Respuesta :
Answer:
The 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter is (0.46, 0.58). This means that we are 99% sure that the true proportion of all U.S. adult Twitter users who get some news on Twitter is between these two values.
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 found that 52% of U.S. adult Twitter users get at least some news on Twitter. The standard error for this estimate was 2.4%
This means that:
[tex]\pi = 0.52, \sqrt{\frac{\pi(1-\pi)}{n}} = 0.024[/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}} = 0.52 - 2.575(0.024) = 0.46[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.52 + 2.575(0.024) = 0.58[/tex]
The 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter is (0.46, 0.58). This means that we are 99% sure that the true proportion of all U.S. adult Twitter users who get some news on Twitter is between these two values.
