8.6 - Orthogonal Contrasts

8.6 - Orthogonal Contrasts

Differences among treatments can be explored through pre-planned orthogonal contrasts. Contrasts involve linear combinations of group mean vectors instead of linear combinations of the variables.

Contrasts

The linear combination of group mean vectors

\(\mathbf{\Psi} = \sum_\limits{i=1}^{g}c_i\mathbf{\mu}_i\)

is a (treatment) contrast if

\(\sum_\limits{i=1}^{g}c_i = 0\)

Contrasts are defined with respect to specific questions we might wish to ask of the data. Here, we shall consider testing hypotheses of the form

\(H_0\colon \mathbf{\Psi = 0}\)

Example 8-5: Drug Trial

Suppose that we have a drug trial with the following 3 treatments:

  1. Placebo
  2. Brand Name
  3. Generic

Consider the following questions:

 Question 1: Is there a difference between the Brand Name drug and the Generic drug?

\begin{align} \text{That is, consider testing:}&& &H_0\colon \mathbf{\mu_2 = \mu_3}\\ \text{This is equivalent to testing,}&&  &H_0\colon \mathbf{\Psi = 0}\\ \text{where,}&&  &\mathbf{\Psi = \mu_2 - \mu_3} \\ \text{with}&&  &c_1 = 0, c_2 = 1, c_3 = -1 \end{align}

 Question 2: Are the drug treatments effective?

\begin{align} \text{That is, consider testing:}&& &H_0\colon \mathbf{\mu_1} = \frac{\mathbf{\mu_2+\mu_3}}{2}\\ \text{This is equivalent to testing,}&&  &H_0\colon \mathbf{\Psi = 0}\\ \text{where,}&&  &\mathbf{\Psi} = \mathbf{\mu}_1 - \frac{1}{2}\mathbf{\mu}_2 - \frac{1}{2}\mathbf{\mu}_3 \\ \text{with}&&  &c_1 = 1, c_2 = c_3 = -\frac{1}{2}\end{align}

Estimation

The contrast

\(\mathbf{\Psi} = \sum_{i=1}^{g}c_i \mu_i\)

is estimated by replacing the population mean vectors with the corresponding sample mean vectors:

\(\mathbf{\hat{\Psi}} = \sum_{i=1}^{g}c_i\mathbf{\bar{Y}}_i.\)

Because the estimated contrast is a function of random data, the estimated contrast is also a random vector. So the estimated contrast has a population mean vector and population variance-covariance matrix. The population mean of the estimated contrast is \(\mathbf{\Psi}\). The variance-covariance matrix of \(\hat{\mathbf{\Psi}}\)ΒΈ is:

\(\left(\sum\limits_{i=1}^{g}\frac{c^2_i}{n_i}\right)\Sigma\)

which is estimated by substituting the pooled variance-covariance matrix for the population variance-covariance matrix

\(\left(\sum\limits_{i=1}^{g}\frac{c^2_i}{n_i}\right)\mathbf{S}_p = \left(\sum\limits_{i=1}^{g}\frac{c^2_i}{n_i}\right) \dfrac{\mathbf{E}}{N-g}\)

Orthogonal Contrasts

Two contrasts

\(\Psi_1 = \sum_{i=1}^{g}c_i\mathbf{\mu}_i\) and \(\Psi_2 = \sum_{i=1}^{g}d_i\mathbf{\mu}_i\)

are orthogonal if

\(\sum\limits_{i=1}^{g}\frac{c_id_i}{n_i}=0\)

The importance of orthogonal contrasts can be illustrated by considering the following paired comparisons:

\(H^{(1)}_0\colon \mu_1 = \mu_2\)

\(H^{(2)}_0\colon \mu_1 = \mu_3\)

\(H^{(3)}_0\colon \mu_2 = \mu_3\)

We might reject \(H^{(3)}_0\), but fail to reject \(H^{(1)}_0\) and \(H^{(2)}_0\). But, if \(H^{(3)}_0\) is false then both \(H^{(1)}_0\) and \(H^{(2)}_0\) cannot be true.

Notes

  • For balanced data (i.e., \(n _ { 1 } = n _ { 2 } = \ldots = n _ { g }\) ), \(\mathbf{\Psi}_1\) and \(\mathbf{\Psi}_2\) are orthogonal contrasts if \(\sum_{i=1}^{g}c_id_i = 0\)
  • If \(\mathbf{\Psi}_1\) and \(\mathbf{\Psi}_2\) are orthogonal contrasts, then the elements of \(\hat{\mathbf{\Psi}}_1\) and \(\hat{\mathbf{\Psi}}_2\) are uncorrelated
  • If \(\mathbf{\Psi}_1\) and \(\mathbf{\Psi}_2\) are orthogonal contrasts, then the tests for \(H_{0} \colon \mathbf{\Psi}_1= 0\) and \(H_{0} \colon \mathbf{\Psi}_2= 0\) are independent of one another. That is, the results of the test have no impact on the results of the other test.
  • For g groups, it is always possible to construct g - 1 mutually orthogonal contrasts.
  • If \(\mathbf{\Psi}_1, \mathbf{\Psi}_2, \dots, \mathbf{\Psi}_{g-1}\) are orthogonal contrasts, then for each ANOVA table, the treatment sum of squares can be partitioned into \(SS_{treat} = SS_{\Psi_1}+SS_{\Psi_2}+\dots + SS_{\Psi_{g-1}} \)
  • Similarly, the hypothesis sum of squares and cross-products matrix may be partitioned: \(\mathbf{H} = \mathbf{H}_{\Psi_1}+\mathbf{H}_{\Psi_2}+\dots\mathbf{H}_{\Psi_{g-1}}\)
Legend
[1]Link
↥Has Tooltip/Popover
 Toggleable Visibility