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The Following MINITAB Output Presents a Multiple Regression Equatior y^\hat { y }

question 19

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The following MINITAB output presents a multiple regression equatior y^\hat { y } =b0+b1x1+b2x2+b3x3+b4x4
The regression equation is
Y=5.5079+1.6552X11.1088X2+1.3981X31.2465X4\mathrm { Y } = 5.5079 + 1.6552 \mathrm { X } 1 - 1.1088 \mathrm { X } 2 + 1.3981 \mathrm { X } 3 - 1.2465 \mathrm { X } 4

 Predictor  Coef  SE Coef  T  P  Constant 5.50790.76401.10020.314 X1 1.65520.70323.19290.002 X2 1.10880.60233.23100.005 X3 1.39810.89701.81370.087 X4 1.24650.82511.14330.354\begin{array}{lllll}\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\\text { Constant } & 5.5079 & 0.7640 & 1.1002 & 0.314 \\\text { X1 } & 1.6552 & 0.7032 & 3.1929 & 0.002 \\\text { X2 } & -1.1088 & 0.6023 & -3.2310 & 0.005 \\\text { X3 } & 1.3981 & 0.8970 & 1.8137 & 0.087 \\\text { X4 } & -1.2465 & 0.8251 & -1.1433 & 0.354\end{array}

 The following MINITAB output presents a multiple regression equatior  \hat { y } =b<sub>0</sub>+b<sub>1</sub>x<sub>1</sub>+b<sub>2</sub>x<sub>2</sub>+b<sub>3</sub>x<sub>3</sub>+b<sub>4</sub>x<sub>4</sub> The regression equation is  \mathrm { Y } = 5.5079 + 1.6552 \mathrm { X } 1 - 1.1088 \mathrm { X } 2 + 1.3981 \mathrm { X } 3 - 1.2465 \mathrm { X } 4    \begin{array}{lllll} \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 5.5079 & 0.7640 & 1.1002 & 0.314 \\ \text { X1 } & 1.6552 & 0.7032 & 3.1929 & 0.002 \\ \text { X2 } & -1.1088 & 0.6023 & -3.2310 & 0.005 \\ \text { X3 } & 1.3981 & 0.8970 & 1.8137 & 0.087 \\ \text { X4 } & -1.2465 & 0.8251 & -1.1433 & 0.354 \end{array}       \text { Analysis of Variance }   \begin{array}{lccccc} \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 4 & 637.5 & 159.4 & 7.1480 & 0.003 \\ \text { Residual Error } & 40 & 893.2 & 22.3 & & \\ \text { Total } & 44 & 1,530.7 & & & \end{array}   Let  \beta _ { 1 }  be the coefficient  X _ { 1 }  Test the hypothesis  H _ { 0 } : \beta _ { 1 } = 0  rersus  H _ { 1 } : \beta _ { 1 } \neq 0 \text { at the } \alpha = 0.05  level. What do you conclude? A)  Do not H<sub>0</sub> B)  Reject H<sub>0</sub>

 Analysis of Variance \text { Analysis of Variance }
 Source  DF  SS  MS  F  P  Regression 4637.5159.47.14800.003 Residual Error 40893.222.3 Total 441,530.7\begin{array}{lccccc}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\\text { Regression } & 4 & 637.5 & 159.4 & 7.1480 & 0.003 \\\text { Residual Error } & 40 & 893.2 & 22.3 & & \\\text { Total } & 44 & 1,530.7 & & &\end{array}

Let β1\beta _ { 1 } be the coefficient X1X _ { 1 } Test the hypothesis H0:β1=0H _ { 0 } : \beta _ { 1 } = 0 rersus H1:β10 at the α=0.05H _ { 1 } : \beta _ { 1 } \neq 0 \text { at the } \alpha = 0.05
level. What do you conclude?


Definitions:

Lateral Moraine

A ridge of debris deposited along the sides of a glacier, marking the previous extent of the glacier's reach.

Terminal Moraine

A moraine deposited at the point of furthest advance of a glacier or ice sheet, marking its maximum extent.

Kettle

A pitlike depression in glacial deposits, commonly a lake or swamp; formed by the melting of a large block of ice that had been at least partly buried in the glacial deposits.

Coastal Dunes

Sand formations located near coastlines, formed by the wind's movement of sand.

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