Quantitative Methods — March/April 2024

KNEC Diploma in Sales and Marketing — Module II. Paper code: 2902/204

Independent worked revision answers. These are not official KNEC marking schemes. Every printed question is covered for revision, including all subparts. Retain unrounded values until the final step.

Question 1(a)

Explain five limitations of using census as a method of data collection. (10 marks)

Answer

  1. High cost: contacting and recording every population unit requires substantial spending on staff, transport, systems and supervision.
  2. Time consuming: covering the entire population and processing all returns may take so long that findings become outdated.
  3. Difficult administration: large or geographically dispersed populations require complex coordination and consistent field procedures.
  4. Non-sampling errors remain: omissions, non-response, inaccurate answers and processing mistakes can affect results even though every unit is intended to be covered.
  5. Unsuitable for destructive or inaccessible populations: testing every item may destroy the stock, while very large, mobile or poorly defined populations may be impossible to enumerate fully.

Question 1(b)

The following is the distribution of profits of companies in two sectors: A and B:

Profits, Ksh.000’s Number of Companies: Sector A Number of Companies: Sector B
40–60 5 2
60–80 30 10
80–100 55 72
100–120 18 14
120–140 18 42
140–160 10 20
160–180 5 16
180–200 2 10
  1. (i) mean profit for each sector;
  2. (ii) Combined mean profit for the two sectors

(10 marks)

Answer

Use the class midpoint x to estimate each class’s profit. All amounts in the working table are in Ksh ‘000’.

Profit class Midpoint x fA fAx fB fBx
40–60 50 5 250 2 100
60–80 70 30 2100 10 700
80–100 90 55 4950 72 6480
100–120 110 18 1980 14 1540
120–140 130 18 2340 42 5460
140–160 150 10 1500 20 3000
160–180 170 5 850 16 2720
180–200 190 2 380 10 1900
Total — 143 14350 186 21900
(i) Mean profit for each sector
Sector Working Mean profit
A ΣfAx/ΣfA = 14,350/143 = 100.34965 thousand Ksh 100,349.65
B ΣfBx/ΣfB = 21,900/186 = 117.741935 thousand Ksh 117,741.94
(ii) Combined mean
Step Working
Total estimated profits 14,350 + 21,900 = 36,250 thousand shillings
Total companies 143 + 186 = 329
Combined mean 36,250/329 = 110.182371 thousand shillings = Ksh 110,182.37

The combined mean weights each sector by its number of companies; it is not the simple average of the two sector means.


Question 2(a)

Outline five advantages of using the simple random sampling method in data collection. (10 marks)

Answer

  1. Every population unit has an equal selection probability under a correctly implemented simple random design.
  2. Random selection limits deliberate selection bias by the researcher.
  3. The selection procedure is straightforward when a complete sampling frame is available.
  4. Standard methods can estimate sampling error and construct confidence intervals from the probability sample.
  5. It avoids the need to divide the population into strata or clusters before selection, making it convenient for a reasonably homogeneous population with an accessible frame.

Question 2(b)

The following data shows the distribution of marks scored by employees in a proficiency test:

marks 20 21 22 23 24 25 26 27 28
Number of employees 24 35 20 80 56 40 8 10 5

Determine the:

  1. (i) Range;
  2. (ii) Standard deviation
  3. (iii) Variance

(10 marks)

Answer

Class / value f x fx fx²
20 24 20 480 9,600
21 35 21 735 15,435
22 20 22 440 9,680
23 80 23 1,840 42,320
24 56 24 1,344 32,256
25 40 25 1,000 25,000
26 8 26 208 5,408
27 10 27 270 7,290
28 5 28 140 3,920
Total 278 — 6,457 150,909

N = 278. The table describes the employees’ distribution, so use the population denominator N.

Part Working Answer
(i) Range Highest observed mark − lowest observed mark = 28 − 20 8 marks
Mean used below Σfx/N = 6,457/278 23.226619 marks
(iii) Variance σ² = Σfx²/N − (Σfx/N)² = 150,909/278 − (6,457/278)² 3.362313 marks²
(ii) Standard deviation σ = √3.362313 1.833661 marks

Question 3(a)

The following data relates to different sectors of an economy:

Sector National Income (Ksh. Billions)
Agriculture 400
Industry 300
Trade 200
Other Sectors 300

Draw a pie chart from the information (10 marks)

Answer

Total national income = 400 + 300 + 200 + 300 = Ksh 1,200 billion. Sector angle = (sector income/1,200) × 360°.

Sector Income (Ksh billion) Percentage Sector angle
Agriculture 400 33⅓% 120°
Industry 300 25% 90°
Trade 200 16⅔% 60°
Other sectors 300 25% 90°
Total 1200 100% 360°
Pie chart of national income: Agriculture Ksh 400 billion (120°), Industry Ksh 300 billion (90°), Trade Ksh 200 billion (60°) and Other sectors Ksh 300 billion (90°).
Pie chart of national income: Agriculture Ksh 400 billion (120°), Industry Ksh 300 billion (90°), Trade Ksh 200 billion (60°) and Other sectors Ksh 300 billion (90°).

Question 3(b)

The following data shows the ages of similar machines in a manufacturing firm and their corresponding annual maintenance costs:

Age (years) Annual maintenance cost (Ksh.000’s)
5 52
2 45
4 50
3 45
7 48
12 47
2 48
1 40
  1. (i) Determine the regression equation of annual maintenance cost on age of the machines.
  2. (ii) Estimate the annual maintenance cost for machine of 10 years of Age.

(10 marks)

Answer

(i) Regression equation
x: machine age (years) y: annual maintenance (Ksh ‘000’) x² xy
5 52 25 260
2 45 4 90
4 50 16 200
3 45 9 135
7 48 49 336
12 47 144 564
2 48 4 96
1 40 1 40
36 375 252 1,721
Step Working
Number of observations n = 8
Slope b [nΣxy − (Σx)(Σy)] / [nΣx² − (Σx)²]
Substitute [8(1,721) − 36(375)] / [8(252) − (36)²] = 0.372222
Intercept a ȳ − bx̄ = 46.875 − (0.372222)(4.5) = 45.2
Regression equation ŷ = 45.2 + 0.372222x
(ii) Ten-year-old machine
Step Working
Substitute x = 10 ŷ = 45.2 + 0.372222(10) = 48.922222 thousand shillings
Estimated annual maintenance cost Ksh 48,922.22

Question 4(a)

Explain four limitations of using observation method in the collection of data. (8 marks)

Answer

  1. It mainly records observable behaviour: motives, attitudes and private experiences cannot usually be established reliably by watching alone.
  2. Observer bias can affect what is noticed, recorded and interpreted, reducing consistency between observers.
  3. People may change their behaviour when they know they are being observed, so the record may not reflect normal activity.
  4. It can be slow and expensive: trained observers may need to wait for rare events or watch activities for long periods, and some settings are inaccessible.

Question 4(b)

A class has 30 girls and 20 boys. two students are picked at random without replacement to represent the class in a contest.

  1. (i) Present the information above in a tree diagram.
  2. (ii) Determine the probability of selecting: (I) 2 girls; (II) 2 boys; (III) a boy and a girl.

(12 marks)

Answer

(i) Tree diagram

There are 50 students initially and 49 after the first selection. Because sampling is without replacement, the second-stage probabilities depend on the first selection.

Probability tree without replacement: first girl 30/50 or boy 20/50; after a girl, girl 29/49 or boy 20/49; after a boy, girl 30/49 or boy 19/49.
Probability tree without replacement: first girl 30/50 or boy 20/50; after a girl, girl 29/49 or boy 20/49; after a boy, girl 30/49 or boy 19/49.
(ii) Required probabilities
Event Working Probability
(I) Two girls (30/50)(29/49) 87/245 = 0.355102
(II) Two boys (20/50)(19/49) 38/245 = 0.155102
Girl then boy (30/50)(20/49) 12/49
Boy then girl (20/50)(30/49) 12/49
(III) A boy and a girl P(GB) + P(BG) 24/49 = 0.489796

Both possible orders must be included for “a boy and a girl”. As a check, P(GG) + P(BB) + P(GB) + P(BG) = 1.


Question 5(a)

Explain four disadvantages of regression analysis in business decision making. (8 marks)

Answer

  1. Dependence on data quality: missing, inaccurate or unrepresentative observations can produce misleading coefficients and forecasts.
  2. Model assumptions may fail: a fitted straight line can misrepresent a nonlinear relationship, and omitted factors or unusual observations can distort the estimate.
  3. Association does not establish causation: a regression relationship alone does not show that changing one variable will cause the predicted change in another.
  4. Forecasting can be unreliable outside the observed range or after structural change: past relationships may not continue under new market conditions.

Question 5(b)

The following data shows the production of wheat in tonnes by a farmer for three consecutive years:

Year 1 2 3 4
2019 37 38 37 40
2020 41 34 25 31
2021 25 27 31 29
  1. (i) Determine the trend values using 3-period moving averages.
  2. (ii) Comment on the results in (i) above.

(12 marks)

Answer

(i) Three-period moving averages

Read the four periods of each year from left to right, then continue into the following year. A three-period moving average is placed at the middle period of its three observations. The first and last periods have no centred three-period value.

Middle period Three observations (tonnes) Three-period total Moving average (tonnes)
2019, period 2 37 + 38 + 37 112 37.3333
2019, period 3 38 + 37 + 40 115 38.3333
2019, period 4 37 + 40 + 41 118 39.3333
2020, period 1 40 + 41 + 34 115 38.3333
2020, period 2 41 + 34 + 25 100 33.3333
2020, period 3 34 + 25 + 31 90 30
2020, period 4 25 + 31 + 25 81 27
2021, period 1 31 + 25 + 27 83 27.6667
2021, period 2 25 + 27 + 31 83 27.6667
2021, period 3 27 + 31 + 29 87 29
(ii) Comment

The smoothed production rises from about 37.33 tonnes to 39.33 tonnes around late 2019, then falls substantially through 2020. It levels near 27–28 tonnes and recovers slightly to 29 tonnes by the third period of 2021. Overall, the later smoothed values are lower than those at the beginning; the pattern is not a steady decline in every period.


Question 6(a)

Explain five limitations of the mode as a measure of central tendency. (10 marks)

Answer

  1. It may not exist uniquely: if no value occurs more often than the rest, there is no single most frequent value.
  2. A distribution can have two or more modes, making one representative central value difficult to choose.
  3. It ignores most observations: it depends on the most frequent value or class rather than using the magnitude of every observation.
  4. It can be unstable between samples or change when grouped-data class boundaries are altered.
  5. It has limited algebraic use: modes cannot generally be combined or manipulated to obtain an overall mode from separate groups using simple totals.

Question 6(b)

The following is the expected net cash inflows from a project:

Year 1 2 3 4 5 6 7 8 9
Cash-in flows (Ksh.000’S) 600 750 500 550 550 550 500 450 550

The project’s discount rate is 12%. Calculate the present value of the cash in flows. (10 marks)

Answer

Assume each cash inflow is received at the end of its stated year. Discount each separately using PVₜ = CFₜ/(1 + 0.12)ᵗ. Amounts in the working table are in Ksh ‘000’.

Year t Cash inflow Discount factor 1/(1.12)ᵗ Present value
1 600 0.892857 535.71
2 750 0.797194 597.90
3 500 0.71178 355.89
4 550 0.635518 349.53
5 550 0.567427 312.08
6 550 0.506631 278.65
7 500 0.452349 226.17
8 450 0.403883 181.75
9 550 0.36061 198.34
Total — — 3,036.02
Result Amount
Present value in table units 3,036.024222 thousand shillings
Present value in shillings Ksh 3,036,024.22

The total is calculated from unrounded discounted cash flows. This is the present value of inflows; no initial outlay is given for an NPV calculation.


Question 7(a)

Outline four advantages of using primary data in the research. (8 marks)

Answer

  1. It can be collected specifically for the research question, improving relevance to the required population and variables.
  2. The researcher controls definitions, sampling, measurement instruments and quality checks.
  3. It can provide current information for the period being studied.
  4. The researcher knows how the information was obtained and can seek clarification or collect additional detail when necessary.

Question 7(b)

The following is the distribution of salaries paid to employees in two different industries:

Salaries, Ksh.000’s Number of Employees: Industry A Number of Employees: Industry B
40–50 15 5
50–60 20 4
60–70 25 15
70–80 14 20
80–90 10 30
90–100 16 36
  1. (i) Calculate the Coefficient of variation of the salaries in each industry.
  2. (ii) Comment on the results in (i) above.

(12 marks)

Answer

(i) Coefficients of variation

Use midpoint estimates and the population standard deviation for each industry: CV = (σ/x̄) × 100%.

Salary class Midpoint x fA fAx fAx² fB fBx fBx²
40–50 45 15 675 30375 5 225 10125
50–60 55 20 1100 60500 4 220 12100
60–70 65 25 1625 105625 15 975 63375
70–80 75 14 1050 78750 20 1500 112500
80–90 85 10 850 72250 30 2550 216750
90–100 95 16 1520 144400 36 3420 324900
Total — 100 6820 491900 110 8890 739750
Industry Measure Working and result
A Mean 6,820/100 = 68.2
A Variance 491,900/100 − (6,820/100)² = 267.76
A Standard deviation √267.76 = 16.363374
A CV (16.363374/68.2) × 100% = 23.9932%
B Mean 8,890/110 = 80.818182
B Variance 739,750/110 − (8,890/110)² = 193.421488
B Standard deviation √193.421488 = 13.907605
B CV (13.907605/80.818182) × 100% = 17.2085%
(ii) Comment

Industry B has the lower coefficient of variation, about 17.21% compared with 23.99% for Industry A. Salaries in B therefore vary less relative to their mean and are more consistent. Industry B also has the higher estimated mean salary, about Ksh 80,818.18 compared with Ksh 68,200 for A.