Let p denotes the population proportion of all adult Americans who watched streamed programming up to that point in time.
Given : 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 devices.
i.e. sample size :n= 2343
[tex]\hat{p}=0.53[/tex]
We know that ,
Critical value for 99% confidence interval = [tex]z^*=2.576[/tex]
a) Confidence interval for population proportion:-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
[tex]0.53\pm (2.576)\sqrt{\dfrac{0.53(1-0.53)}{2343}}\\\\\approx0.53\pm0.0266\\\\=(0.53-0.0265,\ 0.53+ 0.0265)\\\\=(0.5035,\ 0.5565)[/tex]
i.e. Confidence interval : (0.5035, 0.5565)
Interpretation: A person can be 99% confident that the true proportion of all adult Americans who watched streamed programming up to that point in time lies between 0.5035 and 0.5565 .
b) Margin of error : E= half of width of CI= [tex]\dfrac{0.05}{2}=0.025[/tex]
The formula to find the sample size , if the prior estimate of population proportion is known:
[tex]n=p(1-p)(\dfrac{z^*}{E})^2[/tex]
Put all the value in the above formula :
[tex]n=0.53(1-0.53)(\dfrac{2.576}{0.025})^2\\\\=0.2491(103.04)^2\\\\=2644.75488256\approx2645[/tex]
Thus , the minimum sample size = 2645
