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

7.4.3.6.

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: $${\hat{\mu }}_i={\overline{Y}}_{i\cdot },$$ where $${\overline{Y}}_{i\cdot }=\frac{\sum _{j=1}^{n_i}{Y}_{ij}}{n_i}=\frac{Y_{i\cdot }}{n_i}.$$

Variance of the factor level means

The variance of this sample mean estimator is $$s_{{\overline{Y}}_{i\cdot }}^2=\frac{MSE}{n_i}=\frac{{\hat{\sigma }}_e^2}{n_i}.$$

Confidence intervals for the factor level means

It can be shown that: $$t=\frac{{\overline{Y}}_{i\cdot }-{\mu }_i}{s_{{\overline{Y}}_{i\cdot }}},$$ 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: $${\overline{Y}}_{i\cdot }\pm t_{1-\alpha /2,N-k}\sqrt{\frac{{\hat{\sigma }}_{ϵ}^2}{n_i}}.$$

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: $$s_{{\overline{Y}}_{1\cdot }}^2=\frac{MSE}{n_1}=\frac{1.331}{5}=0.2662$$ $$s_{{\overline{Y}}_{1\cdot }}=\sqrt{0.2662}=0.5159.$$ 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: $$\hat{C}=\sum _{i=1}^r{c}_i{\overline{Y}}_{i\cdot }$$ and $$\text{Var }(\hat{C})=\sum _{i=1}^r{c}_i^2\cdot \text{ Var }({\overline{Y}}_{i\cdot })=\sum _{i=1}^r{c}_i^2\left(\frac{{\sigma }^2}{n_i}\right)={\sigma }^2\sum _{i=1}^r\frac{c_i^2}{n_i}.$$ 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 $$\hat{C}=\sum _{i=1}^r{c}_i{\overline{Y}}_{i\cdot }.$$ The estimator of $\text{Var }(\hat{C})$ 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: $$\hat{C}=\frac{{\overline{Y}}_1+{\overline{Y}}_2}{2}-\frac{{\overline{Y}}_3+{\overline{Y}}_4}{2}=-0.5.$$

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.