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.)

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 , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

The margin of error of the interval is given by:

For this problem, we have that:

99% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

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 in this case, so we use , 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

Rounding up

We need a sample size of at least 1842.

62.5 paisa

step-by-step explanation:

4000*70 = 280000

280000-(2000*90) = 100000

4000-2400 = 1600

100000/1600 = 62.5

( give brainliest)

£56

step-by-step explanation: