PubH 6002 Quiz 1-5 & Final Exam – Biostatistics Public Health – (2026) Actual Questions & Answers (GWU) 100% Guarantee Pass

PubH 6002
QUIZ 1-5 & FINAL EXAM
Answer Key
Biostatistical Applications for Public Health

George Washington University

This Document Description:
Complete PubH 6002: Biostatistical
Applications for Public Health QUIZ 1-5 &
Final Exam Answer Key (MCQs with fully
worked solutions)

, PubH 6002:
PubH 6002: Biostatistical
Biostatistical Applications
Applicationsfor
forPublic
PublicHealth
Health


Quiz 11 -- Key
Quiz Key

Student Name:
Name:

Instructions:
Instructions:This
Thisquiz
quizconsists
consistsofof15
15MC
MCquestions.
questions.While
Whilethis quiz
this is is
quiz designed to to
designed take 3535
take
minutes,
minutes, you have
have 22 hours
hours to
to complete
complete it.
it. Work
Work individually!
individually!You
Youmay
mayuse
useyour
yourown
ownformula
formula
sheets containing relevant
relevant hand-written
hand-writtennotes
notes as
as well
wellasasaastandard
standardororscientific
scientificcalculator.
calculator.ToTo
receive full credit,
credit, you
you must
must show
show all
all of
of your
yourwork.
work.Good
Goodluck!
luck!

For questions
For questions1-3, 1-3,refer
refertotothe
thefollowing
followinginformation:
information: Back pain
Back is aismajor
pain a majorhealth problem
health problem
because of
of its
its high
highprevalence
prevalenceand andcosts
costsininterms
termsofofhealth
healthcare
careexpenditures
expendituresand andlost
lost
productivity. Systematic
productivity. Systematicreviews
reviewshave
haveconcluded
concludedthatthatchiropractic
chiropracticspinal
spinalmanipulation
manipulation appears to to
appears
be effective
effective inin some
somesubgroups
subgroupsofofpatients
patientswith
withback
backpain
painand
andthis
thisisisone
oneofof
the
thefew
fewtreatments
treatments
recommended in in clinical-practice
clinical-practiceguidelines
guidelineson onthe
thecare
careofofadults
adultswith
withlowlowback
backpain
painin in
the
the
United States.
States. The
The effectiveness
effectivenessofofphysical
physicaltherapy
therapyfor
forback
backpain
painhashasnot
notbeen
beenwell
wellstudied, and
studied, and
results of comparisons of
the results of physical
physicaltherapy
therapywith
withchiropractic
chiropracticmanipulation
manipulationhave haveconflicted.
conflicted.
Suppose among
among aa large
large group
group ofofpatients
patients with
withlower
lowerback
backpain,
pain,15%
15%visitvisitboth
botha aphysical
physical
therapist and a chiropractor, and 15% 15% visit
visitneither
neitherofofthese.
these.Assume
Assumethe theprobability
probability that
thata a
physical therapist
patient visits a physical therapist is 0.49. Hint:
Hint:Start
Startby
bydrawing
drawingaaVennVenndiagram.
diagram.

Not (PT or
Not or C) = ̅̅̅̅̅̅̅̅̅̅̅
= 𝑃𝑇
ᵃᵃ ᵅᵅ𝑜𝑟ᵃ𝐶
0.15




PT
PT and 𝐶̅
and ᵃ̅ PT and
PT and C C
C and ̅̅̅̅̅
𝑃𝑇
and ᵃᵃ
0.34
0.34 0.15
0.15 0.36
0.36




1. What
Whatisisthe
theprobability
probabilitythat
thata arandomly
randomlychosen
chosenpatient
patientvisits
visitsa achiropractor?
chiropractor?(3(3points)
points)

a.
a. 0.21
0.21
b.
b. 0.49
0.49
c.
c. 0.51
0.51
d.
d. 0.85
0.85
e.
e. 0.15
0.15
-- Define
Define the
the events
events PT
PT==patient
patientvisits
visitsphysical
physicaltherapist
therapistand
andC C= =patient
patientvisits chiropractor.
visits chiropractor.
-- We are
are given
given P(PT
P(PT and
and C)
C)== 0.15,
0.15, P(not
P(notPT PTororC)
C)==.15,
.15,P(PT)
P(PT)==.49
.49
Using
Using the rule, P(PT
the addition rule, P(PT or
or C)
C) = = P(PT) P(C)––P(PT
P(PT)++ P(C) P(PTand andC).
C).
-- Solving
Solving for
for P(C),
P(C), we
we get
get P(C)
P(C)== P(PT
P(PTor C)––P(PT)
or C) P(PT)++P(PTP(PTandandC).
C).
-- By
By the
the definition
definition of
of complements,
complements,P(PT P(PTororC)C)==11- -.15
.15==.85
.85
-- Using
Using substitution,
substitution, P(C)
P(C)== .85
.85 - -.49
.49++.15
.15==.51
.51
-- Alternatively,
Alternatively, since
since (PT
(PT and
and C)
C) isis mutually
mutuallyexclusive
exclusivewithwith (C
(Cand
andᵃᵃ̅)̅̅̅̅
, we
𝑃𝑇 can
), we simply
can add
simply addthese
these
probabilities as
probabilities as P(C)
P(C)==.15
.15++.36
.36==.51 .51


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, 2. What
Whatis isthe
theprobability that
probability a randomly
that a randomlychosen patient
chosen visits
patient a physical
visits therapist
a physical given
therapist that
given that
does not
s/he does not visit a chiropractor? (3 points)
points)

a. 0.31
b. 0.41
c. 0.60
0.60
d. 0.85
e. 0.69

- The complementary
complementary event
event of ̅ whichrepresents
ofCCisisᵀ̅𝐶which representsnot
notvisiting
visitinga achiropractor.
chiropractor.

- This probability
probability isis found
foundfrom
fromP(C)
P(C)as 𝑃(𝐶)̅ )==11––P(C)
as ᵀ(ᵀ̅ P(C)==1 1- .51
- .51==.49
.49

- The probability
probability of
ofvisiting
visitingaaphysical
physicaltherapist
therapistgiven
givennot
notvisiting
visitinga achiropractor
chiropractorisis
𝑃(𝑃𝑇 ∩
ᵀ(ᵀᵀ ∩ ᵀ̅𝐶 )) ̅
𝑃(𝑃𝑇|𝐶̅ )) == ᵀ(ᵀ̅
ᵀ(ᵀᵀ|ᵀ̅ 𝑃(𝐶)̅)


-- The numeratorabove
The numerator aboveis is found
found 𝑃(𝑃𝑇
as as ∩ ∩𝐶̅ )ᵀ̅=
ᵀ(ᵀᵀ ) =𝑃(𝑃𝑇)
ᵀ(ᵀᵀ)−−ᵀ(ᵀᵀ
𝑃(𝑃𝑇 ∩ᵀ)
∩ 𝐶)==.49
.49 − .15
−.15 = .34
= .34
.34
.34
-- Therefore, ᵀ(ᵀᵀ|ᵀ̅̅ )) =
Therefore, 𝑃(𝑃𝑇|𝐶 = .49 = .69
.49 = .69

3. IsIsa apatient
patientvisiting
visitinga aphysical
physicaltherapist
therapistindependent
independentofofa apatient
patientvisiting
visitinga achiropractor?
chiropractor?Why
Why
or why not?
not? (2 points)
points)

Yes,because
a. Yes, becauseP(PT
P(PT and
and C)
C) ≠≠ P(PT)
P(PT) **P(C).
P(C).
No,because
b. No, because P(PT
P(PT andand C) ≠ P(PT)
C) ≠ P(PT) **P(C).
P(C).
c. Yes,
Yes,because
becauseP(PT
P(PTand C)≠≠0.0.
and C)
No,because
d. No, becauseP(PT
P(PTandand C)
C)≠≠0.0.
e. Cannot
Cannotbe bedetermined
determined from
fromthe
thegiven
giveninformation.
information.
- IfIf two
twoevents
eventsPT
PTand
and CCare
areindependent,
independent, then thenP(PT
P(PTand
and C)C)==P(PT)
P(PT)**P(C).
P(C).
- From the given
given information,
information,P(PT
P(PTandandC) C)==0.15
0.15and
andP(PT)
P(PT)==0.49.
0.49.
- From (1), we we found
found that
that P(C)
P(C)==0.51.
0.51.
- Using substitution, P(PT)
P(PT) **P(C)
P(C)==.51*.49
.51*.49==0.2499.
0.2499.
- Since
Since P(PT
P(PT and
and C)C) ≠
≠ P(PT)
P(PT) ** P(C),
P(C), i.e.,
i.e.,.15
.15≠≠.2499
.2499, ,no,
no,visiting
visitinga aphysical
physicaltherapist
therapistis is
not independent of visiting
visitingaa chiropractor.
chiropractor.




2
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, For questions
questions4-8,4-8,refer
refertotothe
thefollowing
followinginformation:
information: The aging
The aging process
processmaymayaffect
affectnormal
normal
reference ranges
reference ranges of some laboratory
of some laboratory results.
results. InIn studies
studies addressing
addressinggeriatric
geriatricsubjects,
subjects, laboratory
laboratory
reference intervals
reference intervals should
should also
also include
include asas aa requirement
requirement for for analysis,
analysis,corrections
correctionsforforage
ageand
and
gender factors,
gender factors, which
which are
are essential
essential for
for separating
separating changes
changes occurring
occurring in inreference
reference ranges
rangesnot not
related to
related to disease.
disease. Disorders
Disorders common
common in in old
old age,
age, and
and the
the high
highfrequency
frequencyof of certain
certain disease
diseasestates
states
in the
in elderly, prevent
the elderly, prevent the
the establishment
establishment of of normal
normal reference
reference ranges
ranges with
with certainty.
certainty. A A study
studywas
was
conducted to
conducted to determine
determine if significant gender
if significant gender differences existed in
differences existed in the
the mean
mean amounts
amounts of of calcium,
calcium,
inorganic phosphorus,
inorganic phosphorus, andand alkaline
alkaline phosphatase
phosphataselevelslevelsininthe
theblood
bloodamong
amongsubjects
subjects65 65years
yearsofof
age and
age and older
older (Boyd,
(Boyd, Delost
Delost and
and Holcomb
Holcomb1998).
1998).The Theresearchers
researchersperformed
performed aa retrospective
retrospective
chart review
chart review of
of laboratory
laboratory procedures
procedures performed
performed in in six
six different physician practices.
different physician practices. TheThe data
data
consisted of
consisted of 178
178 subjects (92 males
subjects (92 males and
and 8686 females)
females) aged
aged 6565 or
or older.
older. AA subset
subset ofof the
the data
data for
for
eight of
eight these subjects
of these subjects is
is provided
provided inin Table
Table 1.1.

Table 1.
Table 1.Data
Datafor
for 88randomly
randomlyselected
selectedsubjects
subjects
OBSNO
OBSNO LAB
LAB AGE
AGEGENDER
GENDER ALKPHOS
ALKPHOSCAMMOL
CAMMOLPHOSMMOL AGEGRP
PHOSMMOL AGEGRP
8 4 68
68 2 153
153 2.45
2.45 1.53
1.53 1
29
29 3 74
74 2 61 2.50
2.50 0.81
0.81 2
39
39 2 76
76 2 89 2.43
2.43 1.14
1.14 3
59
59 2 88
88 1 78 2.35
2.35 0.92
0.92 5
79
79 1 82
82 2 115
115 2.33
2.33 1.45
1.45 4
95
95 1 71
71 2 78 2.50
2.50 1.35
1.35 2
123
123 1 73
73 1 74 2.38
2.38 0.76
0.76 2
176
176 6 67
67 2 84 2.30
2.30 0.99
0.99 1

Questions 4-8:
Questions 4-8:Identify
Identifythe
thetype
typeof
ofeach
eachvariable
variableininTable
Table22below
belowbybyselecting
selectingone
oneofof
the
the
following. Look
following. Look atat the
the data
data in
in Table
Table 11 to
to see
see examples
examplesof of the
the values.
values.Be
Beasasspecific
specificasaspossible!
possible!
(0.5 points
(0.5 points each)
each)

a. Continuous
a. Continuous
b. Dichotomous
b. Dichotomous
c. Discrete
c. Discrete
d. Nominal
d. Nominal
e. Ordinal
e. Ordinal

Table 2.
Table 2.Descriptions
Descriptionsof
ofvariables
variables(Data
(DataDictionary)
Dictionary)
TYPE
TYPE VARIABLE
VARIABLE DESCRIPTION
DESCRIPTION
Lab wherethe
Lab where theblood
blood was
was analyzed
analyzed (1=Metpath,
(1=Metpath, 2=Deyor,
2=Deyor, 3=St. Elizabeth’s,
3=St. Elizabeth’s,
4.
4. d LAB
4=CB Rouche,
4=CB Rouche,5=YOH,
5=YOH, 6=Horizon)
6=Horizon)
5.
5. b GENDER
GENDER Gender
Genderofofthe
thesubject (1=Male,
subject 2=Female)
(1=Male, 2=Female)

6.
6. c ALKPHOS
ALKPHOS Alkaline phosphatase
phosphatase in International Units/Liter
in International (IU/L),
Units/Liter usingusing
(IU/L), wholewhole
numbers only only
numbers

7.
7. a CAMMOL Amount of
Amount of raw
rawcalcium
calcium(mmol/L)
(mmol/L)

8.
8. e AGEGRP
AGEGRP Age
Age group
groupofofthe
thesubject (1=65-69,
subject 2=70-74,
(1=65-69, 3=75-79,
2=70-74, 4=80-84,
3=75-79, 5=85-89
4=80-84, years) years)
5=85-89


3
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