Question

An online poll found that 11% of adults have never travelled beyond the state they were born in. If you were to conduct your own poll, find the sample size need to 96% confident that the error does not exeed 5%

Answer 1

Solution :

Given that,

= 0.11

1 - = 1 - 0.11= 0.89

margin of error = E =5% = 0.05

At 91% confidence level the z is ,

Z/2 = 1.695 ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.695 / 0.05)2 * 0.11 * 0.89

= 112.5=113