Introduction to Biostatistics: Main Exam, 19 November 2018

1.

1. a) Given the data 27.9, 34.7, 54.4, 18.92, 47.6, 39.68. Determine the i. Mean ii. Median (3 marks)
   Model answer

   • Mean: 35.21
   • Median: 34.7
2. b) Determine the geometric mean of the following data 3, 5, 6, 6, 7, 10, 12 (4 marks)
   Model answer

   • Geometric mean = √(3×5×6×6×7×10×12) ≈ 6.77
3. c) State two advantages and two disadvantages of scatter diagrams in studying correlation. (4 marks)
   Model answer

   • Advantages:
   • Shows direction and strength of relationship
   • Easy to identify outliers
   • Disadvantages:
   • Cannot quantify the exact relationship
   • Sensitive to extreme values
4. d) Arrange the following data in the form of stem and leaf diagram 2.97, 3.81, 2.54, 2.01, 3.49, 3.09, 1.99, 2.31, 2.22. (4 marks)
   Model answer

   • Stem‑and‑Leaf:
   • 1 | 9
   • 2 | 01 22 31 54 97
   • 3 | 09 49 81
5. e) A company manufactures a device which contains three components A, B and C. The device fails if any of these components fail and the company offers to its customers a full money back warranty if the product fails within one year. The company has assessed the probabilities of each of the components lasting at least a year as 0.98, 0.99 and 0.95 for A,B and C respectively. The three components within a single device are considered to be independent. Consider a single device chosen at random. Calculate the probability that all three components will last for at least 5 years; the device will be returned for a refund. (5 marks)
   Model answer

   • Probability all last ≥5 years = 0.98×0.99×0.95 = 0.9211
   • Device fails within 1 year = 1−0.9211 = 0.0789
6. f) Plot a scatter plot for this data X | 68 | 64 | 75 | 50 | 64 | 80 | 75 | 40 | 55 | 64 y | 62 | 58 | 68 | 45 | 81 | 60 | 68 | 48 | 50 | 5 (5 marks)
   Model answer

   • Scatter plot shows points at (68,62), (64,58), (75,68), (50,45), (64,81), (80,60), (75,68), (40,48), (55,50), (64,5). Labels: X‑axis: X, Y‑axis: Y
7. g) A random sample of 12 components has a mean weight of 50 grams and standard deviation of 9 grams, determine the 95% confidence interval for the population mean. (5 marks)
   Model answer

   • Degrees of freedom df = 12‑1 = 11
   • t value for 95 % CI (two‑tailed) ≈ 2.201
   • Margin of error = t·s/√n = 2.201 × 9/√12 ≈ 5.72 g
   • 95 % CI = 50 g ± 5.72 g → (44.28 g, 55.72 g)

2.

1. a) The table shows the frequency distribution of 100 items Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 frequency | 12 | a | b | c | 18 Given that the median and the mode are equal and are 50 determine the missing frequencies (10 marks)
   Model answer

   • a = 30
   • b = 35
   • c = 5
   • More answers: any set of integers satisfying a+b+c = 70, a < 38, b > a, b > c and b > 18 (e.g., a=32, b=34, c=4)
2. b) Given the data below Class | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 Frequency | 1 | 3 | 8 | 12 | 9 | 2 Determine i. Mean ii. Standard deviation iii. Modal class (10 marks)
   Model answer

   • Mean = Σf·midpoint / N = 1·4.5 + 3·14.5 + 8·24.5 + 12·34.5 + 9·44.5 + 2·54.5 = 1167.5 g ÷ 35 = 33.36 g
   • Standard deviation
   • Σf·x² = 43 498.75
   • Variance = [Σf·x² – (Σf·x)²/N] / (N‑1) = (43 498.75 – 1 363 056.25/35) / 34 ≈ 134.5
   • SD = √134.5 ≈ 11.6 g
   • Modal class = class with highest frequency → 30‑39 g

3.

1. a) From the following bivariate data, you are required to: Find the regression line and predict Y if x = 20 x | 4 | 12 | 8 | 6 | 4 | 4 | 16 | 8 y | 14 | 4 | 2 | 2 | 4 | 6 | 4 | 12 (11 marks)
   Model answer

   • Σx = 62, Σy = 48, Σxy = 332, Σx² = 612, n = 8
   • Slope (b) = (nΣxy – ΣxΣy) / (nΣx² – (Σx)²) = (8·332 – 62·48) / (8·612 – 62²) = -320 / 1052 ≈ ‑0.30
   • Intercept (a) = (Σy – b·Σx) / n = (48 – (‑0.30)·62) / 8 ≈ 8.36
   • Regression line: Ŷ = 8.36 – 0.30X
   • Predicted Y for X = 20: Ŷ = 8.36 – 0.30·20 ≈ 2.3