Question

A survey was conducted that asked 1003 1003 people how many books they had read in the past year. results indicated that x overbar x equals = 10.5 10.5 books and s equals = 16.6 16.6 books. construct a 99 99​% confidence interval for the mean number of books people read. interpret the interval.

Answer 1

The confidence interval is

$10.5\pm1.35 = (9.15, 11.85)$.  This means that we can be 99% confident that the mean number of books people read lies between 9.15 and 11.85.

To find the confidence interval, we first find the z-score associated with it:

Convert 99% to a decimal:  0.99
Subtract from 1:  1-0.99=0.01
Divide by 2:  0.01/2 = 0.005
Subtract from 1:  1-0.005 = 0.995

Using a z-table (http://www.z-table.com) we see that this is associated with a z-score between 2.57 and 2.58.  Since both are equally far from this value we will use 2.575.

We calculate the margin of error using

$ME=z*\frac{s}{\sqrt{n}} \\ \\=2.575*\frac{16.6}{\sqrt{1003}}=1.35$

This means that the confidence interval is
$10.5\pm1.35$

The lower limit is given by 10.5-1.35 = 9.15.
The upper limit is given by 10.5+1.35 = 11.85