532 CHAPTER 10 Correlation and Regression Use the sample data in Table 10-1 (shown in Example 1). Use Formulas 10-3 and 10-4 to find the equation of the regression line in which the explanatory variable (or x variable) is the jackpot amount and the response variable (or y variable) is the corresponding number of tickets sold. CP YOUR TURN. Do Exercise 13 “Powerball Jackpots and Tickets Sold.” EXAMPLE 2 Using Manual Calculations to Find the Regression Equation SOLUTION REQUIREMENT CHECK The requirements are verified in Example 1. We begin by finding the slope b1 using Formula 10-3 as follows (with extra digits included for greater accuracy). Remember, r is the linear correlation coefficient, sy is the standard deviation of the sample y values, and sx is the standard deviation of the sample x values. b1 = r sy sx = 0.947349# 12.96255 70.50611 = 0.174170 After finding the slope b1, we can now use Formula 10-4 to find the y-intercept as follows: b0 = y - b1x = 26.55556 - 10.17417021214.888892 = -10.87164 After rounding, the slope is b1 = 0.174 and the y-intercept is b0 = -10.9. We can now express the regression equation as yn = -10.9 + 0.174x, where yn is the predicted number of tickets sold and x is the amount of the jackpot. SOLUTION Shown below is the Minitab display of the scatterplot with the graph of the regression line included. We can see that the regression line fits the points reasonably well. Graph the regression equation yn = -10.9 + 0.174x (found in Examples 1 and 2) on the scatterplot of the jackpot>tickets data from Table 10-1 and examine the graph to subjectively determine how well the regression line fits the data. CP EXAMPLE 3 Graphing the Regression Line
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