Chapter 13.3, Problem 11E

To determine

To calculate: The sum of squared differences for the data points $\left(0,10\right),\left(1,9\right),\left(2,6\right),\left(3,0\right)$ and the model $y*=-1.25x^2+0.5x+10$ .

Answer to Problem 11E

The value of sum of squared differences is $0.125$ .

Explanation of Solution

Given information:

The data points $\left(0,10\right),\left(1,9\right),\left(2,6\right),\left(3,0\right)$ and the model $y*=-1.25x^2+0.5x+10$ .

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 $\left(0,10\right),\left(1,9\right),\left(2,6\right),\left(3,0\right)$ and the model $y*=-1.25x^2+0.5x+10$ .

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,

  $\begin{array}{ccccc}x& 0& 1& 2& 3\\ y& 10& 9& 6& 0\\ y*& 10& 9.25& 6& 0.25\\ y-y*& 0& -0.25& 0& -0.25\\ {\left(y-y*\right)}^2& 0& 0.0625& 0& 0.0625\end{array}$

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 $0.125$ .