Biostatistics Assignment: Statistical Measures and Distributions

This document presents a comprehensive solution to a biostatistics assignment, addressing key statistical concepts and data analysis techniques. The assignment includes calculations of sample mean, standard deviation, median, quartiles, and the creation of frequency distribution tables and relative frequency histograms. The solution analyzes cholesterol levels, blood pressures, and baby height measurements, applying appropriate statistical measures to interpret the data. It also addresses the identification of outliers and the selection of appropriate measures of central tendency and dispersion based on the data distribution. The document provides detailed step-by-step solutions to each problem, ensuring a clear understanding of the concepts and methodologies used. The analysis includes the application of statistical formulas and interpretation of results, making it a valuable resource for students studying biostatistics.

Biostatistics
Student Name:
Instructor Name:
Course Number:
10th May 2019

1. A study is run to estimate the mean total cholesterol level in children 2
to 6 years of age. A sample of 9 participants is selected and their total
cholesterol levels are measured as follows.
185 225 240 196 175
180 194 147 223
a. Compute the sample mean.
Answer
Mean=∑ xi
n
¿ 185+225+240+196+175+180+194+147+223
9 = 1765
9 =196.11
b. Compute the sample standard deviation.
Answer
Standard deviation= √∑ ( xi−x )2
n−1
¿ √ ( 185−196.11 )2+ ( 225−196.11 )2 +…+ ( 147−196.11 )2+ ( 223−196.11 )2
9−1 = √ 6728.889
8 = √841.11
c. Compute the median.
Answer
We rearrange the data in ascending order first, so we have;
147 175 180 185 194 196 223 225 240
The median value is the 5th value which is 194
d. Compute the first and third quartiles.
Answer
1st quartile = 3rd value = 180
3rd quartile = 7th value = 223
e. Which measure, the mean or median, is a better measure of a
typical value? Justify.
Answer
Median is the better measure of a typical value since the data
seems not to be symmetrical (follow a normal distribution)
f. Which measure, the standard deviation or the interquartile
range, is a better measure of dispersion? Justify.
Answer
Interquartile is the better measure of dispersion since the data
seems not to be symmetrical (follow a normal distribution)

2. The box-whisker plots in Figure 4–25 show the distributions of total
cholesterol levels in boys and girls 10 to 15 years of age.
 What is the median total cholesterol level in boys?
Answer
The median total cholesterol level in boys is 245
 Are there any outliers in total cholesterol in boys? Justify briefly.
Answer
There are no outliers in total cholesterol in boys. This can be seen from
the plot where there are no values protruding outside the
limits/boundaries
 What proportion of the boys has total cholesterol less than 205?
Answer
The proportion of the boys that has total cholesterol less than 205 is
0.4286 (42.86%)
 What proportion of the girls has total cholesterol less than 205?
Answer
The proportion of the girls that has total cholesterol less than 205 is
1.000 (100.00%)
3. In a study of a new anti-hypertensive medication, systolic blood
pressures are measured at baseline. The data are as follows:
120 112 138 145 135
150 145 163 148 128
143 156 160 142 150
a. Compute the sample mean.
Answer
Mean=∑ xi
n
¿ 120+ 112+ 138+ …+160+142+150
15 = 2135
15 =142.33
b. Compute the sample median.
Answer
We re-arrange the data in ascending order first, so we have;
11
2
12
0
12
8
13
5
13
8
14
2
14
3
14
5
14
5
14
8
15
0
15
0
15
6
16
0
16
3
The median value is the value of the 8th term which is 145
Median=145
c. Compute the sample standard deviation.
Answer

Standard deviation= √∑ ( xi−x )2
n−1
¿ √ ( 112−142.33 )2+ (120−142.33 )2 +…+ ( 160−142.33 )2 + ( 163−142.33 )2
15−1 = √ 2787.33
14 = √199.095
d. Compute the sample range.
Answer
Range=Max – Min
Max=163
Min=112
Range=163−112=51
e. Are there any outliers? Justify.
Answer
Yes there are is at least one outlier in the dataset.
4. Construct a frequency distribution table using the data in the Problem
below and the following categories: less than 35, 35–44, 45–54, 55 or
more.
Problem below
(problem) The following are baby height measurements (in
centimeters) for a sample of infants participating in a study of infant
health:
28 30 41 48 29
48 62 49 51 39
Answer
Frequency distribution table is given as follows;
Class Frequency
< 35 3
35-44 2
45-54 4
>= 55 1
Total 10
a. Compute the sample mean.
Answer
Mean=∑ xi
n

¿ 28+30+41+ …+48+29+ 48+62+49+51+39
10 = 425
10 =42.5
b. Compute the sample standard deviation.
Answer
Standard deviation= √∑ ( xi−x )2
n−1
¿ √ ( 28−42.5 )2+ (30−42.5 )2+…+ ( 51−42.5 )2+ ( 39−42.5 )2
10−1 = √ 1118.5
9 = √124.2778=11.15
c. Compute the median.
Answer
We first re-arrange the data in ascending order, so we have;
28 29 30 39 41 48 48 49 51 62
The median is between the 5th and the 6th values
Median= 41+ 48
2 = 89
2 =44.5
d. Compute the first and third quartiles.
Answer
1st quartile = 30 + 2.25 = 32.25
3rd quartile = 49- ¼ = 48.75
e. Which measure, the mean or median, is a better measure of a
typical value? Justify.
Answer
Me is the better measure of a typical value since the data
seems not to be symmetrical (follow a normal distribution)
f. Which measure, the standard deviation or the interquartile
range, is a better measure of dispersion? Justify.
Answer
Interquartile is the better measure of dispersion since the data
seems not to be symmetrical (follow a normal distribution)
5. Generate a relative frequency histogram using the data in Problem 4.
Answer