In: Statistics and Probability
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
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 * (((( * (1 - )) / n)
= 2.576* (((0.83*0.17) /530 )
E = 0.042