If n=530 and ˆpp^ (p-hat) =0.83, find the margin of error at a 99% confidence level...

If n=530 and ˆpp^ (p-hat) =0.83, find the margin of error at a 99% confidence level

Give your answer to three decimals

Solutions

Expert Solution

Solution :

Given that,

n = 530

Point estimate = sample proportion = = 0.83

1 - = 0.17

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576* ( $\sqrt$ ((0.83*0.17) /530 )

E = 0.042

orchestra answered 2 years ago

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