Question

Of 150 adults selected randomly from one town, 30 of them smoke. Construct a 99% confidence interval for the true percentage of all adults in the town that smoke.

Answer 1

Answer:

The 99% confidence interval for the true percentage of all adults in the town that smoke is (0.1159, 0.2841).

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:

Of 150 adults selected randomly from one town, 30 of them smoke. This means that $n = 150, p = \frac{30}{150} = 0.2$

99% confidence interval

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.20 - 2.575\sqrt{\frac{0.20*0.80}{150}} = 0.1159$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.20 + 2.575\sqrt{\frac{0.20*0.80}{150}}{119}} = 0.2841$

The 99% confidence interval for the true percentage of all adults in the town that smoke is (0.1159, 0.2841).