Question

A poll was conducted that asked 1009 people how many books they had read in the past year. Results indicated that

x bar=13.1 books and s=16.6 books. Construct a 90% confidence interval for the mean number of books people read.

Construct a 90​% confidence interval for the mean number of books people read.

Answer 1

Given that,

= 13.1

s =16.6

n = 1009

Degrees of freedom = df = n - 1 = 1009- 1 = 1008

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,1008= 1.646    ( using student t table)

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

= 1.646* (16.6 / 1009) = 0.8602

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

- E < < + E

13.1 - 0.8602< < 13.1+ 0.8602

12.2398 < < 13.9602

( 12.2398 ,13.9602)