An overview of statistical control, design of experiment, and monitoring and controlling validated processes.

1. Statistical Tools for the Quality Control
Laboratory and Validation Studies:
Session 1
l STEVEN S. KUWAHARA, Ph.D.
l GXP BioTechnology LLC
l PMB #506
l 1669-2 Hollenbeck Avenue
l Sunnyvale, CA 94087-5042 USA
l Tel. & FAX 408-530-9338
l e-Mail: s.s.kuwahara@gmail.com
l Website: www.gxpbiotech.org
IVTPHL1012S1 1

2. NORMAL DISTRIBUTION
2
⎛ X − µ ⎞
e
1 −1/ 2⎜ i
⎟
Y= ⎝ σ ⎠
σ 2Π
IVTPHL1012S1 2

3. IVTPHL1012S1 3

4. IVTPHL1012S1 4

5. IVTPHL1012S1 5

6. NORMAL DISTRIBUTION
PROPERTIES
l The normal distribution has the following
properties:
l Bell-shaped
l Unimodal
l Symmetrical
l Extends from -∞ to +∞ (tails never reach zero
frequency)
l Same value for mean, median, and mode
l This pattern of variation is common for
manufacturing processes.
IVTPHL1012S1 6

7. IVTPHL1012S1 7

8. VARIANCE (S2)
2
ΣX i
2
−
(ΣX i )
S2 = n
n −1
2
S 2
=
(
Σ Xi − X )
n −1
2 2
nΣX i − (ΣX i )
S2 =
n(n − 1)
IVTPHL1012S1 8

9. Averages and Standard Deviations and the
SEM. 1.
l All of the n measurements that go into the mean
() must be measurements of the same thing.
l The mean of fruits and the mean of oranges are
different things unless all of the fruits are oranges.
l But then it is still the mean of oranges not fruits.
l The standard deviation (s) is a measure of the
variation among the n components of  NOT the
variation of  itself.
l Thus the next item (n + 1) from the original population
should have a 95% chance of being within ± 1.96s of 
but not the next average (1).

10. Averages and Standard Deviations and
the SEM. 2.
l The variation in the averages is the standard
error of the mean (SEM) which is: s/√n.
l Thus the next average (1) has a 95% probability
of being within ±1.96(s/√n) or ±1.96SEM of the
original mean ().
l When dealing with single numbers, s is used,
but when dealing with means the SEM is the
number to use.
l It is incorrect to use s to set a specification on a
value that is actually an average.

11. RANGE AND C.V.
l The range can be related to the standard deviation
for n<16.
XL − Xs
s= d 2 is a tabular value.
d2
S
C.V . = X 100 = % RSD
X
IVTPHL1012S1 11

12. F - TEST
2
s2
Fα ,df 1,df 2 = 2 Note : This is slightly different
s1
from the F - test that is used for ANOVA
and factorial experiments.
Note : F0.05,3,3 = 9.28
F0.05,10,10 = 2.98
12

13. Student’s t
x−µ
t= Basic form.
s
n
df = n − 1
x1 − x2
t= Independent averages,
σ 12 σ 2
2
+
n1 n2
known variances.
13

14. t-TEST vs THEORETICAL OR KNOWN
VALUE
l CHON Analysis. 9.55% H calculated.
l Data: 9.17, 9.09, 9.14, 9.10, 9.13, 9.27. n = 6, ! = 9.15,
s = ± 0.0654
l t0.05/2, 5= 2.57, t0.01/2, 5 = 4.032, t0.001/2, 5 = 6.869, p < 0.001
x − µ 9.15 − 9.55
t= = = 14.98
s 0.0654
n 6
14

15. KNOWN VARIANCES, t-TEST OF TWO
AVERAGES
l Karl Fischer H2O. σ = 0.025 from historical data.
l Data: Lot A: 0.50, 0.53, 0.47.
l Lot B: 0.53, 0.56, 0.51, 0.53, 0.50
l n1=3, n2=5, x1=0.500, x2=0.526
l t0.05/2.∞=1.96, df = n1 + n2 – 2 = 6, t0.05/2, 6 =2.447
0.500 − 0.526
t= = 1.424
2 2
(0.025) +
(0.025)
3 5
15

16. t for Unknown and Equal Variances
x − x n n
t = 1 2 1 2
s n + n
p 1 2
x − x n
if n = n t = 1 2
1 2 s 2
p
df = n + n − 2
1 2
16

17. t-TEST, UNKNOWN BUT EQUAL VARIANCES,
1.
l Data (mg/L Fe3+): Lot A: 6.1, 5.8, 7.0.
l Lot B: 5.9, 5.7, 6.1. xA=6.30, sA=0.6245, xB=5.90,
sB=0.2000.
F0.05 / 2, 2 , 2 = 39.00
2
F =
(0.6245) = 9.75
2
(0.2000 )
2 2
2(0.6245) + 2(0.20 )
sP = = 0.4637
(3 − 1) + (3 − 1)
17

18. t-TEST UNKNOWN BUT EQUAL VARIANCES. 2.
l df = n1 + n2 - 2 df = 4
6.30 − 5.90 3 X 3
t= = 1.056
0.4637 3+3
t0.05 / 2, 4 = 2.78
18

19. POOLED VARIANCE
2 2
n1 −1 s1 + n2 −1 s2
( ) ( )
sp =
n1 +n2 −2
19

20. t for Independent Averages with
unknown and unequal variances.
x1 − x2
t =
2 2
s1 s2
+
n1 n2
2 2
⎛ s1 s2 ⎞
⎜
⎜ n + ⎟
⎝ 1 n2 ⎟⎠
df = 2 2
−2
2 2
⎛ s1 ⎞ ⎛ s2 ⎞
⎜
⎜ n ⎟⎟ ⎜
⎜ n ⎟ ⎟
⎝ 1 ⎠ + ⎝ 2 ⎠
n1 + 1 n2 + 1 20

21. t-TEST UNKNOWN AND UNEQUAL
VARIANCES, 1.
l Data:Extension of Previous Fe+3 mg/L study
1 6.1 5.9 l xA = 6.13, sA = 0.3529
2 5.8 5.7 l xB = 5.76, sB = 0.1647
3 7.0 6.1 l nA = nB = 10
4 6.1 5.8
l F0.05/2,9,9 = 4.03
5 6.1 5.7
l F = (0.3529)2 / (0.1647)2
6 6.4 5.6
7 6.1 5.6 l F = 4.59
8 6.0 5.9
9 5.9 5.7
10 5.8 5.6
21

22. t-TEST UNKNOWN AND UNEQUAL
VARIANCES, 2.
t= 6.13 − 5.76
⎜ 0.3529 ⎞
⎛ ⎟
2 ⎜ 0.1647 ⎞
⎛ ⎟
2
⎝ ⎠
+ ⎝ ⎠
10 10
0.37
t= = 3.0044
0.0151664
l t.05/2,17 = 2.110
22

23. t-TEST UNKNOWN AND UNEQUAL
VARIANCES, 3.
2
2 2
⎛ s 1 s ⎞
2
⎜
⎜ n + ⎟
⎟
⎝ 1 n2 ⎠
df = 2
−2
2 2
⎛ s1 ⎞ ⎛ s2 ⎞
⎜
⎜ n ⎟ ⎟ ⎜
⎜ n ⎟ ⎟
⎝ 1 ⎠ + ⎝ 2 ⎠
n1 + 1 n2 + 1
23

24. t-TEST UNKNOWN AND UNEQUAL
VARIANCES, 4.
⎡ ⎤
⎢
df = ⎢
(0.0395799 )2 ⎥
−2
2 2 ⎥
⎢ (0.01245384 ) + (0.0271261) ⎥
⎢
⎣ 11 11 ⎥
⎦
0.0015666 0.0015666
df = =
0.0000141 + 0.0000669 0.000081
df = 19.3 − 2 = 17 rounded to a whole number
24

25. Paired t-Test
d
t= n d = xi1 − xi 2 df = n −1
sd
2
2 (∑ d )
∑d −
sd = n
n −1
25

26. DATA FOR t -TESTS
l Sample New Original d
l 1. 12.1% 14.7% 2.6%
l 2. 10.9 14.0 3.1
l 3. 13.1 12.9 -0.2
l 4. 14.5 16.2 1.7
l 5. 9.6 10.2 0.6
l 6. 11.2 12.4 1.2
l 7. 9.8 12.0 2.2
l 8. 13.7 14.8 1.1
l 9. 12.0 11.8 -0.2
l 10 9.1 9.7 0.6
l ave. 11.60 12.87 1.27
l s 1.814 2.075 1.126
26

27. Paired t-Test Calculation
d 1.27
t= n= 10 = 3.567
Sd 1.126
t 0.05 / 2,9 = 2.26
Therefore a significant
difference exists.
27

28. t-Test for unknown but equal variances.
X1 − X 2 n1n2
t =
Sp n1 + n2
11.60 − 12.87 100
=
1.9488 10 + 10
t = 1.457 df = n1 + n2 − 2 = 18
t 0.05/2,18 = 2.10
l Showing that there is no significant difference?
28

29. Student’s t to a C.I.
x−µ ts
t= Basic form. = x−µ
s n
n
df = n − 1
ts
µ = x± The value of t is taken from a t - table
n
for n - 1 degrees of freedom and the desired confidence.
29

30. CONFIDENCE INTERVAL 1.
C.I . = µ ± 1.96σ
ts
C.I . = X ±
n
t0.05,n −1 t0.05, 2 = 4.30
30

31. DATA SET FOR SETTING SPECS. 1.
} 67.0 65.8 78.1 66.4 69.0 70.5
} 67.5 75.6 74.2 74.5 85.0 81.1
} 76.0 71.9 70.8 67.3 75.0 74.0
} 72.7 68.8 84.9 73.2 74.7 76.6
} 73.1 82.6 72.2 68.7 69.5 64.2
} n = 30, range = 64.2 - 85.0 ! range = 20.8
} Ave. = 73.03 s or σ = 5.4416 SQRT(30) = 5.4772
} t0.995, 29=2.756 99%C.I.(t) = 70.29 - 75.77
IVTPHL1012S1 31

32. DATA SET FOR SETTING SPECS. 2.
SETS OF 3
l 67.0 72.7 71.9 82.6 70.8 66.4 73.2 85.0 69.5 74.0
l 67.5 73.1 68.8 78.1 84.9 74.5 68.7 75.0 70.5 64.2
l 76.0 65.8 75.6 74.2 72.2 67.3 69.0 74.7 81.1 64.2
l Ave.70.2 70.5 72.1 78.3 76.0 69.4 70.3 78.2 73.7 71.6
l s = 5.06 4.10 3.40 4.20 7.77 4.44 2.52 5.86 6.43 6.54
l CV. 7.21 5.82 4.72 5.37 10.23 6.40 3.58 7.49 8.72 9.13
l CI ±29.0 23.5 19.5 24.1 44.5 25.4 14.4 33.6 36.8 37.5
l X3 = 73.03, s = 3.36, C.V.=3.5%, n=10, t0.995,9 = 3.250
l 99%C.I.(ave) = ±3.46 = 69.67 - 76.49
IVTPHL1012S1 32

33. DATA SET FOR SETTING SPECS. 3.
SETS OF 10
} Set A: 67.0 67.5 76.0 72.7 73.1 65.8 75.6 71.9 68.8 82.6
} Set B: 78.1 74.2 70.8 84.9 72.2 66.4 74.5 67.3 73.2 68.7
} Set C: 69.0 85.0 75.0 74.7 69.5 70.5 81.1 74.0 76.6 64.2
} SQRT(10) = 3.162278 t0.995, 9 = 3.250
} A B C
} 72.1 ± 5.13, 7.1% 73.0 ± 5.49, 7.5% 74.0 ± 6.08, 8.2%
} CI.66.8 - 77.37: 5.2 67.4 - 80.6: 5.64 65.7 - 82.2: 8.23
} Ave(10s)= 73.03, s = 0.9300, C.V. = 1.3%, 99%C.I. = ± 5.33
} 99%CI = 67.7 - 78.4. SQRT(3) = 1.7321 t0.995,2 = 9.925
IVTPHL1012S1 33

34. DATA SET FOR SETTING SPECS. 4.
CUMULATIVE
} n Ave. s C.V. 99%C.I. SQRT(n) t0.995,n-1
} 2 67.25 0.35 0.5 15.9 1.4142 63.66
} 3 70.17 5.06 7.2 42.8 1.1731 9.925
} 4 70.80 4.32 6.1 12.6 2.0000 5.841
} 5 71.26 3.88 5.4 8.0 2.2361 4.604
} 6 70.35 4.12 5.9 6.8 2.4495 4.032
} 9 70.93 3.78 5.3 4.2 3.0000 3.355
} 12 72.78 4.97 6.8 4.5 3.4641 3.106
} 18 72.74 5.40 7.4 3.7 4.2426 2.898
} 24 73.13 5.45 7.5 3.1 4.8990 2.807
} 30 73.03 5.44 7.5 2.7 5.4773 2.756
IVTPHL1012S1 34

35. Wilcoxon’s Signed Rank Test 1.
l Nonparametric test for paired test results.
l Does the same thing as the paired t-test but without the
assumption of normalcy.
l First, take your paired data and calculate the
differences, including their signs.
l Second, place the differences in order (low to high)
based on their absolute values.
l Third, assign a rank to the differences and assign to the
rank a sign according to the sign of the original
difference. (continued)
35

36. Wilcoxon’s Signed Rank Test 2.
l Fourth, count the number or positive or
negative ranks, take the group with the smaller
number of members, and sum the absolute
values of the ranks in that group. This will give
a value, Tn, where n = the number of pairs.
l Go to a Wilcoxon table for n pairs and
significance level of at least 95% to obtain a
tabular value of Tn. For significance, the
calculated value must be smaller than the
tabular value for Tn.
36

37. Signed Rank Test: Example
l A minimum of 6 pairs is needed.
l With 6 pairs, all of the differences must have the same
sign. This gives T6 = 0 which is significant at the 95%
level.
l Differences from 19 pairs of test results.
l Diff : +2, -4, -6, +8, +10, -11, -12, +13, +22, -25,
l Rank:+1, -2, -3, +4, +5, -6, -7, +8, +9, -10,
l Diff: -33, +33, +41, -45, +45, +45, +81, +92, +139
l Rank:-11.5,+11.5,+13,-15, +15, +15, +17, +18, +19
37

38. Signed Rank Test: Example: Continued
l There are 7 negative ranks and 12 positive
ranks, so the absolute sum is taken of:
l -2, -3, -6, -7, -10, -11.5, and -15, this gives:
l T19 = 54.5. The tabular value for T0.05, 19 is
46, so the data show no difference between
the groups.
38

39. A Simpler Nonparametric Test 1.
l The following is not as powerful as the Signed
Rank Test, but is faster and easier. It tests the
hypothesis that p = 0.5 for a given sign. It is a Chi-
square (χ2) test.
2
2 ( n1 − n2 − 1)
χ =
n1 + n2
39

40. Simpler Signed Rank Test 2.
l n1 and n2 are the number of positive and
negative differences. From the previous data
there are 12 positive and 7 negative differences
so:
2 2
2 (12 − 7 − 1) 4 16
χ = = = < 1.0
12 + 7 19 19
40

41. Simpler Signed Rank Test 3.
l Usually, Χ2 > 1.0, so this indicates that
there is no significance since the
calculated Х2 should be larger than the
tabular Χ2 for significance.
l This test can be adopted as a rapid and
easy method to decide if further
investigation is required. It is even
possible to have prepared tables for use.
41

42. Basic Statistics for Quality Control and
Validation Studies: Session 2
• Steven S. Kuwahara, Ph.D.
• GXP BioTechnology, LLC
• PMB 506, 1669-2 Hollenbeck Ave.
• Sunnyvale, CA 94087-5402
• Tel. & FAX (408) 530-9338
• E-Mail: s.s.kuwahara@gmail.com
• Website: www.gxpbiotech.org
ValWkPHL1012S2 1

43. Sample Number Determination 1.
• One of the major difficulties with setting the
number of samples to take lies in determining the
levels of risk that are acceptable. It is in this area
that managerial inaction is often found, leaving a
QC supervisor or senior analyst to make the
decision on the level of risk the company will
accept. If this happens, management has failed its
responsibility.
ValWkPHL1012S2 2

44. Sample Number Determination 2.
• The problem is that all sampling plans, being statistical in
nature, will possess some risk. For instance, if we randomly
draw a new sample from a population we could assume or
predict that a test result from that sample will fall within
±3σ of the true average 99.7% of the time, but there is still
0.3% (3 parts-per-thousand) of the time when the result
will be outside the range for no reason other than random
error. Thus a good lot could be rejected. This is known as a
false positive or a Type I error.
• This is the type of error that is most commonly considered,
but there is type II error also.
ValWkPHL1012S2 3

45. Sample Number Determination 3.
• False positives occur when you declare that there is a
difference when one does not really exist (example given in
the previous slide). Sometimes called producer’s risk,
because the producer will dump a lot that was okay.
• False negatives occur when you declare that a difference
does not exist when, in fact, the difference does exist.
Sometimes called customer’s risk, because the customer
ends up with a defective product. It is also known as a
Type II error.
ValWkPHL1012S2 4

46. SIMPLIFIED FORM OF n CALCULATION
n for an ! to compare with a µ
xi − µ ⎛ s ⎞ = x − µ
t= t ⎜ ⎟ i
s ⎝ n ⎠
n
2 2 2 2
ts 2 ts
n
( )
= x−µ =Δ 2
n= 2
Δ
ValWkPHL1012S2 5

47. EXAMPLE OF SIMPLIFIED METHOD
WITH ITERATION
• Δ = 51- 50 = 1 s = ± 2 Z0.025=1.96
• n = (1.96)2 (2)2 / 1 = 3.8416 X 4 = 15.4 ~ 16
• t0.025,15= 2.131 (2.131)2 = 4.541161
• n = 4.54116 X 4 = 18.16 ~ 19
• t0.025,18= 2.101 (2.101)2 = 4.414201
• n = 4.414201 X 4 = 17.66 ~ 18
• t0.025,17= 2.110 (2.110)2 = 4.4521
• n = 4.4521 X 4 = 17.81 ~ 18
ValWkPHL1012S2 6

48. Sample Number Determination 6.
• Because of the need to define risk and consider the
level of variation that is present, sampling plans
that do not allow for these factors are not valid.
• Examples of these are: Take 10% of the lot below
N=200 and then 5% thereafter. The more famous
one is to take :
• in samples. N +1
ValWkPHL1012S2 7

49. DEVELOPMENT OF A SAMPLING PLAN
• Consider a situation where a product must contain
at least 42 mg/mL of a drug. At 41 mg/mL the
product fails. Because we want to allow for the test
and product variability, we decide that we want a
95% probability of accepting a lot that is at 42 mg/
mL, but we want only a 1% chance of accepting a
lot that is at 41 mg/mL.
• For the sampling plan we need to know the
number (n) of test results to take and average.
• We will accept the lot if the average () exceeds k
mg/mL.
ValWkPHL1012S2 8

50. SAMPLING PLAN CALCULATIONS A.
You will need the table of the normal distribution for this.
• Suppose we have a lot that is at 42.0 mg/mL.
•  would be normally distributed with µ=42.0
– And the SEM = s/!n. We want !>k
x − 42.0 k − 42.0
x= >
s s
n n
x = standard normal deviate
From a “normal” table (or “x” with ν = ∞) we want a
probability of 0.95 that “x” will be greater than the
“k” expression.
ValWkPHL1012S2 9

51. SAMPLING PLAN CALCULATIONS A1.
You will need a normal distribution table for this
• x0.95,∞ = 1.645 (cumulative probability of 0.95)
• We know that this must be greater than the “k”
expression.
• We also know that k must be less than 42.0 since
the smallest acceptable  will be 42.0.
• Therefore:
k − 42.0
= 1.645 since k < x
s
n
ValWkPHL1012S2 10

52. SAMPLING PLAN CALCULATIONS B.
• Now suppose that the correct value for the lot is 41.0 mg/
mL. So now µ = 41.0 and we want a probability of 0.01
that !>k. Now:
x − 41.0 k − 41.0
x= > = −2.326
s s
n n
k − 42.0 1.645
= = −0.707
k − 41.0 − 2.326
k = 41.59
ValWkPHL1012S2 11

53. SAMPLING PLAN CALCULATIONS C.
• Going back to the original equation for a passing result
and knowing that s = ± 0.45 (From our assay validation
studies?)
k − 42.0 41.59 − 42.0 − 0.41
= = = −1.64
s s s
n n n
2
[−1.64]s
= (− 0.41) or n =
([− 1.64][0.45])
2
n (− 0.41)
0.544644
n= = 3.24
0.1681
ValWkPHL1012S2 12

54. SAMPLING PLAN
• The sampling plan now says: To have a 95%
probability of accepting a lot at 42.0 mg/mL or
better and a 1% probability of accepting a lot at
41.0 mg/mL or worse, given a standard deviation
of ± 0.45 mg/mL for the test method; run four
samples and average them. Accept the lot if the
mean is 41.59 mg/mL or better.
• Note that the calculated value of n is close enough
to 3 that some would argue for 3 samples.
ValWkPHL1012S2 13

55. SAMPLE SIZES FOR MEANS
• Suppose we want to determine µ using a test where we
know the standard deviation (s) of the population.
• How many replicates will we need in the sample?
• The length of a confidence interval = L
2 2 2 2
2ts 2 4t s 4t s
L= L = n = 2 L = 2Δ
n n L
ValWkPHL1012S2 14

56. Recalculation of Earlier Problem.
2 2
4t s
n= 2
L
L = 2, s = ±2, t0.95,∞=1.960 (two sided)
2 2
4(2) (1.96 ) 61.4656
n= 2
=
(2) 4
n = 15.4 or 16
Iterate : (t )0.95,15 = 2.131 n = 18.16, (t )0.95,18 = 2.101 n = 17.66
(t )0.95,17 = 2.110 n = 17.81 so n = 18
ValWkPHL1012S2 15

57. Sample size for estimating µ
• Note the statement: We are determining the % of drug
present and we wish to bracket the true amount (µ%)
by ± 0.5% and do this with 95% confidence, so L = 2 x
0.5 = 1.0
• We have 22 previous estimates for which s = 0.45
• Now at the 95% level of significance (1–0.95), t0.975,21 =
2.080.
2 2
4(2.080) (0.45 )
n= 2
= 3.5
(1.0)
ValWkPHL1012S2 16

58. POOLED VARIANCE
2 2
n1 −1 s1 + n2 −1 s2
( ) ( )
sp =
n1 +n2 −2
ValWkPHL1012S2 17

59. Calculating the Confidence Interval, Sp
• The results of the four determinations are: 42.37%,
42.18%, 42.71%, 42.41%.
• ! = 42.42% and s = 0.22% (n2 – 1) = 3
• Using the extra 3 df and s = 0.22% we have:
2 2
21(0.45 ) + 3(0.22 )
Sp = = 0.43
21 + 3
ValWkPHL1012S2 18

60. Calculating the Confidence Interval, L
• Sp = s, the new estimate of the standard deviation, so a
new confidence interval can be calculated with 24 df.
t(0.975, 24)= 2.064.
L = 2(2.064)(0.43)
4
L = 0.88752, rather than 1.0.
( )
C.I. = ± L = 0.44376 or ± 0.45
2
C.I. 95% = 42.42 ± 0.45 or 41.97 - 42.87
Note that n = 4 not 25 for calculating L.
ValWkPHL1012S2 19

61. Sample Sizes for Estimating Standard Deviations. I.
• The problem is to choose n so that s at n – 1 will be
within a given ratio of s/σ.
• Examples are found in reproducibility,
repeatability, and intermediate precision
measurements.
• s = standard deviation experimentally determined.
σ = population or true standard deviation. s2 and
σ2 are corresponding variances.
• You will use n to derive s.
ValWkPHL1012S2 20

62. Sample Sizes for Estimating Standard Deviations. χ2
• This is the asymmetric
2
distribution for σ2.
• Now as an example, χ 2
=
(n − 1)s
assume n-1 = 12. At 12 df, n −1 2
χ2 will exceed 21.0261 5% σ
of the time and it will 2 2
exceed 5.2260 95% of the χ n −1 s
time. Therefore 90% of the = 2
time, χ2 will lie between
5.2260 and 21.0261 for 12
(n − 1) σ
df. 2 2
• Check your tables to ⎛ s ⎞ ⎛ χ ⎞ n −1
confirm this. ⎜ ⎟ = ⎜ ⎟
⎜ (n − 1) ⎟
⎝ σ ⎠ ⎝ ⎠
ValWkPHL1012S2 21

63. Confidence interval for the standard deviation.
• Given the data in the previous slide, we know that
(s2/σ2) will lie between (5.2260/12) and
(21.0261/12), or between 0.4355 and 1.7552.
• Thus the ratio of s/σ will lie between the square
roots of these numbers or between 0.66 and 1.32
or 0.66 < s/σ < 1.32. This gives:
• s/1.32 < σ < s/0.66. If you know s this gives you a
90% confidence interval for the standard
deviation.
• Now let’s reverse our thinking.
ValWkPHL1012S2 22

64. Sample Sizes for Estimating Standard Deviations.
Continued. I.
• Instead of the confidence interval, suppose we say
that we want to determine s to be within ± 20% of
σ with 90% confidence. So:
• 1 – 0.2 < s/σ < 1+ 0.2 or 0.8 < s/σ < 1.2
• This is the same as: 0.64 < (s/σ)2 < 1.44
• Since we want 90% confidence we use levels of
significance at 0.05 and 0.95.
• Now go to the χ2 table under the 0.95 column and
look for a combination where χ2/df is not < 0.64,
but df is as large as possible.
ValWkPHL1012S2 23

65. Sample Sizes for Estimating Standard Deviations.
Continued. II.
• Trial and error shows this number to be about 50.
• Next we go to the column under 0.05 and look for
a ratio that does not exceed 1.44, but df is as small
as possible.
• Trial and error will show this number to be
between 30 and 40.
• You must take the larger of the two numbers and
since df = n – 1, n = 51 replicates.
ValWkPHL1012S2 24

66. Do Not Panic. Consider This!
• Instead of the confidence interval, suppose we say
that we want to determine s to be within ± 50% of
σ with 95% confidence. So:
• 1 – 0.5 < s/σ < 1+ 0.5 or 0.5 < s/σ < 1.5
• This is the same as: 0.25 < (s/σ)2 < 2.25
• Since we want 95% confidence we use levels of
significance at 0.025 and 0.975.
• Now go to the χ2 table under the 0.975 column and
look for a combination where χ2/df is not < 0.25,
but df is as large as possible.
ValWkPHL1012S2 25

67. Greater Confidence, But Lesser Certainty
• Trial and error shows this number to be 8.
• Next we go to the column under 0.025 and look for
a ratio that does not exceed 2.25, but df is as small
as possible.
• Trial and error will show this number to be 8. The
same as the other df.
• You must take the larger of the two numbers and
but in this case df = 8 and n = 9.
• You have a greater confidence interval for a
smaller n.
ValWkPHL1012S2 26

68. n for Comparing Two Averages
x1 − x2
tα , df = Δ = x1 − x2 n1 = n2
σ 12 2
σ2
+
n1 n2
2
Δ 2
t α ,df
2
tα .df =
σ 2
σ 2
n
(σ 2
1 )
+ σ 2 = Δ2
2
1 2
+
n n
n=
2
tα , df (2 2
σ1 + σ 2 )
Δ2
ValWkPHL1012S2 27

69. Introduction to the Analysis of Variance
(ANOVA) I.
This method was aimed at deciding whether or not
differences among averages were due to
experimental or natural variations or true
differences among averages.
R.A. Fisher developed a method based on
comparing the variances of the treatment means
and the variances of the individual measurements
that generated the means.
The technique has been extended into the field
known as DOE or factorial experiments
ValWkPHL1012S2 28

70. Introduction to the Analysis of Variance
(ANOVA) II.
• The method is based on the use of the F-test and
the F-distribution (Named after him.)
– The F-distribution, and all distributions related to
errors, is a skewed, unsymmetrical distribution.
2
ns y
F= 2
s pooled
– S2y represents the variance among the treatments and
s2pooled is the variance of the individual results (system
noise).
ValWkPHL1012S2 29

71. Introduction to the Analysis of Variance
(ANOVA) III.
• F increases as the number of replicates increases.
– In simple ANOVA systems n is the same for all
treatments.
– By increasing n you amplify small differences between
the variances of the treatment means and the system
noise.
– An F value of 1.0 or less says that the system noise is
greater than the variance of the means. This suggests
that the differences among the means are due to
experimental or environmental variations.
ValWkPHL1012S2 30

72. Introduction to the Analysis of Variance
(ANOVA) IV.
• Because of the importance of system noise, before
doing an ANOVA or factorial experiment, you
should reduce variation in the system to a
minimum.
– You should remove all special cause variation and
minimize common cause variation.
– Methods such as Statistical Process Control (SPC)
should be used to reduce variations.
• Note: A system where special cause variation has been
eliminated and only common cause variation is left is known as
a system under statistical control.
ValWkPHL1012S2 31

73. Introduction to the Analysis of Variance (ANOVA)
V.
• The F-distribution depends on the number of degrees
of freedom of the numerator and denominator and the
level of type 1 error that you will accept.
– For each level of type 1 error there are different
distribution tables. The exact value of F then depends
on the number of degrees of freedom of the numerator
and denominator.
• If the calculated F exceeds the tabular F, it is then significant at
the1-α level. Where α is the level of type 1 error that you are
willing to accept.
• α is the p value. Most statistical software programs will
calculate the p value. Normally, you want 0.05 or 0.01.
• Type-1 error is where you falsely conclude that there is a
difference. AKA: False positive, producer’s risk.
ValWkPHL1012S2 32

74. Fairness of 4 sets of dice. (Taken from Anderson, MJ and
Whitcomb, PJ, DOE Simplified, CRC Press, Boca Raton, FL, 2007.)
• Frequency distribution for 56 rolls of dice.
Dots White Blue Green Purple
6 6+6 6+6 6+6 6
5 5 5 5 5
4 4 4+4 4+4 4
3 3+3+3+3+3 3+3+3+3 3+3+3+3 3+3+3+3+3
2 2+2+2 2+2+2+2 2+2+2+2 2+2+2+2+2
1 1+1 1 1 1
Mean (y) 3.14 3.29 3.29 2.93
Var. (s2) 2.59 2.37 2.37 1.76
n = 14 Grand Ave. = 3.1625
• Grand average = Total of all dots/56 dice (4X14)
ValWkPHL1012S2 33

75. Fairness of 4 sets of dice. Calculation of F.
Note differences in denominator.
Since F is much less than 1.0 we can assume that there is no
significant difference among the colors even without looking
at an F table.
s2 =
(3.14 − 3.1625)2 + (3.29 − 3.1625)2 + (3.29 − 3.1625)2 + (2.93 − 3.1625)2
y
4 −1
2
s y = 0.029
s 2 = 2.59 + 2.37 + +2.37 + 1.76 = 2.28
pooled
4
2
n * s y 14 * 0.029
F= 2 = = 0.18
s pooled 2.28
ValWkPHL1012S2 34

76. Fairness of 4 sets of dice. How about a loaded
set?
Dots White Blue Green Purple
6 1 3 6 1
5 1 2 5 2
4 1 3 1 3
3 2 4 1 1
2 5 1 0 2
1 4 1 1 5
Mean (y) 2.50 3.93 4.93 2.86
Var. (s2) 2.42 2.38 2.07 3.21
n = 14 Grand Ave. = 3.555
δ= -1.055 0.375 1.375 -0.695
δ2 = 1.1130 0.1406 1.8906 0.4830
Σδ2 = 3.6245 Σδ2/3 = s2y = 1.2082
ValWkPHL1012S2 35

77. Fairness of 4 sets of dice. How about a loaded
set? ANOVA
2 2.42 + 2.38 + 2.07 + 3.21
spooled = = 2.52 df = 4(14 - 1) = 52
4
2
s y = 1.21 df = 3 (4 - 1)
14 *1.21
F= = 6.71 F3,52 = 6.71
2.52
Tabular F3,52 = 2.839 − 2.758 at 5%, p = 0.05
and 4.313 - 4.126 at 1%, and 6.595 - 6.171 at 0.1%.
Range is for F3,40 to F3,60 . Significant at p = 0.001
ValWkPHL1012S2 36

78. Least Significant Difference
Lucy in the Sky with Diamonds (LSD)
• DO NOT EVER USE THIS METHOD WITHOUT THE
PROTECTION OF A SIGNIFICANT ANOVA
RESULT ! ! !
• There are 45 combinations of 10 results taken in pairs. If you focus
mainly on the high and low results, you are almost guaranteed to
encounter a type-1 error.
– This is why you need to use the ANOVA coupled with an LSD
determination.
• The LSD is based on the equations for confidence intervals.
n 2
LSD = ± t(1−α ,df ) × s pooled 2 / n s pooled =
∑s 1 i
n
ValWkPHL1012S2 37

79. LSD for the Current Problem
2.42 + 2.38 + 2.07 + 3.21
s pooled = = 2.52 = 1.59
4
2
LSD = 2.01 ×1.59 = ±1.21
14
at 99% LSD = ±1.333 for t (0.99,df =52 ) ≅ 2.68
• The (1-α) level of the t determines the level of
significance for the LSD.
• n = 14 for replicates, but s2pooled had 4X(14-1)
= 52 df.
ValWkPHL1012S2 38

80. So where are the bad dice?
• Given the LSD = ±1.333, the result can be displayed in
different ways.
• Plot the result as the mean of the average count of the
treatments (colors) ± ½ LSD.
– Then look for overlaps. A significant difference will not have
an overlap.
• Or take the difference between means and compare
them to the LSD.
– In the present case, the white and purple dice are similar, but
the green dice are definitely higher, with the blue dice
different from the white, but not from the green and only
marginally different from the purple.
ValWkPHL1012S2 39

81. White = 2.50 Blue = 3.93 Green=4.93 Purple=2.86
White = 2.50 1.43 2.43 0.36
Blue = 3.93 1.43 1.00 1.07
Green=4.93 2.43 1.00 2.07
Purple=2.86 0.36 1.07 2.07
For 95% confidence, the LSD is ± 1.21 and for 99%, the LDS
is ± 1.33.
So blue and green are different from white, and green is
different from purple and white at the 99% level.
White and purple are the same as are blue and green.
Purple is also similar to blue, but not to green.
All of this holds at the 99% level, thus at p = 0.01 we conclude
that blue and green dice run to higher numbers than white
and purple.
ValWkPHL1012S2 40

82. David.LeBlond@sbcglobal.net

83. “Designing
an
efficient
process
with
an
effec;ve
process
control
approach
is
dependent
on
the
process
knowledge
and
understanding
obtained.
Design
of
Experiment
(DOE)
studies
can
help
develop
process
knowledge
by
revealing
rela;onships,
including
mul;-­‐factorial
interac;ons,
between
the
variable
inputs
…
and
the
resul;ng
outputs.
Risk
analysis
tools
can
be
used
to
screen
poten;al
variables
for
DOE
studies
to
minimize
the
total
number
of
experiments
conducted
while
maximizing
knowledge
gained.
The
results
of
DOE
studies
can
provide
jus;fica;on
for
establishing
ranges
of
incoming
component
quality,
equipment
parameters,
and
in
process
material
quality
aKributes.”
2

84. What
is
it?
The
ability
to
accurately
predict/control
process
responses.
How
do
we
acquire
it?
Scien;fic
experimenta;on
and
modeling.
How
do
we
communicate
it?
Tell
a
compelling
scien;fic
story.
Give
the
prior
knowledge,
theory,
assump;ons.
Show
the
model.
Quan;fy
the
risks,
and
uncertain;es.
Outline
the
boundaries
of
the
model.
Use
pictures.
Demonstrate
predictability.
3

85. Screening
Designs
•
2
level
factorial/
frac;onal
factorial
designs
•
Weed
out
the
less
important
factors
•
Skeleton
for
a
follow-­‐up
RSM
design
Response
Surface
Designs
•
3+
level
designs
•
Find
design
space
•
Explore
limits
of
experimental
region
Confirmatory
Designs
•
Confirm
Findings
•
Characterize
Variability
4

86. Key
Factors
Key
Responses
Cau;on:
EVERYTHING
depends
on
gecng
this
right
!!!
5

87. Fixed
Factors
Responses
Disint
(A
or
B)
Dissolu;on%
(>90%)
Drug%
(5-­‐15%)
Make
Disint%
(1-­‐4%)
ACE
DrugPS
(10-­‐40%)
Tablets
WeightRSD%(<2%)
Lub%
(1-­‐2%)
Day
Random
Factors
6

88. Trial
DrugPS
Lub%
Disso%
1
25
1
85
2
25
2
95
3
10
1.5
90
4
40
1.5
70
Lubricant%
2
95
90
70
1
85
10
40
DrugPS
7

89. Lubricant%
2
95
90
70
1
85
10
40
DrugPS
Disso% = 86.667
+10 × Lub%
−0.667 × DrugPS
+ε
8

90. ž Previous
example
had
only
2
factors.
Ø Factor
space
is
2D.
We
can
visualize
on
paper.
ž With
3
factors
we
need
3D
paper.
Ø Corners
even
further
away
ž Most
new
processes
have
>3
factors
ž OFAT
can
only
accommodate
addi;ve
models
ž We
need
a
more
efficient
approach
9

91. True
response
• Goal:
Maximize
response
• Fix
Factor
2
at
A.
Factor
2
• Op;mize
Factor
1
to
B.
80
E
60
40
• Fix
Factor
1
at
B.
C
• Op;mize
Factor
2
to
C.
A
• Done?
True
op;mum
is
Factor
1
=
D
and
B
D
Factor
2
=
E.
Factor
1
• We
need
to
accommodate
curvature
and
interac/ons
10

92. Response
A
B
C
D
Factor
level
• A
to
B
may
give
poor
signal
to
noise
• A
to
C
gives
beKer
signal
to
noise
and
rela;onship
is
s;ll
nearly
linear
• A
to
D
may
give
poor
signal
to
noise
and
completely
miss
curvature
• Rule
of
thumb:
Be
bold
(but
not
too
bold)
11

93. Trial
DrugPS
Lub%
Disso%
1
10
1
75
2
10
2
100
3
40
1
75
4
40
2
80
2
100
80
Lubricant%
1
75
75
10
40
DrugPS
12

94. Lubricant%
2
100
80
1
75
75
10
40
DrugPS
Disso% = 43.33
+0.667 × DrugPS
+31.667 × Lub%
−0.667 × DrugPS × Lub%
+ε
13

95. ž Model
non-­‐addiKve
behavior
› interacKons,
curvature
ž Efficiently
explore
the
factor
space
ž Take
advantage
of
hidden
replicaKon
14

96. Planar:
no
interac;on
Non-­‐planar:
interac;on
Y = a + b ⋅ X1 + c ⋅ X 2 Y = a + b ⋅ X1 + c ⋅ X 2 + d ⋅ X 1 ⋅X2
15

97. 16

98. 17

99. 18

100. 2
A
B
Trial
DrugPS
Lub%
Disso%
Lub%
1
10
1
C
2
10
2
A
1
C
D
3
40
1
D
10
40
4
40
2
B
DrugPS
B +D A +C A
B
MainEffectDrugPS = −
2 2 C
D
A +B C +D A
B
MainEffectLub% = −
2 2 C
D
C +B A +D A
B
InteractionEffectDrugPS×Lub% = −
2 2 C
D
19

101. Uncoded
Units
Coded
Units
Trial
DrugPS
Lub%
Trial
DrugPS
Lub%
1
10
1
1
-­‐1
-­‐1
2
10
2
2
-­‐1
+1
3
40
1
3
+1
-­‐1
4
40
2
4
+1
+1
• Coding
helps
us
evaluate
design
proper;es
• Some
sta;s;cal
tests
use
coded
factor
units
for
analysis
(automa;cally
handled
by
sotware)
• Easy
to
convert
between
coded
(C)
and
uncoded
(U)
factor
levels
U − Umid
C= ⇔ U = C(Umax − Umid ) + Umid
Umax − Umid
20

102. +1
A
B
Trial
DrugPS
Lub%
DrugPS Disso%
*Lub%
Lub%
1
-­‐1
-­‐1
+1
C
2
-­‐1
+1
-­‐1
A
-­‐1
C
D
-­‐1
+1
3
+1
-­‐1
-­‐1
D
DrugPS
4
+1
+1
+1
B
Disso = a a = (+ A + B + C + D) / 4
+b × Lub% b = MEDrugPS / 2 = (−A + B − C + D) / 4
+c × DrugPS c = MELub% / 2 = (+ A + B − C − D) / 4
+d × Lub% × DrugPS d = IEDrugPS×Lub% / 2 = (−A + B + C − D) / 4
+ε
21

103. Disso = a + b × Lub + c × DrugPS + d × Lub × DrugPS + ε
ž It
is
obtained
through
the
“magic”
of
regression.
ž b
measures
the
“main
effect”
of
Lub
ž c
measures
the
“main
effect”
of
DrugPS
ž d
measures
the
“interac;on
effect”
between
Lub
and
DrugPS
Ø if
d
=
0,
effects
of
Lub
and
DrugPS
are
addi;ve
Ø if
d
≠
0,
effects
of
Lub
and
DrugPS
are
non-­‐addi;ve
ž ε
represents
trial
to
trial
random
noise
22

104. +1
+1
+1
Lub%
Lub%
Lub%
-­‐1
-­‐1
-­‐1
-­‐1
+1
-­‐1
+1
-­‐1
+1
DrugPS
DrugPS
DrugPS
Trial
DrugPS
Lub%
Trial
DrugPS
Lub%
Trial
DrugPS
Lub%
1
-­‐1
-­‐1
1
-­‐1
-­‐1
1
-­‐1
-­‐1
2
-­‐1
+1
2
-­‐1
0
2
-­‐1
-­‐1
3
+1
-­‐1
3
+1
0
3
+1
+1
4
+1
+1
4
+1
+1
4
+1
+1
Inner
product:
+1-­‐1-­‐1+1=0
+1+0+0+1=2
+1+1+1+1=4
23

105. 24

106. Dissolu;on
(%LC)
1%
Lubricant
2%
Lubricant
90
10
40
DrugPS
25

107. y = a + bA + cB + dC + eAB + fAC + gBC + hABC + ε
• Average
Number
of
Number
of
• Main
Effects
Factors
(k)
Trials
(df
=
• 2-­‐way
interac;ons
2k)
• Higher
order
0
1
interac;ons
(or
1
2
es;mates
of
noise)
2
4
3
8
4
16
5
32
6
64
26

108. Main Effects
Trial
I
A
B
C
D=AB
E=AC
F=BC
ABC
1
+
-­‐
-­‐
-­‐
+
+
+
-­‐
2
+
+
-­‐
-­‐
-­‐
-­‐
+
+
3
+
-­‐
+
-­‐
-­‐
+
-­‐
+
4
+
+
+
-­‐
+
-­‐
-­‐
-­‐
5
+
-­‐
-­‐
+
+
-­‐
-­‐
+
6
+
+
-­‐
+
-­‐
+
-­‐
-­‐
7
+
-­‐
+
+
-­‐
-­‐
+
-­‐
8
+
+
+
+
+
+
+
+
y = a + bA + cB + dC + eD + fE + gF + ε
• Can
include
addi;onal
variables
in
our
experiment
by
aliasing
with
interac;on
columns.
• Leave
some
columns
to
es;mate
residual
error
for
sta;s;cal
tests
27

109. Trial
I
A
B
C
AB
AC
BC
ABC
1
+
-­‐
-­‐
-­‐
+
+
+
-­‐
2
+
+
-­‐
-­‐
-­‐
-­‐
+
+
+1 3
+
-­‐
+
-­‐
-­‐
+
-­‐
+
4
+
+
+
-­‐
+
-­‐
-­‐
-­‐
C 5
+
-­‐
-­‐
+
+
-­‐
-­‐
+
+1
B 6
+
+
-­‐
+
-­‐
+
-­‐
-­‐
-1 -1 7
+
-­‐
+
+
-­‐
-­‐
+
-­‐
-1 A +1
8
+
+
+
+
+
+
+
+
y = a + bA + cB + dC
• Create
a
half
frac;on
by
running
only
the
ABC
=
+1
trials
• Note
confounding
between
main
effects
and
interac;ons
• Compromise:
must
assume
interac;ons
are
negligible
• In
this
case
(not
always)
design
is
“saturated”
(no
df
for
sta;s;cal
tests).
28

110. • “I=ABC”
for
this
23-­‐1
half
frac;on
is
called
the
“Defining
Rela;on”
• Note
that
“I=ABC”
implies
that
“A=BC”,
“B=AC”,
and
“C=AB”.
• 3-­‐way
interac;ons
are
confounded
with
the
intercept
• Main
effects
are
confounded
with
2-­‐way
interac;ons
• The
number
of
factors
in
a
defining
rela;on
is
called
the
“Resolu;on”
• This
23-­‐1
half
frac;on
has
resolu;on
III
• We
denote
this
frac;onal
factorial
design
as
2III3-­‐1
29

111. • I=ABCD
for
this
24-­‐1
half
frac;on
is
called
the
Defining
Rela;on
• Note
that
I=ABCD
implies
•
A=BCD,
B=ACD,
C=ABD,
and
D=ABC.
•
AB=CD,
AC=BD,
AD=BC
•
Main
effects
are
confounded
with
3-­‐way
interac;ons
•
Some
2-­‐way
interac;ons
are
confounded
with
others.
We
like
our
screening
designs
to
be
at
least
resolu;on
IV
(I=ABCD)
30

112. Number
of
Factors
2
3
4
5
6
7
8
9
10
11
12
13
14
15
4
Full
III
6
IV
8
Full
IV
III
III
III
Number
of
Design
Points
12
V
IV
IV
III
III
III
III
III
16
Full
V
IV
IV
IV
III
III
III
III
III
III
III
20
III
III
III
III
III
24
IV
IV
IV
IV
III
III
III
32
Full
VI
IV
IV
IV
IV
IV
IV
IV
IV
IV
48
V
V
64
Full
VII
V
IV
IV
IV
IV
IV
IV
IV
96
V
V
V
128
Full
VIII
VI
V
V
IV
IV
IV
IV
31

113. Trial
DrugPS
Lub% Disso%
2 98,102 88,82
1
10
1
76
Lub%
2
10
2
98
3
40
1
73
4
40
2
82
1 76,84 73,77
5
10
1
84
10 40
6
10
2
102
7
40
1
77
DrugPS
8
40
2
88
FiKed
model
is
based
on
averages
SDindividual
SDaverage =
number of replicates
32

114. ReplicaKng
1
measurement
batch
3
batches
per
batch
producKon
Repeated
3
measurements
1
batch
per
batch
measurement
33

115. Trial
DrugPS
Lub% Disso%
ReplicaKon
1. Every
operaKon
that
1
10
1
76
contributes
to
variaKon
is
2
10
2
98
3
40
1
73
redone
with
each
trial.
4
40
2
82
2. Measurements
are
5
10
1
84
independent.
6
10
2
102
3. Individual
responses
are
7
40
1
77
analyzed.
8
40
2
88
RepeKKon
Trial
DrugPS
Lub% Disso%
1. Some
operaKons
that
contribute
variaKon
are
not
1
10
1
76, 84
redone.
2
10
2
98, 102
3
40
1
73, 77
2. Measurements
are
correlated.
4
40
2
82, 88
3. The
averages
of
the
repeats
should
be
analyzed
(usually).
34

116. ž Frac;onal
factorial
designs
are
generally
used
for
“screening”
ž Sta;s;cal
tests
(e.g.,
t-­‐test)
are
used
to
“detect”
an
effect.
ž The
power
of
a
sta;s;cal
test
to
detect
an
effect
depends
on
the
total
number
of
replicates
=
(trials/
design)
x
(replicates/trial)
ž If
our
experiment
is
under
powered,
we
will
miss
important
effects.
ž If
our
experiment
is
over-­‐powered,
we
will
waste
resources.
ž Prior
to
experimen;ng,
we
need
to
assess
the
need
for
replica;on.
35

117. 2 2
⎛ σ ⎞
N = (#points
in
design)(replicates/point) ≅ 4 z1−α + z1−β ( 2
) ⎜ ⎟
⎝ δ ⎠
σ
=
replicate
SD
δ
=
size
of
effect
(high
–
low)
to
be
detected.
α
=
probability
of
false
detec;on
β
=
probability
of
failure
to
detect
an
effect
of
size
δ
α
z1-­‐α/2
β
z1-­‐β
2
0.01
2.58
0.1
1.28
⎛ σ ⎞
0.05
1.96
N ≅ 16 ⎜ ⎟
0.2
0.85
⎝ δ ⎠
0.10
1.65
0.5
0.00
• While
not
exact,
this
ROT
is
easy
to
apply
and
useful.
• Commercial
sotware
will
have
more
accurate
formulas.
36

118. 2 2
⎛ σ ⎞
(
N = (#points
in
design)(replicates/point) ≅ 4 z1−α + z1−β
2
) ⎜ ⎟
⎝ δ ⎠
Disso%
WtRSD
Replicate
SD
σ
1.3
0.1
Difference
to
detect
δ
2.0
0.2
False
detecKon
probability
α
0.05
0.05
z1-­‐α/2
1.96
1.96
DetecKon
failure
probability
β
0.2
0.2
z1-­‐β
0.85
0.85
Required
number
of
trials
N
13.3
8
37

119. Run A B C D E Confounding Table
1 - - - - + I = ABCDE
2 + - - - - A = BCDE
3 - + - - - B = ACDE
4 + + - - + C = ABDE
5 - - + - - D = ABCE
6 + - + - + E = ABCD
7 - + + - + AB = CDE
8 + + + - - AC = BDE
9 - - - + - AD = BCE
10 + - - + + AE = BCD
11 - + - + + BC = ADE
12 + + - + - BD = ACE
13 - - + + + BE = ACD
14 + - + + - CD = ABE
15 - + + + - CE = ABD
16 + + + + + DE = ABC
38

120. ž Sta;s;cal
test
for
presence
of
curvature
(lack
of
fit)
ž Addi;onal
degrees
of
freedom
for
sta;s;cal
tests
ž May
be
process
“target”
secngs
ž Used
as
“controls”
in
sequen;al
experiments.
ž Spaced
out
in
run
order
as
a
check
for
drit.
39

121. Complete
RandomizaKon:
• Is
the
cornerstone
of
sta;s;cal
analysis
• Insures
observa;ons
are
independent
• Protects
against
“lurking
variables”
• Requires
a
process
(e.g.,
draw
from
a
hat)
• May
be
costly/
imprac;cal
Restricted
RandomizaKon:
• “Difficult
to
change
factors
(e.g.,
bath
temperature)
are
“batched”
• Analysis
requires
special
approaches
(split
plot
analysis)
Blocking:
• Include
uncontrollable
random
variable
(e.g.,
day)
in
design.
• Assume
no
interac;on
between
block
variable
and
other
factors
• Excellent
way
to
reduce
varia;on.
• Rule
of
thumb:
“Block
when
you
can.
Randomize
when
you
can’t
block”.
40

122. 41

123. Confounding Table
I = ABCDE
Blk = AB = CDE
A = BCDE
B = ACDE
C = ABDE
D = ABCE
E = ABCD
AC = BDE
AD = BCE
AE = BCD
BC = ADE
BD = ACE
BE = ACD
CD = ABE
CE = ABD
DE = ABC
42

124. StdOrder
RunOrder
CenterPt
Blocks
Disint
Drug%
Disint%
DrugPS
Lub%
11
1
1
2
A
5
1.0
10
2.0
13
2
1
2
A
5
4.0
10
1.0
19
3
0
2
A
10
2.5
25
1.5
15
4
1
2
A
5
1.0
40
1.0
18
5
1
2
B
15
4.0
40
2.0
14
6
1
2
B
15
4.0
10
1.0
20
7
0
2
B
10
2.5
25
1.5
16
8
1
2
B
15
1.0
40
1.0
17
9
1
2
A
5
4.0
40
2.0
12
10
1
2
B
15
1.0
10
2.0
9
11
0
1
A
10
2.5
25
1.5
7
12
1
1
B
5
4.0
40
1.0
1
13
1
1
B
5
1.0
10
1.0
2
14
1
1
A
15
1.0
10
1.0
4
15
1
1
A
15
4.0
10
2.0
3
16
1
1
B
5
4.0
10
2.0
10
17
0
1
B
10
2.5
25
1.5
5
18
1
1
B
5
1.0
40
2.0
8
19
1
1
A
15
4.0
40
1.0
6
20
1
1
A
15
1.0
40
2.0
43

125. RunOrder
CenterPt
Blocks
Disint
Drug%
Disint%
DrugPS
Lub%
Disso%
WtRSD
1
1
2
A
5
1.0
10
2.0
100.4
1.6
2
1
2
A
5
4.0
10
1.0
103.0
2.1
3
0
2
A
10
2.5
25
1.5
88.8
1.6
4
1
2
A
5
1.0
40
1.0
94.3
2.3
5
1
2
B
15
4.0
40
2.0
78.9
1.6
6
1
2
B
15
4.0
10
1.0
102.9
2.0
7
0
2
B
10
2.5
25
1.5
90.9
1.4
8
1
2
B
15
1.0
40
1.0
91.8
2.2
9
1
2
A
5
4.0
40
2.0
76.3
1.4
10
1
2
B
15
1.0
10
2.0
103.4
1.6
11
0
1
A
10
2.5
25
1.5
89.9
1.8
12
1
1
B
5
4.0
40
1.0
91.8
2.2
13
1
1
B
5
1.0
10
1.0
101.2
2.2
14
1
1
A
15
1.0
10
1.0
101.8
2.6
15
1
1
A
15
4.0
10
2.0
102.5
1.4
16
1
1
B
5
4.0
10
2.0
100.3
1.5
17
0
1
B
10
2.5
25
1.5
91.2
1.6
18
1
1
B
5
1.0
40
2.0
76.3
1.3
19
1
1
A
15
4.0
40
1.0
92.4
2.1
20
1
1
A
15
1.0
40
2.0
76.8
1.6
44

126. 45

127. 46

128. 47

129. 48

130. 49

131. Source DF Adj MS F P
Blocks 1 2.21 0.11 0.745
Disint 1 0.30 0.01 0.905
Drug% 1 2.94 0.15 0.707
Disint% 1 0.30 0.01 0.905
DrugPS 1 1174.45 58.93 0.000
Lub% 1 258.61 12.98 0.004
Curvature 1 32.68 1.64 0.225
Res Error 12 19.93
2.179
is
the
1-­‐α/2
th
quan;le
of
the
t-­‐
distribu;on
having
12
df.
50

132. Source DF Adj MS F P
Blocks 1 0.01090 0.51 0.487
Disint 1 0.03751 1.77 0.208
Drug% 1 0.00847 0.40 0.539
Disint% 1 0.08282 3.91 0.071
DrugPS 1 0.00189 0.09 0.770
Lub% 1 2.10586 99.46 0.000
Curvature 1 0.21198 10.01 0.008
Res Error 12 0.02117
51

133. Disso%
• Only
DrugPS
and
Lub%
show
significant
main
effects
• Plot
of
Disso%
residuals
vs
predicted
Disso%
shows
systema;c
paKern.
• The
residual
SD
(4.5)
is
considerably
larger
than
expected
(1.3)
WtRSD
• Only
Lub%
shows
a
sta;s;cally
significant
main
effect
• Curvature
is
significant
for
WtRSD
Therefore
• Only
DrugPS
and
Lub%
need
to
be
considered
further
• The
other
3
factors
can
fixed
at
nominal
levels.
• The
predic;on
model
is
inadequate.
Addi;onal
experimenta;on
is
needed.
52

134. Trial
DrugPS
Lub%
Disso%
1
10
1
C
2
10
2
A
2
A
F
B
3
40
1
D
Lub%
4
40
2
B
G
I
H
5
25
1
E
1
C
E
D
6
25
2
F
10
40
7
10
1.5
G
DrugPS
8
40
1.5
H
9
25
1.5
I
Disso = a + b × Lub% + c × DrugPS + d × Lub% × DrugPS + e × Lub%2 + f × DrugPS2 + ε
Disso = a + b × Lub% + c × DrugPS + d × Lub% × DrugPS + ε
53

135. Response
Factor
54

136. Response surface
design
Factorial or
fractional factorial
screening design
55

137. 56

138. •
“Cube
Oriented”
•
3
or
5
levels
for
each
factor
In
3
factors
Factorial
or
Center
Points
+
FracKonal
Factorial
+
Axial
Points
=
Central
Composite
Design
57

139. 58

140. 59

141. 60

142. Std
Run
Center
Block
Disint
Drug%
Disint%
DrugPS
Lub%
Disso%
WtRSD
Order
Order
Point
11
1
1
2
A
5
1.0
10
2.0
100.4
1.6
13
2
1
2
A
5
4.0
10
1.0
103.0
2.1
19
3
0
2
A
10
2.5
25
1.5
88.8
1.6
15
4
1
2
A
5
1.0
40
1.0
94.3
2.3
18
5
1
2
B
15
4.0
40
2.0
78.9
1.6
…
10
17
0
1
B
10
2.5
25
1.5
91.2
1.6
5
18
1
1
B
5
1.0
40
2.0
76.3
1.3
8
19
1
1
A
15
4.0
40
1.0
92.4
2.1
6
20
1
1
A
15
1.0
40
2.0
76.8
1.6
21
21
-­‐1
3
A
10
2.5
10
1.5
22
22
-­‐1
3
A
10
2.5
40
1.5
23
23
-­‐1
3
A
10
2.5
25
1.0
24
24
-­‐1
3
A
10
2.5
25
2.0
25
25
0
3
A
10
2.5
25
1.5
26
26
0
3
A
10
2.5
25
1.5
61

143. Std
Run
Center
Block
Disint
Drug%
Disint%
DrugPS
Lub%
Disso%
WtRSD
Order
Order
Point
11
1
1
2
A
5
1.0
10
2.0
100.4
1.6
13
2
1
2
A
5
4.0
10
1.0
103.0
2.1
19
3
0
2
A
10
2.5
25
1.5
88.8
1.6
15
4
1
2
A
5
1.0
40
1.0
94.3
2.3
18
5
1
2
B
15
4.0
40
2.0
78.9
1.6
…
10
17
0
1
B
10
2.5
25
1.5
91.2
1.6
5
18
1
1
B
5
1.0
40
2.0
76.3
1.3
8
19
1
1
A
15
4.0
40
1.0
92.4
2.1
6
20
1
1
A
15
1.0
40
2.0
76.8
1.6
21
21
-­‐1
3
A
10
2.5
10
1.5
101.8
1.7
22
22
-­‐1
3
A
10
2.5
40
1.5
84.0
1.7
23
23
-­‐1
3
A
10
2.5
25
1.0
96.7
2.1
24
24
-­‐1
3
A
10
2.5
25
2.0
82.8
1.4
25
25
0
3
A
10
2.5
25
1.5
92.3
1.5
26
26
0
3
A
10
2.5
25
1.5
91.9
1.2
62

144. 63

145. Y = a + b ⋅ DrugPS + c ⋅ Lub% + d ⋅ DrugPS2 + e ⋅ Lub%2 + f ⋅ Drug ⋅ PSLub% + ε
64

146. 65

147. 66

148. 67

149. Source DF Adj SS Adj MS F P
Blocks 2 2.27 1.13 0.48 0.625
Regression
Linear
DrugPS 1 1331.87 1331.87 567.73 0.000
Lub% 1 340.61 340.61 145.19 0.000
Square
DrugPS*DrugPS 1 27.39 27.39 11.68 0.003
Lub%*Lub% 1 0.14 0.14 0.06 0.811
Interaction
DrugPS*Lub% 1 222.98 222.98 95.05 0.000
Residual Error 18 42.23 2.35
Lack-of-Fit 7 25.15 3.59 2.32 0.103
Pure Error 11 17.07 1.55
68

150. Source DF Adj SS Adj MS F P
Blocks 2 0.02341 0.01171 0.41 0.671
Regression
Linear
DrugPS 1 0.00118 0.00118 0.04 0.842
Lub% 1 2.31351 2.31351 80.72 0.000
Square
DrugPS*DrugPS 1 0.04980 0.04980 1.74 0.204
Lub%*Lub% 1 0.09743 0.09743 3.40 0.082
Interaction
DrugPS*Lub% 1 0.00234 0.00234 0.08 0.778
Residual Error 18 0.51589 0.02866
Lack-of-Fit 7 0.28587 0.04084 1.95 0.154
Pure Error 11 0.23003 0.02091
69

151. StaKsKcal
Significance?
Model
Term
Disso%
WtRSD
DrugPS
P
P
Lub%
P
P
DrugPS2
P
P
Lub%2
?
DrugPS
×
Lub%
P
P
Lack
of
Fit
?
Y = a + b ⋅ DrugPS + c ⋅ Lub% + d ⋅ DrugPS2 + e ⋅ Lub%2 + f ⋅ Drug ⋅ PSLub% + ε
70