Respuesta :
Using the z-distribution, considering the standard 95% confidence level, the margin of error is of 0.0486 = 4.86%.
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]
The margin of error is:
[tex]M = 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 by:
[tex]n = 350, \pi = \frac{240}{350} = 0.6857[/tex]
Hence the margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 1.96\sqrt{\frac{0.6857(0.3143)}{350}}[/tex]
M = 0.0486 = 4.86%.
More can be learned about the z-distribution at https://brainly.com/question/25890103
#SPJ1
