 7. Product and Process Comparisons
7.4. Comparisons based on data from more than two processes
7.4.3. Are the means equal?

## Assessing the response from any factor combination

Contrasts This page treats how to estimate and put confidence bounds around the response to different combinations of factors. Primary focus is on the combinations that are known as contrasts. We begin, however, with the simple case of a single factor-level mean.
Estimation of a Factor Level Mean With Confidence Bounds
Estimating factor level means An unbiased estimator of the factor level mean $$\mu_i$$ in the one-way ANOVA model is given by: $$\hat{\mu}_i = \bar{Y}_{i \huge{\cdot}} \, ,$$ where $$\bar{Y}_{i \huge{\cdot}} = \frac{\sum_{j=1}^{n_i} Y_{ij}}{n_i} = \frac{Y_{i \huge{\cdot}}}{n_i} \, .$$
Variance of the factor level means The variance of this sample mean estimator is $$s_{\bar{Y}_{i \huge{\cdot}}}^2 = \frac{MSE}{n_i} = \frac{\hat{\sigma}_e^2}{n_i} \, .$$
Confidence intervals for the factor level means It can be shown that: $$t = \frac{\bar{Y}_{i \huge{\cdot}} - \mu_i}{s_{\bar{Y}_{i \huge{\cdot}}}} \, ,$$ has a $$t$$ distribution with $$(N-k)$$ degrees of freedom for the ANOVA model under consideration, where $$N$$ is the total number of observations and $$k$$ is the number of factor levels or groups. The degrees of freedom are the same as were used to calculate the MSE in the ANOVA table. That is: dfe (degrees of freedom for error) = $$N - k$$. From this we can calculate$$100(1-\alpha)$$ % confidence limits for each $$\mu_i$$. These are given by: $$\bar{Y}_{i\cdot} \pm t_{1-\alpha/2, \, N-k} \,\,\sqrt{\frac{\hat{\sigma}^2_\epsilon}{n_i}} \, .$$
Example 1
Example for a 4-level treatment (or 4 different treatments) The data in the accompanying table resulted from an experiment run in a completely randomized design in which each of four treatments was replicated five times.

 Total Mean Group 1 6.9 5.4 5.8 4.6 4.0 26.70 5.34 Group 2 8.3 6.8 7.8 9.2 6.5 38.60 7.72 Group 3 8.0 10.5 8.1 6.9 9.3 42.80 8.56 Group 4 5.8 3.8 6.1 5.6 6.2 27.50 5.50 All Groups 135.60 6.78

One-way ANOVA table layout This experiment can be illustrated by the table layout for this one-way ANOVA experiment shown below:

 Level Sample j i 1 2 ... 5 Sum Mean N 1 Y11 Y12 ... Y15 Y1. 1. n1 2 Y21 Y22 ... Y25 Y2. 2. n2 3 Y31 Y32 ... Y35 Y3. 3. n3 4 Y41 Y42 ... Y45 Y4. 4. n4 All Y. .. nt

ANOVA table The resulting ANOVA table is

 Source SS DF MS F Treatments 38.820 3 12.940 9.724 Error 21.292 16 1.331 Total (Corrected) 60.112 19 Mean 919.368 1 Total (Raw) 979.480 20

The estimate for the mean of group 1 is 5.34, and the sample size is $$n_1$$ = 5.

Computing the confidence interval Since the confidence interval is two-sided, the entry $$(1-\alpha/2)$$ value for the $$t$$ table is (1 - 0.05/2) = 0.975, and the associated degrees of freedom is $$N$$ - 4, or 20 - 4 = 16.

From the t table in Chapter 1, we obtain $$t_{0.975, \, 16}$$ = 2.120.

Next we need the standard error of the mean for group 1: $$s_{\bar{Y}_{1 \huge{\cdot}}}^2 = \frac{MSE}{n_1} = \frac{1.331}{5} = 0.2662$$ $$s_{\bar{Y}_{1 \huge{\cdot}}} = \sqrt{0.2662} = 0.5159 \, .$$ Hence, we obtain confidence limits 5.34 ± 2.120 (0.5159) and the confidence interval is $$4.246 \le \mu_1 \le 6.434 \, .$$

Definition and Estimation of Contrasts
Definition of contrasts and orthogonal contrasts Definitions

A contrast is a linear combination of two or more factor level means with coefficients that sum to zero.

Two contrasts are orthogonal if the sum of the products of corresponding coefficients (i.e., coefficients for the same means) adds to zero.

Formally, the definition of a contrast is expressed below, using the notation $$\mu_i$$ for the $$i$$-th treatment mean: $$C = c_1 \mu_1 + c_2 \mu_2 + \cdots + c_j \mu_j + \cdots + c_k \mu_k \, ,$$ where $$c_1 + c_2 + \cdots + c_j + \cdots + c_k = \sum_{j=1}^k c_j = 0 \, .$$ Simple contrasts include the case of the difference between two factor means, such as $$\mu_1 - \mu_2$$. If one wishes to compare treatments 1 and 2 with treatment 3, one way of expressing this is by: $$\mu_1 + \mu_2 - 2\mu_3$$. Note that

$$\mu_1 - \mu_2$$ has coefficients +1, -1.
$$\mu_1 + \mu_2 - 2\mu_3$$ has coefficients +1, +1, -2.
These coefficients sum to zero.
An example of orthogonal contrasts As an example of orthogonal contrasts, note the three contrasts defined by the table below, where the rows denote coefficients for the column treatment means.

 $$\mu_1$$ $$\mu_2$$ $$\mu_3$$ $$\mu_4$$ $$c_1$$ +1 0 0 -1 $$c_2$$ 0 +1 -1 0 $$c_3$$ +1 -1 -1 +1
Some properties of orthogonal contrasts The following is true:
1. The sum of the coefficients for each contrast is zero.

2. The sum of the products of coefficients of each pair of contrasts is also 0 (orthogonality property).

3. The first two contrasts are simply pairwise comparisons, the third one involves all the treatments.
Estimation of contrasts As might be expected, contrasts are estimated by taking the same linear combination of treatment mean estimators. In other words: $$\hat{C} = \sum_{i=1}^r c_i \bar{Y}_{i \huge{\cdot}}$$ and $$\mbox{Var } (\hat{C}) = \sum_{i=1}^r c_i^2 \cdot \mbox{ Var } (\bar{Y}_{i \huge{\cdot}}) = \sum_{i=1}^r c_i^2 \left( \frac{\sigma^2}{n_i} \right) = \sigma^2 \sum_{i=1}^r \frac{c_i^2}{n_i} \, .$$ Note: These formulas hold for any linear combination of treatment means, not just for contrasts.
Confidence Interval for a Contrast
Confidence intervals for contrasts An unbiased estimator for a contrast $$C$$ is given by $$\hat{C} = \sum_{i=1}^r c_i \bar{Y}_{i \huge{\cdot}} \, .$$ The estimator of $$\mbox{Var }(\hat{C})$$ is $$s_{\hat{C}}^2 = \hat{\sigma}_e^2 \, \sum_{i=1}^r \frac{c_i^2}{n_i} \, .$$ The estimator $$\hat{C}$$ is normally distributed because it is a linear combination of independent normal random variables. It can be shown that: $$\frac{\hat{C} - C}{s_{\hat{C}}} \, ,$$ is distributed as $$t_{N-r}$$ for the one-way ANOVA model under discussion.

Therefore, the $$1-\alpha$$ confidence limits for $$C$$ are: $$\hat{C} \pm t_{1-\alpha/2, \, N-r} \,\, s_{\hat{C}} \, .$$

Example 2 (estimating contrast)
Contrast to estimate We wish to estimate, in our previous example, the following contrast: $$C = \frac{\mu_1 + \mu_2}{2} - \frac{\mu_3 + \mu_4}{2} \, ,$$ and construct a 95 % confidence interval for $$C$$.
Computing the point estimate and standard error The point estimate is: $$\hat{C} = \frac{\bar{Y}_1 + \bar{Y}_2}{2} - \frac{\bar{Y}_3 + \bar{Y}_4}{2} = -0.5 \, .$$

Applying the formulas above we obtain $$\sum_{i=1}^4 \frac{c_i^2}{n_i} = \frac{4(1/2)^2}{5} = 0.2$$ and $$s_{\hat{C}}^2 = MSE \, \sum_{i=1}^4 \frac{c_i^2}{n_i} = 1.331(0.2) = 0.2662 \, ,$$ and the standard error is $$\sqrt{0.2661} = 0.5159$$.

Confidence interval For a confidence coefficient of 95 % and df = 20 - 4 = 16, $$t_{0.975, \, 16}$$ = 2.12. Therefore, the desired 95 % confidence interval is -0.5 ± 2.12(0.5159) or

(-1.594, 0.594).
Estimation of Linear Combinations
Estimating linear combinations Sometimes we are interested in a linear combination of the factor-level means that is not a contrast. Assume that in our sample experiment certain costs are associated with each group. For example, there might be costs associated with each factor as follows:

 Factor Cost in \$ 1 3 2 5 3 2 4 1

The following linear combination might then be of interest: $$C = 3\mu_1 + 5\mu_2 + 2\mu_3 + 1\mu_4 \, .$$

Coefficients do not have to sum to zero for linear combinations This resembles a contrast, but the coefficients $$c_i$$ do not sum to zero. A linear combination is given by the definition: $$C = \sum_{i=1}^r c_i \mu_i \, ,$$ with no restrictions on the coefficients $$c_i$$.
Confidence interval identical to contrast Confidence limits for a linear combination $$C$$ are obtained in precisely the same way as those for a contrast, using the same calculation for the point estimator and estimated variance. 