Linear Regression : SSR equation The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Help with problem: Estimated Standard Deviation of Regression Equation (Simple Linear)Simple Linear RegressionDoes SSTR (sum of squares for treatments) = SSR (regression sum of squares)?Linear regression: degrees of freedom of SST, SSR, and RSSProve $SST=SSE+SSR$How to show for a simple regression with an intercept and one independent variable $R^2 = r ^2$ , where $r$ is the ordinary correlation coefficient.Linear Regression about SSRProving that SSE and SSR are independentHow to compute SSR with just residuals and Xi?Does $E(sum e_i^2) = sum E(y_i^2) - E(sum hat y_i^2)$ hold true?
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Linear Regression : SSR equation
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Help with problem: Estimated Standard Deviation of Regression Equation (Simple Linear)Simple Linear RegressionDoes SSTR (sum of squares for treatments) = SSR (regression sum of squares)?Linear regression: degrees of freedom of SST, SSR, and RSSProve $SST=SSE+SSR$How to show for a simple regression with an intercept and one independent variable $R^2 = r ^2$ , where $r$ is the ordinary correlation coefficient.Linear Regression about SSRProving that SSE and SSR are independentHow to compute SSR with just residuals and Xi?Does $E(sum e_i^2) = sum E(y_i^2) - E(sum hat y_i^2)$ hold true?
$begingroup$
I've studied SST and the others but I saw the equation that $SSR=(n-1)(β_1)^2(S_x)^2$.
I want to know how the equation is done.
Thnks!!
statistics
asked Dec 22 '16 at 6:58
文奭允文奭允
213
$endgroup$
add a comment |
$begingroup$
I've studied SST and the others but I saw the equation that $SSR=(n-1)(β_1)^2(S_x)^2$.
I want to know how the equation is done.
Thnks!!
statistics
asked Dec 22 '16 at 6:58
文奭允文奭允
213
$endgroup$
1
$begingroup$
I believe you'd want to put $hatbeta_1$ for $beta_1$. Then it should become clear.
$endgroup$
– Sergei Golovan
Dec 22 '16 at 8:03
$begingroup$
what is beta hat..?
$endgroup$
– 文奭允
Dec 23 '16 at 0:55
1
$begingroup$
Beta1 "hat" is the estimate for Beta1
$endgroup$
– Brandon
Dec 24 '16 at 2:08
add a comment |
$begingroup$
I've studied SST and the others but I saw the equation that $SSR=(n-1)(β_1)^2(S_x)^2$.
I want to know how the equation is done.
Thnks!!
statistics
asked Dec 22 '16 at 6:58
文奭允文奭允
213
$endgroup$
I've studied SST and the others but I saw the equation that $SSR=(n-1)(β_1)^2(S_x)^2$.
I want to know how the equation is done.
Thnks!!
statistics
statistics
asked Dec 22 '16 at 6:58
文奭允文奭允
213
asked Dec 22 '16 at 6:58
文奭允文奭允
213
asked Dec 22 '16 at 6:58
文奭允文奭允
213
asked Dec 22 '16 at 6:58
文奭允文奭允
213
asked Dec 22 '16 at 6:58
文奭允文奭允
213
213
1
$begingroup$
I believe you'd want to put $hatbeta_1$ for $beta_1$. Then it should become clear.
$endgroup$
– Sergei Golovan
Dec 22 '16 at 8:03
$begingroup$
what is beta hat..?
$endgroup$
– 文奭允
Dec 23 '16 at 0:55
1
$begingroup$
Beta1 "hat" is the estimate for Beta1
$endgroup$
– Brandon
Dec 24 '16 at 2:08
add a comment |
1
$begingroup$
I believe you'd want to put $hatbeta_1$ for $beta_1$. Then it should become clear.
$endgroup$
– Sergei Golovan
Dec 22 '16 at 8:03
$begingroup$
what is beta hat..?
$endgroup$
– 文奭允
Dec 23 '16 at 0:55
1
$begingroup$
Beta1 "hat" is the estimate for Beta1
$endgroup$
– Brandon
Dec 24 '16 at 2:08
1
1
$begingroup$
I believe you'd want to put $hatbeta_1$ for $beta_1$. Then it should become clear.
$endgroup$
– Sergei Golovan
Dec 22 '16 at 8:03
$begingroup$
I believe you'd want to put $hatbeta_1$ for $beta_1$. Then it should become clear.
$endgroup$
– Sergei Golovan
Dec 22 '16 at 8:03
$begingroup$
what is beta hat..?
$endgroup$
– 文奭允
Dec 23 '16 at 0:55
$begingroup$
what is beta hat..?
$endgroup$
– 文奭允
Dec 23 '16 at 0:55
1
1
$begingroup$
Beta1 "hat" is the estimate for Beta1
$endgroup$
– Brandon
Dec 24 '16 at 2:08
$begingroup$
Beta1 "hat" is the estimate for Beta1
$endgroup$
– Brandon
Dec 24 '16 at 2:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In simple regression, we have
$$
bary = hatbeta_0 + hatbeta_1 barx
$$
by normal equations. Thus,
$$
beginalign
SSR &= sum_i=1^n (haty - bary)^2 \
&= sum_i=1^n left( (hatbeta_0 + hatbeta_1x_i) - (hatbeta_0 + hatbeta_1 barx) right)^2 \
&= sum_i=1^n left( (hatbeta_1)^2 (x_i - barx)^2 right) \
&= (hatbeta_1)^2 sum_i=1^n ( (x_i - barx)^2 \
&= (hatbeta_1)^2S_xx
endalign
$$
answered Mar 31 at 13:32
MoreblueMoreblue
9151317
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
In simple regression, we have
$$
bary = hatbeta_0 + hatbeta_1 barx
$$
by normal equations. Thus,
$$
beginalign
SSR &= sum_i=1^n (haty - bary)^2 \
&= sum_i=1^n left( (hatbeta_0 + hatbeta_1x_i) - (hatbeta_0 + hatbeta_1 barx) right)^2 \
&= sum_i=1^n left( (hatbeta_1)^2 (x_i - barx)^2 right) \
&= (hatbeta_1)^2 sum_i=1^n ( (x_i - barx)^2 \
&= (hatbeta_1)^2S_xx
endalign
$$
answered Mar 31 at 13:32
MoreblueMoreblue
9151317
$endgroup$
add a comment |
$begingroup$
In simple regression, we have
$$
bary = hatbeta_0 + hatbeta_1 barx
$$
by normal equations. Thus,
$$
beginalign
SSR &= sum_i=1^n (haty - bary)^2 \
&= sum_i=1^n left( (hatbeta_0 + hatbeta_1x_i) - (hatbeta_0 + hatbeta_1 barx) right)^2 \
&= sum_i=1^n left( (hatbeta_1)^2 (x_i - barx)^2 right) \
&= (hatbeta_1)^2 sum_i=1^n ( (x_i - barx)^2 \
&= (hatbeta_1)^2S_xx
endalign
$$
answered Mar 31 at 13:32
MoreblueMoreblue
9151317
$endgroup$
add a comment |
$begingroup$
In simple regression, we have
$$
bary = hatbeta_0 + hatbeta_1 barx
$$
by normal equations. Thus,
$$
beginalign
SSR &= sum_i=1^n (haty - bary)^2 \
&= sum_i=1^n left( (hatbeta_0 + hatbeta_1x_i) - (hatbeta_0 + hatbeta_1 barx) right)^2 \
&= sum_i=1^n left( (hatbeta_1)^2 (x_i - barx)^2 right) \
&= (hatbeta_1)^2 sum_i=1^n ( (x_i - barx)^2 \
&= (hatbeta_1)^2S_xx
endalign
$$
answered Mar 31 at 13:32
MoreblueMoreblue
9151317
$endgroup$
In simple regression, we have
$$
bary = hatbeta_0 + hatbeta_1 barx
$$
by normal equations. Thus,
$$
beginalign
SSR &= sum_i=1^n (haty - bary)^2 \
&= sum_i=1^n left( (hatbeta_0 + hatbeta_1x_i) - (hatbeta_0 + hatbeta_1 barx) right)^2 \
&= sum_i=1^n left( (hatbeta_1)^2 (x_i - barx)^2 right) \
&= (hatbeta_1)^2 sum_i=1^n ( (x_i - barx)^2 \
&= (hatbeta_1)^2S_xx
endalign
$$
answered Mar 31 at 13:32
MoreblueMoreblue
9151317
answered Mar 31 at 13:32
MoreblueMoreblue
9151317
answered Mar 31 at 13:32
MoreblueMoreblue
9151317
answered Mar 31 at 13:32
MoreblueMoreblue
9151317
9151317
add a comment |
add a comment |
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1
$begingroup$
I believe you'd want to put $hatbeta_1$ for $beta_1$. Then it should become clear.
$endgroup$
– Sergei Golovan
Dec 22 '16 at 8:03
$begingroup$
what is beta hat..?
$endgroup$
– 文奭允
Dec 23 '16 at 0:55
1
$begingroup$
Beta1 "hat" is the estimate for Beta1
$endgroup$
– Brandon
Dec 24 '16 at 2:08