Question

The Centers for Disease Control reported the percentage of people 18 years of age and older who smoke (CDC website, December 14, 2014). Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .30. a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .02 (to the nearest whole number)

Answer 1

Answer:

A sample of 2017 people should be taken.

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

The margin of error is:

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

Suppose a 95% confidence level:

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

Preliminary estimate of the proportion who smoke of .30.

This means that $\pi = 0.3$

a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .02 (to the nearest whole number)

This is n for which M = 0.02. So

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

$0.02 = 1.96\sqrt{\frac{0.3*0.7}{n}}$

$0.02\sqrt{n} = 1.96\sqrt{0.3*0.7}$

$\sqrt{n} = \frac{1.96\sqrt{0.3*0.7}}{0.02}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.3*0.7}}{0.02})^2$

$n = 2016.84$

To the nearest whole number, 2017.

A sample of 2017 people should be taken.