Question

A poll was conducted that asked 1007 people how many books they had read in the past year. Results indicated that x overbarxequals=12.9 books and

s equals=16.6 books. Construct a 90​%

confidence interval for the mean number of books people read.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 12.9

sample standard deviation = s = 16.6

sample size = n = 1007

Degrees of freedom = df = n - 1 = 1006

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,24 = 1.646

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

= 1.646 * (16.6 / 1007)

= 0.861

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

- E < < + E

12.9 - 0.861 < < 12.9 + 0.861

12.039 < < 13.761

(12.039 , 13.761)