Question

(b) In a study of 120 volunteers who were given nicotine patches, 58 had quit smoking. (i) Construct a 95% confidence interval for p. 1 - a= a = a/2= 20/2= X1 (use at least 4 decimal places for ), ĝ=1 - Ộ = confidence interval for p: Ô Izovo

Answer 1

Solution :

Given that,

n = 120

x = 58

Point estimate = sample proportion = $\hat p$ = x / n = 58/120=0.4833

q =1 - $\hat p$ = 1 - 0.4833=0.5167

At 95% confidence level

$\alpha$ = 1 - 95%  

$\alpha$ = 1 - 0.95 =0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z_{0.025 = 1.96}

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

= 1.96 ($\sqrt$((0.4833*0.5167) /120 )

= 0.0894

A 95% confidence interval for population proportion p is ,

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

0.4833 - 0.0894 < p <  0.4833 +0.0894

0.3939< p < 0.5727

The 95% confidence interval for the population proportion p is : 0.3939,0.5727