Statistical Analysis: Quantitative Methods in Health MATH 1065

This assignment provides solutions to problems related to quantitative methods in health, specifically focusing on statistical analysis. It covers topics such as descriptive statistics, including measures of central tendency and dispersion, normal distribution, probability calculations, and hypothesis testing. The assignment utilizes MINITAB for data analysis and visualization, including histograms, boxplots, and cumulative distribution functions. Problems address real-world scenarios, such as analyzing nap habits across different age groups and moods, evaluating the effectiveness of elderberry in reducing cold days, and assessing glucose levels for gestational diabetes. The solutions include detailed calculations, interpretations of statistical outputs, and explanations of the methods used, such as z-scores and interquartile ranges. This document is designed to aid students in understanding and applying quantitative methods in a health context. Desklib offers additional resources, including past papers and solved assignments, to support student learning.

Running Head: QUANTITATIVE METHODS
Quantitative Methods
Name of the Student
Name of the University
Author Note

1QUANTITATIVE METHODS
Table of Contents
Answer 1..........................................................................................................................................2
Part a............................................................................................................................................2
Part b............................................................................................................................................3
Part c............................................................................................................................................4
Answer 2..........................................................................................................................................6
Part a............................................................................................................................................6
Part b............................................................................................................................................6
Part c............................................................................................................................................7
Part d............................................................................................................................................7
Part e............................................................................................................................................7
Answer 3..........................................................................................................................................8
Part a............................................................................................................................................8
Part b............................................................................................................................................9
Part c..........................................................................................................................................11
Answer 4........................................................................................................................................11
Part a..........................................................................................................................................11
Part b..........................................................................................................................................12
Part c..........................................................................................................................................13

2QUANTITATIVE METHODS

3QUANTITATIVE METHODS
Answer 1
Part a
Age 80+70-7950-6930-4918-29
100
80
60
40
20
0
Percent
Yes
No
Nap
Percent is calculated within levels of Age.
Distribution of Age with Nap
Figure 1: Distribution of Age and Nap
From the above graph it can be inferred that the percentage of people who took a nap and
who did not take a nap in the age group of 80+ is approximately equal. For other groups, 18 – 29,
30 – 49, 40 – 69 and 70 – 79 the percentage of people who did not take a nap is more than the
percentage of people who did take a nap.

4QUANTITATIVE METHODS
Part b
Mood Very happyPretty happyNot too happy
100
80
60
40
20
0
Cumulative Percent Count
Yes
No
Nap
Percent is calculated within levels of Mood.
Distribution of Mood across Nap
Figure 2: Distribution of Mood across Nap
From the above chart it can be inferred that percentage of people who were pretty happy
and very happy when they did not take a nap was more than when they did take a nap.

5QUANTITATIVE METHODS
Part c
Table 1: Distribution Age and Nap
Rows: Age Columns: Nap
No Yes All
18-
29
163 82 245
66.5
3
33.4
7
100.0
0
30-
49
193 106 299
64.5
5
35.4
5
100.0
0
50-
69
280 135 415
67.4
7
32.5
3
100.0
0
70-
79
265 140 405
65.4
3
34.5
7
100.0
0
80+ 40 44 84
47.6
2
52.3
8
100.0
0
All 941 507 1448
64.9
9
35.0
1
100.0
0
Cell Contents
Count
% of Row
From the above table it is found that 52.38% of 80+ age citizens take a nap. 67.47% of
people in the age group of 50 – 69 years’ age do not take a nap.

6QUANTITATIVE METHODS
Table 2: Distribution mood and Nap
Rows: Mood Columns: Nap
No Yes All
Not too
happy
303 267 570
53.1
6
46.8
4
100.0
0
Pretty happy 385 138 523
73.6
1
26.3
9
100.0
0
Very happy 253 102 355
71.2
7
28.7
3
100.0
0
All 941 507 1448
64.9
9
35.0
1
100.0
0
Cell Contents
Count
% of Row
From the above table it can be inferred that of “pretty happy” citizens 73.61% did not
take a nap. Similarly, “very happy” citizens 71.27% did not take a nap. On the other hand, from
“Not too happy” people 53.16% did not take a nap and 46.84% did take a nap.

7QUANTITATIVE METHODS
Answer 2
Part a
1612840
5
4
3
2
1
0
1612840
Mean 4.667
StDev 2.270
N 12
Elderberry
Mean 6.882
StDev 3.533
N 17
Placebo
Elderberry
Cold_Days
Frequency
Placebo
Histogram (with Normal Curve) of Cold_Days by Group
Panel variable: Group
Figure 3: Histogram of Cold Days
Part b
Table 3: Descriptive Statistics for Cold Days across Groups

8QUANTITATIVE METHODS
Part c
PlaceboElderberry
16
14
12
10
8
6
4
2
Group
Cold_Days
Distribution of Cold Days
Figure 4: Distribution of Cold Days across Groups
Part d
The histogram (figure 3) of Cold days of the two groups suggests that the distribution of
cold days is positively skewed. Thus it can be inferred that the mean number of cold days is less
than the median number of cold days.
From the (figure 4) there is no indication of outliers in number of cold days.
From both the histogram and boxplot it can be inferred that the data for cold days is right
skewed. Since the data is not normally distributed, hence median would be the appropriate
measure of central tendency. The median number of Cold days for Elderberry is 4.0 and for
Placebo is 6.0 respectively. The median is used as a measure of central tendency. Thus IQR
would be the ideal measure of dispersion. The IQR for cold days for Elderberry is 3.750 (6.750 –
3.000). The IQR for cold days for Placebo is 4.500 (9.50 – 4.00).
Part e
The mean number of cold days using Elderberry is 4.667 while for placebo is 6.882. The
maximum number of cold days in Elderberry is 9.00 while for Placebo is 15.00. Moreover, the
median number of cold days for Elderberry is 4.00 and for Placebo is 6.00. Thus it can be
inferred that by using Elderberry the number of Cold days is less as compared to Placebo.

9QUANTITATIVE METHODS
Answer 3
Part a
Let X denote the mean glucose level
The mean glucose level is normally distributed with μ=125mg /dl and σ =10 mg/dl
The probability that the mean glucose level is P(X>140mg/dl)
Method 1: Use Standard Normal Tables
z-score for P ( X >140 )
z= 140−125
10 =1.5
Thus, P ( X >140 ) =P ( z >−1.5 ) =1−P ( z←1.5 )
From z-table it is found that P(z = -1.5) = 0.933
P(z >−1.5)=1−0.933=0.067
Therefore, the probability that the patient has gestational diabetes = 0.067
Method 2: Minitab
Cumulative Distribution Function
Normal with mean = 125 and standard deviation = 10
x P( X ≤ x )
140 0.933193
The probability P ( X >140 ) =1−P ( X <140 )=1−0.933193=0.066807

10QUANTITATIVE METHODS
0.04
0.03
0.02
0.01
0.00
X
Density
140
0.06681
125
Distribution Plot
Normal, Mean=125, StDev=10
Part b
Using MINITAB, a normal distributed random sample of size 5 is created.
Sample
121.085
138.256
110.773
138.132
133.129
The mean and standard of the sample is 128.27 and 12.02 respectively.
Method 1: Use Standard Normal Tables
z-score for P ( X >140 )
z= 140−128.27
12.02 =0.975
Thus, P ( X >140 ) =P ( z >0.975 ) =1−P ( z< 0.975 )
From z-table it is found that P(z = 0.975) = 0.83522
P(z >0.975)=1−0.83522=0.16478

11QUANTITATIVE METHODS
Therefore, the probability that the patient has gestational diabetes = 0.16478
Method 2: Minitab
Cumulative Distribution Function
Normal with mean = 128.27 and standard deviation = 12.02
x P( X ≤ x )
140 0.835436
The probability P ( X >140 ) =1−P ( X <140 ) =1−0.835436=0.164564
0.035
0.030
0.025
0.020
0.015
0.010
0.005
0.000
X
Density
140
0.1646
128.3
Distribution Plot
Normal, Mean=128.27, StDev=12.02