

1 2 3 4 5 6 7 8 9 10 11 12

Quarters

Prod

uctio

n (0

00)

(c)

Production

Moving Total 1

Moving Total 2

Trend/ Moving Average

Additive differences

Multiplicative Differences

48 56 70 270 556 69.50 0.50 1.007 96 286 592 74.00 22.00 1.297 64 306 628 78.50 -14.50 0.815 76 322 680 85.00 -9.00 0.894 86 358 724 90.50 -4.50 0.950

132 366 756 94.50 37.50 1.397 72 390 802 100.25 -28.25 0.718

100 412 880 110.00 -10.00 0.909 108 468 188

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MODEL ANSWER TO QUESTION 2 CONTINUED

Or

Quarter Year 1 2 3 4 2004 0.5 22 2005 -14.5 -9 -4.5 37.5 2006 -28.25 -10 -42.75 -19 -4.0 59.5 Average seasonal variation -21.375 -9.5 -2.0 29.75

Quarter Year 1 2 3 4 2004 1.007 1.297 2005 0.815 0.894 0.95 1.397 2006 0.718 0.909 1.533 1.803 1.957 2.694 Average seasonal variation 0.7665 0.9015 0.9785 1.347

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QUESTION 3 (a) State two reasons why samples might be selected from a population instead of the whole

population being examined. (4 marks)

A batch of 500 goods received notes were given numbers in the range 100151 to100650 inclusive. A sample of 10 of these goods received notes is required. Extract from Random Number Tables

56702 93301 76666 0660457354 30875 55880 1120080593 97203 13829 0259198188 69253 75436 9283514876 88932 65734 89277

(b) Using the random number table above, explain briefly how you would select a simple random sample of 10 goods received notes.

(8 marks) The goods received notes referred to above, were received by two departments. 100151 to 100400 were received by department A, and 100401 to 100650 were received by department B. (c) Using this additional information and the random number table above, select a stratified random sample of 10 goods received notes, explaining briefly how you obtained the sample.

(8 marks)

(Total 20 marks)

MODEL ANSWER TO QUESTION 3 (a) A sample may be used in preference to the whole population to save time and costs.

In the case of destructive testing a sample is the only sensible approach. (b) A simple random sample requires that all the population have an equal chance of being

chosen. Using the sequence of numbers which the goods received notes (GRN) already have identifies each item in the population. The initial three digits can be ignored and the number treated as though they run from 151 to 650.From the random number tables three digit numbers are chosen which lie between 151 and 650. If the number chosen lies below 151 or above 650 it is rejected and a further number drawn until 10 numbers in the range 151 to 650 have been selected. For example 567 is chosen, 029 rejected, 330, 176 are chosen, 666 rejected and so on until the 10 GRN have been identified. The following valid random numbers would form a random sample of 10: 567, 330, 176, 430, 558, 593, 382, 591, 375 and 436.

(c) This information means that half the population is from department A and half from department B. The sample will have to reflect this in the numbers drawn above for the random sample. 4: 330, 176, 382 and 375, are from department A and 5: 430, 436, 567, 558 and 593 are from department B. The final random number will have to be rejected and a further number drawn until one in the range 151 to 400 is obtained. This would be 351.

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QUESTION 4 (a) State two benefits of a good quality control system to a business.

(4 marks) A manufacturer uses quality control procedures which set the warning limits at the 0.025 probability point and the action limits at the 0.001 probability point. This means for example that the upper action limit is set so that the probability of a mean exceeding the limit is 0.001. The volume of a can of paint is set at 1025 millilitres with a standard deviation of 20 millilitres. Samples of 8 items are taken from the production line to check the accuracy of the production process. (b) (i) Calculate the values of the warning and action limits and construct a quality control chart to

monitor the production process. (8 marks)

(ii) The results for 6 samples are given below.

Plot these on your quality control chart and comment on the stability of the production process. Sample number 1 2 3 4 5 6 Sample mean ml 1015 1020 1002 1020 1050 1048

(4 marks) (c) If the process mean changed to 1040 millilitres and the standard deviation remained at 20 millilitres calculate the probability that the mean of a random sample of 8 would be below the upper action limit (using the existing control chart).

(4 marks)

(Total 20 marks)

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MODEL ANSWER TO QUESTION 4 (a)

Age x Value y x2 xy 3 104 9 312 7 101 49 707 11 92 121 1012 13 86 169 1118 15 83 225 1245 19 71 361 1349 21 59 441 1239 24 52 576 1248 29 46 841 1334 142 694 2792 9564 Σx Σy Σx2 Σxy

( )( )( )22 ∑∑

∑ ∑∑−

−=

xxn

yxxynb

21422792969414295649

−××−×

=b

496412472−

=b = -2.512

n

xb

ny

a ∑∑ −= 9

142512.29

694−−=a

64.3911.77 −−=a = 116.75 y = 116.75-2.512x (b) (i) 116.75 – 12 x 2.512 = 86.6 (ii) 116.75 – 35 x 2.512 = 28.8

(iii) Comment: (b) (i) is based on interpolation, whilst (ii) is an extrapolation. It is likely that (b) (i) will be more accurate.

(c) Null hypothesis: The correlation coefficient does not differ significantly from zero.

Alternative hypothesis: The correlation coefficient does differ significantly from zero. Degrees of freedom n-2 =9-2 =7 Critical t = 2.37/3.5 2−= nrt = 2998.0 − = 13.03

r−1 2 98.01 − 2

Conclusion: The result is highly significant, the correlation differs from zero

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QUESTION 5 A company uses 2 initial assessment procedures in deciding the suitability of staff for appointments: a psychometric test and an assessment by an interviewing panel. After a six month probationary period for the successful candidates the personnel department obtains an assessment of the suitability of each of the candidates from their manager. The following data show the results for the selection procedures and the managers’ assessment of the suitability of the candidates:

Candidate Psychometric Test Interview Score Manager’s Assessment A 54 28 55 B 72 57 68 C 38 38 43 D 51 38 50 E 66 44 31 F 45 41 51 G 58 53 62 H 51 55 59 I 43 59 65 J 40 40 38

(a) Rank each set of data.

(6 marks) (b) Calculate the Spearman’s Rank correlation coefficient for the Psychometric Test and Interview score.

(8 marks) The Spearman’s Rank correlation coefficient for Interview and Manager’s Assessment is 0.63 and for Psychometric Testing and Manager’s Assessment is 0.33. (c) Write a brief report on your findings obtained in part (a) and the information above.

(6 marks)

(Total 20 marks)

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MODEL ANSWER TO QUESTION 5 (a) Ranking if using equal ranks.

Candidate Psychometric Test Interview Score Manager’s Assessment

A 4 10 5

B 1 2 1

C 10 8= 8

D 5= 8= 7

E 2 5 10

F 7 6 6

G 3 4 3

H 5= 3 4

I 8 1 2

J 9 7 9 Or if using mean ranks.

Candidate Psychometric Test Interview Score Manager’s Assessment

A 4 10 5

B 1 2 1

C 10 8.5 8

D 5.5 8.5 7

E 2 5 10

F 7 6 6

G 3 4 3

H 5.5 3 4

I 8 1 2

J 9 7 9

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MODEL ANSWER TO QUESTION 5 CONTINUED Psychometric

Test Interview d d2

4 10 6 36

1 2 1 1

10 8.5 1.5 2.25

5.5 8.5 3 9

2 5 3 9

7 6 1 1

3 4 1 1

5.5 3 2.5 6.25

8 1 7 49

9 7 2 4 ∑ d 2 = 118.5

Rank Correlation coefficient R = ( )161 2

2

−Σ

−nn

d

R = ( )110105.11861 2 −

×− =

9907111 − = 1-0.718 = 0.282

(c) Brief Report: Topic, To whom addressed, Date

• All three coefficients are positive. • They lie between 0.28 and 0.64. • The relationships between psychometric test and Interview Score, and Psychometric Test and Manager’s Assessment are both weak. • The relationship between Interview Score and Manager’s Assessment is stronger, perhaps showing that only one of “interview score” and “manager’s score” is needed.

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QUESTION 6 (a) Briefly explain two business uses of the Normal Distribution.

(4 marks)

A company makes a product which passes through three departments: Mechanical, Electronics and Assembly. The mean and standard deviation for the time (in minutes) that a product spends in each department is given below. Assume that the times follow a normal distribution.

Minutes Mean Standard Deviation Mechanical 24 8 Electronic 22 4 Assembly 46 8

(b) For Assembly, find the probability that the time involved will:

(i) be less than 34 minutes (ii) exceed 66 minutes (iii) be between 34 and 66 minutes

(6 marks) (c) Find the probability that the total time to make the product will exceed 116 minutes.

(6 marks) (d) Find the time below which 95% of the products will be made.

(4 marks)

(Total 20 marks)

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MODEL ANSWER TO QUESTION 6 (a)

• The basis for mean charts is statistical quality control. • To calculate the proportion of items which falls within given limits, where the population is normally distributed, e.g. physical dimension, heights. • Significance test for product improvement, e.g. drug tests.

(b) (i) Probability less than 34 minutes sd

xxz −=

84634 −

= = -1.5

Table probability = 0.933, required probability = 0.067

(ii) Probability more than 66 minutes sd

xxz −=

84666 −

= = 2.5

Table probability = 0.994, required probability = 1- 0.994 = 0.006

(iii) Probability between 34 and 66 minutes = 1- (0.067 + 0.006) = 1- .073 = 0.927

(c) Joint mean = AEM xxx ++ = 24 + 22 + 46 = 92

Joint standard deviation = 222AEM sdsdsd ++ = 222 848 ++

= 144 = 12

sd

xxz −=

1292116 −

=z 1224

= = 2

Table probability = 0.977, required probability = 1- 0.977 = 0.023 (d) Required z value = 1.64

sd

xxz −=

129264.1 −

=+x

921264.1 −=×+ x = 111.68

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QUESTION 7 (a) Explain why a significant association between two factors does not necessarily mean that there is

cause-effect relationship between them. Give a suitable business example to illustrate your answer.

(4 marks) A random sample of employees of a company was selected and the employees were asked to complete a questionnaire. One question was whether the employees were in favour of the introduction of flexible working hours. The following table classifies the employees by their response and gender, i.e. male or female

Gender Female Male In favour 67 83 Not in favour 43 37 Undecided 20 30

(b) Test whether there is an association between gender and views.

(12 marks)

(b) By combining the data for male and female, find the 95% confidence interval for the proportion of employees in favour of flexible working.

(4 marks)

(Total 20 marks)

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MODEL ANSWER TO QUESTION 7 (a) There may be a third, unknown variable, which is the causal factor. Association only signifies a connectedness. The direction of causation might be uncertain. There may be

spurious or false association with no underlying theoretical basis for the association. There may be other variables which are not isolated. Example: Increases in pay causes an increase in productivity. There may have been an increase in capital equipment. (b) Null hypothesis: There is no association between gender and views on flexible hours. Alternative hypothesis: There is association between gender and views on flexible hours. Degrees of freedom = (R-1)(C-1) = (3-1)(2-1) = 2 Critical X² = 5.99/9.21 Expected Female Male Total In favour 69.6 80.4 150 Not in favour 37.1 42.9 80 Undecided 23.2 26.8 50

Contributions to X² 0.100 0.087

0.924 0.800 0.445 0.386 X² = 2.742

Conclusion: The calculated value of X2 is less than the critical value at the 0.05 level of significance; there is insufficient evidence to reject the null hypothesis. There is no association between gender and views on flexible hours. (c) Number of males and females in favour of flexible working = 67+83 = 150 Proportion in favour = 150/280 = 0.536

95% CI =n

ppp )1(96.1 −± =

280464.0536.096.1536.0 ×

±

CI = 0.478 to 0.594

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QUESTION 8 (a) Explain the circumstances in which a paired t test is used in preference to a two independent

sample test. (4 marks)

The directors of a company were concerned that advertisements for new staff on the internet were attracting too many unsuitable candidates. The format of the advertisements was restyled. After the new format had been used for six months the company examined a random sample of advertised posts before and after the change had taken place. The number of unsuitable candidates for each post was counted. The results were as follows:

Number of unsuitable candidates Before change of advertisement 24 26 18 24 18 28 32 After change of advertisement 18 24 20 12 24 28 18 12 8 (b) Test whether there is a decrease in the mean number of unsuitable candidates

after the change of advertisement. (14 marks)

(c) In part (b), might a type 1 error have occurred? Explain your answer.

(2 marks)

(Total 20 marks)

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Σ ( )2xx − Σ ( )2yy −

MODEL ANSWER TO QUESTION 8 (a) A paired t test is used in preference to the two sample independent tests when a single sample is subject to two treatments, e.g. a before and after test on the same people or items. The two independent sample test is when two different samples are subject to different treatments or the same treatment. (b) Null hypothesis: the number of unsuitable candidates does not decrease after the

change in advertising. Alternative hypothesis: the number of unsuitable candidates decreases after the change in advertising. Degrees of freedom = n + m – 2 = 7+9 –2 = 14 One tail critical to 0.05/0.01 = 1.76/2.62

x y x2 y2 ( )2xx − ( )2yy −

24 18 576 324 0.08 0.05 26 24 676 576 2.94 33.38 18 20 324 400 39.51 3.16 24 12 576 144 0.08 38.72 18 24 324 576 39.51 33.38 28 28 784 784 13.80 95.60 32 18 1024 324 59.51 0.05 12 144 38.72 8 64 104.49 170 164 4284 3336 155.43 347.56 Σx1 Σy Σ x2 Σy2

24.29 18.22

( ) ( )2

22

−+−Σ+−Σ

=mn

yyxxs297

56.34743.155−+

+= = 5.99

mn

s

yxt

11+

−=

91

7199.5

22.1829.24

+

−=

02.306.6

= = 2.01

Conclusion: The calculated value of t is greater than the critical value of t (at the 5% significance level); there is sufficient evidence to reject the null hypothesis. The number of unsuitable candidates does decrease after the change in advertising.

(c) The null hypothesis has been rejected, therefore type 1 error might have occurred.

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