Respuesta :
Answer:
The 95% confidence interval of the true proportion of all Americans who are in favor of gun control legislation is (0.5471, 0.5779).
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 interval [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 zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
A random sample of 4000 citizens yielded 2250 who are in favor of gun control legislation. This means that [tex]n = 4000[/tex] and [tex]\pi = \frac{2250}{4000} = 0.5625[/tex]
Estimate the true proportion of all Americans who are in favor of gun control legislation using a 95% confidence interval
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5625 - 1.96\sqrt{\frac{0.5625*0.4375}{4000}} = 0.5471[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5625 + 1.96\sqrt{\frac{0.5625*0.4375}{4000}} = 0.5779[/tex]
The 95% confidence interval of the true proportion of all Americans who are in favor of gun control legislation is (0.5471, 0.5779).
