Chapter 5.8, Problem 19PPSE

Solution

To calculate : The model that best fits the given data.

The model that best fits the data is $y=-0.025x^2+0.14x+19.595$ .

Given information :

The data is:

U.S Crude Oil and Petroleum

Month (2008)	Products Supplied (millions of barrels/day)
$\begin{array}{l}2\\ 4\\ 6\end{array}$	$\begin{array}{l}19.782\\ 19.768\\ 19.553\end{array}$

Formula used:

  $\left(n+1\right)$ Principle: For any set of $n+1$ points in the coordinate plane that pass the vertical line test, there is a unique polynomial of degree at most $n$ that fits the point perfectly.

Calculation :

Let $x$ represent the months and $y$ represent the products supplied.

$x$	$y$
$\begin{array}{l}2\\ 4\\ 6\end{array}$	$\begin{array}{l}19.782\\ 19.768\\ 19.553\end{array}$

Here, there are three set of points. From the $\left(n+1\right)$ principle, there is a unique polynomial of degree at most $2$ that fits the points perfectly.

To determine the model that fits the data calculate linear, and quadratic model for this data.

For this find the linear, and quadratic regression to get the models using the calculator.

The linear model will be:

  $y=ax+b$

Here, $a=-0.057,b=19.93$ .

So, the linear function will be: $y=-0.057x+19.93$ and $r^2=0.7957$ .

The quadratic model will be:

  $y=a{x}^2+bx+c$

Here, $a=-0.025,b=0.14,c=19.595$ .

So, the quadratic function will be: $y=-0.025x^2+0.14x+19.595$ and $r^2=1$ .

By comparing the values of $r^2$ , the quadratic model best fits the data.

Therefore, the model best fits the data is $y=-0.025x^2+0.14x+19.595$ .