Chapter 4, Problem 10CT

(a)

To determine

To make: The scatter plot of the data and to describe the correlation.

Answer to Problem 10CT

The line of fit describes the positive correlation

Explanation of Solution

Given:

Advertising (dollars) $x$,	$500$	$1000$	$1500$	$2000$	$2500$	$3000$	$3500$	$4000$
Yearly attendance, $y$	$400$	$550$	$550$	$800$	$650$	$800$	$1050$	$1100$

Calculation:

Consider the table shows the amount $x$ in dollars spent on advertising for a neighborhood festival and the attendance $y$ of the festival for several years.

Now make a scatter plot from the above table

  

Conclusion:

The line of fit describes the positive correlation

(b)

To determine

To write: The equation that models the attendance as the function of the amount spent on advertising

Answer to Problem 10CT

The equation that models the attendance as the function of the amount spent on advertising

is $y=\frac{1}{5}x+300$

Explanation of Solution

Given:

Advertising (dollars) $x$,	$500$	$1000$	$1500$	$2000$	$2500$	$3000$	$3500$	$4000$
Yearly attendance, $y$	$400$	$550$	$550$	$800$	$650$	$800$	$1050$	$1100$

Calculation:

Find the slope for the line of fit using the points on the line $(500,400)and(4000,1100)$

  $\begin{array}{l}m=\frac{1100-400}{4000-500}\\ =\frac{700}{3500}\\ m=\frac{1}{5}\end{array}$

Find the slope Now, write the equation of line of fit using the slope $m=\frac{1}{5}$ , and the point $(500,400)$

The point - slope form of the equation of the lines is

  $y-y1=m(x-x1)$

Substitute the values

  $\begin{array}{l}y-400=\frac{1}{5}(x-500)\\ =\frac{1}{5}x-100\end{array}$

Add $400$ on both sides

  $y=\frac{1}{5}x+300$

Conclusion:

The equation that models the attendance as the function of the amount spent on advertising

is $y=\frac{1}{5}x+300$

(c)

To determine

To find: The slope and y - intercept of the line of fit

Answer to Problem 10CT

Slope is $\frac{1}{5}$ and y - intercept is $300$

Explanation of Solution

Given:

Advertising (dollars) $x$,	$500$	$1000$	$1500$	$2000$	$2500$	$3000$	$3500$	$4000$
Yearly attendance, $y$	$400$	$550$	$550$	$800$	$650$	$800$	$1050$	$1100$

Calculation:

The equation of line of fit is $y=\frac{1}{5}x+300$

The slope is $\frac{1}{5}$ .

This means the yearly attendance is increasing by $\frac{1}{5}$ every $500$ dollars

The y-intercept is $300$ which means at $0$ dollars on advertising the yearly attendance is $300$

Conclusion:

The slope is $\frac{1}{5}$ and y - intercept is $300$