Transcript Chapter 3 Experiments with a Single Factor: The Analysis

Chapter 3 Experiments with a Single
Factor: The Analysis of Variance
1
3.1 An Example
• Chapter 2: A signal-factor experiment with two
levels of the factor
• Consider signal-factor experiments with a levels
of the factor, a  2
• Example:
– The tensile strength of a new synthetic fiber.
– The weight percent of cotton
– Five levels: 15%, 20%, 25%, 30%, 35%
– a = 5 and n = 5
2
• Does changing the
cotton weight percent
change the mean tensile
strength?
• Is there an optimum
level for cotton content?
3
An Example (See pg. 61)
• 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.
4
• 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
5
6
7
• Does changing the power change the mean etch
rate?
• Is there an optimum level for power?
• We would like to have an objective way to answer
these questions
• The t-test really doesn’t apply here – more than
two factor levels
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3.2 The Analysis of Variance
• a levels (treatments) of a factor and n replicates
for each level.
• yij: the jth observation taken under factor level or
treatment i.
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Models for the Data
• Means model:
 i  1,2,...,a
yij   i   ij , 
 j  1,2,...,n
– yij is the ijth observation,
– i is the mean of the ith factor level
– ij is a random error with mean zero
• Let μ i = μ + τ i ,  is the overall mean and τ i is the
ith treatment effect
• Effects model:
 i  1,2,...,a
yij     i   ij , 
 j  1,2,...,n
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• Linear statistical model
• One-way or Signal-factor analysis of variance
model
• Completely randomized design: the experiments
are performed in random order so that the
environment in which the treatment are applied is
as uniform as possible.
• For hypothesis testing, the model errors are
assumed to be normally and independently
distributed random variables with mean zero and
variance, σ 2, i.e. yij ~ N(μ + τi, σ 2)
• Fixed effect model: a levels have been
specifically chosen by the experimenter.
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3.3 Analysis of the Fixed Effects
Model
• Interested in testing the equality of the a treatment
means, and E(yij) = μ i = μ + τ i, i = 1,2, …, a
H0: μ 1 = … = μ a v.s.
H1: μ i ≠ μ j, for at least one pair (i, j)
• Constraint:


i
a
i
  i  0
i
• H0: τ1 = … = τa =0 v.s. H1: τ i ≠ 0, for at least one i
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•
n
a
n
Notations: yi   yij , y   yij
j 1
i 1 j 1
yi  yi / n, y  y / N , N  na
3.3.1 Decomposition of the Total Sum of Squares
•
•
Total variability into its component parts.
The total sum of squares (a measure of overall
variability in the data)
a
n
SST   ( yij  y.. )
2
i 1 j 1
•
Degree of freedom: an – 1 = N – 1
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a
n
a
n
2
(
y

y
)

[(
y

y
)

(
y

y
)]
 ij ..  i. .. ij i.
2
i 1 j 1
i 1 j 1
a
a
n
 n ( yi.  y.. )   ( yij  yi. ) 2
2
i 1
i 1 j 1
SST  SSTreatments  SS E
• SSTreatment: sum of squares of the differences
between the treatment averages (sum of squares
due to treatments) and the grand average, and a –
1 degree of freedom
• SSE: sum of squares of the differences of
observations within treatments from the treatment
average (sum of squares due to error), and a(n - 1)
= N – a degrees of freedom.
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SST  SSTreatments  SSE
• A large value of SSTreatments reflects large differences
in treatment means
• A small value of SSTreatments likely indicates no
differences in treatment means
• dfTotal = dfTreatment + dfError
2
2
(
n

1
)
S



(
n

1
)
S
SS
•
1
a
E

N a
(n  1)    (n  1)
• No differences between a treatment means:
variance cane be estimated by
n ( yi  y ) 2
SSTreatments

a 1
i
a  115
SSTreatments
SSE
• Mean squares: MS
, MS E 
Treatments 
a 1
N a
a
n
a
1
1
E ( MS E ) 
E ( yij2   yi2 )   2
N  a i 1 j 1
n i 1
a
E ( MSTreatments )   2  n( i ) /(a  1)
i 1
3.3.2 Statistical Analysis
• Assumption: εij are normally and independently
distributed with mean zero and variance σ 2
• Cochran’s Thm (p. 69)
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• SST/σ2 ~ Chi-square (N – 1), SSE/σ2 ~ Chi-square
(N – a), SSTreatments/σ2 ~ Chi-square (a – 1), and
SSE/σ2 and SSTreatments/σ2 are independent
(Theorem 3.1)
• H0: τ1 = … = τa =0 v.s. H1: τi ≠ 0, for at least one i
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• Reject H0 if F0 > F α , a-1, N-a
• Rewrite the sum of squares:
y2
SST   y ij 
N
i 1 j 1
a
n
1 a 2 y2
SSTreatments   y i 
n i 1
N
SSE  SST  SSTreatments
• See page 71
• Randomization test
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ANOVA Table of Example 3-1
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3.3.3 Estimation of the Model Parameters
• Model: yij = μ + τ i + ε ij
• Estimators: ˆ  y
ˆi  yi  y
ˆ i  yi
• Confidence intervals:
y i ~ N (  i ,  2 / n)
y i  t / 2 , N  a
MS E
MS E
  i  y i  t / 2 , N  a
n
n
y i  y j   t / 2 , N  a
MS E
MS E
  i   j  y i  y j   t / 2 , N  a
n
n
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• Example 3.3 (page 74)
• Simultaneous Confidence Intervals (Bonferroni
method): Construct a set of r simultaneous
confidence intervals on treatment means which is
at least 100(1-): 100(1-/r) C.I.’s
3.3.4 Unbalanced Data
• Let ni observations be taken under treatment i,
i=1,2,…,a, N = i ni,
2
y
SST   y ij2  
N
i 1 j 1
a
ni
a
SSTreatments  
i 1
y i2 y2

ni
N
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1. The test statistic is relatively insensitive to small
departures from the assumption of equal variance
for the a treatments if the sample sizes are equal.
2. The power of the test is maximized if the samples
are of equal size.
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3.4 Model Adequacy Checking
• Assumptions: yij ~ N(μ + τ i, σ2)
• The examination of residuals
• Definition of residual:
eij  yij  yˆ ij ,
yˆ ij  ˆ  ˆi  y  ( yi  y )  yi
• The residuals should be structureless.
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3.4.1 The Normality Assumption
• Plot a histogram of the residuals
• Plot a normal probability plot of the residuals
• See Table 3-6
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• May be
– the left tail of error is thinner than the tail part
of standard normal
• Outliers
• The possible causes of outliers: calculations, data
coding, copy error,….
• Sometimes outliers are more informative than the
rest of the data.
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• Detect outliers: Examine the standardized
eij
residuals,
d ij 
MS E
3.4.2 Plot of Residuals in Time Sequence
• Plotting the residuals in time order of data
collection is helpful in detecting correlation
between the residuals.
• Independence assumption
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• Nonconstant variance: the variance of the
observations increases as the magnitude of the
observation increase, i.e. yij  2
• If the factor levels having the larger variance also
have small sample sizes, the actual type I error
rate is larger than anticipated.
• Variance-stabilizing transformation
Poisson
Square root transformation,
Lognormal
Logarithmic transformation, log yij
Binomial
Arcsin transformation, arcsin
y ij
y ij
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• Statistical Tests for Equality Variance:
H0 : 12     a2 v.s.H1 : abovenot truefor at least one i2
– Bartlett’s test:
q
c
 02  2.3026
a
q  ( N  a) log S   (ni  1) log S i2
2
P
i 1
1  a
1
1 
c  1
  (ni  1)  ( N  a) 
3(a  1)  i 1

a
S   (ni  1) S i2 /( N  a)
2
p
i 1
2
2
– Reject null hypothesis if  0   ,a1
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• Example 3.4: the test statistic is
02  0.43 and 02.05,3  7.81
• Bartlett’s test is sensitive to the normality
assumption
• The modified Levene test:
– Use the absolute deviation of the observation in
each treatment from the treatment median.
d ij  y ij  ~
y i , i  1,2,, a, j  1,2,, ni
– Mean deviations are equal => the variance of
the observations in all treatments will be the
same.
– The test statistic for Levene’s test is the
ANOVA F statistic for testing equality of
means.
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• Example 3.5:
– Four methods of estimating flood flow
frequency procedure (see Table 3.7)
– ANOVA table (Table 3.8)
– The plot of residuals v.s. fitted values (Figure
3.7)
– Modified Levene’s test: F0 = 4.55 with P-value
= 0.0137. Reject the null hypothesis of equal
variances.
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•
•
•
•
Let E(y) =  and y  
Find y* = y that yields a constant variance.
y*  +-1
Variance-Stabilizing Transformations
y and 

= 1 - 
Transformation
y constant 0
1
No transformation
y  1/2
½
½
Square root
y  
1
0
Log
y  3/2
3/2
-1/2
Reciprocal square root
y  2
2
-1
Reciprocal
http://www.stat.ufl.edu/~winner/sta6207/transform.pdf 33
• How to find :
log yi  log   log i
• Use Si   i and yi  i
• See Figure 3.8, Table 3.10 and Figure 3.9
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3.5 Practical Interpretation of
Results
• Conduct the experiment => perform the statistical
analysis => investigate the underlying
assumptions => draw practical conclusion
3.5.1 A Regression Model
• Qualitative factor: compare the difference between
the levels of the factors.
• Quantitative factor: develop an interpolation
equation for the response variable.
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The Regression Model
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3.5.2 Comparisons Among Treatment Means
• If that hypothesis is rejected, we don’t know
which specific means are different
• Determining which specific means differ
following an ANOVA is called the multiple
comparisons problem
3.5.3 Graphical Comparisons of Means
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3.5.4 Contrast
• A contrast: a linear combination of the parameters
of the form
a
a
i 1
i 1
   ci  i ,  ci  0
• H0:  = 0 v.s. H1:   0
• Two methods for this testing.
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• The first method:
a
a
i 1
i 1
Let C   ci y i T henVar (C )  n 2  ci2
a
c y
Under H 0 ,
i 1
i
i
a
~ N (0,1)
n 2  ci2
i 1
a
Hence thest at ist ic,t 0 
c y
i 1
i
i
a
nMSE  ci2
~ t N a
i 1
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• The second method:
a
F0  t02 
(  ci y i ) 2
i 1
a
nMSE  ci2
~F1,N  a
i 1
 a

  ci y i 
MS C SSC / 1

F0 

, SSC   i 1 a
MS E
MS E
n ci2
i 1
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• The C.I. for a contrast, 
a
   ci  i
i 1
σ2
Let C   ci y i . T henVar(C) 
n
i 1
a
a
HenceC.I.  ci y i  t / 2, N  a
i 1
MS E
n
a
2
c
i
i 1
a
2
c
i
i 1
• Unequal Sample Size


  ci y i 
ci y i


i 1
3.SS C   i a1
a
2
2
n
c
MS E  ni ci
 ii
a
a
a
1.  ni ci  0 2. t 0 
i 1
i 1
i 1
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2
3.5.5 Orthogonal Contrast
• Two contrasts with coefficients, {ci} and {di}, are
orthogonal if ci di = 0
• For a treatments, the set of a – 1 orthogonal
contrasts partition the sum of squares due to
treatments into a – 1 independent single-degreeof-freedom components. Thus, tests performed on
orthogonal contrasts are independent.
• See Example 3.6 (Page 90)
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43
3.5.6 Scheffe’s Method for Comparing All Contrasts
• Scheffe (1953) proposed a method for comparing
any and all possible contrasts between treatment
means.
Suppose u  c1u 1   c au  a , u  1,2,  , m
a
C u   ciu y i and S Cu  MS E  (ciu2 / ni )
i 1
i 1
T hecrit icalvalue : S  ,u  S Cu (a  1) F ,a 1, N  a
If C u  S  ,u , thenreject H 0 : u  0
• See Page 91 and 92
44
3.5.7 Comparing Pairs of Treatment Means
• Compare all pairs of a treatment means
• Tukey’s Test:
– The studentized range statistic:
q
y max  y min
MS E / n
, y max and y min are thelargest and smallest
samplemeansout of a group of p samplemeans
MS E
T hecriticalpointis T  q (a, f )
n
or T  q (a, f ) MS E (1 / ni  1 / n j )
– See Example 3.7
45
• Sometimes overall F test from ANOVA is
significant, but the pairwise comparison of mean
fails to reveal any significant differences.
• The F test is simultaneously considering all
possible contrasts involving the treatment means,
not just pairwise comparisons.
The Fisher Least Significant Difference (LSD)
Method
• For H0: i = j
t0 
y i  y j 
MS E (1 / ni  1 / n j )
46
• The least significant difference (LSD):
LSD  t / 2, N a
1

1
MS E   
n n 
j 
 i
• See Example 3.8
Duncan’s Multiple Range Test
• The a treatment averages are arranged in
ascending order, and the standard error of each
average is determined as
S yi  
MS E
, nh 
nh
a
a
1 / n
i 1
i
47
• Assume equal sample size, the significant ranges
are
RP  r  p, f S yi , p  2,3,, a
• Total a(a-1)/2 pairs
• Example 3.9
The Newman-Keuls Test
• Similar as Duncan’s multiple range test
• The critical values:
K P  q ( p, f )S yi
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3.5.8 Comparing Treatment Means with a Control
• Assume one of the treatments is a control, and the
analyst is interested in comparing each of the other
a – 1 treatment means with the control.
• Test H0: i = a v.s. H1: i  a, i = 1,2,…, a – 1
• Dunnett (1964)
• Compute yi  ya , i  1,2,, a  1
• Reject H0 if
y i   y a
1
1 
 d  (a  1, f ) MS E   
 ni n a 
• Example 3.9
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3.7 Determining Sample Size
• Determine the number of replicates to run
3.7.1 Operating Characteristic Curves (OC Curves)
• OC curves: a plot of type II error probability of a
statistical test,
  1  PReject H 0 | H 0 is false
 1  P( F0  F ,a 1, N a | H 0 is false)
50
• If H0 is false, then
F0 = MSTreatment / MSE ~ noncentral F
with degree of freedom a – 1 and N – a and
noncentrality parameter 
• Chart V of the Appendix
a
• Determine
2
2 
n i
i 1
a 2
• Let i be the specified treatments.
Then estimates
a
of i :  i   i   ,     i / a
i 1
2
• For  , from prior experience, a previous
experiment or a preliminary test or a judgment
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estimate.
• Example 3.11
• Difficulty: How to select a set of treatment means
on which the sample size decision should be based.
• Another approach: Select a sample size such that
if the difference between any two treatment means
exceeds a specified value the null hypothesis
should be rejected.
2
nD
2 
a 2
52
3.7.2 Specifying a Standard Deviation Increase
• Let P be a percentage for increase in standard
deviation of an observation. Then
a


i 1
2
i
/a

/ n
1  0.01P
2
 
1 n
• For example (Page 110): If P = 20, then

1.2
2

 1 n  0.66 n
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3.7.3 Confidence Interval Estimation Method
• Use Confidence interval.
y i  y j   t / 2 , N  a
MS E
MS E
  i   j  y i   y j   t / 2 , N  a
n
n
• For example: we want 95% C.I. on the difference
in mean tensile strength for any two cotton weight
percentages to be  5 psi and  = 3. See Page 110.
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3.8 A Real Application
55
56
57
58
3.10 The Regression Approach to
the Analysis of Variance
• Model: yij = μ + τ i + ij
2
•
a
n
a
n
L    ij2    y ij     i 
i 1 j 1
i 1 j 1
L L

 0, i  1,2,  , a
  i
 y
a
n
i 1 j 1
 ˆ  ˆi   0 &   y ij  ˆ  ˆi   0, i  1,2,  , a
n
ij
j 1
59
• The normal equations
Nˆ
nˆ
nˆ
nˆ
 nˆ1
 nˆ1
 nˆ2
   nˆa
 nˆ2

a
 nˆa
• Apply the constraint  ˆi  0
Then estimations are i 1
ˆ  y ,ˆi  yi  y


y
y1

y 2


y a
• Regression sum of squares (the reduction due to
fitting the full model)
2
a
a
y i
R( , )  ˆy  ˆi yi  
i 1
i 1 n
60
• The error sum of squares:
a
n
SSE   yij2  R , 
i 1 j 1
• Find the sum of squares resulting from the
treatment effects:
R( |  )  R(  , )  R(  )
 R(Full Model) - R(Reduced Model)
2
y
y /n
N
i 1
2
i
61
• The testing statistic for H0: 1 = … = a
R( |  ) /(a  1)
F0 
~ Fa 1, N a
a n 2

 yij  R( , ) /( N  a)
 i 1 j 1

62