Question

In an extensive government study conducted this year a simple random sample of 500 households was obtained and 80 had experienced property damage due to a crime. Calculate and interpret a 95% confidence interval for the true proportion of households that have experienced property damage due to crime

Answer 1

Using the z-distribution, the 95% confidence interval is (0.1279, 0.1921), and it means that we are 95% sure that the proportion for the entire population is between these two values.

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

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

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

In this problem:

• A sample of 500 households was obtained and 80 had experienced property damage due to a crime, hence $n = 500, \pi = \frac{80}{500} = 0.16$.

• 95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so $z = 1.96$.

Hence:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.16 - 1.96\sqrt{\frac{0.16(0.84)}{500}} = 0.1279$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.16 + 1.96\sqrt{\frac{0.16(0.84)}{500}} = 0.1921$

The 95% confidence interval for the true proportion of households that have experienced property damage due to crime is (0.1279, 0.1921), and it means that we are 95% sure that the proportion for the entire population is between these two values.

More can be learned about the z-distribution at https://brainly.com/question/25890103