I'm trying to figure out the distribution of this statistic:
$$S=\frac{\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}}}{\sqrt{\hat{\sigma}^2/\sigma^2}}$$
Where:
- $\overline{X}=\frac{1}{n} \sum_{i=1}^n X_i$
- $\hat{\sigma}^2=\frac{1}{n-1} \sum_{i=1}^n (X_i-\overline{X})^2$
And $X_i \sim N(\mu, \sigma^2)$ i.i.d.
The numerator is easy, it is a classical standardization. Thus:
$$\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} = Z \sim N(0,1)$$
For what it concerns the denominator: $$\frac{\hat{\sigma}^2}{\sigma^2}=\frac{\frac{1}{n-1} \sum_{i=1}^n (X_i-\overline{X})^2}{\sigma^2}=\frac{\frac{1}{\sigma^2}\sum_{i=1}^n (X_i-\overline{X})^2}{n-1}=\frac{\sum_{i=1}^n \left (\frac{X_i-\mu}{\sigma}\right )^2}{n-1}$$
Now, since $Z=\frac{X_i-\mu}{\sigma}$ is a standard normal , then $$\sum_{i=1}^n \left (\frac{X_i-\mu}{\sigma}\right )^2=\sum_{i=1}^n Z^2 \sim \chi^2_n$$
This means that:
$$\frac{\hat{\sigma}^2}{\sigma^2} = \frac{\chi^2_n}{n-1}$$
Combining the results above leads me to this final result:
$$\frac{\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}}}{\sqrt{\hat{\sigma}^2/\sigma^2}}=\frac{N(0,1)}{\sqrt{\frac{\chi^2_n}{n-1}}} \sim t_n$$
Where $t_n$ is a Student distribution with $n$ degrees of freedom. This result must be wrong, because the right one should give a Student with $n-1$ degrees of freedom. Something is clearly missing but I cannot understand what it is. Could you please help me?