A public health screening program will estimate the proportion of Perth women ag

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 1,337 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) 243 of the women in the sample were found to be HPV carriers. Calculate a 95% confidence interval for the proportion of Perth women aged 16-24 who have an HPV infection.

Explanation / Answer

(a) p = 0.19 n = 1337

Mean = np = 1337 * 0.19 = 254.03

Standard deviation = (np(1-p)) = (254.03*0.81) = 14.3445.

X = 20% of 1337 = 267.4

Z = (X - ) /

= (267.4 - 254.03) / 14.3445

= 0.9321

P(Z > 0.9321) from tables = 0.8244.

(b) Let the confidence interval be (a,b)

n = 243 = 254.03

z for 95% is 1.96

a = - z / n = 254.03 - 1.96*14.3445 / 243 = 252.2264.

b = + z / n = 254.03 + 1.96*14.3445 / 243 = 255.8336.

The 95% confidence interval is (252.2264, 255.8336).

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