Introduction to Biostatistics: Main Exam, 28 July 2021

1.

1. a) State five types of probability sampling techniques. (5 marks)
   Model answer

   • Simple random sampling
   • Systematic sampling
   • Stratified random sampling
   • Cluster sampling
   • Multistage sampling
2. b) Explain the importance of biostatistics to Health studies. (5 marks)
   Model answer

   • Provides quantitative tools to design studies and determine appropriate sample sizes.
   • Enables objective measurement of disease frequency and risk factors.
   • Facilitates data analysis, hypothesis testing and interpretation of results.
   • Supports evidence‑based decision making and policy formulation.
   • Allows assessment of intervention effectiveness and health outcomes.
3. c) Explain different types of modes. (3 marks)
   Model answer

   • Nominal mode – most frequently occurring category (e.g., blood type O).
   • Ordinal mode – most common rank order (e.g., most patients reporting ‘moderate’ pain).
   • Continuous mode – value(s) with highest frequency in a grouped distribution (e.g., age 30‑35 years interval).
4. d) Distinguish between standard deviation and variance. (2 marks)
   Model answer

   • Standard deviation measures the average distance of each observation from the mean (in original units).
   • Variance is the square of the standard deviation; it expresses dispersion in squared units.
5. e) Using the following values 40, 50, 60, 70,80,90,100,60,65,72 Calculate;
   1. i) Mean (2 marks)
   2. ii) Mode (2 marks)
   3. iii) Median (2 marks)
   4. iv) Mean absolute deviation (4 marks)
   Model answer

   • Mean
   • Sum = 40+50+60+70+80+90+100+60+65+72 = 687
   • n = 10
   • Mean = 687/10 = 68.7
   • Mode
   • Value occurring most often = 60 (appears twice)
   • Median
   • Ordered data: 40,50,60,60,65,70,72,80,90,100
   • Median = (5th + 6th)/2 = (65+70)/2 = 67.5
   • Mean absolute deviation (MAD)
   • Deviations from mean: |40‑68.7|=28.7, |50‑68.7|=18.7, |60‑68.7|=8.7, |70‑68.7|=1.3, |80‑68.7|=11.3, |90‑68.7|=21.3, |100‑68.7|=31.3, |60‑68.7|=8.7, |65‑68.7|=3.7, |72‑68.7|=3.3
   • Sum of absolute deviations = 137.0
   • MAD = 137.0/10 = 13.7
6. f) A machine produces the following number of defects in each successive period of five minutes: 14 21 24 22 27 24 26 5 12 22 23 21 22 28 16 23 23 25 24 5
   Model answer

   • Mean number of defects
   • Total defects = 407
   • Observations = 20
   • Mean = 407/20 = 20.35 defects per 5‑minute period
   • Range
   • Minimum = 5, Maximum = 28
   • Range = 28 − 5 = 23 defects
7. g) Construct a frequency distribution from these data using seven class intervals of equal width.
   Model answer

   • Insufficient data: The raw data values are not provided, so a frequency distribution cannot be constructed.
   • If data were given: Determine the range, divide by 7 to get class width (round up), set lower class limit, then create seven equal‑width intervals and tally frequencies.
   • Example format:
   • Class interval 1: ___ – ___ Frequency = ___
   • Class interval 2: ___ – ___ Frequency = ___
   • …
   • Class interval 7: ___ – ___ Frequency = ___
   • More answers: Use the formula Width = Ceil[(Maximum‑Minimum)/7] and ensure intervals are contiguous with no gaps.

2.

1. a) The following data pertain to Marks obtained by 120 students of class level in biostatistics. Marks 30 - 39 40 - 49 50 - 59 60 - 69 70 - 79 80 - 89 Number of students 1 11 12 43 11 3 Calculate (i) the mean (ii) the median (iii) Mode (iv) the Interquartile deviation (v) standard deviation
   Model answer

   • Mean: 62.3
   • Calculation: Σf·midpoint = 5 044.5; N = 81; Mean = 5 044.5 ÷ 81 ≈ 62.3
   • Median: 63.8
   • Position = (N+1)/2 = 41st value; lies in 60‑69 class.
   • Median = L + [(N/2 – cf_prev)/f]·w = 60 + [(40.5‑24)/43]·10 ≈ 63.8
   • Mode: 64.9
   • Modal class = 60‑69.
   • Mode = L + [(f1‑f0)/(2f1‑f0‑f2)]·w = 60 + [(43‑12)/(86‑12‑11)]·10 ≈ 64.9
   • Interquartile deviation: 5.85
   • Q1 = 56.9 (position 20.25 in 50‑59 class)
   • Q3 = 68.6 (position 60.75 in 60‑69 class)
   • IQR = Q3‑Q1 = 11.7; IQD = 11.7/2 ≈ 5.85
   • Standard deviation: 8.4
   • Σf·x² = 319 980.25; Variance = (Σf·x² / N) – Mean² = 3 950.4 – 3 880.3 ≈ 70.1
   • SD = √70.1 ≈ 8.4