Design of Experiments with Several Factors PDF

JWCL232_c14_551-636.qxd 1/1/70 5:38 AM Page 551 14 Design of Experiments with Several Factors ©Oleksiy Kondratyuk/iStockphoto...

JWCL232_c14_551-636.qxd 1/1/70 5:38 AM Page 551 14 Design of Experiments with Several Factors ©Oleksiy Kondratyuk/iStockphoto Carotenoids are fat-soluble pigments that occur naturally in fruits in vegetables that are rec- ommended for healthy diets. A well-known carotenoid is beta-carotene. Astaxanthin is another carotenoid that is a strong antioxidant and commercially produced. An exercise later in this chapter describes an experiment in Biotechnology Progress to promote astaxanthin production. Seven variables were considered important to production: photon flux density, and concentrations of nitrogen, phosphorous, magnesium, acetate, ferrous, NaCl. It was important to study the effects of these factors, but also the effects of combinations on the pro- duction. Even with only a high and low setting for each variable, an experiment that uses all possible combinations requires 27 ⫽ 128 tests. There are a number of disadvantages of such a large experiment and a question is whether a fraction of the full set of tests can be selected to provide the most important information about the effects of these variables in many fewer runs. The example used a surprisingly small set of 16 runs (16/128 ⫽ 1/8 fraction). The design and analysis of experiments of this type is the focus of this chapter. Such experiments are widely-used throughout modern engineering development and scientific studies. CHAPTER OUTLINE 14-1 INTRODUCTION 14-4 GENERAL FACTORIAL 14-2 FACTORIAL EXPERIMENTS EXPERIMENTS 14-3 TWO-FACTOR FACTORIAL 14-5 2k FACTORIAL DESIGNS EXPERIMENTS 14-5.1 22 Design 14-3.1 Statistical Analysis of the 14-5.2 2k Design for k ⱖ 3 Factors Fixed-Effects Model 14-5.3 Single Replicate of the 2k Design 14-3.2 Model Adequacy Checking 14-5.4 Addition of Center Points to a 14-3.3 One Observation per Cell 2k Design 551 JWCL232_c14_551-636.qxd 1/16/10 9:55 AM Page 552 552 CHAPTER 14 DESIGN OF EXPERIMENTS WITH SEVERAL FACTORS 14-6 BLOCKING AND CONFOUNDING 14-7.2 Smaller Fractions: The 2kⴚp IN THE 2k DESIGN Fractional Factorial 14-7 FRACTIONAL REPLICATION OF 14-8 RESPONSE SURFACE METHODS THE 2k DESIGN AND DESIGNS 14-7.1 One-Half Fraction of the 2k Design LEARNING OBJECTIVES After careful study of this chapter you should be able to do the following: 1. Design and conduct engineering experiments involving several factors using the factorial design approach 2. Know how to analyze and interpret main effects and interactions 3. Understand how the ANOVA is used to analyze the data from these experiments 4. Assess model adequacy with residual plots 5. Know how to use the two-level series of factorial designs 6. Understand how two-level factorial designs can be run in blocks 7. Design and conduct two-level fractional factorial designs 8. Test for curvature in two-level factorial designs by using center points 9. Use response surface methodology for process optimization experiments 14-1 INTRODUCTION An experiment is just a test or series of tests. Experiments are performed in all engineering and scientific disciplines and are an important part of the way we learn about how systems and processes work. The validity of the conclusions that are drawn from an experiment depends to a large extent on how the experiment was conducted. Therefore, the design of the experiment plays a major role in the eventual solution of the problem that initially motivated the experiment. In this chapter we focus on experiments that include two or more factors that the experi- menter thinks may be important. A factorial experiment is a powerful technique for this type of problem. Generally, in a factorial experimental design, experimental trials (or runs) are per- formed at all combinations of factor levels. For example, if a chemical engineer is interested in investigating the effects of reaction time and reaction temperature on the yield of a process, and if two levels of time (1 and 1.5 hours) and two levels of temperature (125 and 150F) are con- sidered important, a factorial experiment would consist of making experimental runs at each of the four possible combinations of these levels of reaction time and reaction temperature. Experimental design is an extremely important tool for engineers and scientists who are in- terested in improving the performance of a manufacturing process. It also has extensive applica- tion in the development of new processes and in new product design. We now give some examples. Process Characterization Experiment In an article in IEEE Transactions on “Electronics Packaging Manufacturing” (2001, Vol. 24(4), pp. 249–254), the authors discussed the change to lead-free solder in surface mount technology (SMT). SMT is a process to assemble electronic components to a printed circuit board. Solder paste is printed through a stencil onto the printed circuit board. The stencil-printing machine has squeegees; the paste rolls in front of the squeegee and fills the apertures in the stencil. The squeegee shears off the paste in the apertures as it moves over the stencil. Once the print stroke JWCL232_c14_551-636.qxd 1/16/10 9:55 AM Page 553 14-1 INTRODUCTION 553 Controllable factors x1 x2 xp... Input Output SMT Process (printed circuit boards) (defects, y) Figure 14-1 The... flow solder z1 z2 zq experiment. Uncontrollable (noise) factors is completed, the board is separated mechanically from the stencil. Electronic components are placed on the deposits and the board is heated so that the paste reflows to form the solder joints. The current SMT soldering process is based on tin–lead solders, and it has been well developed and refined over the years to operate at a competitive cost. The process will have several (perhaps many) variables, and all of them are not equally important. The initial list of candidate variables to be included in an experiment is constructed by combining the knowl- edge and information about the process from all team members. For example, engineers would conduct a brainstorming session and invite manufacturing personnel with SMT experience to participate. SMT has several variables that can be controlled. These include (1) squeegee speed, (2) squeegee pressure, (3) squeegee angle, (4) metal or polyurethane squeegee, (5) squeegee vibration, (6) delay time before the squeegee lifts from the stencil, (7) stencil separa- tion speed, (8) print gap, (9) solder paste alloy, (10) paste pretreatment (11) paste particle size, (12) flux type, (13) reflow temperature, (14) reflow time, and so forth. In addition to these controllable factors, there are several other factors that cannot be easily controlled during routine manufacturing, including (1) thickness of the printed circuit board, (2) types of components used on the board and aperture width and length, (3) layout of the components on the board, (4) paste density variation, (5) envi- ronmental factors, (6) squeegee wear, (7) cleanliness, and so forth. Sometimes we call the uncontrollable factors noise factors. A schematic representation of the process is shown in Fig. 14-1. In this situation, the engineer wants to characterize the SMT process; that is, to determine the factors (both controllable and uncontrollable) that affect the occurrence of defects on the printed circuit boards. To determine these factors, an experiment can be designed to estimate the magnitude and direction of the factor effects. Sometimes we call such an experiment a screening experiment. The information from this characterization study, or screening experiment, can help determine the critical process variables as well as the direction of adjustment for these factors in order to reduce the number of defects, and assist in determining which process variables should be carefully controlled during manufacturing to prevent high defect levels and erratic process performance. Optimization Experiment In a characterization experiment, we are interested in determining which factors affect the response. A logical next step is to determine the region in the important factors that leads to an opti- mum response. For example, if the response is cost, we will look for a region of minimum cost. As an illustration, suppose that the yield of a chemical process is influenced by the op- erating temperature and the reaction time. We are currently operating the process at 155F and 1.7 hours of reaction time, and the current process yield is around 75%. Figure 14-2 shows a view of the time–temperature space from above. In this graph we have connected points of constant yield with lines. These lines are yield contours, and we have shown the JWCL232_c14_551-636.qxd 1/16/10 9:55 AM Page 554 554 CHAPTER 14 DESIGN OF EXPERIMENTS WITH SEVERAL FACTORS 200 Path leading to region of higher yield 190 95% 180 Temperature (°F) 60% 90% 70% 170 82% 58% Current 80% operating 160 conditions 75% 150 Figure 14-2 Contour 56% 69% plot of yield as a func- 140 tion of reaction time and reaction tempera- ture, illustrating an optimization 0.5 1.0 1.5 2.0 2.5 experiment. Time (hr) contours at 60, 70, 80, 90, and 95% yield. To locate the optimum, we might begin with a factorial experiment such as we describe below, with the two factors, time and temperature, run at two levels each at 10F and 0.5 hours above and below the current operating condi- tions. This two-factor factorial design is shown in Fig. 14-2. The average responses observed at the four points in the experiment (145F, 1.2 hours; 145F, 2.2 hours; 165F, 1.2 hours; and 165F, 2.2 hours) indicate that we should move in the general direction of increased temper- ature and lower reaction time to increase yield. A few additional runs could be performed in this direction to locate the region of maximum yield. A Product Design Example We can also use experimental design in the development of new products. For example, suppose that a group of engineers are designing a door hinge for an automobile. The product characteristic is the check effort, or the holding ability, of the latch that prevents the door from swinging closed when the vehicle is parked on a hill. The check mechanism consists of a leaf spring and a roller. When the door is opened, the roller travels through an arc causing the leaf spring to be com- pressed. To close the door, the spring must be forced aside, and this creates the check effort. The engineering team thinks that check effort is a function of the following factors: (1) roller travel distance, (2) spring height from pivot to base, (3) horizontal distance from pivot to spring, (4) free height of the reinforcement spring, (5) free height of the main spring. The engineers can build a prototype hinge mechanism in which all these factors can be varied over certain ranges. Once appropriate levels for these five factors have been identified, an experiment can be designed consisting of various combinations of the factor levels, and the prototype can be tested at these combinations. This will produce information concerning which factors are most influential on the latch check effort, and through analysis of this infor- mation, the latch design can be improved. JWCL232_c14_551-636.qxd 1/16/10 9:55 AM Page 555 14-2 FACTORIAL EXPERIMENTS 555 Most of the statistical concepts introduced in Chapter 13 for single-factor experiments can be extended to the factorial experiments of this chapter. The analysis of variance (ANOVA), in particular, will continue to be used as a tool for statistical data analysis. We will also introduce several graphical methods that are useful in analyzing the data from designed experiments. 14-2 FACTORIAL EXPERIMENTS When several factors are of interest in an experiment, a factorial experiment should be used. As noted previously, in these experiments factors are varied together. Factorial Experiment By a factorial experiment we mean that in each complete trial or replicate of the experiment all possible combinations of the levels of the factors are investigated. Thus, if there are two factors A and B with a levels of factor A and b levels of factor B, each replicate contains all ab treatment combinations. The effect of a factor is defined as the change in response produced by a change in the level of the factor. It is called a main effect because it refers to the primary factors in the study. For ex- ample, consider the data in Table 14-1. This is a factorial experiment with two factors, A and B, each at two levels (Alow, Ahigh, and Blow, Bhigh). The main effect of factor A is the difference between the average response at the high level of A and the average response at the low level of A, or 30  40 10  20 A   20 2 2 That is, changing factor A from the low level to the high level causes an average response increase of 20 units. Similarly, the main effect of B is 20  40 10  30 B   10 2 2 In some experiments, the difference in response between the levels of one factor is not the same at all levels of the other factors. When this occurs, there is an interaction between the fac- tors. For example, consider the data in Table 14-2. At the low level of factor B, the A effect is A  30  10  20 and at the high level of factor B, the A effect is A  0  20  20 Since the effect of A depends on the level chosen for factor B, there is interaction between A and B. Table 14-1 A Factorial Experiment with Table 14-2 A Factorial Experiment with Two Factors Interaction Factor B Factor B Factor A B low B high Factor A B low B high Alow 10 20 Alow 10 20 Ahigh 30 40 Ahigh 30 40 JWCL232_c14_551-636.qxd 1/16/10 9:55 AM Page 556 556 CHAPTER 14 DESIGN OF EXPERIMENTS WITH SEVERAL FACTORS When an interaction is large, the corresponding main effects have very little practical meaning. For example, by using the data in Table 14-2, we find the main effect of A as 30  0 10  20 A  0 2 2 and we would be tempted to conclude that there is no factor A effect. However, when we ex- amined the effects of A at different levels of factor B, we saw that this was not the case. The effect of factor A depends on the levels of factor B. Thus, knowledge of the AB interaction is more useful than knowledge of the main effect. A significant interaction can mask the signif- icance of main effects. Consequently, when interaction is present, the main effects of the factors involved in the interaction may not have much meaning. It is easy to estimate the interaction effect in factorial experiments such as those illus- trated in Tables 14-1 and 14-2. In this type of experiment, when both factors have two levels, the AB interaction effect is the difference in the diagonal averages. This represents one-half the difference between the A effects at the two levels of B. For example, in Table 14-1, we find the AB interaction effect to be 20  30 10  40 AB   0 2 2 Thus, there is no interaction between A and B. In Table 14-2, the AB interaction effect is 20  30 10  0 AB    20 2 2 As we noted before, the interaction effect in these data is very large. The concept of interaction can be illustrated graphically in several ways. Figure 14-3 plots the data in Table 14-1 against the levels of A for both levels of B. Note that the Blow and Bhigh lines are approximately parallel, indicating that factors A and B do not interact signifi- cantly. Figure 14-4 presents a similar plot for the data in Table 14-2. In this graph, the Blow and Bhigh lines are not parallel, indicating the interaction between factors A and B. Such graphical displays are called two-factor interaction plots. They are often useful in presenting the re- sults of experiments, and many computer software programs used for analyzing data from de- signed experiments will construct these graphs automatically. 50 50 40 Bhigh 40 Blow Blow Observation Observation 30 30 20 20 10 10 0 0 Bhigh Alow Ahigh Alow Ahigh Factor A Factor A Figure 14-3 Factorial experiment, no Figure 14-4 Factorial experiment, with interaction. interaction. JWCL232_c14_551-636.qxd 1/16/10 9:55 AM Page 557 14-2 FACTORIAL EXPERIMENTS 557 45 40 35 35 25 30 y 25 y 15 5 20 1 –5 1 15 0.6 0.6 0.2 0.2 10 0 –15 – 0.2 –1 –0.2 B –1 –0.6 – 0.6 –0.6 –0.6 –0.2 –1 B –0.2 0.2 –1 0.2 0.6 0.6 A 1 A 1 Figure 14-5 Three-dimensional surface plot of the data from Figure 14-6 Three-dimensional surface plot of the data from Table 14-1, showing the main effects of the two factors A and B. Table 14-2 showing the effect of the A and B interaction. Figures 14-5 and 14-6 present another graphical illustration of the data from Tables 14-1 and 14-2. In Fig. 14-3 we have shown a three-dimensional surface plot of the data from Table 14-1. These data contain no interaction, and the surface plot is a plane lying above the A-B space. The slope of the plane in the A and B directions is proportional to the main effects of factors A and B, respectively. Figure 14-6 is a surface plot of the data from Table 14-2. Notice that the effect of the interaction in these data is to “twist” the plane, so that there is curvature in the response function. Factorial experiments are the only way to discover interactions between variables. An alternative to the factorial design that is (unfortunately) used in practice is to change the factors one at a time rather than to vary them simultaneously. To illustrate this one-factor-at- a-time procedure, suppose that an engineer is interested in finding the values of temperature and pressure that maximize yield in a chemical process. Suppose that we fix temperature at 155F (the current operating level) and perform five runs at different levels of time, say, 0.5, 1.0, 1.5, 2.0, and 2.5 hours. The results of this series of runs are shown in Fig. 14-7. This figure indicates that max- imum yield is achieved at about 1.7 hours of reaction time. To optimize temperature, the engineer then fixes time at 1.7 hours (the apparent optimum) and performs five runs at different tempera- tures, say, 140, 150, 160, 170, and 180F. The results of this set of runs are plotted in Fig. 14-8. Maximum yield occurs at about 155F. Therefore, we would conclude that running the process at 155F and 1.7 hours is the best set of operating conditions, resulting in yields of around 75%. 80 80 70 70 Yield (%) Yield (%) 60 60 50 50 0.5 1.0 1.5 2.0 2.5 140 150 160 170 180 Time (hr) Temperature (°F) Figure 14-7 Yield versus reaction time with Figure 14-8 Yield versus temperature with reaction temperature constant at 155F. time constant at 1.7 hours. JWCL232_c14_551-636.qxd 1/16/10 9:55 AM Page 558 558 CHAPTER 14 DESIGN OF EXPERIMENTS WITH SEVERAL FACTORS 200 190 95% 180 Temperature (°F) 90% 170 80% 160 150 70% 140 Figure 14-9 Optimization 60% experiment using the one-factor-at-a- 0.5 1.0 1.5 2.0 2.5 time method. Time (hr) Figure 14-9 displays the contour plot of actual process yield as a function of temperature and time with the one-factor-at-a-time experiments superimposed on the contours. Clearly, this one-factor-at-a-time approach has failed dramatically here, as the true optimum is at least 20 yield points higher and occurs at much lower reaction times and higher temperatures. The failure to discover the importance of the shorter reaction times is particularly important be- cause this could have significant impact on production volume or capacity, production plan- ning, manufacturing cost, and total productivity. The one-factor-at-a-time approach has failed here because it cannot detect the interac- tion between temperature and time. Factorial experiments are the only way to detect inter- actions. Furthermore, the one-factor-at-a-time method is inefficient. It will require more experimentation than a factorial, and as we have just seen, there is no assurance that it will produce the correct results. 14-3 TWO-FACTOR FACTORIAL EXPERIMENTS The simplest type of factorial experiment involves only two factors, say A, and B. There are a levels of factor A and b levels of factor B. This two-factor factorial is shown in Table 14-3. The experiment has n replicates, and each replicate contains all ab treatment combinations. The observation in the ij th cell for the kth replicate is denoted by yijk. In performing the experi- ment, the abn observations would be run in random order. Thus, like the single-factor exper- iment studied in Chapter 13, the two-factor factorial is a completely randomized design. The observations may be described by the linear statistical model i  1, 2, p , a Yijk    i  j  12 ij  ijk j  1, 2, p , b (14-1) k  1, 2, p , n JWCL232_c14_551-636.qxd 1/16/10 9:55 AM Page 559 14-3 TWO-FACTOR FACTORIAL EXPERIMENTS 559 Table 14-3 Data Arrangement for a Two-Factor Factorial Design Factor B 1 2 p b Totals Averages y111, y112, y121, y122, y1b1, y1b2, 1 p , y11n p , y12n p , y1bn y1.. y1.. y211, y212, y221, y222, y2b1, y2b2, 2 p , y21n p , y22n p , y2bn y2.. y2.. Factor A o ya11, ya12, ya21, ya22, yab1, yab2, a p , ya1n p , ya2n p , yabn ya.. ya.. Totals y.1. y.2. y.b. yp Averages y.1. y.2. y.b. y... where  is the overall mean effect, i is the effect of the i th level of factor A, j is the effect of the jth level of factor B, ()ij is the effect of the interaction between A and B, and ijk is a ran- dom error component having a normal distribution with mean zero and variance 2. We are interested in testing the hypotheses of no main effect for factor A, no main effect for B, and no AB interaction effect. As with the single-factor experiments of Chapter 13, the analysis of vari- ance (ANOVA) will be used to test these hypotheses. Since there are two factors in the experiment, the test procedure is sometimes called the two-way analysis of variance. 14-3.1 Statistical Analysis of the Fixed-Effects Model Suppose that A and B are fixed factors. That is, the a levels of factor A and the b levels of fac- tor B are specifically chosen by the experimenter, and inferences are confined to these levels only. In this model, it is customary to define the effects i, j, and ()ij as deviations from the mean, so that g i1 i  0, g j1 j  0, g i1 12 ij  0, and g j1 12 ij  0. a b a b The analysis of variance can be used to test hypotheses about the main factor effects of A and B and the AB interaction. To present the ANOVA, we will need some symbols, some of which are illustrated in Table 14-3. Let yi.. denote the total of the observations taken at the ith level of factor A; y.j. denote the total of the observations taken at the jth level of factor B; yij. denote the total of the observations in the ij th cell of Table 14-3; and y... denote the grand total of all the observations. Define yi.., y.j., yij., and y... as the corresponding row, column, cell, and grand averages. That is, Notation for Totals b n yi.. and Means yi..  a a yijk yi..  i  1, 2, p , a j1 k1 bn a n y.j. y.j.  a a yijk y. j.  j  1, 2, p , b i1 k1 an n yij. yij..  a yijk y ij.  i  1, 2, p , a k1 n a b n y... y...  a a a yijk y...  j  1, 2, p , b i1 j1 k1 abn JWCL232_c14_551-636.qxd 1/16/10 9:55 AM Page 560 560 CHAPTER 14 DESIGN OF EXPERIMENTS WITH SEVERAL FACTORS The hypotheses that we will test are as follows: 1. H0: 1  2  p  a  0 (no main effect of factor A) H1: at least one i 0 2. H0: 1  2  p  b  0 (no main effect of factor B) (14-2) H1: at least one j 0 3. H0: ()11  ()12  p  ()ab  0 (no interaction) H1: at least one ()ij 0 As before, the ANOVA tests these hypotheses by decomposing the total variability in the data into component parts and then comparing the various elements in this decomposition. Total variability is measured by the total sum of squares of the observations a b n SST  a a a 1 yijk  y...2 2 i1 j1 k1 and the sum of squares decomposition is defined below. ANOVA Sum of Squares The sum of squares identity for a two-factor ANOVA is Identity: Two a b n a b a a a 1 yijk  y...2  bn a 1 yi..  y...2  an a 1 y.j.  y...2 Factors 2 2 2 i1 j1 k1 i1 j1 a b a b n  n a a 1 yij.  yi..  y.j.  y...2 2  a a a 1 yijk  yij.2 2 (14-3) i1 j1 i1 j1 k1 or symbolically, SST  SSA  SSB  SSAB  SSE (14-4) Equations 14-3 and 14-4 state that the total sum of squares SST is partitioned into a sum of squares for the row factor A (SSA), a sum of squares for the column factor B (SSB), a sum of squares for the interaction between A and B (SSAB), and an error sum of squares (SSE). There are abn  1 total degrees of freedom. The main effects A and B have a  1 and b  1 degrees of freedom, while the interaction effect AB has (a  1)(b  1) degrees of freedom. Within each of the ab cells in Table 14-3, there are n  1 degrees of freedom between the n replicates, and observations in the same cell can differ only because of random error. Therefore, there are ab(n  1) degrees of freedom for error. Therefore, the degrees of freedom are partitioned according to abn  1  1a  12  1b  12  1a  121b  12  ab1n  12 If we divide each of the sum of squares on the right-hand side of Equation 14-4 by the corresponding number of degrees of freedom, we obtain the mean squares for A, B, the interaction, and error: SSA SSB SSAB SSE MSA  MSB  MSAB  MSE  a1 b1 1a  121b  12 ab1n  12 JWCL232_c14_551-636.qxd 1/16/10 9:55 AM Page 561 14-3 TWO-FACTOR FACTORIAL EXPERIMENTS 561 Assuming that factors A and B are fixed factors, it is not difficult to show that the expected values of these mean squares are Expected Values of Mean a b Squares: bn g 2i an g 2j E1MSA 2  E a b E1MSB 2  E a b Two Factors SSA i1 SSB j1 2  2  a1 a1 b1 b1 n g g 12 2ij a b E1MSAB 2  E a b SSAB i1 j1 2  1a  121b  12 1a  121b  12 E1MSE 2  E a b SSE 2 ab1n  12 From examining these expected mean squares, it is clear that if the null hypotheses about main effects H0: i  0, H0: j  0, and the interaction hypothesis H0: ()ij  0 are all true, all four mean squares are unbiased estimates of 2. To test that the row factor effects are all equal to zero (H0: i  0), we would use the ratio F Test for Factor A MSA F0  MSE which has an F distribution with a  1 and ab(n  1) degrees of freedom if H0: i  0 is true. This null hypothesis is rejected at the level of significance if f0 f ,a1,ab(n1). Similarly, to test the hy- pothesis that all the column factor effects are equal to zero (H0: j  0), we would use the ratio F Test for Factor B MSB F0  MSE which has an F distribution with b  1 and ab(n  1) degrees of freedom if H0: j  0 is true. This null hypothesis is rejected at the level of significance if f0 f ,b1,ab(n1). Finally, to test the hy- pothesis H0: ()ij  0, which is the hypothesis that all interaction effects are zero, we use the ratio F Test for AB Interaction MSAB F0  MSE which has an F distribution with (a  1)(b  1) and ab(n  1) degrees of freedom if the null hypothesis H0: ()ij  0. This hypothesis is rejected at the level of significance if f0 f ,(a1)(b1),ab(n1). JWCL232_c14_551-636.qxd 1/16/10 9:55 AM Page 562 562 CHAPTER 14 DESIGN OF EXPERIMENTS WITH SEVERAL FACTORS It is usually best to conduct the test for interaction first and then to evaluate the main effects. If interaction is not significant, interpretation of the tests on the main effects is straightforward. However, as noted in Section 14-3, when interaction is significant, the main effects of the factors involved in the interaction may not have much practical interpretative value. Knowledge of the interaction is usually more important than knowledge about the main effects. Computational formulas for the sums of squares are easily obtained. Computing Formulas for Computing formulas for the sums of squares in a two-factor analysis of variance. ANOVA: Two Factors a b n y2... SST  a a a yijk 2  (14-5) i1 j1 k1 abn a y 2#.. 2 i y... SSA  a  (14-6) i1 bn abn b y.2j. y 2... SSB  a an  (14-7) j1 abn a b y2ij. 2 y... SSAB  a a n   SSA  SSB (14-8) i1 j1 abn SSE  SST  SSAB  SSA  SSB (14-9) The computations are usually displayed in an ANOVA table, such as Table 14-4. EXAMPLE 14-1 Aircraft Primer Paint Aircraft primer paints are applied to aluminum surfaces by ties. A factorial experiment was performed to investigate the two methods: dipping and spraying. The purpose of the primer effect of paint primer type and application method on paint ad- is to improve paint adhesion, and some parts can be primed hesion. For each combination of primer type and application using either application method. The process engineering method, three specimens were painted, then a finish paint was group responsible for this operation is interested in learning applied, and the adhesion force was measured. The data from whether three different primers differ in their adhesion proper- the experiment are shown in Table 14-5. The circled numbers Table 14-4 ANOVA Table for a Two-Factor Factorial, Fixed-Effects Model Source of Sum of Degrees of Variation Squares Freedom Mean Square F0 SSA MSA A treatments SSA a1 MSA  a1 MSE SSB MSB B treatments SSB b1 MSB  b1 MSE SSAB MSAB (a  1)(b  1) MSAB  1a  121b  12 Interaction SSAB MSE Error SSE ab(n  1) SSE Total SST abn  1 MSE  ab1n  12 JWCL232_c14_551-636.qxd 1/16/10 9:55 AM Page 563 14-3 TWO-FACTOR FACTORIAL EXPERIMENTS 563 Table 14-5 Adhesion Force Data for Example 14-1 Primer Type Dipping Spraying yi.. 1 4.0, 4.5, 4.3 12.8 5.4, 4.9, 5.6 15.9 28.7 2 5.6, 4.9, 5.4 15.9 5.8, 6.1, 6.3 18.2 34.1 3 3.8, 3.7, 4.0 11.5 5.5, 5.0, 5.0 15.5 27.0 y.j. 40.2 49.6 89.8  y... in the cells are the cell totals yij.. The sums of squares required The ANOVA is summarized in Table 14-6. The experimenter to perform the ANOVA are computed as follows: has decided to use  0.05. Since f0.05,2,12  3.89 and f0.05,1,12  4.75, we conclude that the main effects of primer type and ap- 2 a b n y... plication method affect adhesion force. Furthermore, since 1.5  SST  a a a yijk 2  i1 j1 k1 abn f0.05,2,12, there is no indication of interaction between these fac-  14.02 2  14.52 2  p tors. The last column of Table 14-6 shows the P-value for each F-ratio. Notice that the P-values for the two test statis- 189.82 2 tics for the main effects are considerably less than 0.05, while  15.02  2  10.72 the P-value for the test statistic for the interaction is greater 18 than 0.05. a y 2.. 2 y... Practical Interpretation: A graph of the cell adhesion force averages 5 yij.6 versus levels of primer type for each ap- i SStypes  a  i1 bn abn plication method is shown in Fig. 14-10. The no-interaction 128.72 2  134.12 2  127.02 2 conclusion is obvious in this graph, because the two lines are  nearly parallel. Furthermore, since a large response indicates 6 189.82 greater adhesion force, we conclude that spraying is the best 2 application method and that primer type 2 is most effective.   4.58 18 b y.2j. 2 y... SSmethods  a an  7.0 j1 abn 140.22 2  149.62 2 189.82 2 6.0    4.91 Spraying 9 18 b y2 yi j 5.0 a ij. y 2... SSinteraction  a a n   SS types  SS methods Dipping i1 j1 abn 4.0 112.82 2  115.92 2  111.52 2  115.92 2  118.22 2  115.52 2  3.0 3 189.82 2 1 2 3   4.58  4.91  0.24 Primer type 18 Figure 14-10 Graph of average adhesion force ver- and sus primer types for both application methods. SSE  SST  SStypes  SSmethods  SSinteraction  10.72  4.58  4.91  0.24  0.99 Tests on Individual Means When both factors are fixed, comparisons between the individual means of either factor may be made using any multiple comparison technique such as Fisher’s LSD method (described in JWCL232_c14_551-636.qxd 1/16/10 9:55 AM Page 564 564 CHAPTER 14 DESIGN OF EXPERIMENTS WITH SEVERAL FACTORS Table 14-6 ANOVA for Example 14-1 Source of Sum of Degrees of Mean Variation Squares Freedom Square f0 P-Value Primer types 4.58 2 2.29 28.63 2.7  E-5 Application methods 4.91 1 4.91 61.38 4.7  E-6 Interaction 0.24 2 0.12 1.50 0.2621 Error 0.99 12 0.08 Total 10.72 17 Chapter 13). When there is no interaction, these comparisons may be made using either the row averages yi.. or the column averages y.j.. However, when interaction is significant, comparisons between the means of one factor (say, A) may be obscured by the AB interaction. In this case, we could apply a procedure such as Fisher’s LSD method to the means of factor A, with factor B set at a particular level. Minitab Output Table 14-7 shows some of the output from the Minitab analysis of variance procedure for the aircraft primer paint experiment in Example 14-1. The upper portion of the table gives factor name and level information, and the lower portion of the table presents the analysis of vari- ance for the adhesion force response. The results are identical to the manual calculations dis- played in Table 14-6 apart from rounding. 14-3.2 Model Adequacy Checking Just as in the single-factor experiments discussed in Chapter 13, the residuals from a factorial experiment play an important role in assessing model adequacy. The residuals from a two-factor factorial are eijk  yijk  yij. That is, the residuals are just the difference between the observations and the corresponding cell averages. Table 14-7 Analysis of Variance from Minitab for Example 14-1 ANOVA (Balanced Designs) Factor Type Levels Values Primer fixed 3 1 2 3 Method fixed 2 Dip Spray Analysis of Variance for Adhesion Source DF SS MS F P Primer 2 4.5811 2.2906 27.86 0.000 Method 1 4.9089 4.9089 59.70 0.000 Primer *Method 2 0.2411 0.1206 1.47 0.269 Error 12 0.9867 0.0822 Total 17 10.7178 JWCL232_c14_551-636.qxd 1/16/10 9:55 AM Page 565 14-3 TWO-FACTOR FACTORIAL EXPERIMENTS 565 Table 14-8 Residuals for the Aircraft Primer Experiment in Example 14-1 Application Method Primer Type Dipping Spraying 1 0.27, 0.23, 0.03 0.10, 0.40, 0.30 2 0.30, 0.40, 0.10 0.27, 0.03, 0.23 3 0.03, 0.13, 0.17 0.33, 0.17, 0.17 Table 14-8 presents the residuals for the aircraft primer paint data in Example 14-1. The normal probability plot of these residuals is shown in Fig. 14-11. This plot has tails that do not fall exactly along a straight line passing through the center of the plot, indicat- ing some potential problems with the normality assumption, but the deviation from nor- mality does not appear severe. Figures 14-12 and 14-13 plot the residuals versus the levels of primer types and application methods, respectively. There is some indication that primer type 3 results in slightly lower variability in adhesion force than the other two primers. The graph of residuals versus fitted values in Fig. 14-14 does not reveal any unusual or diag- nostic pattern. 2.0 1.0 +0.5 zj 0.0 eij k –1.0 0 Primer type 1 2 3 –2.0 –0.5 –0.3 –0.1 +0.1 +0.3 ei j k, residual –0.5 Figure 14-11 Normal probability plot of the Figure 14-12 Plot of residuals versus primer type. residuals from Example 14-1. +0.5 +0.5 eijk 0 Application method ei j k D S ^ yij k 0 4 5 6 – 0.5 –0.5 Figure 14-13 Plot of residuals versus application Figure 14-14 Plot of residuals versus predicted method. values yˆijk. JWCL232_c14_551-636.qxd 1/16/10 9:56 AM Page 566 566 CHAPTER 14 DESIGN OF EXPERIMENTS WITH SEVERAL FACTORS 14-3.3 One Observation per Cell In some cases involving a two-factor factorial experiment, we may have only one replicate—that is, only one observation per cell. In this situation, there are exactly as many parameters in the analysis of variance model as observations, and the error degrees of freedom are zero. Thus, we cannot test hypotheses about the main effects and interactions unless some additional assumptions are made. One possible assumption is to assume the interaction effect is negligi- ble and use the interaction mean square as an error mean square. Thus, the analysis is equivalent to the analysis used in the randomized block design. This no-interaction assumption can be dan- gerous, and the experimenter should carefully examine the data and the residuals for indications as to whether or not interaction is present. For more details, see Montgomery (2009). EXERCISES FOR SECTION 14-3 14-1. An article in Industrial Quality Control (1956, pp. (a) State the hypotheses of interest in this experiment. 5–8) describes an experiment to investigate the effect of two (b) Test the above hypotheses and draw conclusions using the factors (glass type and phosphor type) on the brightness of a analysis of variance with = 0.05. television tube. The response variable measured is the current (c) Analyze the residuals from this experiment. (in microamps) necessary to obtain a specified brightness 14-3. In the book Design and Analysis of Experiments, 7th edi- level. The data are shown in the following table: tion (2009, John Wiley & Sons), the results of an experiment in- (a) State the hypotheses of interest in this experiment. volving a storage battery used in the launching mechanism of a (b) Test the above hypotheses and draw conclusions using the shoulder-fired ground-to-air missile were presented. Three mate- analysis of variance with  0.05. rial types can be used to make the battery plates. The objective is (c) Analyze the residuals from this experiment. to design a battery that is relatively unaffected by the ambient temperature. The output response from the battery is effective life in hours. Three temperature levels are selected, and a factorial ex- Glass Phosphor Type periment with four replicates is run. The data are as follows: Type 1 2 3 1 280 300 290 Temperature (ⴗF) 290 310 285 Material Low Medium High 285 295 290 1 130 155 34 40 20 70 2 230 260 220 74 180 80 75 82 58 235 240 225 2 150 188 136 122 25 70 240 235 230 159 126 106 115 58 45 3 138 110 174 120 96 104 14-2. An engineer suspects that the surface finish of metal 168 160 150 139 82 60 parts is influenced by the type of paint used and the drying time. He selected three drying times—20, 25, and 30 minutes—and used two types of paint. Three parts are tested with each combi- (a) Test the appropriate hypotheses and draw conclusions nation of paint type and drying time. The data are as follows: using the analysis of variance with = 0.05. (b) Graphically analyze the interaction. (c) Analyze the residuals from this experiment. Drying Time (min) 14-4. An experiment was conducted to determine whether Paint 20 25 30 either firing temperature or furnace position affects the baked 1 74 73 78 density of a carbon anode. The data are as follows: 64 61 85 50 44 92 Temperature (ⴗC) 2 92 98 66 Position 800 825 850 86 73 45 1 570 1063 565 68 88 85 565 1080 510 JWCL232_c14_551-636.qxd 1/16/10 9:56 AM Page 567 14-3 TWO-FACTOR FACTORIAL EXPERIMENTS 567 583 1043 590 Copper Content (%) Temperature 2 528 988 526 (ⴗC) 40 60 80 100 547 1026 538 50 17, 20 16, 21 24, 22 28, 27 521 1004 532 75 12, 9 18, 13 17, 12 27, 31 100 16, 12 18, 21 25, 23 30, 23 125 21, 17 23, 21 23, 22 29, 31 (a) State the hypotheses of interest. (b) Test the above hypotheses using the analysis of variance with = 0.05. What are your conclusions? (a) Is there any indication that either factor affects the amount (c) Analyze the residuals from this experiment. of warping? Is there any interaction between the factors? (d) Using Fisher’s LSD method, investigate the differences Use = 0.05. between the mean baked anode density at the three differ- (b) Analyze the residuals from this experiment. ent levels of temperature. Use = 0.05. (c) Plot the average warping at each level of copper content 14-5. An article in Technometrics [“Exact Analysis of and compare the levels using Fisher’s LSD method. Means with Unequal Variances” (2002, Vol. 44, pp. 152–160)] Describe the differences in the effects of the different described the technique of the analysis of means (ANOM) and levels of copper content on warping. If low warping is presented the results of an experiment on insulation. Four desirable, what level of copper content would you specify? insulation types were tested at three different temperatures. (d) Suppose that temperature cannot be easily controlled in The data are as follows: the environment in which the copper plates are to be used. Does this change your answer for part (c)? 14-7. An article in the IEEE Transactions on Electron Devices Temperature (F) (November 1986, p. 1754) describes a study on the effects of Insulation 1 2 3 two variables—polysilicon doping and anneal conditions (time 6.6 4 4.5 2.2 2.3 0.9 and temperature)—on the base current of a bipolar transistor. The data from this experiment follows below. 2.7 6.2 5.5 2.7 5.6 4.9 (a) Is there any evidence to support the claim that either 1 6 5 4.8 5.8 2.2 3.4 polysilicon doping level or anneal conditions affect base 3 3.2 3 1.5 1.3 3.3 current? Do these variables interact? Use  0.05. 2.1 4.1 2.5 2.6 0.5 1.1 (b) Graphically analyze the interaction. 2 5.9 2.5 0.4 3.5 1.7 0.1 (c) Analyze the residuals from this experiment. (d) Use Fisher’s LSD method to isolate the effects of anneal 5.7 4.4 8.9 7.7 2.6 9.9 conditions on base current, with  0.05. 3.2 3.2 7 7.3 11.5 10.5 14-8. An article in the Journal of Testing and Evaluation 3 5.3 9.7 8 2.2 3.4 6.7 (1988, Vol. 16, pp. 508–515) investigated the effects of cyclic 7 8.9 12 9.7 8.3 8 loading frequency and environment conditions on fatigue crack 7.3 9 8.5 10.8 10.4 9.7 growth at a constant 22 MPa stress for a particular material. 4 8.6 11.3 7.9 7.3 10.6 7.4 Environment Air H2O Salt H2O (a) Write down a model for this experiment. (b) Test the appropriate hypotheses and draw conclusions 2.29 2.06 1.90 using the analysis of variance with  0.05. 2.47 2.05 1.93 10 (c) Graphically analyze the interaction. 2.48 2.23 1.75 (d) Analyze the residuals from the experiment. 2.12 2.03 2.06 (e) Use Fisher’s LSD method to investigate the differences 2.65 3.20 3.10 between mean effects of insulation type. Use  0.05. 2.68 3.18 3.24 14-6. Johnson and Leone (Statistics and Experimental Frequency 1 2.06 3.96 3.98 Design in Engineering and the Physical Sciences, John 2.38 3.64 3.24 Wiley, 1977) described an experiment conducted to investi- 2.24 11.00 9.96 gate warping of copper plates. The two factors studied were 2.71 11.00 10.01 temperature and the copper content of the plates. The re- 0.1 2.81 9.06 9.36 sponse variable is the amount of warping. The data are as 2.08 11.30 10.40 follows: JWCL232_c14_551-636.qxd 1/16/10 9:56 AM Page 568 568 CHAPTER 14 DESIGN OF EXPERIMENTS WITH SEVERAL FACTORS The data from the experiment follow. The response variable is 14-9. Consider a two-factor factorial experiment. Develop a fatigue crack growth rate. formula for finding a 100(1  )% confidence interval on (a) Is there indication that either factor affects crack growth the difference between any two means for either a row or rate? Is there any indication of interaction? Use  0.05. column factor. Apply this formula to find a 95% CI on the (b) Analyze the residuals from this experiment. difference in mean warping at the levels of copper content 60 (c) Repeat the analysis in part (a) using ln(y) as the response. and 80% in Exercise 14-6. Analyze the residuals from this new response variable and comment on the results. Anneal (temperature/time) 900 900 950 1000 1000 60 180 60 15 30 4.40 8.30 10.15 10.29 11.01 1  1020 Polysilicon 4.60 8.90 10.20 10.30 10.58 doping 3.20 7.81 9.38 10.19 10.81 2  1020 3.50 7.75 10.02 10.10 10.60 14-4 GENERAL FACTORIAL EXPERIMENTS Many experiments involve more than two factors. In this section we introduce the case where there are a levels of factor A, b levels of factor B, c levels of factor C, and so on, arranged in a factorial experiment. In general, there will be abc p n total observations, if there are n replicates of the complete experiment. For example, consider the three-factor-factorial experiment, with underlying model Yijkl    i  j  k  12 ij  12 ik  12 jk i  1, 2, p , a  12 ijk  μ j  1, 2, p , b (14-10) ijkl k  1, 2, p , c l  1, 2, p , n Notice that the model contains three main effects, three two-factor interactions, a three-factor interaction, and an error term. Assuming that A, B, and C are fixed factors, the analysis of vari- ance is shown in Table 14-9. Note that there must be at least two replicates (n  2) to compute an error sum of squares. The F-test on main effects and interactions follows directly from the expected mean squares. These ratios follow F distributions under the respective null hypotheses. EXAMPLE 14-2 Surface Roughness A mechanical engineer is studying the surface roughness of a periments, we have used Minitab for the solution of this prob- part produced in a metal-cutting operation. Three factors, feed lem. The F-ratios for all three main effects and the interac- rate (A), depth of cut (B), and tool angle (C ), are of interest. tions are formed by dividing the mean square for the effect of All three factors have been assigned two levels, and two interest by the error mean square. Since the experimenter has replicates of a factorial design are run. The coded data are selected  0.05, the critical value for each of these shown in Table 14-10. F-ratios is f0.05,1,8  5.32. Alternately, we could use the P- The ANOVA is summarized in Table 14-11. Since value approach. The P-values for all the test statistics are manual ANOVA computations are tedious for three-factor ex- shown in the last column of Table 14-11. Inspection of these JWCL232_c14_551-636.qxd 1/16/10 9:56 AM Page 569 14-4 GENERAL FACTORIAL EXPERIMENTS 569 Table 14-9 Analysis of Variance Table for the Three-Factor Fixed Effects Model Source of Sum of Degrees of Expected Variation Squares Freedom Mean Square Mean Squares F0 bcn 兺2i MSA A SSA a1 MSA 2  a1 MSE acn 兺2j MSB B SSB b1 MSB 2  b1 MSE abn 兺2k MSC C SSC c1 MSC 2  c1 MSE cn 兺 兺12 2ij 1a  121b  12 MSAB 2  1a  121b  12 AB SSAB MSAB MSE bn 兺 兺12 2ik 1a  121c  12 MSAC 2  1a  121c  12 AC SSAC MSAC MSE an 兺 兺12 jk2 1b  121c  12 MSBC 2  1b  121c  12 BC SSBC MSBC MSE n 兺兺兺12 2ijk 1a  121b  121c  12 MSABC 2  1a  121b  121c  12 ABC SSABC MSABC MSE Error SSE abc1n  12 MSE 2 Total SST abcn  1 P-values is revealing. There is a strong main effect of feed Most