Question

A poll was conducted that asked 1014 people how many books they had read in the past year. Results indicated that x overbar=10.6 books and s=16.6 books. Construct a 90​% confidence interval for the mean number of books people read. LOADING... Click the icon to view the table of areas under the​ t-distribution. Construct a 90​% confidence interval for the mean number of books people read.

Answer 1

Given that,

= 10.6

s =16.6

n = 1014

Degrees of freedom = df = n - 1 =1014 - 1 = 1013

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,1013 = 1.646    ( using student t table)

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

=1.646 * ( 16.6/ 1014) = 0.8581

The 90% confidence interval estimate of the population mean is,

- E < < + E

10.6 -0.8581 < < 10.6+ 0.8581

9.7419 < < 11.4581

( 9.7419 , 11.4581)