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Suppose $x_{n+1}$ is an additional observation, which is distributed as $N(\mu,\sigma^2)$, and is independent of $X_1,X_2,...,X_n$, also $X_1,X_2,...,X_n$ are iid from $N(\mu,\sigma^2)$.

I have tried to change the form of $\frac{X_{n+1}-\bar{X}}{S\sqrt{1+\frac{1}{n}}}$, in order to get N(0,1) on top and Chi-square at the bottom, however, my effort went nowhere. $\frac{X_{n+1}-\bar{X}}{S\sqrt{1+\frac{1}{n}}}$=$\frac{X_{n+1}-\bar{X}}{\sigma\sqrt\frac{S^2}{\sigma^2}\sqrt{1+\frac{1}{n}}}$

Got stuck at dealing with $\frac{X_{n+1}-\bar{X}}{\sigma\sqrt{1+\frac{1}{n}}}$

Can anyone help to take a look at it?

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Note that $$ X_{n+1}-\bar{X}\sim N(0, \sigma^2(1+n^{-1})) $$ where the second parameter is variance since a linear combination of independent normals is normally distributed. It follows that $$ Z=\frac{X_{n+1}-\bar{X}}{\sigma\sqrt{1+n^{-1}}}\sim N(0, 1). $$ Next if $S^2$ denotes the sample variance of $X_{1}, \dotsc, X_n$, we have that $$ W=\frac{(n-1)S^2}{\sigma^2}\sim \chi ^2_{(n-1)} $$ and further $Z$ is independent of $W$ whence $$ T=\frac{Z}{\sqrt{W/n-1}}=\frac{X_{n+1}-\bar{X}}{S\sqrt{1+\frac{1}{n}}}\sim t_{(n-1)} $$ as desired.

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