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If $X_i$, $i =1,...,n$ all follow a normal distribution $N(\mu,\sigma^2)$, and are independent, does $\frac{\sqrt{n}\cdot(\frac{1}{n}\cdot \sum X_i - \mu)}{\sqrt{(\frac{1}{n}\cdot \sum (X_i-\mu)^2)}}$ follow a student t-distribution with n degrees of freedom? I am aware of the classic formula with the n-1 degrees of freedom t-distribution. I just want to understand if this formula is true too.

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If your sums still are $\sum_{i=1}^n$, than yes. The degrees of freedom come from the number of independent $\chi^2_1$-distributed rv in the denominator, the $(X_i-\mu)^2$-terms.

Matrix calculations are enlightning here: In the usual $t_{n-1}$-statistic, $$\sum^n(X_i-\bar{X})^2 = (X_1,\ldots,X_n)P_n (X_1,\ldots,X_n)',$$ where $P_n = I_n - \frac{1}{n}$ is the centering matrix. The rank of this matrix is $n-1$, in particular it has $n-$ times the eigenvalue 1. So after diagonalizing $P_n$ you get that this sum is distributed like $$\sum^{n-1}(X_i-\mu)^2.$$

In your case, $P_n$ is simply replaced by $I_n$. So the rank of the matrix is $n$ and the degree of freedom as well.

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  • $\begingroup$ Hello Horst, thanks a lot for your answer, very clear. Now I am curious, if we take a set of data X1, ... , Xn that we suppose normally distributed and independent. If I compute the let's say 90% confidence interval for the mean with the classic t n-1 formula and then with my t n formula, do I get the same interval? I experimented on R and get slightly different results. The bigger n the number of samples, the smaller the difference. What is the explanation? $\endgroup$ – clotilde Aug 9 '14 at 15:47
  • $\begingroup$ You're right. The $n$ indicates the uncertainty in estimating the variance under the $\sqrt{}$: Small $n$, poor estimation. So this uncertainty "adds" to the normal uncertainty of the mean in the numerator: Extreme values of $\bar{x}$ lead to an even more extreme statistic, if the variance estimator is incidentally extremely small. Hence the quantiles become larger. But this stabilizes: For $n\rightarrow \infty$, $t_n \rightarrow \mathcal{N}(0,1)$ (central limit theorem). You can hence consider the degree of freedom as the penalty for not knowing the variance and thus estimating it. $\endgroup$ – Horst Grünbusch Aug 11 '14 at 8:58

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