Help with line of best fit! (1 Viewer)

Green Yoda · Jun 8, 2015

find the equation of the best fit

ANSWER: y=5x+48.5 (WTF)

Can anyone please do it with full steps?

InteGrand · Jun 8, 2015

Rathin said:
View attachment 32107

ANSWER: y=5x+48.5 (WTF)

Can anyone please do it with full steps?

$Finding the line of best fit is not inside the syllabus, and will require some multivariable calculus.$

$If we have some observed data $(x_1,y_1), (x_2, y_2),\cdots, (x_n,y_n)$, and we want to fit it to the line $y=mx+b$, we essentially are looking for the values of $m$ and $b$ that will give us the lowest sum of squared residuals (this is called the least squares method: \url{http://en.wikipedia.org/wiki/Least\_squares}). In other words, we want the straight line such that at each $x$-value of the data, if we measure the distance of the observed $y$-value to the line, and square it, we want the sum of all these squares to be minimised.$

$Each squared residual is of the form $R_i^2 = (y_i - (mx_i + b))^2$, because the $y$-value on the line at $x=x_i$ is $mx_i + b$, and the observed value is $y_i$. So we want to minimise the following sum by choosing suitable $m$ and $b$: $S(m,b)=\sum _{i=1}^n (y_i - (mx_i + b))^2$. To minimise this, we want the partials of $S$ with respect to $m$ and with respect to $b$ to both be 0, and then solve the resulting linear system for $m$ and $b$. We calculate the partials below.$

$\frac{\partial S}{\partial m}=\frac{\partial }{\partial m}\sum _{i=1}^n (y_i - (mx_i + b))^2$

$$= \sum_{i=1}^{n}\frac{\partial }{\partial m}(y_i - (mx_i + b))^2$ (derivative of a sum is the sum of derivatives)$

$=\sum_{i=1}^{n}[-2x_i(y_i - (mx_i + b))]$

$Also, $$$

$\frac{\partial S}{\partial b}=\frac{\partial }{\partial b}\sum _{i=1}^n (y_i - (mx_i + b))^2$

$$=\sum _{i=1}^n \frac{\partial }{\partial b} (y_i - (mx_i + b))^2$ (derivative of a sum is the sum of derivatives)$

$$=\sum_{i=1}^{n}[-2(y_i - (mx_i + b))]$.$

$As we said, for the minimum, we want these two partials to be simultaneously 0, so we obtain the following system:$

$\begin{cases} \sum_{i=1}^{n}[-2x_i(y_i - (mx_i + b))]=0& (1)\\ \sum_{i=1}^{n}[-2(y_i - (mx_i + b))]=0& (2) \end{cases}$

$(Ignore the $br$'s.)$

$Solve this linear system with two equations and two unknowns ($m$ and $b$) using the given $(x_i,y_i)$ values, and the solutions give you the least-squares optimal line (i.e. the Line of Best Fit).$

$In practice, this will be tedious to do, so it is better to just use some software like Excel.$

leehuan · Jun 8, 2015

Unless this is a Maths General 2 question requiring this formula under Data Analysis...

Formula sheet usable in the General 2 exam

I don't know what a correlation coefficient is though.

Green Yoda · Jun 8, 2015

InteGrand said:
$Finding the line of best fit is not inside the syllabus, and will require some multivariable calculus.$

$If we have some observed data $(x_1,y_1), (x_2, y_2),\cdots, (x_n,y_n)$, and we want to fit it to the line $y=mx+b$, we essentially are looking for the values of $m$ and $b$ that will give us the lowest sum of squared residuals (this is called the least squares method: \url{http://en.wikipedia.org/wiki/Least\_squares}). In other words, we want the straight line such that at each $x$-value of the data, if we measure the distance of the observed $y$-value to the line, and square it, we want the sum of all these squares to be minimised.$

$Each squared residual is of the form $R_i^2 = (y_i - (mx_i + b))^2$, because the $y$-value on the line at $x=x_i$ is $mx_i + b$, and the observed value is $y_i$. So we want to minimise the following sum by choosing suitable $m$ and $b$: $S(m,b)=\sum _{i=1}^n (y_i - (mx_i + b))^2$. To minimise this, we want the partials of $S$ with respect to $m$ and with respect to $b$ to both be 0, and then solve the resulting linear system for $m$ and $b$. We calculate the partials below.$

$\frac{\partial S}{\partial m}=\frac{\partial }{\partial m}\sum _{i=1}^n (y_i - (mx_i + b))^2$

$$= \sum_{i=1}^{n}\frac{\partial }{\partial m}(y_i - (mx_i + b))^2$ (derivative of a sum is the sum of derivatives)$

$=\sum_{i=1}^{n}[-2x_i(y_i - (mx_i + b))]$

$Also, $$$

$\frac{\partial S}{\partial b}=\frac{\partial }{\partial b}\sum _{i=1}^n (y_i - (mx_i + b))^2$

$$=\sum _{i=1}^n \frac{\partial }{\partial b} (y_i - (mx_i + b))^2$ (derivative of a sum is the sum of derivatives)$

$$=\sum_{i=1}^{n}[-2(y_i - (mx_i + b))]$.$

$As we said, for the minimum, we want these two partials to be simultaneously 0, so we obtain the following system:$

$\begin{cases} \sum_{i=1}^{n}[-2x_i(y_i - (mx_i + b))]=0& (1)\\ \sum_{i=1}^{n}[-2(y_i - (mx_i + b))]=0& (2) \end{cases}$

$(Ignore the $br$'s.)$

$Solve this linear system with two equations and two unknowns ($m$ and $b$) using the given $(x_i,y_i)$ values, and the solutions give you the least-squares optimal line (i.e. the Line of Best Fit).$

$In practice, this will be tedious to do, so it is better to just use some software like Excel.$

Umm yeah..can i get a basic solution? Im in year 10 and i have no idea what you did there... The way we are taught is to first get any 2 points on the line and find the gradient of it..then use the point gradient formula to find y=mx+b, thus being the equation..but ive tried like so many different points on the line but the gradient varies from 3:75 to 5:5 but not 5 as the answer shows

leehuan · Jun 8, 2015

Hm, the process of taking two points is strictly defined to be correct for when you have ONLY two points to choose from. If you pick any two points out of all of that, then you can get any random number (between 3.75 and 5.5 as you've stated) as your value for the gradient.

Any of this is possible
(i) - They gathered all the gradients possible and just picked the integer (1, 2, 3, 4, 5, 6, 7) closest to the averages of the gradients
(ii) - A SPECIFIC group of two yields the gradient 5
(iii) - Use of mathematics general 2
(iv) - If you plot the points, m is approximately equal to 5. So 5 must be correct
Or any other reason.

InteGrand · Jun 8, 2015

Rathin said:
Umm yeah..can i get a basic solution? Im in year 10 and i have no idea what you did there... The way we are taught is to first get any 2 points on the line and find the gradient of it..then use the point gradient formula to find y=mx+b, thus being the equation..but ive tried like so many different points on the line but the gradient varies from 3:75 to 5:5 but not 5 as the answer shows

Exactly, for experimental data, unless the data happen to lie exactly on a straight line (which is highly unlikely, due to random errors etc.), the gradients obtained by using two different points will be different for different pairs (in general). The method they taught you won't get you the line of best fit; to find that, we need other methods.

InteGrand · Jun 8, 2015

According to WolframAlpha, the LOBF is different to the one your solution says (assuming I inputted the data correctly): http://www.wolframalpha.com/input/?...,{20.4,152},{23.5,167},{24.3,169},{21.4,156}}

Wolfram also shows the residual plot, $R^2$ value, etc.

Green Yoda · Jun 8, 2015

InteGrand said:
Exactly, for experimental data, unless the data happen to lie exactly on a straight line (which is highly unlikely, due to random errors etc.), the gradients obtained by using two different points will be different for different pairs (in general). The method they taught you won't get you the line of best fit; to find that, we need other methods.

They want us to find the equation of the line of best fit..in the format of y=mx+b

leehuan · Jun 8, 2015

Rathin said:
They want us to find the equation of the line of best fit..in the format of y=mx+b

InteGrand said:
According to WolframAlpha, the LOBF is different to the one your solution says (assuming I inputted the data correctly): http://www.wolframalpha.com/input/?...,{20.4,152},{23.5,167},{24.3,169},{21.4,156}}

Wolfram also shows the residual plot, $R^2$ value, etc.

Compare your textbook answer to the theoretical answer...

Using the point-grad formula was, as already stated, bound to cause error in accuracy.

InteGrand · Jun 8, 2015

Rathin said:
They want us to find the equation of the line of best fit..in the format of y=mx+b

Maybe check your textbook or notes for worked solutions. (What they're doing isn't actually the line of best fit as far as I can tell; it's at best a rough estimate, if you're lucky.)

Help with line of best fit! (1 Viewer)

Green Yoda

Hi Φ

InteGrand

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leehuan

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Green Yoda

Hi Φ

leehuan

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InteGrand

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InteGrand

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Green Yoda

Hi Φ

leehuan

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InteGrand

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