Question

Out of 100 people sampled, 42 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places.

Answer 1

Answer:

The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).

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

Out of 100 people sampled, 42 had kids.

This means that $n = 100, \pi = \frac{42}{100} = 0.42$

99% confidence level

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 - 2.575\sqrt{\frac{0.42*0.58}{100}} = 0.293$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 + 2.575\sqrt{\frac{0.42*0.58}{100}} = 0.547$

The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).