In a sample of 400 students, 60% of them prefer eBooks. A. Find 98% Confidence Interval for the proportion of all students that prefer ebooksb.

b. Find the margin of erro

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).

.mp4 files, because, video

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )