Download presentation

Presentation is loading. Please wait.

Published byPhilomena Watkins Modified over 2 years ago

1
Center for Biofilm Engineering Standardized Biofilm Methods Research Team Montana State University Importance of Statistical Design and Analysis Al Parker July, 2010

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 Design Uncertainty assessment

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

5
Why statistical thinking? Provide convincing results Anticipate criticism Increase efficiency Improve communication

6
Attributes of a standard method: Seven R’s Relevance Reasonableness Resemblance Repeatability (intra-laboratory reproducibility) Ruggedness Responsiveness Reproducibility (inter-laboratory)

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

8
Resemblance 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)

9
Resemblance Example

10
Coupon Density LD cfu / cm 2 log(cfu/cm 2 ) 1 5.5 x 10 6 6.74 2 6.6 x 10 6 6.82 3 8.7 x 10 6 6.94 Mean LD= 6.83 Data: log 10 (cfu) from viable plate counts

11
Resemblance Example Exp control LD Mean LD SD 16.73849 16.820566.832400.10036 16.93816 26.66276 26.739576.714400.04473 26.74086 36.91564 36.745576.852930.09341 36.89758

12
Resemblance from experiment to experiment Mean LD = 6.77 S r = 0.15 the typical distance between a control coupon LD from an experiment and the true mean LD log 10 (cfu/cm 2 )

13
Resemblance from experiment to experiment The variance S r 2 can be partitioned: 69% due to between experiment sources 31% due to within experiment sources log 10 (cfu/cm 2 )

14
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 coupon LD S E = between-experiments variance of control coupon LD n c = number of control coupons per experiment m = number of experiments 2 2 S m E 2 SE of mean control LD =

15
3 Formula for the SE of the mean control LD, averaged over experiments S c = 0.31 x (.15) 2 = 0.006975 S E = 0.69 x (.15) 2 = 0.015525 n c = 3 m = 3 2 2 3 SE of mean control LD =.006975 +.015525 = 0.0771 95% CI for mean control LD = 6.77 ± t 6 x 0.0771 = (6.58, 6.96)

16
Resemblance from technician to technician Mean LD = 8.42 S r = 0.17 the typical distance between a coupon LD and the true mean LD log 10 (cfu/cm 2 )

17
The variance S r 2 can be partitioned: 39% due to technician sources 43% due to between experiment sources 18% due to within experiment sources Resemblance from technician to technician log 10 (cfu/cm 2 )

18
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

19
Repeatability Example Data: log reduction (LR) LR = mean(control LDs) – mean(disinfected LDs)

20
Repeatability Example Exp control LD Mean LD SD 16.73849 16.820566.832400.10036 16.93816 26.66276 26.739576.714400.04473 26.74086 36.91564 36.745576.852930.09341 36.89758

21
Repeatability Example log densitymean log density Expcontroldisinfectedcontroldisinfected log reduction 16.738493.08115 16.820563.293266.832403.135463.69695 16.938163.03196 26.662762.92334 26.739573.034886.714403.056563.65784 26.740863.21146 36.915642.73748 36.745572.660186.852932.708054.14488 36.897582.72651 Mean LR = 3.83

22
Repeatability Example Mean LR = 3.83 S r = 0.27 the typical distance between a LR for an experiment and the true mean LR

23
S n c m c 2 + Formula for the SE of the mean LR, averaged over experiments S c = within-experiment variance of control coupon LD S d = within-experiment variance of disinfected coupon LD S E = between-experiments variance of LR n c = number of control coupons n d = number of disinfected coupons m = number of experiments 2 2 2 S n d m d 2 + S m E 2 SE of mean LR =

24
Formula for the SE of the mean LR, averaged over experiments S c 2 = 0.006975 S d 2 = 0.014045 S E 2 = 0.066234 n c = 3, n d = 3, m = 3 SE of mean LR = 3 3.006975 +.066234 3.014045 + = 0.156 95% CI for mean LR= 3.83 ± t 2 x 0.156 = (3.16, 4.50)

25
How many coupons? experiments? no. control coupons (n c ):23612 no. disinfected coupons (n d ):23612 no. experiments (m) 10.2770.2710.2640.261 20.1960.1910.1870.184 30.1600.1560.1520.151 40.1380.1350.1320.130 60.1130.1100.1080.106 100.0880.0860.0840.082 1000.0280.0270.026 n c m m.006975 +.066234 n d m.014045 + SE of mean LR =

26
Repeats of the same experiment run independently by different researchers in different laboratories produce nearly the same result as indicated by a small reproducibility standard deviation. Requires a collaborative (multi-lab) study. Statistical tool: nested ANOVA Reproducibility

27
Reproducibility Example Mean LR = 2.61 S R = 1.07 the typical distance between a LR for an experiment at a lab and the true mean LR

28
Reproducibility Example The variance S R 2 can be partitioned: 62% due to between lab sources 38% due to between experiment sources

29
S n c mL c 2 + Formula for the SE of the mean LR, averaged over labratories S c 2 = within-experiment variance of control coupon LD S d 2 = within-experiment variance of disinfected coupon LD S E 2 = between-experiments variance of LR S L 2 = between-lab variance of LR n c = number of control coupons n d = number of disinfected coupons m = number of experiments L = number of labs S n d mL d 2 + S mL E 2 SE of mean LR = + S L L 2

30
Formula for the SE of the mean LR, averaged over labratories S c 2 = 0.007569 S d 2 = 0.64 S E 2 =.2171 S L 2 = 0.707668 n c = 3, n d = 3, m = 3, L = 2 SE of mean LR = 3 3 2 3 2.007569 +.2171.64 + = 0.653 95% CI for mean LR= 2.61 ± t 4 x 0.653 = (0.80, 4.42) 3 3 2 2 +.707668

31
How many coupons? experiments? labs? SE of mean LR = n c mL mL.007569 +.2171.64 + n d mL L +.707668 no. of labs (L)111222333444555666 no. control/dis coupons (n c and n d ):235235235235235235 no. experiments (m) 11.1171.0681.0270.7900.7550.7260.6450.6170.5930.5590.5340.5130.5000.4780.4590.4560.4360.419 20.9890.9610.9390.6990.6800.6640.5710.5550.5420.4940.4810.4690.4420.4300.4200.4040.3920.383 30.9420.9230.9070.6660.6530.6420.5440.5330.5240.4710.4620.4540.4210.4130.4060.3850.3770.370 40.9180.9030.8910.6490.6390.6300.5300.5220.5150.4590.4520.4460.4110.4040.3990.3750.3690.364 60.8930.8830.8750.6320.6240.6190.5160.5100.5050.4470.4420.4370.3990.3950.3910.3650.3610.357 100.8730.8670.8620.6170.6130.6090.5040.5000.4970.4360.4330.4310.3900.3880.3850.3560.3540.352 1000.844 0.8430.597 0.5960.4880.487 0.422 0.3780.377 0.3450.344

32
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 numbers of experiments versus more coupons in an experiment).

33
Any questions?

Similar presentations

OK

Statistical Concepts Basic Principles An Overview of Today’s Class What: Inductive inference on characterizing a population Why : How will doing this allow.

Statistical Concepts Basic Principles An Overview of Today’s Class What: Inductive inference on characterizing a population Why : How will doing this allow.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on computer malwares anti Ppt on union budget 2012-13 Ppt on fibonacci numbers examples Ppt on file system in unix grep Ppt on newton first law of motion Ppt on motion force and pressure Ppt on steel structures in india Ppt on c language basics pdf Ppt on panel discussion questions Ppt on airbags in cars