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).

a

step-by-step explanation:

a

2/3x + 7 = 29

step-by-step explanation:

two thirds = 2/3

plus a number = x

at least 29 = 29

there for you should get 2/3x + 7 = 29