Answer:
The 99% confidence level for the proportion of all adult Americans who watched streamed programming up to that point in time is (0.514, 0.566). This means that we are 99% sure that the true proportion of all American adults surveyed said they have watched digitally streamed TV programming on some type of device is between 0.514 and 0.566.
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 z-score that has a p-value of
.
A poll reported that 54% of 2342 American adults surveyed said they have watched digitally streamed TV programming on some type of device.
This means that

99% confidence level
So
, z is the value of Z that has a p-value 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 (0.514, 0.566). This means that we are 99% sure that the true proportion of all American adults surveyed said they have watched digitally streamed TV programming on some type of device is between 0.514 and 0.566.