Question

What sample size is needed to give a margin of error within ±2.5% in estimating a population proportion with 90% confidence? We estimate that the population proportion is about 0.4

Answer 1

Solution :

Given that,

= 0.4

1 - = 1 - 0.4= 0.6

margin of error = E = 0.025

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= (1.645 / 0.025)2 * 0.4 * 0.6

=1039.1136

Sample size = 1040