Question

The table shows (lifetime) peptic ulcer rates (per 100 population) for various family incomes as reported by the National Health Interview Survey.

(a) Make a scatter plot of these data and decide whether a linear model is appropriate.

(b) Find and graph a linear model using the first and last data points.

(c) Find and graph the regression line.

(d) Use the linear model in part (c) to eastimate the ulcer rate for people with an income of \(25,000.

(e) According to the model, how likely is someone with an income of \)80,000 to suffer from peptic ulcers?

(f) Do you think it would be reasonable to apply the model to someone with an income of \(200,000?

Income	Ulcer rate (per 100 population)
\)4,000	14.1
\(6,000	13.0
\)8,000	13.4
\(12,000	12.5
\)16,000	12.0
\(20,000	12.4
\)30,000	10.5
\(45,000	9.4
\)60,000	8.2

Short answer

a. The scatter plot is shown below:

The linear model is appropriate.

b. \(y = - 0.000105357x + 14.521429\). The graph is shown below:

c. \(y = - 0.0000997855x + 13.950764\)

d. 6%

e. The value of y is negative for people with income $200,000, therefore model is not suitable

Step-by-step solution

(a) Plot the data points on a scatter plot

The graph below shows the data points on thescatter plot.

From the scatter plot, the points are primarily close and linear, so the linear model is suitable from the scatter plot.

(b) Find the regression line

The regression line between points \(\left( {4000,14.1} \right)\) and \(\left( {60000,8.2} \right)\), can be calculated as follows:

\(\begin{aligned}y - 14.1 &= \frac{{8.2 - 14.1}}{{60000 - 4000}}\left( {x - 4000} \right)\\y &= - 0.000105357x + 14.521429\end{aligned}\)

The graph below represents the regression line as shown below:

(c) Find the equation of the regression line

Using the following steps in the TI-84 calculator.

1. Hit STAT and then press enter when EDIT appears on the screen.
2. In the \({L_1}\) column enter the data of “Income” and in \({L_2}\) column enter “Ulcer rate”.
3. Hit on STAT and scroll the curser to “LinReg(ax+b)” and then enter.
4. The value of a and b are 0.01879 and 0.30480 respectively.

The equation of regression line is \(y = - 0.0000997855x + 13.950764\).

(d) Find the equation of regression line

Substitute 25,000 for \(x\) in the equation \(y = - 0.0000997855x + 13.950764\).

\(\begin{aligned}y &= - 0.0000997855\left( {25,000} \right) + 13.950764\\ &\approx 11.5\end{aligned}\)

It shows that 11.5 per 100 population.

(e) Find the chances that a person with income of $80,000 to suffer from peptic ulcers

Substitute 80,000 for \(x\) in the equation \(y = - 0.0000997855x + 13.950764\).

\(\begin{aligned}y &= - 0.0000997855\left( {80,000} \right) + 13.950764\\ &\approx 6\end{aligned}\)

Nearly 6% of income with $80,000 to suffer from peptic ulcers.

(f) Check the model for income $200,000

Substitute 200,000 for \(x\) in the equation \(y = - 0.0000997855x + 13.950764\).

\(\begin{aligned}y &= - 0.0000997855\left( {200,000} \right) + 13.950764\\ &\approx - 19.9571 + 13.950764\\ &< 0\end{aligned}\)

As the value of y is negative, therefore model is not suitable for people with income $200,000.