Question

A random sample of 700 voters in a particular city found 266 voters who voted yes on proposition 200. Find a 95% confidence interval for the true percent of voters in this city who voted yes on proposition 200. Express your results to the nearest hundredth of a percent.

Answer 1

Solution :

Given that,

n = 700

x = 266

Point estimate = sample proportion = $\hat p$ = x / n = 266 / 700 = 0.38

1 - $\hat p$ = 1 - 0.38 = 0.62

At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ =1 - 0.95 =0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z0.025 = 1.960

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

= 1.96 ($\sqrt$((0.38 * 0.62 ) / 700)

= 0.036

A 95% confidence interval for population proportion p is ,

$\hat p$ ± E

= 0.38 ± 0.036

= ( 0.344, 0.416 )