Selected Statistics Examples from Lectures Matlab Histograms y
Selected Statistics Examples from Lectures
Matlab: Histograms >> y = [1 3 5 8 2 4 6 7 8 3 2 9 4 3 6 7 4 1 535896246156987534529 659416785429679253196 843679134752985745436 793169567321578531975 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
Probability Examples l Probability that at least one coin will turn heads up from five tossed coins » » l Number of outcomes: 25 = 32 Probability of each outcome: 1/32 complement set Probability of no heads: P(AC) = 1/32 Probability at least one head: P(A) = 1 -P(AC) = 31/32 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:
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: nk = 265 = 11, 881, 376 » Total number of different words containing each letter no more than once:
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
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
Continuous Distribution Example l Probability density function l Cumulative distribution function l Probability of events
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
Matlab: Normal Distribution Example l 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? 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
Discrete Distribution Example l Matlab function: unicdf(x, n), >> x = (0: 6); >> y = unidcdf(x, 6); >> stairs(x, y)
Poisson Distribution Example l l Probability of a defective thin polymer film p = 0. 01 What is the probability of more than 2 defects in a lot of 100 samples? Binomial distribution: m = np = (100)(0. 01) = 1 Since p <<1, can use Poisson distribution to approximate the solution.
Matlab: Maximum Likelihood l 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. 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
Mean Confidence Interval Example l l Measurements of polymer molecular weight (scaled by 10 -5) Confidence interval
Variance Confidence Interval Example l l Measurements of polymer molecular weight (scaled by 10 -5) Confidence interval
Matlab: Confidence Intervals >> [muhat, sigmahat, muci, sigmaci] = normfit(data, alpha) 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. 31 1. 26], 0. 05) muhat = 1. 2580 sigmahat = 0. 0697 muci = 1. 2081 1. 3079 sigmaci = 0. 0480 0. 1273 l
Mean Hypothesis Test Example l Measurements of polymer molecular weight l Hypothesis: m 0 = 1. 3 instead of the alternative m 1 < m 0 Significance level: a = 0. 10 Degrees of freedom: m = 9 Critical value l Sample t l Reject hypothesis l l l
Polymerization Reactor Control Monomers Catalysts Hydrogen Solvent Polymer On-line measurements Lab measurements l Monomer Content Viscosity 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: m 0 = 20%, s 2 = 1 » Control limits: » Sample mean: 21% process is in control
Goodness of Fit Example l Maximum likelihood estimates
Goodness of Fit Example cont.
Matlab: Goodness of Fit Example l Find maximum likelihood estimates for µ and σ of a normal distribution with function mle >> data=[320 … 360]; >> phat = mle(data) phat = l 364. 7 26. 7 Analyze dataset for goodness of fit using function chi 2 gof – usage: [h, p, stats]=chi 2 gof(x, ’edges’, edges) for data in vector x divided into intervals with the specified edges » » If h = 1, reject hypothesis at 5% significance If h = 0, accept hypothesis at 5% significance p: probability of observing given statistic stats: includes chi-square statistic and degrees of freedom >> [h, p, stats]=chi 2 gof(data, ’edges’, [-inf, 325: 10: 405, inf]); >> h = 0 >> p = 0. 8990 >> chi 2 stat = 2. 8440 >> df = 7
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