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Comparing Distributions III: Chi squared test, ANOVA By Peter Woolf (pwoolf@umich.edu) University of Michigan Michigan Chemical Process Dynamics and Controls Open Textbook version 1.0. Creative commons. Unit 1. Unit 2.

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## Comparing Distributions III: Chi squared test, ANOVA By Peter Woolf (pwoolf@umich)

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**Comparing Distributions III:**Chi squared test, ANOVA By Peter Woolf (pwoolf@umich.edu) University of MichiganMichigan Chemical Process Dynamics and Controls Open Textbookversion 1.0 Creative commons**Unit 1**Unit 2 Scenario: You have two parallel processes that carry out the same reaction using very similar equipment. Question: Are these units actually behaving the same or not?**Approach: (1) Gather data on yield from both units**Plot of data does not clearly show any difference**Requires binning**Directly on data Approach: (1) Gather data on yield from both units (2) Perform statistical analysis • Fisher’s exact test • Chi squared test • ANOVA**HIGH**LOW Binning Data: Data reduction by reassigning data into windows**HIGH**LOW Binning Data: Data reduction by reassigning data into windows Choosing a binning strategy: • Assign to bins that naturally appear such as groupings or important thresholds (e.g. yield>50 is profitable, so this is a natural window) • If multiple windows appear, assign multiple bins • If no natural bins appear, choose equally sized bins or above/below average Bin in excel with IF.. THEN statements**HIGH**As mentioned in last lecture, we can use Fisher’s exact to calculate a p-value of the probability of finding this configuration at random LOW For Fisher’s exact and Chi squared tests,create a contingency table. Contingency table Low High 97 150 Unit 1 53 82 68 150 Unit 2 135 165 300**Low**High observed 97 150 Unit 1 53 82 68 150 Unit 2 135 165 300 Low High “more extreme” configuration 98 150 Unit 1 52 83 67 150 Unit 2 135 165 300 “most extreme” configuration Low High 150 150 Unit 1 0 135 15 150 Unit 2 135 165 300**Most likely cases if this were a random sample**Observed case More extreme =0.0005 Less extreme =0.9995 Total area=1.0 Conclusion: • The units are behaving differently IDEA! The distance between observed case and the most likely if random is far, so can we just use that? Probability of configuration # changes away from observed**IDEA! The distance between observed case and the most likely**if random is far, so can we just use that? If this distance is “big” then the observed case is unusual What is this point? Probability of configuration # changes away from observed**Observed case**Low High 97 150 Unit 1 53 82 68 150 Unit 2 135 165 300 Distance between these two cases? Low High 150 Unit 1 150 Unit 2 135 165 300 Chi squared statistic What is this point? Most likely case if random =150*(135/300) =67.5 =150*(165/300) =82.5 But this depends on the magnitude, so normalize it.. =150*(135/300) =67.5 =150*(165/300) =82.5**Low**High 97 150 Unit 1 53 82 68 150 Unit 2 135 165 300 For this case: Low High 150 Unit 1 150 Unit 2 135 165 300 Chi squared statistic Observed case Most likely case if random Okay.. So what? What is the p-value? =150*(135/300) =67.5 =150*(165/300) =82.5 =150*(135/300) =67.5 =150*(165/300) =82.5**For this case:**This can be done in a more automated way in excel using “chitest” Chi squared statistic The chi squared statistic has a known distribution that can be looked up or found in excel using “chidist” with 1 degree of freedom. =chidist(11.33,1)=0.00076 For this case chitest & Fisher’s exact agree**Chi squared test vs. Fisher’s exact**• For a random null, Fisher’s exact will always yield a correct result • Chi squared test is often easier to carry out (the math is easier) • Chi squared will give incorrect results when • fewer than 20 samples are present • if there are between 20 and 40 samples and one expected number is 5 or below Chitest says the result is 2x more significant--error due to small sample effect**Chi squared test vs. Fisher’s exact (continued)**• Chi squared test is easy to do for larger contingency tables and when the expected distribution is not random. • Can be done with a Fisher’s like test, but the math gets much harder. Example: 3 by 3 contingency table with a model for expectations Observed is close to the expected, but far from random**Approach: (1) Gather data on yield from both units**(2) Perform statistical analysis • Fisher’s exact test • Chi squared test • ANOVA Requires binning Directly on data**ANOVA: Analysis of Variance**Method to compare continuous measurements determine if they are sampled from the same or different distributions. For a single factor ANOVA, we assume that each observation in each class can be modeled as: Observation = overall mean + class effect + random error In the study we are following in this class, the class effect would be the effect unit 1 or unit 2. ANOVA analysis can be easily done in Excel using Tools->Data Analysis-> ANOVA**1 way ANOVA**Key value: p-value here tells the probability that both units (each group) are the same.**2 way ANOVA with replicates**Scenario: Testing three units in triplicate, each with three different control architectures: Feedback (FB), Model predictive control (MPC), and a cascade architecture. In each case we measure the yield. Questions: Do the units significantly differ? Do the control architectures significantly differ? Tools->Data Analysis ->ANOVA:Two factor with replication**2 way ANOVA with replicates**Controllers (samples) have a significant effect Columns (units) don’t have a significant effect ?? Looks like an error, and may be why we get a negative F value and no p-value**ANOVA**• ANOVA tells you if factors are significantly related to an outcome according to a linear model • Nonlinear relationships can be strong, but may appear insignificant in an ANOVA analysis. • ANOVA does not tell you the model parameters. • ANOVA, t-test, and z-test all provide similar kinds of information for different kinds of data.**Statistical Analysis**Physical process Experimental Data • Results: • Unit 1 is different from unit 2 • This difference is clearer in the binned data (chi squared and fisher’s<ANOVA) Unit 1 Unit 2**Take Home Messages**• Chi squared tests are analogous to Fisher’s exact tests, but are generally easier to calculate • Chi squared tests fail when sample sizes are small • ANOVA determines if lists of continuous measurements likely the same or different • ANOVA can determine the significance of a set of factors on the measurements**The following pages have additional examples of ChemE**applications of ANOVA analyses**Solution approach:**two factor ANOVA. Factor 1: Farm Factor 2: Shipper See if a factor has a significant p-value**Looking at averages and ranges, it looks like shipper Rex**has a somewhat worse record than Ned. The farms have some variation, but it is small. This said, both shippers will bring wheat with moths, but Rex will bring more.**1) Import data into Excel**2) Select Tools->Data Analysis-> ANOVA: Two factor with Replication Conclusion, the factor “shipper” has a significant Influence on the moth probability with a p-value of 0.03**ANOVA- ChemE examples**How does temperature affect yield?**ANOVA- ChemE examples**Do both temperature and concentration affect yield?**ANOVA- ChemE examples**How can controlling v4 and v2 differently affect process profitability? Example from 2006 controls wiki: http://controls.engin.umich.edu/wiki/index.php/Design_of_experiments_via_taguchi_methods:_one_and_two_way_layouts**DATA**How can controlling v4 and v2 differently affect process profitability? Example from 2006 controls wiki: http://controls.engin.umich.edu/wiki/index.php/Design_of_experiments_via_taguchi_methods:_one_and_two_way_layouts**DATA**ANOVA How can controlling v4 and v2 differently affect process profitability? Example from 2006 controls wiki: http://controls.engin.umich.edu/wiki/index.php/Design_of_experiments_via_taguchi_methods:_one_and_two_way_layouts

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