BUSINESS MATHEMATICS &
QUANTITATIVE METHODS
NOTES:
FORMATION 1 EXAMINATION - APRIL 2011
You are required to answer 5 questions.
(If you provide answers to all questions, you must draw a clearly distinguishable line through the answer not to
be marked. Otherwise, only the first 5 answers to hand will be marked).
All questions carry equal marks.
STATISTICAL FORMULAE TABLES ARE PROVIDED
DEPARTMENT OF EDUCATION MATHEMATICS TABLES ARE AVAILABLE ON REQUEST
TIME ALLOWED:
3 hours, plus 10 minutes to read the paper.
INSTRUCTIONS:
During the reading time you may write notes on the examination paper but you may not commence
writing in your answer book.
Marks for each question are shown. The pass mark required is 50% in total over the whole paper.
Start your answer to each question on a new page.
You are reminded that candidates are expected to pay particular attention to their communication skills
and care must be taken regarding the format and literacy of the solutions. The marking system will take
into account the content of the candidates' answers and the extent to which answers are supported with
relevant legislation, case law or examples where appropriate.
List on the cover of each answer booklet, in the space provided, the number of each question(s)
attempted.
The Institute of Certified Public Accountants in Ireland, 17 Harcourt Street, Dublin 2.
THE INSTITUTE OF CERTIFIED PUBLIC ACCOUNTANTS IN IRELAND
BUSINESS MATHEMATICS &
QUANTITATIVE METHODS
FORMATION 1 EXAMINATION - APRIL 2011
Time Allowed: 3 hours, plus 10 minutes to read the paper.
You are required to answer 5 questions.
(If you provide answers to all questions, you must draw a clearly distinguishable line through the answer not to
be marked. Otherwise, only the first 5 answers to hand will be marked).
All questions carry equal marks.
1.
Two projects are proposed by the Contracts Manager for the consideration of the management team. One is a
capital intensive project requiring a large capital injection immediately with cash flows commencing in year 3. The
second is less capital intensive but shows a smaller revenue stream over it’s life. The cash flows are provided in
the following table. You are required to assess the projects by means of the following:
(i)
Recommend the most appropriate project based on NPV using a discount rate of 10%.
(8 Marks)
(ii)
Derive the IRR for the recommended project.
(8 Marks)
(iii)
Recommend the most appropriate method for your decision.
(4 Marks)
Year
0
1
2
3
4
5
6
7
8
Project 1
Cash Flow
(€000s)
(25)
0
0
8
10
10
12
12
10
Project 2
Cash Flow
(€000)
(8)
4
4
4
4
4
2
2
2
[Total: 20 Marks]
1
2.
The Superior Products Company has obtained information that the co-efficient of variation in the weekly production
of it’s sister company, DIB Ltd, is 12.5%.The following information has been obtained from your Production
Department showing the weekly production by groups of employees. You are required to:
(i)
Derive the mean and standard deviation of weekly production levels; and
(10 Marks)
(ii)
Derive the co-efficient of variation, compare it to DIB Ltd, and explain its significance.
(10 Marks)
Weekly production (units)
350
400
450
500
550
600
650
700
750
800
850
900
950
1000
1050
<
<
<
<
<
<
<
<
<
<
<
<
<
<
<
No. employees
400
450
500
550
600
650
700
750
800
850
900
950
1000
1050
1100
2
5
10
15
20
30
40
30
20
20
12
10
5
4
1
[Total: 20 Marks]
3.
Vodacall is assessing roaming charges for it’s mobile telephone calls. From the company records, roaming calls
have averages of 15 minutes per call. However, in a random sample of 50 calls, the sample mean was 14.10 mins.
per call with a sample standard deviation of 5 mins. The company wishes to test if the call usage has changed. You
are required to:
(i)
Set out the process by which the company can test this hypothesis; and
(8 Marks)
(ii)
Test if there has been a change in the average duration of roaming calls.
(Use a 0.05 level of significance for your test).
(12 Marks)
[Total: 20 Marks]
2
4.
GKN has produced a sales profile for it’s computer network servers over the past 4 years. The company is aware
that there is a major cyclic variation in annual sales. The sales data is shown in the table below. It wishes to use
this data to predict future sales. You are required to:
(i)
(ii)
Outline the steps in the process of time series analysis to calculate the trend and deseasonalise the data;
and
(8 Marks)
Using the method of moving averages, calculate the average seasonal variations.
Year
Qtr. 1
Qtr.2
Qtr.3
Qtr.4
Total
2007
100
55
70
150
375
2008
95
58
65
145
363
2009
99
48
65
155
367
(12 Marks)
2010
103
61
71
148
383
[Total: 20 Marks]
5.
As a Financial Advisor you have been requested to solve the following problems for your client, JB:
(i)
(ii)
JB’s credit card has an outstanding balance of €1,000 per month on which he is charged interest at 2.5%
per month. He has been told that the real rate of interest is higher than this apparent annual rate of 30%.
What is the real rate of interest?
(6 Marks)
JB runs an electrical retail business. He has monitored the prices of three key commodities for the last two
years as shown in the table below and wishes to calculate the increase in prices over the period.
Commodity
TVs
DVD Players
Hi-Fis
2009
€210
€101
€125
2010
€235
€95
€150
He estimates that the TVs contribute 7 times more to turnover than DVD players and that DVD players
contribute 3 times more than Hi-Fis. Using a weighted aggregate index calculate the increase in prices over
the period.
(8 Marks)
(iii)
JB wishes to use the method of ‘reducing balance depreciation’ to reduce the value of the office I.T. system
rather than straight line depreciation which has been used to date. The I.T. system costs €50,000. With a
rate of depreciation of 25% calculate the depreciated value of the I.T. system after 5 years.
(6 Marks)
[Total: 20 Marks]
6.
“Different types of sample may be used depending on the characteristics or attributes of the ‘populations’ to be
sampled and on the objectives of those undertaking the sample”. In the context of this statement outline the key
features of four of the following methods of sampling: (i) Systematic sampling (ii) simple random sampling, (iii)
stratified sampling, (iv) quota sampling, (v) multi-stage sampling, (vi) cluster sampling’.
[Total: 20 Marks]
END OF PAPER
3
SUGGESTED SOLUTIONS
THE INSTITUTE OF CERTIFIED PUBLIC ACCOUNTANTS IN IRELAND
BUSINESS MATHEMATICS &
QUANTITATIVE METHODS
FORMATION 1 EXAMINATION - APRIL 2011
SOLUTION 1
(i)
NPV for both projects.
Cash Flows €000s
Year
Project 1
0
1
2
3
4
5
6
7
8
NPV
(25)
0
0
8
10
10
12
12
10
Discounted Cash Flow
Project 2
Discount Factor
@ 10%
Project 1
Project 2
1.000
0.909
0.826
0.751
0.683
0.621
0.564
0.513
0.455
-25
0
0
6.008
6.830
6.210
6.760
6.156
4.660
11.632
-8
3.636
3.304
3.004
2.732
2.484
1.128
1.026
0.932
10.246
(8)
4
4
4
4
4
2
2
2
2 marks
2 marks
The NPV for Project 1 is €11,632. The decision rule is that the project with the higher NPV should be
accepted particularly if there is a scarcity of financial resources and one project only should be selected.
Therefore project 1 is recommended.
(4 marks)
(ii)
Internal Rate of Return for Project 1. Select another discount factor, say 20%.
Year
Project 1
Discount
Factor @ 10%
Project 1
Cash Flow
Discount
Factor @ 20%
Project 1
Cash Flow
0
1
2
3
4
5
6
7
8
(25)
0
0
8
10
10
12
12
10
1.000
0.909
0.826
0.751
0.683
0.621
0.564
0.513
0.455
-25
0
0
6.008
6.830
6.210
6.760
6.156
4.660
1.000
0.833
0.694
0.579
0.482
0.402
0.335
0.279
0.233
-25
0
0
4.632
4.820
4.020
4.020
3.348
2.330
NPV
11.632
(1.83)
2 marks
5
2 marks
The IRR can be found by interpolation. Since 10% is too low and 20% is marginally too high the IRR = 20%
- (1.83/13.492)x10% = 20% - 1.35% = 18.65%.
(4 marks)
[where the difference of 10% is 13.492]
(iii)
NPV. If the NPV of a project is positive this can be interpreted as the potential increase in cash flow of the
project valued in present day terms. This shows that the principal amount borrowed and the interest can be
repaid from the cash flows leaving a cash balance. This balance is a Net Terminal Value and its present value
is the NPV. The IRR is the discount rate which gives zero NPV. The IRR cannot be found directly – it can be
found by graphical methods or interpolation. Where the calculated IRR is greater than the company’s cost of
capital then the project is acceptable. In the majority of cases where there are conventional cash flows then
the IRR gives the same decision as the NPV. This does not necessarily rank projects in the same order of
attractiveness. In many cases either of these decision criteria can be used successfully but there are
differences in particular situations
-
-
Where projects can be considered independently of each other and where the cash flows are
conventional then NPV and IRR gives the same accept or reject decision
NPV is an absolute measure of the return on a project whereas IRR is a relative measure relating the
size and timing of cash flows to the initial investment. Thus the NPV reflects the scale of a project where
IRR does not.
Mutually exclusive decisions arise where projects are ranked in order of attractiveness and to select the
most profitable. In these cases NPV and IRR may give conflicting rankings.
(4 marks)
[Total: 20 Marks]
6
SOLUTION 2
(i)
Calculation of mean and standard deviation.
Class boundaries
350 < 400
400 < 450
450 < 500
500 < 550
550 < 600
600 < 650
650 < 700
700 < 750
750 < 800
800 < 850
850 < 900
900 < 950
950 < 1000
1000 < 1050
1050 < 1100
∑
F
2
5
10
15
20
30
40
30
20
20
12
10
5
4
1
224
x
375
425
475
525
575
625
675
725
775
825
875
925
975
1025
1075
F(x)
750
2,125
4,725
7,875
11,500
18,750
27,000
21,750
15,500
16,500
10,500
9,250
4,875
4,100
1,075
156,300
2 marks
(x – x)
-323
-273
-223
-173
-123
-73
-23
27
77
127
177
227
277
327
377
Mean, x = ∑fx = 156,300 = €697.8 ≈ €698
∑f
224
Standard deviation, σ =
(ii)
√
(x – x)2
104,329
74,529
49,729
29,929
15,129
5,329
529
729
5,929
16,129
31,329
51,529
76,729
106,929
142,129
2 marks
F(x – x)2
208,658
372,645
497,290
448,935
302,580
159,870
21,160
21,870
118,580
322,580
375,948
515,290
383,645
427,716
142,129
4,318,896
2 marks
(2 marks)
∑f(x –x)2 =
4,318,896 = √19,280.8 = €138.9
∑f
√
224
Co-efficient of variation = σ = 138.9 = 0.199 = 19.9%
x
698
(2 marks)
(4 marks)
The co-efficient of variation measures the spread of the data around the mean - the greater the co-efficient
the greater the spread. In the case of DIB Ltd the spread of the data around the mean is less than Superior
Products showing that there is less of a variation in weekly production. This could imply that the efficiency of
production in Superior Products is less with some group producing more than others and other groups
producing substantially less.
(6 marks)
[Total: 20 Marks]
7
SOLUTION 3
(i)
The procedure followed in hypothesis testing involves stating a null hypothesis (normally designated by H0)
and an alternative hypothesis designated by H1. the null hypothesis assumes that no change has occurred.
The onus therefore rests with disproving the null hypothesis. The alternative hypothesis is accepted if we
reject the null hypothesis. When an hypothesis is tested it is possible to draw both correct and incorrect
conclusions from the analysis.
Therefore to test a hypothesis, the summarised steps are:
state the hypotheses for H0 and H1
state the significance level
state the critical values
calculate the z score, the test statistic [(x – µ)/(σ/√n)], for the sample; the sample standard deviation, s,
is used for σ, providing that the sample size is reasonably large
compare this z sample score with the z critical value/s
come to a conclusion: accept or reject H0
state the conclusion in words, that is, the sample evidence does (or does not) support the null
hypothesis at the stated significance level.
(6 marks)
(ii)
Set out the null and alternative hypothesis for this two tailed test.
H0: µ = 15.0 mins - duration of roaming calls is as expected – no action required
H1: µ ≠ 15.0 mins - duration of roaming calls not as expected – action required
Using a 0.5 significance level is equivalent to saying that there is a 0.05 probability of making a Type 1 error.
This gives a critical value of µ + (Z0.025 x σx).
(2 marks)
The critical value is σx = s/√n; therefore with n = 50, X = 14.10, s = 5,
σx = 5/√50 = 5/7.07 = 0.707.
The critical values are: µ + (Z0.025 x σx)
= 15 ± (1.96 x .707)
= 15 ± 1.385
= 16.385 and 13.615.
(4 marks)
Decision rule: Reject H0 if x < 13.615 or x > 16.385. Otherwise accept H0.
Since the sample mean is 14.10 mins the null hypothesis cannot be rejected. We can therefore say with 95%
confidence that roaming calls are still averaging 15 mins. The sample does not provide evidence that there
has been a change.
(4 marks)
The data and critical values are set out in the diagram.
(4 marks)
[Total: 20 Marks]
8
SOLUTION 4
(i)
Steps in forecasting by using time series analysis.
•
•
•
(ii)
Find the underlying trend of the cycle or data pattern; this can be done by using moving averages. The
length of the moving average should be equal to the length of the cycle or in multiples of the cycle. In
the present case a four year period moving average will be used.
Determine whether these are cyclic or seasonal factors – this is the difference between the points on
the cycle and the underlying trend; the seasonal or cyclic differences can be calculated by determining
the difference between the actual value and the de-seasonalised line calculated in stage 1. the
seasonal values are brought together into a single seasonal factor for each part of the cycle. A simple
average can be taken.
Use the underlying trend and cyclic factors to build forecasts for the future. The calculation of a forecast
can be made by extending the de-seasonalised line (the moving average) into the future. This is the
most up-to-date moving average value. The forecast is completed by adding to this value the seasonal
factor appropriate to the quarter.
(6 Marks)
To derive the seasonal variations by using the above steps.
Year
Qtr. 1
Qtr.2
Qtr.3
Qtr.4
Total
2006
100
55
70
150
375
2007
95
58
65
145
363
2008
99
48
65
155
367
2009
103
61
71
148
383
Year
Quarter
Sales
Running total
Moving average
Seasonal value
2006
1
2
3
4
100
55
70
150
375
93.75
56.25
2007
1
2
3
4
95
58
65
145
370
373
368
363
92.50
93.25
92.00
90.75
2.50
-35.25
-27.00
54.25
2008
1
2
3
4
99
48
65
155
367
357
357
367
91.75
89.25
89.25
91.75
7.25
-41.25
-24.25
64.25
2009
1
2
3
4
103
61
71
148
371
384
390
383
92.75
96.00
97.50
95.75
10.25
-35.00
-26.50
52.25
3 Marks
3 Marks
Getting the average seasonal factor.
Year
Qtr. 1
Qtr.2
Qtr.3
Qtr.4
2006
56.25
2007
2.50
-35.25
-27.00
54.25
2008
7.25
-41.25
-24.25
64.25
2009
10.25
-35.00
-26.50
52.25
Average seasonal factor
6.66
-37.16
-25.91
75.66
(8 Marks)
[Total: 20 Marks]
9
SOLUTION 5
(i)
JB is paying compound interest on the outstanding balance. The rate quoted by the company is the nominal
rate. Since he is making 12 payments per year on the debt of €1,000, the amount at the end of the period is
T = P(1 + r/100m)mn, where m = 12 (no of payments per year), n = 1 year,
r = nominal interest rate.
T
=
1000(1 + 30/100 x 12)1 x 12
=
1000(1.025)12
=
=
1000 x 1.3449
1344.9
(3 Marks)
that is, the true rate of interest is 34.49%.
(ii)
(3 Marks)
The following table sets out the weights, quantities and prices. The weights show the relative importance of
each commodity contributing to the turnover.
Commodity
TVs
VCRs
HI-Fis
∑
Weight
7
3
1
2008 (p0)
€210
€101
€125
2009 (p1)
€235
€95
€150
wp0
1470
303
125
1898
wp1
1645
285
150
2080
(4 Marks)
Simple aggregrate index =
∑wp1 x 100
∑wp0
=
2080
1898
x 100
=
109.5
(4 Marks)
The weighted prices of the group of commodities has increased by 9.5%.
(iii)
The reducing balance depreciated value of an asset can be calculated in a number of ways.
-
Using the formula for reducing balance
D = A(1 - i)n where i = the depreciation rate, A = actual/book value,
n = number of years.
A5
-
= 50000(1 - 0.25)5
= 50000 x 0.755
= 50000 x 0.2373
(3 Marks)
= €11,865
Using a depreciation table.
1st year:
2nd year:
3rd year:
4th year:
5th year:
25%
25%
25%
25%
25%
x
x
x
x
x
50000
(50000
(50000
(50000
(50000
–
–
–
–
12500)
21875)
21093.75)
34181.25)
=
=
=
=
=
12,500
9375
7031.25
5275
3954.68
Total depreciation = €38135.93
Value of asset after 5 years: 50000 - 38135.93 = €11,864
(3 Marks)
[Total: 20 Marks]
10
SOLUTION 6
Sampling Methods. There are a wide range of sampling methods depending on the type of sample required and
the technique being used. A description of four of the following methods is required.
Simple random sampling. A random sample is used when the intention is to give each item in the ‘population’ as
much chance of being selected in the sample as every other item. The key features of random sampling are:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Each item selected has an equal chance of being drawn
Usually adopted when the population is homogeneous, that is, when it is difficult to distinguish between items
Implementation often involves the use of computer-generated random numbers
Selection is unbiased
A major drawback is that the population listing is required and the chosen items need to be located, then
questioned or measured.
This method is used by large marketing companies to obtain a wide geographical spread in the data.
(5 Marks)
Systematic sampling. This is a modified form of simple random sampling. Select subjects from a population list in
a systematic rather than a random way. For example the first person is selected randomly and then a fixed number
are skipped to get the next person, etc. this process makes sample selection very convenient. A further example of
a systematic sampling scheme is to select every nth person until the sample is completed. The main problem
associated with this form of sampling is that it can be expensive to carry out, based on the geographical coverage,
particularly if the sample is based on a national listing.
(5 Marks)
Stratified sampling. This method is used where the population contains distinct groups, that is, is segmented or
heterogeneous, that is, it contains groups with different views about the issues under study or of particular interest.
It is a non-random procedure. The sample can be stratified according to the particular groups so that it has the same
proportions, approximately, as the population. For each stratum the sample members are selected randomly as for
simple random sampling. It is a unbiased method and gives a representative sample. The process of stratification
can be expensive and incurs costs additional to the survey process. The key features are:
(i)
(ii)
(iii)
(iv)
(v)
Used when the population has a number of identifiable attributes
Populations stratified in this way are known as heterogeneous
The composition of the sample must reflect the attributes present in the population
Individuals within each stratum may be selected randomly
It is free from selective bias since it reflects the proportions of any given attribute present in the population as
a whole.
(5 Marks)
Quota sampling. The intention is to introduce selective bias into the samples so that attributes of the members or
items selected will represent the choice of the samples rather than the attributes of the population as a whole. In
this sense there is no attempt to seek a representative or unbiased sample from the population nor is there any
attempt to use random sampling within the quotas selected. Used where interviewing is the main method of data
collection. The key features are:
(i)
(ii)
(iii)
(iv)
Sample includes a specified number or quota of subjects with given attributes
Widely used in market research
Interviewers are often responsible for identification and selection of respondents
A ‘biased’ sample may result but one which may be useful in representing particular customers most likely to
purchase the company’s products.
(5 Marks)
11
Multistage sampling. This is a process used for producing a representative sample form a widespread population.
The key features are:
Used where there is a wide geographical spread of subjects which makes sampling expensive
Involves sampling subjects with a given attribute
The geographical area is divided into regions
A small number of regions is selected randomly
Regions selected are broken into sub-regions from which a random sample is selected
Sub-regions are broken into units from which a random sample is selected (town, streets, or particular
locations)
(vii) Then individuals with a given attribute are identified in specific towns or streets
(viii) Some risk of bias if small numbers of regions/locations are selected.
(5 Marks)
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Cluster Sampling. This method is used when the population items of interest are widely spread and it is desirable
to ensure that the sample elements are grouped together in some way. The key features are:
(i)
(ii)
(iii)
Items are chosen in clusters rather than individually (a cluster might be all the residents in a particular road)
Similar to multi-stage sampling but where the cluster is the single-stage involved
Useful where the population is widely spread geographically but where the various clusters are broadly
representative of that population.
(5 Marks)
Systematic Sampling. This method is similar to simple random sampling. The key features are:
(i)
(ii)
(iii)
(iv)
The data are assumed to be random and every nth member is selected where n is determined by (population
size) / (sample size)
The start point of the sample may be chosen randomly.
It is an inexpensive and easy method to use and is useful if the exact population is unknown.
Often particular steps must be taken to ensure that the sample is not biased.
(5 Marks)
[Total: 20 Marks]
END OF PAPER
12