Chapter 4.5, Problem 18E

a.

To determine

To write: An equation of line of best fit for given data and to further graph the points and plot the line.

Answer to Problem 18E

Equation of the line of best fit is $\text{y= x + 6}\text{.75}$ .

Explanation of Solution

Given Information: Data values of day, $x$ and number of people, $y$ is given.

Calculation:

Entering the data and using the linear regression feature in the graphing calculator,

  $\begin{array}{l}\text{LinReg}\\ \text{ y=ax+b}\\ \text{ a= 1}\text{.0}\\ \text{ b= 6}\text{.75}\\ {\text{ r}}^2\text{= 0}\text{.38356164383561653}\\ \text{ r= 0}\text{.6193235372853324}\end{array}$

As $y=ax+b$ ,

thus, equation of line of best fit is $\text{y= x + 6}\text{.75}$ .

Graph:

  

b.

To determine

To identify: The correlation coefficient and to further interpret it.

Answer to Problem 18E

  $r$ is $0.62$ .

Data values have positive correlation.

Explanation of Solution

Given information: Graphing calculator display from the previous part.

Interpretation:

From the graphing calculator display in the previous part it can be inferred that correlation coefficient $r$ is $0.62$ . Value of r is inclined towards $1$ , this implies that data values have positive correlation and the points lie around a positive slope.

c.

To determine

To interpret: The slope and $y-\mathrm{int}ercept$ of the line of best fit.

Answer to Problem 18E

Slope is positive and is 1.

Line of best fit crosses the $y-axis$ at $\text{6}\text{.75}$ , however, in this case it has no significance.

Explanation of Solution

Given Information: Graphing calculator display from previous part.

Interpretation:

From the graphing calculator display shown in part $(a)$ , it can be inferred that slope $a$ is $\text{1}$ , this implies that slope is positive. It interprets that the number of volunteers at the animal shelter increases by 1 per day.

  $y-\mathrm{int}ercept(b)$ is $\text{6}\text{.75}$ , this means that line of best fit crosses the $y-axis$ at $\text{6}\text{.75}$ . However, as the number of days cannot be zero, $\therefore$ it has no significance in this case.