TV advertising agencies face growing challenges in reaching

TV advertising agencies face growing challenges in reaching audience members because viewing TV programs via digital streaming is increasingly popular. the Harris poll reported on November 13, 2012, that 53% 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 have watched streamed programming.

b. what sample size would be required for the width of a 99% CI to be at most .05 irrespective of the value of p?

Solution

p = 0.53
n = 2343

a.
alpha,a = 1-0.99 = 0.01

standard error, SE = sqrt(p(1-p)/n)
SE = sqrt(0.53(1-0.53) / 2343) = 0.01031

Za/2 = Z0.005= 2.575

confidence interval:
p +/- [Za/2 * SE]
0.530 +/- 0.0266
( 0.5034 , 0.5566 )

b.

error,E = width/2 = 0.05/2 = 0.025

proportion,p = 0.53           
alpha,a = 1-0.99 = 0.01      
.          
Za/2 = Z0.005 = 2.575  
sample size,n >= [Za/2 / E]^2 * p(1-p)          
n >= [2.575 / 0.025]^2 * 0.53(1-0.53)      
n >= 2642.70      
n = 2643