Expected mean squared error and MSR
$begingroup$
In a small-scale regression study, five observations on $Y$ were
obtained corresponding to $X = 1,4,10, 11$, and $14$. Assume that
$sigma=0.6,B_0=5,B_1=3$
a. What are the expected values off MSR and MSE here?
b. For derermining whether or not a regression relation exists, would
it have been better or worse to have made the five observations at $X
= 6,7, 8, 9$, and $10$? Why? Would the same answer apply if the principal purpose were to estimate the mean response for $X = 8$?
Discuss.
$$Y_i=B_0+B_1X_i+epsilon_i$$
$$hat{Y}_i=hat{B}_0+hat{B}_1X_i$$
$$MSR=sum(hat{Y}_i-overline{Y})^2$$
$$MSE=frac{sum (Y_i-hat{Y}_i)^2}{n-2}=frac{sum(B_0+B_1X_i+epsilon_i-hat{B}_0-hat{B}_1 X_i)^2}{n-2}$$
I'm still doesn't understand what they want, they want
$$E(MSE);E(MSR) text{ ?}$$
What they mean by expected values?
statistics self-learning regression mean-square-error
asked Oct 5 '15 at 13:02
RolandRoland
1,35511030
$endgroup$
add a comment |
$begingroup$
In a small-scale regression study, five observations on $Y$ were
obtained corresponding to $X = 1,4,10, 11$, and $14$. Assume that
$sigma=0.6,B_0=5,B_1=3$
a. What are the expected values off MSR and MSE here?
b. For derermining whether or not a regression relation exists, would
it have been better or worse to have made the five observations at $X
= 6,7, 8, 9$, and $10$? Why? Would the same answer apply if the principal purpose were to estimate the mean response for $X = 8$?
Discuss.
$$Y_i=B_0+B_1X_i+epsilon_i$$
$$hat{Y}_i=hat{B}_0+hat{B}_1X_i$$
$$MSR=sum(hat{Y}_i-overline{Y})^2$$
$$MSE=frac{sum (Y_i-hat{Y}_i)^2}{n-2}=frac{sum(B_0+B_1X_i+epsilon_i-hat{B}_0-hat{B}_1 X_i)^2}{n-2}$$
I'm still doesn't understand what they want, they want
$$E(MSE);E(MSR) text{ ?}$$
What they mean by expected values?
statistics self-learning regression mean-square-error
asked Oct 5 '15 at 13:02
RolandRoland
1,35511030
$endgroup$
add a comment |
$begingroup$
In a small-scale regression study, five observations on $Y$ were
obtained corresponding to $X = 1,4,10, 11$, and $14$. Assume that
$sigma=0.6,B_0=5,B_1=3$
a. What are the expected values off MSR and MSE here?
b. For derermining whether or not a regression relation exists, would
it have been better or worse to have made the five observations at $X
= 6,7, 8, 9$, and $10$? Why? Would the same answer apply if the principal purpose were to estimate the mean response for $X = 8$?
Discuss.
$$Y_i=B_0+B_1X_i+epsilon_i$$
$$hat{Y}_i=hat{B}_0+hat{B}_1X_i$$
$$MSR=sum(hat{Y}_i-overline{Y})^2$$
$$MSE=frac{sum (Y_i-hat{Y}_i)^2}{n-2}=frac{sum(B_0+B_1X_i+epsilon_i-hat{B}_0-hat{B}_1 X_i)^2}{n-2}$$
I'm still doesn't understand what they want, they want
$$E(MSE);E(MSR) text{ ?}$$
What they mean by expected values?
statistics self-learning regression mean-square-error
asked Oct 5 '15 at 13:02
RolandRoland
1,35511030
$endgroup$
In a small-scale regression study, five observations on $Y$ were
obtained corresponding to $X = 1,4,10, 11$, and $14$. Assume that
$sigma=0.6,B_0=5,B_1=3$
a. What are the expected values off MSR and MSE here?
b. For derermining whether or not a regression relation exists, would
it have been better or worse to have made the five observations at $X
= 6,7, 8, 9$, and $10$? Why? Would the same answer apply if the principal purpose were to estimate the mean response for $X = 8$?
Discuss.
$$Y_i=B_0+B_1X_i+epsilon_i$$
$$hat{Y}_i=hat{B}_0+hat{B}_1X_i$$
$$MSR=sum(hat{Y}_i-overline{Y})^2$$
$$MSE=frac{sum (Y_i-hat{Y}_i)^2}{n-2}=frac{sum(B_0+B_1X_i+epsilon_i-hat{B}_0-hat{B}_1 X_i)^2}{n-2}$$
I'm still doesn't understand what they want, they want
$$E(MSE);E(MSR) text{ ?}$$
What they mean by expected values?
statistics self-learning regression mean-square-error
statistics self-learning regression mean-square-error
asked Oct 5 '15 at 13:02
RolandRoland
1,35511030
asked Oct 5 '15 at 13:02
RolandRoland
1,35511030
edited Oct 8 '15 at 1:13
Roland
asked Oct 5 '15 at 13:02
RolandRoland
1,35511030
asked Oct 5 '15 at 13:02
RolandRoland
1,35511030
asked Oct 5 '15 at 13:02
RolandRoland
1,35511030
1,35511030
add a comment |
add a comment |
1 Answer
1
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$begingroup$
$$Y_i=B_0+B_1X_i+epsilon_i$$
$$hat{Y_i}=hat{B_0}+hat{B_1}X_i$$
a) $$E[MSE]=E[frac{sum(Y_i-hat{Y_i})^2}{n-2}]=sigma^2=0.6^2$$
$$E[MSR]=E[sum(hat{Y_i}-overline{Y})^2]=sigma^2+B_1sum(X_i-overline{X})^2=1026.36$$
b)
$$sigma(hat{B_1})=sqrt{frac{sigma^2}{sum(X_i-overline{X})^2}}=frac{0.6}{sqrt{sum(X_i-overline{X})^2}}$$
for the case where $X=(1,4,10,11,14)$ we have that $sigma(hat{B_1})=0.05619515$
and for the case where $X=(6,7,8,9,10)$ $sigma(B_1)=0.1897367$, then the first set is better I think.
Is there any difference if it were estimating the mean response for X = 8?
answered Oct 7 '15 at 23:35
RolandRoland
1,35511030
$endgroup$
add a comment |
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$begingroup$
$$Y_i=B_0+B_1X_i+epsilon_i$$
$$hat{Y_i}=hat{B_0}+hat{B_1}X_i$$
a) $$E[MSE]=E[frac{sum(Y_i-hat{Y_i})^2}{n-2}]=sigma^2=0.6^2$$
$$E[MSR]=E[sum(hat{Y_i}-overline{Y})^2]=sigma^2+B_1sum(X_i-overline{X})^2=1026.36$$
b)
$$sigma(hat{B_1})=sqrt{frac{sigma^2}{sum(X_i-overline{X})^2}}=frac{0.6}{sqrt{sum(X_i-overline{X})^2}}$$
for the case where $X=(1,4,10,11,14)$ we have that $sigma(hat{B_1})=0.05619515$
and for the case where $X=(6,7,8,9,10)$ $sigma(B_1)=0.1897367$, then the first set is better I think.
Is there any difference if it were estimating the mean response for X = 8?
answered Oct 7 '15 at 23:35
RolandRoland
1,35511030
$endgroup$
add a comment |
$begingroup$
$$Y_i=B_0+B_1X_i+epsilon_i$$
$$hat{Y_i}=hat{B_0}+hat{B_1}X_i$$
a) $$E[MSE]=E[frac{sum(Y_i-hat{Y_i})^2}{n-2}]=sigma^2=0.6^2$$
$$E[MSR]=E[sum(hat{Y_i}-overline{Y})^2]=sigma^2+B_1sum(X_i-overline{X})^2=1026.36$$
b)
$$sigma(hat{B_1})=sqrt{frac{sigma^2}{sum(X_i-overline{X})^2}}=frac{0.6}{sqrt{sum(X_i-overline{X})^2}}$$
for the case where $X=(1,4,10,11,14)$ we have that $sigma(hat{B_1})=0.05619515$
and for the case where $X=(6,7,8,9,10)$ $sigma(B_1)=0.1897367$, then the first set is better I think.
Is there any difference if it were estimating the mean response for X = 8?
answered Oct 7 '15 at 23:35
RolandRoland
1,35511030
$endgroup$
add a comment |
$begingroup$
$$Y_i=B_0+B_1X_i+epsilon_i$$
$$hat{Y_i}=hat{B_0}+hat{B_1}X_i$$
a) $$E[MSE]=E[frac{sum(Y_i-hat{Y_i})^2}{n-2}]=sigma^2=0.6^2$$
$$E[MSR]=E[sum(hat{Y_i}-overline{Y})^2]=sigma^2+B_1sum(X_i-overline{X})^2=1026.36$$
b)
$$sigma(hat{B_1})=sqrt{frac{sigma^2}{sum(X_i-overline{X})^2}}=frac{0.6}{sqrt{sum(X_i-overline{X})^2}}$$
for the case where $X=(1,4,10,11,14)$ we have that $sigma(hat{B_1})=0.05619515$
and for the case where $X=(6,7,8,9,10)$ $sigma(B_1)=0.1897367$, then the first set is better I think.
Is there any difference if it were estimating the mean response for X = 8?
answered Oct 7 '15 at 23:35
RolandRoland
1,35511030
$endgroup$
$$Y_i=B_0+B_1X_i+epsilon_i$$
$$hat{Y_i}=hat{B_0}+hat{B_1}X_i$$
a) $$E[MSE]=E[frac{sum(Y_i-hat{Y_i})^2}{n-2}]=sigma^2=0.6^2$$
$$E[MSR]=E[sum(hat{Y_i}-overline{Y})^2]=sigma^2+B_1sum(X_i-overline{X})^2=1026.36$$
b)
$$sigma(hat{B_1})=sqrt{frac{sigma^2}{sum(X_i-overline{X})^2}}=frac{0.6}{sqrt{sum(X_i-overline{X})^2}}$$
for the case where $X=(1,4,10,11,14)$ we have that $sigma(hat{B_1})=0.05619515$
and for the case where $X=(6,7,8,9,10)$ $sigma(B_1)=0.1897367$, then the first set is better I think.
Is there any difference if it were estimating the mean response for X = 8?
answered Oct 7 '15 at 23:35
RolandRoland
1,35511030
edited Oct 8 '15 at 21:47
answered Oct 7 '15 at 23:35
RolandRoland
1,35511030
answered Oct 7 '15 at 23:35
RolandRoland
1,35511030
answered Oct 7 '15 at 23:35
RolandRoland
1,35511030
1,35511030
add a comment |
add a comment |
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