Question

A public health screening program will estimate the proportion of Perth women aged 16-24 who have a human papillomavirus (HPV) infection. (There are nearly 200 types of HPV, most of which cause no symptoms in most people.) A random sample of 1,437 Perth women in this age group was selected.

(a) If the population proportion of women carrying HPV were 19%, what is the chance that the sample proportion could exceed 20% in a random sample of this size?

(b) 253 of the women in the sample were found to be HPV carriers. Calculate a 90% confidence interval for the proportion of Perth women aged 16-24 who have an HPV infection.

Answer 1

a) n = 1437

P = 0.19

= p = 0.19

= sqrt(p(1 - p)/n)

= sqrt(0.19 * (1 - 0.19)/1437)

= 0.0103

P( > 0.2)

= P(( - )/ > (0.2 - )/)

= P(Z > (0.2 - 0.19)/0.0103)

= P(Z > 0.97)

= 1 - P(Z < 0.97)

= 1 - 0.8340

= 0.1660

b) = 253/1437 = 0.1761

At 90% confidence interval the critical value is z_0.05 = 1.645

The 90% confidence interval for population proportion is

+/- z_0.05 * sqrt((1 - )/n)

= 0.1761 +/- 1.645 * sqrt(0.1761 * (1 - 0.1761)/1437)

= 0.1761 +/- 0.0165

= 0.1596, 0.1926