Question

A survey was conducted that asked 1011 people how many books they had read in the past year. results indicated that x overbarequals14.8 books and sequals16.6 books. construct a 90​% confidence interval for the mean number of books people read. interpret the interval

Answer 1

Since we do not know the population standard deviation, we will use the t-distribution to construct the 90% confidence interval of the mean number of books people read. The value of $t_{\frac{\alpha }{2}}$ with 1010 degrees of freedom and with alpha equals to 0.10 is 1.64

First, we need to solve for the margin of error, E.
     $E=t_{\frac{\alpha }{2}}\cdot \frac{s}{\sqrt{n}}=1.64\cdot \frac{16.6}{\sqrt{1011}}=0.86$

The lower bound of the confidence interval is 
     $LB=\overline{x}-E=14.8-0.86=13.94$

The upper bound of the confidence interval is 
     $LB=\overline{x}+E=14.8+0.86=15.66$

Therefore, the 90% confidence interval is (13.94, 15.66).

We are 90% confident that the interval from 13.94 books to 15.66 books does contain the true value of the population mean number of books people read.