Respuesta :
Using the z-distribution, as we are working with a proportions, it is found that the 99% confidence interval for the proportion of people who consider themselves athletic is (0.0911, 0.1239).
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, the parameters are as follows:
[tex]z = 2.575, n = 2373, \pi = \frac{255}{2373} = 0.1075[/tex]
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1075 - 2.575\sqrt{\frac{0.1075(0.8925)}{2373}} = 0.0911[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1075 + 2.575\sqrt{\frac{0.1075(0.8925)}{2373}} = 0.1239[/tex]
The 99% confidence interval for the proportion of people who consider themselves athletic is (0.0911, 0.1239).
More can be learned about the z-distribution at https://brainly.com/question/25730047
