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 56% of 2343 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.

(b) What sample size would be required for the width of a 99% CI to be at most 0.05 irrespective of the value of p̂? (Round your answer up to the nearest integer.)

Answer 1

Solution :

Given that,

n = 2343

Point estimate = sample proportion = = 0.56

1 - = 1 - 0.56=0.44

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 (((0.56*0.44) / 2343)

= 0.026

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.56 - 0.026 < p < 0.56 + 0.026

0.534< p < 0.586

The 99% confidence interval for the population proportion p is : (0.534 , 0.586)

solution

width = 0.05

margin of error = E = width / 2 =0.05 / 2 = 0.025

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z_/2 = 2.576

Sample size = n = (Z_/2 / E)² * * (1 - )

= (2.576 / 0.025)² * 0.56 * 0.44

=2616

Sample size = 2616