# In normal hypothesis testing, why do we not use $H_0$ in construction of the estimate of $\sigma^2?$

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 $$\frac{Z}{\chi_{n-1}^2} \sim t_{n-1}$$?

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.
• +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.