1 DESIGN OF EXPERIMENTS by R. C. Baker How to gain 20 years of experience in one short week!

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DESIGN OF EXPERIMENTSby

R. C. Baker

How to gain 20 years of experience in one short week!

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Role of DOE in Process Improvement

• DOE is a formal mathematical method for systematically planning and conducting scientific studies that change experimental variables together in order to determine their effect of a given response.

• DOE makes controlled changes to input variables in order to gain maximum amounts of information on cause and effect relationships with a minimum sample size.

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• DOE is more efficient that a standard approach of changing “one variable at a time” in order to observe the variable’s impact on a given response.

• DOE generates information on the effect various factors have on a response variable and in some cases may be able to determine optimal settings for those factors.

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• DOE encourages “brainstorming” activities associated with discussing key factors that may affect a given response and allows the experimenter to identify the “key” factors for future studies.

• DOE is readily supported by numerous statistical software packages available on the market.

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BASIC STEPS IN DOE

• Four elements associated with DOE:

• 1. The design of the experiment,

• 2. The collection of the data,

• 3. The statistical analysis of the data, and

• 4. The conclusions reached and recommendations made as a result of the experiment.

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TERMINOLOGY

• Replication – repetition of a basic experiment without changing any factor settings, allows the experimenter to estimate the experimental error (noise) in the system used to determine whether observed differences in the data are “real” or “just noise”, allows the experimenter to obtain more statistical power (ability to identify small effects)

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TERMINOLOGY

• .Randomization – a statistical tool used to minimize potential uncontrollable biases in the experiment by randomly assigning material, people, order that experimental trials are conducted, or any other factor not under the control of the experimenter. Results in “averaging out” the effects of the extraneous factors that may be present in order to minimize the risk of these factors affecting the experimental results.

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TERMINOLOGY

• Blocking – technique used to increase the precision of an experiment by breaking the experiment into homogeneous segments (blocks) in order to control any potential block to block variability (multiple lots of raw material, several shifts, several machines, several inspectors). Any effects on the experimental results as a result of the blocking factor will be identified and minimized.

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TERMINOLOGY

• Confounding - A concept that basically means that multiple effects are tied together into one parent effect and cannot be separated. For example,

• 1. Two people flipping two different coins would result in the effect of the person and the effect of the coin to be confounded

• 2. As experiments get large, higher order interactions (discussed later) are confounded with lower order interactions or main effect.

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TERMINOLOGY

• Factors – experimental factors or independent variables (continuous or discrete) an investigator manipulates to capture any changes in the output of the process. Other factors of concern are those that are uncontrollable and those which are controllable but held constant during the experimental runs.

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TERMINOLOGY

• Responses – dependent variable measured to describe the output of the process.

• Treatment Combinations (run) – experimental trial where all factors are set at a specified level.

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TERMINOLOGY

• Fixed Effects Model - If the treatment levels are specifically chosen by the experimenter, then conclusions reached will only apply to those levels.

• Random Effects Model – If the treatment levels are randomly chosen from a population of many possible treatment levels, then conclusions reached can be extended to all treatment levels in the population.

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PLANNING A DOE

• Everyone involved in the experiment should have a clear idea in advance of exactly what is to be studied, the objectives of the experiment, the questions one hopes to answer and the results anticipated

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PLANNING A DOE

• Select a response/dependent variable (variables) that will provide information about the problem under study and the proposed measurement method for this response variable, including an understanding of the measurement system variability

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PLANNING A DOE

• Select the independent variables/factors (quantitative or qualitative) to be investigated in the experiment, the number of levels for each factor, and the levels of each factor chosen either specifically (fixed effects model) or randomly (random effects model).

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PLANNING A DOE• Choose an appropriate experimental design

(relatively simple design and analysis methods are almost always best) that will allow your experimental questions to be answered once the data is collected and analyzed, keeping in mind tradeoffs between statistical power and economic efficiency. At this point in time it is generally useful to simulate the study by generating and analyzing artificial data to insure that experimental questions can be answered as a result of conducting your experiment

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PLANNING A DOE

• Perform the experiment (collect data) paying particular attention such things as randomization and measurement system accuracy, while maintaining as uniform an experimental environment as possible. How the data are to be collected is a critical stage in DOE

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PLANNING A DOE

• Analyze the data using the appropriate statistical model insuring that attention is paid to checking the model accuracy by validating underlying assumptions associated with the model. Be liberal in the utilization of all tools, including graphical techniques, available in the statistical software package to insure that a maximum amount of information is generated

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PLANNING A DOE

• Based on the results of the analysis, draw conclusions/inferences about the results, interpret the physical meaning of these results, determine the practical significance of the findings, and make recommendations for a course of action including further experiments

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SIMPLE COMPARATIVE EXPERIMENTS

• Single Mean Hypothesis Test

• Difference in Means Hypothesis Test with Equal Variances

• Difference in Means Hypothesis Test with Unequal Variances

• Difference in Variances Hypothesis Test

• Paired Difference in Mean Hypothesis Test

• One Way Analysis of Variance

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CRITICAL ISSUES ASSOCIATED WITH SIMPLE COMPARATIVE EXPERIMENTS

• How Large a Sample Should We Take?

• Why Does the Sample Size Matter Anyway?

• What Kind of Protection Do We Have Associated with Rejecting “Good” Stuff?

• What Kind of Protection Do We Have Associated with Accepting “Bad” Stuff?

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Single Mean Hypothesis Test

• After a production run of 12 oz. bottles, concern is expressed about the possibility that the average fill is too low.

• Ho: = 12

• Ha: <> 12

• level of significance = = .05

• sample size = 9

• SPEC FOR THE MEAN: 12 + .1

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Single Mean Hypothesis Test• Sample mean = 11.9

• Sample standard deviation = 0.15

• Sample size = 9

• Computed t statistic = -2.0

• P-Value = 0.0805162

• CONCLUSION: Since P-Value > .05, you fail to reject hypothesis and ship product.

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Single Mean Hypothesis Test Power Curve

Power Curvealpha = 0.05, sigma = 0.15

True Mean

Pow

er

11.8 11.9 12 12.1 12.20

0.2

0.4

0.6

0.8

1

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Single Mean Hypothesis Test Power Curve

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Single Mean Hypothesis Test Power Curve - Different Sample Sizes

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DIFFERENCE IN MEANS - EQUAL VARIANCES

• Ho:

• Ha:

• level of significance = = .05

• sample sizes both = 15

• Assumption: =

• Sample means = 11.8 and 12.1

• Sample standard deviations = 0.1 and 0.2

• Sample sizes = 15 and 15

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DIFFERENCE IN MEANS - EQUAL VARIANCES Can you detect this difference?

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DIFFERENCE IN MEANS - EQUAL VARIANCES

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DIFFERENCE IN MEANS - unEQUAL VARIANCES

• Same as the “Equal Variance” case except the variances are not assumed equal.

• How do you know if it is reasonable to assume that variances are equal OR unequal?

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DIFFERENCE IN VARIANCE HYPOTHESIS TEST

• Same example as Difference in Mean:

• Sample standard deviations = 0.1 and 0.2

• Sample sizes = 15 and 15

• **********************************

• Null Hypothesis: ratio of variances = 1.0

• Alternative: not equal

• Computed F statistic = 0.25

• P-Value = 0.0140071

• Reject the null hypothesis for alpha = 0.05.

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DIFFERENCE IN VARIANCE HYPOTHESIS TEST

Can you detect this difference?

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DIFFERENCE IN VARIANCE HYPOTHESIS TEST -POWER CURVE

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PAIRED DIFFERENCE IN MEANS HYPOTHESIS TEST

• Two different inspectors each measure 10 parts on the same piece of test equipment.

• Null hypothesis: DIFFERENCE IN MEANS = 0.0

• Alternative: not equal

• Computed t statistic = -1.22702

• P-Value = 0.250944

• Do not reject the null hypothesis for alpha = 0.05.

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PAIRED DIFFERENCE IN MEANS HYPOTHESIS TEST - POWER CURVE

Power Curvealpha = 0.05, sigma = 3.866

Difference in Means

Powe

r

-5 -4 -3 -2 -1 0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

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ONE WAY ANALYSIS OF VARIANCE

• Used to test hypothesis that the means of several populations are equal.

• Example: Production line has 7 fill needles and you wish to assess whether or not the average fill is the same for all 7 needles.

• Experiment: sample 20 fills from each of the 9 needles and test at 5% level of sign.

• Ho: =

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RESULTS: ANALYSIS OF VARIANCE TABLE

Analysis of Variance-----------------------------------------------------------------------------Source Sum of Squares Df Mean Square F-Ratio P-Value-----------------------------------------------------------------------------Between groups 1.10019 6 0.183364 18.66 0.0000Within groups 1.30717 133 0.00982837-----------------------------------------------------------------------------Total (Corr.) 2.40736 139

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SINCE NEEDLE MEANS ARE NOT ALL EQUAL, WHICH ONES ARE DIFFERENT?

• Multiple Range Tests for 7 NeedlesMethod: 95.0 percent LSDCol_2 Count Mean Homogeneous Groups--------------------------------------------------------------------------------N7 20 11.786 X N2 20 11.9811 X N1 20 11.9827 X N6 20 11.9873 X N3 20 11.9951 X N5 20 11.9953 X N4 20 12.11 X

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VISUAL COMPARISON OF 7 NEEDLES

N1

N2

N3

N4

N5

N6

N7

Box-and-Whisker Plot

11.5 11.7 11.9 12.1 12.3

Col_1

Col_

2

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FACTORIAL (2k) DESIGNS

• Experiments involving several factors ( k = # of factors) where it is necessary to study the joint effect of these factors on a specific response.

• Each of the factors are set at two levels (a “low” level and a “high” level) which may be qualitative (machine A/machine B, fan on/fan off) or quantitative (temperature 800/temperature 900, line speed 4000 per hour/line speed 5000 per hour).

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• Factors are assumed to be fixed (fixed effects model)

• Designs are completely randomized (experimental trials are run in a random order, etc.)

• The usual normality assumptions are satisfied.

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• Particularly useful in the early stages of experimental work when you are likely to have many factors being investigated and you want to minimize the number of treatment combinations (sample size) but, at the same time, study all k factors in a complete factorial arrangement (the experiment collects data at all possible combinations of factor levels).

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• As k gets large, the sample size will increase exponentially. If experiment is replicated, the # runs again increases.

k # of runs2 43 84 165 326 647 1288 2569 51210 1024

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FACTORIAL (2k) DESIGNS (k = 2)

• Two factors set at two levels (normally referred to as low and high) would result in the following design where each level of factor A is paired with each level of factor B.

RUN Factor A Factor B RESPONSE RUN Factor A Factor B RESPONSE

1 low low y1 1 -1 -1 y1

2 high low y2 2 +1 -1 y2

3 low high y3 3 -1 +1 y3

4 high high y4 4 +1 +1 y4

Generalized Settings Orthogonal Settings

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FACTORIAL (2k) DESIGNS (k = 2)

• Estimating main effects associated with changing the level of each factor from low to high. This is the estimated effect on the response variable associated with changing factor A or B from their low to high values.

2

)(

2

)( 3142 yyyyEffectAFactor

2

)(

2

)( 2143 yyyyEffectBFactor

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FACTORIAL (2k) DESIGNS (k = 2): GRAPHICAL OUTPUT

• Neither factor A nor Factor B have an effect on the response variable.

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• Factor A has an effect on the response variable, but Factor B does not.

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• Factor A and Factor B have an effect on the response variable.

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• Factor B has an effect on the response variable, but only if factor A is set at the “High” level. This is called interaction and it basically means that the effect one factor has on a response is dependent on the level you set other factors at. Interactions can be major problems in a DOE if you fail to account for the interaction when designing your experiment.

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EXAMPLE:FACTORIAL (2k) DESIGNS (k = 2)

• A microbiologist is interested in the effect of two different culture mediums [medium 1 (low) and medium 2 (high)] and two different times [10 hours (low) and 20 hours (high)] on the growth rate of a particular CFU.

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• Since two factors are of interest, k =2, and we would need the following four runs resulting in

RUN Medium Time Growth Rate

1 low low 17

2 high low 15

3 low high 38

4 high high 39

Generalized Settings

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• Estimates for the medium and time effects are

• Medium effect = [(15+39)/2] – [(17 + 38)/2] = -0.5

• Time effect = [(38+39)/2] – [(17 + 15)/2] = 22.5

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• A statistical analysis using the appropriate statistical model would result in the following information. Factor A (medium) and Factor B (time)

Type III Sums of Squares------------------------------------------------------------------------------------Source Sum of Squares Df Mean Square F-Ratio P-Value------------------------------------------------------------------------------------FACTOR A 0.25 1 0.25 0.11 0.7952FACTOR B 506.25 1 506.25 225.00 0.0424Residual 2.25 1 2.25------------------------------------------------------------------------------------Total (corrected) 508.75 3All F-ratios are based on the residual mean square error.

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EXAMPLE:CONCLUSIONS

• In statistical language, one would conclude that factor A (medium) is not statistically significant at a 5% level of significance since the p-value is greater than 5% (0.05), but factor B (time) is statistically significant at a 5 % level of significance since this p-value is less than 5%.

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EXAMPLE:CONCLUSIONS

• In layman terms, this means that we have no evidence that would allow us to conclude that the medium used has an effect on the growth rate, although it may well have an effect (our conclusion was incorrect).

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EXAMPLE:CONCLUSIONS

• Additionally, we have evidence that would allow us to conclude that time does have an effect on the growth rate, although it may well not have an effect (our conclusion was incorrect).

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EXAMPLE:CONCLUSIONS

• In general we control the likelihood of reaching these incorrect conclusions by the selection of the level of significance for the test and the amount of data collected (sample size).

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2k DESIGNS (k > 2)

• As the number of factors increase, the number of runs needed to complete a complete factorial experiment will increase dramatically. The following 2k design layout depict the number of runs needed for values of k from 2 to 5. For example, when k = 5, it will take 32 experimental runs for the complete factorial experiment.

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2k DESIGNS (k > 2)

RUNS

1 -1 -1 -1 -1 -12 +1 -1 -1 -1 -13 -1 +1 -1 -1 -14 +1 +1 -1 -1 -1

5 -1 -1 +1 -1 -16 +1 -1 +1 -1 -17 -1 +1 +1 -1 -18 +1 +1 +1 -1 -1

9 -1 -1 -1 +1 -110 +1 -1 -1 +1 -111 -1 +1 -1 +1 -112 +1 +1 -1 +1 -113 -1 -1 +1 +1 -114 +1 -1 +1 +1 -115 -1 +1 +1 +1 -116 +1 +1 +1 +1 -1

17 -1 -1 -1 -1 +118 +1 -1 -1 -1 +119 -1 +1 -1 -1 +120 +1 +1 -1 -1 +121 -1 -1 +1 -1 +122 +1 -1 +1 -1 +123 -1 +1 +1 -1 +124 +1 +1 +1 -1 +125 -1 -1 -1 +1 +126 +1 -1 -1 +1 +127 -1 +1 -1 +1 +128 +1 +1 -1 +1 +129 -1 -1 +1 +1 +130 +1 -1 +1 +1 +131 -1 +1 +1 +1 +132 +1 +1 +1 +1 +1

2k Design Layouts (k = 2-5)

k = 2

k = 3

k = 4

k = 5

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Interactions for 2k Designs (k = 3)

• Interactions between various factors can be estimated for different designs above by multiplying the appropriate columns together and then subtracting the average response for the lows from the average response for the highs.

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Interactions for 2k Designs (k = 3)

a b c ab ac bc abc

-1 -1 -1 1 1 1 -1+1 -1 -1 -1 -1 1 1-1 +1 -1 -1 1 -1 1+1 +1 -1 1 -1 -1 -1-1 -1 +! 1 -1 -1 1+1 -1 +1 -1 1 -1 -1-1 +1 +1 -1 -1 1 -1+1 +1 +1 1 1 1 1

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2k DESIGNS (k > 2)

• Once the effect for all factors and interactions are determined, you are able to develop a prediction model to estimate the response for specific values of the factors. In general, we will do this with statistical software, but for these designs, you can do it by hand calculations if you wish.

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2k DESIGNS (k > 2)

• For example, if there are no significant interactions present, you can estimate a response by the following formula. (for quantitative factors only)

Y = (average of all responses) + )](*)2

[( LfactorLEVECTfactorEFFE

= BAY BA *)2

(*)2

(

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ONE FACTOR EXAMPLE

• Simple “one factor” example where the factor is the number of hours a student studies for an exam (LOW = 10 HRS, HIGH = 20 HRS) and the response variable is their grade. Estimate the model for prediction a students grade as a function of the number of hours they study.

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ONE FACTOR EXAMPLE

Plot of Fitted Model

#HRS STUDY

GR

AD

E

10 12 14 16 18 2055

65

75

85

95

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ONE FACTOR EXAMPLE

• The output shows the results of fitting a general linear model to describe the relationship between GRADE and #HRS STUDY. The equation of the fitted general model is

• GRADE = 29.3 + 3.1* (#HRS STUDY)

• The fitted orthogonal model is

• GRADE = 75 + 15 * (SCALED # HRS)

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Two Level Screening Designs • Suppose that your brainstorming session

resulted in 7 factors that various people think “might” have an effect on a response. A full factorial design would require 27 = 128 experimental runs without replication. The purpose of screening designs is to reduce (identify) the number of factors down to the “major” role players with a minimal number of experimental runs. One way to do this is to use the 23 full factorial design and use interaction columns for factors.

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Note that * Any factor d effect is now confounded with the a*b interaction* Any factor e effect is now confounded with the a*c interaction* etc.* What is the d*e interaction confounded with????????

a b c d = ab e = ac f = bc g = abc

-1 -1 -1 1 1 1 -1+1 -1 -1 -1 -1 1 1-1 +1 -1 -1 1 -1 1+1 +1 -1 1 -1 -1 -1-1 -1 +! 1 -1 -1 1+1 -1 +1 -1 1 -1 -1-1 +1 +1 -1 -1 1 -1+1 +1 +1 1 1 1 1

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Problems that Interactions Cause!

• Interactions – If interactions exist and you fail to account for this, you may reach erroneous conclusions. Suppose that you plan an experiment with four runs and three factors resulting in the following data:

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Problems that Interactions Cause!• Factor A Effect = 0• Factor B Effect = 0• Factor C Effect = 5• In this example, if you were assuming that

“larger is better” then you would set Factor C at the “high level” and it appears to make no difference where you set factors A and B. In this case there is a factor A interaction with factor B and this interaction is confounded with the factor C effect.

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Problems that Interactions Cause!

Interaction Plot

FACTOR A

5

6

7

8

9

10

RE

SPO

NSE

-1 1

FACTOR B-11

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Resolution of a Design

• The above design would be called a resolution III design because main effects are aliased (confounded) with two factor interactions.

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Resolution of a Design • Resolution III Designs – No main effects are

aliased with any other main effect BUT some (or all) main effects are aliased with two way interactions

• Resolution IV Designs – No main effects are aliased with any other main effect OR two factor interaction, BUT two factor interactions may be aliased with other two factor interactions

• Resolution V Designs – No main effect OR two factor interaction is aliased with any other main effect or two factor interaction, BUT two factor interactions are aliased with three factor interactions.

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Common Screening Designs

• Fractional Factorial Designs – the total number of experimental runs must be a power of 2 (4, 8, 16, 32, 64, …). If you believe first order interactions are small compared to main effects, then you could choose a resolution III design. Just remember that if you have major interactions, it can mess up your screening experiment.

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Common Screening Designs

• Plackett-Burman Designs – Two level, resolution III designs used to study up to n-1 factors in n experimental runs, where n is a multiple of 4 ( # of runs will be 4, 8, 12, 16, …). Since n may be quite large, you can study a large number of factors with moderately small sample sizes. (n = 100 means you can study 99 factors with 100 runs)

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Other Design Issues

• May want to collect data at center points to estimate non-linear responses

• More than two levels of a factor – no problem (multi-level factorial)

• What do you do if you want to build a non-linear model to “optimize” the response. (hit a target, maximize, or minimize) – called response surface modeling

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Other Design Issues• What do you do if the factors levels are

categorical and not quantitative, or some are categorical and some are quantitative?

• What do you do if the structure of you experiment is “nested”? These are called heirarchical designs and will allow you to partition the total variability among the different levels of the design (called variance components)

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Response Surface Designs – Box-Behnken:After screening designs identify major factors –Next step.

• Design class: Response Surface• Design name: Box-Behnken design • Base Design• -----------• Number of experimental factors: 3 Number of blocks: 1• Number of responses: 1• Number of runs: 15 Error degrees of freedom: 5• Randomized: No

• Factors Low High Units Continuous• ------------------------------------------------------------------------• Factor_A -1.0 1.0 Yes• Factor_B -1.0 1.0 Yes• Factor_C -1.0 1.0 Yes

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Response Surface Designs – Box-Behnken

FACTOR A FACTOR B FACTOR C

0 0 0

-1 -1 0

1 -1 0

-1 1 0

1 1 0

-1 0 -1

1 0 -1

0 0 0

-1 0 1

1 0 1

0 -1 -1

0 1 -1

0 -1 1

0 1 1

0 0 0

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Response Surface Designs – Central Composite

• Design class: Response Surface• Design name: Central composite blocked cube-star • Number of experimental factors: 3 Number of blocks: 2• Number of responses: 1• Number of runs: 16 Error degrees of freedom: 5• Randomized: No

• Factors Low High Units Continuous

• ------------------------------------------------------------------------

• Factor_A -1.0 1.0 Yes• Factor_B -1.0 1.0 Yes• Factor_C -1.0 1.0 Yes

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Response Surface Designs – Central Composite

FACTOR A FACTOR B FACTOR C

-1 -1 -1

1 -1 -1

-1 1 -1

1 1 -1

0 0 0

-1 -1 1

1 -1 1

-1 1 1

1 1 1

-1.76383 0 0

1.76383 0 0

0 -1.76383 0

0 0 0

0 1.76383 0

0 0 -1.76383

0 0 1.76383

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Multilevel Factorial Designs• Design class: Multilevel Factorial• Number of experimental factors: 3 Number of blocks: 1• Number of responses: 1• Number of runs: 27 Error degrees of freedom: 17• Randomized: No

• Factors Low High Levels Units• ----------------------------------------------------------------------

---• Factor_A -1.0 1.0 3 • Factor_B -1.0 1.0 3 • Factor_C -1.0 1.0 3

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Multilevel Factorial DesignsFACTOR A FACTOR B FACTOR C

-1 -1 -10 -1 -11 -1 -1-1 0 -10 0 -11 0 -1-1 1 -10 1 -11 1 -1-1 -1 00 -1 01 -1 0-1 0 00 0 01 0 0-1 1 00 1 01 1 0-1 -1 10 -1 11 -1 1-1 0 10 0 11 0 1-1 1 10 1 11 1 1

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Nested Design• Design class: Variance Components• Number of experimental factors: 3 • Number of responses: 1• Number of runs: 27• Randomized: NoFactors Levels Units• -----------------------------------------------• Factor_A 3 • Factor_B 3 • Factor_C 3 You have created a variance components design which will

estimate the contribution of 3 factors to overall process variability. The design is hierarchical, with each factor nested in the factor above it. A total of 27 measurements are required.

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Nested DesignFACTOR A FACTOR B FACTOR C

1 1 11 1 21 1 31 2 11 2 21 2 31 3 11 3 21 3 32 1 12 1 22 1 32 2 12 2 22 2 32 3 12 3 22 3 33 1 13 1 23 1 33 2 13 2 23 2 33 3 13 3 23 3 3

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Response Surface Designs – Box-BehnkenEXAMPLE - RECAP

• Design class: Response Surface• Design name: Box-Behnken design • Base Design• -----------• Number of experimental factors: 3 Number of blocks: 1• Number of responses: 1• Number of runs: 15 Error degrees of freedom: 5• Randomized: No

• Factors Low High Units Continuous• ------------------------------------------------------------------------• Factor_A 10 30 Yes• Factor_B 30 60 Yes• Factor_C 40 60 Yes

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Response Surface Designs – Box-BehnkenREAL MODEL: Y = 40F1+200F2+100F3-10F1F2+9F1F1 + (NOISE)

RUN F1 F2 F3

1 10 45 60

2 30 45 40

3 20 30 40

4 10 30 50

5 20 45 50

6 30 60 50

7 20 45 50

8 30 45 60

9 20 45 50

10 20 60 40

11 10 45 40

12 30 30 50

13 20 60 60

14 10 60 50

15 20 30 60

90


Example: std. dev. of noise = 0

RUN F1 F2 F3 Y0

1 10 45 60 11800

2 30 45 40 8800

3 20 30 40 8400

4 10 30 50 9300

5 20 45 50 9400

6 30 60 50 8300

7 20 45 50 9400

8 30 45 60 10800

9 20 45 50 9400

10 20 60 40 8400

11 10 45 40 9800

12 30 30 50 11300

13 20 60 60 10400

14 10 60 50 12300

15 20 30 60 10400

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Response Surface Designs – Box-Behnken

REAL MODEL: Y = 40F1+200F2+100F3-10F1F2+9F1F1 + (NOISE)


Regression coeffs. for Var_1----------------------------------------------------------------------constant = 0.0A:Factor_A = 40.0B:Factor_B = 200.0C:Factor_C = 100.0AA = 9.0AB = -10.0AC = 0.0BB = 0.0BC = 0.0CC = 0.0

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Optimize Response-----------------Goal: MAXIMIZE Var_1

Optimum value = 13261.4

Factor Low High Optimum-----------------------------------------------------------------------Factor_A 10.0 30.0 10.1017 Factor_B 30.0 60.0 60.0 Factor_C 40.0 60.0 60.0

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RUN F1 F2 F3 Y100

1 10 45 60 11825

2 30 45 40 8781

3 20 30 40 8413

4 10 30 50 9216

5 20 45 50 9288

6 30 60 50 8261

7 20 45 50 9329

8 30 45 60 10855

9 20 45 50 9205

10 20 60 40 8538

11 10 45 40 9718

12 30 30 50 11308

13 20 60 60 10316

14 10 60 50 12056

15 20 30 60 10378

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Regression coeffs. for Var_3----------------------------------------------------------------------constant = 2312.5A:Factor_A = 36.575B:Factor_B = 200.067C:Factor_C = 3.85AA = 9.09875AB = -9.81167AC = -0.0825BB = 0.117222BC = -0.311667CC = 1.10875

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Standardized Pareto Chart for Var_3

Standardized effect0 10 20 30 40

ACBB

B:Factor_BBCCC

A:Factor_AAA

C:Factor_CAB

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Contours of Estimated Response SurfaceFactor_C=60.0

Factor_A

Fac

tor_

B

Var_39300.09500.09700.09900.010100.010300.010500.010700.010900.011100.011300.011500.011700.0

10 14 18 22 26 3030

35

40

45

50

55

60

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• Optimize Response• -----------------• Goal: maximize Y

• Optimum value = 13139.4

• Factor Low High Optimum• -----------------------------------------------------------------------• Factor_A 10.0 30.0 10.1036 • Factor_B 30.0 60.0 60.0 • Factor_C 40.0 60.0 60.0

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Response Surface Designs – Three Level Factorial Design (33)REAL MODEL: Y = 40F1+200F2+100F3-10F1F2+9F1F1 + (NOISE)


RUN F1 F2 F3 Y100

1 10 30 40 8270

2 20 30 40 8272

3 30 30 40 10324

4 10 45 40 9928

5 20 45 40 8520

6 30 45 40 8973

7 10 60 40 11082

8 20 60 40 8377

9 30 60 40 7410

10 10 30 50 9191

11 20 30 50 9331

12 30 30 50 11131

13 10 45 50 10615

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RUN F1 F2 F3 Y100

14 20 45 50 9302

15 30 45 50 9723

16 10 60 50 12088

17 20 60 50 9343

18 30 60 50 8260

19 10 30 60 10313

20 20 30 60 10363

21 30 30 60 12267

22 10 45 60 11763

23 20 45 60 10534

24 30 45 60 10791

25 10 60 60 13281

26 20 60 60 10349

27 30 60 60 9497

101



Regression coeffs. for Y----------------------------------------------------------------------constant = 2887.08A:Factor_A = 36.7028B:Factor_B = 212.806C:Factor_C = -31.0306AA = 8.95833AB = -9.57333AC = -0.190833BB = -0.283704BC = 0.100556CC = 1.30333

102



• Optimize Response• -----------------• Goal: maximize Y


• Factor Low High Optimum• -----------------------------------------------------------------------• Factor_A 10.0 30.0 10.0 • Factor_B 30.0 60.0 60.0 • Factor_C 40.0 60.0 60.0

103



Estimated Response SurfaceFactor_C=60.0

Factor_A

Factor_B

Y

10 14 18 22 26 30 30 35 40 45 50 55 609400

10400

11400

12400

13400

104



Contours of Estimated Response SurfaceFactor_B=60.0

Factor_A

Facto

r_C

Y8200.08600.09000.09400.09800.010200.010600.011000.011400.011800.012200.012600.0

10 14 18 22 26 3040

44

48

52

56

60

105

CLASSROOM EXERCISE

• STUDENT IN-CLASS EXPERIMENT: Collect data for experiment to determine factor settings (two factors) to hit a target response (spot on wall).

• Factor A – height of shaker (low and high)

• Factor B – location of shaker (close to hand and close to wall)

• Design experiment – would suggest several replications

106

CLASSROOM EXERCISE

• Conduct Experiment – student holds 3 foot “pin the tail on the donkey” stick and attempts to hit the target. An observer will assist to mark the hit on the target.

• Collect data – students take data home for week and come back with what you would recommend AND why.

• YOU TELL THE CLASS HOW TO PLAY THE GAME TO “WIN”.

107

CLASSROOM EXERCISE

108

CLASSROOM EXERCISE

MARKER STICK

VERTICAL POLE

1ST OBS 2ND OBS 3RD OBS 4TH OBS MEANSTANDARD DEVIATION

L L -2.750 -4.500 -4.750 -5.000 -4.250 1.021

H L -12.500 -6.750 -4.625 -4.000 -6.969 3.871

L H 3.000 3.250 3.875 6.250 4.094 1.484

H H 4.625 11.250 12.625 14.000 10.625 4.155

MARKER STICK

L = VERTICAL POLE WAS CLOSE TO WALL (MARKER END OF STICK

H=VERTICAL POLE WAS CLOSE TO HAND

VERTICAL POLE

L=SHAKING DEVICE LOCATED LOW ON VERTICAL POLE

H=SHAKING DEVICE LOCATED HIGH ON VERTICAL POLE

109

CLASSROOM EXERCISE

• HOMEWORK: • .Determine the effects “marker stick” and “vertical pole”

have on the mean location of the hit.• .Determine the effects “marker stick” and “vertical pole”

have on the standard deviation of the hit.• .Which factor would you say affects the mean location of

the “hit”?• .Which factor would you say affects the standard

deviation of the “hit”?• OPTIMAL SETTINGS: Where would you recommend

we locate the “vertical pole” and the “marker stick” IF we wish to (a) MINIMIZE THE VARIABILITY OF THE HIT and (b) HIT THE TARGET LOCATED AT “0”?

110

PIN THE TAIL DATA INPUT

111

ESTIMATE OF EFFECTS (MEAN HIT)

• Estimated effects for MEAN• ----------------------------------------------------------------------• average = 0.875 • A:MARKER STICK = 1.906 • B:VERTICAL POLE = 12.969• AB = 4.625 • ----------------------------------------------------------------------• No degrees of freedom left to estimate standard errors.

112


Pareto Chart for MEAN

Effect0 3 6 9 12 15

A:MARKER STICK

AB

B:VERTICAL POLE

113


Main Effects Plot for MEAN

MEA

N

MARKER STICK-1.0 1.0

VERTICAL POLE-1.0 1.0

-6

-3

0

3

6

9

114

INTERACTION PLOT (MEAN HIT)

Interaction Plot for MEAN

MEA

N


VERTICAL POLE=-1.0

VERTICAL POLE=-1.0

VERTICAL POLE=1.0

VERTICAL POLE=1.0

-7

-4

-1

2

5

8

11

115

3-D PLOT OF RESPONSE (MEAN HIT)

Estimated Response Surface

MARKER STICK

VERTICAL POLE

MEA

N

-1 -0.6 -0.2 0.2 0.6 1 -1 -0.6-0.2 0.2 0.6 1-7-4-1258

11

116

CONTOUR PLOT OF RESPONSE (MEAN HIT)

Contours of Estimated Response Surface

MARKER STICK

VERT

ICAL

POL

E

MEAN-4.0-2.00.02.04.06.08.0

-1 -0.6 -0.2 0.2 0.6 1-1

-0.6

-0.2

0.2

0.6

1

117

ANALYSIS OF VARIANCE TABLE (MEAN HIT)

• Analysis of Variance for MEAN• --------------------------------------------------------------------------------• Source Sum of Squares Df Mean Square F-Ratio P-Value• --------------------------------------------------------------------------------• A:MARKER STICK 3.63284 1 3.63284 0.17 0.7511• B:VERTICAL POLE 168.195 1 168.195 7.86 0.2181• Total error 21.3906 1 21.3906• --------------------------------------------------------------------------------

118

ESTIMATED LINEAR RESPONSE MODEL (MEAN HIT)

• Regression coeffs. for MEAN• ----------------------------------------------------------------------• constant = 0.875• A:MARKER STICK = 0.953• B:VERTICAL POLE = 6.4845• ----------------------------------------------------------------------

• The StatAdvisor• ---------------• This pane displays the regression equation which has been fitted to• the data. The equation of the fitted model is

• MEAN = 0.875 + 0.953*MARKER STICK + 6.4845*VERTICAL POLE

119

OPTIMAL FACTOR SETTINGS (MEAN HIT)

• Optimize Response• -----------------• Goal: maintain MEAN at 0.0


• Factor Low High Optimum• -----------------------------------------------------------------------• MARKER STICK -1.0 1.0 0.03311

• VERTICAL POLE -1.0 1.0 -0.139803

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ESTIMATE OF EFFECTS (STD DEV HIT)

• Estimated effects for STD DEV

• ----------------------------------------------------------------------

• average = 2.63275

• A:MARKER STICK = 2.7605

• B:VERTICAL POLE = 0.3735

• AB = -0.0895

121

ESTIMATE OF EFFECTS (STD DEV HIT)

• Analysis of Variance for STD DEV• --------------------------------------------------------------------------------• Source Sum of Squares Df Mean Square F-Ratio P-Value• --------------------------------------------------------------------------------• A:MARKER STICK 7.62036 1 7.62036 951.33 0.0206• B:VERTICAL POLE 0.139502 1 0.139502 17.42 0.1497• Total error 0.00801025 1 0.00801025• --------------------------------------------------------------------------------• Total (corr.) 7.76787 3

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OPTIMAL FACTOR SETTINGS (STD DEV HIT)

• Optimize Response• -----------------• Goal: minimize STD DEV


• Factor Low High Optimum• -----------------------------------------------------------------------• MARKER STICK -1.0 1.0 -1.0 • VERTICAL POLE -1.0 1.0 -1.0

123

INTERACTION (STD DEV HIT)

Interaction Plot for STD DEV

STD

DEV


VERTICAL POLE=-1.0

VERTICAL POLE=-1.0

VERTICAL POLE=1.0

VERTICAL POLE=1.0

0

1

2

3

4

5

124

CONTOUR PLOT OF RESPONSE (STD DEV HIT)

Contours of Estimated Response Surface

MARKER STICK

VERT

ICAL

POL

E

STD DEV1.01.52.02.53.03.54.04.5

-1 -0.6 -0.2 0.2 0.6 1-1

-0.6

-0.2

0.2

0.6

1

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SO, WHAT’S THE ANSWER?

• I WOULD:

• 1. SET THE “MARKER STICK” AT LOW (CLOSE TO THE WALL)

• 2. SET THE “VERTICAL POLE” AT A VALUE THAT WILL HIT THE TARGET.

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SO, WHAT’S THE ANSWER?

• FROM REGRESSION FOR “MEAN HIT”, SET MARKER STICK AT “-1”, HIT AT “0”, AND SOLVE FOR VP

• HIT = .0875 +.953*MS + 6.4845*VP

• 0 = .875 + .953*(-1) + 6.4845*VP

• Resulting in

• VP = .012 and MS = -1

127

Contour Plots for Mean and Std. Dev.

1 DESIGN OF EXPERIMENTS by R. C. Baker How to gain 20 years of experience in one short week!

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Transcript of 1 DESIGN OF EXPERIMENTS by R. C. Baker How to gain 20 years of experience in one short week!