# #StackBounty: #hypothesis-testing #self-study #normal-distribution #t-test #likelihood-ratio Likelihood Ratio Test Equivalent with \$t\$ …

### Bounty: 50

$$newcommand{szdp}[1]{!left(#1right)} newcommand{szdb}[1]{!left[#1right]}$$
Problem Statement: Suppose that independent random samples of sizes $$n_1$$ and $$n_2$$
are to be selected from normal populations with means $$mu_1$$ and $$mu_2,$$
respectively, and common variance $$sigma^2.$$ For testing $$H_0:mu_1=mu_2$$
versus $$H_a:mu_1-mu_2>0$$ ($$sigma^2$$ unknown), show that the likelihood
ratio test reduces to the two-sample $$t$$ test presented in Section 10.8.

Note: This is Exercise 10.94 from Mathematical Statistics with Applications, 5th. Ed., by Wackerly, Mendenhall, and Scheaffer.

My Work So Far: We have the likelihood as
$$L(mu_1, mu_2,sigma^2)= szdp{frac{1}{sqrt{2pi}}}^{!!(n_1+n_2)} szdp{frac{1}{sigma^2}}^{!!(n_1+n_2)/2} expszdb{-frac{1}{2sigma^2}szdp{sum_{i=1}^{n_1}(x_i-mu_1)^2 +sum_{i=1}^{n_2}(y_i-mu_2)^2}}.$$
To compute $$Lbig(hatOmega_0big),$$ we need to find the MLE for $$sigma^2:$$
begin{align} hatsigma^2&=frac{1}{n_1+n_2}szdp{sum_{i=1}^{n_1}(x_i-mu_1)^2 +sum_{i=1}^{n_2}(y_i-mu_2)^2}. end{align}
This is the MLE for $$sigma^2$$ regardless of what $$mu_1$$ and $$mu_2$$ are.
Thus, under $$H_0,$$ we have that
$$hatsigma_0^2=frac{1}{n_1+n_2}szdp{sum_{i=1}^{n_1}(x_i-mu_0)^2 +sum_{i=1}^{n_2}(y_i-mu_0)^2},$$
and the unrestricted case is
$$hatsigma^2=frac{1}{n_1+n_2}szdp{sum_{i=1}^{n_1}(x_i-overline{x})^2 +sum_{i=1}^{n_2}(y_i-overline{y})^2}.$$
Under $$H_0,;mu_1=mu_2=mu_0,$$ so that
begin{align} Lbig(hatOmega_0big) &=szdp{frac{1}{sqrt{2pi}}}^{!!(n_1+n_2)} szdp{frac{1}{hatsigma_0^2}}^{!!(n_1+n_2)/2} expszdb{-frac{n_1+n_2}{2}}\ Lbig(hatOmegabig) &=szdp{frac{1}{sqrt{2pi}}}^{!!(n_1+n_2)} szdp{frac{1}{hatsigma^2}}^{!!(n_1+n_2)/2} expszdb{-frac{n_1+n_2}{2}}, end{align}
and the likelihood ratio is given by
begin{align} lambda &=frac{Lbig(hatOmega_0big)}{Lbig(hatOmegabig)}\ &=szdp{frac{hatsigma^2}{hatsigma_0^2}}^{!!(n_1+n_2)/2}\ &=szdp{frac{displaystylesum_{i=1}^{n_1}(x_i-overline{x})^2 +sum_{i=1}^{n_2}(y_i-overline{y})^2} {displaystylesum_{i=1}^{n_1}(x_i-mu_0)^2 +sum_{i=1}^{n_2}(y_i-mu_0)^2}}^{!!(n_1+n_2)/2}. end{align}
It follows that the rejection region, $$lambdale k,$$ is equivalent to
begin{align} frac{displaystylesum_{i=1}^{n_1}(x_i-overline{x})^2 +sum_{i=1}^{n_2}(y_i-overline{y})^2} {displaystylesum_{i=1}^{n_1}(x_i-mu_0)^2 +sum_{i=1}^{n_2}(y_i-mu_0)^2}&frac{1}{k’}-1=k”\ frac{n_1(overline{x}-mu_0)^2+n_2(overline{y}-mu_0)^2} {displaystyle(n_1-1)S_1^2+(n_2-1)S_2^2}&>k”\ frac{n_1(overline{x}-mu_0)^2+n_2(overline{y}-mu_0)^2} {displaystyledfrac{(n_1-1)S_1^2+(n_2-1)S_2^2} {n_1+n_2-2}}&>k”(n_1+n_2-2)\ frac{n_1(overline{x}-mu_0)^2+n_2(overline{y}-mu_0)^2} {S_p^2}&>k”(n_1+n_2-2). end{align}
Here
$$S_p^2=frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}.$$

My Question: The goal is to get this expression somehow to look like
$$t=frac{overline{x}-overline{y}}{S_psqrt{1/n_1+1/n_2}}>t_{alpha}.$$
But I don’t see how I can convert my expression, with the same sign for $$overline{x}$$ and $$overline{y},$$ to the desired formula with its opposite signs. What am I missing?

Get this bounty!!!

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