Chapter 5.8, Problem 17PPSE

Solution

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

The model that best fits the data is $y=\left(-8.33\times {10}^{-6}\right)x^3+\left(3\times {10}^{-4}\right)x^2-0.02x+2.01$ .

Given information :

The data is:

World Population

Year	Average Growth Rate (%)
$\begin{array}{l}1972\\ 1982\\ 1992\\ 2002\end{array}$	$\begin{array}{l}1.96\\ 1.73\\ 1.5\\ 1.22\end{array}$

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 years after 1970 and $y$ represent the average growth rate.

$x$	$y$
$\begin{array}{l}2\\ 12\\ 22\\ 32\end{array}$	$\begin{array}{l}1.96\\ 1.73\\ 1.5\\ 1.22\end{array}$

Here, there are four set of points. From the $\left(n+1\right)$ principle, there is a unique polynomial of degree at most $3$ that fits the points perfectly.

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

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

The linear model will be:

  $y=ax+b$

Here, $a=-0.02,b=2.019$ .

So, the linear function will be: $y=-0.02x+2.019$ and $r^2=0.9975$ .

The quadratic model will be:

  $y=a{x}^2+bx+c$

Here, $a=-1.25\times {10}^{-4},b=-0.02,c=1.99$ .

So, the quadratic function will be: $y=\left(-1.25\times {10}^{-4}\right)x^2-0.02x+1.99$ and $r^2=0.9995$ .

The cubic model will be:

  $y=a{x}^3+b{x}^3+cx+d$

Here, $a=-8.33\times {10}^{-6},b=3\times {10}^{-4},c=-0.02,d=2.01$ .

So, the cubic function will be: $y=\left(-8.33\times {10}^{-6}\right)x^3+\left(3\times {10}^{-4}\right)x^2-0.02x+2.01$ and $r^2=1$ .

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

Therefore, the model best fits the data is $y=\left(-8.33\times {10}^{-6}\right)x^3+\left(3\times {10}^{-4}\right)x^2-0.02x+2.01$ .