QUANTITATIVE ANALYSIS NOVEMBER 2018 - CPA SOLUTION|REVISION KITS|KASNEB

QUESTION 1a

Q (i) Opportunity loss.

(ii) Expected value of perfect information.

Solution

(i) Opportunity loss

This is the cost of a missed opportunity or a possible profit that was not realized because a decision was made that prevented the investor from realizing that profit.

(ii) Expected value of perfect information

This is the price that a person would be willing to pay to have access to perfect knowledge.

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QUESTION 1b

Q Outline three assumptions ofthe transpotflation model.

Solution

Assumptions of Transportation model

1. Transportation of commodity: The model assumes that all items can be easily moved from one place to another.

2. Certainty of per unit transportation cost: The cost of shipping a product from one place to another is determined.

3. Independent cost per unit: The amount transferred from sources to destinations does not affect the cost per unit.

4. The quantity of units transported determines the route's transportation costs..

5. The goal is to reduce the organization's overall transportation costs.

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QUESTION 1c

Q Determine whether there is any significant difference between the average monthly salaries of employees working in the two departments. (Use a significance level of 5 per cent).

Solution

n₁ = 29

n₂ = 24

x̄₁ = 260,000.

x̄₂ = 310,000

S₁ = 25,000

S₂ = 30,000

S_p

√

(n₁ - 1)S₁² + (n₂-1)S₂²

n₁ + n₂ - 2

√

(29 - 1)25,000² + (24 - 1)30,000²

29 + 24 - 2

√749,019,607.8

27,368.22

∴

S_x̄₁

S_p

√n₁

27,368.22

√29

5,082.15

S_x̄₂

S_p

√n₂

27,368.22

√24

5,586.51

S_{x̄₁ - x̄₂}

√ S²_x̄₁ + S²_x̄₂

S_{x̄₁ - x̄₂}

√ 5,082.15² + 5,586.51²

= 7,552.31

t_calculated

x̄₁ - x̄₂

S_{x̄₁ - x̄₂}

(260,000 - 310,000) / 7,552.31 = 6.62

From T table 5% value with (n₁ + n₂ - 2) = (29 + 24 - 2) = 51 degree of freedom is 2.000

Since calculated value is greater than table value, then there is a significant difference in salaries of the two departments.

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QUESTION 1d

Q (i) Present the above information in a Venn diagram.

(ii) The number of students undertaking all the three courses.

(iii) The number of students undertaking only one course

Solution

(i) Present the above information in a Venn diagram.

Assume accounting represented by A

Assume Computing represented by C

Assume Driving represented by D

Students doing all courses

500 = 329 + 186 + 295 - 83 - 217 - 63 + x

500 = 810 - 363 + x

500 = 447 + x

x = 500 - 447 = 53

A(Accounting)

C(Computing)

A82

C93

164

D68

D(Driving)

Students doing all the three courses = 53

Students doing accounting and computing but not driving = 83 - 53 = 30

Students undertaking accounting and driving but not computing = 217 - 53 = 164

Students undertaking both computing and driving courses but not accounting = 63 - 53 = 10

Students taking accounting but not computing = 329 - 186 = 143

(ii) The number of students undertaking all the three courses

53 students

iii) The number of students undertaking only one course

82 + 93 + 68 = 243 students

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QUESTION 2a

Q (i) The total profit function.

(ii) The maximum profit.

Solution

(i) The total profit function

Total cost function integral of MC

[1 / 2]² - 100x + 250 (integral in "000")

Quadratic function form

Y = a + bx + cx²

1,600 = a + 20b + 400c..........(i)

3,200 = a + 40b + 1,600c.......(ii)

4,800 = a + 60b + 3,600c...... (iii)

1,600 = a + 20b + 400c

a = 1,600 - 20b - 400c............(iv)

Substitute equation (iv) in (ii) and (iii)

3,200 = 1,600 - 20b - 400c + 40b + 1,600c

3,200 = 1,600 - 20b + 40b - 400c + 1,600c

1,600 = 20b + 1,200c... (v)

4,800 = 1,600 - 20b - 400c + 60b + 3,600c

4,800 - 1,600 = -20b + 60b - 400c + 3,600c

3,200 = 40b + 3,200c

1,600 = 20b + 1,200c....x2

3,200 = 40b + 3,200c....x1

3,200 = 40b + 2,400c

3,200 = 40b + 3,200c

1,600 = a + 20b + 400c.......i

3,200 = a + 40b + 1,600c....ii

4,800 = a + 60b + 3,600c....iii

Create equation (iv) by subtracting (ii)-(1)

1,600 = 20b + 1,200c

Create equation (v) by subtracting (iii)-(ii)

1,600 = 20b + 2,000c

800c / 800 = 0 / 800

C = 0

1,600 = 20b + 1,200(0)

20b / 20 = 1,600 / 20

b = 80

1,600 = a + (20 x 80) + (400 x 0)

1,600 = a + 1,600

a = 0

Revenue function = 80x

Profit function = TR - TC

= 80x - (0.5x² - 100x + 250)

= 80x - 0.5x² + 100x - 250

= 180x - 0.5x² - 250

(ii) The maximum profit.

Δπ / Δx = 180 - x = 0

x = 180

= (180 x 180) - 0.5 x 180² - 250

= 15,950 × 1,000

shs. 15,950,000

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QUESTION 2b(i)

Q The market shares of the three companies as at 31 December 2018.

Solution

[

0.90

0.03

0.07

0.10

0.75

0.15

0.05

0.15

0.80

]

[0.33 0.33 0.33]

[

0.90

0.03

0.07

0.10

0.75

0.15

0.05

0.15

0.80

]

x = (0.33 x 0.90) + (0.1 x 0.33) + (0.33 x 0.05) = 0.3465

y = (0.33 x 0.03) + (0.75 x 0.33) + (0.15 x 0.33) = 0.3069

z = (0.33 x 0.07) + (0.33 x 0.15) + (0.80 x 0.33) = 0.3366

Market share 31 December 2018

[0.3465 0.3069 0.3366]

[

0.90

0.03

0.07

0.10

0.75

0.15

0.05

0.15

0.80

]

x = (0.3465 x 0.90) + (0.3069 x 0.10) + (0.3366 x 0.05) = 0.3594

y = (0.3465 x 0.03) + (0.3069 x 0.75) + (0.3366 x 0.15) = 0.2911

z = (0.3465 x 0.07) + (0.3069 x 0.15) + (0.3366 x 0.80) = 0.3396

Market share 31st December 2018

x = 35.94%

y = 29.11%

z = 33.96%

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QUESTION 2b(ii)

Q The long run market shares of the three companies.

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QUESTION 3a

Q (i). Level of significance.

(ii). Region of rejection.

Solution

(i). Levels of significance

➢ Another name for it is Alpha.

➢ It is indicated as α

Significance level refers to the probability that the null hypothesis will be rejected when it is true.

(ii). Region of rejection

This refers to a set of factors that would lead the researcher to reject the null hypothesis.

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QUESTION 3b

Q Summarise three factors that determine the size of the Pearson product moment correlation coefficient.

Solution

Factors that determine the size of the Pearson product moment correlation coefficient

1. Linearity

➢ If a curvilinear relationship exists Pearson r will be sustainability small.

➢ If there is a linear relationship between the two variables under study, then Pearson r will assume a large value.

2. Size of the group

➢ It has zero impact on size of the correlation coefficient

3 . Homogeneity of the group:

➢ As a group under study becomes increasingly homogeneous, then Pearson r decreases.

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QUESTION 3c

Q (i). The least squares regression function relating the direct labour hours to the total overhead cost.

(ii). The coefficient of determination.

(iii). Comment on the results obtained in (c) (ii) above.

Solution

(i). The least squares regression function relating the direct labour hours to the total overhead cost.

y	x	XY	X²	Y²
14,250	856	12,198,000	732,736	203,062,500
13,000	536	6,968,000	287,296	169,000,000
13,000	640	8,320,000	409,600	169,000,000
12,500	600	7,500,000	360,000	156,250,000
13,250	680	9,010,000	462,400	175,562,500
13,750	808	11,110,000	652,864	189,062,500
∑y = 79,750	∑x = 4,120	∑xy = 55,106,000	∑x²=2,904,896	∑y² = 1,061,937,500

nΣxy - ExΣy

nΣx² - (Ex)²

(7 x 55,106,000) - (79,750 x 4,120)

(7 x 2,904,896) - 4,120²

57,172,000

3,359,872

= 17.02

∑y

- b

∑x

79,750

- 17.02 x

4,120

11,392.86 - 10,017.49 = 1,375.37

Regression equation

y = 1,375.37 + 17.02x

(ii). The coefficient of determination.

n∑xy - ∑ x ∑y

√n∑x² - (∑x)² x √ n∑y²+(∑y)²

(7 x 55,106,000) - (4,120 x 79,750)

√(7 x 2,904,896) - 4,120² x √(7 x 1,061,937,500) + 79,750²

= 0.9520

r² = 0.9520²

r² = 0.9063

(iii) Comment

90.63% variation in total overhead costs can be explained by the change direct labour hours.

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QUESTION 4a

Q The adjusted seasonal component for each of the four quarters, using the multiplicative model.

Solution

Year	Quartr	Students y	4 months moving total	4 months moving Average	Annual moving Average	y/t
2015	1	70
	2	100	310	77.5
	3	80	290	72.5	75.0	1.07
	4	60	230	57.5	65.0	0.92
2016	1	50	270	67.5	62.5	0.80
	2	40	290	72.50	70.0	0.57
	3	120	330	82.5	77.5	1.55
	4	80	360	90,0	86.5	0.93
2017	1	90	310	77.5	83.75	1.07
	2	70	270	67.5	72.5	0.97
	3	70	240	60	63.75	1.10
	4	40	270	67.5	63.75	0.63
2018	1	60	330	82.5	75.0	0.80
	2	100
	3	130

Adjusted Seasonal Component
Year/Quater	1	2	3	4
2015			1.07	0.92
2016	0.8	0.57	1.55	0.93
2017	1.07	0.97	1.40	0.63
2018	0.8
Total	2.67	1.54	3.72	2.48
Average	0.89	0.77	1.24	0.83	3.73
Adjustment	1.07	1.07	1.07	1.07
Adjusted seasonal component	0.95	0.82	1.33	0.89

Normalisation ratio = 4/3.73 = 1.07

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QUESTION 4b

Q Estimate the enrollment of students in each of the four quarters of year 2019 using the simple least squares method.

Solution

Simple least square
Year	Quater	x	Students	ASI	y	x²	xy
2015	1	1	70	0.95	73.68	1	73.68
	2	2	100	0.82	121.95	4	243.9
	3	3	80	1.33	60.15	9	180.45
	4	4	60	0.89	67.42	16	269.68
2016	1	5	50	0.95	52.63	25	263.15
	2	6	40	0.82	48.78	36	292.68
	3	7	120	1.33	90.23	49	631.61
	4	8	80	0.89	89.89	64	719.12
2017	1	9	90	0.95	94.74	81	852.66
	2	10	70	0.82	85.37	100	853.7
	3	11	70	1.33	52.63	121	578.93
	4	12	40	0.89	44.94	144	539.28
2018	1	13	60	0.95	63.16	169	821.08
	2	14	100	0.82	121.95	196	1,707.3
	3	15	130	0.33	97.74	225	1,466.1
		∑=120			∑y= 1,165.26	∑x² = 1,240	∑xy=9,493.32

nΣxy - ExΣy

nΣx² - (Ex)²

(15 x 9,493.32) - (120 x 1,165.26)

(15 x 1,240) - 120²

2,568.6 / 4,200 = 0.61

∑y

- b

∑x

1,165.23

- 0.61 x

120

= 77.684 - 4.88 = 72.804

Regression equation

y= 72.804+0.61x

Estimated student enrollment

2019

Quarter 1 2019 = 72.804 + 0.61 x 17 = 83.17 students

Quarter 2 2019 = 72.804 + 0.61 x 18 = 83.784 students

Quarter 3 2019 = 72.804 + 0.61 x 19 = 84.394 students

Quarter 4 2019 = 72.804 + 0.61 x 20 = 85.004 students

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QUESTION 5a

Q Probability theory

(i) Mutually exclusive events.

(ii) Independent events.

(iii) Joint probability.

(iv) Conditional probability.

Solution

(i) Events that are mutually exclusive cannot occur at the same time. For instance, when you flip a coin once, you either get the head or the tail.

(ii) Independent events are events that can happen at the same time, for example, when you flip a coin twice, it's heads or tails.

(iii) Joint probability is the likelihood of two events occurring together and at the same point in time . It's is the probability of event Y occurring at the same time that event X occurs.

(iv) Conditional probability refers to the likelihood of event B happening given that event A has already happened. To calculate conditional probabilities, we employ the Baye's rule.

The formula for conditional probability is:

P(B / A)

P(A / B) x P(B)

P(A)

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QUESTION 5b(i)

Q (i). A network diagram for the project.

(ii). The expected duration of the project,

Solution

(i). A network diagram for the project.

Critical path A - C - D - F - G

(ii) The expected duration of the project,

Expected project duration = 20.2 days

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QUESTION 5b(iii)

Q Simulate the durations of the project on the basis of two runs.

Solution

Activity	Expected time in days
A	3 x 0.25 + 4 x 0.5 + 5 x 0.25 = 4.0
B	0.15 x 4 + 5 x 0.3 + 6 x 0.2 + 7 x 0.2 + 8 x 0.15 = 5.9
C	0.2 x 1 + 3 x 0.65 + 0.15 x 5 = 2.9
D	0.8 x 4 + 0.25 x 5 = 4.2
E	0.15 x 3 + 4 x 0.25 + 0.25 x 5 + 6 x 0.35 = 4.5
F	0.2 x 5 + 0.8 x 7 = 6.6
G	0.5 x 2 + 3 x 0.5 = 2.5

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