THIS IS AN EXAMPLE SINCE CHEGG WONT ALLOW AS WHEN ADDED TO QUESTION GETS TOO LONG

Suppose that we have following information: Bo = 169.59 SE(B) = 16.87 B = 2.7013 SEC ) = 0.5837 B, = -1.6978 SECR2) = 0.6943 S 42.13 RP = 0.06083 Using only this information, and the fact that the sample size is n=398, do the following 1 (a) Construct two different ANOVA tables. In the first table, report the sequential sums of squares with X entered first, and in the second table report the sequential sums of squares with X, entered first. From S= 42.13, RP = 0.06083, n = 398 and p=3, we can show that SSE = (n -pS = 701100.1 SST = SSE/(1 - R") = 746510.3SSR=SST - SSE = 45410.22 The information that we need to divide SSR into components due to X1 and X2 X1, and its components due to X, and X1 X2, comes from the t-statistics which are equivalent to partial F's. The partial F for X, given X is SSR(X2X1) = 5.979691 SE(B) S2 So SSR(X2|X1) = 10613.57. Similarly, the partial F for X1 given X, is SSR(X1|X2) F- = 21.41738 SEB.). S2 So SSR(X1|X) = 38014.49. Source SS 34796.65 X2X1 10613.57 Error 701100.1 (SE)" - (SEC.)" DF 1 1 395 MS 34796.65 10613.57 1774.94 Source DE 1 X1/X2 Error SS 7395.73 38014.49 701100.1 MS 7395.73 38014.49 1774.94 395 (b) Perform five different f tests for this model: (i) the overall for testing the combined significance of X and X2, (ii) the F for testing the significance of X, with X, excluded from the model, (iii) the F for testing the significance of x with X2 included in the model, (iv) the F for testing the significance of x2 with X excluded from the model, and (v) the F for testing the significance of X, with X included in the model. Report degrees of freedom and p-values. For (ii) and (iv), use more stable version of error. (i) The overall Fin 45110.22/2 1774.94 12.792 with 2 and 395 degrees of freedom, and the p-value is 4.14 x 10-6 (ii) The F for X with X2 excluded from the model is 34796.65/1 (10613.57 + 701100.1)/396 with 1 and 396 degrees of freedom, and the p-value is 1.39 x 10-5. (iii) The F for X with X, included from the model is 38011.19/1 1774.94 21.117 with 1 and 395 degrees of freedom, and the p-value is 5.01 x 106. (iv) The F for Xwith X excluded from the model is 7395.73/1 -3.962 (38014.49 + 701100.1)/396 with 1 and 396 degrees of freedom, and the p-value is 0.047. Suppose that we have following information: Bo = 169.59 SE(B) = 16.87 B = 2.7013 SEC ) = 0.5837 B, = -1.6978 SECR2) = 0.6943 S 42.13 RP = 0.06083 Using only this information, and the fact that the sample size is n=398, do the following 1 (a) Construct two different ANOVA tables. In the first table, report the sequential sums of squares with X entered first, and in the second table report the sequential sums of squares with X, entered first. From S= 42.13, RP = 0.06083, n = 398 and p=3, we can show that SSE = (n -pS = 701100.1 SST = SSE/(1 - R") = 746510.3SSR=SST - SSE = 45410.22 The information that we need to divide SSR into components due to X1 and X2 X1, and its components due to X, and X1 X2, comes from the t-statistics which are equivalent to partial F's. The partial F for X, given X is SSR(X2X1) = 5.979691 SE(B) S2 So SSR(X2|X1) = 10613.57. Similarly, the partial F for X1 given X, is SSR(X1|X2) F- = 21.41738 SEB.). S2 So SSR(X1|X) = 38014.49. Source SS 34796.65 X2X1 10613.57 Error 701100.1 (SE)" - (SEC.)" DF 1 1 395 MS 34796.65 10613.57 1774.94 Source DE 1 X1/X2 Error SS 7395.73 38014.49 701100.1 MS 7395.73 38014.49 1774.94 395 (b) Perform five different f tests for this model: (i) the overall for testing the combined significance of X and X2, (ii) the F for testing the significance of X, with X, excluded from the model, (iii) the F for testing the significance of x with X2 included in the model, (iv) the F for testing the significance of x2 with X excluded from the model, and (v) the F for testing the significance of X, with X included in the model. Report degrees of freedom and p-values. For (ii) and (iv), use more stable version of error. (i) The overall Fin 45110.22/2 1774.94 12.792 with 2 and 395 degrees of freedom, and the p-value is 4.14 x 10-6 (ii) The F for X with X2 excluded from the model is 34796.65/1 (10613.57 + 701100.1)/396 with 1 and 396 degrees of freedom, and the p-value is 1.39 x 10-5. (iii) The F for X with X, included from the model is 38011.19/1 1774.94 21.117 with 1 and 395 degrees of freedom, and the p-value is 5.01 x 106. (iv) The F for Xwith X excluded from the model is 7395.73/1 -3.962 (38014.49 + 701100.1)/396 with 1 and 396 degrees of freedom, and the p-value is 0.047