#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}&<k^{2/(n_1+n_2)}=k’\
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-overline{x})^2+n_1(overline{x}-mu_0)^2
+sum_{i=1}^{n_2}(y_i-overline{y})^2+n_2(overline{y}-mu_0)^2}&<k’\
frac{1}{1+dfrac{n_1(overline{x}-mu_0)^2+n_2(overline{y}-mu_0)^2}
{displaystylesum_{i=1}^{n_1}(x_i-overline{x})^2
+sum_{i=1}^{n_2}(y_i-overline{y})^2}}&<k’\
frac{n_1(overline{x}-mu_0)^2+n_2(overline{y}-mu_0)^2}
{displaystylesum_{i=1}^{n_1}(x_i-overline{x})^2
+sum_{i=1}^{n_2}(y_i-overline{y})^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?


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