Answer:
The 99% confidence interval for the fraction of US adult Twitter users who get some news on Twitter is (0.4872, 0.6018). It means that we are 99% sure that the true proportion of US adult Twitter users who get some news on Twitter is between 0.4872 and 0.6018.
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 zs[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex], and s is the standard error.
54% of US adult Twitter users get at least some news on Twitter.
This means that [tex]\pi = 0.54[/tex]
The standard error for this estimate was 2.4%
This means that [tex]s = 0.024[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex]
The lower bound is:
[tex]\pi - zs = 0.54 - 2.575*0.024 = 0.4782[/tex]
The upper bound is:
[tex]\pi + zs = 0.54 + 2.575*0.024 = 0.6018[/tex]
The 99% confidence interval for the fraction of US adult Twitter users who get some news on Twitter is (0.4872, 0.6018). It means that we are 99% sure that the true proportion of US adult Twitter users who get some news on Twitter is between 0.4872 and 0.6018.