Chapter 13.3, Problem 6E

To determine

To calculate: The sum of squared differences for the data points $\left(0,7\right),\left(2,5\right),\left(3,3\right),\left(4,3\right),\left(6,0\right)$ and the model $y*=-1.2x+7$ .

Answer to Problem 6E

The value of sum of squared differences is $1$ .

Explanation of Solution

Given information:

The data points $\left(0,7\right),\left(2,5\right),\left(3,3\right),\left(4,3\right),\left(6,0\right)$ and the model $y*=-1.2x+7$ .

Formula used:

The sum of squared differences is ${\displaystyle \sum _{i=1}^n{\left(y-y_i*\right)}^2}$ where $y_i$ is the y -coordinate of the point and $y_i*$ is the linear model.

Calculation:

Consider the provided data points $\left(0,7\right),\left(2,5\right),\left(3,3\right),\left(4,3\right),\left(6,0\right)$ and the model $y*=-1.2x+7$ .

Construct a table with data points. Substitute the value of each x to compute the value for each $y*$ .

Then compute the difference between each y and corresponding $y*$ . At last compute the value of ${\left(y-y*\right)}^2$ for each point of data set.

Recall that the sum of squared differences is ${\displaystyle \sum _{i=1}^n{\left(y-y_i*\right)}^2}$ where $y_i$ is the y -coordinate of the point and $y_i*$ is the linear model.

Data points are tabulated as,

  $\begin{array}{cccccc}x& 0& 2& 3& 4& 6\\ y& 7& 5& 3& 3& 0\\ y*& 7& 4.6& 3.4& 2.2& -0.2\\ y-y*& 0& 0.4& -0.4& 0.8& 0.2\\ {\left(y-y*\right)}^2& 0& 0.16& 0.16& 0.64& 0.04\end{array}$

Sum up all the values of the last row to obtain the value of sum of squared differences.

Thus, the value of sum of squared differences is $1$ .