Question

2. (-/19 Points) DETAILS DEVORESTAT9 7.E.020. MY NOTES TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streaming is gaining in popularity. A pol reported that 52% of 2344 American adults surveyed said they have watched digitally streamed TV programming on some type of device. (a) Calculate and interpreta 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 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 0.04 irrespective of the value of p? (Round your answer up to the nearest integer.)

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = 0.52

1 - $\hat p$ = 1 - 0.52 = 0.48

Z$\alpha$/2 = Z_0.005 = 2.576

Margin of error = E = Z_{$\alpha$ / 2} * (($\hat p$ * (1 - $\hat p$ ))$\sqrt$ / n)

= 2.576 ($\sqrt$((0.52 * 0.48) / 2344)

= 0.027

A 99% confidence interval for population proportion p is ,

$\hat p$ ± E   

= 0.52  ± 0.027

= ( 0.493, 0.547 )

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

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

$\hat p$ =  1 - $\hat p$ =   0.5

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

= (2.576 / 0.02)² * 0.5 * 0.5

= 4147.36

sample size = n = 4148