Chapter 13, Problem 15CT

To determine

To find the sum of the squared differences using the linear model found in Exercise $14$ .

Answer to Problem 15CT

The sum of the squared differences is $\frac{1}{26}$ .

Explanation of Solution

Given information:

Points $\left(0,2\right),\left(3,4\right),\left(4,5\right)$

Evaluating $y^{\ast }=\frac{19}{26}x+\frac{51}{26}$ ,

$x$	$y$	$y^{\ast }$	$y-y^{\ast }$	${\left(y-y^{\ast }\right)}^2$
$0$	$2$	$\frac{51}{26}$	$\frac{54}{13}$	$\frac{127}{26}$
$3$	$4$	$\frac{1}{26}$	$-\frac{2}{13}$	$\frac{3}{26}$
$4$	$5$	$\frac{1}{676}$	$\frac{4}{169}$	$\frac{9}{676}$

By adding the values in the last column the sum of the squared differences is $\frac{1}{26}$ .