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 2349 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 the true population proportion lies below this interval. We are 99% confident that the true population proportion lies above this 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.

(b)

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,

Point estimate = sample proportion = = 0.59

1 - = 1 - 0.59 = 0.41

Z/2 = Z0.005 = 2.576

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

= 2.576 (((0.59 * 0.41) / 2349 )

= 0.026

A 99% confidence interval for population proportion p is ,

± E   

= 0.59  ± 0.026

= ( 0.564, 0.616 )

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

b) margin of error = E = 0.03/ 2 = 0.015

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

= (2.576 / 0.015)2 * 0.5 * 0.5

= 7373.08

sample size = n = 7374