Chapter 13.3, Problem 9E

To determine

To calculate: The sum of squared differences for the data points $\left(-3,2\right),\left(-2,2\right),\left(-1,4\right),\left(0,6\right),\left(1,8\right)$ and the model $y*=0.29x^2+2.2x+6$ .

Answer to Problem 9E

The value of sum of squared differences is $0.8259$ .

Explanation of Solution

Given information:

The data points $\left(-3,2\right),\left(-2,2\right),\left(-1,4\right),\left(0,6\right),\left(1,8\right)$ and the model $y*=0.29x^2+2.2x+6$ .

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 $\left(-3,2\right),\left(-2,2\right),\left(-1,4\right),\left(0,6\right),\left(1,8\right)$ and the model $y*=0.29x^2+2.2x+6$ .

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,

  $\begin{array}{cccccc}x& -3& -2& -1& 0& 1\\ y& 2& 2& 4& 6& 8\\ y*& 2.01& 2.76& 4.09& 6& 8.49\\ y-y*& -0.01& -0.76& -0.09& 0& 0.49\\ {\left(y-y*\right)}^2& 0.0001& 0.5776& 0.0081& 0& 0.2401\end{array}$

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 $0.8259$ .