Chapter 1.2, Problem 21E

a.

To determine

To make: a scatter plot of these data and check whether a linear model is appropriate. and graph.

Answer to Problem 21E

The scatter plot of the data is shown below.

Explanation of Solution

Given:

The given table shows peptic rules ulcer rates for various rates:

Income	Ulcer rate(per 100 population)
$\$4000$	$14.1$
$\$6000$	$13.0$
$\$8000$	$13.4$
$\$12000$	$12.5$
$\$16000$	$12.0$
$\$20000$	$12.4$
$\$30000$	$10.5$
$\$45000$	$9.4$
$\$60000$	$8.2$

Calculation:

The scatter plot of the data is shown below:

  

b.

To determine

To find: a linear model using the first and last data points and graph the linear model.

Answer to Problem 21E

  $y=-0.000105x+14.52$

Explanation of Solution

Given:

The given table shows peptic rules ulcer rates for various rates:

Income	Ulcer rate(per 100 population)
$\$4000$	$14.1$
$\$6000$	$13.0$
$\$8000$	$13.4$
$\$12000$	$12.5$
$\$16000$	$12.0$
$\$20000$	$12.4$
$\$30000$	$10.5$
$\$45000$	$9.4$
$\$60000$	$8.2$

Calculation:

According to the table, the first point is $\left(4000,14.1\right)$ and the last point is $\left(6000,8.1\right)$ .

The slope of the linear equation is given by $m=\frac{y_2-y_1}{x_2-x_1}$ .

So, $m=\frac{8.2-14.1}{60000-4000}=\frac{-5.9}{56000}=-0.000105$ .

Now, use the given point slope form to write the equation of the line. It is to choose any of the two points to use. Point-slope form of the equation is given by $y-y_1=m\left(x-x_1\right)$ .

Now, if it choose the first point, it get $y-14.1=-0.000105\left(x-4000\right)$ .

In order to solve for y , it must first distribute $-0.000105$ and then add ing $14.1$ to both sides of gives $y=-0.000105x+14.52$ .

The graph of the linear model is shown below:

  

Hence, a linear model using the first and last data points is $y=-0.000105x+14.52$ .

c.

To determine

To find: the least squares regression line and graph the regression line.

Answer to Problem 21E

  $a=-9.978546\times {10}^{-5}\approx -0.0000099785\text{ and }b=13.95076408\approx 13.95076$ .

Explanation of Solution

Given:

The given table shows peptic rules ulcer rates for various rates:

Income	Ulcer rate(per 100 population)
$\$4000$	$14.1$
$\$6000$	$13.0$
$\$8000$	$13.4$
$\$12000$	$12.5$
$\$16000$	$12.0$
$\$20000$	$12.4$
$\$30000$	$10.5$
$\$45000$	$9.4$
$\$60000$	$8.2$

Calculation:

To find the linear regression, use the $\text{LinReg}\left(ax+b\right)$ program on theT1 calculator. The following directions for a T1-84are:

Step 1: Hit STAT and then press enter when Edit is highlighted.

Step 2: Enter the x-coordinate under the $L_1$ column and y-coordinate under $L_2$ column.

Step 3: Hit STAT, use the arrow key to scroll right to CALC and the scroll down to $\text{LinReg}\left(ax+b\right)$ . Press Enter.

Step 4: Press Enter multiples time until the program runs. If X list does not show $L_1$ and Y list does not show $L_2$ , make to sure to fix those before running to program.

Step 5: The program will output values for a and b .

The program outputs the values of $a=-9.978546\times {10}^{-5}\approx -0.0000099785\text{ and }b=13.95076408\approx 13.95076$ .

Hence, the least squares regression line $a=-9.978546\times {10}^{-5}\approx -0.0000099785\text{ and }b=13.95076408\approx 13.95076$

So, the least squares regression line is shown below:

  

d.

To determine

To estimate: the ulcer rate for an income of $\$25,000$ by using the linear model in part (c).

Answer to Problem 21E

  $11.45$ .

Explanation of Solution

Given:

From part (c) $a=-9.978546\times {10}^{-5}\approx -0.0000099785\text{ and }b=13.95076408\approx 13.95076$ .

Calculation:

It is known from part (c), the equation of the linear model according to Least Squares Regression Line approach: $y\approx -0.00000997855x+13.9507640771$

When the income rate is $\$25,000\left(x=25000\right)$ just plug in the linear function equation in order to estimate the ulcer rate:

  $y\approx -0.00000997855\left(25000\right)+13.9507640771\approx 11.4561$

Hence, the ulcer rate for an income of $\$25,000$ is $11.45$ .

e.

To determine

To check: how likely is someone with an income of $\$80,000$ is suffering from ulcer.

Answer to Problem 21E

  $6\%$ .

Explanation of Solution

Given:

The given model is $y\approx -0.00000997855+13.9507640771$

Calculation:

In this case $x=80,000$ , plug the value of x in the equation of the linear model according to Least Squares Regression Line approach:

  $y\approx -0.00000997855\left(80000\right)+13.9507640771\approx 5.96796$

There is a chance of approximately$6\%$for suffering from ulcer when the income is $\$80,000$

f.

To determine

To check: that it is reasonable to apply the model to someone with an income of$\$200,000$.

Answer to Problem 21E

It is not reasonable.

Explanation of Solution

Given:

The given model is $y\approx -0.00000997855+13.9507640771$

Calculation:

It will be not reasonable to apply the linear model to someone with an income of$\$200,000$because the value is far away from the interval where the data is collected.

Besides, apply the equation, it is finding a negative ulcer rate which is an irrational number$y\approx -0.00000997855\left(200000\right)+13.9507640771\approx -6.00623$