Orthogonal Contrasts - Planned Comparisons

Orthogonal Contrast

The analysis of variance method is useful and is a reliable tool for comparing several treatments means.  In comparing t treatments, the null hypothesis states that all treatment means are not different (H₀: μ₁ = μ₂ = ... = μ_t).  If the F test is significant, then HA is accepted, stating that not all treatment means are the same or are one of the mean treatments that are different from the others.  Furthermore, a comparison was made to determine which treatment was different by parsing the Squared Number of Treatments for additional F tests to answer some of the questions that had already been planned.  Contrast or orthogonal methods for separating means require a certain level of knowledge that is a priori, either based on certain scientific considerations or based on the results of previous studies.  Therefore, this method is also called the planned F Test.

If the researcher has a specific question that needs to be answered, the treatment is designed to provide information and statistical tests that will answer the question.  Experienced researchers will select the treatment so that the Squared Amount of Treatment can be parsed to answer several free questions according to the value of the treatment-degree of freedom contained in the analysis of variance.  Consequently, another name for this test is a single degree of freedom test.  If the comparison is mutually free, it is said to be orthogonal.

Definition of Contrast and orthogonality

Contrast is the linear sum in the form:

 $$Q=\sum_{i=1}^{m}{c_i{\bar{Y}}_{i.}}\text{; with } \sum_{i=1}^{m}c_i=0$$

The number of coefficients in each comparison must be zero, ${\bar{Y}}_{i.}$ is the mean of the treatments(it can also be in the form of the number of treatments), and m ≤ t, is the number of treatments to be compared.  For convenience in calculations, the value of the coefficient c_{i is} usually an integer.  Contrast always has a single degree of freedom (degree of freedom = 1).

Suppose  $Q_c=\sum_{i=1}^{m}{c_i{\bar{Y}}_{i.}}$ dan $Q_d\sum_{i=1}^{m}d_i{\bar{Y}}_{i.}$ it is two contrasts.  Both are orthogonal when the number of multipliers of the corresponding coefficients in two or more ratios is zero.

$$\sum_{i=1}^{m}{c_id_i}=0 \text{ or } \sum_{i=1}^{m}{c_id_i/n_i=0} \text{ for unbalanced designs}$$

The main purpose of the orthogonal contrast is to separate the null hypothesis into several combinations of mean comparisons.  The trait of orthogonality is very important for the following reasons: for t-1 orthogonal interchangeable comparators from t of treatment fruits, then the sum of the squares of all such comparators (t-1 pieces) will be equal to the sum of the squares of the original treatment.  This means that the experimental treatment can be broken down into t-1 free and separate experiments, one for each comparison.

Example

Suppose we want to test three treatments, T1, T2 and T3 (control), so that there are 2 degrees free of treatment.  We state the treatment with μ1, μ2,, and μ3.  The null hypothesis: H₀: μ₁ = μ₂ = μ₃.  Since the degree of freedom = 2, there are two types of comparisons that may be made.  For example, to test the hypothesis whether T1 and T2 are not significantly different from controls (T3): μ₁ = μ₃.  and μ₂ = μ₃.

For μ₁ = μ₃ (1μ₁ + 0μ₂ -1μ₃= 0) the coefficient is : c1 = 1, c2 = 0, c3 = -1

For μ₂ = μ₃ (0μ₁ + 1μ₂ -1μ₃= 0) the coefficient is: d1 = 0, d2 = 1, d3= -1

The mean linear combination is a contrast if the sum of all c_{i is} equal to zero:

 $\sum_{i=1}^{m}c_i=0$ : (1 + 0 + (-1) = 0).

However, the contrasts are not mutually orthogonal because:

 $\sum_{i=1}^{m}{c_id_i}\neq0$   (c1d1 + c2d2+ c3d3 = 0 + 0 + 1 = 1).

This suggests that not every pair of hypotheses can be tested using this approach.  In addition, in order for the summation to be 0, the coefficient c_i is almost always an integer value .  For t = 3, for example the pair of coefficients is c1 = 1, c2 = 1, c3 = -2; and d1 = 1, d2 = -1, d3 = 0.  Hypotheses defined based on the orthogonal coefficient are 1) the mean of the two treatments is no different from the control, 2) the μ₁ does not differ markedly from the μ₂.  The two comparisons are contrasting because (1 + 1 + (-2) = 0 and 1+ (-1) + 0= 0) and are orthogonal because (c1d1 + c2d2+ c3d3 = 1 + (-1) + 0 = 0).

Group Comparison

The first application of orthogonal contrast is for comparison in groups.  This is a Analysis of variance on the mean group.  To make it easier to understand the use of orthogonal contrast in group comparisons, we will use the example data in the following Table.  An analysis of its variance is given in the next Table.

Table 1.    Results (mg weight of upper plants, shoot) RAL experiments to determine the effect of seed treatment by various types of acids at the hatchery stage.

* = inorganic, ** = organic

Table 2.     Analysis of variance

For the experiment, researchers may have several questions that have been designed to answer:

1. Does acid treatment decrease seed growth?
2. Are organic acids different compared to inorganic acids?
3. Is there a difference in effect between the two organic acids?

To answer the question, a table of linear comparison coefficients between treatments is presented in the following table.

Table 3.    Orthogonal coefficients for separating the squared number of treatments in the Analysis of variance Table into three mutually free tests.

Coefficients can be used to divide the Sum of Treatment Squares (SST) into three components, each with one degree of freedom to be used in Test F for each comparison.  The critical value of F is based on 1 degree of freedom for numerator and df error for divisor (Test F with single degree of freedom). 

Rules for determining the value of the coefficient on the comparison of groups (Little &Hills p 66).

1. In the mean comparison of the two groups, each group that has the same treatment is given the same coefficient, +1 for one group and -1 for the other group.  Similar cases could be applied to more complex comparisons, more than two groups.
2. In comparing groups whose number of treatments differs, in the first group give a coefficient value equal to the number of the second group, and in the second group give a coefficient of uniform value until the number is equal to the number of groups of the first.  For example, if there are 5 treatments to be compared, the first group has 2 treatments and the second group has 3 treatments.  The coefficient value for the first group should be worth 3 (equal to the number of the second group, 3 treatments) and the coefficient value for each treatment in the second group should be -2 (why -2? the number of the third coefficient of treatment should be equal to the number of the first group, 3+3=6, only the sign should be "-") , so the coefficient for each treatment should be: +3,  +3,  -2, -2, -2. 
3. The coefficient for each comparison is as much as possible the smallest integer for each calculation.  For example, coefficients: +4, +4, -2, -2, -2, -2. it should be simplified (everything is divided by 2) to the smallest integers : +2, +2, -1, -1, -1, -1.
4. Table 4).

Table 4.    Fertilization experiment with 4 treatments, 2 levels for N and 2 levels for P.

The coefficients for the first 2 comparisons are obtained from the first rule.  The coefficient of interaction is the sum of the multiplications between the coefficient values of the two rows.

Note, the number of coefficients on each comparison must be zero and the number of cross products must also be zero.  If both requirements are met, then the comparison is said to be orthogonal, so the conclusions drawn in one of the comparisons do not depend on the comparison of the other groups.

The calculation of the sum of squares for Test F with a single degree of freedom for the linear combination of treatment means is as follows:

$$ SS(Q)=MS(Q)=\frac{(\sum{c_i{\bar{Y}}_{i.})^2}}{1/r\sum c_i^2}\ \ or\ \ \frac{(\sum{c_i{\bar{Y}}_{i.})^2}}{\sum{c_i^2/r_i}}\text{ for unbalanced design}$$

SS (control vs. acid)        = [3(4.19) – 3.64 – 3.73 – 3.87]² / [(12)/5] = 0.74

SS (Inorg. vs. org.)           = [3.64 + 3.73 – 2(3.87)]² / [(6)/5] = 0.11

SS (between orgs.)         = [-3.64 + 3.73]² / [(2)/5] = 0.02

Table 5.    Decomposition of the Squared Number of Treatments on orthogonal Comparisons.

From the above analysis we can conclude that the three acid treatments very markedly reduce seed growth, organic acids cause a greater decrease in seed growth compared to inorganic acids, and there is no difference between organic acid treatments.

The number of single degree of freedoms of group comparison is equal to df treatment.  This will happen if each comparison is orthogonal.  The multiplicity of orthogonal group comparisons is equal to the degree of free treatment.  If the comparison is not orthogonal, the sum of squares for a given group comparison may represent (or be represented by) the sum of the squares of other group comparisons. Therefore, the conclusions obtained from one of the tests may be affected by other comparison tests and the sum of the squares of each comparison will not be equal to the sum of the squares of the treatment.