Respuesta :
Using the z-distribution, the 95% confidence interval for the true proportion of students attending private schools is:
(0.0919, 0.1525).
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 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 the critical value is z = 1.96.
The sample size and the estimate are given as follows:
[tex]n = 450, \pi = \frac{55}{450} = 0.1222[/tex]
Hence the bounds of the interval are:
- [tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1222 - 1.96\sqrt{\frac{0.1222(0.8778)}{450}} = 0.0919[/tex]
- [tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1222 + 1.96\sqrt{\frac{0.1222(0.8778)}{450}} = 0.1525[/tex]
More can be learned about the z-distribution at https://brainly.com/question/25890103
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