Question

A survey was conducted that asked 1008 people how many books they had read in the past year. Results indicated that x overbarequals14.4 books and sequals16.6 books. Construct a 99​% confidence interval for the mean number of books people read. Interpret the interval.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 14.4

sample standard deviation = s = 16.6

sample size = n = 1008

Degrees of freedom = df = n - 1 = 100 8- 1 = 1007

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,1007 = 2.581

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

= 2.581 * (16.6 / 1008)

= 1.35

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

- E < < + E

14.4 -1.35 < < 4.4 + 1.35

13.05 < < 15.75

(13.05 , 15.75)