Introduction to Biostatistics: Main Exam, 27 March 2024

Mount Kenya University · Department of Natural Sciences · Diploma in Dental Technology · DMA1301

Mount Kenya University Introduction to Biostatistics past paper 2023-2024 with model answers for students.

Instructions: ANSWER ALL QUESTION IN SECTION A AND ANY OTHER TWO QUESTIONS IN SECTION B

Section A

1. Distinguish population and sample as used in statistics. (4 marks)

Model answer
  • Population: the entire group of individuals or items that share a common characteristic and about which we want to draw conclusions.
  • Sample: a subset of the population selected for measurement or observation.
  • Population parameters are fixed values (e.g., true mean) that are usually unknown.
  • Sample statistics are calculated from the sample and used to estimate population parameters.

2. Distinguish between ordinal and nominal scale of measurement as used in statistics giving example of each. (4 marks)

Model answer
  • Nominal scale: classifies data into categories without any inherent order (e.g., blood type: A, B, AB, O).
  • Ordinal scale: ranks data in a meaningful order but intervals between ranks are not necessarily equal (e.g., pain intensity: none, mild, moderate, severe).
  • Nominal data can only be counted or expressed as frequencies.
  • Ordinal data allow comparison of relative position but not precise measurement of differences.

3. Explain three limitations of using secondary data. (6 marks)

Model answer
  • Secondary data may be outdated, limiting relevance to current research questions.
  • The original data collection purpose may differ, leading to mismatched variables or definitions.
  • Quality and accuracy cannot be verified, increasing risk of bias or errors.
  • Access restrictions or licensing fees can limit availability.
  • Documentation may be incomplete, making it hard to assess methodology.
  • Lack of control over sampling design can affect representativeness.

4. Explain the following terms as applied in statistics. (6 marks)

  1. i) Random sampling
    Model answer
    • Random sampling involves selecting each member of the population with a known, non‑zero probability, usually equal for all.
    • Selection is performed by a chance mechanism (e.g., lottery, random number table).
    • It reduces selection bias and enhances the likelihood that the sample reflects the population.
    • Enables the use of probability theory to estimate sampling error and construct confidence intervals.
  2. ii) Stratified sampling
    Model answer
    • Stratified sampling divides the population into homogeneous sub‑groups (strata) based on a characteristic (e.g., age, gender).
    • A random sample is then drawn from each stratum, often proportionally to its size.
    • Ensures representation of all key sub‑groups, improving precision of estimates.
    • Reduces sampling variability compared with simple random sampling when strata differ markedly.
  3. iii) Cluster sampling (6 marks)
    Model answer
    • Cluster sampling selects entire groups (clusters) of units, such as schools or villages, rather than individuals.
    • The population is first divided into clusters that are ideally mini‑representations of the whole.
    • A random sample of clusters is chosen, and all members within selected clusters are surveyed.
    • Useful when a complete list of individuals is unavailable but a list of clusters exists.
    • Reduces travel and administrative costs in field studies.
    • Increases sampling error if clusters are internally homogeneous, so larger numbers of clusters are often needed.

5. The Marks obtained by nutrition students in a mathematics test are shown in the frequency distribution table below; Marks 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99 students 4 14 18 28 18 8 6 4

  1. Determine;
    1. i) Mean.
    2. ii) Median.
    3. iii) Third and seventh percentile.
    Model answer
    • Mean 55.5
    • Median 55
    • Third percentile 27.5
    • Seventh percentile 32.1

QUESTION TWO

2.

  1. a) An examination of 11 applicants for an accountant post was taken by a finance company. the Marks obtained by the applicant in the reasoning and aptitude test are given below. Calculate spearman's rank correlation coefficient. applicant A B C D E F G H I J K Reasoning test 20 50 28 25 70 90 76 45 30 19 26 Aptitude test 30 60 50 40 85 90 56 82 42 31 49 (10 marks)
    Model answer
    • Rank the reasoning scores (lowest = 1): A2  B8  C5  D3  E9  F11  G10  H7  I6  J1  K4
    • Rank the aptitude scores (lowest = 1): A1  B8  C6  D3  E10  F11  G7  H9  I4  J2  K5
    • Differences (d) and d²: A 1 (1), B 0 (0), C ‑1 (1), D 0 (0), E ‑1 (1), F 0 (0), G 3 (9), H ‑2 (4), I 2 (4), J ‑1 (1), K ‑1 (1)
    • Σd² = 22
    • Spearman’s ρ = 1 ‑ [6·22]/[11(11²‑1)] = 1 ‑ 132/1320 = 0.90
  2. b) The height of football players in various teams was measured and found to have been distributed with mean μ and standard deviation σ A random sample of 100 students was taken and 95% confidence interval for the population mean was found to be between 160 cm and 170cm. i) Estimate the sample mean. ii) Estimate the standard deviation. 99% confidence interval for the population mean. (10 marks)
    Model answer
    • i) Sample mean = (160 + 170)/2 = 165 cm
    • ii) Estimate of σ: SE = (170‑160)/(2·1.96)=2.55; s = SE·√100 = 2.55·10 = 25.5 cm
    • 99 % CI for μ: 165 ± 2.576·2.55 = 165 ± 6.57 → 158.4 cm to 171.6 cm

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