Question

CNN wants to estimate the percent of people in the US who support a bill in congress. How large a sample will they need if they want a margin of error less than 2.5% at the 90% confidence level?

Answer 1

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 2.5% = 0.025

At 90% confidence level

= 1 - 0.90 = 0.10

/2 =0.05

Z_/2 = 1.645 ( Using z table )

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

= (1.645 / 0.025)² * 0.5 * 0.5

=1082.41

Sample size = 1083