Question

1) A survey was conducted that asked 1002 people how many books they had read in the past year. Results indicated that x=12.9 books and s=16.6 books. Construct a 99​% confidence interval for the mean number of books people read. Interpret the interval.

Answer 1

Given that,

= 12.9

s =16.6

n = 1002

Degrees of freedom = df = n - 1 = 1002- 1 = 1001

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, 1001= 2.581

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

= 2.581* (16.6 / 1002) =1.3535

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

- E < < + E

12.9 -1.3535   < < 12.9+ 1.3535

11.5465< < 14.2535

(11.5465,14.2535)