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 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.) 460

Help with b

Answer 1

given that, n = 2942 = 0.52 To find, Calculate and Interpaet confidence Interval 99%. C.L for popwation of all adult Americans who walehed Streamed posgramming that point In time gg1 C.I is, up to The - (§ + Zanz Jo(16) ) Here, Zoool = 2.576 =) ( 0.52 + 2.576 0.52% 0.48 2342 7 - (0.52 = 0.02659) (0.52 0.52 - 0.02659 0.52 + 0.02659) Ans ) (10.493) 10.547 Ž Iterpart the rest Ans. 0 we 991. confident that this interval contains the toue population pooportion. are

③ given that n=0 > C.L=gg./. E-0.06 ſ = 0.52 1-P = 1-0.52 0.48 La = Lool : 2-58 D n. (Za)? (PCO-)) (E) 2 (2.68)? (0.48X0:52) (0.062 1.6614374 0.0036 461.510 ~ 462 size Ans. Sample 462