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Statistical thinking in antibiofilm research

Statistical thinking in antibiofilm research. Cord Hamilton Al Parker Marty Hamilton. MBL and SBML: 23 October 2008. Topics (presenter). Calculating LR and the within-experiment standard error of LR (Cord)

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Statistical thinking in antibiofilm research

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  1. Statistical thinking in antibiofilm research Cord Hamilton Al Parker Marty Hamilton MBL and SBML: 23 October 2008

  2. Topics (presenter) Calculating LR and the within-experiment standard error of LR (Cord) Using data from repeated experiments to find more reliable LR values in the future (Al) Analysis of dilution series counts (Marty)

  3. Log Reduction (LR) fora Quantitative Assay Vc = viable cell density of biofilm grown in the absence of antimicrobial treatment Vd = viable cell density of biofilm grown in the presence of the disinfectant Log Reduction = log10(Vc) - log10(Vd)

  4. Numerical Example Vc = 107 & Vd = 10 Log Reduction = log10(107) - log10(10) LR = 7 - 1 LR = 6 Interpretation: disinfectant killed 99.9999% of the bacteria

  5. D Calculating LR when there are multiple coupons = mean of control log10 densities = mean of disinfected log10 densities Log Reduction = C D - C

  6. D Example: Mean of logs for 3 disinfected coupons Coupon Density log10Density (i) cfu / cm2 (Di) 1 9.6·104 4.982 2 1.7·104 4.230 3 9.7·1033.987 Mean = 4.400 = mean density = 4.09∙104 log of mean density = 4.61

  7. Example: Control coupons Coupon log10Density (i) (Ci) 1 7.499 2 7.013 3 7.863 = 7.458 C

  8. Calculating LR when there are multiple coupons = 7.458 & = 4.400 Log Reduction = C D - D C = 7.458 - 4.400 LR = 3.058

  9. Within-experiment standard error (SE) of the LR 2 Sc = variance of control log10 densities Sd = variance of disinfected log10 densities nc = number of control coupons nd = number of disinfected coupons SE of LR = (within-experiment) 2 2 2 S S + c d nc nd

  10. Example: Calculating SE for single reactor experiment Sc = 0.181865 and nc = 3 Sd = 0.269272 and nd = 3 SE = 2 2 0.181865 0.269272 + 3 3 = 0.3878

  11. Uncertainty in LR Estimate LR ± SE = 3.058 ± 0.388 or 3.06 ± 0.39 or 3.1 ± 0.4

  12. 3 2 1 0 RDR biofilm: 5 ppm chlorine for 10 minutes Log Reduction ± SE 3 4 5 1 2 Experiment

  13. Experiment repeated 3 times, each using three control and 3 disinfected coupons

  14. Statistical summary for data from 3 experiments, with 3 control and 3 disinfected coupons per experiment

  15. Formula for the SE of the mean LR, averaged over experiments Sc = within-experiment variance of control coupon LD Sd = within-experiment variance of disinfected coupon LD SE = between-experiments variance of LR nc = number of control coupons nd = number of disinfected coupons m = number of experiments 2 2 2 2 2 2 S S S E d c SE of mean LR = + + m nd • m nc • m

  16. Formula for the SE of the mean LR, using estimated standard deviations 2 2 2 0.2574 0.1185 0.0833 SE of mean LR = + + m nd • m nc • m

  17. Choosing the numbers of coupons and the number of experiments. Table cell is the the SE of the mean LR. Shaded SE values are designs requiring 24 coupons.

  18. Dilution series and drop plate technique 10 Counted dilution 32 colonies Source: BiofilmsOnline

  19. Find the fraction of initial beaker volume in each of the dilution tubes 10 0.1 0.01 0.001 0.0001 Beaker: contained all cells from coupon fraction of beaker volume in tube Source: BiofilmsOnline

  20. Estimated number of cells in beaker = cfu count divided by the volume fraction plated 10 Beaker: contains all cells from coupon 10-4 fraction in tube Estimate: 32/(5 x 10-7) = 6.4 x 107 Plated 50 μl from tube; plate contains a fraction 50/10000 = 5 x 10-3 of the volume in the tube. f = (5 x 10-3) 10-4 = 5 x 10-7

  21. Dilution series and filter technique: pooling data from two tubes The 460 cfu corresponds to this fraction of the beaker volume: f = 1.8x10-5 + 2.0x10-6 = 2.0 x 10-5 10 9 ml filtered 10 ml filtered Estimate for beaker = 460/(2.0x10-5) = 2.3 x 107 39 cfu 421 cfu Count 20 fields on each filter; corresponds to 0.02 of filter area f = 0.001 x 0.9 x 0.02 = 1.8 x 10-5 f = 0.0001 x 1.0 x 0.02 = 2.0 x 10-6

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