Question

The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 44 who smoke. Step 2 of 2 : Suppose a sample of 632 Americans over 44 is drawn. Of these people, 462 don't smoke. Using the data, construct the 90% confidence interval for the population proportion of Americans over 44 who smoke. Round your answers to three decimal places.

Answer 1

Answer:

The 90% confidence interval for the population proportion of Americans over 44 who smoke is (0.24, 0.298)

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 $1 - \frac{\alpha}{2}$.

For this problem, we have that:

632 American, 462 don't smoke, 632 - 462 = 170 smoke.

Proportion who smoke, so

$n = 632, \pi = \frac{170}{632} = 0.269$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.269 - 1.645\sqrt{\frac{0.269*0.731}{632}} = 0.24$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.269 + 1.645\sqrt{\frac{0.269*0.731}{632}} = 0.298$

The 90% confidence interval for the population proportion of Americans over 44 who smoke is (0.24, 0.298)