1. 4.5 – Analyzing Lines of Best Fit Today’s learning goal is that students will be able to Use residuals to determine how well lines of fit model data. Distinguish between correlation and causation.

2. Core Concept Residual – the difference of the y- value of a data point and the corresponding y-value found using the line of fit. If the model (line of fit) is good, then the absolute values of the residuals are relatively small.

3. Example 1 – Using Residuals In Example 3 in Section 4.4, the equation y = -2x +20 models the data in the table shown. Is the model a good fit? Week, xSales (millions), y 119 215 313 411 510 68 77 85 Y-value from model y = -2x +20 Residual -2(1) +20 = 1819-18 = 1 -2(2) + 20 = 1615-16 = -1 -2(3) + 20 = 1413-14 = -1 -2(4) + 20 = 1211-12 = -1 -2(5) + 20 = 1010-10 = 0 -2(6) + 20 = 88 – 8 = 0 -2(7) + 20 = 67 – 6 = 1 -2(8) + 20 = 45 – 4 = 1 Notice that all of the residual points are close to zero. Let’s graph them.

4. Example 1 – Using Residuals In Example 3 in Section 4.4, the equation y = -2x +20 models the data in the table shown. Is the model a good fit? Week, x 1 2 3 4 5 6 7 8 Residual 19-18 = 1 15-16 = -1 13-14 = -1 11-12 = -1 10-10 = 0 8 – 8 = 0 7 – 6 = 1 5 – 4 = 1 The points are evenly dispersed about the x-axis. The equation is a good fit.

5. Example 2 – Using Residuals The table shows the ages x and salaries y (in thousands of dollars) of eight employees at a company. The equation y = 0.2x + 38 models the data. (is the line of fit) Is the model a good fit. Ages, xSalaries (in thousands) y 3542 3744 4147 4350 4552 4751 5349 5545 Y-value from model y = 0.2x + 38 Residual 0.2(35) + 38 = 4542-45 = -3 0.2(37) + 38 = 45.444-45.4 = -1.4 0.2(41) + 38 = 46.247-46.2 = 0.8 0.2(43) + 38 = 46.650-46.6 = 3.4 0.2(45) + 38 = 47.052-47 = 5 0.2(47) + 38 = 47.451-47.4 = 3.6 0.2(53) + 38 = 48.649 – 48.6 = 0.4 0.2(55) + 38 = 4945 – 49 = -4.0

6. Example 2 – Using Residuals Let’s graph the residuals. Ages, x Residual 35 -3 37 -1.4 41 0.8 43 3.4 45 5 47 3.6 53 0.4 55 -4.0

7. You Try!  1. The table shows the attendances y (in thousands) at an amusement part from 2005 to 2014, where x = 0 represents the year 2005. The equation y = -9.8x + 850 models the data. Is it a good fit? Step 1: Set up the table Step 2: Calculate the residuals Step 3: Plot the residuals Is the equation a good fit? NO

8. Finding Lines of Best Fit  Graphing calculators use a method called linear regression to find a precise line of fit called a line of best fit.  A graphing calculator often gives a value r, called the correlation coefficient.  Values of r range from -1 to 1.  When r is close to -1, there is a strong negative correlation.  When r is close to 1, there is a strong positive correlation.  When r = 0, there is no correlation.  When r is close to 0, the correlation is weak.

9. Correlation and Causation  Causation – when the change in one variable causes the change in another variable.  When there is causation, there is a strong correlation.  A strong correlation does not necessarily imply causation.

10. Example 5 – Identifying Correlation and Causation  Tell whether a correlation is likely in the situation. If so, tell whether there is a casual relationship (causation). Explain.  a) time spent exercising and the number of calories burned  b) the number of banks and the population of a city There is a positive correlation and a casual relationship because the more time you spend exercising, the more calories you burn. There may be a positive correlation but no casual relationship. Building more banks will not cause the population to increase.

11. You Try!  Is there a correlation between time spent playing video games and grade point average? If so, is there a casual relationship (causation)? Explain your reasoning There may be a negative correlation. A casual relationship is possible because the more time you spend playing video games, the less time you spend studying.