EXST7015 : Statistical Techniques II Geaghan Simple Linear Regression With equations & calculations Page 1 03d-Slr-TreeCalc&Equations.doc Freund & Wilson (1997) : Prediction of weight of wood from trees (Table 8.24) Observation Dbh Weight Dbh*Dbh Wt*Wt Dbh*Wt Predicted Residual1 5.7 174 32.49 30276 991.8 288.42 -114.422 8.1 745 65.61 555025 6034.5 716.97 28.033 8.3 814 68.89 662596 6756.2 752.68 61.324 7.0 408 49.00 166464 2856.0 520.55 -112.555 6.2 226 38.44 51076 1401.2 377.7 -151.76 11.4 1675 129.96 2805625 19095.0 1306.23 368.777 11.6 1491 134.56 2223081 17295.6 1341.94 149.068 4.5 121 20.25 14641 544.5 74.14 46.869 3.5 58 12.25 3364 203.0 -104.42 162.4210 6.2 278 38.44 77284 1723.6 377.7 -99.711 5.7 220 32.49 48400 1254.0 288.42 -68.4212 6.0 342 36.00 116964 2052.0 341.99 0.0113 5.6 209 31.36 43681 1170.4 270.56 -61.5614 4.0 84 16.00 7056 336.0 -15.14 99.1415 6.7 313 44.89 97969 2097.1 466.98 -153.9816 4.0 60 16.00 3600 240.0 -15.14 75.1417 12.1 1692 146.41 2862864 20473.2 1431.22 260.7818 4.5 74 20.25 5476 333.0 74.14 -0.1419 8.6 515 73.96 265225 4429.0 806.25 -291.2520 9.3 766 86.49 586756 7123.8 931.25 -165.2521 6.5 345 42.25 119025 2242.5 431.27 -86.2722 5.6 210 31.36 44100 1176.0 270.56 -60.5623 4.3 100 18.49 10000 430.0 38.43 61.5724 4.5 122 20.25 14884 549.0 74.14 47.8625 7.7 539 59.29 290521 4150.3 645.54 -106.5426 8.8 815 77.44 664225 7172.0 841.96 -26.9627 5.0 194 25.00 37636 970.0 163.42 30.5828 5.4 280 29.16 78400 1512.0 234.85 45.1529 6.0 296 36.00 87616 1776.0 341.99 -45.9930 7.4 462 54.76 213444 3418.8 591.98 -129.9831 5.6 200 31.36 40000 1120.0 270.56 -70.5632 5.5 229 30.25 52441 1259.5 252.7 -23.733 4.3 125 18.49 15625 537.5 38.43 86.5734 4.2 84 17.64 7056 352.8 20.57 63.4335 3.7 70 13.69 4900 259.0 -68.71 138.7136 6.1 224 37.21 50176 1366.4 359.84 -135.8437 3.9 99 15.21 9801 386.1 -33 13238 5.2 200 27.04 40000 1040.0 199.14 0.8639 5.6 214 31.36 45796 1198.4 270.56 -56.5640 7.8 712 60.84 506944 5553.6 663.4 48.641 6.1 297 37.21 88209 1811.7 359.84 -62.8442 6.1 238 37.21 56644 1451.8 359.84 -121.8443 4.0 89 16.00 7921 356.0 -15.14 104.1444 4.0 76 16.00 5776 304.0 -15.14 91.1445 8.0 614 64.00 376996 4912.0 699.11 -85.1146 5.2 194 27.04 37636 1008.8 199.14 -5.1447 3.7 66 13.69 4356 244.2 -68.71 134.71Sum 289.2 17359 1981.98 13537551 142968.3 Sum = 0Mean 6.15 369.34 42.17 288033 3041.9 SS = 670190.732n 47 47 47 47 47EXST7015 : Statistical Techniques II Geaghan Simple Linear Regression With equations & calculations Page 2 03d-Slr-TreeCalc&Equations.doc Intermediate Calculations Sum X = 289.2 Sum Y = 17359 Sum X2 = 1981.98 Sum Y2 = 13537551 Mean X= 6.153191489 Mean Y= 369.3404255 Sum XY = 142968.3 n = 47 Correction factors and Corrected values (Sums of squares and cross-products) CF for X = Cxx = 1779.502979 Corrected SS X = Sxx = 202.4770213 CF for Y = Cyy = 6411380.447 Corrected SS Y = Syy = 7126170.553 CF for XY = Cxy = 106813.2511 Corrected SS XY = Sxy = 36155.04894 Model Parameter Estimates Slope = b1 = 36155.04894 / 202.4770213 =178.5637141 Intercept = b0 = 369.3404255 - 178.5637141 * 6.153191489 = -729.3963003 Regression Line Yi = b0 + b1 * Xi + ei Yi = -729.3963003 + 178.5637141 * Xi + ei ANOVA Table SSTotal =7126170.553 SSRegression 36155.048942 / 202.4770213 = 6455979.821 SSError = 7126170.553 - 6455979.821 = 670190.7322 Source df SS MS F Regression 1 6455979.821 6455979.821 433.4871821 Error 45 670190.7322 14893.12738 Total 46 7126170.553 Standard error of b1 : where t(0.05/2, 45 df) = 2.014103 ; SMSEXbi12=∑ = 8.576401034 P(178.5637 - 2.0141*8.5764 ≤ β1 ≤ 178.5637 + 2.0141*8.5764 ) = 0.95 P( 161.289956 ≤ β1 ≤ 195.8375 ) = 0.95 Testing b1 against a specified value : H0: β1 = 200 versus H1: β1 ≠ 200 tbSHb=−1101β|= (178.5637141 - 200) / 8.576401034 = -2.49945 Note that t2 = F = 6.247251 ; This test would be done in SAS as an F statementEXST7015 : Statistical Techniques II Geaghan Simple Linear Regression With equations & calculations Page 3 03d-Slr-TreeCalc&Equations.doc The variance of a linear combination is given by the sum of the variances plus twice the covariances. e.g. for A = aX + bY + cZ then Var(A) = a2σ2X + b2σ2Y + c2σ2Z + 2(abσXY + acσXZ + bcσYZ) where the covariances are equal to zero if the variables are independent For the linear combination YbbXii=+01, the standard error of Yi is as follows. Standard error of the regression line ( Yi): SMSEnXXXXYiiyx |..=+−−FHGGIKJJ∑122chch The calculation above DOES NOT assume that the covariances of the regression coefficients are independent. However, for the variance of individual points the linear combination is YbbXeYeiiiii=+ +=+01 . For this linear combination the terms for the predicted value and residuals are assumed independent (i.e. Yi is independent of ei). SMSEnXXSMSE MSEnXXSYixxixxyx |..=+−FHGGIKJJ+= ++−FHGGIKJJ11122ch ch Standard error of b0 is the same as the standard error of the regression line where Xi = 0 SQRT(14893.12738(0.021276596 + (0 - 37.8617655) / 202.4770213 = 55.69366336 Confidence interval on b0 where b0 = -729.3963003 and t(0.05/2, 45 df) = 2.014103 P(-729.3963 - 2.0141*55.6937 ≤ β0 ≤ -729.3963+2.0141*55.6937)=0.95 P( -841.5690916 ≤ β0 ≤ -617.223509 ) = 0.95 Estimate and standard error of an individual observation (e.g. the weight of wood for a ten-inch-diameter tree) Y =-729.3963003 + 178.5637141* X = -729.3963003 + 178.5637141 * 10 =1056.240841 se(bx=10) =14893.1274*(1 + 0.02128 + (10 - 14.79794) / 202.4770) = 127.6654 P(1056.2408-2.0141*127.6654 ≤ µx=10 ≤ 1056.2408+2.0141*127.6654)=0.95 P( 799.1094964 ≤ µx=10 ≤ 1313.372185 ) = 0.95 Calculate the coefficient of Determination and correlation R2 = 0.905953594 or 90.59535936 % r =