QUANTITATIVE ANALYSIS NOVEMBER 2019 - CPA SOLUTION|REVISION KITS|KASNEB

QUESTION 1a

Q In the context of the above statement, describe what "business analytics" entails.

A

Solution

Description of 'Business analytics'

Business Analytics - refers to the skills, applications, and practices of continuously exploring data to gain insights that drive business decisions.

Business analysis can help an organization keep up with competition and maintain prospects for future growth in several ways, such as:

(i) Help organizations anticipate and more proactively forecast outcomes and trade-offs.

(ii) Enables businesses to create an innovation-receptive environment to design strategies

(iii)Helps organizations gain a competitive advantage by providing models for decision making purposes such as simulations, decision trees, etc..

(iv) Modern computation applications such as tax analysis e.t.c can help businesses recover taxes on overpaid transactions and prevent future overpayments.

(v) Enables businesses to protect their data with digital watermarking, malware detection measures, and more.

(vi) Prescriptive analytics can also be enhanced by developing optimization techniques for superior business performance.

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QUESTION 1b(i-ii)

Q (i) A binomial distribution.

(ii) A Poisson distribution.

A

Solution

A binomial distribution.

P = probability of being absent
q = probability of not being absent
r = total number of “successes” ( heads or tails etc.)
n = number of consoltant workers
c = Tip: You can use the combinations calculator to figure out the value for _nC_r

P^r = _nC_r x P^rq^n-r

n = 6, Q = 1 - 0.2 = 0.8

P(r ≥ 3) = 1 - [P(r = 0) + P(r = 1) + P(r = 2)]

P(r = 0) = 6C₀ x (0.2)⁰(0.8)^6-0 = 0.2621

P(r = 1) = 6C₁ x (0.2)¹ (0.8)^6-1

= ₆C₁ x (0.2)¹(0.8)⁵ = 0.3932

P(r=2)= ₆C₂x (0.2)²(0.8)⁴ = 0.2458

Thus, P(r ≥ 3) = 1 - (0.2621 + 0.3932 + 0.2458)

=1 - 0.9011

= 0.0989

Therefore, there is a chance or probability of 0.989 that more than two consultants will be absent from work on a given day.

A Poisson distribution.

P(r)

=

e^-λ λⁿ

n!

r is a random variable following a Poisson distribution
n is the number of times an event occurs
P(X = n) is the probability that an event will occur k times
e is Euler’s constant (approximately 2.718)
λ (lambda) is the average number of times an event occurs
! is the factorial function

P(r ≥3) = 1 - [P(r = 0) + P(r = 1) + P(r = 2)]

e = 2.7183

λ = 20% x 6 = 1.2

P(r = 0)

=

2.7183^-1.2 + 1.2^o

0!

= 0.3012

P(r = 1)

=

2.7183^-1.2 + 1.2¹

1!

= 0.3614

P(r = 2)

=

2.7183^-1.2 + 1.2²

2!

= 0.2169

∴P(r ≥) = 1 - (0.3013 + 0.3614 + 0.2169)

=1 - 0.8795 = 0.1205

Therefore, there is a chance or probability of 0.1205 that more than two consultants will be absent from mark on a given day.

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QUESTION 1c

Q The required gross output of each sector in order to meet the new demand.

A

Solution

Production sector		Gross output(units)
x₁	(500+800+200)	1,500
x₂	(600+1,400+400)	2,400

Matrix of input-output coefficient(A)

∴

I - A =

-

(I-A)^-1

=

1

|(I-A)|

[Adjoint(I-A)]

∴ |(I-A)| = (4/5 x 5/12) - (-2/5 x -1/3)

1/5

∴

(1-A)^-1 = 5/1

X

=

[

X₁

X₂

]

D

=

[

d₁

d₂

]

Where;

X= Gross output production matrix

D-Final demand matrix

D

=

[

400

800

]

X

=

[

400

800

]

=

[

2,166.67

4,000

]

X₁ = 2,166.67 units
X₂ = 4,000 units

Hence, the gross output of each sector required to meet the new demand is
X₁ = 2.166.67 units and X₂ = 4000 units

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QUESTION 2a(i)

Q Cumulative profit in the break-even time interval.

A

Solution

Profit = 10x - x² - 5
At breakeven point,
Profit = 0
Thus,
10x - x² - 5 = 0

-x² + 10x = 5

x² - 10x = -5

(x - 5)² = -5 + (5)²

= -5 + 25 = 20

x - 5 = √20

X = 5 + 4.4721

= (5 - 4.4721) or (5 + 4.4721)

= 0.5279 months or 9.4721 months

Therefore, the cumulative profit between 0.5279 months and 9.4721 months is,

Profit ₍₀₀₀₎ = [10(9.4721) - (9.4721)² - 5] - [10(0.5279) - (0.5279)²-5]

= 0.00032159 - 0.00032159

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QUESTION 2a(ii - iii)

Q (ii) The best time to end the production.

(iii) The total profit based on your result in (a) (ii) above.

A

Solution

(ii) The best time to end the production.

The best time to end the production is when the profit reaches the optimum point

At maximum profit:

Marginal profit (MP) = 0

MP= dtp/dx

MP = 10 - 2x

10 - 2x = 0

-2x = -10

x = 5 months

Therefore, the production must be stopped at month 5, because at this point the profit has reached its maximum point.

iii) The total profit based on the result in (a) (ii) above

Profit₍₀₀₀₎ = 10(5) - (5)² - 5 = 20

Thus, profit Sh. 20,000

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QUESTION 2b

Q Test at 95% level of confidence whether the company should accept the new plant.

A

Solution

Since the sample size "n" is very small and the population standard deviation (8) is unknown, a suitable test statistic is the t-test.

Where:

t

=

X̄ - μ

S / √n

μ= 18.5, X̄ = 19.7, n = 25, S = 1.5

∴

t

=

19.7 - 19.5

1.5 / √25

= 0.2 / 0.3 = 0.667(3dp)

Formulating the hypothesis;

HO: u 19.5 That is, the mean reflectometer is less than or equal to 19.5
HI: u > 19.5 that is, the mean reflectometer is greater than 19.5 greater than 19.5

α = 1 - 0.95 = 0.05

The H1, formulated above, implies that the test is one failed.

Thus, t_α,n-1 = t_0.05,25-1

t_0.05,24 = 1.711

The observed to = 0.667 is less then the critical tc = 1.711, and therefore lies in the acceptance region for the null hypothesis (HO). We accept the H0 and conclude that the sample evidence suggests that the mean reflectometer is less than 19.5. the company should therefore reject the new plant.

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QUESTION 2c

Q Explain two advantages and two disadvantages of decision trees as used in decision theory.

A

Solution

Advantages of Decision Trees

Simplicity: Decision trees are easy to understand and interpret, making them a valuable tool for both experts and non-experts in decision-making processes.
Monetary simulation: Allows monetary values to be incorporated into decision trees to help identify and isolate the best alternative course of action.
Visualization: Decision trees provide a visual representation of decision-making processes, which aids in the communication and comprehension of complex decision scenarios.
Handling Non-Linearity: Decision trees can model complex, non-linear relationships in the data, which may be challenging for other algorithms that assume linearity.
No Assumption of Data Distribution: Decision trees do not require assumptions about the distribution of data, making them suitable for various types of datasets.
Doesn't require scaling of data

Disadvantages of Decision Trees

Overfitting: Decision trees are prone to overfitting, especially when they are deep and capture noise in the training data. This can lead to poor generalization on new, unseen data.
Instability: Decision trees are sensitive to variations in the training data, and small changes can result in different tree structures. This instability can impact the reliability of the model.
Data Partitioning Bias: Decision trees may create biased splits in the data, particularly when certain classes are dominant, leading to imbalanced branches and potential performance issues.
Difficulty in Capturing XOR-like Relationships: Decision trees struggle to capture exclusive OR (XOR) relationships between features, as they rely on simple splits based on threshold values.
Complexity: Extensive knowledge and deeper understanding of decision theory is required.
Scaling: As the amount of data increases, building decision trees becomes more complex
Time consuming

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QUESTION 2d

Q (i) The average number of customers in the system.

(ii) The average time spent by a customer in the system.

(iii) The average number of customers in the queue.

(iv) The utilisation factor of the service utility.

A

Solution

(i) The average number of customers in the system.

L _s

=

λ

μ - λ

λ = 2 customers per hour
μ = 3 customers per hour

L _s

=

2

3 - 2

2customer

(ii) The average time spent by a customer in the system.

W_s

=

1

μ - λ

W_s

=

1

3 - 2

1 hour per customer

(iii) The average number of customers in the queue.

L_q

=

λ²

μ(μ - λ)

∴

L_q

=

(2)²

3(3 - 2)

1.3333

(iv) The utilisation factor of the service utility.

e = λ / μ

2 / 3

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QUESTION 3a

Q In the context of time series analysis, describe three differences between "additive" and "multiplicative" models.

A

Solution

Additive model		Mltiplicative model
1	Represents the components of time series i.e. seasonality(s) trends(t) cyclical(c) and residual (r) as The sum: Y = S + T + C + R	Represents the components as the product of time series i.e Y = S x T x C x R
2	Changes in the time series data is usually expressed interms that are absolute	The changes in data are expressed in percentages
3	Is used where seasonal component(s) which over time tends to be constant.	Used where seasonality component (s) tends to change over time.
4	Considers the time series components as being independent of each other.	Treats the components as being interdependent of each other.

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QUESTION 3b(i)

Q Illustrate, using a graph, the number of website hits against the amount spent in website promotion.

Comment on any relationship between website hits and the extent of promotion.

A

Solution

Website promotion (Ksh."000")

Website Hits(No.of visitors)

It can be seen from the graph that there is no clear relationship between click-through rate and the degree of promotion on direct mortgage websites.

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QUESTION 3b(ii)

Q Calculate the correlation coefficient and give an interpretation to its value.

A

Solution

r²

=

nΣxy - ΣxΣy

√(nΣx² -(Σx)²) x (nΣy² - (Σx)²)

X	Y "000"	XY	X²	Y²
25	1	25	625	1
24	1.2	28.8	576	1.44
56	1.6	89.6	3,136	2.56
54	1.4	75.6	2,916	1.96
55	1.2	66	3,025	1.44
58	1.8	104.4	3,364	3.24
∑X = 272	∑Y = 8.2	∑XY = 389.4	∑X² = 13,642	∑Y² = 11.64

n=6

r

=

6(389.4) - 272(8.2)

√[6(13,642) - 272] x [6(11.64) - 8.2²]

r

=

106

√20,456.8

r = 0.7411

The value of "T" above is positive and therefore indicating a strong positive correlation between the website hits and the extent of promotion.

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QUESTION 3b(iii)

Q Determine the regression line.

A

Solution

y = a + bx

a

=

∑y

n

-

b∑x

n

b

=

n∑xy

n∑x²

-

∑x∑y

( ∑x )²

b

=

6(389.4) - 272(8.2)

6(13642) - (272) ²

106/7868 = 0.0135

a = 8.2/6 - (0.0135(272))/6

= 1.3667 - 0.612 = 0.7547

Therefore regression line y = 0.7547 + 0.0135x

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QUESTION 4a(i-ii)

Q (i) Explain the meaning of a "transition matrix'

(ii) Outline two features of a transition matrix.

A

Solution

(i) Meaning of a "transition matrix"

It is a matrix of probability that an element in a markov chain, will move from one state to another.

ii) The features of a transition matrix

➢The sum of the elements of each row of the transition matrix is 1
➢All transition matrices are random i.e. probability
➢ The elements of the matrix are either positive or null, that is to say not negative.

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QUESTION 4b

Q Calculate the output matrix K.

A

Solution

K = T²P^-1

where;

T - is an input matrix

P^-1 - is an inverse of matrix P.

K - is an output matrix.

Given that T

=

[

24

20

]

&

P =

[

24

20

]

T²

=

[

24

20

]

x

[

24

20

]

[

4+88+0

4+08+0

]

T²

=

[

128

48

]

P

=

[

17

04

]

P^-1

=

1

|P|

(Adjoint P)

|P| = (1 x 4) - (0 x 7) = 4 - 0 = 4

P^-1

=

1

4

[

4-7

01

]

K

=

1

4

[(

128

48

)

(

+4-7

0+1

)

]

=

1

4

[

48+0-84+8

16+0-28+8

]

=

1

4

[

48-76

16-20

]

K

=

[

1219

4-5

]

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QUESTION 4c(i-ii))

Q (i) Construct a probability transition matrix of the switching probabilities.

(ii) Construct a vector to represent the initial market share in percentages.

A

Solution

(i) Construct a probability transition matrix of the switching probabilities.

0

M

N

S

[

M

0.6

0.4

0.1

N

0.2

0.5

0.1

S

0.2

0.1

0.8

]

Where;

M = Meta, N = Nzuri, S = Safi

(ii) Construct a vector to represent the initial market share in percentages.

M	N	S
(40	40	20)

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QUESTION 4c(iii)

Q Calculate a new market share a week after the current market share.

A

Solution

(40 40 20)

[

0.6

0.4

0.1

0.2

0.5

0.1

0.2

0.1

0.8

]

(24+16+2 8+20+2 8+4+16) = (42 30 28)

Therefore, the market share will be 42%, 30% and 28% for meta, Nzuri and safi respectively, a week after the current market share.

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QUESTION 4d(i)

Q Predicting business outcome.

A

Solution

Predicting business outcome.

They help managers in evaluating scenarios for forecasting purposes.

Helps organizations implement strategies to improve and enhance employee performance to achieve business goals

Enables organizations to prepare employees for complex business environments through business process modeling.

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QUESTION 4d(ii)

Q Managing business risks.

A

Solution

Managing business risks

The simulation itself is used as a tool to assess and analyze the uncertainty and risk surrounding the business

The simulation model helps managers generate random numbers for effective scenario analysis of risks in the business.

They are used to identify contingencies and facilitate effective cost management.

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QUESTION 5a(i-iv)

Q Advise the businessman on the best alternative under:

(i) Maximin criterion.

(ii) Maximax criterion.

(iii) The Hurwitz criterion, assuming a degree of optimism of 0.6.

(iv) Laplace criterion.

A

Solution

(i) Maximin criterion.

Alternative	Minimum payoff Sh."million"
A	-12
B	-6
C	0	(maximin)

Therefore, the business man should select alternative c resulting in a payoff of sh. 0 in scenario.

(ii) Maximax criterion.

Alternative	Maximum payoff Sh."million"
A	6
B	16	(maximum)
C	12

Therefore, the businessman should choose alternative B resulting in the maximum pay off Sh. 16 million, under this criterion.

(iii) The Hurwitz criterion, assuming a degree of optimism of 0.6.

It is based on weighted payoff where:
Weighted payoff = α (maximum payoff) + 1-d (minimum payoff)

1 - α = 1 - 0.6 = 0.4

Alternative		Minimum payoff Sh."million"
A	0.6(6) + 0.4(-12)	- 1.2
B	0.6(16) + 0.4(-6)	7.2	(maximum)
C	0.6(12) + 0.4(0)	7.2	(maximum)

Therefore, the business man should select either alternative B or C, resultin in the maximum weighted payoff of shs. 7.2 million each.

(iv) Laplace criterion.

It is based on the average payoff

Where;

Average payoff

=

∑(payoffs) for each alternative

Number of states of nature

Alternative		Maximum payoff Sh."million"
A	(6 + 2 - 2 - 12 + 4) ÷ 5	- 0.4
B	(-6 + (-3) + 10 + 16 + 0) ÷ 5	3.4
C	(12 + 8 + 4 + 0 + 6) ÷ 5	6	(maximum)

Therefore, the businessman should select alternative C, resulting in the maximum average payoff Sh. 6 million

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QUESTION 5b(i-ii)

Q (i). Formulate and solve the above Linear programming model graphically.

(ii). Determine the maximum selling price of Product Y at which the solution in (b) (i) above would still remain optimal.

A

Solution

(i). Formulate and solve the above Linear programming model graphically.

Let x and y represent the number of units of products x and y produced and sold respectively.

Objective:

To determine the number of units of each product x and y to be produced and sold in order to maximize profit

Objective function (z)
Z_max = 24x + 32y

Subject to constraints:

1. Material B:

(80/20)x +(80/20)y ≤ 240

x + y ≤ 60

2. Direct labour hours

(64/16)x + (32/16)y ≤ 100

2x + y ≤ 100

3Packaging time

(24/24)x + (48/24)y ≤ 100

x + 2y ≤ 100

4. Non-negativy constraints

x ≥ 0
y ≥ 0

Hence the optimal output is

x = 20 units of product x
y = 40 units of product y

∴ Z_max = 24(20) + 32(40)

1,760

Product Y

ii) Determining the maximum selling price of product y at which the solution in b, i) above would still remain optimal

Let maximum selling price of product y be represented by p"

:. Total sales = total cost + total profit.

Optimal output production of "y" = 40 units as obtained under (b) (i)above

Total cost = 40 (300 + 28)

= 40 x 328 = Sh 13,120

Total profit 40 x 32 = Sh 1,280

∴ Total sales of y = 13,120 + 1,280 = Sh. 14,400

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QUESTION 5c(i-ii)

Q (i) The total cost function using the least squares method.

(ii) If each bread is sold at Sh.50, predict the break-even number of units of bread.

A

Solution

(i) The total cost function using the least squares method.

y = a + bx

X	Y	XY	X²
45	46	2,070	2,025
42	43.2	1,814.4	1,764
55	46.6	2,563	3,025
43	48	2,064	1,849
60	56.4	3,384	3,600
40	44.8	1,792	1,600
48	46.2	2,217.6	2,304
53	50.6	2,681.8	2,809
36.6	40.2	1,471.32	1,339.56
34	33	1,122	1,156
∑x=456.6	∑y=455	∑xy=21,180	∑x²=21,471.56

b

=

n∑xy

n∑x²

-

∑x∑y

( ∑x )²

b

=

10(21,180.12) - (456.6 x 455)

10(21,471.56) - (456.6) ²

4,048.2 / 6,232.04

∴ b = 0.6496

a

=

∑y

n

-

b∑x

n

a

=

455

10

-

0.6496(456.6)

10

45.5 - 29.6607 = 15.8393

∴ y = 15.8393 + 0.6496x

(ii) If each bread is sold at Sh.50, predict the break-even number of units of bread.

Break even units = Total fixed costs/Contribution per unit

∴X(in hundreds) = (15,839.3 x 1,000) / (50 - 0.6496)

15,839.3 / 49.3504 = 320.96

Hence the break even number of units of bread

320.96 x 100 = 32,096 units

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CPA Foundation Leval Quantitative Analysis November 2019 Suggested solutions

Solution

Description of 'Business analytics'

Solution

A binomial distribution.

A Poisson distribution.

Solution

Solution

Solution

(ii) The best time to end the production.

iii) The total profit based on the result in (a) (ii) above

Revision Made Easy. 🚀

Solution

Revision Made Easy. 🚀

Solution

Advantages of Decision Trees

Disadvantages of Decision Trees

Revision Made Easy. 🚀

Solution

(i) The average number of customers in the system.

(ii) The average time spent by a customer in the system.

(iii) The average number of customers in the queue.

(iv) The utilisation factor of the service utility.

Revision Made Easy. 🚀

Solution

Revision Made Easy. 🚀

Solution

Revision Made Easy. 🚀

Solution

Revision Made Easy. 🚀

Solution

Solution

(i) Meaning of a "transition matrix"

ii) The features of a transition matrix

Revision Made Easy. 🚀

Solution

Revision Made Easy. 🚀

Solution

(i) Construct a probability transition matrix of the switching probabilities.

(ii) Construct a vector to represent the initial market share in percentages.

Revision Made Easy. 🚀

Solution

Revision Made Easy. 🚀

Solution

Predicting business outcome.

Revision Made Easy. 🚀

Solution

Managing business risks

Revision Made Easy. 🚀

Solution

(i) Maximin criterion.

(ii) Maximax criterion.

(iii) The Hurwitz criterion, assuming a degree of optimism of 0.6.

(iv) Laplace criterion.

Revision Made Easy. 🚀

Solution

(i). Formulate and solve the above Linear programming model graphically.

ii) Determining the maximum selling price of product y at which the solution in b, i) above would still remain optimal

Revision Made Easy. 🚀

Solution

(i) The total cost function using the least squares method.

(ii) If each bread is sold at Sh.50, predict the break-even number of units of bread.

Revision Made Easy. 🚀

Comments on CPA past papers with answers:

Economics

Introduction to Law and Governance

Quantitative analysis

Comments on:

CPA
Foundation Leval
Quantitative Analysis November 2019
Suggested solutions