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


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 factorlevel 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 oneway 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 \((Nk)\) 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: $$ \bar{Y}_{i\cdot} \pm t_{1\alpha/2, \, Nk} \,\,\sqrt{\frac{\hat{\sigma}^2_\epsilon}{n_i}} \, . $$  
Example 1  
Example for a 4level 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.


Oneway ANOVA table layout 
This experiment can be illustrated by the table layout for this oneway
ANOVA experiment shown below:


ANOVA table 
The resulting ANOVA table is
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 twosided, 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. 

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.


Some properties of orthogonal contrasts 
The following is true:


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 \(t_{Nr}\)
for the oneway ANOVA model under discussion.
Therefore, the \(1\alpha\) confidence limits for \(C\) are: $$ \hat{C} \pm t_{1\alpha/2, \, Nr} \,\, s_{\hat{C}} \, . $$ 

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


Estimation of Linear Combinations  
Estimating linear combinations 
Sometimes we are interested in a linear combination of the
factorlevel 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:
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. 