Question 1157945
the 99%CI half-interval is t(0.995,df=1015)*s/sqrt(n).   {it is a survey}
=2.581*16.6/sqrt(1015)
=1.34
the interval is (12.36, 15.04)
probably should round to (12.4, 15.0) units books

That means while we don't know the average number of books all people have read in the past year, we are 99% confident that the true number lies in the interval (12.4, 15.0)