Question

A survey was conducted that asked

10131013

people how many books they had read in the past year. Results indicated that

x overbarxequals=11.211.2

books and

sequals=16.616.6

books. Construct a

9090​%

confidence interval for the mean number of books people read. Interpret the interval.

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Click the icon to view the table of critical​ t-values.

Construct a

9090​%

confidence interval for the mean number of books people read and interpret the result. Select the correct choice below and fill in the answer boxes to complete your choice.

​(Use ascending order. Round to two decimal places as​ needed.)

A.If repeated samples are​ taken,

9090​%

of them will have a sample mean between

nothing

and

nothing.

B.There is a

9090​%

probability that the true mean number of books read is between

nothing

and

nothing.

C.There is

9090​%

confidence that the population mean number of books read is between

nothing

and

nothing.

Answer 1

sample mean, xbar = 11.2
sample standard deviation, s = 16.6
sample size, n = 1013
degrees of freedom, df = n - 1 = 1012

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.646

ME = tc * s/sqrt(n)
ME = 1.646 * 16.6/sqrt(1013)
ME = 0.858

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (11.2 - 1.646 * 16.6/sqrt(1013) , 11.2 + 1.646 * 16.6/sqrt(1013))
CI = (10.34 , 12.06)

There is 90​% confidence that the population mean number of books read is between 10.34 and 12.06