STA6167 Exercise 2 Solutions: Linear and Quadratic Regression Analysis, Assignments of Statistics

Download STA6167 Exercise 2 Solutions: Linear and Quadratic Regression Analysis and more Assignments Statistics in PDF only on Docsity! STA6167 Exercise 2 (Due July 14, Monday) Please answer all the question as concisely as possible. The answers should be written on standard 8.5”x11” paper and stapled together. 1. a. Sol: See SAS output. The response seems nonlinear. b. Sol: The fitted linear equation: y = −6.96 + 0.279x c. Ans. See SAS output for the plot. No, the residuals are rpm dependent. d. Sol: F = (650.75 − 131.37)/(40− 36) 131.37/36 = 129.795 3.649 = 35.75 ∼ F 4 36 The model does not fit the data well (p < 0.001). 2. a. Sol: The fitted equation is y = 63.36 − 0.71x + 0.0033x2. b. Ans. Yes, the quadratic equation fit significantly better than the linear equation. The quadratic coefficient is significantly different from 0 (p < 0.0001). c. Sol: The lack of fit test F = (149.43 − 131.37)/(39− 36) 131.37/36 = 18.06/3 3.649 = 1.65 ∼ F 3 36 The model has great improvement and passed the no lack of fit test (p > 0.10). d. Sol: The value 43.7 is an outlier at rpm=200. If we take this value away, the fit is much better, but still cannot pass the lack of fit test. Residual plot shows that the data at rpm=120 do no fit into the quadratic equation. They are too large. Thus we may try a square root term. The fit becomes y = −408 − 7.49x + 107 √ x + 0.0111x2. The residual plot for this model looks good and the lack of fit test becomes F = (33.3 − 38.5)/(37 − 35) 33.3/35 = 2.6 0.95 = 2.73 ∼ F 2 35 p > 0.05. We may pass this model and ask the experimenter whether all the terms make sense. 3. a. Sol: See computer output. Basically, the methods are different (p < 0.0001), but the variety (p=0.99) and interactions (p=0.085) are not. b. Sol: Method A is better than B and C. B and C are not significantly different. (At p=0.05 level of testing) c. Sol: The estimate of the difference is 6.89 with standard error 1.04 (t has 75 degrees of freedom). The 95% confidence interval is 6.98 ± 1.96 × 1.043 = (4.94, 9.02) –1–