Question

Help with b) please. It is not 462 nor 460

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 2342 American adults surveyed said they have watched digitally streamed TV programming on some type of device. (a) Calculate and interpret a confidence interval at the 99% confidence level for the proportion of all adult Americans who watched streamed programming up to that point in time. (Round your answers to three decimal places.) 0.493 0.547 Interpret the resulting interval. We are 99% confident that this interval contains the true population proportion. We are 99% confident that the true population proportion lies below this interval. We are 99% confident that this interval does not contain the true population proportion. We are 99% confident that the true population proportion lies above this interval. (b) What sample size would be required for the width of a 99% CI to be at most 0.06 irrespective of the value of p? (Round your answer up to the nearest integer.) 462

Answer 1

Solution :

$\hat p$ = 0.52

1 - $\hat p$ = 0.48

margin of error = E = 0.06 / 2 = 0.03

Z_$\alpha$/2 = 2.576

sample size = n = (Z_{$\alpha$ / 2} / E)² * $\hat p$ * (1 - $\hat p$ )

= (2.576 / 0.03)² * 0.52 * 0.48

= 1841

sample size = n = 1841