For this assignment, I'm asked to find the expected value of the length of a 95% confidence intervial for the population mean from which a sample of size 9 has been drawn. The population is distributed normally, with unknown mean and variance.
I know a t-statistic is required because variance is unknown, and I know that the length of the confidence interval will be 2*(1.397)*(s/3), so the expectation can be simplified to (2/3)*(1.397)*E(s).
The hint I'm given is that E(s) can be rewritten, to be the following:
(pop. st. dev)/(sqrt(n-1)) * E{sqrt[(n-1)*s^2/(pop. variance)]}
(Sorry that's such a mess, it basically is just a coefficient portion outside the expectation, and the inside is distributed as a Chi-Square dist'n with n-1 degrees of freedom.
The problem I'm having is, even though I know the distribution of the interior of the hint's expecation, I don't know how to take the expectation of the square root of the Chi-Square distribution.
How do I evaluate E[sqrt(Chi-Square dist'n, df = 8)]?
Thanks.