A random sample x1, x2, ..., xm from N(0, 0) has sufficient statistics m =(x,-) x=x/m and S = (x-x). 1=1 It is proposed to take a second independent random sample of size n from the same distribution and to calculate from the values y, y2, ..., y, the two statistics = y= yn and S-(-). 1-1 Show that P, S, S) = p(|x, 0)p(S; \0,)(0|S)

d, and hence that p(S;|x, S) = p(S;\S}) = [p(S;\0\)=(0,\S9d02. Using the known expressions for p(S10) (exercise 5.20) and (0|S) (equation 5.3.4) evaluate this integral and show that p(x, S) is such that (m-1)S (n-1)S has an F-distribution on (n-1) and (m-1) degrees of freedom.

The posterior distribution of 0, based on the combined sample of m+n observations will, if m+n is large (so that the f-distribution is almost normal) and the little information about 0, provided by the difference between and is ignored, be approximately normal with mean (mx+ny)/(m+n) and variance (S+S)/(m+n-2) (m+n). Show how to use the above result and tables of the F-distribution to determine the value of n to be taken for the second sample in order to be 90% certain that the 95% confidence interval for 0, obtained by this approximation has width less than some pre-assigned number

c. Obtain also p(x, S).