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The solutions to exercise 2 of the statistics and data analysis (sta6167) course. It includes the answers to various questions related to linear and quadratic regression analysis, including the calculation of regression coefficients, residuals, and tests for model fit. The document also discusses the identification and handling of outliers.

Typology: Assignments

Pre 2010

1 / 2

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–