# Method of Least Squares

• Oct 13th 2008, 10:53 PM
wik_chick88
Method of Least Squares
The following relationship is thought to exist between a response
y and a predictor $x" alt="x" />:

$y$
= $ae^x" alt="ae^x" />.

For a sample (
$x_{1}" alt="x_{1}" />, $y_{1}$), . . . , ( $x_{n}" alt="x_{n}" />, $y_{n}$), use the method of least squares to determine an estimator for a.

• Oct 13th 2008, 10:55 PM
wik_chick88
The following relationship is thought to exist between a response $y$ and a predictor $x$:

$y$ = $ae^x$.

For a sample (

$x_{1}$, $y_{1}$), . . . , ( $x_{n}$, $y_{n}$), use the method of least squares to determine an estimator for a.
• Oct 15th 2008, 07:37 AM
wik_chick88
Please if ANYONE can give ANY help I would be sooo grateful!!!please!!
• Oct 15th 2008, 05:30 PM
mr fantastic
Quote:

Originally Posted by wik_chick88
The following relationship is thought to exist between a response $y$ and a predictor $x$:

$y$ = $ae^x$.

For a sample (

$x_{1}$, $y_{1}$), . . . , ( $x_{n}$, $y_{n}$), use the method of least squares to determine an estimator for a.

$\ln y = \ln \left( a e^x \right) = \ln a + \ln e^x = \ln a + x$.

So get a least squares regression line for $\ln y$ versus $x$ and use it to estimate the value of $a$.
• Oct 16th 2008, 06:45 AM
wik_chick88
Quote:

Originally Posted by mr fantastic
$\ln y = \ln \left( a e^x \right) = \ln a + \ln e^x = \ln a + x$.

So get a least squares regression line for $\ln y$ versus $x$ and use it to estimate the value of $a$.

how do i calculate the least squares regression line?
• Oct 16th 2008, 03:38 PM
mr fantastic
Quote:

Originally Posted by wik_chick88
how do i calculate the least squares regression line?

You go to your classnotes or textbook, find the relevant formulae and then substitute the appropriate data into those formulae. Alternatively, you enter the data into a calculator or software package and let the technology do it for you.
• Oct 16th 2008, 05:45 PM
wik_chick88
Quote:

Originally Posted by mr fantastic
You go to your classnotes or textbook, find the relevant formulae and then substitute the appropriate data into those formulae. Alternatively, you enter the data into a calculator or software package and let the technology do it for you.

thanks so much for your help my classnotes are very unhelpful but i looked it up on the internet and i worked out the least squares regression line to be:

$y = 0.45646x - 0.34665$

can you tell me if this is right?

also, what do i do from here?
• Oct 16th 2008, 06:05 PM
mr fantastic
Quote:

Originally Posted by wik_chick88
thanks so much for your help my classnotes are very unhelpful but i looked it up on the internet and i worked out the least squares regression line to be:

$y = 0.45646x - 0.34665$

can you tell me if this is right?

also, what do i do from here?

If you correctly inputted the correct data into the formula then your line will be correct (you inputted ln y and x, yes?).

From my first post, $\ln a = -0.34665 \Rightarrow a = \, ....$

By the way, things don't look so good ..... Your model predicts that the line of best fit of ln y versus x should have a gradient close to 1 ........ So either:

2. Perhaps you should be using a model $y = a e^{bx}$ ........ In which case your best estimate for b is b = 0.45646 (assuming your line is correct).