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
, and a confidence interval
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the zscore that has a pvalue of
.
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
and ![\pi = \frac{2250}{4000} = 0.5625](https://tex.z-dn.net/?f=%5Cpi%20%3D%20%5Cfrac%7B2250%7D%7B4000%7D%20%3D%200.5625)
Estimate the true proportion of all Americans who are in favor of gun control legislation using a 95% confidence interval
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5625 - 1.96\sqrt{\frac{0.5625*0.4375}{4000}} = 0.5471](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.5625%20-%201.96%5Csqrt%7B%5Cfrac%7B0.5625%2A0.4375%7D%7B4000%7D%7D%20%3D%200.5471)
The upper limit of this interval is:
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5625 + 1.96\sqrt{\frac{0.5625*0.4375}{4000}} = 0.5779](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.5625%20%2B%201.96%5Csqrt%7B%5Cfrac%7B0.5625%2A0.4375%7D%7B4000%7D%7D%20%3D%200.5779)
The 95% confidence interval of the true proportion of all Americans who are in favor of gun control legislation is (0.5471, 0.5779).