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Show work by hand will give a thumbs up

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2. [35 points] A study was performed on wear of a bearing and its relationship to x₁ = oil viscosity and x₂ = load. The following data were obtained. y X1 x2 8² = SSE y'y-B'X'y n-p n-p (a) [10 points] Fit a multiple linear regression model to these data. Round your answers to 3 decimal places. ŷ = Bo + B₁x₁ + B₂x₂ = ( se(Bo)=√√8² Coo = ( se(B₁)=√√8²C₁1 = ( se(B₂) = √√8²C₂2 = ( 293 1.6 851 = 1 2 3 4 5 6 7 8 9 10 11 12 (b) [6 points] Estimate o² and the standard errors of the regression coefficients. Round your answers to 3 decimal places. 230 15.5 816 Percentage Points for of the t Distribution degrees of freedom. .40 325 289 277 271 267 265 263 11 262 261 260 = ( 172 22.0 1058 ) ) ) 700 1.372 260 697 1.363 259 695 1.356 ) + ( 91 43.0 1201 (c) [4 points] Find a 95% confidence interval on the regression coefficient for the variable x₁. Round your answers to 3 decimal places. B₁-ta/2n-pse(B₁) ≤ B₁ ≤ B₁+ta/2n-pse(B₁) ⇒ ( ) )x₁ + ( 113 33.0 1357 .25 .05 .025 005 .10 .01 1.000 3.078 6.314 12.706 31.821 63.657 816 1.886 2.920 4.303 6.965 765 1.638 2.353 3.182 4.541 9.925 5.841 4.604 741 1.533 2.132 2.776 3.747 727 1.476 2.015 2.571 3.365 4,032 718 1.440 1.943 2.447 3.143 3.707 711 1.415 1.895 2.365 2.998 3.499 .706 1.397 1.860 2.306 2.896 3.355 .703 1.383 1.833 2.262 2.821 3.250 1.812 2.228 2,764 3.169 1.796 2.201 2.718 3.106 1.782 2.179 2.681 3.055 )x2 125 40.0 1115 ) ≤B₁ ≤ ( .0005 .0025 .001 127.32 318,31 636.62 23.326 7.453 10.213 14.089 31.598 12.924 5.598 7.173 8.610 4.773 5.893 6.869 4.317 5.208 5.959 4.029 4.785 5.408 3.833 4.501 5.041 3.690 4.297 4.781 3.581 4.144 4.587 3.497 4.025 4.437 3.428 3.930 4.318 ).
(d) [5 points] Find a 95% prediction interval on the y when X₁ = 25 and x₂ = 1,066. Round your answers to 3 decimal places. 90-ta/2n-p√² (1+xo (X'X)-¹x) ≤ Yo ≤o +ta/2n-p8² (1+x (X'X)-¹x0) ) ≤% ≤ ( ⇒( (e) [5 points] Fit a simple linear regression model relating the response variable to the regressor X₁. Round your answers to 3 decimal places. ŷ = Bo + B₁x₁ = ( ) + ( )x₁ (f) [3 points] Find a 95% confidence interval on the slope for the simple linear regression model from part (e). Round your answers to 3 decimal places. B₁-ta/2, n-2se(B₁) ≤ B₁ ≤ B₁ + ta/2, n-2 se (B₁) ⇒ ( Percentage Points for of the t Distribution degrees of freedom. 1 2 3 4 5 6 7 8 9 10 11 12 40 325 289 ) 25 1,000 05 .025 .01 6.314 12.706 31.821 2,920 4.303 6.965 2.353 4.541 3.182 2.132 2.776 3.747 2.571 3.365 1.943 2.447 3.143 1.415 1.895 2.365 2.998 1.397 1.860 2.306 2.896 1.383 1.833 2,262 1.812 2.228 1.372 1.363 1.356 1.796 2.201 1.782 2.179 .10 3.078 816 1.886 277 765 1.638 271 741 1.533 267 727 1.476 2015 265 718 1.440 263 711 262 706 261 703 260 .700 260 697 259 .695 005 63.657 9.925 5.841 4.604 4.032 3.707 3.499 3.355 2.821 3.250 2.764 3.169 2.718 3.106 2.681 3.055 The confidence interval computed in part (c) is ( computed in part (f). Therefore, the ( ) ≤ B₁ ≤ ( .0025 .001 127.32 318.31 14.089 23.326 7.453 10.213 5.598 7.173 4.773 5.893 4.317 5.208 4.029 4.785 3.833 4.501 5.041 3.690 4.297 4.781 3.581 4.144 4.587 3.497 4.025 4.437 3.428 3.930 4.318 0005 636.62 ) linear regression model in this case is preferable. 31.598 12.924 8.610 6.869 5.959 5.408 (g) [2 points] Compare the lengths of the two confidence intervals computed in parts (c) and (f). Which interval is shorter? Does this tell you anything about which model is preferable? ) than the confidence interval )