Chapter 13.3, Problem 7E

To determine

To calculate: The sum of squared differences for the data points $\left(-3,-1\right),\left(-1,0\right),\left(0,2\right),\left(2,3\right),\left(4,4\right)$ and the model $y*=0.5x+0.5$ .

Answer to Problem 7E

The value of sum of squared differences is $6.75$ .

Explanation of Solution

Given information:

The data points $\left(-3,-1\right),\left(-1,0\right),\left(0,2\right),\left(2,3\right),\left(4,4\right)$ and the model $y*=0.5x+0.5$ .

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 $\left(-3,-1\right),\left(-1,0\right),\left(0,2\right),\left(2,3\right),\left(4,4\right)$ and the model $y*=0.5x+0.5$ .

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,

  $\begin{array}{cccccc}x& -3& -1& 0& 2& 4\\ y& -1& 0& 2& 3& 4\\ y*& -1& 0& 0.5& 1.5& 2.5\\ y-y*& 0& 0& 1.5& 1.5& 1.5\\ {\left(y-y*\right)}^2& 0& 0& 2.25& 2.25& 2.25\end{array}$

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 $6.75$ .