Respuesta :
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.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
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 [tex]\pi = 0.54, n = 2342[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.54 - 2.575\sqrt{\frac{0.54*0.46}{2342}} = 0.514[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.54 + 2.575\sqrt{\frac{0.54*0.46}{2342}} = 0.566[/tex]
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.
