1. Chapter 13: Experiments with Random Factors
Design & Analysis of Experiments 8E 2012 Montgomery
STT 511-STT411: Design of Experiments and Analysis of Variance Dr. Cuixian Chen
Chapter 13
Chapter 13: Experiments with Random Factors

2. Chapter 13
Design & Analysis of Experiments 8E 2012 Montgomery

3. Design of Engineering Experiments - Experiments with Random Factors
Text reference, Chapter 13
Previous chapters have focused primarily on fixed factors
A specific set of factor levels is chosen for the experiment
Inference confined to those levels
Often quantitative factors are fixed
When factor levels are chosen at random from a larger population of potential levels, the factor is random
Inference is about the entire population of levels
Industrial applications include measurement system studies
It has been said that failure to identify a random factor as random and not treat it properly in the analysis is one of the biggest errors committed in DOE
The random effect model was introduced in Chapter 3
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

4. Chap 3.9: Random Effects Model in One way ANOVA

5. Review: fundamental one-way ANOVA identity with fixed effect
The total variability in the data, as measured by the total corrected sum of squares, can be partitioned into a sum of squares of the differences between the treatment averages and the grand average, plus a sum of squares of the differences of observations within treatments from the treatment average.
Now, the difference between the observed treatment averages and the grand average is a measure of the differences between treatment means, whereas the differences of observations within a treatment from the treatment average can be due only to random error.
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 3
Chapter 3

6. 3.9 Random Effects Model in One way ANOVA
Text reference, page 116.
There are a large number of possible levels for the factor (theoretically an infinite number).
The experimenter chooses a of these levels at random.
Inference will be to the entire population of levels.
Example: What are the effects of repeated exposure to an advertising message (digital camera)? The answer may depend on length of the ad.
Consider different lengths of the ad. Eg: 1 second, 2 seconds, …, 10 seconds, …, 100 seconds, and so on. (Infinite possibilities)
Now suppose we randomly choose 2 levels of lengths: 15 seconds and 32 seconds. (a=2)
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 3

7. Variance components
Testing hypotheses about individual treatment effects is not very meaningful because they were selected randomly, we are more interested in the population of treatments.
p-value=1-pf(F0, a-1, N-a)
Chapter 3
Design & Analysis of Experiments 8E 2012 Montgomery

8. Covariance structure:
Unlike the fixed effects case in which all of the observations yij are independent, in the random model the observations yij are only independent if they come from different factor levels.
Chapter 3
Design & Analysis of Experiments 8E 2012 Montgomery

9. Observations (a = 3 and n = 2):
Chapter 3
Design & Analysis of Experiments 8E 2012 Montgomery

10. >
Chapter 3
Design & Analysis of Experiments 8E 2012 Montgomery

11. Note: ANOVA F-test is identical to the fixed-effects case.
Chapter 3
Design & Analysis of Experiments 8E 2012 Montgomery

12. Estimating the variance components using the ANOVA method:
Chapter 3
Design & Analysis of Experiments 8E 2012 Montgomery

13. Example 3.11: Random Effect Model
A textile company weaves a fabric on a large number of looms. It would like looms to be homogeneous so that it obtains a fabric of uniform strength. The process engineer suspects that, in addition to usual variation in strength within samples of fabric from the same loom, there may also be significant variations in strength between looms. To investigate this, she selects four looms at random and makes four strength determinations on fabric manufactured on each loom. This experiment is run in random order, and data obtained are shown in Table
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4

14. Example 3.11: Random Effect Model
#install.packages("GAD")
exp3.11<-read.table(" = TRUE);
library(GAD); ## General ANOVA Design ##
## For random effect with any combination of orthogonal/nested and fixed/random ## factors
y<- exp3.11$Strength;
## “as.random” works the same way as as.factor, but assigns an additional class informing that it is a random factor.
gad(lm(y~ as.random(exp3.11$Loom) ));
anova(lm(y~factor(exp3.11$Loom))) ## For regular ANOVA with fixed effect ##
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4

15. Example 3.11: Random Effect Model
#install.packages("GAD")
exp3.11<-read.table(" = TRUE);
library(GAD); ## General ANOVA Design ##
## For random effect with any combination of orthogonal/nested and fixed/random factors ##
y<- exp3.11$Strength;
## “as.random” works the same way as as.factor, but assigns an additional class informing that it is a random factor.
gad(lm(y~ as.random(exp3.11$Looms)));
anova(lm(y~as.factor(exp3.11$Loom))) ## For regular ANOVA with fixed effect ##
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4

16. Example 3.11: Random Effect Model
From the ANOVA, we conclude that the looms in the plant differ significantly.
Q: How to estimate
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

17. Estimators
The ANOVA variance component estimators are moment estimators.
Normality not required to estimate variance component.
They are unbiased estimators.
Finding confidence intervals on the variance components is “clumsy”.
Negative estimates can occur – this is “embarrassing”, as variances are always non-negative.
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4

18. Confidence interval for sigma^2
Chapter 3
Design & Analysis of Experiments 8E 2012 Montgomery

19. Confidence interval for the interclass correlation coefficient
interclass correlation coefficient gives proportion of variance of an observation that is the results of difference b/w treatments.
Chapter 3
Design & Analysis of Experiments 8E 2012 Montgomery

20. Chapter 3
Design & Analysis of Experiments 8E 2012 Montgomery

21. Is there significant variation in the calcium content from batch to batch?
Estimate the components of variance, by POINT estimate method. F
Find a 95% confidence interval for mu.
Find a 95% confidence interval for sigma.

22. Estimate the components of variance. F
Is there significant variation in the Uniformity at different wafer positions?
Estimate the components of variance. F
Chapter 13
Design & Analysis of Experiments 8E 2012 Montgomery

23. Chap 4.1: Random Blocks and Fixed Treatments (page 151)

24. Review: ANOVA for RCBD with fixed treat/block
Suppose single factor, a treatments (factor levels) and b blocks (treat/block are both fixed)
A statistical model (effects model) for the RCBD is
The relevant (fixed effects) hypotheses are
Model Assumption:
Constraints:
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4

25. Review: ANOVA for RCBD with fixed treat/block
The degrees of freedom for the sums of squares in
are as follows:
Therefore, ratios of sums of squares to their degrees of freedom result in mean squares and the ratio of the mean square for treatments to the error mean square is an F statistic that can be used to test the hypothesis of equal treatment means
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4

26. Random Blocks and Fixed Treatments
There are many situations where either treatments or blocks (or both) are random factors.
It is very common to find that the blocks are random.
For conclusions from the experiment to be valid across the population of blocks that the ones selected for the experiments were sampled from.
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4

27. Ch 4.1: Random Blocks and Fixed Treatments
Design & Analysis of Experiments 8E 2012 Montgomery
It is exactly same test statistic we used when blocks were fixed.
Chapter 4

28. Example 1
Q: Find the components of variance for the following example.
Chapter 4
Design & Analysis of Experiments 8E 2012 Montgomery

29. Example 1 Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4
Design & Analysis of Experiments 8E 2012 Montgomery

30. Example 2 Q1: List all the Null and Alternative hypotheses.
Q2: Input all data into R and conduct the ANOVA analysis.
Q3: Based on the outputs, draw conclusions to hypothsis.
Q4: Find the components of variance.
Chapter 13
Design & Analysis of Experiments 8E 2012 Montgomery

31. Example 2 Q1: List all the Null and Alternative hypotheses.
Q2: Draw conclusions?
Q3: Find the components of variance.
Block=c(rep(1, 4), rep(2, 4), rep(3, 4), rep(4, 4), rep(5,4));
Treat=rep(c("A", "B", "C", "D"), 5)
y=c(89, 88, 97, 94, 84, 77, 92, 79, 81, 87, 87, 85, 87, 92, 89, 84, 79, 81, 80, 88); cbind(y, Treat, Block)
library(GAD);
Block = as.random(Block);
Treat = as.fixed(Treat);
gad(lm(y~Treat+Block)); ##anova(lm(y~factor(Treat)+factor(Block)));
Chapter 13
Design & Analysis of Experiments 8E 2012 Montgomery

32. Example 3 Q1: List all the Null and Alternative hypotheses.
Q2: Draw conclusions?
Q3: Find the components of variance.
Treat=c(rep(1, 4), rep(2, 4), rep(3, 4), rep(4, 4));
Block=rep(c("A", "B", "C", "D"), 4)
y=c(-2, -1, 1, 5, -1, -2, 3, 4, -3, -1, 0, 2, 2, 1, 5, 7); cbind(y, Treat, Block)
library(GAD);
Block = as.random(Block);
Treat = as.fixed(Treat);
gad(lm(y~Treat+Block)); ##anova(lm(y~factor(Treat)+factor(Block)));
Chapter 13
Design & Analysis of Experiments 8E 2012 Montgomery

33. Chap 13: Experiments with Random Factors

34. Review: Chapter 5 -- Two-factor Factorial Design
Statistical (effects) model:
Other models (means model, regression models) can be useful.
is overall mean effect, is effect of ith level of row factor A, is effect of jth level of column factor B, is the effect of interaction between and , and
is a random error component.
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 5

35. Review: Chapter 5 -- Two-factor Factorial Design
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4

36. Review: ANOVA Table for Two-factor Factorial Design – Case1: Both are with Fixed Effects
The reference distribution for F0 is the F((numerator df), (denominator df)) distribution
Reject the null hypothesis (equal treatment means) if p-value<0.05,
Where p-value=1-pf(F0, (numerator df), (denominator df))
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 5

37. Design of Engineering Experiments - Experiments with Random Factors
Text reference, Chapter 13
Previous chapters have focused primarily on fixed factors
A specific set of factor levels is chosen for the experiment
Inference confined to those levels
Often quantitative factors are fixed (why?)
When factor levels are chosen at random from a larger population of potential levels, the factor is random
Inference is about the entire population of levels
Industrial applications include measurement system studies
It has been said that failure to identify a random factor as random and not treat it properly in the analysis is one of the biggest errors committed in DOX
The random effect model was introduced in Chapter 3
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

38. Case 2: Experiments with Two Random Effects
Two factors, factorial experiment, both factors random (Section 13.2, pg. 574)
The model parameters are NID random variables.
Random effects model.
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

39. Case 2: Testing Hypotheses - Random Effects Model
Standard ANOVA partition is appropriate. Relevant hypotheses:
Form of the test statistics depend on the expected mean squares:
Test effects of A and B: different from two fixed effects Modes
The reference distribution for F0 is the F((numerator df), (denominator df)) distribution
Reject the null hypothesis (equal treatment means) if p-value<0.05,
Where p-value=1-pf(F0, (numerator df), (denominator df))
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

40. Case 2: Estimating the Variance Components – Two Factor Random model
As before, we can use the ANOVA method; equate expected mean squares to their observed values:
These are moment estimators.
Potential problems with these estimators.
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

41. Example 13.1 A Measurement Systems Capability Study
Statistically designed experiments are frequently used to investigate the sources of variability that affect a system.
A common industrial application is to use a designed experiment to study the components of variability in a measurement system.
These studies are often called gauge capability studies or gauge repeatability and reproducibility (R&R) studies because these are the components of variability that are of interest (for more discussion of gauge R&R studies, see the supplemental material for this chapter).
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

42. Example 13.1 A Measurement Systems Capability Study
Gauge capability (or R&R) is of interest
The gauge is used by an operator to measure a critical dimension on a part
Repeatability is a measure of the variability due only to the gauge
Reproducibility is a measure of the variability due to the operator
See experimental layout, Table This is a two-factor factorial (completely randomized) with both factors (operators, parts) random – a random effects model
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

43. Example 13.1: ANOVA with Two Random Factors
Chapter 13
Design & Analysis of Experiments 8E 2012 Montgomery

44. Example 13.1: ANOVA with Two Random Factors
Chapter 13
Design & Analysis of Experiments 8E 2012 Montgomery

45. Example 13.1: ANOVA with Two Random Factors
Q1: List all Null and alternatives
Q2: If the F values and p-values are missing, may we still find them?
Q3: Find the components of variance.
Q4: The reduced model?
Q5:The model assumptions.
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

46. Example 13.2 Summary Minitab Solution – Balanced ANOVA
When both factors are random:
There is a large effect of parts (not unexpected)
Small operator effect
No Part – Operator interaction
Negative estimate of the Part – Operator interaction variance component
Fit a reduced model with the Part – Operator interaction deleted
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

47. Example 13.1: Reduced model: ANOVA with Two Random Factors
Chapter 13
Design & Analysis of Experiments 8E 2012 Montgomery

48. Example 13.1 Minitab Solution – Reduced Model
Q1: Find the components of variance.
Q2: Compare with the one we have to original model?
Original Model
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

49. Example 13.1 Minitab Solution – Reduced Model
Estimating gauge capability:
If interaction had been significant?
The variability in the gauge appears small relative to the variability in the product.
This is generally a desirable situation, implying that the gauge is capable of distinguishing among different grades of product.
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

50. 2nd example on two random factors model
Q1: The model assumptions.
Q2: Verify the F-value
Q3: Find the components of variance.
Q4: List all Null and alternatives
Q5: Need a reduced model?
Battery<-read.table(" = TRUE);
a=c(5341.9, , ,675.2 )
c(a[1]/a[3], a[2]/a[3], a[3]/a[4])
1-pf(a[1]/a[3], 2,4)
1-pf(a[2]/a[3], 2,4)
1-pf(a[3]/a[4], 4, 27)
## 13.2 Two factor factorial with random factor ##
## Input data into R by hand…
library(GAD);
y<- Battery$Life;
mat<- as.random(Battery$Material);
temp<- as.random(Battery$Temp);
gad(lm(y~mat*temp)); ## anova(lm(y~factor(Battery$Temp)*factor(Battery$Material)));
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

51. 2nd example on two random factors model
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

52. Example 3 Q1: The model assumption. Q2: Verify the F-value
Q3: Find the components of variance.
Q4: List all Null and alternatives
Q5: Need a reduced model?
Q: Fill in the blanks.
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

53. Example 3 Q1: The model assumption. Q2: Verify the F-value
Q3: Find the components of variance.
Q4: List all Null and alternatives
Q5: Need a reduced model?
## 2nd example ##
Tool<-read.table(" = TRUE);
library(GAD);
y<-Tool$Life;
Sp<-as.random(Tool$Speed);
An<-as.random(Tool$Angle);
gad(lm(y~Sp*An))
#SpAn<-as.random(Tool$Speed*Tool$Angle)
#model3.13.2<-lm(y~SpAn) ;
#gad(model3.13.2);
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

54. Case 3: Two-Factor Mixed Model: A fixed & B random
Two factors, factorial experiment, factor A fixed, factor B random (Section 13.3, pg. 581)
The model parameters are NID random variables, the interaction effects are normal, but not independent.
This is called the restricted model. This restriction implies certain interaction elements at different levels of the fixed factor are not independent.
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4

55. Testing Hypotheses - Mixed Model
Once again, the standard ANOVA partition is appropriate
Relevant hypotheses:
Test statistics depend on the expected mean squares:
Test effects of A: different from two fixed effects Modes
The reference distribution for F0 is the F((numerator df), (denominator df)) distribution
Reject the null hypothesis (equal treatment means) if p-value<0.05,
Where p-value=1-pf(F0, (numerator df), (denominator df))
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4

56. Estimating the Variance Components – Two Factor Mixed model
Use the ANOVA method; equate expected mean squares to their observed values:
Estimate the fixed effects (treatment means) as usual
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4

57. Example 13.2: Measurement Systems Capability Study Revisited
Same experimental setting as in example 13.1
Parts are a random factor, but Operators are fixed
Assume the restricted form of the mixed model
Minitab can analyze mixed model and estimate variance components
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

58. Example 13.2 Minitab Solution – Balanced ANOVA
Q1: List all Null and alternatives.
Q2: Verify the F-values.
Q3: Find the components of variance.
Q4: The reduced model?
Q5: The model assumption?
Q6: The difference with 2-factor factor random effects model?
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

59. Example 13.2 Minitab Solution – Balanced ANOVA
Q1: List all Null and alternatives.
Q2: Verify the F-values.
Q3: Find the components of variance.
Q4: The reduced model?
Q5: The model assumption?
Q6: The difference with 2-factor factor random effects model?
When A is fixed and B is random
When both are considered as random
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

60. Example 13.2 The Measurement Systems Capability Study Revisited
When Parts are a random factor, but Operators are fixed:
There is a large parts effect (not unexpected)
Small operator effect
No Part – Operator interaction
Negative estimate of the Part – Operator interaction variance component
Fit a reduced model with the Part – Operator interaction deleted
This leads to the same solution that we found previously for the two-factor random model
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

61. 2nd example on two-factor with one fixed and one random
Q1:The model assumption
Q2: verify the F-value
Q3:Find the components of variance.
Q4: List all Null and alternatives
Q5:the reduced model?
## 13.3 Two factor Mixed Model ##
Battery<-read.table(" = TRUE);
library(GAD); y<- Battery$Life;
temp<- as.fixed(Battery$Temp);
mat<- as.random(Battery$Material);
mod13.3.1<-lm(y~temp*mat);
gad(mod13.3.1);
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

62. Q1:The model assumption Q2: verify the F-value
Q3:Find the components of variance.
Q4: List all Null and alternatives
Q5:the reduced model?
## 2nd example ##
Tool<-read.table(" = TRUE);
library(GAD);
y<-Tool$Life;
Sp<-as.fixed(Tool$Speed);
An<-as.random(Tool$Angle);
mod13.3.2<-lm(y~Sp*An);
gad(mod13.3.2)
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 13

63. Summary: Comparison on different cases with two-factor
Case 1: Both factors of A and B are fixed;
Case 2: Both factors of A and B are random;
Case 3: Factor A is fixed and Factor B is random.
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4

64. Comparison on different cases with two-factor
Case 1: Both factors of A and B are fixed;
## 13.3 Two factor Mixed Model ##
Battery<-read.table(" = TRUE);
y<- Battery$Life;
temp<- Battery$Temp;
mat<- Battery$Material;
anova(lm(y~factor(temp)*factor(mat)));
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4

65. Comparison on different cases with two-factor
Case 2: Both factors of A and B are random;
## 13.2 Two factor factorial with random factor ##
Battery<-read.table(" = TRUE);
library(GAD);
y<- Battery$Life;
temp<- as.random(Battery$Temp);
mat<- as.random(Battery$Material);
gad(lm(y~temp*mat)); ## anova(lm(y~factor(Battery$Temp)*factor(Battery$Material)));
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4

66. Comparison on different cases with two-factor
Case 3: Factor A is fixed and Factor B is random.
## 13.3 Two factor Mixed Model ##
Battery<-read.table(" = TRUE);
library(GAD); y<- Battery$Life;
temp<- as.fixed(Battery$Temp);
mat<- as.random(Battery$Material);
mod13.3.1<-lm(y~temp*mat);
gad(mod13.3.1);
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4

67. Summary: Comparison on different cases with two-factor
Design & Analysis of Experiments 8E 2012 Montgomery
Chapter 4