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Multiple Comparisons and ANOVA Assumptions · Tutorial 879 of 996

Scheffé Adjustments for Flexible Contrasts

Learn to define a contrast, calculate its Scheffé interval, and understand when its broad protection is worth the added conservatism.

Intermediate 10 min read

What You'll Learn

  • Define a contrast of group means and estimate it from sample means.
  • Calculate the standard error for a contrast, including with unequal group sizes.
  • Construct a Scheffé simultaneous confidence interval using the ANOVA error term.
  • Use a Scheffé interval to test a contrast against a specified value.
  • Explain why Scheffé protects flexible contrasts but can be less precise for pairwise questions.

Why Flexible Contrasts Need Simultaneous Protection

After an ANOVA, a useful question may compare two groups. But a scientific question can also compare one group with the average of several others, or compare two different averages. If the comparisons are chosen only after looking at the sample means, ordinary single-comparison intervals do not automatically protect the overall family of conclusions. Scheffé’s method addresses this by providing simultaneous inference for every contrast among the group means.

That flexibility is valuable when the relevant contrasts were not all specified in advance. It also comes at a cost: Scheffé intervals can be wider than intervals from procedures designed for a narrower family, such as pairwise comparisons. The method is not a correction to the group means themselves; it changes the uncertainty threshold used for inference.

Model, Contrasts, and Their Standard Errors

Suppose a one-way ANOVA compares \(k\) groups. Let \(\mu_i\) be the population mean response in group \(i\), \(n_i\) its sample size, and \(N=\sum_{i=1}^k n_i\) the total sample size. The model assumes independent observations, normally distributed responses within each group, and a common population variance \(\sigma^2\) across groups. The pooled within-group variance estimate is the ANOVA mean square error, denoted \(MSE\), with \(df=N-k\) error degrees of freedom.

These assumptions are needed for the standard Scheffé results. They should be plausible given the study design and the data; Scheffé’s adjustment does not repair dependence, serious model problems, or unequal group variances.

Definition: A contrast of the \(k\) population means is a linear combination \(L=\sum_{i=1}^k c_i\mu_i\), where the fixed coefficients satisfy \(\sum_{i=1}^k c_i=0\). Its estimate is \(\hat L=\sum_{i=1}^k c_i\bar{x}_i\), and, under the common-variance ANOVA model, its estimated standard error is \(SE(\hat L)=\sqrt{MSE\sum_{i=1}^k c_i^2/n_i}\).

The coefficients must sum to zero so that a contrast describes a comparison among means rather than their overall level. For example, \(\mu_1-\mu_2\) has coefficients \(1,-1,0,\ldots,0\). The comparison of the average of groups 1 and 2 with group 3 is \((\mu_1+\mu_2)/2-\mu_3\), with coefficients \(1/2,1/2,-1\).

Scheffé’s procedure uses the same estimate and standard error as ordinary contrast inference. Its distinctive feature is the multiplier applied to that standard error. For a familywise error rate \(\alpha\), define \(F_{1-\alpha;k-1,N-k}\) as the \(1-\alpha\) quantile of an \(F\) distribution with \(k-1\) and \(N-k\) degrees of freedom.

Theorem: Under the one-way ANOVA model, a Scheffé simultaneous confidence interval for any contrast \(L\) is
$$ \hat L \ \pm\ \sqrt{(k-1)F_{1-\alpha;k-1,N-k}}\;SE(\hat L). $$
The intervals have simultaneous coverage of at least \(1-\alpha\): the probability that all such intervals cover their respective contrasts is at least \(1-\alpha\).

“Simultaneous” is the key word. The guarantee applies to the whole set of contrasts, not merely to one contrast selected before data collection. It therefore remains valid if a researcher first inspects the results and then chooses a contrast to report. The guarantee is about long-run coverage under the stated model, not a probability assigned to a fixed parameter after the interval has been calculated.

Worked Example: A Difference Between Two of Four Means

Worked Example: A Difference Between Two of Four Means

In a hypothetical experiment, four independently sampled groups each have \(n_i=4\) observations. Their sample means are 12, 17, 14, and 16 response units. The ANOVA gives \(MSE=4\). Assume the design and data support the one-way ANOVA model. Find a 95% Scheffé simultaneous interval for \(\mu_1-\mu_2\).

Here \(k=4\), \(N=16\), and \(df=N-k=12\). The contrast coefficients are \(1,-1,0,0\), which sum to zero. Its estimated difference is

$$ \hat L=12-17=-5\ \text{response units}. $$

Its standard error is

$$ SE(\hat L) =\sqrt{4\left(\frac{1^2}{4}+\frac{(-1)^2}{4}\right)} =\sqrt{2} \approx 1.414\ \text{response units}. $$

For 95% simultaneous coverage, \(F_{0.95;3,12}\approx3.4903\), so the Scheffé multiplier is \(\sqrt{3(3.4903)}\approx3.236\). Using the unrounded values for the margin gives approximately \(3.236(1.414)=4.576\) response units. The interval is

$$ -5\pm4.576=(-9.576,-0.424)\ \text{response units}. $$

The interval is entirely below zero. Under the model and this familywise procedure, the data support a difference in which group 1’s population mean is lower than group 2’s, while maintaining 95% simultaneous coverage across all contrasts. This conclusion is about the population means, not a causal effect unless the study design supports a causal interpretation.

Worked Example: Comparing an Average With Another Group

Worked Example: Comparing an Average With Another Group

In a second hypothetical ANOVA, three groups have sample sizes \(5,4,6\), sample means \(21,12,15\) response units, and \(MSE=9\). Assume independent observations and a plausible common-variance normal model. Estimate the contrast comparing the average of the first two population means with the third, and form its 95% Scheffé interval.

The contrast is \(L=(\mu_1+\mu_2)/2-\mu_3\), with coefficients \(1/2,1/2,-1\), whose sum is zero. Its estimate is

$$ \hat L=\frac{21+12}{2}-15=1.5\ \text{response units}. $$

There are \(k=3\) groups and \(N=15\) observations, so \(df=12\). The standard error accounts for the unequal group sizes:

$$ SE(\hat L) =\sqrt{9\left(\frac{(1/2)^2}{5}+\frac{(1/2)^2}{4}+\frac{(-1)^2}{6}\right)} =\sqrt{9\left(0.05+0.0625+0.166667\right)} =\sqrt{2.5125} \approx1.5851\ \text{response units}. $$

Using \(F_{0.95;2,12}\approx3.8853\), the multiplier is \(\sqrt{2(3.8853)}\approx2.788\). The margin is approximately \(2.7876(1.5851)\approx4.419\), so the interval is

$$ 1.5\pm4.419=(-2.919,5.919)\ \text{response units}. $$

Because zero lies in the interval, this familywise procedure does not establish a nonzero contrast at the 5% level. The estimate is positive, but its direction alone is not enough to establish a population difference.

How the Simultaneous Guarantee Works

The critical multiplier comes from the geometry of the \(k\) group means. The following justification explains why it covers contrasts even when their coefficients are selected flexibly. Let \(d_i=\bar{x}_i-\mu_i\) be the estimation error for group \(i\), and let \(\bar d_w=\sum_i n_i d_i/N\) be its sample-size-weighted average. For any contrast, \(\sum_i c_i=0\), so

$$ \hat L-L =\sum_i c_i d_i =\sum_i c_i(d_i-\bar d_w). $$

Applying the Cauchy–Schwarz inequality to the last sum gives

$$ \frac{(\hat L-L)^2}{\sum_i c_i^2/n_i} \leq \sum_i n_i(d_i-\bar d_w)^2. $$

The right-hand side measures estimation error in the \(k-1\) dimensional space of contrasts. Under the model, its ratio to \((k-1)MSE\) has an \(F\) distribution with \(k-1\) and \(N-k\) degrees of freedom. Thus, with probability at least \(1-\alpha\),

$$ \sum_i n_i(d_i-\bar d_w)^2 \leq (k-1)F_{1-\alpha;k-1,N-k}MSE. $$

Combining the inequalities and taking square roots shows that, at once for every contrast,

$$ \frac{|\hat L-L|}{\sqrt{MSE\sum_i c_i^2/n_i}} \leq\sqrt{(k-1)F_{1-\alpha;k-1,N-k}}. $$

Rearranging this bound gives the Scheffé interval. Since the same bound applies to every coefficient vector summing to zero, it also covers contrasts chosen after examining the data.

Testing a Contrast and Comparing Methods

To test \(H_0:L=L_0\), construct the Scheffé interval for \(L\). Reject \(H_0\) at familywise level \(\alpha\) if the interval does not contain \(L_0\). Equivalently, reject when

$$ \frac{|\hat L-L_0|}{SE(\hat L)} > \sqrt{(k-1)F_{1-\alpha;k-1,N-k}}. $$

For a test against zero, the interval excludes zero exactly when this inequality holds with \(L_0=0\). The corresponding decision is protected as part of the full family of contrasts, rather than as an isolated test.

Different multiple-comparison procedures protect different families. Tukey procedures focus on all pairwise differences, while Bonferroni and Holm adjustments protect a finite set of selected comparisons. Scheffé’s method protects the much broader family of all contrasts. That breadth is useful when a contrast may be formulated after seeing the results, but it generally makes Scheffé intervals more conservative for a pairwise-only question. The method should be chosen to match the inferential family, not simply because it is available.

Common Mistake: Treating Scheffé as a Pairwise Adjustment

A common error is to use a pairwise critical value, or an ordinary single-contrast \(t\) critical value, and then claim protection for any contrast selected from the data. Those thresholds do not provide Scheffé’s guarantee for the full family. The correction is to use \(\sqrt{(k-1)F_{1-\alpha;k-1,N-k}}\) with the contrast’s standard error, using \(k-1\) numerator and \(N-k\) denominator degrees of freedom. Also check that the coefficients sum to zero and use \(MSE\) from the same one-way ANOVA model.

Key takeaway: Scheffé intervals provide simultaneous protection for every contrast among the group means. Their flexibility is especially useful for data-informed comparisons, while their broad coverage can sacrifice precision when the only goal is a prespecified family of pairwise comparisons.

Check Your Understanding

Use the definitions and results above to answer these questions.

  1. Do the coefficients \(1/3,1/3,1/3,-1\) define a contrast? Explain.
  2. In a four-group ANOVA with \(N=28\), what are the numerator and denominator degrees of freedom for the Scheffé critical \(F\) value?
  3. Why does Scheffé’s simultaneous guarantee remain useful if a contrast is chosen after inspecting the sample means?
  4. For a contrast estimate of 3 response units with a Scheffé interval from \(-1\) to 7 response units, what can be concluded about a zero contrast?
  5. Why might a pairwise-focused procedure give narrower intervals than Scheffé for pairwise comparisons?