2 TV advertising agencies face increasing challenges in reac

2. TV advertising agencies face increasing challenges in reaching audience members because TV programs via digital streaming is gaining in popularity. The Harris poll reported on November 13, 2012, that 1242 of 2343 American adults surveyed said they have watched digitally streamed TV programming on some type of device.

(a) Using the approach discussed in class (which involved the addition of 2 successes and 2 failures to the sample thereby increasing the sample size by 4), calculate a 99% confidence interval for the proportion of all adult Americans who watched streamed programming up to that point in time.

(b) Provide an interpretation of this interval.

(c) What sample size, n , would be required for the width (this is half the length of the interval) of the confidence interval to be .025?

Solution

a)
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=1242
Sample Size(n)=2343
Sample proportion = x/n =0.5301
Confidence Interval = [ 0.5301 ±Z a/2 ( Sqrt ( 0.5301*0.4699) /2343)]
= [ 0.5301 - 2.58* Sqrt(0.0001) , 0.5301 + 2.58* Sqrt(0.0001) ]
= [ 0.5035,0.5567]
b)
Interpretations:
1) We are 99% sure that the interval [ 0.5035,0.5567] ~ [ 50.35%, 55.67% ] contains the
true population proportion
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contain the true
population proportion

c)

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.025 is = 2.24
Samle Proportion = 0.5301
ME = 0.026
n = ( 2.24 / 0.026 )^2 * 0.5301*0.4699
= 1848.8965 ~ 1849