Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
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
#SPJ1
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
