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: Analyze the numerator

$$\sqrt{mn}(\bar{X} - \mu_1) = \sqrt{mn} \cdot \frac{(\bar{X} - \mu_1)}{\sigma/\sqrt{n}} \cdot \frac{\sigma}{\sqrt{n}}$$

Since $\bar{X} \sim N(\mu_1, \sigma^2/n)$: $$Z = \frac{\bar{X} - \mu_1}{\sigma/\sqrt{n}} \sim N(0,1)$$

Therefore: $$\sqrt{mn}(\bar{X} - \mu_1) = \sqrt{m} \cdot \sigma \cdot Z$$

Step 2: Analyze the denominator

$$\sum_{i=1}^m (Y_i - \mu_2)^2 = \sum_{i=1}^m \left(\frac{Y_i - \mu_2}{\sigma}\right)^2 \cdot \sigma^2$$

Since $\frac{Y_i - \mu_2}{\sigma} \sim N(0,1)$: $$\sum_{i=1}^m \left(\frac{Y_i - \mu_2}{\sigma}\right)^2 \sim \chi^2_m$$

Let $U = \sum_{i=1}^m \left(\frac{Y_i - \mu_2}{\sigma}\right)^2 \sim \chi^2_m$

Then: $\sum_{i=1}^m (Y_i - \mu_2)^2 = \sigma^2 U$

Step 3: Compute $W$

$$W = \frac{\sqrt{m} \cdot \sigma \cdot Z}{\sqrt{\sigma^2 U}} = \frac{\sqrt{m} \cdot \sigma \cdot Z}{\sigma\sqrt{U}} = \frac{\sqrt{m} \cdot Z}{\sqrt{U}}$$

$$= \frac{Z}{\sqrt{U/m}}$$

Step 4: Identify the distribution

The statistic $W = \frac{Z}{\sqrt{U/m}}$ where $Z \sim N(0,1)$ and $U \sim \chi^2_m$ are independent, follows a t-distribution with $m$ degrees of freedom:

$$W \sim t_m$$

Answer: (A) $W \sim t_m$