Answer :
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:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which:
- [tex]\pi[/tex] 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 [tex]n = 500, \pi = \frac{80}{500} = 0.16[/tex].
- 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].
Hence:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.16 - 1.96\sqrt{\frac{0.16(0.84)}{500}} = 0.1279[/tex]
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.16 + 1.96\sqrt{\frac{0.16(0.84)}{500}} = 0.1921[/tex]
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