Question

A poll is taken in which 349 out of 500 randomly selected voters indicated their preference for a certain candidate.

Find the margin of error for this 99% confidence interval for p.

Answer 1

Solution :

Given that,

n = 500

x = 349

= x / n = 349 / 500 = 0.698

1 - = 1 - 0.698 = 0.302

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 * (((0.698 * 0.302) / 500 )

= 0.053

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.698 - 0.053 < p < 0.698 + 0.053

0.645 < p < 0.751

The 99% confidence interval for the population proportion p is : ( 0.645 , 0.751)