Question

6. A recent poll asked 700 Americans. “How many books did you read in the last year? Results of the poll indicated that t= 12 books and s = 8 books. (a) Construct a 99% confidence interval for the number of books read. (b) Suppose the standard deviation of 8 books is the population standard deviation instead of the sample standard deviation). What would be the confidence interval in that case?

Answer 1

6.

(a)

Formula for Confidence Interval for Population mean when population Standard deviation is not known

s : Tta/2.n-1 XP vn

Sample size : n=700

$\alpha$ for 99% confidence level = (100-99)/100 =0.01

$\alpha$/2 =0.01/2 =0.005

Degrees of freedom = n-1 =700-1=699

t_{$\alpha$/2,n-1} = t_0.005,699 = 2.5829

99% confidence interval for the number of books read

1 – 70.005,699 - = 12 +2.5829 x = 12+2.5829 x 0.3024 = 12 +0.781 = (11.219, 12.781) 700

99% confidence interval for the number of books read = (11.219,12.781)

(b)

Formula for Confidence Interval for Population mean when population Standard deviation is known

士 za/2 X

Population standard deviation : $\sigma$ =8

$\alpha$ for 99% confidence level = (100-99)/100 =0.01

$\alpha$/2 =0.01/2 =0.005

_{Z$\alpha$/2,n-1} = Z_0.005 = 2.5758

99% confidence interval for the number of books read

1 20.005 X = 12+2.5758 x = 12+2.5758x0.3024 = 12+0.7788 = (11.2212.12.7788)

99% confidence interval for the number of books read =(11.2212,12.7788)