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Three independent samples $(X_1,X_2,...,X_n)(Y_1,Y_2,...,Y_n)$ and $(Z_1,Z_2,...,Z_n)$ are drawn from normal population, each having the same unknown variance $\sigma^2$ but, $E(X_i)=m_1,E(Y_i)=m_2,E(Z_i)=m_1-m_2,\forall i$,show that the test of $H_0:m_1=\lambda m_2$ can be carried out by the test statistic $$T=\frac{[(2-\lambda)\bar X+(1-2\lambda)\bar Y+(1+\lambda)\bar Z]\sqrt n}{s\sqrt{6(\lambda ^2-\lambda +1)}}$$ where $s^2$ is the pooled sample variance , $\lambda$ is known constant and $T$~$t_{3n-2}$ , under $H_0$.

First I think $m_1=\lambda m_2$ as $m_1-\lambda m_2=0$ and I can define $T=E(X_i)-\lambda E(Y_i)$ .But since there also given $E(Z_i)=m_1-m_2$ It should be used.Please help me.

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So, under $H_0$

$[(2-\lambda)\bar X+(1-2\lambda)\bar Y+(1+\lambda)\bar Z]$~$N(0,\frac{[(2-\lambda)^2+(1-2\lambda)^2+(1+\lambda)^2]\sigma ^2}{n})$ Now estimate $\sigma ^2$ by the pooled sample variance $s$.

Note that $\bar X$~$N(m_1,\frac{\sigma^2}{n})$,$\bar Y$~$N(m_2,\frac{\sigma^2}{n})$ and $\bar Z$~$N(m_1-m_2,\frac{\sigma^2}{n})$ Similarly $\frac{(n-1)s_1^2}{\sigma ^2}$~$\chi ^2_{n-1}$,$\frac{(n-1)s_2^2}{\sigma ^2}$~$\chi ^2_{n-1}$ and $\frac{ns_3^2}{\sigma ^2}$~$\chi ^2_{n}$ independently.

Hence $[(2-\lambda)\bar X+(1-2\lambda)\bar Y+(1+\lambda)\bar Z]$~$N(0, {6\sigma ^2(\lambda ^2-\lambda +1)}/n)$ and $\frac{(3n-2)s^2}{\sigma ^2}$~$\chi ^2_{3n-2}$

So,$\dfrac{\frac{[(2-\lambda)\bar X+(1-2\lambda)\bar Y+(1+\lambda)\bar Z]\sqrt n}{\sigma\sqrt {6(\lambda ^2-\lambda +1)}}}{\sqrt {\frac{(3n-2)s^2}{\sigma^2}/(3n-2)}}$~$t_{3n-2}$

We get the test statistics $$T=\frac{[(2-\lambda)\bar X+(1-2\lambda)\bar Y+(1+\lambda)\bar Z]\sqrt n}{s\sqrt{6(\lambda ^2-\lambda +1)}}$$

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