Why Orthogonal Contrasts Matter
An analysis of variance can identify whether group means vary overall, but contrasts describe particular patterns among those means. When two contrasts are orthogonal, their estimates carry separate information about the treatment pattern: their covariance is zero, and under normal errors their estimates are independent. This relationship also explains when contrast sums of squares divide the treatment variation into distinct pieces.
There is an important qualification to the phrase “independent contrast tests.” Independent contrast estimates do not automatically make test statistics independent when each test uses the same estimated error variance. The numerator information is independent under the conditions below; the shared denominator can still connect the resulting test statistics.
The sample sizes appear in the orthogonality condition because group means have different variances when group sizes differ. If all group sizes equal \(n\), the condition reduces to \(\sum_i c_i d_i=0\). With unequal sizes, simply checking an unweighted dot product is generally wrong.
From Orthogonality to Independence
The group sample means are independent under the model, with \(\operatorname{Var}(\bar Y_i)=\sigma^2/n_i\). For any two contrasts, linearity of covariance gives the key result.
Thus orthogonal contrasts have uncorrelated estimates. This covariance calculation needs independent observations and the common-variance assumption; it does not require normality. To go from uncorrelated to independent, however, normality matters: the contrast estimates are linear combinations of a jointly normal vector of sample means. A jointly normal vector with zero covariance between two components has independent components. Therefore orthogonal contrast estimates are independent under the normal-error model.
An orthogonal contrast sum of squares is the squared estimated contrast divided by its variance factor, with \(\sigma^2\) omitted:
This quantity is nonnegative and measures the part of treatment variation associated with that contrast. For an individual contrast, its degrees of freedom are one. The definition applies to nonzero contrast coefficients; a zero coefficient vector has no contrast information and cannot be used in the formula.
Why Contrast Sums of Squares Partition Treatment Variation
The treatment sum of squares is \(SS_{\text{Tr}}=\sum_i n_i(\bar Y_i-\bar Y_{\cdot})^2\), where \(\bar Y_{\cdot}=\sum_i n_i\bar Y_i/\sum_i n_i\) is the weighted grand mean. Geometrically, this is the squared length of the centered vector of group means, using weights \(n_i\). A contrast corresponds to a direction within the space of mean patterns whose weighted average is zero. Orthogonal contrast directions are perpendicular in that weighted geometry, so their squared projections add.
The vector with entries \(\bar Y_i-\bar Y_{\cdot}\) has weighted mean zero, so it lies in the treatment-contrast space, which has dimension \(k-1\).
For coefficient vector \(c\), the contrast estimate is \(\hat L_c=\sum_i c_i\bar Y_i\). Its squared projection length is \(\hat L_c^2/(\sum_i c_i^2/n_i)=SS_c\).
For mutually orthogonal contrasts, the projection directions are perpendicular. The squared length of the centered mean vector equals the sum of squared projection lengths when those directions form a basis of the contrast space.
Consequently, \(k-1\) mutually orthogonal contrasts form a complete partition:
If there are fewer than \(k-1\) mutually orthogonal contrasts, their sums of squares account only for the variation in the directions they cover. The remaining treatment variation is not thereby explained or eliminated; it lies in the unexamined orthogonal directions. This partition is an algebraic result for the observed group means. The normal model supplies the additional distributional results used for exact inference.
Worked Example: Two Independent Contrasts in a Balanced Design
A hypothetical study compares mean weekly calibration times for three procedures. Each procedure is applied to four independent units. The observed group means are 12, 10, and 7 minutes. Assume independent normal errors with common known variance \(\sigma^2=16\) minutes squared. Consider \(c=(1,-1,0)\) and \(d=(1,1,-2)\). Find their covariance, establish independence, and verify the sums-of-squares partition.
Method and orthogonality. Both coefficient vectors sum to zero: \(1-1+0=0\) and \(1+1-2=0\), so each defines a contrast. With \(n_i=4\), their orthogonality calculation is \(\sum_i c_i d_i/n_i=(1-1+0)/4=0\). Thus the estimates are uncorrelated; normality makes them independent.
Compute the estimates and variances. The estimates are \(\hat L_c=12-10=2\) and \(\hat L_d=12+10-2(7)=8\). Their variances are \(\operatorname{Var}(\hat L_c)=16(1+1)/4=8\) and \(\operatorname{Var}(\hat L_d)=16(1+1+4)/4=24\). The covariance is \(16(1-1+0)/4=0\). For scale, the standardized estimates are \(2/\sqrt{8}=0.7071\) and \(8/\sqrt{24}=1.6330\), rounded to four decimals.
Compute and check the sums of squares. The contrast sums of squares are \(SS_c=2^2/(2/4)=8\) and \(SS_d=8^2/(6/4)=128/3\). The grand mean is \((12+10+7)/3=29/3\), so \(SS_{\text{Tr}}=4[(12-29/3)^2+(10-29/3)^2+(7-29/3)^2]=152/3\). The contrast contributions sum to \(8+128/3=152/3\), exactly the treatment sum of squares. The common mistake this avoids is using the wrong denominator in a contrast sum of squares: it is \(\sum_i c_i^2/n_i\), not \(\sum_i c_i^2\).
Worked Example: Unequal Group Sizes Require Weighted Orthogonality
A hypothetical assessment records mean completion scores of 10, 12, and 7 for three groups with sizes 2, 4, and 2. Assume independent normal observations with common variance \(\sigma^2=9\). Examine \(c=(1,-1,0)\) and \(d=(1,2,-3)\). Determine whether they are orthogonal and verify their treatment sum-of-squares partition.
Check the contrast conditions and orthogonality. The coefficient sums are \(1-1+0=0\) and \(1+2-3=0\). The weighted inner product is \(1(1)/2+(-1)(2)/4+0(-3)/2=1/2-1/2=0\). They are orthogonal for these group sizes. Their ordinary dot product is \(1-2+0=-1\), which is not zero; treating an unequal-size design as if it were balanced would give the wrong conclusion.
Calculate estimates and covariance. The estimates are \(\hat L_c=10-12=-2\) and \(\hat L_d=10+2(12)-3(7)=13\). Their covariance is \(9[1/2-2/4+0]=0\), and their variances are \(9(1/2+1/4)=27/4\) and \(9(1/2+4/4+9/2)=54\). Because the errors are normal, these estimates are independent.
Verify the partition. The weighted grand mean is \((2(10)+4(12)+2(7))/8=10.25\). Therefore \(SS_{\text{Tr}}=2(10-10.25)^2+4(12-10.25)^2+2(7-10.25)^2=33.5\). The contrast sums of squares are \(SS_c=(-2)^2/(1/2+1/4)=16/3\) and \(SS_d=13^2/(1/2+4/4+9/2)=169/6\). Their sum is \(32/6+169/6=201/6=33.5\), matching the treatment sum of squares. This check confirms that the weighted orthogonality criterion and the weighted contrast sum-of-squares formula agree.
Independent Estimates Are Not Always Independent Tests
In practice, \(\sigma^2\) is usually unknown and is estimated by the pooled within-group mean square \(MSE=SSE/(N-k)\), where \(N=\sum_i n_i\). Under the normal one-way ANOVA model, \(MSE\) is independent of the contrast estimates. For a single contrast, the usual statistic is \(t=\hat L/\sqrt{MSE\sum_i c_i^2/n_i}\), with \(N-k\) error degrees of freedom under the null hypothesis \(L=0\).
For two orthogonal contrasts, the estimates in the numerators are independent, and each is independent of \(MSE\). But the two \(t\) statistics use the same random \(MSE\). Dividing two independent numerators by a shared random quantity generally makes the resulting statistics dependent. The corresponding one-degree-of-freedom \(F\) statistics share that denominator as well. Thus “independent contrast tests” is a useful shorthand for independent contrast information, but it should not be read as a guarantee that all test statistics or decisions are independent.
Under normality, each contrast sum of squares divided by \(\sigma^2\) has a chi-square distribution with one degree of freedom, and sums of squares for orthogonal contrasts are independent. The residual sum of squares is independent of these contrast sums of squares. Nevertheless, individual \(F\) ratios formed by dividing contrast sums of squares by the same \(MSE\) share a random denominator. This distinction matters when interpreting joint test behavior.
Worked Example: Separate Contrast Components, Shared Error Estimate
A hypothetical experiment has three procedures, four observations per procedure, and observed means 9, 7, and 5. Assume independent normal errors with a common variance. The within-group error sum of squares is 36. Use \(c=(1,-1,0)\) and \(d=(1,1,-2)\) to calculate the two contrast tests and identify precisely what is independent.
Find the treatment components. The contrasts are orthogonal because \((1-1+0)/4=0\). Their estimates are \(\hat L_c=9-7=2\) and \(\hat L_d=9+7-2(5)=6\). Their sums of squares are \(SS_c=2^2/(2/4)=8\) and \(SS_d=6^2/(6/4)=24\). The grand mean is 7, giving \(SS_{\text{Tr}}=4[(9-7)^2+(7-7)^2+(5-7)^2]=32\); the two components sum to 32.
Use the pooled error estimate. There are \(N-k=12-3=9\) error degrees of freedom, so \(MSE=36/9=4\). The contrast test statistics are \(t_c=2/\sqrt{4(2/4)}=2/\sqrt{2}=1.4142\) and \(t_d=6/\sqrt{4(6/4)}=6/\sqrt{6}=2.4495\), rounded to four decimals. Their squared values are \(F_c=2\) and \(F_d=6\), respectively.
The contrast estimates are independent under the normal model, and each is independent of the pooled error estimate. The two reported \(t\) statistics and the two \(F\) statistics nevertheless share \(MSE=4\) as their estimated denominator. The common mistake is to infer independence of the test statistics solely from orthogonality of the contrasts; the independence result applies directly to the contrast estimates and their separate sums-of-squares components, not automatically to ratios sharing an estimated denominator.
Exam Checks and Practical Interpretation
When using orthogonal contrasts, check the design before calculating: identify every group size, verify that each coefficient vector sums to zero, and apply the sample-size-weighted orthogonality condition. Then use the matching denominator for each contrast sum of squares. If the contrasts do not form a complete set of \(k-1\) directions, do not claim they exhaust treatment variation. Finally, distinguish the exact normal-theory result from the covariance result: normality justifies independence, while zero covariance alone does not.
These ideas are aligned with the SOA VEE Mathematical Statistics syllabus because they connect the geometry of ANOVA treatment variation with covariance and normal distributions. They are also useful for reading a collection of planned contrasts: orthogonality can make the contrast components interpretable separately, but it does not remove the need to examine how inferential statistics share an error estimate.
Check Your Understanding
For each problem, use the stated model and show the orthogonality or partition calculation before interpreting the result.
- A balanced study has three groups of size 5 and means 6, 4, and 3. For \(c=(1,-1,0)\) and \(d=(1,1,-2)\), verify orthogonality, calculate both contrast estimates and sums of squares, and check their sum against \(SS_{\text{Tr}}\). (Answer: Orthogonal; estimates \(2\) and \(4\); sums of squares \(10\) and \(20\); \(SS_{\text{Tr}}=30\).)
- Group sizes are 3, 6, and 3. Contrast vectors are \(c=(1,-1,0)\) and \(d=(1,2,-3)\). Are these orthogonal under the unequal-size condition? Give the weighted inner product. (Answer: Yes; \(1/3-2/6=0\).)
- Two orthogonal contrasts have estimates with covariance zero. Errors are independent with common variance, but their distribution is not specified as normal. What can be concluded, and what cannot be concluded from orthogonality alone? (Answer: The estimates are uncorrelated under the variance model; their independence does not follow without an additional condition such as joint normality.)
- For four treatment groups, three mutually orthogonal contrasts have sums of squares 4, 9, and 7. The treatment sum of squares is 25. How much treatment variation remains outside those contrasts, and do the three contrasts form a complete partition? (Answer: \(25-(4+9+7)=5\) remains; they do not form a complete partition because the three components sum to 20, not 25.)
- Two orthogonal contrasts are tested using the same pooled \(MSE\). Under normal errors, are the contrast estimates independent? Are the resulting \(t\) statistics guaranteed to be independent? (Answer: The contrast estimates are independent; the \(t\) statistics are not guaranteed to be independent because they share the random \(MSE\) denominator.)