Question

Chapter 9

A survey was conducted that asked 1017 people how many books they had read in the past years. Results indicated that x bar =10.8 books and s = 16.6 books. Construct a 90% confidence interval for the mean number of books people read.

Construct a 90% confidence interval for the mean number of books people read and interpret the results.

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

1.) If repeated samples are taken,90% of them will have a sample mean between ___ and ___

2._ There is a 90% chance that the true mean number of books read is between ____ and ___

3.) There is 90% confidence that the population mean number of books read is between ____ and ___

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 10.8

sample standard deviation = s = 16.6

sample size = n = 1017

Degrees of freedom = df = n - 1 = 1016

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,1016 = 1.646

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

= 1.646 * (16.6 / 1017)

= 0.86

The 90% confidence interval estimate of the population mean is,

- E < < + E

10.8 - 0.86 < < 10.8 + 0.86

9.94 < < 11.66

(9.94 , 11.66)

3.) There is 90% confidence that the population mean number of books read is between 9.94 and 11.66