Chapter 7.2, Problem 56E

To determine

To find:The least squares regression line $y=ax+b$ for the data.

Answer to Problem 56E

The least squares regression line $y=ax+b$ for the data is $y=-0.58286x+5.39048$ .

Explanation of Solution

Given information:

The data is given in the scattered plot below:

  

System of equations to find the values of $a$ and $b$ is:

  $\{\begin{array}{c}6b+15a=23.6\\ 15b+55a=48.8\end{array}$

Concept Used:

To find the least squares regression line $y=ax+b$ for a set of points $(x_1,y_1),(x_2,y_2),\mathrm{....},(x_n,y_n)$ is by solving the system below for $a$ and $b.$

  $\{\begin{array}{c}ab+\left({\displaystyle \sum _{i=1}^n{x}_i}\right)a=\left({\displaystyle \sum _{i=1}^n{y}_i}\right)\\ \left({\displaystyle \sum _{i=1}^n{x}_i}\right)b+\left({\displaystyle \sum _{i=1}^n{x}_i{}^2}\right)a=\left({\displaystyle \sum _{i=1}^n{x}_i{y}_i}\right)\end{array}$

Calculations:

Solving the system of equations using substitution method, we have:

  $\begin{array}{l}6b+15a=23.6\\ \Rightarrow 6b=23.6-15a\\ \Rightarrow b=\frac{23.6-15a}{6}\end{array}$

Substituting the above finding in the second equation, we get

  $\begin{array}{l}15\left(\frac{23.6-15a}{6}\right)+55a=48.8\\ \Rightarrow 59-37.5a+55a=48.8\\ \Rightarrow 17.5a=-10.2\\ \Rightarrow a=-\frac{10.2}{17.5}=-0.58286\end{array}$

Now, using the above finding value:

  $\begin{array}{l}b=\frac{23.6-15a}{6}\\ \Rightarrow b=\frac{23.6-15(-0.58286)}{6}\\ \Rightarrow b=\frac{32.3429}{6}\\ \Rightarrow b=5.39048\end{array}$

Conclusion:

Using the graphing utility, we obtain the following results:

  

Therefore, the results we got are correct.