1. Selected Statistics Examples from Lectures

2. Matlab: Histograms >> y = [1 3 5 8 2 4 6 7 8 3 2 9 4 3 6 7 4 1 5 3 5 8 9 6 2 4 6 1 5 6 9 8 7 5 3 4 5 2 9 6 5 9 4 1 6 7 8 5 4 2 9 6 7 9 2 5 3 1 9 6 8 4 3 6 7 9 1 3 4 7 5 2 9 8 5 7 4 5 4 3 6 7 9 3 1 6 9 5 6 7 3 2 1 5 7 8 5 3 1 9 7 5 3 4 7 9 1]’; >> mean(y)  ans = 5.1589 >> var(y)  ans = 6.1726 >> std(y)  ans = 2.4845 >> hist(y,9)  histogram plot with 9 bins >> n = hist(y,9)  store result in vector n >> x = [2 4 6 8]’ >> n = hist(y,x)  create histogram with bin centers specified by vector x

3. Probability Examples l Probability that at least one coin will turn heads up from five tossed coins »Number of outcomes: 2 5 = 32 »Probability of each outcome: 1/32 »Probability of no heads: P(A C ) = 1/32 »Probability at least one head: P(A) = 1-P(A C ) = 31/32 l Probability of getting an odd number or a number less than 4 from a single dice toss »Probability of odd number: P(A) = 3/6 »Probability of number less than 4: P(B) = 3/6 »Probability of both: »Probability of either: complement set

4. Permutation Examples l Process for manufacturing polymer thin films »Compute the probability that the first 6 films will be too thin and the next 4 films will be too thick if the thickness is a random variable with the mean equal to the desired thickness »n 1 = 6, n 2 = 4 »Probability l Encryption cipher »Letters arranged in five-letter words: n = 26, k = 5 »Total number of different words: n k = 26 5 = 11,881,376 »Total number of different words containing each letter no more than once:

5. Combination Examples l Effect of repetitions »Three letters a, b, c taken two at a time (n = 3, k = 2) »Combinations without repetition »Combinations with repetitions l 500 thin films taken 5 at a time »Combinations without repetitions

6. Matlab: Permutations and Combinations >> perms([2 4 6])  all possible permutations of 2, 4, 6 6 4 2 6 2 4 4 6 2 4 2 6 2 4 6 2 6 4 >> randperm(6)  returns one possible permutation of 1-6 5 1 2 3 4 6 >> nchoosek(5,4)  number of combinations of 5 things taken 4 at a time without repetitions ans = 5 >> nchoosek(2:2:10,4)  all possible combinations of 2, 4, 6, 8, 10 taken 4 at a time without repetitions 2 4 6 8 2 4 6 10 2 4 8 10 2 6 8 10 4 6 8 10

7. Continuous Distribution Example l Probability density function l Cumulative distribution function l Probability of events

8. Matlab: Normal Distribution l Normal distribution: normpdf(x,mu,sigma) »normpdf(8,10,2)  ans = 0.1210 »normpdf(9,10,2)  ans = 0.1760 »normpdf(8,10,4)  ans = 0.0880 l Normal cumulative distribution: normcdf(x,mu,sigma) »normcdf(8,10,2)  ans = 0.1587 »normcdf(12,10,2)  ans = 0.8413 l Inverse normal cumulative distribution: norminv(p,mu,sigma) »norminv([0.025 0.975],10,2)  ans = 6.0801 13.9199 l Random number from normal distribution: normrnd(mu,sigma,v) »normrnd(10,2,[1 5])  ans = 9.1349 6.6688 10.2507 10.5754 7.7071

9. Matlab: Normal Distribution Example l The temperature of a bioreactor follows a normal distribution with an average temperature of 30 o C and a standard deviation of 1 o C. What percentage of the reactor operating time will the temperature be within +/-0.5 o C of the average? l Calculate probability at 29.5 o C and 30.5 o C, then calculate the difference: »p=normcdf([29.5 30.5],30,1) p = [0.3085 0.6915] »p(2) – p(1) 0.3829 l The reactor temperature will be within +/- 0.5 o C of the average ~38% of the operating time

10. Discrete Distribution Example l Matlab function: unicdf(x,n), >> x = (0:6); >> y = unidcdf(x,6); >> stairs(x,y)

11. Poisson Distribution Example l Probability of a defective thin polymer film p = 0.01 l What is the probability of more than 2 defects in a lot of 100 samples? Binomial distribution:  = np = (100)(0.01) = 1 l Since p <<1, can use Poisson distribution to approximate the solution.

12. Matlab: Maximum Likelihood l In a chemical vapor deposition process to make a solar cell, there is an 87% probability that a sufficient amount of silicon will be deposited in each cell. Estimate the maximum likelihood of success in 5000 cells. l s = binornd(n,p) – randomly generate the number of positive outcomes given n trials each with probability p of success »s = binornd(5000,0.87) s=4338 l phat = binofit(s,n) – returns the maximum likelihood estimate given s successful outcomes in n trials »phat = binofit(s,5000) phat = 0.8676

13. Mean Confidence Interval Example l Measurements of polymer molecular weight (scaled by 10 -5 ) l Confidence interval

14. Variance Confidence Interval Example l Measurements of polymer molecular weight (scaled by 10 -5 ) l Confidence interval

15. Matlab: Confidence Intervals >> [muhat,sigmahat,muci,sigmaci] = normfit(data,alpha) l data: vector or matrix of data l alpha: confidence level = 1-alpha l muhat: estimated mean l sigmahat: estimated standard deviation l muci: confidence interval on the mean l sigmaci: confidence interval on the standard deviation >> [muhat,sigmahat,muci,sigmaci] = normfit([1.25 1.36 1.22 1.19 1.33 1.12 1.27 1.27 1.31 1.26],0.05) muhat = 1.2580 sigmahat = 0.0697 muci = 1.2081 1.3079 sigmaci = 0.0480 0.1273

16. Mean Hypothesis Test Example l Measurements of polymer molecular weight Hypothesis:  0 = 1.3 instead of the alternative  1 <  0 Significance level:  = 0.10 l Degrees of freedom: m = 9 l Critical value l Sample t l Reject hypothesis

17. Polymerization Reactor Control l Lab measurements »Viscosity and monomer content of polymer every four hours »Three monomer content measurements: 23%, 18%, 22% l Mean content control »Expected mean and known variance:  0 = 20%,   = 1 »Control limits: »Sample mean: 21%  process is in control Monomers Catalysts Solvent Hydrogen On-line measurements Lab measurements Polymer Monomer Content Viscosity

18. Goodness of Fit Example l Maximum likelihood estimates

19. Goodness of Fit Example cont.

20. Linear Regression Example l Reaction rate data l Sample variances and covariance l Linear regression Confidence interval for  = 0.95 and m = 6 Experiment12345678 Reactant Concentration 0.10.30.50.70.91.21.52.0 Rate2.35.710.713.118.525.432.145.2

21. Matlab: Linear Regression Example >> c = [0.1 0.3 0.5 0.7 0.9 1.2 1.5 2]; >> r = [2.3 5.6 10.7 13.1 18.5 25.4 32.1 45.2]; >> [k, kint] = regress(r’, [ones(length(c), 1), c’], 0.05) k = -1.1847 22.5524 kint = -2.8338 0.4644 21.0262 24.0787

22. Correlation Analysis Example l Polymerization rate data l Correlation coefficient  = 5%, m = 6  c = 1.94 l Compute t-statistic l Reject hypothesis that the hydrogen concentration and the polymerization rate are uncorrelated Experiment12345678 Hydrogen Concentration 00.10.30.51.01.52.03.0 Polymerization rate 9.79.210.710.110.511.210.410.8

23. Matlab: Correlation Analysis Example >> h = [0 0.1 0.3 0.5 1 1.5 2 3]; >> p = [9.7 9.2 10.7 10.1 10.5 11.2 10.4 10.8]; >> R = corrcoef(h,p) R= 1.0000 0.6238 0.6238 1.0000 >> t = ttest(h,p) t=1 (reject hypothesis)