Respuesta :
Using the z-distribution, the 95% confidence interval for the population proportion is (0.5791, 0.7133).
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 other parameters are given as follows:
[tex]n = 195, \overline{p} = \frac{126}{195} = 0.6462[/tex]
Hence the bounds of the interval are:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6462 - 1.96\sqrt{\frac{0.6462(0.3538)}{195}} = 0.5791[/tex]
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6462 + 1.96\sqrt{\frac{0.6462(0.3538)}{195}} = 0.7133[/tex]
More can be learned about the z-distribution at https://brainly.com/question/25890103
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