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).

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Feliz [49]

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N76 [4]

No it is not a function

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