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Sagot :
Using the z-distribution, it is found that the margin of error for the 95% confidence interval is of 0.0889.
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.
The margin of error is given by:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
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 estimate and the sample size are given as follows:
[tex]\pi = 0.71, n = 100[/tex]
Hence, the margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 1.96\sqrt{\frac{0.71(0.29)}{100}}[/tex]
M = 0.0889.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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