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 , and a confidence level of , we have the following confidence interval of proportions.
In which
z is the z-score that has a p-value of .
The margin of error is of:
In a sample of 400 students, 60% of them prefer eBooks.
This means that
98% confidence level
So , z is the value of Z that has a p-value of , so .
Margin of error -> Question b:
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).