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 sample size would be required to estimate the proportion of all hospital referrals to the health center with a margin of error of .04 at 95 % confidence?

Answer 1

Solution :

Given that,

n = 400

x = 80

Point estimate = sample proportion = $\hat p$ = x / n = 0.2

1 - $\hat p$ = 0.8

a)

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.96

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.96 * ($\sqrt$((0.2 * 0.8) / 400)

= 0.039

A 95% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.2 - 0.039 < p < 0.2 + 0.039

0.161 < p < 0.239

(0.161 , 0.239)

b)

sample size = n = (Z_{$\alpha$ / 2} / E )² * $\hat p$ * (1 - $\hat p$ )

= (1.96 / 0.04)² * 0.2 * 0.8

= 384.16

sample size = 385