Data Analysis Examples
This document discusses statistical methods for analyzing temperature fluctuations within a coal gassifier and computing viscosity using a falling bead viscometer. It explores normal distribution characteristics, calculating standard deviation, and provides insights into the probability of exceeding threshold temperatures. Additionally, it covers the impact of measurement uncertainties on viscosity calculations and identifies dominant sources of error. Finally, it delves into hypothesis testing approaches to compare effectiveness between new and old catalysts, offering a comprehensive understanding of statistical analysis in experimental data.
Data Analysis Examples
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Presentation Transcript
Data Analysis Examples Anthony E. Butterfield CH EN 4903-1
#1: The Normal PDF • Your coworker tells you the temperature fluctuations of the outlet temperature from a certain coal gassifier have an average of 1304 K and keep within 12 K of that mean for 95% of her measurements, over months of operation. If we assume the temperature measurements are normally distributed, what is the standard deviation and what are the odds that a temperature measurement would be above 1310 K? • T = 1304 ± 12 K (95% Confidence Level)
Normal Distribution • Probability density function (PDF):
#2: Error Propagation • In a falling bead viscometer, the viscosity may be found by the following equation: • Where r is the bead radius, g is gravitational acceleration, V is the terminal velocity, rB is the bead density and rF is the fluid density. If we find, within a 95% confidence level, that the bead density is 2 ± 0.1 g/cm3, the radius is 3 ± 0.1 mm, the fluid density is 1.1 ± 0.2 g/cm3, and, after terminal velocity is achieved, the bead falls 10 ± 0.2 cm in 12 ± 0.5 seconds. What is the calculated viscosity and the uncertainty in its value? Which measurement is the greatest source of error?
#2: Error Propagation • A couple options:
#3: Log Normal • 2. You find the following particle size distributions from a spray dryer experiment: Table of dataIf we were to assume this distribution of particle sizes is log-normal, what would be the mean and standard deviation for the log-normal pdf? • Nonlinear fitting problem, like #6.
#4: Hypothesis Testing • On a certain stage of a distillation column theory predicts the ethanol concentration should be 27%. You take the following measurements over several runs: • What is the likelihood that your measurements match theory?
#4: Hypothesis Testing • Student’s T-Test. • Mean = 24.52 • StDev = 4.2163 • Degrees of Freedomv = na – 1 = 10 -1 = 9
#4: Hypothesis Testing • T-Statistic:
#4: Hypothesis Testing • Use t-statistic in CDF to find probability. Answer = 9.6%
#5: Hypothesis Testing 2 • You are measuring the effectiveness of a new catalyst on a reaction with a great deal of normally distributed variability. You measure the time to 99% conversion of your reactants with both your new and old catalyst for several experimental runs and find the following data: • Given this data, what is the probability that the new catalyst is more effective than the old? What is the probability that they are equally effective?
#5: Hypothesis Testing 2 • Mean A = 10.25, Mean B = 9.50 • StDev A = 1.071, StDev B = 1.066 • Number A = 22, Number B = 20 • Degrees of Freedomv = na + nb – 2 = 40
#5: Hypothesis Testing 2 • T-Statistic:
#5: Hypothesis Testing 2 • Simple rule: • Greater or less than tests use one tail (two unequal areas) and you can easily know which % you want to use by looking at the means. • Equaltest uses two equal tails. • For T-CDF with v = 40 and at t-statistic of -2.295, P = 2.7%. • P that new catalyst is more effective is a one tail test. • More effective (one tail) = 100% - 2.7% = 97% • Equal (two tail) = 2*2.7% = 5%
#6: Non-Linear Fit • The rate of population growth in a bacteria culture are found to be: • It is thought that this data could be fit to the equation: Rate=b1*sin(b2*t)where b1 and b2 are constants to be determined and t is time. Determine the least squares estimated values for b1 and b2 and give an appropriate confidence interval for a confidence level of 90%. Also, what would you anticipate the rate to be at 24 hr? What would the confidence interval for a 95% confidence level be at 24 hr?
%Anthony Butterfield 2009 %Example of nonlinear fit with CIs clear close all b(1)=1/3; b(2)=1; re=0.1; %random noise strength x=linspace(0,6,20)'; %x data for fitting x2=linspace(0,6,100)'; %x data for plotting n=length(x); y=b(1)*sin(b(2)*x)+re*randn(n,1); %y data for fitting, note the random error added in to make it realistic yt=b(1)*sin(b(2)*x2); %theoretical y data for plotting [beta r J]=nlinfit(x,y,@nlinfitsin,[1 1]); %numerically performs a nonlinear fit bci = nlparci(beta,r,J); %returns the c.i. for the parameters, beta [ypred,delta] = nlpredci(@nlinfitsin,x2,beta,r,J); %returns a predicted y and the c.i. for each y [ypred,delta] = nlpredci(@nlinfitsin,x2,beta,r,J); %returns a predicted y and the c.i. for each y disp('Fit to equation: y = b1 sin(b2 * x)') disp(' x data y data') for i=1:n txt=sprintf(' %5.3f %5.3f',x(i),y(i)); disp(txt) end txt=sprintf('b1 was %3.1f, and is estimated to be: %f ± %f (95%% CL)',b(1),beta(1),abs(beta(1)-bci(1,1))); disp(txt) txt=sprintf('b2 was %3.1f, and is estimated to be: %f ± %f (95%% CL)',b(2),beta(2),abs(beta(2)-bci(2,1))); disp(txt) figure(1) hold on grid on scatter(x,y,10,'r') plot(x2,yt,'Color',[1 0.5 0]) %just wanted to give you an example of how to change the line color to something not preset plot(x2,ypred,'b',x2,ypred+delta,'b:',x2,ypred-delta,'b:') hold off
#6: Non-Linear Fit • nlparci • In “theory” b1 = 0.3; estimated b1 = 0.35 ± 0.05 (90% CL) • In “theory” b2 = 1.0; estimated b2 = 1.04 ± 0.04 (90% CL) • nlpredci • At 24 hr “theory” predicts: • Rate = -0.3019 • Fit predicts: • Rate = -0.1090 ± 0.3839 (95% CL)
