7.4.3.6. Assessing the response from any factor combination

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

_i in the 1-way ANOVA model is given by:

where

Ybar(i.) = {SUM[j=1 to n(i)][Y(ij)]}/n(i) = Y(i.)/n(i)

Variance of the factor level means

The variance of this sample mean estimator is

Confidence intervals for the factor level means

It can be shown that: t = (Ybar(i.) - mu(i))/s(ybar(i.))

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 (1- alpha

)100% confidence limits for each

_i. These are given by: Ybar(i.) =/- t(alpha/2;N-k)*SQRT(sigmahat(e)/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

1-Way ANOVA table layout

This experiment can be illustrated by the table layout for this 1-way ANOVA experiment shown below:

Level			Sample j
i	1	2	...	5	Sum	Mean	N

1	Y₁₁	Y₁₂	...	Y₁₅	Y_1.	_1.	n₁
2	Y₂₁	Y₂₂	...	Y₂₅	Y_2.	_2.	n₂
3	Y₃₁	Y₃₂	...	Y₃₅	Y_3.	_3.	n₃
4	Y₄₁	Y₄₂	...	Y₄₅	Y_4.	_4.	n₄

All					Y_.	_..	n_t

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 n₁ = 5.

Computing the confidence interval

Since the confidence interval is two-sided, the entry alpha

/2 value for the t-table is .5(1 - .95) = .025, and the associated degrees of freedom is N - 4, or 20 - 4 = 16.

From the t table in Chapter 1, we obtain t_.025;16 = 2.120.

Next we need the standard error of the mean for group 1:

s(Ybar(1.))**2 = MSE/n(i) = 1.331/5 = 0.2662

s(Ybar(1.)) = SQRT(0.2662) = 0.5159

Hence, we obtain confidence limits 5.34 ± 2.120 (0.5159) and the confidence interval is

Definition and Estimation of Contrasts

Definition of contrasts and orthogonal contrasts

Definitions

A contrast is a linear combination of 2 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 _i for the i-th treatment mean:

C = c₁

₁ + c₂

₂ + ... + c_j

_j + ... + c_k

where

c₁ + c₂ + ... + c_j + ... + c_k = SUM[j=1 to k]c(j) = 0

Simple contrasts include the case of the difference between two factor means, such as ₁ - ₂. If one wishes to compare treatments 1 and 2 with treatment 3, one way of expressing this is by: ₁ + ₂ - 2₃. Note that

₁ - ₂ has coefficients +1, -1
₁ + ₂ - 2₃ 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.


	₁	₂	₃	₄

c₁	+1	0	0	-1
c₂	0	+1	-1	0
c₃	+1	-1	-1	+1

Some properties of orthogonal contrasts

The following is true:

The sum of the coefficients for each contrast is zero.
The sum of the products of coefficients of each pair of contrasts is also 0 (orthogonality property).
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:

and

Var(Chat) = SUM=1 to r][c(i)**2*Var(Ybar(i.)] =
SUM[i=1 to r][c(i)**2*(sigma**2/n(i))] =
simga**2*SUM[i=1 to r][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 Chat = SUM[i=1 to r][c(i)*Ybar(i.)]

The estimator of is

s(Chat)**2 = sigmahat(e)**2*SUM[j=1 to r][c(i)**2/n(i))

The estimator is normally distributed because it is a linear combination of independent normal random variables. It can be shown that:

is distributed as t_N-r for the one-way ANOVA model under discussion.

Therefore, the 1- alpha confidence limits for C are:

Example 2 (estimating contrast)

Contrast to estimate

We wish to estimate, in our previous example, the following contrast: ((mu1 + mu2)/2) - (mu3 + mu4)/2)

and construct a 95 percent confidence interval for C.

Computing the point estimate and standard error

The point estimate is:

Chat = (Ybar(1) + Ybar(2))/2 - (Ybar(3) + Ybar(4))/2 = -0.5

Applying the formulas above we obtain

SUM[i=1 to 4][c(i)**2/n(i)] = 4*(1/2)**2/5 = -0.5

and

s(Chat)**2 = MSE*SUM[i=1 to 4][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_.025;16 = 2.12. Therefore, the desired 95% confidence interval is -.5 ± 2.12(.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 to 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.