PubH 6002
QUIZ 1 Answer Key
Biostatistical Applications for Public Health
George Washington University
This Document Description:
Complete PubH 6002: Biostatistical
Applications for Public Health Quiz 1 Answer
Key (MCQs with fully worked solutions)
, PubH 6002: Biostatistical Applications for Public Health
Quiz 1 - Key
Student Name:
Instructions: This quiz consists of 15 MC questions. While this quiz is designed to take 35 minutes,
you have 2 hours to complete it. Work individually! You may use your own formula sheets
containing relevant hand-written notes as well as a standard or scientific calculator. To receive full
credit, you must show all of your work. Good luck!
For questions 1-3, refer to the following information: Back pain is a major health problem
because of its high prevalence and costs in terms of health care expenditures and lost productivity.
Systematic reviews have concluded that chiropractic spinal manipulation appears to be effective in
some subgroups of patients with back pain and this is one of the few treatments recommended in
clinical-practice guidelines on the care of adults with low back pain in the United States. The
effectiveness of physical therapy for back pain has not been well studied, and the results of
comparisons of physical therapy with chiropractic manipulation have conflicted.
Suppose among a large group of patients with lower back pain, 15% visit both a physical
therapist and a chiropractor, and 15% visit neither of these. Assume the probability that a
patient visits a physical therapist is 0.49. Hint: Start by drawing a Venn diagram.
̅𝑇
Not (PT or C) = 𝑃𝑜̅𝑟̅𝐶̅
0.15
PT and 𝐶̅ ̅ ̅𝑇̅
PT and C C and 𝑃
0.34 0.15 0.36
1. What is the probability that a randomly chosen patient visits a chiropractor? (3 points)
a. 0.21
b. 0.49
c. 0.51
d. 0.85
e. 0.15
- Define the events PT = patient visits physical therapist and C = patient visits chiropractor.
- We are given P(PT and C) = 0.15, P(not PT or C) = .15, P(PT) = .49
Using the addition rule, P(PT or C) = P(PT) + P(C) – P(PT and C).
- Solving for P(C), we get P(C) = P(PT or C) – P(PT) + P(PT and C).
- By the definition of complements, P(PT or C) = 1 - .15 = .85
- Using substitution, P(C) = .85 - .49 + .15 = .51
- Alternatively, since (PT and C) is mutually exclusive with (C and ̅𝑃̅𝑇̅), we can simply add these
probabilities as P(C) = .15 + .36 = .51
This study source was downloaded by 100000902171055 from CourseHero.com on 12-17-2025 20:44:56 GMT -06:00
https://www.coursehero.com/file/44864797/Quiz1AnswerKeypdf/
QUIZ 1 Answer Key
Biostatistical Applications for Public Health
George Washington University
This Document Description:
Complete PubH 6002: Biostatistical
Applications for Public Health Quiz 1 Answer
Key (MCQs with fully worked solutions)
, PubH 6002: Biostatistical Applications for Public Health
Quiz 1 - Key
Student Name:
Instructions: This quiz consists of 15 MC questions. While this quiz is designed to take 35 minutes,
you have 2 hours to complete it. Work individually! You may use your own formula sheets
containing relevant hand-written notes as well as a standard or scientific calculator. To receive full
credit, you must show all of your work. Good luck!
For questions 1-3, refer to the following information: Back pain is a major health problem
because of its high prevalence and costs in terms of health care expenditures and lost productivity.
Systematic reviews have concluded that chiropractic spinal manipulation appears to be effective in
some subgroups of patients with back pain and this is one of the few treatments recommended in
clinical-practice guidelines on the care of adults with low back pain in the United States. The
effectiveness of physical therapy for back pain has not been well studied, and the results of
comparisons of physical therapy with chiropractic manipulation have conflicted.
Suppose among a large group of patients with lower back pain, 15% visit both a physical
therapist and a chiropractor, and 15% visit neither of these. Assume the probability that a
patient visits a physical therapist is 0.49. Hint: Start by drawing a Venn diagram.
̅𝑇
Not (PT or C) = 𝑃𝑜̅𝑟̅𝐶̅
0.15
PT and 𝐶̅ ̅ ̅𝑇̅
PT and C C and 𝑃
0.34 0.15 0.36
1. What is the probability that a randomly chosen patient visits a chiropractor? (3 points)
a. 0.21
b. 0.49
c. 0.51
d. 0.85
e. 0.15
- Define the events PT = patient visits physical therapist and C = patient visits chiropractor.
- We are given P(PT and C) = 0.15, P(not PT or C) = .15, P(PT) = .49
Using the addition rule, P(PT or C) = P(PT) + P(C) – P(PT and C).
- Solving for P(C), we get P(C) = P(PT or C) – P(PT) + P(PT and C).
- By the definition of complements, P(PT or C) = 1 - .15 = .85
- Using substitution, P(C) = .85 - .49 + .15 = .51
- Alternatively, since (PT and C) is mutually exclusive with (C and ̅𝑃̅𝑇̅), we can simply add these
probabilities as P(C) = .15 + .36 = .51
This study source was downloaded by 100000902171055 from CourseHero.com on 12-17-2025 20:44:56 GMT -06:00
https://www.coursehero.com/file/44864797/Quiz1AnswerKeypdf/
