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 57% 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.) , Interpret the resulting interval. We are 99% confident that this interval does not contain the true population proportion. 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 the true population proportion lies above this interval. Correct: Your answer is correct. (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,

n = 2342

Point estimate = sample proportion = = 0.57

1 - =0.43

a)

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (((0.57 * 0.43) / 2342)

= 0.026

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.57 - 0.026 < p < 0.57 + 0.026

0.544 < p < 0.596

The 99% confidence interval is : (0.544 , 0.596)

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

b)

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

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

= (2.576 / 0.02)² *0.57 * 0.43

= 4066.07

sample size = 4067