Question

4. For a 99% confidence level, how large of a sample size is needed for a margin of error of 0.03 for the es-timate of the population proportion? Past studies are not available

Answer 1

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 0.03

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58 ( Using z table ( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )

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

= (2.58 / 0.03)2 * 0.5 * 0.5

=1849