2. experiments with one factor (ch.3. analysis of variance...

Hae-Jin Choi School of Mechanical Engineering,

Chung-Ang University

2. Experiments with One Factor

(Ch.3. Analysis of Variance ANOVA)

1 DOE and Optimization

Experiments with One Factor

2 DOE and Optimization

Plasma etching process:

3


An engineer is interested in investigating the relationship between the RF power setting and the etch rate for this tool. The objective of an experiment like this is to model the relationship between etch rate and RF power, and to specify the power setting that will give a desired target etch rate.

The response variable is etch rate.

She is interested in a particular gas (C2F6) and gap (0.80 cm), and wants to test four levels of RF power: 160W, 180W, 200W, and 220W. She decided to test five wafers at each level of RF power.

The experimenter chooses 4 levels of RF power 160W, 180W, 200W, and 220W

The experiment is replicated 5 times runs made in random order

DOE and Optimization


4


5

Does changing the power change the

mean etch rate?

We would like to have a systematic way

to answer these questions

Analysis of Variance (ANOVA)


The Analysis of Variance (ANOVA)

In general, there will be a levels of the factor, or a treatments, and n replicates of the experiment, run in random ordera completely randomized design (CRD)

N = an total runs

We consider the fixed effects casethe random effects case will be discussed later

Objective is to test hypotheses about the equality of the a treatment means

DOE and Optimization 6

Statistical Hypothesis


A statistical hypothesis is a statement about the parameters of a

probability distribution. For example, we may think that the mean

values of distributions are equal. This may be stated formally as

(Null hypothesis)

(Alternative hypothesis)

Type I error the null hypothesis is rejected when it is true.

Type II error the null hypothesis is accepted when it is false

0 1 2

1

:

: At least one mean is different

aH

H

Basic Single-factor ANOVA model


the total of the observations for the ith factor level

the average of the observations under the ith factor level

the grand total of all observations

the grand average of all observation

1

= 0a

i

i

. .

1

. = / 1,2, ,n

i ij i i

j

y y y y n i a

.. .. ..

1 1

= /a n

ij

i j

y y y y N

.iy

.iy

..y

..y



.

ij i ijy

i = 1 i = 2 i = 3 i = 4

10


1,2,...,,

1, 2,...,

a parameter common to all factors called an overall mean,

the deviation at the th factor level from the overall mean,

called the th factor effe

ij i ij

i

i ay

j n

i

i

2

ct,

random error, (0, )ij NID

Typically, resulting from measurement error, the effects of variables not

included in the experiment, and so on. (1) normally and independently distributed random variables (2) zero mean value (3) Variance ( ) is assumed constant for all levels of the factor. 2


11

The Analysis of Variance

Total variability is measured by the total sum of

squares:

The basic ANOVA partitioning is:

2

..

1 1

( )a n

T ij

i j

SS y y

2 2

.. . .. .

1 1 1 1

2 2

. .. .

1 1 1

( ) [( ) ( )]

( ) ( )

a n a n

ij i ij i

i j i j

a a n

i ij i

i i j

T Treatments E

y y y y y y

n y y y y

SS SS SS

2..2

1 1

= - a n

ijT

i j

ySS y

anor



12

SSTreatments is the sum of squares due to the factor, that is the sum of squares of differences between factor-level averages and the grand average. It measurers the differences between factor levels

A large value of SSTreatments reflects large differences in treatment means

A small value of SSTreatments likely indicates no differences in treatment means

2

. ..

1

( )a

Treatments i

i

SS n y y2 2

. ..

1

= - a

i

Treatments

i

y ySS

n anor



13

SSE is the sum of squares due to error, that is the sum of square of difference of observations within a specific factor level and the factor-level average. It measures the random error.

If SSTreatments is large, it is due to differences among the means at the

different factor levels.

Thus, by comparing the magnitude of SSTreatments to SSE we can see how

much variability is due to changing factor levels and how much is due to

error.

This comparison is facilitated if we first divide this sums by their number

of degrees of freedom.

or 2

.

1 1

( )a n

E ij i

i j

SS y y = - E T TreatmentsSS SS SS



14

While sums of squares cannot be directly compared to test the hypothesis of equal means, mean squares can be compared.

A mean square is a sum of squares divided by its degrees of freedom:

If the treatment means are equal, the treatment and error mean squares will be (theoretically) equal.

If treatment means differ, the treatment mean square will be larger than the error mean square.

1 1 ( 1)

,1 ( 1)

Total Treatments Error

Treatments ETreatments E

df df df

an a a n

SS SSMS MS

a a n


ANOVA Table

15

The ratio between MS Treatments and MSE follows a F, distribution,

which is completely determined by u, the numerator degrees of

freedom, and v, the denominator degrees of freedom .

In analysis of variance, u is equal to (a-1), the degrees of freedom

of MS Treatments and v is equal to a(n-1), the degrees of freedom of MSE.

The test statistic is

If Fo > F a-1,a(n-1) , we may conclude that the factor-level means are

different

= TreatmentsoE

MSF

MS


16

ANOVA Table

The reference distribution for F0 is the Fa-1, a(n-1) distribution

Reject the null hypothesis (equal treatment means) if

P-value is the probability of Type I error i.e., the probability of rejecting the null hypothesis even if it is actually TRUE

0 , 1, ( 1)a a nF F


17



18



19

Why Does the ANOVA Work?

2 2

1 0 ( 1)2 2

0

We are sampling from normal populations, so

if is true, and

Cochran's theorem gives the independence of

these two chi-square random variables

/ (So

Treatments Ea a n

Treatments

SS SSH

SSF

2

11, ( 1)2

( 1)

2

2 21

1) / ( 1)

/ [ ( 1)] / [ ( 1)]

Finally, ( ) and ( )1

Therefore an upper-tail test is appropriate.

aa a n

E a n

n

i

iTreatments E

a aF

SS a n a n

n

E MS E MSa

F


Residual Analysis


We define a residual as the difference between the actual

observation and the factor-level mean . Therefore, the residual is

The difference between an observation and the corresponding

factor-level mean.

The normality assumption can be checked by plotting the residuals

on normal probability paper.

Residual Plot

Normal Probability Plot

. = - ij ij ie y y

Residual Analysis


Plasma etching process

Residual Plot



Residual Plot


To check the assumption of zero mean and equal variances at each

factor level, plot the residuals against the factor levels and compare

the spread in the residuals.

The independent assumption can be checked by plotting the

residuals against the run order in which the experiment was

performed.

A pattern in this plot, such as sequences of positive and negative

residuals, may indicate that the observations are not independent.



A normal probability paper is a graph of the cumulative normal

distribution of the residuals, that is, graph paper with the ordinate

scaled so that the cumulative normal distribution plots as a straight

line.

To construct a normal probability plot, arrange the residuals in

increasing order and plot the kth of these ordered residuals versus

the cumulative probability point,

on normal probability paper. If the underlying error distribution is

normal, this plot will resemble a straight line

= ( - 0.5) /kP k N

Practical Interpretation of Results


Regression Model

Contrasts

Regression Model



Contrasts


We might suspect at the outset of the experiment that 200W and

220W produce the same etch rate, implying that we would like to

test the hypothesis:

This hypothesis could be tested by investigating an appropriate

linear combination of factor level totals say

3 4

1 3 4

: =

: =

oH

H

3. 4. - = 0y y

3 4 - = 0

Or

Contrasts


If we have suspected that the average of 160 W and 180W did not

differ from the average of 200W and 220W, then the hypothesis,

would have been

which implies the linear combination

1 2 3 4

1 1 2 3 4

: + = +

: + = +

oH

H

1. 2. 3. 4. + - y - y = 0y y

1 2 3 4 + - - = 0

Or

Contrasts


In general, the comparison of factor level means of interest will

imply a linear combination of factor level totals such as

with the restriction that Such linear combinations are

called contrasts. The sum of squares of any contract is

.

1

= ya

i i

i

C c

1

= 0.a

i

i

c

2

.

1

2

1

c y

= 1

a

i i

i

c a

i

i

SS

cn

Contrasts


Plasma etching problem

Hypothesis Contrast

1 2

1 2 3 4

3 4

: =

: + +

: =

o

o

o

H

H

H

1 1. 2.

2 1. 2. 3. 4.

3 3. 4.

+

C y y

C y y y y

C y y

Contrasts



1

2

1

2

2

2

1(551.2) 1(587.4) 36.2

36.23276.10

1(2)

5

1(551.2) 1(587.4) 1(625.4) 1(707.0) 193.8

( 193.8)46,948.05

1(4)

5

C

C

C

SS

C

SS

3

3

2

1(625.4) 1(707.6) 81.6

( 81.6)16,646.40

1(2)

5

C

C

SS

Contrasts



A contrast is tested by comparing its sum of squares to the error mean square. The resulting statistic would be distributed as F with 1 and a(n-1) degrees of freedom.

Contrast is always single degree of freedom since two comparing data and one equation

2. experiments with one factor (ch.3. analysis of variance...

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