Line of Best Fit

Line of Best Fit exercise

• Explain what the line of best fit is.

  Hints

  The blue line seen here is the line of best fit.

  Neither of these lines have a point lying on them.

  A best fitting line means that the sum of the distances of the given points to the line is as small as possible.

  Solution

  The line of best fit is the line that is as close as possible to all points on the graph with as many points above the line as below it.

  The way to figure out which line is the best fitting line is to calculate the residuals of the regarding line. This is the difference of the $y$-values of the line and the corresponding y-coordinate of the given point.

  The lower the sum of the squares of the differences in $y$-values, the better the line fits to the data.

• Identify how to tell which line fits better when two lines are given.

  Hints

  Using this graph, fill in the table from left to right for each given point.

  In the table, we have the $y$-value of the line from substituting the $x$-value $1$ into the equation $y=0.55x+3.2$.

  The difference of the $y$-values is also known as the residual.

  The sum of least squares is needed to determine the line of better fit.

  Solution

  To check if a given line is the better fitting line, we need its corresponding linear equation. For example we can use $y=0.55x+3.2$.

  For each point we proceed as follows:

  1. We record the coordinates of each point in the table. Here the ordered pair $(1,2)$.
  2. We substitute the $x$-value of this point in to the linear equation of the line to get the corresponding $y$-value. Here $y=0.55(1)+3.2=3.75$.
  3. We determine the residual, that means the difference between the two $y$-values. In this case $y_\text{coor.}-y_\text{line} = 2 - 3.75 = -1.75 $
  4. Next, we square the residual. $(-1.75)^2=3.0625$. We'd do steps $1$-$4$ with each point.
  5. Then we'd sum all the squares and compare it to the other lines to find the one with the sum of least squares. That line is the line of best fit.

• Decide which line fits the best when the following residuals are given.

  Hints

  You have to sum the squares of the residuals.

  The lower the sum of the squares of the differences in $y$-values, the better the line fits to the data.

  The sum of squares of the worst fitting line is given by $32.25$.

  Solution

  For each of the given tables, you still have to sum the squares and compare the squares.

  Keep in mind that The lower the sum of the squares of the residuals, differences in $y$-values, the better the line fits to the data.

  So we get the following order:

  $\begin{array}{c} \mathbf{y=x+1}\\\\ \begin{array}{c|c|c|c} (x,y_1)&y_2&y_1-y&(y_1-y)^2\\ \hline (1,3)&2&1&1\\ \hline (2,4)&3&1&1\\ \hline (4,5.5)&5&0.5&0.25\\ \hline (6,6)&7&-1&1\\ \hline &&&\text{sum: }~~ 3.25 \end{array} \end{array}$

  $\begin{array}{c} \mathbf{y=1.5x-1}\\\\ \begin{array}{c|c|c|c} (x,y_1)&y_2&y_1-y&(y_1-y)^2\\ \hline (1,3)&0.5&2.5&6.25\\ \hline (2,4)&2&2&4\\ \hline (4,5.5)&5&0.5&0.25\\ \hline (6,6)&8&-2&4\\ \hline &&&\text{sum: }~~ 14.5 \end{array} \end{array}$

  $\begin{array}{c} \mathbf{y=1.5x+1}\\\\ \begin{array}{c|c|c|c} (x,y_1)&y_2&y_1-y&(y_1-y)^2\\ \hline (1,3)&2.5&0.5&0.25\\ \hline (2,4)&4&0&0\\ \hline (4,5.5)&7&-1.5&2.25\\ \hline (6,6)&10&-4&16\\ \hline &&&\text{sum: }~~ 18.5 \end{array} \end{array}$

  $\begin{array}{c} \mathbf{y=2x-1}\\\\ \begin{array}{c|c|c|c} (x,y_1)&y_2&y_1-y&(y_1-y)^2\\ \hline (1,3)&1&2&4\\ \hline (2,4)&3&1&1\\ \hline (4,5.5)&7&-1.5&2.25\\ \hline (6,6)&11&-5&25\\ \hline &&&\text{sum: }~~ 32.25 \end{array} \end{array}$

• Calculate the sum of squares of the following line.

  Hints

  Substitute the left coordinate of each ordered pair in to the equation $y=0.5x+2$.

  So you get $y_{\text{line}}$.

  Be cautious about the way you order the difference.

  The square of any number is always a positive number. For example $(-2)^2$ is $4$.

  You only have to put in the corresponding results.

  Solution

  To decide if a given line fits the data, we will we have to calculate the residuals, respectively their squares, and then sum those squares. This sum is a measure for the goodness of fit of the line.

  If you have to compare two lines, or more lines, the sum of least squares is the way to find the best fitting line.

  $~$

  $(1,3)$

  • $y_{\text{line}}=0.5(1)+2=2.5$
  • $y_{\text{point}}-y_{\text{line}}=3-2.5=0.5$
  • $(y_{\text{point}}-y_{\text{line}})^2=0.5^2=0.25$

  $~$

  $(2,2)$

  • $y_{\text{line}}=0.5(2)+2=3$
  • $y_{\text{point}}-y_{\text{line}}=2-3=-1$
  • $(y_{\text{point}}-y_{\text{line}})^2=(-1)^2=1$

  $~$

  $(3,3)$

  • $y_{\text{line}}=0.5(3)+2=3.5$
  • $y_{\text{point}}-y_{\text{line}}=3-3.5=-0.5$
  • $(y_{\text{point}}-y_{\text{line}})^2=(-0.5)^2=0.25$

  $~$

  $(4,5)$

  • $y_{\text{line}}=0.5(4)+2=4$
  • $y_{\text{point}}-y_{\text{line}}=5-4=1$
  • $(y_{\text{point}}-y_{\text{line}})^2=1^2=1$

  $~$

  Finally, we sum all those squares to get $0.25+1+0.25+1=2.5$.

• Determine the $y$-value of the given line.

  Hints

  Keep in mind that the first coordinate is the $x$-coordinate.

  Substitute the corresponding $x$-coordinate into the linear equation.

  Looking at an example: $(8,6)$

  The corresponding $y$-value is given as follows:

  $y=0.55(8)+3.2=7.6$.

  Solution

  We need the $y$-values according to each point. So we put the $x$-coordinate, the first or the left coordinate, in the corresponding linear equation.

  • $(1,2)$ $\rightarrow$ $y=0.55(1)+3.2=3.75$
  • $(3,9)$ $\rightarrow$ $y=0.55(3)+3.2=4.85$
  • $(2,7)$ $\rightarrow$ $y=0.55(2)+3.2=4.3$
  • $(7,9)$ $\rightarrow$ $y=0.55(7)+3.2=7.05$

  We will later use these $y$-values to determine the residuals. To check the goodness of fit of a given line, we have to determine the residuals for each point, square those residuals and finally sum those squares.

• Decide which line fits better to the given three points.

  Hints

  Proceed as follows for each point and for each line:

  1. Put the $x$-coordinate in the equation.
  2. Determine the differences of the $y$-values. Those are the residuals.
  3. Square the residuals.
  4. Sum all given squares.

  The lower the sum of the squares of the differences in $y$-values, the better the line fits to the data.

  Both sum of squares lie between $2$ and $3$.

  Keep the signs in mind.

  Solution

  Here we have to decide which of the two given lines fits better to the given points.

  Let's start with the green, one $y=0.5x+0.5$.

  $(1,2)$

  • $y_{\text{line}}=0.5(1)+0.5=1$
  • $y_{\text{point}}-y_{\text{line}}=2-1=1$
  • $(y_{\text{point}}-y_{\text{line}})^2=1^2=1$

  $~$

  $(2,2)$

  • $y_{\text{line}}=0.5(2)+0.5=1.5$
  • $y_{\text{point}}-y_{\text{line}}=2-1.5=0.5$
  • $(y_{\text{point}}-y_{\text{line}})^2=0.5^2=0.25$

  $~$

  $(4,3.5)$

  • $y_{\text{line}}=0.5(4)+0.5=2.5$
  • $y_{\text{point}}-y_{\text{line}}=3.5-2.5=1$
  • $(y_{\text{point}}-y_{\text{line}})^2=1^2=1$

  Lastly, we sum all squares to $1+0.25+1=2.25$ for the green line.

  $~$

  Next, we determine the sum of squared residuals also for the blue line $y=1.2x$.

  $(1,2)$

  • $y_{\text{line}}=1.2(1)=1.2$
  • $y_{\text{point}}-y_{\text{line}}=2-1.2=0.8$
  • $(y_{\text{point}}-y_{\text{line}})^2=0.8^2=0.64$

  $~$

  $(2,2)$

  • $y_{\text{line}}=1.2(2)=2.4$
  • $y_{\text{point}}-y_{\text{line}}=2-2.4=-0.4$
  • $(y_{\text{point}}-y_{\text{line}})^2=(-0.4)^2=0.16$

  $~$

  $(4,3.5)$

  • $y_{\text{route}}=1.2(4)=4.8$
  • $y_{\text{point}}-y_{\text{line}}=3.5-4.8=-1.3$
  • $(y_{\text{point}}-y_{\text{line}})^2=(-1.3)^2=1.69$

  Lastly, we sum all squares to $0.64+0.16+1.69=2.49$ for the blue line.

  $~$

  Because $2.25<2.49$, we can conclude that the green line fits the given three points better than the blue one. However, please note that neither of these lines represent the actual line of best fit because as you can see with the green one all the points are above the line.