Unknown detail about mean R.V.,normal variance

Question

How can I see that for a $N(\mu_0,\sigma^2)$ R.V. $X_1,...,X_n$ this

$$ Z =\frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}} $$

has a standard $N(0,1)$, and not this:

$$ Z =\frac{\bar{X} - \mu_0}{\sigma^2 /{n}} $$?

Answer 1

We assume $X_1,\ldots,X_n$ independent then you have $$\overline{X} = \frac{1}{n}\sum_{k=1}^n X_k$$ and so it follows $$\text{Var}(\overline{X}) = \frac{1}{n^2} \sum_{k=1}^n \text{Var}(X_k) = \frac{\sigma^2}{n}$$ and we get

$$\text{Var}\left(\frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}}\right) = \frac{n}{\sigma^2}\text{Var}(\overline{X}) = \frac{n}{\sigma^2}\frac{\sigma^2}{n} = 1$$ as well as $$\text{Var}\left(\frac{\overline{X} - \mu_0}{\sigma^2 / n}\right) = \frac{n^2}{\sigma^4}\text{Var}(\overline{X}) = \frac{n^2}{\sigma^4}\frac{\sigma^2}{n} = \frac{n}{\sigma^2}$$