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Center for Biofilm Engineering Al Parker, Biostatistician Standardized Biofilm Methods Research Team Montana State University The Importance of Statistical.

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Presentation on theme: "Center for Biofilm Engineering Al Parker, Biostatistician Standardized Biofilm Methods Research Team Montana State University The Importance of Statistical."— Presentation transcript:

1 Center for Biofilm Engineering Al Parker, Biostatistician Standardized Biofilm Methods Research Team Montana State University The Importance of Statistical Design and Analysis in the Laboratory Feb, 2011

2 Standardized Biofilm Methods Laboratory Darla Goeres Al Parker Marty Hamilton Diane Walker Lindsey Lorenz Paul Sturman Kelli Buckingham- Meyer

3 What is statistical thinking?  Data  Experimental Design  Uncertainty and variability assessment

4 What is statistical thinking?  Data (pixel intensity in an image? log(cfu) from viable plate counts?)  Experimental Design - controls - randomization - replication (How many coupons? experiments? technicians? labs?)  Uncertainty and variability assessment

5 Why statistical thinking?  Anticipate criticism (design method and experiments accordingly)  Provide convincing results (establish statistical properties)  Increase efficiency (conduct the least number of experiments)  Improve communication

6 Why statistical thinking? Standardized Methods

7 Attributes of a standard method: Seven R’s  Relevance  Reasonableness  Resemblance  Repeatability (intra-laboratory)  Ruggedness  Responsiveness  Reproducibility (inter-laboratory)

8 Attributes of a standard method: Seven R’s  Relevance  Reasonableness  Resemblance  Repeatability (intra-laboratory)  Ruggedness  Responsiveness  Reproducibility (inter-laboratory)

9 Resemblance of Controls Independent repeats of the same experiment in the same laboratory produce nearly the same control data, as indicated by a small repeatability standard deviation. Statistical tool: nested analysis of variance (ANOVA)

10 86 mm x 128 mm plastic plate with 96 wells Lid has 96 pegs Resemblance Example: MBEC

11 123456789101112 A100 50:NNGCSC B50 50:NNGCSC C25 50:NNGCSC D12.5 50:NNGC E6.25 50:NNGC F3.125 50:NNGC G1.563 50:NNGC H0.781 50:NNGC MBEC Challenge Plate disinfectant neutralizer test control

12 Resemblance Example: MBEC Mean LD= 5.55 Control Data: log 10 (cfu/mm 2 ) from viable plate counts rowcfu/mm 2 log(cfu/mm 2 ) A 5.15 x 10 5 5.71 B 9.01 x 10 5 5.95 C 6.00 x 10 5 5.78 D 3.00 x 10 5 5.48 E 3.86 x 10 5 5.59 F 2.14 x 10 5 5.33 G 8.58 x 10 4 4.93 H 4.29 x 10 5 5.63

13 ExpRow Control LD Mean LDSD 1A5.71 5.550.31 1B5.95 1C5.78 1D5.48 1E5.59 1F5.33 1G4.93 1H5.63 2A5.41 0.17 2B5.71 2C5.54 2D5.33 2E5.11 2F5.48 2G5.33 2H5.41 Resemblance Example: MBEC

14 Resemblance from experiment to experiment Mean LD = 5.48 S r = 0.26 the typical distance between a control well LD from an experiment and the true mean LD

15 Resemblance from experiment to experiment The variance S r 2 can be partitioned: 2% due to between experiment sources 98% due to within experiment sources

16 S n c m c 2 + Formula for the SE of the mean control LD, averaged over experiments S c = within-experiment variance of control LDs S E = among-experiment variance of control LDs n c = number of control replicates per experiment m = number of experiments 2 2 S m E 2 SE of mean control LD = CI for the true mean control LD = mean LD ± t m-1 x SE

17 8 2 Formula for the SE of the mean control LD, averaged over experiments S c = 0.98 x (0.26) 2 = 0.00124 S E = 0.02 x (0.26) 2 = 0.06408 n c = 8 m = 2 2 2 2 SE of mean control LD = 0.00124 + 0.06408 = 0.1792 95% CI for the true mean control LD = 5.48 ± 12.7 x 0.1792 = (3.20, 7.76)

18 Resemblance from technician to technician Mean LD = 5.44 S r = 0.36 the typical distance between a control well LD and the true mean LD

19 The variance S r 2 can be partitioned: 0% due to technician sources 24% due to between experiment sources 76% due to within experiment sources Resemblance from technician to technician

20 Repeatability Independent repeats of the same experiment in the same laboratory produce nearly the same data, as indicated by a small repeatability standard deviation. Statistical tool: nested ANOVA

21 Repeatability Example Data: log reduction (LR) LR = mean(control LDs) – mean(disinfected LDs)

22 ExpRow Control LD Mean LDSD 1A5.71 5.550.31 1B5.95 1C5.78 1D5.48 1E5.59 1F5.33 1G4.93 1H5.63 2A5.41 0.17 2B5.71 2C5.54 2D5.33 2E5.11 2F5.48 2G5.33 2H5.41 Repeatability Example: MBEC 123456789101112 A 100 50:NNGCSC B 50 50:NNGCSC C 25 50:NNGCSC D 12.5 50:NNGC E 6.25 50:NNGC F 3.125 50:NNGC G 1.563 50:NNGC H 0.781 50:NNGC

23 Repeatability Example: MBEC Mean LR = 1.63 ExpRow Control LD Control Mean LDCol Disinfected 6.25% LD Disinfected Mean LDLR 1A5.71 5.554.511.04 1B5.9514.67 1C5.7824.41 1D5.4834.33 1E5.5944.59 1F5.3354.54 1G4.93 1H5.63 2A5.41 3.202.21 2B5.7114.78 2C5.5422.71 2D5.3333.48 2E5.1143.23 2F5.4851.82 2G5.33 2H5.41

24 Repeatability Example Mean LR = 1.63 S r = 0.83 the typical distance between a LR for an experiment and the true mean LR

25 S n c m c 2 + Formula for the SE of the mean LR, averaged over experiments S c = within-experiment variance of control LDs S d = within-experiment variance of disinfected LDs S E = among-experiment variance of LRs n c = number of control replicates per experiment n d = number of disinfected replicates per experiment m = number of experiments 2 2 2 S n d m d 2 + S m E 2 SE of mean LR =

26 Formula for the SE of the mean LR, averaged over experiments S c = within-experiment variance of control LDs S d = within-experiment variance of disinfected LDs S E = among-experiment variance of LRs n c = number of control replicates per experiment n d = number of disinfected replicates per experiment m = number of experiments 2 2 2 CI for the true mean LR = mean LR ± t m-1 x SE

27 Formula for the SE of the mean LR, averaged over experiments S c 2 = 0.00124 S d 2 = 0.47950 S E 2 = 0.59285 n c = 8, n d = 5, m = 2 SE of mean LR = 8 2 2 0.00124 + 0.59285 5 2 0.47950 + = 0.5868 95% CI for the true mean LR= 1.63 ± 12.7 x 0.5868 = 1.63 ± 7.46 = (0.00, 9.09)

28 How many coupons? experiments? n c m m 0.00124 + 0.59285 n d m 0.47950 + margin of error= t m-1 x no. control coupons (n c ): 235812 no. disinfected coupons (n d ): 235512 no. experiments (m) 28.207.807.46 7.16 32.272.152.06 1.97 41.451.381.32 1.27 60.960.910.87 0.84 100.650.620.59 0.57 1000.180.170.16

29 A method should be sensitive enough that it can detect important changes in parameters of interest. Statistical tool: regression and t-tests Responsiveness

30 disinfectant neutralizer test control Responsiveness Example: MBEC A: High Efficacy H: Low Efficacy 123456789101112 A100 50:NNGCSC B50 50:NNGCSC C25 50:NNGCSC D12.5 50:NNGC E6.25 50:NNGC F3.125 50:NNGC G1.563 50:NNGC H0.781 50:NNGC

31 Responsiveness Example: MBEC This response curve indicates responsiveness to decreasing efficacy between rows C, D, E and F

32 Responsiveness Example: MBEC Responsiveness can be quantified with a regression line: LR = 6.08 - 0.97row For each step in the decrease of disinfectant efficacy, the LR decreases on average by 0.97.

33 Summary  Even though biofilms are complicated, it is feasible to develop biofilm methods that meet the “Seven R” criteria.  Good experiments use control data!  Assess uncertainty by SEs and CIs.  When designing experiments, invest effort in more experiments versus more replicates (coupons or wells) within an experiment.

34 Any questions?


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