Question

In the Health ABC Study, 543 subjects owned a pet and 1977 subjects did not. Among the pet owners, there were 293 women; 975 of the non-pet owners were women. Find the proportion of pet owners who were women. Do the same for the non-pet owners. (Be sure to let Population 1 correspond to the group with the higher proportion so that the difference will be positive. Round your answers to three decimal places.)

Answer 1

The proportion of pet owners who were women is 0.540.

The proportion of non-pet owners who were women is 0.493.

Explanation:

The data provided is summarized in the table below.

Pet Owners Non-Pet Owners TOTAL

Men 250 1002 1252

Women 293 975 1268

TOTAL 543 1977 2520

Denote the events as follows:

W = a subject is a women

P = a subject is a pet owner

The probability of an event, say E, is defined as the ration of the favorable outcomes of E to the the total number of outcomes of the experiment.

P (E) = n (E) ÷ N

Here,

n (E) = favorable outcome

N = total number of outcomes

(1)

The conditional probability of an event B given that another events A has already occurred is:

P (B | A) = n (A ∩ B) ÷ n (A)

Compute the probability of a pet owner being a women as follows:

P (W | P) = n (W ∩ P) ÷ n (P)

= 975 ÷ 1977

= 0.5396

≈ 0.540

Thus, the proportion of pet owners who were women is 0.540.

(2)

Compute the probability of a non-pet owner being a women as follows:

P (W | P') = n (W ∩ P') ÷ n (P')

= 293 ÷ 543

= 0.4932

≈ 0.493

Thus, the proportion of non-pet owners who were women is 0.493.