Question

Estimate the minimum sample size needed to achieve the margin of error E= 0.023 for a 95% confidence interval

Answer 1

Solution :

Given that,

= 0.5 ( use 0.5 when not given )

1 - = 1 - 0.5= 0.5

margin of error = E = 0.023

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96 ( Using z table ( see the 0.025 value in standard normal (z) table corresponding z value is 1.96 )

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

= (1.96 / 0.023)² * 0.5 * 0.5

= 1816

Sample size = 1816