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M2 Practice Problem SolutionB

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MPH501 Module 2 Assignment B Solutions

Quantitative Methods for Pub Health Application

Chapter 4 Solutions to Problems

1. A study is run to estimate the mean total cholesterol level in children 2-6 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

X X2
147 21,609
175 30,625
180 32,400
185 34,225
194 37,636
196 38,416
223 49,729
225 50.625
240 57,600
1765 352,865

a. Compute the sample mean.

ΣX 1,765
X̄ = = =196 . 1
n 9

b. Compute the sample standard deviation.

s=

ΣX 2−( ΣX )2 /n
n−1
=

352,865−(1,765 )2 /9
9−1
=
8 √
352,865−(3,115,225/9 )

=
√ 8
=
8√
352,865−346,136 . 1 6,728 . 89
= √ 841 . 1=29 .0

c. Compute the median.

Median=194

d. Compute the first and third quartiles.

Q1 = 177.5 ((175+180)/2) Q3=224 ((223+225)/2)

© 2012 Jones & Bartlett Learning, LLC 1


MPH501 Module 2 Assignment B Solutions
Quantitative Methods for Pub Health Application
e. Which measure, the mean or median, is a better measure of a typical value? Justify.

Check for outliers: IQR = 224-177.5 = 46.5

Lower Limit = 177.5 – 1.5(46.5) = 107.8


Upper Limit = 224 + 1.5(46.5) = 293.8

No outliers, therefore the best measure of a typical value is the sample mean 196.1.

f. Which measure, the standard deviation or the interquartile range, is a better measure of
dispersion? Justify.

Because there are no outliers, the best measure of dispersion is the sample
standard deviation, s=29.0.

4. The following data were collected as part of a study of coffee consumption among graduate
students. The following reflect cups per day consumed:

3 4 6 8 2 1 0 2

X X2
0 0
1 1
2 4
2 4
3 9
4 16
6 36
8 64
26 134

a. Compute the sample mean.

ΣX 26
X̄ = = =3. 3
n 8

© 2012 Jones & Bartlett Learning, LLC 2


MPH501 Module 2 Assignment B Solutions
Quantitative Methods for Pub Health Application
b. Compute the sample standard deviation.

s=

ΣX 2−( ΣX )2 /n
n−1
=

134−( 26 )2 /8
8−1
=
7 √
134−84 . 5
=√ 7 .36=2 .7

c. Compute the median.

Median = 2.5 ((2+3)/2 = 2.5)

d. Compute the first and third quartiles.

Q1 = 1.5 ((1+2)/2=1.5) Q3=5.0 ((4+6)/2=5.0)

e. Which measure, the mean or median, is a better measure of a typical value? Justify.

Q1 = 1.5 Q3=5.0

Check for outliers: IQR = 5.0-1.5 = 3.5

Lower Limit = 1.5 – 1.5(3.5) = -3.75


Upper Limit = 5.0 + 1.5(3.5) = 10.25

No outliers.

No outliers, therefore the best measure of a typical value is the sample mean 3.3.

f. Which measure, the standard deviation or the interquartile range, is a better measure of
dispersion? Justify.

Because there are no outliers, the best measure of dispersion is the sample standard deviation,
s=2.7.

5. In the study of a new anti-hypertensive medication, systolic blood pressures are measured at
baseline (or the start of the study before any treatment is administered). The data are as
follows:

120 112 138 145 135 150 145 163


148 128 143 156 160 142 150

© 2012 Jones & Bartlett Learning, LLC 3


MPH501 Module 2 Assignment B Solutions
Quantitative Methods for Pub Health Application
a. Compute the sample mean.

X̄ =142.3 .

b. Compute the sample median.

Median = 145.

c. Compute the sample standard deviation.

s = 14.1.

d. Compute the sample range.

Range = 163-112 = 51.

e. Are there any outliers? Justify.

Q1 = 135 Q3=150

Check for outliers: IQR = 150 – 135 = 15

Lower Limit = 135 – 1.5(15) = 112.5


Upper Limit = 150 + 1.5(15) = 172.5

The value of 112 is an outlier on the low end (less that the lower limit of 112.5).

Median = 8 and IQR = 5

© 2012 Jones & Bartlett Learning, LLC 4

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