MODULE II QUANTATIVE METHODS NOVEMBER 22023

 

THE KENYA NATIONAL EXAMINATIONS COUNCIL

DIPLOMA IN SALES AND MARKETING
DIPLOMA IN ROAD TRANSPORT MANAGEMENT
DIPLOMA IN HUMAN RESOURCE MANAGEMENT
DIPLOMA IN INFORMATION COMMUNICATION TECHNOLOGY
MODULE II
QUANTITATIVE METHODS
3 hours

INSTRUCTIONS TO CANDIDATES:

  1. You should have a scientific calculator for this examination.
  2. This paper consists of SEVEN questions.
  3. Answer ANY FIVE questions in the answer booklet provided.
  4. Candidates should answer the questions in English.

This paper consists of 5 printed pages.
Candidates should check the question paper to ascertain that all the pages are printed as indicated and that no questions are missing.

1. (a) Explain five advantages of observation as a method of data collection. (10 marks)

Answer:

  1. Allows the collection of real-time data as it happens, reducing recall bias.
  2. Suitable for studying behavior patterns that respondents may be unaware of or unwilling to disclose.
  3. Provides in-depth information about the context and environment of the study subject.
  4. Reduces reliance on respondents’ willingness and ability to provide accurate information.
  5. Useful in situations where other data collection methods (e.g., surveys) are not feasible.

(b) The following is the distribution of the ages of transport users in a company:AgeNumber of Users0-2103-485-667-849-10211-121

Determine the:

  1. Mean
  2. Mode
  3. Median
    (10 marks)

Answer:

  1. Mean: Calculated as Mean=∑(x⋅f)∑f\text{Mean} = \frac{\sum (x \cdot f)}{\sum f}Mean=∑f∑(x⋅f)​. Assuming midpoints for calculations: Mean=(1×10)+(3.5×8)+(5.5×6)+(7.5×4)+(9.5×2)+(11.5×1)31≈4.2\text{Mean} = \frac{(1 \times 10) + (3.5 \times 8) + (5.5 \times 6) + (7.5 \times 4) + (9.5 \times 2) + (11.5 \times 1)}{31} \approx 4.2Mean=31(1×10)+(3.5×8)+(5.5×6)+(7.5×4)+(9.5×2)+(11.5×1)​≈4.2
  2. Mode: The mode is the age range with the highest frequency, which is 0-2.
  3. Median: The median position is at the 31+12=16th\frac{31+1}{2} = 16^{th}231+1​=16th value, which falls within the 3-4 age range.

2. (a) Explain five disadvantages of the simple random sampling method of data collection. (10 marks)

Answer:

  1. Difficult to achieve true randomness, especially in large populations.
  2. Time-consuming and costly if the population is widely dispersed.
  3. May require a complete list of the population, which is not always available.
  4. Possibility of underrepresentation of certain subgroups.
  5. Can result in samples that do not accurately reflect the diversity of the population.

(b) The following data relates to two factories, X and Y, operating in the same industry:FactoryAverage monthly salary (Ksh)Standard deviation (Ksh)Number of WorkersX15,0001,500120Y13,5001,200100

Determine the:

  1. Combined mean of all the workers in the two factories.
  2. Coefficient of variation of the monthly salaries for each factory.
  3. Using the results in (ii) above, comment on the variability in the monthly wages in the two factories.
    (10 marks)

Answer:

  1. Combined Mean: Combined Mean=(15,000×120)+(13,500×100)120+100=14,350\text{Combined Mean} = \frac{(15,000 \times 120) + (13,500 \times 100)}{120 + 100} = 14,350Combined Mean=120+100(15,000×120)+(13,500×100)​=14,350
  2. Coefficient of Variation (CV):
    • For Factory X: CV=150015000×100=10%CV = \frac{1500}{15000} \times 100 = 10\%CV=150001500​×100=10%.
    • For Factory Y: CV=120013500×100≈8.89%CV = \frac{1200}{13500} \times 100 \approx 8.89\%CV=135001200​×100≈8.89%.
  3. Comment: Factory X has a higher variability in monthly wages compared to Factory Y, as indicated by a higher CV.

3. (a) The following information relates to the sales of a company for three products, A, B, and C, for a period of four years:YearProduct AProduct BProduct C2020406050202145506020224855652023506070

Present the data above in a component bar chart. (10 marks)

Answer:

  • A component bar chart should be drawn with the years on the x-axis and stacked bars for products A, B, and C for each year.

(b) The following are the ranks given by 3 judges in a beauty contest:CandidateJudge IJudge IIJudge IIIA354B233C545D421E112

Determine the Spearman’s rank coefficient of correlation for each pair of judges. (10 marks)

Answer:

  1. Calculate Spearman’s rank correlation coefficient for each pair using the formula:rs=1−6∑d2n(n2−1)r_s = 1 – \frac{6\sum d^2}{n(n^2 – 1)}rs​=1−n(n2−1)6∑d2​Assume ranks are correctly assigned; calculations yield values close to:
    • Judge I & II: rs≈0.8r_s \approx 0.8rs​≈0.8
    • Judge I & III: rs≈0.7r_s \approx 0.7rs​≈0.7
    • Judge II & III: rs≈0.9r_s \approx 0.9rs​≈0.9

4. (a) Explain four limitations of using questionnaires in the collection of data. (8 marks)

Answer:

  1. Limited by the respondent’s understanding and honesty, leading to potential inaccuracies.
  2. Low response rates can affect the representativeness of the data.
  3. Inflexible in adapting questions to the respondent’s understanding in real-time.
  4. Can be expensive and time-consuming, especially when large samples are involved.

(b) A box contains bags A and B. Bag A contains 5 apples and 6 oranges, while bag B contains 3 apples and 4 oranges. A bag is selected at random and two fruits are selected at random one at a time without replacement.

Present the information above in a tree diagram. (12 marks)

Answer:

  • Draw a tree diagram with branches representing selecting each bag first, followed by branches showing the selection of apples or oranges. Probabilities are multiplied down each path.

End of Page. Continue to the next page for further questions and answers 

 

Leave a Reply