Factorial Experiments for Random Effects and Mixed Models

14.5 Factorial Experiments for Random Effects and Mixed Models

In a two-factor experiment with random effects, we have the model

Y ijk =μ+A i +B j + (AB) ij +ǫ ijk ,

for i = 1, 2, . . . , a; j = 1, 2, . . . , b; and k = 1, 2, . . . , n, where the A i ,B j , (AB) ij , and ǫ ijk are independent random variables with means 0 and variances σ 2 α ,σ 2 β , σ 2 αβ , and σ 2 , respectively. The sums of squares for random effects experiments are computed in exactly the same way as for fixed effects experiments. We are now

14.5 Factorial Experiments for Random Effects and Mixed Models 589 interested in testing hypotheses of the form

where the denominator in the f-ratio is not necessarily the mean square error. The appropriate denominator can be determined by examining the expected values of the various mean squares. These are shown in Table 14.14.

Table 14.14: Expected Mean Squares for a Two-Factor Random Effects Experiment

Source of Degrees of

Mean

Expected

Variation Freedom

Square

Mean Square

AB (a − 1)(b − 1)

s 2 3 σ 2 + nσ 2 αβ

Error

ab(n − 1) 2 s

Total

abn − 1

From Table 14.14 we see that H ′′

′′′ 0 and H 0 are tested by using s 3 in the de- nominator of the f-ratio, whereas H 0 is tested using s 2 in the denominator. The

unbiased estimates of the variance components are s 2 2 s 2 2 2 2

Table 14.15: Expected Mean Squares for a Three-Factor Random Effects Experiment Source of Degrees of

Expected Variation

Mean Square

A a−1 s 2 1 σ 2 + nσ 2 αβγ 2 + cnσ 2 αβ + bnσ 2 αγ + bcnσ α

B b−1 s 2 2 σ 2 + nσ 2 αβγ + cnσ 2 αβ + anσ 2 βγ + acnσ 2 β

C c−1 s 2 σ 2 + nσ 2 + bnσ 2 + anσ 2 + abnσ 3 2 αβγ αγ βγ γ AB (a − 1)(b − 1)

s 2 σ 2 + nσ 4 2 + cnσ αβγ 2 αβ AC (a − 1)(c − 1)

s 2 σ 5 2 + nσ 2 αβγ + bnσ 2 αγ BC (b − 1)(c − 1)

s 2 6 σ 2 + nσ 2 αβγ + anσ 2 βγ ABC

(a − 1)(b − 1)(c − 1)

s 2 7 σ 2 + nσ 2 αβγ

Error abc(n − 1)

Total abcn − 1 The expected mean squares for the three-factor experiment with random effects

in a completely randomized design are shown in Table 14.15. It is evident from the expected mean squares of Table 14.15 that one can form appropriate f-ratios for

590 Chapter 14 Factorial Experiments (Two or More Factors) testing all two-factor and three-factor interaction variance components. However,

to test a hypothesis of the form

H 0 :σ 2 α = 0,

H 1 :σ 2 α

there appears to be no appropriate f-ratio unless we have found one or more of the two-factor interaction variance components not significant. Suppose, for example,

that we have compared s 2 5 (mean square AC) with s 2 7 (mean square ABC) and found σ 2 αγ to be negligible. We could then argue that the term σ 2 αγ should be dropped from all the expected mean squares of Table 14.15; then the ratio s 2 1 /s 2 4 provides a test for the significance of the variance component σ 2 α . Therefore, if we are to test hypotheses concerning the variance components of the main effects, it is necessary first to investigate the significance of the two-factor interaction compo- nents. An approximate test derived by Satterthwaite (1946; see the Bibliography) may be used when certain two-factor interaction variance components are found to be significant and hence must remain a part of the expected mean square.

Example 14.6: In a study to determine which are the important sources of variation in an industrial process, 3 measurements are taken on yield for 3 operators chosen randomly and 4 batches of raw materials chosen randomly. It is decided that a statistical test should

be made at the 0.05 level of significance to determine if the variance components due to batches, operators, and interaction are significant. In addition, estimates of variance components are to be computed. The data are given in Table 14.16, with the response being percent by weight.

Table 14.16: Data for Example 14.6 Batch

65.2 67.3 67.4 68.7 Solution : The sums of squares are found in the usual way, with the following results:

SST (total)

SSE (error)

SSA (operators)

SSB (batches) = 50.1564,

SS(AB) (interaction) = 5.5161. All other computations are carried out and exhibited in Table 14.17. Since

f 0.05 (2, 6) = 5.14,

f 0.05 (3, 6) = 4.76,

and

f 0.05 (6, 24) = 2.51,

14.5 Factorial Experiments for Random Effects and Mixed Models 591 we find the operator and batch variance components to be significant. Although

the interaction variance is not significant at the α = 0.05 level, the P-value is 0.095. Estimates of the main effect variance components are

Table 14.17: Analysis of Variance for Example 14.6 Source of

Sum of Degrees of

Mean

Computed

Variation Squares

Mixed Model Experiment

There are situations where the experiment dictates the assumption of a mixed model (i.e., a mixture of random and fixed effects). For example, for the case of two factors, we may have

Y ijk =μ+A i +B j + (AB) ij +ǫ ijk ,

for i = 1, 2, . . . , a; j = 1, 2, . . . , b; k = 1, 2, . . . , n. The A i may be independent random variables, independent of ǫ ijk , and the B j may be fixed effects. The mixed nature of the model requires that the interaction terms be random variables. As a result, the relevant hypotheses are of the form

H :σ α 2 0 1 2 =···=B b 0 αβ = 0,

1 : At least one the B j is not zero, H 1 :σ αβ Again, the computations of sums of squares are identical to those of fixed and

H ′ :σ 2 ′′

random effects situations, and the F-test is dictated by the expected mean squares. Table 14.18 provides the expected mean squares for the two-factor mixed model problem.

Table 14.18: Expected Mean Squares for Two-Factor Mixed Model Experiment

Factor

Expected Mean Square

A (random)

σ 2 + bnσ 2 α

B (fixed)

σ 2 + nσ 2 αβ + an

b−1

AB (random) σ 2 + nσ 2 αβ

Error

592 Chapter 14 Factorial Experiments (Two or More Factors) From the nature of the expected mean squares it becomes clear that the test on

the random effect employs the mean square error s 2 as the denominator, whereas the test on the fixed effect uses the interaction mean square. Suppose we now consider three factors. Here, of course, we must take into account the situation where one factor is fixed and the situation in which two factors are fixed. Table 14.19 covers both situations.

Table 14.19: Expected Mean Squares for Mixed Model Factorial Experiments in Three Factors

A Random

A Random, B Random

A σ 2 + bcnσ 2 α

σ 2 + cnσ 2 αβ + bcnσ 2 α

B σ 2 2 + cnσ B j

αβ + acn

σ + cnσ αβ + acnσ β

BC σ 2 (BC) + nσ 2 2 αβγ + an jk

σ 2 + nσ 2 αβγ + anσ 2 βγ

k (b−1)(c−1)

Note that in the case of A random, all effects have proper f-tests. But in the case of A and B random, the main effect C must be tested using a Satterthwaite-type procedure similar to that used in the random effects experiment.

Exercises

14.26 Assuming a random effects experiment for Ex- Operator ercise 14.2 on page 575, estimate the variance compo-

1 2 3 4 nents for brand of orange juice concentrate, for number

Filter

1 16.2 15.9 15.6 14.9 of days from when orange juice was blended until it was

16.8 15.1 15.9 15.2 tested, and for experimental error.

14.27 To estimate the various components of variabil- 16.9 16.3 16.0 14.6 ity in a filtration process, the percent of material lost

16.8 16.5 17.2 15.9 in the mother liquor is measured for 12 experimental

3 16.7 16.5 16.4 16.1 conditions, with 3 runs on each condition. Three filters

16.9 16.9 17.4 15.4 and 4 operators are selected at random for use in the

17.1 16.8 16.9 15.6 experiment.

(a) Test the hypothesis of no interaction variance com- 14.28 A defense contractor is interested in studying ponent between filters and operators at the α = an inspection process to detect failure or fatigue of

0.05 level of significance. transformer parts. Three levels of inspections are used (b) Test the hypotheses that the operators and the fil- by three randomly chosen inspectors. Five lots are used ters have no effect on the variability of the filtration for each combination in the study. The factor levels are process at the α = 0.05 level of significance.

given in the data. The response is in failures per 1000 (c) Estimate the components of variance due to filters, pieces. operators, and experimental error.

Exercises 593 (a) Write an appropriate model, with assumptions.

Operator (b) Use analysis of variance to test the appropriate hy-

1 2 3 4 pothesis for inspector, inspection level, and inter-

Time

1 9.5 9.8 9.8 10.0 action.

9.8 10.1 10.3 9.7 Inspection Level

9.9 9.8 9.8 10.1 Inspector Inspection Inspection Commercial

14.31 A manufacturer of latex house paint (brand A) 5.12 4.87 5.68 would like to show that its paint is more robust to

C 7.70 6.82

the material being painted than that of its two closest 6.42 5.39

competitors. The response is the time, in years, until 5.35 5.01 6.12 chipping occurs. The study involves the three brands

of paint and three randomly chosen materials. Two

14.29 Consider the following analysis of variance for pieces of material are used for each combination. a random effects experiment:

Brand of Paint Source of

A B C Variation

Degrees of

(a) What is this type of model called? AB 3 15 (b) Analyze the data, using the appropriate model.

AC 6 24

BC 2 18 (c) Did the manufacturer of brand A support its claim ABC

6 2 with the data?

Error

Total 47 power setting on the machines used to fill certain types 14.32 A process engineer wants to determine if the Test for significant variance components among all of cereal boxes results in a significant effect on the ac-

main effects and interaction effects at the 0.01 level tual weight of the product. The study consists of 3 of significance

randomly chosen types of cereal manufactured by the (a) by using a pooled estimate of error when appropri- company and 3 fixed power settings. Weight is mea- ate;

sured for 4 different randomly selected boxes of cereal (b) by not pooling sums of squares of insignificant ef- at each combination. The desired weight is 400 grams. fects.

The data are presented here.

Cereal Type 14.30 A plant manager would like to show that the

Power

1 2 3 yield of a woven fabric in the plant does not depend on

Setting

392 392 402 405 machine operator or time of day and is consistently

Low

394 401 399 399 high. Four randomly selected operators and 3 ran-

390 392 404 403 domly selected hours of the day are chosen for the

Current

395 502 400 399 study. The yield is measured in yards produced per

404 406 415 412 minute. Samples are taken on 3 randomly chosen days.

High

401 400 413 415 (a) Write the appropriate model.

(a) Give the appropriate model, and list the assump- (b) Evaluate the variance components for operator and

tion being made.

time. (b) Is there a significant effect due to the power set- (c) Draw conclusions.

ting? (c) Is there a significant variance component due to

cereal type?

594 Chapter 14 Factorial Experiments (Two or More Factors)

Review Exercises

14.33 The Statistics Consulting Center at Virginia results, in kilograms, are as follows: Tech was involved in analyzing a set of data taken by personnel in the Human Nutrition and Foods Depart-

Humidity ment in which it was of interest to study the effects

50% 70% 90% of flour type and percent sweetener on certain physical

Plastic Type

A 39.0 33.1 33.8 33.0 attributes of a type of cake. All-purpose flour and cake

42.8 37.8 30.7 32.9 flour were used, and the percent sweetener was varied

B 36.9 27.2 29.7 28.5 at four levels. The following data show information

41.0 26.8 29.1 27.9 on specific gravity of cake samples. Three cakes were

C 27.4 29.2 26.7 30.9 prepared at each of the eight factor combinations.

30.3 29.9 32.0 31.5 Sweetener

(a) Assuming a fixed effects experiment, perform an Concentration All-Purpose

Flour

analysis of variance and test the hypothesis of no 0 0.90 0.87 0.90

Cake

interaction between humidity and plastic type at 50 0.86 0.89 0.91

the 0.05 level of significance. 75 0.93 0.88 0.87

(b) Using only plastics A and B and the value of s 2 100

from part (a), once again test for the presence of (a) Treat the analysis as a two-factor analysis of vari-

interaction at the 0.05 level of significance. ance. Test for differences between flour type. Test

for differences between sweetener concentration. 14.36 Personnel in the Materials Engineering Depart- (b) Discuss the effect of interaction, if any. Give P- ment at Virginia Tech conducted an experiment to values on all tests.

study the effects of environmental factors on the sta- bility of a certain type of copper-nickel alloy. The basic

14.34 An experiment was conducted in the Depart- response was the fatigue life of the material. The fac- ment of Food Science at Virginia Tech. It was of inter- tors are level of stress and environment. The data are est to characterize the texture of certain types of fish as follows: in the herring family. The effect of sauce types used

Stress Level in preparing the fish was also studied. The response

Medium High in the experiment was “texture value,” measured with

Environment

Low

11.08 13.12 14.18 a machine that sliced the fish product. The following

Dry

10.98 13.04 14.90 are data on texture values:

Hydrogen

10.75 12.73 14.15 Unbleached Bleached

Fish Type

High

10.52 12.87 14.42 Sauce Type Menhaden Menhaden Herring

Humidity

10.43 12.95 14.25 Sour Cream 27.6 57.4 64.0 66.9 107.0 83.9 (a) Do an analysis of variance to test for interaction 47.8 71.1 66.5 66.8 110.4 93.4 between the factors. Use α = 0.05.

Wine Sauce 49.8 31.0 48.3 62.2 88.0 95.2 (b) Based on part (a), do an analysis on the two main 11.8 35.1 54.6 43.6 108.2 86.7 effects and draw conclusions. Use a P-value ap-

proach in drawing conclusions. (a) Do an analysis of variance. Determine whether or

not there is an interaction between sauce type and 14.37 In the experiment of Review Exercise 14.33, fish type.

cake volume was also used as a response. The units are cubic inches. Test for interaction between factors

(b) Based on your results from part (a) and on F-tests and discuss main effects. Assume that both factors are on main effects, determine if there is a significant fixed effects. difference in texture due to sauce types, and deter- mine whether there is a significant difference due

Flour to fish types.

Sweetener

Concentration All-Purpose Cake 0 4.48 3.98 4.42

5.00 4.26 4.34 14.35 A study was made to determine if humidity

75 3.92 3.82 4.06 4.82 4.34 4.40 conditions have an effect on the force required to pull

3.26 3.80 3.40 4.32 4.18 4.30 apart pieces of glued plastic. Three types of plastic

were tested using 4 different levels of humidity. The

Review Exercises 595 14.38 A control valve needs to be very sensitive to (1988), a study on how tire air pressure affects the ma-

the input voltage, thus generating a good output volt- neuverability of an automobile was described. Three age. An engineer turns the control bolts to change different tire air pressures were compared on three dif- the input voltage. The book SN-Ratio for the Quality ferent driving surfaces. The three air pressures were

Evaluation, published by the Japanese Standards As- both left- and right-side tires inflated to 6 kgf/cm 2 , sociation (1988), described a study on how these three left-side tires inflated to 6 kgf/cm 2 and right-side tires factors (relative position of control bolts, control range inflated to 3 kgf/cm 2 , and both left- and right-side tires of bolts, and input voltage) affect the sensitivity of a inflated to 3 kgf/cm 2 . The three driving surfaces were control valve. The factors and their levels are shown asphalt, dry asphalt, and dry cement. The turning ra- below. The data show the sensitivity of a control valve. dius of a test vehicle was observed twice for each level Factor A, relative position of control bolts:

of tire pressure on each of the three different driving center −0.5, center, and center +0.5

surfaces.

Factor B, control range of bolts: 2, 4.5, and 7 (mm)

Tire Air Pressure Factor C, input voltage:

1 2 3 100, 120, and 150 (V)

Driving Surface

Dry Asphalt

C Dry Cement

Perform an analysis of variance of the above data. A 1 B 2 178 171

Comment on the interpretation of the main and in- A 1 B 3 204 190

teraction effects.

A 2 B 1 156 148

14.41 The manufacturer of a certain brand of freeze- A 2 B 2 183 168

dried coffee hopes to shorten the process time without A 2 B 3 210 204

jeopardizing the integrity of the product. The process A 3 B 1 161 145

engineer wants to use 3 temperatures for the drying

chamber and 4 drying times. The current drying time Perform an analysis of variance with α = 0.05 to test is 3 hours at a temperature of −15 ◦

C. The flavor re- for significant main and interaction effects. Draw con- sponse is an average of scores of 4 professional judges. clusions.

The score is on a scale from 1 to 10, with 10 being the best. The data are as shown in the following table.

14.39 Exercise 14.25 on page 588 describes an exper- Temperature iment involving the extraction of polyethylene through

Time ◦ −20 C −15 C −10 C use of a solvent.

9.60 9.61 9.55 9.48 Solvent Temp.

(a) What type of model should be used? State assump-

(a) Do a different sort of analysis on the data. Fit an (b) Analyze the data appropriately. appropriate regression model with a solvent cate- (c) Write a brief report to the vice-president in charge gorical variable, a temperature term, a time term, a

and make a recommendation for future manufac- temperature by time interaction, a solvent by tem-

turing of this product.

perature interaction, and a solvent by time interac- tion. Do t-tests on all coefficients and report your

14.42 To ascertain the number of tellers needed dur- findings.

ing peak hours of operation, data were collected by (b) Do your findings suggest that different models are an urban bank. Four tellers were studied during three

appropriate for ethanol and toluene, or are they “busy” times: (1) weekdays between 10:00 and 11:00 equivalent apart from the intercepts? Explain.

A.M. , (2) weekday afternoons between 2:00 and 3:00 (c) Do you find any conclusions here that contra- P.M. , and (3) Saturday mornings between 11:00 A.M. dict conclusions drawn in your solution of Exercise and 12:00 noon. An analyst chose four randomly se- 14.25? Explain.

lected times within each of the three time periods for each of the four teller positions over a period of months, and the numbers of customers serviced were observed.

14.40 In the book SN-Ratio for the Quality Evalua- The data are as follows: tion, published by the Japanese Standards Association

596 Chapter 14 Factorial Experiments (Two or More Factors) Time Period

any, would be violated?

Teller

(b) Construct a standard ANOVA table that includes 2 16 11 19 14

F-tests on main effects and interactions. If interac- 3 12 19 11 22

tions and main effects are found to be significant, 4 11 9 13 8 10 7 19 8 11 9 17 9 give scientific conclusions. What have we learned? Be sure to interpret any significant interaction. Use

It is assumed that the number of customers served is a your own judgment regarding P-values. Poisson random variable.

(c) Do the entire analysis again using an appropriate (a) Discuss the danger in doing a standard analysis of

transformation on the response. Do you see any variance on the data above. What assumptions, if

differences in your findings? Comment.