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lecture16

# lecture16 - 3.051J/20.340J Lecture 16 Statistical Analysis...

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1 3.051 J / 20 .340 J Lecture 16 Statistical Analysis in Biomaterials Research (Part II) C. F Distribution ¾ Allows comparison of variability of behavior between populations using test of hypothesis: σ x = σ x’ Named for British statistician Sir Ronald A. Fisher. Define a statistic: 2 2 χ 2 = ν 1 S / σ where ν 1 is degrees of freedom. ν 1 χ 2 / ν v 1 then F = χ 2 / ν 1 v 2 2 2 S x F For σ x = σ x’ F = S 2 x ' Procedure to test variability hypothesis: 1. Calculate S x 2 and S x’ 2 (with ν 1 = Ν −1 and ν 2 = N’ -1, respectively) 2. Compute F 3. Look in F -distribution tables for critical F for ν 1 , ν 2 , and desired confidence level P F 1 P < < 4. For F F 1 + P ⇒ σ x = σ x’ 2 2

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2 3.051 J / 20 .340 J Case Example : Measurements of C5a production for blood exposure to an extracorpeal filtration device and tubing gave same means, but different variabilities. Are the standard deviations different within 95% confidence? Control (tubing only): S x’ 2 = 26 ( µ g/ml) 2 , ν 2 =9 Filtration device: S x 2 = 32 ( µ g/ml) 2 , ν 1 = 7 2 1. Calculate S x 2 and S x’ (provided) 2. Compute F 2 S x F = S 2 = 32/26 = 1.231 x ' 3. Determine critical F values from F- distribution chart ν 1 =7 and ν 2 =9 (m, n for use with tables) 1 P = 0.025 F 0.025 = 0.207 2 1 + P = 0.975 F 0.975 = 4.20 2 For 0.207 F 4.20 ⇒ σ x = σ x’ F =1.231 falls within this interval. Conclude σ values for two systems are the same!
3 3.051 J / 20 .340 J D. Other distributions of interest ¾ Radioactive decay Poisson distribution ¾ Only 2 possible outcomes Binomial distribution 3. Standard deviations of computed values ¾ If quantity z of interest is

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