39 nuts are gathered. They have an average size of 3.34 cent and a standard deviation of .45 cent. Find the 95% confidence interval for the mean size of all the nuts
assuming normality this should be a t with n-1 degrees of freedom,
which most people will approximate with a z anyhow, which is 1.96 for a 95 percent CI.
(by the way, is mr fantasy one of the nuts? It would seem so.)
$\displaystyle \bar X \pm t_{n-1,.025}s/\sqrt{n}$
@baldeagle: *Ahem* .... Space 1999 Eagle Transporter - Where All Eagles Come Home
IF we don't have normality, then they want us to use z=1.96 via the central limit theorem. We kind of need to know what the underlying distribution is. Most likley they want the 1.96 percentile from the st normal tables. In any case with n reasonably large (30 according to most books) it doesn't matter which way we go. But one should know where you're sampling from.