?18 BB Sum of Squares Criterion ?In addition to the value of ?R2,?another measurement used to assess the quality of a model is the ?sum of squares of the residuals.?Recall from Section that a residual is die difference between an observed ?y ?value and the value of ?y ?predicted from the model, which is denoted as?y.?Better models have smaller sums of squares. Refer to the data in Table. a.?Find?(?y? ??)2,the sum of squares of the residuals resulting from the linear model. b.?Find the sum of squares of residuals resulting from the quadratic model. c.?Verify that according to the sum of squares criterion, the quadratic model is better than the linear model. Section Table

Solution 18 BB Step 1 : a) Given, Moore’s law : The number of transistors per square inch on integrated circuits will double approximately every 18 months. The data is given below Co 1 1 1 1 1 1 1 1 2 2 2 2 der 9 9 9 9 9 9 9 9 0 0 0 0 yea 7 7 7 8 8 8 9 9 0 0 0 0 r 1 4 8 2 5 9 3 7 0 2 3 7 2 4 7 4 1 3 7 2 1 8 Pop 2 1 2 2 2 1 1 5 0 0 9 ulat . 5 2 7 0 9 8 0 0 0 0 0 ion 3 0 5 0 0 0 0 0 0 0 0 0 0 0 Step 2 : Linear regression equation is obtained in TI-83 by the following procedure. Steps : > Enter the data - Stat - select edit Then we get as follows > enter values of L1 and increase values in L2. > press 2nd and 0 to select catalog. > scroll to to get Diagnostics on, press enter > Press Stat and select Calc then from options select linear regression > Type L1 and L2 separated by comma and then press enter Hence regression results be Y = ax + b Where, a = 13494 and b = -148241 Step 3 : Predicted value is y = ax + b Where x = 1 = 13494*1 - 148241 = -134747 Residual = y - y = 2.3 - (-134747) = 134749.3 Similarly Residual and Square of Residual values are calculated as shown. y - y x y y (y y) 1 2.3 -134747 134749.3 18157373850 4 5 -94265 94270 8886832900 8 29 -40289 40318 1625541124 1 120 13687 -13567 184063489 2 1 275 54169 -53894 2904563236 5 1 118 108145 -106965 11441511225 9 0 2 310 162121 -159021 25287678441 3 0 2 750 216097 -208597 43512708409 7 0 3 420 256579 -214579 46044147241 0 00 3 220 283567 -63567 4040763489 2 000 3 410 297061 112939 12755217721 3 000 3 789 351037 437963 1.91812E+11 7 000 The sum of square value is 2 11 (y y) = 3.66652*10 Step 4: Similarly, for quadratic regression equation y = ax +bx + c Where a = 1249.4 b = - 33961.82685 c = 133198 Residual = y - y = 2.3 - (100486.4) = 134749.3 Residual and Square of Residual values are calculated as shown. x y y y - y (y y)2 1 2.3 100486.4 -100484 10097054353 4 5 17344.4 -17339.4 300654792.4 8 29 -58528.4 58557.4 3428969095 1 120 -944420.4 94540.4 8937887232 2 1 275 -95102 95377 9096772129 5 1 118 -61027 62207.6 3869785498 9 0 2 310 13027 -9927.6 98557241.76 3 0 2 750 127063.6 -119564 14295454445 7 0 3 420 238828 -196828 38741261584 0 00 220 3 W0 325831.6 -105832 11200327559 2 00 3 410 373081.6 36918.4 1362968259 3 000 3 789 587069.6 201930.4 40775886444 7 000 The sum of square value is 2 11 (y y) = 1.422*10 Step 9 : The quadratic regression is better than the Linear regression, residual value indicates the prediction error. If lesser the residual error better is the regression equation. We can observe that the residual value of the quadratic equation is lesser.hence, quadratic equation is better than linear regression.