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Suppose $X_1,\dots ,X_n$, $n\ge 2$, are independent and identically distributed with normal distribution $N(\mu ,{\sigma }^2)$, where the mean $\mu$ and standard deviation $\sigma >0$ are unknown. Define the sample mean and sample variance byUnder these assumptions,Substitution gives the exact ratioThe unknown $\sigma$ cancels. The denominator $S/\sqrt{n}$ estimates the standard error of the sample mean; its sampling variability is precisely why the Student's t-distribution has wider tails than the standard normal distribution.

The crucial independence has a geometric explanation. Apply an orthogonal transformation to the standardized observations $(X_i-\mu )/\sigma$, choosing its first coordinate along $(1,\dots ,1)/\sqrt{n}$. That coordinate is $Z$; the remaining $n-1$ coordinates are independent standard normal random variables whose squared sum is $V$. Their joint probability density factorizes because the transformation preserves the squared length. This also explains the loss of one statistical degree of freedom when estimating the mean. Uncorrelatedness alone would not establish the required independence for arbitrary data.

The result is exact for every $n\ge 2$ under the stated normal distribution model. A small sample size by itself does not justify it: a sample from an arbitrary probability distribution need not have the displayed chi-squared distribution or the necessary independence.

Student's t-test divides a normally distributed estimated effect by its estimated standard error. Under its null hypothesis the statistic has a Student's t-distribution, with degrees of freedom determined by the variance estimate.

For a symmetric interval of the form $[L,U]=[\hat{\theta }-c\mathrm{SE}(\hat{\theta }),\hat{\theta }+c\mathrm{SE}(\hat{\theta })]$, where $c\ge 0$ and the standard error is nonnegative, adding and subtracting the endpoints proves the displayed identities. Thus the endpoints must be ordered and their midpoint must equal the point estimate. Rounding should be assessed using intervals of possible unrounded values before declaring a numerical contradiction.

When this confidence interval is obtained by inverting a matching two-sided test of $\theta ={\theta }_0$, exclusion of ${\theta }_0$ agrees with rejection at the corresponding significance level, with the usual convention at boundary equality. This is a check for a specified symmetric interval procedure, such as the usual t interval for a mean difference; it does not require asymmetric intervals or intervals on transformed scales to have their midpoint at the estimate on the original scale.

For a full-column-rank design matrix in a normal linear model, a fitted mean at covariate vector $v$ is $v^T\hat{\beta }$. The covariance matrix of the least-squares estimator immediately gives the displayed variance. Replacing ${\sigma }^2$ by the independent unbiased residual variance gives its squared standard error and a Student t confidence interval on $n-p$ statistical degrees of freedom. A future response instead contributes an additional error variance ${\sigma }^2$ when its error is independent of the fitted data.

// Target: probability-and-statistics.bigb

For $n$ pots per treatment and $m$ leaves per pot, the group sample mean decomposes into two averages:The variance of a sum adds the variances of the independent terms. Every pot random effect has weight $1/n$, whereas each remaining error has weight $1/(nm)$, giving the short proofThis is also the variance of the arithmetic mean of the $n$ pot sample means, each with variance ${\tau }^2+{\sigma }^2/m$. Equal numbers of leaves per pot make the two averaging procedures identical.

If all $nm$ leaf measurements were falsely treated as independent, the predicted variance would instead be $({\tau }^2+{\sigma }^2)/(nm)$. The ratio of correct to false variance isThus pseudoreplication can underestimate a standard error by the square root of this ratio. Equality holds when $m=1$ or ${\tau }^2=0$; sharing a pot creates an error here precisely when the shared influence has positive variance.

Choose ${\tau }^2=4$, ${\sigma }^2=1$, $n=4$ and $m=3$ as constructed model parameters in squared length units. These are assumptions rather than fitted experimental values. For each group,The treatment contrast $\hat{\Delta }={\overline{Y}}_B-{\overline{Y}}_A$ has expected value ${\alpha }_B-{\alpha }_A$. Since the group errors are independent conditional on the allocation, its variance is twice the group variance, namely $13/6$, whereas the false calculation gives $5/6$. Its correct standard error is $\sqrt{13/6}\approx 1.472$ length units; the false value is $\sqrt{5/6}\approx 0.913$. The variance ratio is $13/5=2.6$.

With four pots fixed, increasing $m$ drives the error contribution ${\sigma }^2/(4m)$ towards zero, but the limit of the group variance remains ${\tau }^2/4=1$. Measuring twelve leaves on each of four pots gives $49/48$. Using the same forty-eight measurements on sixteen pots, with three leaves per pot, gives $13/48$. More independent experimental units reduce both contributions. Extra nested measurements are useful for characterizing each experimental unit, but cannot average away its shared random effect. This is why sample collection effort and the number of independent treatment replicates answer different questions.^[ref]

A Wald interval centers at an asymptotically normal estimator and uses a consistent estimated standard error times a normal quantile.