ST 516
Experimental Statistics for Engineers II
Practical Interpretation
With a quantitative factor, like power in the etch-rate example,
typically use regression modeling:
y = β0 + β1 x + or perhaps
y = β0 + β1 x + β2 x 2 + .
With a qualitative factor, like “method” in the peak discharge rate
example, we typically focus on comparisons among means.
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Single Factor Experiments
Practical Interpretation
ST 516
Experimental Statistics for Engineers II
Comparing Means
A contrast is a linear combination of treatment means:
a
X
i=1
ci µ i ,
with
a
X
ci = 0.
i=1
Examples are:
µ1 − µ2 , where c1 = 1, c2 = −1, c3 = ... = ca = 0;
µ1 − 12 (µ2 + µ3 ), where c1 = 1, c2 = c3 = − 12 , c4 = ... = ca = 0.
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Single Factor Experiments
Practical Interpretation
ST 516
Experimental Statistics for Engineers II
Testing a Contrast
Many hypotheses about treatment means can be written in terms of
a contrast:
a
X
H0 :
ci µ i = 0
i=1
for appropriate contrast constants c1 , c2 , . . . , ca .
We estimate
Pa
i=1 ci µi
by
a
X
i=1
3 / 22
ci µ̂i =
a
X
ci ȳi· = C .
i=1
Single Factor Experiments
Practical Interpretation
ST 516
Experimental Statistics for Engineers II
C is an unbiased estimator, with
a
σ2 X 2
V (C ) =
c ,
n i=1 i
which we estimate by
V̂ (C ) =
a
MSE X 2
c .
n i=1 i
The test statistic is therefore
a
X
C
ci ȳi·
i=1
t0 = q
=q
.
MSE Pa
2
c
V̂ (C )
i=1 i
n
4 / 22
Single Factor Experiments
Practical Interpretation
ST 516
Experimental Statistics for Engineers II
Compare t0 with the t-distribution with df = N − a.
Equivalently, compare F0 = t02 with the F -distribution with
df = 1, N − a.
Confidence interval for C :
a
X
i=1
5 / 22
v
u
a
u MSE X
t
ci ȳi· ± tα/2,N−a
ci2 .
n i=1
Single Factor Experiments
Practical Interpretation
ST 516
Experimental Statistics for Engineers II
Multiple Contrasts
Sometimes we test several contrasts in the same experiment, or
equivalently set up CIs for those contrasts; error rate becomes an
issue.
Control experiment-wise error rate for allppossible contrasts using
Scheffé’s method: replace tα/2,N−a with (a − 1)Fα,a−1,N−a .
Confidence interval becomes
a
X
i=1
6 / 22
ci ȳi· ±
q
v
u
a
u MSE X
t
(a − 1)Fα,a−1,N−a
c 2.
n i=1 i
Single Factor Experiments
Practical Interpretation
ST 516
Experimental Statistics for Engineers II
Pairwise Comparisons
Often the only contrasts we consider are pairwise comparisons
µi − µj .
To control experiment-wise error rate for all pairwise comparisons,
Tukey’s method gives shorter intervals than Scheffé’s: CI is
r
MSE
ȳi· − ȳj· ± qα (a, f )
,
n
where qα (a, f ) is a percent point of the studentized range statistic,
and f = N − a is the degrees of freedom in MSE .
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Single Factor Experiments
Practical Interpretation
ST 516
Experimental Statistics for Engineers II
Fisher’s Least Significant Difference
The basic t-statistic for comparing µi with µj is
ȳi· − ȳj·
t0 = r
MSE n1i +
1
nj
.
So we declare µi and µj to be significantly different if
s
1
1
|ȳi· − ȳj· | > tα/2,N−a MSE
+
.
ni
nj
8 / 22
Single Factor Experiments
Practical Interpretation
ST 516
Experimental Statistics for Engineers II
That is,
s
tα/2,N−a
MSE
1
1
+
ni
nj
is the least significant difference, or LSD.
Notes
In a balanced design, ni = nj = n, so the LSD is the same for every
pair µi and µj .
The LSD method has comparison-wise error rate α; it does not
control experimentwise error rate.
9 / 22
Single Factor Experiments
Practical Interpretation
ST 516
Experimental Statistics for Engineers II
R command
TukeyHSD(aov(EtchRate ~ factor(Power), etchRateLong))
Output
Tukey multiple comparisons of means
95% family-wise confidence level
Fit: aov(formula = EtchRate ~ factor(Power), data = etchRateLong)
$‘factor(Power)‘
diff
lwr
180-160 36.2
3.145624
200-160 74.2 41.145624
220-160 155.8 122.745624
200-180 38.0
4.945624
220-180 119.6 86.545624
220-200 81.6 48.545624
10 / 22
upr
69.25438
107.25438
188.85438
71.05438
152.65438
114.65438
p adj
0.0294279
0.0000455
0.0000000
0.0215995
0.0000001
0.0000146
Single Factor Experiments
Practical Interpretation
ST 516
Experimental Statistics for Engineers II
Graph
plot(TukeyHSD(aov(EtchRate ~ factor(Power), etchRateLong)))
220−200
220−180
200−180
220−160
200−160
180−160
95% family−wise confidence level
0
50
100
150
Differences in mean levels of factor(Power)
11 / 22
Single Factor Experiments
Practical Interpretation
ST 516
Experimental Statistics for Engineers II
Sample Size
Using Operating Characteristic curves: β is the probability of type II
error:
β = 1 − P{Reject H0 |H0 is false}
= 1 − P{F0 > Fα,a−1,N−a |H0 is false}.
Charts plot β against Φ, where
2
Φ =
n
Pa
2
i=1 τi
.
aσ 2
So if we know σ 2 and a set of treatment effects τ1 , τ2 , . . . , τa for
which we want the type II error to be β, we can find the smallest
acceptable n.
12 / 22
Single Factor Experiments
Sample Size
ST 516
Experimental Statistics for Engineers II
Note: often the resulting n is too large for the experiment to be
feasible. The experimenter must accept higher β or larger τ ’s.
Alternative to using OC charts: using length of confidence interval;
for comparing two means, confidence interval is
r
2
ȳi· − ȳj· ± tα/2,N−a MSE × ,
n
If we know σ 2 , can choose n to give desired width.
Still often gives infeasible (too large) n.
13 / 22
Single Factor Experiments
Sample Size
ST 516
Experimental Statistics for Engineers II
The Regression Approach
Recall the “effects” model:
yi,j = µ + τi + i,j ,
i = 1, 2, . . . , a,
j = 1, 2, . . . , ni .
Note that we now allow for unbalanced data (unequal ni ).
14 / 22
Single Factor Experiments
The Regression Approach
ST 516
Experimental Statistics for Engineers II
Suppose we put all the y ’s in a response vector:


y1,1
 y1,2 


 .. 
 . 


 y1,n1 


 y2,1 


 y2,2 
 . 


Y =  .. 


 y2,n2 
 . 
 .. 


 y

 a,1 
 y

 a,2 
 .. 
 . 
ya,na
15 / 22
Single Factor Experiments
The Regression Approach
ST 516
Experimental Statistics for Engineers II
To keep track of the group that an observation belongs to, we also
create a design matrix X.
The row for an observation in group i consists of all zeroes, except
for a single 1 at position i.
For convenience, we add a first column with all 1’s.
16 / 22
Single Factor Experiments
The Regression Approach
ST 516
Experimental Statistics for Engineers II













X=











17 / 22
1
1
..
.
1
1
..
.
0 ...
0 ...
.. ..
. .
0 ...
1 ...
1 ...
.. ..
. .
0
0
..
.
1
1
1
..
.
1
0
0
..
.
1
..
.
0
..
.
1 ...
.. ..
. .
0 ...
0 ...
.. ..
. .
0
..
.
1
1
..
.
0
0
..
.
1 0 0 ...
1
Single Factor Experiments
0
0
0
..
.
1
1
..
.

























The Regression Approach
ST 516
Experimental Statistics for Engineers II
We also put all the parameters µ, τ1 , τ2 , . . . , τa into a parameter
vector:


µ
 τ1 




β =  τ2 
 .. 
 . 
τa
With an error vector constructed like Y, we have
Y = Xβ + .
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Single Factor Experiments
The Regression Approach
ST 516
Experimental Statistics for Engineers II
This is a multiple regression equation.
The least squares estimates µ̂, τ̂1 , τ̂2 , . . . , τ̂a satisfy the least squares
normal equations
X0 Xβ̂ = X0 Y.
If X0 X were non-singular, we could solve these equations:
β̂ = (X0 X)−1 X0 Y.
But it isn’t, so we can’t...
19 / 22
Single Factor Experiments
The Regression Approach
ST 516
Experimental Statistics for Engineers II
The first column of X is the sum of the other columns, which makes
one column redundant.
More formally, if




b=


+1
−1
−1
..
.
−1




,


then
(X0 X)b = X0 (Xb) = 0.
So X0 X is singular.
20 / 22
Single Factor Experiments
The Regression Approach
ST 516
Experimental Statistics for Engineers II
If we leave out one column of X, the remaining columns have no
redundancy, and the reduced X0 X matrix is non-singular.
Leaving out a column means leaving the corresponding parameter out
of the model.
Leave out the first column ⇔ leave out µ ⇔ the means model.
Leave out the second column ⇔ leave out τ1 ⇔ the effects model
with the first level as the baseline.
Leave out the last column ⇔ leave out τa ⇔ the effects model with
the last level as the baseline.
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Single Factor Experiments
The Regression Approach
ST 516
Experimental Statistics for Engineers II
The R function lm() is a general-purpose multiple regression function,
and uses this approach to estimation.
Because the formula EtchRate ~ factor(Power) has a factor on the
right hand side, lm() internally creates a design matrix X with one
column for each level of the factor except the first (thus imposing the
constraint τ1 = 0), and solves the normal equations.
This looks complicated, but it does not require balanced data, and
generalizes easily to models with more than one factor.
22 / 22
Single Factor Experiments
The Regression Approach