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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.)

2 Answers

3 votes

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.

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{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

The margin of error of the interval is given by:


M = z\sqrt{(\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 - (0.01)/(2) = 0.995, so
Z = 2.575.

The lower limit of this interval is:


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

The upper limit of this interval is:


\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.55 + 2.575\sqrt{(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{(\pi(1-\pi))/(n)}


0.03 = 2.575\sqrt{(0.5*0.5)/(n)}


0.03√(n) = 2.575*0.5


√(n) = (2.575*0.5)/(0.03)


(√(n))^(2) = ((2.575*0.5)/(0.03))^(2)


n = 1841.8

Rounding up

We need a sample size of at least 1842.

User Steve Skrla
by
8.6k points
2 votes

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

Explanation:

The 99% confidence interval is given by


\bf p\pm z^*√(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√(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=((z^*)^2p(1-p))/(e^2)

where

e is the error proportion = 0.03

hence


\bf n=((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.

User Dkarzon
by
8.1k points
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