Question

In the Health ABC Study, 546 subjects owned a pet and 1967 subjects did not. Among the pet owners, there were 297 women; 988 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.)

p̂1

=

p̂2

=

(b) Give a 95% confidence interval for the difference in the two proportions. (Do not use rounded values. Round your final answers to three decimal places.)

Answer 1

(a)

1 = Proportion of pet owners who were women = 297/546 =0.544

2 = Proportion of non-pet owners who were women = 988/1967 =0.502

So,

Answer is:

1	=	0.544
2	=	0.502

(b)

Q = 1 - P =0.4887

So,

= 0.0242

= 0.05

From Tabl, critical values of Z = 1.96

So,

Confidence Interval:
(0.544 - 0.502) (1.96 X 0.0242)

= 0.042 0.0474

= ( - 0.005 , 0.089)

So,

Confidence Interval:

- 0.005 < P1 - P2 < 0.089