A survey of 2000 Californians revealed that 1,350 supported stricter gun control laws. Develop a 99%...

A survey of 2000 Californians revealed that 1,350 supported stricter gun control laws. Develop a 99% confidence interval for the true proportion of all Californians who support stricter gun control laws.

Expert Solution

Solution :

Given that,

Point estimate = sample proportion = = x / n = 1350 / 2000 = 0.675

Z_/2 = 2.576

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

= 2.576 * ( $\sqrt$ ((0.675 * 0.325) / 2000)

= 0.027

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.675 - 0.027 < p < 0.675 + 0.027

0.648 < p < 0.702

(0.648 , 0.702)

orchestra answered 1 year ago

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