Question

in a survey of 700 community college students, 481 indicated that they have read a book for personal enjoyment during the school year (based on data from the community college survey of student engagement.) determine a 90% confidence interval for the proportion of community college students who have read a book for personal enjoyment during the school year.

Answer 1

The 90% confidence interval is (0.687, 0.705).

How did we get the value?

In order to construct this confidence interval, we first get our point estimate of the proportion of students who read for personal enjoyment, which we can get through simple division:

p = 481/700

= 0.687

There is need to determine the margin of error, which is the width of the confidence interval. For a 90% confidence interval, we can find the margin of error by solving the equation below, as shown:

`E = 1.64 * \sqrt\frac{{p}(1-p)}n`

`E = 1.64 * \sqrt(0.687(1-0.687))/(700)`

= 0.018

Now, we add and subtract the margin of error from the point estimate to get the boundaries of our confidence interval:

Low = 0.687 - 0.018

= 0.669

High = 0.687 + 0.018

= 0.705

So, the confidence interval is (0.687, 0.705).

The interpretation of this interval is that we are 90% sure the true proportion of college students who read for pleasure is between 0.687 and 0.705.