Question

A survey was conducted that asked 1018 people how many books they had read in the past year. Results indicated that x overbarxequals=11.8 books and s=16.6 books. Construct a 99​% confidence interval for the mean number of books people read. Interpret the interval.

Answer 1

solution

Given that,

= 11.8

s =16.6

n = 1018

Degrees of freedom = df = n - 1 = 1018- 1 =1017

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df= t0.005,1017 = 2.581 ( using student t table)

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

= 2.581 * (16.6 / 1018) = 1.3428

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

- E < < + E

11.8 - 1.3428 < <11.8 + 1.3428

10.4572 < < 13.1428

( 10.4572 ,13.1428)