WRITTEN ASSIGNMENT WEEK 2
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Dec 6, 2023
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Written Assignment:
Measures of Central Tendency and Variability
University of The People
HS 4510:
Biostatistics
Written Assignment:
Week 2
Instructor: Prof Adaugo Eziyi
28 Nov 2023
Table 1
1
Age categories and relative frequencies of mean pulse rate of study
participants(male an female)
Age of individual
participants(years
)
Number of
participant
s
Mean
Pulse rate
Relative Frequencies
10-14
5
70.8
0.2= 20%
15-19
6
74.3
0.24= 24%
20-24
2
64.0
0.08= 8%
25-29
2
71.0
0.08= 8%
30-34
1
73.0
0.04= 4%
35-39
2
84.0
0.08= 8%
40-44
3
79.7
0.12= 12%
45-49
2
65.0
0.08= 8%
2
50-54
1
110.0
0.04= 4%
55-59
1
94.0
0.04= 4%
Totals
25
1 = 100%
What is the proportion of males and females in this sample?
Females: Males=
10: 11
Now there are 4 who were not mentioned their gender. I will assume 2 are male
and 2 are female therefore
Females: Males= 12: 13
a). I would use
Box or Whisker plot and Histograms
. Firstly, the histogram
summarizes discrete or continuous data on an interval scale. Box or Whisker plots
shows continuous numerical data distribution between multiple groups when you want to
compare it.
The data that we have is mutually exclusively that is it does not overlap and
are continuous variables. Histograms show relative frequencies giving a probability
distribution of a continuous variable. Box or Whisker plots would show us the first
quartile up to the third quartile, the median and the outliers data
3. Males: 79, 90, 75, 60, 65, 79, 54, 98, 49
Females: 66, 59, 79, 95, 72, 55, 64, 60, 65
Calculations:
3
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a.
Calculate the
arithmetic
mean
pulse for each gender separately
, and for both
genders together
Males
: 79, 90, 75, 60, 65, 79, 54, 98, 49
Arithmetic mean=1/9x ( 79+ 90+ 75+ 60+ 65+ 79+ 54+ 98+ 49)
=649/9
=72.1bpm
Females:
66, 59, 79, 95, 72, 55, 64, 60, 65
Arithmetic mean=1/9x (66+ 59+ 79+ 95+ 72+ 55+ 64+ 60+ 65)
=615/9
=68.3 bpm
Arithmetic mean for both
=
1/18x (649+615)
=1264/18
=70.2bpm
b.
List the
median
pulse for each gender separately
, and for both genders together
.
Arrange the pulse data in ascending order first. The median divides the data as
50
th
percentile
i) Males: 79, 90, 75, 60, 65, 79, 54, 98, 49
49, 54, 60, 65
, 75, 79, 79, 90, 98
The median for male=
75bpm
ii) Females: 66, 59, 79, 95, 72, 55, 64, 60, 65
55, 59, 60, 64,
65, 66, 72, 79, 95
The median for female=
65bpm
iii)
For both:
49, 54, 55, 59, 60, 60, 64, 65,
65, 66,72, 75, 79, 79, 79, 90 ,95, 98
4
Median for both =1/2 x (65+66)
=131/2
=
65.5 bpm
c.
What are the
degrees of freedom
associated with each gender separately and
with both genders together?
The degrees of freedom are the values that refer to the numbers in a calculation
that are free to vary. We use it when we want to calculate standard deviation(SD).
The n-1 is a better estimate of the population SD than n.
For one sample of mean: df= n-1
For both sample combined: df= (n1 + n2)- 2
Variance in males=
[(49-72.1) ^2+(54-72.1)^2+(60-72.1)^2+(65-72.1)^2+(75-
Ʃ
72.1)^2+(79-72.1)^2+( 79-72.1)^2+(90-72.1)^2+(98-72.1)^2] / (9-1)
=(-23.1)^2 +(-18.1)^2 +( 12.1)^2 +(-7.1)^2+ (2.9)^2 +(6.9)^2+(6.9)^2+
( 17.9)^2+(25.9)^2 /8
=(533.61+327.61+ 146.41+ 50.41+ 8.41+ 47.61+ 47.61+ 320.41+ 670.81) /8
=2152.89/8
=
269.11bpm
Variance in females= (55 -68.2)^2+( 59-68.2)^2+ (60-68.2)^2+ (64 -68.2)^2+( 65-
68.2)^2+(66- 68.2)^2+ (72-68.2)^2+ (79-68.2)^2 +( 95-68.2)^2 /(9-1)
= 174.24 +84.64+ 67.24+ 17.64+ 10.24+ 4.84+ 14.44+ 116.64+ 718.24 /8
=1208.16/8
=151.02 bpm
d.
Calculate the
standard deviation
in each gender separately.
Variance in males=
[(49-72.1) ^2+(54-72.1)^2+(60-72.1)^2+(65-72.1)^2+(75-
Ʃ
72.1)^2+(79-72.1)^2+( 79-72.1)^2+(90-72.1)^2+(98-72.1)^2] / (9-1)
=(-23.1)^2 +(-18.1)^2 +( 12.1)^2 +(-7.1)^2+ (2.9)^2 +(6.9)^2+(6.9)^2+
( 17.9)^2+(25.9)^2 /8
=(533.61+327.61+ 146.41+ 50.41+ 8.41+ 47.61+ 47.61+ 320.41+ 670.81) /8
5
=2152.89/8
=269.11
Standard deviation in males
= √269. 11
=16.40 bpm
Variance in females= (55 -68.2)^2+( 59-68.2)^2+ (60-68.2)^2+ (64 -68.2)^2+( 65-
68.2)^2+(66- 68.2)^2+ (72-68.2)^2+ (79-68.2)^2 +( 95-68.2)^2 /(9-1)
= 174.24 +84.64+ 67.24+ 17.64+ 10.24+ 4.84+ 14.44+ 116.64+ 718.24 /8
=1208.16/8
=151.02
Standard deviation in females
= √ 151.02
= 12.29 bpm
e.
Discuss what the measures calculated in “d” indicate for differences in gender.
Standard deviation shows by how much variation or dispersion the data is from
the mean.
The SD for male is 16.40 and for the females is 12.29. The female SD is low
meaning the data points which were observed are close to the mean (55 to 95)
compared to the higher SD for male which shows the data points for the males
are spread out over large range of values(49 to 98).
f.
Calculate the
range
and the
interquartile range
for both genders together.
Data for both: 49, 54, 55, 59, 60, 60, 64, 65, 65, 66,72, 75, 79, 79, 79, 90 ,95, 98
Range= highest value-lowest value = 98-49= 49
Range=
49
Interquartile Range(IQR)=Q3 – Q1
We have even observations n = 18
6
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Q1 Location = (18+1) /4= 19/4 =4.75
Q3 Location = 3(18+1)/ 4= 3(19) /4= 14.25
4.75 falls between 59 and 60 position
14.25 falls between 79 and 79 position
Estimation of Q1= 59 +3/4 (60-59) =59+ 0.75= 59.75
Estimation of Q3= 79 + ¼ (79-79)= 79+ 0= 79
Therefore, IQR= Q3- Q1 = 79 - 59. 75 =
19.25
Question 4
a.
Calculate the mean length of stay for ‘Intensive care unit’ and ‘Surgery’ together
Weighted mean=
(number of beds in each group x mean of each group)
Ʃ
Total number of beds
= (10x 26)+ ( 20x 12
)
10+20
=500
30
=
16.7 days
b.
Calculate the mean length of stay for all 4 listed departments
Weighted mean=
(number of beds in each group x mean of each group)
Ʃ
Total number of beds
7
=(10x26)+ (20x 12) +( 40x4) +( 50x8 )
120
=260+ 240+ 160 + 400
120
=1060
120
=
8.83 days
References
Illowsky, B. and Dean, S. (2017).
Introductory Statistics
. OpenStax Rice University.
Textbook content produced by OpenStax is licensed under a Creative Commons
Attribution 4.0 International License (CC BY 4.0). Retrieved
from
https://my.uopeople.edu/pluginfile.php/1811982/mod_page/content/3/TEXT
%20introductory-statistics.compressed.pdf
Unit 2. (n.d) CPH451 Biostatistics. Reading Assignments.
PDF
https://my.uopeople.edu/pluginfile.php/1811991/mod_book/chapter/474914/Unit
%202_CPH%204510%20Biostatistics_Reading%20Assignment.pdf
8
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