Question

A recent Gallup poll asked 1006 Americans, “During the past year, about how many books, either hardcover or paperback, did you read all or part of the way through?” Results of the survey indicated that x ¯ = 13.4 books and s = 16.6 books.

Construct a 99% confidence interval for the true mean number of books Americans read either all or part of during the past year.

Answer 1

c )solution

Given that,

= 13.4

s =16.6

n = 1006

Degrees of freedom = df = n - 1 = 1006 - 1 = 1005

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005,1005 =2.581 ( using student t table)

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

= 2.581 * ( 16.6/ 1006) = 1.35

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

- E < < + E

13.4 - 1.35 < <13.4 + 1.35

12.05 < < 14.75

(12.05 , 14.75)