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 58% of 2341 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.)

   ,  

Interpret the resulting interval.

We are 99% confident that this interval contains the true population proportion.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.We are 99% confident that the true population proportion lies below this interval.

(b)

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,

a) Point estimate = sample proportion =   = 0.58

1 - = 1 - 0.58 = 0.42

Z/2 = Z_0.005 = 2.576

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 (((0.58 * 0.42) / 2341)

= 0.026

A 99% confidence interval for population proportion p is ,

± E   

= 0.58  ± 0.026

= ( 0.554, 0.606 )

We are 99% confident that this interval contains the true population proportion.

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

=  1 - =   0.5

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

= (2.576 / 0.02)² * 0.5 * 0.5

= 4147.36

sample size = n = 4148