Question

Let $X_1, X_2, \ldots, X_n$ be a random sample from $N(\mu_1, \sigma^2)$ distribution and $Y_1, Y_2, \ldots, Y_m$ be a random sample from $N(\mu_2, \sigma^2)$ distribution, where $\mu_i \in \mathbb{R}, i = 1, 2$ and $\sigma > 0$. Suppose that the two random samples are independent. Define \[ \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i \text{and} W = \frac{\sqrt{mn} (\bar{X} - \mu_1)}{\sqrt{\sum_{i=1}^{m} (Y_i - \mu_2)^2}}. \] Then which one of the following statements is TRUE for all positive integers $m$ and $n$? 
 

• A. $W \sim t_{m}$
• B. $W \sim t_{n}$
• C. $W^2 \sim F_{1, m}$
• D. $W^2 \sim F_{m, n}$

Hint:

Remember that $(Z/\sqrt{V/m}) \sim t_m$ if $Z \sim N(0,1)$ and $V \sim \chi^2_m$. Squaring converts the $t$-distribution into an $F$-distribution.

Answer 1

The Correct Option is A

Step 1: Distribution of $\bar{X$.}
Since $X_i \sim N(\mu_1, \sigma^2)$, we have \[ \bar{X} \sim N\left(\mu_1, \frac{\sigma^2}{n}\right). \] Thus, \[ Z_1 = \frac{\sqrt{n} (\bar{X} - \mu_1)}{\sigma} \sim N(0, 1). \]

Step 2: Distribution of the denominator.
Each $(Y_i - \mu_2)/\sigma \sim N(0,1)$. Hence, \[ \frac{1}{\sigma^2} \sum_{i=1}^{m} (Y_i - \mu_2)^2 \sim \chi^2_m. \]

Step 3: Combine numerator and denominator.
The numerator $Z_1$ is $N(0,1)$ and independent of the denominator term. Hence, \[ W = \frac{Z_1}{\sqrt{(\chi^2_m / m)}} \sim t_m. \]

Step 4: Square the statistic.
Since $t_m^2 \sim F_{1, m}$, \[ W^2 \sim F_{1, m}. \]

Step 5: Conclusion.
\[ \boxed{W^2 \sim F_{1, m}.} \]