Statistical Tools for the Quality Control Laboratory and Validation Studies

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: [email protected] l  Website: www.gxpbiotech.org

IVTPHL1012S1

2

NORMAL DISTRIBUTION

22/1

21 ⎟

⎠

⎞⎜⎝

⎛ −−

Π= σ

µ

σ

iX

eY

IVTPHL1012S1

3 IVTPHL1012S1

4 IVTPHL1012S1

5 IVTPHL1012S1

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

7 IVTPHL1012S1

8

VARIANCE (S2)

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IVTPHL1012S1

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).

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.

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IVTPHL1012S1

12

F - TEST

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13

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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

98.146

0654.055.915.9

=−

=−

=

nsxt µ

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

( ) ( )424.1

5025.0

3025.0

526.0500.022=

+

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16

t for Unknown and Equal Variances

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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.

( )( )

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75.92000.06245.0

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18

t-TEST UNKNOWN BUT EQUAL VARIANCES. 2.

l  df = n1 + n2 - 2 df = 4

78.2

056.13333

4637.090.530.6

4,2/05.0 =

=+

−=

t

Xt

19

POOLED VARIANCE

( ) ( )211

21

222

211

−+−+−

=nn

snsnsp

20

t for Independent Averages with unknown and unequal variances.

2

11 2

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df

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21

t-TEST UNKNOWN AND UNEQUAL VARIANCES, 1.

l  Data:Extension of Previous Fe+3 mg/L study l  xA = 6.13, sA = 0.3529 l  xB = 5.76, sB = 0.1647

l  nA = nB = 10 l  F0.05/2,9,9 = 4.03

l  F = (0.3529)2 / (0.1647)2 l  F = 4.59

1 6.1 5.92 5.8 5.73 7.0 6.14 6.1 5.85 6.1 5.76 6.4 5.67 6.1 5.68 6.0 5.99 5.9 5.710 5.8 5.6

22


l  t.05/2,17 = 2.110

0044.30151664.037.0

10

21647.010

23529.0

76.513.6

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24


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0015666.00000669.00000141.0

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25

Paired t-Test

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s

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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

27

Paired t-Test Calculation

exists. difference tsignifican a Therefore

26.2

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===

t

nSdtd

28

t-Test for unknown but equal variances.

l  Showing that there is no significant difference?

10.2t182ndf 457.1

1010100

9488.187.1260.11

0.05/2,18

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29

Student’s t to a C.I.

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1

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30

CONFIDENCE INTERVAL 1.

30.4

..

96.1..

2,05.01,05.0 =

±=

±=

− ttntsXIC

IC

n

σµ

}  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

DATA SET FOR SETTING SPECS. 1.

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

DATA SET FOR SETTING SPECS. 2. SETS OF 3

}  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

DATA SET FOR SETTING SPECS. 3. SETS OF 10

}  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

DATA SET FOR SETTING SPECS. 4. CUMULATIVE

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)

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.

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

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.

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.

( )21

2212 1nn

nn+

−−=χ

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:

( )0.1

1916

194

7121712 22

2 <==+

−−=χ

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.

ValWkPHL1012S2 1

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: [email protected]

•  Website: www.gxpbiotech.org

2

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

3


•  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

4


•  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

5

SIMPLIFIED FORM OF n CALCULATION n for an to compare with a µ

( ) 2

222222

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−=⎟⎠⎞

⎜⎝⎛−

=

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xn

stn

sxt ii

µ

µµ

ValWkPHL1012S2

6

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

7


•  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. 1+N

ValWkPHL1012S2

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.

8 ValWkPHL1012S2

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

From a “normal” table (or “x” with ν = ∞) we want a probability of 0.95 that “x” will be greater than the “k” expression.

deviate normal standard x

0.420.42

=

−>

−=

ns

k

ns

xx

9 ValWkPHL1012S2

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:

xk since 645.10.42<=

−

ns

k

10 ValWkPHL1012S2

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:

59.41

707.0326.2645.1

0.410.42

326.20.410.41

=

−=−

=−

−

−=−

>−

=

k

kk

ns

k

ns

xx

11 ValWkPHL1012S2

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?)

( ) [ ][ ]( )( )

24.31681.0

544644.041.0

45.064.1nor 41.0]64.1[

64.141.00.4259.410.42

2

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−

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n

ns

ns

ns

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k

12 ValWkPHL1012S2

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.

13 ValWkPHL1012S2

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

Δ==== 2L 4tn 4L 22

22222

Ls

nst

ntsL

14 ValWkPHL1012S2

Recalculation of Earlier Problem.

L = 2, s = ±2, t0.95,∞=1.960 (two sided)

2

224tnLs

=

( ) ( )( )

( ) ( )( ) 18n so 17.81n 110.2t

17.66n 101.2t 18.16,n 131.2t :Iterate16or 4.15

44656.61

296.124

17,95.0

18,95.015,95.0

2

22

===

====

=

==

n

n

15 ValWkPHL1012S2

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.

( ) ( )( )

5.30.1

45.0080.242

22

==n

16 ValWkPHL1012S2

17

POOLED VARIANCE

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21

222

211

−+−+−

=nn

snsnsp

ValWkPHL1012S2

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:

( ) ( ) 43.032122.0345.021 22

=+

+=pS

18 ValWkPHL1012S2

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. gcalculatinfor 25not 4n that Note42.87 - 41.97or 0.4542.42 C.I.

0.45or 44376.02 C.I.

1.0.n rather tha ,88752.04

43.0064.22

95%

=

±=

±=±=

=

=

LL

L

19 ValWkPHL1012S2

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.

20 ValWkPHL1012S2

Sample Sizes for Estimating Standard Deviations. χ2

• This is the asymmetric distribution for σ2. • Now as an example, assume n-1 = 12. At 12 df, χ2 will exceed 21.0261 5% of the time and it will exceed 5.2260 95% of the time. Therefore 90% of the time, χ2 will lie between 5.2260 and 21.0261 for 12 df. • Check your tables to confirm this.

( )

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−

−

−

1

1

1

21

2

2

221

2

221

ns

sn

sn

n

n

n

χσ

σχ

σχ

21 ValWkPHL1012S2

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. 22 ValWkPHL1012S2

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.

23 ValWkPHL1012S2

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.

24 ValWkPHL1012S2

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.

25 ValWkPHL1012S2

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.

26 ValWkPHL1012S2

n for Comparing Two Averages

ValWkPHL1012S2 27

( )

( )2

22

21

2,

222

21

2df,

22

21

22

.

2121

2

22

1

21

21,

nt

n x

Δ

+=

Δ=++

Δ=

=−=Δ

+

−=

σσ

σσσσ

σσ

α

αα

α

df

df

df

tn

nn

t

nx

nn

xxt

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

28 ValWkPHL1012S2

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.

–  S2y represents the variance among the treatments and

s2pooled is the variance of the individual results (system

noise).

2

2

pooled

y

sns

F =

29 ValWkPHL1012S2

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.

30 ValWkPHL1012S2

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.

31 ValWkPHL1012S2

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

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.

•  Grand average = Total of all dots/56 dice (4X14) ValWkPHL1012S2 33

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

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.

ValWkPHL1012S2 34

( ) ( ) ( ) ( )

18.028.2029.0*14*

28.2476.137.237.259.2

029.014

1625.393.21625.329.31625.329.31625.314.3

2

2

2

2

22222

===

=++++=

=−

−+−+−+−=

pooled

y

pooled

y

y

ssn

F

s

s

s

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 δ = δ2 = Σδ2 =

Grand Ave. -1.055 1.1130 3.6245

= 3.555 0.375 0.1406

Σδ2/3 = s2y =

1.375 1.8906 1.2082

-0.695 0.4830

ValWkPHL1012S2 35

Fairness of 4 sets of dice. How about a loaded set? ANOVA

ValWkPHL1012S2 36

0.001 pat t Significan .F toFfor is 0.1%.at 6.171 - 6.595 and 1%,at 4.126 - 4.313 and

0.05p 5%,at 758.2839.2FTabular

71.6 71.652.2

21.1*14

1)-(4 3 df 21.1

521)-4(14df 52.24

21.307.238.242.2

3,603,40

52,3

52,3

2

2

=

=−=

===

==

===+++

=

Range

FF

s

s

y

pooled

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.

ValWkPHL1012S2 37

( ) ns

nstLSDni

pooleddf∑=×±= −

12

pooled,1 s /2 α

LSD for the Current Problem

•  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

( ) 68.2for t 1.333LSD 99%at

21.114259.101.2

59.152.24

21.307.238.242.2

52df0.99, ≅±=

±=×=

==+++

=

=

LSD

spooled

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

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

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

[email protected]

2

“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.”

3

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.

4

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

5 Cau;on: EVERYTHING depends on gecng this right !!!

Key Factors Key

Responses

6

Make ACE

Tablets

Disint (A or B)

Drug% (5-‐15%)

Disint% (1-‐4%)

DrugPS (10-‐40%)

Lub% (1-‐2%)

Dissolu;on% (>90%) WeightRSD%(<2%)

Fixed Factors Responses

Day

Random Factors

7

Trial DrugPS Lub%

Disso%

1 25 1 85 2 25 2 95 3 10 1.5 90 4 40 1.5 70

DrugPS

Lubricant%

85

95

70 90

10 40 1

2

8

=

+ ×

− ×

+ε

Disso% 86.66710 Lub%0.667 DrugPS

DrugPS

Lubricant%

85

95

70 90

10 40 1

2

9

�  Previous example had only 2 factors. Ø Factor space is 2D. We can visualize on paper.

�  With 3 factors we need 3D paper. Ø Corners even further away

�  Most new processes have >3 factors �  OFAT can only accommodate addi;ve models �  We need a more efficient approach

10

True response • Goal: Maximize response

• Fix Factor 2 at A. • Op;mize Factor 1 to B.

• Fix Factor 1 at B. • Op;mize Factor 2 to C.

• Done? True op;mum is Factor 1 = D and Factor 2 = E.

• We need to accommodate curvature and interac/ons

A

Factor 1

Factor 2

B

C

D

E 80 60 40

11

Response

Factor level A B C D

•  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)

12

Trial DrugPS Lub%

Disso%

1 10 1 75 2 10 2 100 3 40 1 75 4 40 2 80

DrugPS

Lubricant%

75

80

75

100

10 40 1

2

13

DrugPS

Lubricant%

75

80

75

100

10 40 1

2

=

+ ×

+ ×

− × ×

+ε

Disso% 43.330.667 DrugPS31.667 Lub%0.667 DrugPS Lub%

14

� Model non-‐addiKve behavior

›  interacKons, curvature

� Efficiently explore the factor space

� Take advantage of hidden replicaKon

15

Planar: no interac;on

1 2Y a b X c X= + ⋅ + ⋅

Non-‐planar: interac;on

1 2 1 2Y a b X c X d X X= + ⋅ + ⋅ + ⋅ ⋅

16

17

18

19

DrugPS

Lub%

C

B

D

A

10 40 1

2

DrugPS

Lub%

DrugPS Lub%

B D A CMainEffect2 2

A B C DMainEffect2 2

C B A DInteractionEffect2 2×

+ += −

+ += −

+ += −

C B D

A

C B D

A C

B D

A

Trial DrugPS Lub% Disso% 1 10 1 C 2 10 2 A 3 40 1 D 4 40 2 B

20

Trial DrugPS Lub% 1 10 1 2 10 2 3 40 1 4 40 2

Trial DrugPS Lub% 1 -‐1 -‐1 2 -‐1 +1 3 +1 -‐1 4 +1 +1

Uncoded Units Coded Units

•  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

midmax mid mid

max mid

U UC U C(U U ) U

U U−

= ⇔ = − +−

21

Disso ab Lub%c DrugPSd Lub% DrugPS

=

+ ×

+ ×

+ × ×

+ε

DrugPS

Lub%

DrugPS Lub%

a ( A B C D) / 4b ME /2 ( A B C D) / 4

c ME /2 ( A B C D) / 4d IE /2 ( A B C D) / 4×

= + + + +

= = − + − +

= = + + − −

= = − + + −

DrugPS

Lub%

C

B

D

A

-‐1 +1 -‐1

+1 Trial DrugPS Lub%

DrugPS*Lub%

Disso%

1 -‐1 -‐1 +1 C 2 -‐1 +1 -‐1 A 3 +1 -‐1 -‐1 D 4 +1 +1 +1 B

22

Disso a b Lub c DrugPS d Lub DrugPS= + × + × + × × + ε

�  It is obtained through the “magic” of regression.

�  b measures the “main effect” of Lub

�  c measures the “main effect” of DrugPS

�  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

�  ε represents trial to trial random noise

23

Trial DrugPS Lub% 1 -‐1 -‐1 2 -‐1 +1 3 +1 -‐1 4 +1 +1

DrugPS

Lub%

-‐1 +1 -‐1

+1

Trial DrugPS Lub% 1 -‐1 -‐1 2 -‐1 -‐1 3 +1 +1 4 +1 +1

DrugPS

Lub%

-‐1 +1 -‐1

+1

Inner product: +1-‐1-‐1+1=0 +1+0+0+1=2 +1+1+1+1=4

Trial DrugPS Lub% 1 -‐1 -‐1 2 -‐1 0 3 +1 0 4 +1 +1

DrugPS

Lub%

-‐1 +1 -‐1

+1

24

25

10 40

DrugPS

Dissolu;

on (%

LC)

2% Lubricant

1% Lubricant

90

26

Number of Factors (k)

Number of Trials (df =

2k) 0 1 1 2 2 4 3 8 4 16 5 32 6 64

•  Average •  Main Effects •  2-‐way interac;ons •  Higher order

interac;ons (or es;mates of noise)

y a bA cB dC eAB fAC gBC hABC= + + + + + + + + ε

27

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

Main Effects

•  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

28

A

B

C

-1 +1 -1

+1

+1

-1

y a bA cB dC= + + +

Trial I A B C AB AC BC ABC 1 + -‐ -‐ -‐ + + + -‐ 2 + + -‐ -‐ -‐ -‐ + + 3 + -‐ + -‐ -‐ + -‐ + 4 + + + -‐ + -‐ -‐ -‐ 5 + -‐ -‐ + + -‐ -‐ + 6 + + -‐ + -‐ + -‐ -‐ 7 + -‐ + + -‐ -‐ + -‐ 8 + + + + + + + +

•  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).

29

• “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

30

We like our screening designs to be at least resolu;on IV (I=ABCD)

•  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.

31

Number of Factors

2 3 4 5 6 7 8 9 10 11 12 13 14 15

Num

ber o

f Design Po

ints

4 Full III

6 IV

8 Full IV III III III

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

32

Trial DrugPS Lub%

Disso%

1 10 1 76 2 10 2 98 3 40 1 73 4 40 2 82 5 10 1 84 6 10 2 102 7 40 1 77 8 40 2 88

DrugPS

Lub%

76,84

88,82

73,77

98,102

10 40 1

2

FiKed model is based on averages individual

averageSD

SDnumber of replicates

=

33

Repeated measurement 1 batch

3 measurements per batch

ReplicaKng batch producKon

3 batches 1 measurement

per batch

34

ReplicaKon 1.   Every operaKon that

contributes to variaKon is redone with each trial.

2.  Measurements are independent.

3.   Individual responses are analyzed.

RepeKKon 1.   Some operaKons that

contribute variaKon are not redone.

2.  Measurements are correlated. 3.   The averages of the repeats

should be analyzed (usually).

Trial DrugPS Lub%

Disso%

1 10 1 76 2 10 2 98 3 40 1 73 4 40 2 82 5 10 1 84 6 10 2 102 7 40 1 77 8 40 2 88

Trial DrugPS Lub%

Disso%

1 10 1 76, 84 2 10 2 98, 102 3 40 1 73, 77 4 40 2 82, 88

35

� Frac;onal factorial designs are generally used for “screening”

� Sta;s;cal tests (e.g., t-‐test) are used to “detect” an effect.

� The power of a sta;s;cal test to detect an effect depends on the total number of replicates = (trials/design) x (replicates/trial)

�  If our experiment is under powered, we will miss important effects.

�  If our experiment is over-‐powered, we will waste resources.

� Prior to experimen;ng, we need to assess the need for replica;on.

36

( )22

11 2N (#points in design)(replicates/point) 4 z zα −β−

σ⎛ ⎞= ≅ + ⎜ ⎟δ⎝ ⎠

•  While not exact, this ROT is easy to apply and useful.

•  Commercial sotware will have more accurate formulas.

α z1-‐α/2 0.01 2.58 0.05 1.96 0.10 1.65

β z1-‐β 0.1 1.28 0.2 0.85 0.5 0.00

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

2

N 16 σ⎛ ⎞≅ ⎜ ⎟δ⎝ ⎠

37

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

( )22

11 2N (#points in design)(replicates/point) 4 z zα −β−

σ⎛ ⎞= ≅ + ⎜ ⎟δ⎝ ⎠

38

Run A B C D E 1 - - - - + 2 + - - - - 3 - + - - - 4 + + - - + 5 - - + - - 6 + - + - + 7 - + + - + 8 + + + - - 9 - - - + - 10 + - - + + 11 - + - + + 12 + + - + - 13 - - + + + 14 + - + + - 15 - + + + - 16 + + + + +

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

39

�  Sta;s;cal test for presence of curvature (lack of fit) �  Addi;onal degrees of freedom for sta;s;cal tests

� May be process “target” secngs

�  Used as “controls” in sequen;al experiments.

�  Spaced out in run order as a check for drit.

40

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”.

41

42

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

43

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

44

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

45

46

47

48

49

50

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.

51

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

52

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.

53

Disso a b Lub% c DrugPS d Lub% DrugPS= + × + × + × × + ε

DrugPS

Lub%

C

B

D

A

10 40 1

2

Trial DrugPS Lub% Disso% 1 10 1 C 2 10 2 A 3 40 1 D 4 40 2 B

E

F

H G 5 25 1 E 6 25 2 F 7 10 1.5 G 8 40 1.5 H

2 2Disso a b Lub% c DrugPS d Lub% DrugPS e Lub% f DrugPS= + × + × + × × + × + × + ε

I

9 25 1.5 I

54

Respon

se

Factor

55

Factorial or fractional factorial screening design

Response surface design

56

57

•  “Cube Oriented” •  3 or 5 levels for each factor In 3 factors

Factorial or FracKonal Factorial

Central Composite Design

+ +

=

Axial Points Center Points

58

59

60

61

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

62

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

63

64

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

65

66

67

68

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

69

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

70

StaKsKcal Significance? Model Term Disso% WtRSD

DrugPS P P Lub% P P

DrugPS2 P P Lub%2 ?

DrugPS × Lub% P P Lack of Fit

2 2Y a b DrugPS c Lub% d DrugPS e Lub% f Drug PSLub%= + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ ⋅ + ε?

71

•  The simplest model that explains the data is best (Occam’s razor, rule of parsimony)

•  Eliminate “least significant” terms one at a ;me followed by re-‐analysis

•  Always eliminate highest order terms first

•  Don’t eliminate lower order terms which are contained in significant higher order terms

•  Any exis;ng theory or prior knowledge trumps these rules.

2 2Y a b DrugPS c Lub% d DrugPS e Lub% f Drug PSLub%= + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ ⋅ + ε?

72

Estimated Regression Coefficients for Disso% using data in uncoded units Term Coef Constant 105.321 DrugPS -0.478970 Lub% 6.62343 DrugPS*DrugPS 0.0130426 Lub%*Lub% -0.959956 DrugPS*Lub% -0.497745

S = 1.49153 PRESS = 83.4051 R-Sq = 97.76% R-Sq(pred) = 95.79% R-Sq(adj) = 97.20%

73

Estimated Regression Coefficients for WtRSD using data in uncoded units Term Coef Constant 4.66698 DrugPS -0.0293187 Lub% -2.96608 DrugPS*DrugPS 0.000623945 Lub%*Lub% 0.763118 DrugPS*Lub% -0.00161165

S = 0.164211 PRESS = 0.850996 R-Sq = 83.93% R-Sq(pred) = 74.65% R-Sq(adj) = 79.92%

74

Acceptable performance more likely

•  Difficult to do with > 2 factors •  Does not take into account

•  es;ma;on uncertainty •  correla;on among responses •  variability in control of factor levels

•  variability in the underlying true model over ;me

75

76

77

Global Solution DrugPS = 11.2121 Lub% = 1.93939 Predicted Responses Disso% = 100.002 , desirability = 1.000 WtRSD = 1.500 , desirability = 0.117927 Composite Desirability = 0.343404

78

Predicted Response for New Design Points Using Model for Disso% Point Fit SE Fit 95% CI 95% PI 1 100.002 0.621070 (98.7063, 101.297) (96.6316, 103.372) Predicted Response for New Design Points Using Model for WtRSD Point Fit SE Fit 95% CI 95% PI 1 1.49952 0.0683772 (1.35689, 1.64216) (1.12848, 1.87057)

79

1.  Number of trials ≥ Number of model coefficients

2.  Each coded column adds to 0 (balance)

3.  Inner product of any 2 coded columns = 0 (orthogonality)

4.  Use resolu;on V (or at least IV) for screening designs 5.  Factor ranges are bold (but not too bold) 6.  Incorporate process knowledge & sequen;al strategies 7.  Assure adequate sample size (power)

8.  Randomize processing order

9.  Block when you cannot randomize

10.  Incorporate tests for model adequacy (e.g., center points)

11.  Avoid PARC (Planning Ater Research is Complete)

80

1.  Use graphics (picture = 1,000 words) 2.  Always verify model assump;ons (normality, independence,

variance homogeneity)

3.  In model reduc;on, follow rules of hierarchy tempered by prior process knowledge

4.  Use coded factor levels in judging sta;s;cal significance of model coefficients.

5.  Consider predic;on uncertainty when iden;fying op;mal factor secngs

6.  Take advantage of curvature & interac;ons when choosing op;mal factor secngs

7.  Always perform independent trials to confirm predic;ons.

81

Minitab • General purpose stat package • User friendly • Good learning tool JMP • General purpose stat package • Excellent for DOE & SPC • Very advanced features

• Monte-‐Carlo simula;on of DOE models • Good D-‐op;mal design features

• May need sta;s;cal support for some features Design Expert • Exclusive focus on DOE (may want addnl tools) • I have not used but my impression is very good

5

5

10

15

Hard%RSD

MixTime(min)5 7 9 11 13 15MixTime(min)5 7 9

15

20

2.015 17

3.02.5 Water(L)

2.0

3.0

Water(L)

Surface Plot of Hard%RSD

6 11 16

2.0

2.5

3.0

MixTime(min)

Wat

er(L

)

Overlaid Contour Plot of Hardness...Hard%RSD

Hardness

Hard%RSD

19.520.5

07

Lower BoundUpper Bound

White area: feasible region

82

Contour Profiling and overlay for design space idenKficaKon

Monte-‐Carlo SimulaKon to determine effect of poor factor control on future batch failure rate

67

83

• Robust design & Taguchi designs • Mixture (e.g.,gasoline blend) and constrained designs

• D-‐op;mal designs and custom augmenta;on

• Bayesian approaches • Probability of mee;ng specifica;ons • mul;ple correlated responses • incorpora;on of prior knowledge

• Variance component analysis & Gage R&R

• Split-‐plot experiments

84

1.  Box, G. E. P.; Hunter, W. G., and Hunter, J. S. (1978). Statistics for Experimenters: An Introduction to Design, Data Analysis, and Model Building. John Wiley and Sons.

2.  Montgomery D (2005) Design and analysis of experiments, 6th edition, Wiley.

3.  Myers R, Montgomery D, and Anderson-Cook C (2009) Response surface methodology, Wiley.

4.  Diamond W (1981) Practical Experiment Designs, Wadsworth, Belmont CA

5.  Altan S, et al (2010) Statistical Considerations in Design Space Development (Parts I-III) PharmTech Nov 2, 2010. Available on line at http://www.pharmtech.com/pharmtech/author/authorInfo.jsp?id=53118

6.  Conformia CMC-IM Working Group (2008) Pharmaceutical Development case study: “ACE Tablets”. Available from the following web site: http://www.pharmaqbd.com/files/articles/QBD_ACE_Case_History.pdf

7.  ICH Expert Working Group (2008) GUIDELINE on PHARMACEUTICAL DEVELOPMENT Q8(R1) Step 4 version dated 13 November 2008

8.  ICH Expert Working Group (2005) Guideline on QUALITY RISK MANAGEMENT Q9 Step 4 version dated 9 November 2005

9.  FDA CDER/CBER/CVM (November 2008) Draft Guidance for Industry Process Validation: General Principles and Practices (CGMP)

Thank You!!

GMPWkPHL1012S4 1

Statistical Control to Monitor and Control Validated Processes.

Session 4

Steven S. Kuwahara, Ph.D. GXP BioTechnology LLC

PMB 506, 1669-2 Hollenbeck Avenue Sunnyvale, CA 94087-5402 USA

Tel. & FAX: (408) 530-9338 e-Mail: [email protected]

Website: www.gxpbiotech.org

•  All functions mentioned in the GMP are GMP processes. –  Not all GMP processes need to be monitored. –  The processes that introduce variability into the

product or can cause the production of unacceptable product need to be monitored.

–  This goes beyond manufacturing processes. –  QC test methods need to be monitored. –  The annual product review needs to be monitored to see

if batch records are being properly executed. –  Processes that generate critical business information

need to be monitored.

What to Monitor 1.

2 GMPWkPHL1012S4

•  Use information from the process validation and risk assessment to determine what needs to be monitored. (You need proof of your assertion.) –  Even if the process validation shows that a step is very

stable, if the risk assessment shows that that step can create a serious threat to the product or patient, it should be monitored.

–  Steps or processes can be monitored. If there are multiple small steps in a process, the outcome of the process can be monitored in place of each of the steps.

–  If the outcome of a process or step is immaterial to the quality of the product, it does not need to be monitored.

What to Monitor 2.

3 GMPWkPHL1012S4

•  There can be two types of data. –  Variable data forms a continuum and an individual

result can fall anywhere within it. Most statistical procedures are designed to deal with variable data.

–  Attribute data comes in discrete units. Yes/no: red, green, blue, or yellow; pass/fail; high, medium, low.

–  One of the problems with attribute data is that it may be subjective or actually part of a continuum. Is a thing red, reddish, pink, or brown, brownish, rust; what is high or medium; passing?

–  Since attribute data must be discrete, a clear and firm definition is needed.

Data 1.

4 GMPWkPHL1012S4

•  There are statistical methods for dealing with attribute data, especially since they may not follow the normal distribution. –  One technique is to convert attribute data into variable data by

using fractions or percentages, but this is good only with large numbers. With small numbers the discreteness creates large jumps.

–  Attribute data that is in binary (+/-, yes/no) form will follow a binomial distribution.

–  Attribute data is not as strong as variable data so you need more of it.

•  In what follows we will assume that you have variable data.

Data 2.

5 GMPWkPHL1012S4

TOLERANCE vs. CONFIDENCE LIMITS With confidence limits you are trying to determine the value of the “true mean” (µ). The result gives an interval within which you expect to find µ with a certain probability (confidence). The results are means with the same n that was used to calculate the limits. With a tolerance interval you are asking for the values of the next n numbers of results. The result gives an interval within which you have x % confidence that it will contain y % of the results. It assumes that all of the results are from the same population. The result here is a single number. 6 IVTPHL1012S1

TOLERANCE LIMITS FOR A SINGLE RESULT 1.

•  DEFINITION: ± ks. Where k is a value to allow a statement that we have 100(1 - α) percent confidence that the limits will contain proportion p of the population.

•  Requires a normal distribution of the population. •  Allows one to set limits for single determinations

(not averages) with a limited number of replicates available.

7 IVTPHL1012S1

TOLERANCE LIMITS 2. Selected Portions of a Table for Two-sided Limits

•  k 95% Confidence 99% Confidence •  n p=95% p=99% p=95% p=99% •  2 37.67 48.43 188.5 242.3 •  3 9.916 12.86 22.40 29.06 •  4 6.370 8.299 11.15 14.53 •  5 5.079 6.634 7.855 10.26 •  10 3.379 4.433 4.265 5.594 •  20 2.752 3.615 3.168 4.161 •  50 2.379 3.126 2.576 3.385 •  Table XIV, Applied Statistics and Probability for Engineers, D.C.

Montgomery & G.C. Runger, John Wiley & Sons, 1994.

8 IVTPHL1012S1

TOLERANCE LIMITS FOR A SINGLE RESULT 3. Example

•  A product is made at 2.25 mg/mL. n = 10 samples are taken from 5 early lots. The assays show 2.27, 2.25, 2.24, 2.22, 2.26, 2.23, 2.23, 2.24, 2.27, and 2.25 mg/mL. We calculate = 2.246, s = 0.017. From the table at n = 10, 99% confidence for 95% of results k = 4.265 so the Tolerance Interval is:

•  2.246 ± (4.265)0.017 = 2.319 - 2.173. •  We have 99% confidence that 95% of the results from

this population will be in the interval. •  The 95% confidence interval for the mean is ± 0.012,

giving 2.258 to 2.234. 9 IVTPHL1012S1

Statistical Process Control (SPC) Basics. 1.

•  At some level of discrimination, NO TWO THINGS ARE ALIKE. There is always some variation.

•  If things are not being hand made, the variation should be small enough to make each unit interchangeable with any other unit. – A 100 mg tablet may vary by ± 1 mg and be

okay, but not a 1 mg tablet. •  The allowable variation should not be larger than

the variation that causes a unit to be “different.”

10 GMPWkPHL1012S4


•  In the old days, people made many units of product and the “production lot” underwent 100% inspection to segregate “good” units from the “bad.” “Bad” units were scrapped. Thus the name “Quality Control.” –  It cost just as much to make scrap as to make a “good”

unit. –  100% inspection is never 100% effective. –  “Quality Control” was basically a sorting operation.

Quality was really not controlled since the “bad” product had already been made.

11 GMPWkPHL1012S4


•  Walter Shewhart, in the late 1920’s, and, later, W. Edwards Deming basically proposed the idea that making scrap was not cost effective.

•  They proposed the idea that the elimination of scrap would produce the most cost effective manufacturing process.

•  Scrap was the result of excessive variation. •  Variation comes in two forms. In the Shewhart/

Deming terminology there are common cause and assignable cause variations.

12 GMPWkPHL1012S4

Statistical Process Control (SPC) Common Cause Variation.

•  Common (controlled) cause variation is the basic variation that is inherent in a process. It is the sum of all of the small, inherent, random errors associated with the process. Sources of error may be undetectable.

•  Since common cause variation is inherent in a process, it changes only when the process, or the way that the process is managed, changes.

•  Common cause variation, therefore, can only be reduced through the actions of management.

13 GMPWkPHL1012S4

SPC: Assignable Cause Variation.

•  Assignable (special) cause variations result from unusual situations and are not built-in as a part of the process.

•  Because they arise from unusual events, assignable cause variations should be detectable by careful inspection of the system.

•  While management action may be needed, most assignable cause variations can be corrected by line workers themselves, if they have the required knowledge and experience.

14 GMPWkPHL1012S4

X-bar Chart

Time LCL

UCL

X bar x x

x x x x x x x

x x x

x x x x

x

15 GMPWkPHL1012S4

R Chart

•  UCL

Time

Range

LCL

o o o o o o o o o o o o

o

o

o o

o o

o o

o o o

16 GMPWkPHL1012S4

GMPWkPHL1012S4 17

GMPWkPHL1012S4 18

Chart and Process Interpretation

•  Look at R chart first. x bar Chart is meaningless if R has changed significantly or is large.

•  If R has increased, identify any special causes that are responsible for the increased variability and change the process to stop them.

•  If R has decreased, identify special causes responsible and change process to incorporate them.

GMPWkPHL1012S4 19

SETTING CONTROL LIMITS FOR YOUR SPC CHARTS. 1.

•  Control limits for the SPC charts should be incorporated as specifications, unless you are setting your release specifications to be wider than your control limits.

•  The problem here is that your SPC control limits should be based upon what your process is capable of delivering, if it is operating in a state of control. Therefore, if the limits are exceeded, it is an indication that the process is out of control, even if it is within the release limits.

GMPWkPHL1012S4 20


•  First, calculate the average of the averages that you are using. This is:

Now calculate the average range that comes from the ranges of the data that were used for the averages. This is:

N

xx

N

∑= 1

R

GMPWkPHL1012S4 21


•  For “3-sigma limits” the Shewhart formulae are:

7. size subgroupfor 0D that Notenegative.not if

CL

CL

3

3

R4

2

x2

<=

=

==

−=

=+=

RDLCL

RRDUCL

RAxLCL

xRAxUCL

R

R

x

x

GMPWkPHL1012S4 22


•  The constants, A2, D3, and D4 are combinations of other constants and are set for 3σ and change with the subgroup size.

•  The subgroup size (n) is the number of replicates for each average. (Also known as a “rational subgroup.”)

•  When possible, 20 to 30 subgroups should be used to establish the grand average and the average range. Fewer subgroups may be used in the beginning, but the numbers should evolve until 30 is reached.

•  The idea is to reach the point where you are dealing with µ and σ, but recent computer studies have shown that instead of 30, the real number should be around 200.

Western Electric Rules 1. Zones

GMPWkPHL1012S4 23

+1σ

-1σ -2σ

-3σ

+2σ

+3σ

Western Electric Rules 2. •  The process is out of control when any one of the

following happens: •  1. One point plots outside the 3σ control limits. •  2. Two out of 3 consecutive points are beyond the 2σ limits. •  3. Four out of 5 consecutive points plot beyond the 1σ

control limits. •  4. Eight consecutive points plot on one side of the center

line (). •  These are one-sided rules. They only apply to

events happening on one side of the chart. –  If a point is beyond +2σ and the next point plots beyond

-2σ, they are not 2 points beyond 2σ.

GMPWkPHL1012S4 24

Western Electric Rules 3. Considerations

•  Based on the work of Shewhart, Deming, and others when they were with Bell Telephone.

•  Because they are based on the standard deviation zones, these are sometimes known as “zone rules.”

•  These rules enhance the sensitivity of the control charts, but require the use of standard deviations, not ranges.

•  A “point” on a control chart usually represents an average that was calculated for a “rational group.”

GMPWkPHL1012S4 25

Rational Subgroups •  The subgroup that is chosen to represent a

“point” should be chosen so that it properly represents the product unit. –  If you are checking lots over time then the subgroup

units should be randomly chosen from the lot. –  If you are more interested in the different parts of a

lot, you should define the parts and choose from within the “part.”

–  The number of units in the subgroup should be calculated as an “n” value based on your desired confidence interval and the level of risk that you are willing to accept.

GMPWkPHL1012S4 26

CUSUM Charts I.

•  There are CUmulative SUM control charts. –  They are based on the Cumulative Sum of the

Deviations from a target value. •  The CUSUM chart is a relatively new

invention designed to overcome the lack of sensitivity of “Shewhart Charts” and the tendency of the charts to give false alarms, especially when the Western Electric Rules are employed.

•  CuSum chart: A quality control chart with memory

GMPWkPHL1012S4 27

CUSUM Charts II.

•  For deviations greater than ± 2.5σ the Shewhart chart is as good or may be better than the CUSUM chart.

•  CUSUM charts are more sensitive and will detect changes (especially trends) in the ± 0.5σ - ± 2.5σ range.

GMPWkPHL1012S4 28

Types of ‘Time-weighted charts

•  There are 3 types of time-weighted charts. •  Moving Average •  - Chart of un-weighted moving averages •  Exponential Moving Average (EWMA) •  - Chart of exponentially weighted moving averages •  CUSUM •  - Chart of the cumulative sum of the deviations

from a nominal specification

GMPWkPHL1012S4 29

Equations for CUSUM

GMPWkPHL1012S4 30

( )

( )

.deviations thesum you will over which period time theis j

lue. target va theis and jat point CUSUM theis j.at average the

x whereaverages,for S

j.at n observatio theis

x wheredata,point singlefor

jj

j

µ

µ

µ

j

j

ij

j

ijj

Sis

x

xS

∑

∑

−=

−=

Starting a CUSUM I.

•  Specification: Target Value: 100 = µ •  n xj Calculation CuSum value = Sj j = 8 •  1 99 = 99 – 100 = -1 •  2 101 = 99+101-2*100 = 0 •  3 99 = 99+101+99-3*100 = -1 •  4 100 =99+101+99+100-4*100 = -1 •  5 102 = 99+101+99+100+102-5*100 = 1 •  6 101 = 99+101+99+100+102+101-6*100 = 2 •  7 100 = 99+101+99+100+102+101+100-7*100 = 2 •  8 101 = 99+101+99+100+102+101+100+101-8*100 = 3

•  Sj= (Σxj ) – jµ

GMPWkPHL1012S4 31

Starting a CUSUM II.

•  In theory, the sum of the deviations should be zero, if the process is in statistical control.

•  So the cumulative run length (j) should be large enough that you would expect it to be at zero, if the target value (µ) is really the center point.

•  The target value (µ) should be the specification for the measurement.

•  Therefore the center line in a CUSUM chart should be at zero.

GMPWkPHL1012S4 32

Shift then return to Normal

GMPWkPHL1012S4 33

Theory Reality

The V-Mask I. • A statistical t-test to determine if the process

is in a state of statistical control. The V-mask is placed at a distance d from the

last point of measurement. The opening of the V-mask is drawn at an

angle of ± θ. •  First, obtain an estimate of the standard

deviation or standard error (σx) for the value of Sj. This should be known from the specification setting process or from development work.

GMPWkPHL1012S4 34

The V-Mask II. •  Decide on the smallest deviation you want to detect (D).

Calculate: δ = D/σx. If D = σx then δ = 1

•  Decide on the probability level (α) at which you wish to make decisions. –  For the usual ± 3σ level, α = 0.00135.

•  Determine the scale factor (k) which is the value of the statistic to be plotted (vertical scale) per unit change in the horizontal scale (lot or sample number). It is suggested that k should be between 1σx and 2σx, preferably closer to 2σx.

•  Using the value of δ, obtain the value of the lead time (d) from the following table.

GMPWkPHL1012S4 35

The V-Mask III. •  Obtain the angle (ϴ) from the same table by

setting δ = D/k. (Table BB in Juran’s Book) •  Construct the V-Mask from these data.

•  Truncated Table BB.

GMPWkPHL1012S4 36

δ ϴ d 0.2 5o 43’ 330.4 0.5 14o 00’ 52.9 1.0 26o 34’ 13.2 1.8 41o 59’ 4.1 2.0 2.6 3.0

45o 00’ 52o 26’ 56o 19’

3.3 2.0 1.5

V-Masked CUSUM

GMPWkPHL1012S4 37

Action?

Failure

d ϴ

Time

X-bar

CUSUM Rules •  The sample size can be the same as used in an chart. One

recommendation is to use. •  n = 2.25s2/D where s2 is an estimate of the process (lot to

lot) variance. •  The V-Mask is always placed at the last point measured. •  If all points are within the V-mask, the process is under

control. •  Any point out of the V-Mask shows a lack of control. •  The first point out of the V-mask shows when the shift

started even if later points are within the masks.

GMPWkPHL1012S4 38

Moving V-Mask

GMPWkPHL1012S4 39

Failure

References: ISO/TR 7871:1997: Cumulative sum charts --

Guidance on quality control and data analysis using CUSUM techniques

•  http://www.iso.org/iso/iso_catalogue/catalogue_tc/catalogue_detail.htm?csnumber=14804

•  British Standards No. BS 5700 ff SR, 05.03.2009, page 16.

•  Wadsworth, H.M.; “Statistical Process Control,” in Juran’s Quality Handbook 5th Ed.; Juran, J.M. and Godfrey, A.B. (eds.); McGraw-Hill, New York, NY, 1999, Sect. 45, page 45-17. GMPWkPHL1012S4 40

Health & Medicine

Statistical Tools for the Quality Control Laboratory and Validation Studies