Chapter 15 Linear Regression The correlation coefficient tells us the degree to which two variables are related; we can use this coefficient to predict the value of one variable when there is a change in another variable for hypothetical cases. Example: p. 247 (student GPA). Logic of Prediction To predict one variable from another, we need to know their correlation (rxy). Look at Fig. 15.1. To do that, we have to create a regression equation in order to compute a regression line, which will give us our best guess as to what score on the Y variable (college GPA) would be predicted by a score on the X variable (high school GPA). The regression line minimizes the distance between the line and each of the points on the Y axis. Question, on this scatterplot, what is the value of why given X of 3? (Fig 15.3)

The distance between each data point and the regression line is the error in prediction. The larger the error, the lower the correlation (see Figure 15.4). So, what would a correlation of 1 (perfect correlation) look like? Computing the regression coefficient (equation). Remember, Y is our dependent variable and X is our independent variable (or variable of interest or main iv). Y׳=bX + a Y׳ is the predicted value of Y based on value of X b is slope, or direction (sign), of line a is the point at which the line crosses the y-axis

Y׳=bX + a Y׳ is the predicted value of Y based on value of X b is slope, or direction (sign), of line a is the point at which the line crosses the y-axis Also need these for a and b: ΣX ΣY ΣX2 ΣY2 ΣXY

Do example on page 251. What is the regression equation (line)? Using this equation, we can predict what Y will be given a value for X. For example, with our equation Y׳=.704X + .719, we can predict what college GPA would be for a student with a high school GPA of 2.8. How? Plug 2.8 in for X. Answer = 2.69 How good is the fit (prediction)? We calculate the error of the estimate by comparing the predicted value (Y׳) with the actual value (Y) for an observed X. For example, given our equation, we would predict a college GPA of 2.69 for an high school (X) GPA of 2.8, but we know (given data set) that the actual Y value for the person with this high school GPA is 3.5. The difference (3.5-2.8) is .81 (error of the estimate). If you calculate the average error estimate for each X value (all differences between each Y׳ and Y), you get the standard error of the estimate. IV. Significance testing. Just use the computer. We need the t-score, but it is hard to compute. SPSS does it automatically along with significance level.

Regression and SPSS and regression output table (p. 253) Interpretation of coefficient “For every one unit (hour of training) increase in X, there is a _________ unit change in Y (# of injuries) of -.125 (i.e., reduction in the # of injuries).” Interpretation of significance level: Is this a significant finding? Yes, significance level is .011. That is, there is only a 1% chance that the Null is true or that we would commit a Type I error. Interpretation of the Adjusted-R2 value: tells us the percentage of the variance in Y (# of injuries) that is explained by the variance in X (hours training). In this case, 18% of the variation in injuries is explained by the number of hours spent training. Creating a graph with the regression line pp254-255