Statistical Tools for the Quality Control Laboratory and Validation Studies

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

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Statistical Tools for the Quality Control Laboratory and Validation Studies

  1. 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. 2. NORMAL DISTRIBUTION 2 ⎛ X − µ ⎞ e 1 −1/ 2⎜ i ⎟Y= ⎝ σ ⎠ σ 2Π IVTPHL1012S1 2
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  6. 6. NORMAL DISTRIBUTION PROPERTIESl  The normal distribution has the followingproperties:l  Bell-shapedl  Unimodall  Symmetricall  Extends from -∞ to +∞ (tails never reach zerofrequency)l  Same value for mean, median, and model  This pattern of variation is common formanufacturing processes. IVTPHL1012S1 6
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  8. 8. VARIANCE (S2) 2 ΣX i 2 − (ΣX i )S2 = n n −1 2S 2 = ( Σ Xi − X ) n −1 2 2 nΣX i − (ΣX i )S2 = n(n − 1) IVTPHL1012S1 8
  9. 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. 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. 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. 12. F - TEST 2 s2Fα ,df 1,df 2 = 2 Note : This is slightly different s1from the F - test that is used for ANOVAand factorial experiments.Note : F0.05,3,3 = 9.28F0.05,10,10 = 2.98 12
  13. 13. Student’s t x−µt= Basic form. s ndf = n − 1 x1 − x2t= Independent averages, σ 12 σ 2 2 + n1 n2known variances. 13
  14. 14. t-TEST vs THEORETICAL OR KNOWN VALUEl  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 = ± 0.0654l  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. 15. KNOWN VARIANCES, t-TEST OF TWO AVERAGESl  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, 0.50l  n1=3, n2=5, x1=0.500, x2=0.526l  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. 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. 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. 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. 19. POOLED VARIANCE 2 2 n1 −1 s1 + n2 −1 s2 ( ) ( )sp = n1 +n2 −2 19
  20. 20. t for Independent Averages withunknown 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. 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. 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 0.0151664l  t.05/2,17 = 2.110 22
  23. 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. 24. t-TEST UNKNOWN AND UNEQUAL VARIANCES, 4. ⎡ ⎤ ⎢df = ⎢ (0.0395799 )2 ⎥ −2 2 2 ⎥ ⎢ (0.01245384 ) + (0.0271261) ⎥ ⎢ ⎣ 11 11 ⎥ ⎦ 0.0015666 0.0015666df = = 0.0000141 + 0.0000669 0.000081df = 19.3 − 2 = 17 rounded to a whole number 24
  25. 25. Paired t-Test dt= n d = xi1 − xi 2 df = n −1 sd 2 2 (∑ d ) ∑d −sd = n n −1 25
  26. 26. DATA FOR t -TESTSl  Sample New Original dl  1. 12.1% 14.7% 2.6%l  2. 10.9 14.0 3.1l  3. 13.1 12.9 -0.2l  4. 14.5 16.2 1.7l  5. 9.6 10.2 0.6l  6. 11.2 12.4 1.2l  7. 9.8 12.0 2.2l  8. 13.7 14.8 1.1l  9. 12.0 11.8 -0.2l  10 9.1 9.7 0.6l  ave. 11.60 12.87 1.27l  s 1.814 2.075 1.126 26
  27. 27. Paired t-Test Calculation d 1.27t= n= 10 = 3.567 Sd 1.126t 0.05 / 2,9 = 2.26Therefore a significantdifference exists. 27
  28. 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 = 2.10l  Showing that there is no significant difference? 28
  29. 29. Student’s t to a C.I. x−µ tst= Basic form. = x−µ s n ndf = n − 1 tsµ = x± The value of t is taken from a t - table nfor n - 1 degrees of freedom and the desired confidence. 29
  30. 30. CONFIDENCE INTERVAL 1.C.I . = µ ± 1.96σ tsC.I . = X ± nt0.05,n −1 t0.05, 2 = 4.30 30
  31. 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. 32. DATA SET FOR SETTING SPECS. 2. SETS OF 3l  67.0 72.7 71.9 82.6 70.8 66.4 73.2 85.0 69.5 74.0l  67.5 73.1 68.8 78.1 84.9 74.5 68.7 75.0 70.5 64.2l  76.0 65.8 75.6 74.2 72.2 67.3 69.0 74.7 81.1 64.2l  Ave.70.2 70.5 72.1 78.3 76.0 69.4 70.3 78.2 73.7 71.6l  s = 5.06 4.10 3.40 4.20 7.77 4.44 2.52 5.86 6.43 6.54l  CV. 7.21 5.82 4.72 5.37 10.23 6.40 3.58 7.49 8.72 9.13l  CI ±29.0 23.5 19.5 24.1 44.5 25.4 14.4 33.6 36.8 37.5l  X3 = 73.03, s = 3.36, C.V.=3.5%, n=10, t0.995,9 = 3.250l  99%C.I.(ave) = ±3.46 = 69.67 - 76.49 IVTPHL1012S1 32
  33. 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. 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. 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. 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. 37. Signed Rank Test: Examplel  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, +139l  Rank:-11.5,+11.5,+13,-15, +15, +15, +17, +18, +19 37
  38. 38. Signed Rank Test: Example: Continuedl  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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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 0.01that !>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. 53. SAMPLING PLAN CALCULATIONS C.• Going back to the original equation for a passing resultand knowing that s = ± 0.45 (From our assay validationstudies?) 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) 0.544644n= = 3.24 0.1681 ValWkPHL1012S2 12
  54. 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. 55. SAMPLE SIZES FOR MEANS• Suppose we want to determine µ using a test where weknow 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 sL= L = n = 2 L = 2Δ n n L ValWkPHL1012S2 14
  56. 56. Recalculation of Earlier Problem. 2 2 4t s n= 2 LL = 2, s = ±2, t0.95,∞=1.960 (two sided) 2 2 4(2) (1.96 ) 61.4656n= 2 = (2) 4n = 15.4 or 16Iterate : (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. 57. Sample size for estimating µ• Note the statement: We are determining the % of drugpresent and we wish to bracket the true amount (µ%)by ± 0.5% and do this with 95% confidence, so L = 2 x0.5 = 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. 58. POOLED VARIANCE 2 2 n1 −1 s1 + n2 −1 s2 ( ) ( )sp = n1 +n2 −2 ValWkPHL1012S2 17
  59. 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. 60. Calculating the Confidence Interval, L• Sp = s, the new estimate of the standard deviation, so anew confidence interval can be calculated with 24 df.t(0.975, 24)= 2.064. L = 2(2.064)(0.43) 4L = 0.88752, rather than 1.0. ( )C.I. = ± L = 0.44376 or ± 0.45 2C.I. 95% = 42.42 ± 0.45 or 41.97 - 42.87 Note that n = 4 not 25 for calculating L. ValWkPHL1012S2 19
  61. 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. 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. 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. 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. 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. 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. 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. 68. n for Comparing Two Averages x1 − x2tα , df = Δ = x1 − x2 n1 = n2 σ 12 2 σ2 + n1 n2 2 Δ 2 t α ,df 2tα .df = σ 2 σ 2 n (σ 2 1 ) + σ 2 = Δ2 2 1 2 + n nn= 2 tα , df (2 2 σ1 + σ 2 ) Δ2 ValWkPHL1012S2 27
  69. 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. 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. 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. 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. 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. 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. 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 2s y = 0.029s 2 = 2.59 + 2.37 + +2.37 + 1.76 = 2.28 pooled 4 2 n * s y 14 * 0.029F= 2 = = 0.18 s pooled 2.28 ValWkPHL1012S2 34
  76. 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. 77. Fairness of 4 sets of dice. How about a loaded set? ANOVA 2 2.42 + 2.38 + 2.07 + 3.21spooled = = 2.52 df = 4(14 - 1) = 52 4 2s y = 1.21 df = 3 (4 - 1) 14 *1.21F= = 6.71 F3,52 = 6.71 2.52Tabular F3,52 = 2.839 − 2.758 at 5%, p = 0.05and 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. 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. 79. LSD for the Current Problem 2.42 + 2.38 + 2.07 + 3.21s pooled = = 2.52 = 1.59 4 2LSD = 2.01 ×1.59 = ±1.21 14at 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. 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. 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 2.07For 95% confidence, the LSD is ± 1.21 and for 99%, the LDSis ± 1.33.So blue and green are different from white, and green isdifferent 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 concludethat blue and green dice run to higher numbers than whiteand purple. ValWkPHL1012S2 40
  82. 82. David.LeBlond@sbcglobal.net  
  83. 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. 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. 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. 86. Key   Factors   Key   Responses  Cau;on:  EVERYTHING  depends  on  gecng  this  right  !!!   5
  87. 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. 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. 89. Lubricant%   2   95   90   70   1   85   10   40   DrugPS  Disso% = 86.667 +10 × Lub% −0.667 × DrugPS +ε 8
  90. 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. 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. 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. 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. 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. 95. ž  Model  non-­‐addiKve  behavior   ›  interacKons,  curvature  ž  Efficiently  explore  the  factor  space  ž  Take  advantage  of  hidden  replicaKon   14
  96. 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. 97. 16
  98. 98. 17
  99. 99. 18
  100. 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. 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. 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. 103. Disso = a + b × Lub + c × DrugPS + d × Lub × DrugPS + εž  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. 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. 105. 24
  106. 106. Dissolu;on  (%LC)   1%  Lubricant   2%  Lubricant   90   10   40   DrugPS   25
  107. 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. 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. 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. 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. 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. 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. 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. 114. ReplicaKng   1  measurement  batch   3  batches   per    batch  producKon  Repeated   3  measurements   1  batch   per    batch  measurement   33
  115. 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. 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. 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 20.01   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. 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. 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. 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. 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. 122. 41
  123. 123. Confounding TableI = ABCDEBlk = AB = CDEA = BCDEB = ACDEC = ABDED = ABCEE = ABCDAC = BDEAD = BCEAE = BCDBC = ADEBD = ACEBE = ACDCD = ABECE = ABDDE = ABC 42
  124. 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. 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. 126. 45
  127. 127. 46
  128. 128. 47
  129. 129. 48
  130. 130. 49
  131. 131. Source DF Adj MS F PBlocks 1 2.21 0.11 0.745Disint 1 0.30 0.01 0.905Drug% 1 2.94 0.15 0.707Disint% 1 0.30 0.01 0.905DrugPS 1 1174.45 58.93 0.000Lub% 1 258.61 12.98 0.004Curvature 1 32.68 1.64 0.225Res Error 12 19.93 2.179  is  the  1-­‐α/2   th  quan;le  of  the  t-­‐ distribu;on  having   12  df.   50
  132. 132. Source DF Adj MS F PBlocks 1 0.01090 0.51 0.487Disint 1 0.03751 1.77 0.208Drug% 1 0.00847 0.40 0.539Disint% 1 0.08282 3.91 0.071DrugPS 1 0.00189 0.09 0.770Lub% 1 2.10586 99.46 0.000Curvature 1 0.21198 10.01 0.008Res Error 12 0.02117 51
  133. 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. 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 + εDisso = a + b × Lub% + c × DrugPS + d × Lub% × DrugPS + ε 53
  135. 135. Response   Factor   54
  136. 136. Response surfacedesignFactorial orfractional factorialscreening design 55
  137. 137. 56
  138. 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. 139. 58
  140. 140. 59
  141. 141. 60
  142. 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

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