Chapter 4.6, Problem 2GP

To determine

To graph: The residual plot for given function.

Answer to Problem 2GP

The residual plot is shown above.

From the pattern observed in residual plot, it can be implied that the linear model acts as a decent fit to the given data.

Explanation of Solution

Given Information: A table comparing the graduation rates and unemployment rates.

Calculation:

An equation of the best fit line is given by,

  $y=a+bx$ ...... (1)

• Here, a is the intercept and b is the slope.

Finding $xy\text{ and }{x}^2$ ,

$x$	$y$	$xy$	$x^2$
73	6.9	503.7	5329
85	4.1	348.5	7225
64	3.2	204.8	4096
81	5.5	445.5	6561
68	4.3	292.4	4624
82	5.1	418.2	6724

Sum of every column is,

  $\begin{array}{c}{\displaystyle \sum x}=453\\ {\displaystyle \sum y}=29.1\\ {\displaystyle \sum x}y=2213.1\\ {\displaystyle \sum x^2}=34559\end{array}$

Calculate a as,

  $\begin{array}{c}a=\frac{{\displaystyle \sum y\cdot {{\displaystyle \sum x}}^2}-{\displaystyle \sum x\cdot {\displaystyle \sum x}y}}{n\cdot {\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ =\frac{29.1\times 34559-453\times 2213.1}{6\times 34559-{453}^2}\\ =1.46\end{array}$

Calculate b as,

  $\begin{array}{c}b=\frac{n\cdot {\displaystyle \sum xy}-{\displaystyle \sum x\cdot {\displaystyle \sum y}}}{n\cdot {\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ =\frac{6\times 2213.1-453\times 29.1}{6\times 34559-{453}^2}\\ =0.045\end{array}$

Substitute the values of a and b in equation (1),

  $\begin{array}{c}y=a+bx\\ y=0.045x+1.46\end{array}$

Hence, the equation of best line is,

  $y=0.045x+1.46$

Make a table including predicted values.

$x$	$y$ (observed)	$y$ (predicted) $y=0.045x+1.46$	Residual $\left(\text{Observed}-\text{Predicted}\right)$
73	6.9	$\begin{array}{c}y=0.045x+1.46\\ =0.045\left(73\right)+1.46\\ =4.745\end{array}$	2.155
85	4.1	5.285	-1.185
64	3.2	4.34	-1.14
81	5.5	5.105	0.395
68	4.3	4.52	-0.22
82	5.1	5.15	-0.05
73	6.9	4.745	2.155

The graph of residual is,

  

The pattern shown by the residual plot is fairly random. First two residuals are negative, the next one is highly positive, the fourth is also positive, and the last two residuals are negative. This irregular pattern implies that the linear model acts as a decent fit to the given data.