design of experiments

Session 1

Classical Methods of Experimental DesignProf. Shiv G. Kapoor

Introduction

Shiv G. Kapoor Department of Mechanical Science and Engineering University of Illinois at Urbana-Champaign

IE 400 Lecture 1

@ 2006 Dr. Shiv G. Kapoor All Rights Reserved

OUTLINE1. Introduction Role of Experimental Design, Important Concepts in the DOE 2. Review of Basic Statistical Methods and Probability Concepts Discrete and Continuous Probability Distribution Functions, Normal and Sampling Distributions, Tests of Hypotheses 3. Comparative Experiments Comparing Two Treatments 4. Design and Analysis of 2k Factorial Experiments General Factorial Designs and Design at Two levels Calculation and Interpretation of Main and Interaction effe 5. Two-Level Fractional Factorial Designs Rationale for, and Consequences of, Fractions of Two-Level Factorials Concept of Design Resolution

IE 400 Lecture 1


THE NEED FOR STATISTICAL METHODS IN THE DESIGN OF EXPERIMENTS 1. The world around us is not deterministic Variability is part of the natural order of thingsVariation in data is neither totally chaotic nor small enough to be ignored It is real, identifiable, and predictable statistically.

University of Illinois at Urbana-Champaign

2006 by Dr. Shiv G. Kapoor

NATURE OF VARIABILITY IN DATA What is Variability? Let us understand the proper interpretation of variability in data via a dialog between the professor and a young, naive student. Professor: I am interested in the tool life of this tool.

IE 400 Lecture 1


NATURE OF VARIABILITY IN DATA

Conditions Speed: Feed: Workpiece 170 FPM .017 IPR 1018 Steel 6dia, 24 long

Depth of Cut: 0.07 IN

IE 400 Lecture 1



Student: I will go down to the lab and run a test to determine the required tool life.Speed Feed Depth of Cut

Tool Life

Material

IE 400 Lecture 1



Student: I have found that the tool life for the conditions you specified is 15 minutes. Professor: Thats fine-but why dont you go back and run another test. I want to be sure that the tool life is 15 minutes. Student: I told you that the tool life is 15 minutes, but I will run another test if you insist.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved


After running another test he came to Professor and said, Professor, this time I got a tool life of 16.2 minutes. However, I noticed some fluctuations in the machine power. If we install a power regulator, we can probably eliminate this source of variation and get a true tool life value.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved


Professor: Do whatever you need to do and re-run the test.Power Regulator Speed Feed Depth of Cut Tool Life

MaterialIE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

NATURE OF VARIABILITY IN DATA Student: After installing the power regulator, I now get a tool life of 15.6 minutesthis is the true tool life. Professor: Let me ask you if you think that the machine was vibrating? Should you be making the machine more rigid and then rerun the test? Student: You are right, I better make the machine more rigid.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

NATURE OF VARIABILITY IN DATA Student: Now the tool life is 16.2 minutes. I am sure that the true tool life is 16.2 minutes. Professor: How can you be so sure that you have eliminated all variation? What about differences in materials, tools, operator inconsistencies, etc? Student: Well, I suppose they could cause some variation.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved


Professor: What I really want is an average tool life for an average tool cutting a material over average environmental conditions! There will always be some variation in the process-we just do the best we can. Student: OK, Ill run more tests using several tools, material pieces, etc. and find out what the average tool life is.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

NATURE OF VARIABILITY IN DATAStudent: I have run tests over a random sample of tools, Workpieces, times of day, etc. Results of the tests are: Test Tool Life Test Tool Life 1 15 7 15 2 15.6 8 16.7 3 16.2 9 16 4 16.5 10 16 5 16.2 11 16.7 6 16.5 12 16.8 I find the average tool life to be 16.1 min.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

NATURE OF VARIABILITY IN DATA Professor: Thats fine, but how confident are you that 16.1 min is the true tool life? Student: Pretty darn confident! Professor: Thats not good enough. I need more information-you better tell me something like I am x% confident that the true tool life is within a certain range-Doesnt that logical? Student: It certainly is logical-but I dont know the answerIE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

THE NEED FOR STATISTICAL METHODS IN THE DESIGN OF EXPERIMENTS

2. We can easily define many factors of potential

significance but only a few account for the vast majority of the structure/variation in the data.The problem at hand is to screen from a large group of potentially important factors those few, which are worthy of continuing study.

3. The sequential nature of experimentation and the iterative process of building up a knowledge base is central to experimental work.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

THE NEED FOR STATISTICAL METHODS IN THE DESIGN OF EXPERIMENTS4. All processes are subject to identifiable and unidentifiable disturbances which can totally invalidate results. 5. In most physical processes, variables tend not to influence the process are independent of each other. 6. The world around us is non-linear.@ 2006 Dr. Shiv G. Kapoor All Rights Reserved

IE 400 Lecture 1

THE NEED FOR STATISTICAL METHODS IN THE DESIGN OF EXPERIMENTS7. Experimentation is costly and time consuming business. 8. We generally respond to crises, not principles.-There is never enough time to do the job right but always enough time to do the job over.

9. We know much less about what makes things work than we think.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

THE NEED FOR STATISTICAL METHODS IN THE DESIGN OF EXPERIMENTS10. Concepts techniques. drive people people drive

A myth we must dispel everything he is saying sounds logical and probably works fantastically for some people but my specific problem simply doesnt lend itself to this approach and/or simply doesnt need it anymore.

The above statement altogether is made too often.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

THE SEQUENTIAL AND ITERATIVE NATURE OF EXPERIMENTATIONTo optimize and control the process, a logical series of questions must be asked: 1. Which of a list of potentially important variables are worthy of further study? 2. How, specifically, do the important variables tend to influence the results? 3. What levels of the most important variables tend to optimize the process?

IE 400 Lecture 1


THE SEQUENTIAL AND ITERATIVE NATURE OF EXPERIMENTATION

4.

In the context of the above, what characteristics of the experimental environment may be bothersome and, therefore, need to be somehow neutralized? What specific concepts/techniques need to be evoked en route to answering the above?

5.

IE 400 Lecture 1


THE SEQUENTIAL AND ITERATIVE NATURE OF EXPERIMENTATIONThe key to successfully and efficiently embracing these questions is the iterative nature of the learning process. Conjecture: hypothesis about the situation at hand. Design: creation of an exercise/experiment to test/place the hypothesis in jeopardy.

IE 400 Lecture 1


THE SEQUENTIAL AND ITERATIVE NATURE OF EXPERIMENTATIONconduct the experiment and obtain the data. Analysis: examination of data to study its plausibility in light of the hypothesis being tested. This is then followed by modification of the original hypothesis in accordance with the analysis deductions and then designing a second experiment, etc.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

Experiment:

THE SEQUENTIAL AND ITERATIVE NATURE OF EXPERIMENTATION

In initiating an experimental study, we should probably plan in the first experiment to do no more than 25% of the total experiments we have resources for.

IE 400 Lecture 1


THE SEQUENTIAL AND ITERATIVE NATURE OF EXPERIMENTATION One big experiment is not only inefficient but leaves us with no resources left if our initial conjecture was in error. We should begin by looking at many factors in a somewhat superficial fashion and more toward the examination of only the few, most relevant, factors in a more comprehensive fashion.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

DIFFICULTIES WITH EXPERIMENTAL WORK WHICH REQUIRES STATISTICAL METHODS

Three key sources of difficulty confront the experimenter.Experimental error. Confusion of correlation with causation. Complexity of variable effects.

IE 400 Lecture 1



Experimental error1. Composed of many minute disturbances which individually have little effect on the outcome of the experiment. 2. Collectively these small chance occurrences may increase the dispersion or spread of the results to the point where real variable effects are masked.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved


Experimental error3. Composed of more than errors of measurement not all instrumentation oriented. A good measurement system accounts for no more than 10-15% of the total error. 4. Can be a function of both unknown and known sources.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved


Statistical analysis1. Can the results be explained solely by chance causes? 2. How much data is required to reveal the existence of true effects in light of chance error?

IE 400 Lecture 1


DIFFICULTIES WITH EXPERIMENTAL WORK WHICH REQUIRES STATISTICAL METHODS Correlation Versus Causation1. Two factors highly related only because they are related to a third common (often unidentified) factor. Deliberate change in one may not lead to a change in the other- Concept of planned versus passively observed To really find out how changes in some factor affect the output, you have to change that factor deliberately and observe the change.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

2.


Complexity of Variable Effects1. Nice state of affairs: variable effects are linear and additive! Generally not the case!Example: effects of aspirin and coffee on driving reaction time: aspirin increase , coffee reduce 2 . Will one aspirin and one cup of coffee reduce reaction time by 1 ? Additive: Will 10 aspirins and 5 cups of coffee keep reaction time constant? Linear?

2. Need to plan experiments to reveal variable interactions and nonlinear variable effects.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

SOME IMPORTANT CONCEPTS RELATED TO STATISTICAL DESIGN OF EXPERIMENTS External Versus Internal Experimental Comparisons Comparing an experimental result with past data is inviting because data may be there and database may be big! This can be dangerous because prevailing conditions may have changed from time of database development to time of the experiment. the data base may have been passively observed poor control; no active interference with the system.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

SOME IMPORTANT CONCEPTS RELATED TO STATISTICAL DESIGN OF EXPERIMENTS External Versus Internal Experimental Comparisons The presence of extraneous variation factors may yield poor sensitivity of statistical test. It is generally best to stay within the experiment for making comparisons.@ 2006 Dr. Shiv G. Kapoor All Rights Reserved

IE 400 Lecture 1

SOME IMPORTANT CONCEPTS RELATED TO STATISTICAL DESIGN OF EXPERIMENTS Essential Nature of a Significance TestIn the analysis of the results of comparative experiments, it is necessary to assess the magnitude of the test statistic of interest, say a difference in the mean results of two methods, in light of the natural variability inherent in the experiment. The question we are really asking is: could this experimental result--difference--have arisen solely due to chance causes or is there a real difference in the methods?IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

SOME IMPORTANT CONCEPTS RELATED TO STATISTICAL DESIGN OF EXPERIMENTS Essential Nature of a Significance TestTo answer this question we must measure the level of chance variation, usually through genuine test replication, and then determine the probability that we could have observed the result before us or one more extreme if only these chance causes were at work.

This is the role of a statistical test of significance.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

SOME IMPORTANT CONCEPTS RELATED TO STATISTICAL DESIGN OF EXPERIMENTS Blocking to Avoid Nuisance Variation When known sources of extraneous/unwanted variation can be identified, we can design the experiment in such a way as to eliminate their influence and provide more sensitive test of significance test.@ 2006 Dr. Shiv G. Kapoor All Rights Reserved

IE 400 Lecture 1

SOME IMPORTANT CONCEPTS RELATED TO STATISTICAL DESIGN OF EXPERIMENTS Example of BlockingProblem: Two different tool materials are being tested to determine whether or not there is a real difference in their wear characteristics.A=B, Wear?

Material AIE 400 Lecture 1

Material B@ 2006 Dr. Shiv G. Kapoor All Rights Reserved

SOME IMPORTANT CONCEPTS RELATED TO STATISTICAL DESIGN OF EXPERIMENTS Example of BlockingFirst Experimental Design: Twenty tools are made, ten with material A and ten with material B. Twenty machine operators are chosen at random, given the tools, and told to use them in machining as they normally do.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

SOME IMPORTANT CONCEPTS RELATED TO STATISTICAL DESIGN OF EXPERIMENTS Example of BlockingFirst Experimental Design: At the end of the experiments the mean amount of wear is determined based on the ten measurements for each material type. The mean difference is calculated and examined by a statistical test of significance. No real difference is found.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

SOME IMPORTANT CONCEPTS RELATED TO STATISTICAL DESIGN OF EXPERIMENTS Problem with First ExperimentThe twenty operators are very different in terms of their skills and experiences. Hence, a sizeable nuisance variation is introduced within the measurements comprising the mean wear for each material. This nuisance variation markedly increases the level of chance variation and hence may be hiding the presence of a real difference in materials.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

SOME IMPORTANT CONCEPTS RELATED TO STATISTICAL DESIGN OF EXPERIMENTS Solution to the ProblemRe-design the experiment: Give each operator a pair of tools with one tool of material A and one tool of material B. At the end of experiment measure the wear on each tool for each operator and calculate the difference in wear within each operator. Average these differences across all operators and perform a test of significance on this average difference.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

SOME IMPORTANT CONCEPTS RELATED TO STATISTICAL DESIGN OF EXPERIMENTS Solution to the Problem Operator-to-operator nuisance variation is in this way blocked from consideration since only relative differences within each operator-block are examined.

This is an example of the technique of Blocking a very important experimental design concept.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

RANDOMIZATION TO GUARANTEE VALIDITY OF TEST RESULTSUnknown sources of nuisance variation are always present but hard to identify trends from day to day. Systematic conduct of tests may violate basic statistical assumptions for significance tests, i.e.,Normality/Independence. Random sampling imply that the data are independently distributed about their respective means. Block for what you can identify, randomize for what you cannot.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

INTERACTION AMONG VARIABLESOften in experimental work it is assumed that factors influence the results independent of each other the effects are additive. Unfortunately, this is often not the case. More likely the way in which one factor influences the results depends upon the level of one or more other factors; that is, the factors are interdependent on or interact with each other.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

INTERACTION AMONG VARIABLESExample: Two factors: temperature and pressure are thought to affect chemical reaction time:

Mr. Xs Experiment

Mr. Ys ExperimentIE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

SUMMARYExperiments should be comparative. There should be genuine replication: replicates can provide an accurate measure of errors. Whenever appropriate, blocking (pairing) should be used to reduce error. Randomization is needed for homogeneity or independence. Experiments should be designed in such a way as to be able to determine the interaction effects of factors.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

OTHER ISSUES IN PLANNED EXPERIMENTATIONThe purpose of most experimental work is to discover the direction of changes that may lead to improvements in both quality and productivity. A fundamental task in designing an experiment is to select an appropriate arrangement of test points within the space defined by the independent variables and develop a mathematical model.

IE 400 Lecture 1


OTHER ISSUES IN PLANNED EXPERIMENTATIONFor example, if a quadratic relationship between two variables is suspected, an experiment that studies the process at only two levels of these variables will be inadequate. Similarly, an experiment using four levels would be unnecessary and inefficient if the true relationship were linear.IE 400 Lecture 1 @ 2006 Dr. Shiv G. Kapoor All Rights Reserved

Session 2

Review of Basic Statistical MethodsProf. Shiv G. Kapoor

Review of Basic Statistical Methods

Professor Shiv G. Kapoor Department of Mechanical Science and Engineering University of Illinois at Urbana-Champaign@2006 University of Illinois Board of Trustees. All Rights Reserved.




When we collect data, we are interested in finding out how the process is behaving in terms of an output quality characteristic(s), perhaps, in terms of the value of the characteristic on the average and the variation in individual measurements of that characteristic.University of Illinois at Urbana-Champaign 2006 by Dr. Shiv G. Kapoor


In general, we have three choices:1) Observe the process once and use that observation as an absolute reflection of the process behavior. 2) Observe all of the output of the process to get a true reflection of its behavior. 3) Observe part of the output of the process and use it to infer something about the true process behavior.University of Illinois at Urbana-Champaign 2006 by Dr. Shiv G. Kapoor

NATURE OF VARIABILITY IN DATA If we do 1) we are committing a potentially fatal error. If we do 2) we are said to be observing the entire population or universe. The term population, meaning all possible realizations of the process, suggests that all terms in the population constitute a very large number, perhaps in a more abstract sense, an infinite population size.University of Illinois at Urbana-Champaign 2006 by Dr. Shiv G. Kapoor

NATURE OF VARIABILITY IN DATA For any particular circumstance of practical significance, however, a population is generally of finite size though that size could be quite large. It is generally neither practical nor necessary to observe the entire population. Rather, we observe a relatively small subset of it (choice 3 above). When we do this we are said to be sampling the process (population).University of Illinois at Urbana-Champaign 2006 by Dr. Shiv G. Kapoor

NATURE OF VARIABILITY IN DATA How much we sample, when we take the sample, and how we specifically take the sample are critical issues? But the fundamental issue is this: We will draw the sample from the process and use the information contained in the samples to say something about the process. We need an efficient and adequate method to collect and analyze data.University of Illinois at Urbana-Champaign 2006 by Dr. Shiv G. Kapoor

CHARACTERIZATION OF DATA Characterization of Data Three Characteristics: 1) Central Tendency 2) Dispersion or Variability 3) Shape of Distribution of Frequencies



CHARACTERIZATION OF DATA A Measure of Central Tendency Given a sample of n pieces of data (X1, X2,, Xn) taken from a given population of size N, the arithmetic mean of the sample, denoted by is

X = ( Xi ) / ni =1

n

The true mean of the population of size N would be defined as .University of Illinois at Urbana-Champaign 2006 by Dr. Shiv G. Kapoor

CHARACTERIZATION OF DATA Measures of Dispersion or VariabilityThe Range denoted by R is the difference between the largest value and the smallest value of the data. In general, R=Xlargest-Xsmallest The variance is another measure of the variability in data. The sample variance, denoted by s2, is given n by: 2 2

s = [ ( X i X ) ] /( n 1 )i =1

The variance of the population is referred to as 2.University of Illinois at Urbana-Champaign 2006 by Dr. Shiv G. Kapoor

CHARACTERIZATION OF DATA-AN EXAMPLE It has been shown that the average can be used to describe the data characteristics. However, it is possible to find two sets of data to have equal averages, but their degrees of scatter are most likely different, and It has been a common mistake in many cases of applications to put all emphasis on the average but overlooking the scatter of the data.University of Illinois at Urbana-Champaign 2006 by Dr. Shiv G. Kapoor

CHARACTERIZATION OF DATA-AN EXAMPLE Such a mistake usually leads to unnecessary erroneous conclusions which could have been easily avoided if the scatter of the data had been considered. Suppose two different brands, A and B, of cutting tools are used to machine 20 workpieces, 10 with each brand. The surface finish readings taken on the 20 workpieces are shown in the Table 1.University of Illinois at Urbana-Champaign 2006 by Dr. Shiv G. Kapoor

CHARACTERIZATION OF DATA-AN EXAMPLE Table 1 Surface FinishXAI 70 72 71 70 69 70 71 70 68 69 XBI 70 70 70 69 69 70 70 71 71 70(XAi

X A )2

(X

Bi

X B )2

0 4 1 0 1 0 1 0 4 1

0 0 0 1 1 0 0 1 1 0



CHARACTERIZATION OF DATA-AN EXAMPLE The two averages X A and X B are calculated as:

XA = XB =

70 + 72 + .... + 69 700 = = 70 10 10 70 + 70 + ... + 70 700 = = 70 10 10



CHARACTERIZATION OF DATA-AN EXAMPLE Judging from these two averages, it seems that there is no difference in the surface finish reading in as much as which brand of the cutting tool is to be used. However, if we calculate the sample variances for the two samples, we get:SA =2

( X Ai X A )2i =1

10

10 1

= 1.33 , S B =2

( Xi=1

10

Bi

X B )2

10 1

= 0.44



CHARACTERIZATION OF DATA-AN EXAMPLE Hence, the two sets of surface finish data are characterized statistically as

X A = 70 microinches, X B = 70 microinches,S A = 1.15 microinches, S B = 0.66 microinches.



CHARACTERIZATION OF DATA-AN EXAMPLE It would appear that cutting tool B gives a more consistent-less variable-surface finish. Whether the difference we see here in these sample results is real/significant remains to be determined by a test of statistical significance.



Probability Density FunctionFor a continuous random variable, X, the probability behavior is described by a function called the probability density function (pdf), f(x). A pdf must satisfy the following:

f(x) 0, all x and

-

f ( x)dx = 1,

over the interval for X. The corresponding cumulative distribution function (cdf) of a continuous random variable with pdf f(x) is given byF(x) = P(X x) =

-

f(t)dt .

Note thatf(x)= d F(x) dt2006 by Dr. Shiv G. Kapoor


f(x)=

d

F(x)

Mathematical ExpectationThe expectation of a function g(x) of a random variable X is defined as

E(g(x)) =

-

g(x) f(x) dx

g(x) = x : the expectation of X, E(X), is the true arithmetic mean value of X , usually denoted by x. g(x) = (x-x)2: the expectation of (X-x)2 is the true variance of X, which is denoted by 2x. Some useful rules for Expectation: 1. E(cX) c E(X) 2. E(X+Y) = E(X) + E(Y) and E(X-Y) = E(X) E(Y) 3. Var(cX) = c2 Var(X), if c is a constant 4. Var(X+Y) = Var(X) + Var(Y), if X and Y are independent 5. Var(X-Y) = Var(X) + var(Y), if X and Y are independent.



Uniform or Rectangular DistributionThe pdf of a rectangular distribution is given by:

f(x) =

1 axb b-a

= 0 elsewhere The mean and variance of this distribution are: = E(x) =

-

x

1 (b + a ) dx = (b a ) 2 - 2

Var(x) = E (x- ) 2 = (x-) 2f(x)dx = E(x 2 ) [ E(x) ] = E(x 2 ) 2 = (b-a) 2 12



Example 1A wire cutting machine cuts wire to a specified length. Due to certain inaccuracies of the cutting mechanism, the length of the cut wire (in centimeters), say X, may be considered as a uniformly distributed random variable over (11.5, 12.5). The specified length is 12 centimeters. If the length of the cut wire is between 11.7 and 12.2 cm, the wire can be sold for a profit of $0.25. If the length of the cut wire is greater or equal to 12.2 centimeters, a profit of $0.10 is realized. However, if the length of the cut wire is less than 11.7 cm, the wire is discarded with a loss of $0.02. What is the expected profit per wire cut?University of Illinois at Urbana-Champaign 2006 by Dr. Shiv G. Kapoor

SolutionGiven:1 = 1.0 11.5 x 12.5 (12.5-11.5) f(x) = 0 elsewhere Also, let Profit, P be a random variable. It will take the values of f(x) = P = 0.25 11.7 x 12.2 P = 0.10 x 12.2 P = -0.02 x 0 H1 : < 0 H1 : 0


Stepwise Approach to Hypothesis Testing

A statistical hypothesis test consists of the following six steps: 1. State the null and alternative hypotheses. Define the test statistic used to analyze the situation. 2. Determine the significance level, a, at which the test will be made. 3. Collect the data and calculate the test statistic result. 4. Define the reference distribution for the test statistic. 5. Compare the test statistic and its reference distribution under H0. Carry out the necessary analysis of data. 6. Assess the risk.



ExampleTo test the newspaper claim that the mean wage rate of local foundry workers is $16 an hour, 25 foundry workers were randomly surveyed. It was found that the average wage rate for the sample of workers was $14.50. Historical data suggest that the wage rates follow the normal distribution and the standard deviation of wage rates is $3. Can the Union claim that the average wage is not $16 an hour? Assume =0.05.



SolutionStep 1: H 0 : = 16, H1 : 16. The sample statistic is X.Step 2: Given that = 0.05 .

Step 3: X = 14.50.Step 4 : The standard deviation of the normally distributed X is given by 3 x = x = = 0.6. n 25



SolutionStep 5: The critical values for z which place /2 (0.05/2 = 0.025) in each tail of the standard normal distribution with both tails of the distribution (we are concerned with both tails of the distribution, since this is a two-tailed test) are Z0.025 = -1.96 and z0.975 = 1.96. The calculated z value corresponding to the observed sample mean of 14.5 is then given by:

z=

(x- )

=

(14.50 16) = 2.50 0.602006 by Dr. Shiv G. Kapoor


INTRODUCTION TO THE CONFIDENCE INTERVAL CONCEPT Given a random sample of n observations from some process of interest and an estimate of the process mean, it is of interest to make some statement about the goodness of that sample mean, as an estimate of , i.e., the degree of belief or confidence that can be placed on it. One way of approaching this problem is through the concept of the confidence interval.University of Illinois at Urbana-Champaign 2006 by Dr. Shiv G. Kapoor

INTRODUCTION TO THE CONFIDENCE INTERVAL CONCEPT Recall that for random samples of size n drawn from a population, we expect that 95% of all sample means will be within an interval of 1.96 standard deviations of the distribution of the sample mean, i.e.,

1.96 x / nembraces 95% of all the sample means.University of Illinois at Urbana-Champaign 2006 by Dr. Shiv G. Kapoor

INTRODUCTION TO THE CONFIDENCE INTERVAL CONCEPT

x In other words, X 1.96 is called a 95% n confidence interval for the true mean . Graphically, we can show it as in Fig.2 and Fig.3A more general statement is that X z 1 / 2 is a 100(1-)% Confidence Interval for .University of Illinois at Urbana-Champaign 2006 by Dr. Shiv G. Kapoor

xn

INTRODUCTION TO THE CONFIDENCE INTERVAL CONCEPTFig.2 Graphical Representation of 95% Confidence Interval

1.96

xn

1.96

xn



INTRODUCTION TO THE CONFIDENCE INTERVAL CONCEPTFig.3 Graphical Representation of 95% Confidence Interval



INTRODUCTION TO THE CONFIDENCE INTERVAL CONCEPT When the sample size is small and x is unknown, the Confidence Interval is given by

X t v ,1 / 2

s n

where v is the degree of freedom and it equals n-1.



Example

Given that 9 bearings made by a certain process have an average diameter of 0.305 cm and the sample standard deviation of 0.003 cm, construct a 99 % confidence interval for the true mean diameter of bearings made by the process. What is the width of the confidence interval?



SolutionGiven that the sample size, n is small, the sample stat, follows a t-distribution.

X

Hence a 99% confidence interval for the true mean diameter is given by

s s ) X + t ( ) n n 0.003 0.003 0.305-3.355 0.305 + 3.355 9 9 0.30165 0.30835 The width of C. I. is X - t (

2 t

s = 2.(0.00335)=0.0067 . n2006 by Dr. Shiv G. Kapoor


Session 3

Comparative ExperimentsProf. Shiv G. Kapoor

Comparative Experiments

Professor Shiv G. Kapoor Department of Mechanical Science and Engineering University of Illinois at Urbana-Champaign

IE 400 Lecture 3

@2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Two Different Designs of Wood Cutting Saws- An Example Due to the environmental awareness and the enactment of Occupational Safety and Health Act (OSHA), the industrial noise is considered a significant factor in designing the machine and cutting tools. Wudkut Co. manufactures the wood cutting tools. One of their products, the woodcutting circular saw, has traditionally been a high noiseproducing tool.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES It is like a circular plate with cutting blades equally spaced around the periphery. The Director of R & D has known for some time that if the saw blades are perturbed in some fashion, the noise produced by the saw is considerably reduced. In order to confirm these results he developed an optimal perturbation design.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES

Objective:To compare the noise levels produced by two, otherwise identical, circular saws, one designed with equal blade spacing (the conventional design) and another with uneven spacing (the modified design).

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES It was decided to manufacture eight identical circular saws for each of the two designs and send a pair each to eight different customers for test and report. They are to record the noise levels in decibels, db. Furthermore, the decision as to which of the two circular saws should be run first has to be taken by every customer by flipping a coin (Randomization).IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES The results reported by the customers are given in Table 1. The numbers in the parentheses (1 or 2) indicate the testing sequence. Two customers reported that their machines broke down and hence they could not test the conventionally-designed saws. So, in all, there are 14 different results as reported in Table1.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSESTable 1 Noise Level From the Two SawsCustomer Numbers 1 2 3 4 5 6 7 8 TotalIE 400 Lecture 3

Noise from Noise from Conventionally Perturbed Blade Designed Saw, Saw, N m(db) N c(db) 85(2) 77(2) 78(1) 86(2) 80(1) 95(2) 86(1) 74(1) 76(2) 87(1) 77(2) 88(1) 83 78 644

Nc =Nm =

501 = 83.50 6649 = 81.13 8

501


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Strictly speaking, it is a comparison of the two means c and m. But since c, m, and 2 are not available, we would have to base the comparison on the sample statistics N c , N m, Sc2, and Sm2 are the averages, Sc2 and Sm2 are the sample variances.N c and NmIE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES In other words, we are faced with the task of making statistical decisions about the difference of the two means (c - m) on the basis of the sample statistics . We will use the confidence interval approach for the comparison of two means.

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES An Independent t-test analysisWith the 14 test data, a common practice would be to calculate the averages Nc and Nm , and compare these two averages. However, in view of the variation of the tests results, the two means, c and m, cannot justifiably be compared merely from these two statistics, (Nc Nm ) .IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES An Independent t-test analysisHence, for this purpose, we shall hereby introduce the independent t-test as a tool for the comparison of the two designs. Recall that the general expression of a t is

t=

statistics mean of statistics var( statistics)@2006 Dr. Shiv G. Kapoor All Rights Reserved

(1)

IE 400 Lecture 3

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES An Independent t-test analysisIn this problem, the statistics of interest is the observed difference of the two averages Nc and Nm , and the mean of this statistics is the difference between the two means, (c - m). Statistics= (Nc Nm) = 83.50-81.13=2.37, Mean of statistics=c-m (which is unknown) and the estimated variance of the statistics is V(statistics) = V ( N c N m )IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Thus equation (1) can be rewritten ast= ( N c N m ) ( c m ) V (Nc N m )

(2)

The immediate information that is required for equation (2) would be the estimated variance V ( N c N m ) .IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation of V(N c N m ) It can be shown theoretically that the variance of ( N c N m ) isV ( N c N m ) = V( N c ) + V( N m ) - 2 COV( N c N m ) where COV( N c N m ) is called the covariance of N c and N m . (3)

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation of V(N c N m) Assuming that the observations Nc do not affect the observations Nm, or vice versa, that is, Nc and Nm are independent, and equation (3) becomes

V ( N c N m ) = V( N c ) + V( N m )IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

(4)

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation of V(N c N m ) Estimating the variances V( N c ) and V( N m) by the sample variances , Sc2 and Sm2, we have S2 S2 (5) V ( N c N m ) = V( N c ) + V( N m ) = c + mnc nm

Assumption: The two data sets come from the same process (two normal distributions with equal variances, i.e. Sc2=Sm2=S2)IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation of V(N c N m ) The common (Pooled) variance is defined as:

Sp

2

( n 1 )S c + ( n m 1 )S m = c ( nc 1 ) + ( nm 1 )2

2

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Example of Calculation of V(N c N m)Using equation (3) and also referring to Table 1, the sample variances Sc2 and Sm2 are:

Sc =2

(Ni =18

6

ci

N c )2

nc 1mi

=

1 ( 225.5) = 45.10 6 11 ( 212.88) = 30.41 8 1@2006 Dr. Shiv G. Kapoor All Rights Reserved

Sm =2

(Ni =1

N m )2

nm 1

=

IE 400 Lecture 3

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Example of Calculation of V(N c N m )Therefore,

Sp =2

( 6 1)( 45 .1) + (8 1)( 30 .41) ( 6 1) + (8 1) 225 .50 + 212 .88 = = 36 .53 5+7@2006 Dr. Shiv G. Kapoor All Rights Reserved

IE 400 Lecture 3

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Example of Calculation of V(N c N m )Substituting the pooled sample variance Sp2 for Sc2 and Sm2 in equation (5), we obtain:V ( Nc Nm ) = ( 1 1 2 + ) Sp nc n m

1 1 = ( + ) (36.53) = 10.96 6 8V ( Nc Nm )IE 400 Lecture 3

= 3.31@2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalAfter the estimated variance (Nc Nm) has been obtained, the 95% confidence interval for (c - m) can be readily calculated as

(Nc Nm) t V(Nc Nm)IE 400 Lecture 3

(6)


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalA total of 14 tests, hence 14 degrees of freedom, is involved. So far we have used up 2 degrees of freedom in calculating the averages N c and N m as estimates of c and m. Therefore, 12 degrees of freedom are left.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalThe value of t associated with 12 degrees of freedom is t12, .025 = 2.179 Substituting the statistics( N c N m )= 2.37, V ( N c N m ) = 3.31, and t12, .025= 2.179 into equation (6), the 95% confidence interval for (c - m) can be obtained. It is [2.37(2.179)(3.31)]=[-4.84,9.58 ].IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalThis 95% confidence interval extends all the way from -4.84 to 9.58, implying that the true difference between the two designs may be as much as 9.58 db in favor of perturbation design or 4.84 db in favor of conventional design. One might well conjecture that there is really no difference between the performances of these two techniques.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalPutting this conjecture in the form of a question, we might ask, Are these data consistent with

the proposition that the true difference between the two designs is zero?One can get a good understanding of what he is entitled to believe on the basis of the data by seeing where this proposed zero value falls in the 95% confidence interval.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalIt is seen that the value zero falls well within the confidence interval. Hence, this value seems to be consistent with the data. That is, if someone claimed that there was no difference between the two designs, he would not be contradicted by the data.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalHowever, if someone claimed that the perturbed design produced noise 20 db less than what is produced by the conventional design, we could say that his claim does not appear to he supported by the data. This is because the value of 20 lies far outside the confidence interval.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

v

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalThe confidence interval contains, roughly speaking, all the plausible values of the true difference between the two designs. The result of the independent t-test indicates that there is no strong evidence either way in favor of the two designs.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES (An Example)Table 1 Noise Level From the Two SawsCustomer Numbers 1 2 3 4 5 6 7 8 TotalIE 400 Lecture 3

Noise from Noise from Conventionally Perturbed Blade Designed Saw, Saw, N m(db) N c(db) 85(2) 77(2) 78(1) 86(2) 80(1) 95(2) 86(1) 74(1) 76(2) 87(1) 77(2) 88(1) 83 78 644

Nc =Nm =

501 = 83.50 6649 = 81.13 8

501


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Strictly speaking, it is a comparison of the two means c and m. But since c, m, and 2 are not available, we would have to base the comparison on the sample statistics N c , N m, Sc2, and Sm2 are the averages, Sc2 and Sm2 are the sample variances.N c and NmIE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES In other words, we are faced with the task of making statistical decisions about the difference of the two means (c - m) on the basis of the sample statistics . We will use the confidence interval approach for the comparison of two means.

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES An Independent t-test analysisWith the 14 test data, a common practice would be to calculate the averages Nc and Nm , and compare these two averages. However, in view of the variation of the tests results, the two means, c and m, cannot justifiably be compared merely from these two statistics, (Nc Nm ) .IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES An Independent t-test analysisHence, for this purpose, we shall hereby introduce the independent t-test as a tool for the comparison of the two designs. Recall that the general expression of a t is

t=

statistics mean of statistics var( statistics)@2006 Dr. Shiv G. Kapoor All Rights Reserved

(1)

IE 400 Lecture 3

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES An Independent t-test analysisIn this problem, the statistics of interest is the observed difference of the two averages Nc and Nm , and the mean of this statistics is the difference between the two means, (c - m). Statistics= (Nc Nm) = 83.50-81.13=2.37, Mean of statistics=c-m (which is unknown) and the estimated variance of the statistics is V(statistics) = V ( N c N m )IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Thus equation (1) can be rewritten ast= ( N c N m ) ( c m ) V (Nc N m )

(2)

The immediate information that is required for equation (2) would be the estimated variance V ( N c N m ) .IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation of V(N c N m ) It can be shown theoretically that the variance of ( N c N m ) isV ( N c N m ) = V( N c ) + V( N m ) - 2 COV( N c N m ) where COV( N c N m ) is called the covariance of N c and N m . (3)

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation of V(N c N m) Assuming that the observations Nc do not affect the observations Nm, or vice versa, that is, Nc and Nm are independent, and equation (3) becomes:

V ( N c N m ) = V( N c ) + V( N m )IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

(4)

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation of V(N c N m ) Estimating the variances V( N c ) and V( N m) by the sample variances , Sc2 and Sm2, we have S2 S2 (5) V ( N c N m ) = V( N c ) + V( N m ) = c + mnc nm

Assumption: The two data sets come from the same process (two normal distributions with equal variances, i.e. Sc2=Sm2=S2)IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

Test for equal variancesFor independent random samples of size n1 and n2, from the two populations, the F-value for testing 12=22 is the ratioF=S12/s22, where S12 and S22 are the variances computed from the two samples. If the two populations are approximately normally distributed and the null hypothesis is true, the ratio F is a value of the F-distribution with 1 = n1-1 and 2 = n2-1 degrees of freedom.

For the two-sided test, the critical region isFF/2(1,2)

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Example of Calculation of V(N c N m)Using equation (3) and also referring to Table 1, the sample variances Sc2 and Sm2 are:

Sc =2

(Ni =18

6

ci

N c )2

nc 1mi

=

1 ( 225.5) = 45.10 6 11 ( 212.88) = 30.41 8 1@2006 Dr. Shiv G. Kapoor All Rights Reserved

Sm =2

(Ni =1

N m )2

nm 1

=

IE 400 Lecture 3

Test for equal variancesFor our problem: H0: 12=22 H1: 12 22 F = 45.10/30.41=1.48 = 0.10 Critical Region: From the F-dist Table, F0.05 (5,7) =3.97 and F0.95(5,7) =1/F0.05(7,5) =1/4.88=0.20. Since F is greater than 0.20 and less than 3.97, the Null hypothesis is not rejected. Hence 12=22.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation of V(N c N m ) The common (Pooled) variance is defined as:

Sp

2

( n 1 )S c + ( n m 1 )S m = c ( nc 1 ) + ( nm 1 )2

2

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Example of Calculation of V(N c N m )Therefore,

Sp =2

( 6 1)( 45 .1) + (8 1)( 30 .41) ( 6 1) + (8 1) 225 .50 + 212 .88 = = 36 .53 5+7@2006 Dr. Shiv G. Kapoor All Rights Reserved

IE 400 Lecture 3

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Example of Calculation of V(N c N m )Substituting the pooled sample variance Sp2 for Sc2 and Sm2 in equation (5), we obtain:V ( Nc Nm ) = ( 1 1 2 + ) Sp nc n m

1 1 = ( + ) (36.53) = 10.96 6 8V ( Nc Nm )IE 400 Lecture 3

= 3.31@2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalAfter the estimated variance (Nc Nm) has been obtained, the 95% confidence interval for (c - m) can be readily calculated as

(Nc Nm) t V(Nc Nm)IE 400 Lecture 3

(6)


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalA total of 14 tests, hence 14 degrees of freedom, is involved. So far we have used up 2 degrees of freedom in calculating the averages N c and N m as estimates of c and m. Therefore, 12 degrees of freedom are left.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalThe value of t associated with 12 degrees of freedom is t12, .025 = 2.179 Substituting the statistics ( N c N m )= 2.37, V ( N c N m ) = 3.31, and t12, .025= 2.179 into equation (6), the 95% confidence interval for (c - m) can be obtained. It is [2.37(2.179)(3.31)]=[-4.84,9.58 ].IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalThis 95% confidence interval extends all the way from -4.84 to 9.58, implying that the true difference between the two designs may be as much as 9.58 db in favor of perturbation design or 4.84 db in favor of conventional design. One might well conjecture that there is really no difference between the performances of these two techniques.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalPutting this conjecture in the form of a question, we might ask, Are these data consistent with

the proposition that the true difference between the two designs is zero?One can get a good understanding of what he is entitled to believe on the basis of the data by seeing where this proposed zero value falls in the 95% confidence interval.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalIt is seen that the value zero falls well within the confidence interval. Hence, this value seems to be consistent with the data. That is, if someone claimed that there was no difference between the two designs, he would not be contradicted by the data.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalHowever, if someone claimed that the perturbed design produced noise 20 db less than what is produced by the conventional design, we could say that his claim does not appear to he supported by the data. This is because the value of 20 lies far outside the confidence interval.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalThe confidence interval contains, roughly speaking, all the plausible values of the true difference between the two designs. The result of the independent t-test indicates that there is no strong evidence either way in favor of the two designs.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalThe Director of R&D felt, however, intuitively felt that the evidence in favor of perturbation design was stronger than that implied by the 95% confidence interval. One possible source of difficulty, he pointed out, was that the quality of wood and the types of machine tools varied considerably from customer to customer, and this variation was not taken into account in the above analysis.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence IntervalIn addition, he also pointed out that the data of perturbation design vary from a value of 74 db to 88 db, and the data of the conventional design vary from a value of 77 db to 95 db (refer to Table 1). He suspected (correctly!) that the variation from customer to customer was causing the sample variances Sc2 and Sm2 to be inflated, thus resulting in an inflated pooled sample variance Sp2.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Calculation and Interpretation of the 95% Confidence Interval From these observations, he conjectured that Sp2 was not a good estimate of the true variance S2 because Sp2 was estimating, in addition, the variance from customer to customer.

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES What is the Solution? A paired t-test analysis Perform a paired comparison of the two designs for every customer. The difference blocks out the customer-tocustomer effect on variability.

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Paired Design of Comparative Test Analysis It was decided to run 20 additional tests. Twenty identical circular saws for each of the two designs were send to ten different customers for test.

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Paired Design of Comparative Test Analysis The customers were requested to test the two saws. on the same machine, under exactly similar conditions of type and size of wood, cutting parameters, etc. Run tests in a random order. record the noise levels in decibels (db).IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSESTABLE 2 NOISE LEVEL FROM 20 ADDITIONAL TESTSCUSTO M ER NUMBER N O IS E F R O M C O N V E N T IO N A L L Y D E S IG N E D S A W C (D B ) N O IS E F R O M P E R T U R B A T IO N D E S IG N E D S A W M (D B ) d = c - M

1 2 3 4 5 6 7 8 9 10

8 8 (2 ) 8 0 (1 ) 9 3 (1 ) 8 1 (2 ) 7 8 (2 ) 8 3 (1 ) 9 1 (2 ) 8 4 (1 ) 9 0 (2 ) 8 0 (2 )

8 4 (1 ) 7 6 (2 ) 9 0 (2 ) 7 8 (1 ) 7 9 (1 ) 8 3 (2 ) 8 6 (1 ) 7 9 (2 ) 8 5 (1 ) 7 7 (1 )

4 4 3 3 -1 0 5 5 5 3

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Paired Design of Comparative Test Analysis The average difference, d , is therefored =

10

i=1

di

10

=

31 = 3 .1 10

The 95% confidence interval for the difference is given as

d t V(d )

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Paired Design of Comparative Test AnalysisThe 95% confidence interval therefore depends upon three statistics:

d

V (d )t value.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Paired Design of Comparative Test Analysis

d:Table 2 gives the ten differences d1, d2, ...., d10. The average difference is: d = 3.10

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Paired Design of Comparative Test AnalysisV (d ) : d In order to find the sample variance of, ,first we have to find the sample variance of a single difference V(d). The sample variance of d is

10 2 (d d ) i 40.12 i =1 V (d ) = = = 4.458 10 1 9IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES Paired Design of Comparative Test AnalysisThe sample variance d of V (d ) is given by

V (d ) =

V (d ) 4.458 = = 0.4458 n 10

IE 400 Lecture 3


COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSES t value:Here, one degree of freedom has already been used up in calculating the mean, and the value of t associated with nine degrees of freedom and corresponding to a 95% confidence level is t9,.025 = 2. 262. The 95% confidence interval is, from equation (6),

3.1 2.262 0.4458 = [1.6,4.6]Here the confidence interval does not include the point zero.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

COMPARISON OF TWO TECHNIQUES/DESIGNS/PROCESSESThis test indicates that the true difference between the performances of the two different design saws is probably somewhere between 1.6 and 4.6 db. The new design might be as much as 4.6 db or as little as the 1.6 db better than the conventional design. The director may now claim with 95% confidence, that the perturbed (modified) design is in fact better than the conventional design in the sense that the saws made according to modified design produce less noise.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

Blocking

One important lesson to learn from this example is the importance of setting up experiments so that extraneous variation can be eliminated from the results. The extraneous variation in the example was the customer to customer fluctuation in the wood and the types of machine tools. By running tests in pairs, this customer to customer variation was eliminated because direct comparison between the perturbation and conventional design could be made for the same customer. This is called blocking.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

Randomization in Paired and Independent Comparisons

We have seen how on the assumption that a sample of n observations may be treated as if they had been independently drawn from a Normal population with constant mean and constant variance, simple inferential procedures can be developed. Unfortunately, in the real world, these assumptions are usually not exactly justifiable. In fact the observations could be correlated in the sequence, for example, and unexpected changes in both mean and variance of their distribution might occur.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

Randomization in Paired and Independent Comparisons

It was pointed out by Sir Ronald Fisher that it is possible to take a precaution in the actual carrying out an experiment, which guarantees the approximate validity of comparative procedures. This precaution is randomization. In the wood cutting saw example, recall that a coin was flipped by each customer to decide the order of the usage of the saws. This represents a process of randomization and was done for one important reason - to alleviate the unavoidable influences of both the known and unknown variables on the experimental results.IE 400 Lecture 3 @2006 Dr. Shiv G. Kapoor All Rights Reserved

Session 4

Factorial DesignsProf. Shiv G. Kapoor

FACTORIAL DESIGNS (FD)

Professor Shiv G. Kapoor Department of Mechanical Science and Engineering University of Illinois at Urbana-Champaign

IE 400 Lecture 10

2006 Dr. Shiv G. Kapoor All Rights Reserved

Factorial DesignsWhen we wish to compare two different techniques, we may either use the independent t-test, or we may use a paired t-test if a nuisance variable is to be blocked away.

IE 400 Lecture 10


Factorial DesignsSuppose now that we want to compare several techniques, the independent t-tests would no longer be adequate and a k-variable analysis should be used. There are indeed many experimental designs available which may be chosen to suit particular experimental situations.

IE 400 Lecture 10


Factorial DesignsBut these designs, e.g., k-variable analysis, involve certain assumptions and restrictions. We will now introduce a general and effective class of experimental design called the Factorial Design which includes k-variable analysis, etc. as a special case.

IE 400 Lecture 10


Factorial DesignsTo develop a general factorial design, one would select a fixed number of levels for each of a number of variables (factors) and then design tests with all possible combinations. For example, if there are L1 levels for variable 1 , L2 levels for variable 2 .., and Lk levels for variable k, then L1 x L2 x L3 x x Lk is called a factorial design for k variables.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

Factorial DesignsFactorial designs are preferred over other experimental designs because :They require relatively few runs per variable (factor) even though they do not explore the entire region of interest They deal easily with variable interactions They can indicate major trends and help determine a possible direction for further experimentation They can be augmented to form composite designs They form the basis for fractional factorial designs The interpretation of the data produced by the designs can be easily analyzed.

IE 400 Lecture 10


Factorial DesignsVariables vs LevelsA design with 3 variables at 2 levels would require 8 tests for a 2 x 2 x 2 = 23 factorial design A design with 3 variables at 3 levels would need 27 tests for a 3 x 3 x 3 = 33 factorial design. A design with 5 variables at 2 levels would need 32 tests for a 25 factorial design.

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Background of the Study--Welding of High Strength Steel Bars High carbon steel, because of its high strength and low cost, has been known to have a potential for a "good market". However, because of its high carbon content, it is not easy to weld.

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Background of the Study--Welding of High Strength Steel BarsAccording to the code of the American Welding Society (AWS), additional steps of pre-heating and post-heating are required in order to have good quality welds and high strength steel. A user of this steel was interested to study whether or not these additional steps of pre-heating and post-heating were really needed.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Background of the Study--Welding of High Strength Steel BarsAfter a preliminary investigation by manual arc welding tests, it appeared that there were three variables significantly affecting the ultimate tensile stress of a weld. These three variables were: ambient temperature, wind velocity bar size.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Background of the Study--Welding of High Strength Steel Bars The evidence, however, was not decisive, and further experiments were therefore planned.

How many tests should be conducted?IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Based upon available funds and time limit, it was decided that 244 tests be run. In the meantime, an engineering statistician,who was called upon for consultation, suggested that 16 tests be run according to his specified sets of test conditions. These 16 tests formulated a specific statistical experimental design which we shall further discuss.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Experimental Design - 23 Factorial Design The statistical experimental design that was formulated by the engineering statistician for this study was a two-level three-variable factorial design, simply designated as a 23 factorial design. The three selected variables were:ambient temperature, denoted by T wind velocity, denoted by V, bar size, denoted by B.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Experimental Design - 23 Factorial Design Two levels were chosen for each variable based upon desired field conditions to be simulated.One level is called the high level and the other the low level. The high and low levels of the three variables are listed in Table 1.

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN

Table 1: The High and Low levels of the 23 Factorial DesignVariable Ambient Temperature (T) Wind Velocity (V) Bar Size (B) Unit F mph 1/8 inch Low 0 0 4 High 70 20 11

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Experimental Design - 23 Factorial DesignIt is seen that the simulated field conditions would consist of a low of zero degrees Fahrenheit ambient temperature and a high of 70 degrees Fahrenheit. Similarly, the low wind velocity was zero miles per hour and the high wind velocity was 20 mph. Two different bar sizes were used,the smaller one was 4/8 inch which is designated as the low level the larger one was 11/8 inches, designated as the high level.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Transforming EquationsIn order to adopt a notation which will be the same for all two-level factorial designs, we use transforming equations to code the variables such that the high level will be denoted by +1, the low level will be denoted by -1. By so doing, regardless of the physical conditions represented by the two levels, the basic design of any two-level factorial design becomes a simple arrangement of +1 and -1.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN

Transforming Equations For example: The transforming temperature (T) is equation for ambient

X1 =

(T 35) 352006 Dr. Shiv G. Kapoor All Rights Reserved

IE 400 Lecture 10

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Transforming EquationsIn order to code the high and low levels of the ambient temperature into +1 and -1, we simply substitute the two levels, 00F and 700F, of the ambient temperature into the above transforming equation. For the low level, X1 = For the high level, X1 =IE 400 Lecture 10

(0 35) = 1 35

(70 35) = +1 352006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Construction of the 23 Factorial DesignA complete 23 factorial design requires 23 = 8 tests. One systematic way of writing down the eight test conditions in their coded forms is to proceed as follows:For X1, write down the values -1,+1,-1,+1,-1,+1,-1,+1 in a column. The signs alternate each time. For X2, write down the values -1,-1,+1,+1,-1,-1,+1,+1 in a column. The signs alternate in pairs. For X3, write down the values -1,-1,-1,-1,+1,+1,+1,+1 in a column. The signs alternate in groups of four.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Construction of the 23 Factorial DesignIn general, there are 2k sets of test conditions for a 2k factorial design. We can obtain the 2k sets of coded test conditions by writing down columns as follows:For X1, write down a 2k number of -1, +1, -1, +1,.... The signs alternate each time (i.e., 20 = 1, one alternation every time). For X2, write down a 2k number of -1, -1, +1, +1,....The signs alternate in pairs (i.e., 21 = 2, alternate in pairs).IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Construction of the 23 Factorial DesignFor X3, write down a 2k number of -1, -1, -1, -1, +1, +1, +1, +1,.... The signs alternate in groups of four (i.e., 22 = 4, alternate in groups of four). For X4, the signs alternate in groups of eight (23 = 8). Proceed in a similar way for X5, X6, ..., Xk. For Xk, write down a 2k-1 number of 1s, followed by a 2k-1 number of +1's. This method will yield all the 2k distinct sets of coded test conditions without repetitions.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGNTable 2 : The 23 Factorial Design for the High Strength Steel Bar ProblemCoded Test Conditions Test # 1 2 3 4 5 6 7 8 X1 -1 1 -1 1 -1 1 -1 1 X2 -1 -1 1 1 -1 -1 1 1 X3 -1 -1 -1 -1 1 1 1 1 Actual Test Conditions F 0 70 0 70 0 70 0 70 (mph) 0 0 20 20 0 0 20 20 (1/8 in) 4 4 4 4 11 11 11 11

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGNTable 3 Results of the Sixteen Welding Experiments Test Y ai Test Y bi Average Test # X1 X2 X3 Order (kpsi) Order (kpsi) (kpsi)

1 2 3 4 5 6 7 8

-1 1 -1 1 -1 1 -1 1

-1 -1 1 1 -1 -1 1 1

-1 -1 -1 -1 1 1 1 1

6 8 1 2 5 3 4 7

84 90.6 69.6 76 77.7 99.7 82.7 93.7

3 7 5 4 8 1 2 6

91 84 86 98 80.5 95.5 74.5 81.7

87.5 87.3 77.8 87 79.1 97.6 78.6 87.7

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Average Effects of Ambient TemperatureAverage Effect of Ambient Temperature, E1Observe that for Test Nos. 1 and 2 in Table 3, the conditions of wind velocity (X2) and bar size (X3) are the same but the temperature (X1) conditions are different. A high temperature was used for Test No. 2 and a low temperature was used for Test No. 1. Therefore, the difference in these two test results, apart from the intrinsic variation that is present, can be attributed solely to the effect of ambient temperature alone.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Average Effects of Ambient Temperature Similarly, for the pairs of Test # 3 and 4, 5 and 6, and 7 and 8 in Table 3; each pair involved similar test conditions with respect to wind velocity and bar size, but different test conditions with respect to ambient temperature. Thus, the differences in the results within each of these four pairs reflect the effect of ambient temperature alone.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Average Effects of Ambient TemperatureThe differences in the test conditions are:(Test Nos . 1 and 2 ) y 2 y 1 = 87 .3 87 .5 = 0 .2 (Test Nos . 3 and 4 ) y 4 y 3 = 87 .0 77 .8 = 9 .2 (Test Nos . 5 and 6 ) y 6 y 5 = 97 .6 79 .1 = 18 .5 (Test Nos . 7 and 8 ) y 8 y 7 = 87 .7 78 .6 = 9 .1 .IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Average Effects of Ambient TemperatureThe average effect of ambient temperature, designated by E1 is by definition the average of the above four differences and is given as :E1 = 1 / 4[( y2 y1 ) + ( y 4 y3 ) + ( y6 y5 ) + ( y8 y7 )] = 1/4[-0.2 + 9.2 + 18.5 + 9.1] = 9.15 units of 1000 psi = 9150 psi.

Note that the average effect is commonly referred to as main effect.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Average Effects of Ambient Temperature Geometrically, the average effect of ambient temperature, E1, is the difference between the average result in Plane II (high level of ambient temperature) and the average result on Plane I (low level of ambient temperature) as shown in Figure 1 .

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Figure 1. A Geometrical Interpretation of Average Effect of Ambient Temperature

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Average Effects of Ambient Temperature This can be seen by rearranging the average effect equation of E1 as 1 E1 = [(y2 + y4 + y6 + y8 ) (y1 + y3 + y5 + y 7 )] 4 high level of ambient Low level of ambient temperature temperatureIE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Average Effects of Ambient TemperatureThe average effect of ambient temperature tells us that, on the average, over the ranges of the variables in this investigation, the effect of changing the ambient temperature from its low level is an increase of ultimate tensile stress by 9150 psi. But, notice that the individual differences ( -200 psi, 9200 psi, 18,000 psi, and 9100 psi) are actually quite erratic.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Average Effects of Ambient Temperature The average effect, therefore, must be interpreted in conjunction with the intrinsic variabilities that are present in the experimental results.

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Average Effects of Wind Velocity Consider Test No. 1 and 3 in Table 3. For thesetwo tests,the ambient temperature (X1) and bar size (X3) are constant, but for Test #3 the wind velocity is at the high level.

The difference in the two test results, apart from the intrinsic variation that is present, can be attributed to the effect of wind velocity alone.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Average Effects of Wind Velocity Similarly, for Test No. 2 and 4, 5 and 7, and 6 and 8: the ambient temperature and bar size are constant within each pair but the wind velocity is at different levels.

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Average Effects of Wind Velocity Therefore, the average effect of wind velocity, E2, can be obtained by taking the average of the four individual differences, which are in units of 1000 psi:(Test Nos. 1 and 3) y3 y1 = 77.8 87.5 = 9.7 (Test Nos. 2 and 4) y 4 y 2 = 87.0 87.3 = 0.3 (Test Nos. 5 and 7) y 7 y 5 = 78.6 79.1 = 0.5 (Test Nos. 6 and 8) y8 y6 = 87.7 97.6 = 9.9.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Average Effects of Wind Velocity And the average effect is 1 E2 = [( y3 y1 ) + (y 4 y2 ) + (y7 y5 ) + (y8 y6 )] 4 1 = (-9.7 - 0.3- 0.5 - 9.9) 4 = -5.1 ( units of 1000 psi) = -5100.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Average Effects of Wind Velocity Geometrically, the average effect of wind velocity is the difference between the average result on Plane IV ( high level of wind velocity) and the average result on Plane III (low level of wind velocity) as shown in Figure 2.

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGNFigure 2. A Geometrical Interpretation of Average Effect of Wind Velocity

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Average Effects of Wind Velocity The average effect of wind velocity is 5100 psi which tells us that on the average, the ultimate tensile stress decreased by 5100 psi when the wind velocity was changed from 0 mph to 20 mph.

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Average Effects of Bar SizeSimilarly, the average effect of bar size, E3, is the corresponding comparison between Plane VI ( high level of bar size) and Plane V ( low level of bar size) which are indicated in Figure 3. This effect is calculated as follows:E3 = 1 [( y 5 y 1 ) + ( y 6 y 2 ) + ( y 7 y 3 ) + ( y 8 y 4 ) ] 4 1 = (-8.4 + 10.3 + 0.8 + 0.7) 4 = 0.85 ( units of 1000 psi) = 850 psi.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGNFigure 3. A Geometrical Interpretation of Average Effect of Bar Size

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Two-Variable InteractionsThe average effects E1, E2, and E3 represent the individual effects of ambient temperature, wind velocity, and bar size on the ultimate tensile stress. What about the joint effect of two variables, say,ambient temperature and wind velocity on the ultimate tensile stress? or wind velocity and bar size on the ultimate tensile stress?

These joint effects are two-variable interactions.IE 400 Lecture 10

indicated

by

the


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Two-Variable Interactions Physically, what is a two-variable interaction? Let us consider the hypothetical set of data given in Figure 4. The numbers located at each of the four corners represent the hypothetical test results which may be observed at the four sets of test conditions given by the four corners.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGNFigure 4 Hypothetical Results Illustrating the Absence of a Two-variable Interaction(high) 110 120

Wind Velocity

90 (low)IE 400 Lecture 10

100 Ambient Temperature (high)


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Two-Variable Interactions The difference in the test results due to a change of ambient temperature performed at the low level of wind velocity is 100 - 90 =10. At the high level of wind velocity, the difference is 120 - 110 = 10.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Two-Variable Interactions These two differences, which are due to the change in ambient temperature at different levels of wind velocity, are identical. We say, in this case, that there is no interaction between ambient temperature and wind velocity.

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Two-Variable Interactions In other words if the effect of changing ambient temperature is the same at both levels of wind velocity (or, if the effect of changing the wind velocity is the same at both ambient temperature levels), there is no interaction between ambient temperature and wind velocity. In a sense, ambient temperature and wind velocity act independently of one another.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGNTwo-Variable Interactions On the other hand, consider the results given in the following Figure 5.(high ) 110 140

Wind Velocit y 90 (low)IE 400 Lecture 10

Ambient Temperature

100 (high )


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGNTwo-Variable Interactions Here, for the low level of wind velocity the difference in the results due to change in level of ambient temperature is 100-90=10. But at the high level of wind velocity, the difference in the results due to change in level of ambient temperature is 140-110=30. Thus, the effect of changing ambient temperature is not the same at the high and low levels of wind velocity.

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Two-Variable Interactions Likewise, the difference in the results at the low level of ambient temperature due to the change in wind velocity level is 110 - 90 - 20, and the difference in the results at the high level of ambient temperature is 140-100=40. That is, the effect of changing the level of wind velocity is not the same at both levels of ambient temperature.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Two-Variable Interactions In other words, the effect of ambient temperature depends on the level of wind velocity, they do not act independently of one another.

We say, in this case, that an interaction exists between ambient temperature and wind velocity.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Calculation of Two-Variable Interactions There are three two-variable interactions to be calculated, namely:between ambient temperature and wind velocity, denoted by E12, between ambient temperature and bar size, denoted by, E13 between wind velocity and bar size, denoted by E23.

The calculations of the three two-variable interactions will be illustrated in the same order as is given above.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Interaction between Ambient Temperature and Wind Velocity, E12In order to calculate the interaction between ambient temperature and wind velocity, let us go back to Table 3. At the high level of wind velocity, the two differences of results given by (Y 4 Y3 ) and (Y8 Y 7 ) , both reflect individual effect of the change in ultimate tensile stress that could occur due to a change of ambient temperature from the low level to the high level.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGNTable 3 Results of the Sixteen Welding Experiments Test Y ai Test Y bi Average Test # X1 X2 X3 Order (kpsi) Order (kpsi) (kpsi)

1 2 3 4 5 6 7 8

-1 1 -1 1 -1 1 -1 1

-1 -1 1 1 -1 -1 1 1

-1 -1 -1 -1 1 1 1 1

6 8 1 2 5 3 4 7

84 90.6 69.6 76 77.7 99.7 82.7 93.7

3 7 5 4 8 1 2 6

91 84 86 98 80.5 95.5 74.5 81.7

87.5 87.3 77.8 87 79.1 97.6 78.6 87.7

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Interaction between Ambient Temperature and Wind Velocity, E12The average of these two differences, at the high level of wind velocity is 1 [(Y Y 3 ) + (Y 8 Y 7 )] 2 4 Similarly at the low level of wind velocity, the average of the two differences in results due to the sole effect of a change in ambient temperature level is 1 [(Y 2 Y1 ) + (Y 6 Y 5 )]. 2IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Interaction between Ambient Temperature and Wind Velocity, E12 The interaction between ambient temperature and wind velocity is given by the average difference of these two averages, that is,1 1 1 E12 = { [(Y2 Y1 ) + (Y6 Y5 )] [(Y2 Y1) + (Y6 Y5 )]}. 2 2 2 1 = {(Y1 + Y5 + Y4 +Y8 ) (Y2 + Y6 +Y3 + Y7 )} 4IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Interaction between Ambient Temperature and Wind Velocity, E12 Note, therefore, that the interaction between ambient temperature and wind velocity tells us the average change in ultimate tensile stress that would occur due to a change from the low level to the high level in both the ambient temperature and wind velocity.

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Interaction between Ambient Temperature and Wind Velocity, E12 The two-variable interaction between ambient temperature and wind velocity is: E12 = (1/4[(87.5+79.1+87.0+87.7)-(87.3+97.6+ 77.8+78.6)]) = 0.0 psiThat is, no interaction exists between ambient temperature and wind velocity.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Interaction Between Ambient Temperature and Bar Size, E13Similarly, the interaction between temperature bar size is

E13 =

1 1 (Y1 + Y3 + Y6 + Y8 ) (Y2 + Y4 + Y5 + Y7 ) 4 4 = 4650 psi

IE 400 Lecture 10


AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Interaction Between Wind Velocity and Bar Size, E23 Likewise, the interaction between wind velocity and bar size isE23 = 1 1 (Y1 + Y2 + Y7 + Y8 ) (Y 3 + Y4 + Y5 + Y6 ) 4 4 1 1 = (87.5 + 87.3 + 78.6 + 87. 7) (77.8 + 87.0 + 79.1+ 97.6) 4 4 = -0.10 ( units of 1000 psi) = -100 psi.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

AN EXPERIMENTAL STUDY USING A 23 FACTORIAL DESIGN Three-Variable-Interaction, E123Just as a two-variable interaction is a measure of the joint effect of twovariables on a response, a threevariable interaction is indicative of the joint effect of three-variables on a response. The procedure of estimating these effects is similar to estimating the second order effects. However, a simplified procedure that has been developed to estimate all these effects will now be discussed.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

A SIMPLIFIED METHOD FOR MAIN AND INTERACTION EFFECTS

The purpose of going through the average effects and interactions in a rather detailed and descriptive manner, such as above using the geometrical representation of the design, is toprovide a basic understanding better appreciation of the meaning of these effects.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

A SIMPLIFIED METHOD FOR AVERAGE/MAIN AND INTERACTION EFFECTS However, the usefulness of this approach for calculation purposes is somewhat limited since extension to more than three variables is cumbersome. A simplified calculation procedure, which is easily extended for analyzing two-level factorial designs in any number of variables, is therefore needed and is now described.IE 400 Lecture 10 2006 Dr. Shiv G. Kapoor All Rights Reserved

A SIMPLIFIED METHOD FOR AVERAGE/MAIN AND INTERACTION EFFECTS Let us refer again to the design matrix for the High Strength Steel Bar Example:Table 1 Design Matrix

Test 1 2 3 4 5 6 7 8IE 400 Lecture 10

X1 -1 1 -1 1 -1 1 -1 1

X2 -1 -1 1 1 -1 -1 1 1

X3 -1 -1 -1 -1 1 1 1 1

y87.5 87.3 77.8 87 79.1 97.6 78.6 87.7


A SIMPLIFIED METHOD FOR AVERAGE/MAIN AND INTERACTION EFFECTS We now notice that if we are to apply (multiply) the column of 1 values associated with X1 to the data ( y ) (refer to Table 1) and then sum the result and divide the sum by N/2 = 8/2 = 4, we would

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