Chapter 13.3, Problem 21E

To determine

To find: least square regression line for the data and the sum of the squared differences.

Answer to Problem 21E

The least square regression line formed will be $y=1.301x+5.099$ and of squared difference is 0.01247

Explanation of Solution

Given information:

The average annual cost of healthcare

  $\left(12,20.73\right),\left(13,22.03\right),\left(14,23.22\right),\left(15,24.67\right)$

Formula used:

The least square regression line for the set of points $\left(x,y\right)$ can be calculated as $f\left(x\right)=ax+b$

Where $a=\frac{n{\displaystyle \sum _{i=1}^n{x}_i{y}_i}-\left({\displaystyle \sum _{i=1}^n{x}_i}{\displaystyle \sum _{i=1}^n{y}_i}\right)}{n{\displaystyle \sum _{i=1}^n{x}_i{}^2}-{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2}$ and $b=\frac{1}{n}\left({\displaystyle \sum _{i=1}^n{y}_i}-a{\displaystyle \sum _{i=1}^n{x}_i}\right)$

Calculation:

Here number of observations (n) is 4

Construct a table as shown below:

  $\begin{array}{cccc}x& y& xy& x^2\\ 12& 20.73& 248.76& 144\\ 13& 22.03& 286.39& 169\\ 14& 23.22& 370.05& 196\\ 15& 24.67& 370.05& 225\\ {\displaystyle \sum _{i=1}^4{x}_i}=54& {\displaystyle \sum _{i=1}^4{y}_i}=90.65& {\displaystyle \sum _{i=1}^4{x}_i{y}_i}=1230.28& {\displaystyle \sum _{i=1}^4{x}_i{}^2}=734\end{array}$

Now for the regression line

  $\begin{array}{c}a=\frac{n{\displaystyle \sum _{i=1}^n{x}_i{y}_i}-\left({\displaystyle \sum _{i=1}^n{x}_i}{\displaystyle \sum _{i=1}^n{y}_i}\right)}{n{\displaystyle \sum _{i=1}^n{x}_i{}^2-{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2}}\\ =\frac{4\left(123.28\right)-\left(\left(54\right)\left(90.65\right)\right)}{4\left(734\right)-{\left(54\right)}^2}\\ =1.301\\ b=\frac{1}{n}\left({\displaystyle \sum _{i=1}^n{y}_i}-a{\displaystyle \sum _{i=1}^n{x}_i}\right)\\ =\frac{1}{4}\left(90.65-a\left(54\right)\right)\\ =\frac{1}{4}\left(90.65-\left(1.301\right)\left(54\right)\right)\left(\because a=1.301\right)\\ =5.099\end{array}$

Hence the least square regression line formed will be $y=1.301x+5.099$

For the squared difference find $y_1=1.30x+5.099$ for each and every given values of x. Then find $y-y_1$ with each and every value of y, also square the difference

  $\begin{array}{ccccc}x& y& y_1& y-y_1& {\left(y-y_1\right)}^2\\ 12& 20.73& 20.711& 0.019& 0.00036\\ 13& 22.03& 22.012& 0.018& 0.00032\\ 14& 23.22& 23.313& -0.093& 0.00865\\ 15& 24.67& 24.614& 0.056& 0.00314\end{array}$

Now

  $\begin{array}{c}{\displaystyle \sum _{n=1}^4{\left(y-y_1\right)}^2}=\left(0.00036+0.00032+0.00865+0.00314\right)\\ =0.01247\end{array}$

Hence sum of squared difference is 0.01247