Consider the following SAS output for fitting the model Y = ?? + ??x? + ??x? + ??x? + ?. Analysis of Variance Source DF Sum of Squares Mean Square F Value Pr > F Model 3 69.627 23.209 10.73 0.000053 Error 31 67.060 2.163 Corrected Total 34 253.497 Root MSE 1.471 R-Square 0.5095 Dependent Mean 18.791 Adj R-Sq 0.4621 Coeff Var 7.828 Parameter Estimates Variable DF Parameter Estimate Standard Error t Value Pr > |t| Intercept 1 12.8557 2.1563 5.962 1.37e-06 x1 1 2.1596 1.1132 1.940 0.061530 x2 1 1.9515 0.8353 2.336 0.026124 x3 1 1.3437 0.3592 3.741 0.000746 (a) From and only from the output above, get an estimate of the standard deviation of ?. (b) Is the estimate in (a) larger than desirable in regression analysis? Justify your answer with the rule of thumb. (c) From the regression summary, can we conclude that this model will be useful for prediction purposes? Explain why. (d) What can you conclude from the p-value in the ANOVA table at the 0.05 significance level? (e) Consider x?, x? and x?. What can you conclude from each of the individual t-tests for their parameters at the 0.05 significance level? (f) Interpret the numerical value of the parameter estimate for x?.
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Step 1:** Calculate the coefficient of variation (Coeff Var) using the formula: Coeff Var = (Root MSE / Dependent Mean) * 100 ** Show moreā¦
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Suppose that you fitted the model E(y) = Ī²0 + Ī²1x + Ī²2x^2 to n = 20 data points and obtained the following MINITAB printout. Regression Analysis: y versus x, x-sq Analysis of Variance Source DF Adj SS Adj MS F-Value P-Value Regression 2 41227.6 20613.8 981.61 0.000 Error 17 357.0 21.0 Total 19 41584.6 Model Summary S R-Sq R-Sq(adj) 4.58258 99.14% 99.04% Coefficients Term Coef SE Coef T-Value P-Value Constant 12.38 3.40 3.64 0.002 x 9.79 1.49 6.57 0.000 x-sq -2.319 0.138 -16.80 0.000 Regression Equation y = 12.38 + 9.79x - 2.319x-sq What is your estimate of the average value of y when x = 0? (Use the exact values found in the MINITAB output.)
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Researchers at a large nutrition and weight management company are trying to build a model to predict a person's body fat percentage from an array of variables such as body weight, height, and body measurements around the neck, chest, abdomen, hips, biceps, etc. After employing various variable selection techniques, the SAS output for the final model is given below: Dependent Variable: bodyfatpt Source DF Sum of Squares Mean Square F Value Pr > F Model 2 3053.290 1526.645 129.239 <.0001 Error 46 543.379 11.813 Corrected Total 48 3596.670 R-Square Coeff Var Root MSE bodyfatpt Mean 0.8489 9.79360 3.4369 72.4500 Parameter Estimate Standard Error t Value Pr > |t| Intercept -53.954 4.742 -11.379 <.0001 weight -0.162 0.038 -4.230 <.0001 Hips_circum 1.105 0.101 10.912 <.0001
Use the data in CEOSAL. 2 to answer this question. (i) Estimate the model $$\ lsalary = beta_{0}+\beta_{1} \text { lsales }+\beta_{2} \text {lmktval}+\beta_{3} \text { ceoten }+\beta_{4} \text {ceoten}^{2}+u$$ by OLS using all of the observations, where lsalary, lsales, and lmktvale are all natural logarithms. Report the results in the usual form with the usual OLS standard errors. (You may verify that the heteroskedasticity-robust standard errors are similar.) (ii) In the regression from part (i) obtain the studentized residuals; call these stri. How many studentized residuals are above 1.96 in absolute value? If the studentized residuals were independent draws from a standard normal distribution, about how many would you expect to be above two in absolute value with 177 draws? (iii) Reestimate the equation in part (i) by OLS using only the observations with $\left|s t r_{i}\right| \leq 1.96 .$ How do the coefficients compare with those in part (i)? (iv) Estimate the equation in part (i) by LAD, using all of the data. Is the estimate of $\beta_{1}$ closer to the OLS estimate using the full sample or the restricted sample? What about for $\beta_{3} ?$ (v) Evaluate the following statement: "Dropping outliers based on extreme values of studentized residuals makes the resulting OLS estimates closer to the LAD estimates on the full sample."
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