Why a Contrast Has a t Distribution
A contrast summarizes a particular comparison among several population means. For example, it can compare one treatment mean with the average of two other means. The comparison is estimated from sample means, but its variance depends on the unknown common error variance. Replacing that variance with the pooled within-group estimate produces a statistic with a Student t distribution—provided the model assumptions hold.
The exact result is useful because it identifies the reference distribution without relying on a large-sample approximation. Its exactness is not automatic: it depends on independent normal errors with a common variance, and on using the pooled error estimate from that model.
Use the one-way fixed-means model \(X_{ij}=\mu_i+\varepsilon_{ij}\), for \(i=1,\ldots,k\) and \(j=1,\ldots,n_i\). Assume the group sizes are fixed, all observations are independent, and every error has the normal distribution \(N(0,\sigma^2)\) with the same unknown variance \(\sigma^2\). Write \(N=\sum_i n_i\). The pooled within-group sum of squares and its mean square are
The pooled-error degrees of freedom are \(\nu=N-k\): each group mean uses one degree of freedom in estimating its group’s residuals. The contrast’s variance factor is
The group means are independent, with \(\operatorname{Var}(\bar X_i)=\sigma^2/n_i\). Consequently, \(\operatorname{Var}(\hat L)=\sigma^2q\). Replacing \(\sigma^2\) by \(MSE\) gives the estimated standard error \(\sqrt{MSE\,q}\).
Deriving the Exact Student t Result
The key is not merely that the numerator is normal and the denominator estimates a standard deviation. The required result follows because the standardized numerator and the scaled pooled error estimate are independent, and the latter has a chi-square distribution.
Under the model, \(\hat L\) is a linear combination of independent normal sample means, so \((\hat L-L)/(\sigma\sqrt q)\) has the standard normal distribution.
Within each group, residuals are the projection of the normal observations onto the subspace orthogonal to that group’s sample mean. Across all groups, this residual subspace has dimension \(N-k\). Thus \(SSE/\sigma^2\) has a chi-square distribution with \(N-k\) degrees of freedom.
The group means lie in the group-constant subspace, while the residuals lie in its orthogonal complement. For jointly normal observations, orthogonal projections are independent. Therefore \(\hat L\) and \(SSE\) are independent.
The independent standard normal numerator divided by the square root of an independent chi-square variable divided by its degrees of freedom has a Student t distribution.
In detail, set \(Z=(\hat L-L)/(\sigma\sqrt q)\) and \(U=SSE/\sigma^2\). The preceding arguments give \(Z\sim N(0,1)\), \(U\sim\chi^2_{\nu}\), and independence. Since \(MSE=SSE/\nu\),
This is a central Student t result when the numerator is centered at the true contrast \(L\). In particular, if a specified value \(L_0\) equals the true contrast, then \((\hat L-L_0)/\sqrt{MSE\,q}\sim t_{N-k}\). If \(L_0\ne L\), the same statistic instead has a noncentral t distribution with \(\nu=N-k\) degrees of freedom and noncentrality parameter \((L-L_0)/(\sigma\sqrt q)\). It is not central t under that alternative.
Worked Calculations
Worked Example: Comparing One Group With Two Others
A hypothetical study observes three independent normal groups under a common unknown variance. Each group has \(n_i=4\) observations. The sample means are \(\bar X_1=12\), \(\bar X_2=9\), and \(\bar X_3=11\), and the within-group sum of squares is \(SSE=144\). For the contrast \(L=\mu_1-(\mu_2+\mu_3)/2\), calculate the estimated contrast, its standard error, and the standardized statistic relative to \(L_0=0\). State its exact reference distribution if \(L=L_0\).
Method. Use coefficients \(c_1=1\), \(c_2=-1/2\), and \(c_3=-1/2\). They sum to zero, as required. First calculate the estimate:
There are \(N=4+4+4=12\) observations and \(k=3\) groups, so \(\nu=12-3=9\). Thus \(MSE=144/9=16\). The variance factor is
Therefore the estimated variance is \(16(3/8)=6\), and the standard error is \(\sqrt6=2.4495\) (rounded to four decimals). The statistic relative to zero is \(2/\sqrt6=0.8165\) (rounded to four decimals). If the true contrast is zero, this statistic has an exact \(t_9\) distribution.
Check: the contrast coefficients sum to zero, and the degrees of freedom are \(N-k=9\). Common mistake avoided: using \(1/n_i\) without multiplying by \(c_i^2\) would give the wrong variance factor.
Worked Example: Unequal Group Sizes
A hypothetical experiment has three independent normal groups with a common unknown error variance. Their sample sizes are \(6,10,8\); their means are \(8.4,6.9,7.6\); and the pooled within-group sum of squares is \(47.25\). For the contrast \(L=\mu_1-\mu_2\), find the standard error and the statistic relative to \(L_0=0\). Give the exact reference distribution if \(L=L_0\).
Method. The coefficients are \((1,-1,0)\), which sum to zero. The estimated contrast is \(8.4-6.9=1.5\). Here \(N=6+10+8=24\), \(k=3\), and \(\nu=24-3=21\). Hence \(MSE=47.25/21=2.25\). The unequal sample sizes must be retained in the variance factor:
The estimated variance of the contrast is \(2.25(4/15)=0.6\), so its standard error is \(\sqrt{0.6}=0.7746\) (rounded to four decimals). The statistic is \(1.5/\sqrt{0.6}=1.9365\) (rounded to four decimals). If \(L=0\), its exact reference distribution is \(t_{21}\).
Check: the zero coefficient contributes nothing to \(q\), while all three groups still contribute to the pooled-error degrees of freedom. Common mistake avoided: using only groups 1 and 2 to calculate the degrees of freedom would discard the residual information from group 3.
Worked Example: Building the Pooled Error From Observations
A hypothetical instrument study records readings \(4,5,6\) in group A, \(2,5,8\) in group B, and \(6,7,8\) in group C. Assume independent normal observations with a common variance. For \(L=\mu_C-(\mu_A+\mu_B)/2\), calculate \(MSE\), the standard error, and the statistic relative to zero. State its exact reference distribution if \(L=0\).
Method. Each group has three observations. The sample means are \(\bar X_A=5\), \(\bar X_B=5\), and \(\bar X_C=7\). The within-group sums of squares are \(2\), \(18\), and \(2\), respectively: for example, group B contributes \((2-5)^2+(5-5)^2+(8-5)^2=9+0+9=18\). Thus \(SSE=2+18+2=22\). With \(N=9\) and \(k=3\), \(\nu=9-3=6\), so \(MSE=22/6=11/3\).
The coefficients for groups A, B, and C are \((-1/2,-1/2,1)\), which sum to zero. The estimate and variance factor are
The estimated variance is \((11/3)(1/2)=11/6\). The standard error is \(\sqrt{11/6}=1.3540\) (rounded to four decimals), and the statistic is \(2/\sqrt{11/6}=1.4771\) (rounded to four decimals). If the true contrast is zero, the exact reference distribution is \(t_6\).
Check: the three group residual sums add to 22, and the six residual degrees of freedom agree with \(N-k\). Common mistake avoided: using the total variation of all observations instead of the within-group residual variation would not produce the pooled error estimate required here.
When the Result Is Exact—and When It Is Not
The formula is exact for the model stated above, not merely approximately correct because a sample is “large enough.” The assumptions have distinct roles:
- Independence across observations and groups supports the stated variance of the contrast and the independence of group means and pooled residuals.
- Normal errors make the standardized contrast normal and the scaled residual sum of squares chi-square. They also make the two orthogonal projections independent.
- A common variance makes one pooled MSE estimate the variance component used by every group mean.
- Fixed group sizes and coefficients define the variance factor \(q\) used in the result.
- Positive residual degrees of freedom are necessary to estimate the common variance; in this setting, \(N>k\).
If group variances differ, pooling within-group sums of squares does not generally yield the stated chi-square denominator scaled by one common \(\sigma^2\). If observations are dependent, such as repeated observations on the same experimental unit treated as independent, the variance and independence arguments can fail. If errors are substantially nonnormal, the exact normal and chi-square steps are no longer guaranteed. Alternative methods may be appropriate in such settings, but the pooled exact \(t_{N-k}\) result should not be asserted without its assumptions.
A frequent exam trap is to use the right-looking ratio but the wrong variance factor. For example, with coefficients \(1,-1/2,-1/2\) and three group sizes of 4, an incorrect calculation might set \(q=1/4+1/4+1/4=3/4\), forgetting to square the coefficients. The correct calculation is
The incorrect value doubles the variance factor and inflates the standard error by a factor of \(\sqrt2\). Always square each contrast coefficient in \(q\), retain the appropriate \(n_i\) for each group, and use \(N-k\) degrees of freedom for the pooled error.
The exact t distribution is the distributional foundation for inference about a single contrast and is aligned with the SOA VEE Mathematical Statistics syllabus. Its practical value depends on matching the calculation to the model: state the contrast, calculate \(q\), use the pooled MSE and its degrees of freedom, and verify that the assumptions justify the reference distribution.
Check Your Understanding
For each problem, identify the relevant variance factor and degrees of freedom before giving the requested result.
- Three independent normal groups have sizes \(5,5,5\), and \(MSE=9\). For \(L=\mu_1-\mu_2\), the estimated contrast is \(3\). Calculate its standard error and statistic relative to zero, and state the exact reference distribution if \(L=0\). (Answer: \(q=2/5\), standard error \(1.8974\), statistic \(1.5811\), and \(t_{12}\).)
- Three independent normal groups have sizes \(4,6,5\), sample means \(10,8,6\), and \(MSE=4\). For \(L=\mu_1-(\mu_2+\mu_3)/2\), calculate the statistic relative to zero and its exact reference distribution if \(L=0\). (Answer: \(q=41/120\), statistic \(2.5662\), and \(t_{12}\).)
- Four independent normal groups have sizes \(3,4,5,6\), with a common unknown variance. For any fixed contrast of their means, what are the pooled-error degrees of freedom, and what is the exact central reference distribution when the specified contrast value is true? (Answer: \(\nu=18-4=14\), so the reference distribution is \(t_{14}\).)
- For a fixed contrast with variance factor \(q=1/2\), suppose the true contrast is \(L=2\), but the statistic is centered at \(L_0=0\). If the common standard deviation is \(\sigma=3\), give its noncentrality parameter. (Answer: \((2-0)/(3\sqrt{1/2})=2\sqrt2/3\), approximately \(0.9428\); the distribution is noncentral t, not central t.)
- In a four-group study, observations within each group are independent and normal, but the group variances are unequal. Is the usual pooled statistic guaranteed to have an exact central \(t_{N-k}\) distribution for a true contrast value? (Answer: No. The common-variance assumption required for the pooled chi-square denominator and the exact \(t_{N-k}\) result is not satisfied.)