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For a normal sample of variance ${\sigma }^2>0$, define $S^2=(n-1{)}^{-1}{\sum }_i(X_i-\overline{X}{)}^2$. The exact chi-squared distribution identity $(n-1)S^2/{\sigma }^2\sim {\chi }_{n-1}^2$ and the central limit theorem for independent squared standard normal random variables implyThe delta method for $g(v)=\sqrt{v}$, with derivative $g^{\prime }({\sigma }^2)=1/(2\sigma )$, therefore gives the displayed limit for $S$. Exact independence of the normal sample mean and sample variance also makes $S$ independent of $\overline{X}$. The resulting joint limit supports combining location and scale estimates when a model relates their parameters.

Since ${\sum }_{i=1}^n{D}_i=0$, we have $D_1=-{\sum }_{i=2}^n{D}_i$. Squaring and adding the other residual squares givesTherefore $S^2$ is a measurable function of $(D_2,\dots ,D_n)$. By the block independence proved in part i, it is independent of $\overline{X}$. This is the independence of the normal sample mean and sample variance; pairwise independence alone would not justify passing to an arbitrary function of several residuals.

Each before-and-after observation belongs to the same individual, so pairing removes individual baseline variation and allows dependence within a person's two readings. Under the stated independent normal model for the differences,The independence of the normal sample mean and sample variance theorem gives independence of $\overline{D}$ and $S_D^2$, andAlso $(\overline{D}-\mu )/(\sigma /\sqrt{n})$ is standard normal. Dividing it by the square root of the independent ${\chi }_{n-1}^2/(n-1)$ variable givesIts law contains neither unknown parameter, providing the pivotal quantity for a paired t-test.

Write $S^2=Q/(n-1)$ and $S=\sqrt{S^2}$, with $n\ge 2$. Two exact facts about a normal sample areand independence of the normal sample mean and sample variance holds. A chi-squared distribution with $n-1$ statistical degrees of freedom is a sum of $n-1$ independent ${\chi }_1^2$ variables, each with mean one and variance two. The central limit theorem therefore givesFor $g(x)=\sqrt{x}$, the derivative at ${\mu }^2$ is $1/(2\mu )$ because $\mu >0$. The delta method gives the asymptotic normality of a normal sample standard deviation:The exact independence of $\overline{X}$ and $S$ makes their limiting normal distributions independent. Consequently,ThusThis has the same leading asymptotic variance as $\hat{\mu }$. The weights $1/3$ and $2/3$ are those in the optimal asymptotic combination of a normal sample mean and standard deviation. This does not require $S$ to be an exactly unbiased estimator of $\mu$.

For a normal sample $X_1,\dots ,X_n$ with unknown mean $\mu$, positive variance ${\sigma }^2$ and $n\ge 2$, let $S^2=(n-1{)}^{-1}{\sum }_i(X_i-\overline{X}{)}^2$. Thenand an exact $(1-\alpha )$ confidence interval for $\mu$ is $\overline{X}\pm t_{n-1,1-\alpha /2}S/\sqrt{n}$.

The proof uses independence of the normal sample mean and sample variance: $\sqrt{n}(\overline{X}-\mu )/\sigma$ is standard normal, independently of $(n-1)S^2/{\sigma }^2\sim {\chi }_{n-1}^2$. Dividing produces Student's t-distribution. The random scale increases uncertainty relative to a known-variance normal interval. Non-normal samples need an approximation argument; an independent scale estimate alone does not make an arbitrary statistic exactly Student t.