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The common way to determine the degrees of freedom is to subtract 1 from n for each parameter that has been estimated. $$ T_{n} = \sqrt{n} \frac{x̄-\mu}{s} $$ Here I estimate both the mean and the sd to calculate T.

Assuming that the distribution of the data is normal, why do we get n-1 and not n-2 df?

(I understand the process to demonstrate that the distribution with n-1 is correct, I just can't grasp why the rule doesn't seem to hold here, maybe I'm getting something wrong)

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In some sense this is just the definition: a $t_{n-1}$ variable was defined as one where the denominator is the square root of a $\chi^2_{n-1}$.

More helpfully, it's the denominator degrees of freedom that we're talking about, ie, degrees of freedom for estimating $s$, and there we do have just one previously estimated parameter to account for, with $n-1$ df left for estimating $s$.

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