Designed Experimentation

Solutions for HW4

8-1 Suppose that in the chemical process development experiment in Problem 6-7, it
was only possible to run a one-half fraction of the 2 4 design. Construct the design and
perform the statistical analysis, using the data from replicate 1.

The required design is a 24-1 with I=ABCD.

A B C D=ABC
-1 -1 -1 -1 90
1 -1 -1 1 72
-1 1 -1 1 87
1 1 -1 -1 83
-1 -1 1 1 99
1 -1 1 -1 81
-1 1 1 -1 88
1 1 1 1 80

By observing the Pareto chart (below), we see that terms A, AB, AD are the most
significant. We will include those in the model along with B and D to preserve hierarchy.

Pareto Chart of the Effects

(response is C9, Alpha = .05)
22.58
F actor Name
A A
A
B B
C C
AB D D

AD
Term

AC

0 5 10 15 20 25
Effect
Lenth's PSE = 6
Factorial Fit: C9 versus A, B, D

Estimated Effects and Coefficients for C9 (coded units)

Term Effect Coef SE Coef T P

Constant 85.000 1.458 58.31 0.000
A -12.000 -6.000 1.458 -4.12 0.054
B -1.000 -0.500 1.458 -0.34 0.764
D -1.000 -0.500 1.458 -0.34 0.764
A*B 6.000 3.000 1.458 2.06 0.176
A*D -5.000 -2.500 1.458 -1.71 0.228

S = 4.12311 R-Sq = 92.41% R-Sq(adj) = 73.44%

Analysis of Variance for C9 (coded units)

Source DF Seq SS Adj SS Adj MS F P

Main Effects 3 292.00 292.00 97.33 5.73 0.152
2-Way Interactions 2 122.00 122.00 61.00 3.59 0.218
Residual Error 2 34.00 34.00 17.00
Total 7 448.00

Now, by observing that only factor A is statistically significant in the revised model
(p~0.05), we remove the other terms and run the model another time. This is necessary
especially since we see that the ANOVA for ME’s and interactions (highlighted above)
show that neither is statistically significant.

Factorial Fit: C9 versus A

Estimated Effects and Coefficients for C9 (coded units)

Term Effect Coef SE Coef T P

Constant 85.000 1.826 46.56 0.000
A -12.000 -6.000 1.826 -3.29 0.017

S = 5.16398 R-Sq = 64.29% R-Sq(adj) = 58.33%

Analysis of Variance for C9 (coded units)

Source DF Seq SS Adj SS Adj MS F P

Main Effects 1 288.0 288.0 288.00 10.80 0.017
Residual Error 6 160.0 160.0 26.67
Pure Error 6 160.0 160.0 26.67
Total 7 448.0

Now we see that the final model with only factor A is statistically significant. After
verification of NID(0,2) residuals, the final regression model is y=85 – 6 (A).
8-2 Suppose that in Problem 6-15, only a one-half fraction of the 2 4 design could be
run. Construct the design and perform the analysis, using the data from replicate I.

The required design is a 24-1 with I=ABCD.

A B C D=ABC
-1 -1 -1 -1 1.71
1 -1 -1 1 1.86
-1 1 -1 1 1.79
1 1 -1 -1 1.67
-1 -1 1 1 1.81
1 -1 1 -1 1.25
-1 1 1 -1 1.46
1 1 1 1 0.85

By observing the Pareto chart (below), we see that terms C, AC, A and B are the most
significant. We will include those in the model and rerun it. We will also include the AD
interaction, but since B and C are both significant and AD is aliased with BC, we’ll add
the interaction BC instead of AD.

Pareto Chart of the Effects

(response is C9, Alpha = .05)
1.214
F actor Name
A A
C
B B
C C
AC D D

A
Term

AD

AB

0.0 0.2 0.4 0.6 0.8 1.0 1.2

Effect
Lenth's PSE = 0.3225
Factorial Fit: C9 versus A, B, C

Estimated Effects and Coefficients for C9 (coded units)

Term Effect Coef SE Coef T P

Constant 1.5500 0.03432 45.16 0.000
A -0.2850 -0.1425 0.03432 -4.15 0.053
B -0.2150 -0.1075 0.03432 -3.13 0.089
C -0.4150 -0.2075 0.03432 -6.05 0.026
A*C -0.3000 -0.1500 0.03432 -4.37 0.049
B*C -0.1600 -0.0800 0.03432 -2.33 0.145

S = 0.0970824 R-Sq = 97.78% R-Sq(adj) = 92.23%

Analysis of Variance for C9 (coded units)

Source DF Seq SS Adj SS Adj MS F P

Main Effects 3 0.59935 0.59935 0.199783 21.20 0.045
2-Way Interactions 2 0.23120 0.23120 0.115600 12.27 0.075
Residual Error 2 0.01885 0.01885 0.009425
Total 7 0.84940

Now we see that the model looks pretty good (low P values) for all terms and the ME’s
and 2-way interactions. The residuals were checked and the final model is:
Y = 1.55 - 0.1425(A) – 0.1075(B) – 0.2075 (C) – 0.15(A*C) – 0.08(B*C)
8-26 A spin coater is used to apply photoresist to a bare silicon wafer. This operation
usually occurs early in the semiconductor manufacturing process, and the average coating
thickness and the variability in the coating thickness has an important impact on
downstream manufacturing steps. Six variables are used in the experiment. The
variables and their high and low levels are as follows:

Factor Low High

Level Level
Final Spin Speed 7350 rpm 6650 rpm
Acceleration Rate 5 20
Volume of Resist 3 cc 5 cc
Applied
Time of Spin 14 s 6s
Resist Batch Variation Batch 1 Batch 2
Exhaust Pressure Cover Cover On
Off

The experimenter decides to use a 26-1 design and to make three readings on resist
thickness on each test wafer. The data are shown in table 8-37.
Table 8-37
A B C D E F Resist Thick ness
Run Volume Batch Time Speed Acc. Cover Left Center Right Avg. Range
1 5 2 14 7350 5 Off 4531 4531 4515 4525.7 16
2 5 1 6 7350 5 Off 4446 4464 4428 4446 36
3 3 1 6 6650 5 Off 4452 4490 4452 4464.7 38
4 3 2 14 7350 20 Off 4316 4328 4308 4317.3 20
5 3 1 14 7350 5 Off 4307 4295 4289 4297 18
6 5 1 6 6650 20 Off 4470 4492 4495 4485.7 25
7 3 1 6 7350 5 On 4496 4502 4482 4493.3 20
8 5 2 14 6650 20 Off 4542 4547 4538 4542.3 9
9 5 1 14 6650 5 Off 4621 4643 4613 4625.7 30
10 3 1 14 6650 5 On 4653 4670 4645 4656 25
11 3 2 14 6650 20 On 4480 4486 4470 4478.7 16
12 3 1 6 7350 20 Off 4221 4233 4217 4223.7 16
13 5 1 6 6650 5 On 4620 4641 4619 4626.7 22
14 3 1 6 6650 20 On 4455 4480 4466 4467 25
15 5 2 14 7350 20 On 4255 4288 4243 4262 45
16 5 2 6 7350 5 On 4490 4534 4523 4515.7 44
17 3 2 14 7350 5 On 4514 4551 4540 4535 37
18 3 1 14 6650 20 Off 4494 4503 4496 4497.7 9
19 5 2 6 7350 20 Off 4293 4306 4302 4300.3 13
20 3 2 6 7350 5 Off 4534 4545 4512 4530.3 33
21 5 1 14 6650 20 On 4460 4457 4436 4451 24
22 3 2 6 6650 5 On 4650 4688 4656 4664.7 38
23 5 1 14 7350 20 Off 4231 4244 4230 4235 14
24 3 2 6 7350 20 On 4225 4228 4208 4220.3 20
25 5 1 14 7350 5 On 4381 4391 4376 4382.7 15
26 3 2 6 6650 20 Off 4533 4521 4511 4521.7 22
27 3 1 14 7350 20 On 4194 4230 4172 4198.7 58
28 5 2 6 6650 5 Off 4666 4695 4672 4677.7 29
29 5 1 6 7350 20 On 4180 4213 4197 4196.7 33
30 5 2 6 6650 20 On 4465 4496 4463 4474.7 33
31 5 2 14 6650 5 On 4653 4685 4665 4667.7 32
32 3 2 14 6650 5 Off 4683 4712 4677 4690.7 35
(a) Verify that this is a 26-1 design. Discuss the alias relationships in this design.

I=ABCDEF. This is a resolution VI design where main effects are aliased with five-
factor interactions and two-factor interactions are aliased with four-factor interactions.

(b) What factors appear to affect average resist thickness?

Factors B, D, and E appear to affect the average resist thickness.

Pareto Chart of the Effects

(response is Avg., Alpha = .05, only 30 largest effects shown)
28.6
D F actor Name
E
B A V olume
EF B Batch
DE
A BF C Time
BF D S peed
A EF
BE E A cc.
AF
CD F C ov er
BD
A BE
A CD
Term

BC
A
ADE
AB
CE
AC
AE
CF
F
AC E
AD
C
DF
ABD
AC F
ADF

0 50 100 150 200

Effect
Lenth's PSE = 12.9

Estimated Effects and Coefficients for Avg. (coded units)

Term Effect Coef SE Coef T P

Constant 4458.5 9.274 480.76 0.000
Batch 73.6 36.8 9.274 3.97 0.000
Speed -207.1 -103.5 9.274 -11.16 0.000
Acc. -182.9 -91.5 9.274 -9.86 0.000

S = 52.4608 R-Sq = 89.46% R-Sq(adj) = 88.33%

Analysis of Variance for Avg. (coded units)

Source DF Seq SS Adj SS Adj MS F P

Main Effects 3 653998 653998 217999 79.21 0.000
Residual Error 28 77060 77060 2752
Lack of Fit 4 20964 20964 5241 2.24 0.094
Pure Error 24 56096 56096 2337
Total 31 731058
Final Equation in Term of Coded Factors:
y=4458.5+36.8(B)-103.5(D)-91.5(E)

Final Equation in Terms of Actual Factors:

Batch Batch 1
Thick Avg =
+6644.78750
-0.29580 * Speed
-12.19500 * Acc

Batch Batch 2
Thick Avg =
+6718.36250
-0.29580 * Speed
-12.19500 * Acc

(c) Since the volume of resist applied has little effect on average thickness, does this
have any important practical implications for the process engineers?

Yes, less material could be used.

(d) Project this design into a smaller design involving only the significant factors.
Graphically display the results. Does this aid in interpretation?

The cube plot usually assists the experimenter in drawing conclusions.

 Cube Plot (data means) for Avg.

4213.53 4274.98

4404.75 4526.68
7350

Speed 4475.35 4504.35

20

Acc.
4593.28 4675.20
6650 5
1 2
Batch

(e) Use the range of resist thickness as a response variable. Is there any indication that
any of these factors affect the variability in resist thickness?

It appears that no factors are influencing the variation in the response range.

Normal Probability Plot of the Effects

(response is Range, Alpha = .05)
99
Effect Ty pe
Not Significant
95 Significant
90 F actor N ame
A V olume
80 B Batch
70 C Time
Percent

60 D S peed
E A cc.
50
F C ov er
40
30
20

10
5

1
-10 -5 0 5 10
Effect
Lenth's PSE = 4.125
(f) Where would you recommend that the process engineers run the process?

Considering only the average thickness results, the engineers could use factors B, D and
E to put the process mean at target. Then the engineer could consider the other factors on
the range model to try to set the factors to reduce the variation in thickness at that mean.
Project this design into a smaller design involving only the significant factors.
Graphically display the results. Does this aid in interpretation?

Cube Plot (data means) for Avg.

4213.53 4274.98

4404.75 4526.68
7350

Speed 4475.35 4504.35

20

Acc.
4593.28 4675.20
6650 5
1 2
Batch