LCCI Examinations Board ASP M 1140

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Model Answers for

Business Statistics

THIRD LEVEL

Series 2 2002(Code 3009)

1

Business Statistics Third LevelSeries 2 2002

How to use this booklet

Model Answers have been developed by LCCIEB to offer additional information and guidance toCentres, teachers and candidates as they prepare for LCCIEB examinations. The contents of thisbooklet 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 tosee 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 individualquestions or to examination technique

Teachers and candidates should find this booklet an invaluable teaching tool and an aid to success.

The London Chamber of Commerce and Industry Examinations Board provides Model Answers to helpcandidates gain a general understanding of the standard required. The Board accepts that candidatesmay offer other answers that could be equally valid.

© LCCI CET 2002

All rights reserved; no part of this publication may be reproduced, stored in a retrieval system ortransmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwisewithout prior written permission of the Publisher. The book may not be lent, resold, hired out orotherwise disposed of by way of trade in any form of binding or cover, other than that in which it ispublished, without the prior consent of the Publisher.

Typeset, printed and bound by the London Chamber of Commerce and Industry Examinations Board.

2

3

Business Statistics Third LevelSeries 2 2002

QUESTION 1

In the year 2000 a random sample of 96 small companies revealed the following profits and lossesdistribution:

Profits and Losses (£000) Number ofcompanies

-50 up to 0 10 0 and up to 10 1210 and up to 20 2120 and up to 40 2640 and up to 80 1980 and up to 150 8

(a) Calculate the arithmetic mean and the standard deviation of these company profits and losses.(8 marks)

In the previous year a random sample of 150 small companies showed a mean profit of £28,400 with astandard deviation of £28,352.

(b) Test to see if there has been a significant increase in the average profit made by smallcompanies.

(7 marks)

(c) Calculate a 99% confidence interval for the arithmetic mean in the year 2000.(5 marks)

(Total 20 marks)

4

Model Answer to Question 1

(a)f mid pt fx fx2

–50 up to 0 10 –25 –250 6,250Over 0 and up to 10 12 5 60 300Over 10 and up to 20 21 15 315 4,725Over 20 and up to 40 26 30 780 23,400Over 40 and up to 80 19 60 1,140 68,400Over 80 and up to 150 8 115 920 105,800

96 2,965 208,875 f� fx� 2fx�

fxx

�

�=96965,2=x = £30.89 (000) (£30,885)

s = 22

– ��

�

�

��

�

�

�

�

�

�

ffx

ffx s =

2

96965,2–

96875,208

��

���

�

= == 87.221,191.953–78.175,2 £34.96 (000) (£34,955)

(b) Null hypothesis: There has not been an increase in the average levelof profits

Alternative hypothesis: There has been an increase in the average levelof profits

Critical z for 0.05 significance level I tail test 1.64

2

22

1

21

2–1

ns

ns

xxz+

= =

150352.28

9696.34

4.28–89.3022

+09.18

49.2 = 0.59

Alternative answer: 0.58

Conclusions: Accept the null hypothesis, there is insufficientevidence to claim the average level of profits hasincreased.

(c) ci = 96

34.9552.5830.89

n

σ2.58x ±=±

9.2130.893.57x2.5830.89 ±=±=

= £21.68 to £40.1 (000)

5

QUESTION 2

The price of the standard family saloon car and the company market share was recorded for a randomsample of 12 car manufacturers.

Sellingprice £00

137 138 125 142 168 145 135 145 160 146 136 160

Marketshare %

14 15 10 8 9 7 11 5 3 5 7 2

(a) Plot the data on a scatter diagram and comment.(4 marks)

(b) Calculate the product-moment correlation coefficient.(10 marks)

(c) Test to find if the correlation coefficient differs significantly from zero.(6 marks)

(Total 20 marks)

CONTINUED ON NEXT PAGE6

Model Answer to Question 2

(a) Price and Market Share

Comment : Some weak negative relationship.

137 14 18,769 196 1,918138 15 19,044 225 2,070125 10 15,625 100 1,250142 8 20,164 64 1,136168 9 28,224 81 1,512145 7 21,025 49 1,015135 11 18,225 121 1,485145 5 21,025 25 725160 3 25,600 9 480146 5 21,316 25 730136 7 18,496 49 952160 2 25,600 4 320

1,737 96 253,113 948 13,593 x� y� 2x� 2y� xy�

(b) r = ( )( )( ) ( ) �

���

�� ���

���

�� ��

���

2222 ––

–

yynxxnyxxyn

r = ( )( )22 96–948x121,737–253,113x12

96x1,737–13,593x12

r = ( )( )9,216–11,3763,017,169–3,037,356

166,752–163,116

r = ( )( )

0.5506–2,16020,187

–3,636 =

16

14

12

10

8

6

4

2

012,000 13,000 14,000 15,000 16,000 17,000 18,000

Mar

ket S

hare

%

Price £

Series 1

7

Model Answer to Question 2 continued

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

Degree of freedom = n – 2 = 12 – 2 = 10Critical t0.025 = 2.23

t = 2–1

2–

rnr t =

)55.0(–

2–1255.0–21 −

t = 3025.0–161.1–

= 835.061.1–

= –2.086

Conclusions: The calculated value of t is less than the criticalvalue of t. There is insufficient evidence to reject the nullhypothesis.The correlation coefficient does not differ from zero.

8

QUESTION 3

A company is planning the launch of a new product. It estimates the probability of good marketconditions to be 80%. If market conditions are good the probability of a successful launch is 75%, ifmarket conditions are poor the probability of a successful launch is 50%.

(a) Find the probability that the launch is successful.(5 marks)

(b) 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:

Market conditions are good and the product launch is successful £55 millionMarket conditions are good and the product launch is unsuccessful – £13 millionMarket conditions are poor and the product launch is successful £37 millionMarket conditions are poor and the product launch is unsuccessful – £19 million

(c) What is the expected profit from the new product launch?(4 marks)

The company sells an established product that has variable levels of weekly sales with arithmeticmean of £5,000 and standard deviation of £600. You may assume that the sales are normallydistributed.

(d) (i) Find the probability that in one week there are sales of over £6,500

(ii) The sales have to exceed £3,800 in each week for the product to break even, what is theprobability of this happening?

(5 marks)

(Total 20 marks)

9

Model Answer to Question 3

(a) Good market conditions and successful = 0.8 x 0.75 = 0.6Poor market conditions and successful = 0.2 x 0.5 = 0.1

Probability of successful launch = 0.7Accept decision tree

(b) Good and unsuccessful = 0.8 x 0.25 = 0.2Poor and unsuccessful = 0.2 x 0.5 = 0.1

Probability unsuccessful = 0.3

0.330.30.1

ulunsuccessfyProbabilit

ulunsuccessfandPoor==

(c) Expected return

Good market conditions and successful launch 0.8 x 0.75 x £55m = £33m

Good market conditions and unsuccessful launch 0.8 x 0.25 x –£13m = -£2.6m

Poor market conditions and successful launch 0.2 x 0.5 x £37m = £3.7m

Poor market conditions and unsuccessful launch 0.2 x 0.5 – £19m = –£1.9m £32.2m

(d) (i) z = sdx µ– z =

600500,1

600000,5–500,6 =

= 2.5, prop = 1 – 0.994 = 0.006

(ii) z = sdx µ– z = 2

600200,1

600000,5–800,3 ==

Proportion = 0.977

10

QUESTION 4

(a) What are the benefits of an effective quality control system?(4 marks)

A company in its quality control procedures sets the warning limit at the 0.025 probability point and theaction limit at the 0.001 probability point. This means for example that the upper action line is set sothat the probability of the mean exceeding the line is 0.001.

The internal diameter of a bored hole is set at 35 mm with a known standard deviation of 0.1 mm.Random samples of 9 items at a time are taken from the production line to check the accuracy of themanufacturing process.

(b) (i) Construct a quality control chart to monitor the manufacturing process.(8 marks)

(ii) The results for 8 samples are given below. Plot these on your quality control chart andcomment on the graph.

(4 marks)

Sample number 1 2 3 4 5 6 7 8Sample mean (mm) 35.05 34.94 34.89 35.16 35.15 34.95 34.99 35.08

(c) If the process mean changed to 35.02 mm and the standard deviation remained at0.1 mm calculate the probability that the mean of a random sample of 9 components would beoutside the warning limits.

(4 marks)

(Total 20 marks)

11

Model Answer to Question 4

(a) The benefits of a good quality control system are better quality products,fewer rejects and less waste, better customer relations, more sales

Warning limits = 033.0x96.13591.096.13596.1 ±=±=± nx σ

upper w lts 35.06 35.07(2)lower w lts 34.94 34.93(2)

Action limits = 033.0x09.33591.009.33509.3 ±=±=± nx σ

upper act lts 35.10lower act lts 34.90

(b) Quality Control Chart

The process goes out of control twice. It seems unstable.

(c) z = 5.1033.005.0

91.0

07.35–02.35–===sd

x µ

Probability the mean lies outside the upper warning limit = 0.067

7.2

91.0

02.35–93.34 =

Probability the mean lies outside the lower warning limit = 0.004

Total probability = 0.071 or (0.072 if 0.0333 is used)

12

QUESTION 5

(a) In what circumstances is a significance test based on the ‘t’ distribution used in preference to asignificance test based on the normal distribution?

(4 marks)

Data input clerks are sent on an intensive training course to increase their keyboarding speed. Theresults of a before and after test for a random sample of 10 clerks gave the following results. Speed ismeasured in key depressions per minute (kdp).

Clerk a b c d e f g h i j

Before trainingcourse (kdp) 625 598 685 754 658 690 559 840 758 685

After trainingcourse (kdp) 610 620 690 780 690 702 573 851 744 690

(b) Test whether the training course has increased the clerks’ data input speed.(12 marks)

(c) What is meant by a Type I and a Type II error?(4 marks)

(Total 20 marks)

13

Model Answer to Question 5

(a) The ‘t’ distribution is used when n the sample size is small <30 and thestandard deviation is estimated from the sample.

(b) Null hypothesis: The course has not increased the speed of thedata input clerks

Alternative hypothesis: The course has increased the speed of thedata input clerks

Degree of freedom n – 1 10 – 1 = 9Critical t0.05 value = 1.83

625 610 –15 225 615.04598 620 22 484 148.84685 690 5 25 23.04754 780 26 676 262.44658 690 32 1,024 492.84690 702 12 144 4.84559 573 14 196 17.64840 851 11 121 1.44758 744 –14 196 566.44685 690 5 25 23.04

98 3,116 2,155.60

d� 2d� 2

–d ���

���� d

8.91098 === �

ndd sed =

1–

2

nd–d ��

�

����

��

( )1–

–2

2

nndd �

�

= 48.1553.2391–108.155,2 == or

1–101098–116,3

2

t = dse

d 0– =

1048.15

0–8.9 = 9.48.9 = 2

Conclusions: The calculated value of ‘t’ is greater than critical valueof ‘t’ reject the Null Hypothesis the speed of the data inputclerks has increased

(c) A Type I error is the error of rejecting a null hypothesis that is true anda Type II error is the error of accepting a false null hypothesis whichshould be rejected.

If a two sample meanstest is used can score5: nh/ah, conclusions

14

QUESTION 6

(a) (i) What is meant by the standard error of the mean? (4 marks)

(ii) What is the difference between a one and two tail test? (4 marks)

In September 2001, a Travel Agent used a random sample of 36 holiday makers to find out theaverage cost per person of a one week holiday in Ruritania. The following information was found:

September 2001

Mean £372.40Standard deviation £26.10Sample size 36

In the previous year the average cost of each holiday was £356.20.

(b) Test whether the cost of a one week holiday to Ruritania has increased significantly since theprevious year.

(6 marks)

The company based its views on customer satisfaction from the letters it receives. In the previousyear, 68% of the letters it received were of a positive nature.

(c) The company wishes to adopt a more scientific approach to estimating customer satisfaction.What sample size would be needed to estimate the proportion of customers’ views to within 2% ofthe true figure at the 95% confidence level?

(6 marks)

(Total 20 marks)

15

Model Answer to Question 6

(a) (i) The standard error of the mean is the measure of dispersion of the

sample means. It equals nσ

(ii) A one tail test tests the direction of the difference between two statistics.A two tail test tests if there is a difference between the two statistics withoutregard to the direction.

(b) Null hypothesis: There is no difference in the holiday cost betweenlast year and September

Alternative hypothesis: There is an increase in the holiday cost betweenlast year and September

Critical z value = 1.64

z = n

x/–

σµ =

36/1.262.356–4.372 =

35.42.16 = 3.72

Conclusions: The calculated value of z is greater than the critical valueof z. Reject the null hypothesis. There is evidence to suggestthat the holiday cost has increased significantly.

(c)

n0.32x0.68

0.021.96

np)–p(1

0.021.96 >±>±

1.962 >

n32.0x68.0

02.0 2 3.8416 >

n2176.00004.0 n >

0004.02176.0x8416.3 > 2,089.8

n = 2,090

16

QUESTION 7

(a) In what circumstances is the multiplicative model preferable to the additive model in time seriescalculations?

(4 marks)The table below shows the quarterly sales figures for a company:

Quarter 1 Quarter 2 Quarter 3 Quarter 4

1998 48 56 70 96

1999 64 76 86 132

2000 72 100 108 188

(b) By the method of centred moving averages, calculate the trend values for the time series and plotthe trend on a graph.

(10 marks)

The average seasonal components calculated by the multiplicative method are shown below:

Quarter 1 Quarter 2 Quarter 3 Quarter 40.768 0.903 0.980 1.349

(c) Using the average seasonal components provided above and the original sales figures, calculatethe seasonally adjusted sales and plot the results on the graph constructed for part (b). Commenton your results.

(6 marks)

(Total 20 marks)

CONTINUED ON NEXT PAGE17

Model Answer to Question 7

(a) The multiplicative model is preferable when the data has a strong trend with thedifferences between the trend and the original data values varying proportionally tothe trend values rather than absolutely.

(b)

(c) Qtr Sales Total m avg 1 m avg 2Trend

1 482 563 70 270 67.5 69.54 96 286 71.5 745 64 306 76.5 78.56 76 322 80.5 857 86 358 89.5 90.58 132 366 91.5 94.59 72 390 97.5 100.25

10 100 412 103 11011 108 468 11712 188

18

Model Answer to Question 7 continued

Sales Seasonal adjustment Seasonal Adjusted Sales48 0.768 62.5056 0.902 62.0870 0.980 71.4396 1.349 71.1664 0.768 83.3376 0.902 84.2686 0.980 87.76

132 1.349 97.8572 0.768 93.75

100 0.902 110.86108 0.980 110.20188 1.349 139.36

Comment: The trend and seasonally adjusted data values show little difference. This implies there is no change in the underlying trend.

19

QUESTION 8

(a) When might a 2χ test be used?(4 marks)

From its records, a company analyses its sales by size and region. The table below shows the results:

Size of order

Region £0 and up to£1,000

£1,000 and upto £3,000

£3,000 and upto £10,000

£10,000 andover

Northern 40 33 20 15

Midlands 50 45 40 25

Southern 40 32 40 20

(b) Test whether there is a relationship between size of order and region.(12 marks)

(c) Combining the three regions estimate a 99% confidence interval for the proportion of orders thatare valued at less than £1,000.

(4 marks)

(Total 20 marks)

20

Model Answer to Question 8

(a) The Chi-squared test is used to test for association between characteristicsrather than variables, for randomness or changes in proportions.

(b)

Null hypothesis: There is no association between the size of orderand region

Alternative hypothesis: There is association between the size of orderand region

Degrees of freedom = (4 – 1)(3 – 1) = 6

Critical 2χ = 12.59

<£1,000 £1,000<3,000 £3,000<10,000 £10,000+Region 40 33 20 15

50 45 40 2540 32 40 20

Total 130 110 100 60

expected freq 35.1 29.7 27.0 16.252.0 44.0 40.0 24.042.9 36.3 33.0 19.8

0.684 0.367 1.815 0.089contributionto chi 0.077 0.023 0.000 0.042

0.196 0.509 1.485 0.0025.288

EE–O 2)(

�=χ = 5.288

Conclusions: The calculated 2χ is less than the critical 2χ . Thereis insufficient evidence to reject the null hypothesis.There is no association between the size of orderand region.

(c)

99% confidence z = ± 2.58

p = 130/400 = 0.325, (1 – p) = 0.675

ci = nppp /)–1(58.2± = 0.325 ± 2.58 400/675.0x325.0

= 0.325 ± 2.58 x 0.0234 = 0.325 ± 0.0604

0.265 to 0.385

© LCCI CET 200221

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