# Thread: standard error of estimate

1. ## standard error of estimate

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?

Judi

2. 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?

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 $\displaystyle \hat{y}_i = ax_i + b$ or is it the other way around? And you have a formula for a in terms of the data $\displaystyle (x_i, ~ y_i)$ ....?

3. 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,

4. 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:

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

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

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

$\displaystyle = .......... zzzzzzzzzzzz$ uh? uh? ughhh .....

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

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

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

$\displaystyle = .......... zzzzzzzzzzzz$ uh? uh? ughhhh .....

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

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

using the normal equations for linear regression

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

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

$\displaystyle = .......... zzzzzzzzzzzz$ uh? uh? ughhh .....

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

using the normal equations for linear regression again

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

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

$\displaystyle = .......... zzzzzzzzzzzz$ uh? uh? .....

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

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 ......)

5. 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!!

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

7. 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).