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Sagot :
Using the z-distribution, the minimum sample size required by this student is of 385.
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.
We want a margin of error of M = 0.05, with no prior estimate, hence [tex]\pi = 0.5[/tex], then we have to solve for n to find the minimum sample size.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.05 = 1.96\sqrt{\frac{0.5(0.5)}{n}}[/tex]
[tex]0.05\sqrt{n} = 1.96(0.5)[/tex]
[tex]\sqrt{n} = 1.96 \times 10[/tex]
[tex](\sqrt{n})^2 = (1.96 \times 10)^2[/tex]
n = 384.16
Rounding up, the minimum sample size is of 385.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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