Question

An advertising firm states that 27% of households will buy something on the internet next year. how large a sample is needed to be 92% confident that this proportion is within 5% of true proportion.

answer is 242

Answer 1

Solution :

Given that,

= 0.27

1 - = 0.73

margin of error = E = 0.05

At 92% confidence level the z is ,

= 1 - 92% = 1 - 0.92 = 0.08

/ 2 = 0.08 / 2 = 0.04

Z_/2 = Z_0.04 = 1.751

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

= (1.751 / 0.05)² * 0.27*0.73

= 241.72

sample size = 242

The answer is 242