Say I have two groups. I collect $n$ data points from the first ($A$) group ($x_i$) and $m$ data points from the second ($B$) group ($y_j$). The null hypothesis is that the means of the two groups are the same.

$$H_0: \mu_B = \mu_A$$ $$=> \mu_B-\mu_A=0$$ The alternate hypothesis is that the mean of group $B$ is larger than the mean of group $A$ by some effect size, $e$. $$H_1: \mu_B=\mu_A+e$$ $$=>\mu_B-\mu_A=e$$

I'll assume that the variances of the two groups are the same. I want to use the Neyman Pearson lemma to show that the test statistic that will yield the most power is the one used in the two-sample t-test (or the Wald test). I've been unable to do this and wanted to see if someone can help.

My attempt:

In a similar theme to the example provided in the Wikipedia article, I first construct the likelihood function:

$$L(\mu_B, \mu_A | x_i, y_j) \propto \exp\left( -\sum_i(x_i-\mu_A)^2 - \sum_j(y_j-\mu_B)^2\right)$$ $$\propto \exp\left(-\left(\sum x_i^2 + \sum y_j^2\right)-2\left(\mu_A \sum x_i+\mu_B\sum y_j\right) +(\mu_A^2+\mu_B^2)\right)$$

Now it seems like when I take the likelihood ratio for the two hypotheses, everything will cancel and the ratio will depend only on $\sum y_j$. This is obviously not right. Where did I go wrong?

  • $\begingroup$ The likelihood is a function of $(\mu_A,\mu_B)$ not $\mu_a-\mu_B$. $\endgroup$
    – Xi'an
    Dec 13, 2021 at 9:39
  • $\begingroup$ Makes sense, updated. $\endgroup$
    – ryu576
    Dec 13, 2021 at 19:53
  • $\begingroup$ The common (unknown) variance must make an appearance in the formulas for the likelihood. $\endgroup$
    – whuber
    Dec 13, 2021 at 20:06
  • $\begingroup$ Instead of parameterizing the model in terms of $\mu_A$ and $\mu_B$ can you parameterize it in terms of $\mu_A$ and $\theta$ where $\theta\equiv\mu_B-\mu_A$? $\endgroup$ Dec 13, 2021 at 20:11
  • 1
    $\begingroup$ Everything cancels if one uses the same parameter values top and bottom. You have to use different parameters top and bottom, eg $(\mu_B, \mu_A)$ and $(\mu_B^\prime, \mu_A^\prime)$. And maximise top and bottom separately under both hypotheses (for the generalised likelihood ratio test, not Neyman-Pearson). $\endgroup$
    – Xi'an
    Dec 14, 2021 at 5:40


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