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 59% of 2345 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 0.03 irrespective of the value of p̂? (Round your answer up to the nearest integer.)

Answer 1

Solution,

Given that,

=  1 - = 0.5

margin of error = E = 0.03

At 99% confidence level

= 1 - 99%

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005  = 2.576

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

= (2.576 / 0.03)2 * 0.5 * 0.5

= 1843.27

sample size = n = 1844