Question

A recent survey showed that among 700 randomly selected subjects who completed 4 years of college, 151 smoke and 549 do not smoke. Determine a 95% confidence interval for the true proportion of the given population that smokes. 95% CI: to

Answer 1

Solution :

Given that,

n = 700

x = 151

= x / n = 151 / 700 = 0.2157

1 - = 1 - 0.2157 = 0.7843

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z _{/ 2} * [( * (1 - )) / n]

= 1.96 * [(0.2157 * 0.7843) / 700]

= 0.03

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.2157 - 0.03 < p < 0.2157 + 0.03

0.1857 < p < 0.2457

(0.1857, 0.2457)