Introduction to Biostatistics: Main Exam, 29 March 2019

1.

1. a) Explain the following sampling methods i) Quota ii) Simple random iii) Stratified (6 marks)
   Model answer

   • Quota sampling: non‑probability method where the researcher selects participants to fill pre‑set quotas for certain characteristics (e.g., age, sex) to reflect their proportion in the population.
   • Simple random sampling: each member of the population has an equal and known chance of being selected; usually done with a random number table or computer.
   • Stratified sampling: population is divided into homogeneous sub‑groups (strata) and a random sample is drawn from each stratum, often proportionally to its size.
2. b) With the aid of sketches differentiate between the following terms as used in statistics i) Positive and negative skew ii) Positive and negative correlation (6 marks)
   Model answer

   • Positive skew: tail of the distribution extends to the right; mean > median > mode. Sketch: a histogram with a long right tail.
   • Negative skew: tail extends to the left; mean < median < mode. Sketch: histogram with a long left tail.
   • Positive correlation: as one variable increases, the other tends to increase; points cluster along an upward‑sloping line.
   • Negative correlation: as one variable increases, the other tends to decrease; points cluster along a downward‑sloping line.
3. c) A random sample of 12 components has a mean weight of 50 gm and standard deviation of 9 gm. Determine the population mean with a 95% confidence. (4 marks)
   Model answer

   • Standard error (SE) = s/√n = 9/√12 = 9/3.464 = 2.60 gm.
   • For 95 % confidence, t ≈ 2.201 (df = 11).
   • Margin of error = t × SE = 2.201 × 2.60 = 5.72 gm.
   • Population mean 95 % CI = 50 ± 5.72 → 44.28 gm to 55.72 gm.
4. d) 10 % of components produced by a machine are defective. Determine the probability that in a random sample of 8 components at most two will be defective. (4 marks)
   Model answer

   • Use binomial formula P(X≤2)=∑_{k=0}^{2} C(8,k)(0.10)^k(0.90)^{8‑k}.
   • k=0: (0.90)^8 = 0.4305
   • k=1: 8×0.10×(0.90)^7 = 8×0.10×0.4783 = 0.3826
   • k=2: 28×0.01×(0.90)^6 = 28×0.01×0.5314 = 0.1488
   • Total probability ≈ 0.4305 + 0.3826 + 0.1488 = 0.962 (96.2 %).
5. e) The distribution of lifetimes of component are as shown
   Model answer

   • Frequency distribution (example):
   • 0–10 hrs: 5 components
   • 10–20 hrs: 12 components
   • 20–30 hrs: 20 components
   • 30–40 hrs: 15 components
   • 40–50 hrs: 8 components
   • 50+ hrs: 0 components
   • Cumulative distribution: add frequencies cumulatively (5, 17, 37, 52, 60, 60).
   • Mean lifetime = Σ(fi·midpoint)/N = (5·5 +12·15 +20·25 +15·35 +8·45)/60 ≈ 24.2 hrs.
   • Standard deviation = √[Σfi(mid‑point‑mean)² /N] ≈ 10.1 hrs.
   • Interpretation: most components fail between 20–40 hrs; distribution is slightly right‑skewed.

2. (13 marks)

1. a) The data shows the height of 80 malnourished kids in Kiambu county height 10-15 15-20 20-25 25-30 30-35 35-40 40-45 Number of kids 5 x 17 20 y 16 4 Given that the median wage is Ksh 27,000 i) Values of x and y ii) Interquartile range (7 marks)
   Model answer

   • Total frequency = 80.
   • Cumulative frequencies: 5, 5+x, 5+x+17, 5+x+17+20, 5+x+17+20+y, …
   • Median position = (80+1)/2 = 40.5 → lies in the 20‑25 cm class (cumulative up to previous class = 5+x+17). Solve 5+x+17 < 40.5 ≤ 5+x+17+20 → 22 < 40.5 ≤ 42+x → x must be 0 (since 22+20=42). Thus x = 0.
   • Upper half (positions 41‑80) median (Q3) position = 0.75×81 = 60.75 → lies in 35‑40 cm class. Cumulative up to previous class = 5+0+17+20+ y = 42+y. Need 42+y < 60.75 ≤ 42+y+16 → y = 2.
   • Values: x = 0, y = 2.
   • Q1 position = 0.25×81 = 20.25 → lies in 15‑20 cm class. Cumulative up to previous class = 5 → Q1 = lower limit + ( (20.25‑5)/17 )×5 = 15 + (15.25/17)×5 ≈ 15 + 4.49 = 19.5 cm.
   • Q3 = lower limit of 35‑40 cm class (35) + ( (60.75‑(42+2))/16 )×5 = 35 + (16.75/16)×5 ≈ 35 + 5.23 = 40.23 cm.
   • Interquartile range = Q3‑Q1 ≈ 40.23 − 19.5 = 20.73 cm.
2. b) The following distribution represent marks scored by 210 students in bio statistics test marks Number of students 20-29 21 30-39 31 40-49 39 50-59 51 60-69 29 70-79 24 80-89 15 Determine i) Mean mark ii) Standard deviation iii) Interquartile range iv) Mode
   Model answer

   • Mean = 52.5 marks
   • Standard deviation ≈ 17.0 marks
   • Interquartile range ≈ 25.3 marks (Q1 ≈ 40.3, Q3 ≈ 65.6)
   • Mode ≈ 53.5 marks (class 50‑59)

3. (9 marks)

1. a) The cost incurred in maintaining a drug production plant for a period of 8 years is shown Age of plant | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 Cost Ksh 000 | 3 | 5 | 9 | 12 | 14 | 15 | 20 | 28 i) Calculate The Pearson product moment correlation coefficient ii) Interpret the result (10 marks)
   Model answer

   • Pearson r = 0.97 (calculated as 1080/1111.35)
   • Interpretation: Very strong positive linear relationship between plant age and maintenance cost
2. a) Obtain the least squares equation for the data below: x | 4.0 | 4.4 | 4.8 | 4.2 | 5.0 | 5.4 | 5.2 | 5.6 | 5.8 | 6.2 y | 50 | 64 | 66 | 70 | 76 | 82 | 86 | 90 | 94 | 96
   Model answer

   • Least‑squares regression equation: ŷ = -21.2 + 19.5 x

4.

1. a) Explain the following terms as used in statistics i) Point estimate ii) Interval estimate (4 marks)
   Model answer

   • Point estimate: a single statistic from the sample (e.g., sample mean) used to estimate the population parameter
   • Interval estimate: a range of values (confidence interval) that is likely to contain the true population parameter
2. b) In MKU 30% of the students use credit cards. If 12 students are selected at random, determine the probability that i) Exactly two student use credit cards ii) At least ten student use credit cards iii) At most two student use credit cards (8 marks)
   Model answer

   • i) P(X=2) = C(12,2)(0.3)^2(0.7)^10 = 66·0.09·0.0282475 ≈ 0.168
   • ii) P(X≥10) = Σ_{k=10}^{12} C(12,k)(0.3)^k(0.7)^{12‑k} ≈ 0.00021
   • iii) P(X≤2) = P0+P1+P2 ≈ 0.0138+0.0712+0.168 ≈ 0.253
3. c) The diameters of artificial insemination tubes manufactured in a factory are normally distributed with a mean of 70mm and standard deviation of 8mm. Determine i) Probability that a tube chosen at random will have a diameter between 74mm and 80mm. ii) Proportion of test tubes which have more than 68mm as their diameters.
   Model answer

   • i) Z₁=(74‑70)/8=0.50, Z₂=(80‑70)/8=1.25; P=Φ(1.25)‑Φ(0.50)=0.8944‑0.6915≈0.203
   • ii) Z=(68‑70)/8=‑0.25; P(X>68)=1‑Φ(‑0.25)=Φ(0.25)=0.599

5. (10 marks)

1. a) Compute quartile deviation from the following data; ci | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 80-90 f | 12 | 19 | 5 | 10 | 9 | 6 | 6 (10 marks)
   Model answer

   • Quartile deviation
   • Find cumulative frequencies to locate Q1 and Q3.
   • Q1 lies in class 20‑30 (cumulative 12+19=31, 25% of 70=17.5 → 20‑30).
   • Q3 lies in class 50‑60 (cumulative 12+19+5+10+9=55, 75% of 70=52.5 → 50‑60).
   • Midpoints: 25 for 20‑30, 55 for 50‑60.
   • Quartile deviation = (Q3‑Q1)/2 = (55‑25)/2 = 15.
2. b) at an agricultural show 10 high bred bulls were ranked by a qualified judge and by a trainee judge their ranking were given as shown in the table below; Bulls | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Qualified Judge | 2 | 4 | 7 | 8 | 10 | 1 | 5 | 9 | 3 | [not readable in the scan] Trainee Judge | 1 | 3 | 9 | 10 | 7 | 2 | 4 | 8 | 5 | [not readable in the scan] Calculate the spearman's rank Correlation coefficient.
   Model answer

   • Spearman’s rank correlation (ρ)
   • Compute rank differences d for each bull:
   • Bull1: 2‑1=1
   • Bull2: 4‑3=1
   • Bull3: 7‑9=‑2
   • Bull4: 8‑10=‑2
   • Bull5: 10‑7=3
   • Bull6: 1‑2=‑1
   • Bull7: 5‑4=1
   • Bull8: 9‑8=1
   • Bull9: 3‑5=‑2
   • Bull10: (missing) – assume 0 for illustration
   • d² values: 1,1,4,4,9,1,1,1,4,0
   • Σd² = 26
   • ρ = 1‑[6Σd²]/[n(n²‑1)] = 1‑[6×26]/[10(100‑1)] = 1‑156/990 ≈ 0.842.