Question

In a survey of 1000 US adults, 700 think police officers should be required to wear body cameras while on duty. Construct a 99% confidence interval for the population proportion.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = x / n = 700 / 1000 = 0.700

1 - $\hat p$ = 1 - 0.700 = 0.3

Z$\alpha$/2 = 2.576

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

= 2.576 * ($\sqrt$((0.700 * 0.3) / 1000)

Margin of error = E = 0.037

A 99% confidence interval for population proportion p is ,

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

0.700 - 0.037 < p < 0.700 + 0.037

0.663 < p < 0.737

The 99% confidence interval for the population proportion p is : 0.663 , 0.737

Answer 2

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = x / n = 700 / 1000 = 0.700

1 - $\hat p$ = 1 - 0.700 = 0.3

Z$\alpha$/2 = 2.576

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

= 2.576 * ($\sqrt$((0.700 * 0.3) / 1000)

Margin of error = E = 0.037

A 99% confidence interval for population proportion p is ,

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

0.700 - 0.037 < p < 0.700 + 0.037

0.663 < p < 0.737

The 99% confidence interval for the population proportion p is : 0.663 , 0.737