Thread: Width of a Confidence Interval

I am OK with part a, but I am having some troubles with part b.

Attempt for part b:

note: P(T>t_(n1+n2-2),alpha/2)=alpha/2 where T~t distribution with n1+n2-2 degrees of freedom.

Now, as n1 increases and n2 increases,
(i) t_(n1+n2-2),alpha/2 gets smaller
(ii) denominator gets larger
(iii) the ∑ terms gets larger because the upper indices of summation are n1 and n2, respectively

(i) and (ii) push towards a narrower confidence interval, but (iii) pushes towards a wider confidence interval. How can we determine the ultimate result?

Thank you for any help!

[note: also under discussion in sos math cyberboard]

I agree that the "intuitive" answer is that the confidence interval will be "narrower" as the sample sizes increase, but how can we prove this more rigorously? My method above doesn't quite seem to work...
The trouble is that as n1 and n2 increase, the numerator ∑(Xi-Xbar)^2 + ∑(Yi-Ybar)^2 also increases (because now we are summing over more and more non-negative terms as n1 and n2, the upper indices of summation, increase), and this tends to make the CI wider. How can we deal with this matter?