Line of Best Fit

Line of Best Fit y Our objective is to fit a line in the scatterplot that fits the data the best y Line of best fit looks like: Least Square Regression Line ŷ = b0 + b1 x That’s a hat on the y, meaning that it is a prediction not the actual y values. VERY IMPORTANT!!! y The line of best fit we calculate is called the LEAST SQUARES REGRESSION LINE (LSRL) y Called this because it minimizes the sum of the squared residuals 2 ∑ ( y − yˆ ) Slope in the z’s y If we look at the scatterplot of the z-scores we find that the line of best fit must go through (0,0) y The slope of the line that minimizes the sum of squares in the z-scores will always be r. y This tells you that on average for each increase of 1 standard deviation in x there is a change of r standard deviations in y. Example: Square Foot vs. Selling Price for Houses in Boulder, CO (MM Table 2.3) Scatter Plot House Sales 350 300 r = 0.677 250 200 150 100 1500 2000 2500 3000 Square_feet 3500 4000 Here is the scatterplot of the z-scores with the line that minimizes the sum of squares. Scatter Plot House Sales Slope in the z‐scores Ex. zˆy = 0.677zx If a house is 1 standard deviation above the mean for square feet in space we would expect the price to be 0.677 standard deviations above the mean price 3 2 zˆy = 0.677zx 1 0 -1 -2 -1 0 1 2 If a house is 1.4 standard deviations below the mean for square feet in space we would expect the price to be 0.677 x -1.4 = -0.9478 0.9478 standard deviations below the mean price z_sf z_sp = 0.67z_sf Switching Variables y If the explanatory and response variable are switched the correlation remains the same y Therefore the slope remains the same if they are switched as well. zˆy = 0.677zx zˆx = 0.677zy Ex. If a house is 1.2 standard deviations below the mean for price we would expect the space to be 0.812 st. dev. below the mean square feet. Slope in the actual scatterplot y Since the slope of the line in the z-scores compares the standard deviations we include these back to get the slope of the line in the scatterplot of the data. y Thus the slope of the line in the regular scatterplot becomes s b1 = y sx r House sales s price = 45135.6 ssquare _ feet = 640.15 b1 = sy r sx r = 0.677 45135.6 b1 = 0.677 640.15 b1 = 47.734 House sales b1 = 47.734 $ per Sq. Foot Interpretation: On average for every increase of 1 sq. ft. in the size there tends to be an increase of $47.73 in the selling price. Interpretation of Slope On average for every increase of 1 (x unit) in the (x variable) there tends to be an increase/decrease of (Slope) (y units) in the (y variable). Finding intercept, b0 y Now that we have the slope, we only need a point that the line runs through to get the intercept. y We have one: ( x , y ) y So the equation for intercept becomes: b0 = y − b1 x y Interpretation of the intercept is generally meaningless. So be careful! House Sales x = 2627.42 y = 177330 b1 = 47.734 b0 = y − b1 x b0 = 177330 − 47.734(2627.42) b0 = 51912.73 Scatter Plot House Sales 350 300 250 200 LSRL : yˆ = 51912.73 + 47.734 x 150 n price = 51912.73 + 47.734( square feet ) 100 1500 2000 2500 3000 Square_feet 3500 4000 Selling_Price = 51912.73 + 47.734 ( Square_feet) Making Predictions We use the model to make predictions on what the price would be for a certain size of house. Ex. What price would we expect from a house that was 2000 sq. ft? ˆ = 51912.73 + 47.734(2000) price ˆ = $147380.73 price Residuals y Residuals are difference between the actual data and predictions in the data. Or observed minus expected. e = y − yˆ y Positive residuals represent underestimations y Negative residuals represent overestimations y The sum of the residuals should be zero. Ex. From the data the cost of one of the houses that was 2000 sq. ft. was $175,000. n price = $147380.73 e = 175000 − 147380.73 e = $27,619.27 Ch07_Drug_abuse Other_Drugs_ A survey was conducted in the United States and 10 countries of Western Europe to determine the percentage of teenagers who had used marijuana and other drugs. Ch07_Drug_abuse 35 30 25 20 15 10 5 0 0 10 Scatter Plot 1. Describe the scatterplot Positive, Linear, Strong (r = 0.9341) Possible outlier at (55,32) but it’s with in the linear pattern 2. Does the linear model seem appropriate? Yes. There is no obvious curve. 20 30 40 Marijuana_ 50 Marijuana_ 0.9341 S1 = correlation ( ) 3. If you knew a country had a percent of teenagers that used marijuana that was 1.5 standard deviations above the mean, what would you predict about the percent of teenagers that used other drugs? I would expect it to be 1.40 standard deviations above the mean percent of other drug usage. 4. If you knew a country’s percent of teenagers that used other drugs was 1.8 standard deviations below the mean, what would you predict about the percent of teenagers that used marijuana for that country? 60 I would expect it to be 1.68 standard deviations below the mean percent of marijuana usage. Ch07_Drug_abuse Other_Drugs_ Marijuana_ 0.9341 S1 = correlation ( ) 7. If a country had a percent of teenagers that used marijuana at 19%, what would you predict that the percent of teenagers that used other drugs to be for that country? 5. Calculate the equation for the LSRL. 10.2399 b1 = 0.9341 15.5528 b1 = 0.615 b0 = 11.6364 − 0.615(23.9091) yˆ = −3.068 + 0.615(19) yˆ = 8.617% b0 = −3.068 yˆ = −3.068 + 0.615 x x - % of teenagers that use marijuana y - % of teenagers that use other drugs 6. Interpret the slope in the context of the problem. On average for every increase of 1 percent in marijuana usage there tends to be an increase of 0.615 percent in other drug usage. 8. If the actual percent of teenagers that used other drugs for the country in #7 is 8%, what is the residual? Did the model overestimate or underestimate the actual value? yˆ = 8.617%; y = 8% e = 8 − 8.617 e = −0.617% Overestimation 9. If a country had a percent of teenagers that used marijuana of 68%, what would your model predict for the percent of teenagers that used other drugs for that country? How confident are you on that prediction? yˆ = −3.068 + 0.615(68) yˆ = 38.752% I would not be confident in this prediction because the value 68% is far beyond the range of the data. Any prediction made would be an extrapolation.