Question

A poll was conducted that asked 1002 people how many books they had read in the past year. Results indicated that x overbar=10.1 books and s=16.6 books. Construct a 99​% confidence interval for the mean number of books people read.

Construct a 99​% confidence interval for the mean number of books people read.

Answer 1

solution

Given that,

= 10.1

s =16.6

n = 1002

Degrees of freedom = df = n - 1 = 1002 - 1 = 1001

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005, 1001= 2.581    ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 2.581 * (16.6 / 1002) = 1.3535

The 99% confidence interval estimate of the population mean is,

- E < < + E

10.1 - 1.3535< < 10.1+ 1.3535

8.7465< < 11.4535

( 8.7465,11.4535)