15 Linear Regression.ppt

15: Linear Regression Expected change in Y per unit X Introduction (p. 15.1) X = independent (explanatory) variable Y = dependent (response) variable Use instead of correlation when distribution of X is fixed by researcher (i.e., set number at each level of X) studying functional dependency between X and Y Illustrative data (bicycle.sav) (p. 15.1) Same as prior chapter X = percent receiving reduce or free meal (RFM) Y = percent using helmets (HELM) n = 12 (outlier removed to study linear relation) Regression Model (Equation) (p. 15.2) How formulas determine best line(p. 15.2) Distance of points from line = residuals (dotted) Minimizes sum of square residuals Least squares regression line Formulas for Least Squares Coefficients with Illustrative Data (p. 15.2 – 15.3) Alternative formula for slope Interpretation of Slope (b)(p. 15.3) b = expected change in Y per unit X Keep track of units! Y = helmet users per 100 X = % receiving free lunch e.g., b of –0.54 predicts decrease of 0.54 units of Y for each unit X Predicting Average Y ? = a + bx Predicted Y = intercept + (slope)(x) HELM = 47.49 + (–0.54)(RFM) What is predicted HELM when RFM = 50? ? = 47.49 + (–0.54)(50) = 20.5 Average HELM predicted to be 20.5 in neighborhood where 50% of children receive reduced or free meal What is average Y when x = 20? ? = 47.49 +(–0.54)(20) = 36.7 Confidence Interval for Slope Parameter(p. 15.4) 95% confidence Interval for ? = where b = point estimate for slope tn-2,.975 = 97.5th percentile (from t table or StaTable) seb = standard error of slope estimate (formula 5) Illustrative Example (bicycle.sav) 95% confidence interval for ? = –0.54 ± (t10,.975)(0.1058) = –0.54 ± (2.23)(0.1058) = –0.54 ± 0.24 = (–0.78, –0.30) Interpret 95% confidence interval Model: Point estimate for slope (b) = –0.54 Standard error of slope (seb) = 0.24 95% confidence interval for ? = (–0.78, –0.30) Interpretation: slope estimate = –0.54 ± 0.24 We are 95% confident the slope para