Noncentral t Distributions for Exact Power

Why a Contrast Has a Noncentral t Distribution

A t statistic for a contrast compares an estimated linear combination of means or regression coefficients with a hypothesized value, in units of an estimated standard error. Under the null hypothesis, its numerator is centered at zero and the statistic has a central t distribution. Under a fixed alternative, the numerator is shifted: its mean differs from the hypothesized value. The resulting distribution is a noncentral t distribution.

The shift is not determined by the contrast effect alone. It must be measured relative to the contrast’s standard deviation, which depends on both the error variance and the design. Deriving this relationship explains why the noncentrality parameter is a standardized contrast effect and why the variance estimate remains part of the random statistic.

Definition: Let $Z\sim N(0,1)$ and $U\sim\chi^2_\nu$ be independent, where $\nu>0$. The random variable $$ T=\frac{Z+\lambda}{\sqrt{U/\nu}} $$ has a noncentral t distribution with $\nu$ degrees of freedom and noncentrality parameter $\lambda$, written $T\sim t_\nu(\lambda)$. When $\lambda=0$, this is the central $t_\nu$ distribution.

The numerator’s shift is $\lambda$ in standard-normal units. The denominator is random, as it is when a variance is estimated from normal-model residuals. For a test of a contrast $c^\mathsf{T}\beta$ against $\theta_0$, the relevant shift under a fixed true parameter $\beta$ will be $\big(c^\mathsf{T}\beta-\theta_0\big)/\big(\sigma\sqrt{c^\mathsf{T}(X^\mathsf{T}X)^{-1}c}\big)$.

Contrast t Statistics in the Normal Linear Model

Consider the full-rank normal linear model $Y\sim N_n(X\beta,\sigma^2 I_n)$, where $X$ is a known $n\times p$ design matrix of rank $p$, $n>p$, and $\sigma^2>0$. Let $c$ be a nonzero $p$-vector and consider the estimable contrast $c^\mathsf{T}\beta$. The least-squares estimator is $\hat\beta=(X^\mathsf{T}X)^{-1}X^\mathsf{T}Y$. Define $q=c^\mathsf{T}(X^\mathsf{T}X)^{-1}c$, which is positive for nonzero $c$.

Definition: For testing $H_0:c^\mathsf{T}\beta=\theta_0$, the contrast t statistic is $$ T=\frac{c^\mathsf{T}\hat\beta-\theta_0}{S\sqrt{q}}, \qquad S^2=\frac{\|Y-X\hat\beta\|^2}{n-p}, \qquad q=c^\mathsf{T}(X^\mathsf{T}X)^{-1}c. $$ Here $S^2$ estimates the common error variance $\sigma^2$. The degrees of freedom are $\nu=n-p$.

Theorem: Under the stated normal linear model, for any fixed $\beta$ and $\sigma^2>0$, $$ T\sim t_{n-p}(\lambda), \qquad \lambda=\frac{c^\mathsf{T}\beta-\theta_0}{\sigma\sqrt{c^\mathsf{T}(X^\mathsf{T}X)^{-1}c}}. $$ The null hypothesis gives $\lambda=0$; a fixed alternative with $c^\mathsf{T}\beta\ne\theta_0$ gives a nonzero noncentrality parameter.

Proof. Write $P_X=X(X^\mathsf{T}X)^{-1}X^\mathsf{T}$ and $M_X=I_n-P_X$. The least-squares estimation error is $\hat\beta-\beta=(X^\mathsf{T}X)^{-1}X^\mathsf{T}\varepsilon$, where $\varepsilon=Y-X\beta\sim N_n(0,\sigma^2 I_n)$. Consequently, $c^\mathsf{T}\hat\beta$ is normal with mean $c^\mathsf{T}\beta$ and variance $\sigma^2q$. Thus $$ Z=\frac{c^\mathsf{T}\hat\beta-c^\mathsf{T}\beta}{\sigma\sqrt q}\sim N(0,1). $$

The residual vector is $M_XY=M_X\varepsilon$, since $M_XX=0$. The matrix $M_X$ is symmetric and idempotent with rank $n-p$, so $\|M_X\varepsilon\|^2/\sigma^2\sim\chi^2_{n-p}$. In particular, $U=(n-p)S^2/\sigma^2\sim\chi^2_{n-p}$.

It remains to verify independence. The random vectors $P_X\varepsilon$ and $M_X\varepsilon$ are jointly normal, and their cross-covariance is $\sigma^2P_XM_X=0$, because $P_XM_X=0$. Joint normality and zero cross-covariance imply independence. Since $Z$ is a function of $P_X\varepsilon$ and $U$ is a function of $M_X\varepsilon$, $Z$ and $U$ are independent. Finally, $$ T =\frac{c^\mathsf{T}\beta-\theta_0+\sigma\sqrt q\,Z} {\sigma\sqrt q\,\sqrt{U/(n-p)}} =\frac{Z+\lambda}{\sqrt{U/(n-p)}}, \qquad \lambda=\frac{c^\mathsf{T}\beta-\theta_0}{\sigma\sqrt q}. $$ This is the defining representation of $t_{n-p}(\lambda)$. $\square$

Reading the Noncentrality Parameter

The numerator of $\lambda$ is the true contrast’s displacement from the null value. The denominator is the contrast’s true standard deviation, $\sigma\sqrt q$, not the standard error computed with the observed $S$. The design enters through $q$: designs that estimate the contrast more precisely have smaller $q$ and, for the same effect and error variance, a larger absolute noncentrality parameter.

The sign of $\lambda$ records the direction of the contrast relative to $\theta_0$. Reversing the contrast’s direction reverses that sign. The degrees of freedom, by contrast, describe the residual information used to estimate $\sigma^2$; in this model they are $n-p$.

Worked Example: Difference Between Two Independent Group Means

Suppose two independent normal groups share variance $\sigma^2$. The group sizes are $n_1=12$ and $n_2=15$, and the contrast is $\mu_1-\mu_2$, tested against $\theta_0=0$. Under a fixed alternative, let $\mu_1-\mu_2=4$ and $\sigma=6$. Find the noncentrality parameter and degrees of freedom for the pooled-variance t statistic.

For a difference of group means, the design-based variance factor is $$ q=\frac{1}{n_1}+\frac{1}{n_2} =\frac{1}{12}+\frac{1}{15} =\frac{5+4}{60} =\frac{3}{20}. $$ The degrees of freedom are $n_1+n_2-2=12+15-2=25$. The contrast effect is $4-0=4$, so $$ \lambda=\frac{4}{6\sqrt{3/20}} =\frac{4\sqrt{20}}{6\sqrt{3}} \approx 1.721. $$ Thus the statistic has distribution $t_{25}(1.721)$, with the displayed decimal rounded. This distribution describes repeated samples under the specified fixed alternative; it is not obtained by replacing $\sigma$ in $\lambda$ with the observed pooled standard deviation.

The same design calculation gives a useful special case for independent groups. If $\bar Y_i$ is the sample mean in group $i$, with common variance $\sigma^2$, then for a contrast $\sum_i a_i\mu_i$, the estimated contrast is $\sum_i a_i\bar Y_i$. Independence of the group means gives its variance as $\sigma^2\sum_i a_i^2/n_i$. Therefore the pooled-variance t statistic has $\lambda=(\sum_i a_i\mu_i-\theta_0)/\big(\sigma\sqrt{\sum_i a_i^2/n_i}\big)$, with residual degrees of freedom $\sum_i n_i-k$ for $k$ groups.

Worked Example: A Weighted Contrast Across Three Groups

Three independent normal groups have sizes $10,12,8$, respectively, and share standard deviation $\sigma=2.5$. Consider the contrast $\mu_1-\tfrac12\mu_2-\tfrac12\mu_3$, tested against zero. Under a fixed alternative, suppose the group means are $5,3,1$. Find its noncentrality parameter and degrees of freedom.

The true contrast effect is $5-\tfrac12(3)-\tfrac12(1)=5-1.5-0.5=3$. Its design variance factor is $$ q=\frac{1^2}{10}+\frac{(-1/2)^2}{12}+\frac{(-1/2)^2}{8} =\frac{1}{10}+\frac{1}{48}+\frac{1}{32} =\frac{48+10+15}{480} =\frac{73}{480}. $$ The sign of each coefficient disappears when its variance contribution is squared. The standard deviation of the estimated contrast is $2.5\sqrt{73/480}$, so $$ \lambda=\frac{3}{2.5\sqrt{73/480}} \approx 3.077. $$ The pooled residual degrees of freedom are $10+12+8-3=27$. Hence the contrast statistic has distribution $t_{27}(3.077)$, with rounding in the noncentrality parameter.

Worked Example: A Paired-Data Contrast

Suppose $n=16$ paired measurements are reduced to differences $D_i$, each independently distributed as $N(\delta,\sigma_D^2)$. The target is the mean difference $\delta$, tested against $\theta_0=0$. Under a fixed alternative, let $\delta=1.2$ and $\sigma_D=2.4$. Find the noncentrality parameter and degrees of freedom.

The estimated contrast is the sample mean of the differences, with variance $\sigma_D^2/n$. Thus its design standard deviation is $2.4/\sqrt{16}=2.4/4=0.6$. The noncentrality parameter and residual degrees of freedom are $$ \lambda=\frac{1.2-0}{2.4/\sqrt{16}}=\frac{1.2}{0.6}=2, \qquad \nu=16-1=15. $$ Therefore the paired t statistic has distribution $t_{15}(2)$ under this fixed alternative. The model here is for the within-pair differences; its error variance is the variance of those differences, not the variance of either measurement considered separately.

Why the Exact Distribution Matters

A noncentral t distribution retains both sources of variation in the statistic: the shifted normal contrast estimate and the random estimate of the error variance. Its noncentrality parameter summarizes how far the true contrast lies from the null in units of its true standard deviation, while its degrees of freedom describe the variability in the variance estimate. Together they specify the statistic’s distribution under the fixed alternative.

This distribution is the basis for exact power calculations for t tests, including tests with adjusted critical values. To use it correctly, first identify the contrast effect, then calculate its design variance factor, and finally use the model’s residual degrees of freedom. The critical value and rejection region determine how that distribution is used for a particular test.

Common Mistake

A frequent error is to put the observed residual standard deviation $S$ in the noncentrality parameter, for example writing $(c^\mathsf{T}\beta-\theta_0)/(S\sqrt q)$. That quantity is random across samples, so it cannot be the fixed parameter indexing the noncentral t distribution. Under the model, $\lambda$ uses the true $\sigma$; the random $S$ appears in the statistic through the independent chi-square denominator.

A second error is to use a contrast variance factor that ignores the contrast coefficients. For independent group means, the correct factor is $\sum_i a_i^2/n_i$, not $\sum_i a_i/n_i$. Variance contributions involve squared coefficients, even when some contrast coefficients are negative.

Check Your Understanding

Use the definitions and derivation above to answer these questions.

For a normal linear model contrast, what are the noncentrality parameter and residual degrees of freedom?
Why are the standardized contrast estimation error and the residual sum of squares independent in the full-rank normal linear model?
For independent group means, why does the contrast variance factor involve $a_i^2/n_i$ rather than $a_i/n_i$?
What does the sign of the noncentrality parameter indicate?
In the paired-data example, which variance belongs in the noncentrality parameter, and why?