Question

A recent survey showed that among 725 randomly selected subjects who completed 4 years of college, 139 smoke and 586 do not smoke. Determine a 95% confidence interval for the true proportion of the given population that smokes.

Answer 1

Answer:

The 95% confidence interval for the true proportion of the given population that smokes is (0.1630, 0.2204).

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 level of $1-\alpha$, we have the following confidence interval of proportions.

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

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

Sample of 725, 139 smoke:

This means that $n = 725, \pi = \frac{139}{725} = 0.1917$

95% confidence level

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1917 - 1.96\sqrt{\frac{0.1917*0.8083}{725}} = 0.1630$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1917 + 1.96\sqrt{\frac{0.1917*0.8083}{725}} = 0.2204$

The 95% confidence interval for the true proportion of the given population that smokes is (0.1630, 0.2204).