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The Following MINITAB Output Presents a Multiple Regression Equation y^=b0+b1x1+\hat { y } = b _ { 0 } + b _ { 1 } x _ { 1 } +

Question 110

Multiple Choice

The following MINITAB output presents a multiple regression equation y^=b0+b1x1+\hat { y } = b _ { 0 } + b _ { 1 } x _ { 1 } + b2x2+b3x3+b4x4b _ { 2 } x _ { 2 } + b _ { 3 } x _ { 3 } + b _ { 4 } x _ { 4 }
The regression equation is
Y=2.59191.3391X1+0.6212X2+1.6435X3+1.4269X4\mathrm { Y } = 2.5919 - 1.3391 \mathrm { X } 1 + 0.6212 \mathrm { X } 2 + 1.6435 \mathrm { X } 3 + 1.4269 \mathrm { X } 4
 Predictor  Coef  SE Coef  T  P  Constant 2.59190.62691.16680.337 X1 1.33910.67163.51900.002 X2 0.62120.84883.28480.004 X3 1.64350.79341.88210.090 X4 1.42690.76790.98790.345\begin{array}{lllll}\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\\text { Constant } & 2.5919 & 0.6269 & 1.1668 & 0.337 \\\text { X1 } & -1.3391 & 0.6716 & 3.5190 & 0.002 \\\text { X2 } & 0.6212 & 0.8488 & -3.2848 & 0.004 \\\text { X3 } & 1.6435 & 0.7934 & 1.8821 & 0.090 \\\text { X4 } & 1.4269 & 0.7679 & -0.9879 & 0.345\end{array}

S=2.8912RSq=55.3%RSq(adj) =46.4%\mathrm{S}=2.8912 \quad \mathrm{R}-\mathrm{Sq}=55.3 \% \mathrm{R}-\mathrm{Sq}(\mathrm{adj}) =46.4 \%

 Analysis of Variance  Source  DF  SS  MS  F P  Regression 4735.9184.07.73110.003 Residual Error 25594.623.8 Total 291330.5\begin{array}{l}\text { Analysis of Variance }\\\begin{array}{lcccrl}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F P } & \\\text { Regression } & 4 & 735.9 & 184.0 & 7.7311 & 0.003 \\\text { Residual Error } & 25 & 594.6 & 23.8 & & \\\text { Total } & 29 & 1330.5 & & & \\\hline\end{array}\end{array}



Let β3\beta _ { 3 } be the coefficient X3X 3 Test the hypothesis H0:β3=0H _ { 0 } : \beta _ { 3 } = 0 versus H1:β30H _ { 1 } : \beta 3 \neq 0 at the α=0.05 level. \alpha = 0.05 \text { level. } What do you conclude?


A) Do not reject H0H _ { 0 }
B) Reject H0H _ { 0 }

Correct Answer:

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