1. LCCI International Qualifications
Business Statistics
Level 3
Model Answers
Series 3 2010 (3009)
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2. Business Statistics Level 3
Series 3 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
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without prior written permission of the Publisher. The book may not be lent, resold, hired out or
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published, without the prior consent of the Publisher.
3009/3/10/MA Page 1 of 20

3. QUESTION 1
A local authority conducts a random sample of the number of planning applications examined per day
by its planning officers over a period of five years. The data are shown below.
Number of planning Number of
applications examined days
40 and under 80 103
80 and under 120 68
120 and under 160 38
160 and under 200 22
200 and under 300 15
(a) Calculate the mean and standard deviation for the number of planning applications examined
per day.
(9 marks)
(b) Calculate a 95% confidence interval estimate for the mean daily number of planning
applications examined.
(5 marks)
The government has set a standard of 120 planning applications to be examined per day.
(c) Test whether the local authority reaches the government standard.
(6 marks)
(Total 20 marks)
3009/3/10/MA Page 2 of 20

4. MODEL ANSWER TO QUESTION 1
(a)
Number of planning Number of
applications days 2
examined x fx fx
2 f x x
40 and under 80 103
59.5 6128.5 364645.8 215114.5
80 and under 120 68
99.5 6766.0 673217.0 2209.3
120 and under 160 38
139.5 5301.0 739489.5 44706.6
160 and under 200 22
179.5 3949.0 708845.5 121450.8
200 and under 300 15
249.5 ..3742.5 ..933753.8 312337.4
246 25887.0 3419951.6 695818.6
2 2
f fx fx f x x
fx
Arithmetic mean = x
f
= 25887 = 105.2 applications per day
246
2
fx 2 fx
Standard deviation = s
f f
2
3419951.6 25887
s = 13902.2 11073.7 = 2828.5
246 246
or
2
f x x 695818.6
f 246
=
= 53.18 per day 2
f x x
f
2
f x x
f
2
f x x
f
2
f x x
f
==
1743409121
.
165
3009/3/10/MA Page 3 of 20

5. QUESTION 1 CONTINUED
(b) 95% confidence interval z = 1.96
53.18
ci x 1.96 105.2 1.96
n 246
= 105.2 6.65= 98.58 to 111.88
(c) Null Hypothesis: The local authority reaches the government standard.
Alternative hypothesis: The local authority does not reach the government
standard.
One tail test z = 1.64
x 105.2 120
z =
53.18
n 246
= -14.27 = -4.21
3.39
Conclusion: There is evidence to reject the null hypothesis; the local authority has
not reached the government standard.
3009/3/10/MA Page 4 of 20

6. QUESTION 2
A random sample of 12 households records the value of their house and the annual expenditure on
repairs to the house.
Value of house Annual expenditure
(£000) on repairs (£)
215 1850
323 2550
455 3150
329 1850
235 1750
425 1700
327 2600
385 3300
237 2750
155 1300
196 1600
176 1750
(a) Calculate the least squares regression line of expenditure on repairs based on the value of a
house.
(10 marks)
(b) Using the regression equation found in (a) estimate the annual expenditure on repairs to a
house valued at £250,000.
(2 marks)
The value of the coefficient of determination is 0.403 or 40.3%
(c) Explain what the coefficient of determination measures and comment on the accuracy of your
answer in (b) above.
(4 marks)
(d) What factors, other than the value of the house, may affect the expenditure on repairs?
(4 marks)
(Total 20 marks)
3009/3/10MA Page 5 of 20

7. MODEL ANSWER TO QUESTION 2
(a)
Value of Annual
house £000 expenditure
2
on repairs £ x xy
215 1850 46225 397750
323 2550 104329 823650
455 3150 207025 1433250
329 1850 108241 608650
235 1750 55225 411250
425 1700 180625 722500
327 2600 106929 850200
385 3300 148225 1270500
237 2750 56169 651750
155 1300 24025 201500
196 1600 38416 313600
176 1750 30976 308000
3458 26150 1106410 7992600
2
x y x xy
n xy x y
b 2
n x2 x
12 7992600 3458 26150 95911200 90426700
b
12 1106410 34582 13276920 11957764
b= 5484500 = 4.16
1319156
y x 26150 3458
a b 4.16
n n 12 12
a = 981.1
y = 981.1 + 4.16x
(b) Estimated costs of repair and maintenance = 981.1 + 4.16 x 250
= 981.1 + 1040. = 2021.1 (£)
(c) The coefficient of determination measures the change in repairs expenditure due to
the change in house value. Therefore, 40.3% of the change in repair expenditure is
due to the change in house value. The estimated repairs expenditure is not very
accurate.
(d) Age of house, size of family, attitude to house repair, size and value may not
be comparable.
3009/3/10MA Page 6 of 20

8. QUESTION 3
(a) Explain the circumstances in which a paired t test would be used in preference to a two sample
mean test for small independent samples.
(4 marks)
A consumer protection magazine wishes to compare the cost of repairs between two companies. They
used a sample of 9 businesses to request estimates for the cost of a repair by both companies.
Company X Company Y
Business Estimated cost of Estimated cost of
repairs £ repairs £
A 2400 2560
B 2600 2760
C 5696 5952
D 7936 7933
E 4000 4200
F 7040 7245
G 3930 3830
H 6000 5964
I 3576 3620
(b) Test whether the estimated cost of repairs differs between the two companies.
(12 marks)
(c) What practical reasons may the two companies have for giving different estimates of repair
costs?
(4 marks)
(Total 20 marks)
3009/3/10MA Page 7 of 20

9. MODEL ANSWER TO QUESTION 3
(a) The paired t test is used when a single sample is subject to two “treatments”
compared with two samples being compared.
(b) Null hypothesis: The estimated cost of repair does not differ between the two
companies.
Alternative hypothesis: The estimated cost of repair does differ between the
two companies.
Degrees of Freedom = n - 1 = 9 - 1 = 8
Critical t value 0.05 = 2.31: 0.01 = 3.36
Company X Company Y
Estimated Estimated
cost of cost of
repairs £ repairs £ Difference (d d) (d d )2 d
2
d
2400 2560 -160 -61.556 3789.1 25600
2600 2760 -160 -61.556 3789.1 25600
5696 5952 -256 -157.556 24823.8 65536
7936 7933 3 101.444 10291.0 9
4000 4200 -200 -101.556 10313.5 40000
7040 7245 -205 -106.56 11354.1 42025
3930 3830 100 198.444 39380.2 10000
6000 5964 36 134.444 18075.3 1296
3576 3620 ..44 54.444 2964.2 …1936
-886 124780 212002
2 2
Σd (d d) Σd
d -886 = -98.44
d =
9
n
2 2
d2 d 212002 886
sd = = = 117.7
n n 9 9
or
2
d d 124780
sd = = 117.7
n 9
d 0 98.44
t = = -2.36
sd n 1 117.7
9 1
3009/3/10MA Page 8 of 20

10. QUESTION 3 CONTINUED
Conclusion: The calculated value of t is greater than the critical value of t at the
0.05 level; reject the null hypothesis accept the alternative hypothesis, the
estimated cost of repair does differ between the two companies. The results is not
significant at the 0.01 level accept the null hypothesis, the estimates do not differ between
the two companies.
(c) The companies may have different cost structures due to eg labour costs,
error in calculating costs, use of manufacturer’s or generic parts, quality of
the work carried out.
3009/3/10MA Page 9 of 20

11. QUESTION 4
A random sample of workers in the Marketing department of a large company were interviewed and
the degree of job satisfaction and their job status were recorded.
Job Satisfaction
High Medium Low
Job
Management 105 30 165
Status
Skilled 69 32 89
Unskilled 56 38 96
(a) Test whether there is an association between job satisfaction and job status.
(12 marks)
(b) National data show that marketing workers express the following levels of job satisfaction.
Job Satisfaction
High Medium Low
30% 16% 54%
By combining the different job status of marketing workers’ test whether the views
on job satisfaction of workers in the Marketing department differ from the national situation.
(8 marks)
(Total 20 marks)
3009/3/10MA Page 10 of 20

12. MODEL ANSWER TO QUESTION 4
(a) Null hypothesis: there is no association between job satisfaction and job status.
Alternative hypothesis: there is association between job satisfaction and job status.
Degrees of freedom (r - 1)(c - 1) = (3 - 1)(3 - 1) = 2 x 2 = 4
2
0.05 = 9.49; 0.01 = 13.28
Observed frequencies
Job Satisfaction
Job High Medium Low
Status 105 30 165
69 32 89
56 38 96
Expected frequencies
101.47 44.12 154.41
64.26 27.94 97.79
64.26 27.94 97.79
2
Contributions to
0.1228 4.5176 0.7261
0.3489 0.5896 0.7908
1.0629 3.6212 0.0329
chi sq= 11.8127
Conclusion: the calculated Chi-squared is greater than the critical value of Chi-squared
at the 0.05 but not at the 0.01 significance level, there is some association between
job satisfaction and job status.
(b) Null hypothesis: marketing workers have the same levels of job satisfaction
as marketing executives nationally.
Alternative hypothesis: marketing workers do not have the same levels of
job satisfaction as marketing executives nationally.
Degrees of freedom (3 - 1) = 2
2
Critical value of 0.05 = 5.99 0.01 = 9.21
High Medium Low
Observed 230 100 350
Expected 204 108.8 367.2
3.314 0.712 0.806 Chi squared = 4.8313
2 2
Conclusion: the calculated value of is less than the critical value of there is insufficient
evidence to reject the null hypothesis that marketing executives have the same level of
job satisfaction as marketing executives.
3009/3/10MA Page 11 of 20

13. QUESTION 5
A, B and C are candidates for a role as design manager. The Managing Director recommends a
candidate and the company board must ratify the decision. The probability the Managing Director
recommends A is 55%, B 25% and C 20%. The probability that the board ratifies A is 60%, B 30%
and C 10% and is independent of the recommendation made by the Managing Director.
(a) Find the probability that
(i) B is a successful candidate
(ii) No candidate is successful
(iii) Given that one candidate is successful, it is candidate B.
(10 marks)
A container ship terminal can unload standard containers at the rate of 2000 per 8 hour day with a
standard deviation of 200 containers. Assume the rate of unloading is normally distributed.
(b) Find the probability that more than 2360 containers are unloaded in a day.
(3 marks)
(c) A ship carries 6500 containers what is the probability it will be unloaded in less
than 3 days.
(7 marks)
(Total 20 marks)
3009/3/10MA Page 12 of 20

14. MODEL ANSWER TO QUESTION 5
(a) (i) B successful = 0.25 x 0.3 = 0.075
(ii) No candidate is successful
A = 0.55 x 0.4 = 0.22
B = 0.25 x 0.7 = 0.175
C = 0.20 x 0.9 = 0.18..
0.575
(iii) Probability a candidate successful = 1 - 0.575 = 0.425
Probability B successful = 0.075 = 0.176
Probability success 0.425
(b) More than 2360 in a day
x 2000 2360
z z = 360/200 = 1.8
sd 200
Table probability = 0.964 required probability = 1 - 0.964 = 0.036
(c) A joint normal distribution approach is needed.
Joint mean x1 x2 x3 = 2000 + 2000 + 2000 = 6000
Joint standard deviation = sd12 2
sd 2 sd32 = 2002 2002 2002 =346
x 6500 6000
z z = 500/346 = 1.4 (1.445)
sd 346
Table probability = 0.919 required probability = 1- 0.919 = 0.081 (0.075)
3009/3/10/MA Page 13 of 20

15. QUESTION 6
(a) Explain why, when the sample size increases, the sample proportions cluster more closely
about the population proportion.
(4 marks)
A company is concerned about the difference in sick leave between the US factories and the UK
factories. A random sample of 70 US staff showed 11 took more than 10 days sick leave absent,
whilst a sample of 90 UK staff showed 18 took more than 10 days sick leave.
(b) (i) Test whether the proportion of staff taking more than 10 days sick leave is higher in the
UK than the US
(12 marks)
(ii) Explain what is meant by a Type 1 error and whether a Type 1 error may have been
committed in your conclusions in part (b) (i).
(4 marks)
(Total 20 marks)
3009/3/10/MA Page 14 of 20

16. MODEL ANSWER TO QUESTION 6
(a) When samples of a given size are taken, the distribution of the sample proportions
is referred to as the sampling distribution of the proportion .
p (1 p
The formula is se = as n increases for any given proportion the value of se
n
will decrease.
(b) (i) Null hypothesis: the proportion of staff with more than 10 days sick leave in
the US does not differ from the proportion of staff with more than 10 days sick
leave in the UK.
Alternative hypothesis: the proportion of staff with more than 10 days sick leave
in the US is less than the proportion of staff with more than 10 days sick leave in the
UK.
Critical z value for 0.05 significance level = -1.64
p1 = 11 = 0.1571; p2 = 18 = 0.20
70 90
n1 p1 n2 p 2
p
n1 n2
= 0.1571 x 70 + 0.20 x 90 = 11 + 18 = 0.181
70 + 90 160
p1 p2 0.1571 0.20
z
1 1 1 1
p1 p 0.181 0.819
n1 n2 70 90
= -0.0429 = - 0.698
0.0614
Conclusions: the calculated z value is less than the critical value of z. There is insufficient
evidence to reject the null hypothesis. The proportion of staff with more than 10 days sick
leave in the US does not differ from the proportion of staff with more than 10 days sick leave
in the UK.
(ii) A type 1 error is when a true null hypothesis is rejected when it should be accepted.
As the null hypothesis was accepted a type 1 error cannot have been made.
3009/3/10/MA Page 15 of 20

17. QUESTION 7
An investigation was carried out into the number of errors per shift made in dispatching internet orders
from warehouses in London and in Hong Kong. Random samples were taken in both London and
Hong Kong. A summary of the results is given below.
London Hong Kong
Arithmetic mean 225 211
Standard Deviation 31 23
Median 218 217
Sample size 55 47
(a) Calculate a measure of skewness for each of the cities.
(4 marks)
(b) Write a brief memo to the Sales Manager of the company highlighting the main features in the
error statistics between the two cities.
(8 marks)
(c) Test whether the average number of errors in Hong Kong is less than the average number of
errors in London.
(8 marks)
(Total 20 marks)
3009/3/10/MA Page 16 of 20

18. MODEL ANSWER TO QUESTION 7
(a) The measure of skew = 3(mean – median)
Standard deviation
London = 3(225 – 218) Hong Kong= 3(211 – 217)
31 23
= 0.677 = -0.783
(b) Memo layout: Title, Date, From, To and Subject
A range of content is possible. The following is suggested
The arithmetic mean and standard deviation are both larger for London than Hong Kong.
Coefficient of variations 0.138 and -0.109 London is more variable than Hong Kong.
The London Median is greater than the Hong Kong Median. The London Mean is greater than the
Hong Kong Mean.
The London mean must have more extreme high values whilst Hong Kong must have
more extreme low values.
The Hong Kong data is negatively skewed and London data is positively skewed but to
approximately the same degree.
(c) Null hypothesis: there is no difference in the number of errors made in dispatching
internet orders from warehouses in London and in Hong Kong. Alternative hypothesis:
there is a greater number of errors made in dispatching internet orders from warehouses
in London than in Hong Kong.
Critical value of z for 0.05 significance level = 1.64/2.33
x1 x2 225 211
z =
s2
1 s 2
2
312 232
n1 n1 55 47
14 14
z = = 2.62
17.47 11.26 5.23
Conclusions: The calculated value of z is more than the critical value of z. There is evidence
to reject the null hypothesis. There is a greater number of errors made in dispatching internet
orders from warehouses in London than in Hong Kong.
3009/3/10/MA Page 17 of 20

19. QUESTION 8
(a) Explain the impact of a business setting its quality control limits too wide.
(4 marks)
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. The average weight of pre-cut tuna is set at 175
grams with a standard deviation of 5 grams. Samples of 9 items at a time are taken from the production
line to check the accuracy of the manufacturing process.
(b) (i) Draw a control chart to monitor the process
(7 marks)
(ii) 7 samples taken from the production line had the following mean weight (grams):
Weight per piece
178.6 180.8 170.5 168.4 172.5 175.8 175.2
(grams)
Plot these data on your control chart and comment appropriately
(5 marks)
(iii) If the sample mean had been wrongly set at 177 cm, assuming the standard
deviation remains the same and the sample size is 9, what is the probability that
a sample mean lies outside the upper action limit?
(4 marks)
(Total 20 marks)
3009/3/10/MA Page 18 of 20

20. MODEL ANSWER TO QUESTION 8
(a) By setting the quality control limits too wide the number of samples rejected
will decrease and costs of resetting the process will be reduced and form a lower
level of rejected items. The business may gain a reputation for poor quality which
may damage its potential sales.
(b)
5
(i) Warning limits x 1.96 175 1.96 = 175 ± 3.26 = 171.7 to 178.3
n 9
(171.67 to 178.33 for z =2)
5
Action limits x 3.09 175 3.09 = 175 ± 5.15 = 169.85 to 180.15
n 9
(170 to 180 for z = 3)
(ii)
Quality Control Chart
180
Weight per piece grams
178
176
174
172
170
168
1 2 3 4 5 6 7
Sample number
Comment: Samples 2 and 4 lie outside the action limits and the process is unstable.
It needs to be stopped and adjusted.
x 180.15 177
(iii) z z
5
n 9
= 3.15 = 1.9 table proportion = 0.971 therefore answer = 0.029
1.67
3009/3/10/MA Page 19 of 20 © Education Development International plc 2010

21. LEVEL 3
3009/3/10/MA Page 20 of 20 © Education Development International plc 2010

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1517/2/10/MA Page 21 of 12 © Education Development International plc 2010