Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
Answer:
a) The 98% Confidence Interval for the proportion of all students that prefer ebooks is (0.55, 0.65).
b) The margin of error is of 0.05.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In a sample of 400 students, 60% of them prefer eBooks.
This means that [tex]n = 400, \pi = 0.6[/tex]
98% confidence level
So [tex]\alpha = 0.02[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.054[/tex].
Margin of error -> Question b:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 2.054\sqrt{0.6*0.4}{400}}[/tex]
[tex]M = 0.05[/tex]
The margin of error is of 0.05.
A.Find 98% Confidence Interval for the proportion of all students that prefer ebooksb.
Sample proportion plus/minus the margin of error.
0.6 - 0.05 = 0.55
0.6 + 0.05 = 0.65
The 98% Confidence Interval for the proportion of all students that prefer ebooks is (0.55, 0.65).
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
