Question

Many people think that a national lobby's successful fight against gun control legislation is reflecting the will of a minority of Americans. A random sample of 4000 citizens yielded 2250 who are in favor of gun control legislation. Estimate the true proportion of all Americans who are in favor of gun control legislation using a 95% confidence interval

Answer 1

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 $\pi$, and a confidence interval $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

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 $n = 4000$ and $\pi = \frac{2250}{4000} = 0.5625$

Estimate the true proportion of all Americans who are in favor of gun control legislation using a 95% confidence interval

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5625 - 1.96\sqrt{\frac{0.5625*0.4375}{4000}} = 0.5471$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5625 + 1.96\sqrt{\frac{0.5625*0.4375}{4000}} = 0.5779$

The 95% confidence interval of  the true proportion of all Americans who are in favor of gun control legislation is (0.5471, 0.5779).