Prediction Interval Question

[iocal]

Hi guys,

I'm a bit stuck in an exercise and I would appreciate some help.
Let $\displaystyle X_1,X_2,...,X_n,X_{n+1}$ be a random sample of size n+1, n>1, from a distribution that is $\displaystyle N(\mu,\sigma^2)$. Let $\displaystyle \bar{X}=\sum_{i=1}^n X_i/n$ and $\displaystyle S^2=\sum_{i=1}^n(X_i-\bar{X})^2 /(n-1)$. Show that in order for the statistic $\displaystyle c(\bar{X}-X_{n+1})/S$ to have a t-distribution, the constant c must equal $\displaystyle \sqrt{\frac{n-1}{n+1}}$.

Okay, we can view the statistic as a difference in means Confidence Interval. From the definition of the t-distribution we know that $\displaystyle T=\frac{W}{\sqrt{V/r}}$ where $\displaystyle W\sim N(0,1)\ and\ V\sim \chi^2(r)$.
We know that $\displaystyle \bar{X}\sim N(\mu,\frac{\sigma^2}{n})$ and $\displaystyle X_{n+1}\sim N(\mu,\sigma^2)$. Since those two are independent(because of the random sample) we can say that the total variance is equal to $\displaystyle \sigma^2(\frac{n+1}{n})$. Since the means are equal our numerator then becomes $\displaystyle \displaystyle\sqrt{n} \frac{\bar{X}-X_{n+1}}{\sigma\sqrt{n+1}}$.
My problem is now to find the chi squared distributed denominator that will give me the $\displaystyle \sqrt{n-1}$ part of c. Since the variances are unequal I do not think we can use a pooled estimator. Any suggestions are greatly appreciated. Thanks.


EDIT: I have looked it up extensively in the meantime and it seems that everyone uses the coefficient $\displaystyle \displaystyle\frac{\sqrt{n}}{\sqrt{n+1}}$ instead of $\displaystyle \displaystyle \frac{\sqrt{n-1}}{\sqrt{n+1}}$. The first case is the one I am able to show and it conforms with my intuition but I do not know which one is correct. Could it be that my book is mistaken at this point? God, this is confusing indeed.