1. Business
Statistics
Level 3
Model Answers
Series 3 2007 (Code 3009)

2. Business Statistics Level 3
Series 3 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
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1

3. QUESTION 1
Over the past 9 years a company has achieved the following market share
Year Market Share %
1998 22.3
1999 21.6
2000 22.0
2001 21.1
2002 20.8
2003 19.9
2004 19.5
2005 18.4
2006 17.6
(a) Plot the data on a graph and comment on the pattern shown.
(4 marks)
(b) Calculate and state the least squares regression line for Market Share dependent on Year.
(10 marks)
(c) Use the regression line found in (b) to estimate the year in which market share is likely to fall to
15% and comment upon the accuracy of your estimate.
(6 marks)
(Total 20 marks)
3009/3/07/MA 2

4. MODEL ANSWER TO QUESTION 1
(a)
Market Share % v Time
23
22
21
Market Share %
20
19
18
17
16
15
1998 1999 2000 2001 2002 2003 2004 2005 2006
Year
Comment: ‘The market share is decreasing over time.’
(b)
Year Market
Share %
x y x2 xy
1 22.3 1 22.3
2 21.6 4 43.2
3 22.0 9 66.0
4 21.1 16 84.4
5 20.8 25 104.0
6 19.9 36 119.4
7 19.5 49 136.5
8 18.4 64 147.2
9 17.6 81 158.4
45 183.2 285 881.4
Σx Σy Σx2 Σxy
3009/3/07/MA 3 CONTINUED ON THE NEXT PAGE

5. MODEL ANSWER TO QUESTION 1 CONTINUED
n∑ xy − (∑ x )(∑ y )
b=
n∑ x 2 − (∑ x )
2
(9 x881.4) − (45 x183.2)
b=
9 x 285 − 45 2
b = -311.4 = -0.577 (-0.58)
540
a=
∑ y −b∑x
n n If x goes from 1998 to 2006 the
183.2 45 equation is y = 1174.84-0.577x
a= − −0.577 x
9 9
a = 20.36 +2.89 = 23.25
Y = 23.25 – 0.577x
(c)
Y = 23.25-0.557x
15 = 23.25 –0.577x
x = 23.25 – 15
0.577
x = 14.29 Round up to 15
Therefore year is 2012
The answer is an extrapolation and therefore subject to uncertainty. In real life the business might
change strategy to protect its market share. A better estimate would have been obtained from using
the regression line for year on market share.
3009/3/07/MA 4

6. QUESTION 2
(a) Explain what is meant by a 90% confidence interval for a sample proportion.
(4 marks)
A company is concerned about the level of customer satisfaction with the service provided by its
domestic and overseas call centre operations. A random sample of customers calling its domestic call
centre showed that 267 out of 400 were satisfied or very satisfied with the service. A random sample
of customers calling its overseas call centre showed that 350 out of 500 were satisfied or very satisfied
with the service.
(b) Test whether there is a difference in the proportion of customers satisfied or very satisfied with
the service provided at the two call centres.
(12 marks)
(c) Explain what is meant by a type 2 error. Might such an error have been committed in part (b)
above?
(4 marks)
(Total 20 marks)
3009/3/07/MA 5

7. MODEL ANSWER TO QUESTION 2
(a) The 90% confidence interval means that if samples of the same size are taken the sample
proportion will lie within the stated range 90 times out of 100.
(b) Null hypothesis: there is no difference in the proportion satisfied or very satisfied with the
customer service from its domestic and overseas call centres.
Alternative hypothesis: there is a difference in the proportion satisfied or very satisfied with the
customer service from its domestic and overseas call centres.
Critical z value = 1.96/2.58
p1 = 267/400 = 0.6675, p2 = 350/500 = 0.7
267 + 350 617
Pooled value of p = = = 0.6856
400 + 500 900
0.6675 − 0.7 − 0.0325
z= = = - 1.04
⎛ 1 1 ⎞ 0.2156(0.0045)
0.6856(1 − 0.6856)⎜ + ⎟
⎝ 400 500 ⎠
Conclusions: There is insufficient evidence to reject the null hypothesis at the 5% significance level.
There is no difference in the proportion satisfied or very satisfied with the customer service from its
domestic and overseas call centres.
(c) A type 2 error is to accept the null hypothesis when it is false. A type 2 error may have been
committed.
3009/3/07/MA 6

8. QUESTION 3
The consumer price index and the index of average earnings are as follows:
Year Consumer Index of Average
Price Index Earnings
1996 100.0 83.3
1997 101.8 86.8
1998 103.4 91.3
1999 104.8 95.7
2000 105.7 100.0
2001 106.9 104.4
2002 108.3 108.1
2003 109.8 111.7
2004 111.2 116.7
2005 113.5 121.4
2006
116.7 127.0
estimate
(a) Rebase the index of average earnings to 1996 as the base year = 100
(4 marks)
(b) Using the information in the table and your answer to part (a), calculate an index of average real
earnings with 1996 as the base year.
(4 marks)
A worker’s wage has risen, between the years 1996 to 2006, from £350.64 to £420.78.
(c) What further increase would be necessary to enable the worker’s wage to keep pace with the
Real Earnings Index?
(4 marks)
(d)
(i) Explain how you would carry out a Quota Sample and give one advantage and one
disadvantage of the method.
(ii) Explain how you would carry out a Systematic Sample and give one advantage and one
disadvantage of the method.
(8 marks)
(Total 20 marks)
3009/3/07/MA 7

9. MODEL ANSWER TO QUESTION 3
(a) Conversion of Index of Average Earnings to base year 1996 = 100
Divide by 83.3 multiply by 100 throughout
Index of
Average
Index of
Earnings
Year Average
Rebased
Earnings
1996
=100
1996 83.3 100.0
1997 86.8 104.2
1998 91.3 109.6
1999 95.7 114.9
2000 100.0 120.0
2001 104.4 125.3
2002 108.1 129.8
2003 111.7 134.1
2004 116.7 140.1
2005 121.4 145.7
2006
127.0 152.5
estimate
(b) Index of average real earnings
Divide Rebased Index of Earnings by Consumer Price Index
Index of
Average
Consumer Index of
Earnings
Year Price Real
Rebased
Index Earnings
1996
=100
1996 100 100.0 100.0
1997 101.8 104.2 102.4
1998 103.4 109.6 106.0
1999 104.8 114.9 109.6
2000 105.7 120.0 113.5
2001 106.9 125.3 117.2
2002 108.3 129.8 119.9
2003 109.8 134.1 122.1
2004 111.2 140.1 126.0
2005 113.5 145.7 128.4
2006
116.7 152.5 130.7
estimate
(c) £350.64 x 130.7/100 = £458.29
-£420.78 = £37.51
3009/3/07/MA 8 CONTINUED ON THE NEXT PAGE

10. MODEL ANSWER TO QUESTION 3 CONTINUED
(d)
(i) Quota sampling involves taking a sample with a given number of people. Often this is done
by stopping people in the street. Interviewers are given quota controls i.e. the characteristics
that the respondents should have. These often relate to age, gender and income. The
advantages are the sample is relatively cheap; there is no need for a sampling frame and no
need for call-backs. The disadvantages are there may be bias in the choice of respondents,
theoretically the standard error cannot be calculated so significance tests are not valid.
(ii) Systematic sampling involves taking a sample at a given interval. The interval is determined
by the sampling proportion e.g. a 10% sampling proportion would mean items are chosen
every 10 items. The first item to be chosen would be selected randomly between 1 and 10
e.g. 4 and the 4th, 14th, 24th etc item sampled. The advantages are the method is relatively
cheap. The disadvantages are there may be a natural period in the subjects and the method
excludes or includes particular subjects and theoretically the standard error cannot be
calculated so significance tests are not valid.
3009/3/07/MA 9

11. QUESTION 4
(a) Explain the difference between a paired t test and a two independent sample means test.
(4 marks)
Two random samples of twelve car hire companies were taken in countries X and Y. The car rental
costs in £ per 10 day hire for a standard four door 2000 cc saloon are given in the table below.
Car Rental Costs in Car Rental Costs in
Country X Country Y
350 525
525 340
550 430
525 520
390 670
540 470
525 520
570 540
530 650
650 650
540 680
590 450
(b) Test whether the car rentals paid in country X differ from those paid in Country Y.
(12 marks)
(c) Explain what is meant by the sampling distribution of the mean.
(4 marks)
(Total 20 marks)
3009/3/07/MA 10

12. MODEL ANSWER TO QUESTION 4
(a) A paired t test is used when a single sample is subject to two treatments, for example, before and
after: a two independent sample means test is used when the results from two separate samples
are compared in respect of a stated parameter, the mean.
(b) Null hypothesis: there is no difference in the car rentals paid in country X and country Y.
Alternative hypothesis: There is a difference in the car rentals paid in country X and country Y
Degrees of freedom n + m –2 = 12+12 –2 = 22
Critical t values 2.07/2.82
x y x2 y2 (x − x ) (y − y )
2 2
350 525 122500 275625 30189.06 1.5625
525 340 275625 115600 1.5625 33764.06
550 430 302500 184900 689.0625 8789.063
525 520 275625 270400 1.5625 14.0625
390 670 152100 448900 17889.06 21389.06
540 470 291600 220900 264.0625 2889.063
525 520 275625 270400 1.5625 14.0625
570 540 324900 291600 2139.063 264.0625
530 650 280900 422500 39.0625 15939.06
650 650 422500 422500 15939.06 15939.06
540 680 291600 462400 264.0625 24414.06
590 450 348100 202500 4389.063 5439.063
6285 6445 3363575 3588225 71806.25 128856.3
Σx Σy Σx2 Σy2 (
Σ x−x ) Σ (y − y )
2 2
6285 6445
x= = 523.75 y = = 537.083
12 12
− −
( x − x ) 2 + ( y − y) 2 71806.25 + 128856.3 s = 95.5 (95.0)
s= s=
n+m−2 12 + 12 − 2
x− y 523.75 − 537.083 = -0.34
t= t=
1 1 1 1
s + 95.5 +
n m 12 12
Conclusions: There is insufficient evidence to reject the null hypothesis. There is no difference in the
car rentals paid in country X and country Y.
(c) When a large number of samples of a given size are taken from a population the mean values will
vary from sample to sample. If the sample size is large (30 and greater) the sample means will be
normally distributed. If the sample size is small (less than 30) the means will be distributed in
accordance with the t distribution with n-1 degrees of freedom.
3009/3/07/MA 11

13. QUESTION 5
(a) Explain what the difference is between a one tail and a two tail test.
(4 marks)
The records of a company for two manufacturing plants X and Y, contain the following data on order
value for the past year:
Plant X Plant Y
Mean order value £ 760 698
Median order value £ 720 665
Standard deviation £ 145 135
Sample size 35 40
(b) Calculate the coefficient of skewness for both plants and comment on your answers.
(4 marks)
(c) Test whether the mean order value is significantly higher in plant X than in plant Y.
(8 marks)
When the order values of the two plants are combined the mean order value for the company as a
whole is found to be £726.93 and the standard deviation is £139.76.
(d) Calculate the 95% confidence interval for the overall mean order value.
(4 marks)
(Total 20 marks)
3009/3/07/MA 12

14. MODEL ANSWER TO QUESTION 5
(a) A one tail test tests if the difference between two samples varies in a single direction either
greater or smaller; a two tail test tests if the difference between two samples varies in either
direction (greater or smaller).
(b)
Coefficient of Skew = 3(Mean – Median)
Standard Deviation
Plant X = 3(760 -720) = 120/145 = 0.828
145
Plant Y = 3(698 – 665) = 99/135 = 0.733
135
Both distribution are positively skewed, Plant X is more skewed than plant Y.
(c) Null hypothesis: There is no difference in the mean order value between plant X and plant Y.
Alternative hypothesis: The mean order value in plant X is greater than in plant Y.
Critical z value = 1.64/2.33
x1 − x 2 760 − 698 62
z= = = = 1.908
s12 s 2
2
1452 1352 600.71 + 455.625
+ +
n1 n2 35 40
Conclusions: There is evidence to support the alternative hypothesis at the 0.05 level; the mean order
value at plant X is greater than the mean order value at plant Y. At the 0.01 level there is insufficient
evidence to support the alternative hypothesis; the mean order value does not differ between the two
plants.
σ 139.76
(e) 95% confidence interval = x ± 1.96 = 726.93 ± 1.96
n 75
= 726.93 ± 31.63 = £695.30 to £758.56
3009/3/07/MA 13

15. QUESTION 6
A random sample is taken of workers in different sectors of the economy and the number of days they
are absent on sick leave in 2006 are recorded in the table below.
Sector of the Economy
Public Sector Domestic Owned Foreign Owned
Sick Leave Taken Private Private
Less than 5 days sick 35 45 20
leave
5 days and less than 10 95 102 35
days sick leave
10 or more days sick 70 53 45
leave
(a) Test whether there is any association between the sector of the economy and the amount of sick
leave taken.
(12 marks)
In a similar survey carried out 5 years earlier the proportions of people taking sick leave were as
follows.
Less than 5 and less than 10 10 or more days
days sick leave days sick leave sick leave
17% 54% 29%
(b) Test whether the pattern of sick leave taken has changed over the period.
(8 marks)
(Total 20 marks)
3009/3/07/MA 14

16. MODEL ANSWER TO QUESTION 6
(a) Null hypothesis: there is no association between the sector of the economy and the number of
days sick leave taken.
Alternative hypothesis: there is association between the sector of the economy and the number
of days sick leave taken.
Degrees of freedom (R-1)(C-1) = (3-1)(3-1) = 4
Critical Χ2 = 9.49/13.28
Observed
35 45 20
95 102 35
70 53 45
Expected
40 40 20
92.8 92.8 46.4
67.2 67.2 33.6
Contributions to X2
0.625 0.625 0
0.052155 0.912069 2.800862
0.116667 3.000595 3.867857
X2 = 12.00021
Conclusions: The calculated value of X2, 12.00 is more than the critical value of X2 at the
0.05 level: reject the null hypothesis there is evidence to support the alternative hypothesis. There is
association between the work sector and the number of days sick leave taken.
The calculated value of X2, 12.00 is less than the critical value of X2 at the 0.01 level, there is
insufficient evidence to reject the null hypothesis: there is no association between the work sector and
the number of day’s sick leave taken.
(b) Null hypothesis: the pattern of sick leave has not changed.
Alternative hypothesis: the pattern of sick leave has changed.
Degrees of freedom = n-1 = 3-1 = 2
Critical X2 = 5.99/9.21
Observed 100 232 168
Expected 85 270 145
Contributions to X2 2.65 5.35 3.65
X2 = 11.65
Conclusions: The calculated value of X2 is greater than the critical value of X2 at both the 0.05 and
0.01 levels: reject the null hypothesis accept the alternative hypothesis. There is strong evidence to
support the alternative hypothesis, the pattern of sick leave has changed.
3009/3/07/MA 15

17. QUESTION 7
A company’s largest selling product has average weekly sales of 2500 units with standard deviation
250 units. Assume that weekly sales are normally distributed.
(a) Find the probability that weekly sales:
(i) exceed 3050 units
(ii) lie between 3100 units and 2200 units
(8 marks)
(b) Find the level of sales which will be exceeded on 90% of occasions.
(3 marks)
The same company is planning the launch of a new product. It estimates that the probability of good
economic conditions is 70%. If economic conditions are good the probability of the workforce
accepting a new wage offer is 30%; if economic conditions are poor the probability of the workforce
accepting a new wage offer is 50%.
(c) Find the probability that the wage offer is accepted.
(5 marks)
(d) If the wage offer is accepted what is the probability that the economic conditions were poor.
(4 marks)
(Total 20 marks)
3009/3/07/MA 16

18. MODEL ANSWER TO QUESTION 7
(a)
(i) probability that sales exceed 3050 units
x−x 3050 − 2500 550
z= = = = 2.2
σ 250 250
Probability = 1 – 0.986 = 0.014
(ii) probability that sales lie between 3100 units and 2200 units.
Probability of sales between 2500 and 3100
x−x 3100 − 2500 600
z= = = = 2.4
σ 250 250
Probability = 0.992 –0.5 = 0.492
Probability of sales between 2500 and 2200
x−x 2200 − 2500 300
z= = = = 1.2
σ 250 250
Probability = 0.885 –0.5 = 0.385
Probability of sales 2200 to 3100 = 0.492 +0.385 = 0.877
(b)
Nearest table value for 90%/0.90 is z = 1.3
x−x
z= ,
σ
x − 2500
− 1.3 =
250
x = (-1.3 x 250) + 2500 - 325 +2500 = 2175 units
(c)
Probability that the wage offer is accepted.
Good economic conditions and wage offer accepted = 0.7 x 0.3 = 0.21
+
Poor economic conditions and wage offer accepted = 0.3 x 0.5 = 0.15
= 0.36
(d)
Probability poor economic conditions and wage offer accepted = 0.15 = 0.4167 (0.42)
Probability wage offer accepted 0.36
3009/3/07/MA 17

19. QUESTION 8
(a) Explain how the mean and range chart are used in quality control.
(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 weight of a packet of biscuits is set
at 150 grams with a standard deviation of 8 grams. Samples of 8 items at a time are taken from the
production line to check the accuracy of the manufacturing process.
(b)
(i) Calculate the values of the action and warning limits and construct a quality control chart to
monitor the manufacturing process.
(8 marks)
(ii) The results for 7 samples are given below. Plot these on your quality control chart and
comment on the results.
(4 marks)
sample number 1 2 3 4 5 6 7
sample mean (grams) 150.5 156.4 143.9 152.9 148.13 159.5 157.6
(c) Find the probability that a single item taken from the production line lies outside the action limits.
You may assume that the control chart was correctly set up with mean 150 grams and standard
deviation 8 grams.
(4 marks)
(Total 20 marks)
3009/3/07/MA 18

20. MODEL ANSWER TO QUESTION 8
(a) When a product is made to a standard the aim is that average dimension of a sample is close to
the specified value. It is also intended that the variability of the item should remain constant.
The mean chart checks that the average stays within acceptable limits and the range chart
checks that the variability of the product stays within acceptable limits.
σ 8
(b) Warning Limits = x ± 1.96 = 150 ± 1.96 = 144.5 to 155.5
n 8
σ
Action Limits = x ± 3.09 = 141.3 to 158.7
n
Quality Control Chart
165.0
160.0
UAL
155.0 UWL
Weight grams
Mean
150.0
145.0
LWL
LAL
140.0
135.0
1 2 3 4 5 6 7
Sample Number
The process is outside the action limits at sample 6. It needs to be stopped and adjusted.
(c) Probability an item lies above the upper action limit
x−μ 158.7 − 150
z= = z= = 8.7/8 = 1.09
σ 8
Probability for table value for z = 1.09 (1.1) = 0.864
Required probability = 1- 0.864 = 0.136
Probability an item lies below the lower action limit = 0.136
Total Probability = 0.272
3009/3/07/MA 19 © Education Development International plc 2007