advertisement

ENM 500 Regression Exercises 0. Use regression techniques to fit a line to the points (0, 0) and (3, 4) using the format Y 0 4 x 0 3 x2 xy Yhat Y-Yhat (Y-Yhat)2 Write the normal equations. 1. Given the following X = 15 Y = 17 X2 = 55 Y2 = 77 XY = 54 n = 5 Find Sxx Sxy Syy A B SSerror s2 t( = 0) R2 SSexplained Write y-hat and X-hat equations. B = 3/19.2 A = 3 – 3/19.2 * 17/5 = 2.46875 X-hat = 2.46875 + 0.15625y Y-hat = 2.5 + 0.3x 2. Sketch a scatterplot and fit regression line to it using the data : x 1 3 5; y 4 9 8. 3. Use multiple linear regression techniques to fit a plane between the points given below. Y 5 6 0 x1 0 4 1 x2 0 3 -1 x12 x22 x1x2 x1Y Solve your linear equations and write the equation of the plane. 4. Fit a polynomial of degree 2 between three points below. x x2 x2 x4 x3 xY x2 Y 2 2 Y x1 x2 x1 x2 x1x2 x1Y x2Y 5 0 6 4 0 1 Solve your equations and write the polynomial. x2Y 5. Demo of Example 8.25 (ex8.25) (mapcar #' R-sq (list x1 x2 x3 x4 x5) (list-of 5 y)) (mapcar #' MSE-r (list x1 x2 x3 x4 x5) (list-of 5 y)) 1. Solution for #1.X = 15 Y = 17 X2 = 55 Y2 = 77 XY = 54 Find Sxx Sxy Syy A B SSerror s2 t( = 0) R2 SSexplained Y-Hat = 2.500 + 0.300X Sxx = 10 Sxy = 3 SSerror = 18.3 sý = 6.1 Explained Variation = 0.9 Syy = 19.2 R-sq = 0.0469 Sum of Y-squared = 77 For ß = 0, (t = 0.384111 P-VALUE = 0.726505) For = 0, (t = 0.965114 P-VALUE = 0.40568) Sa = 2.590 Sb = 0.781 (-2.168, 2.768) or 0.300 +- 2.468 95% confidence interval for Beta NIL (-5.686, 10.686) or 2.500 +- 8.186 95% confidence interval for Alpha NIL Residuals are: -0.800 0.900 -1.400 3.300 -2. The yhats are (2.8 3.1 3.4 3.7 4.0) The b coefficients in Y's are -0.2 -0.1 0 0.1 0.2 The a coefficients in Y's are 0.8 0.5 0.2 -0.1 -0.4 (-0.091 6.891 ) or 3.400 +- 3.491 95% confidence interval for Y-Hat at x = x-bar NIL (-5.15026 11.95026) or 3.400 +- 8.550 95% confidence interval for Y-Predict at x = x-bar NIL F-ratio = Explain/Error/3 = 0.1475. Source Regression Residual Error Total Analysis of Variance SS df MS 0.9 1 0.9 18.3 3 6.1 19.2 4 F P-value 0.148 0.7265