Question 1: A population study population has identified that 65% of women have normal bone mineral density (BMD score ≥ -1.0), 25% have osteopenia (BMD score between -2.5 & -1.0), and 10% have osteoporosis (BMD score < -2.5).

A. For a randomly selected woman, are the bone health states independent events? Mutually exclusive events? Why or why not?

They are independent because the bone health state of one woman does not depend on the bone health of another. They are mutually exclusive because one woman cannot have more than one bone health state at a given point in time.

B. What is the probability a randomly selected woman does not have osteoporosis?

P(A)= 0.65 P(B)=0.25 P(C)=0.10

P(Osteo-)= P(A)+ P(B)
P(Osteo-)= 0.65+ 0.25
P(Osteo-)= 0.90

C. Assume that older men have a probability of normal BMD of 75%, 20% with osteopenia, and 5% with osteoporosis. Also assume that 40% of the aging population is male and 60% is female. What is the probability that a randomly selected older person has osteoporosis?

P(D)= 0.75 P(E)=0.20 P(F)=0.05

P(Osteo+)= 0.6*P(C)+ 0.4*P(F)
P(Osteo+)= 0.6*0.10 + 0.4*0.05
P(Osteo+)= 0.08

D. Using the info from #3, are gender and osteoporosis status independent? Why or why not?

Assuming that there are no other confounding variable differences between the two groups, gender and osteoporosis status are not independent, since men have a lower probability of having osteoporosis than women.

E. The probability of experiencing a hip fracture during a fall is 4% for women with normal BMD, 9% for women with osteopenia, and 18% for women with osteoporosis. Suppose a woman arrives in the ER with a hip fracture following a fall. What is the probability that the woman has osteoporosis?

P(HF|A)=0.04; P(HF|B)=0.09; P(HF|C)=0.18

P(C|HF)= [P(HF|C) * P(C)] / [P(HF|A) * P(A) + P(HF|B) * P(B) + (P(HF|C) * P(C)]
P(C|HF)=(0.18*0.10)/[(0.04*0.65) + (0.09* 0.25) + (0.18*0.10)]
P(C|HF)=0.3947

Question 1 In a simple linear regression analysis of 19 observatio...