This is a homework which I don't know how to solve. All other examples of pivotal quantities examples I've tackled were relatively easy, this one though seems hard.
Suppose $\rm{X}=(X_1, \dots, X_m)$ and $\rm{Y}=(Y_1, \dots, Y_n)$ are samples from $N(\mu_1, \sigma_1^2)$ and $N(\mu_2, \sigma_2^2)$ respectively. Let $\sigma = \sigma_1^2 = \sigma_2^2$, $\theta = (\mu_1, \mu_2, \sigma)$ and $g(\theta) = \mu_2 – \mu_1$. Prove that pivotal quantity of $g(\theta)$ has the form of
$$
\frac{\overline{Y}-\overline{X}-g(\theta)}{\sqrt{\frac{(m-1)S_X^2+(n-1)S_Y^2}{n+m-2}}\sqrt{\frac{1}{m}+\frac{1}{n}}}
$$
where as usual $S_X^2 = \frac{\sum_1^mX_i-\overline{X_i}}{m-1}$, $\overline{X} = \frac{\sum_i^nX_i}{m}$. Use the pivotal quantity to approximate confidence interval of $g(\theta)$ with confidence $1 – \alpha$.
I've got a couple of hints from my professor:
- for $Z \sim N(0,1)$ and $Y \sim \chi^2_n$ we know $X = \frac{Z}{\sqrt{Y/n}}$ has t-Student distribution with n d.o.f.
- for $X_i \sim N(0,1)$ we have $\sqrt n (\overline X – \mu)/\sigma \sim N(0,1)$ and $(n-1)S^2/\sigma^2 \sim \chi^2_{n-1}$
I tried playing with the equations but that led me nowhere. I assume I should somehow find the forms shown in the second hint and then move on to the first one. Any help greatly appreciated.
Best Answer
Outline: Call your displayed quantity $T$ and call its denominator $SE.$ Then $$T = \frac{(\bar Y - \bar X)-(\mu_y - \mu_x)}{SE} \sim \mathsf{T}(\nu = n+m-2).$$
If $t^*$ cuts probability 0.025 from the upper tail of $\mathsf{T}(n+m-2),$ then $P(-t^* \le T \le t^*)= 0.95.$
Upon pivoting, you get a 95% CI for $\mu_y = \mu_s$ of the form $ \bar Y - \bar X \pm t^*SE.$