Simple Linear Regression

Simple Linear Regression ________________________________________ 1) Discuss conceptual differences between ANOVA and regression. 2) Identify the components of the simple regression equation (1, 0) and explain their interpretation. 3) Demonstrate the Least-Squares method for calculating 1 and 0. 4) Develop a measure for error in the regression model and demonstrate a method for comparing the variance due to error with the variance due to our model. 5) Define and explain the correlation coefficient and the coefficient of determination. 6) Discuss the relationship between correlation and causation. CI vs. ANOVA vs. Regression __________________________________________ Key word for CI: Estimation Key word for ANOVA: t-test Comparison Key word(s) for regression: Prediction Estimation Trivia Wars ______________________________________ Let’s say Amherst declares war on Northampton because Northampton tries to lure Judie's into moving out of Amherst. No one actually wants to kill anyone, so we decide to settle our differences with a rousing game of Jeopardy! You are elected the Captain of Amherst’s team (as if you would be selected instead of me). How are you going to choose the team? Multiple criteria: 1) Knowledge 2) Performance under pressure EX: Cindy Brady 3) Speed Historical roots in WW II Who would be a good ball turret gunner? Regression ______________________________________ What is the relationship between… Grades or Money Relationship Status or Health or …and Life Satisfaction? ______________________________________ How well can I predict a person’s Life Satisfaction if I know their … Grades or Money or Relationship or Health Status ______________________________________ How are we going to do this? • Collect a sample of data, create a scatter plot and devise an equation that will help us do the predicting. General form of Probabilistic (Regression) Models ________________________________________ y = Deterministic Component + Random Error or y = regression line + error or y = variance in y predicted by x + error _______________________________________ E(y) - expected (mean) value of y for each value of x • Regression line connects E(y) for each value of x Simple Regression First-Order Single-Predictor ___________________________________________ y = 0 + 1x +  y = dependent or response variable x = independent or predictor variable criterion E(y) = 0 + 1x (deterministic)  = Random error ______________________________________ 0 y-intercept y = mx + b 1 slope of the regression line (for simple models only) Interpretation of y-intercept and slope ________________________________________ Intercept • Intercept only makes sense if x value can go down to zero. • Regression equation only applies (for sure) to range of data used in analysis. ________________________________________ Slope • Change in y for a unit change in x. o + implies direct relationship o – implies inverse relationship ________________________________________ Most important point: Give me a value for x and the regression equation and I can make a pretty good prediction about the corresponding value of y. An Idealized Regression Line _____________________________________________ Notice that the regression line represents the mean of the distribution of y values at each x in the distribution. ________________________________________ Just like the mean is the best predictor of the value of a single observation drawn at random… ...E(y) is the best predictor of the value of a single observation for a given value of y. Steps to completing a regression analysis (both simple and multiple) ________________________________________ Hypothesize the deterministic component of the model. Step 1 Direct vs. Inverse Relationships Step 2 Step 3 Step 4 Step 5 Use sample data to estimate unknown parameters (0, 1) Specify the probability distribution of the random error term and estimate its SD. Evaluate the usefulness of the model statistically Use the model for prediction, estimation, etc. Fitting a model to our data (Step 2) ________________________________________ Least-Squares method 1) Sum of the vertical distance between each point and the line = 0 2) Square of the vertical distance is as small as possible. When in doubt, think Bribery!! ____________________________________________ You want to determine the relationship between monetary gifts and "BONUS POINTS FOR SPECIAL CONTRIBUTIONS TO CLASS" added to your final average so that you can decide how large a check to write at the end of the semester (though I do prefer cash for tax purposes). Let's say x represents the amount of money contributed by past students, and y represents the number of "Bonus Points" awarded to them. Bonus Points Bribery 10 9 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 Donation 7 8 9 10 Fishing for a regression line ________________________________________ 12 Series1 y=x+1 y=5 Bonus Points 10 8 6 4 2 0 0 1 2 3 4 5 6 7 8 9 10 Donation X Y Gift 4 8 2 6 BP 1 9 5 5 Distance y=5 -4 4 0 0 0 Y=x+1 -4 0 2 -2 -4 Which regression line is better? Is that the ‘best’ regression line? SquaredDistance y=5 y=x+1 16 16 16 0 0 4 0 4 32 24 Formulae for Least Squares Method ________________________________________ 1 = 0 = SP / SSx My – (1* Mx) ______________________________________________ SSx = 2   (  x ) x −   n   SP = (xy ) −  (x )(y )  n  2 Finding the best-fit regression line ________________________________________ x 4 8 2 6 x = 20 SSx = = = = Y 1 9 5 5 y = 20 x2 Xy 16 4 64 72 4 10 36 30 (x2) = 120 (xy) = 116 (x2) – [(x)2 / n] 120 – [(20)2 / 4] 120 – (400 / 4) 120 – 100 = 20 = (xy) – [(x)(y)] / n = 116 – [(20)(20) / 4] = 116 – (400 /4) = 116 – 100 = 16 ________________________________________ SP 1 0 = = SP / SSx 16 / 20 = 0.8 = My – (1* Mx) = 5 – (.8)(5) = 1.0 ________________________________________ The Least-Squares Regression Line ________________________________________ 12 Series1 y=x+1 10 Bonus Points y=5 Least Squares Reg. Line 8 6 4 2 0 0 1 2 3 4 5 6 7 8 9 10 Donation x y E(y) Distance 4 8 2 6 1 9 5 5 4.2 7.4 2.6 5.8 -3.2 1.6 2.4 -0.8 0 SquaredDistance 10.24 2.56 5.76 0.64 19.20 Testing Example __________________________________________ Unbeknownst to you, Biff is the heir to his family’s Widget fortune. For his summer job, Biff was asked to evaluate a group of employees’ widget making ability using a standardized widget-making test. Biff’s boss (Uncle Buck) asks Biff to determine the regression equation that one would use to predict performance on the test from years of service with the company. The data appear below. x (years) 3 4 4 2 5 3 4 5 3 2 x = y (score) 55 78 72 58 89 63 73 84 75 48 y = x2 y2 xy (x2) = (y2) = (xy) = Calculations ______________________________________________ SSx = (x2) – [(x)2 / n] SP = (xy) – [(x)(y)] / n ______________________________________________ 1 = SP / SSx 0 = My – (1* Mx) Widget Test Scatter Plot ________________________________________ 100 Test Score 90 80 70 60 50 40 2 3 4 Experience 5 Assumptions regarding Error () ________________________________________ : essentially vertical distance from regression line _________________________________________ 1) The mean of the probability distribution = 0. Similar to unbiasedness 2) The variance of the probability distribution of  is constant. 3) Distribution of  is normal. 4) Values of  are independent of one another. Factors that contribute to Error ________________________________________ Two types of Error 1) Measurement Error - improper use of measuring instrument EX: incorrect reading of beaker 2) Chance factors not accounted for by our model EX: unusually non/reactive chemical Estimation of Variability due to Error (Step 3) ________________________________________ s2 is analogous to MSE s2 = SSE / dferror SSE = = SSE / n – 2 SSy - 1(SP) SSy = y2 – [(y)2 / n] ________________________________________  s2 s = = SSE / (n-2) = Estimated Standard Error of the Regression Model or = Root MSE MSE Calculate the error ______________________________________ SSy = SSE s2 s y2 – [(y)2 / n] = = = 49,861 – [(695)2 / 10] 49,861 – (483,025 / 10) 49,861 – 48,302.5 = SSy - 1(SP) = = 1558.5 – 11.0(115.5) 1558.5 – 1270.5 = = SSE / (n-2) 288 / (10-2) = 36 = 1558.5 = 288 = 36 (a/k/a MSE) = 6 (a/k/a Root MSE) Important points about error or  ________________________________________ 1. The smaller , the better we can predict y based on x. 2. The smaller , the more tightly packed the individual data points will be around the regression line. 3. A smaller  implies that x is a better predictor of y. Why? Because points fall closer to the regression line. Also, can use this information to develop a sense of how far points should fall off the line. • We can calculate a CI around the regression line. 95% of our points should fall within about 2 RMSEs of the regression line. If not, HMMMM… Evaluate the usefulness of the model (Step 4) ________________________________________ Step 1: Specify the null and alternative hypotheses. • Ho: 1 = 0 • Ha: 1  0 Step 2: Designate the rejection region by selecting . Step 3: Obtain the critical value for your test statistic • t • df = n-2 Collect your data Step 5: Use your sample data to calculate: • 1 SP / SSx • s1 = SE = s / SSx Step 6: Use your parameter estimates to calculate the observed value of your test statistic • t = 1 – 0 / s1 Step 7: Compare tobs with tcrit: • If the test statistic falls in the RR, reject the null. • Otherwise, we fail to reject the null. Calculating whether 1 (slope)  0 ________________________________________ Ho: Ha: 1 = 0 1  0 tcrit 2.306 (df = 8;  = .05) RR Observed t |tobs| > 2.306 = = = = 1 – 0 / (s /  SSx) 11 – 0 / (6 / 10.5) 11 / 1.85 5.94 We would reject the null hypothesis because tobs exceeds the tcrit. In other words, tobs falls in the rejection region. Implication: Correlation Coefficient ________________________________________ Pearson’s product moment coefficient of correlation – a measure of the strength of the linear relationship between two variables. Terminology / notation: • r • Pearson’s r • correlation coefficient ________________________________________ r = SP ( SSx)(SSy) Interpretation: +1 perfect positive relationship (strong positive relationship) 0 no relationship (strong negative relationship) -1 perfect negative relationship r for the Widget Example ____________________________________________ r = SP ( SSx)(SSy) Experience in Years = 115.5 = 115.5 = 115.5 / 127.92 (10.5)(1558.5) 16,364.25 = .90 = .90 Experience in Months = 1386 = 1386 = 1386 / 1535.074 (1512)(1558.5) 2,356,452 Stress and Health ____________________________________________ There is a strong negative correlation between stress and health. Generally, the more stressed a person is, the worse their health is. But, does that mean that stress causes poor health? No... Stress only influences the likelihood of engaging in healthy habits. People under a great deal of stress tend to engage in more unhealthy behaviors. Yes... Stress hormones cause your body continuously to be engaged in "Flight or Fight" mode. This causes your heart, lungs, etc. to work harder, stressing and scarring your blood vessels, which increases the likelihood of heart disease. Coefficient of Determination ________________________________________ r2 represents the proportion of the total sample variability around the mean of y that is explained by the linear relationship between y and x. For simple, linear regression, r2 = r2. ________________________________________ More general formula is as follows: r2 = (SSy – SSE) / SSy = 1 – (SSE / SSy) ________________________________________ SPSS will give us everything we need! Questions about Regression output __________________________________________ 1) What is r? 2) Is this correlation significant? 3) How much of the variance in # of colds per winter can be explained by weekend bedtime? 4) What is the y-intercept? 5) Is it significantly different from zero? 6) What is E(y) if x = 10:00 PM (10)? 7) What is E(y) if x = 2:00 AM (14)? 8) Are your answers to questions 6 and 7 meaningful? SPSS output ______________________________________________ Model 1 Model Summary R R2 Adj R2 SE .204 .041 .034 1.20 ANOVA Sum of Mean Model Squares df Square F Sig. 1 Regression 7.68 1 7.68 5.32 .023 Residual 177.58 123 1.44 Total 185.27 124 Coefficients Unstand Stand Model Coeff Coeff t Sig. B SE Beta 1 (Constant) 5.711 1.69 3.38 .001 bed_we -.266 .12 -.20 -2.31 .023 I just don’t get it ____________________________________________ I know I’m old, but I just don’t get the tattoo thing. I gotta figure that people regret their decision as time passes. The data below represent 100 subjects who had tattoos etched into their skin between 1 and 5 years ago. They rated their satisfaction with their lifetime scar on a scale of 1-10 (10 = extremely satisfied). Is there a relationship between tattoo age and tattoo satisfaction? (x) (y) (x2) (y2) (x)(y) 300 600 1100 3954 1660 Regression Equation SP = SSx = 1 = 0 = (xi)(yi) – [(xi)(yi)] / n xi2 – [(xi)2 / n] SP / SSx My – (1* Mx) Hypothesis Test SSy SSE s2 (MSE) t yi2 – [(yi)2 / n] SSy - 1(SP) SSE / (n-2) 1 - 0 / (s / SSxx) = = = = Correlation Coefficient r = SP ( SS x )(SS y ) Calculating the regression parameters ______________________________________________ SP = (xi)(yi) – [(xi)(yi)] / n SSx = xi2 – [(xi)2 / n] 1 = SP / SSx 0 = My – (1* Mx) Let's do a t-test ______________________________________________ SSy = yi2 – [(yi)2 / n] SSE = SSy - 1* (SP) s2 = MSE s = t = 1 – 0 / (s / SSx) We reject the null and conclude that there is a significant NEGATIVE relationship between tattoo age and tattoo satisfaction. Let's calculate the correlation coefficient ____________________________________________ SSy = yi2 – [(yi)2 / n] r = SP r2 = -.532 ( SS x )(SS y ) = .28 Skipping Class __________________________________________ In a perfect world, the correlation between the number of classes skipped and the percentage of classes skipped should be 1.00. Let's see how well the percentage of classes skipped (x) predicts the number of hours of classes skipped (y). Please calculate the regression line, the correlation coefficient, and the coefficient of determination. (x) (y) (x2) (y2) (x)(y) Regression Equation SP = SSx = 1 = 0 = (xi)(yi) – [(xi)(yi)] / n xi2 – [(xi)2 / n] SP / SSx My – (1* Mx) Hypothesis Test SSy SSE s2 (MSE) t yi2 – [(yi)2 / n] SSy - 1(SP) SSE / (n-2) 1 - 0 / (s / SSxx) = = = = Correlation Coefficient r = SP ( SS x )(SS y )