Transcript Additional Class Notes

Sampling distribution
Random samples
In many statistics problems, the observations in a sample can
be thought as observed values of a sequence of a random
variable.
Def. Let x1, x2, …, xn be a collection of n random variables.
These random variables are said to constitute a random sample
of size n if:
a) the xi’s are independent random variables, and
b) every xi’s has the same probability distribution.
When (a) and (b) are satisfied, the xi’s are said to be
independent and identically distributed (iid).
1
Sampling distribution
Def. The random variables
1 n
x   xi is the sample mean of the
n i 1
random variables x1, …, xn.
n
To   xi is the sample total of x1, …, xn.
i 1
2
Sampling distribution
Law of large numbers/ law of averages
Thm.
If the sample size is large, the probability is high
that the sample mean is close to the mean if the
population from which the sample was drawn.
Key theorem used by inferential statistics in
which sample information is used to make
inferences about the population.
3
Central limit theorem
If X is the mean of a random sample X1, …, Xn, of size n from
a distribution with finite mean  and finite positive variance 2,
X   To  n

then the distribution of: W 
is N(0,1) as n
 n
n
 .
 Important points to notice:
o When n is “sufficiently large” (n>30), a practical
use of the CLT is :
w
1 z2 2
PW  w  
e
dz   w

2
o The theorem holds for any distribution with
finite mean and variance.
4
Central limit theorem
 What it all means:
o If n is “large: and we wish to calculate say
Pa  X  b  or P(a To  b), we need only
“pretend” that X or To is normal, standardize it, and
determine the probabilities from the normal table.
The resulting theorem states that it will be
approximately correct.
5
Central Limit Theorem
• Using the exponential distribution and random number
generator, it is possible to plot the resulting frequency
distributions of data. Notice the trend towards normality.
Frequency
30
20
10
0
0
1
2
3
4
5
6
1 run, lambda = 1
6
Central Limit Theorem
• Continuing,
25
15
10
35
5
30
0
25
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2
5 runs, lambda = 1
Frequency
Frequency
20
20
15
10
5
0
0.7
0.8
0.9
1.0
1.1
1.2
1.3
50 runs, lambda = 1
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T-Distribution
t-distribution or student’s t-distribution
Using s for  in computing standardized z-values to look up on the
standard normal table is not trustworthy for small sample sizes (n<30).
Why? Because the CLT only applies to “large” samples. As a result,
when n is small and/or  is unknown, you must use an alternate
distribution, the t-distribution.
Theorem – When X is the mean of a random sample of size n from a
normal distribution with mean , the random variable
X 
T
~ t-distribution with n-1degrees of freedom.
s n
Note: A t-distribution has 1 parameter called the degrees of freedom,
. Possible values of  are 1, 2, …. Each different value of 
corresponds to a different t-distribution.  df = n-1.
8
T-Distribution
Properties of the t-distribution:
Each t curve is bell-shaped and centered at zero.
As  increases, the spread of the corresponding t  curve
decreases, due to the increasing effect of the CLT.
Each t curve is more diffuse than the standard normal(z)
curve.
As , the t curve approaches the standard normal curve.
Let t, = the point on the t-distribution with  df, such that
the area to the right is .
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T-distribution
• Use of the t-distribution is similar to the
use of the standard normal distribution,
except that the degrees of freedom must be
accounted for. The estimation of the true
process mean μ by the experimental mean
creates the loss of one degree of freedom in
estimating the true process standard
deviation σ by s.
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Confidence Interval
Confidence Interval (CI) estimation for the mean
In many cases, a point estimate is not enough, a range of
allowable values is better.
An interval of the form: Lower (l)    Upper (u) may be
useful.
To construct a CI for a parameter , we need to calculate 2
statistics l and u such that: P(l    u) = 1 - .
This interval is a 100(1- )% CI for parameter.
l and u are lower and upper confidence limits.
1- is called the confidence coefficient.
 = level of significance for Type I error (rejecting valid
hypotheses).
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Confidence Interval
• Interpreting a confidence interval
 is covered by interval with confidence 100(1)%.
If many samples are taken and a 100(1- )% CI
is calculated for each, then 100(1-)% of them
will contain/ cover the true value for .
• Note: the larger (wider) a CI, the more confident
we are that the interval contains the true value of
.
• But, the longer it is, the less we know about ,
due to variability or uncertainty  need to
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balance
Confidence Interval
Confidence interval on mean, variance known
If: random sample of size n: X1, …, Xn
Xi ~ N(, 2) and X ~ N(, 2/n)
Then the test statistic: Z 
X 
 2 /n
~ N(0, 1) by CLT
With a CI, we want some range on , P[-Z/2  Z  Z/2] = 1- ,
X 
P[-Z/2 
 Z/2] = 1- 
/ n
 probability test statistic between 2 points is 1- 
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Confidence Interval
P[-Z/2  / n  X    Z/2  / n ] = 1-   want a range on 
P[- X + (-Z/2  / n ) -  -X + (Z/2  / n )] = 1- 
P[ X + Z/2  / n    X - Z/2  / n ] = 1- 
P[ X - Z/2  / n    X + Z/2  / n ] = 1- 
a 100(1- )% CI (2-sided) on  is:
X
X
- Z/2  / n    + Z/2  / n
X
/ n
or
 Z/2
A CI is a statistic  (table value) x standard error.
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Confidence Interval
CI on mean, variance unknown
Up to now, we have known . But typically we do not know, so
what do we do?
1. If n  30, we can replace  in the CI for the mean with the
sample SD, S.
2. if n < 30, then if X1, …, Xn ~ N(, 2)
the the test statistic t =
X 
S/ n
~ t-distribution with (n-1) degrees of
freedom.
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Confidence Interval
Get a CI for 
P [-t/2, n-1  t  t/2, n-1] = 1-
P [-t/2, n-1 
X 
S/ n
 t/2, n-1] = 1-
100(1- )% CI on  is:
X - (t/2, n-1 S / n )   X + (t/2, n-1 S / n )
or X  t/2, n-1 S / n
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Confidence Interval
• Using the data from the plasma etch
experiment described earlier, develop the 95%
confidence interval for the process mean.
• The mean and standard deviation are estimated
below. Since there are only 9 observations, the tdistribution is used for developing the 95% confidence
interval. (υ = 8, α/2 = 0.025)
• Many of these calculations can be easily done by
spreadsheets automatically.
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Confidence Interval
• Continuing,
9
x
x
i
i 1
9

(570.85  576.86  ...  547.49)
 564.11
9
9
sx 
 (x  x)
i 1
2
i
(9  1)
[(570.85  564.11) 2  ...  (547.49  564.11) 2 ]

 10.747
8
t8,0.025  2.306
95%CI  x  t , / 2
s
 10.747 
 564.11  (2.306)
  {555.85,572.37}
n
 9 
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Hypothesis Tests - Review
• Hypothesis Tests:
– Objective: This section devoted to enabling us to:
•
•
•
•
Construct and test a valid statistical hypothesis
Conduct comparative statistical tests( t-tests)
Relate alpha and beta risk to sample size
Conceptually understand analysis of variance
(ANOVA)
• Interpret the results of various statistical tests:
– T-tests, f-tests, chi-square tests.
• Understand the foundation for full and fractional
factorial
• Compute confidence intervals to assess degree of
improvements
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Hypothesis Tests
• Hypotheses defined
– Used to infer population characteristics
from observed data.
– Hypothesis test: A series of procedures
that allows us to make inferences about a
population by analyzing samples
– Key question: was the observed outcomes
the result of chance variation, or was it an
unusual event?
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– Hint: Frequency = Area = Probability
Hypothesis Tests
• Hypothesis: Definition of terms
– Null hypothesis (H0): Statement of no
change or difference. This statement is
tested directly, and we either reject H0 or
we do not reject H0
– Alternative hypothesis (H1): The
statement that must be true if H0 is
rejected.
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Hypothesis Tests
• Definition of terms
– Type I error: The mistake of rejecting H0 when
it is true.
– Type II error: The mistake of failing to reject
H0 when it is false.
– alpha risk ():Probability of a type I error
– beta risk (): Probability of a type II error
– Test statistic: sample value used in making
decision about whether or not to reject H0
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Hypothesis Tests
• Definition of terms
– Critical region: Area under the curve
corresponding to test statistic that leads to
rejection of H0
– Critical value: The value that separates the
critical region from those values that do not
lead to rejection of H0
– Significance level: The probability of rejecting
H0 when it is true
– Degrees of freedom: Referred to as d.f. or ν,
and = n - 1
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Hypothesis Tests
• Definition of terms
– Type I error: Producer’s risk
– Type II error: Consumer’s risk
– Set so type I is the more serious error
type (taking action when none is
required)
– Levels for  and  must be established
before the test is conducted
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Hypothesis Tests
• Hypothesis: Definition of terms
– Degree of freedom
• Degree of freedom are a way of counting the
information in an experiment. In other words, they
relate to sample size. More specifically, d.f. = n – 1
• A degree of freedom corresponds to the number of
values that are free to vary in the sample. If you have a
sample with 20 data points, each of the data points
provides a distinct place of information. The data set is
described completely by these 20 values. If you
calculate the mean for this set of data, no new
information is created because the mean was implied by
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all of the information in the 20 data points.
Hypothesis Tests
• Hypothesis: Definition of terms
– Degree of freedom
• Once the mean is known, though, all of the
information in the data set can be described
with any 19 data points. The information in a
20th data point is now redundant because the
20th data points has lost the freedom to have
any value besides the one imposed on it by the
mean
• We have one less than the total in our sample
because a sample is at least one less than the
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total population.
Hypothesis Tests
• If the population variance is unknown, use s
of the sample to approximate population
variance, since under central limit theorem, s
=  when n > 30. Thus solve the problem as
before, using s
• With smaller sample sizes, we have a
different problem. But it is solved in the
same manner. Instead of using the z
distribution, we use the t distribution
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Hypothesis Tests
• Using t distribution when:
– Sample is small (<30)
– Parent population is essentially normal
– Population variance () is unknown
– As n decreases, variation within the
sample increases, so distribution becomes
flatter.
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Hypothesis Tests
• Compare the means of two samples:
Steps
– Understand word problem by writing out
null and alternative hypotheses
– Select alpha risk level and find critical
value
– Draw graph of the relation
– Insert data into formula
– Interpret and conclude
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Methods to Test a Statistical Hypothesis
Three methods covered in this course:
Calculate Test Statistics, check if it falls in expected
value range, make conclusion based upon the result
(hypothesis test).
Calculate confidence interval. If H0 :  = 0 falls in
interval, fail to reject the null hypothesis H0
P-value for an event
Reject H0 if p-value   = significant level.
If p-value < , reject H0
If p-value  , fail to reject
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Relationship Between Hypothesis Tests
and Confidence Intervals
For a two-sided hypothesis test:
H0 :  = 0
Ha :   0
Equivalent confidence interval is: (lower-limit, upperlimit)
If is contained within the two-sided interval you will fail
to reject H0
If is not contained within the two-sided interval you will
reject H0
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Relationship Between Hypothesis Tests
and Confidence Intervals
For an upper-tail test:
H0 :  = 0
Ha :  > 0
Equivalent confidence interval is: (lower-limit, )
Use the lower bound interval for comparison.
For an lower-tail test:
H0 :  = 0
Ha :  < 0
Equivalent confidence interval is: (, upper-limit)
Use the upper bound interval for comparison
32
Relationship between Hypothesis Tests and
Confidence Intervals
• Using the data from the plasma etch study,
can a true process mean of 530 angstroms
be expected at a 95% confidence level?
• The 95% confidence interval (developed
earlier in detail) runs from 555.85 to
572.37. Since 530 is not included in this
interval, the null hypothesis of μ = 530 is
rejected.
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Test for comparing two means
i. Test for normal population with known variance
1. Assumptions: independent, normal, known variance,
unpaired, unequal variance.
2. 2 populations: X1 ~ N(1, 12) & X2 ~ N(2, 22)
3. Sample m from X1 & sample n from X2
4. Want to test whether 1= 2
5. H0: 1 = 2
H1: 1  2
X 1  X 2  ( 1   2 ) X 1  X 2

6. Test Statistic: Z 0 
2
2
2
2
1

 2
m
n
1  2

m
n
34
Tests for Comparing Two Means
• A company wanted to compare the production from
two process lines to see if there was a statistically
significant difference in the outputs, which would then
require separate tracking. The line data is as follows:
• A: 15 samples, mean of 24.2, and variance of 10
• B: 10 samples, mean of 23.9, and variance of 20
• 95% confidence
x1  x2
24.2  23.9
0.3
Z0 


; r  runs
2
2
10 20
2.6 6
1  2


15 10
r1
r2
Z 0  0.184  1.96  z0.025
35
P - Values
• The P – value is the smallest level of
significance that leads to rejection of the
null hypothesis with the given data. It is
the probability attached to the value of the
Z statistic developed in experimental
conditions. It is dependent upon the type
of test (two-sided, upper, or lower tail
tests) selected to analyze data significance.
36
P - Values
• Using the data developed from the process line
example, but with line A having a mean of 27.2,
instead of 24.2, the P-value would be:
x1  x2
27.2  23.9
3.3
Z0 


; r  runs
2
2
10 20
2.6 6
1  2


15 10
r1
r2
Z 0  2.021  P  0.0217[~ 1 : 46]
37
Tests for Comparing Two Means
H0: 1 – 2 = 0
H1: 1 – 2  0
Two approaches:
Form a (1 – ) confidence interval on 1 – 2 and
reject H0 if confidence interval does not contain 0.
Perform a “t” test
Reject if the absolute value of t exceeds the critical value
Assumptions
Observations are independent, normally distributed
Variances may be unknown or known, equal or unequal
Observations may be paired or unpaired.
38
Tests for Comparing Two Means
1. Assumptions: independent, normal, unknown, unpaired, equal
variance.
2
2
( R1  1) S12  ( R2  1) S 22
 (Yi  Y1 )   (Yi  Y2 )
2
2. S p 

R1  R2  2
R1  R2  2
3. Var (Y1  Y2 )  Var (Y1 )  Var (Y2 ) 
S p2
R1
4. (1   )confident : Y1  Y2  t / 2, R1  R2  2

S p2
R1
S p2
R2

S p2
R2
( y1  y 2 )
5. t-test: t 
Sp
t-crit = t R  R
1
1 1

R1 R2
2  2,

2
6. Note: Many simulations provide measurements that do not have
equal variance.
39
Tests for Comparing Two Means
• A process improvement by exercising
equipment was attempted for an etch line.
Given that the true variances are unknown
but equal, determine whether a statistically
significant difference exists at the 95%
confidence level.
• “Before”: Mean = 564.108, standard deviation =
10.7475, number of observations = 9.
• “Exercise”: Mean = 561.263, standard deviation =
7.6214, number of observations = 9.
40
Tests for Comparing Two Means
• Since the variances are equal, the pooled variance is used for
creation of the confidence interval. If zero is included, there is no
statistically significant difference.
• There are 16 degrees of freedom, and at the α/2 = 0.025 level, the
critical value for t is 2.120.
s
2
pooled,be
( nb  1) sb2  ( ne  1) se2

( nb  ne  2)
s
2
pooled,be
(9  1)10.74752  (9  1)7.6214 2

 86.7972
(9  9  2)
s pooled,be  9.3165
95%CI be   xb  xe   t0.025,16 s pooled,be
1
1

nb ne
95%CI be  564.108  561.263  (2.120)(9.3165)
95%CI be  {6.466,12.156}
1 1

9 9
41
Test for comparing two means
Unequal variance
 Assumptions: independent, normal, unknown variance,
unpaired, unequal variance.
S12 S 22

 Var (Y1  Y2 )  Var (Y1 )  Var (Y2 ) 
R1 R2
 (1   )confident : Y1  Y2  t / 2,
 
 S2 S2 
 1
2 
R R 
 1
2 

2
 S2 
 1 
R 
 1


R1 1
S12 S 22

R1 R2
2
2

 S2 
 2 
R 
 2


R2 1
42
Tests for Comparing Two Means
• An etch process was improved by
recalibration of equipment. The values for a
determination of statistically significant
improvement at the 95% confidence level are
given as follows:
• “Before”: Mean = 564.108, standard deviation =
10.7475, number of observations = 9.
• “Calibrated”: Mean = 552.340, standard deviation =
2.3280, number of observations = 9.
• The null hypothesis is that μb – μc = 0
43
Tests for Comparing Two Means
• The first task is to determine the number of degrees of
freedom and the appropriate critical value.
 bc 
 bc
 sb2
sc2 

n  n 

c 
 b
2
2
2
2
 s2


 sc

 n  
 b n 
b 
c 



 nb  1
nc  1 




 10.74752
2.3280 2 





9
9



 2  8.936  9
2
2
2
2
  10.7475 
 2.3280  







 
9
9

 
 


10
10






44
Tests for Comparing Two Means
• For 9 degrees of freedom and α/2 = 0.025, the critical
value for t is 2.262
tbc 
( xb  xc )  (  b   c )
sb2 sc2

nb nc

(564.108  552.340)  0
10.74752 2.32802

9
9
tbc  3.210  2.262  t0.025,9
45
Tests for Comparing Two Means
Paired Test
 Assumptions: independent, normal, unknown variance,
equal # of replications
 Case 1: Y1, Y2… YR
Case 2: Y’1, Y’2… Y’R
Different: d1, d2… dR , where di = yi – y’i
2
 di
 (d i  d )
2
d 
Sd 
R
R 1
 H0: 1 – 2 = 0  d = 0
H1: 1 – 2  0  d  0
 (1   )confident : d  t / 2, R 1
 t 
S d2
V (d ) 
R
d
S d2
R
46
Tests for Comparing Two Means
• Two materials were compared in a wear test as a
paired, randomized experiment. The coded data are
tabulated below, as well as the differences for each
pairing.
Run
1
2
3
4
5
6
7
8
9
10
A
13.2
8.2
10.9
14.3
10.7
6.6
9.5
10.8
8.8
13.3
B
14.0
8.8
11.2
14.2
11.8
6.4
9.8
11.3
9.3
13.6
Δ
0.8
0.6
0.3
-0.1
1.1
-0.2
0.3
0.5
0.5
0.3
47
Tests for Comparing Two Means
• The mean, standard deviation, and 95% confidence
intervals are constructed below, with nine degrees of
freedom. If the interval does not contain zero, the null
hypothesis of “no difference” is rejected.
10
d 
d
i 1
i
10

10
s 
2
d
 (d
i 1
i
0.8  0.6  ...  0.3
 0.41
10
 d )2
(10  1)
(0.8  0.41) 2  ...  (0.3  0.41) 2

 0.149
9
sd  0.149  0.386; n  10
95%CI d  d  t0.025,9
sd
(2.262)(0.386)
 0.41 
 {0.134,0.686}
n
10
48
Test for Comparing Two Variances
• The variances can also be used to
determine the likelihood of observations
being part of the same population. For
variances, the test used is the F-test since
the ratio of variances will follow the Fdistribution. This test is also the basic test
used in the ANOVA method covered in the
next section.
49
F - Test
• The F distribution is developed from three
parameters: α (level of confidence), and the
two degrees of freedom for the variances
under comparison. The null hypothesis is
typically one where the variances are
equal, which would yield an allowed set of
values that F can be and still not reject the
null hypothesis.
50
F - Test
• If the observed value of F is outside of this
range, the null hypothesis is rejected and
the observation is statistically significant.
• Tables for the F distribution are found in
texts with a statistical interest. Normally,
the ratio is tabulated with the higher
variance in the denominator, the lower
variance in the numerator.
51
F - Test
• The test statistic and null hypothesis is
given below.
H 0 :  12   22
S12
F0  2
S2
s12
f0  2
s2
52
F - Test
• An etching process of semiconductor
wafers is used with two separate gas
treatment mixtures on 20 wafers each. The
standard deviations of treatments 1 and 2
are 2.13 and 1.96 angstroms, respectively.
Is there a significant difference between
treatments at the 95 % confidence level?
53
F - Test
• 95% confidence level infers α = 0.05, and also since
this is a two-tailed distribution, α/2 = 0.025 is used for
F. There are 19 degrees of freedom for each set of
data. F
 2.53
0.025,19,19
s12 2.132 4.54
f0  2 

 1.182  2.53
2
s2 1.96
3.84
• Therefore the null hypothesis of “no difference in
treatments” cannot be rejected.
54
DOE Overview
• Focus is on univariate statistics (1
dependent variable) vs. multivariate
statistics (more than one dependent
variable)
• Focus is on basic designs, and does not
include all possible types of designs (I.e.
Latin squares, incomplete blocks, nested,
etc.)
55
DOE Overview
• One key item to keep in clear focus while
performing a designed experiment:
• Why are we doing it?
• According to Taguchi (among others) it is
to refine our process to yield one of the
following quality outcomes:
56
DOE Overview
• Bigger is better (yields, income, some
tensile parameters, etc.)
• Smaller is better (costs, defects, taxes,
contaminants, etc.)
• Nominal is best (Most dimensions, and
associated parameters, etc.)
• Remember also that whatever is selected as the
standard for comparison, it must be measured!
57
F - Test
• Compare two sample variances
– Compare two newly drawn samples to
determine if a difference exists between the
variance of the samples.
(Up until now we have compared samples to
populations, and sample means)
– For two normally distributed populations with
equal variances. 12 = 22
– We can compare the two variances such that
s12 / s22 = Fmax where s12 > s22
58
F - Test
• Compare two sample variances
– F tests for equality of the variances and uses
the f-distribution.
– This works just like method used with the t
distribution: critical value is compared to test
statistics.
– If two variances are equal, F = s12 / s22 = 1
,thus we compare ratios of variances.
59
F - Test
• Compare two sample variances
– If two variances are equal, F = s12 / s22 = 1
Thus, we compare ratios of variances
– Large F leads to conclusion variances are very
different.
– Small F (close to 1) leads to conclusion
variances do not differ significantly. Thus for
F test:
– H0: s12 = s22
H1: s12  s22
60
F - Test
• Compare two sample variances
– F tables
• Several exist, depending on alpha level.
– Using F tables requires 3 bits of information.
• Chosen alpha risk
• Degree of freedom (n1 – 1) for numerator term.
• Degree of freedom (n2 – 1) for denominator term.
61
ANOVA
Analysis of Variance
• Employs F distribution to compare ratio of
variances of samples when true population
variance is unknown
• Compares variance between samples to the
variance within samples (variance of means
compared to mean of variances). If ratio of
variance of means > mean of variances, the effect
is significant
• Can be used with more than 2 group at a time
• Requires independent samples and normally
distributed population.
62
ANOVA
• ANOVA
– Anova concept
S
S
2
between means
2
between samples
S
F
S
2
X
2
p
63
ANOVA
• ANOVA Concepts
– All we are saying is:
• Assumption that the population variances
are equal or nearly equal allows us to treat
the samples from many different
populations as if they in fact belonged to a
single large population. If that is true, then
variance among samples should nearly
equal variance within the samples
• H0: 1 = 2 = 3 = 4 =… k
64
ANOVA
• ANOVA Steps
– Understand word problem by writing out null
and alternative hypotheses
– Select alpha risk level and find critical value
– Run the experiment
– Insert data into Anova formula
– Draw graph of relation
– Interpret and conclude
65
ANOVA
• Analysis of Variance (ANOVA) is a
powerful tool for determining significant
contributors towards process responses.
The process is a vector decomposition of a
response, and can be modified for a wide
variety of models.
• Can be used with more than 2 groups at a
time
• Requires independent, normally distributed
samples.
66
ANOVA
• ANOVA decomposition is an orthogonal vector
breakdown of a response (which is why
independence is required), so for a process with
factors A and B, as tabulated below:
Source of
Variation
Grand
Mean
Factor A
Mean
Value
y
 or
Number of
Levels
1

A
Factor B

C = cell
value
N/A
B
Residual
Total
N/A
A * B
67
ANOVA
• The ANOVA values are given by:
Sum of Squares
 (=d.f.) =
Mean Squares (MS)
SSm=
AB
SSa=
B  (   ) 2
1
SSm
A-1
SSa/(A-1)
Factor B
SSb=
A (   ) 2
B-1
SSb/(B-1)
Residuals
SSr=
(c  )
(A-1)(B-1)
SSr/(A-1)(B-1)
Total
SSt=
AB
Source of
Variation
Grand Mean
Factor A
 c 2
2
68
ANOVA
• In this case, we use it to demonstrate how the
deposition of oxide on wafers as described by Czitrom
and Reese (1997) can be decomposed into significant
factors, checking the wafer type and furnace location.
The effects will be removed in sequence and verified
for significance using the F-test. The proposed model
is:
• Y=M + W + F + R, where
• M is the grand mean, W is the effect of a given wafer
type, F is the effect of a particular furnace location,
and R is the residual. Y denotes the observed value.
69
ANOVA
• F-test in ANOVA
• The estimator for a given variance in ANOVA is the
mean sum of squares (MS). For a given factor, its MS
can be calculated as noted before, and the ratio of the
factor MS and residual MS compared to the Fdistribution for significance. To do so, the level of
significance  must be defined to establish the Type I
error limit.
F factor 
MS factor
MS residual
 F / 2 , fa cto r, resid u a l
70
ANOVA
• In this example, the level of significance is selected at
0.10, yielding the following table of upper bounds for
the F-test. In all cases, the higher variance (usually
the factor) is divided by the lower variance (usually
the residual).
Degrees of Freedom
F / 2 , 1 , 2
1 = 2, 2 = 9
4.26
1 = 2, 2 = 6
5.14
1 = 3, 2 = 6
4.76
71
ANOVA
• The set of means observed in the process, broken
down by wafer type and furnace location are (in mils
= 0.001 inch) tabulated below. The grand mean is
92.1485.
Wafer /
Furnace
1
2
3
4
Wafer Means
External
Recycle
92.470
92.157
92.656
93.473
92.6890
Internal
Recycle
91.408
92.524
92.419
93.014
92.3142
Virgin
90.810
91.270
91.381
92.200
91.4153
91.9837
92.1520
92.8957
Location 91.5626
Means
72
ANOVA
• The sum of squares about the grand mean is found by
adding the squares of all of the deviations in the 12
inner cells, and totals 6.7819. One degree of freedom
is expended in fixing the mean, and 11 are left.
Wafer /
Furnace
1
2
3
4
Wafer
Means
External
+0.3215 +0.0085 +0.5075 +1.3245 +0.5405
Internal
-0.7405
+0.3755 +0.2705 +0.8655 +0.1927
Virgin
-1.3385
-0.8785
-0.7675
Location
Means
-0.5859
-0.1648
+0.0035 +0.7472
+0.0515 -0.7332
73
ANOVA
• Determining the sum of squares for the wafer types is
done by multiplying the squared difference between a
type and the grand mean by the number of times that
type appears in the data (e.g. 4*squared differences).
This is done for all types, and totals 3.4764, with 2
degrees of freedom.
Wafer /
Furnace
1
2
3
4
Wafer
Means
External
-0.2190
-0.5320
-0.0330
+0.7840
+0.5405
Internal
-0.9332
+0.1828
+0.0778
+0.6728
+0.1927
Virgin
-0.6053
-0.1453
-0.0343
+0.7847
-0.7332
Location
Means
-0.5859
-0.1648
+0.0035
+0.7472
74
ANOVA
• The “residual” sum of squares totals 3.3145 with nine
degrees of freedom, and indicates that the wafer type
may not be the only significant factor. The
significance of the wafer type is verified using the Ftest:
•
3.4674
Fwafer


2

 4.709  4.26
3
.
3145


9
75
ANOVA
• As noted before, a residual sum of squares of about 50
percent of the total sum of squares may indicate the
presence of a significant factor. The effect of furnace
location is then removed from the data and tested for
significance as before. The furnace location sum of
squares totals 2.7863, with three degrees of freedom.
/
1
2
3
4
Wafer
• Wafer
Furnace
Means
External
+0.3669
-0.3672
-0.0365
-0.0368
+0.5405
Internal
-0.3473
+0.3476
+0.0743
-0.0744
+0.1927
Virgin
-0.0194
-0.0195
-0.0378
+0.0375
-0.7332
Location
Means
-0.5859
-0.1648
+0.0035
+0.7472
76
ANOVA
• The remaining residual sum of squares totals 0.5282
with six degrees of freedom. Repeating the F-tests as
before for both factors yields:
3.4674
2  19.694  5.14
Fwafer 
0.5282
6
2.7863
3  10.550  4.76
Flocation 
0.5282
6








77
ANOVA
• These are very significant values to the  = 0.02 level.
The resulting ANOVA table shows all of the factors
combined:
Source of
Variation
Sum of
Squares
Degrees Mean
of
Squares
Freedom (MS)
1
Ratio to
Residual
MS
N/A
Wafer
Type
3.4674
2
1.7337
19.694
Furnace
Location
2.7863
3
0.9288
10.550
Residual
0.5282
6
0.08803
Grand
Mean
78
ANOVA
• The last task is to verify that the residuals follow a
normal distribution about the expected value from the
model. This is done easily using a normal plot and
checking that the responses are approximately linear.
Patterns or significant deviations could indicate
another significant factor affecting the response. The
plot that follows of the residual responses from all
values shows no significant indication of non-normal
behavior, and we fail to reject (on this level) the null
hypothesis of the residuals conforming to a normal
distribution around the model predicted value.
79
ANOVA
Normal Probability Plot
.999
Probability
.99
.95
.80
.50
.20
.05
.01
.001
-1
0
1
ANOVA Residuals
Average: -0.0000845
StDev: 0.721618
N: 84
Kolmogorov-Smirnov Normality Test
D+: 0.051 D-: 0.059 D : 0.059
Approximate P-Value > 0.15
80