Question

A local health center noted that in a sample of 400 patients 80 were referred to them by the local hospital.

a.


Provide a 95% confidence interval for all the patients who are referred to the health center by the hospital.

b.


What size sample would be required to estimate the proportion of hospital referrals with a margin of error of 0.04 or less at 95% confidence?

Answer 1

Given n=400, p=80/400= 0.2

(a) Given a=0.05, |Z(0.025)|=1.96 (check standard normal table)

So 95% CI is
p ± Z*√[p*(1-p)/n]

--> 0.2 ± 1.96*sqrt(0.2*0.8/400)

--> (0.1608, 0.2392)

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(b) Given E=0.04

So n=(Z/E)^2*p*(1-p)

=(1.96/0.04)^2*0.2*0.8

=384.16

Take n=385