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### lect4-2012

Course: STAT 102, Spring 2012
School: UPenn
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4 Stat102 Lecture - Spring 2012 Chapter 3.1 3.2: Introduction to regression analysis Linear regression as a descriptive technique The least-squares equations Chapter 3.3 Sampling distribution of b0, b1. Continued in next lecture 1 Regression Analysis Galtons classic data on heights of parents and their child (952 pairs) Describes the relationship between childs height (y) and the parents (mid)height...

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4 Stat102 Lecture - Spring 2012 Chapter 3.1 3.2: Introduction to regression analysis Linear regression as a descriptive technique The least-squares equations Chapter 3.3 Sampling distribution of b0, b1. Continued in next lecture 1 Regression Analysis Galtons classic data on heights of parents and their child (952 pairs) Describes the relationship between childs height (y) and the parents (mid)height (x). Predict the childs height given parents height. 75 73 71 child ht Parent ht Child ht 73.60 72.22 72.69 67.72 72.85 70.46 71.68 65.13 70.62 61.20 70.23 63.10 70.74 64.96 70.73 66.43 69.47 63.10 68.26 62.00 65.88 61.31 64.90 61.36 64.80 61.95 64.21 64.96 And more 69 67 65 63 61 63 64 65 66 67 68 69 70 71 72 73 74 parent ht 2 Uses of Regression Analysis Description: Describe the relationship between a dependent variable y (childs height) and explanatory variables x (parents height). Prediction: Predict dependent variable y based on explanatory variables x. 3 Model for Simple Regression Consider a population of units on which the variables (y,x) are recorded. Let y|x denote the conditional mean of y given x. The goal of regression analysis is to estimate y|x . Simple linear regression model: y|x 0 1x 4 Simple Linear Regression Model Model (more details later) y 0 1x e y = dependent variable x = independent variable 0 = y-intercept y x 1 = slope of the line e = error (normally distributed) y x 0 1 x 0 0 and 1 are unknown population parameters, therefore are estimated from the data. Rise 1 = Rise/Run Run x 5 Interpreting the Coefficients e.g.,for each extra inch for parents, the average heights of the child increases by 0.6 inch. The intercept is the estimated mean of y for x=0. However, when working from data this interpretation should only be used when the data contains observations with x near 0. Otherwise it is an extrapolation of the model, and so can be unreliable (Section 3.7.2). 73 71 child ht The slope 1 is the change in the mean of y that is associated with a one unit change in x 75 69 67 65 63 61 63 64 65 66 67 68 69 70 71 72 73 74 parent ht child ht = 26.46 + 0.6 parent ht EXAMPLE 6 Least Squares Regression Line What is a good estimate of the line? A good estimated line should predict y well based on x. We use Least squares regression line: Line that minimizes the squared prediction errors in the sample. Good criterion and easy to compute. Has other nice mathematical properties. 7 The Least Squares (Regression) Line ; Here is a scatterplot of data (n = 4) with two possible lines of fit: Y = X and Y = 2.5 (a horizontal line). Which is a better fit? Sum of squared differences = (2 - 1)2 + (4 - 2)2 + (1.5 - 3)2 + (3.2 - 4)2 = 7.89 Sum of squared differences = (2 -2.5)2 + (4 - 2.5)2 + (1.5 - 2.5)2 + (3.2 - 2.5)2 = 3.99 (2,4) w 4 3 2.5 2 w (4,3.2) (1,2) w w (3,1.5) 1 1 2 3 4 The smaller the sum of squared differences the better the fit of the line to the data. 8 The Estimated Coefficients To calculate the estimates of the coefficients of the line that minimizes the sum of the squared differences between the data points and the line, use the formulas: n b1 (x i i 1 x )( yi y ) The regression equation that estimates the equation of the simple linear regression model is: y b 0 b1x n ( xi x ) 2 i 1 b0 y b1 x 9 Example Heights (cont.) For simple linear regression analysis in JMP: Click Analyze, Fit Y by X; then put child ht in Y and parent ht in X and click OK. Then click red triangle next to Bivariate Fit and click Fit Line. Some commands we will use later can now be found in the red triangle next to Linear Fit 10 Example: Heights (cont) Based on our observations, find b1 and b0 The summary statistics for parent hts and child hts: Child hts Parent hts Mean Std Dev Std Err Mean upper 95% Mean lower 95% Mean N 68.20 2.60 0.084 68.37 68.04 952 Mean Std Dev Std Err Mean upper 95% Mean lower 95% Mean N 68.27 1.79 0.0580 68.38 68.15 952 For the regression line From JMP : b1 = 0.61 b0 = Y - b1 X = 68.20 - 0.61 68.27 = 26.55 The LS equation is y = 26.55 + 0.61x 11 JMP Output Bivariate Fit of child ht By parent ht 75 73 child ht 71 69 67 65 63 61 63 64 65 66 67 68 69 70 71 72 73 74 parent ht Linear Fit child ht = 26.456 + 0.612 parent ht 12 JMP Output (cont) Note the values of b0, b1 in the parameter estimates table The other output entries will be explained later Summary of Fit RSquare RSquare Adj Root Mean Square Error Mean of Response Observations Source Model Error C. Total DF 1 950 951 Term Intercept parent ht 0.177 0.176 2.357 68.202 952 Analysis of Variance Sum of Squares Mean Square F Ratio 1136.50 1136.50 204.59 5277.28 5.56 Prob > F 6413.78 <.0001 Parameter Estimates Estimate Std Error t Ratio Prob>|t| 26.456 (b0) 2.920 9.06 <.0001 0.612 (b1) 0.043 14.30 <.0001 13 Ordinary Linear Model Assumptions Properties of errors under ideal model: y|x 0 1x for all x. yi 0 1 xi ei for all xi The distribution of ei | xi is normal. e1,, en are independent. and Var (ei | xi ) e 2 E (ei | xi ) 0 Equivalent definition: For each xi , yi has a normal 0 1xi and variance e 2 . distribution with mean Also, y1,, yn are independent. 14 Sampling Distribution of b0 ,b1 The sampling distribution of b0 , b1 is the probability distribution of the estimates over repeated samples y1 ,.., yn from the ideal linear regression model with fixed values of 0, 1 and e2 and x1 ,.., xn . Standardregression.jmp contains a simulation of pairs x1 , y1 ,.., xn , yn from a simple linear regression model with 0 1, 1 2, e2 1. AND It contains another simulation labeled x1 , y1 ,.., xn , yn from the same model. Notice difference the in the estimated coefficients calculated from the ys and from the y*s. 15 Bivariate Fit of y By x Bivariate Fit of y* By x 4.0 5 3.5 4 3.0 3 y* y 2.5 2.0 2 1.5 1.0 1 0.5 0 0.0 .0 .2 .4 .6 x Linear Fit y = 1.200 + 1.521 x .8 1.0 .0 .2 .4 .6 .8 1.0 x Linear Fit y* = 1.851 + 0.512 x Two outcomes from standardregression.jmp Each data set comes from the model with b 0 = 1, b1 = 2, s e2 = 1 The values of x1 , ..., x20 are the same in both data sets . See next lecture for more plots and simulations 16 Sampling Distribution (Formulas) b0 and b1 have easily described normal distributions Sampling distribution of b0 is normal with E b0 0 (Hence the estimate is unbiased) 1 x2 1 2 where sx Var b0 e2 xi x 2 2 n 1 i n n 1 sx Sampling distribution of b1 is normal with E b1 1 Var b1 (Hence the estimate is unbiased) e2 n 1 sx2 17 Typical Regression Analysis 1. Observe pairs of data (x1,y1),,(xn,yn) that are a sample from population of interest. 2. Plot the data. 3. Assume simple linear regression model assumptions hold. 4. Estimate the true regression line y|x 0 1x by the least squares line y|x b0 b1x 5. Check whether the assumptions of the ideal model are reasonable (Chapter 6, and next lecture) 6. Make inferences concerning coefficients 0 , 1 and make predictions ( y b0 b1x ) 18 Notes Formulas for the least squares equations: 1. The equations for b0 and b1 are easy to derive. Here is a derivation that involves a little bit of calculus: It is desired to minimize the sum of squared errors. Symbolically, this is 2 SSE b0 , b1 yi b0 b1 xi . i The minimum occurs when 0 SSE b0 , b1 and 0 SSE b0 , b1 . b1 b0 Hence we need SSE b0 , b1 2 xi yi b0 b1 xi and b1 0 SSE b0 , b1 2 yi b0 b1 xi . b0 These are two linear equations in the two unknowns b0 and b1 . Some algebraic manipulation shows that the solution can be written in the desired form xi x yi y and b y b x . b1 0 1 2 xi x 0 19 A NICE FACT thats sometimes useful: a. The least squares line passes through the point x , y . To see this note that if x x then the corresponding point on the least squares line is y b0 b1 x . Substituting the definition of b0 yields y y b1 x b1 x y , as claimed. b. The equation for the least squares line can be re-written in the form y y b1 x x . 3. There are other useful ways to write the equations for b0 and b1 . Recall that the sample covariance is defined as 1 Cov xi , yi xi x yi y S xy , say. n 1 i Similarly, the sample correlation coefficient is S xy R , say. 22 Sx S y 2. 2 2 2 [ S x sx is defined on overhead 18, and S y is defined similarly.] Thus, S xy S y S xy Sy b1 2 R. 22 Sx Sx Sx S y Sx 20 History of Galtons Data: 4. Francis Galton gathered data about heights of parents and their children, and published the analysis in 1886 in a paper entitled Regression towards mediocrity [sic] in hereditary stature. In the process he coined the term Regression to describe the straight line that summarizes the type of relational data that may appear in a scatterplot. He did not use our current least-squares technique for finding this line; instead he used a clever analysis whose final step is to fit the line by eye. He estimated the slope of the regression line as 2 3 Further work in the next decades by Galton and by K. Pearson, Gossett ( writing as A. Student) and others connected Galtons analysis to the least squares technique earlier invented by Gauss (1809), and also derived the relevant sampling distributions needed to create a statistical regression analysis. 5. The data we use for our analysis is packaged with the JMP program disk. It is not exactly Galtons original data. We believe it is a version of the data set prepared by S. Stigler (1986) as a minor modification of Galtons data. In order for the data to plot nicely, Stigler jittered the data. He also included some data that Galton did not. The data listed as Parent height in this data set is actually the average of both parents heights, after adjusting the mothers heights as discussed in the next note. 21 6. Galton did not know how to separately treat mens and womens heights in order to produce the kind of results he wanted to look at. SO (after looking at the structure of the data) he multiplied all female heights by 1.08. This puts all the heights on very nearly the same scale, and allowed him to treat mens and womens heights together, without regard to sex. [Instead of doing this Galton could have divided the mens heights by 1.08; or he could have achieved a similar effect by dividing the male heights by 1.04 and multiplying the female ones by 1.04. Why didnt he use one of these other schemes?] 7. Galton did not use modern random-sampling methods to obtain his data. Instead, he obtained his data through the offer of prizes for the best extracts from their own family records obtained from individual family correspondents. He summarized the data in a journal that is now in the Library of the University College of London. Here is what the first half of p. 4 looks like. (According to Galtons notations one should add 60 inches to every entry in the Table.) 22 Half of p4 of Galtons Journal (note the approximate heights for some records, and the entries tall and deformed) This photocopy, as well as much of the above discussion is taken from Hanley, J. A. (2004), Transmuting women into men: Galtons family data on human stature, The Amer. Statistician, 58, p237-243. Another excellent reference is Stigler, S. (1986) The English breakthrough: Galton in The History of Statistics: The Measurement of Uncertainty before 1900, Harvard Univ. Press. 23
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