Question

In the Health ABC Study, 547 subjects owned a pet and 1971 subjects did not. Among the pet owners, there were 300 women; 982 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.)

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

Solution:

Given:

owned a pet: 547 and number of women = 300 thus n1 = 547 and x1 = 300

Did not owned a pet : 1971 and number of women = 982 , thus n2 = 1971 and x2 = 982

Find the proportion of pet owners who were women.

and

Find the proportion of NON-pet owners who were women.

Give a 95% confidence interval for the difference in the two proportions.

Formula;

where

and

We need to find z_c value for c=95% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.95) /2 = 1.95 / 2 = 0.9750

Look in z table for Area = 0.9750 or its closest area and find z value.

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : Z_c = 1.96

Thus

Thus

Thus we are 95% confident that true difference in the two proportions is: