Question

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streaming is gaining in popularity. A poll reported that 52% of 2343 American adults surveyed said they have watched digitally streamed TV programming on some type of device.

What sample size would be required for the width of a 99% CI to be at most 0.04 irrespective of the value of p̂? (Round your answer up to the nearest integer.)

Answer 1

Solution :

Given that,

= 52% = 0.52

1 - = 1 - 0.52 = 0.48

margin of error = E = width / 2 =  0.04 / 2 = 0.02

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.58

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

= (2.58 / 0.02)² * 0.52 * 0.48

= 4154

Sample size = 4154