Question

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

Answer 1

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.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the z-score that has a p-value of
`1 - (\alpha)/(2)`.

The margin of error is of:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

In a sample of 400 students, 60% of them prefer eBooks.

This means that
`n = 400, \pi = 0.6`

98% confidence level

So
`\alpha = 0.02`, z is the value of Z that has a p-value of
`1 - (0.02)/(2) = 0.99`, so
`Z = 2.054`.

Margin of error -> Question b:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`M = 2.054√(0.6*0.4){400}}`

`M = 0.05`

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).