Discussion
Let \(X_{1}, X_{2}, \ldots, X_{9}, Y\) be independent and identically distributed \(N(\mu, \sigma^{2})\) random variables. Define \(\bar{X} = \frac{1}{9}\sum_{i=1}^{9} X_{i}\), \(S^{2} = \frac{1}{8}\sum_{i=1}^{9} (X_{i} - \bar{X})^{2}\).
The distribution of \(\frac{3}{\sqrt{10}} \left(\frac{\bar{X} - Y}{S}\right)\) is
Answer: C
The statistic has the \(t_{8}\) distribution.
Since \(Y\) is independent of \(\bar{X}\), with \(Var(\bar{X}) = \frac{\sigma^{2}}{9}\),
\[ \begin{aligned} \bar{X} - Y &\sim N\left(0, \sigma^{2}\left(\tfrac{1}{9} + 1\right)\right) \\ &= N\left(0, \tfrac{10\sigma^{2}}{9}\right) \end{aligned} \]
so
\[ Z = \frac{3(\bar{X} - Y)}{\sqrt{10}\,\sigma} \sim N(0,1) \]
Also \(\frac{8S^{2}}{\sigma^{2}} \sim \chi_{8}^{2}\), independent of \(\bar{X}\) and \(Y\). Dividing:
\[ \begin{aligned} \frac{Z}{\sqrt{\frac{8S^{2}/\sigma^{2}}{8}}} &= \frac{Z\sigma}{S} \\ &= \frac{3}{\sqrt{10}}\left(\frac{\bar{X} - Y}{S}\right) \sim t_{8} \end{aligned} \]
The degrees of freedom come from \(S^{2}\), which uses 9 observations and loses one to \(\bar{X}\), so \(t_{9}\) is wrong. A ratio with a normal numerator can be negative, so it cannot be chi-square.
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