Respuesta :
Using the z-distribution, the 99% confidence interval to estimate the population proportion is: (0.2364, 0.4836).
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, we have a 99% confidence level, hence[tex]\alpha = 0.99[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so the critical value is z = 2.575.
The estimate and the sample size are given by:
[tex]\pi = 0.36, n = 100[/tex].
Then the bounds of the interval are:
- [tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.36 - 2.575\sqrt{\frac{0.36(0.64)}{100}} = 0.2364[/tex]
- [tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.36 + 2.575\sqrt{\frac{0.36(0.64)}{100}} = 0.4836[/tex]
The 99% confidence interval to estimate the population proportion is: (0.2364, 0.4836).
More can be learned about the z-distribution at https://brainly.com/question/25890103
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