Question

A supplier of television programs through a satellite service is planning to offer a new package and would like to determine what proportion of the households in their region would purchase this new package. Based on experience it is thought that it is likely 35% of households will purchase the package. If the supplier wants to be 95% confident in the outcome, how many households should they randomly sample to achieve a margin of error no greater than ±0.02 in their estimate?

Answer 1

Solution,

Given that,

= 35% = 0.35

1 - = 1 - 0.35 = 0.65

margin of error = E = 0.02

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_0.025 = 1.96

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.96 / 0.02 )² * 0.35 * 0.65

= 2184.91

Sample size = n = 2185