Respuesta :
Using the z-distribution, we have that:
a) The confidence interval is: (24.33%, 30.37%).
b) The correct option is: No, the confidence interval includes 0.25, so the true percentage could easily equal 25%.
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 90% confidence level, hence[tex]\alpha = 0.9[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.9}{2} = 0.95[/tex], so the critical value is z = 1.645.
The other parameters are given as follows:
[tex]n = 427 + 161 = 588, \pi = \frac{161}{588} = 0.2735[/tex]
Then the bounds of the interval are:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2735 - 1.645\sqrt{\frac{0.2735(0.7265)}{588}} = 0.2433[/tex]
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2735 + 1.645\sqrt{\frac{0.2735(0.7265)}{588}} = 0.3037[/tex]
As a percentage, the interval is: (24.33%, 30.37%).
25% is part of the interval, hence the correct statement is:
No, the confidence interval includes 0.25, so the true percentage could easily equal 25%.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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