standard error of estimate

**Judi** · June 7th 2008, 10:48 PM

standard error of estimate=
square root of sum of squares of difference of y and y hat divided by n-2

how does this become--------->
square root of sum of y square minus a times sum of y minus b times sum of xy and divided by 2

The text book doesn't show progress and just stated they are equivalent.

Would you please show me how to get there?

Thank you everybody in advance,

Judi

**mr fantastic** · June 7th 2008, 11:09 PM

Originally Posted by Judi

standard error of estimate=
square root of sum of squares of difference of y and y hat divided by n-2

how does this become--------->
square root of sum of y square minus b times sum of y minus b times sum of xy and divided by 2

The text book doesn't show progress and just stated they are equivalent.

Would you please show me how to get there?

Thank you everybody in advance,

Judi

It would help the common masses if you said that this was in the context of linear regression.

I can't be bothered working it out ..... Is $\hat{y}_i = ax_i + b$ or is it the other way around? And you have a formula for a in terms of the data $(x_i, ~ y_i)$ ....?

**Judi** · June 8th 2008, 12:36 PM

yes, it is the least squares line yhat=a+bx.
I don't know what you mean by 'a formula for a in terms of the data ....?'
I have just given those(from the first thread)that two formulas are equal to each other. I want to see how one can become the other.
thanks,

**mr fantastic** · June 8th 2008, 07:23 PM

Originally Posted by Judi

yes, it is the least squares line yhat=a+bx.
I don't know what you mean by 'a formula for a in terms of the data ....?'
I have just given those(from the first thread)that two formulas are equal to each other. I want to see how one can become the other.
thanks,

The algebra is quite boring. Consider:

$\sum_{i=1}^{n} (Y_i - \hat{Y}_i)^2$

$= \sum_{i=1}^{n} (Y_i - [a + bX_i])^2$

$= \sum_{i=1}^{n} Y_i^2 - 2Y_i (a + bX_i) + (a + bX_i)^2$

$= .......... zzzzzzzzzzzz$ uh? uh? ughhh .....

$= \sum_{i=1}^{n} Y_i^2 - 2a \sum_{i=1}^{n}Y_i - 2b \sum_{i=1}^{n}Y_i X_i + \sum_{i=1}^{n} a^2 + 2ab \sum_{i=1}^{n} X_i + b^2 \sum_{i=1}^{n} X_i^2$

$= \sum_{i=1}^{n} Y_i^2 - 2a \sum_{i=1}^{n}Y_i - 2b \sum_{i=1}^{n}Y_i X_i + n a^2 + 2ab \sum_{i=1}^{n} X_i + b^2 \sum_{i=1}^{n} X_i^2$

$= \sum_{i=1}^{n} Y_i^2 - 2a \sum_{i=1}^{n}Y_i - 2b \sum_{i=1}^{n}Y_i X_i + n a^2 + b \left(2a \sum_{i=1}^{n} X_i + b X_i^2 \right)$

$= .......... zzzzzzzzzzzz$ uh? uh? ughhhh .....

$= \sum_{i=1}^{n} Y_i^2 - 2a \sum_{i=1}^{n}Y_i - 2b \sum_{i=1}^{n}Y_i X_i + n a^2 + b \left(a \sum_{i=1}^{n} X_i + {\color{red}a \sum_{i=1}^{n} X_i + b X_i^2} \right)$

$= \sum_{i=1}^{n} Y_i^2 - 2a \sum_{i=1}^{n}Y_i - 2b \sum_{i=1}^{n}Y_i X_i + n a^2 + b \left(a \sum_{i=1}^{n} X_i + {\color{red}\sum_{i=1}^{n} X_i Y_i} \right)$

using the normal equations for linear regression

$= \sum_{i=1}^{n} Y_i^2 - 2a \sum_{i=1}^{n}Y_i - b \sum_{i=1}^{n}Y_i X_i + n a^2 + b a \sum_{i=1}^{n} X_i$

$= \sum_{i=1}^{n} Y_i^2 - 2a \sum_{i=1}^{n}Y_i - b \sum_{i=1}^{n}Y_i X_i + n a^2 + a {\color{red}b\sum_{i=1}^{n} X_i}$

$= .......... zzzzzzzzzzzz$ uh? uh? ughhh .....

$= \sum_{i=1}^{n} Y_i^2 - 2a \sum_{i=1}^{n}Y_i - b \sum_{i=1}^{n}Y_i X_i + n a^2 + a \left( {\color{red}\sum_{i=1}^{n} Y_i - na} \right)$

using the normal equations for linear regression again

$= \sum_{i=1}^{n} Y_i^2 - 2a \sum_{i=1}^{n}Y_i - b \sum_{i=1}^{n}Y_i X_i + n a^2 + a \sum_{i=1}^{n} Y_i - na^2$

$= \sum_{i=1}^{n} Y_i^2 - 2a \sum_{i=1}^{n}Y_i - b \sum_{i=1}^{n}Y_i X_i + a \sum_{i=1}^{n} Y_i$

$= .......... zzzzzzzzzzzz$ uh? uh? .....

$= \sum_{i=1}^{n} Y_i^2 - a \sum_{i=1}^{n}Y_i - b \sum_{i=1}^{n}Y_i X_i$ .

zzzzzzzzzzzzzzzzzzzzzzzzzzzzz ......... (Dreaming now. Star Wars: .... "Help me Mr Fantastic, you're my only hope. Help me Mr Fantastic, you're my only hope, everyone else hates statistics ......)

**TheEmptySet** · June 8th 2008, 08:10 PM

Originally Posted by mr fantastic

zzzzzzzzzzzzzzzzzzzzzzzzzzzzz ......... (Dreaming now. Star Wars: .... "Help me Mr Fantastic, you're my only hope. Help me Mr Fantastic, you're my only hope, everyone else hates statistics ......)

This is too funny!!

**Judi** · June 9th 2008, 07:28 PM

Wow~
That's a lot of work. Thank you, Mr fantastic, for spending that much time on this problem
thank you
Judi

**mr fantastic** · June 9th 2008, 07:36 PM

Originally Posted by Judi

Wow~
That's a lot of work. Thank you, Mr fantastic, for spending that much time on this problem
thank you
Judi

Just wait till you get my invoice ......

(Always a pleasure helping someone with such lovely manners).

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