# A recent Gallup poll asked 100

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

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

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 /$\sqrt$n)

= 2.581 * ( 16.6/ $\sqrt$1006) = 1.35

The 99% confidence interval estimate of the population meanis,

– E < < + E

13.4 – 1.35 < <13.4 + 1.35

12.05 < < 14.75

(12.05 , 14.75)

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