Suppose X1, X2, ..., Xn and Y1, Y2, ..., Ym are independent random samples with normal distribution with mean and variance (a1 and 9) and (a2 and 25), respectively. Using the x-bar and y-bar sample mean statistics, find the pivotal quantity for the difference in means a1-a2 !

QUESTION:

Suppose X1, X2, ..., Xn and Y1, Y2, ..., Ym are independent random samples with normal distribution with mean and variance (a1 and 9) and (a2 and 25), respectively.  Using the x-bar and y-bar sample mean statistics, find the pivotal quantity for the difference in means a1-a2 !

ANSWER:

$X_{1},...,X_{n}\sim N(a1,9)$

$Y_{1},...,Y_{n}\sim N(a2,25)$

1. Distribution for $\bar{X}$

By using MGF then:

$\begin{aligned}M_{\bar{X}}&=M_{\frac{\sum_{i=1}^{n}xi}{n}}(t)\\&=M_{\sum_{i=1}^{n}xi}(\frac{t}{n})\\&=\left [ M_{x} \right (\frac{t}{n})]^{n}\\&=\left [ e^{a1\frac{t}{n}+\frac{1}{2}\frac{t^{2}}{n^{2}}9} \right ]^{n}\\&=e^{a1t+\frac{1}{2}t^{2}\frac{9}{n}}\hspace{1cm},\bar{X}\sim N(a1,\frac{9}{n})\end{aligned}$

2. Distribution for $\bar{Y}$

By using MGF then:

$\begin{aligned}M_{\bar{Y}}&=M_{\frac{\sum_{i=1}^{m}yi}{m}}(t)\\&=M_{\sum_{i=1}^{m}yi}(\frac{t}{m})\\&=\left [ M_{y} \right (\frac{t}{m})]^{m}\\&=\left [ e^{a2\frac{t}{m}+\frac{1}{2}\frac{t^{2}}{m^{2}}25} \right ]^{m}\\&=e^{a2t+\frac{1}{2}t^{2}\frac{25}{m}}\hspace{1cm},\bar{Y}\sim N(a2,\frac{25}{m})\end{aligned}$