D4: 3.2b Hw: pg 191 – 43, 46, 48, 53, 63, 71 - 78

1. The Role of r2 in RegressionTarget Goal: I can use r2 to explain the variation of y that is explained by the LSRL. D4: 3.2b Hw: pg 191 – 43, 46, 48, 53, 63, 71 - 78

2. r : correlation coefficientr2 : the coefficient of determination • If the line ŷ is a poor model, the value of r2turns out to be: too small, closer to 0. • If the line ŷ fit the data fairly well, the value of r2turns out to be: larger, closer to 1.

3. What is the meaning of r2 in regression? • Squares of the deviations about ŷ

4. The Role of r2 in Regression The standard deviation of the residuals gives us a numerical estimate of the average size of our prediction errors. There is another numerical quantity that tells us how well the least-squares regression line predicts values of the response y. Least-Squares Regression Definition: The coefficient of determination r2 is the fraction of the variation in the values of y that is accounted for by the least-squares regression line of y on x. We can calculate r2 using the following formula: where and

5. Formula for r2made up of these parts • SST: total sum of squares about the mean y bar. SST = ∑(y – y bar)2 • SSE: sum of the squares for error. SSE = ∑(y – ŷ)2 • r2: coefficient of determination. r2 = SST – SSE SST

6. Ex: Large r2 If , then the deviations and thus the ; in fact, if all of the points fell exactly on the regression line, SSE would be 0. r2 = SST – SSE SST x is a good predictor of y SSE would be small

7. For the data in this example, x: 0 5 10 y: 0 7 8 r2 = SST – SSE = 38 – 6 = SST 38 Conclusion: We say that ____ of the ___________ isexplained by the __________________________. 0.842 84% variation in y least-squares regression of y on x.

8. r2 in Regression The coefficient of determination r2, is the fraction of the variation in the values that are explained by least-squared regression of y on x.

9. The Role of r2 in Regression r 2 tells us how much better the LSRL does at predicting values of y than simply guessing the mean y for each value in the dataset. Consider the example on page 179. If we needed to predict a backpack weight for a new hiker, but didn’t know each hikers weight, we could use the average backpack weight as our prediction. Least-Squares Regression SSE/SST = 30.97/83.87 SSE/SST = 0.368 Therefore, 36.8% of the variation in pack weight is unaccounted for by the least-squares regression line. 1 – SSE/SST = 1 – 30.97/83.87 r2 = 0.632 63.2 % of the variation in backpack weight is accounted for by the linear model relating pack weight to body weight. If we use the LSRL to make our predictions, the sum of the squared residuals is 30.90. SSE = 30.90 If we use the mean backpack weight as our prediction, the sum of the squared residuals is 83.87. SST = 83.87

10. Correlation and Regression Wisdom Correlation and regression are powerful tools for describing the relationship between two variables. When you use these tools, be aware of their limitations Least-Squares Regression Fact 1. The distinction between explanatory and response variables is important in regression.

11. Facts about least-squared regression • Fact 2: There is a close connection between the slope of the least-squared regression line. As the correlation grows less strong, the in response to changes in x. correlationand prediction ŷ moves less

12. Fact 3: Every LSRL passes through • Remember: When reporting a regression, give r2 as a measure ofhow successful the regression was in explaining the response. • When you see a correlation (r), square it to get a better feel for the strength of the association.

13. Exercise: Predicting The Stock Market. Some people think that the behavior of the stock market in January predicts its behavior for the rest of the year. • Take the explanatory variable x to be the percent change in a stock market index in January and the • response variable y to be the change in the index for the entire year. • We expect a positive correlation between x and y because the change during January contributes to the years full change.

14. Calculation from the data for the years 1960 to 1997 gives: x bar = 1.75%, sx = 5.36 y-bar = 9.07%, sy = 15.35% and r = 0.596

15. r = 0.596 a. What percent of the observed variation in yearly changes in the index with is explained by a straight-line relationship the changes during January? The straight-line relationship is explained by r2 = 0.355 or, 35.5% of the variations in yearly changes in the index is explained by the changes during January.

16. What is the equation of the least-squared regression line for predicting full-year change from January change? Find b: , b =1.707 Find a: , a = 6.083%

17. The regression equation is ŷ = a + bx ŷ = 6.083% + 1.707x

18. Predictions • The mean change in January is = 1.75%. Use your regression line to predict the change in the index in a year in which the index rises 1.75% (x bar) in January. Why could you have given this result w/out doing the calculation? Every LSRL passes through (x bar, y bar). Recall y bar = 9.07%, so the predicted change is ŷ = 9.07%.

19. Exercise: Class attendance and grades A study of class attendance and grades among first year students at a state university showed that in general students who attended a higher percent of their classes earn higher grades.

20. Class attendance explained 16% of the variation in grade index among students. What is the numerical value of the correlation between percent of class attended and grade index? r2 r = High attendance goes with high grades so the correlation must be positive. 0.40