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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).
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