Question

A survey was conducted that asked 1016 people how many books they had read in the past year. Results indicated that x overbar = 14.9 books and s = 16.6 books. Construct a 90​% confidence interval for the mean number of books people read. Interpret the interval.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 14.9

sample standard deviation = s = 16.6

sample size = n = 1016

Degrees of freedom = df = n - 1 = 1016 - 1 = 1015

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,1015 = 1.646

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

= 1.646 * ( 16.6/ 1016)

= 0.9

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

- E < < + E

14.9 - 0.9 < < 14.9 + 0.9

14 < < 15.8

(14 , 15.8)