# 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 55% of 2341 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 thr

###### Question:

(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.

(b) What sample size would be required for the width of a 99% CI to be at most 0.03 irrespective of the value of p?? (Round your answer up to the nearest integer.)

## Answers

Answer:a)[0.5235, 0.5765]To interpret this result, we could say there is a 99% of probability that the proportion of American adults who have watched digitally streamed TV programming on some type of device is between 52.35% and 57.65%b) 1,843 American adultsStep-by-step explanation:The 99% confidence interval is given by [tex]\bf p\pm z^*\sqrt{p(1-p)/n}[/tex]where p = the proportion of American adults surveyed who said they have watched digitally streamed TV programming on some type of device = 55% = 0.55[tex]\bf z^*[/tex] the z-score for a 99% confidence level associated with the Normal distribution N(0,1). We can do this given that the sample size (2,341) is big enoughn = sample size = 2,341We can find the [tex]\bf z^*[/tex] value either with a table or with a spreadsheet.In Excel use NORM.INV(0.995,0,1)In OpenOffice Calc use NORMINV(0.995;0;1)We get a value of [tex]\bf z*[/tex]= 2.576and our 99% confidence interval is[tex]\bf 0.55\pm 2.576\sqrt{0.55*0.45/2341}=0.55\pm 2.576*0.0103=0.55\pm 0.265 = [0.5235, 0.5765][/tex]To interpret this result, we could say there is a 99% of probability that the proportion of American adults who have watched digitally streamed TV programming on some type of device is between 52.35% and 57.65%We are 99% confident that this interval contains the true population proportion.(b) What sample size would be required for the width of a 99% CI to be at most 0.03 irrespective of the value of p?? (Round your answer up to the nearest integer.)The sample size n in a simple random sampling is given by[tex]\bf n=\frac{(z^*)^2p(1-p)}{e^2}[/tex]where e is the error proportion = 0.03hence[tex]\bf n=\frac{(2.576)^2p(1-p)}{(0.03)^2}=7373.0844p(1-p)=7373.0844p-7373.044p^2[/tex]taking the derivative with respect to p, we getn'(p)=7373.0844-2*7373.0844pand n'(p) = 0 when p=0.5By taking the second derivative we see n''(p)<0, so p=0.5 is a maximum of nThis means that if we set p=0.5, we get the maximum sample size for the confidence level required for the proportion error 0.03Replacing p with 0.5 in the formula for the sample size we get[tex]\bf n=7373.0844*0.5-7373.044(0.5)^2=1,844[/tex]rounded up to the nearest integer.

Answer:a)The 99% confidence level for the proportion of all adult Americans who watched streamed programming up to that point in time is between (0.524, 0.576).We are 99% confident that this interval contains the true population proportion.b) We need a sample size of 1842.Step-by-step explanation:In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]In whichz is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].The margin of error of the interval is given by:[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]For this problem, we have that:[tex]n = 2341, \pi = 0.55[/tex]99% confidence levelSo [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].The lower limit of this interval is:[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.55 - 2.575\sqrt{\frac{0.55*0.45}{2341}} = 0.524[/tex]The upper limit of this interval is:[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.55 + 2.575\sqrt{\frac{0.55*0.45}{2341}} = 0.576[/tex]The 99% confidence level for the proportion of all adult Americans who watched streamed programming up to that point in time is between (0.524, 0.576).This means that we are 99% sure that the true population proportion is in that interval.So the answer is:We are 99% confident that this interval contains the true population proportion. (b) What sample size would be required for the width of a 99% CI to be at most 0.03 irrespective of the value of p?? (Round your answer up to the nearest integer.)We don't know the value of [tex]\pi[/tex] in this case, so we use [tex]\pi = 0.5[/tex], which is the case for which we are going to need the largest sample size.We need a sample size of n, and n is found when M = 0.03. So[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex][tex]0.03 = 2.575\sqrt{\frac{0.5*0.5}{n}}[/tex][tex]0.03\sqrt{n} = 2.575*0.5[/tex][tex]\sqrt{n} = \frac{2.575*0.5}{0.03}[/tex][tex](\sqrt{n})^{2} = (\frac{2.575*0.5}{0.03})^{2}[/tex][tex]n = 1841.8[/tex]Rounding upWe need a sample size of at least 1842.