Mathematics: 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

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Tv advertising agencies face increasing challenges

Ответ:

lurinha8

08.01.2022

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.