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Forming ANOVA table without observed values.
Ridge Regression: $hat{beta} rightarrow beta$Regression with Mean, Standard Deviation, Range and CorrelationCalculating decreased cost with increasing quantityWhy is $beta$ a linear combination of $epsilon$Best Fit Line with 3d PointsDetermining equation of least-squares line without data pointsQuestion on lines of regressionProportion of explained varianceI need help understanding a few things regarding Least Squares Regression.Use linear regression to find $k$ and $C$ for $PV^k = C$
$begingroup$
Need some help in forming the ANOVA table without observed values.
I only managed to find SST with the first line of hint, but do not know how to proceed to find SSTreatment or SSError.
Any hint/solution will be greatly appreciated.
regression linear-regression
asked Feb 2 at 0:43
jaclynxjaclynx
4916
$endgroup$
add a comment |
$begingroup$
Need some help in forming the ANOVA table without observed values.
I only managed to find SST with the first line of hint, but do not know how to proceed to find SSTreatment or SSError.
Any hint/solution will be greatly appreciated.
regression linear-regression
asked Feb 2 at 0:43
jaclynxjaclynx
4916
$endgroup$
add a comment |
$begingroup$
Need some help in forming the ANOVA table without observed values.
I only managed to find SST with the first line of hint, but do not know how to proceed to find SSTreatment or SSError.
Any hint/solution will be greatly appreciated.
regression linear-regression
asked Feb 2 at 0:43
jaclynxjaclynx
4916
$endgroup$
Need some help in forming the ANOVA table without observed values.
I only managed to find SST with the first line of hint, but do not know how to proceed to find SSTreatment or SSError.
Any hint/solution will be greatly appreciated.
regression linear-regression
regression linear-regression
asked Feb 2 at 0:43
jaclynxjaclynx
4916
asked Feb 2 at 0:43
jaclynxjaclynx
4916
asked Feb 2 at 0:43
jaclynxjaclynx
4916
asked Feb 2 at 0:43
jaclynxjaclynx
4916
asked Feb 2 at 0:43
jaclynxjaclynx
4916
4916
add a comment |
add a comment |
1 Answer
1
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$begingroup$
First set up the error funcion
$$F(beta_0, beta_1) =sum_{i=1}^{N}[y_i-beta_0-beta_1x_i]^2.$$
Differentiate with respect to $beta_0$ and $beta_1$ and set the derivatives equal to $0$ to obtain.
$$sum_{i=1}^Ny_i-Nbeta_0-left[sum_{i=1}^{N}x_iright]beta_1=0$$
$$sum_{i=1}^Ny_ix_i-left[sum_{i=1}^{N}x_iright]beta_0-left[sum_{i=1}^{N}x^2_iright]beta_1=0.$$
The first equation can be rewritten as
$$Nbar{y}-Nbeta_0-Nbar{x}beta_1=0 implies beta_0 = bar{y}-2bar{x}=0.$$
From the second equaiton we can obtain
$$150-15cdot 2cdot 0 - left[sum_{i=1}^{15}x^2_iright]beta_1=0$$
$$implies sum_{i=1}^{15}x^2_i=37.5$$
We know that the correlation $r$ is given by
$$r= dfrac{1/Nsum_{i=1}^{15}x_iy_i-bar{x}bar{y}}{s_x s_y}$$
We have to determine $s_x$ in order to calculate $r$:
$$s_x = dfrac{1}{15-1}sum_{i=1}^{15}(x_i-bar{x})^2$$
$$= dfrac{1}{15-1}left[sum_{i=1}^{15}x^2_i-2bar{x}sum_{i=1}^{15}x_i+sum_{i=1}^{15}bar{x}^2right]$$
$$= dfrac{1}{15-1}left[sum_{i=1}^{15}x^2_i-2cdot 15cdotbar{x}^2+15bar{x}^2right]$$
$$= dfrac{1}{15-1}left[sum_{i=1}^{15}x^2_i-15cdotbar{x}^2right].$$
Finally, we know that $text{SSTotal}=sum_{i=1}^{15}left[y_i-bar{y}right]^2$ and
$$r^2 = dfrac{text{SSTreatment}}{text{SSTotal}}=dfrac{text{SSTreatment}}{sum_{i=1}^{15}left[y_i-bar{y}right]^2}=dfrac{text{SSTreatment}}{(15-1)s^2_y}$$
$$implies text{SSTreatment} =(15-1),r^2,s^2_y$$
answered yesterday
MachineLearnerMachineLearner
4767
$endgroup$
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
First set up the error funcion
$$F(beta_0, beta_1) =sum_{i=1}^{N}[y_i-beta_0-beta_1x_i]^2.$$
Differentiate with respect to $beta_0$ and $beta_1$ and set the derivatives equal to $0$ to obtain.
$$sum_{i=1}^Ny_i-Nbeta_0-left[sum_{i=1}^{N}x_iright]beta_1=0$$
$$sum_{i=1}^Ny_ix_i-left[sum_{i=1}^{N}x_iright]beta_0-left[sum_{i=1}^{N}x^2_iright]beta_1=0.$$
The first equation can be rewritten as
$$Nbar{y}-Nbeta_0-Nbar{x}beta_1=0 implies beta_0 = bar{y}-2bar{x}=0.$$
From the second equaiton we can obtain
$$150-15cdot 2cdot 0 - left[sum_{i=1}^{15}x^2_iright]beta_1=0$$
$$implies sum_{i=1}^{15}x^2_i=37.5$$
We know that the correlation $r$ is given by
$$r= dfrac{1/Nsum_{i=1}^{15}x_iy_i-bar{x}bar{y}}{s_x s_y}$$
We have to determine $s_x$ in order to calculate $r$:
$$s_x = dfrac{1}{15-1}sum_{i=1}^{15}(x_i-bar{x})^2$$
$$= dfrac{1}{15-1}left[sum_{i=1}^{15}x^2_i-2bar{x}sum_{i=1}^{15}x_i+sum_{i=1}^{15}bar{x}^2right]$$
$$= dfrac{1}{15-1}left[sum_{i=1}^{15}x^2_i-2cdot 15cdotbar{x}^2+15bar{x}^2right]$$
$$= dfrac{1}{15-1}left[sum_{i=1}^{15}x^2_i-15cdotbar{x}^2right].$$
Finally, we know that $text{SSTotal}=sum_{i=1}^{15}left[y_i-bar{y}right]^2$ and
$$r^2 = dfrac{text{SSTreatment}}{text{SSTotal}}=dfrac{text{SSTreatment}}{sum_{i=1}^{15}left[y_i-bar{y}right]^2}=dfrac{text{SSTreatment}}{(15-1)s^2_y}$$
$$implies text{SSTreatment} =(15-1),r^2,s^2_y$$
answered yesterday
MachineLearnerMachineLearner
4767
$endgroup$
add a comment |
$begingroup$
First set up the error funcion
$$F(beta_0, beta_1) =sum_{i=1}^{N}[y_i-beta_0-beta_1x_i]^2.$$
Differentiate with respect to $beta_0$ and $beta_1$ and set the derivatives equal to $0$ to obtain.
$$sum_{i=1}^Ny_i-Nbeta_0-left[sum_{i=1}^{N}x_iright]beta_1=0$$
$$sum_{i=1}^Ny_ix_i-left[sum_{i=1}^{N}x_iright]beta_0-left[sum_{i=1}^{N}x^2_iright]beta_1=0.$$
The first equation can be rewritten as
$$Nbar{y}-Nbeta_0-Nbar{x}beta_1=0 implies beta_0 = bar{y}-2bar{x}=0.$$
From the second equaiton we can obtain
$$150-15cdot 2cdot 0 - left[sum_{i=1}^{15}x^2_iright]beta_1=0$$
$$implies sum_{i=1}^{15}x^2_i=37.5$$
We know that the correlation $r$ is given by
$$r= dfrac{1/Nsum_{i=1}^{15}x_iy_i-bar{x}bar{y}}{s_x s_y}$$
We have to determine $s_x$ in order to calculate $r$:
$$s_x = dfrac{1}{15-1}sum_{i=1}^{15}(x_i-bar{x})^2$$
$$= dfrac{1}{15-1}left[sum_{i=1}^{15}x^2_i-2bar{x}sum_{i=1}^{15}x_i+sum_{i=1}^{15}bar{x}^2right]$$
$$= dfrac{1}{15-1}left[sum_{i=1}^{15}x^2_i-2cdot 15cdotbar{x}^2+15bar{x}^2right]$$
$$= dfrac{1}{15-1}left[sum_{i=1}^{15}x^2_i-15cdotbar{x}^2right].$$
Finally, we know that $text{SSTotal}=sum_{i=1}^{15}left[y_i-bar{y}right]^2$ and
$$r^2 = dfrac{text{SSTreatment}}{text{SSTotal}}=dfrac{text{SSTreatment}}{sum_{i=1}^{15}left[y_i-bar{y}right]^2}=dfrac{text{SSTreatment}}{(15-1)s^2_y}$$
$$implies text{SSTreatment} =(15-1),r^2,s^2_y$$
answered yesterday
MachineLearnerMachineLearner
4767
$endgroup$
add a comment |
$begingroup$
First set up the error funcion
$$F(beta_0, beta_1) =sum_{i=1}^{N}[y_i-beta_0-beta_1x_i]^2.$$
Differentiate with respect to $beta_0$ and $beta_1$ and set the derivatives equal to $0$ to obtain.
$$sum_{i=1}^Ny_i-Nbeta_0-left[sum_{i=1}^{N}x_iright]beta_1=0$$
$$sum_{i=1}^Ny_ix_i-left[sum_{i=1}^{N}x_iright]beta_0-left[sum_{i=1}^{N}x^2_iright]beta_1=0.$$
The first equation can be rewritten as
$$Nbar{y}-Nbeta_0-Nbar{x}beta_1=0 implies beta_0 = bar{y}-2bar{x}=0.$$
From the second equaiton we can obtain
$$150-15cdot 2cdot 0 - left[sum_{i=1}^{15}x^2_iright]beta_1=0$$
$$implies sum_{i=1}^{15}x^2_i=37.5$$
We know that the correlation $r$ is given by
$$r= dfrac{1/Nsum_{i=1}^{15}x_iy_i-bar{x}bar{y}}{s_x s_y}$$
We have to determine $s_x$ in order to calculate $r$:
$$s_x = dfrac{1}{15-1}sum_{i=1}^{15}(x_i-bar{x})^2$$
$$= dfrac{1}{15-1}left[sum_{i=1}^{15}x^2_i-2bar{x}sum_{i=1}^{15}x_i+sum_{i=1}^{15}bar{x}^2right]$$
$$= dfrac{1}{15-1}left[sum_{i=1}^{15}x^2_i-2cdot 15cdotbar{x}^2+15bar{x}^2right]$$
$$= dfrac{1}{15-1}left[sum_{i=1}^{15}x^2_i-15cdotbar{x}^2right].$$
Finally, we know that $text{SSTotal}=sum_{i=1}^{15}left[y_i-bar{y}right]^2$ and
$$r^2 = dfrac{text{SSTreatment}}{text{SSTotal}}=dfrac{text{SSTreatment}}{sum_{i=1}^{15}left[y_i-bar{y}right]^2}=dfrac{text{SSTreatment}}{(15-1)s^2_y}$$
$$implies text{SSTreatment} =(15-1),r^2,s^2_y$$
answered yesterday
MachineLearnerMachineLearner
4767
$endgroup$
First set up the error funcion
$$F(beta_0, beta_1) =sum_{i=1}^{N}[y_i-beta_0-beta_1x_i]^2.$$
Differentiate with respect to $beta_0$ and $beta_1$ and set the derivatives equal to $0$ to obtain.
$$sum_{i=1}^Ny_i-Nbeta_0-left[sum_{i=1}^{N}x_iright]beta_1=0$$
$$sum_{i=1}^Ny_ix_i-left[sum_{i=1}^{N}x_iright]beta_0-left[sum_{i=1}^{N}x^2_iright]beta_1=0.$$
The first equation can be rewritten as
$$Nbar{y}-Nbeta_0-Nbar{x}beta_1=0 implies beta_0 = bar{y}-2bar{x}=0.$$
From the second equaiton we can obtain
$$150-15cdot 2cdot 0 - left[sum_{i=1}^{15}x^2_iright]beta_1=0$$
$$implies sum_{i=1}^{15}x^2_i=37.5$$
We know that the correlation $r$ is given by
$$r= dfrac{1/Nsum_{i=1}^{15}x_iy_i-bar{x}bar{y}}{s_x s_y}$$
We have to determine $s_x$ in order to calculate $r$:
$$s_x = dfrac{1}{15-1}sum_{i=1}^{15}(x_i-bar{x})^2$$
$$= dfrac{1}{15-1}left[sum_{i=1}^{15}x^2_i-2bar{x}sum_{i=1}^{15}x_i+sum_{i=1}^{15}bar{x}^2right]$$
$$= dfrac{1}{15-1}left[sum_{i=1}^{15}x^2_i-2cdot 15cdotbar{x}^2+15bar{x}^2right]$$
$$= dfrac{1}{15-1}left[sum_{i=1}^{15}x^2_i-15cdotbar{x}^2right].$$
Finally, we know that $text{SSTotal}=sum_{i=1}^{15}left[y_i-bar{y}right]^2$ and
$$r^2 = dfrac{text{SSTreatment}}{text{SSTotal}}=dfrac{text{SSTreatment}}{sum_{i=1}^{15}left[y_i-bar{y}right]^2}=dfrac{text{SSTreatment}}{(15-1)s^2_y}$$
$$implies text{SSTreatment} =(15-1),r^2,s^2_y$$
answered yesterday
MachineLearnerMachineLearner
4767
answered yesterday
MachineLearnerMachineLearner
4767
answered yesterday
MachineLearnerMachineLearner
4767
answered yesterday
MachineLearnerMachineLearner
4767
4767
add a comment |
add a comment |
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