Chapter 8.1, Problem 93E

To determine

To find:the least squares regression line $y=at+b$

Answer to Problem 93E

The least squares regression line is $\boxed{y=7.5t+28}$

The number of new cases of waterborne disease in 2020 can be estimated as $\boxed{141}$

Explanation of Solution

Given information: $\{\begin{array}{l}12b+66a=831\\ 66b+506a=5643\end{array}$

Concept Involved:

A matrix derived from a system of linear equations (each written in standard form with the constant term on the right) is the augmented matrix of the system.

Elementary Row Operation:

The three operations that can be used on a system of linear equations to produce an equivalent system.

Operation	Notation
1.Interchange two equations	$R_a\leftrightarrow R_b$
2. Multiply an equation by a nonzero constant	$c{R}_a\left(c\ne 0\right)$
3. Add a multiple of an equation to another equation.	$c{R}_a+R_b$

In matrix terminology, these three operations correspond to elementary row operations.

An elementary row operation on an augmented matrix of a given system of linear equations produces a new augmented matrix corresponding to a new (but equivalent) system of linear equations. Two matrices are row-equivalent when one can be obtained from the other by a sequence of elementary row operations.

Row-Echelon Form and Reduced Row-Echelon Form:

A matrix in row-echelon form has the following properties.

1. Any rows consisting entirely of zeros occur at the bottom of the matrix.

2. For each row that does not consist entirely of zeros, the first nonzero entryis 1 (called a leading 1).

3. For two successive (nonzero) rows, the leading 1 in the higher row is fartherto the left than the leading 1 in the lower row.

A matrix in row-echelon form is in reduced row-echelon form when every columnthat has a leading 1 has zeros in every position above and below its leading 1.

Calculation:

Write the system of equation $\{\begin{array}{l}12b+66a=831\\ 66b+506a=5643\end{array}$ in augmented form

  $\left[\begin{array}{cccc}66& 12& ⋮& 831\\ 506& 66& ⋮& 5643\end{array}\right]$

Multiply the 1^st row by 1/66

  $\begin{array}{c}1/66R_1\to \\ \end{array}\left[\begin{array}{cccc}1& 2/11& ⋮& 277/22\\ 506& 66& ⋮& 5643\end{array}\right]$

Add -506 times the 1^st row to the 2^nd row

  $\begin{array}{c}\\ -506R_1+R_2\to \end{array}\left[\begin{array}{cccc}1& 2/11& ⋮& 277/22\\ 0& -26& ⋮& -728\end{array}\right]$

Multiply the 2^nd row by -1/26

  $\begin{array}{c}\\ -1/26R_2\to \end{array}\left[\begin{array}{cccc}1& 2/11& ⋮& 277/22\\ 0& 1& ⋮& 28\end{array}\right]$

Add -2/11 times the 2^ndrow to the 1^strow

  $\begin{array}{c}-2/11R_2+R_1\to \\ \end{array}\left[\begin{array}{cccc}1& 0& ⋮& 7.5\\ 0& 1& ⋮& 28\end{array}\right]$

The matrix is now in reduced row-echelon form. Converting back to a system of linear

equations, you have

  $\{\begin{array}{l}a=7.5\\ b=28\end{array}$

Substituting the values of a, and b in the least square regression line is $y=7.5t+28$

To predict the number of the new cases of the waterborne disease in 2020, we need to substitute $t=15$ in the equation $y=7.5t+28$

  $y=7.5\left(15\right)+28$

  $y\approx 141$

Conclusion:

The least square regression line is $y=7.5t+28$

The number of new cases of waterborne disease in 2020 can be estimated as $141$