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In the normal hypothesis testing we teach in basic stats, why do we not use $H_0 = \mu_0$ in construction of the estimate for $\sigma^2$? E.g., $\frac{1}{n}\sum_{i = 1}^n(x_i - \mu_{0})^2$?

It would seem necessary to do something like this in construction of a likelihood ratio...

Is this a handwaving approximation because of $t$-testing? I.e., $\frac{(n-1)s^2}{\sigma^2} \sim \chi_{n-1}^2$, and $\frac{Z}{\chi_{n-1}^2} \sim t_{n-1}$?

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This is an interesting question! Assuming that the $n$-dimensional vector $x$ has each entry $x_i \stackrel{iid}{\sim} \mathcal{N}(\mu, \sigma^2)$, then it is true that the sample mean $\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ is independent of the sample variance $\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2$, while it is not true that the sample mean $\bar{x}$ is independent of the estimator under the null $\frac{1}{n} \sum_{i=1}^n (x_i - \mu_0)^2$. This independence is required for the numerator and denominator of the $t$-statistic.

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    $\begingroup$ +1 -- but I don't think this is the correct reason. Your logic is that one chooses a test and then chooses a statistic to make the test work, but that's a reversal of what should happen: if a better test can be constructed that accounts for this (small) correlation, then why not use it? The real reason, IMHO, is that basing the estimate of $\sigma^2$ on the assumed value of $\mu_0$ leads to a (substantially) less powerful test. This is intuitive, because under the alternative hypothesis $\mu=\mu_A$ the estimate is increased by $(\mu-\mu_A)^2,$ making it difficult to detect a difference. $\endgroup$
    – whuber
    Commented Sep 27, 2019 at 20:28
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    $\begingroup$ Thanks for including this! I agree that this is putting things in the wrong order and that your stated reason gets to the core of the matter. $\endgroup$
    – user257566
    Commented Sep 27, 2019 at 23:46

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