UWS Autumn 2019 401077 Introduction to Biostatistics Assignment 1

This document presents a complete solution to Assignment 1 for the 401077 Introduction to Biostatistics course, focusing on data analysis and statistical concepts. The assignment covers several key areas, including the identification of categorical variables, graphical representations of data using histograms and boxplots, and the interpretation of distributions. It explores the relationship between variables through scatter plots and analyzes categorical data using pie charts and cross-tabulations. The solution includes calculations for binomial distributions, probabilities, and expected values. Furthermore, it demonstrates the application of z-scores to analyze body temperature data and includes descriptive statistics such as mean, standard deviation, and median. The assignment provides a practical application of biostatistical methods to real-world data, offering a comprehensive understanding of the subject matter.

401077 Introduction to Biostatistics, Autumn 2019
Assignment 1 (Due Sunday March 31, 2019)
Your name: Umar Kareem Mohamed
Your student number: 18758359

Question 1 (2 marks)
The categorical variables in the provided dataset include;
 Sex
 Education level
 Marital status (married)
Question 2 (4 marks)
a)
Figure 1: Histogram and boxplot for age (Data source: project analysis)
b)
The distribution of age seems to be normally distributed since the median mark for the boxplot is at
the centre.
Question 3 (3 marks)
Figure 2: Pie chart of highest education level (Data source: project analysis)

The above figure gives a pie-chart of the highest education level. As can be seen, majority
(37.1%) had Bachelor as the highest education level while minority (9.8%) had post graduate
levels.
Question 4 (4 marks)
Figure 3: Scatter plot of Income vs. log income (Data source: project analysis)
From the plot above, it is clear that there a strong positive non-linear relationship between the two
variables (income and log income).
Question 5 (5 marks)
a)
Education Level
Sex
Male Female
Bachelor 90 (40.72%) 95 (34.30%)
Certificate 50 (22.62%) 118 (42.60%)
Not Tertiary 53 (23.98%) 43 (15.52%)
Post Graduate 28 (12.67%) 21 (75.81%)
b)
More male participants (40.72%, n = 90) have bachelor degree level as compared to the female
participants (34.30%, n = 95). However, majority of female participants (42.60%, n = 118) had
certificate level as their highest education level against 22.62% (n = 50) for the male participants.

c)
Probability of male having post graduate level is 0.1267 (12.67%).
d)
Probability of male having no tertiary level is 0.2398 (23.98%).
Question 6 (5 marks)
a)
This is a binomial distribution and as such we have;
P ( x=k ) =(n
k ) pk ( 1− p )n−k
P ( x=3 ) = ( 10
3 ) 0.093 ( 1−0.09 ) 10−3=( 10
3 ) 0.093 ( 1−0.09 ) 7 =0.045206
b)
P ( x=k ) =( n
k ) pk ( 1− p ) n−k
P ( x=1 )= (10
1 )0.091 ( 1−0.09 )10−1= (10
1 )0.091 ( 1−0.09 )9=0.385137
P ( x=2 ) = ( 10
2 ) 0.092 ( 1−0.09 ) 10−2 =( 10
2 ) 0.092 ( 1−0.09 ) 8=0.171407
P ( x=3 ) = ( 10
3 ) 0.093 ( 1−0.09 ) 10−3=( 10
3 ) 0.093 ( 1−0.09 ) 7 =0.045206
P ( x=4 ) = ( 10
4 ) 0.094 ( 1−0.09 ) 10−4 =( 10
4 ) 0.094 ( 1−0.09 ) 6=0.007824
Thus the required probability is;
P ( 1≤ x ≤ 4 )=0.3851+0.1714 +0.0452+0.0078=0.6096
c)
Probability of O negative=P ¿
Frequency=0.09∗10=0.9 ≈ 1
d)
Expected value= Average=E ( X )=np
¿ 10∗0.09
¿ 0.9

Question 7 (7 marks)
a)
We seek to find the percentage of adults who’s ‘at rest’ body temperature falls below 36.80oC
P( X< 36.80)
We compute the z-score as follows;
Z= x −μ
σ =36.80−36.80
0.41 = 0
0.41 =0
P ( Z <0 )=0.5=50 %
Thus 50% of adults who’s ‘at rest’ body temperature falls below 36.80oC.
b)
Z-scores for Olaf
Z= x −μ
σ =33.00−36.80
0.41 =−3.8
0.41 =−9.2683
Z-scores for Timu
Z= x −μ
σ =38−36.80
0.41 = 1.2
0.41 =2.92683
From the two computed z-scores, Olaf has the most unusual body.
c)
We have the following data points;
37.5,36.3,36.6,36.8,36.5,36.3,36.5,37.5,37.5,36.2,36.4,35.6,35.9,35.6,37.8,36.5,36.8,36.5
Mean=∑ xi
n = 37.5+36.3+…+36.8+36.5
18 =658.8
18 =36.60
Standard deviation= √ ∑ ( xi−x )2
n−1 = √ ( 37.5−36.60 )2+ …+ ( 36.5−36.60 )2
18−1 = √ 6.86
17 = √0.4035=0.4035
d)
Median(50 %)= 9 th+10 th
2 = 36.5+36.5
2 = 73
2 =36.50
First Quartile=5th value =36.30