Chapter 5 Workshop on Fitting of Linear Data
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1 Chapter 5 Workshop on Fitting of Linear Data (Contributed by E.L. Allen, SJSU) 5.0 Learning Objectives After successfully com pleting this laboratory workshop, including the assigned reading, the lab repot sheets, the lab quizzes, and any required reports, the student will be able to: 1. Distinguish between dependent and independent variables. 2. Properly display x-y data in a scatter plot w ith appropriate scaling and axes, both manually and using an Excel spreadsheet with its graphing functions. 3. Recognize from a scatter plot whether data is linear, has some other functional dependence, or has no functional dependence between x and y. 4. Perform a linear regress ion analysis of experimental data, both manually and using an Excel spreadsheet or the linear regression commands in MATLAB (e.g., the Basic Fitting Interface) 5. Correctly display a fitted line and its equations and demonstrate understanding of relationship between fitting parameters and experimental data. 6. Demonstrate understanding of the distinction between raw data and experimental results. 5.1 References 1. C. Chatfield, Statistics for Technology, Chapman and Hall, London, W.J. Palm, "Intro to MATLAB7 for Engineers", McGraw-Hill, 2005, pp Linear Regression Linear regression is a technique for finding the linear relationship between the x and y values of experimental data. It assum es that there is a linear relationship between x and y. This m ay not always be the case; there m ay be a linear relatio nship between x and a function of y; there m ay be a non-linear function, or there may be no relationship at all between the data. We only apply linear regression to data when we have reason to think that there is a linear r elationship. It is necessary to first identify which is the independent variable and which is the dependent variable. Here is a m ethod to follow. Many of you have linear regression progra ms on your engineering ENGR45 Chapter 5: Linear Regression 5-1
2 calculators; it is also embedded in Excel and MATLAB software. However, to understand the m eaning and reliability of f itted data, it is best to perform a manual linear regression at least once in your life. In the exercise that follows, you may use your calculator for adding columns of data only but no t for fitting. W hen you have finished the linear reg ression exercise, you may use your calculator s program to check your result; then you can plot the data in Excel or MATLAB and compare the results obtained with Excel s or MATLAB's algorithm. More on this technique, and other statistical methods, can be found in the references. Before beginning any calculations for a linear regression analysis, consider the following points: 1. Identify x and y, that is the independent and dependent variables in the experiment. 2. Make a scatter diagram, i.e. plot the data on an x-y coordinate system and see if it is roughly linear. 3. If it isn t, think about wh ether you should plot som e f(y) such as log y or y 2 ; this generally requires some notion of what the experiment is about and what is the expected relationship in the data. If you need to transform the x or y data, do so, and create a new set of data (c.f. ENGR25) 4. You now have n pairs of data, which we shall refer to as (, y i ). The x variable is referred to as the control or independent variable; y i is referred to as the response or dependent variable. 5. Assume y is subject to scatter (i.e. random errors) and that s not. 6. Assume a linear relationship exists, as y = a 0 + x (equivalent to: y = mx + b). 7. Our objective is to find estim ates for a 0 and such that the line gives a good fit. The method of least squares is one way to do this. In general, the values of a 0 and are interpreted to have some fundamental physical meaning. This is one way of determ ining the value of physical constants or m aterials prop erties. The values so o btained are only as reliable as the measurement errors. 8. The assum ption of a linear f it in the data pre cludes the possibility that there is a m ore complex relationsh ip be tween them, or th at the relation ship is on ly linear ove r a lim ited range of the independent variable, or that th ere are other factors wh ich inf luence the data which we have not explicitly considered. A ll of these c oncepts m ust be considered in interpreting the data. Now that your data is available for a linear re gression you are ready to begin. The following steps provide a guide to performing a linear regression analysis: Chapter 5: Linear Regression 5-2
3 1. Consider that at, the predicted value of y is: yp = a 0 + a1 2. The difference between the observed y-value (y i ) and the predicted value (yp) is e i = y i ( a 0 + ) where e i is the deviation or residual. There is a value of e i at each data pair. 3. Choose values of a 0 and a1 that minimize the s um of the squares of the e i s at each pair of points. The sum of the squares we will call S. n 2 S = e i = y i a 0 + i=1 S = f a 0, n i=1 ( ) [ ( )] 2 4. To minimize S, we need to find the partial derivatives S a0 and S, set th em equal to zero, and solve the resulting simultaneous equa ˆ a 0 and ˆ tions f or the leas t s quares es timates of. In the pa ges that follow, each sum is assu med to be from i=1 to i=n, where n is the total number of data pairs. S = 2[ y i ( a 0 + )] 1 a 0 ( ) S = 2[ y i ( a 0 + )] ( ) 5. Next set the two partial derivatives equal to zero, and solve for the values of a ˆ 0 and ˆ : S = 0 = a 2[ y i a ˆ 0 a ˆ 1 ] 1 ( ) 0 0 = 2y i + 2a ˆ 0 + 2a ˆ 1 and similarly, for the second partial: y i = n ˆ a 0 + ˆ S = 0 = 2[ y a i a ˆ 0 a ˆ 1 ] x ( i ) = 2 y i + 2a ˆ 0 + 2a ˆ 1 2 y i = a ˆ 0 + a ˆ 1 6. We now have two normal equations in a ˆ 0 and ˆ which need to be solved simultaneously. na ˆ 0 + ˆ ˆ a 0 = y i 2 + a ˆ 1 = y i ENGR45 Chapter 5: Linear Regression 5-3
4 To solve the normal equations, set up a matrix of the coefficients of a ˆ 0 and ˆ. Let X = ˆ and Y = ˆ, and a-f are the coefficients. solving for x and y, we get: where ˆ ax + by = c dx + ey = f x = c by and x = f ey a d c by = f ey a d and by rearranging, eventually we get : y = ˆ = dc af db ae, or y n i ( ) 2 2 n is the slope of the least squares fit. y i a 0 7. Doing a similar calculation for the intercept, we find: a ˆ y i a ˆ 1 0 = n 8. To find the linear fit, then, we simply calculate the terms in a ˆ 0 and ˆ from our data set. Then plot the lin e and conf irm that th ere is reasonable fit to th e data. This explanation does not include further statistical analysis such as ho w good the linear fit is or what the confidence intervals are. These should be investigated further in other courses. ENGR45 Chapter 5: Linear Regression 5-4
5 LAB REPORT SHEET 5.1 Determining a Linear Relationship Key Member (Encourage all m embers of the team to participate, ensure that everyone understands the m aterial, and organize the ta sks and divide them between the tea m members): Graphics Analyst (Generate the needed plots and figures.): Other Group Members: Consider the following sets of experimental data. In each case, determine: Whether there appears to be any functional relationship at all among the data If there is a relationship, determine which is the independent variable and which is the dependent variable. Determine whether there is a linear re lationship between th e data, and why you might expect one. If the relationship is not linear, determine whether there is a way to transfor m the data so that it is linear. 1. You have data from a meteorologist on te mperature for each month of the year, measured in one location using th e sam e therm ometer, on three d ifferent d ays each month. Data is in degrees Fahrenheit. Jan 23, 34,13 Feb 12, 45, 32 Mar 14, 32, 33 Apr 38, 50, 51 May 50, 62, 65 June 70, 75, 79 July 80, 82, 95 Aug 88, 83, 75 Sep 84, 75, 90 Oct 78, 65, 50 Nov 50, 45, 32 Dec 32, 30, 28 ENGR45 Chapter 5: Linear Regression Report Sheet 5.1
6 2. You have data on electrical resistance of an aluminum wire, at various temperatures: Temp (C) Resistance(ohms) E E E E E E E E E E E E E E You have data on the im pact energy of a bu llet, m easured at various take-off velocities. Answer the questions above and determine the mass of the bullet. Velocity (m/s) Energy (J) E E E E E E E E E E E+02 ENGR45 Chapter 5: Linear Regression Lab Report Sheet 5.1
7 LAB REPORT SHEET 5.2 Performing a Linear Regression Key Member (Encourage all m embers of the team to participate, ensure that everyon e understands the material, and organize the tasks and divide them between the team members): Calculations Expert (Work with your calculator to help your team solve the problems.) Graphics Analyst (Generate the needed plots and figures.): Other Group Members: Exercise 1 The following data was record ed in an experiment which measured the variation of the specific heat of a chemical with temperature. In this temperature regime, it was expected that the specific heat (C p ) should depend linearly on the absolute tem perature, T. Two measurements were made at each temperature. Temperature ( O C) Specific Heat (J/mol o C) Note that the two m easurements m ade at the sam e t emperature can be considered independent measurements, so you have 12 pieces of data. 2. Plot the data on a scatter diagram and determine whether a linear relationship exists. 3. Fit a s traight lin e to the data by eye; f ind th e slope and inte rcept o f this line; w rite an equation for this line. 4. Estimate the specific heat of this chemical when the temperature is 75 o C. Exercise 2 1. Perform a linear regression analysis on the data of Exercise 1. Using the regression values for the slope and the intercept, add the fitted line to your plot from Exercise 1, using a different color pencil from the previous line fit to the data by eye. 2. Estimate the specific heat of this chemical when the temperature is 75 o C. 3. What is the percentage difference between the estimated values for 75 o C found in Exercise 1 and Exercise 2? Which value is more accurate? Why? ENGR45 Chapter 5: Linear Regression Report Sheet 5.2
8 ENGR45 Chapter 5: Linear Regression Lab Report Sheet 5.2
GAMINGRE 8/1/ of 7
FYE 09/30/92 JULY 92 0.00 254,550.00 0.00 0 0 0 0 0 0 0 0 0 254,550.00 0.00 0.00 0.00 0.00 254,550.00 AUG 10,616,710.31 5,299.95 845,656.83 84,565.68 61,084.86 23,480.82 339,734.73 135,893.89 67,946.95
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