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 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 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.)

Answer 1

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 $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error of the interval is given by:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

$n = 2341, \pi = 0.55$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.55 - 2.575\sqrt{\frac{0.55*0.45}{2341}} = 0.524$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.55 + 2.575\sqrt{\frac{0.55*0.45}{2341}} = 0.576$

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 $\pi$ in this case, so we use $\pi = 0.5$, 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

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.03\sqrt{n} = 2.575*0.5$

$\sqrt{n} = \frac{2.575*0.5}{0.03}$

$(\sqrt{n})^{2} = (\frac{2.575*0.5}{0.03})^{2}$

$n = 1841.8$

Rounding up

We need a sample size of at least 1842.

Answer 2

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 adults

Step-by-step explanation:

The 99% confidence interval is given by  

$\bf p\pm z^*\sqrt{p(1-p)/n}$

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

$\bf z^*$ 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 enough

n = sample size = 2,341

We can find the $\bf z^*$ 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 $\bf z*$= 2.576

and our 99% confidence interval is

$\bf 0.55\pm 2.576\sqrt{0.55*0.45/2341}=0.55\pm 2.576*0.0103=0.55\pm 0.265 = [0.5235, 0.5765]$

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

$\bf n=\frac{(z^*)^2p(1-p)}{e^2}$

where  

e is the error proportion = 0.03

hence

$\bf n=\frac{(2.576)^2p(1-p)}{(0.03)^2}=7373.0844p(1-p)=7373.0844p-7373.044p^2$

taking the derivative with respect to p, we get

n'(p)=7373.0844-2*7373.0844p

and  

n'(p) = 0 when p=0.5

By taking the second derivative we see n''(p)<0, so p=0.5 is a maximum of n

This means that if we set p=0.5, we get the maximum sample size for the confidence level required for the proportion error 0.03

Replacing p with 0.5 in the formula for the sample size we get

$\bf n=7373.0844*0.5-7373.044(0.5)^2=1,844$

rounded up to the nearest integer.