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2.830J / 6.780J / ESD.63J Control of Manufacturing Processes (SMA 6303)
Spring 2008
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MIT 6.780J 2008: Problem set 8 solutions
Problem 1: example solution courtesy of M. Imani Nejad
a) first we calculate the averages and then use the formula in May and Spanos to find b and b 0 :
x=
y=
1 n
∑ y i = ���������
n i==1
∑ (x − x)(y − y) = 2397.779 = 0.96601
2482.148
∑ (x − x) 2
���������������������������� �����������
y = 35.387 + 0.966x ������������
Fitted Line Plot
y = 35.39 + 0.9660 x
Regression
99% CI
99% PI
90
80
S
R-Sq
R-Sq(adj)
70
9 80695
45.4%
43.5%
y
60
50
40
30
20
10
0
0
5
10
15
20
25
30
35
x
Residual Plots for y
Normal Probability Plot
Versus Fits
99
20
Percent
90
Residual
b 0 = y − bx
50
10
0
-20
-40
1
-40
-20
0
Residual
20
40
Histogram
Frequency
50
Fitted Value
60
Versus Order
20
20
15
Residual
b=
1 n
∑ x i = ���������
n i==1
10
5
0
-20
-40
0
-30
-20
-10
0
10
Residual
20
30
2
4 6
8 10 12 14 16 18 20 22 24 26 28 30
Observation Order
In the above graph the interval is calculated based on the following formula:
⎡1
where α = 99% and V(y) = ⎢ +
y 0 ± t α / 2 V(y)
⎣n
(x − x) 2 ⎤ 2
*s For each point.
2⎥
(x
−
x)
∑
⎦
The regression equation is
y = 35.39 + 0.9660 x
S = 9.80695
R-Sq = 45.4%
R-Sq(adj) = 43.5%
Analysis of Variance
Source
Regression
Error
Total
DF
1
29
30
SS
2316.28
2789.11
5105.39
MS
2316.28
96.18
F
24.08
P
0.000
b) We can see in the above graph that there are 6 points of the 99% confidence interval. So we reduce
the data set and again using the formula and procedure in part (a) we get new results.
b=1.2432
y = 31.84 + 1.2432x
b 0 =31.84
b
y = 31.84 + 1.243 x
Regression
99% C I
99% PI
80
70
S
R-Sq
R-Sq(adj)
y
60
50
40
30
20
0
5
10
15
20
x
25
30
35
2.85224
94.1%
93.8%
Residual Plots for y
Normal Probability Plot
Versus Fits
99
5.0
Residual
Percent
90
50
10
2.5
0.0
-2.5
-5.0
1
-5.0
-2.5
0.0
Residual
2.5
5.0
30
40
60
70
Versus Order
4
5.0
3
2.5
Residual
Frequency
Histogram
50
Fitted Value
2
1
0.0
-2.5
-5.0
0
-4
-2
0
Residual
2
4
2
4
6
8 10 12 14 16 18 20 22 24
Observation Order
Now in the reduced data, we can see that residuals are close to normal. This means that we can rely on the
data more than before. Also we can make an ANOVA test which the result is as follows:
The regression equation is
y = 31.84 + 1.243 x
S = 2.85224
R-Sq = 94.1%
R-Sq(adj) = 93.8%
Analysis of Variance
Source
Regression
Error
Total
DF
1
23
24
SS
2977.78
187.11
3164.90
MS
2977.78
8.14
F
366.03
P
0.000
Also we can see that the standard error is less than before. This is a good indication that the reduced
data is more reasonable. So b and b0 in reduced data is estimated better.
c) Again we reduce the data by deleting the data which are put pf 99% level of confidence.
We get the following results:
y = 32.05 + 1.265 x
b
y = 32.05 + 1.265 x
80
Regression
99% CI
99% PI
70
S
R-Sq
R-Sq(adj)
1.36336
98.3%
98.2%
y
60
50
40
30
0
5
10
15
20
25
30
35
x
Residual Plots for y
Normal Probability Plot
Versus Fits
99
2
Residual
Percent
90
50
1
0
-1
10
-2
1
-3.0
-1.5
0.0
Residual
1.5
3.0
30
40
Histogram
50
Fitted Value
60
70
Versus Order
4
Residual
Frequency
2
3
2
1
0
-1
1
-2
0
-2
-1
0
1
Residual
2
1
2
3
4
5 6 7 8 9 10 11 12 13 14
Observation Order
We can see that the residual are normal. But there is a little difference between new fitting model and
part (b). So we perform ANOVA test and calculate the standard error.
The regression equation is
y = 32.05 + 1.265 x
S = 1.36336
R-Sq = 98.3%
R-Sq(adj) = 98.2%
Analysis of Variance
Source
Regression
Error
Total
DF
1
12
13
SS
1327.56
22.30
1349.87
MS
1327.56
1.86
F
714.23
P
0.000
So the standard error is even less than in the part (b). So if we continue this cycle we get better fitting
model and finally we get a perfect fitting, but we should keep in mind that we fit the curve based on
less data. This procedure can be achieved until the standard error is acceptable value.
Problem 2
This was a hard problem and effort will be generously rewarded.
First: proof of May and Spanos Eqn. 8.38:
Let the true model be:
y = Xȕ + İ
and the estimate for y be:
ǔ = Xb
where b are estimates of model coefficients made using May and Spanos Eqn 8.37. (Note that b should
be b in the text.) Then we have: Var(b) = E[(b – ȕ)T(b – ȕ)]
= E[((XTX)–1XTİ)T(XTX)–1XTİ]
= (XTX)–1XTE(İTİ)X(XTX)–1
= (XTX)–1XTı2IX(XTX)–1
[noise is homoskedastic (has constant variance) and not autocorrelated]
= ı2(XTX)–1XTIX(XTX)–1
= ı2(XTX)–1
= Ȉ b.
Ȉb is the variance–covariance matrix for the estimated parameter set. Then, following the rules for
expressing the variance of sums of variables, the variance of the point estimate ǔ* made at
x* = [1 x1 x12 x2] is:
Var(ǔ*) = Var(x*b) = x*Ȉbx*T
and the confidence interval is
ǔ* ± tĮ/2 [Var(ǔ*)]1/2.
The number of degrees of freedom for the t value is [(number of data) – 4] in this case.
Thanks to Shawn Shen for his excellent work on this problem.
Problem 3 (Drain problem 4) (a), (d)
(b) [courtesy M. Imani Nejad] First line: M = 6, W = 2; second line: M = 2, W = 3: (c)
0.0016
0.00212
0.0016
0.00264
Problem 4 a) From the lecture, for c control parameters and n noise parameters, the
number of tests is given by:
Crossed array: 3c*2n
Response surface: 2c+(c+1)n
b) A half-fraction can be used when main effects are not aliased with twofactor interactions. For 2-level (linear model) experiments, this occurs for
a 24-1 test. The number of tests is thus:
Response surface: 24+(4+1)1=21
Half-fraction response surface: 24-1+(4+1)1=13
Crossed array: 3421=162
c) There must be an interaction between the noise factors and the control
factors; otherwise, the process variation won’t depend on the inputs.
d) Run several replicates at each of the controllable input settings. This will
give two useful response variables: mean and variance. Least-square
models would then be fit for the mean and the variance, and response
surface methods used to find the optimum.