Why a Shared-Control Analysis Is Different
When three experimental treatments are each compared with one control, the three estimated treatment effects are not independent: every contrast contains the same control-arm estimate. Treating the tests as independent, or conducting three unadjusted tests, ignores that dependence and can misstate the probability of at least one false positive. Dunnett’s procedure uses the shared-control correlation to calibrate a common critical value for the planned comparisons.
The aim here is a confirmatory, one-sided analysis in which a larger outcome is prespecified as favorable. The family consists of all three treatment-versus-control hypotheses, whether or not an earlier decision process has nominated a preferred arm. The introductory material on Dunnett adjustment establishes the general procedure; the calculations below show how to obtain the contrast statistics, their dependence, the critical value, and the decisions in a concrete case.
Model, Hypotheses, and Test Statistics
The standard pooled-variance Dunnett model assumes independent random samples across arms, normally distributed outcomes within each arm, and one common within-arm variance. Let \(n_j\), \(\bar{x}_j\), and \(s_j\) denote the sample size, mean, and standard deviation in arm \(j\), including \(j=0\) for control. Define the pooled variance estimate and degrees of freedom by
For each treatment, the estimated contrast is \(\bar{x}_j-\bar{x}_0\), and its standard error is \(SE_j=s_p\sqrt{1/n_j+1/n_0}\). The statistic \(T_j=(\bar{x}_j-\bar{x}_0)/SE_j\) is compared with the same positive critical value \(c_D\). For a one-sided familywise level \(\alpha\), \(c_D\) is chosen so that, when all three contrasts are at their null boundaries,
This is a joint-distribution calibration, not three separate single-test cutoffs. The statistics have a multivariate \(t\) distribution under the boundary configuration: they can be represented as \(T_j=Z_j/\sqrt{W/df}\), where \(W\sim\chi^2_{df}\) is independent of the jointly normal vector \(Z\), whose components have mean zero, variance one, and the correlations derived below.
Why the Test Statistics Are Correlated
Before estimating the common variance, the variance of the contrast \(\bar{X}_j-\bar{X}_0\) is \(\sigma^2(1/n_j+1/n_0)\). For two distinct experimental arms \(j\) and \(k\), their contrast estimates share \(-\bar{X}_0\); the independent treatment-arm means contribute no covariance. Thus their covariance is \(\sigma^2/n_0\), giving the correlation
This correlation is positive. With equal arm sizes it is \(1/2\): the shared control contributes half of each contrast’s variance. With unequal sample sizes, the correlations need not be equal, so the critical value must be computed from the corresponding correlation matrix and degrees of freedom. The common pooled variance estimate also matters: under the stated normal, common-variance model it is independent of the arm means, which yields the joint \(t\) distribution used for calibration.
Worked Example: Three Comparisons With Equal Allocation
Hypothetical data. Suppose a continuous clinical outcome is measured in the same units in a control arm and three experimental arms. Each arm has \(n_j=10\). The four sample means are \(\bar{x}_0=0.0\), \(\bar{x}_1=2.0\), \(\bar{x}_2=1.5\), and \(\bar{x}_3=0.5\). The within-arm sums of squares are 36 in each arm. The favorable direction and one-sided familywise level \(\alpha=0.05\) were specified before confirmatory testing.
The pooled sum of squares is \(4(36)=144\), and the degrees of freedom are \(4(10)-4=36\). Therefore \(s_p^2=144/36=4\), so \(s_p=2\). Each contrast has standard error
The three treatment-minus-control estimates are \(2.0\), \(1.5\), and \(0.5\) outcome units. Their test statistics are
For equal allocation, every pairwise correlation is \(0.1/\sqrt{(0.2)(0.2)}=0.5\). Numerical calibration of the three-variate \(t\) maximum with \(df=36\), this correlation matrix, and one-sided \(\alpha=0.05\) gives \(c_D\approx2.13\). Comparing each statistic with this common cutoff, only \(T_1\) exceeds it. The confirmatory decision is to reject \(H_{01}\), but not \(H_{02}\) or \(H_{03}\). The estimated difference for arm 1 is 2.0 outcome units in the favorable direction; the other estimates do not meet the familywise criterion. A non-rejection is not proof that a treatment has no effect.
Justification of Familywise Error Control
Calibration at the all-null boundary gives equality at \(\alpha\) when all three effects are zero. It also gives strong control when only some of the null hypotheses are true. “Strong” means the error probability is bounded by \(\alpha\) for every configuration of true and false null hypotheses, not only when all three nulls hold.
Let \(I\) be the set of indices whose one-sided nulls are true. False rejections can occur only if at least one \(T_j\), \(j\in I\), exceeds \(c_D\).
At the boundary \(\mu_j-\mu_0=0\) for \(j\in I\), that subvector has the central multivariate \(t\) distribution with \(df\) and the corresponding principal submatrix of the full correlation matrix. Its probability of any exceedance is no larger than that for the full three-statistic vector: the subvector can be viewed as coordinates of that full boundary vector, so its maximum cannot exceed the full maximum.
For effects strictly below zero, the means of the true-null statistics shift downward. In the joint representation the same positive denominator \(\sqrt{W/df}\) divides each component, so decreasing their means cannot increase the probability that any exceeds \(c_D\). Consequently the probability of at least one false rejection is at most the full-family boundary probability, \(\alpha\).
The argument depends on the joint model and on using the full-family cutoff. In particular, it does not follow by comparing the true-null event with the maximum over all statistics under a partial alternative: statistics for false nulls may exceed the cutoff often. The valid comparison is between the true-null subvector at its boundary and the full central boundary vector used to define \(c_D\).
Worked Example: The Strong-Control Argument With Two True Nulls
Suppose, hypothetically, arms 1 and 2 have effects exactly equal to control, while arm 3 has a favorable effect. The true-null set is \(I=\{1,2\}\). At the boundary, \((T_1,T_2)\) has the two-coordinate central \(t\) distribution with the same \(df\) and correlation that those coordinates have in the three-arm model. For a full boundary vector \((T_1,T_2,T_3)\),
If either true effect is strictly below control, its statistic shifts downward under the model, and the chance of a false rejection cannot increase. Arm 3’s favorable effect does not invalidate this argument: the bound is obtained from the distribution of the true-null subvector, not from claiming that the full statistic vector under the alternative has a central distribution.
Another Decision Pattern: A Selected-Looking Winner
Worked Example: The Largest Statistic Still Does Not Pass
In a second hypothetical balanced trial, each of the four arms again has \(n_j=10\), \(df=36\), and within-arm sum of squares 27. The pooled variance is \(4(27)/36=3\), so the common standard error is \(\sqrt{3(0.2)}=\sqrt{0.6}\approx0.7746\). Suppose the means are 1.0 units for control and 2.5, 2.2, and 1.8 units for arms 1, 2, and 3.
The estimated contrasts are 1.5, 1.2, and 0.8 units. Thus
The allocation and degrees of freedom are unchanged, so the same one-sided \(c_D\approx2.13\) applies. None of the three statistics exceeds it. Arm 1 is the largest observed contrast, but being the largest is not itself confirmatory evidence: the familywise decision is that none of the three treatment-versus-control null hypotheses is rejected at 0.05.
Choosing the Family and Checking the Analysis
The Dunnett cutoff is appropriate for a prespecified family of comparisons sharing a control. Its advantages and alternatives differ in how they use that structure.
| Approach | How it treats the comparisons | Practical implication |
|---|---|---|
| Dunnett | Calibrates a common cutoff using the shared-control joint distribution. | Directly targets treatment-versus-control comparisons and uses their correlation. |
| Bonferroni adjustment | Controls the familywise error rate by allocating the error level across tests. | Does not need the contrast correlations, but can be more conservative. |
| Unadjusted tests | Compares each statistic with a single-test cutoff without accounting for the family. | Does not generally control the chance of any false positive at the nominal family level. |
| General maxT procedure | Uses the joint distribution of a maximum across a specified statistic family. | Dunnett is a specialized maxT application for shared-control contrasts. |
Before interpreting results, check that the favorable direction, family of three comparisons, and significance level were specified appropriately; that the analysis uses the correct sample sizes and pooled residual variance; and that the shared-control correlation matches the allocation. For unequal allocation, calculate each correlation from its own sample sizes rather than inserting \(0.5\). Verify the degrees of freedom and use a numerical critical value for the appropriate correlation matrix and sidedness.
Diagnostics should also address whether the common-variance normal model is plausible for the outcome and whether there are severe outliers, influential observations, or marked residual-variance differences. Such problems can make the pooled standard error and its joint \(t\) calibration unreliable. A different variance model or robust method may be needed, but its multiplicity calibration must also be appropriate. Missing outcomes, non-independent observations, or comparisons added after examining results likewise require explicit treatment; the standard Dunnett cutoff does not automatically repair those design changes.
Common Mistakes and Limitations
- Using three ordinary \(t\)-tests. The wrong reasoning is that each test has level 0.05, so the family must also have level 0.05. The correction is that the probability of at least one false rejection can exceed the single-test level; use the calibrated family cutoff.
- Assuming the tests are independent. The wrong reasoning ignores the shared control mean. The correction is to include its contribution to the covariance, and to use the resulting correlation matrix.
- Calling the boundary probability exactly \(\alpha\) under every null configuration. Equality is at the all-boundary configuration for the full family. When some effects are strictly below zero, or only a subset of hypotheses is true, the false-rejection probability is at most \(\alpha\), not necessarily equal to it.
- Changing the family after seeing which arm looks best. The wrong reasoning tests only the apparent winner at the original single-test level. The correction is to use a procedure justified for the actual selection and confirmatory family; a cutoff calibrated for three planned comparisons cannot be casually replaced with an unadjusted winner-only test.
Dunnett’s method is tailored to comparisons against one control. It does not directly test every pairwise treatment difference, does not guarantee good performance under a seriously misspecified variance model, and does not by itself establish that an adaptively selected contrast has the intended confirmatory interpretation. The conclusions concern the specified hypotheses and analysis model.
Worked Example: Unequal Allocation Changes the Correlation
Consider hypothetical arm sizes \(n_0=20\), \(n_1=10\), \(n_2=15\), and \(n_3=20\), with a common variance model. The correlation between the arm 1 and arm 2 contrast estimates is
For contrasts 1 and 3, the calculation is
The degrees of freedom would be \(20+10+15+20-4=61\). The three correlations are not all \(0.5\), so a balanced-design critical value cannot simply be reused. The correct Dunnett cutoff is numerically calibrated for \(df=61\) and the full unequal-allocation correlation matrix. This example illustrates why allocation is part of the multiplicity calculation, not merely a detail in the standard errors.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- With equal sample sizes in all four arms, why is the correlation between two different treatment-versus-control contrast estimates positive?
- In the first worked example, which null hypotheses are rejected, and what is the estimated difference for the rejected comparison?
- Suppose only arms 1 and 2 have true null hypotheses. Why is the probability of falsely rejecting one of them bounded using the true-null subvector rather than the full vector under the partial alternative?
- For \(n_0=20\), \(n_j=10\), and \(n_k=20\), calculate the correlation between contrasts \(j\) and \(k\).
- Why is the largest statistic in a family not automatically a significant confirmatory result?
Solutions. (1) Both contrast estimates contain the same control mean with a negative sign. Their covariance is the variance of that shared term, \(\sigma^2/n_0>0\); with equal arm sizes the correlation is \(1/2\). (2) Only \(H_{01}\) is rejected because \(2.236>2.13\), while 1.677 and 0.5590 are below the cutoff. Its estimated favorable difference is 2.0 outcome units. (3) The true-null subvector has its central joint \(t\) distribution at the null boundary, and the probability its maximum exceeds the full-family cutoff is no greater than the calibrated full-family boundary probability \(\alpha\). Under strict null effects the statistics shift downward, so that probability cannot increase. (4) The correlation is \(0.05/\sqrt{(0.1+0.05)(0.05+0.05)}=0.05/\sqrt{0.015}\approx0.4082\). (5) The cutoff accounts for searching across the whole planned family; the maximum must exceed that familywise-calibrated cutoff, not merely be larger than the other observed statistics.