1. M6141 QUALITY ENGINEERING CONTENTS
1. Introduction
DESIGN OF EXPERIMENTS
2. Fundamentals of Probability and Statistics (cancelled)
A/P Wu Zhang 3. Single Factor Experiments
MAE, NTU
4. Randomized Blocks (cancelled)
Office: N3.2-02-14
Tel: 6790 4445 5. Factorial Experiments
Email: mzwu@ntu.edu.sg 6. 2k Experiments
7. Fractional Experiments
TEXT 8. A Case Study
Montgomery, D. C., (2009), Design and Analysis of Experiments,
John Wiley & Sons.
1-1 1-2
1 INTRODUCTION Example 1.1: Two factors: Machine, Feed rate
Response: Process capability
A designed experiment is a test or series of tests, in which, the
experimenter chooses certain factors for study. He purposely varies Table 1.1
those factors in a controlled fashion, and then observes the effect of Process capability Cp
such factors on the response function. Feed rate (mm/sec.)
Machine 0.2 0.4 0.6
USA 2.1, 2.2 1.7, 1.6 1.2, 1.3
Japan 0.9, 1.0 0.7, 0.6 0.4, 0.3
Conclusion: using USA machine at lowest feed rate will result in
max. process capability.
1-3 1-4

2. Factor (input variable) (3) Run: A run is one set of levels for all factors in an experiment.
In above experiment, there are three factors: (A) temperature,
(1) Types (B) machine and (C) switch. The total number of runs is
Quantitative: Temperature, time: They can be set at any value.
R = 4 × 3 × 2 = 24
Qualitative: Machines, operators: They have different categories
Switch: It has different status For example, [1] A = 10, B = Japan-made, C = on
..........
(2) Levels
[24] A = 40, B = UK-made, C = off
Temperature (oC): 10, 20, 30, 40 [4 levels]
Machine: made by Japan, by USA, by UK [3 levels] For single factor experiments, run is equivalent to level of the
factor. The number R of runs is equal to the number a of levels.
Switch: on, or off [2 levels]
(4) Replicates (n): number of observations at each run. A larger
value of n increases the accuracy of the experimental results, but
makes the experiment more expensive.
1-5 1-6
Run A B C Run A B C Response (or output, observation)
No Temp. Machine Switch No Temp. Machine Switch
1 10 Japan On 13 30 Japan On
Response is dependent on the values of the factors.
2 10 Japan Off 14 30 Japan Off
3 10 USA On 15 30 USA On Types of response:
4 10 USA Off 16 30 USA Off
5 10 UK On 17 30 UK On Larger the better: product yield, process capability, student’s
6 10 UK Off 18 30 UK Off
mark
7 20 Japan On 19 40 Japan On
Smaller the better: product cost, number of errors
8 20 Japan Off 20 40 Japan Off
9 20 USA On 21 40 USA On Nominal the best: diameter of a shaft
10 20 USA Off 22 40 USA Off
11 20 UK On 23 40 UK On By DOE, we can find out which factors have effects on the response,
12 20 UK Off 24 40 UK Off
and the direction of the influence (i.e., whether the increase of a
factor will increase or decrease the response ?)
1-7 1-8

3. Example 1.2: Single factor: A text book (a = 2, n = 3, R = 2) Example 1.3: Two factors: A text book, B studying hours
Response: Students’ marks Response: Students’ marks
Table 1.2 (a = 2, b = 3, n = 2, R = 6)
Students’ marks
Table 1.3
Text 1 76, 74, 78 Ave = 76
Students’ marks
Text 2 48, 50, 52 Ave = 50 1 hr 2 hr 3 hr
Text 1 45, 47 59, 58 76, 75
Text 2 32, 33 51, 54 70, 71
Text book (a qualitative factor)
Text 1: book by Montgomery
Text 2: book by Taguchi
1-9 1-10
The objectives of an experiment may include What Makes Designed Experiment (DOE) Special?
(1) Determining which factors xs are most influential on the response, (1) It conducts the experiments in a systematic and efficient way.
y. Reduce design and development time, as well as the cost.
(2) It presents experiment results in the simplest and clearest way.
(2) Determining where to set the influential x's, so that y is near the
desired requirement. (3) It extracts the maximum amount of information from a given set
of experiments.
(3) Determining where to set the influential x's, so that variability in y
is small. (4) It draws the right conclusions, despite the variability presented in
the data.
Examples: Increase process capability
Improve students’ study
Reduce output variability.
1-11 1-12

4. Example 1.4 Reducing Defect Level in a Process In this application, DOE
A manufacturing engineer is going to apply DOE to a process for
(1) Determines which factors affect the occurrence of defects
soldering electronic components to printed circuit boards, in order to
reduce defect levels even further.
(2) Determines the direction of adjustment for the effective factors to
Factors to be investigated: further reduce the number of defects per unit.
Solder temperature
Preheat temperature (2) Decides whether changing the factors together produces different
Conveyor speed results than just adjusting individual factor separately?
Flux type
Flux specific gravity
Solder wave depth
Conveyor angle
Thickness of the printed circuit board
Types of components used on the board
1-13 1-14

5. 3 SINGLE FACTOR EXPERIMENTS Table 3.1 Tensile strength of paper
___________________________________________________________________________________________________________
Hardwood Observations
3.1 Introduction _____________________________________________________________________________________________
Concen. (%) 1 2 3 4 5 6 Total Aver.
_________________________________________________________________________________________________________________________________
Single factor experiment can be used in the situation where the effect
(influence) of a single factor on the response is dominant. The effects 5 7 8 15 11 9 10 60 10.00
of all other factors are negligible. 10 12 17 13 18 19 15 94 15.67
The procedures and formulae developed for the single factor 15 14 18 19 17 16 18 102 17.00
experiments can be easily modified, and then used for the several 20 19 25 22 23 18 20 127 21.17
factors experiments. __________________________________________________
383 15.96
Study the effect (influence) of hardwood concentration on the tensile
strength of paper.
3-1 3-2
(1) One factor: hardwood concentration. Table 3.3 General data table for a single factor experiment
(2) Four levels: 5%, 10%, 15%, 20%: a = 4. Level Observation Total Average
1 y11 y12 . . y1n y1. y1.
(3) Response is the tensile strength of paper.
2 y21 y22 . . y2n y2. y2.
(4) Each level has six observations or replicates: n = 6.
. . . . . .
(5) Total number of observations: N = an = 24. . . . . . .
a ya1 ya2 . . yan ya. ya.
y.. y..
yij the jth observation taken under the ith level of the factor.
3-3 3-4

6. yi. the total of the observations at the ith level 3.2 The Analysis of Variance (ANOVA)
n
ANOVA is the most important statistical tool used in DOE. Its task
yi. = ∑ yij (3.1)
j =1
is to decide whether and which factors or interactions have
significant effect (influence) on the response.
yi. the average of the observations at the ith level
yi. = yi. / n (3.2) Model of ANOVA
Yij = μ + τ i + ε ij = μi + ε ij (3.5)
y.. the grand total of all observations
a n a
y.. = ∑∑ yij = ∑ yi. (3.3)
μ the overall mean. μ ≈ y.. (mean ≈ average)
i =1 j =1 i =1
μi the ith level mean. μ i ≈ yi.
y.. the grand average of all observations.
τi the ith level effect. τ i = μi − μ ≈ yi. − y..
y.. = y.. / N (3.4)
ε ij the random error. ε ij = yij − μ i ≈ yij − yi.
N = an is the total number of observations
3-5 3-6
Machining Example (1) y1. = 2.1 + 2.0 + 1.9 = 6.0, y1. = y1. / 3 = 2.0
Study the effect of machine on the process capability y 2. = 0.9 + 1.0 + 1.1 = 3.0, y 2. = y 2. / 3 = 1.0
Table 3.4 y.. = (2.1 + 2.0 + 1.9) + (0.9 + 1.0 + 1.1) = 9.0
Machine Process capability Cp
y.. = 9.0 / 6 = 1.5
USA 2.1, 2.0, 1.9
Japan 0.9, 1.0, 1.1 μ1 ≈ y1. = 2.0, μ 2 ≈ y 2. = 1.0
μ ≈ y.. = 1.5
(1) One factor: machine (qualitative).
τ1 = μ1 − μ ≈ y1. − y.. = 2.0 − 1.5 = 0.5
(2) Two levels: USA-made, Japan-made, a = 2.
τ 2 = μ 2 − μ ≈ y2. − y.. = 1.0 − 1.5 = −0.5
(3) Response is the process capability.
(4) Each level has 3 observations or replicates: n = 3.
(5) Total number of observations: N = an = 2 × 3 = 6.
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7. Evaluation of Errors
y
ε11 = 0.1
For instance, if i = 1, j = 1, yij = y11 = 2.1
ε ij = yij − μi μ1 = 2
τ1 = 0.5
ε11 = y11 − μ1 ≈ y11 − y1. = 2.1 − 2.0 = 0.1 μ = 1.5
y11 = 2.1 τ2 = -0.5
Errors for the machining example (1) μ2 = 1
Table 3.5
Machine Errors
USA 0.1, 0.0, -0.1 y11 = μ + τ1 + ε11
Japan -0.1, 0.0, 0.1 = 1.5 + 0.5 + 0.1 = 2.1
Figure 3.2
3-9 3-10
μ1 μ2
μ1
τ1 τ2
τ1
μ1 μ2 μ1 μ2 μ3
μ μ τ1 τ2 τ3 μ
μ3 μ4 μ τ2 τ3 τ4
μ2 μ3 μ4
τ3 τ4
τ4
μ3 μ4 μ4
Figure 3.3
Figure 3.4
3-11 3-12

8. Hypothesis Test: to test if a factor has effect on the response. Null: Ho : the factor has no effect on the response
If a factor has no effect on the response, the response is always the Alternative: H1: the factor has effect on the response
same, regardless the change of the factor,
Ho : τ1 = τ 2 =L = τ a = 0 (|τi| is small, no effect)
μ1 = μ 2 = L = μ a = μ
H1 : τ i = o for at lease one i
/ (|τi| is larger, has effect) (3.6)
( μ1 − μ ) = ( μ 2 − μ ) = L = ( μ a − μ ) = 0
τ 1 = τ 2 = Lτ a = 0 The larger the value of any |τi|, the more effective the factor will be.
Here, the magnitude of τi, rather than its sign, makes sense.
If a factor has effect, the value of the response will be different, along
with the change of the factor. The more the response differs, the Type I error: the null hypothesis Ho is rejected when it is actually
greater the effect of the factor is. true (The factor is concluded effective, when it is not)
τ i = o for at least one i
/ Type II error: the null hypothesis is accepted when it is actually
false (The factor is concluded ineffective, when it is effective).
3-13 3-14
Machining Example (2) y
Table 3.6
Machine Process capability Cp τ1 = τ2 = 0
USA 1.6, 1.5, 1.4 μ = μ1 = μ2 =1.5
Japan 1.4, 1.5, 1.6
Minor variation is caused by
errors, rather than the machines
μ1 ≈ 1.5, μ 2 ≈1.5 μ ≈ 1.5
τ 1 = μ1 − μ = 15 − 15 = 0
. .
τ 2 = μ2 − μ = 15 − 15 = 0
. .
Conclusion: in this example, the factor (machine) has no effect on
the process capability. Because, when using different Figure 3.5
machines, the response is always the same, on average.
3-15 3-16

9. Analysis of Variance (ANOVA) (1) Calculate sum of squares (SS)
Major Steps Three types of differences:
[1] Difference between an individual observation and the grand
(1) Calculate sum of squares (SS). average:
yij - y..
(2) Determine degrees of freedom (D.O.F.).
[2] Difference between a level average and the grand average:
(3) Calculate mean square (MS).
yi. - y.. = τ i
(4) Calculate the test ratio F0. [3] Difference between an individual observation and the
corresponding level average:
(5) Make conclusion. yij - yi. = ε ij
Note: yij - y.. = ( yi. - y.. ) + ( yij - yi. )
3-17 3-18
a n a n a n
Sum of squares is the sum of squares of the above three differences.
∑∑ ( y
i =1 j =1
ij - y.. ) ∑∑ ( yi. - y.. )
i =1 j =1
∑∑ ( y
i =1 j =1
ij - yi. )
SST The total sum of squares, which is a measure of total
a n a n a n variability in the data.
∑∑ ( y - y.. ) ∑∑ ( yi. - y.. ) ∑∑ ( y - yi . )
2 2 2
ij ij
i =1 j =1 i =1 j =1 i =1 j =1 a n
SS T = ∑ ∑ ( yij - y.. ) 2 (3.7)
i =1 j =1
SST = SSF + SSE
SSF the sum of squares due to the factor, that is the sum of squares
of differences between factor level averages and the grand
average (difference between levels).
a n a a
SS F = ∑∑ ( yi. − y.. ) 2 = n∑ ( yi. − y.. ) 2 ≈ n∑ τ i2 (3.8)
i =1 j =1 i =1 i =1
3-19 3-20

10. SSE the sum of squares due to error, that is the sum of squares Alternative Formulae of Calculating the Sum of Squares
of differences between observations and their level
averages (difference within levels). a n
y . .2
a n a n
SS T = ∑ ∑ yij 2 -
i =1 j =1 an
(3.11)
SS E = ∑∑ ( yij − yi. ) 2 ≈∑∑ ε ij
2
(3.9)
i =1 j =1 i =1 j =1 a
yi. 2 y . .2
SS F = ∑ n an
i =1
- (3.12)
SS T = SS F + SS E (3.10) SS E = SST - SS F (3.13)
Formulae (3.7) to (3.9) are used to describe the underlying ideas.
Formulae (3.11) to (3.13) are used for actually computation.
3-21 3-22
Machining Example (1) If SSF is large, it is due to differences among the means at the
different factor levels. See equation (3.8), a large SSF means that τ2i
Table 3.7
(or |τi|) is great. It is an indication that the factor has significant effect
Machine Process capability Cp
on the response.
USA 2.1, 2.0, 1.9
Usually, SSF is standardized by taking SSF / SSE . By comparing SSF
Japan 0.9, 1.0, 1.1
to SSE, we can see how much variability is due to changing factor
levels and how much is due to error.
a = 2, n = 3
y1⋅ = 6 y2⋅ = 3 y⋅⋅ = 9 Machining Examples
y1⋅ = 2 y2⋅ = 1 y⋅⋅ = 15
. Table 3.8
Effective ? SSF/SSE
SST = (2.12 + 2.0 2 + 1.9 2 + 0.9 2 + 1.0 2 + 1.12 ) − 9 2 / 6 = 1.54 Example 1 Yes 37.5
SS F = (6 2 + 32 ) / 3 − 9 2 / 6 = 1.50 Example 2 No 0
SS E = 1.54 − 1.50 = 0.04
3-23 3-24

11. (2) Determine degrees of freedom (D.O.F.). Machining Example (1)
Table 3.9
total number of observations N: an
Machine Process capability Cp
degrees of freedom of SST: N - 1 = an - 1 (3.14) USA 2.1, 2.0, 1.9
degrees of freedom of SSF: a-1 (3.15) Japan 0.9, 1.0, 1.1
degrees of freedom of SSE: R (n - 1) = a (n – 1) (3.16) a = 2, n = 3
where, (n - 1) is the degrees of freedom of errors in a run. total number of observations N: 2×3=6
degrees of freedom of SST: N–1 =6–1=5
Equation of DOF
degrees of freedom of SSF: a–1 =2–1=1
an - 1 = (a - 1) + a (n - 1) (3.17)
degrees of freedom of SSE: a (n-1) = 2 × (3 - 1) = 4
DOF of SST = DOF of SSF + DOF of SSE
Check equation (3.17): 5 = 1 + 4
3-25 3-26
(3) Calculate mean square (MS). (4) Calculate the test ratio F0.
The ratio between a sum of squares and its DOF. MS F
Fo = (3.20)
MS E
SS F
MS F = (3.18)
a − 1 If H0 is true, F0 follows a theoretical F distribution, which is
completely determined by the two parameters ν1 and ν2.
SS E
MS E = (3.19) ν1 = a - 1 the D.O.F. of MSF (numerator)
a ( n - 1)
ν2 = a(n - 1) the D.O.F. of MSE (denominator)
There is no need to calculate MS for the SST
3-27 3-28

12. f(x) (5) Make conclusion.
ν1 = 1
ν2 = 4 (1) Decide the curve from
SS F
ν1 and ν2. MS F a (n − 1) SS F SS
(2) Decide Fα,ν1,ν2 from α Fo = = a −1 = ⋅ = Q⋅ F
MS E SS E a − 1 SS E SS E
a (n − 1)
a (n − 1)
α Where, Q = is a constant.
a −1
SS F
Fα ,ν1 ,ν 2 F0 is equivalent to (the standardized SSF) except for a constant Q.
SS E
Figure 3.6 SS
The density function curve of an example of the F distribution However, while F0 follows the theoretical F distribution, F does
SS E
not.
3-29 3-30
SS F Table 3.10 Control Limit Fα ,ν 1 ,ν 2 found from the F Table
If F0 is large ----> is large ---> SSF is large ----> τ2i is large --->
SS E
|τi| is large ---> the factor has significant effect on the response. α = 0.05
Control limit: Fα ,ν 1 ,ν 2 ν1 1 2 3 4 5 . ∞
ν2
ν1 = a - 1 and ν2 = a(n - 1) are the two parameters of an F 1 161.4 199.5
distribution
2 18.51 19.00
α is the probability of Type I error, which means concluding that
the factor has significant effect on the response when it in fact has 3 10.13 9.55
Examples:
no effect. Usually, set α = 1% or 5% 4 7.71 6.94
F0.05,2,1 = 199.5
If Fo < Fα ,ν 1 ,ν 2 , conclude that the factor has no effect. .
F0.05,1,4 = 7.71
∞
If Fo > Fα ,ν 1 ,ν 2 , conclude that the factor has significant effect.
3-31 3-32

13. Table 3.11 Table 3.12
3-33 3-34
Table 3.13 Machining Example (1) (a = 2, n = 3, N = 6), from Table 3.7
Analysis of Variance for a Single-Factor Experiment
___________________________________________________ Table 3.14
___________________________________________________
Source of Sum of Degrees of Mean Fo Source of Sum of Degrees of Mean Fo
Variation Squares Freedom Square Variation Squares Freedom Square
___________________________________________________
___________________________________________________
Machine 1.50 1 (= ν1) 1.50 150
Factor SSF a-1 MSF MS F Error 0.04 4 (= ν2) 0.01
MS E
Total 1.54 5
___________________________________________________
Error SSE a(n-1) MSE
Total SST an-1
___________________________________________________
3-35 3-36

14. Specify α = 0.05, Control limit Fα ,ν 1 ,ν 2 = F0.05,1, 4 = 7.71 Machining Example (2) (a = 2, n = 3, N = 6),
Table 3.15
Since F0 > Fα ,ν 1 ,ν 2 , machines have significant effect on the process
Source of Sum of Degrees of Mean Fo
capability.
Variation Squares Freedom Square
Sum of squares: 1.50 + 0.04 = 1.54
Machines 0 1 (= ν1) 0 0
Degrees of freedom 1 + 4 = 5
Error 0.04 4 (= ν2) 0.01
Total 0.04 5
___________________________________________________
Specify α = 0.05, Control limit Fα ,ν 1 ,ν 2 = F0.05,1, 4 = 7.71
Since F0 < Fα ,ν1 ,ν2 , machines don’t have significant effect on the
process capability.
3-37 3-38
Table 3.16 Example: Tensile strength of paper 4 6
y.. 2
___________________________________________________________________________________________________________ SS T = ∑ ∑ yij 2 -
Hardwood Observations i =1 j =1 an
_____________________________________________________________________________________________
Concen. (%) 1 2 3 4 5 6 Total Aver. (383) 2
_________________________________________________________________________________________________________________________________ = (7) 2 + (8) 2 + L + (20) 2 - = 512.96
24
5 7 8 15 11 9 10 60 10.00
4
yi.2 y..2
10 12 17 13 18 19 15 94 15.67 SS F = ∑ -
i =1 n an
15 14 18 19 17 16 18 102 17.00
(60) 2 + (94) 2 + (102) 2 + (127) 2 (383) 2
= - = 382.79
20 19 25 22 23 18 20 127 21.17 6 24
__________________________________________________
SS E = SS T - SS F
383 15.96 = 512.96 - 382.79 = 130.17
a = 4, n = 6, N = an = 24.
3-39 3-40

15. Table 3.17 ANOVA for tensile strength of paper SUMMARY: ANOVA for Single Factor Experiment
___________________________________________________________________________________________________________________________________
Source of Sum of Degrees of Mean Fo
Variation Squares Freedom Square 1. Decide the factor to be investigated and the following two
__________________________________________________________________________________________________________________________________ parameters:
Hardwood 382.79 3 (= ν1) 127.60 19.61 1) number of levels a;
concentration 2) replicate n.
Error 130.17 20 (= ν2) 6.51
2. Carry out the experiments, obtain N (= an) observations yij by a
Total 512.96 23 random manner.
______________________________________________________________________________________________________________________________________
Specify α = 0.01, Control limit Fα ,ν 1 ,ν 2 = F0.01,3, 20 = 4.94 3. Check if the residuals (estimates of errors) satisfy the
requirements.
Since F0 > Fα ,ν 1 ,ν 2 , hardwood concentration has significant effect on
the tensile strength of paper. 4. Calculate sum of squares SST, SSF and SSE (3.11, 3.12, 3.13).
3-41 3-42
5. Calculate the degrees of freedom for each of SST, SSF, SSE (3.14, 3.3 Test on Individual Level Means
3.15, 3.16).
ANOVA results in a single parameter F0, which can tell whether a
6. Calculate the mean squares MSF and MSE (3.18, 3.19). factor has effect on the response, or whether the mean response
values will be different at different levels of the factor from an
7. Calculate the ratio F0 between MSF and MSE (3.20). overall viewpoint.
8. Specify a type I error α, and find the control value Fα ,ν 1 ,ν 2 from However, ANOVA cannot decide the direction of influence of the
the F distribution Tables (ν1 = a - 1, ν2 = a (n – 1) ). factors, nor identify which factor level mean is different from another
level mean.
9. Conclude if the factor has significant effect on the response.
3-43 3-44

16. The plots indicate that changing the hardwood concentration has a
strong effect on the paper strength; specifically, higher hardwood
concentrations produce higher observed paper strength.
Box plots show the variability of the observations within a factor
level and the variability between factor levels.
However, a less subjective approach is required to test the individual
level means.
Figure 3.7 Box plots of paper strength example
3-45 3-46
Multiple Comparison Procedures One of the major problems with making comparisons among level
means is that unrestricted use of these comparisons can lead to an
It is the techniques for making comparisons between two or more
excessively high probability of a Type I error.
level means subsequent to an analysis of variance.
When we run an analysis of variance and obtain a significant F0 For example, if we have 10 levels in which the complete null
value, we have shown simply that the overall null hypothesis is false. hypothesis is true (with α = .05)
We do not know which of a number of possible alternative
hypotheses is true. Ho: µ1 = µ2 = µ3 = … = µ10
H1 : µ1 ≠ µ2 ≠ µ3 ≠ µ4 ≠ µ5 A series t tests between all pairs of level means will lead to making at
Or H1: µ1 ≠ µ2 = µ3 = µ4 = µ5 least one Type I error 57.8% of the time. In other words, the
experimenter who thinks he is working at the (α = .05) level of
Multiple comparison techniques allow us to investigate hypotheses significance is actually working at (α = 0.578).
that involve means of individual levels. For example, we might be
interested in whether µ1 is different from µ2, or whether µ2 is
different from µ3.
3-47 3-48

17. The probability of making at least one Type I error increases as we Least Significant Difference (LSD) Test is a useful tool to test the
increase the number of independent t tests we make between pairs of differences between the individual level means.
level means. While it is nice to find significant differences, it is not
nice to find ones that are not really there. We need to find some way The basic requirement for a LSD test is that the F0 for the overall
to make the comparisons we need but keep the probability of analysis of variance (ANOVA) must be significant. If the F0 was not
incorrect rejections of Ho under control. significant, no comparisons between level means are allowed. You
simply declare that there are no differences among the level means
and stop right there.
On the other hand, if the overall F0 is significant, you can proceed to
make any or all pairwise comparisons between individual level
means by the use of a t test.
3-49 3-50
The t distributions To test between the ith level mean (µi) and jth level mean (µj),
Ho: µi = µj
H1: µi ≠ µj (3.21)
we calculate
yi⋅ − y j⋅
t= (3.22)
2 MS E / n
The value of MSE has already been obtained during the overall
ANOVA. The DOF of t is equal to the DOF of MSE, i.e., a(n-1) or
t(30) is a t distribution with a DOF of 30. (N-a).
t(∞) means a standard normal distribution with (µ = 0 and σ = 1). Then, using a two-tailed t test for a specified level of type I error α
(e.g., α = 0.05), we can decide a pair of two-sided control limits
±tα(dof) from the t table.
3-51 3-52

18. If the t value determined by Eq (3.22) falls within the limits ±tα(dof),
we cannot reject the null hypothesis in (3.21). We will therefore
conclude that there is no difference between µi and µj.
On the other hand, if the t value falls beyond the limits ±tα(dof), we
will reject the null hypothesis and conclude that there is a difference
between µi and µj.
3-53 3-54
The Example of Tensile Strength of Paper Problem: if hardwood concentrations 5% and 20% produce the
Table 3.18 same paper strength ?
___________________________________________________________________________________________________________
Hardwood Observations Ho : μ 1 = μ 4
_____________________________________________________________________________________________
Hypothesis:
Concen. (%) 1 2 3 4 5 6 Total Aver. H1 : μ1 = μ 4
/
_________________________________________________________________________________________________________________________________
5 7 8 15 11 9 10 60 10.00 Five Steps:
10 12 17 13 18 19 15 94 15.67
(1) Use Eq (3.22)
15 14 18 19 17 16 18 102 17.00
20 19 25 22 23 18 20 127 21.17 yi ⋅ − y j ⋅ 10.00 − 21.17
t= = = −7.58
__________________________________________________ 2 MS E / n 2 × 6.51 / 6
a = 4, n = 6
(Note: MSE = 6.51 was calculated during ANOVA, which is
y1⋅ = 10.00, y2⋅ = 15.67, y3⋅ = 17.00, y4⋅ = 21.17 always carried out before the test on individual level means)
3-55 3-56

19. (2) Determine the DOF, DOF = a (n – 1) = 4 × (6 – 1) = 20 General Steps to Test the Individual Level Means
(3) Decide a type I error level. We use α = 0.01 as in the ANOVA for (1) Calculate t using Eq (3.22) and find the MSE value obtained
this example. during ANOVA.
(4) Find the limits from the t table: ±tα(dof) = ±t0.01(20) ≈ 3.00 (2) Determine the DOF, DOF = a (n – 1).
(5) Make conclusion: since the calculated t value (= -7.58) falls (3) Specify a type I error α.
beyond the limits ±3.00, we will reject the null hypothesis.
(4) Find the control limits from the t table: ±tα(dof) based on α and
Conclusion: There is a significant difference in paper strength DOF.
between using hardwood concentrations 5% and
20%. Moreover, since t < 0, µ1 < µ4. (5) Make conclusion by comparing the calculated t value against the
control limits.
3-57 3-58
The Bonferroni Procedure To put this in a way that is slightly more useful to us, if you want the
overall family-wise error rate to be no more than (α = 0.05), and you
In a Bonferroni procedure the family-wise error rate (Type I error) is want to run c tests, then each run of them should be at (α' = 0.05/c).
divided by the number of comparisons. To run these tests, you do exactly what you did in Fisher's LSD test,
though you omit any requirement about the significance of overall F.
The basic idea behind this procedure is that if you run several tests
(say c tests), each at a significance level (type I error) represented by
α', the probability of at least one Type I error can never exceed cα'.
Thus, for example, if you ran 5 tests, each at α' = 0.05, the family-
wise error rate would be at most 5(0.05) = 0.25. But that is a too high
Type I error rate to make anyone happy.
But suppose that you ran each of those 5 tests at the α' = 0.01. Then
the maximum family-wise error rate would be 5(0.01) = 0.05, which
is certainly acceptable.
3-59 3-60

20. a ni 2
y..
SST = ∑∑ yij −
3.5 Unbalanced Experiments 2
DOF = N -1
i =1 j =1 N
In an unbalanced experiment, the number of observations taken under
each run is different. a
yi2. y..
2
SS F = ∑ − DOF = a -1
ni is the number of observations taken under the ith run (for i =1 ni N
single factor experiment, it is simply the ith level).
a
a SS E = SST − SS F DOF = ∑ (ni − 1)
N = ∑ ni = n1 + n2 + L + na i =1
i =1
is the total number of observations.
Note, in unbalanced experiments, N ≠ an.
3-61 3-62
Machining Example SST = (2.12 + 2.0 2 + 1.9 2 ) + (0.9 2 + 1.0 2 ) - 7.9 2 / 5 = 1.348
DOET = N – 1 = 5 – 1 = 4
Machine Process capability Cp
USA 2.1, 2.0, 1.9 SSF = (6.02/3 + 1.92/2) – 7.92/5 = 1.323
Japan 0.9, 1.0,
DOEF = a – 1 = 2 – 1 = 1
n1 = 3, n2 = 2, N = 3 + 2 = 5
SSE = 1.348 – 1.323 = 0.025
y1. = 6.0, y2. = 1.9, y.. = 7.9
DOEE = (n1 – 1) + (n2 – 1) = (3 – 1) + (2 – 1) = 3
3-63 3-64

21. ANOVA for Machining Example (a = 2, n1 = 3, n2 = 2, N = 5) 3.7 Guidelines for Designing Experiments
(1) Recognition and statement of the problem.
Source of Sum of Degrees of Mean Fo Fully develop all ideas about the problem and the specific
Variation Squares Freedom Square objectives of the experiment;
Solicit input from all concerned parties -- engineering, quality,
Machine 1.323 1 (= ν1) 1.323 158.8 marketing, customer, management and operators.
Error 0.025 3 (= ν2) 0.0083
(2) Choice of factors and levels.
Total 1.348 4 Choose the factors, their ranges and levels. Process knowledge is
___________________________________________________ required.
Specify α = 0.05, Control limit: Fα ,ν 1 ,ν 2 = F0.05,1,3 = 10.13 Investigate all factors that may be of importance and avoid being
overly influenced by past experience.
Since F0 > Fα ,ν 1 ,ν 2 , machines have significant effect on the process Keep the number of factor levels low (Most often two levels are
capability. used.) in the early stage
3-65 3-66
(3) Selection of the response (6) Data analysis
Most often, the average or standard deviation (or both) of the Analyze the data so that results and conclusions are objective rather
measured characteristic will be the response variable. than judgmental.
Use software packages and simple graphical methods
(4) Choice of experimental design.
Carry out residual analysis.
Decide the factorial fraction experiment, select the replicate
(sample size) and run order for the experimental trials.
(7) Conclusions and recommendations.
Decide whether or not blocking or other randomization methods are
involved. (8) Experimentation is an important part of the learning process.
New hypotheses are formulated based on the investigation of the
(5) Performing the experiment. tentative hypotheses. Don’t design a single, large comprehensive
experiment at the start of a study. As a rule of thumb, the first
Ensure that everything is being done accordingly to plan.
experiment should spend no more than 25% of the total budget of
the study.
3-67 3-68

22. 4 RANDOMIZED BLOCKS Response: coded roughness of a machined surface.
4.1 Randomized Block Design Main factor (factor to be studied in the experiment)
Nuisance factor probably has an effect on the response, but we are Cutter of a machine tool
not interested in that effect.
Known nuisance factor (factor not to be studied in the experiment,
For unknown and uncontrollable nuisance factors, use Complete but we are well aware of the existence of its effect on the response)
Randomization technique to get rid of them.
Different machine tools of the same type
For known and controllable nuisance factors, use Randomized
Blocking technique instead. Unknown nuisance factor (unknown and uncontrollable factors,
which may have some effect on the response)
The variability of the nuisance factor will contribute to the variability
observed in the experimental data. As a result, the experimental error Temperature, humidity
will reflect both the random error of the unknown nuisance factors
and variability of the known nuisance factor.
4-1 4-2
Table 4.1 MS F MS F DOEE
F0 = = = MS F (4.4)
Machine tool (Block) MS E SS E SS E
Cutter 1 2 3 4 DOEE
1 -2 -1 1 5
2 -1 -2 3 4 So, whether F0 will be increased by the addition of the known
3 -3 -1 0 2 nuisance factors depends on which of SSE or DOEE will increase
4 2 1 5 7 more.
If SSE increases more by including the known nuisance factors, F0
SST = SS F + SS known + SSunknown (4.1) becomes less sensitive. Therefore, Randomized Blocking technique
is preferred.
If the effects of the known nuisance factors are included in the error If DOEE increases more, F0 becomes more sensitive. Therefore,
Complete Randomization technique is preferred.
SS E = SS known + SSunknown (4.2)
Note, the increase of DOEE also reduces the control limit Fα ,ν 1 ,ν 2 (ν2
DOEE = DOEknown + DOEunknown (4.3) = DOEE), and therefore, makes F0 relatively more sensitive.
4-3 4-4

23. Randomized Block Design Block 1 Block 2 Block b
It means blocking all the runs according to the different levels of a
known nuisance factor. Within a block, the order in which the y11 y12 y1b
runs of the main factors are tested is randomly determined.
y21 y22 y2b
The blocks form a more homogeneous experimental unit in terms
of the known nuisance factor. .............
y31 y32 y3b
Effectively, this design strategy improves the accuracy of the . . .
comparisons of the responses by eliminating the variability of the . . .
known nuisance factor. . . .
ya1 ya2 yab
Examples of blocks
Equipment, machinery, batches of raw material, people, time slot Figure 4.1
4-5 4-6
The statistical model Hypotheses of interest are
⎧ i = 1,2,L, a H0: μ1 = μ2 = = … μa
yij = μ + τ i + β j + ε ij ⎨ (4.5)
⎩ j = 1,2,L, b H1: at least one μi ≠ μj (4.6)
where, μ is an overall mean, τi is the effect of the ith run, βj is the Or
effect of the jth block, and εij is the usual NID(0, σ2) random error H0: τ1 = τ2 = … = τa = 0
term.
H1: at least one τi ≠ 0 (4.7)
βj will not be studied, because it is the effect of the known nuisance
factor.
4-7 4-8

24. yi. the total of all observations taken under run i (total of row) yi. the average of the observations taken under the ith run
y.j the total of all observations in block j (total of column) y. j the average of the observations in block j
y.. the grand total of all observations
y.. the grand average of all observations.
N = ab the total number of observations.
b
yi. = yi . / b y. j = y . j / a y.. = y.. / N (4.9)
yi. = ∑ yij i = 1,2,L, a
j =1
a
y. j = ∑ yij j = 1,2,L, b (4.8)
i =1
a b
y.. = ∑∑ yij
i =1 j =1
4-9 4-10
Sum of square Degrees of Freedom
SST = SS F + SS Block + SS E (4.10) SST N - 1 = ab - 1
a b
y..2 SSF a-1
SST = ∑∑ yij 2 - (4.11)
i =1 j =1 N
SSBlocks b-1
a 2
1 y..
SS F = ∑ yi.2 - N
b i=1
(4.12) SSE (ab − 1) − (a − 1) − (b − 1) = ab − a − b + 1 (4.15)
1 b 2 y..2
SS Block = ∑ y. j - (4.13)
a j =1 N
SS E = SST - SS F − SS Block (4.14)
4-11 4-12

25. Test Statistic Table 4.2 ANOVA Table
_______________________________________________________
MS F Source of Sum of Degrees of Mean F0
F0 =
MS E Variation Squares Freedom Square
SS F _______________________________________________________
MS F = (4.16)
a −1 SS F MS F
Factor SS F a -1 Fo =
SS E a −1 MS E
MS E =
ab − a − b + 1 SS Blocks
Blocks SSBlocks b–1
We would reject Ho in (4.7), if Fo > Fα,(a-1)(ab-a-b+1). b −1
SS E
Error SS E ab – a – b + 1
ab − a − b + 1
Total SST N -1
_______________________________________________________
4-13 4-14
We may also examine the ratio of MSBlocks (= SSBlocks/(b-1)) to MSE. If Example 4.1
this ratio is large, it implies that the blocking factor has a large effect
Table 4.3
and that the noise reduction obtained by blocking was probably Machine tool (Block)
Cutter 1 2 3 4 yi.
helpful in improving the effectiveness of the experiment. Otherwise,
1 -2 -1 1 5 3
blocking may actually be unnecessary, i.e, the known nuisance factor 2 -1 -2 3 4 4
should be handled as random error. 3 -3 -1 0 2 -2
4 2 1 5 7 16
y.j -4 -3 9 18 y.. = 20
4-15 4-16

26. The sums of squares are obtained as follows: Table 4.4 ANOVA Table
_______________________________________________________
a b
y..2 20 2 Source of Sum of Degrees of Mean F0
SST = ∑∑ yij 2 - = 154 − = 129
i =1 j =1 N 16 Variation Squares Freedom Square
_______________________________________________________
= (3 + 4 + (−2) + 15 ) −
1 a 2 y..2 1 2 20 2
SS F = ∑ yi. - 2 2 2
= 38.5 Cutter 38.50 3 12.83 14.44
b i=1 N 4 16
Machine 82.50 3 27.50
∑ y.2j - N = 4 ((−4) 2 + (−3) 2 + 92 + 182 ) − 16 = 82.5
b 2 2 (Block)
1 y.. 1 20
SS Block =
a j =1 Error 8.00 9 0.89
SS E = SST - SS F − SS Block = 129 − 38.5 − 82.5 = 8 Total 129.00 15
_______________________________________________________
Specify α = 0.05, F0.05,3,9 = 3.86. Since 14.44 > 3.86, we conclude
that the type of cutter affects the mean hardness reading.
4-17 4-18
If the randomized block design is not used, F0.05,3,12 = 3.49, the
hypothesis of equal mean values from the four cutters cannot be
rejected.
Table 4.5 ANOVA Table
Source of Sum of Degrees of Mean F0
Variation Squares Freedom Square
Cutter 38.50 3 12.83 1.70
Error 90.50 12 7.54
Total 129.00 15
Here, the error is actually the sum of error and block in Table 4.4.
4-19

27. 5 FACTORIAL EXPERIMENTS Case one
Two factors: A: Machine, B: Feed rate
5.1 Introduction
Response: Process capability
Factorial experiments study the effects of several factors on the
response. Table 5.1
Process capability Cp
In factorial experiments, all possible runs (combinations of the levels Feed rate (mm/sec.)
of the factors) are investigated. Machine 0.2 0.4 0.6
For example, for two factors A: temperature (with a levels) and B: USA 2.1, 2.2 1.7, 1.6 1.2, 1.3
humidity (with b levels), the factorial experiment has ab possible Japan 0.9, 1.0 0.7, 0.6 0.4, 0.3
runs.
a = 2, b = 3, n = 2, R =ab = 6
If for each run, n observations are taken, then, the total number N of N = Rn = abn = 2 × 3 × 2 = 12
the observations is equal to abn.
5-1 5-2
Case two 5.2 ANOVA for Two-Factor Factorial Experiments
Three factors: A, B and C.
Mathematical model
a = 2, b = 3, c = 4, n = 2, R = abc = 24
Yijk = μ + τ i + β j + (τβ )ij + ε ijk (5.1)
N = Rn = abcn = 2 × 3 × 4 × 2 = 48
μ the overall mean
τi the effect incurred at the ith level of factor A
Number of runs is equal to the product of the levels of all factors.
βj the effect incurred at the jth level of factor B
(τβ)ij the effect incurred at the interaction between the ith
level of A and the jth level of B
εijk the random error (normally and independently
distributed, zero mean and constant variance)
5-3 5-4

28. Interaction indicates the influence of one factor on the effect of Table 5.2 Data Arrangement for a Two-Factor Factorial Design
another factor, and vice versa. Factor B
1 2 … b Total Average
Factors A: temperature 1 y111, y112, y121, y122, … y1b1, y1b2, y1.. y1..
B: pressure Factor A …, y11n …, y12n …, y1bn
2 y211, y212, y221, y222, … Y2b1, y2b2, y2.. y 2..
Response Process yield …, y21n …, y22n …, y2bn
: … … … … … …
If: (1) Increasing temperature alone cannot increase the yield; and
a ya11, ya12, ya21, ya22, … yab1, yab2, ya.. y a..
(2) Increasing pressure alone cannot increase the yield; and …, ya1n …, ya2n …, yabn
(3) Increasing temperature and pressure at the same time does Total y.1. y.2. … y.b. y…
increase the yield Average y .1. y .2. … y .b. y ...
Then: There is interaction between temperature (factor A) and yijk is the kth observation taken at the ith level of factor A and the jth
pressure (factor B) level of factor B.
The total number of observations: N = abn
5-5 5-6
Total and average at the ith level of factor A: Hypotheses
b n
y
yi .. = Σ Σ yijk y i .. = i .. i = 1,2,...,a Factor A:
j =1 k =1 bn (5.2)
H o : τ 1 = τ 2 = ... = τ a = 0 Factor A has no effect on y
Total and average at the jth level of factor B:
a n y. j . H 1 : at least one τ i ≠ 0 Factor A has effect on y
y. j . = Σ Σ yijk y. j. = j = 1,2,..., b
i =1 k =1 an (5.3) Factor B:
Total and average when A at the ith level and B at the jth level: H 0 : β1 = β2 =L = βb = 0 Factor B has no effect on y
n y H 1 : at least one β j ≠ 0
yij . = Σ yijk y ij . = ij . Factor B has effect on y
k =1 n (5.4)
Interaction AB:
Grand total and average
a b n
y... H o : (τβ ) ij = 0 for all i, j Interaction AB has no effect y
y...= ΣΣΣ yijk y...= (5.5)
i =1 j =1 k =1 abn H1 : at least one (τβ ) ij ≠ 0 Interaction AB has effect on y
5-7 5-8