Single Parameter Linear Regression

(10 pts)

For this assignment, your task is to perform a single parameter linear regression for data that's symmetric about the origin, as described in this class note. Your data will be based on your name. Specifically, you'll have 4 points with

• the x-coordinates equal to \pm position of the first character in your first and last name and
• the y-coordinates equal to \pm position of the second character in your first and last name.

For example, my name is Mark McClure. Thus,

• Since M is the 13^{\text{th}} letter of the alphabet, the x-coordinates are \pm13 and \pm13
• Since a is the 1^{\text{st}} letter and c is the 3^{\text{rd}}, the y-coordinates are \pm 1 and \pm3.

Thus, my four points are

Those points, with the regression line, look like so:

Desmos will give me the line, as I've done here. The real task, though, is to use the technique described in the simple linear regression notes to derive and explain the formula for the line.

From my name, the four points are:

Given the symmetry of the data,

To calculate the total squared error,

To find the minimum total squared error, find where the derivative is 0.

which simplifies to

So the line fitting these 4 points is f(x)=1.212x, shown below.

My points are (14,15), (-14,-15), (3,1), (-3,-1).
This is the formula for total squared error:

Here's what I did: ((14a-15)^2)+((-14a+15)^2)+((-3a+1)^2)+((3a-1)^2).
Taking the derivative of this avoids having to expand these terms out. I used the chain rule.

Simplified, it's 56(14a-15)+12(3a-1). Expanded out:

Solving for a, we get 876/820=a or 1.06829.

My points are:

Applying these coordinates in the total squared error looks like this:

To find the smallest value of a, we find E'(a) = 0 :

The solution is a = \frac{80}{41} ; which is approximately 1.9512195122.
The regression line ends up being f(x) = \frac{80}{41}x .
Putting together the four points and regression line looks like this:

My points are:

The formula for the total squared area is:

Plugging in my values, I get:

Taking the derivative gives us:

This simplifies to:

Setting this derivative to 0 and solving gives us:

This is the equation for the line:

This is the graph when the points are plotted with the line:

@User 030 Could you edit that and add your model function? That is, f(x)=\,?

My points are

Plugging in my points to the total squared error formula yields

To find the minimum total squared error, we take the derivative

and then solve for E'(a) = 0 which is at a = \frac{23}{6} meaning our line of best fit is

The graph of my points and the resulting linear regression line looks like so:

My points are: (5,12),(-5,-12),(2,1),(-2,-1)

Applying these coordinates in the total squared error looked like this:

After foiling and simplifying we are left with:

Taking the derivative got me:

Then I solved for a and got:

Thus,

desmos-graph800×800 14.4 KB

From my name, my four points are:

I need to find a in the function

To do this, I need to use the formula for total squared error

So my total squared error equation is:

In order to find the minimum, I need to take the derivative and set it equal to zero

Simplified,

So the formula for my line is:

Here is it plotted:

My points are:

Plugging in my points gets me:

The derivative would be:

Solving for E'(a) = 0 shows a = \frac{848}{788}, this shows the line of best fit at

This is the graph of those points and line:

The four points are:

Symmetry of the data,

To calculate the total squared error

To find the minimum total squared error, find where the derivative is 0.

which simplifies to

So the line fitting these four points is f(x)=\frac{43}{185}x \approx 0.2324x.

This is the graph of line

image1278×554 27.6 KB

Name: BRady JEnkins
Points

Formula:

Set equal to 0 and solve

Siplified

Thus

Graph

For my name, my 4 points are:

(12,21),(-12,-21),(19,1),(-19,1)

My points plugged in:

The derivative of the equation:

E'(a)=2(2(12a-21)(12)+2(19a-1)(19))

Solved for 0:

a = \frac{271}{505} == f(x) = \frac{271}{505} x.

Graph:

image1335×812 67.2 KB

My points are: \{(4,1), (-4,-1), (4,5), (-4,-5)\}

The formula for the total squared error is: \sum_{i=1}^{n} (y_i - \hat{y}_i)^2

To calculate it: E(a) = (1-4a)^2 + (-1+4a)^2 + (5-4a)^2 + (-5+4a)^2

So the line is f(x) = \frac{3}{4}x.

My points:

(18,25),\;(-18,-25),\;(19,20),\;(-19,-20)

Model function:

f(x)=\alpha x

Applying these points to the total squared error:

Finding minimum error:

Simplifies to:

Final regression function:

My points are: (2,18), (-2, -18), (13, 1), (-13, -1)

The formula for the total squared error is: \sum_{i = 1}^n (y_i - \hat{y})^2

Applying these points to the total squared error:

So the line is f(x) = \frac{49}{173}x.

My points are: (10, 15), (-10, -15), (7, 1), (-7, 1).

Using formula:

Plugging in values from the four points:

or

Then taking the derivative with respect to a:

To find the regression line we set E'(a) to zero and solve for a:

So the line of best fit is

On a graph below:

My four points based on my name are:

Applying these points in the total squared error I get using f(x) = ax:

Deriving E(a) I get:

Setting equal to zero I get:

Thus, the regression line of best fit is:

And looks like:

My points are

When putting them into the formula for total squared error, it becomes
E(a) = (5a-26)^2 + (-5a+26)^2 + (3a-1)^2 + (-3a+1)^2
= 2((5a-26)^2 + (3a-1)^2)
The derivative is
E'(a) = 2(5a-26)(5) + 2(-5a+26)(-5) + 2(3a-1)(3) + 2(-3a+1)(-3)
=136a-532
0=136a-532
a = \frac{532}{126}\

So, the line's equation is f(x) = \frac{532}{126}x

My points are
{(13,1) (-13,-1) (12, 9) (-12,-9)}
Given the symmetry of the data
f(x)=ax
to calculate the total squared error

taking the derivative

This simplifies to

Thus,

image1239×834 22.9 KB

My points are (-18,-25), (-9,-26), (9,26), (18,25).

Using the total squared error formula:

My points in the total squared error formula were

In calculating the derivative i found that E'(a) = 1620a-450 or about 5/18
So the line best fitting my 4 points is f(x)=5/18x or about 0.27 repeating.
And the graph looks like: