Question

In: Statistics and Probability

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital...

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 ? (Round your answer up to the nearest integer.)

Solutions

Expert Solution

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


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